8. Testing Differential Protection (87)

TheRelayTest ing Handbook Test ing Di ffer ent ialPr otect ion (87) Chr isW er sti uk Professi onalEngi neer J ourneymanPowerSyst em Electri cian El ectri calEngineer ingTechnologist The Relay Testing Handbook Testing Differential Protection (87) . The Relay Testing Handbook Testing Differential Protection (87) Chris Werstiuk Professional Engineer Journeyman Power System Electrician Electrical Technologist Valence Electrical Training Services 7450 w 52nd Ave.com . M330 Arvada. CO 80002 www.valenceonline. an explanation without intent to infringe. damage. or liability directly or indirectly caused or alleged to be caused by this book. Published in the United States of America .Although the author and publisher have exhaustively researched all sources to ensure the accuracy and completeness of the information contained in this book. neither the authors nor the publisher nor anyone else associated with this publication. Telephone: 303-250-8257 Distributed By: www.A. Neither this book nor any part may be reproduced or transmitted in any form or by any means. or by any information storage and retrieval system. shall be liable for any loss. Image from BigStockPhoto. Flamingo Designs Copyright © 2011 by Valence Electrical Training Services.S. West Hills. The material contained herein is not intended to provide specific advice or recommendations for any specific situation. and recording. CA Cover Art: © James Steidl.com Edited by: One-on-One Book Production. M330. All rights reserved. microfilming.valenceonline. electronic or mechanical. U. without permission in writing from the publisher. Arvada. Trademark notice product or corporate names may be trademarks or registered trademarks and are used only for identification.com Interior Design and Layout: Adina Cucicov. CO. 80002. including photocopying. The Relay Testing Handbook: Testing Differential Protection (87) First Edition ISBN: 978-1-934348-17-8 Published By: Valence Electrical Training Services 7450 w 52nd Ave. and I hope you find this and future additions of The Relay Testing Handbook to be useful. Each chapter uses the following outline to best describe the element and the test procedures. •• Application •• Settings •• Pickup Testing •• Timing Tests •• Tips and Tricks to Overcome Common Obstacles We will review techniques to test differential relays with 3 or 6 channels so that readers can test nearly any differential application with any modern test-set. Real world examples are used to describe each test with detailed instructions to determine what test parameters to use and how to determine if the results are acceptable. Basic electrical fundamentals. This package provides a step-by-step procedure for testing the most common differential protection applications used by a variety of manufacturers. Each chapter follows a logical progression to help understand why differential protection is used and how it is applied. v . A wide variety of relay manufacturers and models are used in the examples to help you realize that once you conquer the sometimes confusing and frustrating man-machine interfaces created by different manufacturers.Author’s Note The Relay Testing Handbook was created for relay technicians from all backgrounds and provides the knowledge necessary to test most modern protective relays installed over a wide variety of industries. Testing procedures are described in detail to ensure that the differential protection has been correctly applied. detailed descriptions of protective elements. and generic test plans are combined with examples from real life applications to increase your confidence in any relay testing situation. Thank you for your support with this project. all digital relays use the same basic fundamentals and most relays can be tested by applying these fundamentals. . Practical Solutions www.Acknowledgments This book would not be possible without support from these fine people Mose Ramieh III Level IV NETA Technician Level III NICET Electrical Testing Technician Vice President of Power & Generation Testing.com vii . CET Magna IV Engineering Superior Client Service.Sc.army. Phil Baker David Snyder Hydropower Test and Evaluation Bonneville Lock and Dam www.nwp. Inc.com Eric Cameron.mil/op/b/ Lina Dennison Roger Grylls. Electrical Guru and all around nice guy.magnaiv.E. B.mantatest.usace. Manta Test Systems www. . Tips and Tricks to Overcome Common Obstacles 37 ix . Restrained-Differential Timing Test Procedure 18 5.Table of Contents Author’s Note v Acknowledgments vii Chapter 1 1 Simple Percent Differential (87) Element Testing 1 1. Application 1 2. Settings 11 A) Enable Setting 11 B) Minimum Pickup (Restrained) 11 C) Tap 12 D) Slope-1 12 E) Slope-2 12 F) Breakpoint 12 G) Time Delay 12 H) Block 12 3. Restrained-Differential Slope Testing 20 A) Test-Set Connections 23 B) Slope Test Procedure 24 6. Restrained-Differential Pickup Testing 13 A) Test-Set Connections 14 B) Pickup Test Procedure 16 4. 1-Phase Restrained-Differential Slope Testing 112 A) Understanding the Test-Set Connections 112 B) Test-Set Connections 114 C) 1-Phase Differential Slope Test Procedure 116 8. 3-Phase Restrained-Differential Slope Testing 86 A) Test-Set Connections 93 B) Pre-Test Calculations 94 C) Post-Test Calculations 103 D) Alternate Slope Calculation 107 E) 3-Phase Differential Slope Test Procedure 108 7. Current Transformer Connections 70 4. Restrained-Differential Timing Test Procedure 84 6. Harmonic Restraint Testing 118 A) Harmonic Restraint Test Connections 119 B) Harmonic Restraint Test Procedure with Harmonics 120 C) Harmonic Restraint Test Procedure with Current 122 D) Evaluate Results 123 9. 3-Phase Restrained-Differential Pickup Testing 75 A) 3-Phase Test-Set Connections 77 B) 3-Phase Pickup Test Procedure 78 C) 1-Phase Test-Set Connections 80 D) 1-Phase Pickup Test Procedure 82 5. Application 40 A) Zones of Protection 40 B) Tap 41 C) Phase-Angle Compensation 51 2. Settings 66 A) Enable Setting 66 B) Number of Windings 66 C) Phase-Angle Compensation 66 D) MVA 68 E) Winding Voltage 68 F) Minimum Pickup (Restrained) 69 G) Tap 69 H) Slope-1 69 I) Slope-2 69 J) Breakpoint 69 K) Time Delay 69 L) Block 70 M) Harmonic Inhibit Parameters 70 3. Tips and Tricks to Overcome Common Obstacles 124 x .The Relay Testing Handbook Chapter 2 39 Percent Differential (87) Element Testing 39 1. Alternate Pickup Test Procedure 133 5. Test-Set Connections 127 3. Timing Test Procedure 136 6. Table of Contents Chapter 3 125 Unrestrained-Differential Testing 125 1. Settings 126 A) Enable Setting 126 B) Minimum Pickup 126 C) Tap 126 D) Time Delay 126 E) Block 126 2. Tips and Tricks to Overcome Common Obstacles 137 Bibliography 139 xi . Simple Pickup Test Procedure 130 4. xii . Table of Figures Figure 1: Simple Differential Protection 2 Figure 2: Simple Differential Protection with External Fault 2 Figure 3: Simple Differential Protection with External Fault 2 3 Figure 4: Simple Differential Protection with Internal Fault 3 Figure 5: Simple Differential Protection with Internal Fault 2 4 Figure 6: Simple Differential Protection with Worst Case CT Error 5 Figure 7: Simple Differential Protection with Worst Case CT Error and External Fault 5 Figure 8: Percentage Differential Protection Schematic 6 Figure 9: Percentage Differential Protection Operating Mechanism 6 Figure 10: Percentage Differential Protection and External Faults 7 Figure 11: Percentage Differential Protection and Internal Faults 7 Figure 12: Percentage Differential Protection and Internal Faults 2 8 Figure 13: Percentage Differential Protection Characteristic Curve 8 Figure 14: Percentage Differential Protection Characteristic Curve with Minimum Pickup 10 Figure 15: Percentage Differential Protection Dual Slope Characteristic Curve 11 Figure 16: Simple 87-Element Test-Set Connections 14 Figure 17: Simple 3-Phase 87-Element Test-Set Connections 15 Figure 18: Simple 3-Phase 87-Element Test-Set Connections with Six Current Channels 15 Figure 19: Pickup Test Graph 16 Figure 20: GE Power Management 489 Analog Input Specifications 17 Figure 21: GE Power Management 489 Differential and Output Relay Specifications 18 Figure 22: GE Power Management 489 Minimum Trip Time 19 Figure 23: GE Power Management 489 Specifications 21 Figure 24: GE Power Management 489 Slope Test Connection Table 21 Figure 25: GE Power Management 489 Slope Test Connections Example #1 22 Figure 26: GE Power Management 489 Slope Test Connections Example #2 22 Figure 27: Simple 87-Element Slope Test-Set Connections 23 Figure 28: Simple 87-Element High Current Slope Test-Set Connections 23 Figure 29: Percentage Differential Protection Dual Slope Characteristic Curve 24 Figure 30: GE 489 Differential Formulas 25 Figure 31: Example Characteristic Curve in Amps with 1st Transition Defined 26 Figure 32: Percentage Differential Protection Dual Slope Characteristic Curve 26 Figure 33: Percentage Differential Protection Dual Slope Characteristic Curve 28 Figure 