Power Systems Control and Stability 2nd Ed by P.M. Anderson & a.a. Fouad

Power System Control and Stability Second Edition P. M. Anderson San Diego, California A. A. Fouad Fort Collins, Colorado IEEE Power Engineering Society, Sponsor IEEE Press Power Engineering Series Mohamed E. El-Hawary, Series Editor IEEE PRESS A JOHN WlLEY & SONS, INC., PUBLICATION Copyright 0 2003 by Institute of Electrical and Electronics Engineers, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470. 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For general information on our other products and services please contact our Customer Care Department within the U.S. at 877-762-2974, outside the U.S.at 317-572-3993 or fax 317-572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print, however, may not be available in electronic format. Library of Congress Caralogng in Publication Data is available ISBN 0-471-23862-7 Printed in the United States of America. 1 0 9 8 7 6 5 4 3 2 1 Contents Preface Part I Introduction P. M. Anderson and A. A. Fouad xi11 ... Chapter 1. Power System Stability 1.1 1.2 1.3 1.4 1.5 Introduction Requirements of a Reliable Electrical Power Service Statement of the Problem Effect of an Impact upon System Components Methods of Simulation Problems 3 3 4 8 10 11 Chapter 2. The Elementary Mathematical Model 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.1 1 2.12 Swing Equation Units Mechanical Torque Electrical Torque Power-Angle Curve of a Synchronous Machine Natural Frequencies of Oscillation of a Synchronous Machine System of One Machine against an Infinite Bus-The Classical Model Equal Area Criterion Classical Model of a Multitnachine System Classical Stability Study of a Nine-Bus System Shortcomings of the Classical Model Block Diagram of One Machine Problems References 13 15 16 20 21 24 26 31 35 37 45 47 48 52 Chapter 3. System Response to Small Disturbances 3.1 3.2 3.3 3.4 3.5 Introduction Types of Problems Studied The Unregulated Synchronous Machine Modes of Oscillation of an Unregulated Multimachine System Regulated Synchronous Machine 53 54 55 59 66 vi i viii Contents 3.6 Distribution of Power impacts Problems References 69 80 80 Part I1 The Electromagnetic Torque P. M. Anderson and A. A. Fouad Chapter 4. The Synchronous Machine 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.1 1 4.12 4.13 4.14 4.15 4.16 Introduction Park’s Transformation Flux Linkage Equations Voltage Equations Formulation of State-Space Equations Current Formulation Per Unit Conversion Normalizing the Voltage Equations Normalizing the Torque Equations Torque and Power Equivalent Circuit of a Synchronous Machine The Flux Linkage State-Space Model Load Equations Subtransient and Transient Inductances and Time Constants Simplified Models of the Synchronous Machine Turbine Generator Dynamic Models Problems References 83 83 85 88 91 91 92 99 103 105 107 109 114 122 127 143 146 148 Chapter 5. The Simulation of Synchronous Machines 5.1 5.2 5.3 5.4 Introduction Steady-State Equations and Phasor Diagrams Machine Connected to an Infinite Bus through a Transmission Line Machine Connected to an Infinite Bus with Local Load at Machine Terminal 5.5 Determining Steady-State Conditions 5.6 Examples 5.7 Initial Conditions for a Multimachine System 5.8 Determination of Machine Parameters from Manufacturers’ Data 5.9 Analog Computer Simulation of the Synchronous Machine 5.10 Digital Simulation of Synchronous Machines Problems References 150 150 153 154 157 159 165 166 170 184 206 206 Chapter 6. Linear Models of the Synchronous Machine 6.1 6.2 6.3 6.4 6.5 Introduction Linearization of the Generator State-Space Current Model Linearization of the Load Equation for the One-Machine Problem Linearization of the Flux Linkage Model Simplified Linear Model 208 209 213 217 222 Contents IX 6.6 6.7 Block Diagrams State-Space Representation of Simplified Model Problems References 23 1 23 1 232 232 Chapter 7. Excitation Systems 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.1 1 Simplified View of Excitation Control Control Configurations Typical Excitation Configurations Excitation Control System Definitions Voltage Regulator Exciter Buildup Excitation System Response State-Space Description of the Excitation System Computer Representation of Excitation Systems Typical System Constants The Effect of Excitation on Generator Performance Problems References 233 235 236 243 250 254 268 285 292 299 304 304 307 Chapter 8. Effect of Excitation on Stability 8.1 8.2 8.3 8.4 8.5 Introduction Effect of Excitation on Generator Power Limits Effect of the Excitation System on Transient Stability Effect of Excitation on Dynamic Stability Root-Locus Analysis of a Regulated Machine Connected to an Infinite Bus 8.6 Approximate System Representation 8.7 Supplementary Stabilizing Signals 8.8 Linear Analysis of the Stabilized Generator 8.9 Analog Computer Studies 8.10 Digital Computer Transient Stability Studies 8.1 1 Some General Comments on the Effect of Excitation on Stability Problems References 309 311 315 321 327 333 338 344 347 353 363 365 366 Chapter 9. Multimachine S’wtems with Constant Impedance Loads 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.1 1 Introduction Statement of the Problem Matrix Representation of a Passive Network Converting Machine Coordinates to System Reference Relation Between Machine Currents and Voltages System Order Machines Represented by Classical Methods Linearized Model for the Network Hybrid Formulation Network Equations with Flux Linkage Model Total System Equations 368 368 369 313 374 317 378 381 386 388 390 X Contents 392 396 397 9.12 Multimachine System Study Problems References Part I11 The Mechanical Torque Power System Control and Stability P. M.Anderson Chapter 10. Speed Governing 10.1 10.2 10.3 10.4 10.5 10.6 The Flyball Governor The Isochronous Governor Incremental Equations of the Turbine The Speed Droop Governor The Floating-Lever Speed Droop Governor The Compensated Governor Problems References 402 408 410 413 419 421 428 428 Chapter 11. Steam Turbine Prime Movers 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 Introduction Power Plant Control Modes Thermal Generation A Steam Power Plant Model Steam Turbines Steam Turbine Control Operations Steam Turbine Control Functions Steam Generator Control Fossil-Fuel Boilers Nuclear Steam Supply Systems Problems References 430 432 435 436 437 444 446 458 46 1 476 480 48 1 Chapter 12. Hydraulic Turbine Prime Movers 12.1 Introduction 12.2 The Impulse Turbine 12.3 The Reaction Turbine 12.4 Propeller-Type Turbines 12.5 The Deriaz Turbine 12.6 Conduits, Surge Tanks, and Penstocks 12.7 Hydraulic System Equations 12.8 Hydraulic System Transfer Function 12.9 Simplifylng Assumptions 12.10 Block Diagram for a Hydro System 12.1 1 Pumped Storage Hydro Systems Problems References 484 484 486 489 489 489 498 503 506 509 510 511 512 Contents xi Chapter 13. Combustion Turbine and Combined-Cycle Power Plants 13.1 Introduction 13.2 The Combustion Turbine Prime Mover 13.3 The Combined-Cycle Prime Mover 513 513 518 527 527 529 53 1 545 555 582 590 614 622 63 1 640 65 1 Problems References Appendix A. Appendix B. Appendix C. Appendix D . Appendix E. Appendix F. Appendix G. Appendix H. Appendix I. Appendix J. Index Trigonometric Identities for Three-phase Systems Some Computer Methods for Solving Differential Equations Normalization Typical System Data Excitation Control System Definitions Control System Components Pressure Control Systems The Governor Equations Wave Equations for a Hydraulic Conduit Hydraulic Servomotors Preface It is well over thirty years since some of the early versions of this book were used in our classes, and it is more than a quarter of a century since the first edition appeared in print. Normally, one would have expected users of the book to almost give it up as old-fashioned. Yet, until very recently the questions the authors were frequently asked explained the rationale for the added material in this edition, especially by new users: When will the Second Edition be out? Over these past thirty years the size of the systems analyzed in stability studies, the scope of the studies (including the kind of answers sought), the duration of the transients analyzed, and the methods of solution may have varied, but central to all is that the proper system model must be used. Such a model must be based on description of the physical system and on its behavior during the transient being analyzed. This book has focused on modeling the power system components for analysis of the electromechanical transient, perhaps with emphasis on the inertial transient. The one possible exception reflects the concern of the time the book came into being, namely analysis of the linear system model for detection and mitigation of possible poorly damped operating conditions. Since the 1970s, several trends made stability of greater concern to power system engineers. Because of higher cost of money and delay of transmission construction because of environmental litigations, the bulk power system has experienced more congestion in transmission, more interdependence among networks, and so on. To maintain stability, there has been more dependence on discreet supplementary controls, greater need for studying larger systems, and analysis of longer transients. Since then, additional models were needed for inclusion in stability studies: turbine governors, power plants, discrete supplementary controls, etc. Thus, the need for modeling the power system components that make up mechanical torque has become more important than ever. The authors think it is time to meet this need, as was originally planned. Now that the electric utility industry is undergoing major restructuring, the question arises as to whether the trend that started in the 1970s is likely to continue, at least into the near future. Many power system analysts believe that the answer to this question is yes. Since the revised printing of this book appeared, the electric utility industry has undergone a significant restructuring, resulting in heavier use of the bulk power transmission for interregional transactions. It is expected that new engineering emphasis will be given to what engineers refer to as mid-term or long-term analysis. We believe that in the restructured environment, this type of analysis will continue be needed because there will be greater emphasis on providing answers about system limitations to all parties involved in the various activities as well as in the interregional transactions. Modeling of mechanical torque will be important in conducting these studies. The material on the “mechanical torque” presented in Chapters 10 through 13 and in Appendices F through J are the work of author Paul Anderson and he should be contacted regarding any questions, corrections, or other information regarding these portions of the book. This material is a bit unusual to include in a book on power system stability and control, but we have recognized that a complete picture of stability and the supporting mathematical models cannot Xlll ... xiv Preface be considered complete without a discussion of these important system components. The models presented here can be described as “low-order” models that we consider appropriate additions to studies of power system stability. This limits the models to a short time span of a minute or so, and purposely avoids the modeling of power plant behavior for the long term, for example, in the study of economics or energy dispatch. P. M. ANDERSON A. A. FOUAD Sun Diego, California Fort Collins, Colorado Part I Introduction P. M. Anderson A. A. Fouad chapter 1 Power System Stability 1.1 Introduction Since the industrial revolution man’s demand for and consumption of energy has increased steadily. The invention of the induction motor by Nikola Tesla in 1888 signaled the growing importance of electrical energy in the industrial world as well as its use for artificial lighting. A major portion of the energy needs of a modern society is supplied in the form of electrical energy. Industrially developed societies need an ever-increasing supply of electrical power, and the demand on the North American continent has been doubling every ten years. Very complex power systems have been built to satisfy this increasing demand. The trend in electric power production is toward an interconnected network of transmission lines linking generators and loads into large integrated systems, some of which span entire continents. Indeed, in the United States and Canada, generators located thousands of miles apart operate in parallel. This vast enterprise of supplying electrical energy presents many engineering problems that provide the engineer with a variety of challenges. The planning, construction, and operation of such systems become exceedingly complex. Some of the problems stimulate the engineer’s managerial talents; others tax his knowledge and experience in system design. The entire design must be predicated on automatic control and not on the slow response of human operators. T o be able to predict the performance of such complex systems, the engineer is forced to seek ever more powerful tools of analysis and synthesis. This book is concerned with some aspects of the design problem, particularly the dynamic performance, of interconnected power systems. Characteristics of the various components of a power system during normal operating conditions and during disturbances will be examined, and effects on t h e overall system performance will be analyzed. Emphasis will be given to the transient behavior in which the system is described mathematically by ordinary differential equations. 1.2 Requirements of a Reliable Electrical Power Service Successful operation of a power system depends largely on the engineer’s ability to provide reliable and uninterrupted service t o the loads. The reliability of the power supply implies much more than merely being available. Ideally, the loads must be fed at constant voltage and frequency at all times. In practical terms this means that both voltage and frequency must be held within close tolerances so that the consumer’s 3 I t might be tempting to say that successful operation requires only that the new state be a “stable” state (whatever that means). they should all remain operating in parallel and at the same speed. failure in a piece of equipment. . Thus it can be accurately stated that the power system operator must maintain a very high standard of continuous electrical service. a drop in voltage of l0-15% or a reduction of the system frequency of only a few hertz may lead to stalling of the motor loads on the system. Furthermore. if a generator is lost. For example.g. Economic power as well as emergency power may flow over interconnecting tie lines to help maintain continuity of service. sudden application of a major load such as a steel mill. I f a generator is separated from the system.” 1.e. the power it was carrying must be obtainable from another source. it must be resynchronized and then loaded. or if a line is lost. successful operation of the system means that these lines must remain in service if firm power is to be exchanged between the areas of the system.. This usually requires a study of large geographical areas since almost all power systems are interconnected with neighboring systems. In any case all interconnected synchronous machines should remain in synchronism if the system is stable. synchronizing forces tend to keep it in step. e. a fault on the network. Random changes in load are taking place at all times. eventually leading to its tripping. The high-voltage transmisssion system connects the generating stations and the load centers. Conditions do arise. If a machine tends to speed up or slow down. or loss of a line or generating unit. If at any time a generator loses synchronism with the rest of the system. i. Synchronous machines do not easily fall out of step under normal conditions. major changes do take place at times.4 Chapter 1 equipment may operate satisfactorily. These problems must be studied by the power system engineer and fall under the heading “power system stability. Synchronism frequently may be lost in that transition period. and small impacts in the system may cause these machines to lose synchronism. the remaining connected generators must be capable of meeting the load demand. in generation. assuming it has not been damaged and its prime mover has not been shut down due to the disturbance that caused the loss of synchronism. I f the perturbation does not involve any net change in power.3 Statement of the Problem The stability problem is concerned with the behavior of the synchronous machines after they have been perturbed. I f an unbalance between the supply and demand is created by a change in load.” such a state never exists in the true sense.. a new operating state is necessary. with subsequent adjustments of generation. Therefore. For example. A second requirement of reliable electrical service is to maintain the integrity of the power network. in which operation is such that the synchronizing forces for one or more machines may not be adequate. significant voltage and current fluctuations may occur and transmission lines may be automatically tripped by their relays at undesired locations. The first requirement of reliable service is to keep the synchronous generators running in parallel and with adequate capacity to meet the load demand. While it is frequently convenient to talk about the power system in the “steady state. Unfortunately. Interruptions in this network may hinder the flow of power to the load. We may look at any of these as a change from one equilibrium state to another. however. the machines should return to their original state. or in network conditions. or growing oscillations may occur over a transmission line. A major shock to the system may also lead to a loss of synchronism for one or more machines. this view is erroneous in one important aspect: it neglects the dynamics of the transition from one equilibrium state to another. ” we now propose a simple nonmathematical definition of the term that will be satisfactory for elementary problems. Since we are concerned here with the tripping of the line. Hence the definition describes a practical specification for an acceptable operating condition. are reflected as fluctuations in :he power flow over the transmission lines. the power fluctuation that can be tolerated depends on the initial operating condition of the system. This problem is termed the stability of the tie line. These oscillations.3. The problem of interest is one where a power system operating under a steady load condition is perturbed. which is of concern in defining system stability. but if the system is stable. This primitive definition of stability requires that the system oscillations be damped. The same thing can be said about tie-line stability. Adjustment to the new operating condition is called the transient period. Definition: If the oscillatory response of a power system during the transient period following a disturbance is damped and the system settles in a finite time to a new steady operating condition. I n fact. . with the subsequent adjustment of the voltage angles. it may be tripped out by its protective equipment thereby disconnecting the two groups of machines. If a certain line connecting two groups of machines undergoes excessive power fluctuations. these oscillations will be damped toward a new quiescent operating condition. The definition also excludes continuous oscillation from the family of stable systems. 1. The perturbation could be a major disturbance such as the loss of a generator. The main criterion for stability is that the synchronous machines maintain synchronism at the end of the transient period. we say the system is stable. or a combination of such events. even though in reality it reflects the stability of the two groups of machines. It could also be a small load or random load changes occurring under normal operating conditions. we will provide a more rigorous mathematical definition. however.1 Primitive definition of stability Having introduced the term “stability. This is a desirable feature in many systems and is considered necessary for power systems. Later. The reason is practical since a continually oscillating system would be undesirable for both the supplier and the user of electric power. A statement declaring a power system to be “stable” is rather ambiguous unless the conditions under which this stability has been examined are clearly stated. causing the readjustment of the voltage angles of the synchronous machines.Power System Stability 5 The transient following a system perturbation is oscillatory in nature. This includes the operating conditions as well as the type of perturbation given to the system. it is considered unstable. a severe (but improbable) disturbance can always be found that will cause instability. If the system is not stable. the disturbances for which the system should be designed to maintain stability must be deliberately selected. it results in the establishment of a new steady-state operating condition. This condition is sometimes called asymptotic stability and means that the system contains inherent forces that tend to reduce oscillations. Therefore. including the line loading and the nature of the impacts to which it is subjected. a fault or the loss of a line. These questions have become vitally important with the advent of large-scale interconnections. The system behavior during this time is called the dynamic system performance. although oscillators are stable in a mathematical sense. If such an occurrence creates an unbalance between the system generation and load. This attitude is adopted in spite of the fact that an artificial separation between the two problems has been made in the past. A more recent and certainly more appropriate name is dynamic stability. Here we would expect the system operator to have scheduled enough machine capacity to handle the load. the stability of the machines will be. Stability depends strongly upon the magnitude and location of the disturbance and to a lesser extent upon the initial state or operating condition of the system. A fault on the highvoltage transmission network or the loss of a major generating unit are examples of large impacts. and smaller and more normal random impacts. while others are beyond the scope of this book. and to prevent self-excitation of machines.2 Chapter 1 Other stability problems While the stability of synchronous machines and tie lines is the most important and common problem. some general comments are in order. including the power-angle curve. First. Stability of synchronous machines Distinction should be made between sudden and major changes. since modern exciters will change the operating curve during the period under study.6 1.e. Some of these problems are discussed in Part 111. the ability of the system to survive a certain disturbance depends on its precise operating condition at the time of the occurrence. In the dynamics of the transition from one operating point to another. which we shall call large impacts. This was simply a convenience to accommodate the different approximations and assumptions made in the mathematical treat- I . i. Those that the system should be designed to withstand must therefore be selected a priori. Without detailed discussion.. network interconnections. dynamic stability tends to be a property of the state of the system. generation schedule. determined by many factors. A change in the system loading. This problem is referred to in the literature as the transient stability problem. these impacts have a finite probability of occurring. I f one of these large impacts occurs. to adjust for load changes. from which the engineer concludes that under given system conditions and for a given impact the synchronous machines will or will not remain in synchronism. Second. or type of circuit protection may give completely different results in a stability study for the same disturbance. other stability problems may exist.. . Let us now consider a situation where there are no major shocks or impacts. The problem of studying the stability of synchronous machines under the condition of small load changes has been called “steady-state” stability. the portion in which the power increases with increased angle. the synchronous machines may lose synchronism. particularly in power systems having appreciable capacitances.3. In such cases arrangements must be made to avoid excessive voltages during light load conditions. Transient stability and dynamic stability are both qoestions that must be answered to the satisfaction of the engineer for successful planning and operation of the system. I t is sometimes incorrect to consider a single power-angle curve. We would also expect each synchronous machine to be operating on the stable portion of its power-angle curve. 133 . but rather a random occurrence of small changes in system loading. I n the United States the regional committees of the National Electric Reliability Council ( N E R C ) specify the contingenciesagainst which the system must be proven stable. Thus the transient stability study is a very specific one. to avoid damage to equipment. I n contrast to transient stability. I n this case the smallest oscillatory adjustments in the large systems are reflected as sizable power oscillations in the tie line. and other factors. energy stored in the rotating masses. These oscillations are reflected in the flow of power in the transmission lines. the dynamic response of the components of the overall interconnected system must be considered. usually several seconds and in some cases a few minutes. The question then becomes. Normally. Performance of dynamic stability studies for such long periods will require the simulation of system components often neglected in the so-called transient stability studies. As these conditions spread. For example. The problem is not only in the tie line itself but also in the two systems it connects and in the sensitivity of control in these systems. the effect of an impact must be studied over a relatively long period. perhaps a few seconds. . The oscillation frequency has an effect on the damping characteristics of prime movers. I f the power in any line is monitored. However. The portion supplied by the different generators under different conditions depends upon electrical proximity to the position of impact. the availability o f high-speed digital computers and modern modeling techniques makes it possible to represent any component of the power system in almost any degree of complexity required or desired. Thus the tie line under study may in effect be connecting two huge systems. governor characteristics. and this is reflected in the power flow over the tie line.4 Tie-line oscillations As random power impacts occur during the normal operation of a system. First.3. Whether they will act to aid stability is difficult to predict beforehand. a 40-MW oscillation on a 400-MW tie is a much less serious problem than the same oscillation on a 100-MW tie. The problem is aggravated if the initial disturbance causes other disturbances in neighboring areas due to power swings. Thus questionable simplifications or assumptions are no longer needed and are often not justified. Second. The electrical strength (admittance) or capacity of the tie cannot be divorced from this problem. These controllers are usually too slow to interfere with the dynamic oscillations mentioned above. Thus different parts of the interconnected system will respond to localized disturbances at different times. this added power must be supplied by the generators. The machines therefore are never truly at steady state except when at standstill. I n support of this viewpoint the following points are pertinent. The situation in a tie line is different in one sense since it connects one group of machines to another. These two groups are in continuous oscillation with respect to each other. To what degree can these oscillations be tolerated? The above problem is entirely different from that of maintaining a scheduled power interchange over the tie line. Each machine is in continuous oscillation with respect to the others due to the effect of these random stimuli. a chain reaction may result and large-scale interruptions of service may occur. in a large interconnected system the full effect of a disturbance is felt at the remote parts some time after its occurrence. To alter these oscillations.Power System Stability 7 ments of the two problems. and perhaps more important. in a large interconnected system. periodic oscillations are observed to be superimposed on the steady flow. The situation may be further complicated by the fact that each machine group in turn is connected to other groups. control equipment can be provided to perform this function. these oscillations are not large and hence not objectionable. 1. e.. Fig. until the input power is adjusted by the machine governors. may be felt. The change in frequency will affect the loads. 1. i. load regulation is 100%. Therefore. The following effects. especially the motor loads.8 Chapter 1 exciters. and we will use the point of impact as our reference point.4 Effect of an Impact upon System Components In this section a survey of the effect of impacts is made to estimate the elements that should be considered in a stability study. there is a minimum size of tie that can be effectively made from the viewpoint of stability. Response of a four-machine system during a transient: (a) stable system. . Our "test" stimulus will be a change in power. the power change will go to or come from the energy in the rotating masses.1. A convenient starting point is to relate an impact to a change in power somewhere in the network. 1. s ) . (b) unstable system. I Time. in whole or in part. The system frequency will change because. A common rule of thumb used among power system engineers is that a decrease in frequency results in a load decrease of equal percentage. The network bus voltages will be affected to a lesser degree unless the change in power is accompanied by a change in reactive power. etc. and frequency to change. rotor angle. Let us now consider a severe impact initiated by a sizable generation unbalance.1 Loss of synchronism Any unbalance between the generation and load initiates a transient that causes the rotors of the synchronous machines to “swing” because net accelerating (or decelerating) torques are exerted on these rotors. they are usually neglected and only the transient effects are considered important. e. a new equilibrium state must be reached before any of the machines experience this condition.4. an appreciable increase in machine speeds may not necessarily mean that synchronism will be lost. I in which the rotor angles of the machines in a hypothetical four-machine system are plotted against time during a transient. Induced currents in the field winding due to sudden changes in armature currents. if the initial transient causes an electrical link in the transmission network to be interrupted during the swing. Induced currents in the damper windings (or rotor iron) due to sudden changes in armature currents. To assure stability. I n case (b) it is evident that the machines are separated into two groups where the rotor angles continue to drift apart. The important parameters here are the kinetic energy in M W .4. 3. Note also that the behavior discussed above depends upon the network impedance as well as the machine parameters. The machine output power will be affected by the change in the rotor-winding EMF and the rotor position in addition to any changes in the impedance “seen” by the machine terminals. until the speed changes to the point where it is sensed and corrected by the governor. If these net torques are sufficiently large to cause some of the rotors to swing far enough so that one or more machines “slip a pole. 2. where the rotor angle is measured with respect to a synchronously rotating reference.” synchronism is lost.1 s and are often referred to as “subtransient” effects. This creates another transient.2 Synchronous machine during a transient During a transient the system seen by a synchronous machine causes the machine terminal voltage. 1. the change in the output power will come from the stored energy in the rotating masses. This is illustrated in Figure I . Since the subtransient effects decay very rapidly. . and the system will be stable if it eventually settles to a new angle. In case (a) all the rotor angles increase beyond K radians but all the angle differences are small. This system is unstable. say excess generation. However. The field-winding voltage will be affected by: I . The important factor here is the angle diference between machines. Thus most of the machine rotor angular velocities will increase. which when superimposed on the first may cause synchronism to be lost. which is twice the stored kinetic energy per MVA. A lesser part will be consumed in the loads and through various losses in the system. The time constants for these currents are usually on the order of less than 0. However. Loss of synchronism can also happen in stages. Change in rotor voltage due to change in exciter voltage if activated by changes at the machine terminal. The major portion of the excess energy will be converted into kinetic energy. The impedance seen “looking into” the network at the machine terminal also may change..Power System Stability 9 1. The time constants for this transient are on the order of seconds and are referred to as “transient” effects. Both subtransient and transient effects are observed.s per u n i t MVA (usually called H) or the machine mechanical time constant rj.g. however. the flow of the tie lines may be altered slightly. The number of power system components included in the study and the complexity of their mathematical description will depend upon many factors. The complexity of the model depends upon the type of transient and system being investigated. 1 linearized system equations If the system equations are linear (or have been linearized). Other important components of the power plant that influence the mechanical torque. The loads and their characteristics.5 Methods of Simulation I f we look at a large power system with its numerous machines. Other supplementary controls. we may tend to think it is hopeless to attempt analysis. Fortunately. 6. The parameters of the synchronous machines. 3. I n addition. The dynamics of the transition period. and loads and consider the complexity of the consequences of any impact. the speeds of all machines change so that they are sensed by their speed governors.10 Chapter 1 When the impact is large. The most common method is to . 7. however. Study of the dynamic behavior of the system depends upon the nature of these differential equations. I n general. The key parameters are the governor dynamic characteristics. 2. and after the transient. The responses of these controls are relatively slow and their time constants are on the order of seconds. Machines under load frequency control will correct for the power change. Supplementary controls are provided to these machines. Thus some machines are assigned the requirement of maintaining scheduled flow in the ties. during. The excitation systems of the synchronous machines. Thus the controlled machines are the ones responsible for maintaining the system frequency. The first step in a stability study is to make a mathematical model of the system during the transient. the basic functions of which are to permit each control area to supply a given load. such as tie-line controls. These components are: 1 . each machine's share will depend on its regulation or droop characteristic. the time constants of the phenomena may be appreciably different. The mechanical turbine and speed governor. Until this correction is made. the techniques of linear system analysis are used to study dynamic behavior. are important. lines. 1S . differential equations are used to describe the various components. however. Thus the basic ingredients for solution are the knowledge of the initial conditions of the power system prior to the start of the transient and the mathematical description of the main components of the system that affect the transient behavior of the synchronous machines. 1. This is appropriate since the scheduled economic loading of machines is secondary in importance to stability. The elements included in the model are those affecting the acceleration (or deceleration) of the machine rotors. allowing concentration on the key elements affecting the transient and the area under study. deemed necessary in the mathematical description of the system. 4. the components of the power system that influence the electrical and mechanical torques of the machines should be included in the model. 5 . The network before. Generally. 2 large system with nonlinear equations The system equations for a transient stability study are usually nonlinear. Both linear and nonlinear equations will be developed in following chapters. . and this is the method usually used in power system stability studies. while they seem to offer considerable promise.2 I . The above methods have been frequently used in studies pertaining to small systems or a small number of machines. Power system security. as discussed in Section I . Power system stability.5. The system inputs are represented by the r vector u. Problems I .4 1. Stability of synchronous machines is usually decided by behavior of their rotor angles. This description has the advantage that A may be time varying and u may be used to represent several inputs if necessary. and Routh's criterion. where A is detined by the equation %=Ax+Bu (1. The type of disturbance.I I . What is a tie line'! Is every line a tie line'! What is an impact insofar as power system stability is concerned! Consider the system shown in Figure P1.Power System Stability 11 simulate each component by its transfer function. b. Usually rirrre sohrions of the nonlinear differential equations are obtained by numerical methods with the aid of digital computers.4. are still in the research stage and not in common use. Such efforts.3 I . frequency domain analysis (Nyquist criteria). Power system reliability. More recently. 1. modern theories of stability of nonlinear systems have been applied to the study of power system transients to determine the stability of synchronous machines without obtaining time solutions.5 Suggest detinitions for the following terms: a.2) is a more difficult task than that of the linearized system of ( 1 .1. b. Determining the dynamic behavior of the system described by (1. 1 ) . The various transfer function blocks are connected to represent the system under study. Here the system is described by a large set of coupled nonlinear differential equations of the form 2 = f(X.U. and these inputs are related mathematically to differential equations by an n x r matrix B.1) where x is an n vector denoting the states of the system and A is a coefficient matrix. c. The system performance may then be analyzed by such methods as root-locus plots. Stability characteristics may be determined by examining the eigenvalues of the A matrix.f) ( 1 . For larger systems the state-space model has been used more frequently in connection with system studies described by linear differential equations. The nature of the detining equations.5 where a mass M is pulled by a driving force f ( f )and is restrained by a linear spring K and an ideal dashpot B. Distinguish between steady-state (dynamic) and transient stability according to a .2) where f is an n vector of nonlinear functions. 12 Chapter 1 Write the diferential equation for the system in terms of the displacement variable x and determine the relative values of B and K to provide critical damping when J(r) is a unit step function.6 Repeat Problem I .5 but convert the equations to the state-space form of ( I .5. I . P1. hf(t Fig. I ). . that in which a positive T. accelerates the shaft. the system is unstable. What quantity or signal. will either return to their original state if there is no net change of power or will acquire a new state asymptotically without losing synchronism. = T.' To N . 2. is a decelerating torque. and T. The principal subject of this chapter is the study of stability based largely on machine-angle behavior. would enable us to test for stability? One convenient quantity is the machine rotor angle measured with respect to a synchronously rotating reference.m2 of all rotating masses attached to the shaft. The dot notation is used to signify derivatives with respect to time.1 Swing Equation The swing equation governs the motion of the machine rotor relating the inertia torque to the resultant of the mechanical and electrical torques on the rotor. when perturbed. The angular reference may be chosen relative to a synchronously rotating reference frame moving with I .m (2. 8 is the mechanical angle of the shaft in radians with respect to a fixed reference.. x = dl .m) acting on the shaft. whereas a positive T. is electrical. . If the difference in angle between any two machines increases indefinitely or if the oscillatory transient is not suficiently damped. is the accelerating torque in newton meters (N. namely. N . Le. Usually the perturbation causes a transient that is oscillatory in nature. Thus x=- .) Since the machine is a generator.m (2. (See Kimbark [ l ] for an excellent discussion of units and a dimensional analysis of this equation. preferably electrical. but if the system is stable.T. The question then arises. is mechanical and the retarding or load torque T..2) which establishes a useful sign convention.chapter 2 The Elementary Mathematical Model A stable power system is one in which the synchronous machines. - . the oscillations will be damped. the driving torque T.etc. dr 13 .1) whereJ is the moment of inertia in kg. Thus we write = J8 T. dx d2x . 6.. for example. in (2. shaft angular velocity in rad/s. = 2 Wk/o. which is designated as in ANSI standards ANSI Y 10.7) in terms of an electrical angle that can be conveniently related to the position of the rotor.P. the number of polepairs. Another form of (2. = 377 rad/s.14 Chapter 2 constant angular velocity wR.6. = M. where W. with the result + J6.) (2.4) or in terms of power. To illustrate: for 60 Hz. = (1 /2) J w i joules. we have J w .’ 0 = (wRr + a) + 6. .3) reference for the rotor angle 6. = ( p / 2 ) 6 . and P is in W. is in rad/s.. which is the angle between the field MMF and the resultant MMF in the air gap. Certainly. which certainly varies during a transient. From (2. It is also the electrical angle between the generated EMF and the resultant stator voltage phasors. . and a 1% change in w. Recalling that the product of torque T and angular velocity w is the shaft power P in watts. in Chapter 4 a particular choice of the 7r/2 + 6. . (See Kimbark [ I ] pp. A constant slip of 1% of the value of w. for one second will change the angle of the rotor by 3.8) = pb. w. pp. it would be more useful to write (2. In relating the machine inertial performance to the network. = Jk. w (2. (In Europe the practice is to write 6.3) where a is a constant. 22-27 and Stevenson [2]. gives a = 1r/2 and 6 = W R f we see that 8may be replaced by&. 336-40 for excellent discussions of the inertia constant. The subscript R is used to mean “rated” for all quantities including speed. The torque angle 6. Hence W R = W I in every case. = P. On the other hand the angular frequency does not change by a large percentage before stability is lost.6) It may seem rather strange to call M a constant since it depends upon w . .7) where M is in J-s. J-s (2. wherep is the number of poles.m (2. is the mechanical (subscript r n ) torque angle in rad with respect to a synchronously rotating reference frame. is called the inertia constant and is denoted by M. is related to the rotor mechanical angle 6. Such an angle is the torque angle . and ‘ is the accelerating torque in N. is the shaft & angular velocity in rad/s.77 rad.4) where J is the moment of inertia in kg. is in rad.5. Then M is computed as Angular Momentum = M = J o . both rotating at synchronous speed.m2. The angle a is needed if 6.77 rad/s. 1968. 6.w.5) The quantity Jw. m. is equal to 3. W (2. Mi. (measured:from a synchronously rotating frame) by 6 = 6. The equation of motion of the rotor is called the swing equarion. .. ~ . which is the same as the electrical angle 6.4) that is sometimes useful is obtained by multiplying both sides the by urn. = P. It is given in the literature in the form of (2.P. w. is measured from an axis different from the angular reference frame.l). where p is WI 2.) It is related to the kinetic energy of the rotating masses W . rad (2. = To N. this would lead to loss of synchronism. Then ( 2 . we compute (J*2ni/900wRSB3)b T.)k pa/sB3 = pan pu (2. will usually be derived from a mixture of M K S and English quantities.1 I ) by (2. The form of the swing equation we use must be in M K S units (or pu) but the coefficients. P U (2. which is always understood to be the electrical angle defined by (2.R = 60S~3/2Tn~ (2. we write (2.13) is in rad/s and T is in pu./TB To. Note that w in (2.ft2. The machine nameplate usually gives the rated shaft speed in revolutions per minute (r/min). particularly the moments of inertia.8) we write (2Mlp). N .2 Units It has been the practice in the United States to provide inertial data for rotating machines in English units. but P is now in pu (noted by the subscript u ) .m (2J/p)$ = (2J/p).10) where M .S.9) which relates the accelerating power to the electrical angle 6 and to the angular velocity of the revolving magnetic field w . the moment of inertia. The U. In most problems of interest there will be a large number of equations like (2.8). 7 )and (2.9). 2.12) and substituting 120fR/nR furp. From ( 2 . 9 )becomes a per unit (pu) equation (2M/pSB])i= (ZM/pSB.13). as a quantity usually called W R 2 . We compute the corresponding M KS quantity as Substituting into (2. The consistent English unit for J is slug-ft' o r W R 2 / g where g is the acceleration of gravity (32. 6.13) where we have Substituted the base system radian frequency wR = 2 T f R for the base .1 I ) NOWnormalize this equation by dividing by a base quantity equal to the rated torque at rated speed: TB = SB~/W.given in units of Ibm.p . practice has been to supply J . one for each generator shaft (and motor shaft too if the motor is large enough to warrant detailed representation). In such large systems problems we find it convenient to normalize the power equations by dividing all equations by a common three-phase voltampere base quantity SB]. frequency. We begin with the swing equation in N . and w are in the same units as before.The Elementary Mathematical Model 15 For simplicity we drop the subscript e and write simply 6.12) where SB] is the three-phase V A rating and nR is the rated shaft speed in r/mind Dividing (2.17398 ft/s2).14) The coefficient of 6 can be clarified if we recall the definition of the kinetic energy Of a .T'= ( 2 M / p ) k = Po w (2. = T.m (2. 311525 x IO-'O)(WR*)n~ MJ Then we write the swing equation in the form most useful in practice: ( 2 H / w ~ ) b= T . Values for Hays is vary over a much wider range. Typical values of J (in MJ) are given in Appendix D. and Crary [71. Recognizing that the angular speed w is nearly constant. Curves for estimating H are given in Figures 2. both steam and waterwheel turbines. Sec. and T is in pu.3 Mechanical Torque The mechanical torques of the prime movers for large generators. The quantities taken from these curves must be modified for use in system studies by converting from the machine base VA to the system base VA. It is particularly used with the classical model of the synchronous machine.18) The quantity H is often given for a particular machine normalized to the base VA rating for that machine.14) may be written as ( ~ W ~ I S B ~ W Tau ) P R ~ U We now define the important quantity (2.17) where H is in s.) However we should carefully distinguish between the case of the unregulated machine (not under active governor control) and the regulated (governed) case. Another form of the swing equation. sometimes quoted in the literature.. 27.2.16) where Sg3= rated three-phase MVA of the system Wk = (2. Note also that the final form of the swing equation has been adapted for machines with any number of poles. Vol. are functions of speed. which we can write as Then (2. . 1. the pu accelerating power Pa is numerically nearly equal to the accelerating torque T. w is in rad/s. 11. Sec. This is convenient since these machine-normalized H quantities are usually predictable in size and can be estimated for machines that do not physically exist.19) The value of Hmaeh usually in the range of 1-5 s. (See Venikov [6].1 and 2. pu (2.3.16 Chapter 2 rotating body wk. A modified (and approximate) form of the swing equation becomes ( 2 H / w ~ ) bZ Pa ! PU (2.. involves some approximation. Thus we compute Hsys = Hmich (SB3mach /SB3sys) s (2. 2. With SB38ya = 100 M V A values of Hays from a few tenths of a second (for small generators) to 25-30 s (for large generators) will often be used in the same study.15) H2 wk/s. Note that w is the angular velocity of the revolving magnetic field and is thus related directly to the network voltages and currents. since all machines on the same system synchronize to the same w R . S (2. For this reason it is common to give the units of w as electrical rad/s. ) 2./Dec. (b) expected future large turbogenerators.. (a IEEE. 56.5r 11 I I I I I I 1 0 20 40 100 Genemtor Rating. Feb. when the machine is not under active governor control) the torque speed characteristic is nearly linear over a limited range at rated speed. This can be verified as follows. Nov. From . Reprinted from IEEE Truns. vol.. Reprinted from E/ecrr Eng. vol. PAS-90. MV A (b) Fig. p.3.The Elementary Maihematical Model 17 i 0 - I 100 I 200 (0 ) 1 300 1 400 500 Generator Rating. The value of the turbine torque coefficient suggested by Crary [7] is equal to the loading of the machine in pu. 1201. 2. 2. No distinction seems to be made in the literature between steady-state and transient characteristics in this respect. MVA '"C 4. Figure 2. including allowance of 15% for waterwheels.1 Unregulated machines For a fixed gate or valve position (Le.0 J 3606 r/min fossil G e n a a b r Rating. (o IEEE. I Inertia constants for large steam turbogenerators:(a) turbogenerators rated 500 M V A and below 13. 1971 .the fundamental relationship between the mechanical torque 4. as shown in Figure 2.2 Inertia constants of large vertical-type waterwheel generators. MVA 60 80 120 140 Fig.3(a) shows that the prime-mover speed of a machine operating at a fixed gate or valve position will drop in response to an increase in load.. 1937). .3(a). /w N . . * 0 WR w d mds e .22) (2. 2.m we compute. = -dw PU (2.. and the . with the result dT. = (I/WR)dPm. = P.24) where all values are in pu. 0 and dT. = ..23) If we assume constant mechanical power input. / u . (b) regulated machine. and power P.R/W:)dW N . This relationship is shown in Figure 2.18 Chapter 2 't . 2 3 2 Regulated machines . dP. In regulated machines the speed control mechanism is responsible for controlling the throttle valves to the steam turbine or the gate position in hydroturbines.. using the definition of the differential.( P m R / w : ) d w N.20) (2.m = (2.3(a). T.21) Near rated load (2.m This equation is normalized by dividing through by TmR = P . (2. (b) Fig.(P. T.3 Turbine torque speed characteristic: (a) unregulated machine. L .2 1 ) becomes dT. This occurs under normal operating conditions and during disturbances.. To be stable under normal conditions.The Elementary Mathematical Model 19 mechanical torque is adjusted accordingly.A = . This means that a load pickup from no load (power) to full load (power) would correspond to a speed drop of 5% if the speed load characteristic is assumed to be linear. P.. = Tm0+ T m A . T. where R is the regulation in rad/ N-mes. We also note that the “effective” regulation in a power system could be appreciably different from the value 0.4 where K = SB/SSB The droop characteristic shown in Figure 2.30) Similarly. then the effective pu regulation is given by RucR = R u ( C S B / C S .28) and (2. = pu mechanical power on machine VA base or Since PmA = P.is the sum of the ratings of all machines. if a system base other than that of the machine is used in a stability study. Le.. = T.25) by w R .... It will be shown in Part I11 that without feedback the speed control mechanism is unstable.3(b). Thus T.w A / R .. and is CS.. the change in mechanical power in pu on the system base PmA.31) A block diagram representing (2. the torque speed characteristic of the turbine speed control system should have a “droop characteristic”. IfCSB the sum of the ratings of the machines under governor control.05 in the United States./Ru PU Ru 9 S B R / W : PU (2.A. and T.Pm0. Finally.05 if some of the machines are not under active governor control. B ) (2. = . .28) where the pu regulation Ru is derived from (2. is usually set at 0. .3(b). R .we can write Let P. / S B R = -wA.3(b) is obtained in the speed control system with the help of feedback. a drop in turbine speed should accompany an increase in load. A typical “droop” or “speed regulation” characteristic is 5% in the United States(4x in Europe).31) is shown in Figure 2. The droop (regulation) equation is derived as follows: from Figure 2.. we should point out that the steadystate regulation characteristic determines the ultimate contribution of each machine to a change in load in the power system and fixes the resulting system frequency error.28) or (2.25) Multiplying (2.. . . Such a characteristic is shown in Figure 2..29) As previously mentioned..(w - wR)/R N-m (2. given by is PmAsu = -(SBwAu/ssBRu) Pu (2.w k w A . therefore. the electrical torque is produced by the interaction between the three stator circuits. For the present we simply note that the electrical torque depends upon the flux linking the stator windings and the currents in these windings. This subject will be dealt with in greater detail in Part 111. Under this transformation both currents and flux linkages (and hence voltages) are transformed into two fictitious windings located on axes that are 90’ apart and fixed with respect to the rotor. the field circuit.a Fig. The flux linking each circuit in the machine depends upon the exciter output voltage.4 Block diagram representation of the droop equation. As the rotor moves. and the loads. Expressions for the electrical quantities such as power and torque are developed in terms of the direct and quadrature axis voltages (or flux linkages) and currents. the terminal voltage is determined in part by the external network. and other circuits such as the damper windings. 2. The other axis lies along the magnetic neutral axis and is called the quadrature axis. Whether the machine is operating at synchronous speed or asynchronously affects all the above factors.20 Chapter 2 w I K = S$S. These flux linkage relations are often simplified by using Park’s transformation. Such a detailed discussion will be deferred to Chapter 4. discussion of the electrical torque can become rather involved. One axis coincides with the center of the magnetic poles of the rotor and is called the direct axis. The speed control mechanism for each machine under active governor control will attempt to adjust its output according to its regulation characteristic. the loading of the magnetic circuit (saturation). and the current in the different windings. If the instantaneous values of these flux linkages and currents are known. Two points can be made here: 1. the correct instantaneous value of the electrical torque may be determined. time lags are introduced by the various delays in the feedback elements of the speed control system and in the steam paths.4 Electrical Torque In general. During transients the discrepancy between the mechanical and electrical torques for the various machines results in speed changes. Thus a comprehensive discussion of the electrical torque depends upon the synchronous machine representation. the dynamic response of the turbine could be appreciably different from that indicated by the steady-state regulation characteristics. 2. If all the circuits of the machine are taken into account. . For a particular machine the regulation characteristic for a small (and sudden) change in speed may be considerably different in magnitude from its overall average regulation. 2. Since the three stator circuits are connected to the rest of the system. In attempting to adjust the mechanical torque to correspond to the speed change. A modified form of Park’s transformation will be used here (see Chapter 4). the flux linking each stator winding changes since the inductances between that winding and the rotor circuits are functions of the rotor position. the other machines. 1.(le''). Positive-sequence damping results from the interaction between the positive-sequence air gap flux and the rotor windings. the synchronous torque and a second component that includes all other electrical torques. the machine reactances. It is usually assumed to be proportional to the slip frequency. While these asynchronous torques are usually small in magnitude. In some studies approximate expressions for the torque are used.. The dc braking is produced by the dc component of the armature current during faults. all the above-mentioned components of the electrical torque will be included. / is the current phasor. By dejnirion the phasor f is given by the transformation 6 where 7 /e9 = /(cos B + j sin e) = 6 [ v f / cos (. which may be used for stability studies. e. divides the electrical torque into two main components. we return momentarily to synchronous power to discuss a simplified but very useful expression for the relation between the power output of the machine and the angle of its rotor.The Elementary Mathematical Model 21 A simpler mathematical model. 2.5 Power-Angle Curve of Q Synchronous Machine Before we leave the subject of electrical torque (or power). We explore this concept briefly as an aid to understanding the generator behavior during transients. Since the negative-sequence slip is 2 .4. Negative-sequence braking results from the interaction between the negative-sequence air gap flux during asymmetrical faults and the damper windings. 2. In general. the network configuration affects the value of the terminal voltage. especially after the first swing.4. The most important effects are the following. the rotor angle. and the so-called quadrature axis EMF. 3 . which is nearly the case for small slips. this effect is beneficial since it tends to reduce the magnitude of the machine oscillations. their effect on stability may not be negligible.5(a). which induces currents in t h e rotor winding of fundamental frequency. which may be thought of as an effective rotor E M F that is dependent on the armature and rotor currents and is a function of the exciter response. For example if / is 'the rms value of the current. particularly the damper wihdings. 2. - .g.2 Other electrical torques During a transient. The most important component is associated with the damper windings. A current 3. It is produced by the interaction of the stator windings with the fundamental component of the air gap flux. A phasor is q complex number related to the ? + corresponding time quantity i ( t ) by i ( t ) (Re (\/I le'"') = cos ( W I + 0) = 6 -'. Their interaction produces a torque that is always retarding to the rotor. Also.1 Synchronous torque The synchronous torque is the most important component of the electrical torque. It is dependent upon the machine terminal voltage. 2.ut e)]. Here we usually make an estimate of the components of the torque other than the synchronous torque. It should be emphasized that if the correct expression for the instantaneous electrical torque is used. the torque is always retarding to the rotor. when considering quasi-steady-state conditions. Its magnitude is significant only when the rotor damper winding resistance is high. A phasor is indic_ated with a bar above the symbol for the rms quantity.' Note that the source V i s chosen as the reference. Consider two sources = V e and E = Ekconnected through a reactance x as shown in Figure 2.s. other extraneous electrical torques are developed in a synchronous machine. 1 Classical representation of a synchronous machine in stability studies The EMF of the machine (i. the voltage corresponding to the current in the main field winding) can be considered as having two components: a component E' that corresponds to the flux linking the main field winding and a component that counteracts the armature reaction.5 A simple two-machine system: (a) schematic representation. the relation between P and 6 is a sine curve.5(b). We say approximately because such factors as magnetic circuit saturation and the difference between direct and quadrature axis reluctances are overlooked in this simple representation. V . which is the infinite bus voltage. V . T = IEf l ow s between the two sources.e. as shown in Figure 2.22 Chapter 2 't (a) Fig. Equation (2. but the former (which corresponds to flux linkage) cannot change instantly. not necessarily the excitation voltage. and x are constant.32) Since E.5. But (2. Note that E can be chosen to be any convenient EMF. At steady state the machine can be represented approximately by the above circuit if V is the terminal voltage of the machine.. and the power output of the machine is a function only of the angle 6 associated with E. (b) power-angle curve.32) is essentially correct for a round rotor machine at steady state. We note that the same power is delivered by the source E and received by the source since the network is purely reactive. constant that we may designate as P to write P = P. We can show that the power P i s given by P = (EV/x)sin6 (2. x is the direct axis synchronous reactance: and E is the machine excitation voltage. 2. . sin 6. Consider a round rotor machine connected to an infinite bus.32) indicates that if E. and x are constant. EV/x is a . which is the E M F along the quadrature axis. 2. The latter can change instantaneously because it corresponds to currents. but then the appropriate x and 6 must be defined accordingly. called 6 is 8.0 + j0. Exciters of the conventional type do not usually respond fast enough and their ceilings are not high enough to appreciably alter .this picture. and the machine initially Solution Using Vas reference.13". Furthermore.6) E@ = 1.1 For the circuit of Figure 2.8 & E .8 = = 1.6 . At no load this time constant is o n the order of several seconds. it has been observed that during a disturbance the combined effect of the armature reaction and the excitation system is to help maintain constant flux linkage for a period of a second or two. Example 2. During the transient the magnitude E is held constant.12 + j0.6 let V operating at P = 0. hence other currents will be induced in the various rotor circuits to keep this flux linkage constant.0/-36. The constant voltage source E f i is determined from the initial conditions. .j0. The model used for the synchronous machine is shown in Figure 2. V = = = I. This period is often considered adequate for determining the stability of the machine. but currents will be produced in the armature. = 1 . Thus in some stability studies the assumption is commonly made that the main field flux linkage of a machine is constant. From the above we can see that for a period of less than a second the natural characteristic of the field winding of the synchronous machine tends to maintain constant flux linkage and hence constant E ' .13" The magnitude of E is 1. . ~~ Fig.j0.6 Representation of a synchronous machine by a constant voltage behind transient reactance.2(0. pretransient conditions.2 pu.The Elementary M a t h e m a t i c a l M o d e l 23 When a change in the network occurs suddenly.. This will be held constant during the transient. although 6 may vary. where x.1314/8.8 pu at 0. while under load it is reduced considerably but still on the order of one second or higher. Both the armature and rotor currents will usually have ac and dc components as required to match the ampere-turns of various coupled coils. while the angle 6 is considered as the angle between the rotor position and the terminal voltage V . 2.6.1314.8 PF. The main field-winding flux is almost the same as a fictitious flux that would create an EMF behind the machine direct axis transient reactance. . The flux will decay according to the effective time constant of the field circuit. x.9" = 0.O& 1. the flux linkage (and hence E') will not change. = 0. The initial value of 6. Le.O pu. is the direct axis transient reactance.16 1. Po + PAe P.) = P. + 6. sin 6. 2. + 6. If the control equipment of the machine is slow or inoperative. cos 6.. when perturbed.’’ 2.(sin bo COS 6. The quantity in parentheses in (2. and since Po P.33) = 1. the machine power as a function of the angle 6 is also given by a power-angle curve. Let us assume that the machine can be represented by a constant voltage magnitude behind a constant reactance. It is important that these oscillations should be small in magnitude and should be damped if the system is to be stable in the sense of the definition of stability given in Section 1. The power is given by (2.2.t = P.34) If 6 is small then.)6. I . is used. This is the case when the machine is connected to a very large power system (infinite bus).. COS &)aA (2. This is the same as having a positive synchronizing power coefficient.5.1314/0. system frequencies.36) Equation (2. These oscillations cause fluctuations in bus voltages.) 6. + cos 6. assuming that Vis held constant..37) (Compare this result with dP. There are also cases where coherent groups of machines oscillate with respect to other coherent groups of machines. while in steady-state stability analysis a saturated steady-state reactance x. 6 = 6. Synchronizing power coefficients Consider a synchronous machine the terminal voltage of which is constant. (which is the same as the angle of the EMF E ) . Thus P For theexamplegiven above P. From (2.. a6 (2.e. i. which corresponds to a rotor angle 6.35) PA = (P. . and the voltage E’ are used. . has several modes of oscillation with respect to the rest of the system.6 Natural Frequencies of Oscillation of a Synchronous Machine A synchronous machine. In dynamic studies x.. Let us assume that 6 changes from its .24 Chapter 2 During the transient period.32) P also changes to P = Po + PA... or (2. sin So. From (2.35) is sometimes written in one of the forms PA = P$i.2 = 5. = (2.32).657.6.)sinb = P. = ap -6 .. 1 and sin 6. and tie-line power flows. Let the initial power delivered by the machine be Po. the differential of P.) I n the above analysis the appropriate values of x and E should be used to obtain P. as shown in Figure 2. it is important that the machine be operating such that 0 I 6 5 7r/2 for the operating point to be stable in the static or steady-state sense. sin 6 . This criterion was used in the past to indicate the so-called “steady-state stability limit. cos 6.Then we may write Po + PA = P.2 = (EV/x.35) we also observe that A P.cos6.sin6 (2. approximately. = + (P.35) is defined to be the synchronizing power coeficient and is sometimes designated p. sin (6. initial value 6 by a small amount 6. is approximately equal to the amplitude of the power-angle curve.. (2. P. especially different machine types. Different machines. system frequency in Hz three-phase machine rating in M V A inertia constant in s synchronizing power coefficient in MW/rad Next. But we let 6 = 6.09 HZ If MKS units are used. w.. = 2 H / w . / M . Let the damping be negligible. + 6. For example. - ..where from (2. From (2...39) where f S.of the synchronous machine. . Therefore. is in pu. we write hsc = ( I /2*) drf(p s / s B 3 H. and H is in s..S.. = 6 . H = = = = P. = 4 2 x 377)/(2 x 8 ) = 6. such that $ = iA and P..Pro = 0 since io= 0. We must also be careful with the units.w.85 rad/s f. the different machines in a power system may have somewhat different natural frequencies. I t should be noted that P. Now P. i.38) indicates that the angular frequency of oscillation of the synchronous machine with respect to the rest of the power system is given by d P s S .. H = 8 . From (2.5(a). . + J 2 ) and the angle is al.37) for small aA we write PrA = PSdA. is the synchronizing power coefficient. is the synchronizing power coefficient in pu (on a base of the machine three-phase rating). and a small change in speed is given to the machine (the rotor is given a small twist). = 2 pu. = Pto + Ped. we should point out that a system of two finite machines can be reduced to a single equivalent finite machine against an infinite bus. is constant./M = P. The voltage E remains constant. We now estimate the order of magnitude of this frequency. The equivalent inertia is J l J 2 / ( J . 8 5 / 2 ~= 1..16) we write MIS. or P.The Elementary Mathematical Model 25 In this section we will illustrate the inherent oscillatory nature of a synchronous machine by the following example. a system having P. Let the resulting angle change be aA. w = wo Solut ion From (2. + r u ( t ) . This frequency is usually referred to as the natural frequency./S. Example 2. P. Then the swing equation may be written as + /B s 3 + Ps6A = 0 c elect rad which has the solution of the form a.e.36) P. I f the initial operating angle 6 is small.10) we write M8/SB3+ Pr = P./2H where P. Then ~Uii'~/Ssj P r A = P. w.6) and (2.2 A two-pole synchronous machine is connected to an infinite bus with voltage through a reactance x as in Figure 2.&) = E 6d3ssB3/M sin (2. have different inertia constants. where u ( t ) is a unit step function. is in rad/s. Compute the change in angle as a function of time and determine its frequency of oscillation. is a function of the operating point on the power-angle characteristic.38) Equation (2.... 5 . To obtain a time solution for the rotor angle. requires the following assumptions: 1. 3. The equation of motion of the rotor of the finite machine is given by the swing equation (2. the resulting motion will be oscillatory and with constant amplitude. stability is decided in the first swing.7 O n e m a c h i n e connected t o a n infinite (b) equivalent circuit. it can be represented by a constant impedance (or admittance) to neutral. each coherent group of machines oscillates with respect to other groups of machines. 2. The mechanical power input remains constant during the period of the transient. (If damping is present the amplitude will decrease with time.10).26 Chapter 2 Thus we conclude that each machine oscillates with respect to other machines. The mechanical angle of the synchronous machine rotor coincides with the electrical phase angle of the voltage behind transient reactance. The synchronous machine can be represented (electrically) by a constant voltage source behind a transient reactance (see Section 2. Fig. A major bus of a power system of very large capacity compared to the rating of the machine under consideration is approximately an infinite bus.5. Consider a power system consisting of one machine connected to an infinite bus through a transmission line. At the start of the transient. A schematic representation of this system is shown in Figure 2. which will be referred to as the classical model. 2. we need to develop expressions for the mechanical and the electrical powers. 4. The inertia of the machines in a large system will make the bus voltage of many high-voltage buses essentially constant for transients occurring outside that system. This model.7(a). Damping or asynchronous power is negligible. the machine loses synchronism and stability is lost. Thus according to this model and the assumptions used. and so on. If it reaches a maximum and then starts to decrease.) .7 System of One Machine against an Infinite Bus-The Classical Model A n infinite bus is a source of invariable frequency and voltage (both in magnitude and angle). The frequencies of oscillations depend on the synchronizing power coefficients and on the inertia constants. and assuming that the impact initiating the transient creates a positive accelerating power on the machine rotor. If the rotor angle increases indefinitely. In this section the simplest mathematical model is used. The period of interest is the first swing of the rotor angle 6 and is usually on the order of one second or less. I). the rotor angle increases.2. If a local load is fed at the terminal voltage of the machine. bus through a transmission line: (a) one-line diagram.7) or (2. but in the classical model there is very little damping. From elementary network theory we can show that the power at node 1 is given by PI = &eEi:or P.7(b).41) is shown in Figure 2.The Elementary M a t h e m a t i c a l M o d e l 27 .33).41) The relation between PI and 6 in (2.. Also shown in Figure 2. The equivalent electrical circuit for the system is given in Figure 2.9. = z. = x.. then Pi = E2Gll + EVYI2sin(6 . y I 2 . The two-port network of Figure 2. and y OI2 .8 is conveniently described by the equation The driving point admittance at node 1 is given by K l = Yil /811 = plz + jjlo where we use lower case y's to indicate actual admittances and capital Y's for matrix elements.8 as nodes I and 2 respectively.and y z o ) . I n Figure 2. which represents the power dissipation in the equivalent network.9. which is determined by the real component of the transfer admittance F2. Note that while three admittance elements are obtained (viz. which is used as reference direct axis transient reactance of the machine series impedance of the transmission network (including transformers) equivalent shunt impedance at the machine terminal. In the special case where the shunt load at the machine terminal is open and where the transmission network is reactive.7 can be eliminated.y) (2. the node representing the terminal voltage E in Figure 2. and horizontally by the angle y. 2.ylo. These are shown in Figure 2.y ) = Pc + PMsin(6 . we can easily prove that Pc = 0 and y = 0.7 we define V = V. . The negative of the transfer admittance vlz between nodes I and 2 defines the admittance = matrix element ( I . we note that the power-angle curve of a synchronous machine connected to an infinite bus is a sine curve displaced from the origin vertically by an amount Pc. Examining Figure 2. 2) or F12 Y12/812 = -yi2. ' 0 EA 0 Fig. The nodes to be retained (in addition to the reference node) are the internal voltage behind the transient reactance node and the infinite bus.8 Equivalent circuit for a system o f one machine against an infinite bus. y z o omitted since it is not needed is in the analysis. In this case the power-angle curve becomes identical to that given in (2.co seII EVYl2cos(eI2 6) + = Now define G I I = YII cosB. including local loads if any By using a Y-A transformation. V&) - = = z.H/2. = terminal voltage of the synchronous machine voltage of the infinite bus. b = EZYl.8 are the admittances obtained by the network reduction. 5 = -j2. Example 2.y) = EVY12 6 = 2Esin sin = 6 Since the initial power is Pco= 0. The inertia constant H = 5 MJ/MVA.1 I initial equivalent circuit of the system of Example 2. the machine is delivering 0.20 pu.0 e.9 Power output of a synchronous machine connected to an infinite bus. The equation of motion of the machine rotor is to be determined.1 1.40 pu. the transformer reactance is 0. Initially.10 pu. 2. All resistances are neglected. 0.10. then E sin 6.0 YlO = V = I .3.3 A synchronous machine is connected to an infinite bus through a transformer and a double circuit transmission line.05 pu.4. O L Fig.8 pu. Pc = 0 and y = 0.10 System of Example 2.1 = -a/2 */2 eI2 = therefore. For this system: - 0 Y12= j2. o== +=E ELL j T l 2 = l/j0. The electrical power is given by P = PI . Solution The equivalent circuit of the system is shown in Figure 2.0 pu. and the reactance of each of the transmission lines is 0. 2.3. . as shown in Figure 2. Fig. The direct axis transient reactance of the machine is 0. 2.28 Chapter 2 Fig. all to a base of the rating of the synchronous machine.8 pu power with a terminal voltage of 1.0 Y I I= -j2. The infinite bus voltage V = 1. = Pc + E V Y I 2sin (6 . . 8 = ( V V .2/90") = = 1.09" = 0. Pro = 0.where admittances are used for convenience.29")(0.0)/0.o = 13.12.0148 + j0.1 1 1 x 1.222sin6) 1 0 rad/s2 From this simple example we observe that the resulting swing equation is nonlinear and will be difficult to solve except by numerical methods.05/13.803/-5. = (( 1.222sin 6 Then the swing equation is given by or -= d26 dt2 377 -(0.909.lII)sin6 = 1.400 = 1.OOO)/j0.30) Bf0 sin sinB. / x )sin OrO = (1. The network 0l now is as shown in Figure 2.05/0.. = 0. Example 2.09" pu Thus E = 1.j0.21"+ (0. Solution By Y-A transformation we compute pI2= -j[(3. We also may write P. (0. The electrical power output = of the machine P.3 (1.367 rad.240 + 0.5 = 0. = I. We have the terminal condition - V = I.O.3= 0.8 .or T = ( q .803/-5.50] sin 6 = 2.O/o PU V.1 1.The Elementary Mathematical Model 29 To find the initial conditions.022 + j0.010sin6 .909 x 1. to neutral. = is now .&l)/j0. and the initial angel is 6o = 21.800 .2l0 I.4 Develop the equation of motion of the system of Figure 21 1 where a fault is applied .29" Then the internal machine voltage is EE= l. we write.OSF.240 .8/3.21" The current is found from = = zi + v.1 I I is a constant that will be unchanged during the transient.1 I 1 /21.160 1. at the sending end (node 4) of the transmission line.O5/13. since resistance is zero.333 x 5)/18.2286 e.909 and since Y.022 + j0. For simplicity we will consider a three-phase fault that presents a balanced impedance of j .V ) / z = (l.333]= -j0. then K2 = j0.074 = 0. PU P..037 + j0.J I 2 . we solve the network of Figure 2.2.I. = 0.8 PU To find the angle of F. We now extend the example to consider a fault on the system. 7(0. At the end of this repeated prediction and correction a final value of S(t + A t ) and w(t + A t ) is obtained.429 pu. From Example 2. 2.45 rad/s2 Now let us assume that after some time the circuit breaker at the sending end of the faulted line clears the fault by opening that line. First. therefore. and the new network (with fault cleared) will have a new value of transfer admittance. Solution The equations for 6 were obtained in Example 2.1.3 and 2. From Example 2.010 x 0.368.4 for the faulted network and for the system with the fault cleared.15 s ) .1. From the calculated values of 6 ( t + A t ) and w(t + At). Tl2= j 1.OIOsin6) rad/s2 = At the start of the transient sin 6o 0. A partial survey of these methods is given in Appendix B. Assume that the fault is cleared in nine cycles (0.8 .3 the equation of motion of the rotor is dt = 37.36. the modified Euler method is used in this example. Le. These are called the predicted values of the variables and are based only on the values of 6 ( t ) . This method is outlined in Appendix B. an estimate is made of the values of these variables at the end of an interval of time A t . The new swing equation will be d26 -= dt 37. and their derivatives.587 sin 6) rad/s2 Example 2. I2 Faulted network for Example 2.4. The process is then repeated for the next interval.8 .70 pu. To illustrate the procedure used in numerical integration. time solutions will be obtained by numerical methods. The procedure is outlined in detail in Chapter IO of [8].4 the initial value of 6 is sin-’0.7(0. and the initial rotor acceleration is given by = dtz = 37.368)] 16. at time t + At. w ( t ) .7[0.P e ( f ) ] (2.30 Chapter 2 4 Fig.5 Calculate the angle d as a function of time for the system of Examples 2.(1.8 . The network now will have a series reactance ofj0. The process can be repeated until a desired precision is achieved. the swing equation is replaced by the two first-order differential equations: 8 E o(t) . and the equation .42) The time domain is divided into increments called At. values of the derivatives at t + A t are calculated. A corrected value of 6 ( t + A t ) and w(t + A t ) is obtained using the mean derivative over the interval. With the values of 6 and w and their derivatives known at some time t .wR & = (wR/2H)[Pm .4 in terms of admittances. These equations are nonlinear.. d26 -H .1.4 0.15 t 2 0.15 The results of the numerical integration of the system equations.43) where Pa is the accelerating power. are shown in Figure 2.7(0.82 s.8 Equal Area Criterion Consider the swing equation for a machine connected to an infinite bus derived previously in the form 2 . The first peak of 48. WR dt' . I 1.800 .7(0.2 I I I I I 0.43) d'6 -= dt2 wR 2H (2.13. 2.The Elementary Mathematical M o d e l n 31 I 0 II 0.800 .2" at t = 0.2 1 7 Fig. for w is given by w = 37. I3 Angle-time curve for Example 2.6 0. performed with the aid of a digital computer.1.5. and the oscillation of the rotor angle 6 continues. From (2.38 s.587sin6) 0 =( t < 0. For the system under study and for the given impact.. The time solution is carried o u t for two successive peaks of the angle 6.P .010sin6) = 37.44) p a . synchronism is not lost (since the angle 6 does not increase indefinitely) and the synchronous machine is stable. = p* P U (2.2" is reached at t = 0. after which 6 is decreased until it reaches a minimum value of about 13.0 1.8 Time.P. 2. .50) can be inter.48) or d6 -= dt 66 Padb Pad6)'" (2.) 5 0. and = 0 (2.46) (2. preted as the area under that curve between & and &. the condition of stability is that a value ...49) gives the relative speed of the machine with respect to a reference frame moving at constant speed (by the definition of the angle 6 ) . For stability this speed must be zero when the acceleration is either zero or is opposing the rotor motion. Thus for a rotor that is accelerating.49) Equation (2.14 Equal area criteria: (a) for stability for a stable system. 2. (b) for an unstable system .. . equation (2. (2..32 Chapter 2 Multiplying each side by 2 ( d s / d t ) .47) Integrating both sides.50) If the accelerating power is plotted as a function of 6.a exists such that Pa(&. (2.45) (2. This is shown in Fig- pa t Pa (t = O+) b) Fig. . 15.. = ratio of the peak of the power-angle curve of the faulted network to PM r. To illustrate the critical clearing angle.8. Then for A .. and in many studies P. 2. is the maximum rotor angle reached during the swing.a z 48". .14(b)./r2PM > */2 = A. < */2 6./P. The conditions for A. The former is the power-angle curve discussed in Section 2. and for critical clearing.80) + r2cOs8. is negative. This situation is shown in Figure 2.14(a) where the net area under the Pa versus 6 curve adds to zero at the angle since the two areas A I and A .8. is added to A . 6.15. and hence the rotor acceleration. I f the accelerating power reverses sign before the two areas A . Therefore. corresponds to the angle 6.. The faulted power-angle curve has zero amplitude.2 Application to a one-machine system The equal area criterion is applied to the power network of Examples 2. = ratio of the peak of the power-angle curve of the network with the fault cleared to PM 6. are equal. Conditions for critical clearing are now obtained (see [ I ] and [2]). Let P M = peak of the prefault power-angle curve r. For this case.5.4-2. coincides with the angle 6 on the power-angle curve with the fault cleared such that P = P . Also at. are equal and opposite. = cOs-'{[1/(r2 ....rl)I[(Pm/PM)(& . is smaller than A .51) Note that the corresponding clearing time must be obtained from a time solution of the swing equation. and the like. and as 6 increases beyond the value where Pa reverses sign again. The angle at t = 0 is 21. This corresponds to the maximum angle obtained in the time solution z shown in Figure 2. is such that = 0 and the areas A . A three-phase fault is applied to the same bus with zero impedance. = sin-' P. . = sin-' P.r1cosbOl) (2.4-2. on the faultcleared power-angle curve. 2. The accelerating power curve could have discontinuities due to switching of the network. Note that the accelerating power need not be plotted as a function of 6.. The prefault and . and A .09" and is indicated by the intersection of P. the critical clearing angle is that switching angle for which the system is at the edge of instability (we will also show that this applies to any twomachine system). = A . The clearing angle 6.a . the area A. We can obtain the same information if the electrical and mechanical powers are plotted as a function of 6. is a constant.1 Critical clearing angle For a system of one machine connected to an infinite bus and for a given fault and switching arrangement. and the results are shown in Figure 2. are equal. synchronism is lost. and A..13.7. The limit of stability occurs when the angle 6. The area A .6"..The Elementary Mathematical Model 33 ure 2. correspond to .a the accelerating power. initiation of faults. a more severe fault is used with the same system and switching arrangement..13) and is about 31. and . The stable system of Examples 2. with the prefault curve. .5 is illustrated in Figure 2. the system is stable and 6. The maximum angle b.. 6 > */2. is obtained from the time solution (see Figure 2. r. = = 0 1.714 6. ' A b Fig. gives = ~ 0 ~ . . 2 6 8 4= 74. 21. 6 .15 Application of the equal area criterion to a stable system.43" 8 This situation is illustrated in Figure 2.73" Calculation of the critical clearing angle.16. using (2.222 = 0.34 Chapter 2 Fig. 2. For this system r.16 Application of the equal area criterion to a critically cleared system. 2.58712.09" 149. . postfault networks are the same as before.5I ) . = = a.' 0 . suggesting load representation by a constant impedance.dSI2= I Ho 6120 0 where 2. 4. The mechanical rotor angle of a machine coincides with the angle of the voltage behind the transient reactance. 2. = H IH 2 / ( H . Classical Model of a Multimachine System The same assumptions used for a system of one machine connected to an infinite bus are often assumed valid for a multimachine system: I .The Elementary Mathematical Model 35 2. 3. . 1 I . Loads have their own dynamic behavior. Passive impedances connect the various nodes and connect the nodes to the reference at load . 131. Thus a load-flow study for pretransient . .6.. are debuses. As in the one-machine system. I n the special case where the resistance is neglected. Values of the damping coefficient usually used in stability studies are in the range of 1-3 pu [9. This model is useful for stability analysis but is limited to the study of transients for only the “first swing” or for periods on the order of one second. . IO.3 Equal area criterion for a two-machine system It can be shown that the equal area criterion applies to any two-machine system since a two-machine system can be reduced to an equivalent system of one machine connected to an infinite bus (see Problem 2. are reported in the literature due to generator damping alone [7. Mechanical power input is constant.. (2. . 5..52) becomes . and the damping effect of electrical loads.. Loads are represented by passive impedances. the initial values of E.8. E . up to 25 pu. both mechanical and electrical. which is usually not precisely known and varies from constant impedance to constant MVA.17.n are the internal machine buses. . Assumption 2 is improved upon somewhat by assuming a linear damping characteristic. This is a subject of considerable speculation. However. Damping or asynchronous power is negligible. This represents turbine damping. Assumption 5 . Load representation can have a marked effect on stability results.J b 1 2 P. We can show that the expression for the equal area criterion in this case is given by (2. The electrical network obtained for an n-machine system is as shown in Figure 2. . much larger damping coefficients. 121. generator electrical damping. = 6. Constant-voltage-behind-transient-reactance model for the synchronous machines is valid. The damping coefficient D includes the various damping torque components. is made for convenience in many classical studies. .52) where a. . Nodes 1.9 H. A damping torque (or power) Dw is frequently added to the inertial torque (or power) in the swing equation. termined from the pretransient conditions. or the buses to which the voltages behind transient reactances are applied.2. + H 2 ) . the major point of agreement being that constant impedance is an inadequate representation. Node 0 is the reference node (neutral).14). = = driving point admittance for node i G.~ ~ ~ co s(o .1 Transmission system r constunt impedance loads . 13. n = EfG. . 2. yii = yi... + EiEj(B.By definition. + j B. 0 Representation of a multimachine system (classical model)....n (2.!?.+ . E... 2 . The admittance matrix of the n-port network. looking into the network from the terminals of the generators..sin(6. P. + 1 3 ~ ) j. 2.. E ~ G . j. I 7 Node. which is the electrical power output of machine i .36 Chapter 2 n -machine system n generators . . C E .I j#i - 13. is given by = (Re.. (2. r .55) . Y . conditions is needed.. i = I .54) The power into the network at node i. + j B... n are held constant during the transient in classical stability studies.. Y i i b = negative of the transfer admittance between nodes i and j = = G.2.53) E. The passive electrical network described above has n nodes with active sources.. is defined by where y has the diagonal elements I=VE (2. .I j#i N i = 1 .) + GVcos(bi..and the off-diagonal elements qj. The magnitudes E.p e.aj)] i = 1. + = .. 0 L-- I I I I n d n +jx' -' I Fig. .t) (2. A classical study will be presented here on a small nine-bus power system that has three generators and three loads.58) where x is a vector of dimension (2n x I ) ..10 Classical Stability Study of a Nine-bus System The classical model of a synchronous machine may be used to study the stability of a power system for a period of time during which the system dynamic response is dependent largely on the stored kinetic energy in the rotating masses.56) I t should be noted that prior to the disturbance ( t n 0-) Pmio P.xo. Generator data for the three machines are given in Table 2. these studies can provide useful information. hence. The equivalent impedances of the loads are obtained from the load bus data.a (2.The Elementary Mathematical Model 37 The equations of motion are then given by 2Hi dwi -wR di j#i E ... since the network changes due to switching during the fault.E. of the generators and to calculate the values of Ei&for all the generators.56) is a set of n-coupled nonlinear second-order differential equations.18.n = 1 (2. This system. These can be written in the form x = f(x. 2. For example. Furthermore. Yijocos (eijo j. E ~ K ~ C O S ( 6.~ i = aj) l . such studies can be conducted in a relatively short time and at minimum cost.. The classical model is the simplest model used in studies of power system dynamics and requires a minimum amount of data. is large enough to be nontrivial and thus permits the illustration of a number of stability concepts and results.19.1 Data preparation In the performance of a transient stability study. while small. A load-flow study of the pretransient network to determine the mechanical power P.1. 2 . they may be used as preliminary studies to identify problem areas that require further study with more detailed modeling. = . A one-line impedance diagram for the system is given in Figure 2. The prefault normal load-flow solution is given in Figure 2. For many power systems this time is on the order of one second or less.I j#i + ). + . Thus a large number of cases for which the system exhibits a definitely stable dynamic response to the disturbances under study are eliminated from further consideration. + €. . 2. the following data are needed: I .. Pmio= E: G.57) The subscript 0 is used to indicate the pretransient conditions..10.. The set of equations (2. and f is a set of nonlinear functions of the elements of the state vector x. Si. This applies to all machine rotor angles and also to the network parameters. 2 -850 $85. + ? 11 56 230 kV 9 7 % LaadB I 8 OS " 2 16.01 (3.70 1. U Y 1.0)-Load C -75.072 v 2 = j0.032 m1.025 13.0 13.7) (-24.013 Fig.1008 13.38 Chapter 2 18 kV 230 kV 0.8 kV 230 kV 24.0119 + jO.0) (15.3) 100. 2. .0745 23OkV 0.0085 ij0.0 (-10.19 Nine-bus system load-flow diagram showing prefault conditions.9 -24.5 k -V-@ E Q Fig. 18kV 230kV (35.0)$ I 85.18 Nine-bus system impedance diagram: all impedances are in pu on a 100-MVAbase.1 (-10.70? ? .8 kV c @ s/2 = j0. all flows are in M W and MVAR.026 0 /4.1645 L 2 II :s h d A 2s 4 3 @ $ 8 0.9) 1. 2. 0608 0.20. = v.89 0.0969 0. All system data are converted to a common base.00 0. 2. b.o 192. + jQ.1813 1.) 2..s 31 MW-S 0 Note: Reactance values are in pu on a 100-MVA base.5 16.60) 3.0521 6.) The equivalent shunt admittance at that bus is given by = PL/VZ .5 1 .0742 5.. reactive power Q. But since E E = jxjK we compute r+ vr* E@' = (V + Qxj/V) + j ( P x i / V ) (2. All time constants are in s. a system base of 100 M V A is frequently used.0 18..- jf?. I969 0. 4I X xt(leakage) 140 Stored energy at rated speed 710 hydro 180 r/min 0.)] = VZ(G. time of switchings. the following preliminary calculations are made: 1. 3.jS.535 640 M W . Let the terminal voltage be used temporarily as a reference. The internal voltages of the generators E. and current & flowing into a load admittance FL = G. System data as follows: a. = ( P .600 2364 M W .0 0. + jI. 0.61) The initial generator angle So is then obtained by adding the pretransient voltage .. Thus if a certain load bus has a voltage F. + jI.2 Preliminary calculations To prepare the system data for a stability study.10.The Elementary Mathematical Model Table 2.8645 0.& are calculated from the load-flow data.1.25 0. 2.85 steam x.8958 0 1198 . These internal angles may be computed from the pretransient terminal voltages V k as follows.85 steam 128. (Several quantities are tabulated that are as yet undefined in this book. Transmission network impedances for the initial network conditions and the subsequent switchings such as fault clearing and breaker reclosings. The needed data for this step are obtained from the load-flow study. Generator Rated M V A kV Power factor Type Speed xd 39 Generator Data 1 2 3 247. then from the relation P + j Q = we have I.0336 8.8 0. If we define 7 = I. + jSL.@ YL = V L ( C ( G L- .jQ)/V. as shown in Figure 2. The loads are converted to equivalent impedances or admittances.3125 0.j(QL/W (2. then P. power P. These quantities are derived and justified in Chapter 4 but are given here to provide complete data for the sample system.0969 0.96 0 3600 r/min 0.s 3600 r/min 1. and the maximum time for which a solution is to be obtained.0 13. .1460 0. The inertia constant H and direct axis transient reactance x j for all generators. The type and location of disturbance.2578 0. V. + Y. or 6 = 6' .62) V matrix for each network condition is calculated. and node i and node j. . . The equivalent load impedances (or admittances) are connected between the load buses and the reference node. The + ff (2.17. 2. c.64) Now the matrices Y and V are partitioned accordingly to get where the subscript n is used to denote generator nodes and the subscript r is used for the remaining nodes. Thus for the network in Figure 2.V. additional nodes are provided for the internal generator voltages (nodes 1. = Y. angle CY to d'. The reduction can be achieved by matrix operation if we recall that all the nodes have zero injection currents except for the internal generator nodes. b. Also. we eliminate all the nodes except for the internal generator nodes and obtain the k matrix for the reduced network. . simulation of the fault impedance is added as required. 4.[ (2. has the dimension (n x 1) and V. Expanding (2. Finally. 2.V. 5 . This property is used to obtain the network reduction as shown below.20 Generator representation for computing 60. has the dimension ( r x 1). .17) and the appropriate values of x i are connected between these nodes and the generator terminal nodes. + Y. 0 = Y. I.40 Chapter 2 + E& Fig..63) = I. v xj Let I where 1 = YV (2.64). The following steps are usually needed: a . All impedance elements are converted to admittances..V. and the admittance matrix is determined for each switching condition. Elements of the matrix are identified as follows: ITi is the sum of all the adis the negative of the admittance between mittances connected to node i. V. n in Figure 2. . The network reduction illustrated by (2.YnrY. Solution The objective of the study is to obtain time solutions for the rotor angles of the generators after the transient is introduced. 2. obtained from (2..65) The matrix (Ynm Y.10.083 s) by opening line 5-7. All impedance data are given to this base.The Elementary Mathematical Model 41 from which we eliminate V. The system base is 100 M V A . The disturbance initiating the transient is a three-phase fault occurring near bus 7 at the end of line 5-7. Values for the generator x i are added to the reactance of the generator transformers.7315" E3/6. The damping torques are neglected.' Y.2. The equivalent shunt admittances for the loads are given in pu as load A: j j L s = 1.'Yrn)Vn (2.5044 load B: pL6= 0.. we are to obtain a solution for the set of equations (2. the identity of the load buses must be retained. step 4. to find I n = (Ynn .2926 load C: pLB = 0. If the loads are not considered to be constant impedances. . the reactance between .57). The initial conditions.1752" The matrix is obtained as outlined in Section 2. For example.0566/2.10. mathematically. and 3 are used to denote the generator internal buses rather than the generator low-voltage terminal buses.2) are: The system base is chosen to be 100 M V A . Example 2. Therefore. the data for which is given in Figures 2. Network reduction can be applied only to those nodes that have zero injection current. = l. For convenience bus numbers I .63)-(2. It has the dimensions (n x n) where n is the number of the generators. Y.8777 .56) are known.2717" E2& = 1. are given by & = 0 and 6.j0." In the classical model the angles of the generator internal voltages behind transient reactances are assumed to correspond to the rotor angles.18 and 2.19 and Table 2.9690 .j0. The fault is cleared in five cycles (0.0502/19. These time solutions are called "swing curves. Make all the preliminary calculations needed for a transient stability study so that all coefficients in (2.2610 . denoted by adding the subscript 0.3391 The generator internal voltages and their initial angles are given in pu by E l k o = 1.65) is a convenient analytical technique that can be used only when the loads are treated as constant impedances. Preliminary calculations (following the steps outlined in Section 2.6 The technique of solving a classical transient stability problem is illustrated by conducting a study of the nine-bus system. for generator 2 bus 2 will be the internal bus for the voltage behind transient reactance.O170/13.56).n) is the desired reduced matrix Y.j0.1. For the purpose of this study the generators are to be represented by the classical model and the loads by constant impedances. 3.0850 0.9690 .0100 0.6. All the coefficients for the faulted network and the network with the fault cleared have been determined in Example 2.1 184 G B Generators* No.8777 0.1823 0. Chapter 2 Prefault Network impedance Admittance Bus no. Plot the angles a.6171 1.0085 0. Example 2.5107 -5.1876 I . A brief survey of numerical integration of differential equations is given in Appendix B.0320 0. 1 No. tThe line shunt susceptances are added to the loads.6980 -9. I700 0.1551 1.0625).1008 0 0 0 -8.56) are available.2399 0. S o ht ion The problem is to solve the set of equations (2.4 and 2.3652 I .01 19 1. The results are shown in Tables 2. We now have the values of the constant voltages behind transient reactances for all three generators and the reduced Y matrix for each network.0720 0. and ti3 are obtained by numerical integration. The resulting reduced Y matrices are shown in Table 2.56) for n = 3 and D = 0.1670 0.604 I . and the corresponding k matrix is given in Table 2.5 respectively.7843 -0. and 4 and their difference versus time.1 198 + 0.10.2634 -0.2820 1.0390 0.5882 .6 calculate the rotor angles versus time.56) is nonlinear.18. R X 0.42 Table 2. the desired time solutions for 6. The fault is cleared in five cycles by opening line 5-7 of Figure 2.9422 1. 2 No.9751 -5. The P matrix for the faulted network and for the network with the fault cleared are similarly obtained. (For hand calculations see [ I ] for an excellent discussion of a numerical integration method of the swing equa- . bus 2 and bus 7 is the sum of the generator and transformer reactances (0.0346 -0. The prefault network admittances including the load equivalents are given in Table 2.0 170 0.1684 0.2. 6. Since the set (2.4459 -5.4855 -4.2275 0.1601 0.I I .1610 0.. 5..13.0920 0.6 for the prefault network.2610 0.2.. Thus all coefficients of (2. Elimination of the network nodes other than the generator internal nodes by network reduction as outlined in step 5 is done by digital computer.7 For the system and the transient of Example 2. a2. and the network with the fault cleared respectively. the faulted network. 3 Transmission lines 1-4 2-7 3-9 4-5 4-6 5-7 6-9 7-8 8-9 0 0 0 0.2835 S h u n t admittancest Load A Load B Load C 5-0 6-0 8-0 4-0 7-0 9-0 ~ ~~ - ~ *For each generator the transformer reactance is added to the generator x i . 7843 j4. 1876 + j5.j19.5882 2.I .5107 + jl0.Table 2.2820 + j5.4855 8 9 9 -1.4.1.8426 . Y Matrix of Network with Fault Cleared 5 .1019 30.5 107 .2820 + j5.1.9422 + j10.j16.I .6171 + j13.6171 +j13.7843 2.1335 -1.5882 .1.6041 2.5107 4.1335 .5882 .4371 .6424 -1.7843 4.6980 -1.1.5.8047 .j18. I684 .6171 + j13.1335 3.I .3652 + jl1.3652 + j I I .6041 0.5882 1.1551 +j9.4855 7 8 9 9 0.6980 3.9559 -1.5882 Table 2.7843 .1684 + j5.6171 .j19.4371 .6041 0.2574 .j16.2820 + j5.9751 .2820 + j5.1019 .8675 -1.1.2574 Table 2.6041 10.j24.9422 + j 10.6424 -1.1551 + j9.7412 .5107 .1876 j4.3937 1.8138-j17.1551 + j9.7843 2.6424 -1.975 I .1019 .1551 + j9. I684 .2574 6 7 j5.1. Y Matrix of Faulted Network 5 6 7 8 9 0.4371 .jl6.1.5882 -1.8426 .1551 + j9.j23.9311 .6980 3.7412 .6171 + j13.3.j23. Y Matrix of Prefault Network 5 6 j5.5107 .3937 1.9422 4.6041 3.3652 + jl1.1.6262 .6980 -1.5107 j4.j23.7412 .2820 + j5.604I 3.I .j19.7843 2.8138 -j17.1.1551 + j9.3937 1.j I I .I .2820 + j5. 5 1 . = 6. .191 + j1.229 0.21 shows the rotor angles of the three machines. Type of network Reduced Y Matrices I Node 2 3 Prefault Faulted Fault cleared I 2 3 I 2 3 1 2 3 0.277 .088 O.6.199 + j1. . Figure 2.486 0.j2.199 + j1.070 + j0.000 + jO.000 0.j3.j2.287 + j1.j2.000 + jO..631 1.j5.796 0.) The so-called transient stability digital computer programs available at many computer centers include subroutines for solving nonlinear differential equations.000 0.816 0.000 0.o TIrne.21 Plot of 61.988 0.079 0.229 0.726 0.OO0 + jO.2 10 + j I .000 .138 + j0.62.213 + jl. I 1 2.6.631 0.138 + j0.726 0.088 0..181 .6.213 + j1.j2. Discussion of these programs is beyond the scope of this book. nine-bus system is made by digital computer for 2.j2. Numerical integration of the swing equations for the three-generator.226 0.j2. and b.368 0.273 .44 Chapter 2 Table 2. = 6. 2.342 tion.j1.174 .210 + j1.657 .724 0.287 + j1.000 + jO.000 0. A plot ofd. .229 0.0 1.513 0.079 0. is shown 0 L cycln I I I 0.5 Fig.513 0.070 + j0.420 .846 .953 0. Also see Chapter IO of [8] for a more detailed discussion of several numerical schemes for solving the swing equation.and 63 versus time.389 .0 s of simulated real time.226 0.191 + j1. 2. The maximum angle difference is about 8 5 ” . Furthermore.5 Fig. If the rotor angles (or the angle differences) reach maximum values and then decrease.o nm4 1. the equipment used for excitation controls was relatively slow and simple.. Today large system interconnections with the greater system inertias and relatively weaker ties result in longer periods of oscillations during transients. To determine whether the system is stable or unstable for the pareither ticular transient under study. particularly modern excitation systems. the system is unstable because at least one machine will lose synchronism. on the supplementary control equipment installed. on the dynamic characteristics of the loads. The machine dynamic response to any impact in the system is oscillatory.5 60 cyclr I eo I I 1 100 120 I 1 0 I 2.43 s. 2. out for two “swings” to show that the second swing is not greater than the first for or &. This is the value of 6 . are extremely fast.22 Plot of 6 differences versus time. Generator control systems. it is sufficient to carry out the time solution for one swing only. Note that the solution is carried . and on the type and settings of protective equipment used. in Figure 2.22 where we can see that the system is stable. This includes the response characteristics of the control equipment on the turbogenerators. If any of the angle differences increase indefinitely. the system is stable’. Thus the classical model was adequate. In the past the sizes of the power systems involved were such that the period of these oscillations was not much greater than one second..0 1 . It is therefore .The Elementary Mathematical Model 45 0 I ’ I 1 I 20 I 40 0. at t = 0.1 1 Shortcomings of the Classical Model System stability depends on the characteristics of all the components of the power system. the longer period that must now be considered and the speed of many modern voltage regulators. Furthermore.46 Chapter 2 questionable whether the effect of the control equipment can be neglected during these longer periods. This assumption is suspect on two counts. it is unrealistic to assume that the mechanical power will not change. Representing loads by constant passive impedance. etc.) 4. with the result that some of these frequencies are quite low (frequencies of periods in the order of 5-6 s are not uncommon). Constant generator mainfield-windingflux linkage. We conclude from this discussion that the constant voltage behind transient reactance could be very inaccurate. In conjunction with this we must: 1. This view is well stated by Ray and Shipley [ 14): We have reached a time when it is appropriate that we appraise the state of the Art of Dynamic Stability Analysis. A large system will have relatively weak ties. If periods on the order of a few seconds or greater are of interest. Another aspect is the dynamic instability problem. Update our knowledge of the response characteristics of the various components of energy systems and their controls (boilers. especially in loss of generation studies [8]. 3. Re-examine old concepts and develop new ideas on changes in system networks to improve system stability. reactors. 2 . 4. The longer period. 5 . It is quite possible that the worst swing may occur at an instant in time when the peaks of some of these nodes coincide. generator regulators. the voltage regulator response could have a significant effect on the field-winding flux. it has also become increasingly important to ensure the security of the bulk power supply. A large system having many machines will have numerous natural frequencies of oscillations. It is therefore necessary in many cases to study the transient for a period longer than one second. Let us now make a critical appraisal of some of the assumptions made in the classical model: 1. In the spring-mass analogy used above. means that the change in the main field-winding flux may be appreciable and should be accounted for so that a correct representation of the system voltage is realized. Consider a bus having a voltage Y to which a load PL j Q L is connected. which may be comparable to the field-winding time constant. where growing oscillations have occurred on tie lines connecting different power pools or systems. As this situation has developed. Constant mechanical power. It is important to account for the various components of the system damping to obtain a correct model that will accurately predict its dynamic performance. Expand our knowledge of the characteristictime response of our system loads to changes in voltage and frequency-develop new dynamic models of system loads. this is a rather poorly damped system. field excitation. The turbine-governor characteristics. Negfecting the damping powers. resulting in loss of synchronism after the system machines had undergone several oscillations. 3. turbine governors. The capacities of most of the tie lines are comparatively small. Indeed there have been recorded transients caused by large impacts. Transient stability is decided in thefirst swing. This has made many engineers realize it is time to reexamine the assumptions made in stability studies. Let the load be represented by the static ad- + . Let us illustrate in a qualitative manner the effect of such representation. 2. and perhaps boiler characteristics should be included in the analysis. Reformulate our analytical techniques to adequately simulate the time variation of all of the foregoing factors in system response and accurately determine dynamic system response. Note that the excitation system also plays an important role in the machine’s mechanical oscillations. In the passive impedance model the load power decreases considerably (since PL a V2).23. In other words.and the increase in frequency does not cause an increase in load power. Such a block diagram is shown in Figure 2. The second control system is the speed control or governor that monitors the shaft speed and controls the mechanical power P. Some excess generation results. This assumption is often on the pessimistic side...1 2 Block Diagram of One Machine Block diagrams are useful for helping the control engineer visualize a problem. = Pa pu (2. Then as we proceed to analyze each system. i. We will be considering the control system for synchronous generators and will do so by analyzing each control function in turn..) To illustrate this.The Elementary Mathematical Model 47 Fig. and the area frequency tends to increase. Initially.P.and that both are frequency independent. In the model used in Figure 2. let us assume that the transient has been initiated by a fault in the transmission network. Three separate control systems are associated with the generator of Figure 2. since it affects the electrical power. At the same time.e. The first is the excitation system that controls the terminal voltage.24 [ 15). this model assumes PL m V z . Thus the model used gives a load power lower than expected during the fault and higher than normal after fault removal. 2. mittances CL = P L / V 2 and B L = Q L / V as shown in Figure 2. QL= V 2 . From the foregoing discussion we conclude that the classical model is inadequate for system representation beyond the first swing. P. Since the first swing is largely an inertial response to a given accelerating torque.24 is (2.17 the change in voltage is reflected in the power and reactive power of the load. TjW = P.23 A load represented by passive admittance. A n increase in system frequency will result in an increase in the load power. the numerical value of which depends on the rotating inertia and the system of units. In real systems the decrease in power is not likely to be proportional to Y 2 but rather less than this. a fault causes a reduction of the output power of most of the synchronous generators. The basic equation of the dynamic system of Figure 2. (There are situations. .18). causing the machines to accelerate. It may be helpful to present a general block diagram of the entire system without worrying about mathematical details as to what makes up the various blocks. 2. we can fill in the blocks with the appropriate equations or transfer functions. . and J has been replaced by a time constant rj. During a transient the voltage magnitude V and the frequency will change. where this assumption can lead to optimistic results.24. This discussion is intended to illustrate the errors implied. while the change in the bus frequency is not reflected at all in the load power. however.66) where has been replaced by G. a transmission network fault usually causes a reduction of the bus voltages near the fault location. the classical model does provide useful information as to system response during this brief period. If a sudden change in w occurs. and TFL. In an isolated system the speed reference is the desired system speed and is set mechanically in the governor mechanism. In addition to the three control systems.. the pu inertia constant. be the accelerating time constant.. It is designed to be quite slow so that it is usually not involved in a consideration of mechanical dynamics of the shaft. Finally. we recognize immediately that the shaft speed w must be accurately controlled since this machine must operate at precisely the same frequency as all others in the system. = TFL. but they are smaller than those in the governor loop. three transfer functions are of vital importance.. Solve the swing equation to find r. time system and specify the units of each quantity (see Kimbark [I]). Le.. in an interconnected system there is a master controller for each system. This to sends a unit dispatch signal (UDS) each generator and adjusts this signal to meet the load demand or the scheduled tie-line power. Hence much effort has been devoted to refinements in excitation control.. Time delays are involved here too.. = 0 and is supplied a constant full load accelerating torque. The generator equations are nonlinear and the transfer function is a linearized approximation of the behavior of the generator terminal voltage C: near a quiescent operating point or equilibrium state. I ) dimensionally using a mass. .48 Chapter 2 Fig. but does so through some rather long time constants. the time required to accelerate the machine from rest to rated speed wR. we have two ways of providing controlled responses to this change. wR.24 Block diagram of a synchronous generator control system. I 2.2 Analyze (2.. can also be related to H . To visualize the stability problem in terms of Figure 2. A second controlled response acts through the excitation system to control the electrical power P. One is through the governor that controls the mechanical power P.24. Let r. the energy source equations are a description of the boiler and steam turbine or of the penstock and hydraulic turbine behavior as the governor output calls for changes in the energy input. These equations are very nonlinear and have several long time constants. length. T. A rotating shaft has zero retarding torque T. Problems 2. Le. The load equations are also nonlinear and reflect changes in the electrical output quantities due to changes in terminal voltage ?. The first of these is the generator transfer function. in terms of the moment of inertia J . Then show that r.. 2. Finally. as will be shown later. Thus in most of our work we can consider the speed reference or governor speed changer (GSC) position to be a constant. .S. and 6.10 In computer simulations it is common to see regulation expressed in two different ways as described below: where P. What are the units of R ? 2. and M.1 I and angle SI.12 A solid-rotor synchronous generator is driven by an unregulated turbine with a torque speed characteristic similar to that of Figure 2.M*I(M. What factor must be used to make the units consistent? 2. where in the region of interest the generator torque is proportional to the shaft angle and the motor torque decreases linearly with increased speed. and angles 6 . 2. and c are constants. Compute the regulation R of this machine.. = mechanical power in pu on SsB Pmo initial mechanical power in pu on SSB = J = system base frequency in Hz R . I4 Derive an expression similar to that of (2.ft2. a . = steady-state speed regulation in pu on a system base = RuSsB/SB s = generatorslip = (uR - w)/2rHz (b) Pm . such a case are exactly equivalent to that of a single finite machine of inertia 2. Find the natural frequency of oscillation for this machine.andJis in Ibm. b.8 A 60-MVA two-pole generator and a 600-MVA four-pole generator are to operate in parallel with other U.0 MJ/MVA is initially operated in A synchronous machine having inertia constant H steady state against an infinite bus with angular displacement of 30 elec.3 2. i. 2.8 if the constant K is to be computed in MKS units rather than pu.7) for an interconnection of two finite machines that have inertia constants M . T . Compute the pu constant K that must be used with these machines in their governor simulations if the system base is 100 MVA. What are the units of 6? 2..3(a).Pmo KIAw PU. = turbine power in pu on SsB fmo = initial turbine power in pu on SsB Kl = SB/RuuRSsB = Au speed deviation.. Relate this time to rr and the slip at speed u . State any necessary assumptions.5 In(2. Find the natural frequency of oscillation and the damping coefficient.1 A 500-MVA two-pole machine is to operate in parallel with other U.33) is written for a salient pole machine to include a reluctance torque term.. 4..deg. 2. 6 is in elec. Show that the equations for . machines. assuming small perturbations from the operating point. Write the equation of motion of the shaft for the following systems: (a) An electric generator driven by a dc motor..The Elementary Mathematical Model 49 2. I 1 and is connected to an infinite bus..S connected to an infinite bus.S.e.4 Solve the swing equation to find the time to reach full load speed wR starting from any initial speed uo with constant accelerating torque as in Problem 2. where P. . 2.. Will this system have a steady-state operating point? Is the system linear? 2.9 Repeat problem 2. = M = M. rad/s Verify the expressions in (a) and (b).bB .6 I n (2. let P = PMsin6 + ksin2S.7) assume that Pis in W and M in J -s/rad. and delivering I . The machine has the same characteristics and operating conditions as given in Problem 2. where in the region of interest the torques are given by T. For this condition find the expression for Pa and for the synchronizing power coefficient. systems and are to share in system governing. .2. assuming small perturbations from the operating point.4) assume that Tis in N-m.O pu power. + M2) 6.. .* . = d2 where a.. (b) An electric motor driving a fan. deg. 2.13 Suppose that (2. Machines A and E are 60 Hz. and angles 6.2) that describe an interconnection of three finite machines with inertia constants M I .50 Chapter 2 2. and M. Fig.E and C are a motor-generator set (frequency changer).17 comprises four synchronous machines.16 The system shown in Figure P2. convert the loads to equivalent passive impedances. The reactance x includes the transmission line and the machine reactances. each represented by a constant voltage behind reactance and connected by a pure reactance. The result of a load-flow study is also given. and calculate the generator internal voltages and initial angles. and the amplitude of its power-angle curve. I5 Derive linearized expressions (similar to Example 2. the mechanical input power. . and show that this system can be reduced to an equivalent one machine against an infinite bus.18 has two generators and three nodes.while machines C and D are 50 Hz. Is there a simple expression for the natural frequency of oscillation in this case? Designate synchronizing power between machines I and 2 as P S l 2etc. faulted. and postfault conditions.1 s by removing line 5 . 2. and 6. A three-phase fault occurs near node 2 and is cleared in 0.16 has two finite synchronous machines. d2. The inertia constants of the two machines are HI and H2s. . Generator and transmission line data are given below. Convert the system to a common 100-MVA base. I6 2. (b) Calculate the Y matrices for prefault. . Write the equations of motion for this system. (c) Obtain (numerically)time solutions for the internal general angles and determine if the system is stable. Give the inertia constant for the equivalent machine. Write the swing equation for each machine. P2. I8 (a) Perform all preliminary calculations for a stability study.18 The system shown in Figure P2..17 The system shown in Figure P2. Fig. M2. Assume that the transmission networks are reactive. P2. 2. 5.20 Repeat the calculations of Example 2. the initial conditions.06 0.01 + j pu. = 0 + j0 pu Case3: PLD = 1.0 0.4 pu P.07 I20 generator transformer reactance Transmission Line Data (resistance neglected) Line number: 3 4 5 6 x p u to 100-MVA base 0. . Voltage Load Angle“ MW MVAR Generator MW MVAR Magnitude pu I 2 3 1. Implementation will require computation of Y.030 1. 7. = xi (PU) xTt (PU) H ( M W-s/MVA) 5 4 Rating (MVA) 0.18 to an equivalent one machine connected to an infinite. but with the following changes in the system of Figure 2.0.. For example.08 0. where R L = 0.21 Repeat Problem 2.19 Reduce the system in Problem 2.2 pu (a) Compute the values of R L D and E and find the initial condition for 6 for each case. Let Y. This is more typical of the arcing re0 sistance commonly found in a fault. This can be accomplished by logical control on some analog computers or by careful hand switching where logical control is not available.0 37.2.5 s. = -0.The Elementary Mathematical Model 51 Generator Data (in pu to generator MVA base) Generator number I 3 tX.25 0. 0.08 50 0.8.. 2.8 2.1 and 0. I 0. (b) Study the damping effect of adding a resistance to the transmission lines of R L in each line where R L = 0. What is the critical clearing angle? 2.5% gives a damping torque of 50% of full load torque.5 0.13 Load-Flow Data Bus no. (a) Use a fault impedance of 2. 0.28 0.0 0.0 80.22 Repeat Problem 2.. To measure the damping. . Apply the equal area criterion to the fault discussed in Problem 2. prepare an analog comp_uter simulation for the system.0 -0. Y (c) Devise a method of introducing additional damping on the analog computer by adding a term K d b in the swing equation. = 0.95.21 using a line impedance of0.20 pu Case 2: P L D = value to give the same generated power as Case 1 P. + jQ.2 + j0. .4 pu.0 -1.0 100.8.0 20.4 j0.018 I .0 23. This will require a means of systematically changing from the fault condition to the postfault (one line open) condition after a measured time lapse. let H = 2. Write the swing equation for the faulted network and for the network after the fault is cleared. 2. + jQ. Consider the effect of adding a “local” unity power factor load R L D at bus 3 for the following conditions: Case 1: PLD = 0.4 r j0.4. + jQ.020 0.18. (d) Make a parametric study of changes in the analog simulation for various values of H.5. Repeat the analog simulation and determine the critical clearing time to the nearest cycle.20 but with transmission line impedance for each line of R L + j0.0 30.0 50.2 pu P.8 pu. = 0.5.06 0. I I . = 0.0 40. bus. . and the potentiometer settings. Estimate the value of Kd by assuming that a slip of 2. 14. New York. Crary. (e) Use the computer simulation to determine the critical clearing angle. 1967.. Young. Wiley. and El-Abiad. 1971. faulted. w.. and Thoits. R.. 6. Narional Power Survey... Concordia. 1964. 7. W.) 13. 9. A S M E J . Transient Phenomena in Electrical Power Systems. 6. Stagg. (Rev. . 8. G. M. 2. 1962. A. Paper 66 CP 709-PWR. Dec. C. D.... I . and for if the fault impedance is Z . L. y12 the prefault. 4. Eng. S . Use the computer for this. Anderson. 1971. Synchronous machine damping and synchronizing torques. Power System Stability. R. and Shortley.. make a phase-plane plot of w. Stability program data preparation manual. 15. 8. PAS-902427. Westinghouse Electric Corp. 2 . Wiley. I E E E Trans. References I . Cottipurer Me1hod. Byerly.. A.01 + j0. New York. J. McGraw-Hill. Vol. Stevenson. H. versus 6. and postfault condition. and Webler. New York.. USGPO. . A I E E Trans.. New York. Also. AIEE Subcommittee on Interconnection and Stability Factors. Pergamon Press. Vol. T. T. Wiley. 1959. Concordia. B. 1975. 1964. 1948. J. ISA Trans. 1968. McGraw-Hill. 1970. and Shipley. C. Economic Control of’lnrerronnected Systels. (d) Perform the analog computer simulation and plot the following variables: T. W. 1959. 1951. H. = 0. and Nanakorn. A new stability program for predicting the dynamic performance of electric power systems. Elements qfPower System Analysis. Am. e. T. 1968. Effects of future turbine-generator characteristics on transient stability. 3. S. 2nd ed. P. 56261 -82. Electr.. presented at the IEEE Winter Power Meeting... Power 81:201 -6. C.31. Federal Power Commission. 14:17-23. C. Venikov. New York.6. Proc. R. Power Con/: 29: 1126-39. 1947. K. B. E.52 Chapter 2 (b) Compute the values of I. 1 1 .. V.s in Power System Analysis. 0. E. Kimbark. Pt. Ray. New York. E. W. D.C. P. M .. writing the admittance matrices by inspection and reducing to find the two-port admittances. 70:73 1-37. Dynamic system performance. Erect of steam turbine reheat on speed-governor performance. 2. Sherman. Macmillan. (c) Compute the analog computer settings for the simulation. A n analysis and comparison of certain low-order boiler models. Compare these results with similar plots with no local load present. Power System Stability. Rept. New York. 70-736. Washington. First report of power system stability. Kirchmayer. D. 5 . 12. P. IO. Lokay. 1937. Eng. Thus system behavior is a measure of dynamic stability as the system adjusts to small perturbations. we would expect that for a temporary disturbance the system would return to its initial state. or a small load added to the network (say 1/100 of system capacity or less). The mathematical models for the various components of a power network will be developed in greater detail in later chapters. For a linear system. with emphasis on the qualitative description of the system behavior. While the power-angle relationship for a synchronous machine connected to an infinite bus obeys a sine law (2. which results in a small change in its rotor angle 6. the system is stable. it was shown that for small perturbations the change in power is approximately proportional to the change in angle (2. modern linear systems theory provides a means of evaluation of its dynamic response once a good mathematical model is developed. the system is unstable. the response of a power system to impacts is oscillatory. A n example of this linearization procedure was given in Section 2. If the system is stable. It is important that synchronism not be lost under these conditions. This small disturbance may be temporary or permanent. In general. It is assumed the system under study has been perturbed from a steady-state condition that prevailed prior to the application of the disturbance.35). while a permanent disturbance would cause the system to acquire a new operating state after a transient period. Under normal operating conditions a power system is subjected to small disturbances at random. Typical examples of small disturbances are a small change in the scheduled generation of one machine. If the oscillations are damped. We now define what is meant by a small disturbance.5.chapter 3 System Response to Small Disturbances 3.1 Introduction This chapter reviews the behavior of an electric power system when subjected to small disturbances. so that after sufficient time has elapsed the deviation or the change in the state of the system due to the small impact is small (or less than some prescribed finite amount). The criterion is simply that the perturbed system can be linearized about a quiescent operating state.33). 53 . If on the other hand the oscillations grow in magnitude or are sustained indefinitely. Here a brief account is given of the various phenomena experienced in a power system subjected to small impacts. In either case synchronism should not be lost. 2 Distribution of power impacts When a power impact occurs at some bus in the network. The method of analysis used to linearize the differential equations describing the system behavior is to assume small changes in system quantities such as b. it is often convenient to assume that the disturbances causing the changes disappear. the power impact is “shared” by the various synchronous machines according to their steady-state characteristics. and power respectively)..2 Types of Problems Studied The method of small changes. each impact is followed by oscillatory power swings among groups of machines to reflect the transition from the initial sharing of the impact to the final adjustment reached at steady state.31. (3. the power impact is shared by the machines according to different criteria. the state variables or the system parameters will usually not change appreciably. resulting in a transient.e. 3. When this transient subsides and a steady-state condition is reached. If these criteria differ appreciably among groups of machines. During the transient period. = 2 AX + BU 3. This is advantageous.7]. In this limited range of operation a nonlinear system can be described mathematically by linearized equations. Equations for these variables are found by making a Taylor series expansion about xo and neglecting higher order terms [4. If the perturbation is small. it will acquire a new operating state. an unbalance between the power input to the system and the power output takes place. it is important to ascertain not only that growing oscillations do not result during normal operations but also that the oscillatory response to small impacts is well damped. since linear systems are more convenient to work with. If the stability of the system is being investigated. Thus the operation is in the neighborhood of a certain quiescent state xo. In examining the dynamic performance of the system. The motion of the system is then free. Under normal operating conditions a power system is subjected to numerous random power impacts from sudden application or removal of loads. The behavior or the motion of these changes is then examined. if the system is described by a set of first-order differential equations. Such behavior can be determined in a linear system by examining the characteristic equation of the system.5. I n other words. This procedure is particularly useful if the system contains control elements. however. PA (change in angle. u.2. the new operating state will not be appreciably different from the initial one.2. which are determined by the steady-state droop characteristics of the various governors [5. If the mathematical description of the system is in state-space form. sometimes called the perturbation method [ 1.54 Chapter 3 3.1 System response to small impacts If the power system is perturbed. voltage. i. is very useful in studying two types of problems: system response to small impacts and the distribution of impacts. Stability is then assured if the system returns to its original state.. These power swings appear as power oscil- . each impact will be followed by power swings among groups of machines that respond to the impact differently at different times.6]. As explained above.2..1) the free response of the system can be determined from the eigenvalues of the A matrix. .2) 60 +6 sin(6 + P A . The equation of motion of a synchronous machine connected to an infinite bus and the electrical power output are given by (2. > 0.18) and (2. however.wR/2H (3. Its response is oscillatory with the frequency of oscillation obtained from the roots of the characteristic equation (2H/wR)s2+ P.4) is marginally stable (Le. = Pm0and using the relationship . 3. It can be analyzed if an adequate mathematical model of the various components of the system is developed and the dynamic response of this model is examined.8) .1.. causing instability.” In large interconnected power systems tie-line oscillations can become objectionable if their magnitude reaches a significant fraction of the tie-line loading. a damping term is added to (3. i. This gives rise to the term “tie-line oscillations.y ) = sin(& . the constant-voltage-behind-transient-reactance model. since they are superimposed upon the normal flow of power in the line. = P. oscillatory) for P.7) where D is the damping power coefficient in pu.e.4) and the new characteristic equation becomes (2H/wR)s2 + (D/wR)s + P.7) are given by (3.2) becomes (3.2.6) If the electrical torque is assumed to have a component proportional to the speed change..System Response to Small Disturbances 55 lations on the tie lines connecting these groups of machines. Letting 6 = = Pc + PMsin(6.. such an answer can be reached by a qualitative discussion of the distribution of power impacts.y) + cos(6o ~ Pe .3) the linearized version of (3. This problem is similar to that discussed in Section 3.) sin(& . If we are interested in seeking an approximate answer for the magnitude of the tie-line oscillations. conditions may exist in which these oscillations grow in amplitude.41) respectively or P. which has the roots s = &jdP. P.3 The Unregulated Synchronous Machine We start with the simplest model possible. = 0. = 0 (3.o (3.4) where The system described by (3. The roots of (3. Furthermore.y) (3.y + 6. Such a discussion is offered here. stator resistance. Mathematically. K 3 is an impedance factor. The system described by (3. The transformer voltage terms in the stator voltage equations are considered negligible compared to the speed voltage terms. Pea = K16A + &EA (3. These results are found in de Mello and Concordia [8] and are based on a model previously used by Heffron and Phillips (91. the response is oscillatory with an angular frequency of oscillation essentially the same as that given by . and the initial conditions. another model of the synchronous machine is used. To account for this effect. If either one of these quantities is negative. 3. (3.3. response lim Ek(t)]6A-o I-- K4 = - 1 .1 1) aA-U(I) v~1-0 . Venikov [4] reports that a situation may occur where the machine described by (3. ri0 is the direct axis open circuit time constant of the machine. This would be the case where there is appreciable series resistance (see [4]. equations for the direct and quadrature axis quantities are derived (see Chapter 4). and the damper windings. the external network. It is not our concern in this introductory discussion to develop the model or even discuss it in detail.4) can be unstable under light load conditions if the network is such that tJo < y. the following s domain relations are obtained.lim K3 1-m EA(r) The constants K I . unstable operation.negative. 3. A negative value of P leads to . Major simplifications are then made by neglecting saturation.6). K 2 is the change in electrical power for a change in the direct axis flux linkages with constant rotor angle. and K4 is the demagnetizing effect of a change in the rotor angle (at steady state). For a machine connected to an infinite bus through a transmission network..2)./o. Sec.9) (3. Le. the rotor angle . used in the simpler machine model of constant voltage behind 1 (3.10) where K .a the field-winding voltage uFArand the voltage proportional to the main field-winding flux EA. Rather. and the roots are complex. the system is unstable. . and K4 depend on the parameters of the machine.7) is stable for P > 0 and for D > 0.16~-0 = K3 = final value of unit step u. Note that K . we will state the assumptions made in such a model and give some of the pertinent results applicable to this discussion. K 2 . as this will be accomplished in Chapter 6.56 Chapter 3 Usually ( D / w R ) 2< 8HP.. Linearized relations are then obtained between small changes in the electrical power Pea.1 Demagnetizing effect of armature reaction The model of constant main field-winding flux linkage neglects some important effects. is the change in electrical power for a change in rotor angle with constant flux linkage in the direct axis. From Chapter 2 we know that the synchronizing power coefficient P is negative if the spontaneous change in the angle 6 is . To account for the field conditions. is similar to the synchronizing power coefficient P. among them the demagnetizing influence of a change in the rotor angle 6. we write Kt = PeA/6AlEb=0 K2 = peA/E. (with D = 0) [ Z s ' + (K.K2K3K4 > 0 and K2K3 K4 > 0. 3. = 0. transient reactance.13) where we can clearly identify both the synchronizing and the demagnetizing components.1 1) are usually positive.4). Thus from Routh's criterion [IO] this system is stable if K.K2K3K4)= 0 (K. . Substituting in the linearized swing equation (3. must be greater than the demagnetizing component of electrical power.9) and (3.2). The change in power due to . we obtain the new characteristic equation. - or we have the third-order system (3. would become negative. (3. and K4 are positive. This would happen because the impedance factor producing the constant K.$ K6 3 0 2H 3.10). s3 1 + -s2 + !Q!K. 2H K3Td0 Note that all the constants (3.3. The second criterion is satisfied if the constants K2. (3. Venikov [4] points out that if the transmission network has an appreciable series capacitive reactance. I Primitive linearized block diagram representation of a generator model. K3.2) we are interested in terms involving changes of power due to changes of the angle 6 and its derivative. The first of the above criteria states that the synchronizing power coefficient K.14) +wR I .12) For the case where V . it is possible that instability may occur.2 Effect of small changes of speed In the linearized version of (3. may be represented by the incremental block diagram of Figure 3. with the initial equation (3. Equations (3.System Response to Small Disturbances 57 Fig.1. 3.+ KiK37. If a transient droop or regulation R is assumed.KZK3K4) = 0 (3. = dsA/dl. causes a change in both electrical and mechanical power.16) From Section 2. was discussed above and was found to include a synchronizing power component and a demagnetizing component due the change in EL with 6.17) where i3Pm/dw]. . The system block diagram with speed regulation added is shown in Figure 3. can be obtained from a relation such as the one given in Figure 2.3 the change in mechanical power due to small changes in speed is also linear PmA = a p m / a w l w ~ W A (3.7) the change in electrical power due to small changes in speed is in the form of PL = (D/WR)WA (3. I. 3.2 Block diagram representation of the linearized model with speed regulation added.58 Chapter 3 6.. The characteristic equation of the system now becomes (3. -L Fig. W.2.o s 1 + (Ki .15) As in (3.19) or + R :[ (D + . we may write in pu to the machine base PmA = . The change in speed.20) .18) which is the equation of an ideal speed droop governor.( ~ / W W A / W R ) PU (3. In this case the new differential equation becomes (3.. .jo sin SijA Y sin tiijo + 6ijA cos Sij0 = .24) dijo is the change in the electrical power of machine i due to a change in the angle between machines i and j . = + bijA.4 . = Gu + jBu is an off-diagonal element of the network short circuit admittance matrix Y Using the incremental model so that 6..(B.2 for one machine connected to an infinite bus.I jzi C (3. . with all other angles held constant. the linearized equations (3. This is left as an exercise (see Problem 3.I j+i n Ei Ej Yij cos (eij .2 I ) where 6 = Si .23) can be derived for a given machine in terms of the voltages at those nodes and their angles.. n .E.= y.i + jBii is a diagonal element of the network short circuit admittance matrix Y y. j.a sin 6 . 3. sin 6.4 Modes of Oscillation of an Unregulated Multimachine System The electrical power output of machine i in an n-machine system is n Pei = E:Gii + + j.22) and (3.(Bijsin 6. we compute = sin Sij0 cos SijA + cos S. Thus we write n peiA = j.22) For a given initial condition sin Sijo and cos bij0 are known.j) . G.System Response t Small Disturbances o 59 Again Routh’s criterion may be applied to determine the conditions for stability. Ei = constant voltage behind transient reactance for machine i . It is a synchronizing power coefficient between nodes i and j and is identical to the coefficient discussed in Section 2. Its units are W/rad or pu power/rad. COS a. for PciA.21) applies to any number of nodes where the voltages are known..5. Thus the concept of the synchronizing power coefficients can be extended to mean “the change in the electrical power of a given machine due to the change in the angle between its internal EMF and . cos 6. and the term in parentheses in (3. 6 cos Sij0 -.Gij sin 6ijo) 6 i j A (3. P.23) where Psij s] 8% = Ei Ej(Bijcos 6 . .I j4i C E. j.Gij sin 6ijo) (3. We also note that since (3.6.I j+i + Gijcos aii) (3.22) is a constant.2). Finally. = E:Gii EiE. ~ the time derivatives of these angles. . Equation (3. &..sin6. . n ..I (3.26) for machine i.i. .1) first-order differential equations.I)-independent equations. .27) Subtracting the n t h equation from the ith equation.-.31) where the coefficients aiidepend on the machine inertias and synchronizing power coefficients. and let x. n = 0.” (An implied assumption is that the voltage at the remote bus is also held constant. jti = 0 i = 1. Thus (3. IJA (3. .2.c$. .6.26) is not a set of n-independent second-order equations.I c n..~os6. Using the inertial model of the synchronous machines. we get the set of linearized differential equations. Let x l .. .28) Equation (3.2.ajnA = o i = 1.) This expanded definition of the synchronizing power coefficient will be used in Section 3.I be th e angles aInA.. x ~ .30) dt2 + j. . We will use the latter formulation to examine the free response of this system.31) represents a set of n ..~ respectively.26) comprises a set of (n ...). .n (3.1 linear second-order differential equations or a set of 2(n . . -2Hi d26iA WR dt2 + 2 j-1 EiEj(B.28) can be put in the form Since 6.25) jti or (3. PS. .26) jti The set (3.I a. .0)6.2.x n .29) can be further modified as = 6inA - (3.60 Chapter 3 any bus... From (3. x 2 .A = 0 i = 1.1 jti (3.G.. since Z b . . . we compute n.A. .I --2Hn j . The system equations are of the be form . .6. with all other bus angles held constant.. n (3.... . I 1 I 0 1. we examine the eigenvalues of the characteristic matrix [ l l . Since the matrix -XU is nonsingular.. Find the modes of oscillation of a three-machine system. ... as follows: -XU I U det _ _ _ _ _ _ _ _ A .35) or I X2U .. I _______--- I O I I I I I I I I I ... A .. This is obtained from the characteristic equation derived from equating the determinant of the matrix to zero.. ..1 vector of the angle changes 6.... . 133...(-1/X)”-IA I I -XU I I (-XU) = (See Lefschetz [ 121.. 010 1 . we compute the determinant of M as IM I = = . .System Response io Small Disturbances 0 .1 vector of the speed changes db..... .. A 12 22 +--------------- 0 4 1 .. p.32)* xn -%+I . The machines are unregulated and classical model representation is used..1 frequencies of oscillations.) The system described by I M I = 0.A I = 0. Thus the system has n .X u ]= d e t M = O [ (3.I! 0 * ’ a 1 .1 complex conjugate pairs. has 2(n . Example 3. which occur in n .. the system equations are given by *See the addendum on page 650.34) where X is the eigenvalue./dt To obtain the free response of the system....A(-XU)-’U I (-l)”-’X“-’ I -XU . Solution For an unregulated three-machine system. 121. 2 ‘ or : XI x2 61 (3.. 0 .A I (3.2 (3.1 I X2U .1) imaginary roots..33) where U XI X2 = = the n = the n the identity matrix ... X2n . det [+ h2 a21 all . the following two equations or The state-space representation of the above system is To obtain the eigenvalues of this system. from the first two. 3 9 . we get (noting that 6 = -aji) and subtracting the third equation I f we eliminate are obtained: by noting that + + = 0./2H. . the characteristic equation is given by det -all =o -a12 -a22 -a21 Now by using ( 3 .62 Chapter 3 Multiplying the above three equations by w. 5563 10.6. Example 3. where C.2936 Note that the 6.. the voltage zngle will exhibit a step change.a .096 (%/2)(Ps3z/H3 .6015 1. 5Bi.0566 I . i = I . Table3.64. and & are constants.524 a21= (0R/2)(Ps3& .6 Ij v i vi I . Assume there are no governors active on any of the three turbines.4598 6. 12 23 31 I .1544 I .System Response to Small Disturbances 63 Examining the coefficients aii. Let these given values be X = i ja.40. 3 respectively. A small IO-MW load (about 3% of the total system load of 315 MW) is suddenly added at bus 8 by adding a three-phase fault to the bus through a 10.0170 1. and 3.226 . 6 .1.0502 1.Psz1/H2) = 33. -180 .556 a2z) da 1 + azz)’ (1 . they will not change at the time of impact. nine-bus system of Example 2. The system base is 100 MVA./H2 + Pa3/H2 + Ps32/H3) = 4(alIa22 153. From Example 2. = C .1 as follows: a12 (WR/2)(Ps12/HI + P s 1 3 / H .18 (load flow) and the computed initial values given in Example 2. Solution First we compute the frequencies of oscillation. Then compare these computed frequencies against those actually observed in a digital computer solution.6 we know H i = 23.17.9035 1. cos (Br + &) + Cz cos ( y t + c$~).01 for i = I .088 1. C.OS02 1. are the values of the relative rotor angles at I = 0-.1. Thus we can compute the values of aijfrom Example 3.0566 Bij 4jti psi.Psiz/Hi) = 59. Also from Example 2. 2 . Observe the system response for about two seconds. For loads that are essentially constant impedance.513 1.555.6 we find the data needed to compute Psij with the results shown in Table 3. 3.we can see that both values of Xz are negative real quantities. operating initially in the steady state with system conditions given by Figure 2.. 2.6 for Ei/66. Synchronizing Power Coefficients of the Network of Example 2.G sin). so these are also the correct values for t = O+.24) Psij = V. Since these .55841)”*] = -77. Assume that the system load after t = 0 is constant and consists of the original load plus the IO pu shunt resistance at bus 8. X = f j y .841 = E a22= (oR/2)(Ps2.0 pu impedance. This is also true of angles at load buses to which appreciable inertia is connected. Compute the frequencies of oscillation that will result from this small disturbance. V.460 Then = = -(1/2)[+1.. From (3. 6.at~l)] f (66336 . + Ps31/H3) = 104. + -(1/2)[-257. %(Si/cos . cos 6.2 Consider the three-machine. are rotor angles.0170 I . . The free response will be in the form 6. however. 3(a) and angle differences relative to 6. As might be expected. (a ) I 2.135 0. Methods have been devised [3. where absolute angles are given in Figure 3.000 2.. are given in Figure 3.807 1.3(b).2 can be transformed to a new frame of reference called the Jordan canonical form.1 Hz.402 0.500 Time.3 Unregulated response of the nine-bus system to a sudden load application at bus 8: (a) absolute angles.500 1.7 s.500 (b) Fig.. s 2. neither of the computed frequencies is clearly observed since the response is a combination of the two frequencies.000 1. a Nine-Bus System Quantity Eigenvalue I 2j8. and a. the variables 6. .500 8. should be observed in the intermachine oscillations of the system.500 1. In the form of equations normally used.713 kj13.000 2.416 13.4 Hz and 2. about 1.1I ] by which a system such as the one in Example 3.01 0.000 1.468 Thus two frequencies. The results of such a solution are shown in Figure 3.500 Time. This can be approximately verified by an actual solution of the system by digital computer.807 Eigenvalue 2 x o rad/s f Hz Ts 8.3.3(b) gives periods in the neighborhood of 0. (or other angle differences) contain ""CI I 24. 3.64 Chapter 3 Now we can compute the frequencies and periods shown in Table 3.0 I 1 I I I 0. Table 3 2 Frequencies of Oscillation of . A rough measurement of the peak-to-peak periods in Figure 3.0 -97..2.0 I I I 1 I 1 0. In Jordan form the different frequencies of oscillation are clearly separated. (b) angles relative to 6 1 .416 2.0 0. using any method [ I .).0 1 D = E-IAE = j13.X.3 Transform the system of Example 3.8854 0.2571 0. We now define the transformation x = E y to compute 2 = E i = A x = A E y j.13831 ! 0. 1 1 1 and call these vectors E.2792 0. where x is defined by and the a coefficients are computed in Example 3.8854 0.OoooO I i -0..83069 .5245 -j3.9221 j1. E.95234 I -0.0 0. = E-’ A E y = D y whereD = diag(X.2312 j3. We now compute the eigenvectors of A.0 0.2792 0.00000 1..0 -j6.2792 -0..2 into the Jordan canonical form and show that in this form the system frequencies of oscillation are clearly distinguishable.7008 j 1 S967 -j1. -j0. Hence we have difficulty observing these frequencies in measured physical variables.9221 r-j13.OOOOO j 1. E E E41 2 .. we compute 1 or E-’= I L -j3.0 0.06266 i E = [E.2.2659 -j I .oooOO -0. Solution The system equations for the three-machine problem are given by .5245 j3.7008 0.2792 0.0 0. E. = j -j0.System Response to Small Disturbances 65 “harmonic” terms generally involving all fundamental frequencies of oscillation. We then use these eigenvectors to define a matrix E.. - 0 or i = A x.0 j6.X.o .X. and E4. Example 3.0 0.5967 0.07543 1 1.23 19 -0.13831 0.95234 where the numerical values are found by a suitable computer library routine. 3.1. 0.2659 0. -0.14523 I 0. Performing the indicated numerical work.0 0.0 00 ..2571 0.14523 . has no time lags. Finally. y A = KS6.5. From (3. The magnitude of the damping. however.24. therefore.38) Substituting in (3. To simplify the analysis. 7. We will consider two simple cases of regulation: a simple voltage regulator with one time lag and a simple governor with one time lag.9) and (3. we return to the model discussed in Section 3.3. is such that the frequencies of oscillation given by the above equations are not appreciably affected. = 0 and the transducer .66 Chapter 3 = Substituting into Dy./6.5 Regulated Synchronous Machine In this section we examine the effect of voltage and speed control equipment on the dynamic performance of the synchronous machine. and the change in the synchronous machine terminal voltage y. uFA depends only upon K modified by the transfer function of the excitation system. a rather simple model of the voltage regulator and excitation system is assumed... 3. 3. Again we are interested in the free response of the system.. This gives the following s domain relation between the change in the exciter voltage u. we note that a change in the field voltage uF. we note that the simple model used here assumes that no damping exists. since the eigenvector computation may be too costly. Analysis of such a system is discussed in Chapter 7.10).36). To use (3.37) where K5 = = y.10). = K6 change in terminal voltage with change in rotor angle for constant E' change in terminal voltage with change in E' for constant 6 The system block diagram with voltage regulation added is shown in Figure 3.. is proor duced by changes in either VREF y . = = regulatorgain regulator time constant To examine the effect of the voltage regulator on the system response.1 Voltage regulator with one time lag Referring to Figure 2.. Such a relation is developed in reference [8] and is in the form v. and E: is needed. These relations are given in (3.. In physical systems damping is usually present. This method of computing the distinct frequencies of oscillation is quite general and may be applied to systems of any size.: 'A F = . If we assume that V.we can compute the uncoupled solution yi = Ciexi' i = 1.36) and (3. For very large systems this may not be practical.4. a relation between 6.1EA = VIA/E6la. the oscillatory response given above is usually damped.2.36) where K.3 for a machine connected to an infinite bus through a transmission network.iKc/(l + 7ts)1 yA (3. we compute . + &E: (3.4 where Ci depends on the initial conditions. however.37) UFA = -[Kt/(l + 7es)l(KS6A + (3. (3.System Response to Small Disturbances 67 'mb REF Fig.41) is of the form s4 + 03S3 + O*S2 + q s + cr. rearranging. .39) and (3. we obtain the following characteristic equation: Equation (3.4 System block diagram with voltage regulation. = 0 (3.42) Analysis of this fourth-order system for stability is left as an exercise (see Problem 3.39) From (3.7). 3.9) peA = Substituting in the s domain swing equation and rearranging. If we assume the simplest model possible. we note that a change in the speed w or in the load or speed reference [governor speed changer (GSC)] produces a change in the mechanical torque T. as shown in Figure 3. .5 Block diagram of a system with governor speed regulation.44) The order of this equation will depend upon the expression used for PeA(s).such as the model given by (3..45) or S3(2HTg/W~) s2(2H/WR) + (Kg + + PsTg)S + P.10). PeA(s)= PSGA(s). Applying Routh's criterion. 3.5.5.9) and (3. *p. the system becomes of fourth order.. For the model under consideration it is assumed that GSCA = 0 and that the combined effect of the turbine and speed governor systems are such that the change in the mechanical power in per unit is in the form (3. Then the linearized swing equation in the s domain is in the form (with wR in rad/s) SSA(S) ( ~ W W R ) ~ ' ~ A-[&/(I + 7sI = S) g) ..d Fig. > 0. the system is stable if Kg > 0 and P. Ifanother model is used for PeA(s).43) where Kg = gain constant = I / R r g = governor time constant The system block diagram with governor regulation is shown in Figure 3..2 Governor with one time lag Referring to Figure 2.24. The amount of change in T. = O (3. depends upon the speed droop and upon the transfer functions of the governor and the energy source.5. characteristic equation of the the system is given by (2H/wR)s2 + [Kg/(I+ T~s)]S+ Ps = O (3..68 Chapter 3 3.Ped($ (3. Its dynamic response will change. Information on stability can be obtained from the roots of the characteristic equation or from examining the eigenvalues of its characteristic matrix.46) The system is now of third order. * n + ~ . Note that the estimates made by the methods outlined below are only approximate. (See also [7.5].(B.7.47) jti. c o s s ~ . This kind of impact is continuously occurring during normal operation of power systems. B~~ ail sin (3. B~~ sii + sin V. an oscillatory transient results before the system settles to a new steady-state condition.System Response to Small Disturbances 69 1 GSCA Fig. the system becomes fifth order. j-l .6 Distributionof Power Impacts In this section we consider the effect of the sudden application of a small load PLA at some point in the network. Thus the scheduled tie-line flows will have “random” power oscillations superimposed upon them.9 and 3. as shown in Figure 3. as is usually the case. 3.6.21) by adding node k. Since the sudden change in load PLAcreates an unbalance between generation and load. If both speed governor and voltage regulation are added simultaneously.4.k For the case of nearly zero conductance pi C j .I n E. 0 - pi = E ~ G ~ ~ + E. the power into node i is obtained from (3.+ ) v ~ ( B ~ai.k . yet they are quite instructive. 3. we also assume that the load has a negligible reactive component. We formulate the problem mathematically using the network configuration of Figure 3.) To simplify the analysis.E. sin ~ + cik cossik) jti. Referring to the (n + I)-port network in Figure 3. sin 6.7 and the equations of Sections 2. Our concern here is to make an estimate of the magnitude of these power oscillations. These oscillations are reflected in power flow in the tie lines. The oscillatory transient is in fact a “spectrum” of oscillations resulting from the random change in loads.6 Block diagram of a system with a governor and voltage regulator. (sin6kjO)6kjA (3.. we compute the linear equations j-l j6i.49) for any k. Note also that the internal angles of the machine nodes d l . J2. The machines are represented by the classical model of constant voltage behind transient reactance.k (3. where the impact P L A is applied. . . do not change instantly because of the rotor inertia.47) and (3.47) and (3. Note that the order k j must be carefully observed since & j = . .48) are nonlinear because of the transcendental functions. We also assume that the network has been reduced to the internal machine nodes (nodes I . v k /6ko + &A.48) Here we assume.70 Chapter 3 n L - 1 (n + I)-port network Fig.j .7 Network with power impact at node k.6. The transcendental functions are linearized by the relations sinbkj cos6kj = = sin(6kjo + 6kjA) cos(6kjO 6kjA) + sin6kj0 + (cos6kjO)6kjA cos6kjO . 3.17) and the node k. 2. or V &becomes .1 Linearization The equations for injected power (3. . and the power into node k (the load bus) is (3.that the power network has a very high X/R ratio such that the conductances are negligible. Substituting (3.48) and eliminating the initial values. 6. The immediate effect (assuming the network response to be fast) of the application of P L A is that the angle of bus k is changed while the magnitude of its voltage v k is unchanged. we linearize these equations to find Pi = P0 i + Pia P k = PkO + PkA and determine only the change variables Pia and P k A . n of Figure 2. Since we are concerned only with a small impact P L A .49) into (3.50) I1 j-l These equations are valid for any time t following the application of the impact. . . 3.6 j k . 51) 0+. Let us consider next the deceleration of machine i due to the sudden increase in its output power Pia.e. At the instant r = O+ we know that a. the stored energy in the rotating masses becomes important. we would like to determine exactly how much of the impact PLA is supplied by each generator P i A . = 0 for all generators because of rotor .54) It is interesting that at the instant of the load impact (i. Note also that PkA = -PLA. Equations (3.6 & = -6&.System Response to Small Disturbances 71 3 6 2 A special case: r = 0’ .(o+) ~ 6&jA = 6&A - 6jA = 6&A(o+) Thus (3.. as shown below. the higher the transfer susceptance Bik and the lower the initial angle 6iko.. Later on when the rotor angles change. In particular.52) and (3.51). Thus the machines electrically close to the point of impact will pick up the greater share of the load regardless of their size.52) and (3. Note that the generator rotor angles cannot move instantly. so the foregoing equations can be written in terms of the load impact as From (3. This isalso evident from the first equation of (3. the source of energy supplied by the generators is the energy stored in their magnetic fields and is distributed according to the synchronizing power coefficients between i and k. which depends upon the reactance between generator i and node k .. In other words. Pia depends upon Psi&or Bik.55) indicate that the load impact PLA at a network bus k is immediately shared by the synchronous generators according to their synchronizing power coefficients with respect to the bus k.53) we conclude that (3.. i = 1.2. at r = O?. Thus we can compute (with both i a n d j indicating generator subscripts) 6”A IJ = 0 6i&A = 6 i A . inertias.we note that at node k (3. hence the energy supplied by the generators cannot come instantly from the energy stored in the rotating masses.52) This is to be expected since we are assuming a nearly reactive network.. The incremental differential equation governing the motion of machine i is given by . We also note that at node i Pia depends upon Bikcos6p. The instant immediately following the impact is of interest..n.the greater the share of the impact “picked up” by machine i.50) becomes n piA(o+) = -psik6&A(o+) Comparing the above two equations at r = p&A(o+) = /-I Ps&j6&A(O+) (3.. each according to its size H i and its “electrical location” given by P. ij A (1/CHi)CWiH1 (3.H. Note that after the initial impact the various synchronous machines will be retarded at different rates. we compute the acceleration in pu to be (3.ik.61) Thus at the end of a brief transient the various machines will share the increase in load as a function only of their inertia constants.57). ) .56). s (l/CH.58) Summing the set (3. given by (3.57) for all values of i.60). 3 6 3 Average behavior prior to governor action ( t = 1.56) Then if PLA is constant for all t . Substituting this value of dwiA/dt in (3. . We now investigate the way in which the impact PLa will be shared by the various machines. Note that while the system as a whole is retarding at the rate given by (3.60). is the time at which governor action begins.)CG. is chosen large enough . The time t . .57) Obviously. (3. the individual machines are retarding at different rates. we compute (3.2. is dependent on the synchronizing power coefficient Psik and inertia H i . The pu deceleration of machine i. let us define an “inertial center” that has angle 8 and angular velocity a. . Each machine follows an oscillatory motion governed by its swing equation. To designate this period simply. t ~ (3.. at t = t . when the transient decays. there will be an overall deceleration of the machines during this period. we refer to time as t l .60). This deceleration will be constant until the governor action begins.60) Equation (3. dwiA/dt will be the same as dGA/dt as given by (3. the shaft decelerates for a positive load P L A . To obtain the mean deceleration. Looking at the system as a whole.60) gives the mean acceleration of all the machines in the system. where by definition. We now estimate the system behavior during the period 0 < t < t.59) (3.72 Chapter 3 2Hi d W i A --+ PiA(t) = 0 W R dt and using (3. and after the initial transient decays they will acquire the same retardation as given by (3. > to..55) i = 1. where t. which is defined here as the acceleration of a fictitious inertial center. Synchronizing forces tend to pull them toward the mean system retardation. although there is no specific instant under consideration but a brief time period of no more than a few seconds. . In other words. we note that immediately after the impact PLA(i.61) implies that the H constants for all the machines are given to a common base. 0.1 and in Example 2.at t = 0+) the machines share the impact according to their electrical proximity to the point of the impact as expressed by the synchronizing power coefficients.6.3. Equation (3. are shown in Figure 3.6.61). Examining (3.19 we determine that V = 1. Solution A nominal IO-MW(0. retaining only nodes 1. chosen system base. and 8.6 with a small IO-MW resistive load added to bus 8 as in Example 3.System Response to Small Disturbances 73 so that all the machines will have acquired the mean system retardation.016 and a. according to their inertias. is not so large as to allow other effects such as governor action to take place..8 for the system operating without governor action. using a library transient stability program. 3. . A matrix reduction of the nine-bus system. the correct powers are obwhere SBs the machine rating and S. -2 1 Fig. From the prefault load flow of Figure 2.7". The resulting power oscillations P. = . 2. gives the system data shown on Table 3.2. 2.8 PrAversus t following application of a 10 M W resistive load at bus 8. is the is tained if H is replaced by HSB3/SsB. Solve the system differential equations and plot PtA and wid as functions of time. If they are given for each machine on its own base.e.4 Consider the nine-bus.1 pu) load is added to bus 8 by applying a three-phase fault through a 10 pu resistance. namely. Compare computed results against theoretical values of Section 3. 3.A..56) and (3.. The prefault conditions at the generators are given in Table 3. Example 3. 3. At the same time t . i = 1. three-machine system of Example 2. After a brief transient period the same machines share the same impact according to entirely different criteria. 5717 19.02 1 2.100 2.6961 3.9370 are computed from (3.745 3.1 9.6955 3.956 2.8 3. k cos 6 i k O .Gi sin 6 i k O ) k These values are tabulated in Table 3.4. the results agree quite well with values measured from the are computer study.OOO 10. .8 at time t = O+ and are due only to the synchronizing power coefficients of the generators with respect to bus 8.5 M W .6 2. Synchronizing Power Coefficients ik 18 28 38 psik psik psik (neglecting G j k ) (with Gik term) c The values of piA(o+) 2.3 Transfer Admittances and Initial Angles of a Nine-Bus System ij Gii Bii b o 1.6392 8.000 Note that the actual load pickup is only 9 1 M W instead of the desired IO MW.2.4.692 2.51242 3. These values are also shown on the plot of Figure 3.028 2.690 9.749 3.00965 2.4752 1-8 2-8 3-8 0. Initial Power Change at Generators Due to IO-MWLoad Added to Bus 8 I 2 3 3. averages about 9.659 2.100 1O.61601 From (3.02 1 4. These oscillations have frequencies that are combinations of the eigenvalues computed in Example 3.7 9. If the computed PIA scaled down by 0.01826 -0. Note that the error in neglecting the Gik term is small.016 4.923 I 2.665 2.958 3.6001 2.55697 2. This is due in part to the assumption of constant voltage v k at bus 8 (actually.5878 2. The results of these calculations and the actual values determined from the stability study are shown in Table 3.55) as where PLA(O+)= 10.74 Chapter 3 Table 3. labeled Z P i A . the voltage drops slightly) and to the assumed linearity of the system.0 M W nominally. Table 3.8 show the oscillatory nature of the power exchange between generators following the impact.03530 -0. Table 3. .03 15 12.5.91.24) we compute the synchronizing power coefficients psik = 6 v k ( B . The total. The plots of P i a versus time in Figure 3.5.6414 8. 8 1:9 2:O %.16 -0.0 1.14 -0.40 + 3. Since the governors have a drooping characteristic.9.- -0.8 is the computed values of PiA(t1) that depend entirely on the machine inertia.15 MW i = 1 = 1.64 + 6.570 rad/s2 = -0..40 .9 Speed deviation following application of a 10 MW resistive load at bus 8.04 0.10 a- -0. -0.System Response to Small Disturbances 75 lime. Note the steady deceleration with all units oscillating about the mean or inertial center.9 and show graphically the intermachine oscillations that occur as the system slowly retards in frequency. It is also apparent that the system has little damping and the oscillations are likely to persist for some time. IO 2(23.64 . This is partly due to the inherent nature of this particular system.8 0. the speed deviation would level off after a few seconds to a constant value and the oscillations would eventually decay.2 1.3 0.5 0. the speed would then continue at the reduced value as .01) pu/s = -0.s 1.1.I 6.2 0.08.-0. '- t Fig. 3.A' % r .1 1..0908 Hz/s The individual machine speed deviations wiA are plotted in Figure 3. Another point of interest in Figure 3. but the same phenomenon would be present to some extent on any system. If the governors were active.01) I = 7. .9. It is fairly obvious that the PiA(t) oscillate about these values of P. The computer program provides speed deviation data in Hz and these units are used in Figure 3.8 as dashed lines. This is computed as &A -I --=PLA - dr = 2 C Hi .9 1.513 x 0.4 1. The second plot of interest is the speed deviation or slip as a function of time. These calculations are made from PiA(tl) = (Hi/CHi)PLA = IOHi/(23.18 .-3. shown in Figure 3.7 0. I 01 -0"'t -0.1 0.A(tl).91 M W i = 3 = lOHi/33.6 0.6 1:7 1.4 0.9 as a straight line.05 and the results are plotted in Figure 3.12 .3 1.09 Hz/s is plotted in Figure 3. The mean deceleration of about 0.94MW i = 2 = 0. . 10 Two areas connected with a tie line. signifying a substantial load increase on the generators. The tie line is carrying a steady power flow of 80 MW from area I to area 2 as shown in Figure 3. Now let a load impact PLA = IO MW (1% of the capacity of one area) take place at some point in area I . 3. Because of the proximity of the groups of machines in area 1 to the point of impact. in area 2. At the end of the initial transient the load power impact PLA will be shared by the machines according to their inertias.7 MW toward area 2.-- z 80.0---- t 0 t=O tl Time.3 MW will appear as a reduction in tie-line flow. the governors would need to be readjusted to the new load level so that additional prime-mover torque could be provided. The two areas are of comparable size. their synchronizing power coefficients are larger than those of the groups of machines = PSI. = Psz. 9Q 80MW - PM = 10 M W Fig.10. Thus 3. = 2ps2. at that instant the tie-line flow becomes 76. Solution Since PSI= 2Ps2. while 1/3 or 3. and determine the distribution of this added load immediately after its application ( I = 0 + )and a short time later ( t = t . I f the speed deviation is great. Example 3. Consider a power network composed of two areas connected with a tie line.3 MW will be supplied by the groups of machines in area 2.7 . In other words.5 Let us examine the effect of the above on the power flow in tie lines. They are connected with a tie line having a capacity of 100 MW. I Fig. If we define CPSikJareaI CPsiklarca2 then let us assume that P. as shown in Figure 3. Let us assume that the machines of area 1 are 2 76..10. ) after the initial transients have subsided. say 1000 MW each.oscillations due to the load impact in area I .1 I Tie-line power. .76 Chapter 3 long as the additional load was present. 3. the instant of the impact 2/3 of the IO-MW load will be supat plied by the groups of machines in area 1. 450 I/f12 + j1. If this load is 200 MW (1. mentioned above is smaller than the time needed by the various controllers to adjust the system generation to match the load and the tie-line flow to meet the scheduled flow.7 MW.12 in pu on a 1000-MVA base. 3. The system data are given in Figure 3.12 Two areas connected by a tie line.000 MW and that of area 2 is 14.112" pu 0. The transition from 76. Example 3.518 pu 0. (a) Find the equations of power for PI and Pz.3 MW (toward area 2). represented by the resistive load P4. at bus 4. (d) Consider a sudden load addition to area 2. while the units of area 2 are of = 2CHi].533/-76.3-M W flow is oscillatory. From the above we can see that in the situation discussed in this example a sudden application of a IO-MW load caused the tie-line flow to drop almost instantly by 3. The time t . (c) Find the synchronizing power coefficients.112" = 0. The capacity of area I is 20. and after a brief transient by 6. (b) Find the operating condition when PI = 100 MW.888" g l o = go = 0 z 0.533/103.I where all H's are on a comlarger inertia constants such that CHIJarea2 mon base. The sharing of the load among the groups of machines will now become 6.System Response to Small Disturbances 77 predominantly hydro units (with relatively small H). The tie-line flow will now become 73.820 = = = Ylz GI1 plz = = = -712 1. The inertia constants of the machines in the two areas are about equal.. and power swings of as much as twice the difference between these two values may be encountered.7 MW contributed from area 2 and 3..875 /76. This would correspond approximately to a 100-MW tie-line flow from area 1 to area 2.128 . find the distribution of this load at f = O+ and t = f l .j0.3 MW. This situation is illustrated in Figure 3.10 are represented by their Thevenin equivalents and the tieline impedance is given.1 I .3 MW from area I . Solution Consider the system as a two-port network between nodes 1 and 2.43% of the capacity of area 2).128 .000 MW.6 We now consider a slightly more complex and more realistic case wherein the area equivalents in Figure 3. Ara 1 eguivalent Tie litm Area 2 equivalent Fig.7-MW flow to 73.e. Then we compute Z12 = 0. 518cos(.128 + 0. + 13.1 pu 0.252" 6240 = 620 .V:GzI = 0 + I.858sin(-0.128 + + = 0.784" .10.875/76.l28co~6~ 0 .009/0.532') = 0.0[0.518cos 10. To complete the problem.52") 9. B12 = 0.009(1.784") = 0.000H + + 14.009)(0.0.548 Vz V4(B24 COS 6240 .000H/(20.533 + j1.518sin6.2[20.2 pu.183 we compute the synchronizing power coefficients P1 s4 = VI Y 4 ( B 1 4 ~ ~ ~.G24 sin 6240) = 1.252" 0.128 = 0.796") 6 1 = 10.128 + 0.533sin(6.128 612 = 61 . V 2 ( G 1 2 ~ ~ + 6B12sin612) V:GI2 ~ 12 = 0 + I.000H)I 0.252")] = 1.G2I sin 6210) 1.2) = 0.784" + 0.183~0~(-0.) + 0.O(-O. smaller than PSz4.128 .l.012)&2 = 1.128sin(.103 sin 10.0.784")] = 0.103 -9.I28c0s6~ .858 + j0.112" = 0. 5 1 8 ~ i n 6 ~ ) 0.O(O.Gl2 sin d120) cos l. P 4 A 200/1000 = 0.K)/Z.280" + (0. while initially area 1 picks up only about .252' 6 4 = 610 .100/19.2)/(Psi4 + Ps24) = (0.1/114 = = = .2)/(Ps14+ P S z 4 ) = (0.1 4 0 6 G14sin6140) = (1.0~)/1.z = From the admittance matrix elements - Y14 Y24 = -1 = y4 . is while the inertia of area 1 is greater than that of area 2.62 = 6.?.784" (4 Pr12 = = PS21 = K W B I Z 6 1 2 0 .640 = -0.532" 10 .1 1765 pu 14.O(-O.0/10.1354 pu = The power distribution according to inertias is computed as PlA(fl) = 0.100 = 0.0646 PU Pz~(o+) Ps24(0.2[14. Thus.2 = -l/Fz4 y4 -0.323)(0.004 = 1.10.677)(0.533sir1(6~ 13.518 PI = V:glo + V.009 + j0.533 cos 10.08235 pu = I In this example the synchronizing power coefficientsPSI.2) = 0. Thus we com- (K K(O-) = 60 = 4 (1.784") + 0.100 + j0.Chapter 3 GI2 = -0..485 = (d) Now add the 200-MW load at bus 4.533sin(dl - 13.532' Ps24 == + 0.796") P2 = V:gzo + VI V 2 ( G 1 2 ~ ~+ 6 2 1 ~ BI2sin .000H)J = 0.128sin 10. we must know the voltage pute f12(0-) = p4at t = 0-.000ff PzA(fl) = 0.150 + P4A Then the initial distribution of PIA(^+) is PS14(o.640 = 10.000ff/(20.796") (b) Given that PI = 0.533 = K VZ(B2I cos 6210 . . the load frequency and voltage characteristics. The swing equation for machine i becomes. These frequencies are determined by finding the eigenvalues X of the A matrix by solving det (A . Thus the conclusions reached above should be considered qualitative and as rough approximations. then after a brief period according to their inertias. If the system is made up of groups of machines separated by tie lines. the fast primary controllers such as some of the modern exciters. Following an impact the synchronous machines will share the change first according to their synchronizing power coefficients.5. Yet these conclusions are basically sound and give a good "feel" for what happens to the machines and to the tie-line flows under the influence of small routine load changes.. 23). For the second transient. i= 0 (3.System Response to Small Disturbances 79 one third of the load P. neglecting PIA. Usually these two frequencies are appreciably different. the effect of the network transfer conductances. The transition from the second to the final stage is oscillatory (see Rudenberg [7).. .62) where R is the regulation and 7. The angular frequency of these oscillations can be estimated as follows. From Section 3.g. Hence they will oscillate with respect to each other during the transient period following the impact. In the above discussion many factors have been neglected. Problem 3.XU) = 0. Ch. the effect of the reactive component of the load impact. 2 picks up the remaining 41%.10 gives another example where the point of impact is in area I (bus 3). they share the impacts differently under different conditions. It is interesting to note the order of magnitude of the frequency of oscillation in the two different transients discussed in this section. the frequency of oscillation is given by Y : ~% 1/2HiRf7. In general. I). The power flow in the connecting ties will reflect these oscillations. For a given machine (or a group of machines) the frequency of oscillation in the first transient is the natural frequency with respect to the point of impact. the initial distribution of a load impact depends on the point of impact.. The characteristic equation of the system is given by s2 + (1/7si)s + 1 / 2 H i R i ~ . where U is the unit matrix and A is defined by ( 3 . the change in the mechanical power PmAis of the form (3. is the servomotor time constant. The analysis given above could be extended to include governor actions. and others. which occurs during the transition from sharing according to inertia to sharing according to governor characteristic.h thes domain. the speed governor droop characteristic. namely.2. which will act to make the load sharing according to an entirely different criterion. The speed change will be sensed by the prime-mover governors. at a later time t = t . it picks up about 59% of the load and area .63) from which the natural frequency of oscillation can be estimated. e. I .1 I Repeat Problem 3. Repeat for K5 = -0.7. P.80 Chapter 3 Problems 3.20 pu. References I . 3.41) has the following data: H = 4.PSz3. Trans. J . T..46).20 + j0. (a) Obtain the synchronizing power coefficients PSlz.46) in state-space form.26. W.1.. G .6. 1968. Macmillan. Then determine the system stability and possible system behavior patterns by sketching an approximate rootlocus diagram. Hore.1967. The internal voltages and angles of the generators are given in Example 2.2 Use Routh’s criterion to determine the conditions of stability for the system where the characteristic equation is given by (3. 12. and determine the conditions for stability using Routh’s criterion. London. 3):692-97. Venikov. 2 . 1945.6 for the three-machine system discussed in Section 2. 6. New York. including the damping term. Englewood Cliffs. T .) f 11. J. Apply Routh’s criterion to (3. 1970. and Concordia. 1947. 4. 2. Mathematical Handbook for Scientists and Engineers. and (b) Obtain the natural frequencies of oscillation for the angles 6 1 2 ~ 6 1 3 ~ .) 8.6.3 I)]for small perturbations about the given operating point. A. Repeat when there is a local load of unity power factor = having Itload 8.60 pu. New York. Adkins and D. New York. Crary. for stability. R.. The transmission line impedance is Zlinc= 0. F. A. 3. 3. I A synchronous machine is connected to a large system (an infinite bus) through a long transmission line. Rabins. K. 1969.8. Y.6 Give the conditions for stability of the system described by (3. and Phillips. 1966. A.6 with the impact point shifted to area I and let P L = 100 MW as ~ before. Hayashi.45). McGraw-Hill. New York. London. Prentice-Hall.0 pu. R. 5. Cambridge.10 and for the given operating conditions. (MIT Press.10 pu. = 0. Routh. 1965. 9. K.P S I ) .. G. 1877.14). 7. 1 . Vols. A. Academic Press. S. New York. Control and Dynamic Systems. 3.5. S. Transient Phenomena in Electric Power Systems. and the corresponding coefficientsaij[see(3. IEEE Trans.10 Repeat Example 3.30. ria = 5. Takahashi. N. and Korn. Power System Stability. Stability of Nonlinear Control Systems.. The infinite bus voltage is 1. Calculate the minimum and maximum steady-state load delivered at the infinite bus (for stability). 1952. . 1964.0. Effect of a modern amplidyne voltage regulator on underexcited operation of large turbine generators. 1967. 3. The synchronous machine is to be represented by constant voltage behind transient reactance with E ’ = 1. Mass.9 The equivalent prefuulr network is given in Table 2. New York.20). de Mello.3. 1950.Kz = 2.10. McGraw-Hill. (Adams Prize Essay. 3.4 Using 1 3 ~ the output variable in Figure 3. = 3.. Reading.8 Write the system described by (3. Find the maximum and minimum values of K. and Auslander. D. Advanced Studies in Electrical Power System Design. Compare the results with those of Section 3. B.1 A system described by (3.. KS = 0.5 Use block diagram algebra to reduce the system described by (3.10 for an initial condition of PLA = 300 MW. Mass. 3. Rutenberg. 3. Dynamics o f a System o Rigid Bodies. V. 3. AlEE Trans.3 Compute the characteristic equation for the system of Figure 3. State-Space Analysis of Control Systems. Heffron. 10. Korn.20. 3. Lefschetz. M. The direct axis transient reactance x i = 0. by B. Pergamon Press. and K b = 0. R. PAS-88:316-29. Concepts of synchronous machine stability as affected by excitation control. Nonlinear Oscillations in Physical Systems. E. Addison-Wesley.. as 3.K3 = 0. M. use block diagram algebra to reduce the system block diagram to forward and feedback transfer functions.1. 1964.2. London. Rudenberg. McGraw-Hill. 71 (Pt.. M. C. C. Compare with the periods of the nonlinear oscillations of Example 2. Ogata. Then determine the system behavior by sketching the root loci for variations in K. Transient Performance of Electric Power Systems: Phenomena in Lumped Networks. Wiley.J.0 pu. K I = 4. Chapman and Hall. Anderson A. A. M.Part II The Electromagnetic Torque P. Fouad . . State-space formulation of the machine equations is used. r is the winding resistance. I I]. The interested reader should consult one of the many excellent references on this subject (see [ 1]-[9]). or flux linkages in terms of the actual winding variables.chapter 4 The Synchronous Machine 4. The transformation used is usually called Park's transformation [ 10. voltages. with certain other features suggested by Concordia (discussion to [12]) and Krause and Thomas [13].1) where X is the flux linkage.1 Introduction In this chapter we develop a mathematical model for a synchronous machine for use in stability computations. It defines a new set of stator variables such as currents. Thus the flux linking each winding is also a function of the rotor position. are discussed. Two models are developed. 83 . The instantaneous terminal voltage u of any winding is in the form. called the direct axis. with positive directions of stator currents flowing out of the generator terminals. The magnetic coupling between the windings is a function of the rotor position. This chapter is not intended to provide an exhaustive treatment of synchronous machine theory.2 Park's Transformation A great simplification in the mathematical description of the synchronous machine is obtained if a certain transformation of variables is performed. which are often used for stability studies.' I . These six windings are magnetically coupled. Park's transformation is developed mathematically as follows. one along the direct axis of the rotor field winding. u = icri A ci.I I ] but is more nearly that suggested by Lewis 1121. called the quadrature axis. Simplified models. The transformation developed and used in this book is not exactly that used by Park [IO. (4. one field winding. The new quantities are obtained from the projection of the actual variables on three axes. a second along the neutral axis of the field winding. The synchronous machine under consideration is assumed to have three stator windings. and two amortisseur or damper windings. and i is the current. one using the currents as state variables and another using the flux linkages. The notation indicates the summation of all appropriate terms with due regard to signs. *x 4. and the third on a stationary axis. The expressions for the winding voltages are complicated because of the variation of X with the rotor position. A multiplier is used to simplify the numerical calculations.. Therefore the machine EMF E is primarily along the rotor q axis. The effect of Park’s transformation is simply to transform all stator quantities from phases a.cosB = + ibsin(B .. i6. It produces an E M F that lags this flux by 90”. which is proportional to the zero-sequence current. For generator . We should remember. ibr i. Park’s transformation uses two of the new variables as the d and q axis components.sinB = idaxis (2/3)[i. 4. be the currents leaving the generator terminals. as shown in Figure 4.cos(B + 2r/3)1 (4. we get the relations iqpxis (2/3)[i. and i.2r/3) + i. We define the d axis of the rotor at some instant of time to be at angle B rad with respect to a fixed reference position.5) The main field-winding flux is along the direction of the d axis of the rotor. and c into new variables the frame of reference of which moves with the rotor.2r/3) + i.2) We note that for convenience the axis of phase a was chosen to be the reference position.. that if we have three variables i. The third variable is a stationary current. however.4) and where the Park’s transformation P is defined as (4.a4 Chapter 4 a axis 4 b axis Fig. Let the stator phase currents ia. I Pictorial representation of a synchronous machine. If we “project” these and currents along the d and q axes of the rotor.3) where we define the current vectors (4.1. Consider a machine having a constant terminal voltage V. otherwise some angle of displacement between phase a and the arbitrary reference will appear in all the above terms. 6 .sin (e + 2 ~ / 3 ) ] + i6cos(B . we need three new variables. Thus by definirion iOdq = Pi& (4. . which means that the transformation P is orthogonal. v B = WRt +6+~ / rad 2 (4. an inverse transformation also exists wherein we may write iabc = P-’iodq The inverse of (4.1 1) . Phase-winding designations s and f refer to “start” and “finish” of these coils. and the two damper windings D-D‘ and Q-Q’. The q axis is located at an angle 6. We prefer the shorter notation used here. Having P orthogonal also means that the transformation P is power invariant. and sc-fc. at the reference axis in Figure 4.5) is unique. (The damper windings are often designated by the symbols kd and kq.8) e sin 8 sin(8 P-I = fl l / & cos(8 .2 r / 3 ) [/G cos(e + 243) + 2*/3) and we note that P .The Synchronous Machine 85 action the phasor gshould be leading the phasor The angle between E and is the machine torque angle 6 if the phasor V is in the direction of the reference phase (phase a ) . These are the three phase windings sa-fa..i.6) where wR is the rated (synchronous) angular frequency in rad/s and 6 is the synchronous torque angle in electrical radians. The d axis of the rotor is therefore located at v. Thus 1 (4.’ = P‘. e.1.’ ~ .3 Flux linkage Equations The situation depicted in Figure 4.) We write the flux linkage equation for these six circuits as stator rotor i 1 Wb turns (4. At t > 0. and the d axis is located at 8 = 6 + u/2. sb-fb.5) may be computed to be 1/dT COS (4.1 is that of a network consisting of six mutually coupled coils.9) p = = = uaia + +.7) If the transformation (4. Expressions similar to (4.10) 4.3) may also be written for voltages or flux linkages.g. = V:briabc = (P-’VOd. the field winding F-F’. At f = 0 the phasor Vis located at the axis of phase a.. vhdqiodq = uoio + udid + uqiq (4. Le.2 ~ / 3 ) sin(t9 . U. and we should expect to use the same power expression in either the a-b-c or the 0-d-q frame of reference. the reference axis is located at an angle uRt with respect to the axis of phase a.)‘(P-’iwq) vhdq(P-’)‘P-Iiwq = v & + ~ P P . VOdq = pvabc AOdq = pxabc (4. L. D = 0 H (4. Lab = L.COS2(8 + * / 6 ) H .3... all rotor self-inductances are constants and. Lbb L. = L. 4.L. with single subscripts are constants in our notation.) 4. according to our subscript convention. 4. Thus L. > L.. From the phase windings to the field winding we write La. L. These inductances may be written as follows 4. L.14) where I M. Prentice [ 141 shows that most of the inductances in (4. COS 2(8 . from phase windings to damper winding D we have . = = = . are constants.~ . LbF = LFb L. C O S ~ (+ 2*/3) H ~ (4. I > L.M.. = LQF = 0 H LDQ = L . Note that signs of mutual inductance terms depend upon assumed current directions and coil orientations.5 = LDF = MR H L. C o q e+ 5*/6) H (4. = LFc = = = MFCOS8 H hfFcos(8 . we consider the mutual inductances between stator and rotor windings.2 ~ / 3 ) H MFCOS(8 + 2 ~ / 3 )H (4. +L . = Lcb L. all of which are functions of the rotor angle 8.3.M . . and both L. and all pairs of windings with 90" displacement have zero mutual inductance.4 Rotor mutual inductances The mutual inductance between windings F and D is constant and does not vary with 8. or M . I I ) are functions of the rotor position angle 8.2 Rotor self-inductances Since saturation and slot effect are neglected. Le. (All inductance quantities such as L.15) Stator-to-rotor mutual inductances Finally.M .3. 4.3 = LF H LD D = LD H LQQ= LQ H (4.3. Note the subscript convention in (4. The coefficient of coupling between the d and q axes is zero.13) Stator mutual inductances The phase-to-phase mutual inductances are functions of 8 but are symmetric.11) where lowercase subscripts are used for stator quantities and uppercase subscripts are used for rotor quantities. and L. Lb.3.12) where L.1 Stator self-inductances The ph ase-w i n di ng self-inductances are given by L. C O ~ H + L.2 ~ / 3 )H + L . = = = L. COS2(8 . = LF.*/2) H ..L. we may use a single subscript notation.16) Similarly.86 Chapter 4 = = where Ljk self-inductance when j = k mutual inductance when j # k and where Ljk = Lkjin all cases. . . =.20) where we have defined the following new constants.LRa = stator-rotor inductances LRR rotor-rotor inductances = Equation (4. We compute where L .19) is obtained by premultiplying (4.' ". = stator-stator inductances La=.2*/3) H LcQ = LQc = MQsin(B + 2*/3) H (4.The Synchronous Machine 87 (4. Ld = Lo L. we compute [.5) applied to the a-6-c partition. . + (3/2)Lm H L.21) .3. since B is a function of time.1 I ) with its time-varying inductances can be simplified by referring all quantities to a rotor frame of reference through a Park's transformation (4. as noted in equation (4.1 I ) by where P is Park's transformation and U3 is the 3 x 3 unit matrix. .2M. + M. = + M. Thus in voltage equations i such as (4.(3/2)Lm q H (4. from phase windings to damper winding Q we have La.L. H L. = MQsinB H LbQ = LQb = MQsin(8 . k = L.19).6 Transformation of inductances Knowing all inductances in the inductance matrix (4. Only four of the off-diagonal terms vanish.17) and finally. operation indicated in (4.3 Performing the Wb turns (4.15). We now observe that (4. we observe that nearly all terms in the matrix are time varying. I ) the h term is not a simple Li' but must be computed as = L i + i . I I). 4.18) The signs on mutual terms depend upon assumed current directions and coil orientation. since the machine is a generator. as the first row and column have only a diagonal term. Note that the stator currents are assumed to have a positive direction flowing out of the machine terminals.2. is completely uncoupled from the other circuits. 1 I]. Mutual inductances are omitted from the schematic for clarity but are assumed present with the values given in Section 4. the circuits are shown in Figure 4. where he let vodq = Qvabr with Q defined as (4. This leads to unnecessary complication when the equations are normalized. 4.22) Other transformations are found in the literature. We also note that the transformed matrix (4. This was not true of the transformation used by Park [ 10.1). Similarly. where coils are identified exactly the same as in Figure 4. Schematically.22) is not a power-invariant transformation and does not result in a reciprocal (symmetric) inductance matrix. Flux linkage A.4 Voltage Equations The generator v.88 Chapter 4 In (4.3.= -ri -X + v. thus conforming with our notation for constant inductances. For the conditions indicated we may write the matrix equation v . .20) is symmetric and therefore is physically realizable by an equivalent circuit.1 and with coil terminations shown as well.2 Schematic diagram of a synchronous machine. 4.oltage equations are in the form of (4. i L F ' o r Fig. The power of Park's transformation is that it removes the time-varying coefficients from this equation.20) Ad is the flux linkage in a circuit moving with the rotor and centered on the d axis. This is apparent since all quantities have only one subscript. A. The transformation (4. It is important also to observe that the inductance matrix of (4. This is very important.20) is a matrix of constants. is centered on the q axis. but these terms can be eliminated by applying a Park's transformation to the stator partition.23) in partitioned form as follows: where Thus (4. RFDQ 0 ] p] [k] k] ~FDQ .25) If r. we may also define Robc = rU3 where U.L .26). This requires that both sides of (4. = rb = rc = r. is the 3 x 3 unit matrix. and we may rewrite (4. as is usually the case.XFDQ (4. For the resistance voltage drop term we compute + VFDQ 'vabc] .The Synchronous Machine 89 or +I.J I -ib ib .24) (4.=-r. 1 [: I lo [: 1I..27) (4.28) ..] v where we define the neutral voltage contribution to vabc as (4. 1 1 1 (4.26) is complicated by the presence of time-varying coefficients in the term.- r.26) be premultiplied by By definition for the left side of (4.23) v.26) (4. The resistance matrix is diagonal.3 1) 0= wAd- V (4. the third term on the right side of (4.26) transforms as follows: (4. we substitute (4.32) Finally. Summarizing.33) where by definition nodq is the voltage drop from neutral to ground in the 0-d-q coordinate system.35) .24).71. Aodq = PA&.28)-(4. as it should. we compute (4.3 1) and (4.recalling the definition (4. For balanced conditions the zero-sequence voltage is zero.26) to write Note that all terms in this equation are known.26) is transformed as (4.33) into (4.AFDQ k ] q ["':"Odj ]: n [ v (4.34) and observe that this voltage drop occurs only in the zero sequence. from which we com- V (4. Using (4. let + + ['-1 VFDQ - - 1.30) We evaluate by. RFDQ ] [?] [ ~FDQ . To simplify the notation.90 Chapter 4 The second term on the right side of (4. and the state variables (through the power network connected to the machine terminals) and (2) a set based onflux linkages as the state variables. But the inductance matrix here is a constant matrix.5 Formulation of State-Space Equations ["I v (4. so we may write h = Li V.r) (4. expanding to full 6 x 6 notation. and the iterm behaves exactly like that of a passive inductance. Let x .35)..as follows.37) is of the well-known form = AX + BU (4. Since these two sets of variables are mutually dependent. Here we will use the formulation x' = [ A d A. . The term has been simplified so that we can compute its value from (4. We may now put this set in the form of (4. Note.35) in terms of one set of variables only. Actually. however.38) Examining (4.20). x' = ( i d i q i F i D i Qwhich has the ]. Le. 4. that (4.35). where L& is the transpose of L. we can replace the terms in X and iby terms in i and . We will mention only two that are common: ( I ) a set based on the currenrs as state variables. i.e.37) or (4.36) XFDQ Recall that our objective is to derive a set of equations describing the synchronous machine in the form x where = f(x. X F X D XQ]. where the particular set to be chosen depends upon how conveniently they can be expressed in terms of the machine currents and stator voltages. advantage of offering simple relations between the voltages u d and u. we can see that it represents a set of first-order differential equations.. and rearranging.The Synchronous Machine 91 Then for balanced conditions (4.35) contains flux linkages and currents as variables.38).35) may be written without the zero-sequence equation as 4. Substituting this result into (4.. in state-space form. which we rearrange in partitioned form. numerous possibilities for the choice of the state variables are available. the set (4. we can eliminate one set to express (4.u.39.37) x = a vector of the state variables u = the system driving functions f = a set of nonlinear functions If the equations describing the synchronous machine are linear.6 Current Formulation Starting with (4. which is usually a good approximation.1. viz. If the speed is assumed constant. the angular velocity. iF..) An examination of the voltage equations reveals the dimensional character shown in Table 4. then (4.39). The price we have paid to get rid of the time-varying coefficients is the introduction of speed voltage terms in the resistance matrix. This problem can be solved by normalizing the equations to a convenient base value and expressing all voltages in pu (or percent) of base. I n any event. Also observe that all the terms in the coefficient matrices are constants except w. [These dimensions are convenient here..7 Per Unit Conversion The voltage equations of the preceding section are not in a convenient form for engineering use. Other possible systems are . This is a considerable improvement over the description given in (4. iqand i. the nonlinearity is never great. and iD. this causes (4. time) system. consisting of WX or wLi products. where all dimensions are expressed in terms of a u-i-i (voltage.39) is linear. A great deal of information is contained in (4. i d .23) in the a-b-c frame of reference since nearly all inductances in that equation were time varying. 4. One difficulty is the numerically awkward values with stator voltages in the kilovolt range and field voltage at a much lower level.92 Chapter 4 (4. These terms. They are similar to those of a passive network except for the presence of the speed voltage terms. as w is usually nearly constant.39) where k = m a s before. Similarly. the q axis speed voltages are due to d axis currents. appear unsymmetrically and distinguish this equation from that of a passive network. Note that the speed voltage terms in the d axis equation are due only to q axis currents. (See Appendix C. current. First. we note that the zero-sequence voltage is dependent only upon io and io.39) to be nonlinear. This equation can be solved separately from the others once the initial conditions on io are given. The remaining five equations are all coupled in a most interesting way. Since w is a variable. To simplify the expression without any loss of generality. let us examine the effect of this choice on the d and q axis quantities.42) Then from ( 4 .39) involve only three dimensions. u. are stator quantities because they relate directly to the a-6-c phase quantities through Park’s transformation.] Observe that all quantities appearing in (4. and base time. let the rms phase quantities be V b V and I& A.The Synchronous Machine 93 FLtQ (force. base current. First note that the three-phast power in pu is three times the pu power per phase (for balanced conditions). Units. and t . Li v = x [vrlil [ 1 /I1 v v = = [I1 radians per second radian (rad) (Ws) Bord dimensionless Choosing a base for stator quantities The variables udr u. Quantity Electrical Quantities. Thus if we choose three base quantities that involve all three dimensions. = VR = stator rated line-to-neutral voltage. To obtain the d and q axis quantities. we may compute base quantities for all other entries. vodq = PvObc or .” we choose the following stator base quantities. For example. To prove this. by combining these quantities according to column 4 of Table 4.sin(O + (Y) = d V s i n ( 8 ub = d V s i n ( O + (Y .1 amperes (A) watts (W) voltamperes (VA) weber turns (Wb turns) ohm (a) henry (H) second (s) [vi1 [v/il [il p = vi ri.40) Before proceeding further. V rms wB = w R = generator rated speed.2 ~ / 3 ) V u.COS((Y 7 ) - (4. Table 4.1.)COS((Y 7) - = ~V. charge) and MLtp (mass.(t) is in the form. = d V s i n ( O + (Y + 2 ~ / 3 ) V + a) v (4. (Also see Rankin [ 151.. time. and Dimensions Units u-i-i Symbol Dimensions [VI Relationship Voltage Current Power or voltamperes Flux linkage v i volts (V) p or S x r Lor M I 0 Resistance Inductance Time Angular velocity Angle 4. Note that exactly three base quantities must be chosen and that these three must involve all three dimensions. The pu power P3* is given by Pj+ (~VI/VBI. if we choose the base voltage. i. we will assume that u.) Using the subscript B to indicate “base” and R to indicate “rated. time. Ad. VA rms V. i. length. permeability). all bases are fixed for all quantities. u.I.. length.4 1 ) where the subscript u is used to indicate pu quantities. 9 . elec rad/s (4. [9] for a discussion of this topic. Lewis [ 121 and Harris et al. and A. we first write the instantaneous phase voltage and currents.1.7. Let SB= SR = stator rated VA/phase. = V. id. The three-phase power is 3 VIcos(a .y) W. = l/wB = 1/wR s V. it is divided by the base quantity of the same dimension. since P is a power-invariant transformation. when there is no danger of ambiguity in the notation. = SB/VB = SR/VR A rms r.46) Similarly.94 Chapter 4 (4. Pj6 = i#dU = + iquuqu 3 I./f.y) P U + cos C Y cosy) (4. uqU= d 3 V U c o s a Obviously. For example.50) Thus by choosing the three base quantities S.cosy (4.KcOS(a .40) and Table 4 1 we compute the following: . the power in the d and q circuits must be the same as the power in the three stator phases. s i n ~ iF = fii. 1. K(sin a sin y = 3f.47) and the pu currents are given by id. ..(A) pu (4. Wb turn = R. = i(A)/f. this subscript is omitted. = d 3 1 .44) (4. = VR/IRWR A. for currents we write i. . They indicate that with this particular choice of the base voltage. (4.51) where we use the subscript u to indicate pu. we can compute base .48) To check the validity of the above. the pu d and q axis voltages are numerically equal to fl times the pu phase voltages.49) We now develop the relations for the various base quantities... values for all quantities of interest. and re.43) In pu udu = ud/VB = &(V/V. From (4. = VB/IB VR/IR Q H (4. To normalize any quantity. Later.r. we can show that if the rms phase current is fly A. = VR/wR = LBf. the corresponding d and q axis currents are given by. Similarly.r. = V.45) (4.)sincu = 6 Ksincu (4. then u: + u& = 3Vt’ The above results are significant. L. V. 000 A. since all other rotor base quantities will then be automatically determined. It can be shown that the choice of a common time base t. care should be exercised in the choice of the remaining free rotor base term. The choice made here for the free rotor base quantity is based on the concept of equal mutualflux linkages. Which is more convenient? To illustrate the above.18. for example.2 Choosing a base for rotor quantities Lewis [ 121 showed that in circuits coupled electromagnetically. i.) is For the synchronous machine the choice of SB based on the rating of the stator. Is one choice more convenient than the other? Are there other more desirable choices? The answer lies in the nature of the coupling between the rotor and the stator circuits. VRB will then be 100. It should be remembered. Assume that its exciter has a rating of 250 V and lo00 A. Then we can show that . depending on the condition of the magnetic circuit.000 V. however. and if we choose VRB = 250 V. and io = I D B be applied one by one with other currents set to zero. but the question is. which are to be normalized. iF = IFB. This means that base field current or base d axis amortisseur current will produce the same space fundamental of air gap flux as produced by base stator current acting in the fictitious d winding. = (LlBL2B)”’. If.The Synchronous Machine 95 4. Referring to the flux linkage equations (4.e. then I R B will be 400. MI.) The choice of equal time base throughout all parts of a circuit with mutual coupling is the important constraint. Therefore. (See Problem 4. If we denote the magnetizing inductances ( 4 = leakage inductances) as (4.. (See Appendix C for a more detailed treatment of this subject.7. Hence some rotor base quantities are bound to be very large. and the time base is fixed by the rated radian frequency. These base quantities must be the same for the rotor circuits as well. it is essential to select the same voltampere and time base in each part of the circuit. that the stator VA base is much larger than the VA rating of the rotor (field) circuits. we choose I R B = 1000 A.. Even then there is some latitude in the choice of the base rotor current. It would seem desirable to choose some base quantity in the rotor to give the correct base quantity in the stator. making the corresponding pu rotor quantities appear numerically small. For example. forces the VA base to be equal in all circuit parts and also forces the base mutual inductance to be the geometric mean of the base self-inductances if equal pu mutuals are to result. There is a choice of quantities. the correct base stator flux linkage or open circuit voltage.20) let id = I. through the magnetic coupling. consider a machine having a stator rating of 100 x lo6 V A / phase.. we can choose the base rotor current to give.52) and equate the mutual flux linkages in each winding. it is not always clear which base quantities are used by the authors.22).5 (idud + lquq).e. gives the same numerical values in pu for synchronous machine stator and rotor impedances and self-inductances as the system used in this book.. Part of the problem lies in the nature of the original Park's transformation Q given in (4.. Note that the modijied Park's transformation P defined by (4. one pu cur- rent and voltage gives three pu power in the system used here and gives one pu power in the other system.- (4. The pu mutual inductances differ by a factor of Therefore. I8). These basic constraints permit us to compute and since the base mutuals must be the geometric mean of the base self-inductances (see Problem 4.5) was chosen specifically to overcome these problems. When the Q transformation is used.96 Chapter 4 - - .54) and this is the fundamental constraint among base currents. While the use of pu quantities is common in the literature.. This leads to the relations fl(Lmd/MF)lB vFB = (3/fl)(MF/Ln1d)~B This choice of base quantities.54) and the requirement for equal S. MFB = kFLB H MDB= kDLB H MQB= kQLB H MRB= kFkDLB H (4. Since the power in the d and q stator circuits is the three-phase power. the mutual coupling between the field and the stator d axis is not reciprocal. Also. 1. i.7. Furthermore. Therefore it is important to understand any major difference in the pu systems adopted. = 1. . From (4. The major differences lie in the following: IFB = a. the three-phase power in watts is given byp. the terms kMF used in this book are numerically equal to M F in pu as found in the literature. synchronous machine data is usually furnished by the manufacturer in pu. the pu system is chosen carefully to overcome this difficulty. The system most commonly used in the literature is based on the following base quantities: SB = three-phase rated V A VB = peak rated voltage to neutral I B = peak rated current and with rotor base quantities chosen to give equal pu mutual inductances. which is commonly used.3 Comparison with other per unit systems The subject of the pu system used with synchronous machines has been controversial over the years. This transformation is not power invariant. we compute .57) 4. = 8660 x 2.79 x io-' H T o obtain M. However. From the air gap line of the no-load saturation curve. and (4..d = 18. denoted by an asterisk.. = kMF/L.3333 MVAIphase VB = lSOOO/fl = 8660. Thus the peak phase voltage corresponds to the product iFwRMF.0.972 Wb turn/phase R. Solution: Stator Base Quantities: S.969 x 52* LO = 5.542 x 52 Power factor = 0.371 52 Ld 6.23). In the system used here uk + u& = 3 V t .4/18.73 x IO-' = 70.854 = 326.40 A t. H MF = 8 6 6 0 d / ( 3 7 7 x 365) = 89.25 V 1.' H Excitation voltage = 375 V kMD = 5. since the manufacturers' base system is so common.(unsaturated) = 0.6526 x s A.ed E From the no-load magnetization curve. the value of the field current at rated voltage is 365 A.779 x IO-' H* Fieldcurrent = 926 A r(125"C) = 1.40 A kMQ = 2.765 kW.sin 8. = 2. where wR is the rated synchronous speed.423 x H* RatedMVA = 160 MVA Rated voltage = 15 kV. there is merit in studying both. = 6158. we use (4.01 x H Then k. = 8660.5595 x lo-. (4. I. viu + u& = The system used here is more appealing to some engineers than that used by the manufacturers [9.1 Find the pu values of the parameters of the synchronous machine for which the following data are given (values are for an actual machine with some quantities.006 x kM. At open circuit the mutual inductance La. Then we compute. 121.5595)10-3 = 5.25/6158. = i+. the value of field current corresponding to the rated voltage on the air gap line is 365 A.854 x 3. = x 89.The Synchronous Machine 97 2.989 x H* Inertia constant = 1.40 = 1.s/hp L .16).2 d = (6. = iFMF cos 8 LaF = MF cos 8 The instantaneous voltage of phase a is u.406 52 H L.006 x = 109.189 H rQ = 18.421 x ioT3 Q* LF = 2.64 A MFB= 18. = 6.where Vuis the pu terminal voltage.730 x L.55)-(4.854.329 x lo-' H .118 x H .782 X w3 H* Statorcurrent = 6158.65 x = 22. = 6158.d = Ld . and the flux linkage in phase a are given by A.341 X w3 H rD 18. while in the other system Vt.11).85 rF(125"C) = 0.341 . Therefore.M. Example 4.57). Y connected = t. = 160/3 = 53. from (4. being estimated for academic study): L Q = 1. = 8660/(377 x 6158) = 3. gives pu values of d and q axis stator currents and voltages that are d’3times the rms values. The basis for converting a field quantity to an equivalent stator EMF is that at open circuit a field current iF A corresponds to an EMF of i F W R M F V peak. favors E4 for this vokage.765( I .98 Chapter 4 V B = (53.001096 0.681326. in a discussion of [17].969 x 10-’/0.406 0. the contribution to the d axis stator flux linkage Ad due to the field current iF is kM& and so on.406 = 0. LB/4 = 0. The International .605 4.68 V F RFB = LFB = 163280.49 0. 2. .933 = 1.73 x = 1. = r = rF = r = D rQ = 6. The authors leave this voltage unsubscripted until a new standard is adopted. The choice of symbol for the E M F due to iF is not clearly decided. 0.. A 4.352 Q Amortisseur Base Quantities (estimated for this example): kMD/L. are defined in Section 4.0540 = 1.0/0.d MDB= kMQ/L.018/1.9 = 0.3413. Electrotechnical Commission (IEC).4 The correspondence of per unit stator EMF to rotor quantities We have seen that the particular choice of base quantities used here.559513.73 = 1. The American National Standards Institute (ANSI) uses the symbol € [ 16).0131 18.70 .526 kMD = kMF = M R = 1. We also note that the coupling between the d axis rotor and stator involves the factor k = and similarly for the q axis..98913.781 10 . = d E in MKS units? m.651 5.0 LB H 2. then i F o R M F = . If the rms value of this EMF is E.55 The quantities L D and LA.64 = 499.7. In synchronous machine equations it is often desirable to convert a rotor current.933 x IO-’ H 2.11813.89 s2 (18..001542/1..326 H L B = LB H D R B = RB s2 D RQB = RBI4 = 0. = LQ = LD = A LA.77915.15 1..000742 0.64 1.18911. A new proposed standard uses Ea 1171.15 kMQ 1.42310. = LQB = Inertia Constant: 5.73 0.730 = 1.0.15 6.\/ZE and i F W R kM.1 1. or voltage to an equivalent stator EMF.37 kW *s/kVA H Ld LF LD {d = 1.73 = 1.64 .326 = 1.0. These expressions are developed in this section. For example.64 = 163280.70 2.78115.3711499.746) = The pu parameters are thus given by: = = = L.782 = 0 5 . flux linkage.33 x 106)/326.351 = 0.845)2 x 3. 60) 0 kMF Lq 0 LF MR 0 0 0 kMQ kMD LD 0 0 ‘Q- where the first three equations are on a stator base and the last three are on a rotor base. At steady-state open circuit conditions A. For the following computations we add the subscript u to all pu quantities to emphasize their dimensionless character. Examine the second equation more closely. or (v. The normalization process is based on (4.we have . as all values of voltage and current will normally be in the neighborhood of unity.The Synchronous Machine 99 Since M . we may normalize the voltage equations (4. Having done this. This in turn corresponds to a peak stator EMF (Uf/rF)wRMF. 0 kMF 0 O kM. = LFiF./rF)wRkMF 4. the stator equations should be numerically easier to deal with. Later this subscript will be omitted when all values have been normalized.59) Normalizing the Voltage Equations Having chosen appropriate base values. corresponds (at steady state) to a field current UF/rF.58) By the same reasoning a field voltage U. We can show that the d axis stator EMF corresponding to the field flux linkage A. M F .51) and a similar relation for the rotor. Thus E is the stator air gap rms voltage in pu corresponding to the field current iF in pu. is given by AF(WRk MF/LF) = flE: (4.39). (4. and setting w = u u w R . the d axis stator EMF corresponds to a field voltage U. which may be substituted into (4. to a corresponding stator EMF. and wR are known constants for a given machine. gives a peak stator voltage the rms value of which is denoted by E:. If the rms value of this EMF is denoted by E.and this value of field current iF. when multiplied by w .39) to give 0 0 r 0 0 0 rD WkMQl O 0 0 rF 0 0 0 0 0 0 0 0 0 0 0 0 kM. We can also convert a field flux linkage A.8 = ~ E F D (4. the field current corresponds to a given EMF by a simple scaling factor. Dividing through by V.. 1 00 Chapter 4 (4.iF.. . If the currents are balanced... - (4.. i ~ pu -. . = r / R B Lqu Ldu = Ld/LB Mu = F MQu E MFWRIFBIVB MQwRIQB/vB = Lq/LB MDu = MDwRIDB/VB (4.. + &+.i. ) .62) may be rewritten with all values except time in pu..+ -iF...iO. The fourth equation is normalized on a rotor basis and may be written from (4.65) where all pu coefficients have been previously defined. "R WR WR * (4. - Lo + 3Ln 10.(r 1 + 3rn).66) "RLB = .. RB .62) We now recognize the following pu quantities.w.kMD.. .ip..kMF. w l VB kMQi. WR WR P U (4.70) ..id. following equations are then obtained.60) may be analyzed in a similar way to write U . the d axis equation (4. kMFu' LFu + . The third equation of (4. : (4. WR ( L .63) Incorporating (4. = rF..69) The The damper winding equations can be normalized by a similar procedure.id.iD.Ld.60) as (4.. The first equation is uncoupled from the others and may be written as uou = - -lo.. RB r + 3r.k * Mw * -i. .iqu.+ -i D ~ id. r.50).0. . (4. . we rewrite (4. + 3 ~ .r. it is easy to show that this equation vanishes.. + U. L.63).67) We now incorporate the base rotor inductance to normalize the last two terms as The normalized field circuit equation becomes uF.61) Incorporating base values from (4. = w. L i. i.61) as UdU = - r .e. I (R + wN)i ..d OR dt d7 (4.76) This equation has the desired state-space form. that accompanies every term containing a time derivative. However.75) where R is the resistance matrix and is a diagonal matrix of constants.74) as v = -(R + oN)i - L i pu (4.39). equation. Incorporating all normalized equations in a matrix expression and -dropping the subscript u since all values are in pu. so we have additional equations to write..76) may be depicted schematically by the equivalent circuit shown in . and have rearranged the equations to show the d and q coupling more clearly.The Synchronous Machine 101 (4. This may be done by normalizing time. This is always possible if base quantities are carefully chosen and is highly desirable. since we are interested in balanced system conditions in stability studies. Equation (4.72) (4.74) is identical in notation to (4. It is important to notice that (4. however. Using matrix notation. N is the matrix of speed voltage inductance coefficients. we may write i = -L..7 I ) These normalized equations are in a form suitable for solution in the time domain with time in seconds. It does not express the entire system behavior. and L is a symmetric matrix of constant inductances.73) is the normalized time in rad. If we assume that the inverse of the inductance matrix exists. we write (4. We do this by setting 1 d _ -. as the same equation symbolically serves both as a pu and a “system quantity” equation.74) where we have omitted the u. some engineers prefer to rid the equations of the awkward 1 / 0 .L-’v pu (4. we write = - Ld kMF kMD 0 pu (4. in pu. and we must treat it as a variable.000742 r. These symbols are quite common in the literature in reference to the damper windings. Note that all self and mutual inductances in the equivalent circuit are constants. This is sometimes confusing to one learning synchronous machine theory because a term XI that appears to be a voltage may be a flux linkage. w is certainly not a constant.3 differ from similar equations found in the literature in two important ways.15 H = 2.2.64 1. = 0. However. Figure 4.55 M R = kMD = 0. The use of X for L or M is based on the rationale that w is nearly constant at 1.0540 = 1.49 r = 0.65 1. These are due to speed voltage terms in the equations. and pu quantities are implied for all quantities. Later. Equation (4. Here the subscript notation k d and kq for D and Q respectively is seen.001096 r F 0.2 Consider a 60-Hz synchronous machine with the following pu parameters: L d = 1. including time.0131 rQ = 0.37s t d = t q .70 Lq = LF = L = D LQ = kMF = 1.3 Synchronous generator d-q equivalent circuit. For convenience of those acquainted with other references we list a comparison of these inductances in Table 4.605 1. Table 4. in a linearized model we will let w be approximated as a constant and will simplify other terms in the equations as well.0 pu so that. Some authors and most manufacturers refer to these same quantities by the symbol x or X . it is a state variable in our equations.74) and the circuit in Figure 4. Note also the presence of controlled sources in the equivalent.2. 4. I n this chapter we use the symbols L and M for self an'd mutual inductances respectively. X = w L L .102 Chapter 4 vQ = r-~$-y-kMQ+-J+ r : + ad - Fig.3. as we shall indicate in the sections to follow. Comparison of Per Unit Inductance Symbols LF Lfl xfl Xkdd Chapter 4 Kimbark [2] Concordia [ I ] Ld Ld xd Lq Lq LD LQ L88 kMF MF MR xJld kMD xakd ~ M Q M8 xakq xq Xkgq xo/ Example 4.526 kMQ = 1. 9 Normalizing the Torque Equations In Chapter 2 the swing equation Jii = (2J/p).00074 R+wN= 0 0 -I . This gives some idea of the complexity of the equations.m (4.2~ .55 I I I I 0 0 O 0 l 0 0 0 0 from which we compute by digital computer 5.0498 -L-I(R 1.534 + wN) = IO- 3.2818 .0 1. The result of this normalization was found to be .3290 I I ______________________LL______________ P U 8 3 7 9 .4~ 5086.55 I .55 1.7433 9 190.55 L = 1.9269 2.414 -5. 3 ~ -8183.75) we have numerically - 0.70 .001 1 0 0 0 0 I I I I 1.5. 5 5 ~ 0.9~ 44.605 I I I _______ 0 PU 0 1.110 -5.060 8.3~ I 5.80 284.7888 - and the coefficient matrix is seen to contain w in 12 of its 25 terms.1 15.7564 8379. U - 3065.7198 66.804 I I 0 I 0 0 I I Then we may compute .1 . = To N .90 I.65 1.49~ 0. -8864.405 -1.3878 -5.001 1 0 1. 5 5 ~ 0.The SynchronousMachine 103 Solution From (4.8 1 8 3 .0131 I I -1.2785 3. I -8504.9~ 2785.55 1. 4.9279 I -8975.70~ .0540 I I 1.90 .55 0 0 0.640 0 0 1.857 -313.869 7.77) is normalized by dividing both sides of the equation by a shaft torque that corresponds to the rated three-phase power at rated speed (base three-phase torque).5.9~ 5598. 9 ~ I..060 -3. 4. This torque is normalized along with other generator quantities . Since (4. pU(34) = the expression used for T..81) transforms this pu torque to the new value T... = 3T. Then = = pu generator electromagnetic torque defined on a per phase V A base Te(N'm)/(SB/wB) Pu (4. while the torque is normalized.m . viz.78) angular velocity of the revolving magnetic field in elec rad/s accelerating torque in pu on a three-phase base H = wR/SB)s and the derivative is with respect to time in seconds. on a basis of S. must be in pu on a three-phase V A base.66) (4. when time is in pu..)k where w T. Tm. V. We compute the generator electromagnetic torque in N . the angular speed w and the time are given in M K S units. Ti = 2HwB (4.85) . which is the pu torque on a three-phase basis. Equation (4. Rewriting (4. to the form of (4.) The procedure that must be used is clear.80) (4.t w. = 3..79) and substituting tu = w. Thus the equation is not completely normalized...82) where all the terms in the swing equation. to give T. We need to couple the electromagnetic torque T.81) Te = TtJ3 ~ ~ ( 3 4 ) ( A similar definition could be used for the mechanical torque. determined by the generator equations. = w / w .78) as ( ~ H / U B ) & T.is normalized on a three-phase basis. = = = T ~ ~ ( 3 6 ) o (4. we must use care in combining the pu swing equation and the pu generator torque equation.79). Thus for a fully loaded machine at rated speed. Beginning with (4.78) is normalized to a three-phase base torque and our chosen generator V A base is a per phase basis. .0.78).. The normalized swing equation is of the form given in (2. including time and angular speed.1 The normalized swing equation In (4.T. Suppose we define (4.78) is the swing equation used to determine the speed of the stator revolving M M F wave as a function of time.#. and t . (4. This normalization takes into account the change in angular measurements from mechanical to electrical radians and divides the equations by the base three-phase torque. T.79) T.. Equation (4. I.. are in pu.104 Chapter 4 (2H/w.84) thus.. we would expect to compute T. Usually.9.83) we have for the normalized swing equation (4.. it is considered wise to convert the generator terminal power and torque to a three-phase base S.. by substituting tu = w.79). While the torque is almost always given in pu.91) where the superscript t indicates the transpose of vob.91) becomes iOk = PO".86) If w and To are in pu (and t in s). the phase-to-neutral voltage. rather than SB3.86) are modified as follows: H dw U 180& dt .. .The Synchronous Machine 105 4. (4.g. in pu on a threephase base. 2HwB dw. we find that .87) If I . P U It would be tempting to normalize the swing equation on a per phase basis such that all terms in (4. (4.9. T0. This is not done here because it is common to express both T. (4. dt 2 H dw = To.8) we may write with a similar expression for the voltage vector... = vbq(P-'Y P-'bq Performing the indicated operation and recalling that P is orthogonal.90) To.79) are in pu based on S.._ = 2H dw 0 . it is often not clear which units of w and f are being used.)h = TO.= dw 90 dt. This could indeed be done with the result that all values in the swing equation would be multiplied by three. P U (4. 4. = uoio + ubib + uric = Vtbciobc pu (4.2 Forms of the swing equation There are many forms of the swing equation appearing in the literature of power system dynamics. w. Therefore. pu dt. T.10 Torque and Power The total three-phase power output of a synchronous machine is given by pou.89) (4. .. is a convenient base to use in normalizing the generator circuits.= ' dt. are in pu (and w in rad/s).79) and (4. by substituting in (4. a summary of the different forms of the swing equation is given in this section. (2H/w.. If We begin with w in rad/s and f in s.88) If w is given in elec deg/s. (4. since a phase voltage of k pu means that the line-to-line voltage is also k pu on a line-to-line basis.79). Note there is not a similar problem with the voltage being based on V. -= dt. to match the basis normally used in computing the machine terminal conditions from the viewpoint of the network (e. and To are all in pu. But from (4.t in (4. even though S . t and T. Then (4. To avoid confusion. do. in load-flow studies). and T..Toll P uH . idxq)W]= iqxd .98) which we recognize to be a bilinear term../ae = aPRd/aW= a/aw [(iqxd . Then by using (4. recalling that the flux linkages can be expressed in terms of the currents.20). is the electrical torque. poul= udid + uqiq + uoio (4.98).97) Then (4.100) where D is a damping constant. the power transferred across the air gap. The machine torque is obtained from the second term. we write from (4.81) and (4. i.106 Chapter 4 the power output of a synchronous generator is invariant under the transformation P. Thusu. and Td is the damping torque.95) can be written as (4.95) The same result can be obtained from a more rigorous derivation.e. = io = Oand poul= udid + u. (balanced condition) (4. the swing equation may be written as . = Ldid + kMFiF + kMDiD xq = Lqiq + kh!fQiQ (4. Now.. the energy in the field is given by 6 wfid = &-I j. is the mechanical torque.idxq PU (4.92) For simplicity we will assume balanced but not necessarily steady-state conditions. and the stator ohmic losses respectively. expressed in pu.99) where T. Starting with the three armature circuits and the three rotor circuits.i.. we can obtain the above relation (see Appendix B of [ I ] ) . T. It is often convenient to write the damping torque as T d = DW pu (4. Then using T = a WRd/a8 and simplifying. T.94) Concordia [ I ] observes that the three terms are identifiable as the rate of change of stator magnetic field energy.I I 5 Cik4 Lkj) (4.36). = aw.96) which is a function of 8. (4. Suppose we express the total accelerating torque in the swing equation as (4.93) Substituting for u d and uq from (4. 1 1 Equivalent Circuit of o Synchronous Machine = f(x.20) with the row for A. .t) as given by (4.102) into (4. the (4.101) that the system is nonlinear. we obtain I I I -L-’(R + wN) ____-kh!fQid I I I I I I I I I I I- + I 37..u. 4.104) M U LD kMQ We may rewrite the d axis flux linkages as 0 ‘Q. Note that the “inputs” are For balanced conditions the normalized flux linkage equations are obtained from (4. omitted.85) and depends on the units used for w and following relation between 6 and w may be derived from (4.37).103) This matrix equation is in the desired state-space form x v and T. t. t d 0 kMF kMD 0 0 LF 0 kM< (4.6). Finally. I I (4.102) 6=w-I Incorporating (4.76).The Synchronous Machine 107 where r j is defined by (4. It is clear from (4.101) and (4. = kMFid.106) . are given in this particular case by A. I IO) where Similarly. = LF - x~ = Ld = x d = kMF = kMD = hf~ (4. . we can see that the pu values of &did. the pu values of Lmd. xD In pu. = L.. and the flux linkage that will be mutually coupled to the other circuits is Ad .e.107) 4.e. pu (4.109) If in each circuit the pu leakage flux linkage is subtracted.59. and AD must be equal. d axis circuits is then L.. in pu. Let iF = i.q = kMD P U (4. AF and AD.. The flux linkage mutually coupled to the othe.4. A. L D L Ld A We can also prove that. to give equal mutual flux. the remaining flux linkage is the same as for all other circuits coupled to it. This can be verified by using (4. The flux linkages in the F and D circuits. = kMDid.&id. (4.108 Chapter 4 where and are the leakage inductances of the d. i. and kMD are equal. LAD LD x d = kM. 4. we can represent the above relations by the circuits shown in Figure 4.7. we Fig.eF. where we note that the currents add in the mutual branch. LQ - XQ = kMQ P U (4. we usually call this quantity LAD.57) and (4. or (Ld . T o complete the equivalent circuit. Thus Ad - x d i d = XF - x F i F = AD - xDiD = A A. A A 4. . Therefore. for the q axis we define L .did.. F. From the choice of the base rotor current.2. the pu q axis mutual flux linkage is given by Following the procedure used in developing the equivalent circuit of transformers.. Ld is the magnetizing inductance Lmd. and A. As stated in Section 4.4 Flux linkage inductances of a synchronous machine.108) Similarly. = 0.xa)id. kMF. and D circuits respectively. XQiQ- .) + iD) (4. Similarly. 4..oxq or ud = -rid LAD(id . F. .11 1) A.5 Direct axis equivalent circuit.6. + iF + i.1 16) (4. A. for the q axis circuits ui = -ri. and D ) are coupled through the common magnetizing inductance L A D .118) to get =0 L~~ iF = (l/xF)(AF - xAD) = (l/xD)(x.iF.(id VD = o = -rDi.i.X. we can show that -rFiF - - LA..1 14) (4. = (l/‘f!d)(Ad We now develop an alternate state-space model where the state variables chosen are From (4.115) The above voltage equations are satisfied by the equivalent circuit shown in Figure 4. The d axis circuit contains a controlled voltage source wA. and A. UQ = 0 = .6 Quadrature axis equivalent circuit. .1 17) + iQ) These two equations are satisfied by the equivalent circuit shown in Figure 4. . .) .xdid -uF = - 4d)id + kMFiF + kM..1 13) Similarly. 4.1 IO) - AAD) but from (4. The three d axis circuits (d. LAQ(iq WAd (4. . AD.L...i. which we can incorporate into (4. 4. - + + + iF i. [(Ld . .1 2 Ad. + iQ) + . consider the voltage equations Vd = = -rid -rid - Ad - WA.(i. - xAD) (4*118) v .The Synchronous Machine 109 Fig. = (id F-z Jk + iF + i D )LAD.(id L. Note the presence of the controlled source in the stator-qcircuit. + i +i q Q t - u* d Fig. which carries the sum of the currents id. id The Flux linkage State-Space Model A. XFiF (4.. .iD) wx.5.Xdid .X. with the polarity as shown. and io. For the dequation ud = -rid - Ad - wX.125) Using (4.123) in matrix form.( ~ F / ~ F ) X F +(rF/tF)h - .wAq - vd or id = -(r/'i!d)Ad + (r/td)AAD = -rFiF OAq - vd (4.36).110 Chapter 4 N o w define then Similarly.127) Substituting for iF b or = .128) b = .1 The voltage equations The voltage equations are derived as follows from (4.118) and (4. (4.36) -uF - XF (4.124) and rearranging. we can show that where we define and the 9 axis currents are given by Writing (4.126) Also from (4.r F ( b / t F .A A D / t F > + UF (4. Ad = -r(Ad/td - AAD/td> . I -1/td I I I (4.124) 0 4.12. If saturation can be neglected the A. Substituting for AAD.. are the .126)-(4.1 18) and rearranging..130) and from the (I axis damper-winding equation. are ud.12.121). only inductances that saturate. in (4. A. are also constant. Using (4. 4. will have constant relationships to the state variables as given by (4. The forcing functions.. We can therefore eliminate A.. as given in (4. Therefore.and T.12.. are constant.102) are in state-space form..121) are needed to relate A. from the machine equations. LA. A. The auxiliary equations (4. appears in the above equations. and A. Equations (4.( D / 7 j ) + ~ (4. to the state variables. = iqAd ... and A. L M D and L M . This form of the equations is particularly convenient for solution where saturation is required.. and 6.3).2 The torque equation From (4. = -(rQ/tQ)AQ (4.120). A. since saturation affects only A. The magnetizing flux linkages A..102).... and LA.134) . I33).The Synchronous Machine 111 Repeating the procedure for the D circuit.133) Finally the equation for 6 IS given by (4. and A. The state variables are A. XD = -(ro/tD)~D + (rD/tD)hAD (4.120) and (4. L/Tj &= -(AAD/td3Tj)Aq + (AAQ/tq3Tj)Ad.3 Machine equations with saturation neglected I f saturation is neglected.13 I). we substitute for the currents to compute = We may also take advantage of the relation t q The new electromechanical equation is given by t d (called t. and A terms can be eliminated (see Section 4.1 2. 4. w . This form is convenient if saturation is to be included in the model since the mutual inductances L A D and LA.+. and (4.uF.. and A.idAq.in many references).129) The procedure is repeated for the q axis circuits. A.. (4.120) and (4.131) Note that AAD or A.95) q.124). (4.. For the uq equations we compute iq = -<r/tq)Aq + ( r / t q ) A A Q + wAd + (rQ/tQ)AAQ uq (4. uq. and this matrix form of the equation is not an appropriate form for solution. The 7 x 7 matrix on the right side of (4.135) Similarly.138) The system described by (4. Solution From the data of Example 4.138) is in the form k = f(x.138) contains state variables in several terms.36) to get (4. I: . Again the description of the system is not complete since u d and uq are functions of the currents and will depend on the external load connections. however.112 Chapter 4 These currents are substituted in the d axis voltage equations of (4. the q axis equations are (4.t).136) and the equation for the electrical torque is given by The state-space model now becomes + (4. Example 4.u.2 for the flux linkage model. serve to illustrate the nonlinear nature of the system.3 Repeat Example 4. It does. 050 3.720 0 0 -5.0.1..286058 X Q ‘h I_ X d (I .980Ad 1.235Ad 0..D pu 1.055 PU 1. either by digital or analog computer.101 PU 1. A..Q = 0.=LMD X F X d 0. this can be accom- .028378 PU +-+- 1 0.%) X F X D 0. + 0 - 4. and only these terms need to be corrected for saturation.15 = 35.278 66.15 +-0.550 = 0.12.=LMD t d ID L M D XD Xd ID LMD “!D X F .330 -o~1030 0 -5.115330 .000349 _-= LuD Xd 0.349Aq -0.eq = 0.005789 L.150 4.2) (I .490 = 0.101 +-= 0.1 15.854 -313.2381 PU L.003743 .605 .743 3.0.028378 PU 1 -1 .005928 . In the simulation of the machine.036 --“ LnQ .000235 LMQ 37j 4 : and we get for the state-space equation for the first six variables..126)-(4.49 0.388 0.1 L.000980 LMQ 7j 2 .005278 1 1.066282 = 3 & ‘& XQ = 0. =o A.2381 pu --.55 1 1 + -0. with D 2..756 .I LMQ 0.4 Treatment of saturation The flux linkage state-space model is convenient for considering the effect of saturation because all the terms in the state equations (4.e: LMD 37j&XF LMD ..133) are linear except for the .002049 0.e.0.000235 = 0.927 1..530 O -0. (I .055 35.235Aq -0.651 .1.000642 = 0.1.526 .550 = 0.308485 = 0.282 0 0 59 : .e.D = 0. These are affected by saturation of the mutual inductances L A D and LA. magnetizing flux linkages A and AAQ. 8 0 = 10-3 44.The Synchronous Machine x d X F 113 = = = .%) -5.%)= X D 0.Q 1..928 ~0x10~ o 0 0 0 284.0.0. 2 X F X F (I .642hq 0.003756 IF LMD --.001387 0. = XQ= 1 -=L.0.036 PU .005927 -MD L37j.044720 . 4. 7-J From (4.120) and (4. are constants to be determined from the actual saturation curve. is a saturation factor determined from the magnetization curve of the machine. and hence K.114 Chapter 4 / / h T iO M ‘MS i Fig.138) we have a set of equations for each machine in the form (4.139).v. F o r a round-rotor machine. 4. be iM. As a practical matter. LAD= UADO = L Q LAQ A O Ks = f(AAD) (4.exp[Bs(XAD - XADT)] XAD > k4DT (4.139) where K. For a given value of the unsaturated magnetizing current is iMo. for the d axis in (4. which is the sum of id + iF + io.iMo.140) To determine K.121) at all times to reflect the state of the mutual inductances. which in turn is a function of X A D . the q axis inductance LA. plished by computing a saturation function to adjust (4. is determined.103) and (4.141) where A. The saturation function K.142) . Thus we may write approximately jMA = A. and B. Knowing iMa for a given value of XAD. seldom saturates. the value of iMs is calculated. The procedure for including the magnetic circuit saturation is given below [ 18).7 Saturation curve for X A D . = f(M + L2p )112 (4. i M A = iMs .7. according to [ 161 LD A = KsLADO L Q = KsLAQO A K. The relation between X A D and i M is given by the saturation curve shown in Figure 4. For salient pole machines.the current increment needed to satisfy saturation. The solution is obtained by an iterative process so that the relation h A D K . The computations for saturated values of these inductances follow.1 3 Load Equations x = f(x. corresponding to L A D O . To calculate the saturated magnetizing current iMs. is a function of this magnetizing current. For flux linkages greater than XADT the current i M A increases monotonically in an almost exponential way. ( X A D ) = LAD& is satisfied. is first calculated. Let the magnetizing current. Note that saturation begins at the threshold value corresponding to a magnetizing current iMT. so it is usually necessary to adjust only XAD for saturation. we compute. the following procedure is suggested. while the saturated value is iMs. Let the unsaturated values of the magnetizing inductances be LADOand LAQO. 144) which we transform to the 0-d-q frame of reference by Park's transformation: VOdq = pvabc = Pvm& + Rei. equations describing the load are required. in terms of the state variables. and up. given the machine terminal voltages.143) In matrix notation (4. These constraints are found by solving the network. the set (4. are known. There are a number of ways of representing the electrical load on a synchronous generator. u. For illustrative purposes here.. Assuming that uF and T. and v is a vector of voltages that includes u d . or some composite of all three. For the present we require a load representation that will illustrate the constraints between the generator voltages. By inspection of Figure 4. . + L e P i a b c v or pu (4.13. to the state variables.8 where a synchronous machine is connected to an infinite bus through a transmission line having resistance Re and inductance L e . w and 6 for the current model. assuming no mutual coupling between Rei. constant power.Next Page The Synchronous Machine 115 where x is a vector of order seven (five currents.145) The first term on the right side we may call v. appearing in the equations.. 4. w and 6 for the flux linkage model). or I I Fig.. Therefore two additional equations are needed to relate ud and u. = u.142) does not completely describe the synchronous machine since there are two additional variables ud and u.1 Synchronous machine connected to an infinite bus Consider the system of Figure 4. the terminal conditions of the machine must be known. 4. To obtain equations for U d and u. These are auxiliary equations.8 Synchronous generator loaded by an infinite bus. we could consider the load to be constant impedance. and angular velocity. or five flux linkages. currents. including loads. which may or may not increase the order of the system depending upon whether the relations obtained are algebraic equations or differential equations and whether new variables are introduced.143) becomes vabc = vmabc + ReULc+ L U L (4. the load constraint is satisfied by the simple one machine-infinite bus problem illustrated below.odq and may determine its value by assuming that vmabc is a set of balanced three-phase voltages. Leia or phases. In other words. constant current. The voltages and current for phase a only are shown.8 we can write u. + + (4. For example. 145) may be computed as follows.32).148) 0 vodq = V .145) may be written as (4. but the order of the system remains at seven.149) which gives the constraint between the generator terminal voltage vodq and the generator current io.149). Note that (4. is the magnitude of the rms phase voltage.147) The last term on the right side of (4. Current model Incorporating (4. .Piobc= io. + Lei.f i -Sin (6 [cos(6 - .Thus Pinbe = iodq . Note also that there are two nonlinearities in (4. 4. (4. we can show that (4.151) . - WL.. in radians. The first is due to the speed voltage term.146) cos(wRt + + (Y 120") where V.150) Thus even this simple load representation introduces new nonlinearities.. Thus (4. we may write -Li = (R + wN)i + (4. From the definition of Park's transformation &dq = Piohr.1 3 2 .79. There is also a nonlinearity in the trigonometric functions of the first term. we compute the derivative iodq = Piobr+ Piobe. Using the identities in Appendix A and using B = W R t + 6 + uf2. the wLei product.Previous Page 116 Chapter 4 (4.PP-'iodq where the quantity PP-'is known from (4.149) into system (4. I:[ -iq v Or PU (4.1 pu or. for a given torque angle 6. The angle 6 is related to the speed by 6 = w .149) is exactly the same whether in M K S units or pu due to our choice of P and base quantities.J CY) + Rei. R.- 1 I 1 0 T" - I o 1 . The function f is a nonlinear function of the state variables and f . The infinite bus voltage V . and u contains T. aild y = 6 .-. (4. determination of its stability depends upon finding a suitable Liapunov function or some equivalent method.3). Ld...1 -- (4. which areAuFAand sion line is incorporated in the matrices R..149) and substituting for id and iq in terms of flux linkages (see Section 4. and Lq terms in L. 4.12.. namely. we may replac? the r . j .3 The flux linkage model From (4. and fi. and (4.. The loading effect of the transmisthe system driving functions.13. where x' = (idiFiDiqiQwS].The Synchronous Machine 117 where K = &V.37). appears in the terms K sin y and K cos y. Because the system (4. and I I I I I I I I I ---Ksin? -OF + I I W i ) 1 I I I I I I O I 0 -----K cos y 0 . u.154) The system described by (4.154) is now in the form of (4.154) is nonlinear. Now let i d = R = r + Re Ld + Le i q = Lq + Le (4. f ) . L.153) Premultiplying by -i-' adding the equations for & and i.152). but rather nonlinear functions of the state variable 6. to Using (4.-_ - I . i d .152) iq obtain the new matrices Lan d (R + w i ) . Note also that these latter terms are not driving functions. and N by Thus -Ksiny -ii = (ii + w h ) i + [ -: j K cos y 0 k. = f(x. This is explored in greater depth in Part I l l .0 .155) . 136) to get (4.159) A.156) with (4.157) and (4. A.138) to give the complete state-space model.118 Chapter 4 (4. we combine (4.139. The resulting equation is of the form Ti wherex‘ = [ A d AF AD = CX + D (4.157) Similarly. (4. .155) with (4.158) Equations (4.156) Combining (4.158) replace the first and fourth rows in (4. w a]. L and 0 0 0 I I 0 0 . premultiply (4.The Synchronous Machine 119 and the matrix C is given by I 0 0 I l I I I I I o o o 0 I i I I I 1. The infinite bus voltage constant K and the damping torque coefficient D are left unspecified. = 0.163) Equation (4. .10 d i q = L q + Le = 2. .3 to include the effect of the transmission line and torque equations.162) If T-l exists.163) is in the desired form.161) D = (4. 7j = 2HoR = 1786. I o] (4.2 and 4. .001096 i d = L + Le = 2.' to get X = T-ICX + T-ID (4. . product nonlinearities and trigonometric functions. i.. I I O 0 0 0 . It contains two types of nonlinearities. f ) and completely describes the system...4 pu.94 rad.-.159) by T . The line constants are Re = 0. .04 . in the form of x = f(x. u.e. Le = 0.4 Extend Examples 4. . Example 4. Solution R r 4- R. 67Kcosy Therefore the state-space current model is given by .00495 -0.71 K s i n y 1.080 I I I 0 -5.550 0 1 I 0 0.591 -1.00183 0.5900 -2.330 I I I I I 1.l [Ki:j 7z: ] [ -Ksin y .040 1.550 1.3.669 2.00187 -0.5470 0.867 -7.00436 0.0769 I 3.550 1.09007 .1.67 vF = .00044 0.0141 -0.490 1.6500 0.:Y. VF 0.286 0 Then - I I 0.001 18 -0.5880 I I -0.00065 -0.12332 i .490 .506~ and we compute 2.0131 I I 2.2060 i-yR + .0960 I -2.6.6090 _____________________-_-_---------------I -3.605 0 0 ______________~-_------I 0 0 0 2.526 -0.00074 0 0 0.669 -1.054oJ I I 0 0 0 0 I I 1.490 0 I I 0 0 0 II I L o 1.08 + 0.591 v. I 0 By digital computer we find 1.1.591 K sin y .A) = -0.710 -1.2020 .550 1.5880 2.87 -1.00187 -0.6500 -2.001 1 0 0.709 0 0 I 1.651 0 1.120 Chapter 4 Then 0.040 1.88 I w .4870 I I 2..1. 00029iq 0 1.12332 I I 0 0 I I 0 0 -_-_____-___________--1----.O0044 121 0.0019 0.3162 0.. .4319 0 0 I ..2365 0.162). = C X by (4.000280id -0.6090 0. I 1.0 I 1 I I I I I I o 0 0 1 0 I .000559 T. and D are given T = 0 0 0 0 I 0 .5.6500 -2.59 K sin y -1.0960 2. 0 L - The matrix C is mostly the same as that given in Example 4 3 except that the w terms ..6678 O I 1 I 1.08 K sin y .1 -0.2020 -0.The Synchronous Machine +O.4810 2. where T.0 I I I I I I O 0 0 0.0769 -0.8810 I I I -0.5880 -O.o _ r0.o 0 1.67Kcosy 1 0.59 uF . C.0 .87 vF 1.67 UP COS y -1.0005590 0 I I I I 0 I 0 0 0 I I 1 0 + i 1 1 7 K sin y .3162 0.00031id 0.Substituting.2060 2.71 K + 6..00183 I I 0 0 0 0 0 0.00029iq 0..547~ 0.0901 -0.00495 1.0141 0. are modified.6500 -2.5880 -3.00436 2.-0. -1 The flux linkage model is of the form Tf.159)-(4. + D.0. 789 0 0 1284......046A9 -1.7~ I 0 I I -5..388 44..705Ad O 2.854 -313...046Aq .720 2..756 -115.....743 3.236 0 . -0._ _ _ .330 -431.927 1.706Aq 1284.278 66..928 0 1 5.954Ad - 0 0 I I O 0 K siny VF + 0....024 -5.705Ad 0 p. Then from (4.17.910A9 I 0..278 66...7) we ... the flux linking the d axis circuit will depend initially on the subtransient inductances...720 -47.8~ 0 I I 1 0 0 0 ~ 667..756 -115...954Ad I -0.1._ ..55960 1 0 0 0 0 28.Chapter 4 .- 1 (4..910A9 I 0..282 3..40 0 I I 0 1 01 I .050 -5..8~ I I I 0 0 0 O 1 I 0 I I ____--_-_____________L_------------- 10000 0 j 188._ .282 -236.388 44.337 -207..766 1..7058A9 1.. and after a few cycles on the transient inductances.I 3162~ -747.. Let the phase voltages suddenly applied to the stator be given by where u(t) is a unit step function and V is the rms phase voltage.5.530 0 I __---____--____-_----------------1--------I I 1I I 0 10-3 -0..733 3..._I.529 .330 -13660 I 1 I 1-3162~ 2 1 1 2 ~ I 0 O I I 0 I 1 I _ _ _ _ _ _ _ _ _ ..14 Subtransient and Transient Inductances and l i m e Constants If all the rotor circuits are short circuited and balanced three-phase voltages are suddenly impressed upon the stator terminals..530 I ..164) 4.1 .854 -313.316 1------------ 2. the flux linkages XF and AD are still zero. Li L.171) (L.(kMF)’/LF Ld . (4. the same procedure will yield (at r = 0+) (4.164) is changed by 90” (ud = fiY sin e). i.2LAD -1 (4.e. but there is no other q axis rotor winding.20) for Ad. since they cannot change instantly.169) where L: is the d axis subtransient inductance. the damper winding current decays rapidly to zero and the effective stator inductance is the transient inductance.167) Substituting in (4. From (4.108).169) = Ld - LD + LF .172) (4. we get (at t 0) ’ LFLD .M i kZM$D io = - kM&F .L:D/L.168) The subtransient inductance is defined as the initial stator flux linkage per unit of stator current. after a few cycles from the start of the transient described in this section.2khfFkMDM~ (4.168) and (4.164) are suddenly applied to a machine with no damper winding. some clarification of the circuits that may exist in the q axis is needed.kMFMR LFLD ..The Synchronous Machine 123 can show that (4. A If the balanced voltages described by (4.174) I n a machine with damper windings.173) where L j is the d axis transient inductance. .M i (4.kMDMR LFLD . If the phase of the impressed voltages in (4. Thus by definition Ad ’ L : i d (4.M i id + LFk2Mi .166) Therefore iF = - kMFLD .LD/L:D) where L D is defined in (4. For such a machine the stator flux linkage after the initial subtransient dies out is determined by es- . ud becomes zero and us will have a magnitude of flV. Before we examine the q axis inductances. with all the rotor circuits shorted (and previously unenergized).165) Immediately after the voltage is applied. Thus at f = O+ = 0 = kMFid + LFiF + MRiD id = = 0 = kM& + MRiF + LDiD id (4. For a salient pole machine with amortisseur windings a q axis damper circuit exists. 124 Chapter 4 sentially the same circuit as that of the steady-state q axis flux linkage. = Lqiq -k kM& A (4. Le. Here the solid iron rotor provides multiple paths for circulating eddy currents. Thus for a salient pole machine it is customary to consider the q axis transient inductance to be the same as the q axis synchronous inductance. = V F u ( t ) .179) We can also see that when i . the 9 axis effective inductance in the “transient period” is the same as L.The voltage equations are given by rFiF -k = vF/F(f) = rDiD + A. Thus for such a machine it is important to recognize that a q axis transient inductance (much smaller in magnitude than L . 4. decays to zero after a few cycles.176) Substituting in the equation for A.182) 0+.L:p/LQ (4. A.181). the subtransient and transient reactances are numerically equal to the corresponding values of inductances in pu.. ) exists.(kMQ)’/LQ = L... Consider a step change in the field voltage. (4. = L..22]. Repeating the previous procedure for the q axis circuits of a salient pole machine.182) in (4. = 0 (4. .181) and the flux linkages are given by (note that id 0) = AD Again at t = = LDiD -k MRiF LFiF -k MRiD (4.178) where L: is the q axis subtransient inductance L : = L. . (4.183) Substituting for the flux linkages using (4.AD = 0 . We should again point out that for a round rotor machine L. . U.177) or A. < L.1 Time constants We start with the stator circuits open circuited. = [L. = -(LD/MR)iD (4.( k M p / L p ) i .1 80) Since the reactance is the product of the rated angular speed and the inductance and since in pu oR = 1.. The situation for a round rotor machine is different. .14. Thus for this type of machine L. < L. which act as equivalent windings during both transient and subtransient periods. or iQ = . L:iq = (4. Such a machine will have efleclive q axis rotor circuits that will determine the (I axis transient and subtransient inductances. This procedure has not been followed in this book but could be developed in a straightforward way [21. . To identify these inductances would require that two q axis rotor windings be defined.( k M p ) 2 / L p ] i . which gives for that instant i. and is given by (see [8]. r.186) Equation (4. approximately. Another time constant is associated with the rate of change of direct current in the stator or with the envelope of alternating currents in the field winding.&L. rFiF + LFiF Ti0 = vFu(t) (4. when the machine is subjected to a three-phase short circuit. 6) r.194) where L2is the negative-sequence inductance.4.192) (4.The Synchronous Machine 125 (4. and 4.e.3. It is denoted open circuit because by definition the stator circuits are open.while > write. Ch. When the damper winding is not available or after the decay of the subtransient current. we can show that the field current is affected only by the parameters of the field circuit.)/2 (4. = 1.195) Typical values for the synchronous machine constants are shown in Tables 4. the corresponding d axis time constants are given by r.. + L.193) For a round rotor machine both transient and subtransient time constants are present. 4.o = L D .185) Usually in pu rD > r F . where = LF/rF (4.M:/LF TD (4. r. = L 2 / r (4. Therefore we can (4.187) This is the d axis open circuit subtransient time constant./L./Ld (4.189) Kimbark (21 and Anderson (81 show that when the stator is short circuited. .oL.186) shows that iD decays with a time constant r. This time constant is r . L D and L F are of similar magnitude. i.188) The time constant of this transient is the d axis transient open circuit time constant r&.190) (4.5.191) = A similar analysis of the transient in the q axis circuits of a salient pole machine shows that the time constants are given by (4. which is given by L* = (L. 2 1.. Typical Hydrogenerator Characteristics Parameter Small units Large units Nominal rating (MVA) Power factor SDeed (r/min) .1000 M W Power factor 0.W.95 Direct axis synchronous reactance xd 140. Inertia constant H. are in the order of 1.. Table 4. Note: All reactances in percent on rated voltage and kVA base. High Synchronous condensers Low Avg.03 0.02 0. .035 0.. No attempt has been made to show kW losses associated with generators.0 . by E.35 0.8 0. U S G M. step-up transformers.2 0.95* 70-200 3.72 3.25 9.3 0.0 0.50 ...4 = T.. .S.S.. I 0-40 0.S% on rated kVA base.' 5.5-4.05 0. Note: All reactances in percent on rated voltage and kVA base.80-0.05 6..o-2. ..5 2.0 T: Ta 0.0 Souree: From the 1964 National Power Survey made by the U.0 2.65 4.0-5.02 0.. Source: From the 1964 National Power Survey made by the U.0 .. @ Wiley. .035 0. 5..50--0. USGPO. since generating plants are generally rated on a net power output basis and losses vary widely dependent on the generator plant design.80-0.95* 70-350 1.5 0.5 Direct axis synchronous reactance xd Transient reactance x i Subtransient reactance x: Quadrature axis synchronous reactance x q Negative-sequence reactance x 2 Zero-sequence reactance x o Short circuit ratio 25-45 20-35 .. High Waterwheel generators Low Avg...01 1.64 0.0 20-35 10-25 I .5 5.4. 23-35 Subtransient reactance x : 15-23 Quadrature axis synchronous reactance x q 150.17 11. 20-45 10-35 I . Table 4.6 0.30 1956.. since generating plants are generally rated on a net power output basis and losses vary widely dependent on the generator plant design. 6. 220 55 170-270 25 45--65 20 35-45 55 100-130 19 35-45 13 15-25 0.10 0. Typical Turbogenerator and Synchronous Condenser Characteristics ~~ ~~ Generators Parameter Range Recommended average Synchronous condensers Range Recommended average Nominal rating 300...5.04 0.35-0..05 1.0 .IkW-s) I IkVA) .8 0.035 0. Turbogenerators Low Avg.3.0 90-1 IO 40-200 0.8 9.0 Inertia constant H.2-l. kW losses for typical synchronous condensers in the range of sizes shown. excluding losses associated with.6 1. 0.15 1.8 0. 3.0-5.5 3.0-8..126 Chapter 4 Table 4.o-2. High 2. Kimbark. +These power factors cocdf conditions for generators installed either close to or remote from load centers. vol..160 18-20 Negative-sequence reactance x t Zero-sequence reactance xo 12-14 Short circuit ratio 0..1 9.180 Transient reactance x. Federal Power Commission. Federal Power Commission. 80-100 20-40 15-30 . Typical Synchronous Machine Time Constants in Seconds Time constant Ti0 7. No attempt has been made to show kW losses associated with generators.80-0..90 60 50-100MVA . Source: Reprinted by permission from Power System Stability.16 0. 40 115 40 20 0. 1 Neglecting damper windings-the F i model The mathematical models given in Sections 4.1 18). Machine with solid round rotor [2].196) .123).12 with io or AD omitted.10 and 4.12 assume the presence of three rotor circuits.120). The mathematical model for this type of machine will be the same as given in Sections 4. 4. usually those nearest the disturbance. i. For example. The simplifications adopted depend upon the location of the machine with respect to the disturbance causing the transient and upon the type of disturbance being investigated. Thus the complete mathematical description of a large power system is exceedingly complex. This assumption assumes that the effect of the damper windings on the transient under study is small enough to be negligible. Often only a few machines are modeled in detail.103) or AD and A.The Synchronous Machine 4.10 and 4.138) the third row and column are omitted.e. The model includes seven nonlinear differential equations for each machine. The underlying assumptions as well as the justifications for their use are briefly outlined. This is particularly true in system studies where the damping between closely coupled machines is not of interest. while others are described by simpler models.1 5 Simplified Models of the Synchronous Machine 127 In previous sections we have dealt with a mathematical model of the synchronous machine. The solid round rotor acts as a q axis damper winding. when the saturation is neglected as tacitly assumed in the current model. taking into account the various effects introduced by different rotor circuits. In addition to these.121) with the D and Q circuits omitted.103) and (4. (4. Another model using familiar machine parameters is given below. the model is also somewhat simplified. and the mechanical torque must be included in the mathematical model. [j= (4. Some simplified models have already been presented. The complete mathematical description of the system would therefore be very complicated unless some simplifications were used. From (4. and simplifications are often used in modeling the system. the excitation system. Some of the more commonly used simplified models are given in this section. Amortisseur efects neglecred. Neglecting the amortisseur windings can be simulated by omitting iD and t g in (4. In general. even with the d axis damper winding omitted. other equations describing the load (or network) constraints. In Chapter 2 the classical representation was introduced. and (4. tion. In this chapter..e. An excellent reference on simplified models is Young [ 191. they are presented in the order of their complexity. In this case the effect of the amortisseur windings may be included in the damping torque.. in (4. Situations arise in which some of these circuits or their effects can be neglected. by increasing the damping coefficient D in the torque equa. In a stability study the response of a large number of synchronous machines to a given disturbance is investigated. in (4. both field effects and damper-winding effects. i.15.138). (4. 59) we define ~ ~ E F RDk M F / r F ) v F W ( v (4. ) A d 4.205) all quantities.:iLF I[] IIL. since the choice of the rotor base quantities is based upon equal flux linkages for base rotor and stator currents.( L A D / L i L F ) A d + (Ld/L. pu.36) and rearranging.)X. -I J PU ~ (4.199).202) and converting to pu f l E & V B = WR(k M F u M F B / L F u L F B ) ( A .203) we compute i d = .208) .204) In a similar way we compute A.201) and (4.197) Therefore.200) Ad = -(r/L. + OX.204) and (4. WR(kMF/LF)AF v (4.36). substituting for i.199) and (4. LFBF B I / f i E & = (k M F u AFu / L Fu ) [WR ( M F B I F B / VB )1 or in pu LADAF/LF = dTE6 P U (4.203) AF/LF = P U (4. from (4. from (4.58) we define 6 E .LF)AFI + i f l E i / L A D F pu (4. UF = r F [ .198) (4. pu (4. - Ud P U (4.203) Now.( r / L . from (4.)Ad + (rLAD/L. i d = -rid - WAq - Vd (4.LF Ld/L.207) Also from (4.Lf E]= L [-L::. From the stator equation (4.199) to write Aq = -(r/L.205) Note that in (4. For the field voltage.200) or from (4. This follows.36) uF = rFiF + A.201) From (4.206) Now from (4.199) 0 0 The above equations may be in pu or in M K S units.LF)AF - WAq- vd pu (4.128 Chapter 4 or (4. ) f i E i - Ox. including lime. are in pu. from (4. and substituting for i F from (4.( r / L . the currents are given by -LAD/L. LAD L + 62 E.21 2). . d Block diagrams of the system equations are found as follows. .2 15) 8.206) we compute 4 * L*D EFD = - L. .. 2 Ad + LF Ld r L d 2f l E .213) we write.W A q 5 P U P U (4. + (r/L.204). .22 I ) Similarly. and (4.95) T. It is a fifth-order system with “free” inputs EFD and T.209). Thus the time constants must be in radians.( I / L i - I/Lq)Addq (4.( r / L q ) A q- Note that in the above equations all the variables (including time) and all the parameters are in pu.2 13) (4.209) From (4.idXq.Li and ri0 = LF/rF.2 19) (4. The signals v and Vq depend upon the external network.2 16) = Now we derive the torque equation.203) and (4.214) (r/Lq)[l + (Lq/r)sIAq= WAd - v q Pu (4.The Synchronous Machine 129 and converting to pu . from (4.205).)E. (4. model. LAD pu (4. ( r / L i ) [ I -t (L. or rPu= tsecwR rad (4.219). From (4./L..1 pu T. (4. = WAd . (4. and (4.f/fi d dq = A q / G vd U d / d vq = vd V q / 6 (4. - PU (4.. From (4.210) Rearranging and using L:D/LF = L d .215). and (4. Pu Substituting for (4.2 14) (4.2‘1 I ) We now define rms stator equivalent flux linkages and voltages A = A. T. in the s domain. \ / ~ E F D u vB ~ E F D = = u WR[(kMFuMFB/rFuRFB)UFu vFBl (k M F u UFu l r F u )(OR pu MFB UFB / vB RFB d E F D = LADUF/rF (4.212) Then (4.207).A.218) describe the E.220) Equations (4. (4.2 18) ~ j b = T.222) .2 1 1) become Ad = -(r/L:)A. From the swing equation = E./r)S]Ad = (r/Li)E.199) we get Tct$ iqXd . from (4.UAq - vd P U (4.DW 6 = 0 . id and i q .220) along with the torque equation (4.217) = (hd/Lq - + (LADAF/LiLF)Aq and by using (4.2 13)-(4. . and E F D are assumed to be known and v and V.9 Block diagram representation of the E. The effect of saturation may be added by computing the additional field current required under saturated operating conditions. From Ad = tdid + L A D i F substituting for id from and (4.199).L i ) / L i ] A d pu (4./Ld)SlE. I O Block diagram representation of (4. d Note that T.223) Now define r i d 4 L. The above equations are repreq sented by the block diagram shown in Figure 4.0(L.10 can be combined to give the block diagram of the complete model.and T.215) (Ld/L. 4.10.130 Chapter 4 I I Fig. The remaining system equations can be represented by the block diagrams of Figure 4. The model developed to this point is for an unsaturated machine. The block diagrams in Figures 4.220). depend upon the load.i)r. 4.)( 1 + 7.. model. i A = L q / r .o Fig. = EFD + [(Ld .9 and 4. 1 . . = & L i / L d .218)-(4.9. and from (4. )/L. from W R M F i F = &E in Section 4.L.[(Ld .2 12).2.4 we can show that iFLAD d E P U (4.1 I Block diagram for generating E .226) (4. I I . Another method of treating saturation is to consider a saturation function that depends upon E. should be modified according to the Figure 4. amounting to a negative feedback term and provides a useful insight as to the effect of saturation (see [20] and Problem 4.= EFD -E (4. (4.12. 4.33).7. and (4.The Synchronous Machine Ld 131 ..5 Determine the numerical constants of the E.[(Ld . This leads to a solution for E. model of Figures 4. with saturation.229) where EA corresponds to the additional field current needed to obtain the same EMF on the no-load saturation curve.4. It is also given that L: = 0.226).185 pu and Li = 0.245 pu.9 and 4.1 and 4.9 that produces the signal E. Young [ 191 suggests the modification of (4.215 ) .228) For the treatment of saturation. (4. ) E i . let EA = f A ( E i ) . T ~ O E . Solution From the given data we compute the time constants required for the model. the portion of Figure 4.)/L.203).229) and (4.225) E ( L d / L .L.IAd Substituting (4.L> Fig. Example 4.227) to the form E = ( L d / L .]Ad + EA (4. This additional current is a function of the saturation index and can be determined by a procedure similar to that of Section 4. then Also. Equations (4. We note that if saturation is to be taken into account.227) Now from (4.. . I I .227) into (4. ) E i .10. Le. using the data of Examples 4.228) can be represented by the block diagram shown in Figure 4. Stator subtransient flux linkages are defined by the equations &' = Ad .64 = 3. + e$ vg = . and L&' = L-i. In addition.231) (See [8] for a complete derivation.) From (4.15. vd = -rid . it is assumed in the stator voltage equations that w E wR.245)= 0.245 = 3. vd = -rid .671 rad The fictitious time constants 7 i d and TAd = T ~ = .446 rad 6. will respond relatively slowly to a change in terminal conditions. Note that (4.230) represents the more general case of (4.593 W R = 0.179) respectively. and respectively.023 s T ~ . + id%" + e : (4. in the equations for v d and v.233) (4. A = wR Ad " ed I1 = A = .1 14 (1. = (4.2 Voltage behind rubtransient reactance-the E" model I n this model the transformer voltage terms in the stator voltage equations are neglected compared to the speed voltage terms [ 19). = -ri. V.170) and (4.iqx" V.245/1.245)(3. 4.7 .230) and (4.593 s = 223. = 7 i 0 L. pared to the terms wX./r L. the field effects and the effects of the damper circuits are included in the machine representation.132 Chapter 4 From this we may also compute the short circuit subtransient time constant as 7.36) the stator voltage equations. -ri. which represents a special case of zero inirial flux linkage.473 4.0. These flux linkages produce EMF'S that lag 90" behind them.718 rad This large time constant indicates that A.&id A" = 9 . = ~A(0.1/1.00447 Note the wide range of gain constants required.230) where L i and L: are defined by (4. under the assumptions stated above.7 = 0. are given by vd = -rid .234) Now from (4./r = = (0.wRiqLi .233).185/0. The various gains needed in the model are as follows: 0.542 x = 3.967 s = 1495. These EM F's are defined by e.wRX.169).08 110.232) Combining (4. the terms icd and i are neglected since they are numerically small com.wRX../L.231) and (4.L"j9 4 (4.245 .w R AO " (4.riq .232). I n other words. = 8. Note that while some simplifying assumptions are used in this model. + uRXd + wRidL: + wRX. are computed as L.73 x 10-~)/1.1 18 x 10--'/1.939 1/0.245)/0.542x = 0. Equations (4. 4.13. under the assumptions used in this model.230). L” d = R L“ q (4.12 Voltage behind subtransient reactance equivalent. In this diagram the q and d axes represent the real and imaginary axes respectively.The Synchronous Machine 133 Fig. E” can be calculated. This EMF is called the volrage behind rhe subtransient reucrance. and if the current is also known.12. From ! . the d and q axis components of which are given by (4. we compute We can show that (4. the above we can see that if at any instant the terminal voltage and current of the machine are known. the relations expressed in (4. the voltage E“ can be determined. . “Projections” of the different phasors on these axes give the q and d components of these phasors. We now develop the dynamic model for the subtransient case.230) into (4.237) q axis “t ri Fig. Its components are E: and E: respectively. Substituting (4.13 Phasor diagram for the quasi-static subtransient case. 4. If quasi-steady-state conditions are assumed to apply at any instant.235) The voltages e: and e: are the d and q axis components of the E MF errproduced by the subtransient flux linkage.234) may be represented by the phasor diagram shown in Figure 4. Also if E: and E: are known.134). -:c E“ X’I R where. the terminal voltage can be determined.234) when transformed to the a-b-c frame of reference may be represented by the equivalent circuit of Figure 4. For example the voltage E” is represented by the phasor ? shown. Lii.244) Similarly from (4.203).251).xx )/(xi . .239) Using (4.. = tdd?FLAD K 2 = -L. L.e.L i ) / ( L . d We can also show that = LiD/LF (4. we can rewrite in terms of E. = (LAQ/LQ)AQ Now from the field flux linkage equation (4.i. (4.L. Then from (4. L M D /X ~ )~h . (4.249) (L. = L. we incorporate (4.247) A. . = 4dtFLAD x. .104). as (L.231). .230) and (4.x i .x : ) i .DLF/xdxFLAD)flE. + LA.~4 ‘ (4. LMDL.250) LADid L A D i F L D i D + + A.246) (4.)i. = 4dtD L: L. . .251) Eliminating if from (4.(xd - Xj)(id +iD)/lA (4. + ( L . : A = (L.K .245) where we define the voltage e d We can also show that A . D (4.x4 (4.231) to compute -(x. . + LAQiQ). we compute in pu e: = [(x: - X~ 11 ( d 3 .242) Substituting in (4.248) From the definition of L.240) Now we can compute the constants K .D) (4.G L M D L F 1 .L.238) Therefore we may write (4.iQ which can be substituted into (4. .226) to compute E = E.X d .x.x = I .240) and using (4.1 74) we can show that L .241) I . .L.L.243) (4. (4.203) and (4.i. = AD) + AD (4.104) in pu. .134 Chapter 4 since by definition (4.104) in pu AD = Xd)’ = LF/(LFLD -L. e: = ( L . = = WRLAQiQ (4.xd x. = = (L. = LADid + LFiF+ L A D i D (4.252) .LMD/?dxF)AF + (LiM D / t d 4 D ) A D L (4.236) as Ai A .E . = L. (4.261) (4.187) and (4.261) becomes T.o Similarly.263) . = e. (4.258). the following differential equations are obtained. and (4. e and .The Synchronous Machine 135 Now substituting (4.260) give the time rate of change AD. the system equations can be reduced to the following: (4. and (4.253) which can be put in the form (4.248). and E. jD -. (4.255) Substituting (4.. (4.250) in (4. (4.id A" d q and if w in pu is approximately equal to the synchronous speed.246). T.99. (4.36) we write rDiD + AD = 0 (4.(xi . (4.258) (4. i.36) V. To complete the model.257) (4.255).x. and (4.203).X $ ) * .259) which may be written as [WRrQ(LAQ/LQ)liQ + [(wR g . From (4.257) we get the differential equation ed / T I90 1 The voltage equation for the field circuit cames from (4. (xi . LC.252).36) we have 'QiQ (4. = rFiF + x.228) E.idA9 = By using (4.248). ' -E which can be put in the same form as (4..230) and recalling that in this model it is assumed that L. Equations (4.: E. The voltage e.254) In addition to the above auxiliary equations. in terms of i D .250) into (4. From (4.245).4Q)/LQliQ = Now from (4. the torque equation is needed. from (4. (4.249).192). (4.260) where E is given by (4. = i9 Ad .243). = E.231). (4. + e$id If saturation is neglected. is calculated from (4.254) relate these quantities to id and iq.)T. e. The auxiliary equations (4.256) + AQ = 0 (4.247). and (4.262) Te4 = i9 A" . which in turn depend upon the load configuration..256).. 241) and (4.. : + KZXD Dw/Ti (4..e:i. as defined in (4./3fi .136 Chapter 4 (4.xt)ld E. and K. E: = -(xq = .264) Now from (4.242) respectively. AD Tj0. / d Then (4.263)-(4. we may write e = d T K .268) To complete the description of the system.266) become (1 E" = e r ' / d = E: + jEj (4.x.267) (4. similar to (4. E. In doing so.270) E'b = +K2A~ + -(x 9 . + Xxdfd KIEi + (4.)fq E..266) (4. 4.14 Block diagram for the E" model. we eliminate the d f r o m all equations by using the rms equivalents.243) and using K . ( X i .2 12)..x") 9 Fig.269) (I (1 + + E: O) :S .ide:/3ri S=w-l The currents id and iq are determined from the load equations. AD = X . The block diagrams for the system may be obtained by rearranging the above equations. we add the inertial equations & = (l/Tj)T. .S) E. Solution The time constant T : ~ = 0.189).90s = 2224.15.14. If saturation is to be included.) Sb = w . s ) ~= T.969 x = 0. From (4.189/0. E = E: Example 4. corresponding to the increase in the field current due to saturation.279 rad which is about twice the d axis subtransient open circuit time constant.(xd . .15 Block diagram for computation of torque and speed in the E" model. We also need the d axis transient open circuit time constant.xi)(id + i D ) / f i (4.371 = 5.1-4..03046 s = 72.192) the q axis subtransient open circuit time constant is 7Y0 = L p / r Q = 1.5 to derive the time constants and gains for the E" model.15. 4. The remaining equations are given by (D + 7 . where we have defined (4.27 3) Use the machine data from Examples 4.14 and 4. - + E:/.149 rad is already known from Example 4.The Synchronous Machine 137 1 D+ 7.271) The block diagram for (4.270) is shown in Figure 4. a voltage increment E. Also the block diagram of the complete system can be obtained by combining Figures 4.6 + E.5.272) The block diagram for equation (4.075s = 28.248).5 U J 1 Fig. Ti0 = LF/rF = 2. is to be added to (4.25 rad Note that this time constant is about 30 times the subtransient time constant in the d .272) is given in Figure 4.423 x 10-3/18.1 (4. For the E" model we also need the following additional time constants. It is computed from (4.. 0. Kd xxd = = = K.274) and the corresponding stator voltages are defined by e.245 .1 5 3 . (4.185 = 1.)i. x. = 0. we can write.185) = 9.L:. To compute the gains. = -wRh.x4 E 1 - From (4.2.150)’ ( x i . A n additional assumption made in this model is that in the stator . vd = -rid .)(x. (4.0.245)(0. = .w ~ . Li i.64 . .)i. while the subtransient effects are neglected [18]. from (4. The stator transient flux linkages are defined by A. The machine will thus have two stator circuits and two rotor circuits. e . 9 .70 .0.x. approximately. The transient effects are dominated by the rotor circuits.275) (4.245 .xiid e .245 .632 (xd . = 1. However.277) (4.245 P = U We can now compute from (4. which are the field circuit in the d axis and an equivalent circuit in the q axis formed by the solid rotor.r i .(1. . we 4.70.271) L.(1.iq + e.150 x. the constant x. is usually small.0.J (xd .x.0.64 ./L.185 pu Then.x.673 (0. two-axis model Ad and 4 for a cylindrical rotor machine-the In the two-axis model the transient effects are accounted for. . 2 A. voltage equations the terms i d and i are negligible compared to the speed voltage terms and that w Y wR = 1 pu. Neglecting .651= 0. It is computed from (..x. algebraic equations). compute the gain.150) o.276) (4.279) Following a procedure similar to that used in Section 4.wRL.138 Chapter 4 axis.536 ) 0.)(x: --~. = Ld K.iq . This means that the integration associated with T : ~ will be accomplished very fast compared to that associated with . e.185 .270). = wRx.70.7) 414: .L.x[ = 1.278) (4. + rid + xii.LiD/LF 1. = vd u. .(l.x:) = (1.55)2/l.j0. = uq + r i .4 .0. + (xi .15. + URLjid + e. I = Aq WA.179)we compute Lc = L.id A A.245)(0.x.526 = 0.49)’/1. or e. or Li is needed. Since the term (xi . the number of diflerentiul equations describing these circuits is reduced by two since i d and k are neglected in the stator voltage equations (the stator voltage equations are now .455 pu. r vd + rid + x.0. . 16.282) with &E: -E: . The d axis : flux linkage equations for this model are Ad = L d i d + LADiF Ad pu XF = LAOid -k L F i F pu (4. Similarly. and e: are the 9 and d components of a voltage e’ behind transient reactance.X i Zq From the Q circuit voltage equation rQiQ + d X Q / d r (4.287) E + xdzd + XAZd = .275).174) and (4. are d and q axis stator voltages.The Synchronous Machine 139 Fig. Rather.(xq - X6)fq (4. = eq = LmiF pu = EA fiE d Ed = ed . as given by (4.203). for uniformity.L A Q ~pu Q (4.288) where. The voltages e. It is interesting to note that since e: and e. = -(LAQ/LQ)XQ pu = (4.281) + LAQiQ pu (LAQ/LQ)xQ XQ = LAQiq + LhiQ pu Pu (4.278) indicate that during the transient the machine can be represented by the circuit diagram shown in Figure 4.282) Eliminating iQ. we adopt the notation .285) We also define fiE We can show that [8].6 E .275) we get e : ’ fif?. .203). we can verify that e.286). We now develop the differential equations for the voltages e and e. = 0.279) and (4. in this model the voltage e’.286) (4.280) By eliminating iF and using (4.283) (4. Also. . is not a constant. for the (I axis X q = Lqiq = a E . and by using (4.XqZq = EA . = L:id P U and by using (4. 4.284) and by using (4.we compute = (Lq - LiQ/LQ)iq (4. e.. For example. it will change due to the changes in the flux linkage of the d and q axis rotor circuits. pu (4. = d E .16 Transient equivalent circuit of a generator.284) and (4. they represent d T tirn e s the equivalent stator rms voltages. Equations (4. which corresponds to the transient flux linkages in the machine. The gains are simply the pu reactances xq Xi 1.287).1-4. = EiId + EiIq . Again saturation can be accounted for by modifying (4.275) to compute T.228) (4. To complete the description of the system.18. The procedure for incorporating this modification in the block diagram is similar to that discussed in Section 4. based on the machine data of Examples 4.140 Chanter A 5 X d - xi 1+7bd 1 E' 9 Fig.290).15. ~ (4.288).287) can be represented by the block diagram shown in Figure 4.292) The block diagram for (4.2. the electrical torque is obtained from (4.289) Similarly.Jq] (4. - ( ~ d . T.291) Example 4.Li) i d l q (4.. Ti0 = TJO = LQ/rQ (4. .DW .[EiId 6=w-I + EiIq .17.LJ)I..274) and (4.6. the block diagram for the complete model is obtained. from the field voltage equation we get a relation similar to (4.17 and 4. In addition we obtain from the manufacturer's data the constant xi = 0.293) where E.(Li .17 Block diagram representation of the two-axis model.64 - 0.455 pu The remaining system equations are given by ~ j h = T. (4.380 pu. By combining Figures 4.260 pu Xd - xi = 1.(Li . and (4.17.380 = 1. E = E.290) Equations (4. = Xdi.Xqid. 4.70 . .. . Solution Both time constants are known from Example 4.0. is a voltage increment that corresponds to the increase in the field current due to saturation (see Young [ 191).which is combined with (4.245 = 1. ) I + E.18.7.93.7 Determine the time constants and gains for the two-axis model of Figure 4.292) is shown in Figure 4..x . 4 1 Block diagram representation of the one-axis model. The component e: is completely determined from the currents and u d .. It is similar to the model presented in the previous section except that the absence of the Q circuit eliminates the differential equation for E.id. - &. T.290) and (4.36) with i d = = E. .275).tq + r t d = Xdiq The torque equation is derived from (4.294) (4.9 .o Fig.The Synchronous Machine 141 b b E' 9 K 1 1 . PU E: = 5 + X. is obtained from (4. .274) and 1.E P U The voltage E.295) 0. the system equations are E 7.15.4 Neglecting amortisseur effects and and iq terms-the one-axis model This model is sometimes referred to in the literature as the one-axis model. changing by the field effects according to (4. 4 1 Block diagram representation of ( . 9 ) .274) and (4. (which is a function of the current i a ) . or e. The voltage behind transient reactance e' shown in Figure 4.oEi = E.293).. Substituting (4. and using (4. 4. - (Xd - x:)td PU (4.16 has only the component e.8 422.o 1 * u) Fig. Thus.99. ..io.296) with the network constraints (to determine the currents) and the condition that E..8. The voltage E.268). the system equations are given by (4. and E are approximately equal in magnitude and that their angles with respect to the reference voltage are approximately equal (or differ by a small angle that is constant).8 The simplified model used in Section 4.296) Thus the remaining system equations are - [EiI. 4. changes at a rate that depends upon ..Iq - (Lq - f--i)ldiq (4. From the assumptions used in the model.20 Network representation of the system in Example 4. w .2 (voltage behind subtransient reactance) is to be used in the system of one machine coiinected to an infinite bus through a transmission line discussed previously in Section 4.263)-(4. This time constant is on the order of several seconds... Under this assumption the voltage behind transient reactance E' or e' has a q axis component E. depends on the excitation system characteristics. is constant.(Lq .228) we note that the voltage E. . This is the constant voltage behind transient reactance representation used in the classical model of the synchronous machine. For the mathematical description of the system to be complete.13. that is always constant.15.i. or e. Solution For the case where saturation is neglected. These equations are obtained from the load constraints.15. which corresponds to the d axis field flux linkage.) remains constant during the transient can be justified. by neglecting the terms in h. .1 PU (4. does not change very fast and if the impact initiating the transient is short. E..L i ) I d I q ] P U 6 = w .297) The block diagram representation of the system is given in Figure 4. A.. This set of differential equations is a function of the state variables e. PU T p rj& = T. and A. A.5 Assuming constant flux linkage in the main field winding From (4. Under these assumptions E' is considered constant. L. in the absence of the Q circuit.Dw = E. 4. . I f E. Example 4. The system equation to be solved is (4... and 6 and the currents id and iq. Equation (4. in some cases the assumption that the voltage E.142 Chapter 4 = noting that. Le. The next step in simplifying the mathematical model of the machine is to assume ' that E..19. (or e. and Aq in the stator voltage equations (compared to the speed voltage terms) and also by as- I - Fig.266) expresses e l as a linear combination o f t h e variables E. The system equations neglecting saturation are to be developed. equations for id and iq in terms of the state variables are needed. This in turn implies the existence of inductances Ld. + P(Vwq (4. L.E. . Another source of concern to the power engineer is that the value of the machine constants (such as L.Li.“)] (4. one d axis amortisseur. the system reduces to the equivalent network shown in Figure 4.a) .E2 = (4. I n some stability studies. . it is found that two rotor circuits (on each axis) are sometimes adequate but the inductances and time constants are not exactly the same as those defined in IEEE Standard No.15. It 7&. 115. ments.263)-(4. and two q axis amortisseurs.10 [16].k f q + i f t I . discrepancies between computer simulation and field data have been observed. 211= V .. + E: + X.14. T o fit the “conventional” view of rotor circuits that influence the so-called subtransient and transient dynamic behavior of the machine..) used in dynamic studies are derived from data obtained from ANSI Standard C42.The Synchronous Machine 143 suming that w wR. which are in common use by power system engineers. (4.a) 1 V.k f d - ittIq E: Vmq= .301) Equations (4.234) are given by Vod = .. L i ..20. The representation of these paths by one discrete circuit on each axis has been questioned for some time.298) I d and Iq are determined Id = (R)2 [+ (P)’ R(V. The procedure for determining the constants for these circuits is to assume equivaI . are based on a classical machine with discrete physical windings on the stator and rotor. L. Data for these circuits can be obtained from frequency tests conducted with the machine at standstill. all of which are intended to define fault current magnitudes and decre. This implicitly assumes two rotor circuits in each axis-the field. These studies show that a detailed representation of the rotor circuits can be more accurately simulated by up to three discrete rotor circuits on the d axis and three on the q axis.268) complete the mathematical description of the system.. sin (6 . Studies have been made to ascertain the accuracy of available dynamic models and data for turbine generators (21-251.. L. As mentioned in Section 4. etc.V.2.1 6 Turbine Generator Dynamic Models The synchronous machine models used in this chapter. 4. cos (6 . and L: and time constants T&.298) where k and = r + R.299) Ve = .300) From (4.301) along with the set (4. T ~ and T. equations (4. the solid iron rotor used in large steam turbine generators provides multiple paths for circulating eddy currents that act as equivalent damper windings under dynamic conditions. By following a procedure similar to that in Section 4. It is now suspected that the reason for these discrepancies is the inadequate definition of machine inductances in the frequency ranges encountered in stability studies.147) and (4. Reprinted from IEEE Trans.s)(l + b .21 Frequency response plot for a 555-MVA turboalternator. Machine data thus obtained differ from standard data previously obtained by the manufacturer from short circuit tests. This comparison is given in Table 4. and c.0- c _p 1. Thus for the d axis we write Ld(s) = Ld (1 (I + a. b. . pu 2.303) and the constants Ld.. If the operational inductance is to be approximated by quadratic polynomials.006 I I IIlIlll ! 