34: Percentage Differential Protection Dual Slope Characteristic Curve in Amps 30 xiii . The Relay Testing Handbook Figure 35: Using Graphs to Determine Pickup Settings 31 Figure 36: GE Power Management 489 Specifications 32 Figure 37: Determine Slope by Rise/Run Calculation 35 Figure 38: Zones of Protection Example 40 Figure 39: 3 Phase Generator Differential Protection 41 Figure 40: 3 Phase Transformer Differential Protection 42 Figure 41: 3 Phase Transformer Differential Protection Using Tap Settings 43 Figure 42: Wye-Delta Transformer Differential Protection Using CT Connections 44 Figure 43: Wye-Delta Transformer Differential Protection Using CT Connections 50 Figure 44: Phase Relationship / Clock Position with Leading Angles 53 Figure 45: Phase Relationship / Clock Position with Lagging Angles 53 Figure 46: Wye-Wye Transformer 54 Figure 47: Wye-Wye Transformer Phasor Diagram 55 Figure 48: Delta-Delta Transformer Connections 56 Figure 49: Delta-Delta Transformer Phasors 57 Figure 50: Wye-Delta Transformer Connections 58 Figure 51: Wye-Delta Transformer Phasor Diagrams 58 Figure 52: Wye-Delta Alternate Transformer Connections 59 Figure 53: Wye-Delta Alternate Transformer Phasor Diagrams 59 Figure 54: Delta-Wye Transformer Connections 60 Figure 55: Delta-Wye Transformer Connections 61 Figure 56: Delta-Wye Alternate Transformer Connections 62 Figure 57: Delta-Wye Alternate Transformer Phasor Diagrams 63 Figure 58: Transformer Nameplate Phase Relationships 65 Figure 59: Common Phase-Angle Compensation Settings 68 Figure 60: Zones of Protection Example 71 Figure 61: CT Connections Example #1 72 Figure 62: CT Connections Example #2 72 Figure 63: GE/Multilin SR-745 Transformer Protective Relay Connections 73 Figure 64: GE T-60 Transformer Protective Relay Connections 73 Figure 65: Beckwith Electric M-3310 Transformer Protective Relay Connections 74 Figure 66: Schweitzer Electric SEL-587 Transformer Protective Relay Connections 74 Figure 67: Schweitzer Electric SEL-387 Transformer Protective Relay Connections 75 Figure 68: Simple 3-Phase 87-Element Test-Set Connections 77 Figure 69: Simple 3-Phase 87-Element Test-Set Connections with Six Current Channels 77 Figure 70: Pickup Test Graph 78 Figure 71: SEL-387E Specifications 79 Figure 72: Simple 87-Element Test-Set Connections 80 Figure 73: Simple 3-Phase 87-Element Test-Set Connections 81 Figure 74: Simple 3-Phase 87-Element Test-Set Connections with Six Current Channels 81 Figure 75: Pickup Test Graph 83 Figure 76: SEL-387E Specifications 83 Figure 77: SEL-387 Differential and Output Relay Specifications 85 Figure 78: SEL-387 Differential Minimum Trip Time 85 xiv . Higher Current 87-Element Test-Set Connections 120 Figure 104: Simple Differential Protection with Worst Case CT Error and External Fault 125 Figure 105: Simple 87U-Element Test-Set Connections 127 Figure 106: Parallel 87U-Element Test-Set Connections 128 Figure 107: Parallel 87U-Element Test-Set Connections with Equal W_CTC Settings 128 Figure 108: Transformer Relay Connections for Single-Phase Differential Testing 129 Figure 109: 1-Phase Differential Test-Set AØ Yd1 Connections 129 Figure 110: SEL-387E Specifications 131 Figure 111: SEL-387 Differential Element Specifications 135 Figure 112: Preferred SEL-387 Output Contact Specifications 136 xv . Table of Figures Figure 79: Common Phase-Angle Compensation Settings 87 Figure 80: Yy12 or Yy0 3-Phase Differential Restraint Test Connections 88 Figure 81: Dd0 3-Phase Differential Restraint Test Connections 89 Figure 82: Dy1 3-Phase Differential Restraint Test Connections 90 Figure 83: Yd1 3-Phase Differential Restraint Test Connections 91 Figure 84: Dy11 3-Phase Differential Restraint Test Connections 91 Figure 85: Yd11 3-Phase Differential Restraint Test Connections 92 Figure 86: Schweitzer Electric SEL-387 Transformer Protective Relay Connections 93 Figure 87: 3-Phase Restrained-Differential Slope Test-Set Connections 94 Figure 88: Percentage Differential Protection Dual Slope Characteristic Curve 95 Figure 89: SEL-387 Slope-1 Differential Formulas 96 Figure 90: SEL-387 Definition of IOP and IRT 96 Figure 91: SEL-387 Slope-2 Differential Formulas 100 Figure 92: Percentage Differential Protection Dual Slope Characteristic Curve in Amps 101 Figure 93: Using Graphs to Determine Pickup Settings 102 Figure 94: Determine Slope by Rise/Run Calculation 107 Figure 95: YDac Transformer Connection 112 Figure 96: Transformer Relay Connections for Single-Phase Differential Testing 113 Figure 97: Schweitzer Electric SEL-387 Transformer Protective Relay Connections 114 Figure 98: 1-Phase Restrained-Differential Slope Test-Set AØ Yd1 Connections 114 Figure 99: 1-Phase Restrained-Differential Slope Test-Set BØ Yd1 Connections 115 Figure 100: 1-Phase Restrained-Differential Slope Test-Set CØ Yd1 Connections 115 Figure 101: Transformer Inrush Waveform 118 Figure 102: Simple 3-Phase 87-Element Test-Set Connections 119 Figure 103: Simple 3-Phase. . 87-Elements are selective and will only operate when a fault occurs within a specified zone of protection with no inherent time delay. trip for faults within the zone of protection and ignore faults outside of the zone of protection. 1 . therefore. and that a difference between the two currents is caused by an internal fault. Application Differential protection operates on the principle that any current entering the protected zone must equal the current leaving the protected zone. 1. There are several methods to apply differential protection but they are all designed to achieve the same basic principal. motor. or other applications where the voltage and CT ratios are the same on both sides of the protected device. We will discuss complex differential protection typically installed for transformers in the next chapter. or a group of equipment which is integral to the electrical system.Chapter 1 Simple Percent Differential (87) Element Testing Differential protection (87) is applied to protect equipment with high replacement costs and/or long replacement/repair times. a generator differential scheme may include other equipment such as PTs and circuit breakers. Percent differential protection is the most commonly applied differential element and this chapter will lay the foundation for understanding most differential protection schemes. the relay will trip. This chapter only discusses the simplified differential protection typically found in generator. Differential CTs are often strategically installed to provide overlapping protection of circuit breakers and. If the difference between currents is larger than the 87-Element setpoint. The zone of protection is defined by strategic placement of current transformers (CTs) that measure the current entering and exiting the zone. All descriptions are made on a single-phase basis and you should keep in mind that identical configurations exist for the other two phases in a three-phase system. The Relay Testing Handbook The following figure displays a simplified version of differential protection that would typically be applied to a generator or bus.0 A I2 = 5. the current enters the non-polarity mark of CT2 and the CT secondary current flows through the Iop coil from top to bottom. The equipment being protected in this example would be connected between CT1 and CT2 which defines the differential zone of protection. If the CTs are operating perfectly and have the same CT ratio.0 A I1 = 5. Simultaneously. The CT secondary current follows the circuit and flows through the Iop coil from the bottom to top. 1000 A 100:5 CT1 100:5 CT2 G FAULT Iop = 0.0 A I1 = 50.0 A Figure 2: Simple Differential Protection with External Fault 2 . 100 A 100:5 CT1 100:5 CT2 G Iop = 0. The current flows from left to right in this system and enters the polarity mark of CT1. If the Iop coil was a simple overcurrent element.0 A I2 = 50. Early differential elements were simple overcurrent devices and the Iop coil would have a relatively low setting. The current entering the zone of protection equals the current leaving the zone and cancel each other out just like the first example.0 A Figure 1: Simple Differential Protection The next figure displays an external fault in an ideal world. the net Iop current would be zero amps because the two currents cancel each other out. nothing would happen because zero amps flow though the Iop coil. 0 A Figure 3: Simple Differential Protection with External Fault 2 The next figure displays an internal fault with Source 1 only. If the CT1 current is greater than the Iop setting.0 A I2 = 50. the Iop element will trip. bottom to top. Chapter 1: Simple Percent Differential (87) Element Testing The next figure shows a fault in the opposite direction. current flows into CT1 and the secondary current flows through Iop. In this scenario.0 A I1 = 50.0 A I1 = 50. 1000 A 100:5 CT1 100:5 CT2 FAULT G Iop = 50. There is no current flowing through CT2 and. therefore.0 A Figure 4: Simple Differential Protection with Internal Fault 3 . nothing to cancel the CT1 current. 