0.s) + a.144 Chapter 4 lent circuits on each axis made up of a number of circuits in parallel.0- - '**I 0.s)(l + b. May/June 1974. a.5 -MVA unit -Test resulk --Adjusted resulk for simulation of hvo rotor windings i n each axis I I11111 0.6 Frequency. b .6.. A n example of the data obtained by standstill frequency tests is given in [24] and is reproduced in Figure 4.06 I I 111111 0. 4.1 0. PAS-93.s)(l + c2s) (4. &(s) becomes (4. .302) where L is the synchronous reactance.2 1.304) are different from those associated with the exponential decay of d or 4 axis open circuit voltages... Both third-order and second-order polynomial representations are given. Thus. Hz I I 1 1 1 1 111111 6 111 l ~ l ~ l d 60 Fig. vol. hence the discrepancy with lEEE Standard No. and N ( s ) and D ( s ) are polynomials in s. for the d axis. a . 115. the constants can be identified approximately with the transient and subtransient parameters. The transfer function for each is called an operational inductance of the form Us) = [N(s)/W)lL (4. Speed. Reference (241 gives a comparison between the two sets of data for a 555-MVA turbogenerator. are determined from the frequency domain response.0006 I Frequency response plots 555. . s ) ( l + c.) (G IEEE.304) The time constants in (4. e. the accuracy of any dynamic machine model is greatly improved when the so-called standard machine data are modified to match the results of a frequency analysis of the solid iron rotor equivalent circuit.and Li are all particu.175 I . The inductance versus frequency plot given in Figure 4.1282 rad/s or 0. ?io.867 0. The transfer functions plotted in Figure 4. of the machine. I .61 0.the computed break frequency is = 1/4.The Synchronous Machine Table 4. s Ld L.2 17 1. the d axis time constant ?&-. denoted by r. Li L. 1974. ~ S ) ] . For example.o s ?o l s Ti0 s 7.1282 = 7.304). Since the amplitude at this frequency is the reciprocal of the d axis transient time constant.76 0. 145 Comparison of Standard Data with Data Obtained from Frequency Tests for a 555-MVA turboalternator Standard data Adjusted data Constants Pu P U pu P U P U pu 4 PU 7.00034 pu.3 = 0.3 0.: = 1/0.305) The break point that gives a better fit of the experimental data corresponds to a frequency of 0.90 0.2326 rad/s = 0.16 7. the break frequencies should give the constants of (4. It was found to be inferior to the more .061 1.00062 pu (4..473 0.213 0.56 0. it is more important in stability studies to use accurate machine data than to use more elaborate machine models. Also.8 s (4.97 0.21 is nothing more than the amplitude portion of the familiar Bode plot with the amplitude given in pu rather than in decibels.8 0. The machine constants thus obtained are given in the third column of Table 4.27 0.30 0. Reprinted from IEEE Trans.6. the machine constants obtained from the standard data are used to obtain the breakpoints for the straight-line approximation of the amplitude-frequency plots.:. given by 7.254 0.074 Source: o IEEE. A study conducted by the Northeast Power Coordinating Council [26] concludes that. If this is used to obtain the first break frequency for log [ 1 /( 1 + T . If. PAS-93.6. however. At the time of this writing no extensive studies have been reported in the literature to support or dispute these results. If this is done. as obtained by standard methods.A larly important in stability studies.16 4. the approximated curve does not provide a good fit to the experimental data.022 0. a comparison of these results and the machine models presented in this chapter are in order. The full model presented here is one of the models investigated in the NPCC study [26] for solid rotor machines. L.306) Reference (241 notes that the proper ajustment of ?. this corresponds to an adjusted value.03 1 0. Finally.3 s. L. is 4.81 0.21 can be approximated by the superposition of multiple first-order asymptotic approximations. in general. vol. 12) by sketching the stator coils as in Figure 4.5) is an orthogonal transformation.2 4.14 4. Problems 4.t + a ..18).22) and is not orthogonal.. Verify (4. are given in (4.24) for the neutral voltage drop.8 we wish to compute the induced EMF in coil sa-fa. D.34) and explain its meaning.10 4. The quantities Ad and A.? (c) Explain (4. 1 I] is that given by (4. improvements can and are constantly being made to provide mathematical formulations that better describe the physical apparatus.13 4. Furthermore.15 Park's transformation P as defined by (4.23). This is not surprising since the added detail due to the extra q axis amortisseur should result in an improved simulation. including the polarity of the induced voltage. Extend Table 4.16)-(4.5 4.13).COS(w. COS (WRI + a) V b ( t ) = V.39).20).2.1 I 4..negative? Why is I M. Verify (4. ~ . Why is the term negative? Consider a machine consisting only of the phase winding sa-/a shown in Figure 4.9) by finding the inverse of (4. (a) Use the rate of change of flux linkages &.32) and compute the speed voltage terms.39). Why? But the transformation Q suggested originally by Park [IO. as in many technical areas. Perhaps more surprising is the fact that the model developed here with F. 1 > L.io and T .. in pu. Can you explain why these inductances are constant? (b) Equation (4.(t) = V.146 Chapter 4 elaborate model based on two rotor windings in each axis. Verify (4. Substitute these quantities into (4. (b) Compute the Blv or speed voltage and the transformer-induced voltage.32). But. Verify (4.. including those developed in this chapter. Sketch a new physical arrangement where the field flux is stationary and coil su-/a turns clockwise. The traditional models. Explain the signs on these equations by referring to the currents given on Figures 4.. Do the results agree? They should! Verify (4.9 4. (Y .1 4.3 4. as done in this chapter.cos(wRf + (a) For the pu system used in this book find the pu voltages ud and u. (b) Repeat part (a) using a pu system based on the following base quantities: SB = threephase voltampere and Vs = line-to-line voltage. Do this by two methods and compare your results.15) in terms of the coefficient of coupling of these coils. ( t ) = V.1 and observing how the inductance changes with rotor position. Use the transformation Q to find voltage equations similar to (4. larger digital integration time steps are possible than with models that use the much shorter time constants 7& and 7t0. This is not to imply that the work of the past is without merit.5).4 4. and Q windings provided practically no improvement over a simpler model with only F and Q windings. Check the computation of PP-'given in (4.8 4.1 and 4. For the new physical machine proposed in Problem 4. with the F-Q model based on time constants . Let ~ . Check your result against (4.2 r / 3 ) + 2r/3) v. Verify the following equations: (a) Equation (4.1 and the field winding F.1 by including the actual dimensions of the voltage equations in an MLffi system. I 4. Verify (4.12 4. are often acceptable. Explain the signs on all terms of (4. Repeat for an FLfQ system. Why is the sign of M. Are these two physical arrangements equivalent? Explain.20). As a general conclusion it is apparent that additional studies are needed to identify the best machine data for stability studies and the proper means for testing or estimating these data. (c) For part (b) find the pu power in the d and q circuits and id and i. as related to the rms voltage V.14).6 4. e. A. = 1.25 Compute the saturation function parameters A .16 147 Using the transformation Q of (4.62) by choosing a pu time T = w R t rad. The M M F wave should be a stepwise sine wave.40 4. but +i.22 Consider a synchronous generator for which the following data are given: 2 poles. 4.28 4. 5/6 pitch..1. inductance. looking i n at the coil ends. Is it radially outward along d or q? 4. 4. and B. Show that the constraint among base currents (4. (start a. (b) The d and q axis circuit power in relation to the three-phase power. 4. find: (a) The d and q axis voltages and currents in relation to the rms quantities. Show the position of N and S salient poles and indicate the direction of pole motion. 4.22) (originally used by Park) and the MKS system of units (volt. = = &.3 is connected to a resistive load of R. and +ib enters sb.. Now assume the machine is operating at 1. i.17 Normalize the voltage equations as in Section 4. where at to 4. Sketch the slots and show two coils of the phase a winding. Label coil I sal-/a. Use the current model.). given two different values ofthe variables A. and finish a . 4.2 I Develop the voltage equations for a cylindrical rotor machine.23 Verify (4..32 and i M O correspond to A. 3 phases.20 Show that the I /wR factors may be eliminated from (4.8 but where the equations are those found from the Q transformation of Problem 4. 2 slots/pole/phase..141). = fi. Hint: Add two state variables related to the voltage (or charge) across the capacitance. Incorporate the load equations for the system of one machine against an infinite bus (shown in Figure 4. I 2 slots. Let A r = 0. entersfc. a machine in which the inductances are not a function of rotor angle except for rotor-stator inductances that are as given in (4.27 4. Repeat Problem 4.r at A.2iM0)/ 1.rgiven that when A. Assume the MMF changes abruptly at the center line of the slot. (iMs .iMO)/iMO = 0.54) based upon equal mutual flux linkages is the same as equal MMF’s in each winding.18 Show that the choice of a common time base in any coupled circuit automatically forces the equality of VA base in all circuit parts and requires that the base mutual inductance be the geometric mean of the self-inductance bases of the coupled windings. 4.. and iMs.. ) and coil 2 sa2-Ju2. Le.16)-(4. defined in (4. The synchronous machine described in Examples 4. = fl and i t s is the saturated current at A. Derive the equations for the state-space current model using uF and T.19 - Plot the MMF as positive when radially outward +in enters sa. etc..2 ~.2 and 4. the direction of currents at time t o . iyo.26 4. 6 slots/pole.29 for a local load simulated by a passive impedance. wherei. The load is to be simulated by a passive resistance.0 PF (internal PF) and note by + and notation.15: (a) Neglecting damper effects..The Synchronous Machine 4. and B.3 I 4. and capacitance. The load has a reactive component. Repeat Problem 4.2iMO = 0.13 1.24 Derive formulas for computing the saturation function parameters A.2 .29 4. ( i b .1.\/5.138). ampere.0 pu. 4. as forcing functions.18).30 4.. coil I beginning in slot I (0”) and coil 2 beginning in slot 7 (180”).. using the data and results of the previous problem.. Obtain the state-space model for a synchronous machine connected to an infinite bus through a series resistance.27 using the flux linkage model Derive the state-space model for a synchronous machine connected to an infinite bus with a local load at the machine terminal.8) in the simplified models given in Section 4.. = 1. 1. Compute the saturation function K.8 m .8. - X'b .34 using the second-order transfer functions for Ld(s) and L.21. C.33.(s). Kimbark. Wiley. 1956. E.sketd Bode diagrams by making straight-line asymptotic approximations and compare with thi given test results. Power System Stability. References I . New York. P4. I . W . 1964. 4. (c) Neglecting damper erects and the terms Ad and A.(s) given in Figure 4.33 Show that the voltage-behind-subtransient-reactancemodel of Figure 4. . 1951. 3.14 can be rearranged to give the model of Schulz [20] given in Figure P4.34 Using the third-order transfer functions for Ld(s) and L. Xi- I I I I I . if the rotor has two circuits on the q-axis. Chapman and Hall. New York. The General Theory ofElecrrical Machines.35 Repeat Problem 4.xi x. London. for a machine with sdid ropnd rotor. 4.304)and substitutini the standard data rather than the adjusted data.148 Chapter 4 (b) Neglecting i d and A. Adkins. B.. 3.35 using the second-order transfer functions of (4. Vols. 4.33 4. Synchronous Machines. Concordia.36 Repeat Problem 4. 2. Wiley.1 Fig. R . P. L. and Stephenson. Pittsburgh.. W. P. Two reaction theory of synchronous machines. B. R. lEEE Trans.. J.vsi. Per unit impedances ofsynchronous machines. Erects of synchronous machine modeling in large-scale syStern studies. 1972. Watson..The Synchronous Machine 149 4. 1965.stribution Keference Book. Direct and quadrature axis equivalent circuits for solid-rotor turbine generators. PAS-92:574 82.1610. Task Force on System Studies. 1929. PAS-93:777 -44. H. London. London. G. W. 1945. C. D. Pa. Young. R. Standard No. B. Park. 1973. 18. 1974. and Thomas. Wiley. Symposium on Adequacy and Philosophy of Modeling: System Dynamic Performance. PAS-88:1593. I . Test procedures for synchronous machines. Simulation ofsymmetrical induction machinery. P. Taylor. System Dynamic Simulation Techniques Working Group. I I . P. 1962. M. IEEE Trm. AI€€ Trans. Prentice. PAS-93:767. H.. IEEE Committee Report. IEEE Trans. Per Unit Systems: With Special Reference IO Electrical Machines. 1933. C. A basic analysis of synchronous machines. W. W. IEE (British) Monograph. 21. R.20. R. M.Digital comparisons with system operating tests. New York. S. Analysis of Synchronous Machines Connected to Power Network.. A I E E Trans. Park. Lawrenson. 1974. D. 1947. Press. Jones. Electrical Transmission and Di..109. PAS-88: 1121-36. 64569-72. P. Dandeno. P. Digital simulation of multimachine power systems for stability studies. P. and Manchur. IEEE Trans.. Schulz. 14. 1969. Dynamic models of turbinc generators derived from solid rotor equivalent circuits. Anderson. 13. P. 9. Synchronous machine modeling.. PAS-92:926-33. PAS84:1038-52. J. Iowa State Univ. 52:352-55. 75 CH 0970-PWR. . 1929. L. scale system studies. H. Press.. 6. 1971. 1945. A.c of a Synchronous Two-Machine System. R. IEE (British) Monograph. / € € E Tran. March. IEEE Trans. 8 . Cambridge Univ. D. Krause.76. Schulz. I): 716. 12. 20. Power System Stability. 1975.PAS-87:73-80. M . B.A. T. Ames. IEE (British) Monograph. Two reaction theory of'synchronous machines. PAS91:99. 24. I . PAS-77:436-55. and Schulz.. L. Westinghouse Electric Corp. 2. Northeast Power Coordinating Council.c. Hauth. 48:716-30. 7. Tensor Ana1. Prabhashankar. P. C. 1968. Jackson. Synchronous machine operational impedances from low voltage measurements at the stator terminals. Pt. Equipment and system modeling for large-scale stability studies. H. Rankin. Analysis of Faulted Power Systems. A I E E Trans. and Ewart. 1965. Stability performance of 555 M V A turboalternators--. and Winchester. Cambridge Univ. IEEE Publ. 22. 5 . Press. and Walshaw. Press. and Dandeno. I 26. Fundamental concepts of synchronous machine reactances. R. London. W. 16. 1973. L. Pt. Kundur. and Janischewskyj. 56 (Suppl. 2. Recommended phasor diagram tor synchronous machines. Final Report. Crary. Lewis. 19. R. Lynn. Etfects of synchronous machine modeling in largeTrans. N. 115. Harris. 1961. 25.839-41. R. A I E E Trans. 1969. Vols. 1958. A I E E Trans. K. 1973. G. C.. I E E E Trans.. NPCC-IO. IO. W. 1950. Pi. 23. Cambridge Univ.. IS. IEEE. 17. I . W. M. lEEE Trans. 1970. flux linkages. In all dynamic studies the initial conditions of the system are required. and EMF’S for the different machine circuits. This is a necessary part of any stability investigation.74) at steady state all currents are constant or. When the machine operates in a steady-state condition. For this situation phasor equations are appropriate. The initial position of the rotor with respect to the system reference axis must also be known. = 0 Then from (4. differential equations are not necessary since all variables are either constants or sinusoidal variations with time.1 Introduction This chapter covers some practical considerations in the use of the mathematical models of synchronous machines in stability studies. The number of these circuits depends upon the model of the machine adopted for the study. id = iF = iD = iq = i. 5. The soGalled “stability study” examines the system behavior following the disturbance. It is common to tacitly assume all machines to be in a steady-state condition prior to a disturbance. From (4. The phasor equations derived here permit the solution of the initial conditions that exist prior to the application of the disturbance. and these will be derived.74) 0 = iDrD 150 0 = iQrQ .chapter 5 The Simulation of Synchronous Machines 5. Both analog and digital simulation<are discussed. and construction of simulation models for the machine.2 Steady-State Equations and Phasor Diagrams The equations of the synchronous machine derived in Chapter 4 are differential equations that describe machine behavior as a function of time. determination of the parameters of the machine from available data. The remainder of the chapter is devoted to the construction of simulation models for the synchronous machine. Among these considerations are the determination of initial conditions. mathematically. Here we show how to obtain the required machine parameters from typical manufacturer’s data. The machine models used in Chapter 4 require some data not usually supplied by the manufacturer. These quantities will be determined from the data available at the terminals of the machine. This includes all the currents. 6) wR..)ejs (5.sin8) (5.' the .\/Z multiplier of (5. i.8) where E is the stator equivalent E M F corresponding to iF.)cos(wRt + 6 + ~ / 2 ) +(-ri.4) From (4. By using the relation j = 1 @in (5.( r i d + wLqi. Then from (5. are dc currents obtained from the modified Park transformation.5) u. The choice of this particular transformation introduced the factor l / d in the equation.12) I . i q vq = -riq + WLdid + kMFUiF + u. expressed as a phasor will have the two rectangular components 1.\/ZE (5. (5. and E are stator rms phase voltages in pu.Simulation of Synchronous Machines 151 or at steady state iD = i.74) as ud = -rid - d .226) we also identify WRhfFiF = . Using phasor notation.3) Using (5.. = (1.5) where by definition 8 = + b + u/2. = m [ . .. while id and i. from (4.7) From (4. Thus if the phasor reference is the q axis. uo = 0. w noted as reactances. and wL products may be de- wLd = xd (5. and I d .6) is conveniently used to define the rms voltage phasor (5.9).5) with balanced conditions.1 1) The stator current i. or wLq = x q = (5. + wLdid + kMpwiF)sin(wRt + 6 + */2)] = -[-(rid + oL.9) we may compute u. We define the phasor d = Aej" as a complcx number that is related to the corresponding time domain quantity a ( r ) by the relation a ( r ) = @ ( t / Z A e J w ' ) = a A cos(wr + a)./G (5. To simplify the notation we define the rms equivalent d and q axis currents as Id ii d / d I.4) and (5.9) where the superior bar indicates a total phasor quantity in magnitude and angle (a complex number).1) we may write the stator voltage equation from (4. + jI. - I . = m ( v d c o s 8 (5.iq)cos(wRt + 6 + r / 2 ) + (-riq + WLdid f kMFfdiF)cos(wRf + A)] WRt At steady state the angular speed is constant. Therefore.10) Note that in this equation V. = 0 (5. 10) and rearranging.0 152 Chapter 5 7 d axis I.13) I.@.. 5. from (5. along the q and d axes respectively.& leads the q axis by 90".2 Location of the q axis from a known terminal current and voltage.Xdld@ id jld@. we note that if the angle 8 is known the phasor diagram can be constructed quite readily.1 Phasor diagram representing (5. Note that the phasor jx.4).1 and (5.1 both V and Id are illustrated as d negative quantities. Also note that in the phasor diagram in Figure 5. This situation is shown in Figure 5.14). The phasorjxdT. EB= and by using E = ED. If the position of the q axis is not known but the terminal conditions of the machine q axis Fig.14) is shown in Figure 5. = (5. makes a 90" angle with the negative d axis since I d is numerically negative for the case illustrated in Figure 5 . + r c + jXqIq@ . Examining Figure 5. / Fig. 5. . 7 = .15). To obtain ud and u. I . and V are the projection of V . we compute the rms stator equivalent voltages d Note that V. Substituting (5. Thus the magnitude of d d is subtracted from x q l q to obtain the magnitude of V.1 [l].1 since lagging current (negative Id) is commonly encountered in practice. and E = E + rTa + jxqTq + j x d L (5.12) and (5.1 I ) in (5.14) The phasor diagram representing (5. 18) Note that Figures 5.17) where X e = w L e . are applicable to both.d V . if V.jV. from (5.2. The remarks concerning the location of the q axis starting from V .a) + Reid + u L e i q WLdid + kMFuiF = v'3VmC0S(6.a ) + Reid + u L e i q ' d V .x .11) and rearranging the above equations. The phasor diagram described by these equations is shown in Figure 5. c o s ( b . then add the voltage drop R.a) . . and the power factor angle are known.2 can be combined since the same q and d axes. + -rid . we compute = = Re)fq 0 + Re)[. Equations (5. I.V . is given by - V. where it is assumed that Vu.a ) + Reiq . I.a) ( r .wL. construction of the phasor diagram requires some manipulation of (5.16).id E VmC0S(6.. L is added (this is a phasor perpendicular to E ) .3 * Machine Connected to an Infinite Bus through a Transmission line To illustrate more fully the procedure for finding the machine steady-state conditions.sin(6 . = V.a ) + R. = R .. + ( x .4) into (5. which is different from that shown in Figure 5. sin(6 . is drawn parallel to 7 drop j x . (4.i. An alternate procedure would be to start with the phasor V. sin (6 .xq)& to the phasor q 4 5.2) is located on the q axis. ) f .2.iq . we solve the simple problem of one machine connected to an infinite bus through a transmission line. I d in the g axis direction and the voltage drop R.15).Id + X. with the q axis as reference.149) becomes u d = u. Then the voltage . The differential equations for one machine connected to an infinite bus through a transmission line with impedance 2.1 and 5.. Again remember that in Figure 5. are also applicable here. and f.149). in Figure 5.v3 V . sin(6 .. here as reference) the voltage drop rT. where the phasor representing the infinite bus voltage V.1.17) represent the components of the voltages along the q and d axes respectively. Thus in Figure 5. .. + X . Starting with E (used . + j u R L r is given by (4.wL.15). As we shall see later.3. it serves well to illustrate the procedure of finding initial conditions for any machine.a) (5. This can be verified by noting that the d axis component of the phasor j x q z is x q < . + jV. which is similar to that shown in Figure 5. in the d axis direction to obtain the phasor E. The end of that phasor (Eqa in Figure 5.. we add the phasor (xd . and the angle between them are known).uL. = .a ) + ( r By using (5.f.7) and (5.3 both I d and V. = . = V . However.I. -ri. cos(6 .3 the machine terminal voltage components Vd and V.3. and the same current I.d are shown as negative quantities.Simulation of Synchronous Machines 153 are given (i. Under balanced steady-state conditions with zero derivatives. Its q axis component however is xq&.(xd + xe)fd (5. this simple problem helps us concentrate on concepts without becoming engulfed in details..e. . an alternate procedure for locating the position of the q axis is illustrated in Figure 5.id (5.16) Substituting for v d and u. the same EMF E..1. can be obtained using (5.. Although this one-machine problem is far simpler than actual systems. Thus to locate the phasor E in Figure 5. 5.. are still applicable except that the currents id and i. /‘ v3 Ild = i.19) I.17). and currents are the same as given by (5.16). These are the d and q axis components of the transmission line current i. and (5. which at steady-state conditions are the same as (5. One convenient method is to add a local load at the machine terminals. = i.. I n other words..4. should be replaced by the currents iIdand i.17). l49).20) The transmission line equations are then given by Fig. with the q axis as a reference.4 Machine Connected to an Infinite Bus with local load at Machine Terminal The equations that relate the infinite bus voltage V.14). . = I. Equations (4.154 Chapter 5 q axis Fig..12) respectively.4. (5. to the stator equivalent EMF E are given by (5. .4 One machine with a local load connected to an infinite bus through a transmission line. are given explicitly.13... where we define (5. r. EMF’S. as shown in Figure 5. Since the terminal voltage is a quantity of considerable d interest./& (5. we seek a solution in which V and V. 5. 5. Note that this form of the equations does not give the machine terminal voltage explicitly.3 Phasor diagram of(5. + j l . For the system shown in Figure 5. the steady-state equations for the machine voltages. Simulation of Synchronous Machines 155 which can be stated in the form (5. IS).22) To obtain a relation between can write the phasor relations 5 and T .vd/RL) + xc(fq V. and (5.4.d =: I d - vd/RL f.25) I. ( x e / R L ) + v. xcfd (5. we refer to Figure 5..a) + Re].xe(Id . (5. = where we define 2.26) into (5. By inspection we (5.a) + R.v .26) Substituting (5.25).24) From (5. COS(b . COS(b .(ld . + jvd Separating the real and imaginary components./RL (5./RL) = v.22).) + Vq(Xe/RL)= .1 Special case: the resistive load.4.a) - RL Now define (5.27) (I + 2 ) ~.vd/RL) Vd(I R. .V.)(R.f.22). (5.27) and rearranging.. 5.Sin(6 .(Y) + R e ( f q.T./R.T. + j X L ) = v.(I + R./RL) v.25) The equations for the q and d axis voltage drops can then be obtained from (5./RL) .24) we can solve for I.28) . Sin (6 .V.V. and t l d r (5. zl = Rl + i0 f. vd = vq = or -v. Substituting (5. + + + X. From (5. c o s ( ~ v = .15) into (5. = RL + jXL. = For this case X L = 0. . (L .CY) .23) iL To . 34) Combining (5./RL)E El = v.(I + r / R L ) + X.CY) + + R./RL) ?d = xe(I + r / R L ) -t X d ( 1 + R .I. Substituting (5. makes an angle y with the 9 axis y = arctan[X. COS(6 - CY) + ReIq (5. sin(6 - CY) z: z: + Reid + X e I . Sin (6 - CY) + R.29) Let us define a phasor E .?. : ( I + Re/RL)E + j(Xe/RL)E where the phasor E .22) and rearranging.32) or Vd(1 . (5./(R. .2 The general case: arbitrary For ZL arbitrary the equations are more complicated.~ . = x.25) into (5. COS(6 - CY) - XdId + . vd(I + RLRe + xLxe) + V q ( R L x e- XLRe = . 5.15).31)is shown in Figure 5.28) can be written as 4- R. + XLXe)/Zt A2 = ( R L X . .5.I+.4.29)--(5. (5.156 Chapter 5 Fig.30) + RL)] (5.33) (RLR. a (5. + R.31) The phasor diagram for (5.XLRe)/ZZ (5.(l Then (5.V .5 Phasor diagram of a synchronous machine connected to an infinite bus with local resistive load./RL)E (1 = = . 5.VdX2 where XI Xi) vqX2 = .ld + xelq XeId + V. k.v. / R L ) ddId A (X. s i n ( 6 .(I + XI) V.33) and (5. xdXZ 5 td X .5.Simulation of Synchronous Machines 157 X2E = . + d x ) / ( V a+ rlr % = .) + j ( x .I. 5 X. 5.. + x d ( l + A . where I.lX)l (5.(I + X I ) + rx2 + R.I. I n either case V.sin(6 . V. (5.38) Fp = COS@ (5.COS(~ a) - fdld (5. and I. sin(6 .) (5.. .. it can be represented by the same phasor diagram in Figure 5. ) + rA2 5 + X. lugs V. + R. + rl.40) The phasor E. in phase with E. ) ( r rl.B in The angle between the q axis and the terminal voltage Figures 5. X.5 Determining Steady-State Conditions The most common' boundary conditions are the terminal voltage V . = (5.V.x d ( l Xl)]Id + [ R .xqX2 + X. as a reference. We also define the power factor Fpas + jf. = I. is the component of I.)E = h= R . x . + j I .cos(6 . and 4 (the power factor angle) are assumed to be known.39) -I.15).XZ r ( l + X l ) ] I q + Again. and the power factor Fp or the generated power P and the reactive power Q (per phase). by defining + + + + (5. (I + XI)E = El 5 (1 sin(6 .11).a) + &Id (I V.a) + [ . can be determined from (5. I . and either the current I. + r(l + XI) . I.xqlx + V .37) Since (5. . Then I.sin 4 where 4 is the angle by which I . i d Re + r(l + A .6.sin(6 .X . (5.36) we may write (5.29).cos$ I. 5 = E + (r + jx.2) is given by 6 - E P = = tan-'[(x.l.. we write Resolving 7 into components with is the quadrature component (which carries its o w n sign). + (I.P + 4) I. .41) (Le.44) The remaining and the rotor quantities id and i.) = + jx. = I.x. = I. I and 5.x.2 is given by E. . ..cos(6 . The currents are obtained from Id = -l. Then we compute Vd ..rXz . I .V.a) [Re + r ( l + XI) . .P ) V.P + 4) (5. ) . the angle 6 .35) in the form X 2 E = . can then be determined from their relationship to Vd and V.P ) and ud and u.xdX2lId [ X .35) + X I ) E + &E.37) is of the same form as (5.42) (5. given by (5.43) The above relations are illustrated in Figure 5.V . in Figure 5.cos(6 . ( l + XI) + rXzll..)E (V. 25). Case 2: Machine terminal conditions V. f . can be determined directly from (5. Case 3: Conditions at infinite bus are known.a are known. I and 5.158 Chapter 5 ' q axis Fig. r. From TL and &. is + determined from (5.7. .2). The terminal current I. I . From K.5. is calculated. The voltage of the infinite bus is then determined by subtracting the appropriate voltage drops to the machine terminal voltage E. is determined from (5.13).a are found. as shown in Figure 5. The infinite bus voltage can then be determined by drawing R d I d + f . we can also determine the power and power factor at the infinite bus. From this information I d . and power factor are known. Then from ' F and Z L we can determine &.25).42). In the case of a synchronous machine connected to an infinite bus the same procedure is followed if the conditions at the machine terminals are given. From (5. I .1 Machine connected to an infinite bus with local load Case 1: V . and the power factor the position of the quadrature axis is determined (see Figure 5..37). r-..37)the phasor E. 5. Thus and the angle 6 . .)..15) we can determine V. and V.. I. the position of the q axis is determined by a procedure similar to the above. If the terminal conditions at the infinite bus are given as the boundary conditions.. Also E can be calculated from (5.6 Phasor diagram illustrating (5. From I. The machined and q axis currents and voltages and the machine terminal voltage can then be determined.. the machine terminal voltage V. can be found. currents and flux linkages can readily be determined once these basic quantities are known. In this case Id and I. Now the conditions at the terminals of the machine are known and the complete phasor diagram can be const ructed . E. can be constructed. This is illustrated in Examples 5. and the machine load angle 6 . parallel to the d axis and R .. . and knowing V .f d I d parallel to the q axis. V. Vd. ) . and V. and the power at the infinite bus is given by 3( VmdIId Vm4f.41) and (5. is found.2.I.36) and (5. and Z . The three-phase power of the machine can be determined from the relation P3+ = 3(VdId + V . 5. Then from (5... from which we can determine Vmdand V-. The current I. 8126 = 39.000 pu 0. power factor. The terminal voltage is 1. = = = xq = r = F.620 x 1.2. = 1.Simulation of Synchronous Machines 159 Fig..788 Then from (5.85 PF lagging conditions (nameplate loading).3 is to be examined at rated power and 0.001096 pu COS^ = 0.6 Examples The procedures described are illustrated by several examples where different initial conditions are given.0/0. I The machine described in Examples 4.B + 4 = 31.640 PU 0.64 .64 + 1. lags the q axis. = Re Le Z.884" = angle by which 1.40 pu impedance to a large system.850 1.44) + 39.4005/87.000 + 0.000 fasin@= -0. Calculate the steady-state operating conditions.40) I. = = 31. Example 5.788".001096 x 0.138" From the given power.02 + j0.00 x 1. .700 PU y. Then from (5. and voltage we compute f. 5.02 PU 0.176 PU The angle I$ is computed from F. = cos-'0. find the infinite bus voltage.001096 arctan 0.00 x 0. as 4 f. = 1. 5.0. = I.7 Construction of the phasor diagram for Case 2.096" arctan and 6 . Solution From previous examples and the prescribed boundary conditions the following data are available: x.620 1.620 From (5. and 4.096 = 70.4 pu 0.85 f.6 (6 .1.85 = = 1. I f this machine is connected by a transmission line of 0. 4.j3) = = 1.42) and Figure 5.0 pu.cosd = 1. = 0.8)] where. I f we subtract the three-phase t 2 r losses.Csin39.667 = 1.344 PU -0.776 PU = U.X d t d + 0.667 = 0.(Y = 27.979) = 1.70(.385 + 1.55(-1. are both zero.979 .209) and (5.788') 1.925 x 1. .828 1-27.O . which should be numerically equal to the three-phase power in pu.994 kMFid + LfiF = 1.651)(2.= E .979 .cos(6 .925) = 1.1 60 Chapter 5 t.rti = 1. We also calculate the infinite bus voltage for this operating condition.(0..1.634 = A.979 PU The currents io and i.776 2.925)(1.935 kMDid M R i F = I.666)/1. = 1.631 pu u = -1.667 pu = -t.09" = V . /a . = 0.000.70 x 1.B) + (.994 = A. iF = &E/L.001096 x = EFD 0.345 + 1.55 pu.i.925 pu From (5.. = = = Ldid + kMFiF = 1.. we confirm the generated power to be exactly P = T.1. .55 x 2.0 . = 0.899" or PU Thus we have V.1 12 at steady state [from (4.49 x 0.788" I . .925) + (1.B - CY) = 39.345 (id + iF)kMF = (2. = 1.995" .55)(2.667 x 1.112 pu id = -1.31.634 Lqiq = 1.OLp. T.@ = 1.55) = 1.4712 155. from Example 4. (fl 2.925) + (1. . The flux linkages are given in pu by Ad = A.385 pu i. = LAD = Now using (5. AD A. = 6= .899" = the angle by which The angle between the infinite bus and the 4 axis is computed as = V./@ = l. and ld = t. I76 /@ 3 1.8) in pu.349" = 0. Then iF = 1. = 1.001 pu. As a check we calculate the electrical torque T.Ze<.094 kMQi.0. = & I. = = i Ad .1 by inspection E = = = V. 0. 0.979) = 1. A. = t.828 pu.64 x 0.004 Then T. ~ 0 ~ 3 4 . and @ . 0 9 " 0.899 = 66..4005 /87. We can write V. Then = V.096 + 27..1. E leads 6 - (Y = (6 ..55(2. A..idA. I38")( I .sin(6 -@ + 4) = -1.43) V.666 + rIq . = = = A...1.092 d pu From Figure 5.B + 4) = 0.094 = 3.l76//3 .. kM. Let V .p v. 620 from which we get 1.620. = 0 .e. i. / 'd q axis REF Fig.85 = 31. 5. + RJX V.021. pu and the component of the current in phase with V.) I + [ 1. The angle p between and 7 is given by an equation similar to (5.)/(l.020 . i. and V . viz. tan 4 = tan (cos-' 0.8.0.2 Let the same synchronous machine as in Example 5.0.788" (Y These angles are shown in Figure 5.1 is needed.020) = 0 .980 .. = 0 . = 1. 9 8 0 ~ ~The angle 6 .e.]/( I . A slight modification of the procedure of Example 5. Solution A good approximation is to assume that the power at the infinite bus is the same as at the machine terminals by neglecting the ohmic power loss in the transmission line (since R.O pu current). is I.0./1. between and V.41. 4 ~ ~ . Using the identity tan @ = (tanp we compute 0.020(0.0 pu.0201.1 be connected to an infinite ~. while the power and reactive power are known at a different point (the machine terminal). the voltage is known at one point (the infinite bus).0 pu at 0. 1.021.392 + 0. used as reference. Let 13 Re = (I. is given by tan0 = 1. with V..85 PF). X J . = = 0.392 + 0.020 = -0.85) = 0. 0 2 ~ ~Then the power at the infinite bus is 0.1. A better approximation is to assume a power loss in the transmission line based on some estimate of current (say 1 . infinite bus voltage is 1. The bus through a transmission line having R. 0 2 ~ and Le = X. . = 0 . tan p = .41. The power factor angle at the machine terminal @ is given by 4 = p +e = COS-'0. The machine loading remains the same as before ( P = 1.) + (0. + Ref. . .41. Then + tan8)/(1 - tanptane) .XJ.42). The boundary conditions given in this example are "mixed".0.8 Phasor diagram of V .Simulation of Synchronous Machines 161 Example 5.0201.020 .392 + 0.)1. is small).021. ..217 pu.00)2(0. .392 . T. 8 ' 243 4 = 19. is given by E = = ( V . 54) With a = 0.) In studies of large systems these boundary conditions are satisfied by iterative techniques.162 Chapter 5 From the known value of I. The position of the q axis can be determined from an equation similar to ( ..7 Ad = 1.46) .020 + 0.1 105 . 1 .cos 4 = 1 . we can now determine 8. = iF = E = ud = U.483 = 31.0.31" pu 0.. For the one machine-infinite bus problem the system may be solved explicitly.980)= 1 . + &Ir)+ j(Xcfr &I. We now consider the bus consisting one machine-infinite bus problem with a local load connected to the of a shunt resistance RL and a shunt capacitance CL.310 + 12.) + 1.X.150 = A . representing the transmission line susceptance. as reference (a! = 0).701 2.1o 93o 8 = tan-'(0. = -1.004 1.106 + j0.213/0. and age and both are used in this book. A.388 = 1. 0.217 = I .676 A. = = 3. local load. voltages.001 I n steady-state system studies (often called load-flow studies) it is common to specify the generator boundary conditions in terms of generated power and terminal voltage are commonly used for the terminal voltmagnitude.980 .5 The apparent power injected at node I may be computed as f.1. (Both V .= P. using a digital computer. + jQI= Frf = ViFE + ~~~~~ (5.148 165 . and flux linkages can then be calculated as in Example 5. The terminal voltage V .003 /. P and q. .I.529 -1.826 2. .591 0.. The results are given below in pu: id = i. Le.4 = 1. with 7.12.004) (1. and line may be conveniently described as a two-port network (Figure 5 9 for which we write.) (5 4 ) .172/19. A.793" which is a good check (see above).48" pu The generator phasor current is = and P = %I. = A . = 194 .OOO I pu (on a three-phase basis).j0. T.082) Also 1 = 1. The system of generator. The currents. 862" = 92.9 One-machine system as a two-port network Then we may compute PI = GllV: + V.73. We then compute or y - Yll = ylo+ jj12 GI.02 + j0.4005/87.. Example 5.cosB + B.Simulation of Synchronous Machines 163 Fig.862 .0 (on a three-phase basis) V. Thus we may solve (5. = = + jX.2.47) PI.V.1 and 5.G. I/Ze = YI2/y -0.788" or = l...2792 /3 = 92.3 Compute the steady-state conditions for the system of Examples 5.074".l7/19. where the given boundary conditions in pu are P = 1. . In doing so. and B. = 1.1347 . GI. are specified.01 pu. .(G.01 pu.01 Solution For the numerical data and boundary conditions given. In (5.4969/92.48) F = COS(? ..47) we define (5.V~)/Y12V. from which P can be found.47) for the angle P. there are limits on the magnitude of PI that can be specified in any physical situation.G1. Then from (5. we compute Z. .4 = 0.01 + jO.. 5. and V.4938 = 2.8) = (PI . = F = + jBll = 0.sinP) (5. = G.48) as (PI .V. V.lV:)/Y12V..OO and where the local load is given in pu as RL = 100 B L = CL = 0.4838 pu We now compute the quantity F defined in (5.138" pu pu -yl.j2. Obviously.47) where we define F. Thus the admittance from L node I to reference is ylo= 0. = Re q. it is convenient to define a constant angle y related to the admittance element = YI2/y.1247 + j2.Vm = 0. + jBkm for all k and m. = 1 .. are known or computed system parameters.17 V. while G . as the cosine function is bounded in (5.788 = 19.074' Then 7-P = COS-' F = 73. = - = 0.862" We are also given that RL = IOOpu and B = 0.48). we note from Figure 5.4 pu.Xe(Kcos@. It is connected to an infinite bus through a transmission line having Re = 0. and voltages are the same as in Example 5.0072 + j0. we compute d-q currents. with the result i d = .2161 pu = 0.2199 radians Then.056 -1.950" 4 B +@ = 30. (Figure 5. The quantity Eqa of Figure 5. > / @ 0.)/Z [R.570 Ad = 1.V. pu = & +& + 90" (T . = 2. XAQ = X .8 PF. .4 pu power at 0. = 1. Now 6 We also write = = = = (V:/RL)E+ ( q / X .500 2. I . A.164 Chapter 5 To find the currents.4 The same machine at the same loading as in Example 5.180 = 3. < = + I'.024".003 = 1. voltages.j0.897 1.sinP z 3 /12.6' or .164/30.V.) + X.2 may be computed as a means of finding 6.1.9 that = + &.446/54. XQ = 1.(V. P. Solution The internal machine currents. Thus.746" The power factor is F .595 PU 0.1 pu and X.9945 1.0149.V. P + jQ = ci.000 Example 5.99056 . as in Figure 5. = cos 30.161 2.672" PU We may now compute.024" PU and 6 = 54.8). in pu.1 1. Thus with a = 0 we compute. as a check.746' = .j0.. = 0.9739 .2012 = 0.163 = = = uq = 1.661 E if = = Xf T.696" With all the above quantities known.859 = 1.9667 .1..Ksin8] + j[R.662 iq = 0. 2.0.746" 6 .cos@ . noting that lies at an angle B from V .6.709 ~d = A.@ +4 = 65.000 + j0.V.794 = 1. Find the conditions at the infinite bus. = IaD= 0. flux linkages. Then we compute 6-0 = 34. and flux linkages in pu as in Example 5.1 has a local load of 0.)] = 0. 666 From the local load information I IL I = 0.j0. PU Therefore IL = 0.3pu.(0. .112)(-0.0.776 x 0.16 + 0. These are the steady-state conditions that exist before the impact. = f d = 2 = .1 2 x 0 1 / 2 0 ’ = 0.8) = 0 5 . Assume that a reference frame is adopted for the power system. From (5.001096 X 1 1 .V.0 1 x 1.)(. = 0.64 = -0.673 V . RL = 16 .16 .673)(-0.839 = From (5.(0. = V .4 . Then the transmission losses are 0..13)(2.303 .6 x 0 1 + 1.16 . in pu.. This reference can .4 + 0.1 12)(2.CY) = (.001096 x 0. = [(0.a) = -(-1.13 .14pu.112 0.303) (0. .385)(2. 5.1.666) = 0.7 Initial Conditions for a Multimachine System To initialize the system for a dynamic performance study.119) + (1.6 Then Rd = R.001096 X 0.501 v.13 X 1.Simulation of Synchronous Machines Id 165 = = = I.25) I.776 x I. A2 = (1.2639 x 0.2 + 0.096” 0. sin ( 8 . terminal voltage.13 = 2.631 2. 01 + .2639 4 0. = vd = E = 0.554 PU The power delivered to the infinite bus is P .385 39.6 x 0.776 -0. which is verified by computing RJ. 0.372 .0 Thus we compute from (5. = . COS (6 .0)2= 0.3 1 7 X 1. 1. 6 - p -1.4)/(2.37) 01 + . If they are not specifically known.673)2+ (0.13 = 2.4/(1. 4 16 .6268 .001096 x 1.7 = -0.6 0. XL = 1.501 = The power delivered to the local load is PL = 0.34) X I = (1.385)(-0. the conditions prior to the start of the transient must be known.631 x 1 6 = -o.385 - + 0.2 ZL = 2.0. = (-0.119 . We can also determine that.1197) .666) = -0.) 0.4 + V .4 pu.6268) + 0.2 x 0.16)(2.d = + -1. power factor. From the knowledge of these conditions we can assume that the power output. = 1.O x V. .1197 .372).631 x 1 2 .501)2]1’2 0. a load-flow study is run to determine them.64 x 1 1 + 0.112 0. and current are known for each machine.4 . 7) i. can also be calculated once the d and q components of I are known. and let the q axis be at an angle 6. which indicates the rotor .’ Negative-sequence = x2 Zero-sequence = xo Armature-leakage = xt . Three-phase MVA Frequency and speed Stator line voltage Stator line current Power factor Parameters: Of the several reactances supplied. = xi = xi = xi’ Subtransient q axis = x. From the load-flow data we can determine for each machine the component I.22) traditionally used by the manufacturers.15) to determine Ei.14) From or (5. vaj 5. can be determined. For a more detailed discussion see Appendix C.3 that the pu self-inductances of the stator and rotor circuits are numerically equal to the values based on a manufacturer’s system. we can determine the angle Si . The following data are commonly supplied.. By using an equation similar to (5. we can determine I. the pu system used here is somewhat different from the manufacturer’s pu system. Note that pi is determined from the load-flow study data. Ratings: m. It was noted in Section 4. but the mutual inductances between rotor and We shall attempt to clarify these matters in stator circuits differ by a factor of this section. while di is the desired initial angle of the machine q axis. I d j .7. vdj.Pi) is the load angle or the position.. and Vqi. The flux linkages .. and with the base rotor quantities chosen to force reciprocity in the nonreciprocal Park’s transformed equations.8 Determination of Machine Parameters from Manufacturers‘ Data The machine models given in Chapter 4 are based upon some parameters that are very seldom supplied by the manufacturer. Then by adding the angle & we get the angle d. I n addition. during the study it will be assumed that this reference frame is maintained at synchronous speed. Then from (5. Consider the ith machine. Let its terminal voltage phasor Vaj at an angle Pi be with respect to the arbitrary reference frame.Pi for this machine.42). Once it is chosen. Furthermore. The difference between these two angles angle between the q axis and the terminal voltage. peak-rated stator current. which can be used in (5. and 6. Reactances (in pu I: Synchronous d axis Synchronous q axis Transientdaxis Transient q axis Subtransient d axis = Xd = x. however. it should not be changed during the course of the study. which is the initial rotor angle of machine i. peak-rated stator voltage to neutral. . with respect to the same reference. They are usually given in pu to the base of the machine three-phase rating. This is necessary because of the choice of Park transformation Q (4.. of the terminal current in phase with the terminal voltage and the quadrature component I. Typical generator data supplied by the manufacturer would include the following. the values of primary interest here are the so-called unsaturated reactances.166 Chapter 5 be chosen quite arbitrarily. ’ Resistances (in Q): Stator resistance at 25°C Field circuit resistance at 25°C Other data: Moment of inertia in Ibm.io = T. LO.eF Li which can be put in the form = 4 d + LADXF/(LAD + XF) PU (5. The field-winding leakage is calculated from Figure S.‘ = 7. the corresponding value of LA. Rotor per unit quantities: Calculation of the rotor circuit leakage inductances is made with the aid of the equivalent circuits in Figure 5. Rotor base quantities: I f & in pu is known. then L A D in pu is determined from L A D = L.. L.. the base mutual inductance M F B is calculated from (4. .X d . L ~L. The base rotor quantities are then determined from (4.50) . L. where iF is the field current that gives the rated voltage in the air gap line.. . L.49) (5.57).Simulation of Synchronous Machines 167 Time constants (in s): Field open circuit Subtransient of amortisseur ( d axis) Subtransient of amortisseur ( q axis) = .IO(a) by inspection: inductance . The mutual field-to-stator inductance MF in H is determined from the air gap line on the no-load saturation curve as d V B = WBhfFiF.55) and (4.10....ftZor WRz (sometimes separate data for generator and turbine are given) No-load saturation curve (at rated speed) Rated load saturation curve (at rated speed) Calculations: The base quantities for the stator are readily calculated from the rating data: SB = V A rating/phase V A VB = stator-rated line-to-neutral voltage V f B = stator-rated current A wB = 27r x rated frequency rad/s The remaining stator quantities follow: Also the stator pu inductances are known from the corresponding reactance values. L.56).and& are known. Thus L d . in H is then calculated. eQLAQ/(&Q + LAQ) (5. + LAD (5. .10(b).& .eF The same procedure is repeated for the q axis circuits.1 I by inspection: L: 44 (5.ed and 4 .54) 4. by inspection of Figure 5. 5.1 I Equivalent circuit of the 9 axis subtransient inductance.168 Chapter 5 6) Fig.52) The self-inductances of the field winding L and of the amortisseur LD are then calcuF lated from L F = .e.e. (b) subtransient inductance.)] (5. .eq = .55) from which we can obtain . a temperature rise of 80-100°C is usually assumed.LF(Li . Similarly.51) from which we can obtain e. IO Equivalent circuit ford axis inductances: (a) transient inductance. is determined from Figure 5. (5. = LADeAL: . and the winding resistance Fig.53) = Lq - where . + LAD LA. = LD 4 . + = LAQ[(L. If this data is not available. 5. .td)/[L.56) and the self-inductance of the q axis amortisseur is given by Resisrances: The value used for the stator winding resistance should be that which corresponds to the generator operating temperature at the rated load.tq>/<Lq- L:)I (5. kMD.70 . We should always ascertain that the parameters thus calculated are self-consistent.60) Since all the inductances in (5. I . 7. Finally. Similarly.and MR. However. The machine parameters are to be calculated and compared to those obtained in Example 4.15 = 1. the engineer may find it convenient to assign values for this data from typical data available in the literature for machines of the same size and type. = = = Ld L. Example 5.60) are known. This section illustrates the procedure that can be used to determine the parameters of the machine.185 = L: PU = Ti0 = 5. from = (L.193) rQcan be found.380 PU = 0. often long before the machine is fabricated./L.WQ/rQ) P U (5.192) and (4. From (4.0.5 + 25)] Q (5.59) where ~i~ is given in pu time. can be determined.190) the d axis subtransient time constant is given by T : = [(LDLF - LiD)/rDLFl(L$/Li) rD pu (5. The latter are usually estimated for a machine of given size and type. We should also differentiate between data obtained from verified tests and those obtained from manufacturers' quotations. (4.61) Again note that T$ and 7. = 0. data supplied by the manufacturer may not be available in the complete form given in this section. are given in pu. L A L.15 pu 1.Simulation of Synchronous Machines 169 is calculated accordingly.o l = 0.245 PU = 0.Similarly. Thus for copper winding the stator resistance for I00"C temperature rise is given by r125 = '25[(234. . This may also explain apparent inconsistencies that may be found in a given set of data. xd xq X : X : = = X.5 The data given by the manufacturer for the machine of Example 4. LI = 1.70 pu xt = k d = 4.58) The same procedure can be used to estimate the field resistance at an assumed operating temperature.023 s . When some of the data is not available.64 PU = 0.187) and (4.189) we compute r~ = W 7 i 0 PU (5. Actual values for several existing machines are given in Appendix D. other information is available to estimate the field resistance.075 s 7.5 + 125)/(234. From (4.55 This is also the same as kMF. The damper winding resistances may be estimated from the subtransient time constants.1 are given below. = 0.24 s Solution We begin by calculating the pu d axis mutual inductance LAD = 1.9 S T$ = 0. 055 PU Also.0.154) and (4.70 .150)/(1.054 pu Then from (5.61) rp = (1. Also.526 PU From the open circuit time constant 5./L. the analog computer can be very useful in representing other nonlinearities such as limiters (in excitation systems) and saturation (in the magnetic circuit).49 PU 0.56) &Q= L.9 Analog Computer Simulation of the Synchronous Machine The mathematical models describing the dynamic behavior of the synchronous machine were developed in Chapter 4.0.49[(0.490 = 0.101 .65 1)(0.15 = 1.64 . = 1.651 .55)(0. = 1. For example (4.245) 0. I . 5.0131 PU From T$ = 0.19 rad 0.15) (1.59) r.O.)?$ = 8. Now.0.l85)] 1. from (5. from (5.550 + 0.50) = kMQ = 1.101 = PU + 1.651/2224. = (L.163) have two types of nonlinearities.651 PU From (5.640 .60) rD= = (1.185 .(1.46 ms = 3. These types of nonlinearities can be conveniently represented by special analog computer components.eD = LD = (1.423 x pu and from (5.25 = 7.605 PU = 0.eF = L .25 rad We compute from (5. Note that the equations describing the machine are nonlinear. = 1.55 x 1.0.185 .036 70 = A = 1.023 x 377)(0.605 x 1.l85/l.52) .9 s = 2224.1.0.170 Chapter 5 LA.19)(0.185 . We begin with the analog simulation. Thus in many ways the analog computer is very well suited for studying synchronous machine problems.651)(0.55 = 1.15) 1.101) .075 s we compute T. .15)/( I . The remainder of this chapter will be devoted to the simulation of these models by both analog and digital computers.55)(0. A brief description of analog computers is given in Appendix B.036 PU + 0. a product nonlinearity of the form xixj (where xi and xi are state variables) and the trigonometric nonlinearities cos y and sin y.55)(0.245 .0.64) = These values are the same as those calculated in Example 4.55[(0.055 = 1.101)(0.245)j 0.526/3.185) (1. .66) & I The mutual flux linkage A.)dt + A. For analog computer simulation (5.. I 18) the d axis and field currents are given by = (l/&. - Ud From (4. For example the state-space equation of the variable xi is ii= J ( x . recall that the state-space model of a synchronous machine connected to an infinite bus is a set of seven first-order. Thus the equations developed in Section 4.) - wx.120) Then from (4.) are also added. r . are the state variables. are the driv- (5. .2.)(x.(O) (5.) and the mechanical torque (for T.) (5. t ) where xi.n.A”.. it is usually limited to problems involving one or two machines with full representation or to a small number of machines represented by simplified models (2. .64) (5.12.63) wherea is the computer time scale factor and wB is required if time is to be in seconds (see Appendix B).5].129) A. .12 are used for the analog simulation..A. u .126) A.128) 1 dt + X. j = 1.69) = (l/tF)(AF - The analog representation of the d axis equations is shown in Figure 5. When the equations for the excitation system (for u.2. Thus while the analog computer is well adapted for the study of synchronous machine dynamics.(O) (5. nonlinear differential equations.4. = >l‘ id iF ‘D (A”.65) and from (4. The model most suited for analog computer representation is the flux linkage model. is computed from (4. Note that all integrand terms are multiplied by wB to compute time in seconds and divided by the time scaling factor a.Simulation of Synchronous Machines 171 To place the matter in the proper perspective.1 Direct axis equalions From (4. k ing functions. Complete representation of only one synchronous machine with its controls would occupy the major part of a large-size analog computer. 5.62) is written as = (5. to avoid differentiation. = 4 l’ [ (A”. . ..9.62) 1. however. . the system is typically described by 14 differential equations.68) (5.. The differential equations will be modified. and u. 3. .X. . 9.2 Quadrature axis equations From (4.12 Analog representation of the daxis equations.70) and from (4. from (4.130) (5. 5.73) Then the q axis current is given by. L -‘ AQ -‘ Q Q ‘AQ -“^Itt 9 Fig.71) The mutual flux linkage is computed from = LQ(hq/&q + hQ/&Q) (5.XAQ) The analog simulation of the q axis equations is shown in Figure 5. I3 I ) XQ = ( A ~ Q XQ)df - + XQ(O) (5. 5. I23). .72) (5. 5.13 Analog sirnulation of the q axis equations. jq = (I/&q)<Xq .13.172 Chapter 5 -a -A ‘AD -A Fig. .if. as shown in Figure 5.76) where if. To generate Vd and u. (5. the following scheme. Howd .74) i. 5.14. + wL.14 the machine terminal voltage and current for phase a are given by u. - Rei. is the phase a current to the infinite bus.) R (5. From Figure 5. The machine is assumed to have a very small resistive load located at its terminal. This load is represented by a large resistance R .3 load equations = In (4.14 One machine-infinite bus system with local resistive load.9.149) a 0 will be used for convenience.Simulation of Synchronous Machines 173 Fig.id]dt + i.75) are useful in generating the voltages u and uq. which should be avoided in analog computer simulation. . cos 6 + u. Therefore. 5. ever. suggested by Krause [2]. . = 2 lo ' [ .\/5 V . if they are used directly.74) and (5. -L Fig.75) Equations (5.15 Analog simulation of the load equations. is used.(O) (5. differentiation of id and i will be required. = (i. 5.. 90) and (4.?!-% Aa wA dt + .4. Vd = (id .74) and (5.77) where i and if.. since d = wB(w.80) The analog computer simulation of (5. From (4.174 Chapter 5 Fig.a and -6 is shown in Figure 5.i. the current if can be resolved into d and q axis components id.78) where T = (i. Following a procedure similar to that used in Section 5. with zero initial conditions and with a time scale factor of a. added as required by Figure 5.78)-(5.we can write (5.6(0) 180 A elec deg (5. and iqf given by (5.and b . The ud and uq signals are obtained from Figure 5. .I ) = W B O A pu.17.17 Simulation ofwA. = (iq .i. 5.17. we compute 6 = !.)R (5.)R u. Equation (5.4 Equations for w and 6 T. with subscript t . Therefore.w.16 and 5. are obtained from (5.78) is integrated with time in seconds to .Xd . 5.80) is shown in Figures 5.79) Note that the load damping signal used is proportional to wA (pu slip).75) respectively. 5.14. compute. Most analog computers require that 6 be expressed in degrees to find sin 6 and cos 6 [6]. with hA = h in pu and = 2HwB.74) and (5. are given by (5.idXq)/3. (5. . The generation of the signals .15. The analog computer simulation of the load equations is shown in Figure 5.73).o- Fig.68) and (5. I6 Simulation of the electrical torque T. requiring appropriate values of D ..9.75). The currents id and i.14 by inspection. The analog repre- &A ” -I -1 1 .99). 13. However. sin 6 and V. and 5.13.12. an electronic resolver is needed. . The complete analog representation of the system is shown in Figure 5. 5. To produce the signals V. cos 6.15-5.17 generate the basic signals needed to simulate a synchronous machine connected to an infinite bus through a transmission line.18. 5.18 Analog computer patching for a synchronous machine connected to an infinite bus through a transmission line. additional multipliers are needed. and whd shown in Figures 5. It is important to - 100 SW 1017 r Fig.12 and 5. other auxiliary signals are needed. For example to produce the signals wX.Next Page Simulation of Synchronous Machines 175 sentations shown in Figures 5. 2 is to be simulated on an analog computer.550 1.18. is 10. = 0.. The system response to changes in U. The time scaling used is 20. The integrator for the speed is kept at hold position. The initial conditions may be calculated from the steady-state equations (as in Examples 5. at zero. In that figure the analog unit numbers and the scaling factors for the various signals are given.055 4.605 = 1.526 = = = r = 0. in operation.9. The integrators for the flux linkages are allowed to operate with the torque T.101 0.666.1-5.12 5. The basic connection diagrams for the analog simulation are given in Figures 5.6 The synchronous machine discussed in Examples 4. Example 5. = 0 and w = w R = constant. the scaling factor for A.001096 rF = rD Le LAD 0. In the first step E F D is applied with T.5.g..02838 = L. including the w integrator.37 &Q= 0.3. I..640 LD .036 Le = 0.0131 1. To initialize the system for analog computation. H T:. Solution The data for the synchronous machine and transmission line in pu is given by: L. = 1. This scaling is best illustrated by an example. This value of E F D with the proper scaling is introduced into the integrator for A. and these values may be used to initialize the integrators. = 2. the analog computer is made to initialize itself by allowing the integrators to reach the steady-state conditions in two steps.1 represent the steady-state conditions. = 0. I . and in Example 5. and 5.490 1. is then applied with the speed integrator in operation. 5.17. The steady-state conditions thus reached correspond to initialization of the system for transient studies. Then T. rQ = 0. This builds the flux linkages to values corresponding to the no-load conditions. = 0. The operating conditions as stated in Example 5. .02 0. LD = = 1.400 = 5. The overall connection diagram is shown in Figure 5. is switched on with all integrators. The load T.00 pu and E F D 2. is to be examined. L. maintaining the speed constant. Note that EFD = E in the steady state.3).1-4.00074 0.0540 L F R = 100.700 1.150 R.6 the scaling is given in detail for the simulation of the synchronous machine. e.828 E The additional data needed is T. and T. The settings of the various potentiometers and the scaling are listed in Table 5 . However.90 s s V.0 4d &F = = = 0. which is given in parentheses. the following procedure is used. the analog computer may be used to compute these initial conditions.Previous Page 176 Chapter 5 note that signals are added by using the appropriate setting for the potentiometers associated with the various amplifiers and integrators scaled to operate within the analog computer rating.e 0. As explained in Section 5.651 = = LA.02836 = 1. 00 r0 0 0 00 r- 2 0 w $4 J Q s 2 I 2 I Q 2 w r w w 0 2 rw - R 3 2 w q 0 I I I 2 I 2 \ 2 I - 9 m - 9 9 z z 2 8 z z z Q Q Q rn a . 0 0 00 “ $ N N II I I 1 I -I-$ “4 1 01 01 2 I 2 2 2 I 4 e 0 m 0 % 4 4 Ei - . v . VI 0 0 X 8 2 'c! m 8 "/f 8 II 9 1 rr: 4 - ro 0 0 00 - 9 ' c ! c ? c ? 00 VI 22 - 0 0 - g o E 3 * o F: F: I= 2 m SI2 B R R a a a a 5s 2 2N 2 N 8 d 8 d g g a SI d . They are compared with the values computed in Example 5. the angular velocity error a.. where all plotted quantities are given in pu.1. zero V. Thus a 10% change in EFDis 0.19-5. ..23. 10% T is 0. the electromagnetic torque T. Figures 5..21 show the following analog computer outputs: the change in the exciter voltage E F D . and the rotor angle 6 . Example 5.. the mechanical torque T. The steady-state conditions reached by the analog computer are listed in Table 5. the stator d axis current id. 5. shown in Figures 5. and .2666...O).1.19-5. corresponds to a terminal voltage V. which is 10% of the nominal value computed in Example 5.180 Chapter 5 Fig. Similarly. The results of the simulation are .1 is used as a base for the computer runs.+.19 Response of a machine initially at 90% load and 90% excitation to a 20% step change in excitation.2. the field flux linkage A.3 pu. the terminal voltage error V. = 1 . of 6 p u (or V. The generator is initially loaded at 90% of rated load (T'+ = 2. The system is initially in exactly the condition calculated in Example 5. This excites a well-damped oscillatory response.. V. id. load (Example 5. and 6 (as well as other variables that are not plotted).io = 5.2. this . However. Figure 5..is the first disturbance. and E F D . w . 5. A good degree of damping is evident.1 conditions) to a 10% increase in T followed by a 10% increase in EFD to assure stable operation..20 shows the system response to 10% step changes in both T. Note that the response to this change in E F D does not excite an oscillatory response except for a small. particularly in T.9 s). The terminal voltage responds nearly as a first-order system with a time constant of about 4 s (.. welldamped oscillation in ob..1 with computer voltages given in Table 5. Fig..19 shows the response of the loaded machine to a 20% change in E F D . A 10% increase in T.20 Response of a machine initially at lOOu/.7). Figure 5.Simulation of Synchronous Machines 181 . .is applied. than the initial value.. The result is a fast movement toward instability. but at a higher A. Repeated runs of the system have indicated that corrective action is required before 6 reaches about 95". followed by a 20% increase in EFDto restore stability. This quickly restores the system to a stable operating state at about the same angle 6 as the initial angle.182 Chapter 5 Fig. which if not arrested will cause the machine to fall out of step. The corrective action chosen was a 10% increase in EFD. The system is initially at 9U% load and 90% EFD(0.21 is similar to 5. and EFD are each 20%. Then a 20% step increase in T.21 Response of a machine initially at 90% load to a 20% increase in T. as evidenced by the rapid increase in 6 and the drop in terminal voltage.20 except that the increments of T.. 5. overload on the system results in a gradual increase in 6 with time.666 = 2.399)..9 x 2. Figure 5. A 20% increase in EFD is . 979 1.42 48.I . the excess load and excitation are removed.10 19.10 1.604 1.634 0. Finally. first very fast and later very slowly.092 1.717 .22 shows a plot in the phase plane. I.Simulation of Synchronous Machines Table 5.and the system is quickly restored to a stable operating state.49 33..a.10 0.90 .1.04 29.935 3.995 *Angle between q axis and infinite bus = 68. Initial conditions of Example 5. .732 . followed by a 10% step increase in EFD(see Figure 5..89 8 .I03 1.13 0.667 2. 1.344 .17 applied at about the time 6 reaches IOO”.39 13.994 1.12 30. 5.67I 1. for exactly the same disturbances as shown in Figure 5.22 Phase-plane plot U A versus 6 for a 10% step increase in’T. The system “spirals” to the right.920 0.66 -44.I .1 -2.997 67. Variable 183 Comparison of Digital and Analog Computed Variables Analog computed values V L W Computed value pu Percent error I .20.85 . or uAversus 6.20).84 0.103 I .003 1.78 .316 -0.94 -2.925 0.345 I .10 39. Figure 5. Just prior to loss of synchronism a Fig.904 2.316 1.094 I .10 .I .004 66.I2 52.I .60 .1.2.97 33..63 -38.60 -0. following the 10% increase in T.29 0. 10 Digital Simulation of Synchronous Machines Early efforts in solving synchronous machine behavior by digital computer were simply digital applications of the constant-voltage-behind-transient-reactance model.184 Chapter 5 Fig. governors. following along the lower trajectory. A 10% increase in T.9.9]has been aimed at finding the best machine model for system dynamic studies. 04 versus 6 for a 10% step increase in T. IO?.22 and 5. usually in the form of one of the simplified models of Section 4. As larger and faster computers became available. = 0. 5. A comparison of Figures 5. causes the system to oscillate and to seek a new stable value of 6.23 Phase-plane plot EFD = 2. A simple flow chart of the process is shown . increase in EFDcauses the system to return to about the original 6.23 shows a n example of a stable phase-plane trajectory. This caused an expansion in power plant modeling to include exciters.666.1 computed value of E F D . or 2. and turbines.666. I t also introduced more detailed synchronous machine models into many cornputer programs. Figure 5. the time domain is broken up into discrete segments of length the equations solved for each segment. The system is initially at 90% load but with 100% of the Example 5. using a step-by-step solution method similar to that of Kimbark [7].15. with initial conditions T. 5.23 shows the more rapid convergence to the target value of 6 in the stable case. More recent research [8. Le. All digital computer simulations must solve the differential equations in a discrete and manner. engineers quickly realized that the digital computer was a powerful tool for handling very large system of differential equations.. i iMD we estimate saturation.8 pu .10. is a function ofi. as shown clearly in the analog computer representation of Figure 5. which gives a new AAD.12 to illustrate a digital program for calculating synchronous machine behavior in a numerical exercise. i. This is difficult because saturation is an implicit function. and so on. Our principal concern is the mathematical model used in the simulation. 5.. some of which are presented in Appendix E.. which flows in the magnetizing inductance L A D ..1 Digital computation of saturation One of the problems in digital calculation of synchronous machine behavior is the determination of saturation. and i. = id + iF + i.24 Flow chart of digital integration. From these A’s we compute ..24.. It is based upon computing the offset from the air gap line in pu based on the field current required to produce rated open circuit voltage. A. The exponential estimate is often used since exponentials are easy to compute.Simulation of Synchronous Machines 185 nonlinearities + t= t c t results Fig. Actually. 5.e. and this gives new currents. in Figure 5. by a polynomial approximation. Our concern in this book is not with numerical methods. A. = f ( A A D ) . The first requirement in computing saturation is to devise some means of determining the amount of saturation corresponding to any given operating point on the saturation curve. Each integra. shown in Figure 5.But the currents id. Usually it is assumed there is no saturation at 0. There are several proven methods for performing the actual numerical integration. For this procedure the saturation curve is represented by a table of data of stator EMF corresponding to given field current. iF. depend upon AAo. From tion step gives us new A’s by integration. although this is important. A number of models are given in Chapter 4. We shall use the flux linkage model of Section 4.25 as iFO.12. or by an exponential estimate. these values can be substituted into (5.82) to solve for the and saturation parameters A G and BG.4BG) = A G (5. 5. we compute In(SGI/AG) Then 0.1. From (5. ~ S G . = d .81) Then any saturation may be estimated as an exponential function of the form SG = AGf?8GvA (5. If sGI S G 2 are given. = 1 .84) Rearranging. = V.2B~ I ~ ( ~ . Since at open circuit X A D tion in terms of X A D .86) This result may be substituted into (5. ..82) we write S G ~ Ace = 0.2iFO (5.25 Estimating saturation as an exponential function.2s~~ = Ace 0. A 01 pu Fig.82) where V.2sG2 /AsGI) (5.186 Chapter 5 Field Current. (5.2SG2 (5. Although we define saturation to be zero for V.81 This is appealing since X A D = (id + iF + iD)LAD LAD the only inductance that and is saturates appreciably.O and 1.2 respectively. actually SGassumes a very small posi- . IF. / A G ) In(SG. < 0. The function SGis always positive and satisfies the defined values SG. /AG)’ = = 0.48 G (5.85) to compute BG = 5 In (1.83) SG = A c e x p [ ( X A ~ / d ). We then compute the normalized quantities SGI = iFI im iF0 sG2 = iF3 .286 1.8 pu.85) or AG = s l/I a . ~ S G ~ /0.8.iF2 iF2 iF3 .2iFo 1.4BG = In( I .0. voltage. and s G 2 at r/.0.81) and (5.87) Appendix D shows a plot of SGas a function of V. can also compute saturaV we . If the air gap line has a slope (resistance) R we have V.27397 Then from (5.27397) = 0. Then.9315 5.7 Determine the constants A G and B G needed to compute saturation by means of the exponential definition.K)/K ) from which we may write the nonlinear equation = Rip . given the following data from the saturation curve. V. we usually assume that saturation has a similar effect under load. = RkiFo. Example 5. Include in the program a treatment of saturation that can be . including the effect of saturation..2(0.81) we compute in pu S G ~ 30/365 = 0. Solution From (5.81) we can write for any voltage level. it reduces the terminal voltage by an amount V. we are ready to update the integrands in preparation for numerical integration.kiFO)/kiFO (5. SG = (iF .2(0.Simulation of Synchronous Machines 187 tive value in this voltage range. The exponential function thus gives a reasonably accurate estimate of saturation for any voltage.SGfrom the unsaturated value. = 1.8 Prepare a FORTRAN computer program to compute the integrands of the flux linkage model for one machine against an infinite bus using the machine data of the Chapter 4 examples.10. = 1. from (5. From (5.RkiFo)/RkiFo = (Rip .0 on the air gap line is iFo = 365 A.89) where Rip is the voltage on the air gap line corresponding to field current i F .. Example 5. = 1.08219 = S G = 120/1.27397/0.08219] = 6.SG where SG is a function of V.2 Updating the integrands After computing the new value of saturation for each new time step. Le.86) AG (0.2(365) ~ = 0. Equation (5.88) where iF is the field current required to produce an open circuit voltage V. the actual terminal voltage is not Rip but is reduced by an amount V.87) BG = 5 In [1.K s ~ ( 4 ) (5.2 P U SGZ= 120 A The field current corresponding to V. This process is illustrated by an example.0205 and from (5. Because of saturation.08219)2/1. However.81) s ~ ( 5 = (RiF ..0 PU S G l = 30 A V.89) describes only the no-load condition. Fig. .26 CSMP program for computing initial conditions. 5. 5.188 Chapter 5 Chapter 5 C C N NUGUS S Y S T E M M O D E N G PROGRAM * * * * C C N T I T I N U G U S S Y S T E M MODELLIIN 6 P R O G R A M * * * * VERSION 1.3 0 Fig.26 CSMP program for computing initial conditions. 5.Machines Simulation of Synchronous Machines 189 189 Fig.26 (continued) . 5.1 90 Chapter 5 Fig.26 (continued) . Note that the statement of the problem does not give any explicit numerical boundary condition. 3. Use the Continuous System Modeling Program (CSMP) [IO] for solving the equations and plotting the results.. problems with different boundary conditions but of the same type can be solved with ease. 2. this computation depends upon the boundary conditions that are specified.27 coincide.3. X F . . or until points A and B of Figure 5. computer program: once it is written and verified.91) we compute an estimate of the new currents.27 Saturation curve for the magnetizing inductance L A D .1-5. we compute the saturation function SGD= f ( X A D ) in the Fig. we must find the correct new X A D iteratively. This estimate is not exact because the value of X A D used in (5. V. = 1.XAD>/‘?!d (AD .17(VT). but to some new point B. Since X A D is a function of the currents and of saturation.26give P = 1. Include a local load on the generator bus in the computation. I .OO(VINF). From the equations id = ( A d iD = .00 (PGEN).X A D ) / { D i~ iMD = (XF . Thus iMD computed by (5.X A D ) / ~ F = id + + iF iD (5.91) does not correspond to point A of Figure 5. and AD are the integrated new values. The boundary conditions chosen for this example are those of Example 5. P and at the generator terminals.3.27. whereas the flux linkages Ad.26 called INITIAL. Solution An essential part of the computer program is a routine to compute the initial conditions.91) is the value computed at the start of the last A t . The boundary conditions specified in Figure 5. The FORTRAN coding for this section of the program is included in the portion of the program listing in Figure 5. Make a preliminary estimate of XAD (AAD is named WADS in the program. W being used for X and S meaning “saturated”). Compute the new currents. This is one of the advantages of a. As noted in Examples 5.Simulation of Synchronous Machines 191 executed prior to integration at each time step. viz.and V . To estimate the new XAD. We do this by changing our estimated XAD slightly until iMDagrees with X A D on the saturation curve. 5. = I. . 5.28 CSMP program for updating integrands.192 Chapter 5 Fig. Then we compute A. we will know both the new current and the new saturated value of A A D .Simulation of Synchronous Machines 193 usual way. and A N .01. typically.28.40.27. The computed output of several variables to a step change in T. using (5..3. we esti- AAD . finding new currents. we compute AAD I < ? f where E is any convenient precision index. defined as G A D = XAD + AA. Le.A. As the process converges. and the error mea- + SGD) Now define a new A.83). In both cases. GAD = 4.Ao)/(I + S G D )= I GAD A L A D ~ M D / ( + SGD) ~ to see if it is significantly different from A A D . etc.D = FAD = If the test fails. Now we test GAD AAD to be GAD... the step input is applied at t = TSTART = 0.29-5. .2 s. Then we compute + ( A N .AAD) where h is chosen to be a number small enough to prevent overshoot. Now the entire procedure is repeated.h ( G A D . such as mate a new A A D from neWA. defined in Figure 5. h = 0. A0 = AAD(1 + SGD) AN = L A D i M D Then the error measured on the air gap line is X E sured on the saturation curve is approximately AA = = AN . The computer program for updating the integrands is shown in Figure 5. The second part of the program computes the integrands of all equations in preparation for integration (integration is indicated in the program by the macro INTGTL).. and E F D is shown in Figures 5. returning to step 1 with the XAD = F A D . Computer mnemonics are given in Table 5.. 5735 I I I I l I l l l l l I . I - --. l l l l l l l --- l l --I l l I l l I I I I I I I I l l l l l l l l l l l l l l l l I l l I l l I l l I l l I l l I l l l l l l l l l l l l l l l l l l ... --l l I l l l l l l .5 2.ma- Response to a 10% step increase in T..6182 2 A % 1.I I I I I I I I I I l l I l l l I l l l l I l l l --I 0 0. .- - --I l l I I I I l l I l l I l l I l l . l l .6826A -- I . I I I I I I I I I I I I I l I l I l l l I l I l l l l 1 1 l l . il % 1.I .5 1. step increase in EFD 1.5 Fig. 1 1 ..0 2. s 1.6406 1.I I I I . ..0 2. .5999 .29 d axis flux linkages Ad.5 1. . I I I I I I I I I I 0 Time.I 1..-I I I 1 l I I l I l ! I 1 I I I 1 I I I l 1 1 l 1 l I l l I l l ccI l l I l l I l l I I .6629 -.---- --. I l l 1 1 I I I I l .. I I I I I I I I I I I I I I I I I I Response to a 57. .o Time. I I l I l I l I I I I I I I 1 I l l l l l l l l 1 l l l l l l l l 1 L. a 18 2. 5. 5 2. 1 . 2.o The.* c I I 1 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I cc4-. * * .16117 777.1 1 1 I I I I I I I I 1 1 I I I 1 I I I 1 I 1 l I l I I I 1 I 1 l I 1 I I I 1 I 1 l I 1 I I I 1 I 1 l I 1 I I I 1 I 1 l I 1 I I I 1 I 1 l I 1 I I I 1 I 1 l I l I I I 1 I 1 l I 1 I I I 1 I 1 .- I I I I I I I I I I I I I I I I I I I I I I I I I I I .5 I -.. 0:s Fig.1000 -1 I I I 1 . I ..-.-.----.. I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I 1 I I 1 .-. I -: I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I 1 1 1 1 1 1 1 2. . I . .. IOOO-L I I I I 1 I I I I I I I I I 1 I I I I I I 1 I I I I I I l I I I I I I I I I I I I I I I 1 I I I I I I I I I I I I I I I I I I I I I 1 1 1 I I I I I I I I I I I I I I I I I I I I I I I I I I I I l I I I I I I 1 I I 1 1 i 1 1 -.--r---*LI -I I I I I I I I I 1 1 2. I I I I I 1 1 I C . ~2. 1. -r. s 1.I .2548 a d i L 2.1750--: I . I I I I I I I I I I I I I I 1 I I I 1 I I I I I I I I I I I I I 1 I I I I I I I I 1 . .0 The. I 1. .. I ..0 2.... .- I I I 1 I I I I 1 I I I 1 I I I I I 1 I I I I 1 I I 0 0..2. .3 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I . s 30 Field flux linkages XI.I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I 25 . e .1473-! I I I I I 1 I I I I I I I 1 I I I I I I I I I I------ -. ... ~.2026 .I l I -. I I I l I I I I I I I 1 I I I I I t . a d I I 1 1 1 1 1 1 1 I I I I I I I I I I I I I I I I L C C .5 2.. .I I I i i i i i i + 1 1 1 1 1 1 l . .---.e I I 1 I I I c c c c c - I I I I I I I I I I I I I .C C l .#-.-I-...5 Response to'a . 1 I I I I I I 1 I I I I i i l+ l .. I 1 I I I 1 I l *. . .:I I I I . .2287' 2.a L I I I I I I I I I I I I I L 1 1 I I I I I 1 6 I I .1)88-... I I I I I I I I I I I I I I I 1 I I I I I t I I . I I I I I I I I I I I I I I 1 I I I .---1 .- 1.2..+ I I .-- 1 I 1 1 I 1 I I I 1 I 1 1 I 1 I I I 1 I 1 1 I 1 I I I 1 I 1 1 I 1 I I I 1 I l 1 I 1 I I I 1 I 1 1 I 1 I I I ---I-- 1 1 1 1 l I I I I : l I I I I I I I I I I I I I 2. . ..2026-L L L2 L I I I 1 I I I 1 I I I 1 I I I 1 I I I 1 I I I 1 .-. .. c 1 . -I I 1 I I + 1 1 . ?! f!f !! ! ! !! ! !! ! ! ! . I I I I I I I I I I I I I I I I Response to a 5':d step increase in EFD I .---I 1 1 1 1 1 1 1 1 1 1 1 I I I I I I 1 :o lime. I . I I 1 I 1 I I I I I I 1 I 1 I I I l I I 1 I 1 I I I l I I 1 I 1 I I I 1 I I 1 I I I I I 1 e .- + I + I 1 * I l l + * l l l I 1 1 1 l I I 3----- 1.. I .-5 Fig. -.@tJ&.5 I 1 ..* * .. r ! ! . ! ! ! * * .8679--j iiii 1 I l l I 1 l I I 1 I 1 1 I 1 1 I I I I I I I M .c . 1 I I I I 1 l l I 1 I I I I I I I I I I I I 1 I I I I I I I I I 1 I I I I I I I 1 I I I I I 1 I I I I I I I I I I I I. I .8375-...89827 . 1 I 1 I I ccccc.. . I 1 I I I 1 I 1 I 1 I 1 I I I 1 I - c .5 I 1..5 I 2. I + + i i i i i i i ! ! 1 1 1 . s 1. !! !! . . P 1 I I I I 1 I I I 1 I l l I 1 l I I I I 1 1 I I 1 I I -. .. I I 1 I I I I I 1 e . 1 I I I 1 I I I 1 1 8 1 I I I I I I I I I t -I -. c c c c c I I I 1 I 1 i i i i i I I I 1 I 1 I I I I I 1 I I I l I 1 I I I I I 1 I I I I !!!:! i i i i i I I I I I I I I I I I I I I I I I I I I I I I I I I . I .31 daxis amortisseur flux linkages AD.- 1 1 1 1 l I I I I I I I 1 I I I 1 1 I I I I I I I 1 I I I I I 1 I I I I I 1 I I I I I I I I I ----I I I I I 1 1 1 1 I I I I I I 1 l 1 1 1 I---- I I I I I I I I I I 1 1 1 1 I 1 1 1 1 1 1 1 1 1 I I 1 1 I 1 .1 . 1..5 210 2 .. . . 196 . .II 1 I . .I l I I I I I 1 1 1 1 * I I I I I 1 1 1 1 1 I I I I I I I I 1 1 1 1 1 1 1 I l l 1 1 I I I I I I I I I I I I 1 I I I I I I I I I I 1. I 1.. 8 0 0 . 1 1 1 1 1 1 1 1 1 1 0 I . . 1 . . . . 5. 1 i I 1 1 I I I I I 1 l 1 1 1 l I 1 1 I 1 l I 1 1 I 1 l I 1 1 I I 1 0 I 0. I I I I I I 1 I I I I I 1 I I I 1 I 1 I I I 1 I 1 I I I 1 :.- . + .0 I s 2. . 3 . ..o lime.1 a d ..e ..!. ..!. .. 9193 < -I I I r I I I I I I I I I -. e .8ooo-L L L L l l 1 I I I I I 1 I I I I I I I 1 I I I 1 I I I 1 I I I 1 I I ---I- 1 0 0.9404 -- b b 1. . 1 I I I I I I I I I I I 1.0 2. -I--- I 1 I I I I I I I I I I I I I I I I I 1 1 1 I I I 1 l I I 1 l I I . I I I I 1 . 197 .o 2..32 Saturated d axis mutual flux linkages A A ~ s ..8374-: w e .5 2. 8 0 ~ ~ 0.5 Fig. . 1-11 *I--- 1. r i i i i i r ..8982-. I I I I 1 I I .s 1 Io Time... 4 1. 5.- l l 1 I I I I I I I I 1 1 1 1 I I I I I I I I I I I I I I I I I I I I 1 I . . I I I I I I I I 1 I I I I 1 l l I 1 I I I I 1 I I I I I I 1 I I I I I I I I 1 I I I I I I I l I I 1 I I I I I I 1 I I I I I I I l I I 1 I I I I I I 1 I I I I I I I 1 I I 1 I I I I I I 1 I I I I I I I 1 I I 1 + I I . s 2. a . I I I I I l l 1 1 1 1 1 * c c c c .. I 1. 1.o lime. I I I I 1 l l I I I I 1 l l I I I I 1 1 1 I t I I .Response to a 10% step increase in T.8679--: n I I I I I I I I I l l 1 I I C I I I I 1 -: : I l l I 1 I I I I 1 1 1 I 1 I I I I 1 I l I l 1 I l I l 1 I 1 I 1 1 I I I I I I I I I I 1 1 1 1 l I I I I I I 1 I I I I 1 . - b k .-5 1.. I I I I 1 I I I . l l I l l .5 1 . 0. I I Response to a 5'2. t I 1.17446 ..5 I Fig. s 2:o 215 *.15546-! I I I I I 0.--e- 0.0 Time. I I I I I I I I I I I I I I I I I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I I 1 ----.19283 - -.2O- 0.C C e .33 d axis saturation function SGD. I l .5 I 2.0 I 2. I I I I l I I I l I I I . I 0...--.1m :I - I 0 :A A .--I 1 I I 1 1 I I I 1 1 I I 1 1 I I I 1 1 I I 1 1 I I I 1 1 I I 1 1 I I I 1 1 I I 1 1 I I I 1 1 I I 1 1 I I I I I I I I I 1 ------I I I I I I I I I I I I I I I I I I I I I 1 I 1 I I I I 1 1 I 1 1 I I 1 1 I I I 1 I 1 I I I I 1 l I 1 I 1 I I I I 1 1 I 1 I 1 I I I I . 1 I 1 I 1 I I I I 1 1 I 1 I 1 I I I I 1 1 I 1 I 1 I I I I 1 1 I 11---1- 1 1 1 l 1 1 1 I I I I I I I 1 1 1 l 1 1 1 1 1 1 1 1 1 1 0. 198 .13245--f I I I 01Oo: . I l 1 . 5.21119 Response to a lax step increase in T.-------L. L: ! 015 1 io Time...+ + + + + i i I I I I I I I I I I I I I I I 1 1 1 1 1 1 1 1 1 1 1 1 1 r . I I . step increase in EFD L .C e - I I I I 1 1 I I I I I I I 1 1 I I I I I I I I I I l I 1 1 I I I I l I 1 1 I I I I l I 1 1 I I .5 I 1. L. v) 3 0. . I1 I . s 2.I c . s 1. 1 1 I ...1 1 1 I I I 1 1 I I I 1 1 I I I ! I 1 I 1 1 I I 1 I 1 1 I I 1 1 1 1 1 1 1 1 1 1 1 1 .5 2.9800 0 0.I .9800 0 015 1 Time..I I I c + 1 I 1 I 1 I 1 I I I I 1 I 1 I 1 I 1 I I 1 I 1 I 1 I 1 I I 1 I 1 I 1 I 1 I 1 I 1 1 I I I 1 I 1 I 1 I 1 I 1 I 1 1 I I I 1 I 1 I 1 I 1 I 1 I 1 1 I I i i I 1 1 I 1 I! * !!!!!: 1 I I I i i i i 1 l I l 1 l I l 1 1 I 1 !!!:! 1 I I I 1 l I I 1 l I I I I I I I I 1 l 1 1 I I l I I I 1..0766- .Response to a 10% step increase in T. 199 .5 Response to a 5% step increase in EFD 1. a M i .1 I 6 7 I- ..5 1 .L C c . C C . . 5...” C L C .0355- 0. .0 2.34 Line current i. I 1 1 I I I I 1 I I 1 I I I - 1...5945 0. 1 1 1 1 c c i i I . .I I I l l ..o Time. .o Fig. 1 . I C C C .W14 0.5945 0. I I l l + a 1. c . 0322 - 3. 3.0616- 0 0 1 :o lime.0 I 2:s Fig.00757 . E 2. 5.. .1W3- Response to a 10% step increase in T. s 15 .! L I 0 0 I 1 :o llme. 2:o Response to a 5% step increase in EFD I I 3.oooo-!.3. 3.. . .35 200 Field current if. I I I I I 1 + i i + I I I I I I l l l l 1 1 1 ...36 d axis amortisseur current iD.!.c t ---.. .. 8 4.....o Time..... c e c Response to a 10% step increase in T..- I 1 I I I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I I I 2... I I I I I 1 1 1 1 1 I I 1 I I I 1 I I I 1 I I l 1 I I l 1 I i i I I I 1 I I I I I I I I I I I I -0.. 1 1 I 1 I I 1 1 I I I I 1 1 1 I 1 I I 1 1 I I I I 1 1 1 I 1 I I 1 l I I I I 1 1 1 I 1 I I 1 1 I I I I 1 1 1 I 1 I I 1 1 I I I I 1 1 1 I 1 I I 1 I l I 1 l I I I l I I l I I I I I I I I I I 1 I I l I I I I I I I I I I 1 I I l I I I 1 1 ... .... 0......* I I I I I 1 I I l l I I I I I I I I I I I I I I I I I I I I I I I I e... c... 1 1 1 .. I I l + ... ... l l * I l l I l l I l l I l l I l l I l l I l l * 1 1 I I I 1 I 1 I 1 I 1 1 I I I 1 I 1 l 1 I 1 1 I I I 1 I 1 l 1 I 1 1 I I I 1 I 1 1 1 I I 1 I I I I I I I I .. ... I I I 1 * I I I 1 1 1 I 1 I I 1 1 I *---------1 I I 1 1 1 I I 1 I I 1 1 1 I I 1 I I 1 1 1 I I 1 I I 1 1 1 I I 1 I I 1 1 1 I I 1 I I l 1 1 I I 1 I I 1 1 1 I I 1 I I 1 1 1 I I 1 I I l 1 1 I I 1 1 1 1 1 1 1 1 1 1 1 I I I I I I I l l l l l 1 I 1 1 1 1 1 1 1 1 1 1 I 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 I 1 1 1 1 1 1 l 1 1 1 1 I I I I I I I I l l l l I 1 1 1 1 1 1 1 l 1 1 1 1 I I I I I I I I I I I 1 1 I I 1 1 1 I I 1 I I 1 l 1 I I 1 I I 1 I 1 1 l I . .. i i I t I 1 I I I ..... I I I I l ..... I I 1...01866c i . .---- 1 1 I 1 I I 1 1 I I 1 I I I I I I I I I I I -I 1 I I 1 1 1 1 I 1 I I I I I I I I I I I 1-11- I I I I I I I I I I I I -c I I I I l I I I 1 I -0*02000 -L!.00343..... I 1 1 1 l I I I I I I I I I I I I I I I I I I I 1 0 l I I I I I I I I 1 I I I I I I I 1 I I I I I I I 1 1 I I I I I I I I 1 I I I I I I I l I I I I I I I l I . 5.....! 0.. .....02000 -I I 0 I 0. .00687 -1 I ii . ....... . .... I.....5 t 2..!.5 I 1 ..........!.5 I 1 ..l I l + I l l I a .....00657CI I l l * I l l I I I I I l l 1 I I I 1 I 1 I I I 1 I I I 1 I I n I I I I I I I I I I I I I I I 1 I I I . ...0...!.. ... . c * * .....""'A'.. I I I I I I I I I I I I I I r I .. 1 1 1 1 1 l 1 1 1 l I 1 I 1 1 I 1 1 l I 1 I 1 1 I 1 l 1 I 1 I 1 1 1 1 1 1 I 1 I 1 1 1 1 1 1 I 1 I 1 1 1 1 1 1 I 1 I 1 1 1 1 1 1 I 1 I 1 1 1 I 1 1 I 1 I 1 I 1 1 F.. n -0.5 I Fig. . I I 1 I I I I I I 1 I I I I ... s 1..L I I I I I 1 I l I I I I 1 I I I I I I 1 I * I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I I 1 I I I I I 1 0 I !....... .c ...C C - .. -i m I I I . 20 1 . I I I I I 1 I I l l c w -I -------..!... ..... * * I I I I I I 1 I I I I _I I I l I I 1 l I l I I I l I I 1 l I l 1 . .. * I * I I I 1 * I I I 1 .EFD 0..5 t Response to a 5% step increase in ..01762-! -0. ....03075 - 1 . .........o The..... ... .LL..5 2.. ....... ...c b ... * * ..-........... I I 1 1 I 1 I I I I I I . .... I I I I I I I >- a . I ..-I I I I I I I I I I --.. . .I I I I I I I I O I I I I I I I I I I I I I I I I I I I I I I I I I I I I -------- .5 rime. .-I.C e C .. I I I l l I l l I l l I - -.. I826 --. I I I l I 1 I I I I l l I I I I I I I I I I I I I I l l I I I I I I I I I l l 1 I l l I I I I I I I I I I I I I l l I I I I I I I I l I I I l l I I I l 1 I I I 1 I I I I I l l l l l l l l l l I I I I I I I I I I I I I I I I I I $ I l I I I I I I I I I I I I I I I I I I I I l l 1 1 1 1 1 I 1 1 1 -I ..5 I 2. I I I . .I .! ..0 I . : I I I I I I I O I I I I I I I I I I I I I I 1 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I .. . ......-..-.o Time. ---.-I I I I I I I I I I I I I I I I I I I I I I I I ..- I I I 1 I I I I I I I I I 1 . t I I I I I I I I I I I I I I I Response to a 57. a . s I l l I I I I l l I l l 1 ...e .- I I I I I I I I I I I I I I I I I I I I I I I . I I I I I I I I I I I I I I I I I ..- 4-1-1 1...1500-L ! .e - I l l I l I I I I I l 1 1 I I I I l l l l l l l l I I I I I I I I I I I I I I I I I I 1 l I I 1 l I I I I I I I I I I I I I I I l l I l l I l l I l l -c I I I I I I I I I I ..I .. .! ! ! ! .I I I 0 05 : 1 :o -"-I he .r .Response to a 10% step increase in T. I I I I I I . . step increase in EFD I I I I I *. I I I I I I I I I 1 1 I I I I I I 1 I I I I -I. i i i --I * . .-.._ J I I I I I I I I I I r I I - I -*. 3 I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I c I I n I I I I I I I I I I I I I I I I I I I I I I 1 1 1 1 1 1 1 1 ..nI I I I I I I I l I I I 1 I I I I I I I I I :::::::. & . .* I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I -..-- I I I I I l l 1 I I I I I 1 1 1 1 I I I I I I I I I I I I I I I I l l 1 --.. I I l l I l l I I I l l .-.---I l l 1 I I I I I I I l l 1 1 I l l 1 I I I I I I I I I I I I l l l I l l l l I l l l l l l I l l l l I l l l l I l l l l l l I l I I I I I -.-.. 1-. 1 .----_ I I I I I . * * -. I - >- I I I I I I I I I I 1 1 1 1 1 1 1 1 1 I --------e- 1 1 1 1 1 1 1 1 1 I ) I t 1 I I I I I I I I I I I I I I I I I I I I I I I I I I I I l I I I I I I l I I 11 6: . I I I I l l I l l I l l I l l . I I I I I I I I I I I I I I I I I I -c ---MI I 2..e .....- I I I I I I I I I I I I I I I I I I I I I I I I 1 1 1 1 I I I I I I I I I 1 ' 1 I --I- I 1 --. l .. -.0 I 2.-1 . 1. I I I I I I I I I I I l l IiiI I I I 1 1 1 I I I 1500-A I I I 1 I I I I 1 I l l I l l I l l I l l I l l 1 I I I I 0 ---- . .--.----.1642-1I I I I I I I I I L eoI I I I I I I I I I I I I I I I I I ..-. 1 I I 1 I I I I I I . .5 I Fig.. 1708- 1..17631 . A L A ! .- I I I I I I I I I I I I I I I I I I I I I I e I I I I . 1. 5..I I I 1 I I I I 1 I I I l 1 l I I I I I I l l I I I I l l I I l l I I I l l I 1 I l l I I I l l I I I l l 1 l .37 Terminal voltage V. 1 I I I I I I I 1 I I I I I I I 1 I I I I I I I 1 I I I I I I I 1 I I I I I I I 1 I I I I I .-. -_ I 1 I I I I 1.- l I I l l 1 l 1 1 I l l 1 I l l I I I I I I I l l I I I I I I I l l I I l I l l 1 I I I I I I I I I 1 1 1 1 1 I I 1 I l I I 1 1 I I I I I 1 1 I I I I I 1 1 I I I I I 1 1 I I I I -. . . I I I 1 I I l l 1 I I I I I I I I l l 1 I I I I I I I I I I --I l l l l -4. I I I I I I r I I I I I l l l l l l l I 1 I I I I L l 1700-. o 0 0.Response to a 10% step increase in T.38 Torque angle 6 in degrees. s Fig.5 2.0 2.o Time.5 1 .o 1. 0 1 . 5. .5 2.5 Time. s 1. . 5. .00147 4.5 ub 2.00160 . 3 n . .00102 ! I .0 Fig. . .00064 -0..39 Speed deviation in pu.. .Response to a 10% step increase in T. .00185 - I Response to a Sg step increase in EFD I 1 8 0.00019 -0. P 0. 0 t 1l o Time. I 1 ..O Time. -----.. s 0. .*+**.4236 - ... 2.W)oO-’ I I I ~ 1 1 1 I I I I I I ~ I I I 1 1 1 1 I 1 1 1 1 I I l l 1 1 1 1 1 I I I I l l l l l I I I I I I I I I I I I II I I 1 1 1 1 1 1 1 1 1 1 l I l I 1 1 l I l I 1 1 I 0 --. I I I I I 1 l 1 * * .- 0. I-@ I 1 I I I I I l I 1 1 1 1 1 1 1 1 1 I l l 1 I I I I I 1 I l l l l l l 1 1 1 1 1 1 1 1 1 1 1 1 I l I . .1432e- I I I 1 I I I 1 I 1 I I I I I 1 2.5638- . .- 3. 25 . I I I I I I I I I I I .2834-. .. .*.1 d .-_-....o I .:.o Time.. 5....1 3 - 2.5 I 1 . 1 . 2.. . s 1..8000 I .**.^-I-I l l l I l l I l l 1 I I I I I l l 1 I . step increase in EFD 3. i 0:5 1 . I 20 . .-+------ I I I I I I I 1 1 1 1 1 1 1 I I I I I I I I I I 1 I 1 1 1 I I--. step increase in T...3JoOo--’ I I I 1 I 1 + I I 1 1 . . I 3.9000-. l Response to a loo/. ..5 Fig. I 3..I I I I I I 1 I 1 I I 1 1 1 1 I l l 1 l I 1 1 1 1 l I I I 1 l 1 1 1 l l 1 l I 1 1 1 1 l I I I 1 l 1 1 1 l 1 1 1 I 1 1 1 1 1 I I I 1 1 1 1 1 1 1 I I l l I I 1 l 1 I I 1 1 I I 1 1 I I I I I I I I I I I I I 1 I I I I I I I I I I I I I I I I 1 I I1 I I I l 1 I 1 I I I I I I I --e--- .5 20 .. * * * * I l l 1 +. . .40 Electromagnetic torque Tcb .---- 1 1 l 1 1 1 l 1 I 1 1 l 1 1 1 l 1 I 1 1 1 1 1 1 1 1 I I 1 I I I I I I 1 1 I I I I I I I I 1 I I I I 1 I I I I 1 I I I I I I I I I I I I I I I I 15 .- . 1 1 1 1 1 1 1 1 1 I I I I I I I 1 1 I I I I I I I I I I I I I I I I I I 1 I I I I I I 1 I 1 1 I 1 1 1 I 1 I 1 I I l l I I 1 l .- 1 1 1 1 1 1 I I I I I I I a 2. Response to a S?. ! . . .+. lime. . f I I l 1 I I I 1 I I I t + I I I I I I I I I I I I I I I I I I 1 1 1 1 l 1 l 1 l l l l .. Electr.30 5. identify the amplifiers and potentiometers in Table 5.9 PF lagging.6) the synchronous machine is to be represented by the simplified model known as the one-axis model given in Section 4. and the angle 4. currents. the potentiometer settings. M. D. The angle between the q axis and the terminal voltage is 45". The same synchronous machine connected to the same transmission line. Analog computer representation of a synchronous machine. Note: In the load equations. thesis. v. The infinite bus voltage is 1.1. I n the system of one synchronous machine connected to an infinite bus through a transmission line (discussed in Examples 5. Publ.39 5. Prepare a complete analog computer simulation of this system. = 0. and the amplifier gains.3. B.8 The synchronous machine discussed in Examples 5.4 using the two-axis model of Section 4. Eng. Handbook of Analog Compurarion.1.1 and 5. C .3 I 5..13 the signal to the resolver represents the infinite bus voltage.34 5. . Repeat Problem 5.C. identify the potentiometer and amplifier settings that need adjustment. flux linkages. References I .29 5. i = lei.0 pu.. as i n Examples 5. IEEE Trans.15.9 pu at 0. PAS-88:1593-1610. Princeton.S. which is represented by a resistance R = 10 pu. 1967. PAS-75: I I78--84. Find the steady-state operating condition: the d and q axis voltages.2 with the machine output power being 0. N. Purdue Univ. Analogue computer representations of synchronous generators in voltage regulator studies. VT DLD DOMU TE Problems 5.6 5.J. Simulation of a single machine--infinite bus system.2 is operating at rated terminal voltage. F. D . If the time scaling is changed to (lo). Unpubl.0001-3. Repeat Problem 5. and its output power is 0. 3. 5. Schroder. 15. 1967.7 5. . Find the operating condition ofthe machine. 00800. Krause.2 5.35 5. 4. Recommended phasor diagram for synchronous machines. C..38 5. 6.206 Table 5. IEEE Committee Report. Ames.4 using the voltage-behind-subtransient-reactance model of Section 4. Paper C 73 313-4. West Lafayette.36 5.2.5 5. Ind. I f the level of this signal is reduced by a factor of 2 while the level of all the other signals are maintained. M. Electronic Associates. I n Figure 5.3 5. 2.37 5. 1968.9 PF lagging.15. Canada. Figure Chapter 5 Computer Mnenomics of Output Variables Variable Ad AF Computer mnemonic 5. I that will be affected. 1973. 1956. and 5. AI&& Trans... Iowa State Univ.40 WD WF WKD WADS SG D IA AD XADS SCD ia iF iD IFF IKD 6 (in degrees) W A (in pu) Tt.2. M . assume that ~ . Mimeo notes. Inc. the amplitude and time scaling.4 5. and Anderson.33 5. In the analog computer simulation shown in Figure 5.1 and 5.. the time scaling is (20).9 pu at 0. Buckley. I 5. presented at the IEEE Summer Power Meeting. Repeat Problem 5. Vancouver. 1969.1. 5.. has a local load of unity power factor. Compensation of synchronous machines for stability. Indicate the signal levels for the operating conditions of Example 5. P. The power at the infinite bus is 0. 2nd ed.32 5. Dept. Riaz. P..80 pu.13 and Table 5. Simulation of Synchronous Machines 207 7. 1973. 1973. Hauth. PAS-92:926-33. and Ewart. IO. IEEE Trans. Dandeno. GH2O-0367-4. System/360 Continuous System Modeling Program Users Manual.. L. and Schulz. International Business Machines. Jones. D.. P. E. R . Schulz. ElTects of synchronous machine modeling in large-sale system studies. 1967. R . 1948. W. Vol. IBM Corp. D. New York. P. PAS-92:574-. Power System Stahiliry. I .. . Kimbark. 8. Wiley.. N. 9. R .. lEEE Trans. P. W. Dynamic models of turbine generators derived from solid rotor equivalent circuits. L.82. any bounded input will give a bounded and therefore a stable output. The synchronous machine models developed in Chapter 4 have two types of nonlinearities: product nonlinearities and trigonometric functions. it tends to acquire a new operating state. 208 . The first-order approximations for these have been illustrated in previous chapters and are outlined below. Both the forced response and the free response are decided by the roots of this equation. I f the two states are such that all the state variables change only slightly (i.1) We note that xjo and xio are known quantities and are treated here as coefficients.chapter 6 Linear Models of the Synchronous Machine 6. the variable x i changes from xio to xio + x i Awhere x i A is a small change in x i ) . Let the state variables x i and x j have the initial values xio and x j o . consider the product x i x i . which is assumed to be negligibly small. The initial state may be considered as a quiescent operating condition for the system. The new value becomes (xi0 + XiA)(xjO + x j A ) = XjOXjO + XjOXjA + xjoxjA + XjAxjA The last term is a second-order term. The dynamic response of a linear system is determined by its characteristic equation (or equivalent information). the system equations are linearized about the quiescent operating condition. During the transition between the initial state and the new state the system behavior is oscillatory. while x i Aand x j Aare “incremental” variables. It is shown that when the system is subjected to a small load change. the system is operating near the initial state. Initially their product is given by x i o x j o .. From a point of view of stability the free response gives the needed information. Let the changes in these variables be x i Aand x j A . Thus for a first-order approximation. the change in the product x i x j is given by (xi0 + xiA)(xjO + XjA) - XiOXjO = x j O x j A + XiOxjA (6. By this we mean that first-order approximations are made for the system equations. To examine the behavior of the system when it is perturbed such that the new and old equilibrium states are nearly equal. The new linear equations thus derived are assumed to be valid in a region near the quiescent condition.e. If it is stable.1 Introduction A brief review of the response of a power system to small impacts is given in Chapter 3. As an example of product nonlinearities. l i n e a r Models of the Synchronous Machine 209 The trigonometric nonlinearities are treated in a similar manner as COS ( 6 0 + 6.) are different for every new initial condition.5).9) The elements of the A matrix depend upon the initial values of the state vector xo. although we would expect it to be continuous and relatively smooth.2 + 6. i.f) (6.) - sin 6 0 (COS~O)~~ (6.t) + A(xO)XA + B ( x ~ ) u + B(XO)U (6.9) are determined from the nature of the eigenvalues of the A matrix.cos60 EZ (6.g.4) XA = [boi F o io0 iqo ieo wo 601 t = At the occurrence of a small disturbance. Therefore.. but we do require that it be known. The state space may be thought of as an n-dimensional space. The dynamic properties of the system described by (6. Thus x = X O to". the states will change (6.e. e.. COS(^^ + 6.the incremental variable is bA and its coefficient is -sin J0.7) In expanding (6. reduces to i o -k XA = f(x0 + xA.r) (6. Being nonlinear. Let the state-space vector x have an initial state xo at time t rent model is used..) = C O S ~ ~ C OA 6S . i.6) which.7) becomes Xo + *A f(X0. we can show that the incremental change in the term sin 6 is given by sin ( 6 0 6. For a specific dynamic study it is considered constant. i. The quiescent operating point xo and the functions A(xo) and B(x.5) + XA Note that xo need not be constant..e. after slightly from their previous positions or values.) . Similarly. with dx % xA .7) all second-order terms are neglected.sin 6 0 sin 6 A (-sin60)6. We may also compute the A(xo) by finding the total differential d x at xo with respect to all variables. and the operating conditions constrain the operation to a particular surface in this n space. if the cur(6..e.8) from which we obtain the linearized state-space equation XA = A(xO)XA (6. by using (6. The state-space model is in the form x = f(x. the surface is not flat. terms of the form x i A x j 4are assumed to be negligibly small. The system (6.2) The incremental change in cos 6 is then (-sin 60)6A. with COS bA E I and sin 6A % J A .3) linearization of the Generator State-Space Current Model = t o . for the q axis voltage change we write \ (6.210 Chapter 6 where the quantity in brackets defines A(xo).1 1) Similarly. For the first equation (of the d circuit) we write Expanding the product terms and dropping the second-order terms.14) The linearized damper-winding equations are given by (6.15) (6.16) From (4.17) . proceeding one row at a time. We begin by linearizing (4.10) which is equal to (6.12) which is equal to (6.101) the linearized torque equation may be established as (6.13) For the field winding we compute (6. The quantity in parenthesis on the right side is exactly equal to udo. (6.74). Rearranging the remaining quantities. M-' .20) or in matrix form v = . we may write these equations in the following matrix form: (6.102) may be written as 8.( 1 / 3 ) [ ( L d i q o = kh!fgi. the torque angle equation given by (4.M-'v pu (6.21) Note that the matrix M is related to the matrix L of equation (4.linear Models of the SynchronousMachine 21 1 which can be put in the form 7 j h ~ TmA .18) Finally.74) by Assuming that M . If we drop the A subscript. is = . the state equation for the synchronous generator.22) .MX PU (6.19) are the linearized system equations for a synchronous machine (not including the load equation). = (&A (6.11)-(6. not including the load equations.19) Equations (6. since all variables are now small displacements.K x .' exists.oiDA - XqO)idA - (Ad0 - Lqid0)iqA - kMFiqoiFA -t k M ~ i d o i ~ ~D]W b - (6. .550 1.o 0._ _ .0011 1 I I I I I . Solution The matrix M is related to the matrix L of Example 4.. determine the matrices M and K for the generator described in Examples 4. Let rj = 2HwR = 1786..2 as follows Then we write b.0007 0 K = L o 0 0 I I 0 0 I I -1 01 When the machine is loaded..640 1.3.0 0 0 I I I I I- 1..506 .490 1.550 1. Ld.23) Example 6.. For example. r .20) 0.1 As a preparation for later examples involving a loaded machine.21 2 Chapter 6 which is the same form as X = AX + BU (6.700 1.64 0 1..550 I 1._ _ ._ _ _o _ _ .. I I 0 1.49 I I X. certain terms in these matrices change from the numeric values given to reflect the impedance of the connecting system.. and L.490 I I 1. change .550 1.r--------- I The matrix K is defined by (6.1-4.550 I I I I I I I I I 1 0 M = lI .526 I I .651 1. when loaded through a transmission line to a large system.550 1..94 rad. Equation (4.25) into (6.26) Rearranging (6. The same procedure followed previously is used to linearize this equation.28) with (6.linear Models of the Synchronous Machine 213 to 8 .12). (6.14)--(6. I8). and iq noted in Section 4.25) Substituting (6.149) is repeated here for convenience: (6. after dropping the subscript A.19). Other terms are load dependent (such as L.26) and making the substitution (6..27) we get.16). tem equations . i.11) and (6. and (6.13. L d . 6 3 linearization of the load Equation for the One-Machine Problem . and LY is the angle of V. with the result (6..24) where K = V .28) Combining (6.. we get for the linearized sys(6. (6. as the currents and flux linkages) and must be determined from the initial conditions. and this change is usually accompanied by damped oscillations of the variables. In matrix form (6.29) are the field voltage uFA and the mechanical torque T m A . I t is convenient to compute A as follows. the machine is spinning at synchronous speed and is delivering some known power to the infinite bus. will cause the system to seek a new operating point. and assuming that M-' exists..29) is a linearized set of seven first-order differential equations with constant coefficients. taking into account the load equation. Find the new expanded A matrix.31) Note that the only driving functions in the system (6. I -L_ I I I 0 1 I Equation (6. X = .. Initially. Let Then (6.214 Chapter 6 I 0 I 0 I I I AI I -LI I I I I I (6.2 Complete Example 6.29) I 1 .M-I K X - M-Iv = AX + BU (6.2.29) becomes v = -Kx . Example 6. . Note that the new matrices M and K are now expanded to include the transmission line constants and the infinite bus voltage.M i . Assume D = 0.1 for the operating conditions described in Example 5.30) where A = -M-'K. A change in either uF or T. CU) = ~ T ( ~ 0 ~ 5 3 .040 1.605 0 I I I I I I I 0 M = 0 1 I I I I I I 2.150 + (-1.64 1.550 1. id0 X40 I I 0 I o = = 1.400 = 2.(-kMDi@) 3 1 .100 = i d = Lq = 1.55 X 3 0.550 I I I I I I I I 1.550 1.526 j I I I I I 0 -1786.014 (XqO - LdiqO) = .1.397 1 1.676 + 1.(-Ado 1 = = -1.Linear Models of the Synchronous Machine 215 Solution From Example 5.70 x 0.591) 3 = -1.021 I 1.700 + 0. .490 1.591)(0.040 1.150 .640 + 0.701)(0.362 X 3 + L&o) -(1.400 = 2.490 1.039 + (0.a) = v'T(sin 53.676 1.550 1.651 1.020 = 0. .9 0 1 0 We also compute.701 = -0.430 3 KCOS(&.550 1.428 The matrix K is given by K = The new A matrix is given by A = -M-'K. 7 3= " ) 5 1.025 K sin (6.735") = 1.001 1 + 0. or with D = 0.2 we compute ff The matrix M is given by 2.100 1. in pu.701 3 = -0.4) = 1.550 0.4) = 1. 9983 A s = -0.66 s.439 -4.0 0.. Note also that the real parts of all the eigenvalues are negative..0.43 fI .72 2649.064 -2587.86 1608.0007 A4 = -0.- 0.0359 ...071 f -2327.1106. .0 0. namely small perturbation about a quiescent operating condition.776 .. The results are given below...439 -4.0 f I I 982..72 -2587..- A = 3589.. Solution To perform the computation of the eigenvalues for the A matrix obtained in Example 6.0078 -0.01 .7I ..18 76.216 - Chapter 6 .-- ---------------L--------------l------- -- -0.320 0.218 I I I I I 1776.0075 -0.062 0.2331.0 .68 1543.63 1751.857 -96.472 22.. This is the 60-Hz component injected into the rotor circuits to balance the M M F caused by the stator dc currents.86 1206....062 12.1217 All the eigenvalues are given in rad/rad.18 -2547.72 I -36..54 -331.-..0 Example 6.2204.36. -123.69 958.0 0.142 76. Solution A procedure similar to that followed in Examples 6. which is damped at a much faster rate.50 -605.950 4...605.0 ....._ .37 ---- .46 I I .01 804. Examine the stability of the system..36..064 35.776 --- -------------------L_--_--_---_-_-L__-__- 11I I I 1206.36.0 0..2..123.2027 I -0.017 I I I I I I I ..39 -- 12.0 - 0.0289 = -0. I - ...356 14.2444..320 I-959. A2 A.072 I I I I I . Generator loading is that of Example 5.60 .472 22._ ..54 -2587..0 I I 1000 0.3 gives the following results: .40 10-3 I .95 2649.950 4.2.2..01 2202.1929 0..66 2257.J.- 2387. _ -I_ ..218 -0. I I -0.1929 0.0 14. .0016 + j0. they are damped with a time constant of 1/(0..54 -0..78 1469.0I 1608. .5351 0....._ I_ ..142 1-3487..9983 A6 = -0.j0.0016 .98 ... -0.8399 00 .43 90.I I I I I -3487.------- 0..0 0..01 880....70 IO-’ 35. a digital computer program is used.33- 845..72 .54 -0.4 Repeat the above example for the system conditions stated in Example 5.-.0 I I 0.857 -96...j0.7993 . The pair A s and A6 correspond to frequencies of approximately I .0991 A 7 = -0.. .. Note that there are two pairs of complex eigenvalues.3 Find the eigenvalues of the A matrix of the linearized system of Example 6.3505..95 ..0 0.72 -3505.01 880.63 2202.....0359 + j0.. This complex pair and the real pole due to A.0016 x 377) or 1.70 ------- 2649....73 Hz.-. = Example 6. The other complex pair corresponds to a very fast transient of about 60 Hz.017 -2547.0289 = -0.... A.54 ..70 -2581...10 A = 3589.356 2649.2027 -0..0 1000 L 0.. which means that the system is stable under the conditions assumed in the development of this model. dominate the transient response of the system..63 90.2 and 6.1735..4422 I I I .. I ... 0248 -0.34) Similarly the q axis equation (4.( I &D 2) ADA (6.32) (6. 4 F rD F A.j0. A.33) .137) becomes (6.0359 + j0.1230 A.136) can be linearized to give (6.0991 -0. following a procedure similar to that used above for the current model. A.135) we can compute the linear equations (6. = rD .. A.j0.95 s) compared to the previous example.0248 -0.37) Similarly.Linear Models of the Synchronous Machine 217 and the eigenvalues are given by A.9983 -0.d . 6.49 Hz) and a greatly increased time constant (2.4 Linearization of the Flux Linkage Model We now linearize the flux linkage model of a synchronous machine. From (4.9983 -0.36) The torque equation (4. A. = = = = -0. = = = -0.0005 Note that this new operating condition has a slightly reduced natural frequency ( I .35) (6.A . the swing equation becomes . A. LMD A 4 D t d + rDAL h 4.0009 . Thus damping is substantially reduced by the change in operating point.0009 + j0.0359 . 42) .34).38) For a system of one machine connected to an infinite bus through a transmission line. we can show that the matrix T is exactly the same as (4.158).157) and (4. the load equations are given by (4.37)-(6. several interesting observations can be made.218 Chapter 6 (6. I62). and D are similar to those defined in Section 4.. = uA.161). I59)-(4. If we write C as dFD qQ wb (6. and (6. The linearized equations of the system are (6. but not exactly the same as (4. where and d = r + Re and K = 2/? V . First.160).13.36).3 for the nonlinear model. (6. C.41) where the matrices T. The matrix C is similar.33). These are then linearized to give [I + 2 I)" (I - 4 4 A. If the state equations are written out in the form of (6.40) and 8.41) and compared with the nonlinear equations (4.In matrix form we write TA = CX +D (6. (6. . and C.45) D = [0 UFA O O O Tm.. Example 6.42) by T .5 Obtain the matrices T. we note 01' (6.l i n e a r Models of the Synchronous Machine 219 with partitioning as in (4. we can premultiply both sides of (6..d I _-_----___----- ("" I I 7) iLMDAqO I LMDAqO I 1-. Since the choice of the state variables is arbitrary.a) 1 (6..I ' . and C. We may write matrices C. we can observe that C.47) The matrices A and B will have constant coefficients. Finally.46) which is of the form k = AX + BU (6. Submatrices C.. and C. C. and C. and C.161) if w is replaced by w.44) I where X A D o and A A Q o are the initial values of AAD and A. C. each measured from and the arbitrary reference. is the initial angle of the q axis. The order of the system does not change.. there are many other equations that could be written. and C.. = '[ L 3.COS(6.43) L where a is the angle of vm 6. Solution Machine and line data are taken from previous examples in pu as: . are exactly as in (4. however. now become [ f . = 0 0 .' to obtain = T-'CX + T-ID (6. Submatrices C. c. are considerably changed.' &V..r------- 0 0 : O J (6. Note that the matrices A and B will not be the same here as in the current model. and A of the flux linkage model for the operating conditions discussed in the previous examples. which were formerly zero matrices./7j Assuming that the inverse of T exists. the new D matrix to be respectively. are exactly the same as in the nonlinear equation. as C.j(.161).. C... C. however. which are dependent upon the quiescent operating conditions... and there are still seven degrees of freedom in the solution. o .3656 I I I I 0 O 0 0 0 -2.025 a) = 1.2: A.397 The matrix C corresponding to Example 5. - a) = 1. 1..914 d T V ..2364 0.1625 0 0 0 0 I o I I I I I 0 0 0 0 I O 0 0 - 0 I o .o O 1.200 1.0 I O I I 0 0 0 0 0 0 1.1.1118 1.7478 1 . COS($ sin(6.220 Chapter 6 - 3. the following data is obtained from the initial operating conditions as given in Example 5.0 0 0 I I I Io 0 L I I j I 0 1 - I I 0 I I O To calculate the matrix C.3 162 0.43 I8 1 I I I I .0 O I 0 0 0 0 I I I I o I I I ------_-----_---__I------------l------ 0 0 3.676 2. AQo Ado AFO = = = = = A.150 1.045 1.2 loading is then calculated to be . 1.1622 0 0 T = 0 -0.0 0 1.0 I 0 I - r0. _ -0.437 -5. .7322 0 1 - 0 0 0 I 1000 0 - The eigenvalues of this matrix are the same as those obtained in Example 6. A = 1.76 72.1.330 .3816 I I I -114.9867 I 1 I .388 44.3246 I I I 574.53 2111. .____-- 999.345..378 )I 1039.022 3. . in this example are somewhat different from those in Example 4. = 0.530 1.1.80 0 j I 154.055 284. and K sin ( 6 . . .388 44.278 66.16 0 -0.63 441. ._ . .094.756 .115..142 1 I 328. .756 I -3162.I. ADO = 1.278 66.282 -236.4009 -0. A.7155 111.1 1 1024. - 0 0 0 0 0 and the matrix A is given by . A..48 0 0 1000 0 0 0 -0.Linear Models of the Synchronous Machine -114..854 .530 ) 1..438 -5.68 10-3 C = 3162.9867 -174. ..59 0 0 -313. = 1.634.58 0 I I I -3162.854 .1 we obtain the same matrix T.75- 0 0 I I o O 0 0 1320.120 A = -5. . .6969 I I I I j .055 284.278 66. I o o 0 0 0 0 10-3 -----_--.147 284.282 -747.035 1.32 1396.16 0 . .55 0 0 1 I -313.2. . .5607.1.53 0 2 I I I . .1365.0285 I I .994. For the operating condition of Example 5.3207.44 0 3. The A matrix is given by 1.7322 0 I I I 0 1000 0 1 0 0 0 - Note that some of the elements of the matrices C. .I_ . .330 -1365.78 0 0 -1361. Kcos(6..1430. .76 0 72. . .9790 0 -0.3 and correspond to the loading condition of Example 5.530 0 0 -115.88 0 -431.022 3..388 44. .--__________L______________1----_-____--~--_--.115.6503 I I 0 0 ..378 -313. .854 -1. The matrix C for the operating conditions of Example 5.4009 0 -0._ _ .4 since the resistance is not the same in both examples. . . For this operating condition the initial conditions in pu are given by A = 1.a) = 0.58 I I 0 0 1 I 0 ------______________-L__----- ____-__I____________- 3162. .114. . .330 I I 0 0 0 0 I . . . .720 C = 39.756 I .30 560.035 1.6503 0 -________________-__l______________l____---_---I I 1.78 0 1 I ._ _ _ _ _ .a) = 1. . .0285 -0.530 1.1 is given by - 114.935. and C.720 39.282 -747. -1.. . = &E.80 -174. and ox. - - 0 0 I I I j __ 0 lo00 0 0 - 0 The eigenvalues obtained are the same as those given in Example 6._ .33 . be approximately equal to w R X . I ..278 66. ..L ... _ _ .222 Chapter 6 - 16.51) . .12 667...50 177.1.6969 I 0 0 ...48) Eliminating i...189).. . _. The i d and A.. _ . --- I I 0 154....1 The E' equation From (4...3246 ~ 181. Let the following assumptions be made: 1 .4 and correspon to the loading condition of Example 5. Also let E F D be the stator EMF that is produced by the field current and corresponds to the field voltage v.. 2.. .74) and (4._ .85 -1. Also using the above definition for E. Le. 3.49) in (1 Ti0S)E..141 3. The terms w X in the stator and load voltage equations are assumed to. be the stator EMF proportional to the main winding flux linking the stator.330 -.3816 -0.. f i E 6 = U R k M F X F / L F .. terms in the stator and load voltage equations are neglected compared to the speed voltage terms wX. 6. 5.422 I .388 39. ..) can be developed (see references [ I ] and (21).. 6. ~ E F =D R k M F v F / r F O Using the above definitions and main EFD .756 I -1000._ 999. = @RkkfFiF/d + (xd - xj)ld = E + (xd - xi)ld (6. . .. .83 I l I -430.7155 0 . Balanced conditions are assumed and saturation effects are neglected. .88 0 .104) the field equations are given by VF = rFif + AF + AF = LFiF + kMFid (6... 0.53 . 4.50) where I d = i d / G and s is the Laplace transform variable.15 284._ .._..5.49) Now let e.76 417. Amortisseur effects are neglected.282 ~ 0 -- o 0 o o 0 0 44. Under the assumptions stated above the equations describing the system are given below in pu.848 -26.5.5 Simplified linear Model A simplified linear model for a synchronous machine connected to an infinite bus through a transmission line having resistance R .9790 -0.. i s . we get from (6.io defined by (4... Stator winding resistance is neglected.... and inductance Le (or a reactance X.48) to give E.- -236. 0 -__ . .60 10-3 0 0 I ..720 A = -115. we can arrange the second equation in (6. we get UF = (rF/LF)XF - (rF/LF)kMFid (6.14 -313. 1..44 -431.. the s do- = - (Xd - xi)ld (6. .53) where K = f i V .5 1) and (6.W R L .60) Electrical torque equation The pu electrical torque T. = .& = + [ v.50) to compute EFDA = ('l/K3 r& s ) E b A + + K4 6 . From (4. and K4 is related to the demagnetizing effect of a change in the rotor angle.( x q + X. + + .(xq + x e ) i q ~(Kcos(60 .) K4 = V-K/(Xd .a) (6. Therefore. (xi a) K' + X.a)] (6.) From (6.a ) + Reiq .)sin(6. Le.4 d u.58) and (6.X. R.10) because of the introduction here of E.+ R e i d + wRL.51) are linear.a) .52). .~.57) where we define (in agreement with [2]) I/K.54) + xc>rqA [ vm cos Solving (6. sin(& .57) we get the followings domain relation (6.)(X. .50) and (6. .55) where K/ 1/[Rf + (xq + Xe)(xi+Xe)I (6.5. ) ] S A (6. = WRLdid W R kMFiF = + v.2 --]SA + 1 EbA = constant (6. d v . .54) for I d A and IqA.) R.149) and (4. d we compute v and uq for infinite bus loading to be u = . Note that (6.R .cos(6.a) + R. and V. is numerically equal to the three-phase power. rather than uF. K4 = K 3 6. T. = . ) [ ( X . is the infinite bus voltage to neutral.53).74) and from the assumptions made in the simplified model.52) Linearizing (6.X . 0 0 = -RciqA = -Reid& + (xd + x e ) i d A + URkkfFiFA + [Ksin(6. C O S ( 6 0 (6.~. Rearranging (6.7.j .)cos(6.59) differs from (3. = (I/j)(UJd Uqiq) = (&Id + PU (6.59) we can identify that Kl is an impedance factor that takes into account the loading effect of the external impedance.61) where under the assumptions used in this model.sin(6. Sin (6 . -(xi = E. cos ( 6 . we compute [t.IqA + (xq - a)]6A (6.4..linear Models of the Synchronous Machine 223 where E is as defined in Section 4..58) Then from (6.56) We now substitute I d into an incremental version of (6.wRL.] [ = -(xq + X.a) . + Xe)sin(6.( . = 1 K/(X. X.59) [Note that (6.( Y ) ] 6 A + + x e ) I d A + R. i d CY) (6. .V.67) and the synchronizing power coefficient discussed in Chapter 2 and given by (2. If the field flux linkage is constant.51) t o write the initial condition = + (xd - xq)Id EO (xd - xq)IdO x.a) (xi Xe)cos(6. = K / V .66) Substituting (6.(x.E.a ) l ) a A ~ + + + K/irqOIR: K. .224 Chapter 6 Using (6.2 as Eqa = E with E taken from (6.R . K.65) where we have used the q axis voltage E. 6. . + X. in (6.36). c o s ( ~.. is the change in electrical torque for a small change in rotor angle at constant d axis flux linkage.51) in the second equation of (6.64) Linearizing (6.62).64).R. is the change in electrical torque for small change in the d axis flux linkage at constant rotor angle We should point out the similarity between the constant K . . + vi .a)] + Iq0(x.63) and (6.x ..)sin(6. cos(6..e.3 Terminal voltage equation From (4. = v9 = xd' ld + E i (6.)'qO'dA (6.~a)] ~ .)IdO = E~o (xd - xi)IdO + (xd - xq)IdO = .67) Where K .)Id]f9 (6. .sin(6.5.61) T. ) [ ( X . . defined in Figure 5.a) . .)[(x.55) and (6.x. + + ( x q + X~)zl + EqaORe)E6A K. E6 will also be constant and K .a) .6.(x. i.41) the synchronous machine terminal voltage is given by v: or in rms equivalent variables = (l/3)(u. (6. The model is reduced to the classical model of Chapter 2. + u:) (6.{Eqa. = v. - a) + ( x i + X . IEqa0[R.x. = 0.. ) C O S ( .)]I + Iq0(x.sin(6. .65). we compute the incremental torque to be T.68) v.63) [ E : .[R. we compute T ~ A= IqOE6A + EqaO'qA (xq (xq xi)IdOIIqA - (xq - xi)lqO1dA = 19JiA + - - x. b = -x919 From (6. - (6. the synchronizing torque coefficient = K. + xq) sin (6.56) into (6. 59). from (6. To complete the model. (6.90) is used.63) in (6. Solution We can tabulate the data from Example 5. Le. and the infinite bus voltage. but with the. or for a small change in the d axis flux 6.TeA (6. armature resistance set to zero.7 I ) where K. for a small change in rotor angle at constant d axis flux linkage.. K.linear Models of the Synchronous Machine 225 This equation is linearized to obtain (6. the linearized swing equation from (4.69).6 Find the constants K .70) (6.67).73) The angle 6..69) (6.5. .7 1) are the basic equations for the simplified linear model.4 Summary of equations Equations (6. Example 6.1..69) Substituting (6. is the change in the terminal voltage V.. K.1 as follows. T is the total torque to a base quantity of the three-phase machine power. K. in radians is obtained by integrating on cbA twice.70) and Substituting for lqA I. or and K6 is the change in the terminal voltage linkage at constant rotor angle.A .. through K6 of the simplified model for the system and conditions stated in Example 5. I n the above equations the time is in pu to a base quantity of 1/377 s. and 7j = 2Hw.55).. K4.72) We note that the constants K. (6. and K6 depend upon the network parameters. and (6. (6. (6.. the quiescent operating conditions. 7jLjA = T. = = 1.828)(0.02) 1.395(2.70 .7598)(0.(0. = 1.828 x 0.979 Id0 = f. .a)] + I@(xq .7598{0. cos(6.)(X.) sin (. .9205 + 0.455)(2. = V. K.7124 We then calculate K .2578 K2 = = = K5 and K6 are calculated from (6.5995(0. Synchronous machine data: 0.a)]) 0.V.226 Transmission line data: Chapter 6 Re Infinite bus voltage: = 0.645)(0. Also. . .58) = = [ l + (1/1.04))-1 = 0.a)] = [(0. ..02 x 0.a = 66. From (6.0.a) + (x: + X. cos (6.5995 x 0. . K4 = = Rt + (x.700 pu x.651] = 0.1 12(1.631/1.7598 x 1.7598 x 0.. = 0.04)(0. 0.9205 .5995 I/K.64)(-0..776 We can calculate the angle between the infinite bus and the q axis to be 6.55)2/1. .(2.64) = 2. .828)(1.40 PU = V.645 x 0.0.) [(x.(K.3072 0.[(1. K.X.9205 .a) .66) we compute = 1.3908) .02 x 0.R.%. = E.) = 1.(x.02 X.a)] 6 .245 PU Also.3908) = 1.1.995".385 0.3908.1.Vmx.245)(0.[Re sin (6.776/ 1. + X.112 -0.9205)] .828 xd X .67). = v.)cos (6.V.3908) + (0. -1.1 iFo = 2. . cos(d.828[2.385[(0..71): K. + X.X.a) + ReSin(60 .0)][(0.7598 + X.a) = 0./ qo)[R.0409 .9205)] = -0.02)(0.0)[(0. .04 x 0.7598)(0.Z + (xq + X 1 1 + EqaoRe1 e2 0. &. from Example 5.55 x 2. + X e ) sin (6.0755 K/{Iqo[R.a) .02)2 + (2.3162)(1.455(2. Then sin (6.04 x 0..631 1.9205.000 %o = 0.3162 Then we compute from (6. and K2 from (6. = (K.979/dT .a) = 0.VdO/C/rO)[(X~+ Xe)cOs(60 .3908) + 0.3853 x 1. = = = K.04)2] + 2.02 x 0.3908)] 1. K.700 .640 pu 1.02)(0. 813~)]6.1925 Then K3= (I + 2. = 1.7598)( 1.5915)]) = 1.4738 l / K l = R f + ( x .~I~S)]EF.63 1)(0.4047 PU KO= 0.8125 s (0.77611 . = 1.55 x 2 .497 1 ( %o/ Therefore at this operating condition the linearized model of the system is given by EiA T. + 0.9185) = 2.1.) = 1.0755 6.316 = o.8259 -0.02 x 0.497 1 E . cos(6.395(-0. .4047[(0.a) = 0. From this data we calculate E.6 for the operating conditions given in Example 5. Solution From Example 5. = = 1.7598)( 1 . + A'.a) = 0. 7 Repeat Example 6.455 x 0. K 2 = 0.9185 -0.645)(0.805 The effective field-winding time constant under this loading is given by K37i0 = 0.( Go/ Yo)K.xqRe 0.3072 x 5.9670 PU v.7598)(0. and Eqno 1.O) ((2. The greater value in this example is indicative of a lower loading condition or a greater ability in this case to transmit synchronizing power. = 0. + 1.8063 .2578 E d A -0.736" and sin(6..04 x 0.5915.64)(0.245)(2.04)2] + (2.3072 = K4 T.90 s (2.2.59 15)] + (0.02)(0.4479 = We note that for this example the constant K . y b = = = [0. i 5 0 = Id0 = = Yo = 2.3072/(1 + I.000 pu 60 - LY = 53. The constant K . A + 1.)(x.8 2 6 /d % .9 K.Linear Models of the Synchronous Machine 227 K6 = = Yo)[1 .02) = 0.3174 .02)2 + (2.0409 6. is greater in magnitude than in Example 6.A [0.474)(0..474)[(0.7598 (0.2 .K1xXx9 + Xt)l . corresponds to the synchronizing power coefficient discussed in Chapter 2.9185 Eqno= 1.5915) 1.395)[(2.04)] + (0.02)(0.8063) .7598 E60 = = 1.(0.0.8063.4047)( 1. + A'. ..455)-' 1.6.8063) + (0.04)(0.3162 5.02)) = 1.6628 1.1925 .1.(0.172 Z9.3162 Kl = 0. Example 6 .04 x 1.5261/(1 1. 7 its value is 0..228 Chapter 6 K. = (0. the system damping is affected by the constant K. = To illustrate the demagnetizing effect of the armature reaction. K4.245)(2. 2..:i8)[(0. + K37A0s)16A (6.5.. the voltage regulator decreases the natural damping of the system (at that operating condition). The constants K .64)(0. + I .02) 1. substituting in the expression for K A = IK.67).02946. If this constant is negative. The cases studied in the above examples represent heavy load conditions.172 ) The linearized model of the system at the given operating point is given in pu by E.6 and 6. let EFDA 0.. - K2K3K4/(1 The bracketed term is the synchronizing torque coefficient taking into account the effect of the armature reaction.7 we note that the demagnetizing effect of the armature reaction as manifested by the E. From Examples 6.o = -(K4K6/7h)6A (6. we note the following: I . is reduced by a factor K2K4/ TiO. has reversed sign.02)(0.0409.9670 - - (0. Initially. K.5 Effect of loading Examining the values of the constants K .A].7598)(1.77) .and K6 are comparable in magnitude in both cases.5257 EA .5915) + (0. This is usually compensated for by the use of supplementary signals to produce artificial damping.74) (6. + 0.813~)]6A 6. + 1. we TeA = iK. and (6. This effect is more pronounced in relation to the change in the terminal voltage.04 1 )] 0. From (6. while K. In the heavier loading condition of Example 6.4479 6 .3174 E:.8063)] (1.0)(0. then E6A = [K3K4/(1 + K37iOs)18A + K37hls)16A (6.8063)] = = 0. and in Chapter 8 it will be pointed out that in machines with voltage regulators.04)(0.245) 0’9670 [(0. Similarly.172) - (0. TeA K A = = = [0.7598)(0.57) and (6.02)(0. 0.7598)( 1. - K3K4K6/(I KA.. has a value of -0.58). and inifially the change in the terminal voltage is given by ‘.64) (-t.O)( 1.6. the coefficient K .5915) . .75) and substituting in the expression for TeA get. The constant K 3 is the same in both cases.3072/(1 + 1.58) we note that K3 is an impedance factor and hence is independent of the machine loading.0294 K6 = (Xb 0. and in the less severe loading condition of Example 6.(2. From (6.A dependence is quite significant. This is rather significant. (6.645)(0.5546/(1 1.0294.. Certain effects are clearly demonstrated.5257 ( -o‘6628 (0. through Kb for the loading conditions of Examples 6.76) The second term is usually much larger in magnitude than K.7.6 and 6. K.7598)(I .[0.813 s)]EFnA .71) we note that these constants depend on the initial machine loading. P 02 .o 1 Real Power. . (o IEEE.8 06 .1 Variation of parameters K .1 I 01 0 2 .6 Comparison with classical model The machine model discussed in this section is almost as simple as the classical model discussed in Chapter 2.4 06 . Thus (6. The results are shown in Figure 6./Oct.21. Z .. (c) K4 versus P and Q . (e) K 6 versus P and Q .1- re-0.6 08 . K2. and K6 are studied in reference [3] for a one machine-infinite bus system very similar to the system in the above examples except for zero external resistance. Sept. R e a l Power.0y" 0. 1 1 1 04 . 0.0 1. . (d) K5 versus P and Q . I n (6. . power) as parameter.K6 with loading: (a) K I versus P (real power) and Q (reactive . (b) K2 versus P and Q . . o 01 0 2 . PAS-92. P -.) The effects of the machine loading on the constants K . 0.8 1 . P 0.-o . . 0.80. . 1 . Rml Power.linear Models of the Synchronous Machine 229 1. .o 1 Real Paver.. Z constant.4 00 . except for the variation in the main field-winding flux.7- 0. The classical model does not account for the demagnetizing effect of the armature reaction.5.. 1973.72) the classical model will have K6 = 0. P xe = 0.8 1 . 6.67) in the classical model would have K2 = 0. 0m21 01 .4 Q = 0.o Fig.K4. K. 06 . vol.4 R e a l Power.8 1 .4 06 .4 0. 6. manifested as a change in E:.9- 0.5 011 I 0 1 02 .0 r e = 0.0 xe = 0.1. 1 1 1 0.59) the effective time constant is assumed to be very large so that E.o 0. . 01 0 2 . . Reprinted from IEEE Trans. P 0. I t is interesting to compare the two models. 06 . . Also in (6. Example 6. Note that the angle 6 here is not the same as the rotor angle 6 discussed previously.1. . E = = E & = 1 + jO.62 sin 6. solve the system of Example 6. the same system in Example 6.) 6 . To obtain the linearized equation for VI.) = ( E V .0 0.are the machine terminal voltage and the infinite bus voltage respectively..02 x To compare with the value of K.826 X 1.2.0 I .)] 0.8177 cos 6)’ 2 VI.980 ..OOO [(1.645 X 0.645 + jO.+ X . Vf = (0.4164 (0.V I . in Example 6.217) The synchronizing power coefficient is given by P.7.230 Chapter 6 To illustrate the difference between the two models.(0. based on the three-phase power. The phasor E = E Lis the constant voltage behind transient reactance.0 + j0.7 we note the difference in the pu system. The phasors 7 and 7. + (0.020 + j0.3798 + 0. Le.8 Using the classical model discussed in Chapter 2.G. Therefore. K.3186cos6 .43” + (0..7 is solved by the classical model.neglecting R .(B.8177)’ sin’ 6 or V I A= .4761) = 1. For convenience we will use the pu system used (or implied) in Chapter 2. / Z 2 ) [ ( x . = .00) + j1. it is the angle of the fictitious voltage E.j0. Thus the classical model gives a larger value of the synchronizing power coefficient than that obtained when the demagnetizing effect of the armature reaction is taken into account. we get fa - = VI = 1 .sin6.645)(0.40 Substituting. X’ d e ‘ Re Fig. Solution The network used in the classical model is shown in Figure 6.1261 6 .cos6.sin6.8794 + 0. ) C O S + ~ ~ R.7. = 1.2 Network of Example 6. = 6-60 .3186sin6]/j0.3 I86 /28.1.448.3186 “I = EV.0. 6. . we get for the magnitude of V. .59). = -0. 6. By designating the state variables as and and the input signals as E. 6 .5 the system equations are given by K3Thktj6~+ E:. K. The block diagram representation of (6.. .78) Eliminating V.67).K3K4 6 .7 is given by KA]... and T.7 State-Space Representation of Simplified Model From Section 6. K 6 / ~ . = = = = v... T.3 Block diagram of (6. 7 s - ~ 6 elec rod Fig. In both diagrams the subscript A is omitted for convenience. and T.72) the resulting block diagram is shown in Figure 6. 6. is shown in Figure 6. The output is the terminal voltage change V .T. The diagram and its equations show that the simplified model of the synchronous machine is a third-order system.73)... such as E:. E. .. and Fig 6.6 Block Diagrams - ( K .-~+ = 6.4.Linear Models of the Synchronous Machine 231 The corresponding initial value in Example 6. This block diagram “generates” the rotor angle 6. . from the above equations.. Figure 6. and (6. and 6. namely...73) and the equation for 6. K..EiA K. (6..aA K~E:A T.3.4 has two inputs or forcing functions.4 Block diagram of the simplified linear model of a synchronous machine connected to an infinite bus. T. I ..6 . w . Note that Figure 6.EFDA . + = @A (“ 5 (6.4 is similar to Figure 3. 0 ) 6 . K. When combined with (6.12526. Other significant quantities are identified in the diagram. R. Dynamic system stability.rated terminal voltage and with constant turbine output.. W. 6. I References I . PAS-88:316-29. A.3 using the flux linkage model. I using the flux linkage model 6. compute the eigenvalues of the A matrix with the damping D = I . can you draw from the results? Let D = 0. D . 6.. Note in particular the system damping as compared to the analog computer results of Chapter 5 . and 3.. The former will be discussed in Chapter 7 while the latter is discussed in Part 111. the above equation is in the desired state-space form k ! = AX + BU where (6. C. for the classical model and compare with the corresponding values obtained by the simplified linear model. and Concordia. 1952.1. Find the synchronizing power coefficient and V. 2.. 6. 71:692-97. 3.8 Examine the linear system (6.2 using the flux linkage model.4 Repeat Problem 6.10 Repeat Example 6. if any. M. IEEE Trans.G. PAS-92:1538-46. M.1. IEEE Trans..3 Using the data of Problem 6.5 Repeat Problem 6..6..7 Make an analog computer study using the linearized model summarized in Section 6. Phillips.4. P.8 for the system of Example 6. Pt.9 For the generator and loading conditions of Problem 6. Compute the current model A matrix for these three power factors.. How many elements of the A matrix vary as the power factor is changed? How sensitive are these elements to change in power factor? Use a digital computer to compute the eigenvalues of the three A matrices determined in 6. de Mello..81) In the above equations the driving functions E. N E E Trans. F.2 is loaded to 75% of nameplate rating at. Find the sensitivity of the eigenvalues to this parameter. through K6 for the simplified linear model.5.2 Problem 6. are determined from the detailed description of the voltage regulator-excitation systems and the mechanical turbine-speed governor systems respectively. What conclusions. and T.79) and write the equation for the eigenvalues of this system. 6. and Mehta. Find the characteristic equation and see if you can identify any system constraints for stability using Routh’s criterion. Heffron. Concepts of synchronous machine stability as affected by excitation control. Effect of a modern voltage regulator on underexcited operation of large and turbine generators. calculate the constants K. . Problems The generator of Example 5. 6. I .80) (6. 2.6 Repeat Problem 6. The excitation is then varied from 90% PF lagging to unity and finally to 90% leading. 6.1 at 90%PF lagging.232 Chapter 6 T. 6. El-Sherbiny. as a function of 6 . 6. Determine a value of D that will make the linear model respond with damping similar to the nonlinear model. K. 1969. 1973. Under this assumption all power received as steam must leave the generator terminals as electric power. This simplified view is expressed diagramatically in Figure 7.1. namely. which serves to orient our thinking from the problems of represenlalion of the machine to the problems of confrol. governor.1 Principal controls of a generating unit.2 where. I . and exciter. This is not a bad assumption when total losses of turbine and generator are compared to total output. the stator is represented in its simplest form.e at pressure. Assume that the generating unit is lossless. REF v Fig. 7. The excitation system controls the generated EMF of the generator and therefore controls not only the output voltage but'the power factor and current magnitude as well.P Firing control Governor Excitation RE. Refer to the schematic representation of a synchronous machine shown in Figure 7. Here 233 .chapter 7 Excitation Systems Three principal control systems directly affect a synchronous generator: the boiler control. V Turbine Generator -+. The amount of steam power admitted to the turbine is controlled by the governor. for convenience. A n example will illustrate this point further. let us examine briefly the function of each control element. h Power at voltage.I+ Power setpoint +PI+p3.1 Simplified View of Excitation Control Referring again to Figure 7. leaving the consideration of governors and boiler control for Part 111.1 is nothing more than an energy conversion device that changes heat energy of steam into electrical energy at the machine terminals. P Enthalpy. Thus the unit pictured in Figure 7. Sa -t m . 7. In this chapter we shall deal exclusively with the excitation system. by an EMF behind a synchronous reactance as for round rotor machines at steady state. 234 I-&-+, \ E 9- Chapter 7 I ' + Excitation Fig. 7.2 Equivalent circuit of a synchronous machine. the governor c n rols the torque or the shaft power input and the excitation system controls E,, the internally generated EMF. Example 7.1 Consider the generator of Figure 7.2 to be operating at a lagging power factor with a current I, internal voltage E,, and terminal voltage V. Assume that the input power is held constant by the governor. Having established this initial operating condition, assume that the excitation is increased to a new value E;. Assume that the bus voltage is held constant by other machines operating in parallel with this machine, and find the new value of current I ' , the new power factor cos 0: and the new torque angle 6: Solution This problem without numbers may be solved by sketching a phasor diagram. Indeed, considerable insight into learning how the control system functions is gained by this experience. The initial operating condition is shown in the phasor diagram of Figure 7.3. Under the operating conditions specified, the output power per phase may be expressed in two ways: first in terms of the generator terminal conditions P = v~cose (7.1) and second in terms of the power angle, with saliency effects and stator resistance neglected, P = (E, VIA') sin 6 (7.2) (7.3) ( 7.4) In our problem P and V are constants. Therefore, from (7.1) ICOS~, = k where k, is a constant. Also from (7.2) E, sin 6 = k, where k, is a constant. Fig. 7.3 Phasor diagram of the initial condition. Excitation Systems 235 p + II I I A-- E Fig. 7.4 Phasor diagram showing control constraints. Figure 7.4 shows the phasor diagram of Figure 7.3, but with k, and k, shown graphically. Thus as the excitation is increased, the tip of Eg is constrained to follow the dashed line of Figure 7.4, and the tip of I is similarly constrained to follow the vertical I dashed line. We also must observe the physical law that requires that phasor T and phasor Tlie at right angles. Thus we construct the phasor diagram of Figure 7.5, which shows the “before and after” situation. We observe that the new equilibrium condition requires that ( I ) the torque angle is decreased, (2) the current is increased, and (3) the power factor is more lagging; but the output power and voltage are the same. By similar reasoning we can evaluate the results of decreasing the excitation and of changing the governor setting. These mental exercises are recommended to the student as both interesting and enlightening. I‘ Fig. 7.5 Solution for increasing .Ep at constant P and V Note that in Example 7.1 we have studied the effect of going from one stable operating condition to another. We have ignored the transient period necessary to accomplish this change, with its associated problems-the speed of response, the nature of the transient (overdamped, underdamped, or critically damped), and the possibility of saturation at the higher value of E,. These will be topics of concern in this chapter. 7.2 Control Configurations We now consider the physical configuration of components used for excitation systems. Figure 7.6 shows in block form the arrangement of the physical components in 236 Input torque Drime mover Chapter 7 rd Generator I Output voltage d I current I I Auwi I iary Fig. 7.6 Arrangement of excitation components any system. I n many present-day systems the exciter is a dc generator driven by either the steam turbine (on the same shaft as the generator) or an induction motor. An increasing number are solid-state systems consisting of some form of rectifier or thyristor system supplied from the ac bus or from an alternator-exciter. The voltage regulator is the intelligence of the system and controls the output of the exciter so that the generated voltage and reactive power change in the desired way. I n earlier systems the “voltage regulator” was entirely manual. Thus the operator observed the terminal voltage and adjusted the field rheostat (the voltage regulator) until the desired output conditions were observed. In most modern systems the voltage regulator is a controller that senses the generator output voltage (and sometimes the current) then initiates corrective action by changing the exciter control in the desired direction. The speed of this device is of great interest in studying stability. Because of the high inductance in the generator field winding, it is difficult to make rapid changes in field current. This introduces considerable ‘‘lag’’ in the control function and is one of the major obstacles to be overcome in designing a regulating system. The auxiliary control illustrated in Figure 7.6 may include several added features. For example, damping is sometimes introduced to prevent overshoot. A comparator may be used to set a lower limit on excitation, especially at leading power factor operation, for prevention of instability due to very weak coupling across the air gap. Other auxiliary controls are sometimes desirable for feedback of speed, frequency, acceleration, or other data [I]. 7.3 Typical Excitation Configurations To further clarify the arrangement of components in typical excitation systems, we consider here several possible designs without detailed discussion. 7.3.1 Primitive systems First we consider systems that can be classified in a general way as “slow response” systems. Figure 7.7 shows one arrangement consisting of a main exciter with manual or automatic control of the field. The “regulator” in this case detects the voltage level and includes a mechanical device to change the control rheostat resistance. One such directacting rheostatic device (the “Silverstat” regulator) is described in reference [2] and consists of a regulating coil that operates a plunger, which in turn acts on a row of spaced silver buttons to systematically short out sections of the rheostat. In application, the device is installed as shown in Figure 7.8. In operation, an increase in generator output voltage will cause an increase in dc voltage from the rectifier. This will cause an increase in current through the regulator coil that mechanically operates a solenoid to insert exciter field resistance elements. This reduces excitation field flux and voltage, thereby lowering the field current in the generator field, hence lowering the generator Excitation Systems Commutator Exciter 237 I Field I Exciter field rhecntat I * T PT‘s Manual control Fig. 7.7 Main exciter with rheostat control. voltage. Two additional features of the system in Figure 7.8 are the damping transformer and current compensator. The damping transformer is an electrical “dashpot” or antihunting device to damp out excessive action of the moving plunger. The current compensator feature is used to control the division of reactive power among parallel generators operating under this type of control. The current transformer and compensator resistance introduce a voltage drop in the potential circuit proportional to the line current. The phase relationship is such that for lagging current (positive generated reactive power) the voltage drop across the compensating resistance adds to the voltage from the potential transformer. This causes the regulator to lower the excitation voltage for an increase in lagging current (increase in reactive power output) and provides a drooping characteristic to assure that the load reactive power is equally divided among the parallel machines. The next level of complication in excitation systems is the main exciter and pilot Generator Fig. 7.8 Self-excited main exciter with Silverstat regulator. (Used with permission from Efecrricul Trammission and Distribution Reference Book, 1950, ABB Power T & D Company Inc., 1992.) 238 && ’ e? Chapter 7 Main exciter Canmutator Commutator Slip T I breaker I I JI T Fig. 7.9 Main exciter and pilot exciter system. exciter system shown in Figure 7.9. This system has a much faster response than the self-excited main exciter, since the exciter field control is independent of the exciter output voltage. Control is achieved in much the same way as for the self-excited case. Because the rheostat positioner is electromechanical, the response may be slow compared to more modern systems, although it is faster than the self-excited arrangement. The two systems just described are examples of older systems and represent direct, straightforward means of effecting excitation control. I n terms of present technology in control systems they are primitive and offer little promise for really fast system response because of inherent friction, backlash, and lack of sensitivity. The first step in sophistication of the primitive systems was to include in the feedback path an amplifier that would be fast acting and could magnify the voltage error and induce faster excitation changes. Gradually, as generators have become larger and interconnected system operation more common, the excitation control systems have become more and more complex. The following sections group these modern systems according to the type of exciter 131. o Fig. 7.10 Excitation control system with dc generator-commutator exciter. ( IEEE. Reprinted from l E E E Trans., vol. PAS-88, Aug. 1969.) Example: General Electric type NA143 amplidyne system 141. Excitation Systems 239 I I' Fig. 7.1 I Excitation control system with dc generator-commutator exciter. (w IEEE. Reprinted from IEEE Trans.. vol. PAS-88, Aug. 1969.) Example: Westinghouse type W M A Mag-A-Stat system [ 6 ] . 7.3.2 Excitation control systems with dc generator-commutator exciters Two systems of U S . manufacture have dc generator-commutator exciters. Both have amplifiers in the feedback path; one a rotating amplifier, the other a magnetic amplifier. Figure 7.10 [3) shows one such system that incorporates a rotating amplifier or amplidyne [5] in the exciter field circuit. This amplifier is used to force the exciter field in the desired direction and results in much faster response than with a self-excited machine acting unassisted. Another system with a similar exciter is that of Figure 7. I I where the amplifier is a static magnetic amplifier deriving its power supply from a permanent-magnet generator-motor set. Often the frequency of this supply is increased to 420 Hz to increase the amplifier response. Note that the exciter in this system has two control fields, one for boost and one for buck corrections. A third field provides for self-excited manual operation when the amplifier is out of service. 7.3.3 Excitation control systems with alternator-rectifier exciters With the advent of solid-state technology and availability of reliable high-current rectifiers, another type of system became feasible. I n this system the exciter is an ac generator, the output of which is rectified to provide the dc current required by the generator field. The control circuitry for these units is also solid-state in most cases, and the overall response is quite fast [3]. An example of alternator-rectifier systems is shown in Figure 7.12. In this system the alternator output is rectified and connected to the generator field by means of slip rings. The alternator-exciter itself is shunt excited and is controlled by electronically adjusting the firing angle of thyristors (SCR's). This means of control can be very FdSt 240 Exciter Chapter 7 power Fig. 7. I 2 Excitation control system with alternator-rectifier exciter using stationary noncontrolled rectifiers. (G IEEE. Reprinted from I E E E Trans.. vol. PAS-88, Aug. 1969.) Example: General Electric Alterrex excitation system 171. since the firing angle can be adjusted very quickly compared to the other time constants involved. Another example of an alternator-rectifier system is shown in Figure 7.13. This system is unique in that it is brushless; i.e., there is no need for slip rings since the alternator-exciter and diode rectifiers are rotating with the shaft. The system incorporates a pilot permanent magnet generator (labeled PMG in Figure 7.13) with a permanent magnet field to supply the (stationary) field for the (rotating) alternator-exciter. Thus all coupling between stationary and rotating components is electromagnetic. Note, however, that it is impossible to meter any of the generator field quantities directly since these components are all moving with the rotor and no slip rings are used. Rotating elemenk Other inputs ----Fig. 7. I3 Excitation control system with alternator-rectifier exciter employing rotating rectifiers. (o IEEE. Reprinted from IEEE Trans., vol. PAS-88, Aug. 1969.) Example: Westinghouse type WTA Brushless excitation system 18.91. Excitation Systems IConhollebl 24 1 Fig. 7.14 Excitation control system with alternator-SCR exciter system. ((c> IEEE. Reprinted from IEEE Trans., vol. PAS-88, Aug. 1969.) Example: General Electric Althyrex excitation system I 1 11. The response of systems with alternator-rectifier exciters is improved by designing the alternator for operation at frequencies higher than that of the main generator. Recent systems have used 420-Hz and 300-Hz alternators for this reason and report excellent response characteristics [S, IO]. 7.3.4 Excitation control systems with alternator-SCR exciter systems Another important development in excitation systems has been the alternator-SCR design shown in Figure 7.14 [3]. In this system the alternator excitation is supplied diLinear reactor Fig. 7. I5 Excitation control system with compound-rectifier exciter. (o IEEE. Reprinted from IEEE Trans., vol. PAS-88, Aug. 1969.) Example: General Electric SCTP static excitation system [12,13]. 242 Chapter 7 rectly from an SCR system with an alternator source. Hence it is only necessary to adjust the SCR firing angle to change the excitation level, and this involves essentially no time delay. This requires a somewhat larger alternator-exciter than would otherwise be necessary since it must have a rating capable of continuous operation at ceiling voltage. I n slower systems, ceiling voltage is reached after a delay, and sustained operation at that level is unlikely. 7.3.5 Excitation control systems with compound-rectifier exciter systems The next classification of exciter systems is referred to as a “compound-rectifier’’ exciter, of which the system shown in Figure 7.15 is an example [3]. This system can be viewed as a form of self-excitation of the main ac generator. Note that the exciter input comes from the generator: electrical output terminals, not from the shaft as in previous examples. This electrical feedback is controlled by saturable reactors, the control for which is arranged to use both ac output and exciter values as intelligence sources. The system is entirely static, and this feature is important. Although originally designed for use on smaller units [ 12, 131, this same principle may be applied to large units as well. Self-excited units have the inherent disadvantage that the ac output voltage is low at the same time the exciter is attempting to correct the low voltage. This may be partially compensated for by using output current as well as voltage in the control scheme so that (during faults, for example) feedback is still sufficient to effect adequate control. Such is the case in the u n i t shown in Figure 7.15. 7.3.6 Excitation control system with compound-rectifier exciter plus potentialsource-rectifier exciter A variation of the compound-rectifier scheme is one in which a second rectified outp u t is added to the self-excited feedback to achieve additional control of excitation. Auxi I iory power input far start-up Fig. 7. 16 Excitation control system with compound-rectifier exciter plus potential-source-rectifier exciter. (@ IEEE. Reprinted from lEEE Trans., vol. PAS-88, Aug. 1969.) Example: Westinghouse type WTA-PCV static excitation system [ 14). Excitation Systems 243 This scheme is depicted in Figure 7.16 [3]. Again the basic self-excited main generator scheme is evident. Here, however, the voltage regulator controls a second rectifier system (called the “Trinistat power amplifier” in Figure 7.16) to achieve the desired excitation control. Note that the system is entirely static and can be inherently very fast, the only time constants being those of the reactor and the regulator. 7.3.7 Excitation control systems with potential-source-rectifier exciter The final category of excitation systems is the self-excited main generator where the rectification is done by means of SCR’s rather than diodes. Two such systems are shown in Figure 7.17 and Figure 7.18 (3). Both circuits have static voltage regulators that use potential, current, and excitation levels to generate a control signal by which the SCR gating may be controlled. This type of control is very fast since there is no time delay in shifting the firing angle of the SCR’s. 7.4 Excitation Control System Definitions Most of the foregoing excitation system configurations are described in reference [3], which also gives definitions of the control system quantities of interest in this application. Only the most important of these are reviewed here. Other definitions, including those referred to by number here, are stated in Appendix E. All excitation control systems may be visualized as automatic control systems with feed forward and feedback elements as shown in Figure 7.19. Viewed in this way, the excitation control systems discussed in the preceding section may be arranged in a general way, as indicated in Figure 7.20 and further described in Table 7.1. Note that the synchronous machine is considered a. part of the “excitation control system,” but the control elements themselves are referred to simply as the “excitation system.” The type of transfer function belonging in each block of Figure 7.20 is discussed in reference [ 151. The reference to systems of Type 1, Type 3, etc., in the last column of Table 7. I also refers to system types defined in that reference. This will be discussed in greater detail in Section 7.9. Our present concern is to learn the general configuration Auxiliary power UII u inputfor stort-up&Te& Fo[$jField breaker power potential transfanner ’ CT PT’S regulator -----_I Fig. 7. I7 Excitation control system with potential-source-rectifier exciter. (@ IEEE. Reprinted from / € E € Trans., vol. PAS-88. Aug. 1969.) Example: General Electric SCR static excitation system [14]. 244 Auxiliary Rawer itput tor start-ur, Slia Chapter 7 : L5lment5 ---I . buildup !__ ~ \ J PT's rm Trinistat power amplifier I --II Exci t a t im power potential transfoner I 1 I Excitation power ~ Regulator power control) --1 I rifol;ag7 t I . current I Reguloting system Fig. 7.1 8 Excitation control system with potential-source-rectifier exciter. (c: IEEE. Reprinted from / L E E Trans.. vol. PAS-88. Aug. 1969.) Example: Westinghouse type WTA-Trinistat excitation system. of modern excitation control systems and to become familiar with the language used in describing them. 7.4.1 Voltage response ratio A n important definition used in describing excitation control systems is that of the defined in Appendix E, Def. 3.15-3.19. This is a rough measure of how fast the exciter open circuit voltage will rise in 0 . 5 s if the excitation control is adjusted suddenly in the maximum increase direction. I n other words, the voltage reference in Figure 7.20 is a step input of sufficient magnitude to drive the exciter voltage to its ceiling (Def. 3.03) value with the exciter operating under no-load conditions. Figure 7.2 I shows a typical response of such a system where the voltage u, starts at the rated load field voltage (Def. 3.21) that is the value of u,, which will produce rated response ratio Reference Directly control led Feedback signal (Def 3.30) I (Def 2.05) Fig. 7.19 Essential elements of an automatic feedback control system (Def. 1.02). (E. IEEE. Reprinted from / € E € Truns.. vol. PAS-88. Aug. 1969.) Note: In excitation control system usage the actuating signal is commonly called the error signal (Del. 3.29). (See Appendix E for definitions.) Table 7.1. I Components Commonly Used in Excitation Control System 7 Type of exciter mplifierl 2 Pre- Power amplifier Power sources 1 Manual control Signal modifiers 1 See note 6 dc Generatorcommutator exciter Alternatorrectifier exciter ;ee note Rotating, magnetic. 3 thyristor Rotating. thyristor Self-excited or separately excited exciter Compensated input to power amplifier. Selfexcited field voltage regulator Exciter output voltage regulator Self-excited MG set MG set. synchronous machine shaf Synchronous Synchronous machine machine shaf shaft. MG set. alternator output Alternatorrectifier (controlled) exciter Compoundrectifier exciter Compoundrectifier exciter plus potential-soun:e rectifier exciter Potential-source rectifier (controlled) exciter V Thyristor Alternator output Synchronous machine shaf Magnetic, thyristor Thyristor Synchronous machine terminals Synchronous machine terminals Synchronous machine terminals machine terminals Compensated input to power amplifier Thyristor Synchronous machine terminals Synchronous machine terminals Exciter output voltage regulator. Compensated input to power amplifier 7 I Source: c IEEE. Reprinted from / € E € Truns.. vol. PAS-88. Aug. 1969. 2. Primary detecting element and reference input: can consist of many types of circuits on any system including dil son. intersecting impedance. and bridge circuits. 3. Preamplifier: Cons.ists of all types but on newer systems is usually a solid-state amplilier. 6. Signal modifiers: (A) Auxiliary inputs-reactive and active current compensators: system stabilizing signals pro (B)Limiters-maximum excitation. minimum excitation. maximum V/Hz. 8. Excitation control system stabilizers: can consist of all types from series lead-lag to rate feedback around any ele *IEEE committee report [15]. 246 Chapter 7 Power source [regulator) ( regulator (exciter) power source ** Synchronous machine - i t" I Power I I I stcbi lizer I (Def 2.14) (Def 2.11) (Def 2.03) (Def 2.13) lnpuh i i I - I Fig. 7.20 Excitation control systems. (,q IEEE. Reprinted from IEEE Trans., vol. PAS-88. Aug. 1969.) Note: The numerals on this diagram refer to the columns in Table 7.1. (See Appendix E for definitions.) generator voltage under nameplate loading. Then, responding to a step change in the reference, the open-circuited field is forced at the maximum rate to ceiling along the curve ab. Since the response is nonlinear, the response ratio is defined in terms of the area under the curve ab for exactly 0.5s. We can easily approximate this area by a straight line ac and compute Response ratio = cd/(Oa)(0.5) pu V/s (7.5) Kirnbark [I61 points out that since the exciter feeds a highly inductive load (the generator field), the voltage across the load is approximately u = k d $ / d t . Then in a short time A t the total flux change is Ac$ = 1 JA' = area under buildup curve udt k (7.6) 0 Time, s 0.5 Fig. 7.21 Delinition of a voltage response ratio. 5 s is almost meaningless. 3.5 was chosen because this is about the time interval of older “quickresponse” regulators between the recognition of a step change in the output voltage and the shorting of field rheostat elements. Briefly.5s interval Oe in Figure 7. This is discussed in reference [3]. In dynamic operation where the interest is in small. however. (z IEEE. Aug.21 by an interval Oe = 0. I . 1969. 0 1 2 3 9 101112 Response Rotio (Def 3.22 [3] and shows how close the 0.16) is the time required to reach 95% of ceiling]. Reprinted from /&E€ Trans. and a new definition is introduced (Def. A comparison of three systems. such as a fault condition.18) 4 5 6 7 8 Fig. with still larger machines being equipped with 350-V. 7. It is reasonable that an exciter with a high ceiling voltage will build up to a particular volt- .21) . 171.1-s response is to the ideal system. that modern fast systems may reach ceiling in 0. it may be helpful to state certain numerical values of exciter ratings offered by the manufacturers (see [2] for a discussion of exciter ratings). The voltage rating and the ceiling voltage are both important in considering the speed of response [ I . each attaining 95% ceiling voltage in 0.4. Reference [ I ] tabulates the pattern of ceiling voltages for various response characteristics in Table 7. and extending the triangle acd out to 0. 375-V.1 I d hoving on expomntiol response Synchronous machine mted Imd field voltoge (Lkf 3. It has been recognized for some time. say 10 MVA and below.1 s or less. or 500-V exciters.1 s. . which shows the improved response for higher ceiling voltage ratings (and the lower ceiling voltage for solid-state exciters).. Buildup rather than build-down is used because there is usually more interest in the response to a drop in terminal voltage.05) that replaces the 0. say up to l00MVA. such as t h e one shown in Figure 7.2. Larger units usually have 250-V exciters.1 s for “systems having an excitation voltage response time of 0. exciters are usually rated at 125 V for small generators.) The time A t = 0. 1. is given in Figure 7. .1 s or less” [the voltage response time (Def.2 Exciter voltage ratings Some additional comments are in order concerning certain of the excitation voltage definitions. fast changes. a step function. 7.5) is an adequate definition if the voltage response is rather slow.21. build-down may be equally important.Excitation Systems 247 Syl t m otfoining 95% ceiling in 0. vol. PAS-88.22 Exciter ceiling voltage as a function of response ratio for a high initial response excitation system. Equation (7. First. and pu synchronous internal voltage are all equal under steady-state conditions with no saturation.I . the rated load field voltage.. The IEEE [3] recommends the use of system 9. (B) rated load field voltage. (C) rated air-gap voltage (the voltage necessary to produce rated voltage on the air gap line of the main machine in the case of a dc generator exciter). Consider. Table 7.5 I . an exciter rated at 250V. One useful description.o I . and the like. damping. of Figure 7.23 Per unit voltages for a 250-V exciter.65 I . System D is not illustrated in Figure 7. It is generally agreed that the quantity of primary interest is the exciter voltage-time characteristic in response to a step change in the generated voltage of from 10 to 20% [18. The problem is how to state in words the various possible slopes. I . overshoots.I . often used in control system specification. is that based on the . 7. This is an important consideration in comparing types and ratings of both conventional and solid state exciters as shown in Table 7.00-2.o 1. pu field current. I n adopting a pu system for the exciter. System C is often convenient since.2. System B is often used.25. and (D) no-load field voltage.19]. Some possibilities are (also see [2]): (A) exciter rated voltage.70.55. there is no obvious choice as to what base voltage to use. delays. System performance could be measured under any number of forcing conditions.4. IO *Based on rated exciter voltage..30.I .3 Other specifications Excitation control system response should be compared against a suitable criterion of performance if the system is to be judged or graded. The pu system A pu System Fig. pu exciter voltage. as an example.248 Chapter 7 age level faster than a similar exciter with a lower ceiling voltage simply because it saturates at a higher value.0 .23.50 I . with rated air gap voltage as a base.35 1. Typical Ceiling Voltages for Various Exciter Response Ratios Response ratio Per unit ceiling voltage conventional exciters* SCR exciters 0.5 2 .23 has little merit and is seldom used.20.23 and is seldom used.40 1.SO 4.45.1 .2. 7.55 2.40.I .25 1. For this rating some typical values of other defined voltages are given in Figure 7.I .I .20 1. The natural resonantfrequency w. i.f2)]sinw. In this case the voltage rise is more “sluggish.25.7) and is related to the values a . . 7.24 Time domain specifications 1221. Here the overshoot is zero. the system is oscillatory. such as in Figure 7. Then determine the area under this curve for 0.Excitation Systems 249 time Tim. is also of interest and may be given as a specification. it has very little overshoot (about 5%). (i.24. based on that of a second-order system.t + [{/(I f2)”Z . w:) (7. the time for the response to settle within k of its final value). the settling time is T.e.O. a . and a2 of two successive overshoots [23]. is a reasonable one on which to base time domain specifications since many systems tend to exhibit two “least-damped” poles that give a response of this general shape at some value of gain [20. and the rise time is TR.” as shown in Figure 7.8) w. the system is often “overdamped”.7. The damping ratio is that value for a second-order system defined by f in the expression G(s) = K/(s’ +2 f ~+~ . such as the terminal voltage. cu’rve shown in Figure 7.9].e. For newer. In the case of the second-order system (7.7). and the settling time.tI (7. f > I .21]. the rise time. Sometimes it is given as the time required to arrive at the final value after first overshooting this value. The settling time is the time required for the response to a step function to stay within a certain percentage of its final value. Here the curve is the response to a step change in one of the system variables. fast systems reference [3] suggests simulation of the excitation as preferable to actual testing since on some systems certain parameters are unavailable for measurement [8.25. pu. The rise time is the time for the response to rise from IO to 90% of the steady-state response.‘{cosw.9) When f = 0. Three quantities describe this response: the overshoot. This response. the response to a step change of a driving variable is c(r) where = 1 - e-f**. The overshoot is the amount that the response exceeds the steady-state responsein Figure 7. = w. In dealing with an exciter being forced to ceiling due to a step change in the voltage regulator control. Reference (191 suggests testing an excitation system to determine the response.. The first definition is preferred. (I - (7.. Critical damping is said to occur when { = 1 . I Fig.24..5 s and use this as a specification of response in the time domain. when f = 0. 2. Thus we need to be very critical of this important system component. stiction. is sluggish and includes deadband and backlash due to mechanical friction. In the late 1930s and early 1940s several types of regulators.4 0. no matter how small the deviation. In such a system the voltage reference is the spring tension against which the solenoid must react. As machines of larger size became more common in the 1930s the indirecracring rheostatic regulators began to appear. such as the Silverstat [2] and the Tirrell(241. to hydrogenerators.2 0.1 Electromechanical regulators The rather primitive direct-acting regulator shown in Figure 7.26]. This is the device that senses changes in the output voltage (and current) and causes corrective action to take place. and loosely fitting parts. some dating back to about 1900. I f the regulator is slow. (Reference [24] gives an interesting tabulation of the progress of these developments. the system will be a poor one. has deadband or backlash. it will not alter its response until instructed to do so by the voltage regulator.3 I 0.) 7. it is necessary that the voltage regulator be a continuously acting proportional system. These devices use a relay as the voltagesensitive element [24]. Direct-acting regulators.9. however. This regulator is limited in its speed of response by various mechanical delays.1 & 0. usually connected between the pilot exciter and the main exciter.250 Chapter 7 1 k L V e . This type of system was therefore studied intensively and widely applied during the 1940s and 1950s. or is otherwise insensitive. the direct-acting and the indirect-acting. thus the reference is essentially a spring. to . then to turbine generators.8 is an example of an electromechanical regulator. and large errors are to receive stronger corrective measures than small errors.12). This relay operates to control a motor-operated rheostat.25 Response of an excitation system 7. Thus no deadband is to be tolerated.5. and finally.5 Time.5 Voltage Regulator In several respects the heart of the excitation system is the voltage regulator (Def. Fig.25]. 7. 0 0. Such devices were widely used and steadily improved. in the early 1950s. while maintaining essentially the same form. N o matter what the exciter speed of the response. Once the relay closes. Two types of electromechanical regulators are often recognized. as in the direct-acting device. were developed and tested extensively [24. have been in use for many years. electronic and static. This means that any corrective action should be proportional to the deviation in ac terminal voltage from the desired value. beginning with application to synchronous condensers. The response. It is reliable and independent of auxiliaries of any kind. These tests indicated that continuously acting proportional control “increased the generator steady-state stability limits well beyond the limits offered by the rheostatic regulator” [24. as in Figure 7. In addition to high reliability and availability for maintenance. The . coupled with mechanical backlash. Generally speaking.3 Rotating amplifier regulators In systems using a rotating amplifier to change the field of a main exciter. The development of rotating amplifiers in the late 1930s and the application of these devices to generator excitation systems (28. and much larger field currents can be controlled than with the direct-acting regulator. has inherent deadband and this.2 Early electronic regulators About 1930 work was begun on electronic voltage regulators. constitutes a serious handicap. Such circuits supplied the field windings of the rotating amplifiers.10. A special set of contacts closes very fast with rapid rotor accelerations that permit faster than normal response due to sudden system voltage changes. as in Figure 7. Here we take the view that the rotating amplifier is the final. and this was at least partly justified in later analyses [25]. 7. The contact type of control. such static circuits were designed to exclude any electronic active components so that the reliability of the device would be more independent of component aging. These devices were considered quite successful. The electronic exciters or pilot exciters were high-power dc sources usually employing thyratron or ignitron tubes as rectifying elements. A contact assembly attached to the rotor responds by closing contacts in the rheostat as the shaft position changes. which were connected in series with the main exciter field. depending on polarity. This torque is balanced against a spring in torsion so that each value of voltage corresponds to a different angular position of the rotor. The operation of a typical rotating amplifier regulating system can be analyzed by reference to Figure 7. The response of this type of regulator is fairly fast. This scheme has the feature that the rotating amplifier can be bypassed for maintenance and the generator can continue to operate normally by manual regulation through a field rheostat. and nearly all large units installed between about 1930 and 1945 had this type of control.5. high-speed relays are used to permit faster excitation changes. The generator is excited by a self-excited shunt exciter. these early electronic devices provided “better voltage regulation as well as smoother and faster generation excitation control” (241 than the competitive indirectacting systems. This connection is often called a “boost-buck” connection since. They never gained wide acceptance because of anticipated high maintenance cost due to limited tube life and reliability.10. the rotating amplifier is in a position to aid or oppose the exciter field. 291 have been accompanied by the development of entirely “static” voltage sensing circuitry to replace the electromechanical devices used earlier. high-gain stage in the voltage regulator. I n such a device the output torque is proportional to the average three-phase voltage. devices employing saturable reactors and selenium rectifiers showed considerable promise. and electronic pilot exciters used in conjunction with a conventional main exciter (24. Another type of indirect-acting regulator that has seen considerable use employs a polyphase torque motor as a voltage-sensitive element [27]. the response is quite fast. electronic exciters. electronic voltage regulators were of two types and used either to control electronic pilot exciters or electronic main exciters [25]. Usually. as in Figure 7. This is due to the additional current “gain” introduced by the pilot excitermain exciter scheme. it is not altogether clear whether the rotating amplifier is a part of the “voltage regulator” or is a kind of pilot exciter. In some cases.10. For example.Excitation Systems 25 1 short out a rheostat section. Many are still in service. In general. 7. 251.5. however. g. Under this condition the rotating amplifier voltage is zero.5 (e. and sturdy construction. The speed of response is due largely to the main exciter time constant. This reduces the exciter voltage. Magamp of the Westinghouse Electric Corporation and Amplistat of the General Electric Company.g. The control characteristic may be better understood by examining Figure 7. 7. 30. The field rheostat is set to intersect the saturation curve at a point corresponding to rated terminal voltage. field circuit can be controlled either manually by energizing a relay whose contacts bypass the rotating amplifier or automatically. the magnetic amplifier consists of a saturable core reactor and a rectifier. The ceiling voltage is an important factor too. The voltage rating of the rotating amplifier in systems of this type is often comparable to the main exciter voltage rating.5.252 Chapter 7 Field > u Y e 0 V m > u . see Appendix D).. Magnetic amplifier regulators Another regulator-amplifier scheme capable of zero deadband proportional control is the magnetic amplifier system [6. a static amplifying device [32. but this is no drawback in power applications. e. which in turn reduces if. replaces the rotating amplifier. Thus the shaded area above the set point in Figure 7. 331. the generator field current.4 . long life.26 is called the buck voltage region. zero warm-up time.26. i. A similar reasoning defines the area below the set point to be the boost voltage region. Usually.) I n this system a magnetic amplifier. with the amplifier providing a feedback of the error voltage to increase or decrease the field current. 7.e. Rotating amplifier systems have a moderate response ratio. the exciter voltage required to hold the generated voltage at rated value with full load. The voltage sensing circuit (described later) detects this rise and causes the rotating amplifier to reduce the field current in the exciter field. It is restricted to low or moderate frequencies. It is essentially an amplifying device with the advantages of no rotating parts... often quoted as about 0. 311.e.26 V-l characteristic defining boost and buck regions.. exciters with higher ceilings having much faster response than exciters of similar design but with lower ceiling voltage (see [ 171 for a discussion of this topic). which is much greater than the amplidyne time constant. (We use the generic term “magnetic amplifier’’ although those accustomed to equipment of a particular manufacturer use trade names. i. and the voltage swings of the amplifier change rapidly in attempting to regulate the system [24]. Now suppose the generator load is reduced and the generator terminal voltage begins to rise.U Exciter Shunt Field Current Fig. however. and hence the average load current.27 [33]. with 1-mW input signals. The second stage is driven to maximum output when the input stage is at half-maximum. It is shown in [34] that for static magnetic amplifiers it is equal to one-half the ratio of power output to stored magnetic energy. The figures of merit (341 are about 200/cycle for the input stage and 500/cycle for the output stage. since . The power amplifier has a figure of merit of 1500/cycle with an overall delay of less than 0. the magnetic amplifiers and reference are usually supplied from a 420-Hz system supplied by a permanent-magnet motor-generator set for maximum security and reliability. has several features to distinguish it from the previous example. however.1 1 (61. The current Rowing through the load is basically limited by the very large inductance in the saturable core main windings.Excitation Systems 253 u Supply oc Sotwoble core Laad Fig. Basically. The fact that this amplifier is very nonlinear is of little concern. The experience indicates that. One type of regulator that uses a magnetic amplifier is shown in block diagram form in Figure 7. The main exciter also has a self-excited. and its transient response is also about three cycles. (The figure of merit of an amplifier has been defined as the ratio of the power amplification to the time constant. only two units of which are regulated.10 [4]. Note. is the essence of any amplifier. The rotating amplifier is located in series with the exciter field in the usual boost-buck connection. the current jumps to a large value limited only by the load resistance. of controlling a large output current by means of a small control current. The power amplifier supplies the main exciter directly in this system. can respond to maximum output in three cycles of the 420-Hz supply. rheostat-controlled field and can continue to operate with the magnetic amplifiers out of service.1 I consists of a two-stage pushpull input amplifier that. making this type of regulator favorable to remote (especially hydro) locations. shown in Figure 7. The magnetic amplifier in the system of Figure 7. Another application of magnetic amplifiers in voltage regulating systems. that the exciter must have two field windings for boost or buck corrections since magnetic amplifiers are not reversible in polarity.) Reference [6] reviews the operating experience of a magnetic amplifier regulator installation on one 50-MW machine in a plant consisting of seven units totaling over 300 MW. 7.27 Magnetic amplifier. One important feature of this system is that the magnetic amplifier is relatively insensitive to variations in line voltage and frequency. This compares with about 500/s for a conventional pilot exciter. the magnetic amplifier is similar to that shown in Figure 7. This feature. we control the firing point on each voltage (or current) cycle. As the core becomes saturated. Here the magnetic amplifier is used to amplify a voltage error signal to a power level satisfactory for supplying the field of a rotating amplifier.01 s. First. By applying a small (lowpower) signal to the control winding. particularly reactive load. In fact.6 Exciter Buildup Exciter response has been defined as the rate of increase or decrease of exciter voltage when a change is demanded (see Appendix E. we have A. self-excited.. however. In the case under study. the machine terminal voltage was regulated to i0. such as the complete shorting of the field resistance. it causes the machine on which it is installed to absorb much of the swing in load. separately excited. and boost-buck) are of interest. Def.6. Various configurations are used depending on the manufacturer. They must be analyzed independently. The future will undoubtedly bring more applications of solid-state technology in these systems because of the inherent reliability. The best way to determine the exciter response is by actual test where this is possible. and Dah1 [35] to which the reader is referred for additional study. reveals that both exciter voltage and line currents undergo rapid fluctuations when regulated but are nearly constant when unregulated. 0 current. (Portions of this analysis parallel that of Kimbark [16]. but all have generally fast operation with no appreciable time delay compared to other system time constants.1 The dc generator exciter I n dealing with conventional dc exciters three configurations (Le.28.5 Solid-state regulators Some of the amplification and comparison functions in modern regulators consist of solid-state active circuits (31.10) = i = up = flux linkages of the main exciter field. Usually we interpret this demand to be the greatest possible control effort. the voltage buildup must be computed. ease of maintenance. V . a “build-down’’ curve can also be recorded. A pilot exciter voltage. Curves thus recorded do not differ a great deal from those obtained under loaded conditions.. Rudenberg [20].whereas a i 1% variation was observed with the regulator disconnected [ 6 ] . where A. we compute the response under no-load conditions. This serves to satisfy the terms of the response ratio definition and also simplifies the computation or test procedure. Then a step change in a reference variable is made. This is to be expected since the regulation of machine terminal voltage to a nearly constant level makes this machine appear to have a lower reactance. Since the exciter response ratio is defined in terms of an unloaded exciter (Def. This is called a “buildup curve. close observation of operating oscillograms.254 Chapter 7 the magnetic amplifier regulator is so much faster than the primitive rheostatic regulator. 3. driving the exciter voltage to ceiling while the voltage is recorded as a function of time. Wb turns main exciter field resistance.) Consider the separately excited exciter shown in Figure 7.5. R = -k Ri = vp (7. 7. because the equations describing them are different. Summing voltage drops around the pilot exciter terminal connection. 7.25:(. If it is impractical to stage a test on the exciter.” In a similar way. hence it absorbs changes faster than its neighbors. We now turn our attention to this problem. The exciter is operated at rated speed (assuming it is a rotating machine) and with no load. and low initial cost of these devices.19).15). when operating with an arc furnace load. 3. 7. we replace (7. with relative magnitudes also depending on saturation. W.1 1) is nonlinear. I I ) The voltage of the pilot exciter up may be treated as a constant [ 161. vol. is proportional to the air gap flux 4. Furthermore. (7. by E..Excitation Systems iF 0 = . I t does not link all N turns of the pole on the average and is usually treated either as proportional to or proportional to i . 1956. Kimbark. The field flux has two components. compares with &.12) The problem is to determine how 4. its voltage U .c i 255 a- Pi lot exciter Ypcmbcbr Main exciter Fig. (Reprinted by per.. then 44 where C is a constant. (see [I61 for a more detailed discussion). leakage flux and armature flux. comprising 10-20% of the total.1 1) by a term involving the voltage ordinate u. 0 . = (7. mission from Power Sysiem Siabiliry. Let us assume that r$4 is proportional to 4. in Since magnetization curves are plotted in terms of U . versus i.13) (7. and field flux @ E = 9 + 4 4 . we have N& + Ri = up (7. 7. leakage flux 44. It is helpful to think in terms of the field flux & rather than the field flux linkages. as shown in Figure 7. Le.. Assuming the main exciter to be running at constant speed. If we assume the field flux links N turns. the flux & has two components. (7.14) = 4a + 44 Fig. Also. traverses a high-reluctance path through the air space between poles.28 Separately excited exciter.29. Thus we have an with all other terms constant. since @E = c4. 7. The leakage component. 3 . Therefore..29 Armature of air gap flux &. o Wiley.) . The problem is that i deequation in terms of i and pends on the exact location of the operating point on the saturation curve and is not linearly related to u. and (4) analog or digital computer solution.3 I Self-excited exciter with a rotating amplifier (boost-buck).17) i =O P Fig. An empirical relation. We usually assume up to be a constant. as U. These are (1) formal integration.18) for the self-excited case where rEis the same as in (7. usually expressed graphically by means of the magnetization curve. and i.16). the Frohlich equation [35] -y+ R Fig.19) Kimbark [I61 suggests four methods of solution for (7. be known explicitly.256 Chapter 7 we have 4E = (1 + C)4 = r J 4 (7. I to 1. (3) step-by-step integration (manual). 7.15) where u is called the coefficient of dispersion and takes on values of about 1 .30. is not a linear function of i.2. (7.1 I ) . we may write the nonlinear equation rEbF + Ri = V. where we have hE + Ri = U. This equation is still nonlinear. (7.31.15) into (7. and where we usually assume u to be a constant. Formal integration requires that the relationship between v. (2) graphical integration (area summation). . 7.19). rECF + Ri = up (7.16) where r E = ( N a / k ) s. In a similar way we establish the equation for the self-excited exciter with boostbuck rotating amplification as shown in Figure 7. + V. Writing the voltage equation with the usual assumptions. Following the same logic regarding the fluxes as before. or N& + Ri = vF (7. rECF+ Ri = U. In a similar way we may develop the differential equation for the self-excited exciter shown in Figure 7.30 Self-excited exciter. Substituting (7.16)-(7. however. I n this chapter the buildup of a dc generator will be computed by the formal integration method only. (See Kimbark [ 161. For convenience we shall use the Frohlich equation (7. This method. Method 2. Solution By examination of Figure 7. i UF A V O 0 1 30 2 60 3 90 4 116 5 134 6 147 7 156 8 164 9 172 IO 179 Since there are two unknowns in the Frohlich equation. b. which may be solved f or i to write i = buF/(a .. and c must be found by cut-and-try techniques. Example 7.22). Approximate this curve by the Frohlich equation (7.v. although somewhat cumbersome. It is unlikely. However. Table 7 3 Exciter Generated Voltages and Field Currents . in general. makes use of the saturation curve to integrate the equations. a nonlinear equation is necessary to represent the saturation curve. and accurate. we select two known points on the saturation curve. easy to use. substitute into (7. is quite instructive. . To illustrate this. an analog computer solution and a digital computer technique are outlined in Appendix B. and solve for a and b. To use formal integration.22). The actual methods of computation are many but. that anyone except the most intensely interested engineer would choose to work many of these problems because of the labor involved. graphical integration.20) or (7.21) can be tried. I n either case the constants a. we will select two pairs of points and obtain two different solutions.35]).22) We illustrate the application of (7.20) may be used.) Method 3.20).22) by an example. the step-by-step method (called the point-by-point method by some authors [ 16.3.32.32 we make the several voltage and current observations given in Table 7. or the so-called modified Frohlich equation vF = d/(b + i ) + ci (7.) (7.Excitation Systems 257 V. I n this method. however. with the value during the interval being dependent on the value at the middle of the interval. If this is reasonably successful. = ai/(b + i) (7.2 A typical saturation curve for a separately excited generator is given in Figure 7. the time derivatives are assumed constant over a small interval of time. the equations can be integrated by separation of variables. Method 4 is probably t h e method of greatest interest because digital and analog computers are readily available. One experienced in the selection process may be quite successful in obtaining a good match. is a manual method similar to the familiar solution of the swing equation by a stepwise procedure [36]. Rudenberg [20]. and Dah1 [35] for a discussion of this method. nonlinear functions can be handled with relative ease and with considerable speed compared to methods 2 and 3. . i.9 V b.9 V a2 = 279. 7.32 Saturation curve of a separately excited exciter.258 181 Chapter 7 16 14 12 c P i i p . = 315. = 7.65 A .vF = 172 Then the equations to solve are 90 172 = = 3a/(3 9a/(9 + b) + b) 116 164 = = 4 ~ / ( 4+ b) 8 ~ / ( 8+ b) for which the solutions are a.uF = 90 i = Solution #2 9. Solution # I Select i = 3.53 A b2 = 5. enperer 4 Fig.d e 0 c u' 10 6 4 i 1 I I I 8 2 1 1c 6 Exciter Field Current. Separately excited buildup by integration.24) Example 7.95) + 48i (7.) (7.25) Equation (7. Integrating (7. / ( a .hU~)]du~ (7.(abR/h’)In[(aU. solution 2. and IO.Excitation Systems 259 Both solutions are plotted on Figure 7.53~.000 UFO = . (7. b.95 48./(279.hUFO)] This equation cannot be solved explicitly for u.22).25) is not plotted on Figure 7. = (I/h)(uF - UFO) Example 7.4 Using the result of formal integration for the separately excited case (7. = 359i/(i . Assume that the following constants apply and that the saturation curve is the one found in Example 7.)/(avp .26) This equation may be solved by separation of variables. versus t relationship for values of I from 0 to Is and find the voltage response ratio by graphical integration of the area under the curve.) (7.23) and for solution 2 U.27).uF) - = up (7.2 up = 125 V R = 34 S? 90 V k = 12. Then. compute the U.28) .0 This gives us the modified formula U. For simplicity. Select values of i = 2 .28).91/(5.27) where we have defined for convenience.9 . .)/(UUp . let the saturation curve be represented by the Frohlich equation (7. For solution 1 U. Rearranging algebraically.9i/(7. N u = = 2500 turns 1. so we leave it in this form. 5. and c. Solution i = 2 i = 5 i = IO 60 = 2 ~ / ( 2 6) 134 179 = 5a/(5 = + + 2~ + b) + 5c 10a/(10 + 6) + 1Oc c = Solving simultaneously for a../(315. = 279. we write dI = [TE(U U.32.65~.32 by a modified Frohlich equation. u = 359 b = -21. TEOF + b R ~ . h (1 to)/TE = up + bR. substituting for the current in (7.32 but is a better fit than either of the other two solutions.3 Approximate the saturation curve of Figure 7.2.u.9 .53 + i) + i) or i or i = 7.hu. = 315.21.65 = 5.u. I ) 6. 000 0.40 109. = ai.65 + 3.62.50 0.65 0.94 108.08 110.87 107. the initial voltage uFois 90 V . 2 2 ) we compute io = 5. Then from the Frohlich equation ( 7 .2)/12. From (7. 9 V .5) with cd = 2 7 . Then from (7. Since up is 125 V.4.45 0.21 110.29) .68 109.33.33. i.675)/(5.675) = Using the above constants we compute the uF versus I relationship shown in Table 7.22).55 109. For a self-excited machine whose saturation curve is represented by the Frohlich approximation (7.96 1 10. Assume that we completely short out the field rheostat.675 A 110.20 0.675 A This means that there is initially a total resistance of R.75 0.30 0.3 V Then.4 and illustrated in Figure 7.5) = 0. the response ratio = 27.65(90)/(280 . changing the resistance from 46.we have T&F + bRUF/(a - uF) = UF (7.90) = 2. Buildup of Separately Excited uF for Example 7. from the Frohlich equation the ceiling voltage is uF.00 95.35 106. IO 0.80 0.35 0.9 280 b = 5.9/90(0.55 0.05 0. which has the same area as that under the uF curve abd. Table 7. we compute the final values of the system variables. From the field circuit.7 D of which all but 34 52 is in the field rheostat. = v p / R = 125134 = 3.85 100.) = 280(3.65 Now.7 to 34 0 at t = 0.15 0./(b + i.30 From Figure 7.2 a = 279.79 109.25 110.25 0.260 Chapter 7 S o h I ion First we compute the various constants involved. = 12512.12 103. from the given data.675 = 46. from Example 7.00 0. by graphical construction we find the triangle acd. as shown in the figure.28 110.25 s = Also.16) rE = Na/k = (2500)(1.16 110.18 105.60 0. Self-excited buildup by integration.19 0.70 0.4 90.85 0. 4.3 0. 7. Solving (7. 0.30) This equation can be integrated from with the result (7.34. I Fig.4 ( 06 .68 = 30 9.bR. Example 7. Solution The ceiling voltage is to be 110. .Excitation Systems 26 1 I4 2 c C A 0 0.5) = 0.bR)VF . This is recognized to be identical to the previous case except that the term on the right side is U. Then the resistance must be R = 110. Compare the two buildup curves by plotting the results on the same graph and by comparing the computed response ratios.31) whereK = a .68 A (from the Frohlich equations).31) with this value of R and using Frohlich parameters from Example 7.342 for the self-excited case.33 Buildup of the separately excited exciter for Example 7.4/90(0.5 and the solution curve of Figure 7. at which point the current in the field is 3. instead of up. we have the results in Table 7. The response ratio = 15.7 Time.5 Compute the self-excited buildup for the same exciter studied in Example 7. Change the final resistance (field resistance) so that the self-excited machine will achieve the same ceiling voltage as the separately excited machine.3 V.1 0. Again we rearrange the equation to separate the variables as dt = vF)dvF (a .4.2 0.3/3.V k to to t (7.4. uR .33).52 101. 103.25 0.7 0.M In ( M Q ( M - QQ- 2u~)(M Q .00 91.87 93. Integrating (7.10 99.61 95.80 0.05 0.55 0.u.36 107. or T&.65 0.34 Buildup of the self-excited exciter for Example 7.5. The equation for the boost-buck case is the same as the self-excited case except the amplifier voltage is added to the right side.6 0. + U.35 0.85 105.15 0.4 Time.uX) (7.00 0.57 102. I 0.1 0.47 106.8 0 Fig.03 106.2 03 ..75 0.20 0.I5 104./(a - OF) = U. - (7.38 104..)du. we may separate variables to write dt = TE(a .5 0.5 I VF I VF 0. we compute -= t TE to 2a .. Table 7.70 0.33) where A = auR and M = a . ui) (7.u:) + MU. 7.85 0.52 106.23 96.60 0. .90 .96 107. .45 90.30 0. . 0. + bRu.71 .73 98.34) whereQ = d4A + M2.5.40 0.~UFO) + ~UFO)(M Q . Buildup of Self-excited up for Example 7.~ U F ) 1 + -In 2 (A (A + MU.32) Rearranging./(A + Mu.37 100.52 0.262 Chapter 7 0 0.10 0.50 0.b R . Boost-buck buildup by integration.. 23 97.12 110. Repeat for an amplifier voltage of 100 V.5 I 109.6 Compute the boost-buck buildup for the exciter of Example 7.14 109.65 0.82 109.32 100. = 100) = 2c'd/Oa = 2(29)/90 = 0.55 0.00 110.68 A so that R may be computed as R = 160.6. = 50) = 2cd/Oa = 2(24.80 0.3 V and an amplifier voltage of 50V.68 109. = 100 V gives a second set of data.83 110. Note that increasing he amplifier voltage has the effect of increasing the response ratio. other possibilities may be worth investigating. Using this R in (7.35.24 110.45 0. I Buildup of Boost-Buck UF for Example 7.90 90. = 0 Ri.90 107.36. from 50 to 100 V gives a result that closely resembles the separately excited case.68 = 43. maintains its .00 96.Excitation Systems 263 Example 7.3.20 I 10.00 94.537 R R ( u .60 0.27 110.6.3 V .19 109.30 110.40 0.72 104. Compare with previous results by adjusting the resistance until the ceiling voltage is again 110.00 0. in which R = 57.84 106. This requires that i .70 0.33 Th e r ults re plotted in Figure 7.84 103.96 I10. = 160.47 105.05 110. = uF + U.34 108.2 Linear approximations for dc generator exciters Since the Frohlich approximation fails to provide a simple uF versus t relationship.50 102.15)/90 = 0. Solution With a ceiling voltage of 110. value of 50 V. also tabulated.47 109.3/3.2 Q.20 0. again be equal to 3. IO 0. In this case changing U.98 106. This equation applies as long as U.645 7.6.72 108.70 100.12 107.15 0.05 0.6 VR = U F for 50 UF for U R = I00 0.17 90. Table 7. where . One method that looks attractive because of its simplicity is to assume a linear magnetization curve as shown in Figure 7.32 110. This value of R will insure that the ceiling voltage will again be 110.12 110.31 110.3 V .75 0.85 0.50 0.84 109.30 0.4 where the amplifier voltage is assumed to be a step function at I = to with a magnitude of 50V.35 0.25 0.56 109. we compute with 6.34) results in the tabulated values given in Table 7.6 Q. I n each case the response ratio (RR)may be calculated as follows: RR(u.56 108. Repeating with U. vF = mi +n (7.(f?/m)(vF .35) Substituting (7. 7. .6. i.35) into the excitation equation we have the linear ordinary differential equation TEi)F = v . self-excited uF + vR boost-buckexcited Exciter Field Current.36) where v = = = up separately excited v. 7. amperes Fig.36 Linear approximation to a magnetization curve.n) (7.264 Chapter 7 Fig.35 Buildup of boost-buck exciters for Example 7. 37 Solution of the linear equation./k2)(l - e-k2‘)u(t) + u. if the comparison is made by analog computer.38) is solved by the analog computer connection shown in Figure 7. s Fig. I n this process.Excitation Systems 265 This equation may be solved by conventional techniques.m we set u = up so that (7. 7.37) u. one controlling the slope and one controlling the intercept. both the linear and nonlinear problems are solved simultaneously and the solutions compared on an oscilloscope. The procedure will be illustrated for the separately excited case.9.36) for each case and then systematically trying various values of m and n to find the best “fit. which is a graph made directly by the computer.38 Analog computer comparison of linear and Frohlich models of the separately excited buildup.38. Adjusting potentiometers k. = ( I / T ~ ) ( u ~ + n R / m ) Solution of (7.26) given in Appendix B.(f) = (k. I n a similar way linear approximations can be found for the self-excited and boost-buck connections. for the best fit.k2uFwhere .e-kZ’u(f) (7.36) becomes 6 = k.38) Equation (7. The question of interest is. Having adjusted k. In the separately excited case R/r. . quickly provides the “best fit” solution shown in Figure 7. will quickly and easily permit an optimum choice of these parameters. if any. A simple manipulation of two potentiometers.37) gives k2 = (7.37 and compared with the solution of (7. shown in Figure B.” This extremely laborious process becomes much less painful. “FO L4-J Fig. k. and k. lime. . Linear approximation of the separately excited case. 7. or even fun. will give solutions close to the actual nonlinear solutions? This can be resolved by solving (7. What values of m and n. . the potentiometer settings are read and the factors m and n computed. and k. 266 7. with i. with a load impedance connected to the stator.19.io where Ti0 = LF/r. = 0. In modern solid-state circuits this effect will usually be small. But. I n systems of this type a small delay may be required for the amplifiers and other circuits involved. Kimbark [ 161 has observed that the current in the dc field winding changes much more slowly than the corresponding change in the ac stator winding.14. If the exciter is loaded. the ac exciter voltage . The rate of change of field current depends a great deal on the external impedance of the stator circuit or on the load impedance.40) to be This linearized result does not include saturation or other nonlinearities. Using relation (7.e. One way to solve this system is to assume that U. the load current will affect the terminal voltage of the exciter U. mitting a step change in u. For such fast systems the time constants are so much smaller than others involved in the system that assuming a step change in U. changes linearly to ceiling in a given time delay of t d s. Appendix E) we may assume that the ac exciter is open circuited.6. 3. much lower than that of the large 60-Hz generator that is being controlled. in the Laplace domain. the effective inductance seen by the field current is smaller and the time constant is smaller. taking into account the major time constants and ignoring other effects.4 Solid-state exciters Modern solid-state exciters. the field voltage may be computed from (7.3 The ae generator exciters Chapter 7 As we observed in Chapter 4.6. there is no simple relationship between the terminal voltage and the field voltage of a synchronous generator. since the terminal voltage is proportional to i (neglecting saturation). i.6. Including all the detail of Chapter 4 in the analysis of the exciter would be extremely tedious and would not be warranted in most cases. will change approximately as fast as its field current changes. using the response ratio definition (see Def. but does include the major time delay in the system. such as the SCR exciter of Figure 7.5 Buildup of a loaded dc exciter Up to this point we have considered the response characteristics of unloaded exciters. can go to ceiling without any appreciable delay. 7. should be fairly accurate.. I f the regulator output experiences a step change of magnitude D at t = to. The field voltage may then be assumed to depend only on this delay. ) where uF(s)is the Laplace transform of the open circuit field voltage and u ~ ( s is the transform of the regulator voltage.39) This will give the most conservative (pessimistic) result since. by an amount depending upon the internal impedance of the exciter.T s (7. 7. An ac exciter designed for operation at a few hundred Hz could have a very reasonablei&. We therefore seek a reasonable approximation for the ac exciter voltage. Therefore. amounting to . This is nearly the same as per. I n this case the field current in the exciter changes according to the “direct-axis transient open circuit time constant” .39) we write. where t d may be very small. = 1. 7.0 pu. Kimbark [ 161 treats this subject thoroughly and is recommended to the interested reader. (Reprinted by permission from Power System Stability.39. Thus..e. Kimbark. if the machine has interpoles. certain complications must be ignored if the solution is to be manageable. W.23. To facilitate analysis. has a constant value. .6.39 No-load and load saturation curves. 7. i.) We can analyze the effect of load current in a dc machine as follows. the net effect of load is in the resistance drop (including brush drop) and in the decrease in flux due to cross-magnetizing armature reaction. As in all engineering problems. We will ignore the loading effect in our analysis in the interest of finding a reasonable solution that is a fair representation of the physical device. we assume the load current i. First. This is the pu system designated as C in Figure 7.O pu corresponds to V. designated i in our notation. amperes Fig. 3. and at the relatively slow rate of buildup to be experienced this voltage drop is negligible. we do have to estimate the effect of cross-magnetizing armature reaction. 9 Wiley. i. since in addition to the i F R drop there is also the brush drop.Excitation Systems 267 essentially a small series i. the drop due to armature reaction. Thus at no load and with no saturation. I n rotating dc machines the effect is greater. = I . This means the i. we may neglect demagnetizing armature reaction." This curve shows the voltage generated by air gap flux at this value of i . 1956. (Dah1 [35]provides an exhaustive treatment of this subject and Kimbark [I61 also has an excellent analysis. a typical result of which is shown in Figure 7. by E. However. exciter voltage that produces rated no-load terminal voltage with no saturation.) saturation curve is added the resistance drop to obtain a fictitious curve designated "distortion curve. To the load ~~ Exciter Field Current. E.6 Normalization of Exciter Equations The exciter equations in this book are normalized on the basis of rated air gap voltage. Furthermore.R drop is constant. The combined effect is determined most easily by test.R drop. vol. we recognize that the armature inductance is small. The magnitude of this difference is greatest near the knee of the curve. which causes a net decrease in the air gap flux.. as a function of i. and the armature reaction effect depends on the value of current in the field. and the drop due to armature inductance. and it differs from the no-load saturation curve by an amount due to armature reaction. . reducing x until the lower contact L is made. will reduce V. The required relationship is given by (4. many of which are still in service.. Dividing through by VFB we have the pu equation ~ ~ = f (f u F Ui ) .16) we have. Simply stated. I t is necessary. This will give us a feel for the equations that describe these systems and will illustrate the way a mathematical model is constructed. as we have seen. etc. This.. used an electromechanical means of changing the exciter field rheostat to cause the desired change in excitation.0 pu E F D is not the same base voltage as that chosen for the field circuit in normalizing the synchronous machine. the rheostat setting will change at an assumed constant rate until the maximum or minimum setting is reached. . the arm attached to the plunger will find a new position x for each voltage V.55) we have VFB= VB~B/]FB = SB/lFB V This base voltage is usually a very large number (163 k V in Example 4. the exciter base voltage and the synchronous machine base for the field voltage differ. Note that the reference is the mechanical setting of the reference screw. closing the rheostat motor contact and moving the rheostat in the direction to increase R. Note that there is n o corrective action at all until a contact is closed. across the regulating coil and a given coil current i. A n y given level of terminal voltage will. The base voltage for E F D . As noted in Figure 7. High values of V.268 Chapter 7 The slip ring voltage corresponding to 1. D / f l r F to convert to an equation in EFD. Thus. 7. including the voltage regulators. For example.59). Mathematically.. we can describe this action as follows.. to always maintain the "gain constant" & r F / L A D between the exciter E F D output and the up input to the synchronous machine. = f(uF) V. for example). for the separately excited arrangement. and a change of base between the two quantities is required. The purpose of this section is to compute the response of typical systems.. and pull the arm to the right. This constant is the change of base needed to connect the pu equations of the two machines.40. 7. on the other hand. will increase the coil voltage u.20. From (7.1. would be on the order of 100 V or so. This constitutes an intentional dead zone in which no control action is taken. of course. result in a given voltage u. and then multiplied by L . This current flowing in the regulating coil exerts a pull on the plunger that works against the spring K and dashpot B. for the dc generator exciter we have an equation of the form T E f i . Multiplying by ~ ~ L A D / d r fwe write the exciter equation 7 E E F D u = f ( E F D u ) .42) Thus any exciter equation may be divided through by VfB to obtain an equation in u. after rectification. reducing x. A typical scheme is shown in Figure 7. From (4.1 Noncontinuously regulated systems Early designs of voltage regulating schemes.7 Excitation System Response The response of the exciter alone does not determine the overall excitation system response. This causes current to flow in the coil L. the excitation system includes not only the exciter but the voltage regulator as well. that pulls the arm slowly to the right. which may be explained as follows. which can be written as EFD = (LAD/firF) UF pu EFD = (WBkMf/flrF) ' F (7.. depending on the reference screw setting. Now imagine a gradual increase in V. Once control action is begun.7. . It results in a slow excitation change. In operation the beam position x is changed continuously in response to variations C . responding to a change in V. Note that any change in x from the equilibrium position is a measure of the error in the terminal voltage magnitude.4 I RH versus f for the condition 1 x 1 > K. and spring constants respectively.41 where the choice of curve depends on the rnagnitude of x being greater than the dead zone f K.t s a a c RH t lime.42. = MR + B i + KX (7. -Time delayed raise d lower contack -0pemting coils Fig. damping. But R changes as a function of time whenever the arm position x is greater than some threshold value K. . > 0. This condition is shown in Figure 7. 7.44) where xo is the reference position.. rECF = up . B . I f the beam mass is negligible. the right side of (7.. A block diagram of this control action is shown in Figure 7. where the rheostat motor steadily changes the rheostat setting. is the plunger force. This control action is designated the “raise-lower mode” of operation.F. and M. 7.40 A noncontinuous regulator for a separately excited system.43) and in this case the regulating is accomplished by a change in R. and K are the mass.Ri (7. = K(x~ +. large enough to exceed the threshold K.44) can be simplified.Next Page Excitation Systems 269 -Quick raise 6 lower COntock .e) . The scheme illustrated is a simplified sketch similar to the Westinghouse type BJ system (21. F. 4 is the unstressed length of the spring. s ” Fig. The balanced beam responds to an accelerating force F. K.. I n (7. Any change in V.t (raise) R. If the x deflection is largeenough to make the QL or Q R contacts.. or RQR are switched into or out of the field respectively.43 as an added quick control mode to the original controller. < ( R . The foregoing discussion pertains to the raise-lower mode only. the position x returns to the threshold region 1 x 1 < K.Previous Page 270 Chapter 7 Plunger PT & r u t Fig. is the value of R .. large enough to cause 1 x 1 2 K.45) Thus the exact R depends on the integration time and on the direction of rotation of the and Figure 7. . H At any instant the total resistance R is given by R = RQR + R ..43 Block diagram of the combined raise-lower and quick-raise-lower control modes.z) < R. The quick raise-lower mode is initiated whenever I x 1 > K. = = RQR + Ro . and the motor stops.. This value is constrained by the physical size of the rheostat so that for any time t . initiating a quick response in the exciter. with the resulting action described by KL I Balanced beom Raise.R.a second possible mode of operation is recognized.41. leaving R at the value finally reached..lower threshold Plunger PT 6 rect Fig. results in the rheostat motor changing the setting of R H .40. Referring again to Figure 7. retained following the rheostat motor..45) last integration. f K..lower threshold Rheostat motor -1 Fc 1 - R I 'min Quick raise. in V. + Ro + K M t (lower) (7. the fixed field resistors R. 7. This control scheme is shown in Figure 7.42 Block diagram of the raise-lower control mode. As the rheostat is reset. 7. R. The resulting change in R affects the solution for uF in the exciter equation (7.45.7. The complete excitation system is the combination of Figures 7. a more realistic solution results.45)-(7.48) The function S E is nonlinear and can be approximated by any convenient nonlinear function throughout the operating range (See Appendix D).43 and 7.47) Then we can show that = (I + SE)fB E = (1 A + SE)EB (7. > K.43) the exciter equation is ~ E i ) p= up .2 Continuously regulated systems Usually it is preferable for a control system to be a continuously acting.43 operates to adjust the total field resistance R to the desired value.RGvF . (quick raise) = + R. although the raise-lower mode will also be operational when 1 x I > K. x > K . sE = ('A - 'B)/'B (7.Excitation Systems 27 1 R = R.45.RGvFSE A block diagram for use in computer simulation of this equation is shown in Figure 7.46). R... x < K. this control mode will be initiated only for large changes in V.. Thus.44 Exciter saturation curve. we can write the total (saturated) current as i = GuF(I + S. and will provide a fast response. I f saturation is added.Ri = up .46) If we set K. Mathematically.) = Gu.. Saturation is often treated as shown in Figure 7.44. + GuFSE (7. we can describe the complete control action by combining (7. proportional system. 7.50) Substituting (7. 7. The controller of Figure 7. where we define the saturation function Fig. Le. it will probably not have time to move appreciably before x returns to the deadband. (quick lower) (7.43). the control signal is always present and exerts an effort proportional . where the exciter voltage is converted to the normalized exciter voltage EFD.49) (7. If the air gap line has slope l / G .49) into (7. We analyze each block separately.10 where the feedback signal is applied to the rotating amplifier in the exciter field circuit. Reduced to its fundamental components. Le. called the error voltage.272 Chapter 7 Fig. . The actual delay in this system is small. Voltage regulator and reference (comparator). The second block compares the voltage V. which is proportional to the difference. however. 7. If we let the average rms voltage be represented by the symbol I(. and we may assume that 0 < T R < 0.1). to the system error (see Def. Potential transformer and rectifier.45 Exciter block diagram. 2. where the potential transformer secondaries are connected to bridged rectifiers connected in series. Consider the system shown in Figure 7. the familiar boost-buck system.46 Simplified diagram of a boost-buck system. There is often an objection.we may write (7. (7. against a fixed reference and supplies an output voltage K. and the associated electronic power supplies because of reliability and the need for replace- Fig. transistors. 7.52) This can be accomplished in several ways. to using active circuits containing vacuum tubes. Thus the output voltage GCis proportional to the sum or average of the rms values of the three phase voltages. since it is typical of this kind of excitation system.47. this is shown in Figure 7. Here we shall analyze one system.06 s. One way is to providk an electronic difference amplifier as shown in Figure 7.46.. One possible connection for this block is that shown in Figure 7.5 I ) where KR is a proportionality constant and 7 R is the time constant due to the filtering or first-order smoothing in the transformer-rectifier assembly.12.48. where the time constant of the electronic amplifier is usually negligible compared to other time delays in the system. Most of the excitation control systems in use today are of this type. 48 Electronic difference amplifier as a comparator: (a) circuit connection.53) The operation of the bridge is better understood by examination of Figure 7. But for a deviation u. which is also displaced equally above and below i R E F . and u. Under this condition the currents ia and ib are equal. (7. Le.47 Potential transformer and rectifier connection.49.. (7. = 0. > k.Excitation Systems 273 t I 'de Fig. V.i + ib.53). + kLU. is always equal to & the applied voltage. 7.k U . or i. is V.55) to write k(V& - &c) (7. we assume that the voltage gain is large and that the input current is negligible. Note that the nonlinear resistance shown is quite linear in this critical region. above or below VREF results in a change i. = which may be incorporated into (7.u. + R .kN)UA = . This difficulty could be overcome by having a spare amplifier with automatic switching upon the detection of faulty operation. = -(kL . V. Then the output voltage V . = . in the total current. Here the input current idc sees parallel paths io and ib or id. Another solution to the problem is to make the error comparison by an entirely passive network such as the nonlinear bridge circuit in Figure 7. (7. + kNUA V L = V. Combining (7. 7.. the operation in the neighborhood of VREF is essentially linear.55) = VREF + u. .54) where k. ment of aging components. Fig. (b) block diagram.54) and (7. But since the output is connected to an amplifier. where we set 7 = 0 for the passive circuit. = u.48(b). Since ia = ibrthe sum of volt. seen by io and ib is also given.56) We note that (7. age drops u. we compute V. U N = U.56) has the same block diagram representation as the difference amplifier shown in Figure 7. a deviation U. Thus we may write for a voltage deviation u.. ..50 where the u-i characteristics of each resistance are given and the characteristic for the total resistance R.. If we choose the nonlinear elements carefully. this circuit makes an inexpensive comparator that should have long life without component aging.57) V = KAK/(l + AS) R As with any amplifier a saturation value must be specified.49 Nonlinear bridge comparison circuit.5 1 .59) The generator. Since the exciter is a boost-buck system. (7. The exciter output voltage is a function of the regulator voltage as derived in (7. The generator voltage response to a change in uF was examined in 'dc / R ~ + R ~ 'REF vc "REF 'REF -v A V Fig.58) where KE=RG.50 The u versus i characteristics for the nonlinear bridge. I f nonlinear resistances of appropriate curvature are readily available. The amplifier portion of the excitation system may be a rotating amplifier. These conditions are both shown in the block diagram of Figure 7.1 (7. a magnetic amplifier.45. A closer study of Figure 7. 7. we can write the normalized equation EFD = (VR - EFDsE)/(KE + TES) (7. I n any case we will assume linear voltage amplification K A with time constant T ~or . The exciter. or conceivably an electronic amplifier.274 Chapter 7 Input to amplifier Fig. The amplifier. . is a convenient reference and that two identical gang-operated potentiometers in the bridge circuit would provide a convenient means of setting the reference voltage. What circuit element constitutes the voltage reference? Note that no external reference voltage is applied.50) and with block diagram representation as shown in Figure 7. The nonlinear bridge circuit has the obvious advantage of being simple and entirely passive. such as VRmin< VR < VRmax. 7. The major difference between that case and this is in the definition of the constant KE. A natural question to ask at this point is.50 will reveal that the linear resistance R . it could be done by employing the same technique as used for the exciter.1 0. 2.05 = 40 KG = 1. there is no need to consider saturation of the generator since its output is not undergoing large changes.51 Block diagram of the regulator amplifier. saturation. Let us designate this value as 7 and the gain as Kc to write.0 3. 7.7. when unloaded and ~jwhen shorted. Construct the block diagram of the system described in Section 7. Sketch a root locus for this system and discuss the problem of making the system stable. I f saturation must be included. Chapter 5.7 1 . Looking at the problem heuristically. . I n the region where linear operation may be assumed.05 rR = KE KA = -0. we would expect the generator to j 0 respond nearly as a linear amplifier with time constant . the system transfer function is 1231 Y/%F = KG(s)/[I + KG(s)H(s)l where. Block diagram of the excitation control system.Excitation Systems 275 Fig.52. Solution 1 The block diagram for the system is shown in Figure 7. 7. 0. Find the open-loop transfer function for the case where 7" = = 0.44. Example 7.52 KR v. we have t+rRI= Fig.5 TG = 10 . If we designate the feed-forward gain and transfer function as K G and the feedback transfer function as H. where a saturation function S. neglecting saturation and limiting. neglecting . with the actual time constant being load dependent and between these two extremes.1 and compute the system transfer function. would be defined as in Figure 7. 4 ~ ) ( K+ € 7S( €)I + TcS)(I + T R S ) + KAKGKR and the system is observed to be fourth order. 7.9 . we have the rootlocus plot shown in Figure 7.9s' + = 177s + K' 400K .$)(1 -k 7~S)(l -k 7 ~ s ) Using the values specified and setting K 400 KAKRKG.0.9s' + 226.43: 1/3.5 0. ( I ) Center of gravity = ( C P .278 E 0.4: right breakaway at -0.16.# Z ) (2) Breakaway points (by trial and error): = -(30. we have K Ol( .91 1.53 1. where we compute [22] crossing -10 origin Fig. .)S KGH = (s + IO)@ - + (pen) I)(s + 20) (reg) (amp) (ex4 Solution 3 Using the open-loop transfer function computed in Solution 2. or KGH = KAKGKR ( 1 -k 7.4 + 1/15.276 Chapter 7 or .53.C Z ) / ( # P .v.75 left breakaway at .57 + 1/9.4TAS)(K€f = 7.0)/4 = -7.52.53 Root locus for the system of Figure 7. Solution 2 The open-loop transfer function is KGH. - KAKG(l f TRs) VREF ( 1 f 7 .6 = 1/6.57 + 1/0.20 and K KAKRKG = 40KR.57 = 1/0.89 (3) Gain at j w axis crossing: From the closed-loop transfer function we compute the characteristic equation +(s) = S4 where K' = + 30.281 1/19.4 + 1/16. which requires that R in (7.20 < (177/0.59) have a greater value.7 illustrates the need for compensation in the excitation control system. 2 ~ ~1266 + = 0 s2 = -5. This can take many forms but usually involves some sort of rate or derivative feedback and lead or lead-lag compensation. 7. Excitation system compensation.Excitation Systems 277 Then by Routh's criterion w e have s4 s3 s2 s' so 226.2 K' 177 .9 30. Of these options only (c) is of practical interest.05 K' = 400K .21 We may also compute the point of j o axis crossing from the auxiliary polynomial in s2 with K' = 1266.0.9 I77 221. and (c) adding some kind of compensation that will bend the locus to a more favorable shape in the neighborhood of the j o axis.14) = 1266 K < 3.S Fig. . Thus the response is governed largely by the generator and exciter poles that are very close to the origin.54 Block diagram of a typical compensated system. (b) moving the generator pole to the left.14K' 0 K' 1 K' For the first column we have: From row so K' From row s ' = 400K . We note that for any reasonable gain the roots due to the regulator and amplifier excite response modes that die out very fast and will probably be overdamped.20 > 0 K > 0. (It is Olhcr KG I+rGs I - "t KR 1 +Tp. Even modest values of gain are likely to excite unstable modes in the solution.73 s = +j2. This can be improved by (a) moving the exciter pole into the left half of the s plane. or 2 2 1 .4 An examination of the root locus reveals several important system characteristics. Example 7. which would need to be done as part of the generator design rather than afterwards. 2. Repeat part 2 using an analog computer solution.8 I . . The effect of compensation will be demonstrated by an example.55 Excitation system with rate feedback neglecting S.) This can be accomplished by adding the rate feedback loop shown in Figure 7.7 for the system shown in Figure 7. 7. 7F and KF to mini- KA 1 b KG T I 1 I Vt KF’ 1 + T G I K K (1 + ~ K ~ T ~ S ) ”.54. where time constant T~ and gain KF are introduced.54. Repeat Example 7.278 Chapter 7 interesting to note that Gabriel Kron recognized the need for this kind of compensation as early as 1954 when he patented an excitation system incorporating these features [37]. and limiter: (a) original block diagram. T ~ s ) ( K ~<)(1 + + z K+ (1 +lG4 + KR KG (1 + T+) - Fig. Example 7. (b) with rate feedback take-off point moved to V. Use a digital computer solution to obtain the “best” values for mize the rise time and settling time with minimum overshoot. (c) with combined feedback.. Also notice that provision is made for the introduction of other compensating signals if they should be necessary or desirable. Such a compensation scheme can be adapted to bend the root locus near the j w axis crossing to improve stability substantially. 3. 7.61).56 Locus o f zeros for the open loop transfer function o f (7. 5 ~m Fig.61) A given T~ fixes all poles of (7. KG = 1. as shown in Figure 7. Figure 7.05. The locus of the roots for this system will have a branch on the real CaseI: 0 < a < 1 -10. = 0 and with limiting . The forward loop has a transfer function KG(s) given by KG(s) = K A KG ‘ATETG (s I ~/TA)(S K E / ~ E ) ( ~ 1/76) and the feedback transfer function H ( s ) is given by H(s) = (KF7G/KG7F>s<s + + 1/7R) + (KR/7R)(s (s + 1 / 7 F ) ( s + 1 / 7 R ) 1/7G)(s + 1/7F) The open loop transfer function is thus given by KGH = KAKF .5 - case 111 : a > 20 .62). and Case 111. 0 < a < 1. From the numerator we write S(S + I)(s + 20) + 20(7.05.62) where we let K = 20(rF/KF)and a = 1 / ~ ~ .Excitation Systems 279 Solution 1 The system transfer function can be easily computed for S. . Case 11.61)./KF)(s + + 1 / ~ ~= ) 0 o= I + 20(7.I / ~ F ) ( S + 1/7~) Substituting the values and KR = 1. By using block diagram reduction.61) will have a pole at 0.O.55(b). The locus therefore falls between the origin and .)(s ss ( I( )s + + 20) 1/7F) = + ss ( K ( s + a) + I)(s + 20) (7. the takeoff point for the rate feedback signal is moved to V. 7.a . This means that (7..s(s + TA7€TF 1/7G)(s + 1/7R) + (KRKGTF/7RTGKF)(S l/TF) (s 1/7~)(s = + KE/TE)(S + 71 1 1/7~)(S 4. Then the shape of the locus depends on the location of the zeros. These cases are shown in Figure 7.m is the location of the asymptote. ignored.0./K. KGH = KF 20KA TF (s + 20) 1O)(s .61).a .55(c).5<m<-10 CaseII: l < a < 2 0 l O c m < -0.62). Thus we examine the zeros of (7.l)(s 4- 20(7~/K~)(s I / l p ) + + l)(s + 1/7p)(s + 20) (7. The locus of the roots of (7.1.0 . then the two feedback signals are combined in Figure 7. I)(S = 0.0 KE = -0.1.0. = 1.$6(a).61) would have a zero on the real axis near the origin. value of a = 1 / ~ ~ There are three cases of interest (note that a > 0): Case I. where a zero falls on the negative real axis at .56 where .5. which gives the zeros of (7.1 and a zero on the real axis at . 1 < a < 20.a . Case I is sketched in Figure 7.55(a) shows a block diagram of the system with S = 0 and without the limiter. which is between the origin and . T~ S(S 0. depends upon the . a > 20. T~ = 0. Thus the open loop transfer function of (7. 62) have branches that. are more likely to be located further to the left of the imaginary axis in Case I1 than in Case 111. Thus we will concentrate on Case IIB for further study. Cases I1 and I11 are shown in Figures 7.280 Chapter 7 Case 1 A Case 1 B Case I1 A Case I1 B x-x X -1 X Case 111 A Case Ill B Fig. two may appear as a complex pair. with the proper choice of the ratio K. (Also see (381 for a further study of this subject. 7. I n both cases.O.57 provides a pictorial summary of all six possibilities. Its dynamic response will be sluggish.o l ( . in Case I1 the loci approach the asymptotes to the left of the imaginary axis. A further examination of the possible loci of zeros in Figure 7.56 reveals that for the three zeros.61)the open loop transfer function is given by KGH = KF s3 + 21s2 + 20(1 20KAT F (S + lO)(S . Only in Cases I I B and IIIB is there any hope of pulling this dominant root away from the origin. Futthermore. and of these two.61). However. Thus there are two situations of interest: (A) all zeros real and (B) one real zero and a complex pair of zeros.56(b) and (c).) From (7.I)(s + a ) axis near the origin. these are the zeros for the system described by (7. . The position of the roots of (7. Case IIB is clearly the better choice.57 Root loci of KGH = 20K" ( K F / 7 F ) [ ~ ( I)(s + ~ (s + 20) + 20(s + a)] + 20)(s + IO)(s + I)(s .61). both conditions can appear in all cases. while for Case 111 the loci approach the asymptotes to the right of the origin.63) where 1 < I / T ~< 20. Figure 7. give a pair of complex roots near the imaginary axis.)S + T ~ / K ~+) s 0 / K F 2 + l)(S + 20)(S + l / r F ) (7. and the system dynamic performance will be dominated by this root.62) and hence the zeros of (7. Again. the root-locus plots of (7. In all but two cases the system response is dominated by a root very near the origin. 01.Excitation Systems 281 20 ’ T F =0.5 and 0.6. and 0.58(a) ElTect of variation of K F on dynamic response: T F = 0.58.03 F 15 x a a . K F = 0.2 8 lo E 5 2 .03 respectively.IO E E C 0 5 0 -20 -15 Reo1 -10 -5 Fig.40.W 0. 00. 2 15 1.02 while T F is varied between 0. 7. a root-locus program. E f 5. 0 -io -is -io Real r -5 ’ 00 .20. .7. .01 and 0.. 0 1 F 15.6.58(a) r F is held constant at 0.IO. I n Figure 7.40. The programs used are a root-finding subroutine for polynomials to obtain the zeros of equation (7. \ P 7 0. KF = 0. The root-locus plots and the time-response for the system are repeated.58(b).00 oh0 1:M) 2140 3:a lime. Plots of the loci of the roots are shown for the three cases.€ : .03. is varied between 0.6 while K. 0. KF is held constant at 0. The effect of increasing r F is to reduce the overshoot. .0 0- I 20 r = 06 K = 0 0 F .63). In Figure 7. Solution 2 The above system is studied for different values of rF and K F with the aid of special digital computer programs. The most obvious effect of reducing KF is to reduce the settling time.0 0.02. .. Type I excitation system..0. along with the timeresponse for the “rated” value of KA. F . and a timeresponse program. I 2c r = 0. K ~ 0 .6. Two sample runs to illustrate the effect of rF and KF are shown in Figure 7. K =0.00 0.02. There is. and 0. Type I excitation system.58(b) we can see that the values of T .80 Time. K =0. and T.7.5.02 F 0. From Figure 7.58(b) Effect of variation O f T F on dynamic response: K F = 0. 7 F =0.6. Solution 3 An engineer with experience in s plane design may be able to guess a workable location for the zero and estimate the value of K ..80 1. 1.60 I 2.02 F 15. 0. which gives a reasonably good dynamic response.02 F I T F =0.00 0. = 0. a variety of choices of K.5.5.6. K -0.02 seem to give the best results. and we shall consider this technique as an alternate design procedure. TF = 0. and KF significantly influence the dynamic performance of the system. For this particular system.20 Time.6 and K.02 F I qF -0.40 3. however. K =0. the analog computer can be a great help in speeding up the design procedure. T .60 I 2. with V. 7. = 0.58(a) and 7. For most engineers. 0 -20 -15 -10 1 Real 1 T F = 0. that will give satisfactory results.20 m. .282 Chapter 7 20 7 F =0. K =0. From Figures 7.6. = 0.02 F 15 . Fig.54 we write.40 3.E E - : IC : 5 0 Real 0.7 respectively. and the equation may then be factored. can be rewritten as 4 = (KF/7F)EFD . Run 00 = Summary of Analog Computer Studies for Example 7.65) = (l/S)[(KA/TA) v.66) Using (7.2 0-90% rise time.05 8. we may construct the analog computer diagram shown in Figure 7. In each case both the forward loop gain and feedback gains may be o ptim ized .Excitation Systems 283 Fig.30 0.35 1. The constants in these studies may be used to compute the cubic coefficients (7.16 0.65) by an integrator with feedback.16 50 50 50 I .37 0. s 0.oo 0.75 0.16 0. I f the roots are known. Table 7.0 70.64)--(7.64) For the amplifier block of Figure 7. Then we may systematically move the zero from s = 0 to the left and check the response.64) may be represented on the analog computer by a summer and (7.16 50 very long 0. All other blocks except the derivative feedback term are similar to (7.59 Analog computer diagram for a linear excitation system with derivative feedback.05 I .65).05 2.20 . For the derivative feedback we have 4 = sKFEFD/(I+ 7 F ~ )which .215 I 2 3 4 1.8 42. a root locus Table 7. and other gains have been chosen to optimize V.62).16 50 ISO I .0 22. - ‘R1 Equation (7. In all cases KR has been adjusted to unity. (7. 7.7.59.7 shows the results of several typical runs of this kind.8 KF TC I KA Settling time.(I/TpT)Vj (7. s Percent overshoot 9.25 1 .0 5 0. in a qualitative sense. may be (7.25 0.75 0.66).54 we have VR = K A V e / ( 1 rearranged as ‘R + T ~ s ) which . Onesecond timing pulses are shown on the chart.60. . The actual analog computer outputs for run 2 are shown in Figure 7. which in physical systems would both be limited by saturation. Nevertheless. may be plotted and a comparison made between this and the previous uncompensated solution. This system is tuned to optimize the output which responds with little overshoot and displays good damping.8 are intended to give us some feeling for the derivative feedback of Figure 7. even in linear simulation. Note. Inclusion of saturation is a practical necessity. however.54.60 Analog computer results for Example 7.8. A study of the eigenvalues of a synchronous machine indicates that a first-order approximation to the generator voltage response is only approximately true. that this requires excessive overshoot of EFDand v. These studies also suggest how further improvements could be realized by adding series compensation.Chapter 7 Fig. Solution 2. 7.7 and 7.. but this is left as an exercise for the interested reader. Examples 7. The plot is made so that 20 such pulses correspond to 1 s of real time. Visualizing the root locus of the control is helpful and shows clearly how the compensated system can be operated at much greater gain while still holding a suitable damping ratio. v. making this simplification helps us to concentrate on the characteristics of the excitation system without becoming confused by the added complexity of thegenerator. In rewriting (7. t'. we set x8. = S .. ( E F D ) is a nonlinear function of point to write we linearize at the operating where we define the coefficient SL to describe saturation in the vicinity of the initial operating point. = (KR/TR)y - (1/7R)6 (l/TF) = ( K F / T F ) iFD - V. If the system equations are linearized about a quiescent operating state.- 6 . V.8 State-Space Description of the Excitation System Refer again to the analog computer diagram of Figure 7. vR > VRmin - [(sE + KE)/7ElEFD (7.V. x.69) where u. From (4. EFD = = (KA/7A)ve 'R (1/7A)vR < VRmax.67) to eliminate E F D in the second equation we observe that.and xll to correspond to the variables VI.67) EFD.. thus (7. By inspection we write the following equations (including saturation) in per unit with time in seconds. Arbitrarily. when per unit time must be divided by wR for the system of units to the consistent. + u:) (7.69) V: = (I/~)(u.68) In equation (7.46) or (6. the product (rFrE) The preliminary equations are obtained: O 1 + O I 0 (7..= VREF+ V. a linear relation between the change in the terminal voltage y. and the change in the d and q axis volt- . Suppose we arbitrarily assign a state to each integrator associated with the excitation. vR r'.68) the term ( K R / 7 R ) is a function of the state variables.Excitation Systems 285 7. and u. xlo.69) is nonlinear. are functions of the state variables. Since S. V3. is used.59.Vl2 and E F D . 5. / L A D ) EFD.74) (7.286 Chapter 7 ages U d A and u./T..1 Simplified linear model A simplified linear model can be constructed based on the linear model discussed in Section 6.( K ~ / T .70) The linear model is completed by substituting for U d A and UqA in terms of the state variables and from (6. . 6 = W (7.72) = T.73) and (7.( D / T ~ ) ) w and from the definition of q..69) and repeated here: (7. From (7. V.0) E. The complete state-space description of the system is given by x' = [Eiwd (7.72)-(7..20) and by setting u. E: t = -(l/K37. E F D ] .71) T.72) (7.68) and (7.71) + KbE.8. and T. 7.75) V. . .(K1/7.(K4/?:0) 6 + (1/d0) EFD From the torque equation (6. = ( f l r ..73) (7. The driving functions are V.E. assuming that V.77) . V.. = K . The linearized equations for the synchronous machine are given by (the A subscripts are dropped for convenience) (7.) 6 .. is obtained..76) The system is now described by (7.68) is zero. 6 -I K2E: = KS6 (7. in (7. The state variables are V.76). Such a relation is given in (6. using (6. rF LAD The matrices M and K are thus given by the defining equation v = -Kx .20).8..Excitation Systems 287 7. V3 VR E. where id iF iD i. M-'K. Before this is done. Note that in (7.Mk.20) to obtain the complete mathematical description.80) the state variable for the field voltage is E F D and not u. Thereu.68).M-’K. The remaining equations in (7. U. (7. The equations introduced by the exciter (for V. These are repeated here (with the A subscript omitted). the equation for the field current is adjusted accordingly. must be expressed in terms of the state variables. .78) and using we get (7.79) Substituting in the first equation in (7.78) From (7.70). M is given by . In this equation the term uF is changed to ( & r F //L A D ) ) EFD. V.25) and (7. the excitation system equations are added to the system of (6.2 Complete linear model By using the linearized model for a synchronous machine connected to an infinite bus developed in Chapter 6. i. The new A matrix for the system is given by A = . = 0) will thus become This set of equations is incorporated in the set (6. fore.68) will be unchanged. w d V. .. . .. 1(Aqo . ........ ... .. .. I 0 0 0 0 O I I I I I I I rn V....... ......81) And the matrix K becomes :I : id iF iD 4 0 r.L. 1 I I I I I Lf M R I l I o ! I I I I 0 I I I I I I I I 0 i" kM...... 0 . .....I . ..... . ......... 0 . 0 0 I I 71 ......288 Chapter 7 i..... 'R 0 0 O 0 I I I I - -KRf . k M.. -I K... .. 4 0 P -wokMF -wokMD l ...... l o r. -o TR 0 1 I I I I I t 0 0 0 I 0 0 0 I 1 I 0 0 O I 0 0 : I o 0 0 I ] (7.. .I 1 rQ -v3vwo 0 0 0 K.. . K - . .. ............. . .r' 0 I 0 0 . . I 0 0 0 I O I I 0 I O I 7f 1 I 1 I o I 0 S.... 0 O l I I WJ. l o . I 0 0 0 I I O 0 0 rD . .. .2 doL.. ... -(-Ado L.. ...... 1 1 R 0 0 .... ....1 kM& 3 3 - 1kMDiVoI 3 O I I + kMQido j -D j o I O I I I 1 0 &I 0 0 0 0 0 0 ....82) . 0 I I I I I L I 0 0 K I I I . .:.. ..iqo) .. 0 0 . I O I 0 i o 0 I -_ 76 + Kr 78 (7.. V" 0 0 0 0 0 ...i...-A. .. 1 I 0 ...... .. .) 1 3 I 0 K4 8 I o I o ! 0 0 I . . . . 77)(-0..476 x 0. 1..0053 = 0.85 PU 78 = K R = 1. A l / r A = 1/18.148 1.= aE .80).555 x 2.0037 -O&F/TFTE = -0.4 = -0. 2.0 TA = 0.148/1.55 = -0.5 x 188.265 = 1/71.0039 exp (1.476 x 1.85 = 0. The linear saturation coefficient at the initial operating point is SEI = .0561 (1.2.1.122 = K.0/3.1 = 3..4) = 0.5) = -2.053 1 / =~ 0.4762 i..326 x 0.967 X K A / ~ = 40/18.o) = id0 = = .0523 -(1.967 x X 0.4 + 0. / W R kMF = C3(0.9 Expand Example 6.70 + 0. = 0.02) = 0.04 Let the exciter saturation be represented by the nonlinear function Solution From the initial conditions UdO = ug0 = 5 0 = -1.715 = 269.70 & V.397) = 0.5 PU -0.675 1.05 0.05 s K A = 40 KE KF = 7.85 = 2.0053 ~ ( A + K E ) / ~ = 0.172 (I/3)(uqo/v.000796 S E C 3 r 1 .0.555 [0.77)(-0.59)0.529 -(1/3)(1.025 EFDO = do 40 = (l/3)(Ud0/vro) = = 0..7 X lo+ -wRKF(SA K E ) / T F T E = 2.1.326 x 1.025 + 0.476 x 0.172) = 0.326 x 0. Assume that the machine is operating initially at the load specified in Example 6.04 X 377/(269.dO = = . = 1 I= 0. The excitation system parameters are given by TR = 00 S .675/1.77 P U 18.0321 -(1.Excitation Systems 289 Example 7.0/3.5 s = 188.0751 I / T ~ 0.2 to include the excitation system using the mathematical description of (7.529)] = 0.000829 + The new K matrix is given by .397 1.0/3.326 x 0.476 x 1.15 X 0.000742)/ 1.26 = 7.3264 (1/3)(1.3095 The exciter time constants should be given in pu time (radians).02 .0/3.59 &v. The new terms in the K matrix are -(1.77)(-0.77)(-0.55 P U 0.172) = -0. 0235 -0.. .526 I I O I M = 0 I I I I 0 I I O . _ ...2967 -5.._ _ _ _ _ . ..0 -2331.1735...040 1. .142 1-3487.6 I 1751..50 124.550 1.46 I _ _ _ _ _ _ . .32 I .100 1.550 1.. ...550 1.3 -605. ._ .1 -2122._ I 4._ _ _ _ _ _ 4 I -0...0 880....8673 0 0 .5 -2587.7 I -36.79581 0 -3.. .064 1544..-.490 I I I O I I I 1.4 2649....2027 -0.36..2 76.605 0 I I O I I I I I I I I O I I I 0 I I I 2....472 -4.019 0 0 1 1 0 0 0 .7993 -3505._-.. .550 I I 1.550 1.4388 14.0078 -0.5 1 35.7 -2587.361 134.072 I 2387.. .651 1.0 -2444...4 I t 0 0 A _ _ _ _ _ _ .02 -108. .017 1 2202. _ _ I 0 0 25.0 0 o ! 0 O I io o o 1 I I iJ The A matrix is given by L .. _ _ _ .290 Chapter 7 K - The new M matrix is given by 2.86 1608.490 1. .26 56..776 0 0 0 0 0 0 0 0 0.65'1 -265._ .. . _ .1 0 .5317 -4.-I..... . .0 90. .7 2649...2027 t -0.062 22... I I 845..4422 I I 1 0 0 1 I 0 0 0 o i o 0 0 0 0 0 0 I loo0 I .0 -1106..... . .... -2122.1 1776. 12.3557 -96.7 0 0 0 ~ -0.7099 0 0. _ * _ .218 .J .... .4904 5.. .-.._ .15 211. ... _ _ _ .7 3590.. _ _-_ __ _ _ _ I..052 0 0 1 I 0 0 0 0 0 ' 0 I o 0 1 .---.. lo-' 1! 0 0 0 0 1 I 0 0 1 0 O 0 0 i 0.550 1.077 -0.857 I -2547..9503 1 1206..6 .123.4 -0.3052 53.. _ _ .394 55. 02139 j0.03594 -0. We describe the four IEEE models below.09804 -0. standardized the representation of excitation systems in four different types and identified specific commercial systems with each type.07870 -0. the IEEE formed a working group in the early 1960s to study standardization.OOl85 .02468 .86637 - j0.j0. Thus.12302 -0.j0.86637 + j0.00353 .02444 .12315 -0.j0.10. For this exciter a frequency of approximately 50 Hz is obtained.00340 -0.265 x .02139 + j0. The eigenvalues associated with the exciter parameters did not change significantly with the machine loading.26525 -0.02468 + j0.16664 -0.9 Computer Representation of Excitation Systems Most of the problems in which the transient behavior of the excitation system is being studied will require the use of computers. anything from a very simple linear model to a more complex nonlinear model may be formulated by following these generalized descriptions.00177 -0.2 and for the same exciters.j0. The additional eigenvalues obtained in Example 7. Recognizing this fact.265 x -0.292 Chapter 7 Table 7. .j0.9 are for the same machine and loading condition as used in Example 6.10 and not previously present are comparable in magnitude except for one complex pair associated with the W Low rE Brushless exciter.00071 -0.99827 IO2 + j0. depending upon the available data or importance of a particular exciter in a large system problem.03594 -0.07870 -0. which presented its final report in 1967 [15].j0.1) Exciter type W Brushless W low rE Brushless W TRA -0. I2299 -0.99826 IO2 + j0. The results obtained indicate that only one pair of complex eigenvalues change with the machine loading.00076 -0. we note that two pairs of complex eigenvalues and two real eigenvalues are essentially present in all the results.03594 -0. These models allow for several degrees of complexity.03912 + j0.99826 - j0.07300 -0. Eigenvalues for System of Example 7. This group.00082 -0.OOl85 - -0.0. We can conclude that these eigenvalues are identified with the parameters of the machine and are not dependent on the exciter parameters.02536 -0.00249 - j0. several models have evolved for such systems.02444 + j0.99827 x j0.03594 -0. The same example was repeated for the loading of Example 5.02536 -0. 7. Then each manufacturer can specify the constants for the model that will best represent his systems.00082 -0.03594 -0. I t is therefore recognized that the solution of systems can be greatly simplified if a standard set of mathematical models can be chosen.03594 -0.09763 -0.02444 + jO.00249 -0.00076 -0. As the use of computers has increased and programs have been developed that represent excitation systems.99826 + J0.99826 IO2 . Comparing the results of Examples 6.00353 The results tabulated in Table 7. Actually.00177 + j0.03912 .00447 -0.00071 -0. This pair is one of the two complex pairs associated with the machine parameters.02444 .4 and 7.16664 -0.00447 + j0. and the data acquisition problem will be simplified for the user.4 except for the addition of the exciter models. IO (Loading of Example 5.9. the differences in these representations was more in the form of the data than in the accuracy of the representation. which might be introduced by the extremely low exciter time constant.jO.00340 + j0. Note that if we have no filter and the rate feedback is zero (KF = 0). 0 ~ ~ . Type 4 regulator output voltage maximum value of VR minimum value of VR VRH = generator terminal voltage regulator gain exciter constant related to selfexcited field regulator stabilizing circuit gain current circuit gain in Type 3 system potential circuit gain in Type IS or Type 3 system fast raise/lower constant setting. Reprinted from IEEE Trans. 7. = generator terminal current K.61. K. (c IEEE.= = regulator reference voltage setting field rheostat setting - Note: Voltages and currents a r e s domain quantities.10. EFD = 1 . 1968. = s..83) .9. Usually rR is very small and is often approximated as zero. Type 4 system exciter saturation function auxiliary (stabilizing) input signal I. Table 7. and its output is limited by VRmax and VRmin.0 pu exciter voltage is that voltage required to produce rated generator voltage on the generator air gap line (see Def. Symbol Excitation System Model Symbols Symbol Description Description exciter output voltage IF = generator field current v.) The amplifier has time constant T. Table 7.20 in Appendix E).. = VREF - r: (7. vol. with a filter time constant of r R . Note that provision is made for first-order smoothing or filtering of the terminal voltage V.10 gives a list of symbols used in the four I E E E models. = K.1 Type 1 system-continuously The block diagram for the Type 1 system is shown in Figure 7. 3..61 Type I excitation system representation for a continuously acting regulator and exciter.= regulator amplifier time constant exciter time constant regulator stabilizing circuit time constant same as T~ for rotating rectifier system regulator input filter time constant rheostat time constant. changed slightly to conform to the notation used throughout this chapter. acting regulator and exciter 7.O pu generator voltage is the rated generator voltage and 1.Excitation Systems 293 The excitation system models described use a pu system wherein 1 . This means that at no load and neglecting saturation. 0 ~ ~ gives exactly = 1 . Fig. and gain K. the input to the rotating amplifier is the error voltage v. PAS-87. v. = = = = K. KF K. we can observe the obvious similarity. to give a new effective value of pR. 1968. and this voltage is small.. (7. domain to TEEFD = -KEEFD Equation (7.) IEEE..88) Comparing with (7.SEE. + TES) (7. = (A .S E E F D bRVF/(U (7.. V R = V . T.. Note the that for sufficiently small EFD system is nearly linear (S. vol..62 by the relation s . where we computed TECF = VF + VR - .87) which includes the nonlinear function SEE.SEE.. However.viz..85) for we have c.. 7. From the block diagram we write EFD = f R / ( K .s).B)/B (7.84) and is thus a function of E. Reference [IS] suggests taking .85) This altered value is operated upon linearly by the exciter transfer function.. The saturation function is defined as shown in Figure 7. + TEsEFD = -KEEFD + VR .. Note also that the exciter transfer function contains a constant K.294 Chapter 7 . but finite in the steady state.VF) with the nonlinearity approximated by a Frohlich equation. Fig. for example. Reprinted from l E E E Trans. = 0).87) corresponds in the time + VR . This transfer function vR G(s) = l/(K. R (7. The exciter itself is represented as a first-order linear system with time constant T.62 Exciter saturation curves showing procedure for calculating the saturation function S . This alters the amplifier voltage VR by an amount SEE.86) is not in the usual form for a linear transfer function for a first-order system (usually stated as 1/(1 + T S ) . PAS-87. and substituting (7. a provision is made to include the effect of saturation in the exciter by the saturation function S. that is nonlinear.32). 7. are computed from saturation data.) e .. usually the exciter ceiling voltage and 75% of ceiling.92) with the constraint VRmin< VR < VRmax. A state-space representation of the Type IS system can be derived by referring to (7. and 7 F are respectively the gain constant and the time constant of the regulator stabilizing circuit. Reprinted from IEEE Trans.. with . Some engineers approximate the saturation function by an exponential function. . and S. 'Rrnax = 7. This time constant introduces a zero on the negative real axis. vol. setting V = E F D and eliminating (7. and BE.Excitation Systems KE 295 = S E I E ~ ~ ( 0= ) f lEFD(0)I (7. i. then (7.. +" * E ~ ~ I r j Fig. PAS-87. Finally we examine the feedback transfer function of Figure 7.%'EfD) (7. = 0.e. 1968. and E F D are specified at two points.61 H(s) = K.90) The coefficients A.94) Such systems have almost instantaneous response of their main excitation components such that in Figure 7. I n this case the maximum regulator voltage is not a constant but is proportional to V .61 K. = 1 .89) which corresponds to the resistance in the exciter field circuit at t = 0.2 Type 1S system-controlled rectifier system with terminal potential supply only This is a special case of continuously acting systems where excitation is obtained through rectification of the terminal voltage as in Figures 7. See Appendix D. = 0. (! IEEE.91) introduces both a derivative feedback and a first-order lag.18.65).90) is easy to compute and provides a simple way to represent exciter saturation with reasonable accuracy. Note that (7. Le. Reference [ 151 points out that the regulator ceiling VRmar and the exciter ceiling EFDmax interrelated through S.S/(l + TFS) (7.17 and 7.9. 'Rmax = KP< (7.. 7.63 Type IS system. and K. Under steady-state conditions we compute are VR = KEEFD + (7.. where S. This system is shown in Figure 7.91) where K.63. The function (7. sE = ft E f D ) = exp (BE.67) (written for the Type 1 system).93) (KE + SEmar)EFDmax Thus there exists a constraint between the maximum (or minimum) values of EfDmax and 'Rmax tEFDmm and R n ' mi ). 96) where the f coefficients are constants. vol. - (7. 7.] x2 1 we can show that <= 0 hEFD -k k-I h x k (7. function 'REF V.) rectifier system before 1967. the excitation voltage is not available to feed back. 1 Fig. = ( K F / ~ F ) ~ F D (1/7F) - E F D = (K. incorporates damping loops that originate from the regulator output rather than from the excitation voltage [39] since.9.4/TA) < v 3 'Rrnaxr > vRrnin v.97) where 7 Note that only three states are needed in this case.v.0 TR 0 + K 0 0 -F TF 0 0 1 (7... Note that two time constants one of which approximates appear in the damping loop of this new system. For the linearized system discussed in Chapter 6 where the state variables x' = r: [idi. + v.. we can express as a function of the state variables. Re- .296 Chapter 7 the result pi = (KR/~R) v. iD iq iQ w 61 = [xI x3x4x5x6x. . The IEEE description of this system is shown in Figure 7.( ~ / T R ) vi 6 (l/TA) f.64 Type 2 excitation system representation-rotating printed from IEEE Trans.. where the damping feedback loop is seen to be different from that of Figure 7.79) and substituting for id and iq.95) By using (7. PAS-87.3 Type 2 system-rotating rectifier system Another type of system. being brushless. 1968.61. the rotating rectifier system of Figure 7. r F . Rearranging. = v. 7.13.64. (m IEEE. we write 1 . and rF2. . for the terminal current we may write i. 7.IF (7. To formulate a linearized state-space representation." V .68).298 Chapter 7 If: A > 1 . Multiplying V . . Vc KFE.s/(I + + 7p~) EFD= VB/(K. + K. i. are proportionality factors indicating the proportion of the "Thevenin voltage. we may write the self-excitation components as Vc = Kl V. Obviously.103) to the following form: (7. vol. (a IEEE. + K21.66 Type 3 excitation system representation-static with terminal potential and current supplies.101)-(7. be the term on the right. is a signal proportional to I. V B ' o Fig. = MdXl 4- MqXq (7.. Also. ..104) Note that u.. = Mdid + Mqi. .) Vc represents the self-excitation from the generator terminals. Reprinted from I € € € Trans. which accounts for variation of self-excitation with change in the angular relation of field current (IF) and self-excitation voltage ( V T H[)151. due to potential and current information.101) TRS) But we m a y write the terminal voltage in the time domain as 1 (7.. systems of this type are nonlinear. 1968. PAS-87.E 7 . Constants K. and i . (7. and K.100) Then we write for the entire system VB = 4 = v..103) If we define the states as in (7. ~ ~ 1 VR + = KRy/(I + = [KA/(1 + TAs)] V. we reduce (7.102) where for brevity we let u. are all linear functions of xI-x. . Because the Type 4 system is so nonlinear.time constant of rheostat travel = T R H .~” VRH limited 7. typically set at 5%.5 which.g.5. The RR of modern fast systems are often in the range of 2.7. where K . although typical..7. in addition to the system representations.1 1 are for a system with a response ratio of 0. to the higher of the two input quantities. If V. I I and. that controls the fast-change mechanism on the rheostat. This difference is due to the different choice of base voltage for V .. 7. they have dead zones in which the system operates essentially open loop. there is no advantage in representing it in state variable form. while a small-error voltage will cause the segments to be shorted one at a time. Westinghouse BJ30 or General Electric GFA4 regulated systems) often have two speeds of operation depending upon the magnitude of the voltage error. The equations for the Type 4 system are similar to those derived for the electromechanical system of Section 7. the rheostat setting is changed by motor action with an integrating time constant of 7 R H .9. to VRmal . 7.e. is certainly not fast by today’s standards. a great many systems are of an earlier design similar to the rheostatic system of Section 7.67 Type 4 excitation system representation---noncontinuously setting regulator.0-3. and VRmin given in Table 7. by the different exciter manufacturers and does not necessarily imply any marked difference in the regulator ceilings or performance.10 Typical System Constants Reference [ 151 gives. For any real system all quantities should be obtained from the manufacturer. A comparison of these two systems is recommended. although common. Therebase voltage of V . These data are given in Table 7.Excitation Systems 299 I I Fig.1 1 are unity in column I Note that the values of VRmax and higher values in columns 2 and 3. Also note that the values in Table 7.. Note: and VRmax.. they are generally characterized as slow due to friction and inertia of moving parts. Type 4 systems (e. Thus a large-error voltage may cause several rheostat segments to be shorted out. between VR. is below this limiting value of K. Changing the affects all the other constants in the forward loop.1. a table of typical constants of physical systems.67.1 and are noncontinuous acting. is the raise-lower contact setting.5 Type 4 system-noncontinuous acting The previous systems are similar in the sense that they are all continuous acting with relatively high gain and are usually fast acting. However. An “auctioneer” circuit sets the output V . The computer representation of a system is illustrated in Figure 7. do not necessarily represent any physical system accurately. In addition to this. i. o ..3 -0.03 I . 8 ABB Power T & D Company Inc.28 4.9 I .5 -0..06 25-SO* 0.50 3. caution must be used in comparing gains.5 0.5 0. As experience has accumulated in excitation system modeling.06-0.76 0.I 4.95 0.22 0.o 1 .1 0.5 ... Since these constants are specified on a normalized basis.17 0.5 0.35. and limits for systems of different manufacture.5 0.30 3..5 -3.o .22 0..105 .50 1.o 0.85 . 0.I 7 0.5) Self-excited exciters.03 1 . 20 .I7 0..5 -4.02 7. ..25 0. 4.86 0..05 400 0.95 0. .0 400 0..0 0.3 -7.95 1.. 1992.22 0..04 400 0. (Pu)* (Pu)* 1 . 8.2 -8.05 'mx Ra 'm Rm KF TF KE TE SErnax S E 75max 0 0 -0.08 .o 0 K" TRH . fore.5 0.02 3.5 -3. *Values given assume up (full load) = 3.5) Mag-A-Stat Rotating-rectifier BJ30 Rototrol Silverstat TRA Excitation system type TR I I 0.. Symbol Westinghouse Excitation System Constants for System Studies (excitation system voltage response ratio = 0.. 3600 r/min I800 r/min 'a Rx m 'i Rn m sErnax SE. I5 give examples of excitation system parameters that can be used for estimating new systems or for cases where exact data is unavailable. commutator...lSmax 7E (s) 3.300 Table 7. ..25 4.22 0. multiply * values by ud3.3 0. Since the formation of the National Electric Reliability Council (NERC) a set of deTable 7. or silicon diode with amplidyne voltage regulators (1) Symbol Self-excited commutator exciter with Mag-A-Stat voltage regulator (2) Rotating rectifier exciter with static voltage regulator (3) TR KA TA 0..028 0.50 0..20 1 .2 0. 0.22 0. they can often be used with reasonable confidence on other simulations where data is unavailable. Chapter 7 Typical Constants of Excitation Systems in Operation on 3600 r/min Steam Turbine Generators (excitation system voltage response ratio = 0.074 3.o 0.02 200 0.. .95 0.5 1.95 0.028 0....70 -4.04 1 .3 0.2 11 .5 1. Dec.267 0..5 0..5 0.0 400 0.05 200 0.2 -0.0-0.80 0.5 0.12-7.0 (s) KA TA (s) (Pu)* (Pu)* EFDrnax EFDmin KE KF T F (SI 0.86 0.50 Source: Used with permission from Stability Program Data Preparation Manual.I7 0.3 0.O I -0..5 .12.0 0.95 0.. 0.0 .. . time constants.o -0..0.0 400 0.0 pu.8 8.. Advanced Systems Technology Rept. 70-736.17 .95 0.1.95 0. If not. the manufacturer and utility engineers have determined excitation system parameters for many existing units.11.05 4. ..05 ..0 -0. 1972.05 0.3 -7. Tables 7. . 4 I I 1 0.22 0.50 *For generators with open circuit field time constants greater than 4 s.5 1 7 -3.3 3. 1. 8 1. Philadelphia Electric Co. .555 1.5 I .76 0.25 0.05 0. governors.04 0 -1. 1971. . This has caused an enlarged interest and concern in the accuracy of modeling all system components.2 -1.555 0 1. if known.o I .~SE/KA rE/ KA 4TE/ K A 87E/KA * * 0 0 0 0 0 I .95 0.056 0.o 0.05 0 0 0.0 0.. 1971.15.3 8.05 0.8 -8.13.0039 0.5 3.o I .1 Source: Used by permission from Power System Stability Program User’s Guide.2 8.21 T&& I .5 I .25 0.o * * 400 400 0. particularly the generators.059 1.02 0.0039 0.o 1 .03 I .5 -1.2 1.3 0.555 1. Thus it is becoming common for the manufacturer to specify the exciter model to be used in system studies and to provide accurate gains and time constants for the system purchased. *Data obtained from curves supplied by manufacturer.17 A EX BEX Mag-A-Stat (Type I SCPT* (Type 3) BJ30 (Type 4) Rototrol (Type I ) Silverstat (Type I ) T R A (Type 1) GFA4 (Type 4) Brushless (Type 2) 3600 r/min I .03 Source: Used bv .o Brushless (Type 2) 1800 r / m i n 0.17 0.0 0 1.2 0. * K p = 1.555 I .ma. Table 7.o -0. Typical Excitation System Constants Type of regulator TR KA TA “. .Excitation Systems Table 7.02 I .35 0 0.465 0.14.o rj0/ 10.o 1 .2 0.~EFDFL 1 ?High-speed contact setting.17 -0.05 200 200 400 20 0.5 0. and loads.0052 0.056 0 I I .2 0.o -3.3 3.ost 0 0.05t 0.05lt 1 .85 0. sign criteria has been established specifying the conditions under which power systems must be proven stable.5 0.17 -0.o I .19 -sin(cos-’Fp) “Emax = [ + ap] [ study M V A base generator MVA base ~.855 Brushless (Type 2) 1800 r / m i n I . if known.06 400 120 0.0 .= 1.o 1. “Rmin KF/TF TF 301 Mag-A-Stat (Type 1) SCPT (Type 3) BJ30 (Type4) Rototrol (Type I ) Silverstat (Type I) TRA (Type I ) G FA4 (Type4) NAlOl (Type I ) Amplidyne N A 108 (Type 1 ) Amplidyne N A 143 (Type 1) Amplidyne < 5 k W NA143 (Type I ) Amplidyne > 5 kW Brushless (Type 2) 3600 r/rnin 0 0 20. exciters.04 0.5 -0.0039 0.05 -0.o 7.084 0. permission from Power System Stability Program User’s Guide.2 3.00105 0.8 -3.555 1. tHigh-speed contact setting.o I . Philadelphia Electric Co.06 0.10 o..0039 0 0.19 K.5 3.o 0.12 0.5 0. Typical Excitation System Constants Type of regulator KE -0. For typical values see Appendix D and Table 7.1.0 -7.o I.45 1 .0 0.15 0 0.0 -1.5 0. P .302 Chapter 7 .68 Full model generator response of lo"(. Initial loading of Example 5. 7..1.. and 0 e 0 0 &FD.- x E .- C a2 0 C L L Fig. with no exciter and no generator saturation. step increase in T. sA = 0. KR = 1.. V R = ~ ~ ~ -7. Exciter parameters (Westinghouse Brushless): KA . and 5% step increase in V R E Fwith initial loading of Example 5.93. = 3.0. 7... V. T R = 0.3. KF = 0.8. no gen-. TE = 0.t "/\ .400.0.03. Fig.. = 7.69 Full model generator response to 10% step increase in T. TF = 1. r .1.0.3.. eratm o exciter saturation.02. K E = 1.0. 1. As a result.5 I . $See (7. that the output power is held constant by the governor.304 Chapter 7 Table 7.0/3 IO~io/3 25 25 2OT.0108 1. is changed. This causes the change in 6 to be more stable. The exciter modeled for this illustration is similar to the Westinghouse Brushless exciter. .5 2. I n the regulated machine a 5% step increase in VREFis made. nearly constant when T. improves the system response dramatically. we can construct a computer simulation of a generator with an excitation system.15. Typical Excitation System Constants for Exciters with Amplidyne Voltage Regulators ( N A I O I ..69. and with a decrease in 6 to just below the initial value. N A 1 0 8 .69 with the exciter operative. Adding the excitation system.0171 0. For the purpose of illustration.1 Consider thegenerator of Figure 7.1 1 The Effect of Excitation on Generator Performance Using the models of excitation systems presented in this chapter and the full model of the generator developed in Chapters 4 and 5. both in the transient and dynamic modes of operation.o -0. as shown in Figure 7.0/3 0.0333 0. We conclude that for the load change observed.o 10~20 0. The phase plane plot shows a stable focus at the new 6.25 0.0833 25~.90). NA143) 0.0/22 25 25 50 50 17~.0093 0. The results of this simulation are interesting and instructive and demonstrate clearly the effect of excitation on system perform ance.898 0.69 show the response of the system to a 10% step increase in T. 1971. tFor NA143 over 5 k W .18.1 but assume that the machine is located at a remote location so that the terminal voltage 4 increases roughly in proportion to Eg.~0/3 50 50 10~. Philadelphia Electric Co. Following the increase in torque the system is subjected to an increase in EFD. This increase in 6 is most clearly shown in the phase plane plot./l3 25~. Both Figures 7.0445 -0..1. beginning with the full-load condition of Example 5.v Program User's Guide.0058 0.2 as analyzed in Example 7. has been added to the generator analog simulation of Figure 5. Assume.465 1. * F o r a l l N A l O l .68 for constant EFDand Figure 7. a n d N A 1 4 3 5 k W orless.5 1.06 0. the exciter has a stabilizing influence due to its ability to hold the flux linkages and voltage nearly constant. however.1 level. NA108. this torque increase causes a monotone decay in both A. The results are roughly the same with increases noted in A.0 -0. Problems 7.. 7. and an increase in 6 that will eventually cause the generator to pull out of step. Repeat Example 7. This is accomplished by switching the unregulated machine E F D from 100% to 110% of the Example 5. In Chapter 8 we will consider further the effects of excitation on stability. Note that the exciter holds AF and V.0016 0. For the generator with no exciter. The results are shown in Figure 7. 6 is increased to its new operating level in a damped oscillatory manner.68 and 7.0240 -0..1428 20~. Appropriate switching is arranged so the simulation can be operated with the exciter active or with constant EFD.79 Source: Used by permission from Power Sysrem Sra6ilir.61. and V. a Type 1 excitation system similar to Figure 7. and V. The system is shown in Figure P7.. Write the differential equations that describe the system. Consider the separately excited exciter E shown in Figure P7.1 except that instead of increasing the excitation. Assume linear variations where necessary to establish your arguments. By means of a phasor diagram analyze the change in 6.7 1.4.2.2 305 7. P7.e.1 Consider the generator of Example 7.9. The initial current in the generator field is p when the exciter voltage uF = ko. Le. to restore the power factor to its original value? Repeat Example 7.3 7. At time t = a a step function in the voltage uF is introduced. I.6 7. + k .1 connected in parallel with an infinite bus and operating with constant excitation. Following the change described in Problem 7. and in what amount. holds. for up = u ( t ) . +p-LqLF Fig. What assumptions must be made for the above relation to be approximately valid? Compute the current i2 due to a step change in the pilot exciter voltage.9 . Fig. what action would be required.8.7. i. u(f . in particular.1 and Problems 7. Plot the current function i n the s plane. the change i n 6 and 8. P7. Consider the exciter shown in Figure P7.4 7. L f / r F is 7.Excitation Systems 7. decrease Ex to a magnitude less than that of V. Observe the new values of 6 and 8 and. Sketch this result for the cases where the time constant both very large and Lery small... and 8 when the governor setting is changed to increase the power output by 20%. Note particularly the change in 6 in both direction and magnitude.a). Discuss the operation of this device and comment on the feasibility of the proposed design. can you make any general statement regarding the sensitivity of 6 and 8 to changes in P and ER? Establish a line of reasoning to show that a heavily cumulative compounded exciter is not desirable.1-7. uF = k .8 ComDute the current i F .5 1. where the main exciter M is excited by a pilot exciter P such that the relation uF = k'wc z ki.9 A solenoid is to be used as the sensing and amplification mechanism for a crude voltage regulator. Comparing results of Example 7. I2 has a magnetization curve as given in Table 7. .1.10 Chapter 7 A n exciter for an ac generator. Write the equations for this system and show that. Assume no load on the exciter.14 Assume that the constants r A .01.3. uFI = 120 V.1 I 7. I 2 (a) Determine the buildup curve beginning at rated voltage. is driven by a separate motor with a large flywheel. Assume uF1 = 40 V. Analyze the system given in Figure P7. Let r R take the values of 0.12 The separately excited exciter shown in Figure P7. P7. 0. a degree of stabilization is achieved. uF2 = 180 V. I f this forcing voltage is held constant. and 0....2 R = 8 s2 in field winding k = 12. r. K.000 uF = 120 V (rated) 6 +--’+ 2 Fig. compute the buildup.13 Given the same exciter of Problem 7. r E . What are the initial and final values of resistance in the field circuit? (b) What is the main exciter response ratio? 7. 7. with parameters carefully selected. K. 7.306 7.12.. Consider the motor to have a constant output torque and write the equations for this system. Le. and KA are the same as in Example 7..1 I to determine the effectiveness of the damping transformer in stabilizing the system to sudden changes. particularly for large values of R. Other constants of interest are N = 2500 UP = 125V u = 1.001. Find the effect of rR on the branch of the root locus near the imaginary axis. consider a self-excited connection with an amplidyne boost-buck regulation system that quickly goes to its saturation voltage of +IO0 V following a command from the voltage regulator.7. instead of being driven from the turbine-generator shaft. IO. A . R. Brushless excitation system. Belgium. A I E E Trans. Experience with automatic voltage regulation on a 115-megawatt turbogenerator. D a t a for the excitation system is given in Table 7. IEEE Committee Report. K. 1964. S. W. Chicago. A new regulator and excitation system.14 with rR = 0. P. 12. Presented at Association des IngCnieurs Electriciens de I’lnstitute Electrotechnique Montefiore. F.1961 13.14.. Mass. New York. IO.05 and 7F = 0. 1968. Rec. Alternator-rectifier exciter for Cardinal Plant. Control and Dynamic S-vstems. M . PAS-80 1072--77. PAS-80: 1066--72.. I E E E Trans. Excitation voltage response definitions and significance in power systems. PAS-76:1491-96. M..19 For the excitation system described in Example 7. PAS-87: 1460-64. 3.22 7.I. E.. A I E E Trans. Rogers.1..6. trial and utility steam turbine-generators. J . P. and Temoshok. M . I I . (MIT Press. Wiley. Recent developments in amplidyne regulator excitation systems for large generators.~85. Pa. 19. Rubenstein. K. Southwest I E E E Con/: (SWIEEECO).. E. 5 . PAS-88: 1248-58... G. C. B. 1965. Repeat Problem 7. and Bobo. E.Edwards.20 eigenvalues.. I E E E Trans.. F. Vol. and Bobo.. Rubenstein.. 1950.v. 1952. Rubenstein. Dandeno. A. J. Carlson. Show how the choice of base voltage for the voltage regulator output VR affects other constants i n the forward loop. Barnes..1 I . Proc.9 and for the machine model and operating conditions described in Example 6. A I E E Trans. and Temoshok. J. C.M. D.. F.02 a n d 0. Alexanderson. S. Rotating rectifier exciters for large turbine-driven ac generators. A n electric utility brushless excitation system. Concordia.16 Obtain the loci of the roots for the polynomial of (7.. Lane. 1953. E_xcitationsystems-DesigFs and practices in the United States. Domeratzky. M. 1970. S. J. 1969. Repeat Example 7. Transienf Performance o/’ Electric Power Systems: Phenomena in Lumped Networks. and Bowman.. A. 9. 125. and McClymont.. H.. M . Liege. PAS-80 1077. 1957.23 References Generator excitation systems and power system performance. The amplidyne generator-A dynamoelectric amplifier for power control. Cornelius. M. Kimbark. Whitney. 3. F. M. A new excitation system and a method of analyzing voltage response. Also include the e r e c t of saturation in the simulation. S. M. E. 0. J. S. presented at the IEEE Summer Power Meeting. A static excitation system for industrial and utility steam turbine-generators. Computer representation of excitation systems. J. and Temoshok. 17. Westinghouse Electric Corp. PAS-87:I 189-98. 1957.. Proposed excitation system definitions for synchronous machines. L. 14. 1967. and Cory. L. Myers. PAS-78:1821-24.. A. R. Temoshok.. H.. 7.3. and Temoshok. Mass. Rabins. Addison-Wesley. A I E E Trans. 7.19 for the conditions of Example 6. Cawson. and Rothe. A.17 Obtain (or sketch) a root-locus plot for the system of Example 7. New York.15 Repeat Problem 7. 1950. McGraw-Hill. Power Con/:. H . 1956.sfein Stability. P. Rubenstein. S. Excitation system response: A utility viewpoint.. M. Lee. 8. 6. 27. 7.. A. I E E E Trans. W. Am. 18.05 a n d 0. Reading. 4.7. 7. Hoover. = 0. 1959.. 1966. I. and Kedy. I E E E Trans. C.2 I 7. and Auslander.05 a n d f a r values of 7” = 0. 21. P. 20. Power S. 7. Paper 3 I CP 67-536. 0.63) for T~ = 0. Pittsburgh. A. Elecfrical Transniission and Distribution Re/erence Book. 2. Conv. M .Excitation Systems 307 7. 1966. Chambers. A I E E Trans. T.. Use a Type 1 exciter. C. D. I E E E Int. A. Oliver.. and Temoshok.2. PAS-72:175-83. Proc. A I E E Trans. Assume the usual bases for a n d E. E.. P. 1940.. F. Vol.1961. W. H. 43: 104-6. Myers. 0.8 for K. and Vance. A .9 (with the same operating condition) using a Type 2 excitation system. Rudenberg. IEEE Committee Report.9 for the operating condition of Example 6. H. obtain the A matrix of the system and find the 7. and Horton. Portland. 15. A I E E Trans.. F. . W. PAS-’II:184-87. H. Oreg. Cambridge. L.18 Complete the analog computer simulation of the system of one machine connected to a n infinite bus (given in Chapter 5) by adding the simulation of the excitation system. 16. 1968. General Electric Rev. Design and tests of a static excitation system for indus1961. Bobo. D. W. Takahashi. 7. PAS-76: 1497-1501. 1967). Repeat Example 7. K..3 and for values of KF between 0. Ferguson. I. Storm. M. 1966. A. Herbst. 30. PAS71~239-45. 58:838-44. Theoryand Application. Kimbark. Power Sysreni Stability. G. Brown. 1952. and Horton.S. McClure. Pt. Pt.. G.. Harder. Ewart. 1965. Iowa State Univ. R. 1972. Patent No. 28.... McGraw-Hill. 1938. J. A I E E Trans. Crenshaw. A. 2. and Kinghorn. W. The development of modern excitation systems for synchronous condensers and generators.. 1945. F. A. E. GET-2980. New York. Elerrric Power Circuits. E. Oplinger. and Temoshok. Ph. and Miller. Modern excitation systems for large synchronous machines. Regulating system for dynamoelectric machines. 1954. R.. Hand.1954. N. W. Analytical studies of the brushless excitation system. C. Concordia. C.. Elentents of Srability Calculations. 33. T.s. Eng. Hunter. 2.308 Chapter 7 22. A I E E Trans. . PAS-71:894 -900. Feb. PAS-73:486-91. F. R... Development of a modern amplidyne voltage regulator for large turbine generators.. L. F. 37. 1950.. 0.692. J. 1939. F. 2. 39. E. The amplistat and its applications.. Vol. McClure. J. 31. Rothe. 38. W. J. A I E E Trun. 1959. Vol.. M. Power system stability program. Performance of new magnetic amplifier type voltage regulator for large hydroelectric generators. T.. P.1952. A I E E Trans. McGraw-Hill. Genewl Electric Rev. Ogle. S. Static voltage regulator for Rototrol exciter.. Bull. K.967. and Nilsson. A I E E Trans. C. 1946. and Valentine. Paper CP 65-208. W. PAS-7 1:201-6.. Dahl. Wiley. Effect of boost-buck voltage regulator on steady-state power limit. U. Ames. A I E E Trans. G. Kron. Amplidyne regulator excitation systems for large generators. 34. Oct. Whittlesley. C. Effects of system nonlinearities on synchronous machine control. C. and Hartman. 64: 601. P. 1962. 0 . Research Rept. PAS-78:1815-21. R.. D. Recent developments in generator voltage regulation. H. 41. W. 25. E. 3. Power System Planning Div.. 36. Westinghouse Electric Corp. 1950. M. A I E E Trans. E. H. J. C. Carleton. S. N. New York. Magamp regulator tests and operatand ing experience on West Penn Power System. Stability program data preparation manual. Users Guide U6004-2. Wiley. Elecrr. Basic Feedback Control Sysrennl Design. A static excitation system for steam turbine generators. J. The figure of merit of magnetic amplifiers. 1952. PAS-65:939-45. Carleton. and Valentine. I. PAS691380-84. 1971. A I E E Trans. Savant. ERI-71130.. I .D. General Electric Co. M. E. J r . and Todd. F. 8. 23. Aug. J . Bobo. R . PAS-65: 1070-27. Lane. presented at the IEEE Winter Power Meeting. L. New York. Oyetunji. 29. G . 1948. W. 1971. M. A.. 70-736. 26. thesis. J. 32. 27. J . Porter. 40. 24.. Inrroducrion to Linear Systents Analysis. W. H. 1958. Advanced Systems Technology Rept.. 1946. New York. Patent Office. A I E E Trans. 42... . Philadelphia Electric Co. Mendel. Hanna.: Pt. New York. E. M. K. 35. Unpubl. Kallenback. L. L. and Dandeno. chapter 8 The Effect of Excitation on Stability 8. Some authors call the dynamic stability problem by the ambiguous name of “steadystate stability. then.. Engineers learned that the system damping could be enhanced by artificial signals introduced through the excitation system. it was assumed that a large control effort could be expended through excitation control with a relatively small input of control energy. Early investigators realized that the socalled “steady-state” power limits of power networks could be increased by using the then available high-gain continuous-acting voltage regulators [ I ] . is to determine this limit. Furthermore. to find the exciter design and control parameters that can provide good performance at reasonable cost [ 141.1 Introduction Considerable attention has been given in the literature to the excitation system and its role in improving power system stability.” Other variations are found in the literature. While basically sound. In the first few cycles these requirements may be significantly different from those needed over a few seconds. The success of excitation control in improving power system dynamic performance in certain situations has led to greater expectations among power system engineers as to the capability of such control Because of the small effective time constants in the excitation system control loop. i. It should be noted that this terminology is not universally used. but usually the two problems are treated separately as noted. In the 1960s large interconnected systems experienced growing oscillations that disrupted parallel operation of large systems [3-121. the transient (short-term) problem and the dynamic (long-term) problem. It was also recognized that the voltage regulator gain requirement was different at no-load conditions from that needed for good performance under load. The subject of excitation control is further complicated by a conflict in control requirements in the period following the initiation of a transient. 309 . This suggests the separation of the excitation control studies into two distinct problems. and stabilizing feedback circuits came into common use (21. This scheme has been very successful in combating growing oscillation problems experienced in the power systems of North America. A part of the engineer’s job. In the early 1950s engineers became aware of the instabilities introduced by the (then) modern voltage regulators. It was discovered that the inherently weak natural damping of large and weakly coupled systems was the main cause and that situations of negative damping were further aggravated by the regulator gain [ 13). it has been shown that the best control effort in the shorter period may tend to cause instability later.e. this control is limited in its effectiveness. these induced changes causing their own interaction with neighboring machines (see Section 3. Note that if V. sets up all kinds of oscillatory responses and the system “rings” for a time with many frequencies present. is the machine terminal voltage and V . Thus the one change in load. Consider a multimachine system feeding a constant load (a condition never met in practice). when the fault is removed and the reactance x of (8. thus improving the chances of holding V . at a reasonable value. and electrical location with respect to the load. Now how will this load change manifest itself at the several machines in the control group? Since it is a load increase. Let us assume that at a given instant the load is changed by a small amount. the most beneficial attributes the voltage regulator can have for this situation is speed and a high ceiling voltage. and speeds will be different for each machine in the control group because of differences in unit size. is reduced. Let us examine the behavior of the machines in the time interval prior to the governor action. rotating energy [( 1/2)mu2]is used to supply the load requirements until the governors have a chance to adjust the power input to the various generators.1) is increased due to switching. then somewhat later. there is an immediate increase in the output power requirements from each of the machines. the usual approximation for the power transfer is given by P = (V.1. currents. . this increased power requirement will come first from stored energy in the control group of machines.1) where V. The machines nearest the load electrically will see the largest change. and those farther away will experience smaller and smaller changes until the change is not perceptible at all beyond the boundary of the control group.V. This interval may be on the order of 1 s. Thus energy stored in the magnetic field of the machines is released. and the requirements on the excitation system are also different. Prevention of this reduction in P requires very fast action by the excitation system in forcing the field to ceiling and thereby holding V . Thus each unit responds by contributing its share of the load increase. Each unit has its own natural frequency of response and will oscillate for a time until damping forces can decay these oscillations. Also. By dynamic stability we mean the ability of all machines in the system to adjust to small load changes or impacts. usually a fault. a step change. Assume further that this change in load is just large enough to be recognized as such by a certain group of machines we will call the control group.6). The dynamic stability problem is different from the transient problem in several ways. design. another fast change in excitation is required. In this time period the changes in machine voltages. If we consider the one machine-infinite bus problem. with its share being dictated by the impedance it sees at its terminals (its Thevenin impedance) and the size of the unit. say by the energizing of a very large motor somewhere in the system. P is reduced by a corresponding amount. Indeed. at the needed level. is the infinite bus voltage. Since step changes in power to turbines are not possible.310 Chapter 8 8. These violent changes affect the machine’s ability to release the power it is receiving from the turbine./x)sinb (8. which is maintained for a short time and causes a significant reduction in the machine terminal voltage and the ability to transfer synchronizing power.1 Transient stability and dynamic stability considerations In transient stability the machine is subjected to a large impact. These changes are effectively controlled by very fast excitation changes. (This problem was introduced and analyzed by Concordia [ 11. In the older electromechanical systems there was a substantial deadband in the voltage regulator. although somewhat slowly. and several investigators have shown examples wherein systems are less oscillatory with the voltage regulators turned off than with them operating [7. either as a perceptible change in terminal voltage. and linearization about this normal or “quiescent” point is possible and desirable. This system lag then is a detriment to stable operation. The small impact or dynamic stability problem is different.1. 121. since as the speed voltage changes. or both. The excitation system has one major handicap to overcome in following these system oscillations: this is the effective time constant of the main exciter field which is on the order of a few seconds or so. . and each action or reaction is accompanied by an excitation change. causing a new excitation adjustment to be made. terminal current. This is a nonlinear problem.2 Effect of Excitation on Generator Power limits We begin with a simple example.1 One machine-infinite bus system. During this delay time the state of the oscillating system will change. These systems recognize the change in load immediately. Thus from the time of recognition of a desired excitation change until its partial fulfillment.1 Consider the two-machine system of Figure 8. The machines closer to the load change would recognize a need for increased excitation and this would be accomplished. Here we are concerned with small excursions from normal operation.Effect of Excitation o n Stability 31 1 Now visualize the excitation system in this situation. we may study the response using the tools of linear systems analysis. in this way not only can we analyze but possibly compensate the system for better damping and perhaps faster response. the oscillating control group machines react with one another. where we consider one machine against an infinite bus. and we examine the response in building up from normal excitation to ceiling excitation. the terminal voltage also changes. the excitation of these machines would remain unchanged.) The power output of the machine is given by P 6 = = [EIEz/(XI 6 + 62 1 + X2)] 6 sin Fig. and unless the generator was relatively close to the load change. Moreover. Example 8. and the shape of the magnetization curve cannot be neglected. Thus each oscillation of the unit causes the excitation system to t r y to correct accordingly. Our approach to this problem must obviously depend upon the type of impact under consideration. For the large impact. as we have seen. 8. there is an unavoidable delay. 8. Newer excitation systems present a different kind of problem. the purpose of which is to show that the excitation system can have an effect upon stability. Having done this. we are concerned with maximum forcing of the field. such as a fault. 