1000 A 100:5 CT1 100:5 CT2 FAULT G Iop = 0. The current-in equals the current-out and cancel each other out. The Iop element does not operate. and the secondary current with +10% error equals 5. This is a worst case scenario with nominal current.0A × 0. differential current is a simple overcurrent relay connected between two or more CTs as shown in previous figures.5A ( 5.0A + 5.0A or the relay will trip under normal load conditions assuming a worst case CT mismatch. 1000 A 1000 A 100:5 CT1 100:5 CT2 FAULT G Iop = 100. 100A flows through CT2. In fact. The Iop element must be set larger than 1.10 ) flowing through Iop from bottom to top. 4 .0 A Figure 5: Simple Differential Protection with Internal Fault 2 In an ideal world. 100A flows through CT1.The Relay Testing Handbook The next figure displays an internal fault with sources on both ends.0 A I2 = 50. protection CTs typically have a 10% accuracy class which can cause problems with the protection scheme described previously.0A × 0. Current flows into CT1 and its secondary current flows through Iop from bottom to top.0A. and the secondary current with -10% error equals 4. Current flows through CT2 and its secondary current flows through Iop from bottom to top also. The difference between the two CT secondaries is 1.10 ) flowing through Iop from top to bottom.0A − 5.5A ( 5. we do not live in an ideal world and no two CTs will produce exactly the same output current even if the primary currents are identical. The news gets worse when you also consider that CT accuracy can jump to 20% when an asymmetrical fault is considered. Unfortunately.0 A I1 = 50. In this case the currents add together instead of canceling each other out and the Iop element will trip if the combined currents are larger than the Iop trip setting. The difference between CT operating characteristics is called CT mismatch and the effects of CT mismatch are displayed in the following figures. As in our previous example. This element would have to be set greater than 10A to prevent nuisance trips if a fault occurs outside the zone of protection.5 A I2 = 5. This setting does not appear to limit equipment damage very well and a new system was developed to provide more sensitive protection and prevent nuisance trips for external faults. Chapter 1: Simple Percent Differential (87) Element Testing 100 A 100:5 CT1 100:5 CT2 G Iop = 1. the CT1 secondary current with -10% error is 45A flowing through the Iop from bottom to top.0 A I2 = 55. The CT2 secondary current with +10% error is 55A flowing through Iop from top to bottom. 1000 A 100:5 CT1 100:5 CT2 G FAULT Iop = 10. This setting is very high and any internal fault would need to cause greater than 200A (2x the rated current for a single source system) or 100A (dual source system) of fault current to operate the differential element.5 A Figure 6: Simple Differential Protection with Worst Case CT Error The next figure displays an external fault with the worst case CT mismatch.0 A I1 = 45.0 A I1 = 4.0A.0 A Figure 7: Simple Differential Protection with Worst Case CT Error and External Fault 5 . The differential current is 10A and the Iop element would incorrectly trip for a fault outside the zone using the previously defined setting of 1. The Iop coil attempts to pull the trip contacts together and the coil force is directly related to the Iop current.5A) flows through the Iop coil as before. The CT1 secondary current (4. The 6 .5 A I2 = 5. The two coils are designed so the Iop coil will be able to close the contacts if the ratio of Iop to Ir exceeds the relay’s slope setting. This Ir coil provides counter-force in electro-mechanical relays as shown in Figure 9 which pulls the trip contacts apart. The Ir coil force is directly related to the average of the I1 and I2 currents. 100 A 100:5 CT1 100:5 CT2 G Iop = 1.0 A I1 = 4. Therefore. the designer chooses a slope setting instead of a fixed current and the relay will adjust the pickup setting to the system parameters.5A) flows through Iop in opposition and the 1.0 A Figure 8: Percentage Differential Protection Schematic Ir Iop Figure 9: Percentage Differential Protection Operating Mechanism With this new system.The Relay Testing Handbook This new system adds a restraint coil (Ir) along with the existing operate coil (Iop)in the circuit as shown in Figure 8. The CT2 current (5. any slope setting is a ratio of Iop to Ir.5 A)/2 Ir = 5.5 A + 5. We will define current flowing through the Iop coil as "operate current" and current flowing through the Ir current as the "restraint current" for the rest of this book.5 A Ir = (4. Figure 10 displays normal operating conditions. Our example slope setting will be 25% for the following figures which repeat the previous fault simulations with the new design.0A differential current energizes the Iop coil which tries to close the trip contact. 5A of CT1 current and the other half has 5.0 slope = 20% < Slope setting 25% NO TRIP Figure 10: Percentage Differential Protection and External Faults The next figure displays an internal fault with only one source.5 A slope = 100 x Iop / Ir slope = 100 x 45.5 A)/2 Ir = 5. In this case the average of the two CT currents is 5.2 × 100 = 20% ).0 / 22. 100 A 100:5 CT1 100:5 CT2 G Iop = 1.0 A I1 = 45. The pickup 50A setting of this relay is 25% so the restraint coil is applying more force than the operate coil and the relay will not trip.0 A slope = 100 x Iop / Ir slope = 100 x 1.0 / 5. Chapter 1: Simple Percent Differential (87) Element Testing trip contact does not close because one-half of the restraint coil has 4.5A of CT2 current.5 slope = 200% > Slope setting 25% TRIP Figure 11: Percentage Differential Protection and Internal Faults 7 .5 A + 5.2. 1000 A 100:5 CT1 100:5 CT2 FAULT G Iop = 45.0 A Ir = (45 A + 0. The relay trips in this scenario due to the 200% slope. The average of these two currents creates the Ir force holding the contacts open.0A.5 A I2 = 5.0 A I1 = 4.0 A)/2 Ir = 22. 0. The ratio of 10A operate coil current and restraint coil current is 20% ( = 0.5 A Ir = (4. The Relay Testing Handbook The next figure displays an internal fault with two sources.5 0.5 Operate Current (Iop) 2.0 6.0 7. The relay will trip if the ratio of Iop to Ir is plotted above the line and will not trip if the ratio falls below the line.0 12.5 1.0 A Ir = (45 A + 55 A)/2 Ir = 50 A slope = 100 x Iop / Ir slope = 100 x 100 / 50 slope = 200% > Slope setting 25% TRIP Figure 12: Percentage Differential Protection and Internal Faults 2 The characteristic curve of this element is shown in the following figure.0 A I2 = 55. The relay trips in this scenario due to the 200% slope. Differential Protection 25% Characteristic Curve 3.0 A I1 = 45.0 0.0 8.0 Restraint Current (Ir) Figure 13: Percentage Differential Protection Characteristic Curve 8 .0 Restraint Area 0.0 5.0 Trip Area 1.0 11.0 2.0 1.0 2.0 3. 1000 A 1000 A 100:5 CT1 100:5 CT2 FAULT G Iop = 100.0 4.0 10.0 9. 16A 100A We can combine the two formulas to determine the calculated Tap setting in one step. If we were to zoom in on the origin of the graph in Figure 13. if the Tap settings exist. and we do not want the 87-Element to operate under normal conditions. Chapter 1: Simple Percent Differential (87) Element Testing Most relays have a Tap setting that defines the normal operating current based on the rated load of the protected equipment. Power = P-P Volts × Amps × 3 600000W =Amps = 83. One common situation occurs when the transformer is energized with no connected load. 4160V. we can determine that the rated primary current is 83. 600 kW generator.27A 5A × 83. and the CT ratio.27A ( 4160V × 3 ) The secondary current at full load using 100:5 CTs would be 4. CTSEC × Power AmpsSEC = P-P Volts × 3 × CTPRI The Tap setting could also be considered the per-unit current of the protected device and. 9 . imagine a 3-phase. but it is a normal condition. is the basis for all differential calculations. This is the definition of a differential fault.16A CTSEC AmpsSEC = CTPRI AmpsPRI 5A AmpsSEC = 100A 83. There will be some excitation current on the primary side of the transformer and no current on the secondary side. we would see that it takes very little operate current to trip this element at low power levels. Using the standard 3-phase power formula. the primary voltage.27A.27A AmpsSEC = = 4. or excitation current could cause this element to trip under these normal conditions. Any induced current. noise. For example. 0 0. 10 .0 4.3 x Tap current).0 2.0 1.0 1. The transition between the two slope settings is called the breakpoint or knee and is pre- defined by the manufacturer in some relays or user-defined in others. the chance of CT saturation or other problems occurring dramatically increase which can cause up to a 20% error during asymmetrical faults.4 Operate Current 1.0 7. Also notice that the Operate and Restraint currents are plotted as Multiples of Tap and not amps.0 8.The Relay Testing Handbook A new setting called Minimum Pickup is introduced to ensure that the 87-Element will trip under fault conditions and ignore normal mismatch at low current levels. The differential characteristic curve with a Minimum Pickup and two slope settings is shown in the following figure.4 0.8 0.8 1. and the differential protection will be less sensitive during high-level external faults. we can add a second slope setting to be used when the restraint current exceeds a manufacturer or user- defined level of current.6 1. This setting sets the minimum amount of current that must be present before the 87-Element will operate.6 Restraint Area 0.0 3. The breakpoint is set in multiples of Tap.2 0. Notice the flat line between 0 and 1. This setting is sometimes in secondary amps for simple differential applications.0 0.0 6.00 Restraint Current at the beginning of the curve which represents the Minimum Pickup setting (0.0 Restraint Current (Ir) x TAP Figure 14: Percentage Differential Protection Characteristic Curve with Minimum Pickup As the fault current increases.0 5.3 times the Tap setting. The differential characteristic curve changes when the Minimum Pickup setting is applied as shown in Figure 14. Differential Protection 25% Characteristic Curve 2. This second slope has a higher setting. but it is more commonly set as a percentage or multiple of the Tap setting and is typically set at 0. Because digital relays use programming instead of components for their pickup evaluations.2 Trip Area (Iop)xTAP 1. 00 8.00 Slope 2 (Iop)xTAP 1.00 1.50 Slope 1 1. the setting should be disabled or OFF to prevent confusion.00 2. This setting should be set around 0. If the element is not used.00 7.00 6. 11 . Unlatched indicates that the relay output contacts will open when the trip conditions are no longer present.00 0.00 5. A Latched option indicates that the output contacts will remain closed after a trip until a reset is performed and acts as a lockout relay.50 Trip Area Operate Current 2. Settings The most common settings used in 87-Elements are explained below: A) Enable Setting Many relays allow the user to enable or disable settings. Make sure that the element is ON.00 3. This minimum operate current is used to prevent nuisance trips due to noise or metering errors at low current levels.00 0.00 2. B) Minimum Pickup (Restrained) The Minimum Pickup setting is used to provide more stability to the differential element by requiring a minimum amount of current to flow before the differential element will operate.00 Restraint Current (Ir) x TAP Figure 15: Percentage Differential Protection Dual Slope Characteristic Curve 2. Some relays will also have “Latched” or ‘Unlatched” options. Chapter 1: Simple Percent Differential (87) Element Testing Differential Protection 25%/40% Characteristic Curve 3.50 Restraint Area 0.3 times the nominal or Tap setting of the protected device.00 4. or the relay may prevent you from entering settings. Always verify correct blocking operation by operating the end-device instead of a simulation to ensure the block has been correctly applied. Verify the correct Tap setting using the following formula. CTSEC × Power AmpsSEC = P-P Volts × 3 × CTPRI D) Slope-1 The Slope-1 setting sets the ratio of operate current to restraint current that must be exceeded before the 87-Element will operate. and the CT ratio. The Breakpoint is defined as a multiple of Tap and if the restraint current exceeds the Breakpoint setting. This setting is used as the per-unit operate current of the protected device and most differential settings are based on the Tap setting. If enabled. E) Slope-2 The Slope-2 setting sets the ratio of operate current to restraint current that must be exceeded before the 87-Element will operate if the restraint current exceeds a pre-defined or user-defined break point between Slope-1 and Slope-2. H) Block The Block setting defines a condition that will prevent the differential protection from operating such as a status input from another device. make sure the condition is not true when testing. G) Time Delay The Time Delay setting sets a time delay between an 87-Element pickup and trip. This setting is rarely used.The Relay Testing Handbook C) Tap The Tap setting defines the normal operate current based on the rated load of the protected equipment. F) Breakpoint This setting defines whether Slope-1 or Slope-2 will be used for the differential calculation. 12 . The slope setting is typically set at 20-30%. but can be set as high as 3 cycles for maximum reliability on some relays. This is typically set at the minimum possible setting. the primary voltage. the 87-Element will use Slope-2 for its calculation. Restrained-Differential Pickup Testing Performing simple differential pickup testing is very similar to the test procedure for overcurrent protection with a few extra phases to test.1 x CT •• Differential Trip Slope-1 = 20% •• Differential Trip Slope-2 = 80% •• Differential Trip Delay = 0 cycles The actual Minimum Pickup setting for this relay is 0.0A in North America and 1.16A and the Pickup setting is 0. The expected Minimum Pickup is 0. Apply a current in one phase. 13 . ( 4. If the relay does not have a Tap setting. and raise the current until the element operates.1 × 5. you must determine what the expected result should be before performing any test.5 A ) If the pickup test results are significantly higher or lower than the expected result. Repeat for all affected phases. refer to the next chapter’s instructions for pickup testing because the algorithm for the relay under test probably uses a complex equation for restraint current and correction factors will apply.3. Record the Pickup and Tap settings.0 A = 0.248A ) We will use a GE Power Management 489 relay with the following Phase Differential settings for the rest of the tests in this chapter. ( 0. •• Phase Differential Trip = Unlatched •• Assign Trip Relays = 1 •• Differential Trip Minimum Pickup = 0.3 = 1. The CT designation indicates the nominal CT secondary current (5. Chapter 1: Simple Percent Differential (87) Element Testing 3. The expected pickup current for this example would be 1. Multiply the Pickup and Tap settings to determine the Minimum Pickup in amps.248A.0A in Europe). The Tap setting from our previous example is 4.16A × 0.1 x CT because this relay does not have a Tap setting. As usual. refer to the manufacturer’s literature to determine if the relay uses some other nominal setting such as rated secondary current (5A usually).1 × CT = 0.5A. move the test leads from Winding-1 A-Phase amps to B-Phase amps and perform the test again. Repeat until all enabled phases are tested on all enabled windings. After the Winding-1 A-phase pickup test is performed. RELAY INPUT RELAY AØ PU RELAY TEST SET WINDING 1 A Phase Input=Pickup Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps AØ Test Amps 0° Test Hz + + B Phase Amps C2 Amps + + C Phase Amps C3 Amps WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection DC Supply - + Element + + Timer - Element Timer Output Input Output + Input Figure 16: Simple 87-Element Test-Set Connections You can also connect all three phases to one winding and change output channels instead of changing leads. After all Winding-1 tests are completed.The Relay Testing Handbook A) Test-Set Connections The most basic test-set connections use only one phase of the test-set with a test lead change between every pickup test. move the test leads to Winding-2 and repeat. 14 . Chapter 1: Simple Percent Differential (87) Element Testing TEST #1 RELAY INPUT TEST #2 RELAY INPUT TEST #3 RELAY INPUT AØ PU BØ PU CØ PU RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps AØ Test Amps 0° Test Hz + + B Phase Amps C2 Amps BØ Test Amps 0° Test Hz + + C Phase Amps C3 Amps CØ Test Amps 0° Test Hz WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection DC Supply - + Element + + Timer - Element Timer Output Input Output + Input Figure 17: Simple 3-Phase 87-Element Test-Set Connections If your test-set has six available current channels. TEST #1 TEST #2 TEST #3 TEST #4 TEST #5 TEST #6 RELAY INPUT RELAY INPUT RELAY INPUT RELAY INPUT RELAY INPUT RELAY INPUT A1Ø PU B1Ø PU C1Ø PU A2Ø PU B2Ø PU C2Ø PU RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps A1Ø Test Amps 0° Test Hz + + B Phase Amps C2 Amps B1Ø Test Amps 0° Test Hz + + C Phase Amps C3 Amps C1Ø Test Amps 0° Test Hz WINDING 2 + + A Phase Amps C4 Amps A2Ø Test Amps 0° Test Hz + + B Phase Amps C5 Amps B2Ø Test Amps 0° Test Hz + + C Phase Amps C6 Amps C2Ø Test Amps 0° Test Hz Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 18: Simple 3-Phase 87-Element Test-Set Connections with Six Current Channels 15 . you can use the following connection diagram and change the output channel for each test until all pickup values have been tested. 