31 2 Chapter 8 Fig. 8.2 Phasor diagram for Example 8.1. This equation applies whether or not there is a voltage regulator. Determine the effect of excitation on this equation. Solution We now establish the boundary conditions for the problem. First we assume that XI = X 2 = 1.0 pu and that V, = 1 .O pu. Then for any given load the voltages E , and E2must assume a certain value to hold at 1.0 pu. If the power factor is unity, E, and E2 have the same magnitude as shown in the phasor diagram of Figure 8.2. If E, and E2 are held constant at these values, the power transferred to the infinite bus varies sinusoidally according to (8.2) and has a maximum when 6 is 90". Now assume that E, and E2are both subject to perfect regulator action and that the key to this action is that V, is to be held at 1.0 pu and the power factor is to be held at unity. We write in phasor notation E, = 1 + jf = dmej*/z E2 = I - jf = dme-j6/2 Adding these equations we have E, + E2 = 2 = 2 r n C O S 6 / 2 I I I I I I I I I I I I Angle 6, degrees Fig. 8.3 Comparison of power transferred at unity power factor with and without excitation control. Effect of Excitation on Stability 313 or El = E2 = I cos 612 (8.3) Substituting (8.3) into (8.2) and simplifying, we have for the perfect regulator, at unity power factor, P = tan612 (8.4) The result is plotted in Figure 8.3 along with the same result for the case of constant (unregulated) E l and E 2 . In deriving (8.4), we have tacitly assumed that the regulators acting upon E l and E 2 do so instantaneously and continuously. The result is interesting for several reasons. First, we observe that with this ideal regulation there is no stability limit. Second, it is indicated that operation in the region where d > 90" is possible. We should comment that the assumed physical system is not realizable since there is always a lag in the excitation response even if the voltage regulator is ideal. Also, excitation control of the infinite bus voltage is not a practical consideration, as this remote bus is probably not infinite and may not be closely regulated. Example 8.2 Consider the more practical problem of holding the voltage E2 constant at I .O pu and letting the power factor vary, other things being the same. Solution Under this condition we have the phasor diagram of Figure 8.4 where we note that the locus of E2 is the dashed circular arc of radius 1.0. Note that the power factor is constrained by the relation e, = a2/2 (8.5) where 8, = IT - 8 and 6 = 6, 6,. Writing phasor equations for the voltages, we have + e Fig. 8.4 Phasor diagram for Example 8.2. 314 Chapter 8 Toque Angle, 4, degrees Fig. 8.5 System parameters as a function of 6 2 . El = I + jT = I - [sine + j l c o s e = ~ , e ' " E2 = 1 - j r = I + Isin6 - jfc o s e = E2e-jb2 (8.6) where 6, el, A I , and a2 are all measured positive as counterclockwise. Noting that E2 = 1, we can establish that I sin 6 = = 2sin0, 2sin6, E , sin6 = = 2sinJ2 sinb2/(2 - cos&) 03-71 Thus once we establish J2. we also fix 0, I, 6, and 6 , , although the relationships among these variables are nonlinear. These results are plotted in Figure 8.5 where equations (8.7) are used to determine the plotted values. We also note that tan6, P = V,~COS~ = (8.8) sin a2 or (8.9) but from the second of equations (8.6) we can establish that I cos 8 P = sin& so 62 also establishes P. Thus P does have a maximum in this case, and this occurs when 62 = 90" (E' pointing straight down in Figure 8.4). In this case we have at maximum power E, e = 2 + jl = 2.235/26.6" I = 1.414 = -450 6 = 116.6" The important thing to note is that P is again limited, but we see that 6 may go Effect of Excitation o n Stability 315 0 90 Torque Angle 180 b, degrees Fig. 8.6 Variation of P with 6. beyond 90" to achieve maximum power and that this requires over 2 pu E , . The variation of P with 6 is shown in Figure 8.6. These simple examples show the effect of excitation under certain ideal situations. Obviously, these ideal conditions will not be realized in practice. However, they provide limiting values of the effect of excitation on changing the effective systey parameters. A power system is nearly a constant voltage system and is made so because of system component design and close voltage control. This means that the Thevenin impedance seen looking into the source is very small. Fast excitation helps keep this impedance small during disturbances and contributes to system stability by allowing the required transfer of power even during disturbances. Finally, it should be stated that while the ability of exciters to accomplish this task is limited, other considerations make it undesirable to achieve perfect control and zero Thevenin impedance. Among these is the fault-interrupting capability. 8 3 Effect of the Excitation System on Transient Stability . In the transient stability problem the performance of the power system when subjected to severe impacts is studied. The concern is whether the system is able to maintain synchronism during and following these disturbances. The period of interest is relatively short (at most a few seconds), with the first swing being of primary importance. In this period the generator is suddenly subjected to an appreciable change in its output power causing its rotor to accelerate (or decelerate) at a rate large enough to threaten loss of synchronism. The important factors influencing the outcome are the machine behavior and the power network dynamic relations. For the sake of this discussion it is assumed that the power supplied by the prime movers does not change in the period of interest. Therefore the effect of excitation control on this type of transient depends upon its ability to help the generator maintain its output power in the period of interest. To place the problem in the proper perspective, we should review the main factors that affect the performance during severe transients. These are: 1. The disturbing influence of the impact. This includes the type of disturbance, its location, and its duration. 2. The ability of the transmission system to maintain strong synchronizing forces during the transient initiated by a disturbance. 3. The turbine-generator parameters. The above have traditionally been the main factors affecting the so-called first-swing transients. The system parameters influencing these factors are: 316 Chapter 8 1. The synchronous machine parameters. Of these the most important are: (a) the inertia constant, (b) the direct axis transient reactance, (c) the direct axis open circuit time constant, and (d) the ability of the excitation system to hold the flux level of the synchronous machine and increase the output power during the transient. 2. The transmission system impedances under normal, faulted, and postfault conditions. Here the flexibility of switching out faulted sections is important so that large transfer admittances between synchronous machines are maintained when the fault is isolated. 3. The protective relaying scheme and equipment. The objective is to detect faults and isolate faulted sections of the transmission network very quickly with minimum disruption. 8.3.1 The role of the excitation system in classical model studies In the classical model it is assumed that the flux linking the main field winding remains constant during the transient. If the transient is initiated by a fault, the armature reaction tends to decrease this flux linkage [15]. This is particularly true for the generators electrically close to the location of the fault. The voltage regulator tends to force the excitation system to boost the flux level. Thus while the fault is on, the effect of the armature reaction and the action of the voltage regulator tend to counteract each other. These effects, along with the relatively long effective time constant of the main field winding, result in an almost constant flux linkage during the first swing of 1 s or less. (For the examples in Chapter 6 this time constant K37j0is about 2.0 s.) It is important to recognize what the above reasoning implies. First, it implies the presence of a voltage regulator that tends to hold the flux linkage level constant. Second, it is significant to note that the armature reaction effects are particularly pronounced during a fault since the reactive power output of the generator is large. Therefore the duration of the fault is important in determining whether a particular type of voltage regulator would be adequate to maintain constant flux linkage. A study reported by Crary [2] and discussed by Young [ 151 illustrates the above. The system studied consists of one machine connected to a larger system through a 200mile double circuit transmission line. The excitation system for the generator is Type 1 (see Chapter 7) with provision to change the parameters such that the response ratio (RR)varies from 0.10 to 3.0 pu. The former corresponds to a nearly constant field voltage condition. The latter would approximate the response of a modern fast excitation system. Data of the system used in the study are shown in Figure 8.7. A transient stability study was made for a three-phase fault near the generator. The sending end power limits versus the fault clearing time are shown in Figure 8.8 for different exciter responses (curves 1-5) and for the classical model (curve 6). From Figure 8.8 it appears that the classical model corresponds to a very slow and weak excitation system for very short fault clearing times, while for longer clearing times it approximates a rather fast excitation system. If the nature of the stability study is such that the fault clearing time is large, as in “stuck breaker” studies [IS], the actual power limits may be lower than those indicated when using the classical model. In another study of excitation system representation [ 161 the authors report (in a certain stability study they conducted) that a classical representation showed a certain generator to be stable, while detailed representation of the generator indicated that loss of synchronism resulted. The authors conclude that the dominant factor affecting loss Effect of Excitation on Stability Exciter 317 Fault Generator: xd = 0.63 pu x = 0.42 P U Xd = 0.21 pu 7 T ~ O = Regulating system: Pz = 20 H = 5.0 s 5.0 s P,,= 4 r, = 0.47 s E,,, = 2.25 PU E,i, = -0.30 PU x, = 0.10 pu Line: x = 0.8 Il/mi/line r = 0.12 Q/mi/line y = 5.2 x mho/mi/line System: x = 0.2 pu , H = 50.0 s System damping: Fault on rdll Td12 Td21 rd22 Fault cleared 4 I 0 15 3 3 18 Fig. 8.7 Two-machine system with 200-mile transmission lines. of synchronism is the inability of the excitation system of that generator, with response f ratio o 0.5, to offset the effects of armature reaction. 8 3 2 Increased reliance on excitation control to improve stability .. Trends in the design of power system components have resulted in lower stability margins. Contributing to this trend are the following: I . Increased rating of generating units with lower inertia constants and higher pu reactances. 2. Large interconnected system operating practices with increased dependence on the transmission system to carry greater loading. These trends have led to the increased reliance on the use of excitation control as a -0 i* .a a ::I\ 1.05 Curve r? e RR 3.0 2.0 2 L 1.00 ? -. 50 9 I 2 3 0.02 0.04 0.06 0.08 Fault Clearing Time, I 0 0.10 4 5 6 0.042 s 0.17 s 0.68 s 2.70 s 11.0 s I .o 0.25 0.10 Classical model Fig. 8.8 Sending-end power versus fault clearing time for different excitation system responses. 318 Chapter 8 00 . 1.0 Time, I 20 . 3.0 6) lime, s (C 1 Fig. 8.9 Results of excitation system studies on a western U.S. system: (a) One-line diagram with fault location, (b) frequency deviation comparison for a four-cycle fault, (c) frequency deviation comparison for a 9.6-cycle fault: A = 2.0 ANSI conventional excitation system; B = low time constant excitation system with rate feedback; C = low time constant excitation system without rate feedback. (@ IEEE. Reprinted from IEEE Trans.. vol. PAS-90, Sept./Oct. 1971.) means of improving stability [ 17). This has prompted significant technological advances in excitation systems. As an aid to transient stability, the desirable excitation system characteristics are a fast speed of response and a high ceiling voltage. With the help of fast transient forcing of excitation and the boost of internal machine flux, the electrical output of the machine may be increased during the first swing compared to the results obtainable with a slow exciter. This reduces the accelerating power and results in improved transient performance. Effect of Excitation on Stability 319 Modern excitation systems can be effective in two ways: in reducing the severity of machine swings when subjected to large impacts by reducing the magnitude of the first swing and by ensuring that the subsequent swings are smaller than the first. The latter is an important consideration in present-day large interconnected power systems. Situations may be encountered where various modes of oscillations reinforce each other during later swings, which along with the inherent weak system damping can cause transient instability after the first swing. With proper compensation a modern excitation system can be very effective in correcting this type of problem. However, except for transient stability studies involving faults with long clearing times (or stuck breakers), the effect of the excitation system on the severity of the first swing is relatively small. That is, a very fast, high-response excitation system will usually reduce the first swing by only a few degrees or will increase the generator transient stability power limit (for a given fault) by a few percent. In a study reported by Perry et al. [I81 on part of the Pacific Gas and Electric Company system in northern California, the effect of the excitation system response on the system frequency deviation is studied when a three-phase fault occurs in the network (at the Diablo Canyon site on the Midway circuit adjacent to a 500-kV bus). Some of the results of that study are shown in Figure 8.9. A one-line diagram of the network is shown in Figure 8.9(a). The frequency deviations for 4-cycle and 9.6-cycle faults are shown in Figures 8.9(b) and 8.9(c) respectively. The comparison is made between a 2.0 response ratio excitation system (curve A ) , a modern, low time constant excitation with rate feedback (curve B) and without rate feedback (curve C). The results of this study support the points made above. 8.3.3 Parametric study Two recent studies [ 17,191 show the effect of the excitation system on “first-swing’’ transients. Figure 8.10 shows the system studied where one machine is connected to an infinite bus through a transformer and a transmission network. The synchronous machine data is given in Table 8. I . The transmission network has an equivalent transfer reactance A, as shown in ’ Table 8.1. xd Machine Data for the Studies of Reference [ 191 T;O T;O = 1.72 pu = X; = = = = X : = XE 0.45 0.33 PU PU = 7;o = 7p 6.3 s 0.033 s 0.43 S 0.033s 2 1.68 PU 0.59 pu 0.33 PU = = H 4.0 s Fig. 8.10 System representation used in a parametric study of the effect of excitation on transient stability. (e IEEE. Reprinted from IEEE Trans.. vol. PAS-89, July/Aug. 1970.) 320 Chapter 8 Figure 8.10. A transient is initiated by a three-phase fault on the high-voltage side of the transformer. The fault is cleared in a specified time. After the fault is cleared, the transfer reactance X , is increased from x , b (the value before the fault) to X,, (its value after the fault is cleared). The machine initial operating conditions are summarized in Table 8.2. Table 8.2. Prefault Operating Conditions, All Values in p u 0.2 0.4 0.6 0.8 1.0 1.0 1.0 1.0 0.94 0.90 0.91 0.97 0.90 0.90 0.90 0.90 0.39 0.45 0.44 0.44 With the machine operating at approximately rated load and power factor, a threephase fault is applied at the high-voltage side of the step-up transformer for a given length of time. When the fault is cleared, the transmission system reactance is changed to the postfault reactance X,, and the simulation is run until it can be determined if the run is stable or unstable. This is repeated for different values of X , until the maximum value of X,,, is found where the system is marginally stable. Two different excitation system representations were used in the study: 1. A 0.5 pu response alternator-fed diode system shown in Figure 8.1 1. 2. A 3.0 pu response alternator-fed SCR system with high initial response shown in Figure 8.12. This system has a steady-state gain of 200 pu and a transient gain of 20 pu. An external stabilizer using a signal V , derived from the shaft speed is also used (see Section 8.7). “REF I 1 -0.0445 + 0 . 5 I I ‘FD - I 1 0.16s + I U Fig. 8.1 I Excitation block diagram for a 0.5 R R alternator-fed diode system. (c IEEE. Reprinted from IEEE Trans.,VOI. PAS-89, July/Aug. 1970.) From the data presented in [ 191, the effect of excitation on the “first-swing” transients is shown in Figure 8.13, where the critical clearing time is plotted against the , transmission line reactance for the case where X = X , b and for the two different types of excitation system used. The critical clearing time is used as a measure of relative stability for the system under the impact of the given fault. Figure 8.13 shows that for the conditions considered in this study a change in exciter response ratio from 0.5 to 3.0 resulted in a gain of approximately one cycle in critical clearing time. Effect of Excitation on Stability "REF 321 t . pu 49 I Fig. 8.12 Excitation block diagram for a 3.0 RR alternator-fed SCR excitation system. printed from IEEE Trans., vol. PAS-89, July/Aug. 1970.) (@ IEEE. Re- 8.3.4 Reactive power demand during system emergencies A situation frequently encountered during system emergencies is a high reactive power demand. The capability of modern generators to meet this demand is reduced by the tendency toward the use of higher generator reactances. Modern exciters with high ceiling voltage improve the generator capability to meet this demand. It should be recognized that excitation systems are not usually designed for continuous operation at ceiling voltage and are usually limited to a few seconds of operation at that level. Concordia and Brown [ I71 recommend that the reactive-power requirement during system emergencies should be determined for a time of from a few minutes to a quarteror half-hour and that these requirements should be met by the proper selection of the generator rating. 8.4 Effect of Excitation on Dynamic Stability Modern fast excitation systems are usually acknowledged to be beneficial to transient stability following large impacts by driving the field to ceiling without delay. However, these fast excitation changes are not necessarily beneficial in damping the oscillations that follow the first swing, and they sometimes contribute growing oscillations several seconds after the occurrence of a large disturbance. With proper design and compensation, however, a fast exciter can be an effective means of enhancing stability in the dynamic range as well as in the first few cycles after a disturbance. Since dynamic stability involves the system response to small disturbances, analysis as a linear system is possible, using the linear generator model derived previously [ 1 I]. For simplicity we analyze the problem of one machine connected to an infinite bus E 6 0 2 L 0.2 0.4 06 . 0.8 xeo = Xeb, PU Fig. 8.13 Transient stability studies resulting from studies of [19]: A = 0.5 RR diode excitation system; E = 3.0 pu RR SCR excitation system. (Q IEEE. Reprinted from IEEE Trans., vol. PAS-89, July/Aug. 1970.) 322 Chapter 8 through a transmission line. The synchronous machine equations, for small perturbations about a quiescent operating condition, are given by (the subscript A is omitted for convenience) T, E; V, = = = T ~ W S= Kl6 K2E; [K,/(I K~T;oS)]EFD [ K i K 4 / ( 1 K56 + K6E; T,,, - T, + K~T;oS)]~ (8.10) (8.1 1) (8.12) (8.13) where is the direct axis open circuit time constant and the constants K , through K6 depend on the system parameters and on the initial operating condition as defined in Chapter 6. In previous chapters it was pointed out that this model is a substantial improvement over the classical model since it accounts for the demagnetizing effects of the armature reaction through the change in E; due to change in 6. We now add to the generator model a regulator-excitation system that is represented as a first-order lag. Thus the change in EFD is related to the change in V, (again the subscript A is dropped) by E F D / V ,= -Kc/(I where K , is the regulator gain and 8.4.1 T, + 7,s) (8.14) is the exciter-regulator time constant. Examination of dynamic stability by Routh’s criterion To obtain the characteristic equation for the system described by (8. l0)-(8.14), a procedure similar to that used in Section 3.5 is followed. First, we obtain r - From (8.13) for T,,, = 0, s26 = - ( W ; / T , ) T , = - ( w R / ~ H ) T , (8.16) By combining (8.15) and (8.16) and rearranging, the following characteristic equation is obtained: s4 + as’ + ps2 where C Y + ys + 7 = 0 (8.17) = = = I/?, [(I + l/K3rAO p + K3K6Kc)/K3T;OTr] + KI(WR/2H) .=-[ wR (Ki/Tr 2H WR + Ki/&T;O - K2&/Th) K2K4 2H KI(I + K3K6Kr) K3 4 0 Tf Applying Routh’s criterion to the above system, we establish the array Effect of Excitation on Stability 323 where b2=O c ~ = u ~ = v (8.18) According to Routh's criterion for stability, the number of changes in sign in the first column ( I , a, al , b l , and c I ) corresponds to the number of roots of (8.17) with positive real parts. Therefore, for stability the terms a,a l , l , and cI must all be greater than b zero. Thus the following conditions must be satisfied. 1. a = 1/7, + l/K37;0 > 0, and since 7, and ri0 are positive, dO/T, > - 1/K3 (8.19) K3 is an impedance factor that is not likely to be negative unless there is an excessive series capacitance in the transmission network. Even then 72,)/T, is usually large enough to satisfy the above criterion. 2. 01 = p - y/a > 0 (1 + K3K6Kt K3 T;O 7 , + K~ z)- K3 4 2H K3T;O 0 1, + T, 2 k1 K ~ T ; o T ,) - q] ( Tt + K3dO K2K4 , 0 or (8.20) This inequality is easily satisfied for all values of constants normally encountered in power system operation. Note that negative K , is not considered feasible. From (8.20) K , is limited to values greater than some negative number, a constraint that is always satisfied in the physical system. We now recognize the first expression in parentheses in the last term of (8.21) to be the positive constant CY defined in (8.17). Making this substitution and rearranging 324 Chapter 8 to isolate K , terms, we have (8.22) The expressions in parentheses are positive for any load condition. Equation (8.22) places a maximum value on the gain K, for stable operation. 4. c , = q > o Since KIK6 - K2K5 > 0 for all physical situations, we have This condition puts a lower limit on the value of K,. Example 8.3 For the machine loading of Examples 5.1 and 5.2 and for the values of the constants K, through K6 calculated in Examples 6.6 and 6.7, compute the limitations on the gain constant K , , using the inequality expressions developed above. Do this for an exciter with time constant 7, = 0.5 s. Solution In Table 8.3 the values of the constants K, through K6 are given together with the maximum value of K, from (8.22)and the minimum value of K, from (8.23). The regulator time constant 7, used is O S s , 7 j 0 = 5.9s, and H = 2.37s. Case I is discussed in Examples 5 . I and 6.6; Case 2, in Examples 5.2 and 6.5. From Table 8.3 it is apparent that the generator operating point plays a significant Table 8.3. Computed Constants for the Linear Regulated Machine Constants Case I (Ex. 5.1) Case 2 (Ex. 5.2) Kl K2 K3 K4 K S K6 a K2 K3K47t K372cl + Tt K37207t, K2K41aTdo K45 a KS7d0 K47237, 1/7, Kt K. < ’ 1.076 1.258 0.307 1.712 -0.041 0.497 2.552 0.33 1 2.313 0.906 0.143 0.85 1 -0.616 5.051 4.000 - 2.3 269.0 I .448 t.317 0.307 1.805 0.029 0.526 2.552 0.365 2.313 0.906 0. I58 0.949 0.442 5.325 4.000 -3.2 1120.2 4./l(1 + K.5)./VREF can be obtained by inspection.) + ~ ( 7 + d o ) + do7. and K . we usually have T A ~ > T .24) Equation (8. Computer based methods are available to determine the behavior of such systems and are recommended for the more complex cases [20. As shown in Example 8. This change is in the direction to lower the permissible maximum value of exciterregulator gain K. i. Fig.10)-(8.’) (8.22) the left side tends to decrease while the right side tends to increase. more than the other constants. by introducing a lead-lag network with the proper choice of transfer function. > 1. At heavier loads the values of these constants change such that in (8.S + W. than that for the less severe Case 2. See Problem 8.25) where K = K . 2 { w . In [ 1 I ] de Mello and Concordia point out that the same dynamic performance can be obtained with higher values of K . < K 3 ~ j O )I./~RW = K. V. K. but the resulting equations become complicated to the point that one is almost forced to find an alternate method of analysis.14 Block diagram representing the machine terminal voltage at no load. ~ / ~ T . 8.23) does not apply. 0 7 .V* + 2{W. Changes in this latter voltage follow the changes in EFD with a time lag equal to 7A0. The analysis here has been simplified to omit the rate feedback loop that is normally ar! integral part of excitation systems.2 Furtfrer conridemtionr of the regulator gain and time constant At no load the angle 6 is zero.e. One special case of the foregoing analysis has been extensively studied [ I I]. For this condition we can easily show that the machine terminal voltage V.Effect of Excitation o n Stability 325 role in system performance. W: = ( 1 + K. For the problem under study. .n this special case certain simplifications are possible. for good damping characteristics.K. Routh’s criterion is a feasible tool to use to find the limits of stable operation in a physical system. = ( I / T * + 1/7i0). A block diagram representing the machine terminal voltage at no load is shown in Figure 8.4.. From that figure the transfer function for V.K. Rate feedback could be included in this analysis. and the 6 dependence of (8. 211. a reasonable value of { is I a For typical values of the gains and time constants in fast exciters / .S2] . < 8.14. We can show then that for good performance > > T . is the same as the voltage E:. This is usually lower than the value of gain required for steady-state . . This is left as an exercise (see Problem 8. the heavier load condition of Case 1 allows a lower limit for K. the results are dependent upon both the system parameters and the initial operating point..24) can be put in the standard form for second-order systems as v /VREF K/(.3. / 7 . For good dynamic performance. and K . This analysis assumes high regulator gain (K..)/T&T. = (8. performance. > I ) and low exciter time constant > ( 7 . The loading seems to influence the values of K. 13) we compute the torque as a function of angular frequency to be TJ6 = K .31) Note that the damping torque Td will have the same sign as K S . which is reduced by the demagnetizing effect of the armature reaction.30) is approximately given by T.26) is the synchronizing torque component.26). we write the expression for Tc/6as which at a frequency w can be separated into a real component that gives the synchronizing torque T.3 Chapter 8 Effect on the electrical torque The electrical torque for the linearized system under discussion was developed in Chapter 3. is given by TsZZ Kl . From (3.40) the change of the electrical torque with respect to the change in angle is given by _ -K Te - 9 . This point is discussed in greater detail in [ I I]. . A t very low frequencies the synchronizing torque T. .K 2 K s / K 6 (8.13) gives the change in the electrical torque for the unregulated machine as a function of the angle 6. At very low frequencies (8. Equation (3.4. s ) in the numerator is very small compared to the term K 2 K S K . With use of the linear model. . the electrical torque in pu is numerically equal to the three-phase electrical power in pu.J w K 3 ~ i 0 ) 1 (8.27) In the unregulated machine there is positive damping introduced by the armature reaction.-K2 K4 s + (]/re + K5K/K4Te) I t can be shown that the effect of the terms K 2 K 4 (1 + 7 . which is given by the imaginary part of (8. Using this simplification.27).6).32) which is higher than the value obtained for the unregulated machine given by (8. This corresponds to the coefficient of the first power of s and is therefore a damping term. The same relation for the regulated machine is given by (3.326 8. These components are given by (8. and into an imaginary component that gives the damping torque Td. In this case the regulator reduces the inherent system damping. EZ K I .K2KJK4 (8.[K2K3K4/(I+ w 2 K : ~ 2 ) ] ( . I n the regulated machine we may show the effect of the regulator on the electrical torque as follows. This latter quantity can be negative at some operating conditions (see Example 6.26) The real component in (8.40). From (3. 8. while the exciter is represented in detail. 8. In this section a more detailed representation of the exciter is adopted.10)-(8.16. Fig. In Section 7. it reduces the inherent system damping when K 5 is negative. a very simple model of the generator is used. along with the simplified linear model of the synchronous machine that takes into account the field effects.7). To simulate the damping effect of the damper windings and other damping torques.4 the exciter model used is a very simple one.15. The excitation system model used here is similar to that in Figure 7.5 Root-locus Analysis of a Regulated Machine Connected to an Infinite Bus We have used linear system analysis techniques to study the dynamic response of one regulated synchronous machine. The system to be studied is that of one machine connected to an infinite bus through a transmission line.16 Block diagram of the simplified linear model of a synchronous machine connected to an infinite bus with damping added. The signal V .D w is added to the model as shown in Figure 8. Therefore. is the stabilizing signal that can be derived from any convenient signal and processed through a power system stabilizer network to obtain the desired phase relations (see Section 8. a common condition for synchronous machines operated near rated load. This model is shown in Figure 8.3 and is based on the linearized equations (8. In this figure the function G&) is the rate feedback signal.13). whereas the regulator improves the synchronizing forces in the machine at low frequencies of oscillation.54 except for the omission of the limiter and the saturation function S E .15 Block diagram of a linearized excitation system model. The combined block diagram of the synchronous machine and the exciter is given in Figure 8. a damping torque component .17 (with the subscript A omitted for convenience). . 8.Effect of Excitation on Stability 327 I l c r r “1 Fig. In Section 8.8. This model used for the synchronous machine is essentially that given in Figure 6. 17 Combined block diagram of a linear synchronous machine an . 8.KR l + T S R Fig. I8 is solved by linear system analysis techniques. The constants K I through K6 in pu for the operating point to be analyzed are KI = : 1.7.74 s. KE = -0. Bode diagram of the open loop transfer function. For a given operating point we can obtain the loci of roots of the open loop system and the frequency response to a sinusoidal input as well as the time response to a small step change in input.5257 Example 8. The common takeoff point desired is the terminal voltage V . The exciter data = 0. The results of the linear computer analysis are best illustrated by some examples. Root-locus plot.0294 = 0. is usually very small at heavy load conditions). 211.3072 K2 = 1.3174 K4 = 1.4479 KS = 0. = 400. 4.0 pu and 7A0 = 5.9 s.8052 KS K6 = 0. The resulting block diagram is shown in Figure 8.05. and 6 is analyzed for the loading condition of Example 6. and feedback loops to be studied are the regulator and the rate feedback GF(s). D = 2. KR = 1.mKzK3K5 (1 K ~ T A ~ s ) ( W S DS + Kim) . Bode diagram of the closed loop transfer function.4 Use a linear systems analysis program to determine the dynamic response of the system of Figure 8.q K z K & f ~ + (8.18 with and without the rate feedback.0 and T~ = 0. is omitted ( K . Time response of VA to a step change in VREF.95.17. A number of computer programs are available that are capable of solving very complex linear systems and of displaying the results graphically in several convenient ways or in tabular forms [20. . In the analysis given in this section. 2. constants are 2H = 4.Effect of Excitation on Stability 329 KA ]1+T + A I ‘e) & % N(4 vt - I KR 1 + 7 s R - To study the effect of the different feedback loops. are K. 3. This is done by standard techniques used in feedback control systems [22]. The machine = 0.33) Note that the expression for N ( s ) can be simplified if the damping D is neglected or if the term containing K. using the digital computer. The system of Figure 8. The following graphical solutions are to be obtained for the above operating conditions: 1.18. the machine discussed in the examples of Chapters 4. I n that figure the transfer function N ( s ) is given by N(s) = K3K6(2HS2 + DS + Kim) . . we manipulate the block diagram so that all the feedback loops “originate” at the same takeoff point.5. 21 Bode plots of the closed loop transfer function: (a) GF = 0 (b) GF z 0.20 Time response to a step change in V R E F : GF(s)= 0. (b) with rate feedback.Fig. . 8. .17: (a) without rate feedback. (a) Fig. Fig.19 Root locus of the system of Figure 8. 8. Ib) GF(s)# 0. 8. j10. 8.04 .j10.19-8.69170 .ooooo . Compute these graphical displays for two conditions: (a) GI.4 Zeros Poles (a) K F = 0 -0.83244 -0.35020 . The pair that causes instability is determined by the field Table 8. the system dynamic response is dominated by two pairs of complex roots near the imaginary axis. with KF = 0.00000 -0.1.72620 . = 0 (b) G F + 0.72620 -0.19724 + j0.04. Condition Root-Locus Poles and Zeros of Example 8.72620 -0.7.j0.45130 (b) K F = 0.35020 + j10.35021 + j10. With G&) = 0.20.I . I7894 -0.0. I n each figure.19--8.27324 -20. part (a) is for the result without the rate feedback and part (b) is with the rate feedback.(S) = 0 (b) GAS) = sK.69170 -0.j10. and T~ = 1.j10.Effect of Excitation on Stability 33 1 Fig. .22 for the different plots.19724 .40337 .22 Bode plots of the open loop transfer function: (a) GI.27324 -0.4.1.40337 + j10.0 s Solution The results of the computer analysis are shown in Figures 8.17894 -0.45130 -0 21097 .21097 + j10.83244 .20 show clearly that the system is unstable for this value of gain without the rate feedback./(l + T#).0. Note the basic problem discussed in Example 7. Figures 8.00000 .72620 .35021 . 332 Chapter 8 winding and exciter parameters.74)'12 = 10. The rate feedback modifies the root-locus plot in such a way as to make the system stable even with high amplifier gains. z K s / K ~ ) KK The computer output for K 5 = 0 is essentially the same as that of Example 8.2 1-8.13910 + j10.35021 i j10. (a) For the case of D = 0 it is found (from the computer output) that the poles and zeros affected are only those determined by the torque angle loop.5 Repeat part (b) of Example 8.23-8. The damping coefficient D primarily affects the roots caused by the torque angle loop at a frequency near the natural frequency w .23 Root locus of the system of Example 8. The net effect is to move the branch of the root locus determined by these poles and zeros to just slightly away from the imaginary axis.72550 (instead of -0. The second pair of roots. (b) It has been shown that K 5 is numerically small. . These roots occur near the natural frequency w. A very significant point to note about the two pairs of complex roots that dominate the system dynamic response is the nature of the damping associated with them. (b) K S = 0. The examples given i n this section substantiate the conclusions reached in Section 7.22. gives a somewhat lower fre- Fig.73 rad/s. Except for the situations where K 5 becomes negative. determined by the field circuit and exciter parameters. 8. w2n = ( w R / ~ H ) (.7 concerning the importance of the rate feedback for a stable operation at high values of gain. The poles and zeros obtained from the computer results are given in Table 8. .4479 x 377/4.. These poles now become -0.72620). The root-locus plot and the time response to a step change in VREFfor the cases of D = 0 and K 5 = 0 are displayed in Figures 8. = (1.4. its main effect is to change 0 to the value .4 with (a) D Solution = 0 and (b) K 5 = 0.24. The effect of the pair caused by the torque angle loop is noticeable in the Bode plots of Figures 8.5: (a) D = 0.4. Example 8. These effects contribute the electrical torque components designated T. (b) K S = 0.25 Approximate representation of the excitation system..1 Approximate excitation system representation The approximate system to be analyzed is shown in Figure 8.25 where the exciter and the generator have been approximated by simple first-order lags [ 1 I]. The resulting reduced system is composed of two subsystems: one representing the exciter-field effects and the other representing the inertial effects.17 is such that 0) or near steady state (f a). . - - 8. quency and its damping is inherently poor. and T. 8.5: (a) D = 0. and stabilizing signals are often needed to improve the system damping (Section 8. We recognize that the effect of the rate feedback G&) in Figure 8. In this frequency range the system damping is inherently low. The feedback path through K4 provides a small positive damping component that is usually considered negligible [ 1 I].7).6 Approximate System Representation I n the previous section it is shown that the dynamic system performance is dominated by two pairs of complex roots that are particularly significant at low frequencies. 8.Effect of Excitation o n Stability 333 for Fig.6. 8. This is an important consideration in the study of power system stabilizers. it can be neglected at low frequencies (s = j w We have already pointed out that K S is usually very small and is omitted in this approximate model..24 Time response to a step change in VREF the system of Example 8. respectively. A straightforward analysis of this system gives 4 Gx 0) - Fig. Here we develop an approximate model for the excitation system that is valid for low frequencies. is usually small and the system is poorly damped. The damping f. venin equivalent impedance as “seen” at the generator terminals.2. 2 3 .(s) is to monitor the terminal voltage while injecting a sinusoidal input signal at the voltage regulator summing junction [8.(s). connected to the machine under study by a series impedance. The one machine-infinite bus system assumes that the generator under study is connected to an equivalent infinite bus of voltage Vm/cr through a transmission line of impedance z = Re + jX. the gains and time constants may not be precisely known.. 8.36) We are particularly concerned about the system frequency of oscillation as compared to w . The use of these constants. 12. First. The function G. the theoretical model based on the constants K I through K6 is not only load dependent but is also based on a one machine-infinite bus system.6.334 Chapter 8 (8.(s) must be determined either by calculation or by measurement on the physical system..36) to calculate the approximate parameters for G. requires that assumptions be made concerning the proximity of the machine under study with respect to the rest of the system. Therefore.2 Estimate of G. It should be emphasized that this procedure has some serious drawbacks. where & subtracting the drop &Ze is the generator current.(s) The purpose of this section is to develop an approximate method for estimating K Ithrough K6 that can be applied to any machine in the system. These constants can be used in (8. The resulting amplitude and phase (Bode) plot can be used to identify G. . Second.i at the generator terminal node i is known. we must estimate the parameters of G. (8. if the driving-point short circuit admittance E. we assume that z.(s) in (8.6.35) where W . = I / F i The equivalent infinite bus voltage vm calculated by is from the generator terminal voltage cedure is illustrated by an example. Lacking field test data. then. I 2 . This equivalent impedance is assumed to be the The. The pro- .241. (8. is given in Section 8.(s) by calculations derived from a given operating condition. A proven technique for measurement of the parameters of G. and the use of estimated values may give results that are suspect [IO. 2 4 . A procedure based on deriving an equivalent infinite bus.37) E. is the undamped natural frequency and { is the damping ratio: . 2 5 ] .35). ) ] = 1/0.4 We can establish the terminal conditions from the load-flow study of Figure 2. ( X d .j2.)(X.0550 + j0.~ / q~ ) 620 .029" Then we compute from (6. + X.( & 0 . - + jV.0521 ~ i o 2= 6.0 X42 = H2 = 6. ~ ~ .280 = 61.634 12 = f 2 / . + X . Xd2 = = 0.820 = 1. since it is connected to the internal node by xi2.1476 0.1 198 Xq2 = xi2 = 0.789 P U From Table 2. X .592 /-54.339" pu From Figure 5.156"and V2 = V2/& .630 . However.098" Then - - P2 + d2 = 54.2 = (6.025 /-51.6 the following data for the machine are known (in pu and s).156" = f 4 2 = + . = 0.290 pu Neglecting the armature resistance. = 0. using the equivalent infinite bus method outlined above.592/-2.280". can be obtained by using the approximate relation Z .Effect of Excitation o n Stability 335 Example 8.3177 + + = 2.p = 51.2388 = 0.5084 We can compute the infinite bus voltage . z.6 the driving-point admittance at the internal node of generator 2 is given by - Y2.025 1.j Q z ) / h = (1. ~ -( xV 2 f X 2= 1.j1.818 + 9. ) = 3.806 pu jZ.420 . ) ( X .jxi2.j0.932 .272 .19: 1 2 k & = = 1.2450/77. Solution From Example 2.724 pu The terminal voltage node of generator 2 had been eliminated in the reduction process.066)/1.8645 0. = 0. = 0.8958 0. r 0.39925 1 + K .X . Note that the three-machine system is certainly not considered to have an infinite bus. = 620 51.56) KI 1/K3 = = pu K3 = 1/[RZ (x.1969 0.6. = l/Fz2. The exact reduction process gives z.j0.6 t a n ( ~ 5 ~P 2 ) = X .2 + J1.P2 + &) = 1..818" = V. through K6 for generator 2 of Example 2.6 Compute the constants K . and the results might be expected to differ from those obtained by a more detailed simulation. ~ ~ 2 1 4 2 = Xd21dz Eqoo = E20 = v42 5 2 - = 1.749 PU 1.818" 2 But from the load flow P2 = 9. 3898 $. From Example 3.ZJ.I .E.4) calculate the excitation control system phase lag.O)Ix.318 KS = 0.R.I .2450 /77.592 /6.475 K = 3.027 K6 = 0.92 to convert to the machine base. This is not valid for low frequencies.o + I. . Lacking such measurement.)']l = 3.4 Hz or w.0 x 0.siny (xi + X.2226 = 0.95 s. The constants K.0)/(2 x 5.4750 K.x.) Solution From (8.Next Page 336 Chapter 8 - v.CY = 61. + X./a = v 2 - 0.)cosy + R.028 Iw2 + j0.507 Note that these constants are in pu on 100-MVA base whereas the machine is a 192MVA generator. = d( jw) = I .93) = 5.12.967 x 0. For the system natural frequency (see Example 3.(-12.0.661 . It is judged that the latter is important at the low frequencies of interest.064 K1 = 2.941") V.029")( I .j0.. erroneously.4 the dominant frequency of oscillation is approximately 1. It is sometimes assumed.35): 5.0265 (K.{R.8 rad/s. = d(0. Compute the parameters of G.094 2 K4 = 2.)[(x. 1.~qo[ReCOSy .x.o)K.3898/..0640 + X.)cosy] + I.(s) is obtained by substituting s = j w in the denominator of the first expression in (8.0 x 0. I761 + j0.914" The angle required in the computations to follow is y = 620 K.(xd.95 + 0.x q VdO[(x.)siny .967 rad/s (0.7 The exciter for generator 2 of the three-machine system has the constants K.ag = tan-' (0.. + X.132 and the excitation system is poorly damped.318 x 6.[RZ + (x.O/KO)[I K. = 0.0443~ At the frequency of interest ( w = 8.098 .(s).)siny . We use the regulator gain and the exciter time constant. At any frequency the characteristic equation of G.914) = 74.318 x 6.R. This point is a source of some confusion in the literature.Vm/V.(0.siny]l = 0.o(x. I76 I ) = I6 1.507 x 400)/(6.02519.)[(x.. = 400 and r .x. = = = .95) = 0.xqRe = 0.cosyll K6 = (V.9706 .012" K.Vm(Eqao[R. a judgment is made as to which parameters should be used.280" . z 8.(xq + XP)I .(xq + Xe)sinyI .(VO/V. (Here again we emphasize the need for actual measurement of the system parameters.0941 VmK. Example 8. that the regulator time constant is to be used when the excitation system is represented by one time constant.8 rad/s) we have d( jwosc) = . = = Kz = K4 = KS = .36) we have w. + X. and K2 should be divided by 1.5070 Summary: KJ = 0.cosy] = 2.9958 I. + 2. with u = 8. give curves ofphase shift as a function of normalized frequency. U 02 .17. Many textbooks on control systems.6. such as [22]. 8.26(b) that the phase lag is great.7 is rather large.7). I 06 ..26.967 = 1. 1 ' 0 -1 5 ' -30 -45 -60 s -75 f = damping ratio j -90 .03 0. (a) 0 4 06 .8/5. In the above example. The excitation system phase lag in Example 8. For the case where damping is present.Previous Page Effect of Excitation o n Stability 337 12 6 c = domping ratio . it is apparent from Figure 8. as shown in Figure 8.06 01 .26 Characteristics of a second-order transfer function: (a) amplitude. (b) phase shift. .040. and rx is small. I 1 I I 3 6 I 1 0 I 30 I 60 !OO I U (b) Fig. For small damping the phase changes very fast in the neighborhood of w.Bo ' 4 >- f Q " -18 2 3 4 5 00 .47 and 5 = 0. (where ding = 90").13.060.1 I I I 03 . 8. and phase compensa> tion is likely to be required (see Section 8..02 0. . The phase lag is large because oorc w.1 0.3 The inertial transfer function The inertial transfer function can be obtained by inspection from Figure 8. u = w/w.l5 -o -120 -135 -150 -165 -180 0.01 I 0. 0 0 2 1 4 ~ + At the system frequency of oscillation w = = w.27. 8.31 ) indicates that the voltage regulator introduces a damping torque component proportional to K 5 . For example. = + 0..0183/(-0. To offset this effect and to improve the system damping in general. Use D = 2 pu. O. 0 1 3 7 ~ ~J 0 .1 Block diagram of the linear system We have previously established the rationale for using linear systems analysis for the study of low-frequency oscillations. Such an arrangement is shown schematically in Figure 8. .= re2 Chapter 8 -6 s2 + = = wR/2H = D K ~ w R s2 -s + . d ( j w ) = 1 .39) The damping of the inertial system is usually very low.2 H 2H + ~J.38) Where onis the natural frequency of the rotating mass and 5.6)..156s + 72. = 0. I/KIwR/ZH D/4Hwn = D / 2 d 2 H K I w R (8. Example 8.. in the excitation system shown in Figure 7.4.6 we compute d ( s ) = s2 w.538 rad/s {.‘ wR/2H (8. 8 x 2.3 that under heavy loading conditions K 5 can be negative.0.894 = = 8. We have also shown in Section 8.. We noted in Section 8. tan-’ [0.009 8. Solurion From the data of Examples 2. the signal usually obtained from speed or a related signal such as the frequency.0222 .8 Compute the characteristic equation.3” 8. {.s+ u.0604)] = 163.7.0 . and the damping factor of the inertial system of generator 2 (Example 2.7 Supplementary Stabilizing Signals Equation (8.54 the stabilizing signal is indicated as the signal K .338 . For any generator in the system the behavior .w. is processed through a suitable network to obtain the desired phase relationship. the undamped natural frequency...975 x 377)”2] .8 rad/s. Stabilizing signals are introduced in excitation systems at the summing junction where the reference voltage and the signal produced from the terminal voltage are added to obtain the error signal fed to the regulator-exciter system. These are called “supplementary stabilizing signals” and the networks used to generate these signals have come to be known as “power system stabilizer” (PSS) networks. artificial means of producing torques in phase with the speed are introduced.6. Thus it can often be assumed that the voltage regulator introduces negative damping. T o illustrate. = 2 / [ 2 ( 1 2 . These are the situations in which dynamic stability is of concern.2 that the excitation system introduces a large phase lag at low system frequencies just above the natural frequency of the excitation system.6 and 8. is the damping factor. 40) is a reset term that is used to “wash out” the compensation effect after a time lag 7 0 .40) The first term in (8. In terms of a Bode or frequency analysis (see [22]. This is difficult to show quantitatively in the complete system because of its complexity. The second term in G. with typical values of 4 s [ I I] to 20 or 30 s [ 121. Examining Figure 8.6 for an approximate method to determine these constants) but may be considered constant for small deviations about the operating point. for example) the system is likely to have inadequate phase margin.27 Schematic diagram of a stabilizing signal from speed deviation. requiring a slight modification of the inertial transfer function using standard block diagram manipulation techniques.6 and the results obtained in that section.36) and (8.2 Approximate model of the complete exciter-generator system Having established the complete forward transfer function of the excitation control system and inertia. through K 6 are load dependent (see Section 8.(s) is a lead compensation pair that can be used to improve the phase lag through the system from VREF to u. 8. This block diagram is shown in Figure 8.0pu.28. 8. we may now sketch the complete block diagram as in Figure 8.. The PSS is shown here as a feedback element from the shaft speed and is often given in the form [ I I ] (8. w.0--3. The system time constants. We therefore take advantage of the simplified representation developed in Section 8. and w. The damping constant D is usually in the range of 1.39) respectively. Qualitatively. Thus the PSS network must provide lead compensation. can be conveniently characterized and the unit performance determined. . are defined in (8.7.29. we can recognize the existence of a potential control problem in the system of Figure 8. We note that a common takeoff point is used for the feedback loop. The parameters rx.Effect of Excitation on Stability 339 Fig. cn. the power system stabilizer must compensate for much of the inherent forward loop phase lag.28 due to the cascading of several phase lags in the forward loop. and inertia constants are obtained from the equipment manufacturers or by measurement. We also note that the output in Figure 8. The use of reset control will assure no permanent offset in the terminal voltage due to a prolonged error in frequency. gains.29 we can see that to damp speed oscillations. at the power system frequency of oscillation.29 is the negative of the speed deviation. from the linear block diagram of that generator. such as might occur in an overload or islanding condition. The constants K . 8.L ”1 r 1 1 Fig.28 Block diagram of a linear generator with an exciter and power syste . 30(b)..30(a).41) The transfer function has the pole zero confguration of Figure 8. UL T GSk) 8. Ei (0) (b) Fig. where the zero lies inside the pole to provide phase lead.7. + 7 ~ S ) [ 1+ ( T c + T D ) S ] + (?A + TB)S (8. 8..Effect of Excitation on Stability Kz K/ Ten Pt2C x x u s t o = x - Kd r=t2fwrto= n n n --w 341 AU Fig. the transfer function of this circuit is . = stabilizing circuit time constant < rC < RB/(RA + RB) RD/(Rc + R D ) Eo/Ej = Approximately.= Ei (i/a)(i + aTSj 1 + 7s where (8.29 Block diagram of a simplified model of the complete system.3 Lead compensation One method o providing phase lead is with the passive circuit of Figure 8. then ( 1 -t TAS)/(I + TCS) = (I + U T S ) / ( ~+ T S ) (8.“ I: . = T~ T~ = = = = T~ Kl Kz = K I R C I = lead timeconstant R IC I = noise filter time constant < T. For this simple network the magnitude of the parameter a is usually limited to about 5 . 8. For this circuit we compute E= -O I (1 E. < K z R C z = lag time constant R C .3 I [26].43) wherea = K l C I / K 2 C 2 1. (b) pole zero configuration.30 Lead network: (a) passive network. > .42) where T. Another lead network not so restricted in the parameter range is that shown in Figure 8. . f I f loaded into a high impedance. 45) tan+.o(l/aT*) = log.a7 . ~ ) ]= = .48) Now. height (a .1 ) / 2 f i (8.45) From trigonometric identities (tan x - tan y ) / ( I + tan x tan y ) W.o(l/T4i) Then w. ~ ) ] = tan -lw. occurs at the geometric mean of the corner frequencies. u ~ ) ( w .W.y ) = = x A - y (8.e. . = arg[(l + jw.46) Therefore. = I / T f i (8.1)/(1 + aw. asymptotic approximation is illustrated [22].44) The magnitude of the maximum phase lead 4 is computed from . using (8.uT)/(l + j w .1) and hypotenuse 6 . = (w.46) in (8. i. 8. The maximum phase lead 4 occurs at the median frequency w.tan-lw.a ~ . where the .44) to compute tan#. where w.. Ioglowm = (1/2)[loglo(l/a7) + Io ~ I o ( I /T ) I = (1/2)log. Fig. 6 .~’) (8.T(U (8.31 Active lead network. I I I db I I I I/. For any lead network the Bode diagram is that shown in Figure 8.32 Bode diagram for the lead network ( I + UTS)/( I + TS) where a > I . (a . 8.47) This expression can be simplified by using (8.. visualizing a right triangle with base 2 f i .T tan (x .342 Chapter 8 R Fig.T)/[I + ( w .32. Thus = I/w.7.50) These last two expressions give the desired constraint between maximum phase lead and the parameter a.8" = a = (1 + sin 80.Effect of Excitation on Stability 343 wecompute bZ = (a .40) as G. so that the stabilizer output will never dominate the terminal voltage feedback.9 Compute the parameters of the power system stabilizer required to exactly compensate for the excitation control system lag of 161. and is usually field adjusted for good response. we assume that w.. Example 8.9) = 9..(s) = [K~T~s/(I + T O ~ ) l [ (+ a T s ) / ( ' l + i = T s ) ~ (8.037 UT = 0..T. say 0. In many practical cases the phase lead required is greater than that obtainable from a single lead network.3488 Thus G.48 This is a very large ratio.6/2 = 80. and it would probably be preferable to design the compensator with three lead stages such that 4..9'.50) = 161. Then a = (I + sin 53. S o h ion Assume two cascaded lead stages.I)* + 4a = (a + = I)*or sin& (a .(s) = [K. Prepare a table showing the phase lead and the compensator parameters as a function of a. = 53. Solurion As before.G= 0.49) This expression can be solved for a to compute a = (I + sin&)/(l .I)/@ + I) (8.44) T 8.037~)]' A suitable value for the reset time constant is r0 = IO s. . The gain KO is usually modest [26]. determines the time constant T from (8.8 rad/s.s/(~ + TOS)][(~ + 0.. I n this case two or more cascaded lead stages are used.I From (8.28..9)/( I - sin 53. as shown in Figure 8.. = which is a reasonable ratio to achieve physically..8)/( I .10 Assume a two-stage lead-compensated stabilizer.. Then the phase lead per stage is $ .42 = w.6"computed in Example 8. The procedure then is to determine the desired phase lead qjm. Thus we often write (8. The natural frequency of oscillation of the system is w.1 < KO < 100.sin&) (8. = 8..8 rad/s.50)..349~)/(1+ 0. from (8. Knowing both a and the frequency w.8) 154. It is also common to limit the output of the stabilizer.51) where n is the number of lead stages (usually n 2 or 3).sin 80.44). EJcample 8. This fixes the parameter a from (8. 05 64.935 2. 3.272 I .80 122.62 109.5.83 34.79 67.81 54.6 -4 Rea I -2 7 Rea I (a 1 (b) Fig. Furthermore. Example 8.32 and Problem 8.4401 0. The results of the linear computer analysis are best illustrated by an example.783 2.760 These results show that for a large a or large 4.0508 0.28 is solved by linear system analysis techniques using the digital computer (see Section 8. the corner frequencies wHi and wLo must be spread farther apart than for small 6. (a) no power system stabilizer and (b) a two-stage lead stabilizer with a = 25: 0 .28: 1. 8.254 1 0. 8.3593 0.344 Chapter 8 Table 8..0359 0.08 39. 4.33 Root locus of the generator 2 system: (a) no PSS.1 1. In this section the system of Figure 8.76 0.0254 0. .00 0.90 61. l / U r n 6 WHi = 117 a7 WLO = ]/a7 5 10 15 20 25 41. a Lead Compensator Parameters as a Function of a ?'= 9m 29 . Root-locus plot Time response of wA to a step change in VREF Bode diagram of the closed loop transfer function Bode diagram of the open loop transfer function.5). 2.5082 0.8 .68 27.5682 3.0227 19.58 134. compute these graphical displays for two conditions.38 83.968 I .0293 0.35 44. I I Use a linear systems analysis program to determine the following graphical solutions for the system of Figure 8. (b) with the PSS having two lead stages with a = 25. See Figure 8.8 linear Analysis of the Stabilized Generator I n previous sections certain simplifying assumptions were made in order to give an approximate analysis of the stabilized generator.10 129. Effect of Excitation on Stability i 345 I Fig. thereby increasing the damping. (b) with the PSS having two lead stages with : a = 25.35 Frequency response (Bode diagram) of the closed loop transfer function: (a) no PSS.34 Time response to a step change in V R E F(a) no PSS. The results are shown in Figures 8. The root locus shows clearly the effect of lead compensation and has been used as a basis for PSS parameter identification [27]. 8.36 for the four different plots. (b) with the PSS having two lead stages with a = 25.28 except that the PSS limiter cannot be represented in a linear analysis program and is therefore ignored.0227~)]* The system constants are the same as Examples 8.33-8. forcing the locus to loop to the left and downward. In the root-locus plot (Figure 8. Note that \ 6) Fig.8.7 and 8. 8. . Solution The system to be solved is that of Figure 8. G ~ ( s = [ IOS/( I ) + IOS)] [( I + 0 .33) the major effect of the PSS is to separate the torque-angle zeros from the poles. 5 6 8 S ) / ( I + 0. part (a) is the result without the PSS and part (b) is with the PSS. In each figure. 33(a)].467 rad/s.000 -0.j0.1.000 -45.289 .452 + j8. (b) with the PSS having two lead stages with u = 25.000 + j24.500 .010 + jO.500 + jO.OOO + jO.000 -20. the locus near the origin is unaffected by the PSS.944 + j0.533 . but the locus breaking away vertically from the negative real axis moves closer to the origin as compensation is added [this locus is off scale in 8.000 + jO.102 + jO. in Figure 3.100 + jO..000 -0.533 -0.439 -0.. Table 8 6 Root-Locus Poles and Zeros .1 .3. 8.36 Frequency response (Bode diagram) of the open loop transfer function: (a) no PSS.959 -0.j7.34 shows the substantial improvement in damping introduced by the PSS network.6.538 rad/s computed in Example 8.000 + jO.j8.346 Chapter 8 la) (b) Fig. From this table we note that the natural radian frequency of oscillation is controlled by the torque-angle poles with a frequency of 8.000 0.955 -0.955 + j7. From the computer we also obtain the tabulation of poles and zeros given in Table 8. Note the slightly decreased frequency of oscillation in the stabilized response.959 -0.955 .289 + j8.439 -45.000 -0.289 .533 .000 . Figure 8.000 0.8 using the approximate model and also checks well with the frequency of b..j8.955 -0.100 + jO.467 -0. This agrees closely with w.847 -45.941 .289 + j8.000 -0. = 8.j8.944 .847 .452 .000 + jO.533 -0.467 -0.000 -0.j0.I79 + jO. Condition Poles Zeros No PSS WithPSSa = 25 -20.j24.jO.000 -0.179 + jO.941 + j0.000 -0.000 -45. . 9 for the three excitation systems described in Table 7.18. . With the generator equipped with a W TRA exciter. and W Low T & Brushless.7... Lead compensation improves the phase substantially in this region.37.69 and 8.. The analog computer representation of the excitation system is shown in Figure 8.8 is extended to include the effect of the excitation system. Potentiometer Calculations for a Type 1 Representation of a W TRA Exciter ( a = 20) 0.37. With a variety of compensating schemes available to the designer and with each having many adjustable components and parameters. in Figure 5 .. W Brushless.9.36 show the frequency response of the closed loop and open loop transfer functions respectively. The synchronous machine used is the same as in the examples of Chapter 4 with the loading condition of Example 5.. .35 and 8.8.7. The results with W Brushless and W Low T~ Brushless exciters are shown in Figures 7. the response due to a increase in T.. The new "free" inputs to the combined diagram are VREF and T.. and the terminal marked u.38. IO00 lim 800 .1 Effect of the rate feedback loop in Type 1 exciter As a case study. Example 5.8O00 0.. Note that the output of amplifier 614 (Figure 8. V R m a x= 3. Saturation is represented by an analog limiter on VR in this simulation.8.37) connects to the terminal marked E. .5L' ... 8.8 and 8 . Three IEEE Type I exciters (see Section 7.. . .18. .. The potentiometer settings for the analog computer units are given in Tables 8. thus improving gain and phase margins. 8.5 pu = 3. The parameters for these exciters are given in Table 1.Effect of Excitation on Stability 347 Figures 8. ..5pu = -3. in Figure 5..5 u VRmin = -3. this can be done using the complete nonlinear model of the synchronous machine. 8.. Table 8..1) are used in this study: W TRA. .9 Analog Computer Studies The analog computer offers a valuable tool to arrive a t an optimum setting of the adjustable parameters of the excitation system.1 are shown in Figure 8.. comparative studies of the effectiveness of the various schemes of compensation can be conveniently made. The uncompensated system has a very sharp drop in phase very near the frequency of oscillation..and 5% change in VREFand the phase plane plot of wA versus aA for the initial loading condition of Example 5. .9. This system is added to the machine simulation given in Figure 5.. I8 connects with switch 4 2 1 in Figure 8.. I . . .39 respectively.4994 0. 800 . . Furthermore.. 8.Fig.37 Analog computer representation of a Type I excitation syste . out potentiometer Calculations for a Type I Representation of a W Brushless Exciter ( a = 20) Int.. gain I I 100 Pot.7217 Comparing the responses shown in Figures 8.0 KF/+F = 0.1000 0. Amp.0625 0..08 I / u 7 1/20 = 0.5033 0.5 0.0625 I/TF= 1/1.15 0. set. The dynamic response comparison is based on observing the rise time.. I no. .5 I / a r E = 1/(20)(0. .00 0..0 I / u = 1/20 = 0.02) = 1000 I / a s A = I/(20)(0.5000 0. VREF vREF VR so REF io0 50 5'!. R ' 50 I 0.8) = 0. IO 50 I/sF= 1/0.5 = 2.00 10.00 0. At the same time. v.. by comparing the dynamic response due to changes in the mechanical torque T.25 0..39 we can see that the steady-state conditions are reached sooner with the exciter present.5 \/(IT€ = I ::EI ::: I 20.02 KA/arA= 400/(20)(0..2000 0. . .03/1.1 1.5.1 0.6671400 = = 1. .69. . .9.0 = IO I I I I I I 801 801 IO 50 50 100 50 IO 100 10 50 802 802 810 V.2 Effectiveness of compensation A detailed study of the effectiveness of four methods of compensation is given in (281.. and the change in electrical power output PeA is observed.9. The terminal voltage.02) 1000 X00 VR -EFD -EFD V.0066 Pot.20.00 1. -V.. . 600 601 800 701 601 601 800 800 801 801 50 50 I REF REF 100 100 0.05 l / V T = 0.69. . KF/rF= 0. we note that without the exciter the slow transient is dominated by the field winding effective time constant.10 . 40 2. .00 1. in a given transient. .OfP6Ol I + 2. 803 v. and 8.50 5". 0 1 P60 I I + 2.0625 KE/arE= 1/(20)(0.00 1.5033 0. and percent overshoot of either PeA or V.0 0.1000 0.50 5.96 U ~ ~ ~ = -6.0252 0. . .0252 0..1 IO IO0 801 VR -EFD I IO 1/(20)(0.6250 0. v.V. .1 0.0066 0.50 5.00 VR -EFD I 10 IO K A / a r A= 400/(20)(0. .. .1000 0. I I 1. From Figures 8.5773 0. the response is more oscillatory.96 PU = 6.96 .50 0. vR VR -EFD 0.3333 = = 703 802 810 812 803 lim 800 . . settling time.1500 . Table 8.3333 0. a 10% increase in the reference torque is made. 7.5 I I 100 0.3333 802 802 810 803 800 v..015) 33.4000 I 1 0. and the reference excitation voltage VREF at various machine loadings.0 = 1.50 0.38.. 7. 8.0 2.E ~ D 1. .25 0.03 . the field flux linkage.0 -EFD V. 100 -EFD V. set.39 with that of Figure 5. I out VREF VREF VR constant 1.72 I 7 . 601 601 800 701 800 L~ 50 I I IO In REF L~ 100 L ~ / L ~ 0.00 I / a r A = I/(20)(0.7217 0.2500 0.2000 0.Effect of Excitation on Stability 349 Table 8. .5773 V VRmin 100 50 40 2. .. v. .04/0.7217 LJ .. For example.38. .. (L~'Li)C no.8. -vv V.50 C = constant cap. Pot.02 1. Potentiometer Calculations for a Type 1 Representation of a w Low 7 E Brushless Exciter ( a = 20) C' = I I Amp. and the rotor angle are slow in reaching their new steady-state values. .3333 0.6250 0. .8) = 0.0625 0. 0. no.0 1.96 PU = -6.4 and 8.50 0.02) = 2. Amp. 50 50 V. 0.. .015) 3.333 = 3.00 10.667/400 = I -v. V. . R = 6.5 = 0.0252 0.05 I / Y 3 = 0. .. and 8.5033 20. 1..2500 0.02) = 2. The machine data and loading are essentially those given in the Examples 8.00 KE/arE 1/(20)(0.0 2. . h E Fig. generator equipped with U m C . 8.- .38 System response to a step change in 7. and V R ~ ~ . 8.Fig.... and VRE.39 System response t o a step change in T. generator equipped with a . 86 = 7p = 7R = K R = = 0.0 s 7. two-stage lead-lag and speed Power system stabilizer 0.0 5 .6 0.04 SEmax 0.04 0.22 0.O 100. = = 21 rad/s 2 r = 0.2 I 86.42 10.50 VRmax 8.20 0.10 for the initial operating condition of Tm0 = 3.352 Chapter 8 However.0 1.015 s KF = 0.3).10.22 0. The data of the exciter are: KA 7E = TA = 400 PU 0. the frequencies of these oscillations are usually on the order of 0. When growing oscillations occur in large interconnected systems.2 s r2 = 0.56 0.28 0.0 = 0. 1.3 Hz.)/(s2 + nons + u.2-0. This is particularly valuable in studies to improve the system damping. Rise time Settling time Overshoot ”/.02 s 0.05 0.with other frequencies superimposed upon them.05 s 0.05 s A sample of data given in reference [28] is shown in Table 8.0 0.1 Power system stabilizer: C 7 = 3.0 SE27Smin K. Table 8. .37 0. The excitation system used is Type 2.26 = VRmin= -8.9.06 0.85 PF lagging.0 0.05 73. Case Rise time Comparison of Compensation Schemes Settling time Overshoot ”/.0 pu at 0.0 33. the machine is fully represented on the analog computer. Thus it is important to know the dynamic response of the synchronous machine under these conditions.4 82. Uncompensated Excitation rate feedback Bridged-T only Bridged-T. Other valuable information that can be obtained from analog computer studies is the response of the machine to oscillations originating in the system to which the machine is connected.0 60. s + w.20 0.2 I 0.) w. = 0. This can be simulated on the analog representation of one machine connected to an infinite bus by modulating the infinite bus voltage with a signal of the desired frequency. a rotating rectifier system (see Section 7.2 I 0.23 5-10 Source: Schroder and Anderson (281.23 0.06 0.6 80.98 0.26 = EFDmax = 4.45 The methods of compensation used are: 7Fs) Rate feedback: sKF/(1 Bridge-T filter with transfer function: + C/R n = (s2 + m w .60 4. The impedance diagram of the system (to 100-MVA base) and the prefault conditions are shown in Figures 2.98 peak. When the stabilizer is reinstated at point C . is shown here. and C are represented by .19 respectively.02 and 0.18 and 2. At point D the modulation is removed.40 shows the effect of the PSS under these conditions. but operating under the heavy loading condition of Example 5. Figure 8.. causing growing oscillations to build up especially on P. In this study the loads A . has its bus voltage modulated by a frequency of one-tenth the natural frequency.1 0 Digital Computer Transient Stability Studies To illustrate the effect of the excitation system on transient stability. The transient is initiated by a three-phase fault near bus 7 and is cleared by opening the line between bus 5 and bus 7.10. The generator data are given in Table 2. The PSS is removed at B.1 rad/s is added. B.1.40 Synchronous machine with PSS operating against an infinite bus whose voltage is being modulated at one-tenth the natural frequency of the machine. At time A the modulating signal of 2.Effect of Excitation o n Stability 353 Fig. A sample of this type of study.1.. transient stability studies are made on the nine-bus system used in Section 2. The modulating signal varies the infinite bus voltage between 1. the oscillations are quickly damped out. 8. 8. The same machine discussed above. which would simulate tie-line oscillations. Note also that the frequency of these oscillations is near the natural frequency of the machine. taken from [28]. 0 Note: See Figure 8.o 0.05 0....044 0. the model used for the synchronous machine is the so-called "one-axis model" (see Section 4..60 1. 3.5 ..15 0. I .5 -3..78 0. .00I6 I .62 1.555 . For generator 2.1. 1.. ..465 1 . Excitation Systems Data Parameter Amplidyne MagA-Stat SCPT -0.02 1 . . (It is based on a program developed by the Philadelphia Electric Co.o -1. constant voltage behind transient reactance..95 0. generators I and 3 are represented by classical models.Jdf 7 - 1 constant impedances. an additional value EA is added due to saturation Table 8.05 0....06 0.35 0. .o . When the machine EMF E (corresponding to the field current) is calculated...0039 0.50 0.o 120 1 .20 0.2 2.2 ..41 for BBC exciter parameters..o 400 -0. A modified transient stability program was used in this study. Le.19 2.11. provision is made for the excitation system representation.0805 25 .354 Chapter 8 (1 + TA1)(l t ?$) (1 t O - K. Kz 7 A I)(1 t- K? Kzo T I ) + + F e .17 0.4) with provision for representing saturation.0 I .o 0 0.04 I . with modifications to include the required new features.) When the excitation system is represented in detail...15. . 0 pu response.0.3 Time.141)].Effect of Excitation on Stability 355 l 120. .U al f . the results of generator 2 data are shown in Figures 8. 3 O C 110- . The types of field representation used with generator 2 are: 1. amplidyne NAlOl exciter (see Figure 7.50. This is given by EA = A. IEEE Type I .66).1 Effect of fault duration Two sets of runs were made for the same fault location and removal. 5.47-8.5 pu response. 4.0 pu response (see Figure 7. 8.(El .61). 2.3 v d w B d 90- // 1 ---- --- BBC exciter -constant E Type 1 . Similar results for a six-cycle fault are shown in Figures 8. IEEE Type 3.5 Fig. Mag-A-Stat exciter (see Figure 7. 2. Classical model. 8.0.61).1 1.10. For a three-cycle fault. The breaker clearing times used were three cycles and six cycles. and BEare provided for several exciters [see (4. SCPT fast exciter. 3.52) The constants A. effect and based on the voltage behind the leakage reactance E l .46.41).2 821 0. 0 RR Classical model - I 0 I 0.5 RR 80- P - Mag-A-Stat BBC exciter 2 .42 for various exciters with a three-cycle fault.42-8.100t . Brown Boveri Company (BBC) alternator diode exciter (see Figure 8.0. but for different fault durations.1 I 0.exp[B. The excitation system data are given in Table 8. 2.4 I 0. IEEE Type 1.8)] (8. I I 0. The action of the exciter and the armature reaction effects are clearly displayed in Figure 8.-0. We note. From Figure 8.6ium 0 0 -.0- 0. Figure 8. a slow IEEE Type 1 exciter. however.1 I 0. Again it can be seen that the different models give essentially the same power swing for this generator. 8. 0 response. while all the others give the position of the 9 axis.. a large swing was obtained.5 0.2 Chapter 8 1.. Results with three-cycle fault clearing.42 shows a plot of the first swing of the angle d. Note that the classical run gives the angle of the voltage behind transient reactance..44.4 I 0. Figures 8. The effect of the armature reaction is dominant in this period.) and minimum variation of the terminal voltage after fault clearing. A run with constant E F D is also added.43 V.42 that for a three-cycle clearing time the classical model gives approximately the same magnitude of a.43 we conclude that the slow exciter gives the nearest simulation of a constant flux linkage in the main field winding (and hence constant E.4 / / I 0 1 0. The plot shows that the first swing is the largest. for different field representations. and E. In Figure 8.. for the first swing as the different exciter representations.2.8 = -2--z---- ---<----_---- ./ e .45 shows a time plot of P2 for this transient..46 seem to indicate that for this fault the system is well below the stability limit. since the magnitude of the first swing is on the order of 60".46 the rotor angle d2. with the subsequent swings slightly reduced in magnitude. is hardly affected by the value of E. I I 0. that the minimum swing is obtained with the slow exciter while the maximum swing is obtained with the classical model.356 1. as seen by the EMF E. however. We conclude from the results shown in Figure 8. All generator .E$ 2 0.3 Time.2 I 0. Figure 8.6 Fig.5RR Type1 ----/ I >* 0. It is interesting to note that the actual field current. and a relatively fast exciter with 2 . When the exciter model was adjusted to give constant E F D . is plotted for a period of 2.BBC exciter .0 RR Mag-A-Stat -_.42-8. for most of the duration of the first swing after the fault is cleared. for various exciters with a three-cycle fault.0 s for the classical ~ ~ model. 3 1 0. The magnitude of the first swing for the cases where the excitation system is represented in detail is significantly larger than for the case of the classical representation. For the case of a six-cycle clearing time. We also note that the classical model does not accurately represent the generator response for this case. . the plot of the angle d2.4 I 0. The effect of the power system stabilizer on the response is hardly noticeable until the second swing. The Type 1 exciter gives the highest swing.Effect of Excitation on Stability 357 4 3 Q Y W a n Lu B 2 1. and the system is perhaps close to the transient stability limit. Fig. Here the swing curves for the generator with different field representations are quite different in both the magnitude of swings and periods of oscillation. 2 models give approximately the same magnitude of rotor angle and power swing and period of oscillation.44 EFD and E for various exciters with a three-cycle fault.2 Time.1 I 0. The swing curves indicate that this is a much more severe fault than the previous one.47 for the classical model and for two different types of exciter models. is shown in Figure 8. 8.0pu response exciter is pronounced after the first swing.46 and 8. I 0.5 0 . I I 0. The effect of the 2. we note that for this severe fault the rotor oscillation of generator 2 depends a great deal on the type of excitation system used on the generator. Comparing Figures 8.47. Results with six-cycle fault clearing. 8.5 Fig. 0.45 Output power P2 for various exciters with a three-cycle fault. Fig.1 02 . .358 Chapter 8 C h i c a l model BBC exciter 2.4 0. I 03 .46 Rotor angle 621 for various exciters with a three-cycle fault.5 RR - \ - t B im 80 0 0. 8. Time.0 RR Mag-A-Sa t Type 1 0. 2 0. The generator acceleration will thus decrease. 1 s and stays fairly constant thereafter. Thus after the first voltage dip.8 2.47 Rotor angle 621 for various exciters with a six-cycle fault.0pu exciter and the Type 1 exciter respectively. It would appear that for slightly more severe faults the classical model may predict different results concerning stability than those predicted using the detailed representation of the exciter.8 lime.2 I 1.0 Fig. 0. and for the duration of the first swing it is fairly constant.0 pu exciter.48 for different exciter representations. follow the changes in the field voltage EFDwith a slight time lag. The first swing after the fault seems to be dominated by the inertial swing of the rotor. the swings in V.6 1.0 pu exciter than with the Type 1 exciter. and E. Figures 8. with the action of the exciter dominating the subsequent swings in y . The output power of generator 2 is shown in Figure 8.49 and 8.50 show plots of the various voltages and EMF’S of generator 2 for the case of the 2. The oscillations of terminal voltage V. Again the recovery of the terminal voltage is faster with the 2.4 06 . 8. the flux linkage in the main field winding (reflected in the value of E:) drops only slightly (by about 5%). The curves for E: show that although the fault is near the generator terminal. recovers slowly and continues to increase steadily. respocse. While the general shape of these curves is the same. The excitation system increases the output power of the generator after the first swing.4 I 1. I I 1. This effect is not noticed in the classical model. some significant differences are noted. The faster recovery occurs with the 2. . reaches a plateau at about I .Effect of Excitation on Stability 359 I 0 I I 1 I 0. For the Type 1 exciter E. We also note that the excitation system introduces additional frequencies of oscillation.0 I 1. are somewhat complex. which appear in the V. causing the rotor swing to decrease appreciably. for example.7s. The component of E due to the armature reaction seems to be dominant because the field circuit time constant is long. 8. In the first 0. Since the initial rotor swing is largely an . the changes in E . however.10. is due to the effects of both E . The general shape of the EMF plot. for a more severe fault or for studies involving long transient periods. the excitation representation is not critical in predicting the system dynamic responses. and the armature reaction.360 Chapter 8 The plots of E clearly show the effect of the armature reaction. However. it is important to represent the excitation system accurately to obtain the correct system dynamic response. are reflected only in a minor way in the total internal EMF E. the PSS is helpful in damping oscillations caused by large disturbances and can be effective in restoring normal steady-state conditions. From the data presented in this study we conclude that for a less severe fault or for fast fault clearing.2 Effect of the power system stabilizer For large disturbances the assumption of linear analysis is not valid. However. I Fig.9 i P 3. >i.10 P 10 .I.5 R R exciter.w e 0. 1.50 Voltages of generator 2 with Type I 0. . 0.9 d P 0. 8.Effect of Excitation on Stability 361 4.0 - - : a Llw 0.10 $ 2.8 I Time.7 . Fig.8 . Y b W I a e m u 1. 8. -3.49 Voltages of generator 2 with BBC exciter. 2 1. 4 -3 Y -B Y 2 a e - s. 8. however. = Fig.5 20 . s I I 1. some transient stability runs are made for a threephase fault near bus 7 applied at t = 0.25 I 05 .52 Exciter voltage EFD with and without a PSS.25 05 .51 Torque angle 621 for a three-phase Fault near generator 2.5 15 .oo Time.0 I 1.0167 s ( 1 cycle) and cleared by opening line 5-7 0 0.9 rad/s.75 1 .25 15 .362 Chapter 8 I5 2 I Ii 0 I I!II I I I I 0.75 I 2 . wOEC 8. the effect of the stabilizer is quite pronounced.oo Time.0 I 0.0 07 . I 12 .1 Fig. the stabilizer has little effect on this first swing. On subsequent oscillations. PSS with a = 25.0 17 .5 1 . . To illustrate the effect of the PSS. 8. inertial response to the accelerating torque in the rotor. OS 4..5 1 and 8.52. .6” 3. For exciters to perform this function.10s (6 cycles). warns that “we cannot expect to continue indefinitely to compensate for increases in reactance by more and more powerful excitation systems. they need high gain. shown in Figure 8. Note that this exciter is not particularly fast (RR = O S ) .. has been . however. The discussion in Section 8.. . = a.. Generator 2 is equipped with a Type 1 Mag-A-Stat exciter with constants similar to those given in Table 8. Stability runs were made with and without the PSS..12 (a = 25) with a limiter included such that the PSS output is limited to *O.1 1 Some General Comments on the Effect of Excitation on Stability In the 1940s it was recognized that excitation control can increase the stability limits of synchronous generators. defined as 6 . of Second Swing a 25 16 Limit = +0.52. If the excitation system is slow and has a low response ratio. is interesting. However. Regarding the dynamic performance. From the plot of d. The phase of E. The improvemew in the angle A.4” 8. A comparison of several runs is shown in Table 8.51 note that while the change in the first peak (due to the PSS) is very small. and the voltage E. This results in a delayed E. which tends to limit the downward 6 excursion by retarding the building in T.8” 5... optimistic results . .7” 5.pss).. data for the angle 6.. In the first place. for more severe transients or for transients initiated by faults of longer duration. The above summarizes the situation regarding the so-called steady-state stability or power limits. Also. This is felt to be important in view of the trends toward higher capacity generating units with higher reactances.lOpu.10 seem to indicate that for less severe transients. ramp as 6 swings downward. modern excitation systems play an important part in the overall response of large systems to various impacts. (no pss) . From the stability runs.3 and the studies of Section 8. The comparison in E. 6 . investigated for different PSS parameters.6. changes when the PSS is applied to produce a field voltage that is almost 180” out of phase with &. both in the so-called transient stability problems and the dynamic stability problems.12. and the response tends to be a ramp up and then down. the effect of modern fast excitation systems on first swing transients is marginal. Modern exciters are faster and more powerful and hence allow for operation with higher series system reactance. I t is found that this angle improvement is sensitive to both the amount of lead compensation and to the cutoff level of the PSS limiter. Improvement at Peak .. an accurate representation of the field flux in the machine is needed. Series compensation makes it possible to have a high dc gain and at the same time have lower “transient gain” for stable performance. for faults near the generator terminals it is important that the synchronous machine be modeled accurately. are taken with and without the PSS. Table8.. Another way to look at the same problem is to note that fast excitation systems allow operation with higher system reactances. The PSS constants are the same as in Example 8.. these modern exciters can have a more pronounced effect. Concordia [ 171. the improvement in the peak of the second swing is significant.10 Limit = *O. in Figure 8.” A limit may soon be reached when further increases in system reactance should be compensated for by means other than excitation control. The results are displayed in Figures 8.1 1.Effect of Excitation o n Stability 363 at t = 0. if the transient study extends beyond the first swing.12. 54. 13. may be obtained if the classical machine representation is used. Transient studies are frequently run for a few swings to check on situations where circuit breakers may fail to operate properly and where backup protection is used.5.05 pu. 9. approximate methods of analysis can be used to obtain preliminary design data. I f such measurements are not possible. The parameters of the PSS for a particular generator must be adjusted after careful study of the power system dynamic performance and the generator-exciter dynamic response characteristics. For example. 23. As shown in Section 8. This has been repeatedly demonstrated in computer simulation studies and by field tests reported upon in the literature [8. 8. the effect of excitation system compensation on subsequent swings (in large transients) is very pronounced. These signals must be introduced with the proper phase relations to compensate for the excessive phase lag (and hence improve the system damping) at the desired frequencies [ 321.05. The excitation system used is the Brown Boveri exciter shown in Figure 8. while the effect on the magnitude of the first swing was hardly noticeable. but in the cases where the system exhibits negative damping characteristics. Usually the PSS parameters are optimized over a range of frequencies between the natural mode of oscillation of the machine and the dominant frequency of oscillation of the interconnected power system. Supplementary signals to introduce artificial damping torques and to reduce intermachine and intersystem oscillations have been used with great success. to obtain these characteristics. As indicated in Section 8. causing eventual loss of synchronism. and the swing curves obtained with and without the PSS (for the same fault) are shown in Figure 8. in a stability study conducted by engineers of the Nebraska Public Power District. This is not too surprising in large interconnected systems with numerous modes of oscillations. 30. 1 3 = 0. field measurements are preferred.6. Large interconnected power systems experience negative damping at very low frequencies of oscillations. Ked = 0. with provision for the adjustment of the PSS parameters to be made on the site after installation. . limit = j~0.10. the voltage regulator usually aggravates the situation by increasing the negative damping. 1 5 = 0. = IO. I t is not unlikely that some of the modes may be superimposed at some time after the start of the transient in such a way as to cause increased angle deviation. Their effect on damping torques are small.53 Block diagram of the PSS for the BBC exciter with a 2.41.53. 1 2 = 0.364 Chapter 8 Shaft speed Terminal volta +=@a = Fig.0 RR: KQI = Kp3 T . 311. the effect of the PSS on damping the subsequent swings was found to be quite pronounced. Voltage regulators can and do improve the synchronizing torques.5. 29. 14 = 0. I t should be mentioned that several transients have been encountered in the systems of North America where subsequent swings were of greater magnitudes than the first.05. The PSS used is shown schematically in Figure 8.KQz = I. using the numerical data given in Example 8. Repeat the determination of stable operating constraints developed in Section 8. These equations are solved for the gain constants. 33-39].14). Whether the stabilizing signal derived from speed provides the best answer is an open question. Here signals derived from the various “states” of the system are fed back with different gains to optimize the system dynamic performance. Most of these studies concentrate on the use of a signal derived from speed or frequency deviation processed through a PSS network to give the proper phase relation to obtain the desired damping characteristic.) Recently many studies have been made on the use of various types of compensating networks to meet different situations and stimuli. It would seem likely that the principle of “optimal control” theory is applicable to this problem.19. What is the order of the system? Use block diagram algebra to reduce the system of Problem 8. with the following assumptions (see [ I I]): .54 Effect of the PSS on transient stability. Determine the open loop transfer function for the system of Problem 8. This approach seems to concentrate on alleviating the problem of growing oscillations o n tie lines [ 1 I .4 Construct a block diagram for the regulated generator given by (8. This optimization is accomplished by assigning a performance index.1.3. Problems 8. 8. This index is minimized by a control law described by a set of equations.1 8. arranged as in Figure 7. However. This subject is under active investigation by many researchers [40-441.3 8.2.10)-(8. 26.1 to a feed-forward transfer function KG(s) and a feedback transfer function H ( s ) .Without PSS i n operation d 0 With PSS in operation Time. 13.2 8. 30. (Obtained by private communication and used with permission.Effect of Excitation on Stability 165 365 . 24. in a large interconnected system it is possible to have a variety of potential problems that can be helped by excitation control. for (a) Case 1 and (b) Case 2. Find the upper and lower limits of the gain K. cycles Fig. 14.4. . Byerly. 2):834-44. Hunkins. IO Gs 8. and Hattan. PAS-88:316-29. I E E E Trans.24). A I E E Trans. Damping for the northwest-southwest tieline oscillations-An analog study. 7. In (8. PAS-87:1306. 5.6. and a Bode plot for Example 8.. McClymont. Hattan. E.7 8. Goodwin. Proc. E.. J.. 12. I E E E Trans.13 8. W. Am. (b) A triple lead compensator with Q = IO.. C.16 where w. J.5 + + 8. D. Hanson..10. H. B.4. A.12 8. Long distance power transmission. R. 1969. H.3. Dynamic stability of the Peace River transmission system.8 8.3. Karas.15 8. F. de Mello. Influence of excitation and speed control parameters in stabilizing intersystem oscillations. Skooglund. using the numerical constants K . D. F. F.. W. I E E E Trans. and White. 0.14 8..13. T. A I E E Trans. C o m p u t e the constants K . L.1 I with .2 that a higher value of regulator gain K. A.14 represents the machine terminal voltage a t n o load. 1966. n = 2 a n d r = 0. shown in Figures 8. and & > 7 . F o r each of the cases > in Example 8. PAS-85:586-600.9 8. Effect of high-speed rectifier excitation systems on generator stability limits. 1944. 1967.. J. 1969. IO. I E E E Trans. J.366 Chapter 8 Recompute the gain limitations.. Dandeno. and Dandeno. 69. and Gish. IEEE Trans. F. 1950. F. J. / K 3 . Hardy. Schleif. ( a ) A dual lead compensator with a = 15.29 for a stabilizing signal processed through a bridged Tfilter: 8. I E E E Trans. W. and Td a s functions of w between w = 0. PAS-63:215-20. and Blythe.1 I = (s2 + mw. Analyze the system in Figure 8. L. W.10. 11. K.05 s. Modify the block diagram of Figure 5 .. PAS-85: 1239-47. Concepts of synchronous machine stability as affected by excitation control. and Concordia. E.31) assume that K6K. Schleif. Blythe. E. Schleif.6 8. G.. L. Schleif. and Angell. Perform a transient stability run. time response t o a step change in V. R. F... W. >> I/K3. P..1 rad/s and w = 10 rad/s (use semilog graph paper). R. choose 71 a n d sen. N. and Watson.. I8 showing the analog computer simulation of the synchronous machine to allow modulating the infinite bus voltage. plot T. A. through K6 given in Table 8.. With the help of the field voltage equation (vF = rFiF AF). 1968. = 0. Martin. H.. 6. Hunkins. It is stated in Section 8. Plot E and a s functions of time and comment on these results. Control of rotating exciters for power system damping: Pilot applications and experience.s + w:)/(s2 + nuns + w i ) . T h e s domain equation for y / VREFis given by (8. If the transfer function of such a network is (1 such that the value of the gain can be increased eight times. . Steady-state stability of synchronous machines as affected by voltage regulator characteristics. PAS-86:438-42. 4. H. discuss the plots of EFD. Crary S. Sketch root loci o f e a c h case. Sketch Bode diagrams of the several lead compensators described in Example 8. 1966. 1967.1. M. a n d E. L. (Pt.42 shows a larger swing than the other field representation. 6. PAS-87:3 15-22. M. Use a linear systems analysis program (if o n e is available) t o compute root locus. I E E E Trans.. PAS-87: 1426-34. 8. R. IEEE Trans.43 and 8. 2. R.7 if a low time constant exciter is used where K.. and Keay. R. C.44. T h e block diagram shown in Figure 8.30) and (8. Ellis. C o m p u t e the open loop transfer function of the system of Figure 8. Field tests of dynamic stability using a stabilizing signal and computer program verification. Damping of system oscillations with a hydrogenating unit. 9. E. and Skooglund. W. = 400 and 7. 8.28 both with and without the stabilizer. Shier. PAS-88: 1259-66. E. Control of generator excitation for improved power system stability. R. 1968. R. Excitation control to improve power line stability. using a computer library program to verify the results of . P. Determine the excitation control system phase lag of Example 8.. G. Martin. through K6 for generator 3 of Example 2. PAS-87: 190-201. R.. P. 1968. Section 8. E. 3. Power ConJ 29:lOll-1022. + References I . Explain why the curve for constant EFD in Figure 8. 1968. Concordia. E. can be used if a suitable lead-lag network is chorls)/(l q s ) .. C. is the natural frequency of the machine. I E E E Trans. Effect of Excitation o n Stability 13. Klopfenstein. A. 367 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. Experience with system stabilizing controls on the generation of the Southern California Edison c o . /€€€ Trans. PAS-90:698-706, 1971. de Mello. F. P. The effects of control. Modern concepts of power system dynamics. IEEE tutorial course. IEEE Power Group Course Text 70 M 62-PWR. 1970. Young. C. C. The art and science of dynamic stability analysis. IEEE paper 68 CP702-PWR, presented at the ASME-IEEE Joint Power Generation Conference, San Francisco, Calif., 1968. Ramey, D. G.. Byerly, R. T., and Sherman, D. E. The application of transfer admittances to the analysis of power systems stability studies. / E € € Trans. PAS-W.993-Ip00, 1971. Concordia, C.. and Brown, P. G. Effects of trends in large steam turbine generator parameters on power system stability. / € € E Trans. PAS-90221 1-18. 1971. Perry. H. R.. Luini. J. F.. and Coulter, J. C. Improved stability with low time constant rotating exciter. /€€€ Trans. PAS-902084-89. 197 I . Brown. P. G.. de Mello. F. P., Lenfest. E. H., and Mills. R. J. Effects of excitation. turbine energy control and transmission on transient stability. / E € € Trans. PAS-89 1247-53. 1970. Melsa, J. L. Cottipurer Progranis for Cottiputatiowl Assistance in the Studr of' Linear Control Theory. McGraw-Hill. New York, 1970. Duven. D. J. Data instructions for program LSAP. Unpublished notes, Electrical Engineering Dept., Iowa State University. Ames. 1973. Kuo. Benjamin C. Autonraric Control Systettrs. Prentice-Hall. Englewood Cliffs. N .J., 1962. . Gerhart. A. D., Hillesland. T.. Jr.. Luini. J. F.. and Rocktield, M. L . Jr. Power system stabilizer: Field testing and digital simulation. / € € E Trans. PAS-902095-2101, 1971. Warchol, E. J., Schleif. F. K., Gish, W. B. and Church, J. R. Alignment and modeling of Hanford excitation control for system damping. / E € € Trans. PAS-90714-25, 1971. Eilts. L. E. Power system stabilizers: Theoretical basis and practical experience. Paper presented at the panel discussion "Dynamic stability in the western interconnected power systems" for the IEEE Summer Power Meeting, Anaheim, Calif., 1974. Keay. F. W.. and South, W. H. Design of a power system stabilizer sensing frequency deviation. / € E € Trans. PAS-90707-14. 1971. Bolinger. K.. Laha. A.. Hamilton. R., and Harras. T. Power stabilizer design using root-locus methods. / E € € Trans. PAS-94: 1484-88. 1975. Schroder. D. C.. and Anderson, P. M. Compensation of synchronous machines for stability. IEEE paper C 73-3 13-4, presented at the Summer Power Meeting, Vancouver, B.C., Canada. 1973. Bobo. P. 0.. Skooglund, J. W., and Wagner, C. L. Performance of excitation systems under abnormal conditions. I€€€Trans. PAS-87547-53, 1968. Byerly. R. T. Damping of power oscillations in salient-pole machines with static exciters. / E € € Trans. PAS-89:1009-21. 1970. McClymont. K . R., Manchur. G.. Ross, R. J., and Wilson, R. J. Experience with high-speed rectitier excitation systems. /€€€ Trans. PAS-87: 1464-70. 1968. Jones. G. A. Phasor interpretation of generator supplementary excitation control. Paper A75-437-4, presented at the IEEE Summer Power Meeting. San Francisco, Calif.. 1975. El-Sherbiny. M . K.. and Fouad. A. A. Digital analysis ofexcitation control for interconnected power systems. / E € € Trans. PAS-90441 -48. 1971. Watson. W.. and Manchur. G. Experience with supplementary damping signals for generator static excitation systems. /€€E Trans. PAS-92: 199-203. 1973. Hayes. D. R.. and Craythorn. G. E. Modeling and testing of Valley Steam Plant supplemental excitation control system. /€€€ Trans. PAS-92:464-70, 1973. Marshall. W. K.. and Smolinski. W. J. Dynamic stability determination by synchronizing and damping torque analysis. Paper T 73-007-2. presented at the IEEE Winter Power Meeting, New York. 1973. El-Sherbiny. M. K., and Huah. Jenn-Shi. A general analysis of developing a universal stabilizing signal for different excitation controls, which is applicable to all possible loadings for both lagging and lerding operation. Paper C74-106-1. presented at the IEEE Winter Power Meeting, New York. 1974. Bayne. J. P.. Kundur, P.. and Watson. W. Static exciter control to improve transient stability. Paper T74-521-1, presented at the IEEE-ASME Power Generation Technical Conference, Miami Beach, Fla.. 1974. Arcidiacono. V.. Ferrari. E., Marconato. R.. Brkic,T., Niksic, M.. and Kajari. M. Studies and experimental results about electromechanical oscillation damping in Yugoslav power system. Paper F75-460-6 presented at the IEEE Summer Meeting. San Francisco, Calif., 1975. Fosha. C., E.. and Elgerd. 0 . I. The megawatt-frequency control problem: A new approach via optimal control theory. / E € € Trans. PAS-89563-77. 1970. Anderson, T H.The control of a synchronous machine using optimalcontrol theory.Proc. IEEE-5925-35, . 1971. Moussa, H. A. M., and Yu. Yao-nan. Optimal power system stabilization through excitation and/or governor control. / E € € Trans. PAS-91: 1166-74. 1972. Humpage, W. D., Smith, J. R . and Rogers, G . T. Application of dynamic optimization to synchro. nous generator excitation controllers. Proc. /€€(British) 120:87-93. 1973. Elmetwally, M. M.. Rao. N. D. and Malik. 0 . P. Experimental results on the implementation of an optimal control for synchronous machines. / € € E Trans. PAS-94: I 192-1200. 1974. chapter 9 Multimachine Systems with Constant Impedance Loads 9.1 Introduction In this chapter we develop the equations for the load constraints in a multimachine system in the special case where the loads are to be represented by constant impedances. The objective is to give a mathematical description of the multimachine system with the load constraints included. Representing loads by constant impedance is not usually considered accurate. It has been shown in Section 2.1 1 that this type of load representation could lead to some error. A more accurate representation of the loads will be discussed in Part I11 of this work. Our main concern here is to apply the load constraints to the equations of the machines. We choose the constant impedance load case because of its relative simplicity and because with this choice all the nodes other than the generator nodes can be eliminated by network reduction (See Section 2.10.2). 9.2 Statement of the Problem In previous chapters, mathematical models describing the dynamic behavior.of the synchronous machine are discussed in some detail. In Chapter 4 [see (4.103) and (4.138)] it is shown that each machine is described mathematically by a set of equations of the form ir = ~ ( X , V T,,t) , (9.1) where x is a vector of state variables, v is a vector of voltages, and T, is the mechanical torque. The dimension of the vector x depends on the model used. The order of x ranges from seventh order for the full model (with three rotor circuits) to second order for the classical model where only w and d are retained as the state variables. The vector v is a vector of voltages that includes u d , uq, and up If the excitation system is not represented in detail, uF is assumed known; but if the excitation system is modeled mathematically, additional state variables, including up, are added to the vector x (see Chapter 7) with a reference quantity such as V, known. In this chapter ,, we will assume without loss of generality that uF is known. Consider the set of equations (9. I). In the current model developed in Chapter 4, it represents a set of seven first-order differential equationsfor each machine. The number of the variables, however, is nine: five currents, w and 6, and the voltages ud and uq. Assuming that there are n synchronous machines in the system, we have a set of 7n differential equations with 90 unknowns. Therefore, 2n additional equations are 368 Multimachine Systems with Constant Impedance Loads 369 needed to complete the description of the system. These equations are obtained from the load constraints. The objective here is to derive relations between udi and uqi,i = 1, 2, . . . , n, and the state variables. This will be obtained in the form of a relation between these voltages, the machine currents i9i and i d i , and the angles d i , i = I , 2, . . .,n. In the case of the flux linkage model the currents are linear combinations of the flux linkages, as given in (4.124). For convenience we will use a complex notation defined as follows. For machine i we define the phasors and 5 as - Vi = Vqi + j Vdi Ii - = Iqi + jIdi (9.2) where (9.3) and where the axis qi is taken as the phasor reference in each case. Then we define the complex vectors and f by v (9.4) Note carefully that the voltage and the current 8 are referred to the q and d axes of machine i . I n other words the different voltages and currents are expressed in terms of different reference frames. The desired relation is that which relates the vectors andT. When obtained, it will represent a set of n complex algebraic equations, or 2n real equations. These are the additional equations needed to complete the mathematical description of the system. v 9.3 Matrix Representation of a Passive Network Consider the multimachine system shown in Figure 9.1. The network has n machines and r loads. It is similar to the system shown in Figure 2.17 except that the machines are not represented by the classical model. Thus, the terminal voltages y., i = I , 2, . . . , n, are shown in Figure 9.1 instead of the internal EMF’S in Figure 2.17. Since the loads are represented by constant impedances, the network has only n active sources. Note also that the impedance equivalents of the loads are obtained from the pretransient conditions in the system. By network reduction the network shown in Figure 9.1 can be reduced to the n-node network shown in Figure 9.2 (see Section 2.10.2). For this network the node currents and voltages expressed in phasor notation are 4, &, . . . , and V,, 6 , . . . , Vn respectively. Again we emphasize that these phasors are expressed in terms of reference frames that are different for each node. At steady slate these currents and voltages can be represented by phasors to a com- < 370 Chapter 9 rn c IFig. 9.1. Multimachine system with constant impedance loads. mon reference frame. To distinguish these phasors from those defined by (9.2). we will use the symbols ii and vi, i = 1. 2, . . . , n, to designate the use of a common (network) frame of reference. Similarly, we can form the matrices i and 6. From the network steady-state equations we write (9'3) where and is the short circuit admittance matrix of the network in Figure 9.2. 9.3.1 Network in the transient state -"I ... V" (9.6) Consider a branch in the reduced network of Figure 9.2. Let this branch, located between any two nodes in the network, be identified by the subscript k. Let the branch +n la--- * ', 1 t -2 -1 + va ' n *O _____I, Multimachine Systems with Constant impedance Loads 371 resistance be rk, its inductance be t k , and its impedance be T,. The branch voltage drop and current are vk and i,. I n the transient state the relation between these quantities is given by v k = tk;k + rkik k = 1,2, ..., b (9.7) where b is the number of branches. Using subscripts abc to denote the phases abc, (9.7) can be written as vablk = kiahrk + rkiabck k = 1, 2, . . ,b (9.8) This branch equation could be written with respect to any of the n q-axis references by using the appropriate transformation P. Premultiplying (9.8) by the transformation P as defined by (4.5), Vabrk = t k p iobrk + rkp iabr& (9.9) Then from (4.3I ) and (4.32) jabr = ;Odq - [-:I (9. IO) Substituting (9.10) in (9.9) and using (4.7). (9.1 I ) which in the case of balanced conditions becomes (9.12) I t is customary to make the following assumptions: (1) the system angular speed wR and (2) the terms 4; are does not depart appreciably from the rated speed, or w negligible compared to the terms u t i . The first assumption makes the term @&(& approximately equal to x k i k ,and the second assumption suggests that the terms in ik are to be neglected. Under the above assumptions (9.12) becomes (9.13) Equation (9.13) gives a relation between the voltage drop and the current in one branch of the network in the transient state. These quantities are expressed in the q-d frame of reference of any machine. Let the machine associated with this transformation be i. The rotor angle Oi of this machine is given by ei = oRt + 7r/2 + ai (9.14) where ai is the angle between this rotor and a synchronously rotating reference frame. 372 Chapter 9 Reference frame at synchronous speed) Fig. 9.3. Position of axes of rotor k with respect to reference frame From (9.13) multiply both sides by l / d ; and using (9.3), v ki q () rktqk(iJ - xktdk(ij Vdk(i) = rktdk/iJ + xktqk(iJ (9.15) where the subscript i is added to indicate that the rotor of machine i is used as reference. Expressing (9.15) in phasor notation, - ‘kliJ = = ‘qk(iJ +j - ‘dk(il Xktdk(iJ) (rktqk(iJ + ‘j(rktdkliJ + XktqkliJ) = (rk + jxk)([qk -k jtdk) or (9.16) Equation (9.16) expresses, in complex phasor notation, the relation between the voltage drop in branch k and the current in that branch. The reference is the q axis of some (hypothetical) rotor i located at angle bi with respect to a synchronously rotating system reference, as shown in Figure 9.3. 9.3.2 Converting to a common reference frame To obtain general network relationships, it is desirable to express the various branch quantities to the same reference. Let us assume that we want to convert the phasor = Vqi + j Vdito the common reference frame (moving at synchronous speed). Let the same voltage, expressed in the new notation, be = VQi + jVDi as shown in Figure 9.4. From Figure 9.4 by inspection we can show that < VQi + j VDi = (VqiCOS bi - vdi sin S i ) + j( Vqisin bi + Vdi COS Si) or pi = vejai (9.17) Now convert the network branch voltage drop equation (9.16) to the system reference frame by using (9.17). pke-j*i = 2 k k i e-jai 1.2. ..., b (9.18) or pk = zkjk k = where b is the number of branches and 2, is calculated based on rated angular speed. Comparing (9.18) and (9.5) under the assumptions stated above, the network in the transient state can be described by equations similar to those describing its steady-state Multimachine Systems with Constant Impedance loads 373 Vdi - -- 4 Fig. 9.4. T w o frames of reference for phasor quantities for a voltage Vi. behavior. The network (branch) equations are in terms of quantities expressed to the same frame of reference, conveniently chosen to be moving at synchronous speed (it is also the system reference frame). Equation (9.18) can be expressed in matrix form v b = ?.bib (9.19) where the subscript b is used to indicate a branch matrix. The inverse of the primitive branch matrix 4 exists and is denoted sib, thus ib = y b v b (9.20) Equation (9.20) is expressed in terms of the primitive admittance matrix of a passive network. From network theory we learn to construct the node incidence matrix A which is used to convert (9.20) into a nodal admittance equation i where A = = (A'ybA)v yv (9.21) v is the matrix of short circuit driving point and transfer admittances and [a,,] = 1 if current in branch p leaves node q = -1 = if current in branch p enters node 9 0 if branch p is not connected to node 9 (9.22) withp = 1,2, . . . . b a n d q Since V-' Z exists, = 1 ,2,..., n. v = v-lj 2 zi (9.23) where z is the matrix of the open circuit driving point and transfer impedances of the network. (For the derivation of (9.21)-(9.23), including a discussion of the properties of the and matrices, see reference [ I ] , Chapter I I .) v z 9.4 Converting Machine Coordinates to System Reference Consider a voltage V,bei at node i. We can apply Park's transformation to this voltage to obtain vdqi. From (9.2) this voltage can be expressed in phasor notation as y , using the rotor of machine i as reference. I t can also be expressed to the system reference as vi,using the transformation (9.17). 374 Chapter 9 Equation (9.17) can be generalized to include all the nodes. Let eJ6I = ;e.:; 1' 0 ... ::: ... (9.24) (9.25) Then from (9.2), (9.14), (9.17), and (9.25) V = TV (9.26) Thus T is a transformation that transforms the d and q quantities of all machines to the system frame, which is a common frame moving at synchronous speed. We can easily show that the transformation T is orthogonal, Le., T-1 Therefore, from (9.26) and (9.27) Similarly for the node currents we get = T* (9.27) V = T*3 (9.28) i 9.5 = Tf I = T*P (9.29) Relation between Machine Currents and Voltages From (9.22) f = YV. By using (9.29) in (9.22), T f = VTV Premultiplying (9.30) oy T - ' (9.30) I where = (T-IYT)V & FiV M 2 (T-IYT) (9.31) (9.32) and if R-l exists, - = (T-lvT)-'i = (T-'ZT)i V (9.33) Equation (9.33) is the desired relation needed between the terminal voltages and currents of the machines. It is given here in an equivalent phasor notation for convenience and compactness. It is, however, a set of algebraic equations between 2n real Id17 Id,,. voltages VqI, v d l , . . . , V n , Vdn, and 2n real currents IqI, . . . ,Iqnr Example 9. I Derive the expression for the matrix for an n-machine system. Multimachine Systems with Constant Impedance Loads 375 Solution The matrix of the network is of the form (9.34) and from (9.24) (9.35) From (9.34) and (9.35) y , ,,J(@ll+ a , ) YI2,J(@12+62) ... y In eJ(@in+6n) eJ(82n+6n) 2n VT = y2,,;(e21 +61) 22 eJVz~+6~i ... ... yn l e J ( @ n i + 6 ~ ) yn2e J(un2+62) ... ... . . . Y,, e J ( U , + ~ , ) 1 ... y and premultiplying by T - ' , we get the desired result To simplify (9.36), we note that yikeJ(@ik-6ik) = (Gi, cos 6 , + Biksin 6 , ) + j(Bikcos 6, Gik COS aik - Giksin d i k ) Now define FG+B(dik) FB-G(Bik) = FG+B = = FB-G + Biksin 6, (9.37) (9.38) = Bikcos dik - Gi, sin Then the matrix %Tis given by - M = H + j S where H and S are real matrices of dimensions ( n x n ) . Their diagonal and offdiagonal terms are given by hii Example 9.2 = Gji hik = FG+B(6jk) = Bii sik = FB-G(6i&) (9.39) Derive the relations between the d and q machine voltages and currents for a twomachine system. 376 Chapter 9 Solution From (9.3 I ) and (9.38) = ( H V , - SVd) + j ( S V , + HVd) (9.40) For a two-machine system the q axis currents are given by and the d axis currents are given by We note that a relation between the voltages and currents based upon (9.33) (i.e.. giving V , , . V,,, Vdl, and Vd2in terms of I q l .I,,, Idl. and I d , ) can be easily derived. I t would be analogous to (9.40) except that the admittance parameters are replaced with the parameters'of the matrix of the network. z Example 9.3. Derive the complete system equations for a two-machine system. The machines are to be represented by the two-axis model (see Section 4.15.3), and the loads are to be represented by constant impedances. Solution The transient equivalent circuit of each synchronous machine is given in Figure 4.16. A further approximation, commonly used with this model, is that x i x i 2 x'. The network is now shown in Figure 9.5. The representation is similar to that of the are not constant. classical model except that in Figure 9.5 the voltages E; and The first step is to reduce the network to the "internal" generator nodes 1 and 2. Thus the transient generator impedances rl + j x ; and r, + j x ; are included in the net= ESI j E j l and E; = Ei2 jEj2, worky ( or Z ) matrix. The voltages at the nodes are and the currents are 6 = IqI + j I d l and & = I,, + j f d z . The relation between them is + + + Fig. 9.5. Network of Example 9.3. Multimachine Systems with Constant Impedance Loads Multimachine Systems with Constant Impedance Loads 377 377 given equation similar The equations for each given by an equation similar to (9.40). The equations for each machine, under the assumption that x x the two equations sumption that x ii zz x;,i , are the two axis equations of Section 4.15.3. (9.41) (9.41) and (9.41) completely describe the replaced with Equations (9.40). with system. Each machine represents a fourth-order system, with state variables E& Eii, wi. and 6,. The complete system equations are given by z', (9.42) The system given by (9.42) is nor an eighth-order system since the equations are not independent. This system is actually a seventh-order system with state variables E i , , E ; , , Ei2, Ei2. w , , w 2 , and 6,2. The reduction of the order is obtained from the last two equations a,, = WI - w2 Furthermore, , 7 = D2/rj2 D / T (or Furthermore, if damping is uniform; i.e.. if D 1 / r jj,l = D2/rj2 = D/7i ~ if damping is not present) then the system is further reduced in order by one, and the two torque equations equations can be combined in the form 9.6 System Order In Example 9.3 it was shown that with damping present the order of the system was reduced by one if the angle of one machine is chosen as reference. It was also pointed system out that if damping is uniform, a further reduction of the system order is achieved. We now seek to generalize these conclusions. We consider first the classical model with zero transfer conductances. We can show that the system equations are given by 378 Chapter 9 TJlbr D,w, = + 2 I- El~Bll(sin6; sin6,) - I Jfl 6,=w, i = 1 , 2 ,..., n (9.43) where the superscript s indicates the stable equilibrium angle. Defining the state vector x, the vector I J , and the function f by x' = Iw,, (Jz, &(ak) = . . . % ( 6 , - a;),(& - 6 % . . . ,(a" - 691 k = l , 2 , ...,m E,,E,B,,,(sin(u, + 6iq) - sin6;J 3 m = n(n - 1)/2 and u = C x where C is a constant matrix. The system (9.43) may then be written in the form i = A X - Bf(a) (9.44) where A and B are constant matrices. The order of the system (9.44) is determined by examining the transfer function of the linear part (with s the Laplace variable) W(S) = C(s1 - A)-' B (9.45) This has been done in the literature [2, 3). Expanding (9.45) in partial fractions and examining the ranks of the coefficientsobtained, the minimal order of the system is obtained. It is shown that the minimal order for this system is 2n - I . For the uniform damping case, i.e., for constant D,/T,,, the order of the system becomes 2n - 2 (see also (41). The conclusions summarized above for the classical model can be generalized as follows. If the order of the mathematical model describing the synchronous machine i is k,, i = I , 2 , . . . ,n , and if damping terms are nonuniform damping, the order of the system is (E?=ki - 1). However, if the damping coefficients are uniform or if the damping terms are not present, a further reduction of the order is obtained by referring all the speeds to the speed of the reference machine. The system order then becomes (2?=1 - 2). kj The above rule should be kept in mind, especially in situations where eigenvalues are obtained such as in the linearized models used in Chapter 6. Unless angle differences are used, the sum of the column of 6's will be zero and a zero eigenvalue will be obtained (see Section 9.12.4). 9.7 Machines Represented By Classical Methods In the discussion presented above, it is assumed that all the nodes are connected to controlled sources, with all other nodes eliminated by Kron reduction (see Chapter 2, Section 2.10.2). The procedure used to obtain (9.31) assumes that all the machines are represented in detail using Park's transformation. For these machines we seek a relation, such as (9.3 I). between the currents y and the voltages 9 . The former are either among the state variables if the current model is used, or are derived from the state variables if the flux linkage model is used (see [5]). If some machines are represented by the classical model, the magnitudes of their internal voltages are known. If machine r is represented by the classical model, the angle 6, for this machine is the angle between this internal voltage and the system reference axis. In phasor notation the voltage of that node, expressed to the system refer- Multimachine Systems with Constant Impedance Loads 379 ence, is given by V, = VQ, + j VD, = E, cos 6, + j E , sin 6, (9.46) At any instant if 6, is known, VQrand V,,, are also known. Since the voltage E, is considered to be along the q axis of the machine represented by the classical model, we can also express the voltage of this machine in phasor notation as V, = E, + j0 r = 1,2, ..., c (9.47) Also from where c is the number of machines represented by the classical model. (4.93) on a per phase base Pe* = V,id + uqiq pu Dividing both sides by three changes the base power to a three-phase base and converting to stator rms equivalent quantities. divides each voltage and current by fl, Thus we have pe = bfd + pu(3d) and using (9.47), Per = IqJr ~ ~ ( 3 4 ) (9.48) Note that E, is in per unit to a base of rated voltage to neutral. Assuming that the speed does not deviate appreciably from the synchronous speed, P, and from the swing equation (4.90) on a three-phase base then T, h = r (I/~jr)(Tm, - EJqr) - (Dr/Tjr)wr 8, = wr - 1 (9.49) A machine r represented by the classical model will have only w, and 6, as state variables. I n (9.49) E, is known, while I,, is a variable that should be eliminated. To do this we should obtain a relation between I,, and the currents of the machines represented in detail. Similarly the voltages Vgi and bi of the machines represented in detail should be expressed in terms of the currents f q i and fdj of these machines and the voltages E, of the machines represented classically. To obtain the above desired relations, the following procedure is suggested. Let m be the number of machines represented in detail. and c the number of machines represented by the classical model; Le., A m + c = n Let the vectorsTand v be partitioned as A = - V = (9.50) E,, + j0 R.380 Chapter 9 Then from (9.lI R.51) can be rearranged with the aid of matrix algebra to obtain [.. MI..iv21M~liv12 I..37) and using Fl2 = Y2.5 1) M1 I 2 M*2 E]=[ R . -M.52) is the desired relation between the voltages of the machines represented in detail along with the currents of the machines represented classically... Solution From (9.--_--------- FiZIR.-z-]E. (9. = -6. We note that the matrices Mil. .54) (9.] MI1 a 1 2 (9.lmlz -------!. Example 9. and 4.52) Equation (9..55) .-I= [z-. as functions of the current variables of the former machines and the known internal voltages of the latter group.50) and (9. R2. are functions of and the angle differences as well as the admittance parameters.53) by inspection (9.31) where in (9.4 Repeat Example 9.51) the complex matrix 11.53) - and from (9. I .2 assuming that machine 1 is represented in detail by the twoaxis model and machine 2 by the classical model. Now since Mrl1exists. (9.[] (9.T is partitioned. w. ..E2[FG+E(621) 6. Note that the variables needed to solve for the swing equations are only and Iq2.4) and (9.)[GllEjl + FB-G(aIZ)EZI .8 . + aI2) Id1 (9.51).sin ( 2 4 . 2 . and the transient impedances of both generators are included in the network (or E) matrix.Multimachine Systems with Constant Impedance l o a d s 38 1 or + Id2 = [y. mv.41) for generator I . For the two-machine system this is the same as (9.5 Repeat Example 9..32)... Solution Again the nodes retained are the “internal” generator nodes.57) linearized Model for the Network From (9..26) V = TV.40).3.FE-G(aZl)E:l + G22E21 .56) 1 Yz: YZ2 e. YI I ollj E. (9.(xql .3 I ) .x. it is convenient to use (9.. and E: as state variables.49) for generator 2. and an additional set of algebraic equations relating the node currents to the node voltages.D2w2 .. Vdl.2 .xi) EFDl TmI d o l Eil = 7jl‘l ‘jZLjZ = = TmZ WI .FG+B(612)E21 + [Bll(xdl . = 9. Since the two-axis model retains E.c0se. with machine 1 represented mathematically by the two-axis model and machine 2 by the classical model.sin(d1. and v r. (9. Example 9. Also from (9. The equations needed to describe this system are (9. %. 7601 % ‘ = IBii(xql . with replacing VI and = E. - ell + cos(OI2- el.24) and 57 and V are defined by (9. sin YI I ell)] E. [I:: .IIE:I . wI.x.[Gll(Ej: + + FE-G(a12)E~IE2 + FG+E(aIZ)EiIEZI . The state variables w The complete system equations are given by for this system are E i l ..31)T = where A is given by (9. + j 0 replacing The system is now fifth order. .cos(2e12 y:. E:.17). Linearizing (9.iEil . where T is defined by (9.xl) + (xdl .)[G. . sin biiA GijA. . . -- MAVO = -j k-' I (9. and ... 1. ..58) becomes [ . .60) Therefore the general term in M A given by is mijA Y . .62) n . ..... IJ IJo (9. I (9.382 Chapter 9 where a.a i = 1.~ I ~ O ...... with all the diagonal terms equal to zero.. of the vector V. . ...~ I ~ L \ ) vois the initial value ... I and the linearized equation (9... Then the matrix R becomes YlIeJel1 ~ .* n I A ) .... ) 'IA .6 n l L l . n... .e = IJ j(9.. IJ IJo )e-j6ijA Using the relation cos 6...2.... yIn ...-6...63) . (9.Y.j Y i j e - j(e.-a. .. yn2 eJ(@n2-6n20-6nZJ) . vkO Y n k e . we get for the general term (1 .59) The general term mij of the matrix Rl is of the form Yije -8. yn l e J ( e n l . . ej(eij+6ij~) (9. is evaluated at the initial angles . ) 'J0 6.. .j&jA) ij - y.61) Thus MAhas off-diagonal terms only. j(s.-a. Let di = ai0 + ai... ~ J ( ~ I Z .thus z.. We now formulate (9. Thus from % + MA we compute = Ho i Mo + MA = (No + NA)y (To+ TA).63) can be derived relating to 1. and 6 i j A ..63) is that needed to complete the description of the system. show.Ti1VT06. = 68 diag(blA..64) 2N No + NA to compute We can Note carefully that T-I # Tcl + T i 1 and that (TA)-' # (T-I)A :NA.. Also . The network elements involved in this case are elements of the open circuit impedance matrix Z.63) in a more compact form. Neglectingsecond-orderterms. however.&.24) let T = To + T to A compute vA T A Similarly. - MA = -J(T. that (To)-' = (T-')o = No.A) (9. From (9.) ('9. . A similar equation analogous to (9. ..'bAPTO ..Multimachine Systems with Constant Impedance loads 383 The set of equations (9. .66) From matrix algebra we get the following relations.. &A 1 = r 6 I A . we let T-I = jTo6... . J]v0 (9.70) and the network equation is given by - 1. for a two-machine system.l ~ (9.wecan either manipulate (9.6 Derive the relations between 'iTA and 1.75) .j[6A - mc16Am.Qo6A)& Example 9.j(aAQ0.66). (9.mO6A] mO6A]vO (9.70) is the same as (9. To obtain a relation between and TA.73) VA = Q o L .74) (9.67).68) M A = -j[6Amo . Solution From (9.j[6AMO Note that (9.-'TA .70) to obtain - vA V A = mcl.72) (9.69) (9.63). and (9. Define Q We can then show that m-1 = ~ .1 y .53) we get for M o (9. = M O T A .71) or follow a procedure similar to the above.o From (9.384 Chapter 9 * e-J*. 48) we get = h)A ~ j & . tions between l q l A . Iq2. By separating the real and the imaginary terms in equation (9.(x.80) 6. d i A . in (9.x')IqA T ~ O E ~= E F D A A T~&A = EiA + (xd . and I d z A and CIA.OE~A = .41) we get T.48).3.E i A . I d l a ..x ' ) l d A TmA . we get four real equaV .DoA 8. Vq2.76) Substituting (9.76). = WA (9.41) and the classical model as given by (9. Solution From (9.DwA . .= T m A .78).EIqA .70). From (9.Multimachine Systems with Constant Impedance Loads 385 (9.(IdoEiA Iq0EiA - + + EioldA + E6oiqA) (9.8 I ) . V d 2 ~ and 6126. (9.77).7 Linearize the two-axis model of the synchronous machine as given by (9. These are given below: Example 9. We note again (as per the discussion in Section 9. m + c = n. It is of the form Ax + Bu. From the initial conditions. Then from (9. (9.2 .9 Hybrid Formulation Where a combination of classical and detailed machine representations exists.lo. is 4 + 2 .80). Solution From (9. The system order.79).5.6) that the order of the mathematical description of machine 1 is four. and b120and from the network V matrix all the coefficients of the A matrix of (9. .where (9.82) is a set of five first-order linear differential equations. Stability analysis (such as discussed in Chapter 6) can be conducted. 9. the variables wI and w 2 can be combined in one variable w I 2 .8 Linearize the two-machine system of Example 9.84) can be determined.1 = 5 . and (9. x = Equation (9.386 Chapter 9 Example 9. however. a hybrid formulation is convenient. One machine is represented by the two-axis model.81) and dropping the A subscripts for convenience. l. which determine EAlo.58). and c machines represented classically. If the damping terms are not present. and the second is represented classically.E ~ I o . Let rn machines be represented in detail. that of machine 2 is two.84) E IAIo. 78) and (9.VI0 Y1ze j ( k + * n o )a .63). By comparing (9.86) (9.(SA) is a (C x I ) vector. (9..Multimachine Systems with Constant Impedance Loads 387 From (9.89) Example 9.9 Obtain the linearized hybrid formulation for the two-machine system in Example 9.70) (9.85) and (9.86) where Rm(aA)is an (m x I ) vector and if.88) from which we get (9.I e M Iz- * 120) _______-_----- (9.85) and (9.87) Therefore (9. Solution From Example 9.90) 21A . e j(@ + * 120) yl yz2 "22 e Gnd from (9.2 yI1 ejell 12 y.86) .85) where the subscript m indicates a vector of dimension m.4.8 120) a 1 2 j = 1 [""""""':. From (9. . for the various machines are not state .94) This is a complex equation of order n. or 2n real equations.. in terms of and 8..33) and (9. These complex equations represent four real equations: TI.388 Chapter 9 Substituting in (9.89) or and (9. J = Qi (9.92) Equations (9. and I.72).92) are the desired relations giving K A and 72. If the flux linkage model is used. I. 9.10 Network Equations with Flux linkage Model The network equation for the flux linkage description is taken from (9..91) and (9. ) sin - . For machine i we have Equations (9.. Together with the machine equations they complete the description of the system........sin(&... nl .12.aril) .. (9..&. . These equations are obtained from Section 4. Therefore. it is exceedingly complex to implement.. xnn aIn] [ ...99) and substituting (9. .... . Rnn ZInsin(8.97) XI1 Znls i n ( h ... .100) a . While the above procedure appears to be conceptually simple. . UQIA"] [ UdlAdl + UFIAFI . z nl c os( e n....aril) XI I --- ..... ..96) The complex vector ithus becomes - I = uqn + UQn A Q n Now the matrix 0 in (9.. --- + =[ [ ] Iql + jld.Multimachine Systems with Constant Impedance Loads 389 variables.94) is of the form .. . Z I n (eln .. -t ~ DIADI UdnAdn + UFnAFn + UDnADn I (9... X"" ] (9. (QR + jQI)(Iq (QRIq .... Zn.. To simplify the notation. auxiliary equations are needed to relate these currents to the flux linkages.94) and (9. Rll ..99). .. 1qn + jbn c q i Aqi + . ..95) is put in the form Iqi Idi F UqiAqi UdiAdi + UQiAQi + UFiAFi + U D i A D i i = 1.. Iq2 + jId2 ..414) + j(QIIq + ZlnCOS(flIn (9...94). Rnn = QR + jQI = = Expanding (9.2.n (9.. + uFIAFI dl Adl . This is illustrated below. + + uDIADl udnAdn + uDnAD (9.. ... ....hnl) .. V q + jvd Rll ..98) .... - . .97) into (9..95) are the desired network equations. The whole system is of the form X = f(x.104) k (see [ 5 . This allows the direct use of a relation similar to (9.. Assuming that VFk and Tmk.73). 6. Much of the labor is reduced when some of the simplified synchronous machine models of Section 4. the following relations apply ik = &k = dk = -L.v..103) V& = and the matrices Rk.T.2.n. 7. and (9.31) provides a constraint between all the stator voltages and currents (in phasor notation) as functions of the machine angles.97). For example.. The network is reduced to the generator internal nodes. Lk and Nk are where ik = [idki~ki~ki~ki~k]'.Q/o6a)Id0] (9.t) (9. 9. is used. The linearized equations for the flux linkage model are obtained from (9.(~AQRo QRo8a)Iqo i ( ~ A Q I O . become state variables.100) and (9.are known.15 are used. This has been illustrated in some of the examples used in this chapter. . = = VA q + jVdA = (QRO JQ/o)(IqA + [ Q R o l q A .' defined by (4.QioIdA I ~~ J1dA) .97).iQfo) .2. 8. The vector x includes all the staror and rotor currents of the machines. if the constant voltage behind subtransient reactance is used.. The above illustrates the complexity of the use of the full-machine flux linkage model together with the network equations. .n + + 3Tmk) (9.IOl). These equations are also nonlinear. (9.I k = 1.. and the vector v includes the stator voltages plus the rotor voltages (which are assumed to be known). = 1..390 Chapter 9 (9.31) to complete the mathematical description of the system model. the voltages Eli and E. which is linear.101) Equations (9..J [ a A ( Q R o -t . . I38).. Iqo..Li'vk (l/3Tjk)(-&kiqk Xqkidk . [vdk -vFk 0 vqk 0.1 1 Total System Equations From (4.100) and (9. such as given in (4. Following a procedure similar to that used in deriving (9.(QRO+ ~ Q / o ) ~ A -k (Jho) o (aAQio .'(Rk + WkNk)ik .103) for each synchronous machine and hence for each node in Figure 9. and 91). we expand (9. and Id0 are substituted for by the linear combinations of the flux linkages given by (9. The set (9.102) - The terms in I.Q ~ o J A ) I ~ (~AQRO Q R O ~ A ) ~ ~ O ] o + j [ Q d q p -k QRoIdA .3DkWk wk .74).104) represents a set of 7n nonlinear differential equations.73) into real and imaginary terms as follows: 5 .101) are needed to eliminate Ci and vdi in the state-space equations when the flux linkage model. .2. I. . ) l .2.40). This creates difficulties in the solution of these equations and is referred to in the literature as “the interface problem” [IO]. The system has ten differential equations. . The network equations would then give relations between the currents lqiand l&.31) we note the following: The differential equations describing the changes in the machine currents. six auxiliary machine equations.103) only. Finally. This problem is mentioned here to focus attention on the system and solution complexities. and four algebraic equations for the network (or two complex equations).Idi E” qi di + KdiADi gi = wi .I pi..10 Give the complete system equations for a two-machine system with the machines represented by the voltage-behind-subtransient-reactance model and the loads represented by constant impedances.105) and T&E$ = -E$ T&iA. Also in (9. = KI& (9. and the machine equations are given in Section 4.Aoi + (xi.1 i = 1. and E$.138). 2. if the flux linkage model is used (for the case where saturation is neglected). i = I . The machine equations are obtained from (4. This problem will be discussed further in Part 111 of this work. Solution The network constraints for this system are given in complex notation in (9.i T&iEii T.2 (9. 2. The other phase is the solution of the differential equations of (9. .lqiA’y + E$ i Gi = -rilqi + IdiiX: + E.6.101).Multimachine Systems with Constant Impedance loads 39 1 By examining (9. One phase deals with the state of the network. They are E..103) and (9.31) are functions of the angles of all machines.x Ql‘ . assuming “known” internal machine quantities. .&. The solution alternates between these two phases.40).270).40). E. and angles are given in terms of the individual machine parameters only. The voltage-current relationships (9. in terms of node voltages and currents. and E: should replace Vgi and Vdi.31) or in real variables in (9. the system equations will be (4. and w . Some of the computational labor can be reduced if the subtransient reactances of the generators are included in the network matrix (or Z matrix).x&i)ldi EFDJ ( 1 + K d i ) E .106) The network constraints are obtained from (9. and (9. .6.E”.w2 as state variables. As per the discussion in Section 9.i = l . rotor speeds. Example 9. (9. Again the “interface problem” and the computational difficulties are encountered. IOO). + KZiADi b = --rildi . 1: 1 (X .15. and the voltages E. i Xdifdi -k Tmi. . The nature of the system equations forces the solution methods to be performed in two different phases (or cycles). qi .234) and (4. some differential equations can be eliminated by using 6 . The auxiliary equations for bi and <i can be omitted. l ql = = = . 561 5 0. + j I.1 (some of which are repeated below for convenience).613 .9320 .392 9.63) with B replaced with 6 and for a three-machine system (using = -631).4777 0. Generator 3 (two-axis) Table 9.1312 I .S = V.6940 0.27 1' 7 1 .1447 0. The system under study comprises three generators and three loads. Generator 1 (classical) V / @ .1: (r I.6900 6.8404 - elec deg 0. Three-Machine System Data Quantity S PU PU PU S PU S PU PU PU PU PU PU PU bU Unit Generator 2 (two-axis) 23. are used 0) to obtain the data in Table 9. is given in Figure 2. 9.jQ)/V % . The initial operating system condition. 9. The network is assumed to include the transient reactances of the generators. + jV.2 linearized network equations e. The synchronous machine models to be used are as follows: classical model for generator I .4000 4825. Linearized machine equations are to be used.6668 0.2872 I .1 2 Multimachine System Study Chapter 9 The nine-bus system discussed in Section 2.0392 -0. and the q axis be located at angle 6.p + 4) = Iq + jI.7882 -0.8900 2220.9467 0.1.4865 1.7760 0. The loads are to be simulated by constant impedances based on the initial operating conditions.1 2. Data for the three generators are given in Table 2.6000 226.xqI.8057 61.OS66 6. = I ~ = E: = tan(S .I400 0.1900 5.0852 0.0100 2269.. At these nodes the voltages are and E.1.0361 0 0 8. The network is reduced to the generator internal nodes.1 2..18.9600 3377. ]/-(a . All angles are measured from reference. 4 2 = 421..2902 0.p) = xqIr/(V .6661 -0.7791 54.0412 2.0000 226 1.6780 -0. A one-line impedance diagram is given in Figure 2. The generator current lags the terminal voltage by the power factor angle 4.I&: E: ZZ 6 + ZqX.I43 1" . From (9. indicating the power flows and bus voltages. derived in Section 5.1 Preliminary calculations Let the generator terminal voltage be V @ .10.19. The following relations.4863 0.0975" 3.) (P .7679 -0.10 is to be examined for dynamic stability at the initial operating point given in Section 2.6336 -0.0765 0.5350 201.6400 17824.5. and the two-axis model for generators 2 and 3.6194 -0. . EA. It relates the incremental currents to the incremental state variables Ei2. = Ut (9.DIU.108) Separating real and imaginary parts and dropping the subscript A for convenience. .3 Generator equations From Example 9. EA3.107) and using a = 6. (9. .4.. 9.Yl3eJ~@I3-*IM) y23eJ(@23-d230) j[El yzleJ(@12 +*120) -jEio Y23eJ(*23-62)o) j [ E. Y12eJ(@12120) -6 Y13eJ(B13 130) - ElA = 0 and Elo + j0 = -* = E . (a constant).107) With generator 1 represented classically.E..7 we obtain the following generator equations (again the subscript A is omitted): Generator I (classical) ~ j l h l= T.109) Equation (9.12.109) is the desired linearized network equation. YI]eJ(# I3 + + Yl2eJ(@23 D O ) +* yl3eJ@33 Yz3eJ(@23 -*ZM)] -jEio Y23eJ(e23* B O ) + * 130) ] + y23eJ(e23+*2M)] "IA- (9. and with node E. as the arbitrary reference node = E. A. . aI2. . 1 Substituting in * -j Yl.eJ(d12 120) -j~&.Iq. E&. I IO) . .Multimachine Systems with Constant Impedance loads 393 (9.. and 613. (9. 3681 0. t d 2 .109).0800 .1458 0. = wi i = 2. the last equations in (9.3 (9.7238 0. (9.2871 + j1.1.oiE. ) l d i ~ j i & .0288 .9859 (9. Table 9.2770 .9883 = 2 3 = 0. The resulting system comprises nine linear first-order differential equations.6 as the prefault repeated here in Table 9.11 1) Again we recall that.110) and (9.2871 + j1.3681 = The coefficients of (9.1. ~ .j2. The main system equations are given below.0800 1.0288 -0.0541 ..10-51q1 - 6 1 = w) . The state variables are E i 2 .4200 .394 Chapter 9 Generators 2 and 3 (two-axis model) 7.2096 + j1.5129 1. The incremental currents Iqiand Idi are calculated from (9.2642 0.Jqi = (x.w 2 . Iq3. Node Reduced 7 Matrix for a Three-Machine System 2 = I 3 = I 0.109). IO86 /78. I q 2 ..2770 2. and I.oiE.5399 /79. = w.1 12) By using (9.3 (9.Ebi. E.2133 + j1.6062 0.. 9.1.3434 2.{ + . T.2256 1.30" 0. - wi i = 2.9216 -0.1086/78. is given in Table 2.*.1.4200 2.i -Eii - a'.4414 1.4 Development of the A matrix The V matrix of the network.2256 1.8455 .8347 0.5399 f79.368 1 -0.30" 0.1 I 1) are then calculated.9279 .1058 0.E.Diwi .!)f.9666 1.0879 1.1484 0.3434 1.1 13) The generator differential equations are: Generator 1 (classical) (5. w 3 .i .2.9216 0.2434 /80.25" 0.3.I .1 10) and .4200 0.i = E F D .wl.4914 0.1. .1 1 1).I.Iqio E.8347 I . It is the generator transient reactance.25" 0.8160 -0. to obtain an independent set.0879 1. Data for the terms in (9.j2.2133 + j1. I q 1 .1891 1.2770 -2.Eii0Idi .2096 + j I .5129 1.x .2.jioE:i . - 1.E.6104 x 10-5D101 5.I458 .1 2.i (xdi .E i 3 .6104 x 10-5T.1058 .0265 0. I d l .109) and (9.7239 0.2434 /80. reduced to the internal generator nodes and including matrix. tS12.9 I = 0.i ~ 7.5. = 5..x.91" 0.8305 .5805 -0.i . and 6 .1. are eliminated from (9.1 1 1 ) are combined to give a.109) are calculated and given in Table 9.0541 0.j2. 5919 I .9084 6. &2 = = = -4.6923 x 10-31q2 4.91 I7 -4.3836 X 10-41q3 w3 (9. I I5).3434 I .4 D 3 ~ 3 2..0288 20.3 + .4383 x 10-41d2 . we obtain the linearized differential equations for the three-machine system.1714 0.0802 1 1 1 1 yijc0~(eij aijo) .054 I 85.5675 .4210 X 1 0 . (9..aij0) I .5035 E.4333 8. the currents are then eliminated in (9.1829 .8259 aijo e.4307 X 10-41d2 2.92 I6 28.2952 -5 I .4063 x 1 0 .21.4210 x 10-’Ei3 .] E.6163 0.4111 0 0 0 0 -0.3161 . using the relation 6 = 6i ..9581 x 10-3E.8629 0.4063T. .2434 80.9380 X 1 0 .9540 0.1 14) By using (9.4.loo00 0 0 0 0. 01 612 611 0.5981 0 0 0 0 -13.7247 0 0 .4 E ~ ~ 34.Multimachine Systems with Constant Impedance loads 395 Table 9.4285 eij + aijo 1.7592 x 10-’tq3 4.1 14).loo00 - 4.2.6793 0.5375 I ..0944 X 10-41d3 4.0723Tb2 + lo-‘ 4.3.58.8349 -5.6.7292 X 10-4Ed3 + 2.. + aiio) 0.4592 -4.2156 1.0723 x 1 0 . I058 Generator 2 (two-axis) E.9205 2. Combining terms and . 0 .4 f d 3 .9552 .4 E ~ ~ 24.156.4.9545 7 I .4.0723 x 10-4Tm2 2. The results are shown in (9.4023 5.5399 79. E63 & 3 = = = 63 = -4. 0 0 0 .. I554 0 0 0 0.1 170 68.4076 52.9314 x 10-4Ei2 + 2..16.4409 .12..87 I4 132..9520 -0.2010 0 0 .0 138.5472 42.1.71 16 .6099 1.1666 -0.3 .4210 X i0-4E62 + 3. I458 Yijsin(Oij . 0 0 - .4238 I .2309 -38.4948 3.0723b1 -2.56ior.5463 1.1 13)..4431 Y ~ ~ c o+s 6(i j~ )~ ~ o Yij sin (e.1 15) 4.2 + 1.5035 X 1 0 .0935 0.6270 3.150.1. Nodes Yij eij Preliminary Calculations for Three-Machine System 1-2 1-3 2-3 I .6736 x 10-4E.1781 38.1.6238 0 0 .2 .5035 X 10-4E63 + 5.3. IO86 78.10.3385 10..4 D z ~ 2 1.3. .a.421OEFD2 2.4063 x 10-4T. which is o f the form WI E. €21 W2 E.9766 0 -2.4063 D 0 10.6334 x 10-41q2 62 = w2 Generator 3 (two-axis) rj.0800 I .7494 -3.4741 x 10-4E.2544 .8347 0. while the long period frequency has disappeared. voltages.15.000129 All the eigenvalues have negative real parts. for each ma.. neglecting the amortisseur effects for the synchronous machine represented in detail (Section 4.o.000622 + j0.j0. = 0.7 Develop (9.016659 0 = X7 As X9 = = = -0. Calculate the initial position of the q axes.000529 j0.40) for a two-machine system with G.15. i z j . and the angles 6120 and 6130. Problems matrix of the network.1 If the 90". The eigenvalues are XI A2 = = = = X3 A4 As = -0. Thus if we plot P I 2 from the data of Figure 3.4 Hz.3.j0.000199 .022984 -0. They are XI X2 X3 X4 = = = = = D3 = -0. 9. which checks with the data obtained here.j0.034636 Since this is a special case of uniform damping ( D / s j = 0). the system order is reduced by one. .4 Repeat Example 9.5.000529 . .1). 9.] = E F DT~ 2 E F D Tm3I m ~ The eigenvalues of the A matrix are obtained for the case of D I = D2 1 .000281 -0.002664 . in terms of V and V. is such that 8. I&).5. 9.4 Hz respectively. E ~ oand E.000458 -0.1.8.016644 = X7 X8 X9 = = = -0. using the results of Problem 9. 9.002664 -k j0. a. 9. VdO. Obtain the A matrix and examine the system eigenvalues for stability.4). I.j0.010373 -0..022984 -0. obtain the real matrices for I. c. = 9. and the system is stable for the operating point under study. = G 2 .034636 -0.2 For the conditions of Problem 9.002459 .000622 .022983 -0.010366 -0.3 Repeat Example 9. Compare with (9. .j0. A similar run was obtained for the same data except for D I = D2 = D3 = 0. 9.. using the synchronous machine model called the one-axis model (see Section 4.1 Hz and 1.000129 -0.O pu. using a library computer program. and I. b.034648 -0. The frequencies corresponding to the electromechanical oscillations are almost unchanged. These frequencies are the rotor electromechanical oscillations and should be very similar to the frequencies obtained in Example 3.8. chine.89) for a three-machine system with zero transfer conductances. we find that the dominant frequency is about 1. reduced to the generator nodes.022983 -0.002459 + j0. V.000199 + j0.O.5 Linearize the voltage-behind-subtransient-reactancemodel of the synchronous machine. IO the dynamic stability of the postfault system (with line 5-7 open) is to be examined. 