5A ELEMENT PICK-UP 4A 3A PICKUP SETTING 2A 1A STEADY-STATE PICKUP TEST Figure 19: Pickup Test Graph 4. Compare the pickup test result to the manufacturer’s specifications and calculate the percent error as shown in the following example. Make sure pickup indication operates. (“Accuracy: @ < 2 x CT: ± 0. For our example. We can immediately consider it acceptable as it falls within the metering accuracy specifications in Figure 20.500A ). output contact. The following graph displays the pickup procedure.5% of 2 x CT” = 0. (Chattering contacts or LEDs are not considered pickup. set the fault current at 0.5% × 10A =0.5A setpoint.525A for an element with a 0. 3.508A. Slowly raise current until pickup indication is fully on. etc…See previous packages of The Relay Testing Handbook for details.) 2. The difference between the test result and setting is 0. front panel display.05A ). Looking at the “Output And Neutral End Current Inputs” specification in Figure 20.008A ( 0. 16 . The 87-Element pickup setting is 0.The Relay Testing Handbook B) Pickup Test Procedure Use the following steps to determine pickup: 1.) Record the pickup values on your test sheet. Determine how you will monitor pickup and set the relay accordingly (if required). (Pickup indication by LED.5A and the measured pickup was 0.05A. we see that the acceptable metering error is 0. Slowly lower the current until the pickup indication is off.508A − 0. Set the fault current 5% higher than the pickup setting. Actual Value . Use “Accuracy: @ < 2 x CT: ± 0. Chapter 1: Simple Percent Differential (87) Element Testing OUTPUT AND NEUTRAL END CURRENT INPUTS Accuracy: @ < 2 x CT: ± 0.5A 10% Allowable Percent Error The measured percent error can be calculated using the percent error formula below.500A × 100 = Percent Error 0. Maximum Accuracy Tolerance × 100 = Allowable Percent Error Setting 0.5% × (2 × CT) × 100 = Allowable Percent Error 0.5A 0.5% of 2 x CT @ > 2 x CT: ± 1% of 20 x CT OUTPUT RELAYS Operate Time: 10ms PHASE DIFFERENTIAL Pickup Accuracy: as per Phase Current Inputs Timing Accuracy: +50ms @ 50/60 Hz or ± 0.Expected Value × 100 = Percent Error Expected Value 0.05A × 100 = Allowable Percent Error 0. we can calculate the manufacturer’s allowable percent error.500A 1.005 × 10A × 100 = Allowable Percent Error 0.5A 0. the allowable percent error is 10%.6% Error 5.508A .5A 0.5% of 2 x CT” from the specifications in Figure 20 because the applied current is less than 2x CT (20A). Repeat the pickup test for all phase currents that are part of the differential scheme.005 × (2 × 5A) × 100 = Allowable Percent Error 0.0. 17 .5 % of total time Figure 20: GE Power Management 489 Analog Input Specifications Using these two sections. 500 0.500 4. Restrained-Differential Timing Test Procedure The timing test procedure is very straightforward.20 1.60 W3 PICKUP NA NA NA 0.00 1.508 0. 18 . Use Figure 21 to determine the acceptable tolerances from the manufacturer’s specifications.6 cycles. Apply a single-phase current 10% greater than the Minimum Pickup setting into each input related to the percent differential element and measure the time between the applied current and the relay trip signal. we first must discover what an acceptable result is. However.500 1.508 0.508 0.The Relay Testing Handbook DIFFERENTIAL TEST RESULTS PICK UP 0.60 1.500 0. there are software and hardware delays built into the relay that must be accounted for as shown in Figure 22 that state that the maximum allowable time is 60ms or 3.60 1. As with all tests.1 TIME DELAY 0 SLOPE 1 10% SLOPE 2 20% MINIMUM PICK UP TESTS (Amps) A PHASE B PHASE C PHASE MFG % ERROR W1 PICKUP 0.506 0. Figure 21: GE Power Management 489 Differential and Output Relay Specifications The time delay setting for our example is 0 cycles which means that there is no intentional delay.505 0.00 W2 PICKUP 0. Connect the test-set input(s) to the relay output(s) that are programmed to operate when the restrained-differential relay operates.45 2.600 OK OK OK W3 NA NA NA NA NA 19 .500 3. 5.48 3. Choose a connection diagram from Figures 16-18 on pages 14-15.55A 2.490 2.55A 4. 2. Record the test results. RESTRAINED DIFFERENTIAL TIMING TESTS (cycles) WINDING TEST A PHASE (cy) B PHASE (cy) C PHASE (cy) MFG (cycles) % ERROR W1 0.600 OK OK OK W2 0.410 2.6 cycles Figure 22: GE Power Management 489 Minimum Trip Time Perform a timing test using the following steps: 1. Set a single-phase current at least 10% higher than the Minimum Pickup setting. Configure your test-set to start a timer when current is applied and stop the timer and output channels when the appropriate input(s) operate.55A 2. Minimum Pickup × 110% = 0. Run the test plan on the first phase related to restrained-differential. Repeat on all phases related to the restrained-differential.1 = 0. Chapter 1: Simple Percent Differential (87) Element Testing Reason Delay Phase Differential Timing Accuracy (Software) 50ms Output Relay Operate Time (Hardware) 10ms Total Time 60ms or 3.5A × 1. 3.44 2. Let’s look at the manufacturer’s recommended connection diagram shown in Figure 23 for the GE Power Management 489 relay in our example. A simple differential slope test is performed by choosing a phase (AØ for example) and applying current into that phase for two windings that are part of the differential element. a good understanding of the differential relay’s operating characteristics. This is where we will connect the first current from our test-set. Follow the other side of the CT to terminal H6 which is where we will connect the second current-channel from the test-set. Follow the other phases to determine these connections when testing B or C phases. Connect the neutral of the test-set current channel to G3 by following the other side of the CT to its relay terminal. refer to the Restrained-Differential Slope Testing section of the next chapter for more details. The instructions for this section use single-phase test techniques that will only work when applied to simple differential relays with identical CT ratios and no phase-angle shift between windings. and information from the manufacturer regarding the relay’s characteristics. If your test results do not fall into an acceptable range with this test procedure. 20 . Follow the AØ primary buss through the neutral CTs (Phase a) then follow the secondary CT to terminal H3. Keep following the primary buss through the generator to the other CT and then follow the secondary to terminal G6. If the results still do not match the expected results. The following test plan also assumes a simple slope calculation as described previously. but different relays may use different methods to determine slope. Restrained-Differential Slope Testing Differential slope testing is one of the most complex relay tests that can be performed and requires careful planning. refer to the manufacturer’s literature for the relay’s slope calculation and apply the manufacturer’s formula to the results. This is the neutral of the CTs so we will connect the test-set’s second current-channel-neutral to G6.The Relay Testing Handbook 5. G6 489 .H8 Channel 2 .H7 489 . Chapter 1: Simple Percent Differential (87) Element Testing Figure 23: GE Power Management 489 Specifications Test-set A phase B Phase C Phase Channel 1 + 489 . 489 . 489 .G7 489 .G5 Channel 2 + 489 .G4 489 .G8 Figure 24: GE Power Management 489 Slope Test Connection Table 21 .G3 489 .H3 489 .H4 489 .H6 489 .H5 Channel 1 . 0A / 5. This quick fix isn’t always possible when testing installed relays so we change the Channel 2 phase-angle to 180º to achieve the same result.0 A I2 = 5. the easy solution would be to switch the H3-G3 or H6-G6 connections.0 A H3 H6 Ir = (5 A + 5 A)/2 Slope = Iop / Ir Ir = 5 A Slope = 0. We would not be able to perform a slope test because the slope will always be 200% as shown in Figure 25.0 A G3 G6 I1 = 5.0 A H3 H6 Ir = (5 A + 5 A)/2 Slope = Iop / Ir Ir = 5 A Slope = 10.The Relay Testing Handbook Notice that the polarity marks of the CT’s are in opposite directions.0A / 5.0 A = 0% Figure 26: GE Power Management 489 Slope Test Connections Example #2 22 . all of the applied current would be differential or operate current. CHANNEL 1 CHANNEL 2 5A @ 0º 5A @ 180º G FAULT Iop = 0. If we were to apply both currents into the relay at the same phase-angle.0 A G3 G6 I1 = 5. If you look at Figure 26 carefully. CHANNEL 1 CHANNEL 2 5A @ 0º 5A @ 0º FAULT G Iop = 10.0 A = 200% Figure 25: GE Power Management 489 Slope Test Connections Example #1 If this test was performed on a bench. you’ll see that this is the normal running configuration for the GE 489 relay.0 A I2 = 5. Some tests require large amounts of current which will require the connection in Figure 28. Chapter 1: Simple Percent Differential (87) Element Testing A) Test-Set Connections The test-set connections for a simple slope test are as follows.00A 0° Test Hz WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection + + DC Supply - + Element Timer - Output Input Element Timer Output + Input Figure 27: Simple 87-Element Slope Test-Set Connections W1 RELAY INPUT W2 RELAY INPUT C1 Amps C2 Amps C3Amps RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps W1 Test Amps / 2 0° Test Hz + + B Phase Amps C2 Amps W1 Test Amps / 2 0° Test Hz + + C Phase Amps C3 Amps W2 Test Amps 180° Test Hz WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection + + DC Supply - + Element Timer - Output Input Element Timer Output + Input Figure 28: Simple 87-Element High Current Slope Test-Set Connections 23 . W1 RELAY INPUT W2 RELAY INPUT C1 Amps C2Amps RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps W1 Test Amps 0° Test Hz + + B Phase Amps C2 Amps W2 Test Amps 180° Test Hz + + C Phase Amps C3 Amps 0. 00 0.00 5. Then we can calculate the transition point between the Minimum Pickup and Slope-1 operation.00 6.1 x CT •• Differential Trip Slope-1 = 20% •• Differential Trip Slope-2 = 80% •• Differential Trip Delay = 0 cycles Differential Protection 20/80% Characteristic Curve 3.50 Trip Area Operate Current 2. •• Phase Differential Trip = Unlatched •• Assign Trip Relays = 1 •• Differential Trip Minimum Pickup = 0.00 2. Apply equal current into Winding-1 and Winding-2 with a 180º phase shift between windings.00 3. We can start by re-plotting the characteristic curve shown in Figure 29 using the following relay settings.1 × CT ) as we calculated earlier in this chapter. and raise one current until the relay operates. Slope-1 will be enabled when the operate current is greater than the minimum pickup.00 Slope 2 (Iop)xTAP 1.00 8.The Relay Testing Handbook B) Slope Test Procedure The test procedure seems straightforward.00 7. 24 .00 0. The difficult part of this procedure is determining what starting current to apply and what the expected pickup should be.50 Restraint Area 0.50 Slope 1 1.00 4. The Minimum Pickup is 0.50A ( 0.00 Restrain Current (Ir) x TAP Figure 29: Percentage Differential Protection Dual Slope Characteristic Curve We first need to change the restraint and operate current values to amps instead of multiples of Tap.00 1. Different relays have different slope calculations and we can refer to the GE 489 relay instruction manual to determine the relay’s differential calculation as shown in Figure 30.00 2. The minimum pickup value sets an absolute minimum pickup in terms of operate current. 0.501 Iop > k × Ir 20% = 100 × Ir 0. but rides through normal operational disturbances. Slope 2 if IR > = 2 X CT) IA = phase current measured at the output CT Ia = phase current measured at the neutral end CT Differential elements for phase B and phase C operate in the same manner.505 A 25 .5 A) defined previously.501 20% 20 =Ir = 2.501 50. The delay can be fine tuned to an application such that it still responds very fast.501 = 20% × Ir 100 × 0. and CT performance may produce erroneous operate signals. The bars on either side of each current indicate that we use 2 absolute values when calculating the restraint current. Chapter 1: Simple Percent Differential (87) Element Testing FUNCTION: The 489 percentage differential element has dual slope characteristic. our restraint current must be greater than 2.505 0. The bars above each current indicate a vector sum calculation IW 1 − IW 2 and we can redefine the Iop calculation as IW1 − IW 2 because our vectors will always be 180º apart.Ia IRestraint = 2 where: IOperate = operate current IRestraint = restraint current k = characteristic slope of differential element in percent (Slope 1 if IR < 2 X CT. Based on these calculations.505 0.2 Ir = 2. Figure 30: GE 489 Differential Formulas I A + Ia Notice that the Ir current (IRestraint) is defined as which can be translated into 2 IW 1 + IW 2 for our terminology. The Differential element for phase A will operate when: IOperate > k × IRestraint IOperate = IA . Use the first IOperate formula to calculate the transition between Minimum Pickup and Slope-1 where Slope-1 begins when Iop exceeds the Minimum Pickup setting (0.Ia IA . more than 2 x CT.1 Ir = = = 2. Notice that our generic formula on the right also works for this relay.505A to test Slope-1. This allows for very sensitive settings when fault current is low and less sensitive settings when fault current is high. We can drop the bars and re-define Ir as IW1 + IW 2 . The Iop current (IOperate) is defined as I A − Ia which can be translated into 2 for our calculations. Slope 2 if IR > = 2 X CT) IA = phase current measured at the output CT Ia = phase current measured at the neutral end CT Differential elements for phase B and phase C operate in the same manner. A quick review of the manufacturer’s operate description (Figure 32) of the manual informs us that the breakpoint is pre-defined as 2 times CT. where: IOperate = operate current IRestraint = restraint current k = characteristic slope of differential element in percent (Slope 1 if IR < 2 X CT.0 Restraint Area 5.0 Min 0. We can use the slope formula ( Slope(%) Iop ) = 100 × Ir to calculate the restraint current (Ir) at the transition point between Slope-1 and Slope-2 but we first need to define the transition point.The Relay Testing Handbook Figure 31 displays the revised characteristic curve so far. This is usually the Breakpoint setting.0 Slope 20. but this setting is not available in the GE 489 relay.0 1 15.0 Transition 1 10.0 30.0 Trip Area Operate Current (Iop in Amps) 25. Figure 32: Percentage Differential Protection Dual Slope Characteristic Curve 26 . Differential Protection 20% Characteristic Curve 35.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Restraint Current (Ir in Amps) Figure 31: Example Characteristic Curve in Amps with 1st Transition Defined We also need to determine the maximum restraint current before we accidentally start testing Slope-2 when testing Slope-1. 99 = IW 2 27 .222 8. This time we’re going to expand the formula to determine the actual W1 and W2 currents instead of operate and restraint currents. Iop > k × Ir Iop  I − IW 2  Slope(%) = 100 × = 100 ×  W 1  I + IW 2  Ir  IW 1 + IW 2   IW 1 − IW 2  =0.98 = 2.2IW 1 + 0.2IW 2 2I − 2IW 2  = 100 ×  W 1 Slope(%)  IW 1 + IW 2  2IW 1 − 2IW 2 = 0.222W 2 2 IW 1 19.98 = 2.222IW 2 + IW 2 9.8 20IW 1 + 20IW 2 = 200IW 1 − 200IW 2 IW 1 = 1.99 = IW 2 is defined as Ir<10A IW 1 + IW 2 Ir = Slope 1 expected test 2 currents can be defined by: 1.2 ×  W 1   2   2    0.2IW 2 20 ×  IW 1 + IW 2  = 100 × 2IW 1 − 2IW 2  2.2IW 2 + 2IW 2 1.2IW 2 Slope(%) ×  IW 1 + IW 2  = 100 × 2IW 1 − 2IW 2  2IW 1 − 0.222IW 2 The maximum amount of Ir 8.99 = IW 1 = 1.222IW 2 2 19.2IW 1 + 0.8IW 1 = 2.2I + 0.222IW 2 20IW 1 + 20IW 2 − 200IW 1 = −200IW 2 The maximum amount of Ir 20IW 1 − 200IW 1 = −200IW 2 − 20IW 2 is defined as Ir<10A −180IW 1 = −220IW 2 I + IW 2 Ir = W 1 220IW 2 2 IW 1 = 180 1.99 = IW 1 = 1.2IW 2 20IW 1  + 20IW 2  = 200IW 1 − 200IW 2  IW 1 = 1.2IW 2 2 ×  IW 1 − IW 2   IW 1 − IW 2  =W 1 Slope(%) = 100 × 2  IW 1 + IW 2  2 ×  IW 1 − IW 2 = 0.2IW 1 = 0. Chapter 1: Simple Percent Differential (87) Element Testing You can use the same Minimum Pickup formulas to calculate the transition from Slope-1 to Slope-2.222IW 2 + IW 2 9.222IW 2 IW 2 = 1. 0 20. Differential Protection 20/80% Characteristic Curve 50.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Restraint Current (Ir in Amps) Figure 33: Percentage Differential Protection Dual Slope Characteristic Curve 28 .0 Trip Area 40.0 Transition 1 15.0 Operate Current (Iop in Amps) 30.0 Restraint Area 5.0 Min 0.0 Transition 2 Slope 2 35.0 Slope 1 10.The Relay Testing Handbook The new characteristic curve will look like Figure 33.0 45.0 25. 333IW 2 280IW 2 IW 1 = 120 IW 1 IW 2 = 2.8IW 1 + 0.333IW 2 29 .2IW 1 = 2. Iop > k × Ir Iop  I − IW 2  Slope(%) = 100 × = 100 ×  W 1 Ir  IW 1 + IW 2   I + IW 2   2   IW 1 − IW 2  =0.8IW 2 = 100 ×  W 1 Slope(%)  IW 1 + IW 2  2IW 1 − 2IW 2 = 0. Chapter 1: Simple Percent Differential (87) Element Testing Now we can define the expected test currents when Slope-2 is required.8IW 2 Slope(%) ×  IW 1 + IW 2  = 100 × 2IW 1 − 2IW 2  2IW 1 − 0.8IW 1 + 0.2 80IW 1 + 80IW 2 = 200IW 1 − 200IW 2 IW 1 = 2.8IW 1 = 0.333IW 2 80IW 1 + 80IW 2 − 200IW 1 = −200IW 2 80IW 1 − 200IW 1 = −200IW 2 − 80IW 2 Slope 2 expected test currents can be defined by: −120IW 1 = −280IW 2 IW 1 = 2.8 ×  W 1     2  2 ×  IW 1 − IW 2  0.8IW 2 Slope(%) = 100 ×  IW 1 − IW 2  =W 1  IW 1 + IW 2  2 2I − 2IW 2  2 ×  IW 1 − IW 2 = 0.8IW 2 + 2IW 2 80 ×  IW 1 + IW 2  = 100 × 2IW 1 − 2IW 2  1.8IW 2 80IW 1  + 80IW 2  = 200IW 1 − 200IW 2  2.333 IW 1 = 2.8I + 0.8IW 2 IW 1 = 1. IF ((IW 2 * 1. 30 .0A from the x-axis to the characteristic curve.333) Differential Protection Characteristic Curve 40 35 Trip Area Slope 2 30 25 Operate Current (I W1 in Amps) 20 15 10 Restraint Area Slope 1 5 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Restraint Current (I W2 in Amps) Figure 34: Percentage Differential Protection Dual Slope Characteristic Curve in Amps You could use this graph to determine the expected pickup current for a Slope-1 test by choosing a W2 current greater than the Minimum Pickup and less than Slope-2.222). IW 2 * 1.0A) then IW 1 = 2.IW 2 < 0. then follow the crossover point back to the other axis to determine the expected pickup.0A as shown in Figure 35. IW 2 * 2.5A) then IW 1 = 1. Find that current on the x-axis.222IW 2 •• If (IW 2 > 10.0A to both windings and increase the W1 current until the relay operates. follow 12. we could test Slope-1 by applying 5.99A and Iop < 0. and follow it to the y-axis. The expected pickup current for the Slope-2 test is 28A as per Figure 35.5.222) .333IW 2 •• •• The spreadsheet calculation for these equations could be: IW 1 =IF (IW 2 < 8.5 + IW 2 . The relay should operate at approximately 6. For example.99A and Iop > 0.99. Test Slope-2 by choosing a current below the Slope-2 portion of the graph (12. follow the current up until it crosses the line.5 + IW 2 •• If (IW 2 < 8. 