9.396 Chapter 9 i where x' = [wI Ed2 Ei2 w 2 Ed3Ei3 w3 U' = [ T m I ' = AX + BU a12a]. The generator powers are the same as those of prefault conditions. The dominant frequencies are about 2. derive the general form of the matrixm.4.034648 -0. From a load-Row study obtain the system Rows.8 For the nine-bus system of Section 2.000455 -0.6 Repeat Example9. and angles. 70 M62-PWR. 9.. Laughton. K. and Murthy. 2. 5. Van Ness. W. J . IEEE Power Engineering Society Tutorial. pp. New Liapunov functions for power systems based on minimal realizations. Ames... Undrill. 1968. Digital simulation of multimachine power systems for stability studies. Janischewskyj. IEEE Trans. W. 1974. IEEE Trans. PAS-87:80-84. IO. 1972. Anderson. Int. IEEE Trans. P.. Willems. A. Pai. and Goddard. Evaluation of concepts for studying transient stability. Int. 1974. 8.and Janischewskyj. Iowa State University Press. Analysisof Faulted Power Sysiems. Pal. A partial stability approach to the problem of transient power system stability. 7. Simulation of the nonlinear dynamic response of interconnected synchronous machines (in two parts). Publ. Prabhashankar. Prabhashankar. W. 1968. Proc. F. J. M. E. 1968. Control 19:l-14. G. 53-60. Conrrol 19401-15. J.. M. K. F. 1973. A. State-space representation of multimachine power systems. Matrix analysis of dynamic stability in synchronous multi-machine systems. IEEE Paper C 74 396-8. J . PAS-87:73-80. Dynamic stability calculations for an arbitrary number of interconnected synchronous machines. W. Spec. PAS-87:835-44. 1974. K. 1970. . presented at the Summer Power Meeting. IEE (British) I13:325-36. 3. 4. PAS-91:2064-77. IEEE Trans. Paul M. Calif. Tinney. 1966.Multimachine Systems with Constant Impedance l o a d s 397 References I. and Dandeno. Anaheim. M. Formation of the coefficient matrix of a large dynamic system. P. 6. M. L. J. . Anderson . M.Part 111 The Mechanical Torque Power System Control and Stability P. . By the end of the 19th century. L. C. E. Watt’s governor is extensively treated in the literature of that era and even some elementary quantitative analysis is evident in works prior to 1850 [2]. the governing is accomplished by a speed transducer. J. have been in use since the late 1700s. The speed governor in the figure is a speed transducer. 21. the dynamic speed control problem had been thoroughly documented in the technical literature. I. A. Thompson (Kelvin). During the 19th century interest in speed governing intensified and a number of scholarly papers were written on the subject. B. and was even the subject of historical studies [2]. A. was presented in textbooks and handbooks. Over 100 references on the subject are given in the Royal Society of London Catalogue of Scientific Papers. Hammond [5] notes that J. such as the float valve regulator of French design. Routh (1877). M. Maxwell. The dynamic problems associated with speed governing almost certainly provided the incentive for establishing the mathematical theory underlying automatic control. B.chapter 10 Speed Governing Prime mover governors. James Watt first applied a centrifugal governor to a steam engine in about 1788. Pontryagin [4] refers to the work of the Russian engineer Vishnegradski (published in 1876) as of “complete clarity and simplicity” and credits him as being the originator of automatic controls (in Russia). I. These include C. In a steam or hydraulic turbine-generator system. Maxwell (1868). but it is placed in a modern setting and is attacked with the tools of the control engineer developed in this century. a comparator. Vyshnegradski. 1800-1900 [3]. Hurwitz (1895). and A. Vyshnegradskii (1876). The mechanical flyball governor of Watt and Boulton came into wide use during the early 19th century and easily outstripped competing devices. Figure 10. J. J.1 shows the system block diagram for a steam turbine generator. C. C. identified the instability of an early governor design as being due to a positive eigenvalue [6]. the control dynamic problems inherent in feedback systems were not recognized until the second half of the 19th century. and I. Many of the prominent engineers and scientists of that era made contributions to the description and analysis of governors. Stodola (1893/94). The treatment in this book is therefore the restatement of a very old problem. the output of which is typically the position (stroke) of a rod that is therefore pro- 401 . Airy (1840/5l). Siemens. these works consisted of attempts to solve the differential equations by classical methods and did not present a generalized theory of feedback control. and one or more force-stroke amplifiers. There is evidence that he considered a patent application for his governor and probably decided against it because of earlier patents for similar centrifugal devices used to regulate the speed of water wheels and windmills in the milling industry [l. J. Lyapunov (1892). Maxwell. However. writing in 1868. W. especially centrifugal flyball governors. W. Foucault. Mostly. Mayr [2] lists the earliest contributors to this quantitative theory as G. This controls the pressure downstream from the orifice. In writing the governor equations. Such devices are not widely used for central station speed governors.1 The Flyball Governor Consider the flyball governor shown in Figure 10. The centrifugal flyball governor has historically been used for this purpose. The newest governor designs use high-speed electronic logic to control electrohydraulic force-stroke amplifiers. However.1 Block diagram of steam turbine control system from [I I ] with permission. All three have the same essential components: the flying weights (flyballs). and because they are still widely used. Speed sensing may also be done electromechanically by coupling a small generator to the shaft. the output voltage or frequency of which is speed dependent.402 Chapter 10 Load Reference Position Fig. These electrohydraulic systems have high sensitivityand fast response.2 shows three examples of flyball governors as conceived by different designers. the output pressure of which is only one-fifth or so that of the main pump pressure.1. This approach is used because mechanical devices are easy to understand and analyze. If we assume that the gravitational force is negligible compared to the centrifugal force F. portional to speed. This same figure also describes a hydraulic turbine control system if the valve position is changed to a gate position and the steam valve block is considered the wicket gate and hydro turbine system.4 [4. The force that controls this position error is small and must be amplified in both force and stroke. hydraulic. and the subsequent amplification of the error. It must measure shaft speed and provide an output signal in an appropriate form (position.3. the governor pump pressure varies as the square of the speed. similar equations can be derived for other types of governors. Figure 10. a main oil pump supplies high-pressure hydraulic fluid that flows through an orifice to the governor oil pump. pressure. or electrical device. the restraining spring (speeder spring). 10. Our motive is not to present any given system as being superior to others but to derive a typical mathematical model that will increase our understanding of the governor as a control system component and allow us to analyze systems similar to that of Figure 10. Examples are given in Appendix F.. or voltage) for comparison against the reference.3 (also see Figure A. it will be convenient to use several of the control system component descriptions and formulas given in Appendix A. 61. then there are two forces acting . The analysis followed here is based largely on the mechanical flyball governor. This stroke is mechanically compared to a preset reference position to give a position error proportional to the speed error.27 of Appendix A). 10. In Figure 10. The speed transducer is the heart of the governor system and may be a mechanical. This is the purpose of the two amplifiers labeled speed relay and servomotor. including the penstock. which is used to control the throttle setting through a hydraulic control system. The amount of governor oil flow is determined by the pressure produced by the governor oil pump. In most cases. and a mechanical linkage that changes a shaft or collar position as the speed changes. An example of a hydraulic governor is shown in Figure 10. . 10.2 Examples of centrifugal flyball governors.Speed Governing 403 Arms Governor Travel Fig. the downward force of the spring. . and the radial displacement R of the mass m: Fig.the peripheral velocity v.404 Chapter 10 Turbine Shaft Main Oil Pump I Governor Fig. 10. and a downward spring force Fsacting on the throttle rod.3 A hydraulic governor. The total outward force Fc on the two flyballs depends on the mass m. The reference position r is adjusted to correspond to the desired speed.4 A flyball governor. 10. on the flyball-crankann system: an outward centrifugal force Fc acting on the masses. we can relate Fi to Fs. (10. both the applied at the ballhead.1) as (10.. o is in radians per second.C. = bla is the lever ratio constant.5) where C.Speed Governing 405 (10. = Z Forces = . using Figure 10. By simple geometry.2) *=No we can write (10.7) where Fi is the force due to the spring and BXl is the force required to overcome friction. m is in kg. = 2mN2(d. Equating moments about the pivot. 10. Then the ballhead force Fc may be written in terms of F. relating the governor speed to the turbine shaft speed through the gear ratio N.d = --x b a = -C x R=d-C.3) Fc = 2mMRo2 N (10. as follows: -Fib sin a = -Fsa sin a 1 2 1 2 (10.6) mx. we s u m the forces on the ballhead using Newton’s law: (10.5.4) where Fc is in Newtons.x)oZ NOW.1) Using the familiar relation between peripheral velocity v and shaft angular velocity JI we can write v=RJI Now. and R is in meters. .8) Fig.5 Crank arm geometty and forces.FC Fi 2 2 (10.x x as (10. we can relate R to x1 and x as follows: xl = R .. actual spring force.BXI . -F s-. Equation (1 0.9) about a steady-state operating point (subscript 0 ) from which we will study small deviations (subscript A). In order to gain a better grasp of governor behavior. The ballhead force Fc acts in the x . .16) .x ) Cr = 2mN2(d . Substituting (10.9) where K: is the spring constant of the speeder spring.C r x ) d = 2mx + 2Bx + K? (d Cr (10. particularly K.C. Thus. This force creates an equivalent force in the -x direction. escent condition = io 0.1 1) shows clearly the nonlinearities of the system.406 and solve for F i with the result Chapter 10 a 1 F i = -F s-.xA + K.11) where we define an effective spring constant K..15) Fib = K.13) Now. + 2Bk..9) into (10.. (10.12) At the quiescent points. which introduces nonlinearities that are not continuous functions of x.C r x ) d (10. and their derivatives. direction (see Figure 10. but the coefficients. from (10.1 1) must still be = satisfied. we write (10.11) and using (10. From Figure 10.12) into the system equation (10. This gives the qui- (10. and wA. we linearize (10. there may be backlash in the gears and dead zones in the pivots or other mechanical connections.xA.XO)OA 2mX~ 2 B x ~ (K. r .5) on the total mass 2m. This is rm justified since the speed will deviate f o its rated value only by small amounts in normal operation.Ki(r . Not only are the products and quadratic terms nonlinear.Upon linearization we can write (10. w.6) and (10. - Kl(r .5 we readily compute Fd = CrFc = 2mPCr(d .5) x I = -Crx and the entire equation can be written in terms of the variables x and w with the result 2mP K . with x.x ) Cr Cr (10. = KYCZ. and B. which we shall call F. we compute KJA - 4mN2wo cr (d .10) Now.2mN2&)X~ = + + (10. Furthermore.C r x ) d Clearly. can not be expected to be linear over a large range of x and x.7) we have 2mx. substituting (10. Fd is a function of both x and o.oA (10.13) to simplifl the result.14) which is a linear differential equation in the variables rA. we know that the solution of (10...19) is commonly used to represent governor behavior in power system simulations. it is obvious that the system is unstable when K..rA .. which relates to the throttle rod motion..20) Furthermore..14) the system equation becomes K.17) Using these defined constants in (10.wA = 2 m X ~ ~BXA (K. from which we compute (10. The assumption of linearity is justified since deviations from synchronous speed are small. a minimum spring stiffness exists for satisfactory operation. is always positive. can be applied directly to the throttle valve or.K. as an estimate.18) Taking the Laplace transform of this linear equation we can visualize the computation of x.. we will neglect the governor oscillatory behavior to write.18) will reach steady state much faster than the turbine shaft. Since the governor is physically small and it controls a massive turbine. The linear equation (10. 1 2ms2+2Bs+(Ks-Kx) Fig.6 Block diagram of a linear speed governor. the increase in x.K J x A (10. .6. = . should be in a direction to further open the throttle valve. Therefore. or a well-"tuned" governor may respond in a critically damped mode. In any event.Speed Governing 407 where we define the positive constants The Ballarm Scale K. Therefore. even for large disturbances such as faults on the generator terminals. This equation is algebraic and specifies that a reduction in speed results in an immediate increase in x. from the block diagram of Figure 10. We are interested primarily in the motion of the turbine shaft. Since a reduction in speed would normally accompany an increase in load on the turbine. The variable x. It determines the natural frequency of oscillation of the governor from (10. We would expect a response that is probably oscillatory when a step change is made in or OJ. . more commonly.18).19) which will be sufficiently accurate in the longer time span of interest. 10. (10.. is an important parameter in governor design. The linear equation of motion of the governor is a second-order equation. applied first to a force-stroke amplifier that drives the throttle.18).= 4mN2Cr(d Crxo)o. and K. < K. The spring constant K. - aFi dw 1 0 The Ballhead Scale (10. the frequency of oscillation and the damping ratio are determined by the coefficients on the right-hand side of (10. is in fact the speed re$ erence.17) this means that d > Cp. what is needed is a force-stroke amplifier to magnify the stroke and exert a sufficient force to manipulate the valve.408 Chapter 10 Also note that the system is designed for correct operation with K . which is recommended for further reading on the subject.2 The IsochronousGovernor The flyball governors similar to those shown in Figure 10. From (10. the hydraulic reaction force due to the spool valve.acting through the spring constant K. and a transient component that is proportional to i.. A detailed discussion of (10. Therefore. 10. *Portions of the development here and in subsequent sections are similar to the treatment in Raven [7]. .* The flyball governor equation is the same as (10. and a piston that is capable of exerting a large linear force. the force available to move a throttle mechanism in the x direction is small and the displacement is usually small as well.4 and (10. a steady-state component that is always proportional to x and acts in a direction to close the valve orifice. Adding these forces to (IO.. note carefully that rA.and may be either a stabilizing (closing) or a destabilizing (opening) force [7]. must be added. will cause o to seek a new steady-state value. This hydraulic reaction force. has two components. we need to represent only the steady-state hydraulic reaction force.19). which we write as simply Fh = KhXA (10.. Finally. 10. or Bernoulli force. a spool (pilot) valve. A simple manipulation of this position will cause a change in x and eventually.2 are capable of sensing changes in speed and responding by making a small change in a displacement or stroke (x) according to Equation (10.7 The isochronous governor. which is recommended for further reading. Consider the system shown in Figure 10. the governor-plus-spool valve equation can be written as Pilot J Valve 1 Flow Control Valve Fig. Since the valve transient period is very short compared to the turbine response time. > 0. as the shaft responds.5) that this inequality always holds.21) where the hydraulic reaction scale Kh depends on the orifice area gradient and the pressure drop across the orifice. but we see from Figure 10. which consists of a flyball governor.19) except that a new force.19). However.21) is given in Appendix F.7. This is accomplished by means of a hydraulic amplifier or servomotor (see Appendix E). 26) The leading coefficient is interpreted as the inverse of a time constant T~ in seconds (the may be simplified by perreader may wish to veri@ the dimensions).)xA + KhXA KgxA (1 0.K.OR= RwR rads Substituting into (10. we define the per-unit (pu) quantities.22) into (10.27) (10.29) This is the same result as that discussed in Section 2.22) where Kg = K.28).26) can be determined from (10.26) can be simplified to the normalized form (10. Note that the spool valve-piston combination is in fact an integrator since the output y continues to increase as long as a positive x displacement exists.Speed Governing 409 KsrA . The hydraulic piston moves in the +y direction as long as there is a positive x displacement of the spool valve. It is convenient to normalize (10. Assume the system is initially in the steady state QA 0) and at rated full load (reference) condition (rA= rR)when the load is suddenly dropped.53).25) Then (10. Substituting (10.7) since Fhproduces a reaction in the -xA direction for an acceleration in the +xAdirection]. we write K f l A = aIYA (10. . To do this.28) Then the coefficient of oAU (10. (10.24) on the basis of the full load rating of the generator.24) we compute (10.K.K. . From Appendix J. This is designated hereafter by a subscripted capital R. The coefficient of wAU = forming the following conceptual test.24) may be written in the Laplace domain as (10. Thus (10.24) and we see clearly the integrating effect of the hydraulic piston.30) . with subscript u as follows. we have (10. + Kh.23) and solving for the piston displacement. with the result in (10. Equation (J.23) where Kq is the spool valve volumetric flow per unit of valve displacement and a l is the piston area. The new governor equation is basically the same as before except the xAcoefficient is largis er since the hydraulic reaction force is in opposition to the displacement [Fh subtracted from the right-hand side of (10. causing a change in speed of *A= o .oA = (K.3. which corresponds to the valve position.41 0 Chapter 10 Fig. A block diagram of the isochronous governor is shown in Figure 10. we can estimate the behavior by taking partial derivatives in this neighborhood.3 IncrementalEquations of the Turbine In order to study the performance of the governor. Command Signal 1 . The control transfer function and feedback function are.9 can be constructed [7.9 General Block Diagram of a Control System [7. 10. Note that the comparator is due to the flyball governor and the integration is due to the hydraulic servomotor. the turbinegenerator system. where 7. The output of this control is the “manipulated variable” M(s) = yA(s). 10.O in this problem. As in many control system problems. 101. In the preceding section we developed the equations and the block diagram for the control section corresponding to an isochronous governor. Therefore.9].30) is called an isochronous governor since it attempts to integrate the speed error until the error vanishes. This variable would correspond to the steam valve stroke (or valve area) for a steam turbine or the wicket gate position (or gate area) for a hydro turbine. it will be useful to develop the system and control equations such that a block diagram similar to Figure 10.3 1) as noted in Figure 10. respectively. in this case.1 Fig. It is not necessary here to provide a detailed analysis for large excursions since we are interested in the system behavior only in the neighborhood of the steady-state operating point. The input transfer function A(s) = 1. 1 (10.8.8. it is desirable to develop the incremental (linear) equations of the controlled plant. KgWR - 10. = KSKfR The integrating governor system described by (10. .8 Block diagram of the isochronous governor.9. so the command signal U(s)and the reference R(s) are identical. T.10.9 can be evaluated in terms of the steady-state gain of each block [7]. . W=k2P Then the incremental flow can be written as = kyP (10. in the steady state. which in turn is a function of the valve stroke y. Finally. from (5. For the purpose of this elementary model. In a steam turbine. we can construct the system block diagram shown in Figure 10. but is approximately 0.9 we can write. we write the swing equation.(O) Kp = Gp(0) K N = N(0) KH = H(0) (10.38) . Then.35) and (10..8.33) For the analysis in this chapter we will consider the pressure to be constant such that we may write wA = kyYA (10.35) where we combine the two constants Kt and Ky into the single positive constant K . (10. we determine the gain of each block with s replaced by zero. where the output speed C(s) = o is controlled by the governor. .36) which describes the inertial behavior due to any upset in torque. The relationship between Wand the developed mechanical torque ky is a direct one since all working fluid entering the turbine produces torque with no appreciable delay [lo-121.6 in steam turbines due to valve nonlinearities [ 111. there is a lag associated with the control valve steam chest storage and another greater lag associated with the reheater (see Chapter 11). Combining the plant equations (10.7 admits steam (water for a hydro turbine or fuel mixture for a combustion turbine) as a function of valve area.37) K = A(0) A that is. Usually. The steady-state operation of the general control system block diagram of Figure 10. the valve is designed such that valve area is linearly related to stroke (see Appendix F.36) with the control equations of Figure 10. we include a simple firstorder lag T~for the turbine control servomotor system to write (10.32) (10.Speed Governing 41 1 We now seek a general relationship for the plant transfer function Gp(s) the disturbance and function N(s) for a turbine..7. The fluid flow rate W through the valve is proportional to the product of valve area A and fluid pressure P. Suppose we define for this purpose the steady-state gain functions K.34) where ky is a positive constant. from Figure 10. The flow control valve in Figure 10.DoA P = U (10. = G. There are also lags in hydro turbine systems (see Chapter 12). The term DwA is added to account for electrical load frequency damping and turbine mechanical damping.78): 2 H h ~ TmA. K1would be expected to have a normalized value of unity. function generators). 10. >r. changes.s) -m - (10. for the isochronous governor K.40). = lim s-a rls(l + 7.11 Steady-state operating characteristic of the isochronous governor. where the manipulated variable T. but the transient response also needs to be considered.41) The where we define the constants b = l/rs. For the system of Figure 10. This is a desirable steady-state characteristic.T I r. following any deviation in speed. root locus plot is sketched in Figure 10. I Fig. For the isochronous system we can write OLTF= rls(l + rp)(D+ 2Hs) cg - K s(s + b)(s + c) (10.40) and the steady-state o is a constant for any T. the steady-state performance equation for zero error becomes wss -rss = cg 1 = Rrss (10.. A I Fig..39) Since K. The system is Tm . the controller will drive the system until rAand CgwAare equal. Indeed. o . T i means that.For each setting of the reference.is plotted against the controlled out. the error E must be zero for steady-state operation. is infinite.12 for a typical small value of c and a larger value for b. constant from (10. and K = K1Cg/2Hrlrs.412 Chapter 10 Z.10 System block diagram for the isochronous governor. The transient response of the isochronous governor can be evaluated by plotting the roots of the open-loop transfer function or OLTF on the complex plane. even if the load torque T.1 1.10. 10. this is the hs unique characteristic of any integral control system. >r. .. c = D/2H. Another view of the steady-state operating characteristic of the isochronous governor is shown in Figure 10. Now. is put o. or the steady-state speed is independent of load torque. rather than integral. as (10. for small displacements. and becomes unstable for low values of gain. stable for small values of the gain K but will have a sluggish response since two roots are very near the origin. although having good steady-state characteristics. KqxA = alYA (10.Wp (10.K.42) + K. A better control scheme for this application is to use proportional. . -+ Kh)XA-k K. Substituting into (10.” as shown in Figure 10.wA =0 (K.12 Root locus plot for the isochronous governor. The mechanical feedback transforms the hydraulic integrator into an amplifier.3 and (10. which is used to increase the force and stroke of the governor throttle rod position.43) For the summing beam. we can again write.22).13. .Speed Governing 41 3 Fig. control. Furthermore. the system is unstable. Using the notation of Section 10.46) .44) where L =a + b. is sluggish in its transient response. 10.. Using Kg = K. 10. we can write the displacement equation.. This governor is called a “speed droop governor” or a regulated governor.23).xA or -k KhXA + K.K.43) we get (10.45) For the hydraulic piston.K. it the damping D is zero.XA = -K. This can be accomplished by using mechanical feedback in the form of a “summing beam. this equation can be written as (10. from (10. We conclude that the isochronous governor has a desirable steady-state operating characteristic.4 The Speed Droop Governor The isochronous governor. we s u m forces in the x direction to write K.is very nearly unstable and with sluggish response for reasonable values of gain.(xA + X i ) . Combining (10. in the steady state . in For the second test..47) we compute (10. in the s domain (1 0.48) To determine the normalized coefficients in Equation (1 0. we remove the load. but with the reference held at the same position.45) and (10.48) is unity.49) Substituting (10.41 4 Chapter IO Flow Control Valve Fig.46) we have (10.50) which means that the coefficient of rAu (10.49) into (1 0.13 The speed droop governor.e. The conditions for this test are. 10.47) This equation is normalized and rearranged to write. allowing the speed to increase.48) we perform two conceptual tests. i. (10. The first test is conducted at full (rated) load with the system operating at steady-state rated speed. 51) where we recognize that the speed change in going from full load to no load is. the steady-state speed is reduced.54) Then (10. Note that r1can be adjusted by changing the ratio d L .56) Fig. is increased. Dropping the u subscript. The steady-state performance of the speed droop governor is analyzed from (10. The governor block diagram is shown in Figure 10. the steady-state speed is now a function of both the reference setting rss and the generator load T. In the steady state. RwR. we write the per-unit speed droop governor equation as (1 + T ~ S ) Y A = rA . the mechanical torque applied to the shaft. by definition.55) for the speed droop governor..aKs "R -- where r1= alK&/aK&.14.52) YR K ~ L R Thus. we see that the isochronous integrator l h l s has been transformed into the amplifier 1/(1 + 7.. In order to analyze the performance of the speed droop governor. we interface the system of Figure 10. . Substituting (10.14 with a single turbine representation using one-time lag.8 for the isochronous case. Comparing this diagram with Figure 10.50).53) _ .. together with the inertial torque equations derived in the previous section. IO.s) by means of mechanical feedback through the summing beam. The result is the system of Figure 10.C G W ~ (10.CguA (10.15. to be Tmss = K1 ~ s= s K I(rss . The manipulated variable for this system is T..14 Block diagram of the speed droop governor. we compute (10.37) using the factors (10. the coefficient of wAuin (10.47) and using (10. Clearly.48) is Cg= 1/R as in the isochronous case.Speed Governing 415 (10.5 1) into (10. Note that the integral control of the isochronous case has been replaced by an additional lag in the control system. we can compute T. We will now examine the steady-state and transient performance of this system. In particulary as T. The characteristic of Figure 10. and K = K. = 27. is the steady-state error. we would expect to find r1< r.Cg/2Hrlrs. from (10. the steady-state operating characteristic may be visualized as the family of curves shown in Figure 10. and c are exactly the same as for the isochronous case.16 should be carefully compared with the operating characteristic shown in Figure 10. Thus. where E. 10.56).57) where a = llr.. Thus. This equation describes a family of parallel straight lines in the Tm-oplane. that the error E. the steady-state speed is dependent on the shaft load T. and that the higher loads cause a greater reduction in speed.16 Steady-state operating characteristic of the speed droop governor. Compare this plot with that of Figure 10.17. for each setting of the reference.C.41 6 -------------I------- Chapter 10 Fig. Note that the eigenvalues of the speed droop governor have much larger negative real parts than can be achieved for the isochronous governor. the root locus takes the form of Figure 10. c = D/2H.11 for the isochronous governor. each with Tmintercept K I and with slope -K. IO. whereas it was always integrated or reset to zero for the isochronous governor. A positive steady-state error signal is characteristic of a proportional control system..Note that K.16.. This means that the system can be satisfac- Tm f I Fig. I5 Typical system application block diagram. b. Note that. being about typical [l 11.. . is always greater than zero. Also note.1 1 for the isochronous governor. b = llr-. The transient response of the speed droop governor may be analyzed by plotting the root locus of the open-loop transfer function (OLTfl: OLTF = K1 c g (1 + r1s)(1 + r$)(D + 2Hs) - K (s + a)(s + b)(s + c) (10. In most physical systems. with r. c > 0 K>O (a + b + c)(ab + bc + CU) .58) s3 1 (a (ab + bc + ca) s2 S I + b + c) m (abc + K ) so (abc + K ) 0 0 Then the necessary conditions for stability are found to be a.59) K < (a + b)[c2+ (a + b)c + ab] (10.as (10. the performance of the speed droop governor is preferred because of its better transient response. for the isochronous governor case.( U ~ C K ) + m= >O a+b+c The latter of these constraints may be simplified by converting into the form (10.17 Root locus for the speed droop governor.60) .Speed Governing 417 '\ \ Fig. Figure 10. Overall. torily operated at much higher values of gain and with improved damping and smaller settling time. T. The improvement in transient response is accomplished by moving the pole which is well to the leR in at the origin. We can analyze the closed-loop governor behavior by writing the closed-loop transfer function for a given electromagnetic torque. b. 10.17. to s = -a = -UT]. . wA = KJIs s3 + (a + b + c)s2 + (ab + bc + ca)s + (abc + K ) (10.47) ating point. the speed droop governor does not hold the frequency exactly constant. The droop or slope of the locus is rarely changed in operation.61) for the case where D = 0 to compute 3 < 2 4 $ + +) R (10. we compute the response of the system to a step increase in reference rA(or a step decrease in TeA). but as the load cycles up and down. substituting gains and time constants and simplifying. The gain K1 is a function of the control valve and turbine design and is fixed for a given system.13.66) The response to a step increase in the reference rAis shown in Figure 10. as a limiting case @A(w) = (10.13. r2.50). if D = 0.63) Thus. the net error is usually small. In the root locus plot of Figure 10. increasing a1L increases R and decreases which increases the stability margin. Finally. This increases the negative feedback. this increase in alL moves the pole at s = -a farther to the left.7.shown in Figure 10.65) or. we get (10.58) we have. Because of the change in speed that takes place with changes in load.17. The quantities R and r1 and (10. we note that increasing the ratio alL moves the flyball connection with the summing beam to the right. although it may vary slightly with the opervary with the lever ratio alL since we define. with rA= A h . increases the droop.61) Since the damping D is always a stabilizing force.62) Now T~and H are fixed positive constants.41 8 Chapter 10 or. From (10. we examine (10. Hydro turbines often use a special kind of speed droop governor discussed in Section 10. and reduces the governor time constant. Frequency corrections can be made by adjusting the reference thumbscrew T.18 for two different values of the regulation R (ignoring any oscillatory behavior).or r3)to a new position in Figure 10. This thumbscrew is usually driven by a governor speed changer (GSC) electric motor. from (10. The speed droop governor is widely used for governing steam turbines and combustion turbines. (10. Each new setting of the reference moves the torque-speed curve (labeled r l .64) From the final value theorem we write (10.16. From Figure 10. with L = a + b. Here. -LFG + bFp = 0 (10.wA = (Ks .67) where P is the pressure of the hydraulic supply.13. KJA . The force FG acting at point G on the walking beam is that produced by the governor and is positive for a drop in speed or an increase in the reference position.67). 10. Summing moments about R in the clockwise sense.69) Now we can also write the summing beam displacement equation and the hydraulic servomotor equations in the usual way. we write.Speed Governing 41 9 0 Fig.)xA + bKh L (10.68) or.EYA Kqyb = a l L A Combining (10.A positive R movement in y i produces an upward force F due to the hydraulic piston acting at R.70) . the effect is the same as the design of Figure 10. that is. However. b a Y b = F x A . This results in a positive change in y i with its associated hydraulic reaction force of the pilot valve acting on the point P.K.5 The Floating-Lever Speed Droop Governor Another speed droop governor design is the floating lever governor shown in Figure 10.70) we get (10. The equations of motion of the floating lever governor are determined as follows.18 Step response of the speed droop governor. substituting from (10.K. These forces are computed in the usual way to write (10.19(a).71) (10. 10.69) and (10. the mechanical feedback acts directly on the servomotor pilot valve rather than on the speeder spring. 69) is normalized in the usual way to write rAu .420 Chapter 10 (a) Schematic diagram Flow Control Valve (b) Free body diagram of the walking beam Fig. where Equation (10.72) .19 The floating-lever speed droop governor.TIS)YAu (10.cgoAu = (l $. 10. . is the compensated governor shown in Figure 10.3.Speed Governing 42 1 where and Cg= 1 R f. that the time constant T~ is deis 1.72) identical with ( 0 5 ) Note. Equation (10. fined differently for the two governor designs. We have observed that the speed regulation provided by proportional (drooping) control is important in providing good response and also contributes to the stability of the prime mover Increase ROW Decrease Wow\ \ Flow Control Valve Fig.This governor incorporates an added feedback that gives it a unique operating characteristic.6 The Compensated Governor Another important governor design that is widely used. 10. however.20.20 The compensated governor. 10. particularly in the control of hydraulic turbines. This is particularly important on isolated systems where only one. Suppose that walking beam c-d were disconnected. Then. The time required to change from the temporary to the permanent droop is also adjustable.20. and of the speed o. without lever c-d the compensated governor would act isochronously. u. In doing this.(xA G + x A) + KGA .22(a). The first subsystem is the flyball governor system shown in Figure 10. The resulting speed deviation is gradually removed by slowly correcting the speed back to a second (relatively low) value of droop. Note. the governing is a stable isochronous operation. the larger droop provides stability and the smaller droop provides good speed regulation in the long term.K.73) The force at G’ is equal and opposite to this force.” which is a governor with two values of regulation. we can write equations for the forces acting at G and G’ as functions of the displacements x and X I . for which we write both a displacement and a force equation. or (10. The compressed spring on piston u3 will slowly force this piston downward.74) F&= K. the governor provides a temporary droop characteristic. As piston ul moves in the +y direction. Thus. The added feedback involves a floating lever system a-b connecting the speeder rod (x). following a disturbance. but it would do this in a special way. but is isochronous in the long term. each of which is adjustable. This gives the governor both a permanent and a temporary droop characteristic. As long as the piston location z remains at its steady-state equilibrium position. Still. This need is satisfied by the “compensated governor. machines control the frequency. Since the hydraulic fluid in the chamber connecting pistons a2 and u3 is trapped. tending to close the pilot valve. If the smaller value of droop is zero. however. The principle of operation is to provide a given (relatively large) droop in response to fast load changes. Using the methods developed in previous sections.(xA + x i ) . The mechanical feedback provided by the summing beam c-d provides a temporary droop exactly as in the design of Figure 10.13. or a very few. it causes transmitting piston u2 to be displaced downward. which is held in its steady-state position by a spring. the flyweights must also be in their equilibrium position if the pilot valve is held closed. it would be desirable to have the governor hold nearly constant speed (frequency) if possible. To analyze the compensated governor.oA (10. the pilot valve (u). and y. Thus. the speed of entry or escape depending on the needle valve orifice area. the ballhead would return to the same position when the receiving piston (z) returns to equilibrium. an increase in load would cause the governor to respond to positive displacements in x. if there were no permanent droop through lever c-d. These objectives are met in the compensated governor design of Figure 10.21. increasing the turbine power gradually and restoring the flyweights to their normal positions. this will cause receiving piston u3 to move upward.KJA + KmwA The second subsystem is the upper summing beam shown in Figure 10.A22 Chapter IO system. and a receiving piston of area u3. Thus. that the hydraulic chamber also contains a needle valve that will allow hydraulic fluid to move in or out of the chamber slowly. we can write (10. For incremental displacements. This means that.75) . it is essential that the defined positive directions of all variables be used in summing forces or moments. The two values of droop are called the “temporary” and “permanent” droops and are both adjustable within certain limits. pushing against its spring. it is helpful to break the system into subsystems and write the force and displacement equations for each subsystem.Thus the force acting at G can be written as F = -K. ' (a) Upper Summing Beam 1' $ A 4 (b) Pilot Valve Summing Beam (c) Compensator Summing Beam Fig. Summing moments about R in the clockwise sense. 10. where LI= c + d. Then summing moments about G in the clockwise direction we have XMG = 0 = aFp + L2FB or (10. .(xA + xA) .78) (10.22(b) we can write.oA + LlFS A (10.22 Mechanical beams of the compensated governor. 10.79) c G' h d S R Y F.21 The flyball governor subsystem.C K ~+ cK. for incremental displacements (10.76) For the pilot valve beam of Figure 10.77) where L2= a + b. we compute 8MR= 0 = cF6 + LIFs = cK.Speed Governing 423 Fig. is the needle valve constant in ft5/slbf or in3/spsi. summing moments in the clockwise sense about N.and the load mass is small compared to this force. and other quantities are as previously defined.82) with the result (10. we write the equations for the forces acting at B and E as (10.81) and (10.24.75) and (10. Since the available force Fs is usually much greater than the load FV. may be calculated from (10.80) and. Here. If load force and mass are important considerations.a3ZA (10.424 Chapter 10 The compensator beam of Figure 10.81) where Psis the supply pressure behind the hydraulic ram and a l is the ram area. The final subsystem is the hydraulic piston or ram shown in Figure 10.23 on an enlarged scale.82) The equation for the volumetic discharge rate of fluid through the needle valve is C$'A =a3k .22(c) is nothing but a lever for which we can write the displacement equation e Yd = -YA f (10. .76) we compute (10. From (10.84) for this subsystem. we write only the integrator equation KquA = aIYA (10.23 The compensator system.83) where P A is the incremental pressure change in the chamber in Ibf/ft2. (10. We now collect the equations necessary to describe the total system behavior. the complete equations for the piston should be written (see Appendix E).86) Fig.85) But F. C. This completes the subsystem equations. The compensator system is shown in Figure 10. 10. Note that (10.Substituting this result into (10.24 The hydraulic piston subsystem.92) where we define r2= CdK r3= But r3may also be Written as (1 0.91) may be written in the form 7 y = ZA 3b + 72iA (10.90) from which we can find P A as a function of zAand yA. .93) where 6' is defined as the coefficientmultiplying r2. Substitutinginto (10.80) we compute (10. we have From compensator equation (10.Speed Governing 425 Fig.79) we can write (10. 10.94) a 3 ea I a3L2Kq + afa 1c&h fa:L2CdK&q (10.85) and rearranging. from (10.83) and beam equation (10.91) which is the desired equation for the compensator.88) and simplifyingwe have (10.88) Also.84) we compute (10.89) and using (10.82) and (10. 96) that whenever y A = 0.95) 6 'r2$A =ZA + 72ZA (I 0. = yR.96) are determined from full-load and no-load steady-state tests.99) YR and the ?-&.K.100) Using (10.102) may be written in a slightly improved form by defining a new variable vA = K?A rR _ _-- (10.87) bcKs 2fa3Ks Then the system equations (10. In performing these tests.100) we compute (10.96) Equation (10. we note from (10. from (10. where 6' is given by (10.97) becomes Cg = 1/R as before.97) is unity.96) can be normalized in the usual way to write (10. Now.94).96) may be written as (10.98) or (10.101) and the coefficients of wAu in the normalized equation (10.96) becomes (10. then we also have zA = ZA = 0 as well. if we arbitrarily let Z.97) The coefficients of (10.87) and (10.96) becomes (10. equation (10.426 Chapter 10 We may also define.99) in (10.92) may be summarized as K= aLI(Ks.) --ea2L:Kz (10. Now. coefficient of (10. and that this always holds in the steady state. we can write .102) Equation (10.103) If we multiply the compensator equation by K and define 6 = KS'. if the load is removed and the reference is held at rR the speed will reach wA = RwR at steady state and (10.then (10. At full (rated) load and rated speed at steady state. 14 in the long term.s I ( 10. If (10. The transient performance of the compensated governor is not easily analyzed using the manual root locus or Routh techniques because of the added compensation. This is the desired system description.141. The block diagram helps clarify the role of the compensation feedback and the derivative effect of the temporary droop 6. This form of representation is instructive as it directly parallels the permanent (R) and temporary (R6) droop factors and also shows the integrating effect of the servomotor in the absence of droop.26 is analyzed using (10. The result is the composite system shown in Figure 10. P IVA 1+z. .26. This result was anticipated as the compensation signal vA goes to zero in the steady state. the system block diagram is that given in Figure 10.105) This is exactly the same result obtained for the speed droop governor with no compensation. A computer root lo- Fig. 10.26 Alternate form of compensated governor representation [I 3. 12.104) is written in the s domain.27.Speed Governing 427 0. The steady-state performance of the system shown in Figure 10.15. we again apply the governor as the controller in the system of Figure 10.25. Note that the signal vAwill always return to zero in the steady state and the system tends toward the speed droop governor similar to Figure 10.104) Fig. To analyze the performance of the compensated governor. Another form of the compensated governor derived by Ramey and Skoogland [ 13.37) with the result (10.25 Block diagram of the compensated governor. 141 is shown in Figure 10. 4 Evaluate the function 1 . ~ ~ = O . Hammond. and Rhys Jenkins. As an instructive alternative.32) that " m \ sin cpol Based on this premise. P.. The Origins of Feedback Control (translation of Zur FrGhgeschite der Technischen Regelungen).27 Typical system application block diagram for the compensated governor. 1962. cus method can be used for numerical results. 4. 10. 3 s 2H=4. Macmillan.1 1). lrs = 0 .428 Chapter 10 Control ' I T A I Plant Fig.2 Verify that the dimension of the leading coefficient on the right-hand side of (10. W. 10.1 Verify the development of equation (1 0. 1900. Cambridge.L.26) is in inverse of a time constant in seconds. Give a physical explanation for the resulting effective spring constant of K. Prob1ems 10. Oxford. Use the following constants for all simulations.. CataIog of Scientific Papers. Otto. James Watt and the Steam Engine. New York. 2. and compensated governors. 1927. 1970.'/C?. speed droop. References 1. H. Plot the results. Dickinson. v.MA. 10. S. Feedback Theory and its Applications.Opu S Cg=20pu Determine suitable settings for the gain K . Subject Index. 10. 1800-1900. Ordinary Differential Equations. pp. 1958. 5. 10. 136-137. 11. in all governors and for the parameters S and r2in the compensated governor. May-r. Royal Society of London. MIT Press. but this requires a cut and try procedure to optimize the variable parameters in the system.74s D=2. one can use an analog computer or digital simulator to determine suitable values for all parameters and then examine the behavior in the s plane for further insight into the design optimization. Pontryagin. and for various positive values of +o between 0 and 75 degrees. Mechanics. find the expression for stability of the system. = K. Cambridge. 3.(sin 4Jsin 40)for values of C#J~ = 10. H. and 30 degrees. . Boston.20. Addison-Wesley.3 From Appendix E we find the mathematical statement in (C.5 Perform a computer simulation of the isochronous. 9.. 14. G. Project Report. 12. Auslander. Wiley. Hydro unit transfer functions. Eggenberger. 6. Hydraulic Control Systems. Ramey. May 1969.M.pp.. Sixth PICA Conf. IEEE Tutorial Course. 13. D.. 1970.. C. Pacific Gas and Electric Company. 1868.. 1972. 490-501. Michael J.. Eggenberger. San Francisco. Modeling Thermal Power Plants for Dynamic Stability Studies. Proc. pp. Francis H. Boston. Rabins.. Herbert E. 11. D. A.. 1968. November 27-December 2. 1970.. New York. Skoogland. Yasundo. The Role of Prime Movers in System Stability. Addison-Wesley. 7.Y. J. and J. Merritt. Proc. General Electric Co. A. 10. 34-39. Anderson. 1970. McGraw-Hill. Takahashi.. Introduction to the basic elements of control systems for large steam hrbine-generators.M. W. ASME Winter Annual Meeting. and David M. Raven. . Royal Society of London. IEEE pub. N. 16. On Governors.Detailed hydro governor representation for system stability studies. G. Paper 60-WA-34. A simplified analysis of the no-load stability of mechanical-hydraulicspeed control systems for steam turbines. Automatic Control Engineering. 1967.Speed Governing 429 Maxwell. pp. New York. v. P. 8. 70M29-PWR. Control andDynamic Systems. GET-3096B. 270-283.1960. Ramey.M. New York. computing a unit dispatch signal (UDS). where the governor input is set to hold constant speed or frequency. The system control center measures the power produced by all generators and the interchange power with neighboring systems. while still maintaining prudent reserves to assure adequate generation if unforeseen unit outages should occur. which can be interpreted as a speed error. we concentrate on steam turbines and develop models that can be used to represent this type of machine in computer studies of the power system. thereby assuring constant long-term system frequency. which we write per unit as (11. which are based on the economic dispatch of generation considering individual unit generation costs. Note that the control center does not measure the system loads. It compares the tie line flows with their scheduled values.chapter 11 Steam Turbine Prime Movers 1 1. The measurement of system frequency is used to assure adequate total generation to meet load and maintain rated speed. The governor compares the speed reference or load control signal against the actual speed and drives the governor servo amplifiers in proportion to this difference. Figure 11. Following this general overview of prime movers.1 Introduction We begin this chapter with some general considerations of prime movers and how they are controlled. as shown in Figure 11. The system dispatch computer sets the governor input signal to control the mechanical torque of the prime mover. Note that this control is different on an isolated system.which indicates the position of the turbine control or throttle valves. and these flows are coordinated with neighboring utilities. The servomotor output is a stroke or position YsM. = the mechanical torque output of the turbine in per unit Te = the electromagnetic torque or load of the generator in per unit Ta= the accelerating torque in per unit 430 . as the system load varies. Then.1) where 7j = a time contant related to the unit moment of inertia in seconds w = shaft angular velocity in radians per second T.1 shows on overview of a large power system and the generation control structure. The control center receives measurements of all generator outputs and compares these values with desired values. the control center can change the generation dispatch to economically meet the demand in the most efficient manner. Other types of prime movers are discussed in Chapters 12 and 13.2. The fast dynamics of the generation of each unit is the solution of Newton’s law. The system dispatch computer determines the desired generator output and sets the governor input signal to control the mechanical torque of the prime mover. if the system is to be studied over a long time period.. the boiler firing rate. There exists a cross-couplingto the torque output T. this block can be represented by a constant and in others it may be a simple first-order lag. . the turbine should be represented in greater detail as an energy source transfer function.. Finally. the prime mover term in Figure 11. 1 I . In general. In some cases. the energy source controller receives feedback signals from several points. and the condensate pumping rates. The excitation system is used primarily as a voltage controller and acts much as a single-input. single-output system with V.but this effect is secondary. which indicates the position of the turbine control or throttle valves. to control simultaneously the turbine valve position. as the output. In some modem thermal units.. 1 Power system generationcontrol. which can be interpreted as a speed error. The governor compares the speed reference or governor speed changer (GSC) signal against the actual speed and drives the servomotor amplifiers in proportion to this difference.Steam Turbine Prime Movers 431 r-----l SYSTEM TRANSMISSION NETWORK I I I 1 .2 is a transfer function that relates the turbine control valve position to the mechanical (shafi) torque. 1 Generator v Generation Unit Generated \ # ’ Syste. The servo motor output is a stroke or position Y. for example. including the generated power (or load control signal) and the turbine throttle pressure. Loads System Tie Line Power Tie Line Frequency Reference Fig. where the unit will respond only to changes in its own firing and pumping rates.2. Turbine following is also used in some modem complex systems when the boiler capability is limited for some reason.1 The turbine-following control mode The control system shown in Figure 11. Thus the system response is very slow. even for a rapidly responding boiler.2 Block diagram of a generating unit. This is true because the two units. the pressure changes at the throttle (the turbine control valves). Then a back-pressure control on the turbine changes to hold the throttle pressure constant. In this control mode. the two subsystems must operate in unison under both steady-state and transient conditions. although it is sometimes referred to as “base boiler input” and “admission pressure control” systems (the latter mostly in Europe). the different control modes commonly used by the industry are presented and compared. since limited energy storage is possible in the boiler-turbine system. turbine following is seldom used because of its slow response and its failure to use the heat storage capability of the boiler in an optimal manner to aid in the transition from one generator load level to another. In this section. monotonic. a load demand signal is used to adjust the boiler* firing rate and the fluid pumping rate. In general.3 is usually called the “turbine-following” control. generator and turbine. 11. It is often used in start-up or initial stages of unit operation. *The term “boiler” used here should be taken in a general way to indicate a steam generator and that receives its thermal energy from either a fossil fuel or nuclear energy source. 1 1. operate together to provide a given power output and. 1 1. This back-pressure control is very slow. such as a fan or pump outage. and very stable.432 Chapter 1 1 PTs and VREF I + Tie Line Flqws xcitation System Fig. As the boiler slowly changes its energy level to correspond to the demand signal. Turbine following may be used on a base-load unit.2 Power Plant Control Modes The controls of the steam generator and turbine in a power plant are nearly always considered to be a single control system. . Boiler-following control has the disadvantage that pressure restoration is slow and the control is nonlinear. This type of system simultaneously adjusts firing rate. . and is generally stable under constant load [ 11. pressure. stable value. This control mode is shown in Figure 11. 1 1. If a change in demand exceeds the boiler stored energy. There also may be troublesome interactions between flow. The response is an immediate change in generator load due to a change in turbine valve position and the resulting steam flow rate. Boiler-following control is widely used as the normal control mode of many thermal generating units.2. Many newer units employ a more complex control system in which all control functions are integrated into one master control. and temperature variations. the result may be an oscillation in steam flow and electric power output until the pressure reaches a final. Throttle Pressure Fig.4. utilizes stored boiler energy effectively. 1 1. 11.3 The coordinated control mode Most modern thermal generating units employ a control scheme that is usually called an integrated or coordinated control system. The boiler “follows” this change and must not only “catch up” to the new load level. but even in these more complex controllers. This control mode is sometimes called the “conventional mode” or (in Europe) the combustion control mode.Steam Turbine Prime Movers 433 Fig.2 The boiler-followingcontrol mode A more conventional mode of boiler control is called “boiler-following” mode.4 The boiler-following unit control mode [I].2. boiler following is offered as an optional control mode that may be required if there are limitations in turbine operation. This type of control responds quickly. particularly the older drum-type boiler units. 11. This control scheme divides the control function such that the governor responds directly to changes in load demand. but also must account for the energy borrowed or stored in the boiler at the time the change was initiated.3 The turbine-followingunit control system [I]. A comparison of the three control methods described above is shown in Figure 11. 9 -.5. pumping rate.6 i I 1 THROTTLE PRESSURE *-e----- * . e * .. Pressure deviation is controlled as a function of both the thermal storage and the generation error.."h. I I I I 1 3 4 Time in minutes 5 6 7 Fig. 11..5 The coordinated control mode. +--. 2 . . In this manner.434 rinng Chapter 1 1 Load P. both pressure and generated output are fed back for the control of both boiler and turbine. and turbine throttling in order to follow changes in load demand. Such a coordinated control mode is shown in Figure 11. 1 1.-Sl Loaa IBoiler Fig. Both pumping and firing rates are made proportional to the generation error so that these efforts are stabilized as the load approaches the required value. In this type of control.____ -e---- set Point COORDINATEDCONTROL SYSTEM d .6 Comparison of the results of different control methods [2]. This is accomplished by making maximum use of the available thermal storage in the boiler. . it is possible to achieve the stable and smooth load changes of the turbine-following mode and still enjoy the prompt response of the boiler-following mode. The size of these generating units has increased over time. Usually. railroad ties. and tar oil.12 0. diesel.08 Petroleum (2) Natural gas (3) Nuclear Hydro. A more descriptive way to compare these results is by plotting the numerical values. propane. Percent 53. biomass (wood.186 129. sulfur.76 2.1. wood waste.430 628. agricultural byproducts.581 72.3 Thermal Generation The most universal method of electric power generation is accomplished using thermal generation. butane. kerosene. petroleum coke. batteries. landfill gases.1 Net Generation.702 328.23 17. peat.831 92.61 9.11 -0.104 544.99 10.72 3.040 3.867 -4. The second largest in order of size is nuclear generation. We will not belabor these concepts here as our primary motive is to study the system operation and control. (4) Includes geothermal.conventional Other (4) Pump storage (5) Other (6) (1) Includes coal. (3) Includes natural gas.872. waste coal. but this may gradually change as the sources of fossil fuels are depleted or become more expensive to recover and process than nuclear fuels. methane. wind. natural gas. variations of the straight Rankine cycle are used.Electric Power Industty by Energy Source in GWh Energy Source Coal (1) 1997. culm. such as the Pacific Northwest. liquid propane.763 -4. however.S.727 497.7. liquid butane. Fossil generation uses primarily coal. straw. this could change. Fossil fueled plants generate the majority of the electrical energy. wood liquors. pitch wood sludge. GWh 1.27 2. and photo voltaic.949 73. By “thermal” generation we usually mean a system that operates on the physical principle of the vapor power cycle or Rankine cycle. it is clear that coal is by far the largest energy source used in the United States. which summarizes data compiled by the U. and the most common machine for this production is the steam turbine. chemicals.137 1998. and fish oils). anthracite. Here. The role of hydro generation is rather small taken on a national basis. Nuclear generation uses fission reactors that operate by breakup of high-mass atoms to yield a high energy release that is much greater than that produced from chemical reactions such as burning. (6) Includes hydrogen.765 673. waste heat. waste gas. and lignite waste. hydro is very important in certain regions. This is true in many parts of the world. as shown in Figure 11.843. U.12 0.S. (2) Includes petroleum. (5) A more complete designation of this source is hydro pumped storage. solar.05 18. GWh 1. oil waste. with the largest units now being over 1200 MW.57 15. and purchased steam. at least for the time period represented. As coal becomes depleted or more costly to extract. municipal solid waste. which is more dependent on this energy source. coke breeze. . 1998. and other gas.01 -0. but a thorough understanding of this important subject is available through many fine refer- Table 11.09 Percent 51. where the predominant energy generation depends on available local natural resources.644 358.Steam Turbine Prime Movers 435 1 1. with two important innovations being the reheat cycle and the regenerative cycle. Department of Energy for the years 1997 and 1998. The steam used in electric production is produced in steam generators or boilers using either fossil or nuclear fuels as primary energy sources.08 2. and oil as fuels. fine coal. bituminous gob.478 2. In the United States over 85% of all generation is by powered by steam-turbine-driven generators [l]. tires. The prevalence of thermal energy production in the generation mix of the United States is shown in Table 11.65 14.905 1997. . 1 1.. i E: i.. .....1 ' .. Our objective here is to study the physical design of thermal power plants with the intention of understanding how these plants work and respond to controls...............4 A Steam Power Plant Model Steam power plants are of two general types: those fueled by fossil fuels such as natural gas or coal....... Figure 11............... The overall unit control is largely independent of the source of energy.5 . etc...7 Net generation by type of energy source.... ences on the subject [2-51.... 2 i 3 i 4 ....... .............. and those fueled by nuclear energy produced in a thermal reactor.. ........8 The control system for a thermal generating unit.........' 1 coal 3 6 8 1. 0. in which the source of thermal energy is a steam generator Fig........... 11............................. etc Pumped Storage Hydrogen. ........... as both types of plants must have a means of controlling the power output as well as the frequency...8 shows a block diagram of the controls for a thermal power plant... k 0... .0 I I I I 1 2 3 4 5 6 7 8 Fig.......436 Chapter 11 15x10 ......o 8 5 6 7 8 Petroleum NaturalGas Nuclear Hydro Geothermal. 1998 (top line) and 1997.... 1 1. relatively high efficiency. If there is interest in extending the studies to several minutes. Thus. and the steam turbine and turbine controls.9 (a). Among these advantages are the balanced construction. the U D S is hand set by the plant operator. In transient stability studies of 1-10 seconds duration. The speed governor acts as a continuous. The degree of detail required for computer simulation of the power system depends on the length of time required in the simulation. such as the generator. impulse and reaction blading. it is common to consider the generator. to be exhausted. The two moving stages on . automatic adjustments to unit speed in response to a speed error. the steam expands and its pressure drops as it passes through a nozzle. ease of maintenance. Studies of several minutes would usually require some consideration of the steam generator and steam system controls. and it may be necessary to consider the dispatch computer as well. The unit speed is used by the speed governor as a first-order control on this parameter. network. then it is probably necessary to add at least a simple boiler model to the simulation. Internally. the generated power (PGEN). exciter. In impulse blading. The generated power of the unit is fed back to the control center so that any error in generated power can be corrected. To accomplish this goal. For very long periods of interest. which is usually superheated. This is due to the many advantages of the steam turbine over reciprocating engines. the type and arrangement of turbine blading is important in extracting all possible energy from the steam and converting this energy into the mechanical work of spinning the turbine rotor and attached electric generator. the steam turbine consists of rows of blades designed to extract the heat and pressure energy of the steam.5 Steam Turbines A large portion of the conversion of thermal to electrical energy occurs in steam turbines. Studies of system performance of a few seconds. the more system components that might enter into consideration. The boiler control inputs are the unit demand signal (UDS). Note also that the boiler controller can be turbine following (adjusting firing rate according to desired power).9 (b). giving the shaft a torque due to the unbalanced forces acting on the blade intake and exhaust surfaces. as illustrated in Figure 11. The input from the dispatch computer is optional and is not used when the unit is on local control. the fastest responding components might be represented in a very simple manner and may not be required at all. proportional controller to make fast. This kinetic energy is converted into mechanical energy as the steam strikes the moving turbine blades and pushes them forward. it is seen that the longer the desired simulation. and convert this energy into mechanical energy.10. and the speed or frequency (w). for example. few moving parts. and availability in large sizes. The term “boiler” is used here to designate any type of steam generator. and speed governor. usually to a condenser. Reaction blading operates on a different principle. This mechanism is much faster than the governor speed changer (GSC) adjustment of the boiler controller. The unit demand signal is set by the system dispatch computer based on the method of dispatch and on the level of load to be served. Two types of turbine blading are used. In that case. Thus. or a completely integrated or coordinated control that does both simultaneously. 1 1. need consider only those system components with response times of a few seconds.Steam Turbine Prime Movers 437 that could utilize either fossil or nuclear fuel. Here the “nozzle” through which the steam expands is moving with the shaft. at relatively low pressure and temperature. boiler following (adjusting firing rate to hold throttle pressure).8 would be applicable to these longer-duration studies. A somewhat more realistic picture of the combined impulse-reaction blading is shown in Figure 11. The general block diagram of Figure 11. leaving the nozzle at high velocity as shown in Figure 11. and may require some consideration of the dispatch system. high-pressure steam is admitted through a set of control valves and allowed to expand as it passes through the turbine. 430 Chapter 11 Fig. 1 1.9 Two types of turbine blading. the left of the figure are impulse stages, whereas those on the right are reaction stages. In many turbines, impulse stages are used at the high-pressure, high-temperature end of the turbine and reaction blading at lower pressures. This is because there is no pressure drop across impulse stages and hence there is little tendency for the high-pressure steam to leak past these stages without doing useful work. As the steam expands in passing through the turbine, its volume increases by hundreds of times. At the lower pressures, reaction blading is used. Here, the steam expands as it passes through the blading and its pressure drops. The steam velocity increases as it passes through fixed blading as shown in Figure 11.10, but it leaves the moving blades at a speed about equal to the blade speed. The impulse stage nozzle directs the steam into buckets mounted on the rim of the rotating disk and the steam flow changes to the axial direction as it moves through the rotating disk. In reaction blading, the stationary blades direct the steam into passages between the moving blades and the pressure drops across both the fixed and moving blades. In impulse blading, pressure drops only across the nozzle. In the velocity compound stages, steam is discharged into two reaction stages. The velocity stage uses a large pressure drop to develop a high-speed steam jet. Fixed blades then turn the partially slowed steam before it enters the second row of moving blades, where most of the remaining energy is absorbed. Because of the tremendous increase in the volume of steam as it passes through the turbine, the radius of the turbine is increased toward the low-pressure end. In many turbines, the steam flow is divided into two or more sets of low-pressure (reaction) turbines. Figure 10.1 1 shows several typical tandem compound configurations and Figure 11.12 shows several typical crosscompound designs. In some designs, the steam is reheated between stages to create a reheat cycle, as noted in the figures, which increases the overall efficiency. In other designs, a portion of the steam is exhausted from the various turbine pressure levels to preheat water that is entering the boiler, which is called a regenerative cycle system. The various valves that control the turbine operation are shown in Figure 11.12 and will be discussed in the order encountered by the steam as it moves through the system. Steam leaves the main steam reheater of the boiler at high pressure and is superheated, in most cases, to high superheat temperature. For example, a large fossil fuel unit uses superheated steam at 2400 psi and 1000°F for a 1.0 GW unit [15]. A modem 750 M W nuclear design uses 850 psi saturated (0.25 percent moisture) steam [16]. The steam heaters contain steam strainers Steam Turbine Prime Movers 439 Steam Pressure Fixed t Fixed t t Fixed Fig. 1 1.10 Combined impulse and reaction blading [ 6 ] . to catch any boiler scale that could damage the turbine. A typical steam generator and turbine system is shown in Figure 11.13 [7]. The main stop valve or throttle valve (#2 in Figure 11.13) is one means of controlling the steam admitted to the turbine. It is often used as a start-up and shut-down controller. During startup, for example, other inlet valves may be opened and steam admitted gradually through the stop valve to slowly bring the turbine up to temperature and increase the turbine speed to nearly synchronous speed, at which point the governor can assume control of the unit. This mode of control is known as full-arc admission. The main stop valve is also used to shut off the steam supply if the unit overspeeds. The unit may be under automatic or manual control, but is usually controlled automatically through a hydraulic control system. A typical example of the several valves controlling a large steam unit is presented in Figure 11.13 [7]. This system is typical of many large steam power plants, having both superheater and reheater boiler sections and three separate turbines, representing high pressure (HP), intermediate pressure (IP), and low pressure (LP) units. The admission or governor valves, also known as control vaZves (#3 in the figure), are located in the turbine steam chest and these valves control the flow of steam to the high-pressure turbine. In large units there are several of these valves, and the required valve position is determined by the governor (D in the figure). An overview of the turbine control for a typical steam power plant is shown in Figure 11.14. Steam is admitted through the main stop valves to a set of control valves and admission of steam into the high pressure turbine is regulated by a set of nozzles distributed around the periphery of the first stage of turbine blading. If only a few of the control valves are open, the A40 Chapter 11 Single-Casing Single-Flow t Single-Casing Opposed-Flow t Two-Casing Double-Flow Reheater Two-Casing Double-Flow-Reheat Reheater Three-Casing Tripple-Flow-Reheat Reheater Four-Casing Quadruple-Flow-Reheat Fig. 1 1 . 1 I Typical tandem compound steam turbine designs with single shaft [6]. steam is said to be admitted under partial arc of the first stage rather than through all 360 degrees of the circumference. This is called “partial arc admission.” Two types of overspeed protection are provided on most units. The first is the normal speed control system, which includes the control valves and the intercept valves. The second type of overspeed control closes the main and reheat stop valves, and if these valves are closed, the unit is shut down. Two types of control valve operation are used. In one type, the control valves are opened by a set of adjustable cum Zijlers, as shown in Figure 11.15. In this arrangement, the valves can be opened in a predetermined sequence as the cam shaft is rotated. In response to a load increase, the flow of steam to one input port may be increased and a closed port may simultaneouslybe cracked Steam Turbine Prime Movers 44 1 Reheater Reheater t Two-Casing Double-Flow Two-Casing Double-Flow-Reheat 1 t 1 Four-Casing Quadruple-Flow-Reheat I r""l I Reheater Four-Casing Quadruple-Flow-Reheat Five-Casing Sextuple-Flow-Reheat Six-Casing Sextuple-Flow-Double-Reheat Six-Casing Octuple-Flow-Reheat Fig. I 1.12 Typical cross-compound steam turbine designs with multiple shafts [ 6 ] . 442 Chapter 1 1 1rlll Fig. 1 1.13 Example of a large boiler configurationshowing major system components and controls 171. Steam Generator I- - - - - - I Main stop Valve Crossover F -, , , ,-I Overspeed Trin - -_ ' I I High -b Pressur -. Intermediate Pressure I I I Low Pressure-Turbines n Generator -- Jr ! Valves L-----2&' - J . Load ' Intercept Valve Condenser Reheat -- Reheater Fig. 1 1 . I A reheat turbine flow diagram. 4 Steam Turbine Prime Movers 443 Fig. 1 1.15 Cam lift steam turbine control valve mechanism. open. This distributes the steam around the periphery of the first stage, assuring a uniform temperature distribution and controlling the power input. The cam shaft is controlled by the governor acting through a power servomotor, as shown in Figures 11.13 and 11.14. The other type of steam admission control is called the “bar lift” mechanism. This type of valve control is shown in Figure 11.16; each valve in a line of valves is lifted using a bar, but each valve is a different length so that the valves open sequentially. As load is added to the turbine, the bar is raised and steam flow is not only increased to the first-opening valve, but additional valves are also opened. The separate valves feed steam to different input ports around the periphery of the first-stage blading and thus increase the power input to the turbine. The bar lift is actuated by the governor servomotor through a lever arrangement. Fig. I 1 .I6 Bar lift steam turbine control valve mechanism [2]. 444 Chapter 1 1 The high-pressure turbine receives steam at high pressure and high temperature, and converts a fractionfof the thermal energy into mechanical work. As the steam gives up its energy, it expands and is cooled. Steam is also bled from the turbine and piped tofeedwater heaters. This has proven economical in reducing the boiler size and also reducing the size required at the low-pressure end of the turbine. The turbine extraction points vary in number from one to about eight, the exact number being dictated by design and economy. In the reheat turbines, shown in Figure 11.14, the steam exhausted from the HP (high-pressure) turbine is returned to the boiler in order to increase its thermal energy before it is introduced into the intermediate-pressure (IP) turbine. This reheat steam is usually heated to its initial temperature, but at a pressure that is somewhat reduced from the HP steam condition. Following the reheater, the steam encounters two valves before it enters the IP turbine, as shown in Figures 1 1.13 and 11.14. One of these is the reheat stop valve and serves the function of shutting off the steam supply to the IP turbine in the event the unit experiences shut-down, such as in an overspeed trip operation. The second valve, the intercept valve, shuts off the steam to the IP turbine in case of loss of load, in order to prevent overspeeding. It is actuated by the governor, whereas the reheat stop valve is actuated by the overspeed trip mechanism. The IP turbine in Figure 1 1.13 is similar to the HP turbine except that it has longer blades to permit passage of a greater volume of steam. Extraction points are again provided to bleed off spent steam to feedwater heaters. The crossover, identified in Figure 1 1.14, is a large pipe into which the IP turbine exhausts its steam. It carries large volumes of low-pressure steam to the low-pressure (LP) turbine@). Usually, the LP turbine is double or triple flow as shown in Figures 11.11 and 11.12. Since a large volume of steam must be controlled at these low pressures, doubling or tripling the paths available reduces the necessary length of the turbine blades. The LP turbines extract the remaining heat from the steam before exhausting the spent steam to the vacuum of the condenser. It is desirable to limit condensation taking place within the turbine, as any water droplets that form there act like tiny steel balls when they collide with the turbine blades, which are traveling at nearly the speed of sound. We previously specified that the HP turbine extracts a fractionfof the thermal power from the steam. Then the IP and LP turbines extract the remaining 1 - f of the available power to drive the shaft. Usually,fis on the order of 0.2 to 0.3. For example, in a certain modern 330 MW turbine,fis determined to be 0.24. This is a rather typical value. 1 1.6 Steam Turbine Control Operations The controls for a steam turbine can be divided into those used for control of the turbine and those used for the protection of the turbine. It is difficult to sketch a “typical” control system for a steam turbine since these controls depend on the age of the unit and the type of controls available at the time of unit installation. Since power plants operate for many years, there are likely to be many different controls, using different technologies, on any given power system. However, we can summarize the most common controls as being either “traditional” or “modem,” with those terms also having a somewhat variable meaning due to the steady advance in control technology. The control operations that are usually considered to be “traditional” are listed in Table 11.2. These are controls that have been required for many years and that require only the very basic technologies for their operation. It is apparent that plant control systems become more complex due to the demands of interconnected operation and the availability of more modem methods of control. The newer controls provide many functions that were not considered necessary for older units, and some that were not available due to limitations of the available technology at the time of manufacture. Steam Turbine Prime Movers Table 11.2 Traditional and Modem Steam Turbine Generator Controls 445 Traditional Controls Speed control, near rated speed Overspeed protection Load control-manual or remote Modem Controls All traditional controls and protections Long-range speed (zero to rated speed) Automatic line speed matching Load control; automatic load setback Admission mode selection Automatic safety and condition monitoring On-line testing of all safety systems Fast or early valve actuation Interface to the plant computer Interface to area generation control system Basic control and protection Initial pressure Vacuum Vibration Others, as needed Many of the plant controls are hydraulic, using high-pressure oil supplied by a shaftmounted main oil pump. These high pressures are practical for the operation of power servomotors for control purposes. For example, many control valves are actuated by hydraulic means. In modem plants, many systems also use electric controls as well. The control functions for the turbine include the servomotor-driven control or governing valves and the intercept valves, which control the amount of steam admitted to the turbine. Positioning intelligence for these valves comes primarily from the speed governor, the throttle pressure regulator, or from an auxiliary governor. There is also an interlocking protection between the control and intercept valves so that the control valves cannot be operated open when the intercept valves are closed. The protective controls include the main stop valve (throttle valve) and the reheat stop valve. The reheat stop valve is always either fully open or fully closed, and is never operated partially open. The main stop valve may operate partially open when used as a startup control. Both valves are under control of a device that can rapidly close both valves, shutting down the turbine on the occurrence of emergency conditions such as overspeed trip, solenoid trip, lowvacuum trip, low bearing oil trip, thrust bearing trip, or manual trip. During normal operation, both of these stop valves are completely open. A primary function of the main stop valve is to shut off the steam flow if the unit speed exceeds some predetermined ceiling value, such as 110% of the rated value. Steam turbine blading experiences mechanical vibration or oscillation at certain frequencies. The turbine designer assures that such oscillations occur above or below synchronous speed, with a generous margin of safety. Also, with the longer blades traveling at nearly the speed of sound, destructive vibration levels may be reached if the speed is permitted to increase substantially beyond rated speed. Thus, speed control on loss of load is very important and is a carefully designed control function. [9]. The operation of a steam turbine on loss of load is approximately as shown in Figure 11.17. It is assumed that the generator breaker opens at t = 0 when the unit is fully loaded. On loss of load, the turbine speed rises to about 109% in about one second. As the speed increases, the control valves and intercept valves are closing at the maximum rate and should be completely closed by the time the speed reaches 109% of the rated value, at which time the turbine speed begins to drop. At about 106%, the intercept valves begin to reopen so that a no-load speed of 105% might be achieved. If the speed changer is left at its previous setting, the unit will continue to run at 105% speed on steam stored in the reheater. There is usually sufficient steam for one to three minutes of such operation. Once the reheater steam supply is exhausted, the speed will drop to near 100% and the governor will reopen the control valves. The definition of what constitutes an emergency overspeed [IO] is a figure agreed upon by 446 Chapter 1 1 110-1 lo! 11 0 Intercept Valve starts iliary Load I lo 1012 1 \ Remaining on Generator \ -. - on Generator 0 1 Time in minutes 2 rm Fig. I I . 17 Estimated speed versus time following sudden reduction f o a maximum load to the values noted. turbine manufacturer and purchaser, but may be in the region of 1 10 to 120%of the rated value. If the speed reaches this range, an emergency overspeed trip device operates. Usually the overspeed trip mechanism depends on centrifugal force or other physical measurements that are not dependent on the retention of power supply. Some devices include an eccentric weight or bolt, mounted in the turbine shaft, with the weight being balanced by a spring. At a predetermined speed, such as 1 1 1%, the centrifugal force overcomes the spring force and the bolt moves out radially far enough to strike a tripper, which operates the overspeed trip valve. 1 1.7 Steam Turbine Control Functions We now investigate the transfer functions that describe the operation and control of a typical steam turbine.* The system under investigation is the reheat steam turbine of Figure 1 1.13, with controls as described in the preceding sections. The block diagram for this system is shown in Figure 11.18 [lo], with controls as described in the preceding paragraphs. Our immediate concern is with the thermal system between the control valves, with input q2and turbine torque T. The symbols used in Figure 11.17 represent per-unit changes in the variables, as defined in Table 11.3. For the present, we will accept the transfer functions of governor and servomotor and reserve these for later investigation. Let us examine the functions between qz and T in Figure 1 1.18 more carefully. The control valve transfer function is nearly a constant and would be exactly 1 .O were it not for nonlinear variations introduced by control valve action. This is due to a combination of nonlinearities. First of all, the steam flow is not a linear function of valve lift, or displacement, as shown by the right-hand block of Figure 11.19. It is, in fact, quite nonlinear, exhibiting a definite saturation as the valve opening increases. One way to counteract this nonlinearity is to introduce a nonlinearity in the valve lifting mechanism, as shown in the left block of Figure 11.19. This is accomplished with a cam lift mechanism, as shown in Figure 11.20. Here, the cam acts as a function generator providing an output *We follow closely here the excellent reference by the late M. A. Eggenberger [lo] who did significant work in this field. The authors are indebted to Mr. Eggneberger for having shared his work, some of which is unpublished. Steam Turbine Prime Movers 447 . Fig. 1 1.18 Block diagram of mechanical reheat turbine speed control [lo]. L =f(v2, L) (1 1.2) in which the output L is a function not only of q2but also of L. In this way, the transfer function of the two blocks taken together are nearly linear for any given valve. Still, a small nonlinearity exists in the overall transfer function, as shown in Figure 11.18, due to “valve points,” as this phenomenon is known in the industry. This refers to the point at which one valve, or set of valves, approaches its rated flow and a new valve (or valves) begins to open. Table 11.3 Definition of Per-Unit Change Variables Per Unit Change Variable Speed of rotation Developed torque Load torque Steam flow Servomotorstroke Speed relay stroke Speedlloadreference Speed governor stroke Speed error signal Valve steam flow HP turbine torque Reheat pressure IP + LP torque Accelerating torque Defining Equation (T= Remarks N R = Rated speed NA NR q-= TmA TmR TmR = Rated full load torque A=& TeR TeR = Rated electrical torque QA P=- Q R = Rated steam flow in Ib/sec QR 172 = Y2R Y2A YzR= Servomotorposition for steady rated load Y I R Speed relay stroke for full load = RR= Reference position at rated load and rated speed XR= Speed governor stroke for 5% speed change 711 = Y,R RA RR YIA P= - l =XA XR E EL” 7HP +R q-IP&LP 7, Speed relay input Control valve output HP turbine output variable Reheater output variable IP + LP turbine torque Chapter 1 1 ,&{ lift k (11.3) (11.4) (11.5) Fig. 1 1.19 Block diagram for camshaft and valve function generators [IO]. This causes the transfer function to consist of a series of small curved arcs, as shown in Figure 11.18. To compute the transfer knction of steam flow versus servomotor stroke, we write K3= Pv If it were not for valve points, the curve expressing the function K3 would be a constant with value of unity, with the incremental regulation at the operating point the same as that of the governor (usually 5%). If we define incremental regulation Rias [ 101 du R.= ' dP where u is the per-unit speed, P is the per-unit power, and Riis evaluated at the operating point. If we let Rs be the steady-state regulation or droop L Valve Lift Fig. 1 I .20 Mechanical function generator (cam-operated control valve). Steam Turbine Prime Movers 449 then we have K3= RS (11.6) Eggenberger [lo] points out that Riis often between 0.02 and 0.12 over the range of valve strokes and may be taken as 0.08 as a good approximate value. Using this value, we would compute for a typical case 0.05 K3 = -= 0.625 0.08 (11.7) From Figure 11.8, we see that the steam is delayed in reaching the turbines by a bowl delay T3,expressed in terms of servo stroke and turbine flow parameters as (11.8) where T3 is the time it takes to fill the bowl volume VB(ft3) with steam at rated initial conditions, with specific volume initially of v (lbdsec), or [ 101 T3 = -seconds VQY VB (11.9) Typical values of T3are given as 0.05 to 0.4 seconds. For a straight condensing turbine with no reheat, the torque versus servomotor stroke is given by (1 1.8). This situation is illustrated in Figure 11.21 and is accomplished mathematically by replacing p T in (1 1.8) by 7. This is equivalent to setting the fractionfof torque provided by the HP turbine to unity. For a reheat turbine, there is a large volume of steam between the HP exhaust and the IP inlet. This introduces an additional delay in the thermal system. From Figure 11.18 with elementary reduction, we have [ 101 ( 11.10) Fig. 1 1.21 Torque production as controlled by servomechanism stroke. 450 Chapter 1 1 wheref is the fraction of the total power that is developed in the high-pressure unit and is usualR ly between 0.2 and 0.3. The parameter T is the time constant of the reheater and is defined in a manner similar to (1 1.9) or (11.11) where VR= volume of reheater and piping, ft3 QR,.= full load reheater steam flow, lbdsec v, = average specific volume of steam in the reheater, Et3/lbm R Since the reheat temperature is not constant, computation of T involves taking averages, but it is usually in the neighborhood of 3 to 11 seconds. This long time constant in the reheater causes a considerable lag in output power change following a change in valve setting. In HP turbines, there may be a delay of up to 0.5 seconds, depending upon control valve location. A much larger delay occurs in the IP and LP sections, however. This is due to the large amount of steam downstream of the control valves, and this steam must be moved through the turbines and reheater before the new condition can be established. These delays are both shown in Figure 11.22, where the control valve is given a hypothetical step change and the power output change is plotted [lo]. A five second value for TRis assumed. The speed-torque transfer function is given in Figure 11.18 as [101 -=-. 0 r 1 T4s (11.12) The time constant T4 is the total time it would take to accelerate the rotor from standstill to rated speed if rated torque, T,, is applied as a step function at t = 0. At rated speed, the kinetic energy in the rotating mass is 1 Wk = -J 2 w ~ (11.13) 70% Control Valve Position 60% 0 1 2 3 4 5 Time, seconds 6 7 8 9 Fig. 1 1.22 Reheat turbine response to a control valve change. )(s+ l/TR) + + (11.165 x 109)Pr so that T4 = where P. Referring to Figure 11. = rated power in MW WR2 = rotor inertia in lbm-fi2 N = rated speed in rpm R Another useful constant is the so-called specific inertia of the turbine-generator [lo]: JSP= (11. These constants are shown in Table 1 1.14) for constant torque gives (11.21) . Additional insight into the control of the steam turbine system is gained through an evaluation of system performance by the root locus method [12].) + ~/T. as the losses are very small.98 Jspseconds (1 1. A set of typical constants for all values shown in Figure 1 1.)(s 1/T2)(~ ~/T. 12).15) the rated value of torque. we may write the open-loop transfer function as KG(s) = S(S K(s + llfT.13) we can compute m T4 = -seconds Pr where the units must be consistent. as the turbine speed increases..14) (11.16) since T R= Pr/wR.18 that relates (T to T by a feedback system wherein a portion of the speed increase is fed back as a negative torque [ 101.4.18 is given in [ 101 and is valuable for making comparisons of the various system lags under consideration. this is usually neglected. the load torque increases and the loss torque varies as some power of the speed.19) and this is convenient since it usually turns out to be nearly unity.18) (F)( &) WR2 N 2 x lbm-ft/MW (11. We usually compute wk (11. T4 = 5. Eggenberger [101 shows that this can be accounted for by replacing the single block in Figure 1 1.3) through (1 l .14 and equations (1 l .83(WR2)N. However.16) and (1 1.Steam Turbine Prime Movers 45 1 and the differential equation of motion is Jh = Ta = a constant where we take Ta = TmR (11.From (1 1. Solving (1 1.17) - 0.2 MWs 3600 x lo6 (WR2)N? seconds (2. In terms of this constant.20) Actually. 303 9. Other component values affect the response as well.14 s 0.30 s 0. servomotor.4 Typical Values of Constants Used in Steam Turbine Analysis Parameter C . Table 11.333 MaximUln 12.3 s - 0. speed relay.55 3.05 to 0.6 to 0.00 0.5 has some influence on system behavior.24. where poles of a nonreheat turbine are plotted as a band of values rather than as a point in the s plane.00 Reheat Minimum 5.8 0.27 Maximum 12. For large disturbances. This means that the reheater pole and zero will always be relatively close to the origin and will.33 2. and Gains Item PoleIZero Pole Symbol 1/T.25 s 0. 11T2 l/T3 1/TR 1JJTR K Nonreheat Minimum 7. Here.3 5to12s where Considering the range possible for each variable as shown in Table 11. even for small disturbances. which may vary from 0.50 6.00 3. have a great influence on the system dynamic response. this system may be designed with a wide range of response characteristics.08 to 0.33 - Zero Gain - - 1. as shown in Figure 11. we have a range of pole-zero locations and gains as shown in Table 11. A similar plot for the reheat turbine is shown in Figure 11.15 to 0. since the system response depends on these pole locations. therefore. which may be quite close to the origin.667 1600 46. the four poles due to the inertia.2 to 0. This is especially true for the valve bowl delay. the problem is greatly complicated because the reheater should then be treated as a nonlinear model to account for the spatial distribution of flow and pressure in both reheater and piping.05 to 0.50 0.67 20.3 seconds [IO]. one for the straight condensing (nonreheat) turbine and one for the reheat turbine will illustrate the method. Non-reheat Turbine 20 Reheat Turbine 20 TI T2 K3 T3 TR f T4 Normalized speed governor constant (5% regulation) Speed relay time constant Servomotor time constant Valve gain at no-load point Valve bowl time constant Reheater time constant Load on HP turbine per unit Turbine characteristictime 0. especially the servomotor pole.5 Range of Values for Poles. and valve bowl are far enough from the origin to be offscale for the scale chosen for this figure.5.625 0.3 5340 . The range of values shown in Table 11.05 to 0.452 Chapter 11 Table 11. A convenient method of analyzing steam turbine systems is to use the root locus technique [12].15 4.18 s 0. It is obvious that.50 6.4 s 3tolls 6t012s 0.4.091 0. Two examples.67 20. Zeros.15 to 0.08 to 0.23. 625 and C = 20.0667 s 3 T = 10.1 Prepare a root-locus plot for a nonreheat turbine with the following constants: T = 0. Example 11. Solution The block diagram for this system is that shown in Figure 11. . roots if K3 = 0.-0.T2T3T4 = 937.5 Fig. we can compute the gain K as KG T.25.5 -1 -0.22) For the constants given in this example.1s I T = 0.Steam Turbine Prime Movers 453 I P w I 8 z mzAw I s 0 (I . -. 1 1.\I Ilr I U \)U n I AA M \ u 0 \ -20 -15 -10 valve bowl delay -5 “0 - ---5 Fig.0 s 4 Determine the damping ratio and undamped natural frequency for the two least damped .5 (1 1. The open-loop transfer function is KG(s) = K s(s + 5)(s + lO)(s + 15) K= - K s4+ 30s3 + 225s2 + 750s (1 1.23) < I 1 Zero Range I F 1 I V I I -2 -1.5 llR ** U 0 - M Range -.+5 fi . 11.24 Pole and zero for the reheater.2 s 2 T = 0.23 s Plane plot of poles for the nonreheat turbine. 26) In our case.24) 3.Z = 4 .XZ = -=-7. The locus “breaks away” from the negative real axis at points kl and k2 defined by the equations . Write the polynomial XP . The asymptotes lie at angles of e. *I350 P-z (1 1.9K K 275 750 3K 0 K 0 For the first column in this array to be positive. which are required in order to construct the root locus plot: 1.27). we require that K 5 6167 The auxiliary polynomial [131 is 740s2+ 3(6167) = 0 or s = *j5 (11. 1 1. = (2’+ 1)”O0 = *450.5 -30 P-z 4 D(s) + KN(s) = 0 (1 1.25 Block diagram for the nonreheat turbine We also compute the following constants.0 = 4 2. The center of gravity is located at C.G. we construct the Routh’s table [131to find the critical value of gain and the point of the w-axis crossing: s4 s3 S2 S‘ SO 1 30 740 55500 . The excess of poles over zeros = P .25) (1 1.454 Chapter 11 Fig. = 4.28) 5.27) From (1 1. we have s(s + 5)(s + lO)(s + 15) + K = s4 + 30s3+ 275s2 + K (1 1. 30) k2 = 15 .Next Page Steam Turbine Prime Movers 455 1 -_- k l 1 5-kl +-+.24) to (1 1. We can also locate the point corresponding to the assumed gain of 937. the damping ratio is s = 0.1. .26. (11.29) by trial and error to find k.91) and. With this value of gain. by symmetry.29) -=- 1 15-k~ 1 1 +-k2-5 + -k2 We solve (1 1.09 (11.26 Root locus for a nonreheat turbine system. we construct the root locus diagram shown in Figure 11. E 1. 1 I .7 (1 1.3l).32) / \ \ / / Fig.1 10-kl 1 kz-10 1 15-kl (1 1.91 = 13.91 (actually -1.31) 6. Incorporating information accumulated in equations (1 1. In our case. By comparison.36) (1 1.2 root locus plot that the poles are labeled to remind us of the reason for their existence. being located at approximately -13. corresponding to a gain of 937. Thus.26.2 and 0. 4' 2 0.6%)and oscillationsring down for almost four seconds.7.2 TR=5s Then the open-loop transfer function becomes KG(s) = and the normal value of K is K(s + 1) s(s + 5)(s + lO)(s + 15)(s + 0.37) . If some oscillation can be tolerated.-sin(w.2 is desirable as this corre0. what we would like.7 there is very little overshoot.2) (11.2 If the system of Example 11.27 shows a typical second-order response for values of 5 of 0.5 in the negative-real direction. We would hope to have the damping factor 5 be fairly large for good damping and to prevent an overshoot or too long an oscillation. Example 11. They can be moved by changing the appropriate design parameters. Note that when 5 = 0.54 (1 1.2 radiansls (11.Previous Page 456 Chapter 1 1 and the undamped natural frequency is wn = 2.1 is a reheat system. Certainly. Two of these solutions correspond to responses that are very quickly damped out. there are actually four solutions.6%). Note that.t + 4) (11. the fractionf of power generated by the HP turbine and the reheater time constant T R must be specified. with l= there is practically no overshoot and the system is very well damped. generally speaking. this system could be operated at a higher gain.2 the overshoot is about 50% (actually 52.35) k where k= 4 = tan-*- k w. indicated by the dots on the locus. Suppose we let f = 0. our system will respond approximately as a second-order response [ 141: e-bnt a(t) = u(t) . this system could be operated at a higher gain. Also note in L.34) and we can neglect the quickly damped solutionswith very little error. but with 5 = 0. Figure 1 1.7 sponds to about 50% overshoot (actually 52.33) These values are indicated in Figure 11. We now recognize the significance of the solutionjust obtained. the least damped roots are located at -50. If some oscillation can be tolerated. = -1. = kwn u(t) = unit step function 5 This response is a damped oscillatory response and this is. corresponding to an overshoot of about 25%. as shown in Figure 1 1.28 Block diagram of a reheat turbine system. however. we observe that for a gain of about 187.28. Fig. which would improve the product by a factor of three or four and the oscillations would decay much faster as we see from the exponent of (1 1.Steam Turbine Prime Movers 457 I I 1.5 or 2.2 I - I I I I I Fig. 11. The principal dynamic components that effect the time lag of delivered mechanical power are the speed relay. . The block diagram for this new system is shown in Figure 11.6 1 I I I I I -0.14. 1 1. In normal operation. and low-pressure turbines on a single shaft. depending on the scheduled generation output of the unit.30 by solid lines. intermediate-pressure.29. that the system gain could be increased substantially with practically no change in 5 up to a frequency of about 1. These dynamic components are connected in the system diagram of Figure 11. the intercept valve is fully open.2 = (11.and the feedwater heaters.35).38) is much less than for the straight condensing turbine. This more detailed model consists of high-pressure. the damping ratio is about 0. the drum. From this plot. control valves. Thus the product --&" -0.0.4. but the control valve may be only partially open.30. Note also. driving a generator and excitation system. steam bowl. The root locus plot is shown in Figure 11.5 radians per second. The block diagram of a more detailed dynamic model of a reheat steam turbine system is shown in Figure 11. and the undamped natural frequency is about 0.27 Step response of a second-ordersystem. This system will initiate fast turbine control and intercept valve closure in the event of a load rejection. measured by the generator current. This protection will operate if the difference between these measured power values becomes greater than a preset value. 1 1. such as faults. The recovery time of boiler pressure following a sudden change in turbine control valve setting is measured in minutes for systems of conventional design. The control logic operates by comparing the turbine power. typically about 40% of full load. This provides overspeed protection for the generating unit that might follow a loss of load. the boiler-turbine system is operating with . The steam generator is such a component. 1 1. Thus. which is determined by measuring cold reheat pressure.8 Steam Generator Control The expansion of power system interconnections has necessitated more precise control in order to hold the fiequency stable and to control disturbances.4 \ b0 \ \ \ \ \ \ \ * -1 5 bowl delay speed relay \ \ \ /' servo / / / / / / K = 187 - / / / / / / / / / / \ \ \ \I \ \ / / / /' Y' \ \ A Fig. During this period. system components that are usually thought of as quite slow in response must be investigated for possible behavior that might be detrimental to system damping. Steam generators can be either fossil or nuclear fuel systems. The dashed lines in Figure 11.458 \ Chapter 1 1 \ \ c= 0. It has also introduced a new class of stability problems that are not so much concerned with system recovery following major impacts.29 Root locus for a reheat steam turbine system. as with the control and damping of sustained oscillations over periods of several minutes duration. but here we shall concentrate on fossil-fueled boilers. and the rate-of-change in generator current is also greater than a set point value.30 show the connection of an overspeed protection system. and the generated power. 30 Typical turbine control dynamic for a reheat steam tur . 11.Fig. The size of the off-diagonal terms. with the result that faster response and more precise control are being realized. Any response to such an error will.460 Chapter 1 1 Table 11. However. as shown in Figure 11. requires a more sophisticated control. For example. G&). in most boilers. . i Zj. This kind of system model causes cross coupling between variables. Repeating for other components of m determines G completely. because of its thermal design.31. so that several input variables can actuate a number of actuators simultaneously. 11. This increased interest in boiler control has affected later designs for drum-type boilers too.3 1 Block diagram of a coupled two-variable process. Traditionally. One alternative to this situation is the use of one multivariate controller [15. The introduction of the once-through boiler in the late 1950s also focused attention on boiler control.6 [151.is an indication of the cross coupling that exists in the system. each responding in its own way. 161. In this kind of control. Such controllers should force the system toward the new steady-state position in a much more optimal manner. the usual single-variable controls are those shown in Table 11. in most cases.6 Normal Boiler Single Variable Controls IndependentVariable Desuperheatingspray Firing rate Burner tilt Feedwater flow Controlled Variable Main steam temperature Output (drum) pressure Reheat temperature Drum level its open-loop gain changing and possibly oscillating slowly. With this type of system.3 1. as indicated in Figure 11. The output x corresponding to this component of m determines one column of the transfer function G. a chain reaction of controlled responses follows the change in one error and may unbalance the system for several minutes while all systems readjust themselves.39) Each element of G(s) may be found by setting all inputs m to zero except one. but they can hardly be considered to be beneficial. Thus. cause errors to appear in other variables. the design of a multivariable controller requires the use of an accurate model of the Fig. This type of boiler. which respond to an error in a single variable. How these low-frequency oscillations will affect the overall system behavior is not always clear. the control system for a boiler has been accomplished by using analog devices. a step change in any of the independent variable references or in load will cause a readjustment of all variables. the outputs x are related to all inputs m by a matrix G(s) in the equation x(s) = G(s)m(s) (11. 33 Multivariable control. .32 A multivariable process. The once-through (or once-thru. the drum boiler employs a large drum as a reservoir for fluid that is at an evaporation temperature. as it is often called) design has no drum and the fluid passing through the system changes state into steam and then into su- Fuel Air Tilts spray reedwater ' Process 'r. two distinct types of fossil-fueled steam generators have been designed and are widely used. As suggested by its name. controlled plant and this is not available for many problems.32 and 11.33.Steam Turbine Prime Movers 46 1 Throttle Pressure * I 4 VJ 0 5 8 P 8 A Tilt Feedwater Turbine Valve. - Boiler Main Steam Temperature Reheat Steam Temperature > Turbine' Drum Level > System Steam Flow Rate ~ 3 li > u 8 '53 Excess Air Fig. drum-type boilers and once-through boilers. Applying this concept to a steam generator system. 11.9 Fossil-Fuel Boilers As the technology has evolved. we can construct the system model as shown in Figures 1 1. 11.34. 1 1. I 1 Pressure SP! Trottle Temp SP Including >. A simple comparison of these two types of boilers is illustrated in Figure 1 1. Actuators -1 I Controller Matrix k%' Fig. a very elastic connection as the drum is not an “infinite bus” of thermal energy. The once-through design contains less fluid than the drum-type design and generally has faster transient response.35. It also serves as a storage reservoir to receive energy following a sudden load rejection. however. 11. In such a system. Since the fuel firing and pumping systems lag behind the drum demand by several seconds. New York. 1960. A.9.Water Steam --+-Flue Gas + _ _ _ a _ Line Types Tube Waterwall Sections Superheater Section Evaporator Section E D Drum F F WC 0 Feedhmp Water Circulating Pump Steam Output to Turbine Fig.) .34 Drum and once-through boiler configurations. the drum serves as a reservoir of thermal energy that can supply limited amounts of steam to satisfy sudden increases in demand. G . Some of the major control systems for the drum-type boiler are the following [16]: (a) Combustion control-he1 and air control (b) Burner and safety control (c) Boiler temperature control-burner tilt. perheated steam.- 0: It I I I I I E FP I I I P Drum-Type Boiler T S Once-Thru Boiler Legend --.462 6 Chapter 11 I :o I . It is. 1 1.1 Drum-type boilers A simplified sketch of the working fluid path in a drum-type boiler is given in Figure 11. Figures adapted from similar items in Power Station Engineering andEconomy. Vopat. Skrotski. gas recirculation (d) Feedwater control (e) Superheater temperature control-desuperheating (f) Reheat temperature control-gas recirculation . Bemhardt. the drum serves as a buffer between the turbine-generator system and the boiler-firing system. and W. McGraw-Hill. A. 35 A drum-type boiler arrangement. Some other control systems are: (a) Feedwater heating system control (b) Air heater temperature control (c) Fuel oil temperature control (in an oil fired boiler) (d) Turbine lubricating oil temperature control (e) Bearing cooling water temperature control (f) Mill temperature control (in a coal burning boiler) .Steam Turbine Prime Movers 463 Fig. 1 1. If we linearize about a quiescent operating point. it should provide at least a rough idea of the system behavior and permit us to study various control arrangements without becoming burdened by system complexity. Some valuable work [17-191 has added to our knowledge of boiler behavior as an element in a dynamic system. A certain mass of steam is stored in the boiler and any change in this mass affects the boiler pressure. We also recognize that the pressure at the drum is not the same as pressure at the control valves because of the pressure drop across the superheater.36 (b) Referring to the linear circuit of Figure 11.40) Drum Pressure I Throttle Pressure Superheaters Turbine Steam How (a) Schematic of Boiler-Turbine System v (b) Electric Analog of Boiler Pressure Phenomenon Fig. Still. which vanes as the square of steam flow rate.boiler pressure depends on steam flow.36 (a) and is easily studied by means of an electric analog as shown in Figure 11. This simplified model assumes that feedwater effects can be neglected and that the feedwater control satisfies the drum requirements. = drum pressure Z. we define the following analogous quantities: VRT throttle pressure = V.36 A simplified boiler-turbine representation [20]. Such is the approach presented in [20].464 Chapter 11 These controls are usually single-variable control loops. which is actually a huge distributed parameter system. however. the change in pressure drop is proportional to the change in flow rate and we are justified in using the linearized model of Figure 11.= steam generated Z2= steam flow to turbine R = friction resistance of the superheater R = resistance of the turbine at a given valve opening T (1 1. Thus. it is necessary to have an adequate mathematical model of the process. Such changes result from transient effects wherein the steam generated and the steam demanded by the turbine are unbalanced.36 (b).36 (b). In order to apply advanced control concepts. 1 I . . One boiler representation [20] considers the drum as a lumped storage element as shown in Figure 11. It also ignores the geometry of the boiler. PT. waterwall time constant. we can write PDA ( ~ K Q o ) Q A = where Qo is the steady-state flow rate and QA is the change in flow rate. With this configuration. The steam generated by the boiler is proportional to the heat released in the furnace. but lags behind this heat release by 5 to 7 seconds. may be represented by the lumped parameter model shown in Figure 11. linearized about a quiescent operating point. Q = KVPT Linearizing. a change in control valve opening is represented by a change in RT. the change in valve and the throttle pressure VTR opening.41) and solving for I z A we get (1 1. and may be anything from zero to about 30 seconds [20]. Q . for small perturbations. The delay time constant TFis typically about 20 seconds and the dead time Td depends on the type of fuel system. All of the above relationships. we write the pressure drop from drum to throttle as PD(in lb-mass) or. the system can be arranged as shown in Figure 11.38.43) where K is the friction coefficient and Q is the steam flow rate in l b d s .42) will experience a drop proportional to RTA.45) and is a function of Qo as noted. To study the control of the boiler dynamics.37. is a function of the throttle pressure. is a function of load level. as an estimate [20]. it is possible to investigate the nature of the control system and also to optimize the effect of both .e.Steam Turbine Prime Movers 465 In this model. We may then write v c = H2 + R T I ~ V O + VCA = R(I20 + ZZA) + (RTO + RTA)(z20 + z2A) C (11.44) R = 2KQo (1 1.. Finally. the The boiler storage effect is an integration with capacitance (or thermal inertia or time constant) C. This gives the needed relationship between the net unbalance in boiler steam flow to the drum pressure. In terms of system quantities. then we can think of the generated steam as being delayed by a time constant Tw. In the analog.and a coefficient Kv proportional to the valve opening. at constant firing rate: Po = KQz (1 1.47) where K . Then. The value of R is a function of the quiescent point of operation (the load level).If we let Qw be the flow of steam from the boiler. The steam flow to the turbine.46) (1 1. the fuel system dynamics can be represented by a delay and dead time. i. we write (1 1. (11. etc.. Increased steam flow tends to decrease temperature. the exact operating point plus conditions of soot.38 is recognized to be a “boilerfollowing’’ control arrangement. 1 1.+ * Boiler Fig. an increase in firing rate will always produce an increase in pressure. These are primary or dominant effects and their sign is always the same.38 Typical control system configuration for a drum-type boiler. Generation \ Generation Combustion Control output . will effect the response and its direction. are opposing.. in a boiler..y C . and so on. on the other hand. Some effects. I I pressure and flow changes.. an increase in desuperheat spray will always decrease throttle temperature. an increase in fuel increases steam pressure and this tends to increase steam flow. whereas the increase in fuel input would ordinarily increase temperature.... The configuration of Figure 11.466 Chapter 1 1 I I I I I I Fuel I pBr Air . slag.B o i l e i System Fig.....37 Block diagram of a lumped parameter drum-type boiler. Thus. Thus.Control Desired steam 0 output -Generator . . an increase in air flow will always decrease boiler pressure. 1 I .. Thus.. Multivariable controllers have an additional problem not usually present in single variable controllers-the consistency of results [ 191. 1 1. This kind of control is designed to perform the following operations: 1. Reduce operating level (runback) to safe operating level upon loss of auxiliaries. Computer outputs are generated to the combustion and governor controllers. The usual approach to the solution of these equations is to break the space continuum into a series of discrete elements and convert the partial differential equations into ordinary differential equations in the time domain [18.39 is shown in block diagram form in Figure 11. shown in Figure 1 1. the system does not simultaneously adjust all possible variables. Compare Figure 11. the desired unit demand signal (from the automatic load control device). 3. It consists of two components: the “boiler-turbine governor” and the “unit coordinating assembly.19]. . Figure 1 1. Finally. The controller of Figure 11.” The boiler-turbine governor produces a “required output” set point that takes into account the capa- Desired Unit Generation Actual Unit Generation V Direct Energy Balance Control System Y Combustion Control A 1 Governor Control Y * f Boiler Main Steam Pressure Generator Fig. Observe load limit capabilities of boiler. The references cited will be helpful to one who wishes to pursue the subject further.38 to see the difference between the two types of controls.39 with Figure 11. is called a “Direct Energy Balance Control System’’ [21] by its manufacturer.39.39 A multivariable control system [2 11. and generator. and desired steam pressure are all input quantities to the controller. but it does deal with the primary variables. and these equations -are nonlinear partial differential equations in space and time.39 displays the major components of this type of system.40. This system. Referring to the figure. The equations of the system are those of mass flow and heat transfer in superheater and reheater tubes. These equations may be solved by digital computer.Steam Turbine Prime Movers 467 One of the problems in designing an appropriate controller is that of starting with a good mathematical model of the system. before leaving the subject of drum-type boiler control we note one type of multivariable control that has been used on both drum-type and once-through boilers. This is especially difficult in boiler systems because of the difficulty in modeling a distributed parameter system and also because of the nonlinear character of steam properties. 2. Models of this kind have been studied but are beyond the scope of this book. Adjust both boiler and turbine-generator together. main steam pressure. as required by automatic or manual controls. Thus. actual unit generation. turbine. This unit compares the required output for the unit against the actual unit generation and computes an error signal from which the governor and fuel-air systems are controlled. When operating under automatic load control.. In this mode. and the turbine follows the boiler.- Generation I I I I Boiler Turbine Governor I I I I I I I I I I Frequency Bias (Rates of Change) I I I I I I I I I I I (Limits) (Runbacks) I I To Combustion Control . the governor adjusts the pressure automatically. the control just described may be operated in any one of the following four modes. but in opposite directions. It also fixes the rates of change according to a preselected setting and provides for emergency runbacks and limits.. as shown in Figure 11. The “unit coordinating assembly” is shown in greater detail in Figure 11.7._ * * . In this mode. In practice. the unit automatically achieves the new setting at a preset maximum rate of change. the operator adjusts the boiler inputs and the turbine governor manually. This is because the governor (control) valves and fuel-air systems have opposite effects on pressure.3. properly biased when system frequency is other than normal. a signal is received from the load control unit. bilities of all components-boiler. Base input-turbine follow. The overall effect of the control is to take appropriate action for changes in both load and pressure as noted in Table 11.41.* . turbine. Base input control. 11..40 Block diagram ofa controller [21]. the measured pressure is compared against a desired pressure set point and this produces a pressure error that is used to bias both the governor and fuel-air action. a selector switch provides an input signal from a manual setting. The operator selects the operating mode he wishes to use..468 Chapter 1 1 Generation . This fixes the desired generation for this unit._ . and auxiliaries. Both of these blocks are described in greater detail below.. taking limits into account as noted. When not on automatic control. The “boiler-turbine governor” is shown in greater detail in Figure 11. 1. The operator runs only the .:sid Pressure 61 Miin Steam Pressure To Governor Control Fig. 2. For any size step change in the manual output setter. At the same time.. The unit coordinating assembly coordinates the combustion control with the turbine-governor control. an increase in governor setting tends to reduce the pressure but an increase in fuel-air setting tends to increase it.42. 1 1. boiler inputs. This mode is the normal operating mode for this type of control and is the mode for which the system was designed. 3. except that use is made of the “required output” signal.34 for a simple description of the two types of boilers. 4. In the once-through boiler. It frees the operator from having to watch both the boiler and the turbine. and waterwall risers.9. Instead of these features.4. This mode is like the “conventional” mode as illustrated in Figure 11.35 is the absence of the drum. limiting and runback actions. either automatically or manually. and fixed rates of change. water from the boiler feed pump passes through the economizer. This mode is often used during startup and certain unusual operating conditions. down comers. which provides several advantages over conventional boiler-follow control. 11. and superheater to reach the turbine.Steam Turbine Prime Movers 469 Generation Setter Other I Runback Actions I I I of Change Setter Min. Direct energy balance automatic control.43. Fuel Min. Air Max. Fuel Max. See Figure 11. an increasing number of large boilers installed have been of the “once-through” design. passing from liquid to vapor along the way. such as providing frequency bias. Automatic control-boiler follow.41 Boiler-turbine governor control unit [19]. It also couples the governor and the fuel-air controls to provide an anticipatory boiler signal to accompany governor changes due to a load change. The striking difference between this type of boiler and the conventional drum-type boiler of Figure 11. Air Limit Max. Feedwater Governor Open Limit High Deviation Required Output To Unit Coordinating Assembly Fig.2 Once-through boilers Since the late 1950s. . furnace walls. This “automatic boiler-follow mode” is shown in Figure 1 1. and has lower operating costs.44 [221. A simplified flow diagram of a typical once-through boiler is shown in Figure 11.the control must equate flow into and out of the tube. 11.42 The unit coordinating assembly [21]. holding steam tem- Table 11. however. the pumping rate has a direct bearing on steam output as well as the firing rate and turbine governing. because of the absence of the drum. The heat absorbed (Btu/hr) divided by throughput (lbm/hr) gives the enthalpy (Btu/lbm). require a more intelligent control system. It also costs less. The once-through boiler has a significantly smaller heat storage capacity than a drum-type boiler of similar rating. heat is absorbed by the fluid at a constant rate and the steam temperature is a function of the boiler throughput (pumping rate).7 Net Control Action by the Unit Coordinating Assembly [I91 Steam Pressure High Low Low High Generator output High Action Applied To Governor Difference = Zero Difference= Decrease Difference= Zero Difference = Increase Action Applied To Fuel and Air Inputs Sum = Decrease Sum = Zero Sum = Increase Sum = Zero High Low Low .470 Chapter 1 1 Required Unit Pressure Control System To Turbine Governor Fig. In operation. It does. Thus. If the pressure is constant. A valve at the discharge end can be used to control the pressure. for steady-stateoperation. since it contains much less fluid. the once-through boiler is much like a single long tube with feedwater flowing in one end and superheated steam leaving at the outlet end. .. &Finishing Enclosures urbine Throttle Valve I I I I 0?-Lower Furnace F 'd Air I Superheater aid Reheater D m e s apr Reheat r---- th Feedwater 1 Heating System I Boiler Feedpump Economizer . 1 1. . Thus. when load is increased. L I Fig...Steam Turbine Prime Movers Desired Unit tieneration 471 * Boiler Turbine Governor Actual Unit tieneration Pressure Error Generation Error - v Combustion A A Governor Control Main Steam Pressure v 4 Turbine Generator - output perature at the desired value by maintaining the correct ratio of heat input (fuel and air) to throughput (flow rate).44 Fluid path for a once-through boiler [22]. the pumping rate must be increased to satisfy the increased load and provide greater energy storage.. and heat input must simultaneously be increased to match load and the increased storage level [23]. Transient conditions are difficult to control because of the limited heat storage in the fluid. but such analysis is necessary if a control system is to be designed accurately. Also. Heat Transfer Firing Rate 4Air Flow < By-Pass Damper Position . Assuming operation in the neighborhood of a quiescent point results in a linearized system of equations that may be numerically integrated by known digital techniques. A flow diagram of the iterative process is shown in Figure 11. there is little of the “cushioning effect” that exists in drum-type boiler designs. Having eliminated the spatial parameter by lumping. but it is also more accurate for larger excursions from the quiescent point.472 Chapter 1 1 Partly because of the lower storage of the once-through design. temperature and f o relations as well as lw pump characteristics - - Density Specific Heat Flow Rate Profile and Transport Delays Metal Heat Storage Heat Transfer Metal T----. Another report describes the use of 36 lumps to describe a large boiler used to supply a 900 MW generating unit [26].-d. the response to sudden load changes is much faster than that of the drum-type boiler.. This method is more time consuming than the linearized model. is a difficult problem. like the drum-type boiler. Rigorous analysis of the once-through boiler. The time required for water to pass through the boiler and be converted to superheated steam is only two or three minutes compared to six to 10 minutes for the dnun-type designs [24].. flow rate. pressure-temperature-density Spray Valve Position steam table relations. give the boiler open-loop re- Iterated Pump Speed Presssure. Convection. This method has been used on a supercritical unit for a 191 M W unit in which the analysts divided the boiler into 14 sections or lumps [25].45. turbine pressure. Comparison of such results with field tests have generally been quite good [25. A common approach is to lump the spatial variation and waste heat transfer equations for each lump.-- I Gas to Metal Heat Flux Profile i Gas Path Energy Balance Radiation. Turbine Valve Position continuity. Another approach to this problem has been pursued [22] in which the boiler is lumped into 30 or so sections and the nonlinear equations for each lump are solved iteratively by digital computer. and density profile from iterative solution of pressure drop. The solutions obtained by this process.26].. since the pumping rate is directly coupled to the steam produced. the resulting ordinary differential equations are nonlinear. The authors of [22] present variations to the basic control scheme of Figure 11. PA the pressure error. Detailed mathematical models of these systems have been constructed and are used by system designers and control experts. we investigate various innovations that may improve response. spray flow. This scheme has been used for many once-through boiler installations. it is clear that large fossil-fuel boilers are large complex systems.Steam Turbine Prime Movers 473 sponses to step changes in turbine valve position. at most.46.KVPA = KvPo (1 1. we examine the significance of combining MW error into the control scheme. and heat flux. MW the megawatt level. and KY a constant proportional to the valve opening. then. . The control system of Figure 11. Considering this control scheme. but shown in block diagram form.48) This difference is proportional to the load level and is interpreted as the turbine valve opening. Referring to Figure 11. a few minutes. If we let Po be the pressure set point. Basically. pump speed. from [l 13 MW = KvP = Kp(f'0 or i-PA) MW .46 Coupling of turbine load controls with boiler controls [22]. These results have been used in the synthesis of a control philosophy and control hardware. However.46 is basically the direct energy balance system of Figure 11. a portion of which is described below.3 Computer models of fossil-fueled boilers From the foregoing discussion. the problem is to design an adaptive control system that has the ability to alter its control parameters to satisfy the changing. these large detailed models are not appropriate for use in power system stability analysis.9. nonlinear needs of the system at various load levels and to do this in the shortest possible time.39. I I . 1 1. Our interest is simply in the ability of the boiler to maintain steam pressure and flow for a few seconds or. Frequency Speed MW Position Control I I Demand For: Feedwater Firing Rate I Pressure Anticipatory Feed Forward Action From Desired MW I IBoif aria besy d y Etc Fig.46. 1 1. These reports focus especially on the dynamics of prime movers and energy supply systems in response to power system disturbances such as faults. on the other hand. in turn. Later. The mechanical shaft power is the primary variable of interest as it drives the generator. labeled PT in the figure.37]. Clearly. However. Models of these system components are needed in order to provide an adequate dynamic model of the mechanical system. affected by throttle pressure. This diagram is instructive as it links the boiler-turbine systems to the controlled turbine-generator system and the external power system.48. what is needed for stability analysis is a low-order model that will correctly represent the steam-supply system for up to 10 to 20 seconds.48. and system separations. but they simply are far too detailed. improved models of a steam turbine system. Steam flow through these valves is. where the boiler-turbine system is shown within the dashed lines. There are several types of turbine systems of interest in a power system study. These large detailed models are too detailed and too cumbersome for power system stability analysis. not that they are incorrect.474 Chapter 1 1 Boiler control. This variable is directly affected by the turbine control valve (CV) and intercept valve (ZV). have been developed and are shown in a general way in Figure 11. The IEEE Power Engineering Society has been particularly active in documenting appropriate model structures and data for proper representation and two excellent reports have been issued as a result of these efforts [29. Their inclusion would greatly retard the solution time and the added complexity is unwarranted. The relationship between the prime mover system and the complete power system are shown in Figure 11.both of which admit steam to the turbine sections. but only the essential steam supply and pressure at the throttle valve. which shows how the boiler and turbine models are linked to other power system variables and controllers.47 shows the elements of the prime mover control model that was developed by the IEEE working group. This pressure is directly affected by the boiler performance. it is also not correct to assume that the boiler is an “infinite bus” of steam supply under all conditions. . The prime mover energy supply system is shown inside the dashed box in Figure 11. including the effects of the intercept valve. It is a complex nonlinear system.48 [38]. These generic models are described in [37]. Figure 11. We can see that the prime mover responds to commands Load Reference Load Demand LD L A Turbine L4-L Speed Load IV ~ cv > Turbine Including Reheater Fig. loss of generation or loads. This problem has been investigated for many years and is well documented in the literature [26-371. involves the analysis of system performance over many minutes and analysis of various subsystems within the control hierarchy. The stability analyst is not concerned with the many control loops within the boiler.47 Elements of a prime mover system [37]. and these manufacturers should be consulted to determine the best way to model their equipment. This boiler model is shown in . for generation changes from the automatic generation control system. these effects are modeled linearly as a first-order lag. These experts can also provide appropriate numerical data for the model parameters. An appropriate low-order boiler model has also been recommended by the IEEE committee responsible for the above speed-governing system model. The resulting mechanical power responds to changes in main steam pressure and turbine valve positions. The effect of intercept valve operation is that portion of the figure within the dashed box. I I I I I I I I I I Main Steam Pressure I I I I I I I I I I I I i i I I i I i I i I Fig.Steam Turbine Prime Movers 475 i InterchangePower Electric System Automatic I Generators Generation Frequency Control Network Loads v- 1 Desired Unit Generation Angle Unit Electric Power 1 Turbine/ k I --- I . or from manual commands issued by the control center. In some studies it is also desirable to provide a model of the boiler.49. The output variable of primary interest is the unit mechanical power that acts on the turbine inertia to accelerate or decelerate the inertia in accordance with Newton’s law. The older units operated under a mechanical-hydraulic control system. A generic model of this type of control system is shown in Figure 11. In many cases. This model is believed to be more accurate as it accounts for the valve limits. The manufacturers of speed-governing equipment have their own special models for speed governors of their design.48 Functional block diagram of prime mover controls [38]. where the intercept valve opening or area is represented by the “IV” notation.50. The steam turbine speed and load controls are of two types. The turbine-boiler control also responds to changes in speed. A more detailed model of a generic turbine model is shown in Figure 11. 1 I . The control valve position is shown as “CV” in this figure. This is true of studies that extend the simulation time for long periods where boiler pressure may not be considered constant. in a chain reaction.5 1 and features a lumped volume storage of steam at an internal pressure labeled here as drum pressure.49 Generic turbine model including intercept valve effects [38]. C. 1 1. is required to slow down the neutrons and thereby enhance the probability of fission. or a rate of mass flow). principally water. The output of the model is the steam flow rate to the high pressure turbine. connected through an orifice representing the friction pressure drop through the superheater and piping. C.476 Chapter 1 1 Fig.representing a derivative with respect to time.both of which contain saturated steam and water. The steam generation process is a distributed one and this is approximated in the model by two lumped storage volumes for the drum. 11. or graphite. In once-thru boilers. (note carefully the dot over the m. the major storage is in the transition region.10 Nuclear Steam Supply Systems Nuclear power plants generate steam by utilizing the heat released in the process of nuclear fission. Figure 11. and with steam leads and their associated friction pressure drops. in series with a superheater. In the so-called "thermal" reactors a moderator. and the superheater. The major reservoir for energy storage is in the waterwalls and the drum. or the splitting of the heavy uranium atom by the absorption of a neutron. 1 1. The energy input to the boiler represents heat released by the furnace.50 Approximate representation of control valve position control in a mechanical-hydraulic speed governing system [38]. heavy water. This heat generates steam in the boiler waterwalls at a mass flow rate of rh.. rather than by a chemical reaction as in a fossil-fuel boiler. Speed Relay Position 1 TSM Rate Limits Limits 1 S & 4 Servo Motor - Fig. . The nuclear reactor controls the initiation and maintenance of a controlled rate of fission. gas-cooled. In the BWR. 4.. graphite-moderated reactors In the PWR. These reactors use heavy water under pressure and utilize natural uranium as a fuel. Our treatment will focus on the BWR and PWR types.Steam Turbine Prime Movers 477 Turbine Equivalent Orifice n Control VllVPc HP Turbine Drum and Water Walls vuyu. The major systems in use are the following: 1. Boiling water reactor (BWR) Pressurized water reactor (PWR) CANDU reactor Gas-cooled. since they are so common in the United States. the heat generated in fuel assemblies is removed by carbon dioxide. which is used to produce steam that is carried to steam generators. There are several distinct types of nuclear steam supply systems that have been designed and put into service in power systems. 2. the reactor is cooled by water under high pressure. 3. 11.51 A computer model of boiler pressure effects 1381. In these reactors. The high-pressure water is piped to heat exchangers where steam is produced.the water coolant is permitted to boil and the resulting steam is sent directly to the turbine. In Europe.VULur and Steam Leads (a) The Physical System Turbine Valve Water Wall Lag (b) The S s e Model ytm Fig. . graphite-moderated reactors have been developed. . The CANDU reactors have been developed in Canada. 1 1 .1 Boiling water reactors The major components in a BWR nuclear reactor are shown in Figure 11. .S2 Major components of a BWR nuclear plant 1391. This is a low-order model for such a complex sys- Fig.47% Chapter 1 1 Fig. Note that the steam produced by the reactor is boiled off the water surface and fed directly to the turbines. A block diagram for the boiling water reactor is shown in Figure 11. The variables noted in the figure are defined in Table 11. 11.8. 1 1.52 [39] and these components should be included in a dynamic model.10.53 Block diagram of a reduced-order BWR reactor model.53 [40]. 1 1. One model for the PWR is that shown in Figures 11.2 Pressurized water reactors The major components in the pressurized water reactor are identified in Figure 11. which is not specified here.54 and the major subsystem interactions are shown in Figure 11. where the high.and low-pressure valve positions are unspecified or are unchanging. . tern. One can also LLl-I Bypass Pw Rod Position Rod position Regulator ~ - - PRW Reactor Fig.56. 11. where it is important to keep models reasonably simple.55 and 11.55.55 Interaction of P W R subsystem models 1411. 11.10. These positions are functions of the speed governor model. and was constructed for use in power system stability analysis. The model of the P W R nuclear reactor and turbine are rather complex. but is similar to other speed governor models.54 Major components of a PWR nuclear reactor model [39].Steam Turbine Prime Movers 479 1nput Signal Control Rod Fig. model the turbine bypass system [41]. 1 1.1 by working through each step of the problem and plotting the root locus diagram.1 by performing a Bode plot. s LR = Turbine load reference PR = Reactor Pressure MT = Turbine Steam flow MB = Bypass steam flow Ms= Total steam flow R. Repeat for a longer bowl delay using T3 = 0. Verify the results of Example 11. per Unit LD = Load demand PT = Throttle pressure Ks = Steam flow pressure drop factor T = Oscillation period. Problems 11.1. What is the gain margin? The phase margin? Table 11.25. = Speed regulation Ao = Speed error .2. = Oscillation rate TC. but that option is not pursued here and the total bypass flow is assumed to be a zero input in the reactor model. s Tp = Power response TC. 11.480 Total Chapter 1 1 PWR Reactor Model I 1 I . s 5 = Oscillation damping factor T.8 Variable Identification. Locate the points for which the gain is approximately 937. Several other PWR models have been presented and these are recommended for study [42-46]. Examine the stability of the open-loop transfer h c t i o n of Example 11.56 PWR reactor and turbine model [41]. Turbine Model I + 9 Fig. ” IEEE Publication 600. instead of the short bowl delay used in the example.” paper 83T12. 13.” from “Reheat Turbines and Boilers. 1962. Power Plant Theory and Design. W. New York. September 1952. 1964. 11. pp. Savant.6. a Power Magazine special report. Pmin ‘ IPower Governor steam System Dynamics A governor. G. 1963. New York.’’ American Society of Mechanical Engineers Publication. J.. W. Introduction to Linear Systems Analysis.25 s. P.1 and find the gain margin and phase margin. and J. AZEE. and reheat steam turbine system 11. using the values given for the various parameters. 10. “Recommended specification for speed governing of steam turbines intended to drive electric generators rated 500 MW and larger. 1965. M.Steam Turbine Prime Movers 48 1 1 1. January and February. Locate the points for which the gain is about 187. Find the state-space model for the governor and boiler system shown in the following figure. 