0.5A) then IW 1 = 0.The Relay Testing Handbook The actual characteristic curve of this 87-Element using Winding-1 and Winding-2 currents can be plotted where W2 (Restraint Current) is a group of arbitrary numbers and W1 (Operate Current) uses the following formulas: If (IW 2 < 8.0A for example). (Pickup indication by LED. Chapter 1: Simple Percent Differential (87) Element Testing Differential Protection Characteristic Curve 40 35 Trip Area Slope 2 30 25 Operate Current (IW1 in Amps) 20 15 10 Restraint Area Slope 1 5 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Restraint Current (I W2 in Amps) Figure 35: Using Graphs to Determine Pickup Settings It can be more convenient (and accurate) to use the formulas to determine the expected pickup current as shown below.222(5.11A IW 1 = 27.333(12. The W2 current for our example will be 5.0A) IW 1 = 6.) 2.0A) IW 1 = 2. Slope-1 Slope-2 IW 1 = 1. 31 . The pickup indication should be ON because we have applied a 200% slope as per the calculations earlier in this chapter.996A Follow these steps to test the Differential Slope settings: 1. Determine the Slope-1 restraint current by selecting a value between the Minimum- Pickup and Slope-2 transition points. front panel display. Apply the restraint current through W2. etc…See previous packages of The Relay Testing Handbook for details.222IW 2 IW 1 = 2.0A. Determine how you will monitor pickup and set the relay accordingly.333IW 2 IW 1 = 1. output contact. if required. ) 7. This is the Slope-1 pickup.11A using the Slope-1 formula when W2 = 5. 4. Looking at the “Output And Neutral End Current Inputs” specification in Figure 36. You can switch to the pulse method to reduce the possibility of equipment damage.05A per applied current or 0. Some organizations want more than one test point to determine slope. We measured the pickup to be 6. Apply identical current 180º from W2 into W1.0A.065A ( 6. Raise the W1 current until the element operates.175A − 6. 6.The Relay Testing Handbook 3. we see that the acceptable metering error is 0. OUTPUT AND NEUTRAL END CURRENT INPUTS Accuracy: @ < 2 x CT: ± 0.5% of 2 x CT” = 0.5% × 10 A =0.065A ) and the relay passes the test based on this criteria.05 A ).11A = 0.5% of 2 x CT @ > 2 x CT: ± 1% of 20 x CT OUTPUT RELAYS Operate Time: 10ms PHASE DIFFERENTIAL Pickup Accuracy: as per Phase Current Inputs Timing Accuracy: +50ms @ 50/60 Hz or ± 0. Repeat steps 2-4 with another restraint current between the Minimum-Pickup and Slope-2 transition points until the required number of tests are completed.5 % of total time Figure 36: GE Power Management 489 Specifications 32 . (Review the Instantaneous Overcurrent (50-Element) pickup procedure in previous packages of The Relay Testing Handbook for details. 5. we calculated that the 87-Element should operate when W1 = 6.175A. Slope-1 In our example. The pickup indication should turn OFF because there is no operate current as the two applied currents cancel each other out. Test Slope-2 by repeating steps 2-4 with a restraint current greater than the Slope-1 to Slope-2 transition level. The difference between the test result and calculated pickup is 0.175A. The measured pickup for our example is 6.10A including all applied currents (“Accuracy: @ < 2 x CT: ± 0. Compare the pickup test result to the manufacturer’s specifications and calculate the percent error as shown in the following example. Remember that any test current greater than 10A can potentially damage the relay and should only be applied for the minimum time possible. 712A.5% × (2 × CT) × 100 = Allowable Percent Error 6.11A × 100 = Percent Error 6.0% of 20x CT”=1% x 20 x 5A = 1.11A 2 × 0. 33 .005 × 10A Allowable Percent Error × 100 = 6.11A 0.5% of 2 x CT” from the specifications in Figure 36 because the applied current is less than 2x CT (20A). Looking at the “Output And Neutral End Current Inputs” specification in Figure 36.11A 1.11A 2 × 0. Maximum Accuracy Tolerance × 100 = Allowable Percent Error Setting 2 × 0. Chapter 1: Simple Percent Differential (87) Element Testing We can also calculate the manufacturer’s allowable percent error. the allowable percent error is 1.005 × (2 × 5A) × 100 = Allowable Percent Error 6.Expected Value × 100 = Percent Error Expected Value 6. Actual Value .00A per applied current or 2.996A using the Slope-2 formula when W2 = 12.6%. We measured the pickup to be 28.716A ( 28.0A. we calculated that the 87-Element should operate when W1 = 27. we see that the acceptable metering error is 1.00A).175A .996A = 0. The difference between the test result and calculated pickup is 0.10A × 100 = Allowable Percent Error 6.00A including all applied currents (“Accuracy: @ > 2 x CT: ± 1.716A ) and the relay passes the test based on this criteria.06% Error Slope-2 In our example. Use “Accuracy: @ < 2 x CT: ± 0.6.712A − 27.11 A 1.6% Allowable Percent Error The measured percent error can be calculated using the percent error formula below. 1%.01 × (20 × 5A) × 100 = Allowable Percent Error 27. Use “Accuracy: @ 2 x CT: ± 1.Expected Value × 100 = Percent Error Expected Value 28. change all connections and references to W2 to the next winding under test (W2 becomes W3 for W1-W3 tests). If more than two windings are used. 8.56% Error Rule of Thumb Remember that there are other factors that will affect the test result such as: •• Relay close time: The longer it takes the contact to close the higher the test result will be if ramping the pickup current.0% of 20 x CT” from the specifications in Figure 36 because the applied current is greater than 2 x CT (20A). Actual Value . 34 .996A × 100 = Percent Error 27.712A . Some of these error factors can be significant and a rule-of-thumb 5% error is usually applied to test results.996 A 2.996A 7.1% Allowable Percent Error The measured percent error can be calculated using the percent error formula below.The Relay Testing Handbook We can also calculate the manufacturer’s allowable percent error. the allowable percent error is 7. Repeat the pickup test for all phase currents that are part of the differential scheme. •• Test-set sensing time.0% × (20 × CT) × 100 = Allowable Percent Error 27. •• Test-set analog output error. Maximum Accuracy Tolerance × 100 = Allowable Percent Error Setting 2 × 1.27.996A 2 × 0. Remember to use the IOPERATE ( IW1 − IW 2 ) and IRESTRAINT ( IW1 + IW 2 ) formulas.06 1.29 W1 NA W3 NA NA NA NA SLOPE 2 TESTS (Amps) RESTRAINT TEST A PHASE (A) B PHASE (A) C PHASE (A) MFG (A) % ERROR W1 12 A W2 28.902 9.712 27.175 6.11 3.712 28.11 W1 NA W3 NA NA NA NA Alternate Slope Calculation There is another way to test slope that does not require as much preparation. Determine slope using the rise-over-run graphical method as shown in Figure 37 and the following formulas. Chapter 1: Simple Percent Differential (87) Element Testing SLOPE 1 TESTS (Amps) RESTRAINT TEST A PHASE (A) B PHASE (A) C PHASE (A) MFG (A) % ERROR W1 5A W2 6.899 9.085 34.083 36.996 2. Follow the same steps as the previous test and test 2 points on each slope and record the results.56 2.082 36.23 1.06 1.175 6.06 W1 8A W2 9.56 W1 15 A W2 36.11 3.712 28.995 3.896 9. 2 Differential Protection Characteristic Curve 40 35 Trip Area S lope 2 Rise2 30 25 R un2 Operate Current (I W1 in Amps) 20 15 10 S lope 1 R is e1 5 R un1 R es traint A rea 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Restraint Current (I W 2 in Amps) Figure 37: Determine Slope by Rise/Run Calculation 35 .56 2.110 1.26 1.175 6.776 1. 