3.” ASME. A. 67.. A. B.. R. McGraw-Hill Encyclopedia of Science and Technology.” Trans. Basic Feedback Control System Design.. A. “Overspeed trip systems for steam-turbine generator units. Brown. use a long bowl delay of T3= 0. Examine a turbine control system similar to that of Example 11.. 5. and E. Evans. 4. Initial Power I m ‘a Auxiliary Signal ’r I I . Verify the results of Example 11. “Recent development of the reheat steam turbine. A. Nilsson.4. Kure-Jensen. 15. 11. Power Station web site. “Plant dynamics of a drum-type boiler system.7. 9. Eggenberger. 11. B. Examine the pressure control systems of Figures B. Steam Turbines. R. 1960.. P. PA. A. 2. Reynolds.1 except that. H. Ronald Press.5. References 1. Scranton. ASME Power Test Codes.gov/. IEEE. 1959. General Electric Company publication GET 3096A.2. IEEE Report. A. General Electric Company. J. H. 8. Compare these results with those of the previous problem. Zerban. 7. McGraw-Hill. 12. IEEE. and W. 1959. Skrotzki. McGraw-Hill. Power Test Codes 20. (Associate Editor). 11. Philip J. for example: http://www.. International Textbook Co. reprinted from Mechanical Engineering. 6. Wiley. “Control of large modem steam turbine-generators. Introduction to the Basic Elements of Control Systems for Large Steam Turbine Generators.8. 1967. Power Station Engineering and Economy. 1992. McGraw-Hill.1. Prepare a Nyquist diagram for the system of Example 1 1. Potter.. and B.10 of Appendix B by root locus. Nye. then under “search EIA using” enter “power station” and hit GO.2 by working through each step of the problem and plotting the root locus diagram. . 1958. G. 1-7.7. June 1961. R. F. 547-551. 1983. 14.” Trans. B. Vopat. PAS-82.9. New York. 1948. 7th Edition.. boiler. Jr. Sketch the root locus and find the normal operating point for K3and Cgas given in Example 11. C. deMello. 1952 and the May 1952 Transactions o the f ASME. New York. Power Plants. pp. New York. “Graphical analysis of control systems.3. New York. Skrotzki.fmtgov. =I. Novmec. “Simulation of Bull-Run Supercritical Generating Unit. 34. B. Detroit. deMello.” IEEE Trans. May 1991. 1967.Nov. 1980. Younkins. 35. 9.” CanadianElectrical Association. U. “Dynamic models for steam and hydro turbines in power system studies.1986.. J. IEEE Trans. February 1965. Minneapolis. Thompson. PAS-84. Clar.. 3. . January 30-February 3. Gainesville. C. 1973. F. PAS-85.mec. Stanton. Philadelphia. IEEE Committee Report. 1966. A. IEEE Working Group on Power Plant Response to Load Changes. “Long Term Power System Dynamics. 39. New York. E. Louis. C. 6. IEEE Committee Report. deMello. L. on Power Apparatus and Systems. PWRS 2. Jan. and F. 28. Schulz. 27. and J.” IEEE Trans.” IEEE Trans. D. IEEE Committee Report.“Steam turbine fast valving: Benefits and technicalconsiderations. “Bibliography of literature on steam turbine-generator control systems. 1964.” IEEE Trans.” paper presented at the Fourth Winter Institute on Advanced Control. 1959. 38. S...Instrument Societyof America.” paper presented at the 6th National ISA Power Instrumentation Symposium.. 1983. 30. “MW response of fossil-fueled steam units. Chen. Munro. 1904-1915. PAS-92. 3. T. 24. and D. 23. “An integrated combustion control system for once-through boilers. 2. 18. July 13-18. C. and P. and P. Washington.2. PWRS-1.. presented at the IEEE Winter Power Meeting. 32. Ahner. 7.. 17. A.National Power Survey. May 1987. PA. R. N. July 1966. “Dynamic models for steam and hydro turbines in power system studies.University of Florida. Scutt. Turner.. May 16-18. E. government Printing Ofice. L. P. 1973.” IEEE paper 31PP67-12.” EPRI Report RP90-7. “Boiler pressure control configurations. T. “Effect of prime-mover Response and Governing Characteristicson System Dynamic Performance. C. P. J. January 1968and February 1968. al. “A dynamicmodel of a drum-type boiler system. Schweppe. 1994. D.6.. pp. IEEE Committee Report. 1963. 1988. Ewart. Concordia.” ZEEE Trans. IEEE Task Force on Stability Terms and Definitions. 1963. February 20-24.. 20. P. 31. Hirsch. June 1974and v. March 2426. Summer. deMello.. Federal Power Commission.” Proc. “The application of Direct Energy Balance Control to Unit 2 at Portland Station. PositionPaper ST 267. PAS-82.“A technique for developing low order models of power plants.” IEEE Trans. on Power Apparatus & Systems. Florida. P. Bachofer. August 1986. EC-3.” Proc. Jr. P. “Conventions for block diagram representation. “Once-through boiler control.. 37.. “Analysis and design of controls for a once-through boiler through digital simulation. F. Spanbauer.” Power Engineering. E. PAS-109. 22. presented at the IEEE Power Engineering Society Summer Meeting. 21.1967. and T. and F. T. Oct. “Dynamic models for fossil fueled steam units in power system studies. Adams.S. 33. Kundur. Beaulieu. 26. Littman. May 13-1 5 . Schulz.New York. Kundur. on Power Apparatus and Systems. PWRS-6. Dyer. 1973. pp. 1974.” IEEE Trans. 1904-1915. D. v. on Energy Conversion. N. 92.” IEEE Paper 94 WM 187-5 PWRS. R.C. Kirchmayer. 29. R. F. 36. American Power Conference. Imad.” IEEE Trans. American Power Conference. 1966.. et. P. J. F. Michigan. T. and D. “Nuclear plant models for medium to long-term power system stability studies. “Mathematical modeling of once-through boiler dynamics. D.” IEEE Paper 80SM598-3. P.482 Chapter 1 1 16. R. C. R. E. IEEE Committee Report. J..” IEEE Trans. deMello. D. 2. Ichikawa. on Power Apparatus and Systems. and V. P.” paper presented at the Ninth Annual Power Instrumentation Symposium.. Starbuck. R. 25. and R. 1. F. Whitten. 28. 19. “Update of bibliography of literature on steam turbine-generator control systems..” IEEE Trans. P. Inoue. “Computer control of power plants. presented at the IEEE Power Engineering SocietyMeeting. Kenny. L.. “Fast valving with reheat and straight condensing steam turbines.” IEEE Trans. 1963.. “Plant dynamics and control analysis. K. Morris. L. C. P. 92. PAS-82. 29-Feb. Inoue. A. K. 44. IEEE Power Engineering Society. New York. 47. Robinson nuclear plant. New York. F. R. PWRS-3. May 1988.” paper presented at the IEEE Symposium on Prime Mover Modeling.Steam Turbine Prime Movers 483 40. E. W. and P. PAS102. Ichikawa.J. Winter Meeting. T. Van de Meulebroeke. and J. P. M.. E.to long-term power system stability studies. 30-Feb 2. Jr. February 1977. “Long term power system dynamics.” IEEE Paper 94 WM 187-5 PWRS. 1979. and P. 3.“Reduction of program size for long-term power system simulationwith pressurized water reactor. 1992. January 30. pp. 42. K t . New York. and M. “Modelling of a PWR unit. “Modeling of CANDU nuclear power plants for system performance stud- 45. on Power Apparatus and Systems. March 1983. Moret. D. “A reduced order dynamic model of a boiling water reactor. R. Palo Alto.. Thakkar. Kerlin. T.. Winter Meeting. Di Lascio. CA.” Nuclear Technology. Culp. Jan. Strange. A. Hirsch. 48. B.. 463-71. Dar. and T. 1992. ies. phase I1 final report. on Power Systems. Kundur. McGraw-Hill. New York. and A. P. M..” Project EL-367. Inoue. Principles o Energy Conversion. 41.Winter Meeting. New York.” paper presented at the IEEE Symposium on Prime Mover Modeling. T.” IEEE Trans. “Nuclear plant models for medium. 46. . Ichikawa.” IEEE Trans.. T. September 1976. W. Kundur. Turner. Electric Power Research Institute.” paper presented at the IEEE Symposium on Prime Mover Modeling. Poloujadoff. presented at the IEEE Power Engineering SocietyMeeting.. 1992. “Theoreticaland experimental dynamic az analysis of the H. G. 30. T. E. 43. “Light water reactor plant modeling for power system dynamic simulation. 1994. P. January 30. IEEE Power Engineering Society. IEEE Power Engineering Society. f Schulz. Younkins. January 30.. 1 Inhuduction The generation of hydroelectric power is accomplished by means of hydraulic turbines that are directly connected to synchronous generators. it is not advisable to use the nozzle to cut off the water jet abruptly. Thus.2 The Impulse Turbine The impulse or Pelton wheel is generally used in plants with heads higher than 850 feet (260 meters). several nozzles are directed toward each wheel. steady changes in water flow and power input. and the propeller or Kaplan turbine. although some installations have lower heads. Occasionally. The turbine wheel is spun by directing water from nozzles against the wheel paddles and using the high momentum of the water to drive the wheel.chapter 12 Hydraulic Turbine Prime Movers 12. Four types of turbines or water wheels are in common use. The three most common are the impulse or Pelton turbine. A stripper. This arrangement is shown in Figure 12. The type of turbine used at a given location is based on the site characteristics and on the head or elevation of the stored water above the turbine elevation. Figure 12. which combines some of the best features of the Kaplan and Francis designs.2 (a) and is seen to be similar to the familiar garden hose nozzle. also shown in Figure 12. All of these types make use of the energy stored in water that is elevated above the turbine. A fourth and more recent development is the Deriaz turbine. Impulse turbines are often installed on a horizontal shaft with the generator mounted beside the turbine.1 shows a double-overhung unit with a single nozzle for each wheel. Speed regulation of the impulse turbine is accomplished by adjusting the flow of water through the nozzle by means of a needle that can be moved back and forth to change the size of the nozzle opening. thereby increasing the efficiency of the unit. One plant. Water to power the turbines is directed to the turbine blading through a large pipe orpenstock and is then discharged into the stream or tailrace below the turbine. However. is used to clear water from the bucket as it moves upward. This needle adjustment is used to make small. since the impulse wheel is used in plants having high heads and long penstocks. another means must be found to divert the water stream away from the wheel while the nozzle is closed slowly. has a static head of 2575 feet (785 m) and another in Switzerland has a head of over 5800 feet (about 1800 m). at Bucks Creek in California. The reason for this is that a sharp cut-off in flow causes a pressure wave to travel back along the penstock causing possible damage due to water hammer. the reaction or Francis turbine.1. One way this is accomplished is by mechanically deflecting the water stream by means of a jet deflector as . Some designs have two turbines on a shaft with a generator between them and are called “double-overhung” units. 12. All of the energy input to the shaft is in the form of kinetic energy of the water. . the governor of an impulse wheel will control the nozzle for normal changes. 12. shown in Figure 12. the water veloc- Fig.Hydraulic Turbine Prime Movers 485 Fig.1 A double-overhung impulse wheel. and this energy is transformed into the mechanical work of driving the shaft or is dissipated in fluid friction. In an impulse turbine. Ideally then.2 Impulse wheel nozzle and deflector arrangements. the total drop in pressure of the water occurs at the stationary nozzle and there is no change in pressure as the water strikes the bucket.2(b). Thus. but must recognize a load rejection by quickly moving the jet deflector. 12. however.. we can compute Q =AVft3/s (12. ft C = nozzle coefficient.5) 12. the head.wHQ hp 550 (12.3 The Reaction Turbine In the impulse turbine. Since so much of the turbine blading is active in this energy transfer. and a nozzle coefficient. = power availble at the nozzle. In these turbine designs. ft3/s H = static or total head.3) where A =jet area.where k is a constant. The quantity of water depends on the water velocity. water under pressure enters a spiral case surrounding the moving blades and flows through fixed vanes in a radial inward direction.= -hp 8. and transfers both pressure energy and kinetic energy to the runner blades. the shaft power may be written as HQT. there is only a partial pressure drop at the nozzle.4 lbm/ft3 Q = quantity of water. usually 0. exert- . = . Francis.1) where P. Most reaction turbines in use today are of a radial inward-flow type known as the “Francis” turbine after James B. hp W = weight of one cubic foot of water = 62.4) h=kH for a given situation. is the turbine efficiency. The power available at the nozzle is given by the formula P . Thus. f t Recall that 550 l b d s is equal to one horsepower.2 ft/s2 h = net head at nozzle entrance.8 (12. The water then falls through the runner. water completely fills the cavity occupied by the runner. Actually.2) where the maximum efficiency is usually 80 to 90% [ 13. P. the remainder taking place in the rotating runner. a small kinetic energy remains and is lost as the deflected water is directed downward to the exit passageway. In the reaction turbine. who designed the first such water wheel in 1846. then we may write Ps= k. the diameter of the reaction turbine is smaller than an impulse turbine of similar rating. the high pressure in the penstock at the nozzle is changed to momentum so that no pressure drop is experienced at the turbine.A86 Chapter 12 ity is reduced to zero after it strikes the turbine buckets. It is also restricted by the mean river or stream flow.H3I2 (12. ft2 V = jet velocity. flows across this pressure drop. For a given design. If 77.98 If we assume that (12. which is dictated by nature. ft/s Then v=cv?@ft/s where g = 32. with the vertical turbines being the most common. the water flows perpendicular to the shaft. Positioning these vanes can cause the water to have a tangential velocity component as it enters the runner. The control for a reaction turbine is in the form of movable guide vanes called wicket gates through which the water flows before reaching the runner. thereby reducing the kinetic energy losses at discharge. Reaction turbines are classed as radial flow. however. This low pressure tends to increase the pressure drop across the turbine blading and increase the overall efficiency. In some designs. At any other wicket gate setting. usually at 80 to 90% of wide open. the runner will operate at maximum efficiency. In axial flow the stationary vanes direct the water to flow parallel to the shaft. or mixed flow according to the direction of water flow.3 is the draft tube.3 A typical vertical shaft reaction turbine arrangement. Although the wicket gates are close-fitting. this kinetic energy would be lost. Mixed flow is a combination of radial and axial flow. axial flow. The draft tube is an integral and important part of the reaction turbine design. a large butterfly valve is often installed just ahead of the turbine case for use as a shut-down valve.6) Fig. a partial vacuum is formed due to the fast-moving water.3. a portion of the energy is lost due to less efficient angling of the water streamline. In radial flow. being applicable to installations with heads as high as 800 feet (244 m) and as low as about 20 feet ( G 6 m). they usually leak when fully closed and subject to full penstock pressure. 12. . The large tube with the 90" bend just below the runner in Figure 12. The generator is usually directly connected to the runner shaft as shown in Figure 12. Reaction turbines are installed either in a horizontal or vertical shaft arrangement. this energy may be as high as 50% of the total available energy. One of the important empirical formulas used in waterwheel design is the specific speed formula. Thus.Hydraulic Turbine Prime Movers 407 ing pressure against these movable vanes and causing the runner to turn. It is a versatile design. Without the draft tube. The importance of the draft tube is evident when the energy of water leaving the runner is considered. With the draft tube constructed air-tight. It allows the turbine runner to be set above the tailwater level and it reduces the discharge velocity. (12. For one such position. It serves two purposes. 1) to (1 2. As a general guide.5 10 to 100 80 to 200 10 to 100 max Ns 10 150 250 where N = speed in rpm H = head in feet Ps= shaft power in hp This quantity is the speed at which a model turbine would operate with a runner designed for one horsepower and at a head of one foot. The force required to move this assembly is very large and two servomotors are often used to rotate the ring. as shown in Figure 12.4 Wicket gate operating levers and position servomotors.5) used in conjunction with the impulse turbine also apply for the reaction turbine.4. Department of the Interior. Electric Power Branch. U. Figure courtesy F. R. the value of C is about 0. high-capacity turbine. For (12. low-capacity (in water volume) turbine and the reaction turbine is a high-speed.1 Typical Specific Speeds for Watenvheels Type of Wheel Impulse Reaction Propeller Deriaz N S 0 to 4. These are deflected simultaneously by rotating a large “shifting ring” to which each gate is attached.1 are applicable. 12. The control of a reaction turbine is through the movable wicket gates. Bureau of Reclamation. The same formulas (12. Schleif.488 Chapter 12 Table 12. Fig.8 and this value usually decreases for turbines with higher values of Ns.S. It serves to classify turbines as to the type applicable for a certain location.6 to 0. .4). Under this classification. W. then. USBR photo by C. an impulse turbine is a low-speed. we say that the specific speeds given in Table 12. Avey. This permits optimization of turbine efficiency over a wide range of head and load conditions. This design has the advantage of fairly high efficiency over a wide range of head and wicket gate settings. Compared to the Francis turbines. Surge Tanks. We will assume that a reservoir of water exists and is large enough in capacity that. Nagler. as shown in Figure 12. Wicket gates are generally not used with a Deriaz turbine and control is maintained by blade adjustment only. Thus. That is to say.6 Conduits. It operates at a high velocity and operates efficiently only for fixed head and constant flow applications. These differences are illustrated in Figure 12. leaving the runner with a fast swirling motion. but are beyond the scope of this text. as indicated in Table 12. Three types of propeller turbines can be discussed. 12.4 is one of the generators at the Grand Coulee Dam Powerhouse in Washington State.If load is rejected and the wicket gates are driven closed very quickly by the governor servomotor. in 1919.6.4 Propeller-Type Turbines The propeller-type turbine is really a reaction turbine since it uses a combination of water pressure and velocity to drive the shaft. a relatively level section . this design is well suited for situations requiring large variations in loading schedules. This prevents the large momentum of penstock water from hammering against the closed wicket gates. thereby changing the wicket gate position of all gates.Hydraulic Turbine Prime Movers 489 The machine shown in Figure 12. In some cases. The fixed blade or Nagler type was developed in 1916 by F. The selection of sites and construction of dams. It employs water velocity to a greater extent than the Francis turbine. It shows the wheel pit of a 165.5 The Deriaz Turbine The Deriaz turbine is a more recent development in reaction turbine design and incorporates the best features of the Kaplan and the mixed-flow Francis designs. The blades are contoured similar to the Francis blading and are set at 45 degrees to the shaft axis rather than 90 degrees as in the Kaplan turbines.1. spillways. A second control device used in reaction turbines is a large bypass valve. which is actuated by the shgting ring.8. The two rods are connected to power servomotors and operate to rotate the shifting ring. during periods of interest for control analysis. Thus. Kaplan developed the adjustable blade propeller turbine shown in Figure 12.7. A few years later. The pressure regulator then closes slowly to bring the water gradually to rest. The Deriaz turbine has the capability of operating at high turbine efficiency over a wide range of loadings. Many excel1 lent references are available that discuss these important items [5. and flows to the turbine as shown in Figure 12. It also has a higher specific speed. where the blades are identified by the letters A and the direction of water flow by the letter W. the water source is an infinite bus. From the reservoir. Adjustments of wicket gate setting and blade angle can both be made with the unit running. the pressure regulator is caused to open and does so very rapidly.000 horsepower turbine generator. It is essentially a propeller turbine with adjustable blades. 12. Kaplan turbines are used at locations with heads of 20 to 200 feet (about 15 to 150 m). and Penstocks It is assumed that any hydroelectric generation site has a supply of elevated water from which water may be drawn to power the turbine. A. 12. the Kaplan units operate at higher speeds for a given head and the water velocity through the turbine is greater.5. and the like are important. water is drawn from an area called the forebay into a couduit or large pipe. the draft tube design is important in Kaplan turbine applications. 6 . the head is constant. Under steady-flow conditions. feet H = static head. this head loss at the turbine is h =H .h L where hL= head loss. 12.490 Chapter 12 Fig. called the conduit. a head loss develops.5 The Kaplan propeller turbine. measured in feet.7) . feet k = a constant corresponding to pipe resistance = kQ" (12. The hydraulic gradient in Figure 12. As the water flows through this conduit and penstock at a steady rate. is necessary to move the water to a point where it begins a steep descent through the penstock to the turbine. of pipe. feet h = effective head at the turbine. as a hnction of distance from forebay to turbine. similar to the voltage drop in a nonlinear resistor.8 represents the approximate profile of the head. Hydraulic Turbine Prime Movers 491 (a) The Francis Runner (b) The Kaplan Runner W (c) The D & e Runner Fig. This causes a pressure wave to travel along the penstock. caused by sudden changes in the rate of water flow [ 6 ] . to the positive water-hammer gradient. The turbine-governor reacts to this change by quickly moving the wicket gates toward the closed position and. Water hammer is defined as the change in pressure. the governor will react by opening or closing the wicket gates. This supernormal pressure is not stable. Creager [ 6 ] gives a graphic example of this phenomena as shown in Figure 12. 12. the head loss will be directly proportional to the length of pipe. .9. where 1 5 n 5 2 Thus. One of the serious problems associated with penstock design and operation is that of water hammer. above or below normal pressure. because of the momentum built up by the penstock water. following a sudden change in load. Suppose the load on the turbine is dropped suddenly. as indicated in the figure. the hydraulic gradient to changes from the normal full load gradient A-C. gradient A-D swings to A-E and oscillates back and forth until damped by fhction to a new steady-stateposition. A-D. when the flow is steady.6 Comparison of reaction turbine runners. Q = flow rate. possibly subjecting the pipe walls to great stresses.Thus. and once the wicket gate movement stops. ft3/s n = a constant. 8 A typical conduit and penstock arrangement. = 100 Fixed Propeller 0 ‘ 0 I 1 I I I > 20 40 60 % of Full Load 80 100 Fig. Each step change causes a positive pressure wave to travel up the penstock to the forebay and. reveals that it is much like the distributed parameter transmission line.” it is reflected back as a negative ----_ \ Static Hydraulic Gradient -- - Tailrace Fig. not only must the penstock be well reinforced near the turbine.7 Turbine efficiency as a hnction of load.492 100 Chapter 12 4 Denaz Impulse Kaplan Francis N. Thus. . = 50 Francis N . 12. but it must be able to withstand these shock waves all along its length. Examining this phenomenon more closely. The (closing) wicket gate can be thought of as a series of small step changes in gate position. A sudden increase in load. upon reaching this “open circuit. accompanied by wicket gate opening has just the opposite effect. 12. feet a = pressure wave velocity. inches e = pipe wall thickness.Hydraulic Turbine Prime Movers 493 \\ Penstock Wicket Gates Fig. W S g = acceleration of gravity. Note that to keep water hammer to a low value. feet vA = change in velocity.seconds 2L a (12. The change in head due to water hammer produced by a step change in velocity has been shown to be [6] (12..9) where d = pipe diameter. p. which is defined as p = .Ii I f ne _ i Tailrace pressure wave of almost the same magnitude.9 Hydraulic gradient following a loss of load. ft/s For steep pipes.10 is the hdamental equation for water hammer studies. ft/s2 and a is the pressure wave velocity as previously defined.10) where hA= change in head. inches Pressure wave velocities of 2000 to 4000 feet per second are not uncommon. 12. the wave velocity is approximately a= 4675 ft/s 1 + (d100e) (12.8) where L = length of penstock. vA must be . wlk. Equation 12. The time of one “round trip” of this wave is called the critical time. from a distance below this equal to the friction head for stopping) we have [ 5 ] y= (gA+ P y 2 aLv% (12. The pressure regulator is helpful in controlling positive water hammer as it provides relief for the pressure buildup due to closing of the gates. the surge tank should be as close to the turbine as possible but. can be a serious problem in penstock design.11) where a = conduit area. both positive and negative. it is often placed at the top of the steep-descent portion of the penstock. Sometimes an “equalizing reservoir” is constructed to serve as a surge t n for large installations and may actually be cheaper and more beneficial.10 Conduit and penstock with a surge tank.494 Chapter 12 kept small either by using a pressure regulator or by introducing intentional time lag in the governor. Usually. Suppose. This is due to the general rule that the larger the tank area. Surge tank dimensions are important. The t n must be high enough so that in no case is air ak drawn into the penstock. The introduction of time lags are particularly troublesome for interconnected operation as this contributes to tie-line oscillation [7]. It may also cause violent pressure oscillations. that the gate is opened by only a small amount. For this reason. To be most effective. the time for closure of the wicket gates of a hydraulic turbine is much greater than p of equation (12. the pressure rise can be greater than that due to closure from full gate to zero. It may require that penstocks be built with much greater strength than would ordinarily be necessary. In such a case. the smaller the pressure variation [6]. However. ft Surge Tank -I . ft2 L = conduit length. such that it can be closed in a time p . as shown in the ak figure. p is usually considered the critical governor time. From the above. since it must also be high enough to withstand positive water hammer gradients without overflowing. 12.10. Letting y denote the maximum surge up or down in feet (measured from the reservoir level for starting. A device often used to relieve the problems of both positive andnegative water hammer is the surge tank. -Forebay Tailrace Fig. . however. a large tank usually located between the conduit and penstock.8). it is of no help in combating negative water hammer. as shown in Figure 12. we see that water hammer. which can interfere with turbine operation. Note the relatively long period (about 300 seconds. Ws g = 32. This surge would be due to a sudden increase in load. The damping effect due to the added friction of the differential surge tank is shown in Figure 12. ft2 Barrow [ 5 ] also gives a formula for the time interval that elapses between turbine load change and the occurrence of the maximum surge as (12. ft A = area of surge tank. Since damping is desirable.Hydraulic Turbine Prime Movers 495 v. Because of this restriction. These heads are determined by water in the riser tank. This is done in two ways: by placing a restricted orifice between the t n and the penstock. Note that an accelerating head is created. The diameter of the differential surge tank is about one-half that of a simple surge tank. which increases steadily for about 80 to 85 seconds. where the surge is compared for two types of tank design [6].11.1 1 The differential surge tank. which acts like a simpler surge tank with small diameter.2 fus2 F = friction head. it is sometimes advantaak geous to add hydraulic resistance at the surge t n opening to produce a choking effect..12) where c = coefficient of fiction cv2 = q = flow in ft3/s The factor F in (12. shown in Figure 12. . or by conak structing a “differential surge tank. or five minutes) of the surge. 12. where the turbine wicket gates are opened at time t = 0.___- Tailrace Fig.12. the water level in the outer tank is independent of the accelerating head and the head acting on the turbine. at which time the flow Surge A - Riser I I . consists of two concentric tanks: an inside riser tank of about the same diameter as the penstock and an outer or surge tank of larger diameter with a restricted passage connecting it to the penstock.10) is important since it represents the friction that eventually damps out oscillationsfollowing a sudden change.”The differential surge tank. = velocity change. The riser diameter is usually the same as that of the penstock. If the effect of water wheel speed on flow is neglected. the turbine can be simulated by G or GA.496 Chapter 12 5 Differential: v1 Q 5 15 20 25 -.13 Electric analog of the hydraulic system. : 0 I I I I I I I I I I I I I I I I I I I I I 1 I I I I 1 % 50 100 150 200 250 300 350 Time in seconds Fig. and the turbine is represented by the variable conductance. (How could a differential surge tank be represented?) Conduit Penstock V I I I Fig. of water from that tank ceases. . penstock. Figure 12. 12. In the differential tank. respectively (series resistance could be added to represent hydraulic resistance). volumetric flow is analogous to current. With water being considered incompressible. In the discussion of a technical paper [SI. deMello suggests a lumped parameter electric analog of the hydraulic system. and turbine [9].12 Comparison of surges in simple and differential surge tanks. 12.the accelerating head is established very fast. the inertia of water in the penstock and conduit are represented by inductances L. where head is analogous to voltage. G. but not so fast as to prevent the governor from keeping up with the change.where a change in gate setting is under consideration.and L2. surge tank. The surge tank behaves much like a capacitor as it tends to store water (charge) and release it when the head (voltage) at the turbine falls. including conduit.13 shows this analog. (12. we get ilA = 2(GA/GO)v10 2vo s(L. + L.19) may be written as PA = 1 +-s TW (12.L.19) Ro= 7 10 VI0 (12. assume a change in turbine power at constant efficiency or PA = vIOiIA + ilOVIA Po$(?- s(L.16) Combining.18) reduces to the so-called waterhammer formula Po?( 1 +-s where 1- ks) Jh 2O R (12.21) 2 .14) From the square root relationship between flow and head Q=GG we write il = G (12.20) Then (12.17) Now. C is large and (12.15) G (12.)(1 + LCS2) i0 1 + L2C2S2 + (12.18) When the surge tank is very large.1 3) where Also (12.L2Cs3) vo -+ i0 +L2) +voL2cs2I L1L2cs3 2 2 10 (12. for the head at the reservoir described in the s domain.Hydraulic Turbine Prime Movers 497 If linearized equations about a quiescent operating point are written we have. + L2) + ?L2CS2 10 . As shown in the previous section. For the uniform pipe. This is easily seen when the partial differential (wave) equations for a uniform pipe with negligible friction are examined. ft h = head.v + GV ax dt (12.490 where [9] Chapter 12 .7 Hydraulic System Equations The hydraulic system and water turbine transfer functions have been thoroughly analyzed by Oldenburger and Donelson [8].22) Furthermore.23) These results are not greatly changed by considering the conduit and penstock as a distributed parameter system.19) becomes (12.24) =a constant = p g ( k + X) p = density of fluid g = acceleration of gravity K = bulk modulus of elasticity of fluid r = internal pipe radius f = pipe wall thickness E. or L. when the tunnel inertia is great.'Z = water starting time = 1 second (12. R -cy (12. then (12. the flow of water through a conduit is analogous to an electric transmission line in which head is analogous to voltage and volumetric flow rate is analogous to current.=--cy. 12.=-gdx dt where u = water velocity. as pointed out by deMello [9]. which can be written as follows: a --d i= C.dh dx at du dh . is large. . This excellent description is based on a rigorous mathematical analysis and is supported by substantial experimental evidence to testify to its validity.= Young's modulus for the pipe Equation (12. we write du .25) The similarity for the lossless case should be obvious.24) should be compared to the equations of the transmission line. Ws x = distance along pipe. and K2 as they depend on boundary conditions.. This helps in evaluating the constants C. dx dH . c cash .31) and (12. C2.28) This result can be written in hyperbolic form as U = C.31) or (12. = U. x) = L[h(t.27) The solution of (12.= _ -1 su h g (12. we may write C. 11.32) we may subscript all x's with a numeral (I. With this simplification. or i) to indicate the particular section under study. = ~ .Sda (12. cosh -XI a S + S sinh -4 a S S C2= .6 g H . assuming zero initial conditions. ~ . cosh .27) may be shown to be u = K e-sx/a + K2e+sda I H = K3e-sda + K4e+. or any arbitrary cross section i.+ C4sinh a sx (12.14.U.31) or U = CI cosh .24) with the result. For example.x)] u = U(S..C2sinh a a sx sx H=-- G i sx c. in (12.x ) ] (12..1 a (12.30) These results may be simplified by eliminating of the arbitrary constants subscripted by 3 and 4.32) to any cross section of pipe such as I or I1 of Figure 12. let us define the following: H = H(s.s x +a~ ~ ~ + s x / a y / (12.K.x ) = L[u(t.+ C2 sinh a a a sx sx sx H = C3cosh .. cosh -X.26) We may write the Laplace transform of (12.we have [8] .33) .Hydraulic Turbine Prime Movers 499 Now.29) where a= = wave velocity (12.32) 6sinh~ sx Note we may apply (12. Thus.. sinh -XI a a (12. ft2 then we may write or.37) where L T. cosh -XI cosh -&. sinh -X. let x.= u. a S . we convert the U equation to a Q equation and rewrite (12. C2= --HI (12.32) as. cosh -XI . sinh -X. simply Q=AU (12. = U. a a a (12.37) as ..36) and Ulr= U. (12.34) Now.500 Chapter 12 Fig. 12.= elastic time a (12. a a S S S S + *HI a cosh -X.35) c. sinh -XI.U. Then. Thus.sinh a S S S -X.=o X = L = length of pipe .39) where q = volumemetric flow rate. We may then write (12. R3/s A = pipe cross sectional area. sinh Tp U I HI. cosh Tp .42) and this applies at any section such as I or 11.40) (12.33) and (12. U. since q=AU (12.34) become (12. for the section at 11.38) Now.sinh Tp + HI cosh Tp (12.14 A view of an arbitrary pipe section selected for study. = -. = .6 g H .agH. We can see that (12. we also L[(cosh T.45) We readily conclude that the Laplace transform of the following differential equation may be written: L[(sinh T. From these relations.sinh Tp Z O HII= -ZoQI sinh Tp + HI cosh Tp where 1 --= (12.44) From the time-domain translation theorem of Laplace transform theory we write e-bsF(s)= L[u(t .51) 1 = (sech . Similarly. we rearrange (12.-(sinh Tep)hI ZO hII = -Zo(sinh T.b)] (12. t ) = hI. t) = 0 for t < T. t ) = h.p)f(t)] = F(s) cosh Tes (12.p)qI + (cosh T. Tep)hI . t) = 0 for t > T.s Z O HI = ZoQIIsinh T.Z~(tanh TePkll (12.46) can be rearranged and hyperbolic identities used to write QI= QIIcosh Tes + -HIIsinh T.s + HIfcosh Tes and in the time domain this equation pair becomes 1 (12.. write = 0 when = F(s) sinh Tes (12.(si& TeP)h.48) where 4x0.(O.(L. we conclude that the second item in (12.(L.p)hI 411 = (cash 1 (12.50) to write the hybrid equation pair 41 = (cash TeP)q/I + Z..bMt . Now note that (12.43) is the Laplace transform forf(t) when t < T.49) (12..50) where qI.43) z O- A 6 the “characteristic” impedance (12.p)f(t)] for T > 0 andf(t) .Hydraulic Turbine Prime Movers 501 I QII= QI cosh Tp .43) is the Laplace transform of the equations TeP)qI .49) and subsequently (12. Now.46) t < Te and where we use the notationp = d/dt.47) forf(t) = 0 when t < T. 55) and write new equations in terms of the A-quantities.51).57) in per-unit terms by dividing through by a base quantity. we may add more terms.51) and (12.5 1) = Tep converge rapidly.59) .z o ( d TeP)qIl. or hIIA = (sech Tep)h.5 1) are linear in both q and h such that. if we define (12.A. f o (12. we may write for the first of equations (12. we may expand the hyperbolic terms by the expansions If these sequences in u (12.502 Chapter 12 Equations (12. we may write approximately e'ft -&() = (1 .51) becomes 41 = (cash TeP)qII + -(si& 1 Z O Tep)hll (12. The head loss due to friction has been shown to be proportional to q2.58) We may also write (12. in per-unit terms.Tep)f(t) (12.51) may be evaluated by expanding the hyperbolic differential operators in an infinite series. (12.56) This nonlinearity is removed by the approximation (12.k2q11A (12.53) or. the new equations will be identically the same as (12. if more accuracy is require.55). the head equarm tion is. We recall that (12. In a similar way.k. 4110 2 (12. Let Base q = qo Base h = ho Then.57) k = 2k.Thus.zO(tanh where TeP)qIIA.51) and including a friction-lossterm 2 (sech TeP)hI . qk (12.54) We also note that equations (12.52) and if this series converges rapidly. = riser tank head.7 to typical hypothetical situations and derive transfer functions for the hydraulic system. = elastic time for the conduit hw = forebay head. = ho 41 per unit q1= 40 per unit qII= 40 41 1 per unit Zo= Z .57) applied to the conduit (from forebay to surge t n ) we have ak (12. All values are per unit. per unit Experimental runs verify this assumption [8]. Thus.). 2. 12. or h. = h. we are interested in dynamic oscillations about some quiescent operating point. Also. The results of this section and the assumptions made have been verified for at least one physical case as recorded in [8]. In so doing.. From (12. per unit . = surge t n head. the levels in the riser tank and surge tank are practically the same. wherein the wicket gates are oscillated at a range of frequencies and measurements taken to determine the system Bode diagram. we will assume: 1.Hydraulic Turbine Prime Movers 503 where we define h I per unit hI = h0 hII per unit h. Verification was checked by the frequency-response method [8. when the frequency of oscillation is high. All flows and heads are deviations from the steady state. per unit ak h. In what follows.60) We need not use any special symbol to indicate whether these are per-unit or system quantities as the equations are identical (except for Zoand Z. but we will avoid using the A subscript for brevity. Partial derivatives of nonlinear relationships are assumed to be derived at the quiescent point or Q-point. 101. It has been observed in physical situations that when the wicket gates are oscillated at low frequencies. we assume that the levels in riser and surge tanks are identical.62) where T. =- zo40 h0 (12. We will not dwell on this technique except to acknowledge that experimental verification has been checked by others. the levels in both tanks are practically constant as the water inertia prevents it from responding to rapid changes.61) (12. where h.8 Hydraulic System Transfer Function We now apply the equations of Section 12. 66).64) and taking the Laplace transform with zero initial conditions. we have (12. qp. the hydraulic system up to the penstock is completely described by (12.65) where T. 12. from Figure 12.62) and (12. + 4r + 4 p (12. Combining (12.15 Notation for changes in flow and head (all values are considered deviations from the quiescent values). that the per-unit flow rate at the surge-tank end of the conduit is 4c = 4. I --- Tailrace Fig. = surge tank riser time. h. and penstock flow rate.63) since there will be no change in head at the forebay.15.. we may write hw= 0 (12. We now observe that.normalized conduit impedance 4 =zocqO -h0 qc = conduit flow rate near surge tank. In other words.depends only on the conduit and surge-riser tank characteristics and not on the characteristics of any component following the surge tank.504 Chapter 12 . . per unit 4Jc = friction coefficient for conduit If we assume that the reservoir is large.64) We can further describe the flow into the two tanks by the differential equation Ttht = 4 + 4 r 1 (12.67) This equation is especially interesting since it indicates that the relationship between surge tank head.66) where (12. surge-riser tanks. not in the way the turbine acceleration is restrained by shaft load.68) where qp = friction coefficient of penstock Te= elastic time of penstock zp=-= normalized impedance of penstock ‘Oq0 h0 and all h’s and q’s are defined in Figure 12. we apply equations (12.4pq 1 qp = (cosh Tep)q+ -(sinh ZP T. and not on the turbine characteristicsas determined by partial derivatives in (12. . All values defined as a’s in (12.15.70) where T.64). Combining equations (12.Hydraulic Turbine Prime Movers 505 For the penstock.73) which gives a relation between the per-unit turbine flow rate and the turbine head.z (12.= -h dh dT.n + -zdz dn = aZlh + aZ2n+ a2. T.63) and (12. and conduit. (12.72) where (12. we can write = -h dh dq 34 + -n + -z dn aZ = allh + aI2n+ a132 (12.69) dT.69) and (12. We note that it depends only on the characteristicsof the penstock. T.63) and (12. These values will be read from curves of turbine characteristics.70) are not constants but are nearly constant for any operating quiescent point. we have J.71) Here we assume no electrical torque as we are interested only in the relationship between the variables.7 1).57) to write h = (sech TeP)ht .65) we can write (12.zp(bh TePh . nor on the turbine inertia as given by (1 2. = turbine starting time dn dt = T.51) and (12. we may write the following equation [8]: q where n = per-unit turbine speed z = per-unit gate position Also.p)h (12. is the per unit turbine mechanical driving torque. For the turbine. Also from Newton’s Law. + aT.where J = per-unit mechanical inertia . We may. and (12. Now.76) and (12. (12.9 Simplifying Assumptions It is quite apparent that the transfer functions (12.69).69). (12.77).78) and Figure 12. a complex hyperbolic function is represented by an infinite series and then higher-order terms can be deleted as an approximation. Another approach to simplification is through a combination of mathematical manipulation and physical reasoning.79) In block diagram notation. Using equations (12. This requires a certain amount of experience and intuition.16 (a). lump these characteristics and use only (12.70). drawing generously from the recorded thoughts of Old- .78) Finally. This is a purely mathematical approach and is quite acceptable as long as the deleted terms are small. between (12. and (12.76). 12.78) are very difficult to work with and that some simplification would be helpful. One approach is suggested at the end of Section 12.75) and (12. and (12. 12.78) we deduce that (12.p 1p Hydraulic Supply Water Turbine (a) Hydraulic Components Fig.72) we get (12.74). (12. In this approach. and should be verified by staged tests on a physical system. we have the representation of Figure 12.79).74) Equation (12. however. . combining (12.506 Chapter 12 .75) where (12. I6 Hydraulic System (b) Hydraulic System Block diagrams of a hydraulic system.76) and (12.8. we can express the hydraulic system as shown in Figure 12.72). Our approach is this latter method.77) where F6= "23 (12.16. we can write (12. (12.74) is not yet in the desired form.70).16(b). Combining (12. the resistance head-loss term we so carefully added in equation (12.Hydraulic Turbine Prime Movers 507 enburger and Donelson. Under this condition. It is noted that. We will not bother to remove the term in all expressions.65) we have qc = 0 = (41+ qr) + qp Ttht = qr + qr = -qp (12. 1. In the simplified expression for F3(s)from (12. but note that little error would result from doing + so.64) and (12. Thus. These approximations are not only those devised by experienced engineers. from (12.62) we set the conduit flow to zero.81) and the surge tank acts as an integrator.79) have been validated by experiment. One possible simplificationis that of neglecting the conduit portion of the hydraulic system and assume that the surge tank isolates the conduit from the penstock.e.82) Both this assumption and the assumption on the isolation of the conduit (12.84) Using this approximation. the error in neglecting the hydraulic resistance term is negligible.83) with the result (12. + Zp tanh T p ZP = - 1 Zp tanh T p (12. The first approximation noted is that concerning the hydraulic resistance..82) we can set. in equation (12. although present in F.. +c = 0. but tested extensively to prove their validity.85) . i. as an approximation. We now examine certain approximations suggested by Oldenburger and Donelson [8]. A second simplification involving F3 is possible from experience with physical systems. we compute (12. tanh T p 2 T p (12. which provide several degrees of simplification.80) or 1 Fl = Tts (12. This says that the water flow in the conduit does not change and the conduit is essentially closed. as presented in [8]. We write Fl l+-hnhTp F3(s)= +p + F . and all other factors (note +c and +p). F3.56) is not needed in the small-disturbance case. Thus. 3. finally. the function F4is F4= 5th degree polynomial 6th degree polynomial and is much more detailed than the previous case. TcTp2+ 4cTp + 1 (12. by letting F1= ZCT2 and F by 3 1 (12.86) F6 = (e3s3 e2s2+ els+ eo)(c2s2 + c1s+ co) + - 5th Order Polynomial 5th Order Polynomial (12. Simplify F .az2(b1s b0)(e3s3 e2s2+ els + eo) + + + + a23(e3s3 e2s2+ els + e0)(c2s2 CIS + cg) + + (12. then equation (12.87) 2.93) .=Tc4c+4Ac In this case. In this case. with (12.89) and. All of the above should be compared to the classical water-hammer formula based on a lumped system: (12.91) FI = T 2 + (6.90) This results in a more complex model that is undoubtedly more accurate. F1 becomes a second order function: (12.s + eo a2&i2s2+ dls + d0)(c2s2 cIs + co) . such detail is not necessary.62) becomes [8] h. for all except the most careful experiments. 4.508 Chapter 12 - d2s2+ dIs + d o e3s3+ e2s2+ e. If the water in the conduit is assumed to be rigid.92) and the other transfer functions also become higher order.88) (12. Experiments have indicated that.-h. 71). the speed is governed by the prevailing system frequency and the setting of the reference p determines the load that will be assumed by this machine. They note that a second-order representation of F4 is adequate unless very accurate studies are to be performed. This gives a second-order representation for F4. but having a much greater mechanical force to drive the wicket gates. This would be the case in an isolated system. the wicket gates are very large and massive. This operation also usually introduces a delay or lag. although the hydraulic system is quite complicated.We use the There..Hydraulic Turbine Prime Movers 509 Load Torque Te Penstock Error Ref Signals Servo Stroke Position Gate . 12. which is compared (usually mechanically) against a reference position p. In verifLing these approximations experimentally. surge tank. the gate position is fed back mechanically as a means of adjusting the droop or speed regulation. which is shown in Figure 12. 12. We can analyze the hydro system operation in a general way as follows. which is amplified by a control or servo amplifier to produce a servo stoke Y. it may be represented adequately for control purposes by a linear model in which all transfer functions are ratios of polynomials. *It is common to represent the torque by the symbols Tor M.17. This means that the servo amplifier must also be very large and capable of exerting large driving forces for moving such a large gate in a timely manner. In many hydro installations.1 Turbine Head I Shaft Speed Speed Governor ~ Fig. . scroll case. Any change in speed is changed by the speed governor into a change in position or displacement x. The servomotor stroke Y repositions the wicket gates to produce a new gate position 2.I7 Block diagram of a hydro turbine speed control system. In hydro turbines. They verified that hydraulic resistance may be neglected without serious error. which depends on the design of the servomotor. Any difference in these positions produces an error signal cl. In an interconnected system. Oldenburger and Donalson conclude that the hydraulic system consisting of conduit. penstock. where Tw is the so-called “water starting time” (about one second). Thus. The assumption that the surge tank isolates conduit and penstock systems is also verified. proportional to E.10 Block Diagram for a Hydro System In considering the problem of controlling a hydro station. it is convenient to think of the system block diagram. the electrical torque* is a constant and the speed N will be that set by the speed reference p. For a given steady load on the turbine T. riser tank. but recognize that this symbol is also used for time constants. and draft tube can indeed be represented by a single transfer function relating Q to H a s in (1 1. Thus. usually at times of high system loading. where there is a continual flow of water past the dam. where an elevated pool can be built above the plant site. Obviously. then the overall economics of constructing such a facility may be quite attractive. This is accomplished using a design of generator that can be operated efficiently as a motor and utilizing a turbine that can be operated as a pump.1 1 Pumped Storage Hydro Systems The hydro systems described above assume a storage reservoir of water that is elevated in a configuration that will permit the water to be directed through a system of penstocks to hydro turbines that are situated at a lower elevation. In some cases. . a minimum river flow might be necessary to support navigation or other uses of the water downstream. The operating modes of a pumped storage system are shown in Figure 12. and the generation provided by the pumped-storage plant is of high value. In the pumped-storage system there are two reservoirs. there must be generation available for pumping that can be obtained at a cost differential that will make the entire facility operation an economic success. This is true of stations that use a storage system fed by high-altitude streams. but must be created by forcing the water into the elevated storage reservoir. the elevated water is not provided by nature. Thus. some portion of which might be directed through hydro turbines to produce electric energy. confined behind a dam. There is a cost associated with providing the pumping power. If the pumping energy is available at a reasonable cost. The confined water is held in storage until power output from the station is needed.510 Chapter 12 12. since peaking load usually requires the scheduling of peaking generation with higher operating costs. even if the generators are unavailable for some reason. A pumped storage hydro power plant is different from the run-of-river system.18 The two operating modes of a pumped storage power plant. Such a variation of energy value on a daily basis is not uncommon.18. at which time it is used to power hydro turbine generators. one at a high elevation into which water is pumped for release later. Pumped storage plants require a suitable topology. which must be performed at off-peak times when excess generation is available. This requires the ability to pump power at a reasonably modest cost and a higher energy value during the generating cycle. the economic parameters must be carefully evaluated in considering the construction of a pumped-storage facility. Aside from this physical restriction. there is an interesting economic tradeoff between the cost of providing the pumped storage facility and the availability of off-peak capacity to operate the pumps. This type of system is also used for a run-of-river system. 12. Fig. (F.. including a wide range of heads and physical features.10 A21 = 1.25 s Jm=8s The base quantities are: Torque: Gate: Speed: Head: Flow rate: f4.75) and (12.-+ 3 15 and finding the transfer functions for each F. The system under study in [8] has the following constants: Tec= 13 s Te = 0. the head. . Use the data from problem 3.3. and F6 by using the approximation (a) tanh(Ts) = TS (b) tanh( Ts) = TS . F3.. and (c) and determine.009 s 4 . 12.5.16 (b). Document the sources of your research and prepare a brief report on your findings. 12..13 A. T2) tanh F =h(F3) 4 12. = 1. .2.4.18 A22= -0. 40 MW at 225 rpm 8 inches (at 80% of servomotor stroke) 225rpm 428 feet (headwater-tailwater) 1600 fi3/s The turbine constants per unit are: All = 0.5 Use approximation (1 2..35 A23 = 1. Find the transfer function of the hydraulic system shown in Figure 12.Hydraulic Turbine Prime Movers 51 1 Problems 12.1. etc. Select a hydroelectric site of interest to you and record the physical features of the plant including the type of turbine. 12. Examine the effect of nonlinearity on the transfer functions F. respectively. the installed capacity. (b).57 A21 = -0. F. Use an approximating technique to factor the truncated polynomials of (a).001 s =4 z =z .70) and compute the following: Fl ="us) F3 =f. by pole-zero plots.- (Ta3 3 ( a 2(TQ5 T 3 (c) tanh(Ts) = TS.79). where the hydraulic supply and water turbine transfer functions are given by (12. = 0. Prepare a list of at least 10 hydroelectric sites. how the addition of extra terms in the series changes the system response.= 0. AZEE. H. Section 10. B. McGraw-Hill. Washington. AZEE. 3. R. 750-758. 7. Justin. The Coordinationof Hydraulic Turbine Governors for Power System Operation. July 1966. 418419. 1950. 1964 Federal Power Commission. and J. New York. 4. pp. Wiley.R. 6. Hydroelectric Handbook. F. 5. New York. 10. Washington.New York. New York. P.. U. Wilbor. Oldenburger. Nuclear Energy and the US.S. D. pp. 1941. PAS-85. “Dynamic response of a hydroelectricplant. 403419. and A. 8. deMello. Knowlton. F. P. 81.C. Discussion of reference 8. 1943. P. R. . National Planning Association. Craeger.W. E. Barrows. Schleif. GovernmentPrinting Office. 1962. n. Prime Movers. 1962. D. Standard Handbookfor Electrical Engineers. National Power Survey... Trans. Oldenburger. A. D. Oct. Tietelbaum. 7. 81. 1955-1980. 1964 Notes on Hydraulic Turbines. 1956. 9..” Trans. v. D.. Los Angeles Department of Water and Power. p. and J. Fuel Economy. Frequency Response.51 2 Chapter 12 References 1. 2.IEEE Trans. K.C. Part ZZZ. Macmillan. Oct. Part ZZI. Donelson.. Water Power Engineering. McGrawHill.1964.. Private Communication. They can be delivered new in a relatively short time and are quickly installed compared to the complex installations for large steam turbine units.1 Introduction Two additional types of generating unit prime movers that are growing in importance are the combustion turbine and combined-cycle units. They are also widely used in industrial plants for driving pumps.2 The Combustion Turbine Prime Mover Combustion turbines. may be considered a reasonable alternative to steam turbine generating units. The early units were not large. and have a low cost per unit of output. ready to accept load. in some cases. and electric generators. 13. They are also subjected to fewer environmental controls than other types of prime movers [I]. Combustion turbines have many advantages as a part of the generation mix of an electric utility. A more recent addition to the available types of generating units is the combined-cycle power plant. compressors. but later units have become available in larger sizes and. perhaps most notably in powering jet aircraft. are used in a wide variety of applications. and can come up to synchronous speed. the combustion turbine is widely used as fast-startuppeaking units. The combination of low capital cost and low efficiency dictates that combustion turbines are used primarily as peaking units. they can operate on a rather wide range of liquid or gaseous fuels. In utility applications. compared to steam turbines. even by remote control. being dependent on the Brayton cycle. Combustion turbine units were once considered as generating additions that could be constructed quickly and were reliable units for rapid start duty. Moreover. in which the prime mover duty is divided between a gas or combustion turbine and a heat recovery steam turbine. with each turbine powering its own generator. They are relatively small in size. This makes these units desirable as peaking generating units. The major disadvantage of combustion turbines is their relatively low cycle efficiency. Combustion turbines are quickly started. 513 . limited to about 10 MVA. which makes combustion turbines undesirable as base-load generating units. The dynamic response of combined-cycle power plants is different from that of conventional steam turbine units and they must be studied carefully in order to understand the dynamic performance of these generatingunits. often called gas turbines. in a short time. Another disadvantage is their incompatibilitywith solid fuels.chapter 13 Combustion Turbine and Combined-Cycle Power Plants 13. 13. ( \ # EXHAUST TEMPERATURE A SPEED REFERENCE AIR 2 BURNER 3 - Fig. and exhaust temperature inputs. in practice the single-shaft design is the most common [ 11. which in turn. Auxiliaries that could reduce unit power capability are the AUXILIARY ATOMIZING AIR SYSTEM FUEL DEMAND AUXILIARY FUEL HANDLING SYSTEM \ # > MAIN FUEL SYSTEM I\ # CONTROL SYSTEM .514 Chapter 13 Combustion turbines can be provided in either one. The model is intended for the study of power system disturbances lasting up to a few minutes. The combustion turbine model presented here represents the power response of a singleshaft combustion turbine generating unit [2].The control system develops and sends a fuel demand signal to the main turbine fuel system. in some cases connected to the turbine shaft through a gear train. to obtain economical computer execution times. Figure 13. load. The axial-flow compressor (C) and the generator are driven by a turbine (T). It is assumed that the model is to be used in a computer simulation in which. The model is intended to be valid over a frequency range of about 57 to 63 Hz and for voltage deviations from 50 to 120% of rated voltage. The generator may be on a separate shaft. Air enters the compressor at point 1 and the combustion system at point 2. The model is a rather simple one. but it should be adequate for most studies since the combustion turbine responds rapidly for most disturbances. the timestep of the model might be one second or longer. Hot gases enter the turbine at point 3 and are exhausted to the atmosphere at point 4. the speed.regulates fuel flow to the burner. However. In the two-shaft design.1 shows a simple schematic diagram of a single-shaft combustion turbine-generator system with its controls and significant auxiliaries [2]. the second shaft drives a low-pressure turbine that requires a lower speed.or two-shaft designs. based on the unit set point.1 Combustion turbine schematic diagram [2] . These ranges are considered to be typical of frequency and voltage deviations likely to occur during a major system disturbance. A nonlinear droop characteristic may be used in some cases. Then the frequency error is divided by the per-unit regulation to determine the input demand. The frequency governing characteristic is often characterized as a normal linear governor “droop” characteristic. An alternative input KM represents a manual input that is used if the generator is not under automatic generation control. = reference temperature in “C Linear or Nonlinear Frequency Governing Characteristics ~ ‘1’“ S Off-Nominal Voltage and Power out > 1+&s 0 AGCPS Limit Effects on Power Output Nonwindup Load Nonwindup Magnitude Demand Magnitude Limit Limit Governor Speed Changer Position (GSCP) Fig. temperature. AGCPS.2 shows a block diagram of a single-shaft combustion turbine-generator control system. The output of this model is the mechanical power output of the turbine. 13.1 [2]. (13. The load demand upper power limit varies with ambient temperature according to the relation Pr.2 as GSCP. The auxiliary fuel handling system transfers fuel oil from a storage tank to the gas turbine at the required pressure. is the power signal from the automatic generation control (AGC) system.2. 13. The input signal.1) where A = (the per-unit change in power output per per-unit change in ambient temperature) T = ambient temperature in “C T.1 Combustion turbine control Figure 13. and flow rate. . = 1 +A( 1 - 6) = 1 + 0.2 Combustion turbine model block diagram [2]. The load demand signal shown in the diagram is the difference between the governor speed changer position and the frequency governing characteristic.Combustion Turbine and Combined-Cycle Power Plants 515 atomizing air and fuel handling systems shown in the figure.1 1( 1 - ). Typical data for the parameters shown in Figure 13. in perunit power per second.2 are provided in Table 13. is the integral of the AGC input. The atomizing air system provides compressed air through supplementary orifices in the fuel nozzles where the fuel is dispersed into a fine mist. noted in Figure 13. The power is expressed in the system MVA base [2]. The governor speed changer position variable. and the thermocouple time constants. l).516 Chapter 13 Value 0. The lower power limit corresponds approximately to the minimum fuel flow limit. per-unit MW/s on given base Conversion. 13.00278 - Table 13. see Figure 13. the transport times.O per-unit power at a reference ambient temperature of 15 "C. 0 ' 0 I I I I I > 0.1 Typical Combustion Turbine Model Parameters [2] Constant KM Description Manual rate. There are three different off-nominal voltage and frequency effects.1 0. per-unit fi-eq/puMVA Alternate regulation.04 R1 R2 0. These are defined in the next section. Notice the fast response characteristicof the unit to its new power level. The power limit is increased for temperatures below the reference and is decreased for ambient temperatures above the reference. .4 0. This limit is necessary to prevent the blowing out of the flame and corresponds to zero electric power generated. The turbine response will vary by several tenths of a second for other models or when using other fuels. using liquid fuel [3].3 shows the approximate computed response of a General Electric FS-5.Model N.2 0.3 CT response to a step change in setpoint from no load to rated load [3].4 4 UL 0.unit basekystem base GSCP upper position temperature Combustion turbine time constant.5 Fig. Figure 13.01 According to (13. The analytical model used to compute this response included the effects of the controls. s Normal regulation. heat soak effect of turbine components in the hot gas path.25 0. the turbine will provide 1.3 Time in seconds 0. single-shaft combustion turbine in response to a step change in setpoint from no load to full load.11 Tc 0. the utility practice.4) where DPA is the per-unit change in unit output due to a per-unit change in frequency from the base point frequency oBP . 01 (13. and the site variables. the particular installation limitations.B 1(DPF)(0 B p . bypasses the multiplier effect osYssystem frequency = wBP= system frequency when unit exceeds its power limit The RPFE is one of the possible limiting effects noted by the limitation block on the righthand side of Figure 13.max[DPA(oBp.~) = Reduced power frequency effect multiplier where (13. This represents another limiting function that is referred to in the literature as the auxiliary equipment voltage effect. which represents the limiting multiplier on power demand when the unit is running on an exhaust temperature limitation. the combustion turbine power output capability decreases as the frequency drops. A basic characteristic of the combustion turbine is that the air flow decreases with shaft speed and the fuel flow must also be decreased to maintain the firing temperature limit.2.2 Off-nominal frequency and voltage effects The power supply for the governor system is usually provided by the station battery that can provide power for at least 20 minutes and is..2) B. if the frequency declines 3 Hz or 5% on a 60 Hz system. or AEVE [2]: AEVE = 1 .2 The invocation of this limitation depends on the initial power level of the generating unit and the change in frequency during the transient.o~. such as the fuel system.max[DPV( VBp . For example. This is shown in equation (13.3) Another unit limitation is based on a reduction in system frequency. and air handling equipment. then the power capability of the unit will be reduced by 2% for each 1% reduction in speed after the power limit is exceeded. 01 where DPV = per-unit change in unit output per unit change in voltage VBp voltage level above which there is no reduction in unit output = VT= generator terminal voltage (13.Combustion Turbine and Combined-Cycle Power Plants 517 13. heaters. Off-nominal voltage and frequency both have an effect on the system auxiliaries. =( 01 when power demand < power limit when power demand > power limit DPF = per-unit change in unit output per-unit change in frequency = 0 if data not available.VT). If the power demand exceeds the power limit. These effects vary depending on the unit design. The amount of the air flow decrease is on the order of 2% in output capability for each 1% drop in frequency.osYs). A unit operating initially at full load would reach the power limit immediately and the output of the unit would be decreased by 10%.2). The shaft-driven main fuel and lubrication oil systems can be considered as unaffected by ac system voltage deviations. RPFE = 1 . This limit in defined as 121 AEFE = Auxiliary equipment frequency effect = 1 . unaffected by the voltage and frequency of the ac power system [3]. therefore. it should be noted that the nonlinear droop characteristic was suggested as one device for improving the system response to very large disturbances. and power plants that use a combination of power cycles can have higher efficiencies that those dependent on a single power cycle.3]. An example of a nonlinear droop characteristicis shown in Figure 13.3 Nonlinear governor droop characteristic In some cases. which is not linear over a wide range. The gas exhausted from the gas tur- . one might devise a continuous nonlinear curve to represent a range of frequencies and power responses.4 Nonlinear governor droop characteristic [l]. Given adequate data.2. This system utilizes a combination of a gas turbine Brayton cycle and a steam turbine using a Rankine cycle. nor is it clear that the center frequency in the R1 range should be exactly at the center between o. One typical combined-cycle turbine model is shown in Figure 13. 13. However.and %.3 The Combined-Cycle Prime Mover There are a number of ways in which a combination of power cycles can be used in the generation of electricity. This is only one type of droop characteristicthat might be examined.4 [ 1. the droop characteristic of Figure 13.2.5.which create large upsets in power plants as well as loads.51 8 Chapter 13 f f I R1 ---c \ I I I I JC-AlJFig. Some studies are not intended to accurately represent the power system under such extreme conditions. but tends to saturate for large excursions in speed or power. for example. For example.and low-frequency ranges. I I I I 0 AP * All of the foregoing limiting functions apply to the limiter block on the right-hand side of Figure 13. it is desirable to include in simulations a nonlinear governor droop characteristic rather than the simple 4% or 5% linear droop characteristic often assumed. 13. it is not entirely clear that the slopes labeled R2 need to be equal in the high. lacking better data. in which case the single droop Characteristic may be adequate. 13.4 probably represents an improvement over the single droop characteristic so often used. in providing an accurate model of the speed governor characteristic. This might be necessary. Finally. supplementary firing is used. but all systems can be conceptually reduced to the configuration of Figure 13. Moreover. The output is the fuel demand signal.5. which is described in greater detail below.. and the speed deviation.7. The inputs are the load = \demand. . which shows only the basic components. Their detailed model of the combined-cycle unit is shown in Figure 13. FD.6 shows the schematic diagram for a combined-cycle power plant with a heat recovery boiler (HRG) [ 11. which in turn provides the working fluid for the steam turbine. In some large-system designs. which could significantly increase the steam supply and therefore the power production of the steam subsystem.8 shows the interactions among the subsystems of the combined-cycle system [6]. which may cause the steam turbine to achieve a rating greater than that of the gas turbine. Many combined-cycle power plants are more complex than that shown in Figure 13.9.5 A typical combined-cycle power plant arrangement [3]. bine contains a significant amount of sensible heat and a portion of this heat is recovered in a steam generator. the steam turbine may have a lower rating than the gas turbine. V. In some designs. and identifies the input and output variables of each subsystem and the coupling among these submodels.Combustion Turbine and Combined-Cycle Power Plants 519 Fig. The speed and load controls are described in block diagram form in Figure 13. hN. 13.5. A descriptive technical paper on combined-cycle power plants has been prepared by the IEEE Working Group on Prime Mover and Energy Supply Models for System Dynamic Performance Studies [6]. More practical systems are described below. Figure 13. Figure 13. This structure is convenient for mathematical modeling of the combined-cycle power plant. there may be more than one HRG. 6 Schematic flow diagram of a combined-cycle heat-recovery boiler [l]. 13.520 Chapter 13 Combustion Chamber Air Compressor Gas Turbine = Generator 1 Air Fuel Optional * Supplementary Firing System i Steam SU = Superheater B = Boiler EC = Economizer I i - - Steam Turbine Generator 2 \/ \/ Condenser Deaerating Heater Boiler Feed Feedwater Heater I Fig. . 13. . FueL Gas Turbine ~ ~ z Power ~ Exhaust Temperature Gas Turbine Flow Rate Steam Turbine b Steam Turbine Mechanical Power Fig. + Deviation I SpeedLoad Control Controls + . 13.Combustion Turbine and Combined-Cycle Power Plants 52 1 Stack Steam Turbine Generation Cooling Water J Condensate Pump Gas Turbine Generation ITreatmentl Fig.8 Subsystems of the combined-cycle power plant [ 6 ] .7 Two-pressure nonreheat recovery feedwater heating steam cycle generating unit (HRSGwith internal deaerator evaporator) [ 6 ] . This is accomplished by scheduling air flow with the load demand FD and setting the turbine exhaust temperature reference TRto a value that is calculated to result in the desired load with the scheduled air flow at constant turbine inlet temperature. The exhaust temperature reference is calculated from the following basic gas turbine thermodynamic relations (taken from reference [6]).1 Fuel and Air Controls The gas turbine fuel and air controls are show in block diagram form in Figure 13.9 Combined-cycle speed and control [ 6 ] .522 Chapter 13 MAX f- AN MIN Fig. 13.3. (13. and are active over a limited range.10 [6]. This allows maintaining high turbine exhaust temperatures.5) T R FD Fig.10 Gas turbine fuel and air flow controls [6]. improving the steam cycle efficiency at reduced load.In this control scheme. The fuel and guide vanes are controlled over the load range to maintain constant gas turbine inlet temperature. . 13. the inlet guide vanes are modulated to vary the air flow. 13. These are determined by the solution of (13. The block identified as A in Figure 13. These performance effects have been incorporated into equivalent compressor and turbine efficiency values [61. The actual air flow W is shown as a product of desired air flow and shaft speed.6) with T. specific heat changes. Desired values of WD and T are functions of FD (the desired values of turbine output from speed/load controls) and ambient temperature T.0. R . Also KO= 3413 kWo WO f QCP gT + (13. The reference exhaust temperature T is given by (13.set equal to unity. and at a specified per-unit ambient temperature Tp The .7) and (13..Combustion Turbine and Combined-Cycle Power Plants 523 where T = reference exhaust temperature per unit of the absolute firing temperature at rated condiR tions Also x=(pR)(rWY= ( p R o j q p l Y Y where PRO design cycle pressure ratio = PR PROW isentropic cycle pressure ratio = = y = ratio of specific heats = cJcv We also define the following W = design air flow per unit q3 = turbine efficiency T = turbine inlet temperature per unit of design absolute firing temperature f (13. Equations (13.8) and where we define kWo = base net output per unit WgO =base net flow per unit Tfo = turbine inlet temperature per unit of design absolute firing temperature Cp= average specific heat = compressor inlet temperature per unit of design absolute firing temperature qc = compressor efficiency The combustor pressure drop.7) where kW is the design output in per unit.8) with appropriate limits on WD and TR.6) Then the per-unit flow required to produce a specified power generation at the given gas turbine inlet temperature T is given by the turbine power balance equation f (13.10 represents the computation of the desired air flow WD and the reference exhaust temperature over the design range of air flow variation by means R of vane control.8) determine the air flow Wand pressure ratio parameter Xfor a given per-unit generated power in kW.7) and (13. reference exhaust temperature T is given by (13. range. The air flow must be R subject to the control range limits. The vane control response is modeled with a time constant T and with nonwindup limits corresponding to the vane control R .6) by setting T= 1. and the detailed treatment of cooling flows have been deleted for purposes of illustration of the general unit behavior. The gas turbine output is a function of the computed turbine inlet temperature Tf.. 13.1 per unit.1 1 Gas turbine mechanical power and exhaust temperature model [6]. + Wr. through the low-select (LS) block.1 1. (13.7)for the computation of X. 1.2 The gas turbine power generation A block diagram of the computation of gas turbine mechanical power PMG the exhaust and E temperature T is shown in Figure 13. giving a fuel flow signal W.6).as another input to the gas turbine model.which is a function of the turbine air flow Wj.3.-= design air flow per unit The gas turbine exhaust temperature T is determined by equation (13. Normally. If T should exceed TR. K2= . T is less than TR.substituting T E E for T and using (1 3. The equations used in the development of the gas turbine mechanical power PMG shown are 31 in Figure 1 . . The fuel valve positioner and the fuel control are represented as given in [7]. the controller will come off limit and integrate to the point where the its output takes over as the demand signal for fuel V.524 Chapter 13 The measured exhaust temperature T is compared with the limiting value TR and the error E E acts on the temperature controller.= per-unit combustor temperature rise AT Fig.which causes the temperature E controller to be at the maximum limit of about 1. The mechanical power PMG a function of the R is turbine inlet temperature and the flow rate of combustion products W. 13.9) where Tfo TcD = compressor discharge temperature per unit of absolute firing temperature W. T / n ) and where Qeconl.and low-pressure steam generators. which can be approximated.15) = w ~ g 2 ( T '. The transient heat flux to the high.Tml) + (Qeconl + Q'econl) (1 3.Combustion Turbine and Combined-Cycle Power Plants 525 13.T m 2 ) + (Qecon2 + Qeconl) where &icon I = ~ e c ~ ( 7 "' tecon2 = TW~HP 9~ + 77ec2(Tt' . rlgl = Tex. Temperatures Tmland Tm2are the average metal temperatures in the HP and IP evaporators. The exhaust gas and steam absorption temperatures through the HRSG are indicated in Figure 13.12 Steam energy exhaust gas temperature versus heat absorption [6].14) (13. The gas heat absorption by the HRSG section can be computed as follows [ 6 ] .12) (13.11) = Wqgl(Tex . respectively.3.13) (13. 13.T' Tex (13.TI' (13. QgHp QgLp =- T' . 100 .10) . % Fig.This heat is delivered to the high. are I Heat Absorption.3 The steam tvrbine power generation The heat recovery steam generator (HRSG) system responds to changes in the exhaust flow from the gas turbine Wand its exhaust temperature TE. and Q'econl the HP and IP economizer heat fluxes.12. Qecon2.12.and are computed as follows [6].Tml rlg2 T -Tm2 where T' and T" are the gas pinch points shown in Figure 13.and low-pressure steam generation sections can be approximated using the relations for constant gas side effectiveness. 13. The steam flows.h L p m L P + h/wmLpJw (1 3.1 1) through (13.17) where K T = throttle valve flow coefficient K' = admission point flow coefficient Steam pressures PHpand PLpare found by integrating the transient energy equations.h h p m H P + hJWmHP + hJWmHPJW =Qgw . which are given as DIIPPHP DLPPLP = QgHP .16) Then equations (13.19) Fig.. = M H P * AEHP ' mLP ' AELP 3413 (13.526 Chapter 13 The economizer heat absorption is approximated using the constant effectiveness expressions. The steam turbine power in kilowatts is computed as kW. . as follows [6]: (1 3.13 Steam system model.17) are solved to find the temperature and heat flux profiles. and Tm2are determined by integration of the gas and steam side heat flux as shown in Figure 13. mHpand mLpare computed by the pressurehlow relationship at the throttle and admission points as follows: ~ H = KTPHP P mHP+ mIp= K'PIp (13.13.1 8) The HP and LP metal temperatures T. New York. Make a sketch of how such a closed-cycle system might be configured.. Sketch how such a two-shaft unit might be configured and compare with the single-shaft design.Combustion Turbine and Combined-Cycle Power Plants 527 Fig. The detailed model should be considered for studies of disturbances in the vicinity of the combined-cycle plant. 13.and the IP evaporator metal temperature Tm. It is noted in reference [6] that the total contribution to mechanical power from the two pressure boilers can be approximated with a simple two-time constant model. especially studies in which the major disturbance of interest is far removed from the combined cycle power plant. and may be quite satisfactory for may types of studies. A. 13. or from design heat balance calculations for rated and partial load conditions. . Powerplant Technology. if needed. Moreover. M.1 is a single-shaft design. Note that the heat transferred from the high pressure boiler QG. in general. These simplificationswill result in a low-order model as shown in Figure 13. 1984. P. this simple model could be “tuned” by comparing it against the more detailed model of Figure 13. Other combustion turbines are designed to employ two different shafts. Schulz.14 [6]. .1 The combustion turbine presented in Figure 13.2 The single-shaft combustion turbine shown in Figure 13.the HP evaporator metal temperature T. Explore this cycle using appropriate references on thermodynamic cycles and sketch both the P-V and the T-S diagrams for this cycle.. The gain between the gas turbine exhaust energy and the steam turbine output will. User’s Guide to the LOTDYS Program. and R. References 1. April 1978.McGraw-Hill. Research Project 764-2. What are the advantages of a two-shaft design? Hint: Consult the references at the end of the chapter.is a function of the exhaust gas temperature TE. are the steam actual available energies [6].. be a nonlinear function that can be derived from steady-state measurements through the load range. 2.1 is called a “direct open cycle” design since it exhausts its hot exhaust to the atmosphere.13. Final Report. The dynamic relations for the HRSG and steam turbine are shown in Figure 13. Long Term Power System Dynamics. CA. A different design is called a “closed-cycle” system. M.14 A simplified steam power response model [6]. Electric Power Research Institute. New York. E. where AErip and AE. Palo Alto. which recycles the exhaust back to the air input port. Such a low-order model would be very simple to implement in a computer simulation.3 It has been noted that the ideal cycle for the gas turbine is the Brayton cycle. 13. El-Wakil.13. From [6] the values of the time constants for this simplified model are given as TM= 5s T5-=20~ Problems 13. Turner. volume 1. Turner. 1983. Pier. MA. October 1983. ASME. Bailie. Long Term Power System Dynamics. Reading. IEEE Working Group on Prime Mover and Energy Supply Models for System Dynamic Performance. June 1974. 9. Energy Conversion Engineering. J. Ewart. N. 1698. and D.. Chairman. 6. R. E. 105 (l). 1978. 3 . A. August 1994. 72-EU-2099. p. Addison-Wesley. Summary and Technical Report. I. . R. Series A. W. EPRI Report 90-7-0 Final Report. 1972. Bednarski. F. “Simplified mathematical representations of heavy-duty gas turbines. P. 7. pp.” General ElectricCompany Report. deMello. 5. B.“Dynamic models for combined cycle plants in power system studies. and S.” ZEEE Transactions Power Systems. 865-869. Schulz.Chapter 13 3. Journal of Engineeringfor Power.. 4. P. Rowen.” Trans. “A simplified single shaft gas turbine model for use in transient system analysis.. C. 1 / 2 s i n e c o s e . sin(@f 120) = -1/2sinB f ~ ' 3 / 2 c o s 8 cos(8 sin2(e f 120) cos2(e f i = = (A.120) = 1/4sin28 .12O)cos(8 = = sinecost9 . the engineer encounters a large number of trigonometric functions involving the angles f 120".appendix A Trigonometric Identities for Three-phase Systems I n solving problems involving three-phase systems.I / ~ C O S 'F ~ v'3/2sin8 ( A .v'3/4cos2e .v 3 / 4 sin8cosB + 120) sin (e .I / 2 sin e cos e -1/4sin 28 f d 3 / 2 cos2e f + fl/4cos28 fl/4 sin(e + 120)cos(e + 120) = . Although the symbol (") has been omitted from angles i 120".v'3/4 sinZe .lO) e sin (e + I 20) = = = .1) 120) = .3/4cos28 = -1/4 .120) sin(@.120) cos(e .1/4 COS 28 'F v T / 4 sin 28 -1/2sinBcosB =F v'3/2sinZ8 .4) (A.12) (A.1/2cos28 .6) I /4 sin2e + 3/4 cos2e r d 3 / 2 sin e cos e 1/2 + I / ~ C O SF ~v'3/4sin28 = ~ I /4 cos2e + 3/4 sin2e i v T / 2 sin e COS e I /2 .1/4 sin 28 f d / 4 cos 20 =F d 3 / 4 sin8cos(e COS = = (A. it is always implied.7) (A.v?/4 = 1/2sin28 .1/4 + 1/4 cos 28 f 4 / 4 sin 28 e cos(B f 120) 120) = = .120) sin(t9 = = + G / 4 = 1/2sin28 + f l / 4 (A.11) .2) (A.&/4 cos 28 + vT/4sin2e sin(@+ 12O)cos(B .5) (A.1 /4 sin 20 + f l / 4 COS 28 (A.8) (A.1/4 sin 28 .1 / 2 sin e cos e + d / 4 cosze .1/4 COS 28 i &/4 sin 28 = 120) = = sinesin(e f 120) COS = -1/2sin2e f t/S/2sin8cose .9) (A.13) 529 + 120)sin(e .1/2 COS' 0 =F ~ ' 3 1 2sin e COS e .1/4 . Some of these are listed here to save the time and effort of computing these same quantities over and over.3) (A. 15) (A.1/2 -1/4 + I / ~ C O S ~ (~ ~ .l4) sin (28 f 120) sin 28 f .sin28 = cos28 = sinZ8= I 1/2sin28 cos28 ( I + cos2e)/2 (1 .16) (A.~ 0 ~ 2 8 ) / 2 (A.120) + sin (8 + 120) cos (8 + 120) = 0 sin8 sinZ8+ cos28 = sinBcos8 = cos28 .530 cOs(8 Appendix A + I ~ O ) C O S-B120) ( = I / ~ C O S ~3/4sin2B = ./rl2 cos 28 7 (A. the following commonly used identities are often required: .20) + sin2(8 + 120) = 3/2 cosze + cos2(e .120) + cos(e + 120) = o sin28 + sin2(8 .120) + cos2(e + 120) = 3/2 sin 8 cos 8 + sin ( 0 .17) COS(^^ f 120) = .1/2 cos 28 4 / 2 sin 28 + sin(8 .21) In addition to the above.~ = .18) (A.120) + sin(8 + 120) = 0 case + cos(e .120) cos(8 .19) ( A .20) (A. and other special functions. As such it may be a valuable aid to the understanding of some of the text material and may be helpful in attempting an actual analog simulation. multiplication. Usually V4 is limited to 100 V (IO V on some computers). with hybrid systems as a combination of the two. where both the analog symbol and the mathematical operation are indicated. we will connect several components to solve a simple differential equation. The summer. Computer solutions fall into two categories.1. Hand computation of such large systems of equations is exceedingly cumbersome. A short bibliography of references on analog and digital solutions is included at the end of this appendix. It should be used as a supplement to the many excellent books on the subject. This material is divided into two parts: analog computer fundamentals and digital computer solutions of ordinary differential equations. Later. 6. Note that the amplifier inverts (changes the sign) of the input sum and multiplies each input voltage by a gain constant k. On most computers ki may have values of I or IO. the engineer who attempts an actual simulation will surely need the instruction manual for the computer actually used. The first important component is the summer or summing amplifier shown in Figure B. The purpose of this appendix is to reinforce the material of the text by providing some of the fundamentals of computer solutions. but some models have other gains available.1 Analog computer components Here we consider the most important analog computer components. division. integration.1.1 Analog Computer Fundamentals The analog computer is a device designed to solve differential equations. Some physical systems. multiplication by a constant. These include summation. and computer solutions are usually called for. This is done by means of electronic components that perform the functions usually required in such problems. such as power systems. The purpose of this appendix is to acquaint the beginner with the basic fundamentals of analog computation. analog and digital. 53 1 . I n particular. We discuss these components using the common symbolic language of analog computation and omit entirely the electronic means of accomplishing these ends.appendix B Some Computer Methods for Solving Differential Equations The solution of dynamic systems of any kind involves the integration of differential equations. are described by a large number of differential equations. 6. selected by the user. On some analog computers it is necessary to use high-gain amplifiers to simulate certain operations such as multiplication. although it should be mentioned that this symbol is not used by all manufacturers of analog equipment. The functionfmust be single valued. usually being greater than IO4 and often greater than IO6. B.2. The summer. *v Fig. integration may be done rapidly and very reliably by electronic means. = -(k1 V I + k 2 V2 + k . VO I Fig.3. The integrator.5. V. This feature makes it possible to simulate with reasonable accuracy certain nonlinear functions such as generator saturation. O 5 k 1. The function generator. Note that the gain of the amplifiers is very high.I.4. Gain constants ki are chosen by the operator and are restricted to values available on the computer. Function generators are represented by thk “pointed box” shown in Figure B.532 Appendix B Fig. as shown in Figure 9. and this function is set according to the instructions for the particular computer used. The function generator is a device used to simulate a nonlinear function by straight-line segments. V2 = k V l . The potentiometer. 8. It is necessary to be able to perform integration if differential equations are to be solved. where Vo is the initial value (at I = 0) of the output variable V4. This . The high-gain amplifier. = -1 V. The potentiometer. Fortunately. V2 = / ( V I ) .. The output voltage is limited. where the signal is implied as going from left to right. Potentiometers are usually IO-turn pots and can be reliably set to three decimals with excellent accuracy.)dr). The function generator. B. The symbol usually used for this is shown in Figure B. V I ) . usually to 100 V. The integrator. Fig. 8.2. The potentiometer is used to scale down a voltage by an exact amount as shown in Figure B.4 where the function f is specified by the user. + l‘ ( k l V I + kzVz + k 3 V. usually 1 and IO. V.3. The multiplier inverts and divides the result by 100 (on a 100-V corn pu ter) . These specialized devices are left for the interested reader to discover for himself. and f is a nonlinear function of v and t . Suppose we write where T = r / T .( v . V. Time scaling can be illustrated by means of a simple example. one proportional to u. which in turn are squared and subtracted.s Fig. I n an analog computation the integration time must be chosen so . The high-gain amplifier. A > lo4. = . say VI.6. or one quarter of the difference of the squared signals. Then we form sum and difference signals.6. To do this. p = vi. Consider the first-order equation dv T .VI V2 PU. .1. T is a constant.e.VI V2/100 V = .t ) dt where u is the dependent variable that is desired. V . but in its simplest form it may be represented as shown in Figure B. The multiplier. The multiplier used on modern analog computers is an electronic quarter-square multiplier that operates on the following principle. Thus replacing the constant T by unity as in (B. The symbol used for multiplication varies with the actual components present in the computer multiplier section.( v 2 . Other components. Le.i ) 2 = ( v 2 + 2vi + i 2 ) . The constant T would appear to be merely an amplitude scale factor. = . i. Note that it is usually necessary to supply both the positive and negative of one signal.2vi + i2) = 4vi and p = ( 1 / 4 ) M .2) amounts to time scaling the equation.A V . .Appendix B 533 . B. time scaling and amplitude scaling. A + Fig. The multiplier. we begin with two voltages.= Y(v.. Most full-scale analog computers have other components not described here. the other proportional to i .2 Analog computer scaling Two kinds of scaling are necessary in analog computation. B S . 8. Suppose v and i are to be multiplied to find the instantaneous power. including certain logical elements to control the computer operation.. M = (v + i ) 2 . but such is not the case. means that the input voltage of such amplifiers is essentially zero since the output is always limited to a finite value (often 100 V). 534 Appendix B that the computed results may be conveniently plotted or displayed.. 8. = assigned output level.1 Suppose the integrator in Figure 8. Also define L as the level of a particular variable in volts. The scaling procedure follows: 1 . its maximum safe value.7): PG = KL. corresponding to 1. ifa = 100..0 pu. One reason for this is that the electronic integrator is unable to tell the difference between the two scale factors.l) ordinarily does not go f .7 is to integrate -6 (in pu) to get the torque angle 6 in radians. the computer should be time scaled to plot the results more slowly than this limit. Then we write . 0 ~ I~ the computer is rated IOOV.7. Let the time scaling constant a be defined as follows: T = computer time t = real time a = T - I = computer time real time (B. 0 5 P 5 1 G = amplifier or integrator gain K = physical constant computed for this potentiometer Lo. Lin tntegrator Er sunmer Fig. It also means that 100 s on the output plotter corresponds to I s of real time. this means that it will take the computer 100 times as long to solve the problem as the real system would require. Choose a time scale a that is compatible with plotting equipment and will give rea- sonable computation times (a few minutes at most). if the output plotter has a frequency limit of 1 . Actually..O kHz. it is convenient to scale time and amplitude simultaneously. For example. Apply the following formula to all potentiometer settings (see Figure B. V Lin= assigned input level. the amplifier would reach 100 V . Choose levels for all variables at the output of all summers and integrators.1. suppose the variable u in (B. We begin with the following definitions. This requires that the user estimate the maximum value of all variables to be represented and scale the values of these variables so that the maximum excursion is well below the computer rating. 2. (B. Then if u goes to 5.3) For example. 3. we could set L = 2 0 V on the amplifier supplying u.3 Analog computation Example B.0 pu of that variable. above 5 . Time and amplitude scaling. Moreover. this makes one equation suffice for both kinds of scaling. Analog computers must also be amplitude scaled so that no variables will exceed the rating of the computer amplifiers (usually 100 V).5) where a = time scale factor P = potentiometer setting. V B./aL. For example. B.503 = potentiometer Example B..5) is wR. Use numerical data from Examples 7.5. Also estimate d. Then compute PG = KL...745 x 50 < 100). i in S o h ion Let a = 50. (1. up when separately excited when self-excited + U.. (1. = 1.8. I n our example let wR = 377. Solution diagram for dc exciter buildup.7 and (B.Ri)/T (B.which is required to convert from pu to i in rad/s./aLin = (377 x 50)/(50 x 75) = = = 5. are constants. = 75 V. and 7. . An alternate solution utilizing the Frohlich approximation to the magnetization curve is described by the equation Solving this equation should exactly duplicate the results of Chapter 7 where this same equation was solved by formal integration.2 Compute the buildup curve of a dc exciter by analog computer and compare with the method of formal integration used in Chapter 7. so let Lo. = 100" = 1.= 50 V. when boost-buck excited where both up and U. Then the levels are computed as fol!ows: 6.8..25 pu. Thus the analog computer diagram is that shown in Figure B. IO = gain of integrator and P 0.03 Since 0 5 P S I let G setting.Appendix B 535 JO Thus the constant K in Figure B.7) = U. Solution For this problem we have the first-order differential equation bF where u = = (u . so let L .4. where uF0 = ~ ~ ( 0 ) . 7.745 rad. = U.6... Fig.25 x 75 < 100).. from Table 7. = v .9.6 we have Separately excited: u = up = 125 V Selfexcited: v = U.25 s a = 279. and the potentiometer settings are given in Table B. 7.9) where R and v depend on the type of system being simulated. Suppose we choose a base voltage of 100 V. B.0565 R~FI(2.0s-' Then the factor 0. the same computer setup solves the separately excited. C .65 RvFI(279. = u . and boost-buck buildup curves respectively. Solution diagram for Frohlich approximated buildup.0. Voltage levels are assumed for R Switch Code R = Separately excited C = Self-excited L = Boort-buck excited ( ) = Voltage level of 1 . Rewriting equation (8.799- VF.25 in front of (B. I . The analog computer solution for (B.9 h = 5.4 we have T~ = 0.lO) A convenient time scale factor is obtained by writing o r a = T / t = l / r E = 4.536 Appendix B Using numerical data from Example 7. we have 0. selfexcited.F.lO) becomes unity.REF Fig.O pu . Also.25 L.3 V in all cases.65 The values of R and u depend upon the type of buildup curve being simulated.9) by the base voltage we have the pu equation 0. . R = 30 R Boost-buck excited: u = vF + 50 V R R = = 34 Q 43.) (B.5.5 we note that the derivative of uF can be greater than 100 V/s. Then dividing (B. and 7. and L.4. By moving the three switches simultaneously to positions R .8)with numerical values.O s of real time. This will help us scale the voltage level of fiF.9.lO) is shown in Figure B. and 4 s on the computer corresponds to 1 . From Examples 7.9 - uF) V (B.6 52 and these values will give a ceiling of 110.25 CF where uF and u are now in pu.5. 20 0. 8. uFo bR (separately) 6 R (self) bR (boost-buck) Scdk UP UR I . The nonlinear differential equation here is VF.o 1 . Since the solution is known. I25 0.25 0.5.Ri) (B.6. a 2. One effective method of introducing a subject is to turn immediately to a simple example that can be solved without getting completely immersed in details.050 0. These values are substituted into (B. our numerical exercise will serve as a check on the work of Chapter 7. the real reason for choosing this example is that it is a scalar (one-dimensional) system that we can solve numerically with relative ease.6.384 x 1 Other table entries are similarly computed.92 I .2..050 0.. Potentiometer Potentiometer and Gain Calculations for Figure B. 7. I .25 = 0. Such problems are generally called “initial value problems” because the dependent variable vF is known to have the initial value (at r = 0) of u.199 0.492 0.2 Digital Computer Solution of Ordinary Differential Equations The purpose of this section is to present a brief introduction to the solution of ordinary differential equations by numerical techniques. All methods divide the time domain into small segments A t long .492 0.5) to compute the PG products given in Table B.0/4)(50/10) PG (1. Larger n-dimensional systems of equations are more challenging.384 0.=-= . Table B.4 -7.o 1 .40 0..50 1 . which was solved by integration in Examples 7./Lin)= (1.384 = 0. but the principles are the same.I. We shall use this technique. However.46 1 .o . 8..1 Brief survey of numerical methods There are several well-documented methods for solving the initial value problem by numerical integration.25 0. for potentiometer 5 PG = (K/u)(LOu.Appendix B 537 each amplifier and are noted in parentheses. The treatment here is simple and is intended to introduce the subject of numerical analysis to the reader who wishes to see how equations can be solved numerically. dr l -(u 7E .45 1.40 0.20 0.56 ..(O) = v.339 0.o 0. dv.695 2.125 x IO or for potentiometer 7 = = 0.1 I ) which we will solve by numerical techniques using a digital computer. 1 10 I I I I The computed results are shown in Examples 7. I25 I I I I 9 IO 11 0.92/1)(10/50) = 1. I25 0. and 7.9 Function K PG P G I 2 3 4 5 6 7 8 scale scale time scale initial value. Our sample problem is the dc exciter buildup equation from Chapter 7. For example.45 0.56 0.339 0.384 0.20 0.20 I .4. A complete analysis of every method in Table B. Crane . at the end of each segment.I) i = f(X.. Instead. a given computation scheme may start the integration using one method and then change to another method for increased speed or accuracy. A brief outline of some known methods of numerical integration is given in Table B. x. trapezoidal rule. Some methods are self-starting and others are not.12) we may write .I) (B.2 Modified Euler method Consider the first-order differential equation fi = f(u. slow Start by Runge-Kutta or Taylor series Imposes maximum condition on A t for stable solution Varies size of S t to control error (W2 (W3 (At)' (A05 . we will investigate only the modified Euler method in enough detail to be able to work a simple problem. However. the speed of computation.. and the generation of errors.2.t) (B.14) . Speed is important because..2. Table B.. choosing Af too small may greatly increase the cost of a computed result and may not provide enough improvement in accuracy to be worth the extra cost. = f .. it is easily shown that any nth-order equation can be written as n first-order equations. d ' or in matrix form x'2 = fi(U. Thus.13) Thus we concern ourselves primarily with the solution of a first-order equation. any process that generates a great deal of computation may be expensive..2 is beyond the scope of this appendix and the interested reader is referred to the many excellent references on the subject. Thus instead of (B... therefore. 8. for example. although the digital computer may be fast. I n doing this there are three problems: getting the integration started.I derivatives to solve for u(") Self-starting Self-startingpredictor-corrector Self-starting. Simpson's rule Euler Modified Euler (Heun) Runge-K u tta Milne Hamming Af Must known .538 Appendix B and solve for the value of u... ( u . Note that the form of equation is given in each case as an nth-order equation...2. Some Methods of Numerical Integration of Differential Equations Method ~~~ ~ Form orequation Order oferrors Remarks Direct integration... But boAt is the rectangular area shown in Figure B. Now define which gives the initial slope of the u versus f curve. ) . we compute Now approximate the true area under the 6 versus t curve between 0 and A t by a trapezoid whose top is the straight line from 6.16) which is an extension of the initial slope out to the end of the first interval. as P ( u l ) = U. Calling this value P ( f i . + .Appendix B 539 V Fig.At (B. is too large [also see Figure B. 14). Next a predicted value for u at the end of the first interval is computed.lO(b)]. Suppose we now approximate the value of fiI by substituting P(u.)into the given differential equation (B. as shown by the dashed line in Figure B. I f we define u = u1 when t = At. . ) . where u is known for t = 0 (the initial value). we compute a corrected value of u I . Suppose the curves for u and 6 are as shown in Fig.10. so we conclude that P ( u l ) viously larger than the true area under the .'to P ( c l ) .lO. which we call C ( u .lO(a) and is obversus t curve. we compute the predicted value u. Graphical interpretation of the predictor-corrector routine: (a) versus 1. as shown in Figure B. B. (b) o versus 1.lO(b). IO(a). Using this area rather than the rectangular area.. B. where the time base has been divided into finite intervals A t wide. by the same method.)k - C(U. )differ from one another by less than some prescribed precision index or until C(U. Let us proceed using the latter of the two methods.4.25) where a and b may be found as in Example 7. = 500 - 282. ) . Use numerical values from Example 7.2 we have a = 279. Thus using linear interpolation.u.18) the corrector equation.23) where i as a function of u. we use it as the starting point to find u. We could store the data of Table 7. Once u I is determined as above. for example).)/(u.l. S o h t ion The equation requiring solution is rECF = up . ./(a - 0) .) (8. + (i. + ~ ) ui = + {[Ci + P(Ci+l)]/2}Af + Lji(At) (B.23) becomes (B. Now we substitute the corrected value of u.. ) .C ( u . and substitute in (B../(279.5 ~.9 - UP) (B.2 1) (B. A n alternative method is to use an approximate formula to represent the nonlinear relationship between U.i . we have for any value of u between uI and u2 i = i.u.3 Solve the separately excited buildup curve by the predictor-corrector method of numerical integration. .23) to find C. where from Example 7. We could proceed in two different ways at this point. ) ( u . i = bU.18) rather than P(6.3.65 Thus (B.)"-' 5 6 (B.20) where k is the iteration number and e is some convenient. small precision index (IO-('. At] We now repeat this operation.3 in the computer and use linear (or other means) interpolation to compute values of i for U.27) . is known from Table 7.19) C(ci 1 = f[C(uI). using C(Cl) in (B. (B.9 b = 5.This is done over and over again until successive values of C ( u . by the Frohlich equation.22) 8. Thus.540 Appendix B We call (B.26) or 6.)to obtain an even better value for C ( u .R i (B. (B. The general form of predictor and corrector equations is P ( u ~ += ) ~ ui C ( V .3 Use of the modified Euler method Example B.24) I n this way we can compute the value of i corresponding to any U. and i.2.into the original equation to get a corrected r.2. between given data points. we drop the subscript on uF. WRITE W =V t VDOT’ DELTA T. V. The derivative may not be needed. VDOT T=T OLD = W CVDOT= W D O T v = cv t DELTA VDOT = CVDOT I 1 COMPUTE CV = V t 0. Computer flow diagram.12.4 for values of t from 0 to 0.5 (VDOT + CVDOT)* DELTA I COMPUTE OLD = C V Fig. but it is known and can just as well be printed. The computer flow diagram is shown in Figure B. B. separately excited case. represent U. V . and replace 7 by T to write U = W / T . The FORTRAN coding is given in Figure B. .8 s.28) The data that must be input to begin the solution is shown in Table B. by a constant W . The solution is printed in tabular form in Table B.u)] (B. VDOT El J=J+l COMPUTE B. The computed results agree almost exactly with the results of Example 7.( R b / T ) [ u / ( a. Note that both uF and bF are given.3 with certain additional variables that must be defined.4 and are therefore not plotted. To avoid confusion in programming.I I .Appendix B 54 1 v READ DATA A COMPUTE 1 7 WRITE T.II for the separately excited case. F10.5* (VDOT+CVDOT)*DEtTA IF(CV-OLD-EPS) 107.2 X X X X X X X X X X X X PV DOT CVDOT PV cv T F5.PV) OLD=PV CVDOT = PVDOT CV = V + 0.R.2 F5.VO.3.F5. Symbol UP Data and Variable Symbols.12.542 Appendix B VDOTl(W.EPS 101 FORMAT(F5.3 X .2.7 F5.VDOT DO 200 1= 1. Table B.F10.2 F6.V) WRITE(3.2.3.TEE.3 F5. Names.3.0 VDOT = VDOT 1 (W.0 PV = 0.KEND.F4.0 cv = 0.3 T vo DELTA KEND EPS V V DOT F5.F6. 8.F5.2 F4.106 CVDOT-VDOTl(W.A.I lO)T.0 T=0.1 lO)l.F7.V) = (W-R‘ B’V/(A-V))/TEE READ( 1.2. and Formats N amr W TEE R B A Format Constant X X X Variable F5.0 CVDOT 50.107.F5.V.2 F5.CV) OLD = CV GO TO 104 T=T+DELTA v=cv VDOT=CVDOT WRITE(3.3 F6.3. KEND PV=V+VDOT‘DELTA PVDOTsVDOTl (W.F5.I3.4 13 F7.DELTA.3.2.7) v=vo VDOT = 0. FORTRAN coding for the separately excited case.B.V.VDOl - 105 102 13 0 104 106 107 110 FORMAT(”.2) 200 CONTINUE STOP END Fig.0 PVDOT 0.lOl)W.4.F10. 00 70.23 367.24 139.540 0. Brace and World. R .66 147.32 299.00 132. Analog Computation and Simulation: Laboratory Approach.96 167.4.26 184.50 68.730 0. lnlext Educational Publ.20 77.23 72.65 177. .470 0. L.73 59. J . 1965.490 0.72 104.73 154.350 0.85 0.070 0. Boston. Prentice-Hall.52 14.15 110.03 375.78 405.1I 166.430 0.56 149.94 27. New York.410 0.58 92.130 0. Johnson.08 103. A .640 0.14 119.520 0.680 0.02 73.79 263. Smith.480 0.83 66.5I 74.440 0.3 IO 0. R . 1969.210 0.50 48.9 I 9.530 0. 1963.290 0.64 176.5 I 32.20 452.59 42.72 123.10 116..56 39.30 155.90 446.J.760 0.240 0.03 34.20 390.450 0.74 272.06 96. C . McGraw-Hill. Scranton.60 172.. L.45 177.270 0.97 I13.320 0.63 143.7I 89.23 48.85 19. C.36 I5 I .060 0. I50 0.030 0.92 85.46 176.300 0.090 0. Introduction to Analog Computation. M .360 0.52 10.85 52. Englewood Cliffs.010 0.630 0.420 0.55 159.09 152.84 4 12.82 75.040 0. and Wolford.780 0.95 124.09 121.510 0. I40 0.16 163. M.41 130. .38 12.160 0.600 0.670 0. Pa.94 228.45 160.82 117..88 236. 1965. I9 8.76 145. New York.05 81.00 44.280 0.50 134.65 28 1.650 0.570 0. . New York.720 0.95 177.6 I 100.590 0.24 164.580 0. I2 22. I90 0. Analog and Analog/Hybrid Computer Programming.0 0.09 74.750 0. International Textbook C o .68 9.79 157. G .500 0.41 11. t Separately Excited Results in Tabular Form 6F "F I VF .49 171.86 176. I5 29.340 0.29 359.00 16.50 426.30 57.21 2 10. Pa.52 98.100 0.330 0.52 Analog Computation Ashley.800 158.52 290.230 0.1 I O 0.F 0.55 440. Harcourt.05 219.80 176. N.34 177.220 0.7 I 37. Hausner.1 I 175.73 130. I97 I .87 87.740 0.92 137..560 0.09 177. I O 433.51 169. Analog Computer Siniulation of Engineering Systems.OO 63.26 165. Allyn and Bacon.70 3 17.790 0. Allyn and Bacon.43 73.370 0. 2nd ed.690 0.550 0.93 53.21 166. Wiley. James. R .080 0.76 168.22 177.48 141.400 0.610 0.87 170.1 1 72. -.17 74. 1971.050 0.80 193.68 55.46 106.46 20 I ..7 IO 0. I20 0.21 169.76 109.620 0.41 13.60 6 I .24 325.01 342. J . J . Blum.84 245.75 4 19.08 176.380 0.57 398.5I 152. Analog and Digital Computer Methods in Engineering Analysis.63 175.87 25.69 145.93 24. I70 0.72 15.20 137.700 0.38 175..770 0.28 176.57 168.82 45.02 162.390 References 40. Boston.42 93.200 0.69 383.260 0.05 308.21 351.82 254.Appendix B 543 Table B. 1964.660 0. Analog Computer Techniques.72 127. Jennass.06 171.180 0.68 334. J .42 82.460 0.82 161.020 0.1 I 78. Scranton.37 18.55 177.43 20. 1963.82 74.84 176. introduction to Analog Coitrputation.58 73.250 0. Analog Computation and Sitnulation. and Korn.r Method. 1965. F. H. G .s /i. New York.. 1963. 1971. James. Harcourt Brace Jovanovich. and Wolford. and El-Abiad. J. G . 1962. C'oniputer Methud. Scranton.. New York. G. Englewood Cliffs. Cornpuler Simulation for Engineers. Prentice-Hall.. 1964. Pipes.v in Power S ~ I ~ I P J Analysis. H. McGraw-Hill. New York. Analog and Digital Cottrputer Method. M. Stephenson. Smith. International Textbook Co. H. A .r the Phyhpical Sciences.. McGraw-Hill. Matri.. 1968. Introduction to Nuttierical Analysis. M.. Wiley.544 Appendix B Digital Coniputation Hildebrand. Pennington.s in Engineering Analysis. Korn. wilf.J. 1968. Macmillan... S. A. New York. L. New York. Marhematics Handbook for Scientists and Engineers. N. 1956. W. B. McGraw-Hill. New York. T. A. Matheniutic.vJur Engineering. C. S t a g . Pa. R. L. R. M. E. . lnrroducrory Computer Methods and Numerical Analysis. The normalization scheme used by U. 4. For the study of system dynamic performance it is important to choose a normalization scheme that provides a convenient silnulation of the equations. manufacturers does not satisfy requirement 2. All mutual inductances muSt be capable of representation as tee circuits afier normalization. as given by (4. the authors have adopted the following guidelines against which any normalization system should be measured. 545 .S. Other pu impedances must be related to and easily derived from the data supplied by the manufacturer. Thus both before and after normalization we may write p = ku'i (C. 3. the pu system is to be developed so that the same pu stator and rotor impedance values are obtained. The system voltage equations must be exactly the same whether the equations are in pu or M K S units. The system power equation must be exactly the same whether the equation is in pu or M K S units. C. This means that power is invariant in undergoing the normalization.. 2 . First we write t h e equations in M K S quantities. The major pu impedances traditionally provided by the manufacturers must be maintained in the adopted system for the convenience of the users.appendix C Normalization There are many ways that equations can be normalized. Having carefully considered a number of normalization schemes for synchronous machines and weighed the merits of each. which is different from the transformation used in this book.l Normalization of Mutually Coupled Coils Consider the ideal transformer shown in Figure C .3]. This requirement is included to simplify the simulation of the pu equations. and no one system is clearly superior to the others [ I .1) and k is the same both before and after normalization. I . amperes. However. as given by (4. I .22).2 and either comply wholly or provide a clear tran1 sition to a new system. volts. ohms.2. The manufacturers use the original Park's transformation. This means that the equations are symbolically always the same and no normalization constants are required in the pu equations. At the same time it is also important to consider the traditions that have been established over the years [ I .5). Le. and henrys. we compute vlBllB /tlB f2B E L2I I l B / V2BflB = V2B12B/t2B or SIB/tlB = S2B/t2B . Schematic diagram of an ideal transformer. C .i. R2i2 di2 + L. using the subscript u to clearly distinguish pu quantities.. Ljk = PrnNjNk and f o r j .e.. 02 = R.. u. in terms of the mutual permeance ern the coil turns N. Now choose base values for voltage. I . i. we require that L I2 I2B / and since L12= La. we have. and time in each circuit.546 Appendix C ?I "I [Te-. Ideal I Fig. we have the pu (normalized) voltage equations We can define Now examine the mutual inductance coefficients. To preserve reciprocity.+ L2I dt di2 di. k = 1. For circuit I: V l B I l B I l B For circuit 2: 12Bf2B Then since any quantity is the product of its per unit and base quantities.2. V where.s N. Dividing each equation by its base voltage. di + LIZdr v dr = + L22 . current. H. Appendix C 547 The ideal transformer is also characterized as having the following constraints on primary and secondary quantities: n = i2/il = uI/u2 (C. we have n = 12Bi2u/llBilu = &BUIu/hBU2u Thus the pu turns ratio nu must be nu = i z U / i l u= ~ ~ I B / =I U I . C. From (C. Xm2 = L Z l l l BWbturns (C.8) we compute = v2B/&B or VIB1IB = v2B12B SIB = S2B ’ SB (C.. Rewriting in terms of base and pu values.9) (C. / N 2 . 12) From the flux linkage equations we write (in M K S units) Injecting a base current in circuit 1 with circuit 2 open.IIIB = tIBand i2 = 0.7) where n = N .8) and base quantities are often chosen to make nu 11B/12B I . and the first requirement is satisfied.BU L . L I I= 41+ L m l L22 = 4 2 + Lm2 H (C.4) become Then the voltage equation is exactly the same in pu as in volts. 16) Equal pu mutual flux linkages require that Xmlu = Xm2u ( C . we divide the coil inductances into a leakage and a magnetizing inductance: Le.2 Equal Mutual Flux Linkages To adapt the voltage equations to a pu tee circuit.. the power is also invariant and the second requirement is also met. i. IO) Combining with (C.14) In pu these flux linkages are Xmlu Xm2u = hml/hlB = LmlIIB/LIBtlB = Lml/LIB = hmZ/X28 = L2lIlB/L2BI2B (C.17) .it is apparent that we must have t l B = 128 = IB A and the mutual inductance terms of the voltage equation (C. ~ / = = ~ ~ B / V I B (C.e. with i l gives the following mutual flux linkages XmI = L.6). Furthermore. if this identical relationship exists between currents and voltages. 15) (C. 22). uIU = R l u i l u + teilu + Lmu(ilu + t 2 k izU) uzU = RZui2. L Fig.25) (C.N:I:. @.22). (C.21) Comparing (C.548 or Appendix C LmI/LlB = Lmlu = L21tlB/L2B12B (C. SB(Lml / L I B ) = SB(LmZ/L2B) (LmZ /L2B )(I:BLZB) Lm2I:B @. with i2 = IzB i l = 0) gives the following pu flux linkages: and knlu = L12lZB/LIBIIB Xm2u = LmZ/LZB (C.23) in the voltage equation (C. Now using (C. (C....20).4).19) Again equal pu flux linkages give Lm2/L2B = Lm2u = Ll2IZB/LlBllB (C. 18) Following a similar procedure.20) and (C.2.26) (C.12).21) = IAL2B (C. 2 .27) (C.28) or in terms of the mutual permeance S . C .24) which is represented schematically by the tee circuit shown in Figure C. = or N i I f B = N$IZB or in terms of M M F = F2B . Tee circuit representation of a transformer An interesting point to be made here is that the requirement for equal pu mutual flux linkages is the same as equal base MMF's. = SIB IfBLiB and from (C. + + L m u ( i l u + bu) (C.18) and (C. Thus the third requirement is satisfied. we can show that injecting a base current in circuit 2 with circuit 1 open (Le.20) FromS. and (C.N$f$B or (Lml /LIB ) ( I : B LIB ) LmI I : B = = (C. and hence this choice of stator base quantities does n o t meet requirement number 1. Because of a certain awkwardness in the original Park’s transformation resulting from the fact that the transformation is not power invariant. . a system of stator base quantities is used by U. we get u.146) ud = . 2. IO).30) are not identical. Since the transformation used in this book is power invariant. 3. Such a choice.17). All circuits must have the same time base (C. \ / ~ v c o s ( ~ a) = v (C. the other three requirements stated in the previous sections may not be satisfied. it would appear that adoption of stator base quantities of rated rms line voltage and . Choosing VI. The factor of 4 appearing in the d and q axis equations of Chapter 4 would be eliminated... however.3 Comparison with Manufacturers’ Impedances We now select the base stator and rotor quantities to satisfy the fourth requirement.)cos(d .28). For example. if the S phase voltage u.23). would reveal that the requirement of having the same identical equation hold for the M K S and the pu systems would be violated. The choice of the stator base voltage VI.2.. For this reason it is customary to use a stator base voltage equal to the peak line-to-neutral voltage and a stator base current equal to the peak line current. This requires equal pu mutual flux linkages (C. = ~ V C O( q t + a).9). C.\/5 times line current would be attractive. and the stator base current f l B determines the base stator impedance. A variety of possible stator base quantities can be chosen to satisfy the condition of having the same pu stator impedances as supplied by the manufacturers.4 V s i n ( d .a) pu (C.29) where V = rms voltage to neutral.29) and (C. For example. Note that in all these choices the base stator impedance is the same. leads to a rotor VA base equal to the threephase stator VA base.9). to give the same pu impedances as those supplied by the manufacturers.S. rms line-to-neutral voltage and rms line current.Appendix C C.30) Note that (C. manufacturers that facilitates the choice of rotor base quantities.the d a n d q axis voltages are obtained by a relation similar to that of (4. which in turn requires that the base MMF be the same in all circuits (C.. or rms line voltage and fl times rms line current. namely. the awkwardness referred to above is not encountered.1 Summary 549 The first three normalization specifications require that 1. All circuits must have the same VA base (C. &VLN (rated). = .6). The requirement of a common pu tee circuit means equal pu magnetizing inductance in all circuits (C. However. To illustrate. among the possible choices for the stator base: peak line-to-neutral voltage and peak line current (same as the manufacturers). along with the requirement of equal base ampere turns (or equal pu mutuals). = (V/V. (C. Careful examination. and (C.a) u. 32) and (C.38) Thus the value of the pu d axis mutual inductance of any rotor circuit is the same as the pu magnetizing inductance of the stator. Note that the base inductances and resistances are the same in both systems. kMF.).. From (C. B fl (C....S (C. - VIB (k) = ~FLIB (C. A IIB = I/%. manufacturers is given in the Table C. V A = VIE = rated rms voltage to neutral.33) or where kF = k MF/Lm.36) The base for the mutual inductance is obtained from (C.32) (The subscript 2 is used to indicate any rotor circuit. Thus in M K S units. llBL..11) and (C. t B .34) From (C. and FB (or A.l = f2~kMF k = a (C.34) for the rotor resistance base R ~ = B v2~/12~ = k$(v1B/llB) E ~ : R .35) The inductance base for the rotor circuit is then given by L2B = VZBtB/12B = (kMF/Lmd2(V1B/IlB) () i = kzFLlB (C.37) (Lml/kMF)&3 The pu d axis mutual inductance is then given by (C. = kMD. The same derivation applies to a field circuit or to an amortisseur circuit.550 Appendix C In this book the stator base quantities selected to meet the requirements stated above are S I B rated per phase voltampere. = MRu = Lmlv (C.) Equal mutual flux linkages require that the mutual flux linkage in the d axis stator produced by a base stator current would be the same as the d axis stator flux linkage produced by a d axis rotor base current.33)and (C. Equal V A base gives vIBIIB = v2B12B VA (C.33) we obtain for the rotor circuit base voltage v2B = vlBIIB/12B = kFvlB (C.33) = Ll2B -12B V I B ~ B -. V f l B = rated rms line current. I .39) A comparison between the pu system derived in this book and that used by US.31) The rotor base quantities are selected to meet the conditions of equal SB.. . manufacturers* Quantity/system C.185 0.100 (C. The machine used for this data is the I60-MVA. = 0.150 = X . Ratings: 160MVA Unsaturated reactances in pu: xd = 136MW 0. There . = 0.5. two-pole machine that is used in many of the text examples. The following data is provided by the manufacturer (this is actual data on an actual machine with data from the manufacturers bid or “guaranteed” data).24 = 0.075 (C.5.40) xq = X . Starting with the pu impedances supplied by the manufacturer.85PF 15kV (C .023 T. = 1. = 0.185 0.001 113 rF = 0.8 of the text.185 X..64 0. 55 1 Comparison of Base Quantities Per unit system used In this book By US. we provide a consistent set of data for a typical synchronous generator. = 0. the base quantities are derived and all the impedance values are calculated.Appendix C Table C.70 1.9 7. The data given and results computed are the same as in Example 5.4 Complete Data for Typical Machine To complement the discussion on normalization given in this appendix. One problem not mentioned there is that of finding the correct value of field resistance to use in the generator simulation.245 0.380 x ~ . The method used is that of Section 5.. = X : = = x2 XO = 0.43) Resistances in ohms at 25°C: r.2687 (C. 0.I.42) Excitation at rated load: UF = 345 V iF = 926 A (C. Computations here are carried to about eight significant figures using a pocket “slide rule” calculator.4 I ) Time constants in seconds: = 5.44) Computations are given in Example 5. 59). r F = 345/926 = 0. Thus.8C.46) 3.5. This is obtained from the air gap line of the magnetization curve provided by the manufacturer.781800664 mH = 18. Compute from (5.406 250 22. Compute from ((2. which is a reasonable result.2.652 582 384 326.37257 s2 (C.333 333 333 163 280.9 = 0.660 254 036 2.1 I5 4415 1.972 0373 3. no such data is given for any of the amortisseur circuits. Compute from (C.43). and frequency. we have 0.730 193 98 53.44) at an assumed operating temperature of 125°C: rf = 0.5 + 25)) = 0.2687[(234.. The given values are easily identified since they are written to three decimals.885 8653 433.677 2. voltage.0)/(234.333 333 333 8.O = 2. The base quantities for all circuits are given in Table C. Working backward to compute the corresponding operating temperature.2 and the pu values from Example 5..372245 Q (C. using LF from Table C. IFB RFB AFB LFB MFB IBl k ~ VFBIIFB VFB~B XFBIIFB G 53.325 988 441 0. Stator base values are derived from nameplate data for voltamperes.48) or the operating temperature is 0 = 123.402 872 1..49) Note that a key element in determining the factor k F . at operating temperature.0102349 mH/5. = 109. Table C. we may construct Table C.4 of q axis parameters. = kM.371097 s2 The value computed from L. and hence all the rotor base quantities./L. Unfortunately.3 of d axis parameters and Table C.5 + 25.652 582 384 6158. while the pu values of the various amortisseur elements can be determined. is the value of M F (in H). Using the base values from Table C.85402857 (C. Circuit Stator Base quantity Base Values in MKS Units Numerical value Formula Units SB V B lB SB. XBI~B SB VBI~B VB~B SBI~FB 1 ..070 329 184 MV A / ph ase kVLN ms A R Wb mH MVA/phase V ms A R Wb H H .1 where we compute k. their corresponding M K S data are not known. The method of relating stator to field base quantities through the constant kF is shown in Example 4.371097 (C.0)] = 0.5 + 8)/(234.5 + 125. must be used if the correct time constant is to result./T.552 are three possibilities: Appendix C I .3 r F LF/T./~ VLLI d3 112~60 sB/ B ‘ I B RB AB Field LB SFB VFB IF.189475/5.2.45) 2.635 915 499.2687[(234. 550 IS O 0.Appendix C Table 553 C. Quadrature Axis Parameters in pu and M K S pu value M K S value Units Symbol L.096 0.099 90. Ti0 7 .550 0.216 I .371 097 586 0. Direct Axis Parameters in pu and M KS pu value M K S value Units Symbol Ld L.096 463 455 x .559 529 097 I .490 0.055 0.185 1.477 2224.028 378 3784 0.90 0.030 459 0.557 989 025 0.791 1.341 329 761 mH I . 1 13 1..075 8.134 800 664 529 097 475 759 282 084 193 675 mH mH H H H LD -e.265 5697 1.559 2.550 1.791 607 397 x IO-’ 1.442 11.605 4 I6 667 1. 25°C r.541 901 734 0.380 (not used) 0.651 202 749 I .035 I .274 3.eF 5.541 901 734 0.550 0.055 416 667 1.089 006 484 0.575 28.F .550 0.189 808 581 579 905 357 607 463 955 204 333 482 4715 397 x 455 x 165 IO-’ IO-’ 1.490 0. Lmd x d 1. eL:i.670 364 135 868 599 450 945 295 x 90 x IO-’ 44 1 .460 365 85 mQ mi2 S S ‘9 T7? T7? Tq 89 785 ms . 25°C r.245 0.525 808 581 5. 125PC 2. I50 1.85 mR mn n S S S S 52 1 69 195 726 0.265 5691 LF L.742 13. L.3.Q LQ MQ kMQ LMQ r. I50 1. MF kMF MD kMD 0.053 203.4.2687 (not used) 0.117 518 122 mH mH mH 0..490 6.640 0.482 8.700 0. Tk 7.101 202 749 I .24 5. 125°C rF 25°C rF H O I TD 70 1.54 0. 1.113 1.189 2.028 0.109 010 235 H H MR LMD r.185 6.023 S Table C.247 320. 0.781 0. 1EE Monogr. I. A I E E 7runs.Y . Power system stability. 4. Generdl Electric Co. J.554 References Appendix C I . AIEE Trans. R. . Lewis. A . Ser. 1973. A. Electric Utility Engineering Seminar. W. Press. Lawrenson. Cambridge Univ. Per unit impedances of synchronous machines. Harris. Schenectady. 77:436 56. M. P. 2.. 1970. Rankin. N. A basic analysis of synchronous machines. 1958.. W. J . 3. and Stephenson.. 64569 841. 4. Section on Synchronous Machines. M. 1945. f e r Unit Systrmr: With Specin/ R e f m w c c to Electrical Machines. Pt. and these are often the only known data for a machine. Table references on these items are given in parentheses following the identifying symbol. (See Tables D. data are provided for typical transmission lines of various voltages. the exciter. This is helpful in estimating simulation constants that can be used to represent other typical medium to large units. Thus it is often necessary for the engineer to estimate or calculate the missing information. it is often helpful to have access to typical system constants. Such constants help the student or teacher become acquainted with typical system parameters. A rather complete set of data is given for various sizes of machines driven by both steam and hydraulic turbines. For example. ( 1) Short circuit ratio The SCR is the “short circuit ratio” of a synchronous machine and is defined as the ratio of the field current required for rated open circuit voltage to the field current required for rated short circuit current [I].I. 555 . Referring to Figure D. The items included in the tabulations are specified in Table D. data taken from manufacturers’ bids are limited in scope.8 the end of this appendix. Finally.2) z I/xd pu where x. An explanation of these referenced items follows.appendix D Typical System Data In studying system control and stability. and the power system stabilizer. is the saturated d axis synchronous reactance.) at D.l Data for Generator Units Included here are all data normally required for dynamic simulation of the synchronous generator.1) (D. and they permit the practicing engineer to estimate values for future instal la t ions. 1 ---D. we compute SCR It can be shown that SCR = l B / l s PU (D. the turbine-governor system. Certain items in Table D.l.1 require explanation. I n most cases such an accumulation of information is not available without special inquiry. The data given here were chosen simply because they were available to the authors and are probably typical. Data are also provided that might be considered typical for certain prime mover systems. I we arbitrarily define 51 = (fF2 .3) is valid for any point yl [ 2 . We compute it is common to specify two values of saturation at V. From Figure D. full load.l F l ) / f F l where (D.2 pu.the pu saturated magnetizing inductance. which is defined in terms of the open circuit terminal voltage versus field current characteristic shown in Figure D.556 Appendix D ‘A IB ‘s I F ‘C Field Current. These values are given under open circuit conditions so that V. 31. Thus we can easily determine two saturation values from the generator saturation curve to use as the basis for defining a saturation function.3) and will use these two values to generate a saturation function. Fig.. .I . sG at (D. D. is actually the voltage behind the leakage reactance and is the voltage across LA. Open circuit.O and 1. With use of this definition.2. and short circuit characteristics of a synchronous generator (2) Generator saturation Saturation of the generator is often specified in terms of a pu saturation function SG. = I . I Suppose that measurements on a given generator saturation curve provide the following data: S.8) SGI. .o/AG) 0. = ~ ~ ~ ~ ~ . IF Fig. (D.26. explicitly.2SGl.20)f/1. Construction used for computing saturation.04167 B. l.0 = c Rearranging and taking logarithms.10). 2 ~ c 1 .2(0. (SGl. In(Scl.8/0.6) (D.B.o/AG)2 = I. = ~ ~ 1 .Appendix D 557 Field Current.e0.2SG1. we can solve for A.8 (D.10) 2 O ~ In( I . 2 / ~ G l .2. is usually less than 0. 1 where we define s.4 BG = (D.2 BG = Then.6) contains two unknowns and the quantities S.843 This gives an idea of the order of magnitude of these constants.9) Example D. I and B. ~ ~ G I .2/AG) 0. one of which is given in Section 5 .z = 0. From the given data we write A e0. IO. using (D.20) = 7.80 Then we compute. where = A.20 S. . 8 ~ ~ .o = 0.2 x 0.7) Va = is the difTerence between the open circuit terminal voltage and the assumed saturation threshold of 0 .0.*/& or A.enGvA r/.2 A. = (0.l. 0 / ~ . = 51n(1. D. Since (D. and VAare known at two points.80) = 0.2SGl. is usually between 5 and IO. and B.l. A. There are several ways to define a saturation function.4nG = (D. A. 4.15) describes how V.6.I = V. 2 Other methods of treating saturation are found in the literature (I. Note that S. 12) vi = RIF2 . is the drop in voltage due to saturation. l . The method of introducing the damping is by means of a speed or slip feedback term similar to that shown in F i g ure 3.3. Referring again to Figure D . 4.4. The approximate saturation function. 2. > 0 for any V.SGV. < 0.15) where S. we write the voltage on the air gap line as V.8 is negligible.12) V.3. 5. 4 .2 . determined above may be used to compute the open circuit voltage (or flux linkage) in terms of the saturated value of field current (or MMF).8. current but only produces VFlror K2 = RIF2 (D. . Equation (0. (D.14) Then from ( D . But from Figure D. is clearly a function of yI. so the saturation computed for V.. The computed saturation function has the shape shown in Figure D./Kl (D.2. and 4. is reduced by saturation below its air gap value RI. = R I F Z R = I'. we assume a similar reduction occurs under load./(IFz - - V.15 and is treated in the literature [8-121. S. = 0. 2. ( 3 ) Damping It is common practice in stability studies to provide a means of adding damping that is proportional to speed or slip. Note that A .9.V. where we assume that saturation begins. at no load.13) (IF2 - IFI)/IFI = K/RIFI = I'. Note that the exponential saturation function does not satisfy the definition (D. Usually. D. When saturation is present. = RI..10. and the approximation solution is considered adequate in the neighborhood of 1 . The error is small. $G =AGeGA B V *G1 I I * 't Fig.3) in the neighborhood of V. where D is the pu damping coefficient used to compute a damping torque Td .. IF2 (D.558 Appendix D The value of S.3) we write SG E = IFI) (D..3.7]. This concept is discussed in Sections 2. where V. is usually a very small number.2 tan0 From (D. however.11) does not give Refer to Figure D..O pu voltage. The models proposed by the IEEE committee in 1968 [3] have been largely superseded by newer systems and alternate models for certain older systems. Then (D. up to 25 pu is sometimes used as a representation of amortisseur damping if this important source of damping is omitted from the machine model.16) where all quantities are in pu. fR is the base frequency in Hz. i. WMA.18) J A where S.pU. is the three-phase MVA base. Silverstat. For example. T I "bin i S ' Other signals SE + KE - -I+rF' i Fig. 18) is D' = PG/fR is MW/Hz (D. KA 1 I -1 . a damping of 1-3 pu is often used to represent damping due to turbine windage and load effects (21. and the slip in Hz.D)WA~ MW (D. Type A-continuously acting dc rotating excitation system. The approach used here is the alphabetic labeling adopted by the Western Systems Coordinating Council (WSCC)..19) wnere Pc is the scheduled power generated in M W for this unit. A much higher value.4-D.17) becomes Td = (sB3D/fR)fA = 'YA MW (D.4. both in design and in the models for representing the various designs. Then with the slip wA in pu Td = (SB. provided through private communication. Representative systems: (1) TR = 0: General Electric NA143. NA 108: Westinghouse Mag-A-Stat. Westinghouse Rototrol./sB.e. This corresponds to D = P. Excitation systems have undergone significant changes in the past decade.Appendix D defined as Td = 559 DwAVP U (D. The need for expanded modeling and common format for exchange of modeling data is under study by an IEEE working group at the time of publication of this book. In some simulations the torque is computed in megawatts. . A value sometimes used for D' in (D. 1 1 . D. Allis Chalmers Regulux: (2) TR# 0 General Electric N A 101. fa Hz. The value used for D depends greatly on the kind of generator model used and particularly on the modeling of the amortisseur windings.16). (4) Voltage regulator type The type of voltage regulator system is tabulated using an alphabetical symbol that corresponds to the block diagrams shown in Figures D. The value of D also depends on the units of (D.. TRA.17) I t is also common to see the slip computed in hertz. V B = O 'FD I Fig.Westinghouse pre-1967 brushless. D.7.HEV)Z KE t- If: A > 1. D. Type D-SCPT system. KA 1 1 - -f EFDmin = s ' Other signals KFs . Type C-.' 1 1 __ 1 +T I R c 'Rmin 1 T I -43 'FDmin s' Oher signals / Stabilizer - ~ SK F -1 + T S F .5.6.I t T S Stabilizer F Regulator Stabilizer "Rmin E~~ SE t K E - S ~ t K ~a Fig. D.\ -'E ' KE -1 I+T FI S Fig. .78XL1FdV.Westinghouse brushless since 1966 'REF \ 1 Exciter \ 1 t j S ' E~~ Other signals 2 trp 1 - 'Rmin __1 A = (0. Type B--. V R = V Rmin - ‘Fhax EFDmin Fig.K V . D.Appendix D 56 1 “REF If: hax 1 ‘E’ ‘Rmin “to b V t + .8. Representative systems: General Electric GFA4. D.I t T S F . Type E--noncontinuously acting rheostatic excitation system.9. D. “REF l+r I AI - “bax 1 +IA2‘ 1 ‘Rmin “I signals Fig. Integrating regulator / / Exciter - A K ( I + TAS) “Rmin Other rigmlr .1 KFs .10. Type F--Westinghouse continuously acting brushless rotating alternator excitation system. Westinghouse BJ30.SE+KE Fig. + ’E KE 47p EFDmin = 0 EFCmax . Type G--General Electric SCR excitation system. choosing a different base affects the constant S .B ) / B SE75max = ( E . 0.D)/D (D. + KE)EFD.75 of ceiling and full load. D.20) Exciter Field Current Fig. = (C ..562 Appendix D 1 t T AI S - 1 A2 ‘Rmin -fa EFDmin E~~ “5 Other si gna Is Stabilizer SE t K E F’ 1 t T S F - Note that the regulator base voltage used to normalize V. . and also the gain K. A dc exciter saturation curve. and K. SEmrx= ( A .F)/F S.12. we define the following constants at ceiling. may be chosen arbitrarily.. of ceiling.(S. Since the exciter input signal is usually VR . (5) Exciter s a t u r a t i o n The saturation of dc generator exciters is represented by an exponential model derived to fit the actual saturation curve at the exciter ceiling (max) voltage (zero field rheostat setting) and at 757. Referring to Figure D.12. 75EFDmaa.( D.F)/DEFDmaa (D. EFDmaa = EFDmar(V)/EFDFL(V)= B / D pu (D.7SmaX respectively.22) we can write SEmar (D. At E D = E D a F F mx SE = S E = ~ A .26) simultaneously to find (6) Governor representation Three types of governor representation are specified in this appendix: a general governor model that can be used for both steam and hydro turbines. and a hydraulic governor model. The appropriate model is identified by the letters G. any convenient base may be used). for U.25) and (D. units. The governor block diagrams are given in Figures D..24) which gives the approximate saturation for any EFD.. These values are called SEmpa and cal values of saturation at EFDmaa SE. Suppose we are given the numeriand 0.Appendix D 563 Then in pu with EFDFL as a base (actually.13-D. Combining (D..21) or = DEFDmaa We can also compute B/F = 413 = DEFDmaa/F or F' = O. .15.~~DEFD.B)/DEFDmaa AExe ( ~ ~ = B E E X FDmax (D.F)/o*75DEFDmaa = (4/3)(E .75mai (E - F)/F = (E . Using these two saturation values.22) = = ( A .20)--.23) Now define the saturation function SE 5 AExe BEXEFD (D.25) We then solve (D. The regulation R is the steady-state regulation or droop and is usually factory set at 5". C .S.B)/B = ( A - B)/DEFDmaa SE. we compute the two unknowns AEX and BEX as follows. and H in the tabulation. a cross-compound governor model. Cross-compound governor block diagram. D. . Fig.564 Appendix D Fig. P DdTd' 7 ~ I Fig.13.14. General purpose governor block diagram. D. D. Hydroturbine governor block diagram.15. 88: Apr. 12. K. 8. Fig. B. Eng. Still there is some merit in having approximate data that can be considered typical of stabilizer settings. Crary. Elecfric Machin~rI~. R. A . T. Power Srsfenr Sfabdit. D. 70731 -37. Eng. Power Con/: 29: 1126. B. the dynamic characteristics of the system. Table D. ed. Rept.. Computer representation of excitation systems. = accelerating power = Pa. 1967. Saturated synchronous reactance. 1968.. quency deviation = fA.Appendix D KQV I + T 565 I vs "s iim 1 t T KQS = -V lim = wA. Electr. D.8 for estimating the impedance of transmission lines. A m . References I . Wiley. Kingsley.: 124. Concordia. Concordia.r. Usually. Prubhashankar. D. L. and McCauley.. 2.. New 3rd York. Jr.: 300 305. A / € € Trans.-50... Westinghouse Electric Corp. Kilgore. Stabilizer types: ( I ) V. M.Z--D. 4. . K. C. Kingsley.32: discussions. Econoniic Control ojlnrerconnected Systents. 1935. Fitzgerald.2 Data for Transmission lines Data are provided in Table D. 9. 1950. Mar. and March. Elecfr. rotor slip (2) Vx = fre- (7) Power system stabilizer The constants used for power system stabilizer (PSS) settings will always depend on the location of a unit electrically in the system. L. 2.16. lEEE Trans.: 484. Values given in Tables D.) 3. Elrect of steam-turbine reheat on speed-governor performances. Stability program data preparation manual. 1971. 5 . Power.. L. Equivalent reactance of synchronous machines. PAS-87:73-40. Eng. McGraw Hill. E. IEEE Working Group. based on actual utility line design information. A. and Kusko. P. Byerly. R. May: 545. A. The PSS block diagram is given in Figure D. 6. EnR. C. S. T. 7. 1935.Vol. L. Dec. Jan.8 provides data for making rough estimates of transmission line impedances for a variety of common 60-Hz ac transmission voltages. Kirchmayer. S . C. Young. Mar. A new stability program for predicting dynamic performance of electric power systems. Sherman. 1972. PAS-87: 1460 64. and the dynamic characteristics of the unit. Elecrr. / E € € Trans. 16. ASME J . 1951. 70 736. C. Proc.39. Jr. 1968. 1934. IO. 1959. Digital simulation of multi-machine power systems for stability studies. accurate data are available for transmission circuits. New York.. Synchronous machine damping and synchronizing torques. (3) V . Wiley. C. Erects of saturation on machine reactances. (Rev. Crary.. New Y o r k . Apr. I I . and Janischewdkyj. W. C.. Shildneck. Power system stabilizer block diagram. M. and Webler. 1970. A. E.: 603 7. 1947.5 are actual settings used at certain locations and may be used as a rough estimate for stabilizer adjustment studies.: 201 -6. 566 Table D. C = cross-compound.1. or K .4 s pu TA o r 7.or hydro gute time constant (type G) or dashpot time constant (type H ) Steam valve bowl timeconstant (iero for type G hydrogovernor) o r ( r w / 2 for type H 1 Steam reheat timeconstant or I /I hydro water starting time constant (type C o r G)o r minimum gate velocity in M W / s ( t y p e H ) p u shaft output ahead of rehealer o r -2. Y b' "d Xb' xb xq 'a x. r o maximum gate velocity i n M W / s (type H 1 STABILIZER PSS (7) (7) PSS feedback: f = frequency.4 I s 'A 2 s Excitation system type Excitation system name Exciter response ratio (formerly ASA response) Regulator input filter time constant Regulator gain (continuous acting regulator) or fast raise-lower contact setting (rheostatic regulator) Regulator time constant ( # I ) Regulator timeconstant (62) KQV k' QS (7) s s 'Q rQl PSS speed gain.Opu voltage i n pu (2) Machine saturation at I .2 EFDFL D EXCITER VR Type Name RR Arbitrary reference number Machine-rated M V A : base M V A for impedances Machine-rated tcrminal voltage in kV: base k V for impedances Machine-rated power factor Machine short circuit ratio Unsaturated daxis subtransient reactance Unsaturated J axis transient reactance Unsaturated d axis synchronous reactance Unsaturated 4 axis subtransient reactance Unsaturated 4 axis transient reactance Unsaturated 4 axis synchronous reactance Armature resistance Leakage o r Potier reactance Negative-sequence resistance Negative-sequence redclance Zero-sequence reactance d axis subtransient short circuit time constant d axis transient short circuit time constant d axis subtransient open circuit time constant d axis transient open circuit time constant 4 axis suhtrunsient short circuit time constant 4 axis transient short circuit time constant 4 axis subtransient open circuit time constant 9 axis transient open circuit time constant Armature time constant M W * s Kinetic energy ofturbine + generator atratedspeedinMJorMW. pu PSS reset time constant First lead time constant First lag timeconstant Second lead time constant Second lag time constant T h i r d lead time constant T h i r d lag time constant PSS output l i m i t setting.O sG I. starting at full load tield voltage Exciter self-excitation at full load lield voltage Exciter time constant Rotating exciter saturation at 0. pu 'Ql . P = accelerating power PSS voltage gain. - ' R h'& mi" 'E SE.e2 'Q2 rQ3 s s s 'PI 'Slim s s pu . Rated M V A Rated kV Rated PF SC R .t.orxP '2 x2 XO 'b' 'b 1% 'bo 1. GENERATOR Appendix D Definitions of Tabulated Generator Unit Data EXCITER (conrbrued) ' R max Unit no. starting at full load tield voltage Minimum regulator output.75max SEmar A EX BEX EFDmax EFDmin KF ' F o r 'FI 'F2 Maximum regulator output.0 for hydro units(types C o r 6 ) .75 ceiling voltage.s I1 Machine lield resistance in II (2) Machinesaturation at 1.2 pu voltage i n pu (2) Machine full load excitation i n pu (3) Machine load damning coetticient . or lip for SC'PT exciter Derived saturation constiint for rotiiting exciters Derived saturation constant for rotating exciters Maximum field voltage o r ceiling voltage. H = hydraulic Turbine steadystate regulation setting o r droop Maximum turbine output in M W Control timeconstant (governor delay) o r governor response timc(type H ) Hydro reset time constant (type G ) or pilot valvetime(typeH) Servo time constant (type G o r C ) . for SCPT exciter Rotating exciter saturation at ceiling voltage. pu Minimum field voltagt: Regulator stabilizing circuit gain Regulator stabiliring circuit time constmt ( # I ) Regulator stahiliiing circuit time constant(d2) T U R B I N E-GOVERNOR tiov R Pmax (6) (6) MW s 'I '2 s s r3 '4 s '5 z F (6) Governor type: G = general. ' S = speed. pu (4) (4) R ' K. 'b '1 'bo 'a *R 'F sCI. 5 0..0I8 2.79 13.385 0.000 REGULUX 0.835 .930 0.05 I 4.137 OS60 0.000 I .ow ..003 0.000 IXO.... 107. .940 I .1507 2. H9 86.9185 A A A A 3.306 0..080 2.135 .2 (2) (2) 1.400 ‘do T4 Tb 1’’ ...320 -5.20 0.0032 0.035 1.224 .680 0.3127 0..130 0. .408 0...623 1.235 0.120 0.040 0.200 25.150 2.380 1.990 0..5 0.090 0... Rated M V A Rated kV Rated PF SC R Xb‘ HI 9 .297 0. H2 17.507 I .940 I .000 O.610 0..00 0..OO0 0.269 0.258 0.00 13.300 .33 I 0..199 0.000 I.1861 2.660 0. .060 0.340 0.3 I2 0..400 0.950 0..540 0.0022 .000 NA108 WMA I . I30 0.099 Name RR TR KA ‘ A Or ‘ A I ‘A 2 E AJ23 0.004 I 0.. . .000 0. .000 ‘FZ 0.90 0.329 0. 5 ~ 0.500 . .000 0.050 20..174 0.3.80 0.000 0. I70 0..00 0.064 I.18 0. 0.000 ..00 13.550 .000 1. X.320 0..90 0.0042 0.9 70 0.5 0.018 2...200 0..607 I.5 0.410 -1. 0.00 13.. .330 ....000 0. 0.190 .310 ...570 0.240 3..670 0.240 0.0062 0.019 0.00 xb Xd X9 6. 1...000 0.615 0. 0.480 ...200 0.2 1 0 “ R max “ R min KE ‘E .90 1..000 0..000 0.000 .. 3.100 1. ..000 2.030 0..7059 3.5 0.948X 1.0015 0.410 -0.096 oano 5.057 1.450 1.ooo I. .1.0027 1..00 13.340 0.017 rqO ? MW..210 0.80 0.950 0.000 0.760 0.130 2. ..615 0. H5 40. I 50 0.000 .O .100 0.495 0..I.80 0...50 . 1.. .80 H4 35..000 I ..140 0..155 0.030 1.660 0.5 0.050 20.035 1.90 1.035 2.328 0.2.. .7375 2... . 0. .060 0.200 0..9185 3.100 Unit no.210 0.055 1. . GENERATOR 567 Typical Data for Hydro ( H ) Units H3 25.. 1. 524.103 I .000 E GFA4 0..00 13.220 0.620 0. .000 0.167 0.646 0.850 ‘? .00 WR ‘F SGI.95 2.000 0. ’a x x or ..0027 sE.120 0..035 2.5 0.000 0.000 65.000 ..050 1. 183..000 NA108 0.000 0. 0.000 0..125 H6 54. I30 .. 0.000 I 2.000 0.064 1.000 0..000 E WMA 0.80 0. ..070 I...000 4.000 242. 0.033 .600 ...570 -2..375 0.550 -2..327 0.050 1.000 0..0017 1.OOO 0.000 3.220 0.90 1.018 2..074 rb’ ‘f‘0 . . 0.000 -3.lXO . 176. O..00 13. 7.150 ..000 0.IX00 23.460 2. 5..000 0.320 2.607 .301 0. 0.332 0.33 0.06J SG..550 0.5738 5..306 0.95 2..000 37. . ..75 max SEmax A EX BEX EFD max &FDmin KF ?For l F I 0.020 0.000 0.95 1.000 0.288 0.S il (2) .80 0.320 1.900 HX 75. 1.5..480 0..650 0. 5.150 I .260 0..286 233.260 0..3480 0.x5 0.0645 1. 0. .000 0.90 1.000 0. 5.200 .500 0..490 0.80 0.924 0. 0.260 .36 0...050 0..000 1.320 -0.0885 0.700 .500 ...0 I 4 0.687 0.270 0.1219 2.020 ..ooo .000 0.00 ... I20 0..sp ‘2 .300 ..000 0.91 I .130 2..180 0.193 ..50 7.88 0. 4.320 2 ..200 0..000 5.. I I I -0.Appendix D Table D.000 0. I.000 0.2. ..120 0.318 0.140 0. .680 ti 7 65.. ...000 0.035 1..044 2.175 0..760 0.035 0.340 0..446 2.176 0... . I 70 0.000 -0.90 I ..360 0.650 0. 254.. 117.000 0.550 0. I827 0.3 IO xb xq 0. 1. .00 0.. 0.390 0.......700 0. 0.245 0.000 4..000 EFDFL D EXCITER VR type (3) . 168.000 .250 0.440 1.030 0.904 2.000 A NA143 0..160 0.670 0.000 0..300 0.0029 0. 7.. .000 .18 0. ..0059 0.012 1.00 0.000 5.00 .050 20.000 0...7412 3.770 2.000 0.540 0..240 -3.620 0...000 ~E RHEO 0. I30 0.0357 1.0121 I . I30 2.050 20..670 x2 XO .. 0..2 IO I ..280 0.264 0.00l6 I ...004 0.195 0.685 2.000 -0.2100 ..3566 3..580 0 . 050 52. .050 40.. ..000 I..920 0. ...2. . .100 0.....568 Table D..300 -2.000 0.300 .2.. (conr. ...60 48. 0...000 0.. .600 2.000 2.800 0. . . 0... . .ow 0...000 0.000 G 0....000 0.430 -2.. 7QI ..400 0..100 PU .000 0.500 0 . .. .000 G 0...000 0...050 ..785 -2.050 86. .... ... .2... . .000 0.000 0 500 0.XO 16.400 0.00 0..000 0....030 0.000 0..500 0.300 ....050 14.50 25.000 0. ...020 0. ‘Q I ‘02 TQZ .050 0. QS lQ (7) (7) S S S S . 0.......700 0.000 0.000 30.......050 23. ..000 0..000 0.920 0.545 ..Ooo I ...850 . . F F F . ..000 G 0.. . .....50 0. .000 0. .579 -2..050 90.. ... .350 ...500 0...000 G 0.OO0 G 0.) . .000 G 0. .000 65..000 3.000 0.000 0. .440 4.00 16.2.2... ... ... .. ..150 IO. .. .... ... ..758 0. . .000 2.700 0..000 G 0..500 0.....400 0.00 20. .000 30..500 0..000 0.050 71 S S 73 74 75 S S f (6) (7) 8.000 0.634 0. .. ..000 4..920 0.2.095 Tuble D.030 0..000 O..... ..000 0...000 G 0..000 3. 0. . .00 16.00 12. .000 0.000 0.056 40.2 (continued) TURBINE-GOVEKNOR Appendix D ~~ ~~ GOV R Pmax TI (6) (6) MW S G 0. ... ...020 0. .100 S S S ‘Q3 ‘Q3 YSlim . ...000 STABlLlLEK PSS KQV . ..... . ..000 2. .. .758 0... ...000 4. 000 0..020 0..680 .XY79 0. I50 0.3.20 0.2847 0..250 0.... .120 -0.1 13 0.030 .00 0.000 -0.600 2.0236 I.027 0.405 0...120 -4. 0.310 0.10 13..0096 1.175 0.000 50.1.770 0.182 0. 392.560 1.143 0.219 (2) 0..000 54...000 -0.478 1.870 .00 ..008 0.0245 0. 0.‘d xi ..270 ..510 0.000 1.80 0.1Sh 0..00 13.1274 0.000 0.870 5.184 -0.220 0..360 1.535 0.3. I20 1..975 .. 0. 0.070 0..080 0.310 .03 I 7 0..220 0.000 0. 0...000 0.950 1..15max SEmax AEX BEX EFDmax &FDmin KF TFor‘FI ‘F2 WMA NAI43A SC K WMA I .0195 0.OO0 400...000 0. ..o I. 0.040 ‘bo 1 .000 0.060 0.000 0.330 0.80 0..04 I 7.029 5..163 .9227 3.278 M W .573 HIS 158.040 0..7 I O -0. 0.140 0.00 13. 0.200 . 0. .000 0.000 0.080 0.646 0.0043 1. 0.320 0.000 100.90 .050 0.050 0.0049 0...00 WR ‘F sGI.375 0.260 0.300 0.039 6..870 -3.850 -2.725 2.o 0..000 O.000 0.vi .870 3.295 0.r 7.000 0.850 2..024 1.290 0.686 0.010 0.os 0.120 I..050 0..060 .00 ..000 WMA NA143 0.0237 0.960 .269 0.020 3.221 0.480 1. 1 . .280 0.985 0.000 0. 1.273 0.000 400..00 14. . 469.O sGl.. .000 ..000 1.034 .200 3.200 .000 7..315 I .21 I 0.. I70 0. ..000 0.... .850 2.200 0.00 12..730 .000 0.000 0.1461 3. I70 HI 4 145.0769 0.2 .732 0...000 0.300 1. .85 1.90 I. 0.90 1.000 0. I80 458....990 -5.0377 0.000 0.330 0.340 .040 8..000 0..280 .000 0.002 I 0.700 .568 0.Oh0 0.00 15..610 0.734 (2) (2) 2.002 0.646 .000 0. 0.20 0. ..00 I8.230 2...000 . 0.410 0.Appendix D Table D.80 0. 7.155 0.5. I 3 5 0. I30 0. ..000 0.147 0.178 0.220 2..030 .. I20 0.09 0... 2. 0.5 0.00 I1 0. 0.255 0.000 0.00 0..035 1... H 18 615. I642 0.730 -4.960 3.200 786.0 0.150 HI3 131.. 0. .000 .00 0. 0..~396 ..000 0.245 0..rp ‘2 .004 0.000 2m . .160 .330 EFDFL D EXCITER VK type Name RR 7R . ..484 0.062 I.080 .438 1. .000 I .990 -0.08 I 2 0.030 7.00 O..i ‘ 2 HI0 100.. .030 2..200 0.243 -0.90 1.000 .180 0. lq0 la .161 HI2 125.. HI1 115.5 0.000 .592 2. .000 7. x .000 0.040 0.100 0.953 0..000 3.0014 0.270 0.014 0...00 13...131 1. 1603.940 . . ..000 .000 5.120 .5 I.330 439.01 I 0.000 1.290 0.50 0. I20 0.200 0.995 0.282 I . (4) A A A G A A KA ‘ A Or ‘ A I ‘A 2 “Rmax “R KE TE min SE.07 I . .000 A SIEMEN I ...000 .3..610 0..000 ..020 0.00 200.000 0.5 I . 0. l ? .0276 1..314 1.308 0. S 312..000 A J ASEA I .12 0.320 0..0023 0.85 I.950 2.206 0..07 I .00 0.020 0..1 ..258 0.332 0...400 1.985 -2.300 0..20s 0.800 0..6303 2. Rated M V A Rated kV Rated PF SC R Xb‘ 569 .000 -3.020 0.00 0. I55 0.730 4.230 0. .480 0.2 (conhuedl GENERATOR Unit no.060 0..312 0.... 6..000 .r4 or .600 0....510 0. 3 166..00 0. 2.229 (3) 2.60 13.710 -0.600 0...287 0...573 0.000 0. 9.90 I . 0.0024 0..0648 1..6Xh 0...rb ...080 .060 0.326 .. .5612 1.333 0..40 0.200 .80 0. I 2 7 0.302 0. 0.000 0.200 0.379 0.407 0..40 0. ..028 HI6 23 I . 0.360 502.195 0.000 17.o 0.612 2.000 0.0303 0.550 .010 1.88 2.000 0. . ..020 HI7 250. 0.2995 O..400 ‘b ‘20 .000 4. 0.181 0.000 0..3.050 0..105 0.300 0..950 0.402 0.930 0. .920 0.243 0.690 0... . ..80 0.533 0.000 272.330 0.570 0.568 0.ooo 0..600 0.% ‘a .570 0.770 0.04s 0.2IX 0. 0...990 2.000 276.000 0..95 1. . .020 0.000 0. -2.050 267.000 0.000 0.300 3.200 0.. . .040 0.000 0..051 115.000 3.520 0... .100 0.. .000 0. . ..300 10... .120 0.900 -2...000 STA B I LlZER PSS KQV QS (7) (7) (7) S F F F .....020 0.000 0.000 15. .OOO 30..000 0.000 G 0.000 0.00 4..000 0.431 0.00 G 0..000 -2.. .... ...oO0 0.050 . .000 G 0.000 0.000 0. . .053 0..100 4..393 0.500 0. . . .050 0..020 0. ..00 27. ..380 0.500 0.000 .00 52.00 124. ...240 6.000 0.000 G 0.30 36.700 0. ..415 -2.. F 'Q r QI S S 'e I '3 Q '2 4 7e2 S S S S 'Q3 "slim P" 0.000 10..000 0..100 .000 .050 603.00 31... .O 0.590 0. ..000 0. ..250 0.000 .oO0 10.020 0.500 65.00 160..000 0..050 250..... ... .380 0. .020 1.000 G G 0...100 0.. .000 0..000 0.020 0O O . .090 F o.000 0. .. . 0.00 G 0.000 8.020 0.....000 0.....030 133.000 6.00 30.000 0. ...600 0.050 0.520 0. ..500 0.500 0. .000 0...700 0.600 0.050 171.000 0. -2. .000 0.650 -2.000 4. ..000 0.000 0.800 0.470 8..000 0.000 IO.000 1.515 -2.740 -2.oO0 F o....020 0...570 Table D.. . ...431 0. .000 55.000 0.000 1.053 0..000 G 0.. .. ..000 0.038 120.498 -2... . 0.2 (continued) TURBINE-GOVERNOR GOV Appendix D R Pmax TI (6) (6) MW S S S S S 72 13 74 75 F (6) G 0. ..050 155.050 5. .145 0.537 0.209 I ..003 I 0.500 0.372 F3 51.100 0.099 0.. ' A Or I A I E BJ30 0..042 5..470 43 I ..134 0. I ..80 0.210 1.400 .....035 .724 2. F5 Typical Data for Fossil Steam (Fl Ilnits 100....80 0.80 F2 35...750 0.000 .120 0. .. 0. .000 57.232 1. .os0 0...216 0.00 .1026 0.060 0. 0.3.000 0.060 0.2 EFDFL D EXCITER ~~ ~~~~ (2) (2) (3) FI 25.140 498. GENERATOR Unit no.670 2..4044 2...00 0.. 6. 0...250 0.600 5....033 8. ..882 F7 147.886 2. I30 0.0933 0.300 0..000 0..90 0.2 I6 0.134 .900 ..390 596.210 0. I34 0.300 0. . . . 0.078 0...0082 0.50 0..038 6.093 0.520 0..085 0.050 20.00 13.050 GFA4 0.092 0.070 0. 0.80 0.80 09 .980 0.29 13.. ....0072 . .80 0.000 0.035 0. . .. 0.364 2.220 0.070 0. 0.640 . I30 0..177 125..000 260...105 0. 0.00 13. .Table D.023 . I74 I . 0.220 1. .250 1.220 0. . I20 2.50 0.145 0..077 F4 75.50 0.80 I .85 0.108 .290 0.016 0.020 0.075 0.3928 2.1 16 0. .270 0..059 4.4320 2.300 0.100 .50 0.805 3.02 I5 0.000 .850 I . I85 I .133 0.970 0.00 0.64 0.004 .050 K" .40 0.140 464.200 WMA 0.976 I .017 0.1 I6 0.000 0.500 2.118 0. 154.279 0.85 0...00 0.50 0..715 I .. .292 2 .160 0.90 0..240 F6 125..000 400..057 0.065 0.80 0...50 NA143A 0.. 0. ...220 2..2 I8 1.050 20.00 0.0034 0. 0.295 0.299 I .050 ..000 sG I .23 I 1.000 0.500 0.50 0...10 15.00 13.80 0.00 15.140 S NAlOl 0. .. I34 0..0035 0.360 0.000 25.000 0.095 0.0284 0.118 0.000 WMA I .023 1.. .180 0.3 IO 2..90 0.375 0.120 0.105 0.00 I4 0.882 0. . I20 0.000 0.216 0.80 0..050 0.000 175.0484 4..2067 0.50 0.. .500 0.380 I ..000 0. .070 0..50 0.50 0.85 0..200 25.035 .20 13.80 0..500 .000 2. 0.000 VR type Name RR TR A A E (4) (4) S DU A A A NAlOl 0.280 0.0 0.215 0. ...... .... 2 0.2667 0.000 0.3604 0.000 O.....000 0.. . . ...000 .438 0.090 0.000 -4.6758 0..000 -0.Ooo 0. .50 0.0684 0..000 I ..050 121..Ooo F G 0..050 5 .... .200 0. .020 0. ‘Q3 .200 O.6 1... .083 0..0895 0. ...0769 1. . .000 0.000 6.0137 0... .000 I ....OO0 1. ..300 0.020 0. .300 0...2096 4.5833 348 . ... .812 I ....7128 3. .. .I 0 3 -4...2254 0.Ooo TURBINE GOVERNOR GOV R Pmax S S (6) (6) MW G 0.Ooo 0.105 0.200 0. ..370 0 1I 0 .000 4.300 0..000 0..000 I . ..000 I ..300 0.. .300 0.395 I . ...8807 357 .... ..Ooo 0.00 0. .000 0. .200 O.‘A2 0. .090 ‘I ‘2 ‘3 S s S 0.Ooo 0. . 4im ...Ooo G 0.350 0... .6544 0.00I 6 1. .OO0 -1.6774 4..000 1.200 0....200 0.... .500 0. . . . .00 0.Oo0 0.200 0....... ...0027 1... ‘Q I ‘Q I ‘Q2 ‘Q2 ‘Q3 S . ... .10 0..0445 0.0967 0.120 -0.3 -3. ..I 0 3 0..414 0.090 0.108 0..280 I o...000 I . ... ..950 0.050 36.908 0.000 0...Ooo 0...000 G 0..000 0..300 0.OO0 -0.Ooo I .0582 0..120 -3. Qs ‘Q (7) (7) (7) S S S .000 0. ..500 -4.6130 -0.Ooo 0.417 0.030 EFDmin KF ‘ForrFl ‘F7 0.ooo 0.Ooo O. .500 0.000 0. .050 22.OO0 .....OO0 -1.952 0.080 I .050 75. .220 0. . .000 0.0012 I .. 0. ..00 0.000 I .180 0. ..170 0..250 STABILIZER PSS KQV . .Ooo G 0.Ooo I .6349 3.Ooo -1.Ooo 0.0392 0.. .. ...Ooo G 0.000 0.75max SEmax AEX BEX EFDmax 0.700 10.700 0.330 -3.349 0.000 I .350 0.3774 0. ..000 -0.4628 4. S S S PU .980 0. ....00 0.. .6I30 -0...Ooo ‘Rmax ‘Rmin KE ‘E SE..0 0.0015 I . .Oo0 14 ...Oo0 0. .....0016 1..000 I .300 0.078 53. O.. .330 0.. .380 O.000 0.0601 0.0924 0.000 3.. ‘5 F (6) G 0.040 0. 0. . .300 0.200 0..100 .090 0.090 0.050 105..050 120 3. 50 0.20 .. .Table D.023 0..90 0.50 0.00 24.so 0..042 ..255 I ..000 C A WMA G BRLS A NA143A A BBC 0..s 'F sG I ..470 1..500 .000 3.0054 0.060 .2738 I .000 0. .70 1347. 1 518.58 I 0.85 F15 448.000 3010.035 0.778 0.160 0.000 ALTHYREX I .50 0. I .032 3.000 0. .7911 0..000 4 0 6 0 0.00 0.000 2. .550 .0047 .650 rf .480 . 834 I 0.00 24...50 0.000 ..00 I9 0. 0. .580 0. 0... 0.00 0.000 0.OOo 0.. 0.000 0.800 0..90 0.1 I2 0. I72 0.0910 0.....140 0.580 0...7469 0..700 ..50 4OO.07 1 I 0.000 50.258 I..247 1.000 0...1930 0..00 0.650 .400 0.004 0..00 0.210 0..508 3....7895 2. .199 0. 6.3 (continued) GENERATOR FI2 330... I346 .00 22.000 30.270 I .3950 0.020 0. I245 0.230 Ti +io +a S S .000 3.050 O.500 0.0104 1.00 0.400 2.035 0.090 0.. 1. . .120 I..290 WR MW.2632 0. .. .os I 1. .. I357 0.042 5.580 X. . Rated MVA Rated kV Rated PF SC R (1) 0.600 0.85 F17 552. 0.950 xi Xd Xb' F16 512... I50 0..0043 0.2239 1.2284 0.023 I..920 ...198 0..000 C BRLS 0.000 200...I320 0.000 400.700 xb xQ a ' PU PU PU PU PU PU .00 ... Tb . 1.070 0.000 0.. I59 0..2261 0.324 I . .265 1..90 0.90 F14 410. 3.670 0.450 1006.000 0.082 0. 2. 992.205 0. .2 EFDFL D 0.500 1..O n (2) (2) (2) (3) . 0.535 I 2. .042 5. I.200 0.580 F13 384... 0....000 0.167 0..162 0..150 I 190. 0 I58 .770 0..0014 0.2374 0.no 0. 0.. .00 20.. ...432 XZ .00 24.030 0. P U P" P U P" PU .. .518 sGI. XO 7.470 0.00 0.460 1.013 0.317 I .. 3.870 2 .50 0.00 0...400 2.....000 . x t or x p ' 2 .i ...00 0. . 0.7668 0.OO0 0.000 400.260 0...060 0...020 0...053 2.175 0. TfO S S S S S S S .205 0..000 0.1 I I 0.00 24...580 Unit no.580 0.700 2. .000 1. ‘Q3 ‘Q3 S .000 0. 0.000 O..OO0 .050 0.3910 3.890 0...300 0. .000 TURBINE GOVERNOR GOV G G G 0.Oo0 290.050 497. .500 -4. .050 1...Oo0 -1.270 0.. I 30 -8.... I 0 5 0.290 0.....000 0...I572 0.500 0. .520 0. .. S . ...OO0 1. .Ooo F (6) STABILIZER 0..000 .. P” YIlim ..Oo0 0..ooo .050 .050 I0.Ooo I .000 0.400 0.100 0.1658 0.250 0.050 0... 0..300 460.Ooo O.300 0.000 0.000 4.010 0..350 0.656 0.5465 4.810 -0.130 0.050 8.0465 0...000 0.600 0.7~max SEmax AEX 3. ‘Q I ‘Q2 TQ2 S . 0.. .270 I .3857 O.910 0...Oo0 G G 0...1 I 4 5 0. 0. . I50 0.100 S n ...250 0..459 0.920 0.00 0.. .0012 O.030 I . .560 %in KE I .040 0.Oo0 0.000 0..5... ..060 I.170 0. .990 -0.000 0... .0 000 0..2639 4.230 ...270 8.2909 4.. .OOO 10.840 -3...“Rmax .950 0.. .000 0..Ooo O. ..000 0.. ..810 -3...050 0.00 0.. 26.890 -4. .... ..I80 O.278 0.220 0.OO0 5. .Oo0 TI?..0635 I .840 I .320 0.. ...840 *EX €FDmax €FDmin KF ‘ForrFl 4.050 347.812 0.840 0.OOO 0..100 O.3.050 367. .050 360..730 1.267 0.220 0.OO0 0..I 0 5 0.. .000 0.250 0.200 0. .00 0.270 KO00 0.00 0...000 .320 -4..000 3 0.Oo0 O.100 R Pmax (6) (6) MW S ‘I ‘2 F S 0.000 0..00 0.OO0 3.100 0.130 5....050 G 0. . . 75 0. ..5.435 0. .020 0.300 PSS KQV .990 .0027 -0.050 0.020 0.. . .300 O.OO0 ‘E S~.ooo 0. ....250 8.260 73 ‘4 5 .000 0.. . ... KQs (7) (7) (7) S F S S ‘Q ‘Q I ..000 0.071 0...00 0.0832 I. S S .040 1.Oo0 3.950 0. oo Rated MVA Rated k V Rated PF SCR 0.171 305... I50 0.00 0.00 I8 .047 5..229 0.043 5 .0905 0.023 0. 0.023 0.171 0.175 0.50 VR type Name RR T? i WMA WMA NA143 O.64 0. Typical Data for Cross-Compound Fossil Steam (CF C'F2-HP 192.0 sG I .800 0.....023 I .000 .225 0.50 0.00 . I50 I.092 0.224 0.205 0.224 0...00 0.326 0.000 2 000 EXCITER A A A A A NAlOl (4) (4) S NAlOl WMA WMA 0. .00 0.0039 0.390 596.90 0.330 14in.640 0.225 0.023 0.640 2. I35 0.600 0.101 .590 0.500 'GI.252 0.070 0.670 0. . 0.390 464.171 0.292 0.140 0.Oo0 KA PU 0.304 : 0.680 0.141 0.85 0.3 I5 I .570 0..390 650.00 13.043 5 .650 0.610 0. I86 0.00 0.0036 0.032 4.840 2.675 0.000 245.53 I 0.00 0.610 2.58 0.24X 0.730 0.291 0.150 639.034 5.380 I.85 X.Z EFDFL D 0..570 2.023 1.500 0.50 0.70 0.224 0.000 0. I50 1.141 0.00 22.000 0.229 C'F3.Table D.90 0..224 0.000 275.50 0..50 0.OM 592.4161 2.70 0..130 0..0982 0.00 C'FI-LP 128.0043 0.000 0.500 2.020 12 . 0.028 0.70 20. .000 0.640 0.260 1.648 0.023 0..4.58 0..500 0.650 0.64 0.00 .31 I 1.955 I .205 0.250 0.000 0.979 1.80 0...028 0.000 0..460 I..L P 22 I . GENERATOR Unit no.64 0.023 1.369 1.000 A A 2.121 0.000 245.58 I 0.660 0. 1.c Or xp '2 x2 XO C'FI-HP 128. I249 0.004 0.500 0.. 0.000 275.80 I 8 .249 .390 0..50 0.890 0. 12 C'F4-HP 445.4161 2.23 I 0.095 0. .64 0. I35 1.023 0.000 .820 0.020 120 TdO 1.0982 0. .022 0.0024 0.0036 0.400 0.250 0.64 0.840 2. .500 0.oo C'F2-L P 192.8 15 0.150 1. 0.3 I5 1.640 2.250 . 0..003 0.85 0..080 0.029 0.30 20.1122 0.00 .000 .345 2.000 2.101 CF3-HP 278.000 0...000 0.060 25.. GO TbO la WR 'F 0.00 13.85 0. I86 0.232 1.958 1.670 0..958 1.4 I3 2.100 0. I40 0.026 0.060 0.820 0.590 0.028 0.433 2.410 0.060 25.171 0.50 0.i xd Xd Xb' xb 'a x.500 0.205 787. 0.053 5. 0..000 0..565 1.020 0.037 5.. .90 0.320 1.470 0. 100 G 0.570 -3.OO0 1.150 0.Oo0 O.1.000 0.88 0.000 12.100 O.750 0.260 -4.3.960 0.330 2.0.OO0 O.Oo0 0.Oo0 O.350 I .606 G 0.020 O.0027 1.020 0.984 -0.771 5.000 8.00 0.OO0 0.00 0.450 0.091 0.490 0.000 2.050 0.050 0.960 .033 0.400 S S S 0.000 0.020 O.000 0.160 0.0.000 0.050 F 0.600 10.0667 I .080 0.060 .455 0.000 0.3035 0.033 0.75max SEmax A EX BEX (5) (5) (5) (5) P (5) U P U PU 0.200 0.050 O.000 €FDmax €FD min KF TForsFl 'F2 O.0013 1.000 0.0307 0. O OO F 0.200 0.OO0 0.OO0 0.5331 4.780 -0.50 0.220 0. 10 '2 73 '4 75 O.540 STA El LlZER (7) (7) (7) S S S S S S 0.560 0.260 0.570 0.050 0.Ooo O.490 0.OoO 0.780 .OO0 DU O.050 172.250 G 0.OO0 -0.720 0.000 0.020 O.960 .OO0 -0.050 107.048 0.OO0 0.639 3.984 .230 0.700 0.2978 0.060 O.000 I o.2978 0.Oo0 13.000 TURBINE GOVERNOR GOV R G 0.455 IO.300 10.OO0 O.265 0.000 I .OO0 IO.020 0.000 F 0.250 0.150 0.950 0.570 0.050 267.50 0.05 I 'E SE.OO0 1.13.020 0.3750 3.330 O.020 0.050 107.310 -5.OO0 O.5685 0.05 K€ PU S .OO0 0.050 172.100 G 0.000 0.950 0.220 0.160 0.100 8.000 0.094 3.020 0.265 0.3750 3.0667 I . I688 0.OO0 0.050 213.150 0.020 0.000 0.Ooo 0.3035 0.OO0 0.OO0 0 .370 0.OO0 I .350 0.OO0 10.591 0.300 10.020 O.2.OO0 0.260 0.09 I 0.OO0 0.OO0 I .000 .020 O.050 12.000 0.0778 0. I50 0.040 s S 0.230 0.780 -0.3.000 O.OOo 0.260 -4. 1.170 1.5 I2 I .000 10.50 0.310 0.053 'A2 S O.570 .250 Pmax 'I G 0.370 0.0027 I .OO0 0.000 O.00 I3 1.050 41 1.1688 0.000 F 0.000 0.5685 0.300 12.200 0.Oo0 0.00 0.300 4.549 0.OO0 O.250 0.OO0 O.50 0.OO0 O.050 V R max ' R min P (4) U P (4) U .OO0 0.' A Or ' A I S 0.100 I 7 0.000 -0.960 0.OoO 0.080 0.050 O.020 0.170 1o.OO0 O.050 0.639 3.070 I .600 I o.020 S S 10.984 - O.040 I .3.780 -2.000 0.5331 4.05 I 0.170 .300 4.984 -0.Oo0 I .060 0.020 0.000 0.0307 0.OO0 O.0778 0.000 0. 95 0..580 0..000 0. .000 0.320 1.3814 3..201 I .500 Ti0 Ti0 .060 1.2932 0. .500 0.230 0..3055 2.000 I.144 1.230 . .90 0...90 0.000 A NAlOl 0.0714 0.000 0.000 6.0027 I S966 3.. I44 I ....002 I 0.340 2.. I29 0.3900 0.480 0.090 -4.800 0..693 0.281 0.460 O.043 6.000 6.0489 0.0863 0... I20 0.038 7. I 52 1.0576 0.710 2.000 0.780 .2 EFDFL D .310 -4.. .00 0.756 0.Appendix D Table D.004 I 0.960 I.... N2 245..000 25. ..85 0.220 .4 I 0. .517 2.00 0..5630 0. 0..vb NI 76.550 0..68 I8 5. I296 0..660 0.00 I 3 1..000 0..020 1..228 ..1455 0.028 0.150 .035 1.060 25..743 0.195 N7 1300.00 I 6 1.360 .739 0.565 1.790 0.000 10.100 . 0. 6..636 0.034 9.07 I 4 0.073 0.630 0.029 0.2200 0.000 A WMA 0.000 0..660 0.665 0..000 -0..200 0..270 2.200 0.6.0900 0.881 0.070 0..020 0.500 0. G E N ERATOR Unit no.029 0.75max SEmar A EX BEX E F D ~ ~ ~ EFDmin KF 'ForTFI T F7 A NAlOl 0.. 0.782 0..7750 BRLS 2.093 0..00 I 6 1.000 0. I 80 4690.640 0. ..228 . I 2 5 0.000 I .08 I 6 0.090 0.000 256..85 0.1.500 .000 NAIJ3 0. I24 1.000 .933 . .0244 0.4015 3.0019 0.4 0. 577 Typical Data for Nuclear Steam ( N ) Units N3 500.9500 0.70 .000 400.. .1 IO 0..380 0.210 I 136.0794 0.000 0.450 .084 I .2 t 0 Tb' 0.00 0.650 0..0857 0.375 1.150 0..2 IO 0.. ..1.040 0.040 0.650 . 0.3520 2.I.251 N8 1340.OOo .032 4. .312 0.000 WR 'F I990.00 0. 0.000 0.0979 0.99 I 1.050 0.460 0.3 IO N4 920... 4580..881 -3.000 -0.170 2.000 c' BRLS 2.oo 0..100 120 ...000 . SGI.205 0.OO0 0..533 I 2. 0.000 -0.00 0.90 0.. . I309 0. 0.350 0. .3 I00 3.570 I .015 0...10.00 .275 0... .015 0.035 0...020 0.50 0.. .00 0.000 ..80 13..50 0.650 1. 0.160 0...00 0.2 I5 0.000 .4475 4.....0029 0.00 2s .665 -3.444 1.020 0..308 1...0056 0. .3933 2.000 A C' c BRLS 2 ..467 1.020 . .000 0. 4.0032 0... P .607 0.040 I..032 7. 281.00 .277 1.870 .0769 0.000 0.6120 4.35 18..000 0..5600 0..120 .vb' .060 25.960 -6..00 0.000 50.025 0. 0... 4698. .2148 0.5330 4.000 0. .5.730 ..0761 0.512 0. 2.00 18.23 0.000 ..000 .ooo 0..020 0.050 0.190 0. . .660 Id0 19 T b ..055 0.358 2. .0516 0.0048 0.000 2.052 6.850 0.195 NS 1070..2784 0.000 1.283 0.510 1.oo 0.5 14.so 0.27s 0..00 0.. 0..00 22. 0.000 1.350 0. 0.275 0.. .3 I j 0.945 2. 0.000 400. 1.90 0.050 0.280 Rated M V A Rated kV Rated PF SC R .467 2.580 .0464 0..000 0. 3312.3244 2.0779 0.00 0.. .0576 0.00 25.v4 or x '2 -.210 0.0901 0.579 0.. N6 I280.500 0.oOU 400..710 0.000 I .3093 0.284 .320 1.50 0.237 0..200 0. A EXCITER VR type Name RR TR KA Or T A I ' A2 " R max V R min KE 'E SE..350 0.000 0..1.355 1.860 0. 0.000 0. 0.217 0.000 ..088 0..346 I...vi "d Xq .900 .000 .281 0.059 0. 0.074 0.00 0.000 -0.708 2.000 0.00 22.4100 2..125 0..858 -2.000 . .90 0..0 sc1.580 Ti .237 0.310 0.055 6.480 0.. 0. .0752 0.....19 3464.020 0.50 0...470 Q 'a I.. .0233 0.000 EA210 I .215 0.3400 0.858 -0.522 0.8 0. ..... I ..100 0.150 0.000 1. .675 G 0.300 0...000 0..00 0..040 0. ..100 .050 450... . .Oo0 7YO. ..050 0...000 0. ... .000 20.020 1.000 1.300 PSS KQV (7) (7) S O..180 0. . STABILIZER .... ...340 F o...050 Y 5 I.020 0.000 0..000 1o.000 1o. .280 2..200 5.. .OOo 0...000 0.000 0. 5 I0 0. . .OO 0.000 1.00 G 0.000 0..300 G 0.100 F 0... .180 0. ..050 65...050 208.000 0.300 0..330 0.000 0.. PU .530 3. ..300 5.814 6.000 0...020 0.. ..100 .200 10.000 0. .0 000 O.330 5 .OOo 0.OO 1216.. .. ‘Q2 ‘Q3 ‘Q3 YSlim S S S 0.5..000 0... .. . I80 0. 0.....210 0. . .200 5..000 0..050 IOYO. .. ..000 O. ..OO0 ‘e Qs (7) S S S 0.000 0.000 0..200 0. G G 0...578 Appendix D Table D.050 0. ...000 0.320 . .Oo0 0. ..00 0.250 0..000 0.000 0..000 0. S 0... ..050 G 0..020 0.. ..330 ........460 0.. .OX0 0...Ooo F 0.030 o. .050 0..150 0. .I8 ...00 0..... . ..OO0 G 0.. . 10. TURBINE GOVERNOR (continued) G 0.. . .040 0.050 1205.020 0. ..100 0..100 ‘Ql ‘0I ‘Q2 0..000 20. .020 0.000 0.080 0. . . .6. .000 Or l F I TF2 6.180 0.. I50 0. 0. 0.477 0.80 0. A WMA A WMA RR 71 1 KA ' A 0.0047 0.I I .050 0.23 I 0.180 0..630 -0..0025 0.0 I53 1 .000 'E SE. I70 0.oo 0.274 0.855 0.0027 0.790 0.000 A WMA A N A I43 0.950 0.04 I 3.220 0.304 I .oo ..040 1.141 0....220 0.058 12..290 60.769 0..000 0.000 0.0669 0. 0..80 0.800 0.257 0.6884 8.235 .4782 7.00 12.ooo I .279 0. I59 rF SG I .ooo 0.000 -0.560 60.476 0. .199 0.. A .040 I .ooo 0.000 I . .070 1.000 400. .75max SEmax A EX EFDmax EFDmin KF 4.. I77 0...OO0 400.0634 0. 3.000 I . ...5795 0.200 0..000 0.500 -8.500 0.373 1. 0.200 I05.270 . G EN E R A T 0 R 579 Typical D a t a for Synchronous Condensor (SC) U n i t s ~ ~~ U n i t no..3956 14.201 0.160 .0027 0.1512 0.00 x : xd xi xq Xq ra xc or xp r2 x2 XO 1.730 .350 0..5.000 0. 0. 0.98 0.165 0.000 -0.2 EFDFL D EXCITER V R type Name 30.720 0.0987 0.000 1 .558 0.O sGl.0024 0.000 5...70 0.3 IO 2.14..850 -0.000 Or ' A I O.230 0.407 -4.80 0.6.0138 0.000 18. .00 13.0151 0.4407 0.00 0.666 3.270 0.000 0.146 I .60 0.708 4..225 0.776 4.050 0. I32 0.ooo 2 .80 0.000 I I ..950 0. sc2 40.50 0.00 13.00 sc1 .. 25.650 .1 IS 0..0356 5. ...244 I .040 1 .600 ..039 16. 0.I ..ooo 0.007 I 0.80 0.540 400.00 5c5 75. 0. 0..850 -5.304 0.343 2.630 .0873 0. 3..170 0... I28 0.00 I .050 0.950 0.035 .385 2.ooo 0.172 0.560 0.058 0.858 0.950 0.288 89.000 0.Appendix D Table D.180 0.0027 0. .500 0.950 0.006 0.180 .170 1 ..ooo 0.500 0.000 .950 0.063 I 0.7.188 0..050 0..200 .950 0.083 0..650 0.0473 .000 200.407 -0.85 0.041 0.220.7802 7.. Rated M V A Rated kV Rated PF S CR X.ooo 0...170 0.320 I .0525 8.500 -7. I045 0.26 I 1.170 0.185 0.540 .295 0.172 1.000 0.220 0..oo 0..00 13.035 Ti T& .004 sc4 60.ooo .000 0.00 13.2035 0. .00 I7 0.720 0.00 0.050 6. 0.790 .058 1 I .224 .00 sc3 50. . I70 0.150 0.0027 1..180 1.338 0.. . .610 'bo If 62.000 0.75maa SEmax .2681 2.300 1.05 I 0...4348 2..500 0. 'b G O .102 0.0870 0.. GENERATOR TvDical Data for Combustion Turbine (CT) Units <.500 C BRLS 0. .. .000 400.85 0.Appendix D Table D 7 ..500 0.80 0. Rated MVA Rated kV Rated PF SC R X.50 0...054 7.025 0. I59 1.19' 2.80 0...i xb Xd Xf CT I 20. x2 XO . .580 0.225 1.. EXCITER Uni1 no.640 'I '2 5 0. .000 .. ..000 0.000 G 0. 4.350 7 13.025 0.00 0.. .020 0..000 120.020 0.000 7. 2..575 0. ..500 VR type Name RR 'R KA ' A Or 7 A I l A2 D SC PT ..65 13.000 1..OO0 0... .0 . 1..000 0.040 82.. 'i . 0._ .000 Fuel: Oil G a s 0... xi x9 '0 .000 0.100 0..850 CT2 .261 0..500 1.640 0.300 -7..300 0.000 0.113 0.000 I .580 0..860 0.740 x 4 or x p 12 .250 0.. .50 0. 0. 1...050 17..000 0.000 . I55 0. A EX BEX EFDmax €FDmin KF 'ForrFl 'F2 .100 EFDFL D '3 74 S '5 S S F (6) 0..102 0.I .200 .50 13..000 0..30 TURBINE GOVERNOR GOV WR 'F sG I . ..85 0..46 I .O SGI.035 0.55 0..2972 7..700 0..034 0.306 1.030 I . .253 VRmaa VRmin KE 'E SE.035 0...700 0. 183.188 0. 0.200 1.730 0. .5 0.107 I .352 0..2 R Pmaa (6) (6) MW S G 0.32 ~~~ 0.000 ~~ 'b '50 'q0 '0 ... . .....050 0.100 0.0983 0. 03 0.09 0.500) ( I .4694 0.5 477. I529 0.6 70. spacing I2 14 16 18 Flat phase spacing Ut) Geometric mean distance illmi Xd .44 I 0.5622 0.03 0.4694 0.4694 0.805 I 0. Transmission Line Data ACSR Conductor area (Ordiam) 60-Hz inductive reactance Line-toline voltage (kV) kCM (in.420 0.1677 0.382) 22 28 28 38 38 38 38 56 56 15.3480 0.10 0.3789 0.430 0.750) ( I .2922 0.4694 0.4325 0.3641 0.00 ..5166 0.5 ( I .6002 0.9 14) ( I .3 47.6 0.1 17.0988 .02 0..9 47.Table D 8 Typical 60-Hz .01 0.09 0.3 35.7 35.4325 0.4030 0.5166 0.8089 0.7761 0.9 47.10 0.6223 0.Vu + xd 0.) (ft) Conductors per phase 8 1 8in.602) (1.06 0.3336 0. I 22.750) ( I .6 20. y 69 I I5 I38 161 230 345 345 500 500 500 500 735 735 I I I I I I 2 I 2 3 4 3 4 226.7944 0.9 70.3294 0.0456 0.10 0.45 I 0.8 336.5278 0.5950 0.8230 0.7990 0.0584 0.7 27.4 397.246) (2.465 0.165) (0.01 0.5682 0.0784 0.0 556.9 47.7616 0.07 0. Finally. 1 . and the supplement to C85. A feedback control system which op- 1. 1. The definitions that follow are those proposed by the 1969 report. feedback. The first [ I ] appeared in 1961 under the title “Proposed excitation system definitions for synchronous machines” and provided many definitions of basic system elements. PAS-88. added the new definitions required by technological change and attempted to make all definitions agree with accepted language of the automatic control community. 1. o IEEE.appendix E Excitation Control System Definitions There are two important recently published documents dealing with excitation control system definitions.1 [ 5 ] . ANSI Standard C85. Proposed IEEE Definitions 1.. Reprinted with permission from IEEE Trans. automatic feedback.01 Control system. A control system which operates to achieve prescribed relationships between selected system variables by comparing functions of these variables and using the difference to effect control. using the first report as a starting point. vol. and manual control.04 Excitation control system (new).1 second or less. erates without h u m a n intervention.10 on rotating machines [3]. 582 . definition 41. The second report [2] was published in 1969 under the same title and.05 High initial response excitation system (new). which defines certain time constants and gain factors used in excitation control systems. A feedback control system which includes the synchronous machine and its excitation system. An excitation system having an excitation system voltage response time of 0.’ Reference is also made to the definitions given in ANSI Standard C42. 1. reference is made to the IEEE Committee Report “Computer representation of excitation systems” [6).0 Systems 1. regulator.1 on automatic control [4].03 Excitation system [ l . 1969.02 Control system. The source of field current for the excitation of a synchronous machine and includes the exciter. feedback. It creates a droop in generator voltage proportional to reactive current and equivalent to that which would be produced by the insertion of a reactor between the generator terminals and the paralleling point.04 Control. Line drop compensators modify generator voltage by regulator action to compensate for the impedance drop from the machine terminals to a fixed point. A feedback element of the regulator which acts to compensate for the effect of a variable by modifying the function of the primary detecting element.03 Compensator [l. A reactive current compensator is a compensator that acts to modify the functioning of a voltage regulator in accordance with reactive current. An element or group of elements associated with a feedback control system by which adjustment of the level of a controlled variable can be made. definition 441. The voltage drops of the resistance and reactance portions of the impedance are obtained respectively in pu quantities by an “active compensator” and a “reactive compensator. Notes: I . Those elements in the excitation control system which provide for manual adjustment of the synchronous machine terminal voltage by open loop (human element) control. Reactive droop compensation is the more common method.” 2. 2. terms such as “equalizing reactor” and “cross-current compensator” have been used to describe the function of a reactive compensator. Those elements in the controlling system which change the feedback signal in response to the directly controlled variable. These terms are deprecated. They function in the following two ways. The difference current for any generator from the common series current creates a compensating voltage in the input to the particular generator voltage regulator that acts to modify the generator excitation to reduce to minimum (zero) its differential reactive current. manual (new). Action is accomplished by insertion within the regulator input circuit of a voltage equivalent to the impedance drop. A device whose output is an enlarged reproduction of the essential features of an input signal and which draws power therefore from a source other than the input signal. It is obtained by a series differential connection of the various generator current transformer secondaries and reactive compensators. b.Appendix E 583 2. .05 Elements. definition 401. a. Examples are reactive current compensator and active current compensator. Historically. 3.01 Adjuster (1. 2. An active current compensator is a compensator that acts to modify the functioning of a voltage regulator in accordance with active current. Reactive differential compensation is used where droop in generator voltage is not wanted. 2. Reactive compensators are generally applied with generator voltage regulators to obtain reactive current sharing among generators operating in parallel.0 Components 2. 2.02 Amplifier. 4. power reactor. synchronous machine [ 1 definition 8 .10 Exciter. The source of all or part of the field cur1 rent for the excitation of another exciter. and a given regulator may not have all the items included. I t is exclusive of input control elements. definition 7. or the shaft of the synchronous machine.91 a dc generator.07 Element. The exciter includes the power potential transformers. forward. auxiliary inputs. 2. 1 regulator couples the output variables of the synchronous machine to the input of the exciter through feedback and forward controlling elements for the purpose of regulating the synchronws machine output variables. Definition 151.12 Regulator. primary detecting. An exciter whose energy is derived from. It is exclusive of input control elements. The exciter includes the power transformers (current and potential). power amplifier. The rectifiers may be stationary or rotating with the alternator shaft. The source of all or part of the field current for the excitation of an electric machine. the regulator is assumed to consist of an error detector. An exciter whose energy is derived from an alternator and converted to dc by rectifiers.20. Those elements situated between the actuating signal and . 2 0 Exciter [l. acts to limit a variable by modifying or replacing the function of the primary detector element when predetermined conditions have been reached. 2 0 . including gate circuitry.94 a stationary ac potential source and converted to dc by rectifiers.09 Exciter. The source of all or part of the field current for . 2 0 . these regulator components are assumed to be self-explanatory.6 the controlled variable in the closed loop being considered. power rectifiers which may be either noncontrolled or controlled. That portion of the feedback elements which first either utilizes or transforms energy from the controlled medium to produce a signal which is a function of the value of the directly controlled variable. A n exciter whose energy is derived from . prime mover. pilot (1. power rectifiers which may be either noncontrolled or controlled. main [ l . The exciter may be driven by a motor. I t is exclusive of input control elements. Potential source rectifier exciter. or by the shaft of the synchronous machine. where used. 21 1 Limiter [ 1. The exciter includes a dc generator with its commutator and brushes. An exciter whose energy is derived from the .584 Appendix E 2 0 Elements. As shown in Figure 7.09.2 Alternator rectifier exciter.93 currents and potentials of the ac terminals of the synchronous machine and converted to dc by rectifiers. 2. . Functional regulator definitions describing types of regulators are listed below. 2 0 . 2. including gate circuitry. stabilizers. 2. including gate circuitry. Note: In general. 3. The exciter includes an alternator and power rectifiers which may be either noncontrolled or controlled. DC generator commutator exciter. and limiters. The alternator may be driven by a motor. 2. prime mover. preamplifier. Compound rectifier exciter.definition 51. exclusive of another exciter. The term "dynamic-type" regulator has been omitted as a classification [ I . A synchronous machine . definition 4 1 A feedback element of the excitation system which . It is exclusive of input control elements. definition 51.8 the excitation of an electric machine. . exciter nominal [ 1 definition 2 1 Nominal exciter ceiling voltage is the ceiling voltage of an exciter loaded with a resistor having an ohmic value equal to the resistance of the field winding to be excited and with this field winding at a temperature of I . 2. the effect of such . An indirect-acting type of regulator is a rheostatic type that controls the excitation of the exciter by acting on an intermediate device not considered part of the regulator or exciter. 3. power system (new). frequency. frequency.13 Stabilizer. lating function by mechanically varying a resistance. Quantitatively.6 action) intended to improve performance with respect to some specified characteristics. Note: In control usage this characteristic is usually the system deviation. ponent system output voltage that is able to be attained by an excitation system under specified conditions. 2 1 Stabilizer. 2. 3. . definition 2 1 Exciter ceiling voltage is the maxi4. 5. One that requires a sus. excitation control system (new). 2 1 . I00"C for field windings designed to operate at rating with a temperature rise greater than 60°C.23 . A direct-acting type of regulator is a rheostatic type that directly controls the excitation of an exciter by varying the input to the exciter field circuit. 2. Compensa- .22 tained finite change in the controlled variable to initiate corrective action.0 Characteristics and performance 3 0 Accuracy. m u m voltage that may be attained by an exciter under specified conditions. 2 1 .definition 111. Definitions 13. definition 2 1 The maximum dc com. rheostatic type regulators have been further defined as direct-acting and indirect-acting. 3 0 Compensation. 3. excitation control system (new). 3 0 Ceiling voltage. synchronous machine electrical power and other. ambient temperature. 4: Note [ I . 75°C for field windings designed to operate at rating with a temperature rise of 60°C or less.definition 1 1 One that initiates a correc. 1 1 Historically. Rheostatic type regulator [ 1 definition 1 1 One that accomplishes the regu. A n element or group of elements which pro.4 vides an additional input to the regulator to improve power system dynamic performance. and supply voltage variations. it is expressed as the ratio of difference between the controlled variable and the ideal value. Noncontinuously acting regulator [l.05 Ceiling voltage. 3.21 0.3 1 6.02 Air gap Line. tive action for a sustained infinitesimal change in the controlled variable. Continuously acting regulator [I. excitation system 1 . The degree of correspondence be.1 tween the controlled variable and the ideal value under specified conditions such as load changes.04 Ceiling voltage.Appendix E 585 2 1 . A number of different quantities may be used as input to the power system stabilizer such as shaft speed. humidity. A modifying or supplementary action (also. exciter [l. An element or group of elements which modifies the forward signal by either series or feedback compensation to improve the dynamic performance of the excitation control system. The extended straight line part of the no-load saturation curve. This line is determined by establishing the area acd equal to area abd.11 Duty.08 Deviation. definition 361. which rate if maintained constant. drift is determined as the change in output over a specified time with fixed command and fixed load. The duty cycle will include the action of limiting devices to maintain synchronous machine loading at or below that defined by ANSI C50.14 Error. On a practical system. accuracy.1. which change is unrelated to input. A system having an excitation system voltage response time of 0. . transient. 2. The instantaneous value of the ultimately controlled variable minus the command. definition 211. Note: ANSI C85 deprecates use of the term as the negative of deviation.I . Referring to Fig. The starting point for determining the rate of voltage change shall be the initial value of the excitation system voltage time response curve. 3.09 Disturbance. Notes: 1 . 3.586 Appendix E tion is frequently qualified as “series.. Note: The duty cycle usually involves a three-phase fault of specified duration located electrically close to the synchronous generator. excitation system (new). The instantaneous value of the ultimately controlled variable minus its steady-state value.1 s or less is defined as a high initial response excitation system (Definition l . The rate of increase or decrease of the excitation system output voltage determined from the excitation system voltage-time response curve. precision in ANSI C85. Note: The change is a plus or minus variation of short periods that may be superimposed on plus or minus variations of a long time period. system. environment. or load.12 Drift [l.07 Deviation.” “feedback. An undesired variable applied to a system which tends to affect adversely the value of a controlled variable. Its primary purpose is to specify the duty that the excitation system components can withstand without incurring maloperation or specified damage. to indicate the relative position of the compensating element. E. excitation system (new).13-1965. . Similar definitions can be applied to the excitation system major components such as the exciter and regulator.An indicated value minus an accepted standard value. 3. See also 3. Those voltage and current loadings imposed by the synchronous machine upon the excitation system including short circuits and all conditions of loading.05).15 Excitation system voltage response [ l . with specified environmental conditions.10 Duty. A n initial operating condition and a subsequent sequence of events of specified duration to which the excitation system will be exposed. 3. 3. 3.13 Dynamic.” “parallel. 3. A n undesired change in output over a period of !ime. the excitation system voltage response is illustrated by line ac. Referring to a state in which one or more quantities exhibit appreciable change within an arbitrarily short time interval. would develop the same voltage-time area as obtained from the curve for a specified period. or true value. 3.” etc. s Fig. The excitation system output voltage expressed as a function of time.15). under specified conditions. E. Note: A similar definition can be applied to the excitation system major components: the exciter and regulator separately. measured over the first half-second interval unless otherwise specified. and then suddenly establishing circuit conditions that would be used to obtain nominal exciter ceiling voltage. in volts per second. The time in seconds for the excitation voltage to reach 95 percent of ceiling voltage under specified conditions. Referring to Fig. The main exciter response ratio is the numerical value obtained when the response.17 Excitation system voltage time response [ I . which is obtained when the excitation system voltage response in volts per second.19 Exciter main response ratio. response should be determined at rated speed. 3. formerly nominal exciter response. where ao = synchronous machine rated load field voltage (Definition 3. unless otherwise specified. Unless otherwise specified. the same excitation voltage-time area as attained by the actual exciter. . with the exciter voltage initially equal to the rated-load field voltage. 3.21) and oe = 0.1 the excitation system voltage response ratio = (ce . if maintained constant. For a rotating exciter.1 . Exciter or synchronous machine excitation system voltage response (Def.ao)/(ao)(oe). the excitation system voltage response ratio shall apply only to the increase in excitation system voltage. definition 2 1 The numerical value 3. definition 191. which response.5 I 00 =Synchronous machine rated load field voltage e - 0 Time. 3.5 second. would develop. in one half-second. This definition does not apply to main exciters having one or more series fields (except a light differential series field) nor to electronic exciters. Note: The response is determined with no load on the exciter. is divided by the rated-load field voltage. 3.Appendix E 587 I oe = 0.18 Excitation system voltage response ratio [I. E.16 Excitation system voltage response time (new). is divided by the rated-load field voltage of the synchronous machine. 3. base (new). and terminal voltage and with the field winding at 25°C. output.29 Signal. band of [I.26 Regulated voltage. 1. rated speed. on the speed of a motor. definition 391. No-load field voltage is the voltage required across the terminals of the field winding of an electric machine under conditions of no load. band or zone.24 Limiting.20 Field voltage.31 Signal. A signal applied to a system or element.19). 75°C for field windings designed to operate at rating with a temperature rise of 60°C or less. expressed in percent of the rated value of the regulated voltage. X = input transform.Next Page 588 Appendix E 3. Nominal band of regulated voltage is the band of regulated voltage for a load range between any load requiring no-load field voltage and any load requiring rated-load field voltage with any compensating means used to produce a deliberate change in regulated voltage inoperative. Illustration: Y = =tPX where P = proportional gain. input.g.30 Signal. 3. 3. 3. 3. proportional.27 Regulated voltage. feedback.32 Signal. 3. 100°C for field windings designed to operate at rating with a temperature rise greater than 60°C. within which the excitation system will hold the regulated voltage of an electric machine during steady or gradually changing conditions over a specified range of load. . Note: This defines one pu excitation system voltage for use in computer representation of excitation systems [ 6 ] . the signal resulting from subtracting a particular return signal from its corresponding input signal (Figure 7. actuating. no-load [ l . nominal band of. 1. definition 381.. The intentional imposition or inherent existence of a boundary on the range of a variable. 3. formerly nominal collector ring voltage. 75°C for field windings designed to operate at rating with a temperatux rise of 60°C or less. definition 3 1 Band of regulated voltage is the 7. rated-load [ 1. error. I00"Cfor field windings designed to operate at rating with a temperature rise greater then 60°C. The decrease of controlled variable (usually speed or voltage) from no load to full load (or other specified limits).28 Signal. The ratio of the change in output due to proportional control action to the change in input. Rated-load field voltage is the voltage required across the terminals of the field winding or an electric machine under rated continuous-load conditions with the field winding at one of the following. 3. load.19).23 Gain. That return signal which results from the reference input signal (Figure 7. 3.22 Field voltage.21 Field voltage. 3. 3. The synchronous machine field voltage required to produce rated voltage on the air gap line of the synchronous machine at field temperat ures. 2. e. and Y = output transform. 3. 2.25 Regulation. The reference input signal minus the feedback signal (Figure 7.19). 3. In a closed loop. A signal delivered by a system or element. American National Standards Institute. New York. 3. Proposed excitation system definitions far synchronous machines.34 Signal. IEEE Trans. American National Standards Institute. rate.e. or amplitude. Note: It should be recognized that under some system conditions it may be necessary to use power system stabilizing signals as additional inputs to excitation control systems to achieve stability of the power system including the excitation system. 589 A signal that is responsive to the rate of change of an input signal. the property that the output remains bounded. Definitions of electrical terms. such as value. PAS-88: 1248-58. Note: It may describe a condition in which some characteristics are static. reference input. 3. For certain nonlinear systems or elements. Terminology for automatic control. rate (new). American National Standards Institute. AIEE Committee Report. the property such that its output is asymptotic. 3. References I . In a control loop. That in which some specified characteristic of a condition. ANSI Standard C85. 1961.38 Stability. 6.la-1966. IEEE Committee Report. Computer representation of excitation systems.39 Steady state. 3.41 Variable. 3. AIEE T ~ u ~ PAS-80:173-180. The ability of the excitation system to control the field voltage of the principal electric machine so that transient changes in the regulated voltage are effectively suppressed and sustained oscillations in the regulated voltage are not produced by the excitation system during steady-load conditions or following a change to a new steady-load condition. 1963. In a variable observed during transition from one steady-state operating condition to another that part of the variation which ultimately disappears. A condition of a linear system or one of its parameters which places the system on the verge of instability. New York. and transmitted by the loop and to be subtracted from that input signal. i. rotating machinery (group 10). in a limit cycle of continued oscillation. IEEE Trans. 4. 1963. ANSI Standard C85. In a closed loop. Proposed excitation system definitions for synchronous machines. excitation system.37 Stability limit. IEEE Committee Report. . 1969. For a feedback control system or element.. PAS-87: 146064. ANSI Standard C42.40 Transient. Note: ANSI C85 deprecates using the term to mean the total variable during the transition between two steady states. periodicity. that variable whose value is sensed to originate a feedback signal. New York. Supplement to terminology for automatic control (285. IO. One external to a control loop which serves as the standard of comparison for the directly controlled variable. 3. within the linear range and without continuing external stimuli. s.35 Signal. the signal resulting from a particular input signal. exhibits only negligible change over an arbitrarily long interval of time.g.33 Signal. 3.l-1963. directly controlled. 3. 5. will ultimately attain a steady-state.36 Stability. when the input is bounded. 1968. 3. 2. 1957. e..1-1963. return. others dynamic.Previous Page Appendix E 3. integration. We use linear superposition to evaluate C1withy fixed and C2with x fixed. F. .appendix F Control System Components The electrical engineer is usually acquainted with common control system components u used in all-electric or electromechanical systems.l Summation A summer is a device that adds two or more quantities with due regard for algebraic sign. O r goal here is to introduce mechanical and hydraulic components and.Y ) (Fa For small displacements.-El z (2 + dz =fix. summing the currents entering the summingjunction. . R2. By similar triangles. we have y=c. . to compare these with electric components that perform a similar function. not forces. or E. of x and y with the displacement z being proportional to some function of x and y. 590 . Eggenberger and his excellent paper “Introduction to the Basic Elements of Control Systems for Large Steam Turbine Generators” [l]. differentiation. A.Rnto the input of an operational amplifier with feedback resistor RP as shown in Figure F.x+c2y *Many of the ideas illustrated here are due to the late M. . this is easily done by adding as many connections as desired through resistors R1. we can write -E2 + . I.2. in some cases.* The purpose for doing this is to enable one to recognize basic functions such as summation. and amplification when performed either electrically or mechanically. Such familiarity is an obvious aid to both analysis and synthesis of control systems. Electrically. The object is to sum displacements. Therefore. + -En) R f R f R2 Rn A mechanical summer can be built using a “floating lever” or “walking beam” as shown in Figure F. = . we assume a linear approximation z=x+- dx Ir Ir where the bar-r notation means the derivative is evaluated at a reference position. where the voltage is practically zero because of the high gain A. . then the s u m is X+Y+W Z = 3 Another way of adding more than two quantities is to add them to the same beam. (F.2 Mechanical summer (floating lever). F. Unfortunately. F.5) and (F. changing one of the coefficients also changes the others. If the wobble plate is an equilateral triangle. but is reasonably accurate if the tilt angle is less than 30".3. z= b X+- a+b a a+bY For the special case where a = b we have z=X +Y 2 Obviously. In a similar way.l Electric summer using resistors and op amp. so this must be studied for each individual case.3) includes a term for each component.as shown in Figure F.6) should not be used if the beam becomes tilted. in which case (F. we can use a wobble plate to add three displacements.Control System Components 591 Feedback output Inputs Summing Junction 4 I % K Practically Ground Potential Fig. . X Z V Fig. Therefore. F. F. adding currents entering the summingjunction we have E. This is due to the wide-band amplifying capability of the operational amplifier and the fact that s = S +j w is in the numerator. Since all electronic equipment generates a certain mount of noise. this circuit will not perform well due to the amplification of noise. this circuit is not practical and is usually avoided. which may lead to serious error. the addition of mechanical hardware can cause problems of friction and backlash. Unlike the electronic summer.8) where a separate beam is used for each partial sum and still another beam for the total. Still another way of adding more than two quantities to break up the sum into partial sums. Feedback C I I v output 'Summing Junction Fig. Angular addition of two quantities can be performed by a mechanical differential gear arrangement. and resistance networks. difference amplifiers.2 Differentiation Differentiation would seem to be possible in an electric network by using the technique shown in Figure F.g. = -(RfCs)Ei (F. e. Other electric summers include transformers.4 An electronic differentiator.592 Appendix F Y Fig. where 1 2. any high-frequency noise (large o)available at the input is amplified at the output. = cs Then. 10) which is obviously a differentiation of Eimultiplied by a negative constant.3 A mechanical summer for three variables (wobble plate).4.3] F. . However. Therefore.9 z = u + v f X + y = (u + v) + (x + y ) P. Many of these schemes are described in the literature [2. Various electrical and electromechanical circuits for approximate differentiationhave been proposed [2].: 3 Ts 1 + Ts .Control System Components 593 0 Fig.5 Mechanical position differentiator (for low frequency). 11) Ys () -=Xis) IfTs4 1 . The transfer function of this device is found from the differential equation Mj. we get . Usually. we can solve the system equation by integration rather than differentiation and this is recommended. In fact.K y which.6.3 Integration Integration involves none of the problems of noise amplification present in the circuit of Figure F. The usual way of doing this is by means of the circuit of Figure F.j ) .Ts (F. = B ( i . One method of strictly mechanical differentiationat low frequencies is the dashpot. with M sz 0 and T = B/K becomes (F. Adding the currents entering the summingjunction. integration tends to smooth any input disturbances and is an operation ideally suited for electronic simulation.5. 14) F. 13) and y(t) = Ti@) (F.4. 12) Y(4 (F. F. shown in Figure F. Obviously.6 An electronic integrator. this slight movement x1 will cause the piston to continue its motion. By graphical integration. although not as a linear function of x. if the pilot valve is opened a greater amount. and has a gain of 1/RiCp A good example of a mechanical integrator is the combination of a pilot valve and a piston. as indicated by the minus sign. 14) we see that Fig. Suppose the pilot valve is lifted an amount x1 above its neutral position. in the s-domain Kx(s) Ys = () S (F.7. Thus. F. traveling at some given speed. where Kx. is the velocity. except for small displacements. the piston will travel a distance Ay = Kxldt. Ri Summing Junction Fig. as shown in Figure F. the high-pressure hydraulic fluid will flow through this pipe and push against the piston. F. Its operation is explained as follows.15) This integrator is inverting. in each increment of time dt.7 Mechanical integrator. Unless the piston reaches a stop. we have y(t) = Kfx(t)dt 0 (F. as shown in Figure F. As this opens the port to the pipe connecting the pilot valve to the piston. El3 output E.= - -Ei RiCfs (F.14) Rearranging (F. compressing the piston spring.594 Appendix F Feedback c (k r \ E. the velocity will be increased. 13) or. .8. say from 10” to The circuit for doing this is shown in Figure F. A mechanical stroke amplifier is shown in Figure F. Here.9 where (F. Using a high-gain operational amplifier.19) Fig. F. . from which we can write b Y(s)= -X(s) a (F. 15) or the speed of y is proportional to the displacement of x . 16) we have (F. F. We can write Jcj = T. it is desirable to produce gain in mechanical devices. is the s u m of all torques acting to accelerate the shaft. 16) where T. the moment of inertia is the gain constant. Another familiar example of a mechanical integrator is a rotating shaft such as a turbine. W ) Y(0 = (F. it is quite easy to produce gains over several orders of magnitude.4 Amplification The amplifier is a common device in electrical technology.17) Another example of an integrator is a steam pressure vessel in which the steam pressure in 1 the vessel is the integral of the algebraic s u m of steam flows into the vessel [1 . 18) In many cases. (F.9 A de voltage amplifier. Transforming (F. F.Control System Components 595 0 Fig.8 Graphical integration. 10. only the stroke or displacement.12 [4]. 11 will typically amplify the energy level by 1000:1 or so and can be used to drive substantial loads. A mechanical power amplifier. We may analyze the system of Figure F. which also uses power from an auxiliary (+B) supply. The device in Figure F.19) By inspection of Figures F. 12 we write G . The output Y follows a change in Xposition with a time lag. Usually. 10 A mechanical stroke amplifier. is usually called a servomotor or a mechanical-hydraulic amplifier.8 and F.(in) Fig.. which amplifies both stroke and force. = -R . under pressure from an auxiliary power source. 11 is called double-acting since the two control “lands” of the pilot valve simultaneously control fluid flow to and from the opposite sides of the piston. as shown in Figure F. 11./E = X The pilot valve transfer hnction is b XY=O a + b 5- (F. Note that theforce is not amplified in this device. such as oil. F.. uses hydraulic fluid.ll A mechanical-hydraulic power amplifier or servomotor.20) G . The servomotor pictured in Figure F.21) w. This is analogous to an electronic amplifier. Such a device. . the mass of the moving parts is low compared to the force available such that the response is quite fast. F.11 according to the block diagram of Figure F.596 Appendix F Fig. By inspection we write [2] (F.Q” =Q 2E 0 (F. This relationship is illustrated in Figure F. F.22) VO -SAP@) 2B or = Qxs) (F. QV is the valve flow in cubic inches per second. -- AP (F. The leakage coefficient of the valve is defined as the change in flow per unit change in pressure [4].Control System Components 597 Fig.13.24) where AP is the change in pressure on either side of the piston in psi.23) (F. F. B is the bulk modulus of elasticity of the fluid in psi. Vois the fluid volume at zero pressure differential in in3 and Q. Calling this leakage coefficientL. . where Qo is the average flow gradient for small displacements.12 block diagram of the power servomotor. for constant E. and E is the valve displacement in inches. is the compressibilityflow. inches Fig. we have. =L-in3/s psi Transfer function G3can be derived fkom the fluid compressibilityequation [4] H 2- Q.13 Valve flow curve for a pilot valve. 30.11 and F. we compute H I which gives the relationship between valve displacement and piston velocity at zero feedback [4] or AY = QP or HI=- QP Y =AS (F.20) through (F.30) where the servomotor gain is K = -GI . 11. we compute.25) G --=- Y 4- AP A Ms2 (F. where the oil force on one side of the servomotor is replaced by a strong spring. 14.33) and . A servomotor can also be constructed as a “single-acting” unit. by inspection F Combining (F.29) If the mass M is small.27) From Figures F.26) Finally. but in this case -b K=a (F. as we have assumed here. and Qo = Wp+as in Figure F.32) with A = A. then (F. Then My = F = A .12.28) we get ac (F. Y has the opposite direction of X The transfer function for this configuration is given by equation F.bd = H3 ac and Tis the servomotor time constant (F. as shown in Figure F. (F.598 Appendix F We find G4 from Newton’s Law.28) Y(s) GJH. In this figure.3 1) (F. Consider a force F acting on an area A with a small change in pressure AP. Af’ or (F. 30 and may be represented by the curves of Figure F. 14 and the mechanical integrator in Figure F.35) I I I I T > Fig. Note that this is not the response for the electronic amplifier in equation F.7. I5 Step response of the servomotor. The difference is clearly the presence of the mechanical feedback linkage such that the amplifier finds a new equilibrium position corresponding to a new input position x.14 A single-acting servomotor. F.9 with a parallel R-C combination such that a 4 (F.15.9 slightly to obtain a first-order delay similar to Figure F. If we replace the feedback resistor in Figure F. We may change the electronic amplifier of Figure F. T= a + b 44 Note carefully the difference between the force-stroke amplifier of Figure F.34) 5=1 + RCs R (F.Control System Components 599 5 Fig. where there is no delay indicated. Recall that the integrator continues to drive the piston for any pilot valve displacement until the pilot valve is returned to its neutral position.15.17. . The response of the servomotor amplifier is given by equation F. F. 16. then (F. F.36) which is comparable to (F. F. 16 An electrohydraulic amplifier. with the device response shown in Figure F. 100% Output Step Input(e.600 Appendix F Input Fig. Such a device is shown in Figure F.) / 0 100% 4 Fig.17. . this is a higher-order response than the first-order lag shown in Figure F. Clearly.17 Response of the electrohydraulic amplifier.15.30) Eggenberger [ 11 also gives an example of an electrohydraulic amplifier that can be used to drive large loads such as steam valves. as shown in Figure F. Here.19 Response of the circuit of Figure F.20 Response of the circuit of Figure F. 18.19.Control System Components 601 Ei(+) . F.20. if either.18 for EL < 0. will be the greater (less negative) of either EL(-) or -(Rf/Rl)E.5 Gating A gate is a device that makes a decision as to whether a signal should be passed or not. Ri Fig.(+). which illustrates a “low-value gate” device. should pass the gate. or that chooses between two eligible input signals to determine which. . Reversing the diode and the polarity of ELgives the response shown in Figure F. Thus. . 18 with diode reversed and EL > 0. F. it is assumed that El is positive and E L is negative. “auctioning off’ the output to the highest (or lowest) bidder. F. Then E.18 Electrical low value gate. t” Fig. F. it is seen that this circuit has the ability to select between Ei and EL. tEo Fig. This can be accomplished in an electric circuit by a scheme such as that shown in Figure F. can be used to control Y providing that X. a transducer is useful over a limited range and these limits must be compatible with the normal operating range of the quantity to be measured. Many other gating circuits are possible and such circuits often contain diodes. .X. X controlling) (b) Mechanical Overriding Device (Double-acting relay.21. In some applications.is between its maximum and minimum limits. which behaves in a certain way up to a limiting value. to the measured quantity. where both inputs X.21 Mechanical gating devices. Another useful device is the comparator. If X. F.. the value of EL is fixed and the circuit is called a limiter.19 and F.is outside these limits. Usually.20 are possible.602 Appendix F (a) Mechanical Overriding Device (Single-actingrelay. F. Both limiters and comparators could be used as ovemding gates in the sense intended here. Xcontrolling) Fig. see [5] and [6]. Such a system is shown in Figure F. then XI has no control over the variable Y. or some other nonlinear elements.g.6 Transducers A transducer is a device that measures some quantity and produces an output that is related. in a useful way.Other circuits with characteristics similar to Figures F. Gating can also be accomplished using hydraulic-mechanical controls. and X. Many references in the analog computer field give examples of such circuits.can be either control signals or limit signals. e. Zener diodes. In both systems. then changes state and acts in a different manner. 22. when changes in speed are small. since the generated emf (the rms value) varies directly with speed. The “output” will usually be a mechanical position or a voltage.37) Actually. as shown in Figure F.23 is single-valued in the range of interest (n > 0) so that the use of (F. our treatment will be confined to components used in power system control. An (F. Probably the oldest and best method know for measuring shaft speed is the flyball governor shown in Figure F.23 (also see Appendix C).Control System Components 603 In many cases. Moreover. if within specified limits. by the expression Ax -=K. Space does not permit an exhaustive survey of all known transducers. An example of an electromechanical speed transducer. We can approximate the transfer function of this device. the characteristic is not linear. F. F.37). which is convenient is some cases.26. but quadratic. However. is the permanent magnet ac generator as shown in Figure F. the transducer will be designed such that its output varies linearly with the measured quantity. for small parameter changes. the characteristic of Figure F. as shown in Figure F. One advantage of this device is its linearity.24. A com- Limit Fig.6. the error in assuming linearity is not great and the approximation of (F.25. .22 A mechanical speed governor. An electromechanical scheme is the magnetic pickup device shown in Figure F.37) is adequate. even though technically incorrect. will always generate an error signal of the correct polarity. Here.1 Rotational speed transducers (tachometers) It is very important to have a simple and reliable measure of the angular velocity of the generator shaft so that frequency can be closely monitored and controlled. 26 Another important speed transducer is the shaft-mounted oil pump. If a gear-type pump is used. bination of these last two devices is also possible.604 Appendix F Speed n (units) Fig. a square root characteristic exists between flow and pressure drop.40) Permanent - Fig. Thus. or Q = kP < .38) When discharged through an orifice.24 Permanent magnet generator speed transducer.23 Characteristicsof the mechanical speed governor.39) P = k2n2 (F. The oil discharge of the pump is directed through an orifice or needle valve. the flow of oil will be directly proportional to speed. F. as in Figure F. or Q = kln (F. F. wherein a frequency of the PM generator is sensed and converted to a voltage. we have the relationship between speed and pressure (F. . 26 Magnetic pickup speed transducer.25 Characteristicsof permanent magnet generator speed transducer. . F. F. Fig.Control System Components 605 0 1 Fig. F. Magnetic --kP pi u c 1 &J ~ Voltage Converter Wheel %Tooth Fig.27 An oil pump speed transducer used as a governor. 606 Appendix F Fig. two secondary windings. which moves axially. F. which is located in the center (axially). One common way is to use a potentiometer.2 Position Transducers It is often desirable to convert a mechanical position into an electrical signal. The magnetic circuit is excited by the primary winding. and a movable magnetic core. but is a stair-step function. depending on the potentiometer design.27. which we can linearize for small changes. F. The windings are concentric about the cylindrical core. as shown in Figure F. F. the resolution is finite and this may be a problem for some applications. There are many ways of doing this. The movable core provides a flux path Fig. Another useful position transducer is the linear variable differential transformer (LVDT) shown in cross-section in Figure F. This technique can be used to indicate translational or rotational position and can be linear or nonlinear. The main advantage of this type of device is its simplicity and low cost. If the potentiometer is wire-wound.29 Cross-section of a linear variable differential transformer (LCDT).28. This device consists of a primary winding. A typical oil-pump governor arrangement is shown in Figure F. In this case.28 A potentiometer used to indicate position. the transfer function is not a straight line as in Figure F. .29 [8].6. as indicated in the figure.28. and e2 with polarity such that the connection shown gives the difference. F.Control System Components 607 + J O T ..) - xN e.30 LVDT demodulator and filter [8]. we require a demodulator.Moving the coil toward one end increases the coupling to one secondary and.. . which is proportional to displacement. Thus.: F:+ U Fig. e. i. el = e2. .e..3 1 LVDT transfer function. each secondary is equally coupled to the primary and the induced voltages in the secondaries are equal.e. reduces the coupling to the other.. rectifies e. simultaneously. To convert the secondary voltages to dc. F. i. This device. movement of the core to the right will result in el > e2.41) 7-"I Valve 9 e A LVDT Core Position (in.eb = -Ki (F. When the core is exactly in the center. in Figure F. shown in Figure F.29.e o dt+Dm d ao k . for magnetic flux to link the primary and secondary coils. (volts) I I I -$ I -4 I 1 I 1 I * o \ ' +h 1 $8 Fig.30. . change in capacitance with change of plate spacing. and many others.32. piezoelectricity. For example.3 1 shows the LVDT transfer function. (F.32 A mechanical pressure transducer (for low pressure). i.e.43) where P is the pressure. F. A Ax=-AP G (F. magnetostricton. change in resistance with strain. F.608 Appendix F The final stage in Figure F.3 Pressure transducers Pressure transducers can be either mechanical or electrical. An electrical pressure transducer makes use of the LVDT shown in Figure F. say 100K. A is the effective bellows area. For small displacements.. the output of which is loaded into a loading resistor. Some of these devices are useful over a very small range of displacement [3]. Other translational and angular position transducers are available that utilize different principles.30 is a low-pass filter.6. and G is the spring gradient. Our concern here has centered on devices usable over relatively large changes in displacement. that is. Note the linearity of the device and the fact that the resolution is infinite. the output can be either a position or a voltage. the change in output Ax is proportional to the change in pressure. where the output voltage change may be written as AV=KAP (F.42) Figure F. A common mechanical pressure transducer is the spring-loaded bellows shown in Figure F.44) where K is a constant depending on both the LVDT characteristic and the Bourdon tube charac- Fig. where X is indicated as a steam turbine valve position and shows typical values of parameters used.33. such that ep = -K. 35. This nonlinear function generation can also be accomplished in the feedback path.34. Thus. and permits the linearizationof steam flow using a nonlinear compensator. in Figure F. There are many mechanical function generators in the machines of industry. We compensate for this by opening the valve faster at large values of stroke Y2. the stoke Y2 opens the valve according to the curvature of the cam. A cam is a function generator as it determines the position and velocity of a valve as a function of time or as a function of the control stroke. F.33 An electrical pressure transducer (for high pressure). where a given nonlinear characteristic is duplicated by an electronic simulation. However. down to almost zero pressure. In this particular case. teristics. F. the steam' flow saturates for large values of valve lift. as shown in Figure F. .7 Function Generators Function generators are rather common in analog computer work. This gives the valve lift L a nonlinear characteristic. A few examples will illustrate the use of such function generators in the control scheme of a steam turbine. F. it must be mounted where vibration will not produce noise in the output.36. This transformer is very linear. as 0 0 0 0 0 0 " output Fig.Control System Components 609 Fig.34 A cam as a function generator. as shown in Figure F. Fig. F. F. Yz Fig.36 Camshaft and valve function generators.35 Typical valve lift vs. Fig. stroke nonlinearity.37 Mechanical function generator in feedback (intercept valve relay).610 Appendix F Valve Stroke. F. . . F. F. Final (Next Valve Starts Opening) Actual Valve Characteristic ~ 0 Valve Lift 100% Fig.38 Block diagram of mechanical intercept valve flow control using a feedback function generator Valve Steam Valve (Electrical Cam) Demodulator RAM - Steam Flow .40 Approximation of valve characteristic by electrical function generator (utilizing two slopes).L IL----I Fig. PlV Feedback Cam Fig.j .Control System Components 61 1 Intercept *. v aive InterceDt Valve Relay Servo Motor .39 Electrohydraulic valve flow control with feedback function generator. F. 37. (the output in Figure F. The nonlinear feedback path tends to linearize the p. but one easy way is that shown in Figure F. There are several ways to do this electrically. the nonlinear “valve” characteristics must be simulated electrically. where the feedback signal is electrical rather than mechanical. such as diodes.41 Example of an electrical function generator in a feedback circuit. publicationGET-3096A. 1958. An electro-hydraulic valve controller is shown in Figure F. General Electric Co. Note that the feedback cam has the same nonlinear characteristic as the intercept valve. giving the flatter characteristic of Figure F.39. . New York. M.. Suppose the desired curve is similar to that shown in Figure F. As the input stroke Y.32). but not linearly.4 1. A. If greater accuracy is required.“ versus Y. The notations in Figure F.40.LVDT .= POSITION RAM Fig. In block diagram form.40 and the representation is to be as shown. where two straight lines are used to approximate the curve. the valve is an intercept valve that is operated by stroke Y. (a negative value) the current flows through the initial slope and final slope resistors in parallel.61 2 Appendix F SERVO AMPLIFIER - TO SERVO VALVE ( DEMODULATOR . 1967.38 refer to Figure F. shown in Figure F. J. 2. this situation behaves as shown in Figure F. Introduction to the Basic Elements o Control Systems for Large Steam Turbinef generators.. several break points can be incorporated so that the straight-line segments become shorter and the functional representation more precise. McGraw-Hill. These same ideas can be used in electromechanical systems in which an electronic simulation of the nonlinearity replaces the cam. calling for additional output Y. Basic Feedback Control System Design. Until the voltage Ed-) becomes as negative as the value set as the break point. once the break-point voltage is reached flows through the initial slope resistance R2. C . Savant.38. This is usually done using several straight line segments and nonlinear elements.37. Jr. References 1. all current However. the feedback position F is increased. Eggenberger. Thus. Here. F... increases. Ramo.Unpublished technical notes. Introduction to Analog Computation. J. and H. 1962. New York. 1961.Control System Components 613 3. 4. 6. 1967. E. Bragge.. E. 5. HandbookofAutomation. E.. S. M. Design of Hydraulic Control Systems. Systems and Components. McGraw-Hill. Westinghouse Electric Corp. E. New York. Servoactuators. McGraw-Hill. New York. E. 8. Fundamentals of Turbine Speed Control. Ashley. Private communication. Lewis. . Stem. Computation and Control. 3. Elliott Company. Shigley.Wiley.1963. J. Woolridge. New York.Wiley. Simulation of Mechanical Systems: An Introduction. 7. Bulletin H-21A.. and D. R. v. Feedback lever L . In doing so.l(b) are related to physical quantities shown in Figure G. 1. q. chanical differentiation and is called reset control. pp = 0 *This analysis follows closely that of Eggenberger and Callan.2. I). The multiplier of Figure G. and the temporary feedback. it will be convenient to refer to a typical physical system that exhibits some of the features under discussion.l(a) will be eliminated by mathematical manipulation. 614 . 7. pp. Pressure Regulator. for zero reference.29. The transfer functions for Figure G. This amounts to a mepreset needle valve KNv.l(b) will be derived. where system variables are defined both by name and by symbols. such as the turbine-following system of Figure 11. The variables defined in Figure G. 1. The constant 8 is a droop constant fed back mechanically to stabilize the system.* The block diagram for such a control system is shown in Figure G. We compute.appendix G Pressure Control Systems Pressure control systems. vl8. error is applied and with no feedback. Such a physical system is shown in Figure G. C. ref. including the pressure-sensing bellows. have been analyzed from a control viewpoint.3. go through full or unit stroke if a rated pressure . This relay piston operates the force and stroke amplifier to obtain the stroke np (not shown). which operates the pilot valve input to the relay piston integrator (see Appendix F). Feedback lever L2 produces a transient droop that is gradually reduced to zero by controlled leakage through a which equalizes the initial pressure difference.3. produces the steady-state droop by acting in opposition to E (negative feedback) with the droop adjusted by changing the lever arm as noted. It consists of a summing beam B (see Appendix F) on which several forces act.All forces are summed with the correct algebraic sign to provide an output. the reference. E. Three transfer functions for pressure regulation are used: (a) Proportional control is represented by the block diagram of Figure G.l(a). the steady-state feedback. where 1/G is to the time it would take the output (stroke) 7. Thus. where the system is arranged to slowly reset itself.2. G..4. 1 A turbine-following representation.2 is inactive for proportional control and the needle valve is open. KNy= 03 (b) Proportional plus reset control is represented by the block diagram of Figure G. where K = GT. TL is fairly large (a few seconds) and is adjusted by setting the needle valve KNy Figure G. Le. in We compute which we simplify to GR= We have defined G(1 + TLs) ~ (+ GTL6 + TLs) l . TR = 1IG6p The temporary feedback loop in Figure G.Pressure Control Systems 615 Pressure Regulator Servo Motor - Steam Vessel + (a) Identification of System Variables (b) Identification of System Transfer Function Fig. (c) Proportional plus partial reset control is represented by the block diagram of Figure G.5.5) Here. TL is defined as before and two new time constants are defined as follows: Fig. G.3 Block diagram of a regulator for proportional control.. where the transfer function is given as G .2 A typical pressure regulator.61 6 Appendix G Fig.711 = . G.s) TLs2+ [ 1 + G(Sp + SJTLS]+ G S p G(l ((3.$ + T. C TL = KNV where C is a mechanical constant and KNvis the flow factor (in3/sec-psi) for the damping needle. . . and CM We begin by assuming the flow through the valve is proportional to the product of the equivalent valve area and the pressure: Fig. and is defined as 3.Pressure Control Systems 617 Fig. By proper choice of the several parameters. Hydraulic Servomotor.4 Block diagram of a regulator for reset control. this type of regulator is adaptable to many applications.5 Block diagram for a regulator with proportional plus partial reset control. G. is shown in Appendix F. Steam ValveSteam Flow. 13. TR2 = . Gh The transfer function of a hydraulic servomotor of a force and stroke amplifier. 2.1 OR2 where the two frequencies are defined according to the choice on the sign of the second term. G. e. 4. 12) GAGM(P. We represent the steam-volume portion of the system by the block diagram of Figure G. imately equal to pi.6.H=O and the loop time constant is Fig. 13) where Tvis the characteristictime of the steam volume. we write the differential (G. The steam vessel or drum ahead of the control valves acts as an integrator.lO) Since under normal operation the pressure is at nearly rated value. pi.H=l p=O. for p=l. 11) (G.. being fed into the steam volume. . P = PRand the first term in (G.9). Thus. is constant and is independent of pressure.618 Appendix G M =A P l b d s (G.9) This assumes that the equivalent valve area has been linearized in the valve drive cams or in the valve itself. where the feedback hiction H(a)is approxi. We would like to eliminate this multiplication and to linearize equation (G. By definition Since a unit change in q2produces a unit change in a I . Steam Volume We assume that the steam flow. 10) may be evaluated at rated pressure. where (G. any flow in that is not balanced by flow out of the drum will increase the pressure at a rate given by the integrator gain G. G.6 Block diagram for flow-volume-pressure relationships. l(b).)= 1 The change is Ap caused by d$l can be introduced at the summing point as shown in Figure G. To do this. GA = 1 Therefore (G. 16) Combining all of the above. .14) (G.8 Bode diagram of a proportional-pressure control system. the block diagram for a turbine-following system with proportional control is given by Figure G. or (G.7.17) +90 Fig.29 solves this system using Bode diagrams with the result shown in Figure G. A larger regulation..8 for typical values of the time constants. T V T= H For 0 < H < 1 the transfer fhction is given by (G. Reference 11.29) points out that. makes the system more stable. (3.15) Reference (1 1. S. but results in a greater steady-state error. Recalling that the steady-state error is defined as [26] Kv = lim sKG(s) = lim s s+o s-to K3/& Tv =. G. we may assume this to be an integration. The quantity most easily changed is S f .K3 ~ (+lTp~)(1 T ~ s ) SpTv + (G. in most cases.7 Block diagram for proportional initial-pressure control with a large steam vessel (Tv % 1).Pressure Control Systems 619 P P ++ EP +4 Wl l/ijp 1 TRs I Fig. respectively.0 (b) Bode Diagram for Proportional Plus Reset Control Fig. (a) Block Diagram of Proportional Plus Reset Control +40 +90 e 2 +0 +45 j . Kv. G. 2 0 0 9 3 h s -45 g-20 40 0. the results are changed as shown in Figure G. 17).& 4i. These systems could also be analyzed by root locus and this method is recommended to the interested reader. Stability of depends on the gain.9 Proportional plus reset pressure control.33 'eo a.0 10.ET z .r a d s 5.lO.9 and G.620 Appendix G The system is type 1 [26] and has a steady-state position (pressure) error of zero. If either proportional plus reset control or proportional plus partial reset control are used. where typical values of constants are used. (G. . Pressure Control Systems 62 1 (a) Block Diagram of Proportional Plus Partial Reset Pressure Control Diagram +90 B 4-45 -8 0 [ d 'g E -45 a J -90 a. . G .10 Proportional plus partial reset pressure control system.radiands (b) Bode Diagram of Proportional Plus Partial Reset Pressure Control Diagram Fig. the balls are thrown out. where two flyballs are held to a rotating shaft by rigid arms L. 1. H. 1. It is presented as a background for the material for Chapter 10 and forms a basis from which simplifying assumptions may be made for physical systems.1 The equilibrium equations To analyze the forces acting on one of the flyballs of Figure H.1) *The development here follows closely that of Pontryagin [8]. holding the ball of mass M. Consider the flyball governor shown in Figure H. 1 The Flyball Governor. some of it quite elementary [l-71. 622 . and L2and further restrained by a spring K. = Spring Force due to Spring K The difference between centrifugal force and spring force is the net outward force Fc or Fc= Fd . As the flyball rotates at angular velocity o (radians/second). This appendix explores the governor equations in greater detail than is usually needed for linearized control.appendix H The Governor Equations Considerable literature exists on governors. Only a few references provide a more rigorous analytical treatment [8. r ure H. the vertical position of the collar C from some stationary reference is a measure of the angular velocity and a mechanical linkage attached to C could be used to control the throttle of the prime mover.9]. the ball is acted upon by three forces: Fg = Gravitational Force (weight) Fd = Centrifugal Force (weight) F.Fs From classical dynamics we write (H. As the angular velocity of the shaft increases. Under these conditions. refer to the sketch in Fig. such that the collar C slides upward on the shaft.swings out to radius R and assumes an angle 4 with the vertical rotating shaft. Thus. providing the force available is sufficient to move the throttle lever. the a m of length L. 1.2. H. 1 A simple flyball governor. H. Fs f Fc Fig. H. .2 Forces acting on the flyball.The Governor Equations 623 Fig. then we may rewrite (H.7) we get Fc = mLo6 sin 4 .. = 2K(R . where we define a = mw&-2K b = -2K sin $ .) where R. I).7) F. mo&.(mL sin L sin L sin 4 Also. In terms of the angle c$ we note that R = L sin 4 or F&= mv2 = -. (H.mg (H.2KL)cos r#~ . we can write (H..12) -2K sin 4.6) (H. is the angle corresponding to R. The forces permendicular to the arm L are defined as Fp.2KL sin C#J + 2KL sin 4.624 Appendix H where v is the peripheral velocity of the ball. is awkward to solve. &)os we have F = mLo&sin 4 d For the spring force. Then cos 4 = a (H..lO) as Fp= mLwi sin 4 cos 4 . cos 4-mgsin 4 If the system is in equilibrium.lO) COS 4+ 2KL sin 4. is the unstressed length of the spring.R. cos C#J for which the equilibrium condition is tanf#l= (H.3. where Fp=FcCOSfp-FGSin or Fp=mLw$ sin +cos 4-2KL sin 9 (H. unfortunately. and (H. 11) which. For the force FG we have the familiar expression for the weight of an object FG m g = where g is the acceleration of gravity. If the spring is quite stiff and it overpowers the gravitational effect. Combining (H. (mLw&.15) . then Fp= 0 and we compute the relationship tan f#J= -2KL sin c$. writing v as a function of R and )oG. 14) v2TP (H.2KL sin 4 cos 4 + 2KL sin 4.mR20& . 13) This can be viewed as a right triangle as shown in Figure H. (Ha where 4.6).2K (H. = 0 (at then this simplifies the equilibium condition for (H.. 18) In any case. 17) If there is no spring at all. +1 4 P m2 + sin2 4.-- cos 4 = 4Kw. 2K cos 4 = w. we assume that the spring has an unstressed length R..4.3) for R and defining A =L sin 4 we have (H.(L cos 4 + L d A 2 .sin2 4) .The Governor Equations a 625 b Fig.-- 2K m 2K m2 (H. we obtain 4 as a function of w.) 1'" (H.lO) such that (H. g (H. we get w. then K = 0 and we have cos 4 = LO.l we note that an angular displacement 4 results in a linear displacement of the collar c.. From Figure H. 15) or w. 16) c$u = 0) If. H. on the other hand.20) where R Substituting (H. or. This is shown in Figure H. 14) 2K i -sin 4.. where we note that X = d .3 Definition of the angle 4. -- wi-- 2K m (H.2 1) = LJL x = d . by trigonometric maniputation cos 4= Factoring the numerator.(a + b) 04-19) or x =d- (L cos 4 + W) (H. .23) + X A = d . Furthermore.26) . = L this becomes x =d-2L Thus.626 Appendix H Fig. . a displacement force due to the position of the mass (or the angle 4) at any time.6 = mLw2 sin 4 cos 4 . Any acceleration of the mass M is governed by Newton’s laws and the equations describing the system behavior are differential equations.. There is also a viscous friction where B is the viscous force acting to retard the motion and this force is usually depicted as Bq% constant. 1. given by Fp in ( .T. and electrical (load) torque T.2L COS(+^ + 4 ~ ) XA = (2L sin &J+A from which we compute (H. we must include all forces actH 1 ) is ing on the mass M. Combining all forces we write the following equation.mg sin 4 . this is a dynamic problem..25) Now. m. we have concerned ourselves with the “static” governor equations. for a turbine angular velocity o we can write J h = T. Then. the equations derived for cos small displacements XO COS 4 (H.. that is.2KL sin 4 cos 4 + 2KL sin 4. For (H.22) 4 may be used as a proportional measure of x.Bq% (H. H. = T. If L.24) H.2 The dynamic equations Up to this point.The force (perpendicular to L). the equations based essentially on constant speed. (H. of course. cos 4 . 0 . considering m to be a point mass.4 Relationship between 4 and x. Actually. suppose the turbine-generator has moment of inertia J with mechanical driving torque T. the rotor speed o must be constant. i. we note that the governor stroke.COS 4 0 ) (H. Also. cos 4 m m (H. T. T. ignoring any delays in converting governor stroke to mechanical torque.3 1) These equations are the state equations for the system. is the accelerating torque.The Governor Equations 627 where T.24) and (H.T. a state of equilibrium exists where +. Thus.\ 1 Fig.L 2ru +. T. However. . (H. This explanation ignores the delays in servos and steam systems. H.22). We now define a constant F as follows: F = T. is a function of cos Thus.. Note that. as 4 increases..e. thus giving constant governor speed @Nandconstant governor angle Thus.o + kcos 4 0 (H. write +. *= (i. (H.27) From (H.. as the speed decreases.. is increased by the admission of more steam as shown in Figure H.. the 4 direction to be $ i.30) we have a normalized system of equations as follows: $ = n2LoZ sin $cos 2K. the mechanical torque must be proportional to cos If we assume an operating angle +o at which point the we torque is Tm0.5 Mechanical torque as a function of angle I$. for convenience. WG = NO (H. there is a simple gear ratio N relating o and w ~ . I I 4 90’).o + COS .30) Combining (H. +.32) t I 0I 1 \ \ \ \ I 1 I I .28) decreases and vice versa (also note that 0 where k > 0 is a constant. we define the angular speed in .-sin +cos ++ -sin 4.e. = T.. decreasing 4.5. x.29) which is dependent on the load torque T... When operating at a constant load T. g sin + 2KL m + +A) .-(sin 2KL c $ ~.sin 2KL cos2 4 0 cos 4o m sin qj0 #o .sin r$Jcos +o 0 = ~ c o s 4 0 -F J (H.37) Equation (H. such as those involving squared variables. A23 = " 0 2gsin #o 4KL + -(sin .sin 4u)cos #o (H.32).3 I). Also (H. COS(C$~ 4 A ) ..35) to write (H.36) + -sin +..m+A B (H. etc.. cos #o m (H.3 1) by the substitution 4=40°+4A G=GO+$A w= "0 + (H.39) "Om The result is a linear system that is restricted to small deviations from the initial states.33) [ cos +o = F/k 2KL sin 4.38) where A21 = - gsin2 +o --. 0 = $0 0 = W-4jsin 4ocos #o .36) must be examined for higher-order terms.34) may be incorporated to give the result (H.33) we compute .g sin +o [ From (H. we learn more about the state of equilibrium by setting the left-hand side to zero and substituting (H.628 Appendix H From (H.34) We now linearize (H. which may reasonably be neglected. Recall from (H. ul‘2 ’ a3u0 (H.A2 + a. . by definition -Azlh kj423 + -sin+o J (H. defined by R.42) and incorporating (H.34) with the result (H. Rearranging (H. Now..45) This corresponds to the slope at a given point (wo.45) is a positive quantity. (H. therefore.The Governor Equations 629 It is instructive to examine the stability of the linear system (H. We call the system matrix A and compute P(A) = det A .42) Note that uj > 0.28) that F is a constant for a given value of 0.A1 = 0 where 1is the unit matrix.46) “t Fig. H 6 Location of the operating point on a torque-speed curve. if stability is to be assured.A + a. Thus we have the result (H.38). the incremental regulation computed by (H. but also.44) corresponds to a particular operating point on the torque speed characteristic of the prime mover.= ‘ dw - dF (H. The derivative (H.6. not only must all a’s be positive. This is the sufficient condition for stability [8]. . T.40) B P(A) = A3 + -A2 m or. These incremental changes on the torque-speed curve are referred to as the “incremental regulation” (incremental droop) of the prime mover.43) where these coefficients are defined above. by Routh’s criterion.44) where F is proportional to the load torque.33) we compute (H.. To) on the torque-speed curve as shown in Figure H. we require that. the right-hand side of (H.45) may be computed from (H. Since the slope is usually negative.41) P(A) = a3A3+ a. Rockford.46). Elliott Company Publication H-21A. f Elliott Company. . pared by Westinghouse Electric Corporation engineers. 9. 2. Feedback Theory and its Applications. Addison-Wesley. we have the result Ri (H. New York. Recommended Specifcation for Speed Governing o Steam Turbines. PA. Eggenberger. U. Private Communication. Jeanette. Macmillan. P. As seen from (H. 3. 6. Unpublished notes on Hydraulic Turbine Governors and Turbine (Steam) Lubrication Systems.. Part ZV.630 from which we compute . Governors. f 1959. The Controlled System. Pontryagin. 8. 2. 1962. Hammond. 7. Part IZZ. Woodward Governor Company PublicationPMCC 66-1. MA. A. References 1. 3. S. For 4u= 0. Floor. f Speed Governor Fundamentals. Introduction to the Basic Elements o Control Systemsfor Large Steam Turbine f Generators. Parallel Operation o Alternators. H. 1958. IEEE Publication 600. The Control o Prime Mover Speed: Part Z. City of Los Angeles. Illinois. Parts Z and ZZ. this depends on the values of K and and the spring can be either beneficial or detrimental.General Electric Company Publication GET-3096A.44). 4. Fundamentals o Turbine Speed Control.. Reading. Part ZI. J. Unpublished notes pre- 5. From (H. 25031. L. Department of Water and Power. B (friction) is essential for stability Large J(inertia) is beneficial to stability Large m (flyball mass) is detrimental to stability Stability is increased by increasing the regulation or droop of the torque-speed characteristic For a given system with fixed B. Governors and Governing Systems. and Supervisory Instruments. Mathematical f Analysis. Publication No.48) we may summarize our findings as follows: 1.. Woodward Governor Company.47) By factoring BJ/m from the left side of (H.7 Appendix H 2F wo 1 Ri Ri w 2K m (I-%)] iP BJ -> . The Elliott Company. IEEE. 4.. The Controlled System. and m.48) This is an important result and is the sufficient condition for stability. Private Communication. M. Ordinary Diperential Equations. the only control we have on stability is through the regulation. a large K is detrimental to stability.1 m '[ (H. are given in Table I. which is recommended for further reading on the subject.2) Pc = r[(H + dH) . then the fluid at face C has area A + dA.3).Z) lbf/ft2 where H i s the head and Z the height at B. 1.(Z + dZ)] lbf/ftz where we compute and dZ=-sinadxft Substituting into (1.= g.1 Dynamic Equation of Equilibrium The dynamic equilibrium condition (F = mu) for an element of water dx long may be derived as follows. these equations are often called the "water hammer" equations. where The pressure also is different at the two faces. l. we get (1.appendix I Wave Equations for a Hydraulic Conduit The purpose of this appendix is to derive the equations for head and velocity of fluid in an elastic conduit.6) 631 . At C the pressure is (1. since they describe mathematically the traveling pressure waves in a conduit.together with the variable names. The derivation used here follows closely that of Parmakian [11. The resulting equations are very similar to the familiar wave equations used by electrical engineers to describe the voltage and current at any point along a transmission line. If the fluid at face B has area A. the acceleration of gravity Y P 1. In hydraulic applications. It will be convenient to recognize that . Consider two faces or sections along the conduit labeled B and C in Figure I. All variables used in this derivation. At B the pressure is PB = HH . 1 Variable Names Symbol A R Variable Pipe inside area Pipe inside radius Pipe inside diameter Pipe length Pipe wall thickness Modulus of elasticity of pipe material Bulk modulus of water Specific weight of water Fluid (water) mass density Acceleration of gravity Angle of slope of conduit Distance along pipe in direction of flow Head at any point x and at any time t Height above conduit outlet or gate Velocity of fluid Longitudinal stress in pipe wall Circumferential stress in pipe wall Pressure at any point x Force of fluid Dimenson ft2 ft A ft D L e E K g P g a! X H = H(x. we analyze the forces at faces B and C caused by the pressure acting over a given area. .Z) lbf (1. t ) z f t Ib/ft2 Ib/A2 Ib/ft3 Ibm/A2or lbf-s2/ft2 W2 S radians ft ft f t V U I a2 WS Ib/A2 Ib/ft2 Ibf/ft2 Ibf P F Finally. I.7) and at C Fc = y(A + Z d r ) [ ( H . At face B FB = ?A(H .632 Appendix I Table 1. 1 Sketch of conduit showing element of length dw between faces B and C [ 11.Z) + ($+ sin CY)&] lbf Hydraulic Gradient for Gate Closure Fig. 10) of which a fraction.2 Area. 15) Finally then (I. Pressure. as indicated in Figure I.Fc (I. Ibf/R2 Value at Face B A Y V . 14) (I. 12) so that we can write the approximate solution dH Fa = -YAP& ax (1. 1. and Force Quantities on a Differential Length dx of Fluid Quantity Area. there is also a force due to gravity.dV g or dt (I.3 {(H- Value at Face C dA A+-& dx + (g+ sin a)&] .8) are summarized in Table 1.13) But Fa = (mass) x (acceleration) = yA -&. the accelerating force may be written as Fa = (FB+ Fg sin a). In addition to the forces due to fluid pressure. 16) Table 1. Then (1.2. Thus. ft2 Pressure. which acts on the center of gravity of the element.Wave Equations for a Hydraulic Conduit 633 The quantities computed in (1. along the pipe longitudinalaxis.Fg sin a acts to the right.2)-(1. Calling this downward force Fg we compute Fg = YAc& where we take the area at the center of gravity to be A + (1/2)dA. 11) It is often assumed that A- dH r3x % (H-2)- dA ax lbf (I. Thus. .2. The element boundaries are B and C at time t as shown in (a) but have moved to D and F. This condition requires that all space inside the boundaries of the element be occupied by water at all times. 17) and the velocity at face F is VF= Vc + dVc = avc dV v + -& + -& + -dt avc dx dx dt dx dV =V+-&+dx ” (V + . Consider the element of water dx long as shown in Figure 1. 1. respectively.634 Appendix I which is one of the wave equations for the conduit and is derived from the equations of dynamic equilibrium for an element of water. V. in time dt. dX (a) At timet B D C F I ----av av V+-BD+-dt ax at (b) At time t +dt Fig. At time t + dt we may compute the velocity at face D as VD= VB+ dVB = dV V + -&+ dx dV V + BDdx dV -dt df = + -dt dV dt (I.2 The change in length of dr in time dt [ 13. and t is derived from the continuity condition.& c F)+ . x. 1. at time t + dt. B moves to D and C moves to F.18) These velocities are shown in Figure 1.2.( : V+5dx)dt (1.2 The Continuity Condition The second equation relating H. C F = ---dwdt The change in length computed by (1. a. if dL > 0. 2.3. and p as Poisson’s ratio. Now.22) 1. causing a further change (decrease) in length. 1.CF (I. the average velocity of face B in moving to D in time dt is 1 1 dV 1 dV =V+--BD+--dt 2dx 2 dt dV (1.Wave Equations for a Hydraulic Conduit 635 The change in length of the element is dL = BD . . as the circumferential stress.2. 1. then the change in radius may be computed as (1.3 A sequent of the pipe shell of length dx [2].21) Then. we can compute.as the longitudinal stress. Note that these two effects are additive. neglecting higher-order terms dL = B D . Since the water is compressible.22) is caused by two factors.20) The distance that face B moves in time dt is (1. dV dx (1. 19) where we note that. the element becomes shorter or compresses because of the way that dL is defined. a change (increase) in pressure causes a change (decrease) in the volume of water within the element. If we define a. The change (increase) in pressure causes the pipe shell to expand and causes dx to shrink in order to contain the same volume of water.1 Deformation of the shell A small segment of the pipe shell is shown in Figure 1.23) Stwss Center line axis of pipe Fig. .25) dL. It is apparent that. Three cases that are sometimes of interest are shown in Table 1. for Three Cases of Interest 1. then we can write new volume ./A)AcT~] + (1.27) The exact solution of (1.3 for steel pipes. We may also compute the change in length due to stressing of the pipe material as SX = -(A01 Cix E .91. the A quantities are changes in stress due to a change in pressure. Pipe with expansion joints throughout its entire length 2e yDdH yDdH yDdH ( 1.p2)dX Ee 0 yDdH 2e Ee . For example. and .23) and (I.29) Parmakian [11 gives examples to show that the results are nearly the same for all values of C .95.26) by incorporating (1. (2 . free at the other yDdH yDdH 4e PAUZ 2e 2.636 Appendix I where el2 is negligible relative to R. with ~ l= 0.3.L)Au.= -[(lE ~/. Pipe anchored at one end.old volume old area .n(R + AR)’(dx + &) . in any case.wR2& TR2 AR =Sx+2---& (1. we may compute the new volume of the element as New Volume = 7T(R+ AR)2(Cix + &) If we define the change in length due to change in stress as dL. we get dx dL. Table 1. we compute CI to have values of 0.. Expanding (1.24).0. = R (1.24) In both equations (1.26) with higher-order terms neglected. Knowing AR and 8X. we may write dL.23) and (I.3 Evaluationof dL.28) (1.27) depends on exactly how the pipe is anchored. Pipe anchored throughout its entire length 3.PAUZ) ft (1.24).= C 1 where yDdH dx Ee (1. in general.yDdHdx eE dHdx .30) This change in volume causes a change in length dL. we can set the two expressions equal and write (1.38) .22). C. dx (1.+ v- dH dt dH =--- 1 dV dx K .35) or (1.Wave Equations for a Hydraulic Conduit 637 0.(nR2)( ydH)dx R~ K K (1.37) Using this expression. Thus.ydHdx -t -- =(++A ‘ID K (1.36) Now define (1. 1.32) But His a function of both x and t so that dH = dH -dx dx -t dH -dt dt = dH dx --dt dx dt + dt (1. equal to (1.2 Compressibilityof the water The change in volume of the original length dx of water due to water compressibility under pressure change ydH is AV= (force)& .85 for the three cases. we could take C .3 1) Then the total change in length is dL = dL. + dL.9. we can write (1.(area x pressure)& .34) Since the change in length is also computed in (I.33) Then we may write (1. to be a constant somewhat less than unity.2.36) as . or about 0. 39) where (1. We compute v dV -dx V v dH =*--dH -dx a dt Now. the constant “a” may be evaluated for a given physical system and will typically have a value of fiom 2000 to 4000.a2 dV dt g dx + x=fat+k (1. It is sometimes convenient to write (1.638 Appendix I which is the second of the wave equations.42) dH The solution to these equations is well known and may be thought of as two waves traveling in the +x and -x directions at a velocity of a feet per second. This is 100 times or so the value expected for V.42).45) - =-gax (1.42) to write dV at dH -=--.40) Then we can write dH dH . so both quantities (1.46) . We conclude that -svdt dH -s v dt and we can neglect the second terms on the left side of (1.44) av dV dx dH dx dH (1. this one being derived from the continuity of water inside the pipe.+v-=--dt a* dV (1. This being the case. we may write This simple relationship helps us analyze the second terms on the left side of (1.38) in a slightly different way. Suppose we let (1.41) gdx The wave equations then may be written as dV dt v - dV dx =-gx (1.43) =f - VdV a at (1.44) have multipliers V/a that are very small. J. Prentice-Hall. WaterhammerAnalysis. The solution may be thought of as an incident wavef+ and a reflected wavef.( t + v. . 1955.or H-Ho=f+(t. Parmakian.vo = a Reference qf+( a) t- a) +f-(t + i)] (1.i ) + f .. New York.47) 1.Wave Equations for a Hydraulic Conduit 639 This is the more familiar form of wave equation and corresponds to a lossless transmission line. e. and a line to the load-a three-way configuration. 1. four-way spool valve shown in Figure J. Many valves. A = orifice area. These devices have two main mechanical components: a control valve and a piston. The newer designs incorporate electromechanicalelements to improve the speed and accuracy. H. The purpose of this appendix is to write the basic equations that describe the behavior of these two components and of the servomotor system. Consider a three-land. ft3/s C = dimensionless discharge coefficient . The flow past the spool orifices are given by Bernoulli’s equation* Q. Actual devices might be analyzed using different dimensions for convenience.appendix J Hydradic Servomotors The hydraulic servomotor. R2 *Dimensions of all quantities are given in a consistent set of units. such as the valve shown in Figure J. is a class of control devices that are used to move large loads with precision and speed. using A in square inches or metric units..g. Our analysis follows closely that of Merritt [I]. All are analyzed in a similar way. such as the mechanical integrator described in Appendix A. All valves require at least a supply line. are four-way valves. =CAI/= QZ = WZ/= = Q3 P where Q = volumetric flow rate.l. which is recommended for further study. 640 . This valve is described by four sets of equations that describe the flow and pressure relationships. a return line. 1 Control Valve Flow Equations The control valve or spool valve is usually described in terms of the number of spools or lands and the number of ways the hydraulic fluid can enter or leave the valve. often using the R-lbm-s system. Qz (5.Q4 = Q3 . The first simplification is to assume matched symmetrical valve orifices: Matched A1 =A3 A2 = A4 (J.(x. Thus. J..5) Symmetrical: A&) A3W = A2(-x) =A 4 6 4 We also define the neutral position area . 1-J.Hydraulic Servomotors 641 4 Fig. must be solved simultaneLe. we can write Finally. J. we note that the pressure drop across the load is given by PL=PI -P 2 (5.2) and these relationships are readily verified by examining the Wheatstone bridge equivalent of the spool valve in Figure J.4.l A three-land. The orifice area in each case is a function of the displacement x. lbf/fiz p = mass density of fluid. P. as a function of x and PLY Q. ously to give Q. = Q.). and P = pressure. lbm/ft2 or lbf-sz/fV The flow to the load can be written as QL = QI . 1.with appropriate simplifications. four-way spool valve [I].4) These four equations. i.2. we write or P . . Now.642 Appendix J Usually./2 V . we compute (J. el2 e P.e..8) = Q.10) (J.. and using (JS). A = wx where w is the width of the slot in the valve sleeve in ft2/ft (or in2/in).2 Graphical illustration of pressure division for matched symmetric orifices.11) These relationshipsare shown graphically on a pressure scale in Figure 5. el2 6 I $. From (J.4). we assume that orifice area varies linearly with valve stroke so that only one defining equation is required. 12) P. Drop Across 2 V Fig. for matched symmetrical valves Qi (5. = O J.9) Q2 = Q4 From the first equality.2) we also compute (J. (J.10) with (5. +P2 Combining (J. Across 1 Drop 4 P. 5. = P . We can use a Taylor's series expansion to write (J. however. 16) is obviously not linear. we can write (J. where the spool is displaced a small amount in the +x direction. as equation (J. J. Consider the spool valve shown in Figure J. 18) where Kq = the flow gain = K. =C ./F) (. 17) Thus (J.13) If leakage is neglected.Hydraulic Servomotors 643 Also. = the flow-pressure coefficient = -(J. Qs= Q I+ Q2 (5. we can write A ~ ( x ) = A ~ ( .18) is the desired relationship and will be used in eva.CA . 1. even though much of the operating range is reasonably linear. for any x we can write (J.16) Now.15) = wx (5. our goal is to determine a linear equation for Q L . from Figure J. Continuity requires that QI = Q2 =C ~ O /. There are obvious limitations that should be kept in mind.x)= lAll 11 x Thus.19) Equation (5.2 Control Valve Force Equations The equations giving the forces acting on the spool valve are developed for either a steadystate or a transient condition.3.14) For a symmetrical valve.Jating the sma signal behavior of the system.0 52) . The magnitude is given by . Memtt. The effective area. where we have defined the discharge coefficient as the product c. . where we have defined C. The direction of this force tends to close the valve./= (5. 0 1967 by John Wiley & Sons.. If the fluid element is accelerated in the direction of flow. Inc.15) to express A. as a linear function of x. Herbert E. From Hydraulic Control Systems. .26) This is a steady-state (Bernoulli) force that always acts in a direction to close the orifice. = c.23) The steady-state force acting on the spool valve is given by (5. for small x (J.22) Thus.3. we have devined A.25) Using (J. or in the -x direction in Figure J. The transient flow force is derived by considering the forces produced by accelerating the element of fluid shown in Figure J.= velocity coefficient = 0. due to flow contraction is given by [ 11 A2 = CJO (J. we write QI = Q2 = C+42. or the pressure at face a exceeds that at face b.3 in reacting with the face area of the spool. the pressure on the left must exceed that on the right. 5 3 Flow forces on a spool valve due to flow leaving the valve chamber. to be the orifice area.pl) ( P ~ COS Fs= K ~ x A e (J.24) which is a force normal to the plane of the vena contracta.98 (J.c. by . The force normal to the spool is given by F~ = F ~ C O S e = ~ C .644 Appendix J Fig. < 1.O) C. we write.6 < C.21) Also. = contraction coefficient (0. C ~ . 5.20) with the area expressed as a linear function of x.3 The Hydraulic Valve Controlled Piston A hydraulic valve controlled piston or linear servomotor is shown in Figure 5. The quantity L is called the damping length and is the axial length of fluid between incoming and outgoing flows. This is similar to the mechanical-hydraulic integrator described in Appendix F and reference 2. that equal pressure Fig. 8 . = Ma = pLA- 4QiW = pL-~ Q I dt dt (5. J.Hydraulic Servomotors 645 F.4 A hydraulic-valve-controlledpiston [I].27) Using Ql from (J.4. it is customary to ignore the transient force ( . . The second term is usually neglected.28) where P A= PI .Pz.This is simply in recognition of the fact that the valve transient period is very short compared to the load transient period. 52) In power system control analysis. we assume that the valve orifices are matched and symmetrical. Merritt [l] observes that the first term on the right side of (J. In our analysis. we compute (5.28)is the more significant as it represents a damping term. 646 Appendix J drops exist across the valves. ft/s2 p = density. fi5/s-lbf . . Thus. connecting line. from (J.33) may be written as dV V dP Z Q i n . and that the supply pressure. that the valves have equal coefficients.for small deviations.33) (5.+ g v dt dt But we can also write the weight flow rate as (5. Ibf/s2 g = accelerationof gravity.30) we can write dV dp ZK. fi3 Cip= internal cross port leakage coefficient of piston. we can write the continuity equation (5.-XWou. p. ft5/s-lbf C = external leakage coefficient of piston.=gp.30) where W = weight flow rate. For the piston chambers.Z Q o u t = . is constant. Then.. If we consider a contained volume V of mass m and density p.+ -dt pe dt which is a convenient form of continuity equation for this problem [ 11. is the effective bulk modulus (lbf/ft2)and P is the pressure. ft3 From (5.KJ'L (5.31) can be written as dV Vdp ZQin-ZQout=.18). and piston volume. QL = K$ . lbm/ft3 (or lbf-s2/ft") v = volume. We can also write a continuity equation for the weight flow rate in and out of the contained volume. we write the continuity relations (5.35) (5. ft3 V2= total volume of return chamber. at constant temperature Po p=po+-P Pe (5.+ -dt p dt Now. (5.P2 is the pressure drop across the load or across the piston.32) Then (5.34) where po is the density at zero pressure.36) where V . Ps.31) (5.29) where PL = P I . = total volume of forward chamber including valve. = v. VO2 initial volumes.. we can show that the last term on the right side of (5.(5.43) M f y = -Ky .39) (5. lbf-s/ft K = spring constant. we have three equations that describe the servomotor behavior. = load force.37) and substituting into (5. we subtract these equations and divide by two to write Using (5.=2vo Taking derivatives of (5. ft3 = and assume that [I] v.=v.. (5.+v.. lbf/ft F. using PL = P I . Also.B p j -FL + A$L (5. these equations are QL=K~x-KPL (5.. lbf In summary.42) c.Hydraulic Servomotors 647 Now.36) we get (5. let (5.P2.44) where Mt = total mass of piston and load... ft2 Val.40) Now.38) v. lbf-s2/ft B.45) . 1l). In the sdomain. = viscous damping coefficient of piston and load. V.= v.37) where A. Also note that the total volume.41) can be written as QL= where we define =C 7f $ . then.= piston area.p cip = + C"P 2 We now apply Newton's law to the forces acting on the piston to write (5.+ -dt 2pe dt (5. + A Qi + Q2 dY vo dPL ...40) is zero. is constant. Le. = hydraulic natural frequency 4=% !% VIM (5. f Cip+ % (5. + C. .51) Pressure Limits Fig.48) where we have incorporated the assumption that [11 We also have defined the following parameters: q-= K = .50) Note that (5.46) to Y= (5. the spring force is missing and K = 0.648 Appendix J These equations are easily combined to write where we define the new coefficient Kce= K. = K. This changes the form of (5. the time constant is small. In most applications.e.47) Equation (5. but with K = 0.45) can be arranged in the block diagram form shown in Figure 5.48) has a pure integration.5 Block diagram of servomotor position y as a function of control valve position x and load force FL.5. the mass Mt of the piston and load is negligible.46) where the spring was included. which is not present in the system (5..6. In some systems.lag time constant WAC. i. J. or M I -e1 B P (5. The block diagram for this system is the same as Figure 5. In such a case. . References 1. FL -e AppL (5. Merritt.6. Hydraulic Control Systems. Herbert E. New York. the intercept valve for a large steam turbine may weight three or four tons. Wiley. Eggenberger. the load force can be neglected entirely and the transfer function for the servomotor becomes (5. the valve-controlled piston is approximated as an integrator.Hydraulic Servomotors Pressure Limits 649 F I I Fig. 5... such as the speed governor servomotor for a steam turbine can be modeled as a system similar to Figure 5.5 becomes simply an integration. M. When this assumption holds.6 Servomotor with negligible load mass and small lag time constant.6. For example.53) or the entire system becomes an integrator with integrating time AJKq. It should be noted that (5. A. Many practical systems. 2.. 1970. it may not be a good assumption to write (5. the output transfer function in Figure 5. 1967.53) unless the piston area Ap and pressure drop PLare both very large such that the acceleration can be very fast compared to the turbine response. If we also assume that time constant 7 is small.52) In this case. General Electric Company Publication GET-3096 B.52): K=O v.54) and when these assumptions hold. In summary. Q 4PeKce FL Fp (5. Introduction to the Basic Elements of Control Systems. the following assumptions have been used in deriving (5. Le. Another assumption that is commonly made is that the load force FL is small compared to the piston force Fp. the system reduces to that of Figure 5.53) may not be an adequate mathematical model if the piston load is massive. This is the form often assumed for the power servomotor. =I 2Hi j#l OR -Psni 2Hn OR where n is the number of machines and a machine n is the reference.32: A i i = .Addendum Page 61. 3. general formula for the A’s in Eq.C -p . 650 . mean value Admission valve Admittance matrix: defined primitive reduction Air gap line A matrix See also Eigenvalues including excitation system American National Standards Institute (ANSI) Amplidyne Amplification Amplifier: as analog computer component defined figure of merit magnetic rotating transfer function Analyzing steam turbine systems Analog computer simulation: differential equations excitation control system excitation system Links 72 439 36 373 40 248 11 219 287 14 239 595 532 451 253 239 239 274 452 531 302 257 307 265 347 282 535 252 251 584 65 221 290 98 251 209 232 307 143 318 581 212 386 214 394 40 370 This page has been reformatted by Knovel to provide easier navigation. .INDEX Index Terms A Acceleration. exciter voltage Bypass valve. P. in voltage regulator Bar lift Base quantity.G. Armature reaction. choice Bode plot: compensated excitation system lead compensator machine inductance regulated synchronous machine Boiler configuration (large) -follow control. Buildup. (automatic) -following mode -turbine representation (simplified) storage effect Boiling water reactors Boost-buck Braking: dc negative sequence Brown Boveri Corp.) synchronous machine Anderson. for hydro turbines 21 21 354 321 247 489 358 361 364 329 342 144 329 233 442 471 433 464 465 478 250 268 274 305 334 334 366 339 344 366 238 443 93 550 95 104 147 167 250 This page has been reformatted by Knovel to provide easier navigation. . P.Index Terms Analog computer simulation (Cont.M. Brown. demagnetizing effect Automatic control Links 170 125 56 401 352 228 229 326 B Backlash. steam turbine control Ceiling voltage. nin–bus system Clearing angle. Root locus current excitation system excitation system. critical Clearing time. lead network linear analysis Compensator Compressibility of water Links 443 23 295 402 401 26 35 45 22 37 33 33 519 518 513 515 513 514 421 422 422 422 422 237 277 339 344 451 637 321 341 341 344 584 363 366 422 320 316 316 55 355 358 247 311 260 562 263 584 266 This page has been reformatted by Knovel to provide easier navigation.Index Terms C Cam lift. See also Bode plot. critical Combined cycle power plant Combined cycle prime mover Combined cycle units Combustion turbine control Combustion turbine units Combustion turbine schematic diagram Compensated governor analysis of principle of operation permanent droop temporary droop Compensation. . exciter Centrifugal flyball governor Centrifugal governors Classical model: defined multimachine system shortcomings synchronous machine Classical stability study. of combustion turbines Computer methods. in hydraulic conduits Control: generating unit optimal system Control system: for a boiler components Control valve Control valve flow equations Control valve force equations Control valve operation Control valve position control Coordinated control mode. Critical time.G. B. Conduits Constant flux-linkage assumption Constant voltage behind transient reactance See also Classical model Continuous system modeling program (CSMP) Continuity conditions. C. excitation system Links 516 531 56 321 23 142 188 634 234 365 581 560 590 411 640 643 440 476 433 316 493 237 439 83 325 489 46 102 363 494 106 311 D Dahi. for a thermal unit Crary.Index Terms Computed response. of a hydro unit Current compensation. O. Damping: critical effect. . S. of a hydro unit 249 495 254 257 267 This page has been reformatted by Knovel to provide easier navigation.C. differential equations Concordia. for hydro units Drift Links 377 297 5 21 249 309 21 339 237 250 635 32 14 56 585 495 592 537 257 184 353 92 20 256 267 53 484 487 585 584 22 23 84 325 35 558 306 268 311 46 106 326 334 336 337 46 558 527 This page has been reformatted by Knovel to provide easier navigation.) effect on system order excitation system generator unit oscillation positive sequence ratio system oscillation torque (Dω) Damping transformer. . machine equations Direct axis Dispersion. Deviation Differential surge tank Differentiation. coefficient Distortion curve Disturbance Double-overhung hydro units Draft tube. F. as excitation stabilizer Deadhand. in voltage regulator Deformation of the shell. in hydro conduits Delta: maximum value mechanical (σm) de Mello. P. in control systems Digital computer simulation: differential equations excitation system synchronous machine transient stability Dimensions.Index Terms Damping (Cont. excitation system Dynamic Dynamic equation of equilibrium. for a hydraulic conduit Dynamic equations of governors Dynamic system performance Links 10 19 461 585 454 631 626 5 46 325 58 563 E E(EMF proportional to iF). in steam turbines Equal area criterion two-machine system Equal mutual flux linkage Equivalent circuit. in steam turbine control Emergency trip conditions. A matrix Electric analog of a hydro system Electrical angle (δc) Electrical load frequency damping Electrohydraulic systems. defined Economic control Eigenvalues: A matrix A matrix with linear exciter effect of uniform damping Eigenvectors. synchronous machine 11 216 291 378 65 496 15 411 402 445 31 35 95 107 547 54 222 307 61 232 79 284 209 396 98 99 99 152 10 129 128 This page has been reformatted by Knovel to provide easier navigation.Index Terms Droop Droop characteristic Drum-type boilers Duty. defined EFD (EMF proportional to vF). defined Eq′(EMF proportional to λF). defined Eqa. . . pu d-q quantities See also Stator Error Euler method. modified Excitation control: equivalent alternator–rectifier system alternator–SCR systems brushless compound rectifier configurations dc generator–commutator systems potential source rectifier rheostatic Routh’s criterion simplified view Excitation control system See also Excitation control analog computer solution boost-buck response comparison with classical representation complete linear model definitions linear analysis of compensation linear numerical example simplified linear model Excitation systems approximate representation compensation See also Bode plot Links 129 585 30 532 239 241 240 242 236 239 243 236 11 271 233 236 282 275 316 287 243 344 288 286 431 333 277 583 296 583 244 583 583 268 57 277 584 59 322 68 232 304 compound rectifier plus potential source rectifier 242 581 This page has been reformatted by Knovel to provide easier navigation.Index Terms Equivalent stator. B Types C.) compensation. computer representation Types A. D Types E. G Type K Type 1 Type 1S Type 2 Type 3 Type 4 damping defined duty effect on power limits effect on stability high initial response normalization primitive rate feedback response continuously regulated noncontinuously regulated rheostat self-excited saturation separately excited stabilizer state–space description thyristor typical constants Links 292 559 560 591 562 293 355 295 296 297 299 297 243 605 311 309 247 248 236 277 268 271 268 236 237 271 238 237 285 239 299 241 266 294 305 306 338 307 562 247 268 325 585 352 581 267 299 246 581 307 355 304 359 307 316 347 This page has been reformatted by Knovel to provide easier navigation. .Index Terms Excitation systems (Cont. F. .) voltage response Exciter: boost-buck transfer function ceiling voltage voltage rating Exciter build–down Exciter buildup ac generator exciter analog computer solution dc generator exciter digital computer solution formal integration linear approximation loaded exciter response ratio solid-state exciter Links 585 274 23 295 247 254 247 266 535 254 540 256 263 266 268 266 585 259 306 254 247 311 260 562 263 584 266 F Faults. synchronous machine network 85 388 248 248 253 352 35 366 46 315 320 587 587 3 19 244 16 244 355 309 329 352 This page has been reformatted by Knovel to provide easier navigation.Index Terms Excitation systems (Cont. effect on transient stability Feedback Feedback control system Field voltage: base rated load Figure of merit. amplifiers Filter. bridged T First swing stability Flux-linkage: equations. in control systems General Electric Co. . in combustion turbine unit Fuel system dynamics. 0dq) Frequency: natural resonant (undamped) oscillation Friction head.) mutual subtransient transient Flyball governor subsystem Fossil-fueled boiler computer models low-order model prime mover control model block diagram of prime mover controls Fossil-fueled steam generators drum-type boilers once-through boilers FORTRAN Frame of reference. in steam turbines Function generators Links 95 132 138 401 423 473 474 474 475 461 461 461 187 84 249 24 495 256 540 522 465 609 263 265 294 535 310 541 402 408 622 416 134 G Gain Gas turbine power generation Gating.Index Terms Flux-linkage (Cont. in a hydro penstock Frohlich equation Fuel and air controls. 587 524 601 238 560 240 243 252 299 This page has been reformatted by Knovel to provide easier navigation. (abc. in U. Head. in steam turbines Grand Coulee Dam Links 430 430 431 435 432 10 407 424 427 427 406 417 546 563 10 518 622 622 68 439 489 58 48 68 233 H H. power systems Generating unit block diagram Governor analysis: Ballarm scale compensator system block diagram transient performance behavior closed-loop block diagram computer representation droop characteristic. nonlinear. 16 17 126 93 493 490 56 This page has been reformatted by Knovel to provide easier navigation. . R. in combustion turbine system equations equilibrium equations linear synchronous machine values.S. M. change of base estimating curves typical values Harris. G.Index Terms Generation control isolated system network system Generation mix. change in Head loss Heffron. W. Index Terms Hybrid formulation. linear n-machine system Hydraulic gradient Hydraulic reaction force. n-port Incremental variables Inductance. in governors Hydraulic servomotor: general description transfer function Hydraulic system equations Hydraulic system transfer function Hydraulic turbine prime movers adjustable blade propeller turbine Deriaz turbine Francis turbine impulse turbine Kaplan turbine Nagler turbine Pelton turbine propeller type turbine reaction turbine Hydraulic valve controlled piston Hydro system. synchronous machine leakage defined 54 8 6 501 373 208 108 86 111 122 383 53 69 546 This page has been reformatted by Knovel to provide easier navigation. block diagram Links 386 492 408 640 618 498 503 484 489 484 484 484 484 489 484 484 484 645 509 489 487 489 489 490 489 486 493 I Ideal transformer Impact: distribution effect large vs small Impedance. . characteristic Impedance matrix. in steam turbine systems Interface. in control systems Integrator. .Index Terms Inductance. ISO Institute of Electrical and Electronics Engineers (IEEE) Integration. synchronous machine leakage (Cont. analog computer component Intercept valve. defined See also H M. defined units Infinite bus Initial conditions: examples stability study.) negative sequence magnetizing table transient and subtransient 0dq. between system differential and algebraic equations International Electrotechnical Commission (IEC) Instability. defined Inertia constant: effect on stability H. dynamic Isochronous governor Links 125 108 126 123 87 317 14 14 15 26 159 165 143 321 593 532 444 522 98 46 408 238 347 292 355 316 581 319 16 115 143 J Jordan canonical form 64 This page has been reformatted by Knovel to provide easier navigation. A. synchronous machine Lefschetz. Limiter. M. in combustion turbines Lmd and Lmq. W. A. Kinetic energy of rotating mass.Index Terms K Kimbark. defined L0. Wk Krause. Lewis. defined LMD and LMQ. S. Kron. E. Liapunov. defined Leakage inductance. Gabriel Kron reduction See also Matrix reduction Links 13 184 14 83 278 378 14 246 16 173 48 254 102 256 125 266 L LAD and LAQ. P. defined Lq. W. constant impedance 115 213 114 35 38 35 46 368 37 162 108 87 517 108 110 87 87 108 61 83 117 583 53 53 10 587 321 70 60 208 386 381 93 95 111 This page has been reformatted by Knovel to provide easier navigation. C. defined Limiting effects. nonlinear equations Linearized system equations Load equations: infinite bus form linear current model for one machine synchronous machine Load-flow study nine-bus system Load representation. defined Ld. . dynamic stability Linearization. defined Linear analysis. in governor control Links 237 439 10 40 402 9 35 533 467 250 445 13 378 13 165 255 305 583 15 368 N National Electric Reliability Council (NERC) Nebraska Public Power District Network equations: based on flux-linkage model. . linearized n-machine form n-machine system Nine-bus system: defined linearized solution load-flow study oscillation stability simulation swing curve n-machine system hybrid formulation network equations system equations 37 392 38 61 353 44 35 386 369 377 381 386 396 388 381 36 369 300 364 This page has been reformatted by Knovel to provide easier navigation. elementary See also Classical model Matrix reduction Mechanical flyball governor Moment of inertia Multimachine systems Multiplier. Excitation Main stop valve Mathematical model.Index Terms M Main exciter See also Exciter. analog computer component Multivariable control system. system equations. .Index Terms Node incidence matrix Nonlinearities. machine equations Nonlinear system equations Nonreheat turbine block diagram Normalization: comparison of pu systems guidelines swing equation synchronous machine equations time torque equations Northeast Power Coordination Council n-port network Nuclear steam supply systems Numerical methods. in combustion turbine systems Omega: mechanical (ωm). integration Nyquist. n-machine system Oscillation: generator unit modes natural frequencies system 5 59 24 309 46 364 310 558 14 14 26 469 377 386 115 153 311 517 This page has been reformatted by Knovel to provide easier navigation. H. defined One machine-infinite bus solution Once-through boiler Order. Links 373 107 208 11 454 96 545 15 92 101 103 145 370 477 537 11 103 99 545 550 116 119 170 185 O Off-nominal frequency and voltage effects. defined rated (ωR). H.) three-machine. Park. H. Phillips. .Index Terms Oscillation (Cont. Pilot exciter See also Exciter. in d-q reference frame reference frame definition reference frame transformation relation between system d-q quantities Philadelphia Electric Co. in steam turbines Overspeed trip Links 61 7 249 440 445 319 20 20 371 489 319 54 96 92 103 21 152 374 370 372 374 301 56 238 84 614 272 250 255 305 583 304 354 372 151 102 550 83 494 88 115 146 55 P Pacific Gas and Electric Co. R. A. R. Park’s transformation Penstocks Peny. in control systems Potential transformer. transfer function This page has been reformatted by Knovel to provide easier navigation. nine-bus system tie-line Overshoot Overspeed protection. synchronous machine equations. Excitation P matrix Position transducers. R. Perturbation method Per unit: comparison of various systems conversion torque Phasor: defined diagram. typical Pressure transducers Pressurized water reactors Prime mover governors Proportional plus partial reset pressure control Proportional plus reset pressure control Pumped storage hydro systems Links 532 15 157 85 311 24 21 10 338 359 30 86 614 616 608 479 401 620 620 510 343 562 534 345 584 352 357 33 414 32 33 Q Quadrature axis Quiescent operating point 20 209 84 311 R Rankin. analog computer component Power: accelerating factor invariance limits. 93 435 277 46 325 This page has been reformatted by Knovel to provide easier navigation. excitation system Ray. J. Rankine cycle Rate feedback. Pressure regulator. J. B. W. R. effect of excitation synchronizing See also Synchronizing power coefficient Power-angle curve Power system components.Index Terms Potentiometer. in stability study Power system stabilizer (PSS) Predictor-corrector method Prentice. in control systems Pressure regulator. A. . typical synchronous machine Reheat steam turbine system block diagram Reheat stop valve Reheat turbine flow diagram Reheat turbines Reheat vapor cycle Reliability Reset control. excitation system Rise time. direct-axis: synchronous. See also Voltage regulator continuously acting proportional control. in hydro systems Root locus compensated excitation system Rotor angle Rotational speed transducers Links 22 23 486 321 370 13 566 250 616 66 459 444 442 444 435 3 617 617 244 306 236 249 495 11 276 366 13 603 372 371 271 583 584 585 457 248 316 247 260 320 268 263 357 299 363 281 306 327 344 This page has been reformatted by Knovel to provide easier navigation. xd transient. . emergency demand Reference frame: phasor synchronously rotating Regenerative vapor cycle Regulator. defined Riser tank. xd′ Reaction hydro turbine Reactive power.Index Terms Reactance. block diagram proportional plus reset control Response ratio Rheostat. C. response 54 53 53 402 249 46 555 587 236 10 294 255 185 271 114 285 20 533 352 148 239 320 113 355 556 294 186 307 593 562 563 257 267 271 251 Simplifying assumptions. R. defined Silverstat regulator Simulation methods Small disturbances: defined response See also Linear analysis Small impacts.Index Terms Routh’s criterion applied to excitation control system Rudenberg. Links 11 271 79 57 277 254 59 322 257 68 232 S Saturable reactor Saturation: computer representation of exciters dc generator exciter digital calculation excitation systems exponential function linearized exciter synchronous machines Scaling. Schulz. . D. Short circuit ratio Signal. P. for hydro transfer functions 506 This page has been reformatted by Knovel to provide easier navigation. R. analog computer solutions Schroder. SCR See also Thyristor Servomotor Settling time Shipley. R. B. Index Terms Specific inertia of steam turbine generator Speed droop governor block diagram eigenvalues floating lever root locus Speed reference. governor Speed regulation See also Droop Speed relay. statement of simulation in nine-bus system steady-state synchronous machines transient See also Transient stability Stabilizer See also Power system stabilizer Links 451 413 416 416 419 417 408 19 402 402 90 5 5 6 304 317 588 35 33 3 5 4 353 6 6 6 327 46 338 309 584 315 24 309 46 588 315 320 13 53 309 588 310 321 49 57 563 This page has been reformatted by Knovel to provide easier navigation. synchronous machine Stability: asymptotic defined dynamic effect of excitation effect of inertia constant excitation system first swing limit power systems primitive definition problem. . governor sensing Speed voltage. synchronous machine Steady-state stability Steam plant control functions Steam power plant model fueled by fossil fuels fueled by nuclear energy Steam turbine Steam turbine control operations control of protection of Steam turbine generator controls Steam turbine power generation.Index Terms Stabilizing signal. supplementary. . for excitation system State-space equations: current form. in combined cycle plants Steam valve–steam flow Links 338 91 285 217 118 111 109 209 117 231 83 390 129 369 4 150 6 446 435 436 436 430 444 444 444 445 525 618 437 136 379 588 157 24 309 139 151 154 91 107 296 368 This page has been reformatted by Knovel to provide easier navigation. synchronous machine excitation system flux-linkage form: linear loaded machine neglecting saturation synchronous machine linear current form loaded machine simplified linear machine synchronous machine total system Stator equivalent. rms pu quantities Steady state Steady-state equations. in control systems Summer. loss Synchronizing power coefficient (ps) Synchronous machine: analog simulation block diagram classical model constant field flux–linkage model digital simulation E′ model Links 618 14 9 132 132 123 590 400 413 423 489 41 44 13 16 37 13 16 103 29 4 24 230 170 47 22 142 184 127 57 55 67 231 340 9 59 71 224 227 111 46 79 494 495 134 135 This page has been reformatted by Knovel to provide easier navigation. D. analog computer component Summing beam. W.Index Terms Steam volume Stevenson. . in speed droop governors compensator Surge tank Swing curve: defined nine-bus system Swing equation approximate. synchronous machine Summation. Subtransient: effects EMF flux linkage inductances. in pu power classical n-machine system defined most useful form normalized form simple nonlinear form Synchronism. . unregulated load equations local load normalization equations one-axis model operational inductance parameters. regulated linear.) E′′ model equivalent tee circuit flux–linkage equations governor inductance linear models linear. from manufacturers’ data phasor diagram regulated saturation simplified model simulation solid roror dynamic models speed voltage stability state-space equations steady-state equations subtransient inductance time constants two-axis model typical parameters unregulated unsaturated flux–linkage model voltage equations Links 132 107 85 68 86 56 327 55 114 154 92 141 144 166 152 66 20 56 150 143 90 6 83 150 123 125 138 126 55 111 88 110 552 91 157 134 143 107 109 368 329 113 127 334 355 222 556 551 99 354 545 108 60 111 208 122 322 143 This page has been reformatted by Knovel to provide easier navigation.Index Terms Synchronous machine (Cont. reheater Time constants. Throttle valuve. in steam turbine systems Thyristor. H. (Dω) dc braking electromagnetic or electrical mechanical regulated 13 21 21 339 21 13 13 18 20 16 105 46 111 326 35 558 46 106 326 14 125 126 250 143 104 106 3 435 14 77 83 439 239 7 450 241 55 266 85 This page has been reformatted by Knovel to provide easier navigation. . excitation system Tie-line oscillations Time constant. tabulation of typical values Links 250 581 584 566 271 584 T T eφ defined derived from field energy Tesla.Index Terms System: continuously acting. proportional control noncontinuously acting System data. defined Thevenin equivalent Thomas. Nikola Thermal generation Theta (θ). synchronous machine derived table Tirrell regulator Torque: accelerating asynchronous damping. C. synchronous machine reactance Transducers Transfer functions. for three-phase systems Turbine blading.Index Terms Torque (Cont. typical data Trigonometric identities. . of steam turbine control Transient stability: defined digital simulation effect of excitation effect of faults first swing steps in problem solution Transmission line equations Transmission lines. steam impulse blading reaction blading Links 17 21 103 103 21 6 14 235 588 9 138 138 123 22 605 446 6 353 315 316 46 41 498 4564 398 437 437 437 438 438 355 315 319 309 85 326 This page has been reformatted by Knovel to provide easier navigation.) unregulated negative sequence braking normalization equations per unit synchronous Torque angle defined effect of excitation Transient: defined effects EMF’s flux linkage inductance. . hydro systems Turbine-following control mode Turbine mechanical damping. of steam turbine systems Links 492 432 411 19 27 451 U Units: English inertia constant MKS 15 15 15 V Vapor power cycle Venikov. in torque equations Turbine torque-speed characteristics Two-port network Typical constants. synchronous machine Voltage reference: nonlinear bridge circuit transfer function Voltage regulator See also Amplifier backlash boost-buck deadhand direct-acting electromechanical electronic indirect-acting linear synchronous machine magnetic amplifier models of physical systems rotating amplifier 238 251 250 250 250 251 250 66 239 559 251 272 252 268 305 250 256 268 262 311 272 273 272 236 250 560 435 56 88 110 This page has been reformatted by Knovel to provide easier navigation. V. A. Voltage equations.Index Terms Turbine efficiency. in hydro penstocks Western Systems Coordinating Council (WSCC) Westinghouse Electric Corp. defined xd′′. C. defined xq.) solid-state Voltage response ratio See also Response ratio Links 254 244 W Water hammer formula Water starting time Wave equations for a hydraulic conduit Wave velocity.Index Terms Voltage regulator (Cont. rotor 487 15 239 297 561 242 299 252 304 269 347 509 491 494 X xd. 127 131 140 316 This page has been reformatted by Knovel to provide easier navigation. . defined xp. defined 151 166 166 151 166 151 166 166 166 166 166 166 Y Young. 484 497 498 631 493 560 237 291 349 Wicket gate WR. defined xd′. defined xd′′. C. defined xd′. defined x0. defined x2.