90A − 8A) − (6.896A − 1.175 6. The Relay Testing Handbook Rise %Slope2 = 100 × Rise Run %Slope1 = 100 × Run IOPERATE 2 − IOPERATE1 %Slope2 = 100 × I 2 − IOPERATE1 IRESTRAINT 2 − IRESTRAINT 1 %Slope1 = 100 × OPERATE IRESTRAINT 2 − IRESTRAINT 1 (IW 12 − IW 2 2) − (IW 11 − IW 21) %Slope2 = 100 × (I 2 − IW 2 2) − (IW 11 − IW 21)  IW 12 + IW 2 2   IW 11 + IW 21  %Slope1 = 100 × W 1  −   IW 12 + IW 2 2   IW 11 + IW 21   2   2   2 − 2      (36.71A − 12A) %Slope2 = 100 × (9.5875A 4.896 9.712 W1 15.085 84.712 28.082 36.902 21.28 7.3 84.5 21.175 6.08A + 15A   28.3% SLOPE 1 TESTS (Amps) RESTRAINT TEST A PHASE (A) B PHASE (A) C PHASE (A) SLOPE (%) % ERROR W1 5.3 5.3 84.3605 %Slope2 = 100 × 0.899 9.843 Slope1 = 21.356A %Slope1 = 100 × 8.541A − 20.0 W2 28.175A+5A   2   2   2 − 2      21.185 %Slope1 = 100 × 3.175 W1 8.37 5.90A + 8A   6.71A+12A  %Slope1 = 100 ×  −   9.083 36.6 7.712 28.67 8.5 21.5% Slope2 = 84.721 5.175A − 5A)  36.0 W2 9.712A %Slope2 = 100 × 1.39 W1 NA W3 NA NA NA 36 .35 5.082A − 16.948A − 5.37 %Slope2 = 100 × 0.0 W2 6.0 W2 36.175A 25.08A − 15A) − (28.07 W1 NA W3 NA NA NA SLOPE 2 TESTS (Amps) RESTRAINT TEST A PHASE (A) B PHASE (A) C PHASE (A) SLOPE (%) % ERROR W1 12. Chapter 1: Simple Percent Differential (87) Element Testing 6.5. ¾¾ Are you applying the currents into the same phase relationships? ¾¾ Did you calculate the correct Tap? ¾¾ Don’t forget to use the entire operate and restrained calculations. Tips and Tricks to Overcome Common Obstacles The following tips or tricks may help you overcome the most common obstacles. ¾¾ Before you start. the relay probably is applying correction factors and the test procedures in the next chapter should be used. apply current at a lower value and review the relay’s measured values to make sure your test-set is actually producing an output and your connections are correct. 37 . ¾¾ 180º apart? ¾¾ If the pickup tests are off by a 3 . or 0. 1.577 multiple. . ¾¾ The relay should trip if we close into a fault.Chapter 2 Percent Differential (87) Element Testing The previous chapter discussed differential protection on a single-phase basis. There are several new problems to be resolved when differential protection is applied to transformers which add complexity to our testing procedure: ¾¾ The simple definition of differential protection (current in equals current out) no longer applies (now power in equals power out) because the primary and secondary sides of the transformer have different voltage levels. ¾¾ Transformer differential relays are designed to ignore the large inrush current that occur when a transformer is initially energized. but most digital relays use Wye-connected CTs and the phase shift is compensated by relay settings. which is fine when discussing the simplest differential protective elements typically used for generator or motor protection. 39 . Additional protection is applied to operate quickly to clear the fault. ¾¾ The situation becomes more complex with Delta-Wye or Wye-Delta transformers. The phase-angle on the primary and secondaries of the transformer are no longer 180 degrees apart. Electro-mechanical relays solved this problem by changing the CT connections to compensate for the phase shift. Different voltage levels create different CT ratios and we now have to compensate for different Tap values on either side of the protected system. CT7. •• Relay 87OA is an overall differential relay with 4 inputs that will operate for any fault within the plant as defined by the locations of its input CTs. 40 . and CT11. GCB-1 GCB-2 CT1 CT2 CT3 87T1 GSU XFMR T1 87OA CT4 CT8 CT5 CT9 G AUX 87T2 87G GEN XFMR T2 G1 CT6 CT10 CT7 CT11 Figure 38: Zones of Protection Example There are four differential relays in Figure 38 that protect each major piece of equipment and the overall plant.The Relay Testing Handbook 1. Application We will start with a description of zones of protection typically used in protection schemes. CT1. The following figure depicts a typical electrical system with multiple zones of protection. A) Zones of Protection Zones of protection are defined by strategic CT placement in an electrical system and there are often overlapping zones to provide maximum redundancy. CT2. The differential protection should operate when a fault occurs inside the zone and ignore faults outside the zone. we were able to isolate one phase at a time to perform our testing. CT5. Conversely. Chapter 2: Percent Differential (87) Element Testing •• GSU XFMR T1 is protected by differential relay 87T1 that will operate if a fault is detected between CT3. 100 A 100:5 CT 100:5 CT A Phase a Phase IA Ia B Phase b Phase IB Ib C Phase c Phase G IC Ic Iop1 IrA Ira Iop2 IrB Irb Iop3 IrC Irc Figure 39: 3 Phase Generator Differential Protection 41 . the gap between CT8 and CT9 means that a fault in that gap will only trip the 87OA relay and will be difficult to locate. and CT8. The overlap between CT4 and CT5 provides 100% protection between the two transformers. Because there is no voltage difference or phase shift between windings. This is the same protection but displayed on a 3 phase basis. The CT ratios on both sides were identical and we could ignore the Tap setting as well. •• 87T2 protects AUX XFMR T2 between CT4 and CT6. B) Tap i) Simple Differential Protection The previous chapter demonstrated differential protection on a single-phase basis which should be applied to all three phases as shown in Figure 39. 98A Iop2 =0. If we apply the percent-slope formula from the previous chapter.4 A.500V.78A@-120º Ib=4.19A@-60º Ir3=3.000V to 34.98A Figure 40: 3 Phase Transformer Differential Protection A quick review of Figure 40 (that does not apply Tap settings) shows that the Iop current under normal conditions in each phase is 0.The Relay Testing Handbook Tap settings are very important when more complicated differential schemes are applied such as the 115. 151 A XFMR 1 503 A 30MVA 115kV:34.78A@120º Ic=4.98 Remember that these are ideal conditions and there is an additional 10% allowable CT error under symmetrical conditions and 20% error under asymmetrical conditions.19A@-60º Iop1 =0.4A IrA Ira IA=3.5kV 200:5 CT1A 600:5 CT1a A Phase Y:Y a Phase IA=3.19A@60º Ir2=3.19A@60º 200:5 CT1C 600:5 CT1c C Phase c Phase IC=3. We can apply Winding Taps to compensate for the different nominal currents.05% IRestraint 3.4 %Slope =100 × =100 × =10. 30 MVA transformer depicted in Figure 40.78A@0º Ia=4.19A@180º Ir1=3.98A Iop3 =0. we discover that there is approximately a 10% slope under normal operating conditions.4A IrB Irb IB=3.78A@120º Ic=4.78A@-120º Ib=4. 42 .19A@180º 200:5 CT1B 600:5 CT1b B Phase b Phase IB=3. IOP 0.78A@0º Ia=4.4A IrC Irc IC=3. 78A Actual 4.19A@-60º Actual 3.01% IRestraint 1.00 Figure 41: 3 Phase Transformer Differential Protection Using Tap Settings Now the measured slope is less than 1% under normal conditions in an ideal situation.99@0º Ira=1.2A =0.80A Tap setting on Winding-1 and 4.00 Notice that all of the numbers in the slope calculation do not include A or amps because these numbers are now related in percent of Tap or per-unit (PU).2A on Winding-2. 151 A XFMR 1 503 A 30MVA 115kV:34.01 %Slope =100 × =100 × =0. so it is now easier to use % of Tap or per-unit for all of our differential calculations. 43 .99@120º Irc=1.00 Iop3 =0.78A@-120º Ib=4. Understanding the Tap settings and correctly applying them to your differential calculations is vital when testing more complicated differential protection. Different Tap settings are applied.00@180º Ir1=1.00@60º Ir2=1.01 IrC Irc IrC=0.01 IrA Ira IrA=0.8A Tap 4.78A@120º Ic=4.19A %Tap = = = 0.19A@60º 200:5 CTC 600:5 CTc C Phase c Phase IC=3. Chapter 2: Percent Differential (87) Element Testing Figure 41 displays the same protection with a 3.00 Tap 3.78A@0º Ia=4.5kV 200:5 CTA 600:5 CTa A Phase Y:Y a Phase IA=3. IOP 0.00@-60º Ir3=1.01 IrB Irb IrB=0.99 Iop1 %Tap = = = 1.00 Iop2 =0.99@-120º Irb=1.19A@180º 200:5 CTB 600:5 CTb B Phase b Phase IB=3. 01 IrA Ira IrA=0. most transformers do not have Wye-Wye connections and many have Delta-Wye or Wye-Delta Connections.2A IrC Irc IrC=0.00@150º Ir1=1.78A@240º 200:5 CTH3 600:5 CTX3 H3 X3 @120º @120º 291A 151A IC Ic Ih3=3.00 Actual 6.00 Tap 6.99@210º Irb=1. Remember that we will discuss much simpler methods that require less analysis later in the chapter. The High and Low side currents (and corresponding CT ratios) entering and leaving the transformer have not changed 44 .The Relay Testing Handbook ii) Transformer Differential Protection with E-M Relays Unfortunately.01 Tap 4.55A@90º H0 Iop1 =0.19A %Tap = = = 0.55A@210º Ih3-Ih2=6. We will go through all of the changes in Figure 42 on a step-by- step basis to help you understand all of the challenges and it does get quite involved. 151 A XFMR 1 503 A 30MVA 115kV:34.99@330º Ira=1.78A@0º 200:5 CTH2 600:5 CTX2 H2 X2 @240º @240º 151A 291A IB Ib Ih2=3. The Wye-Delta transformer in Figure 42 has essentially the same characteristics (Voltage.19A@150º Ix2=Ib-Ia=4.19A@270º Ix1=Ia-Ic=4.99@90º Irc=1.01 IrB Irb IrB=0. MVA.55A Iop3 Actual 4. etc) but the transformer secondary has been reconfigured for a delta connection which completely changes the differential protection scheme.00 Figure 42: Wye-Delta Transformer Differential Protection Using CT Connections Let’s start with the primary connections.60A =0.00@30º Ir2=1.55A@330º Ih2-Ih1=6.99 %Tap = = = 1.00@270º Ir3=1.5kV 200:5 CTH1 600:5 CTX1 H1 Y: X1 151A 291A @0º @0º IA Ia Ih1=3.19A@30º Ih1-Ih3=6.78A@120º Ix3=Ic-Ib=4.00 Iop2 =0. we had to relabel many of the measured currents because of the new connections.78@ 0° + 3. and Ih3 to indicate that they are the secondary currents of the bushing CTs.78Amp 21 0 3.78@ 0°) 0 24 0 0 270 30 3.78@240° + (3.78@240° + −(3.78@120° + −(3.55@ 90° 45 .78@120° + (3.78@240° + 180°) 21 0 33 0 24 0 0 270 30 3. If you follow the circuit back from IrA.78@ 0° + 180°) 3. (CTH1 = IA. Ih2.78@180° 12 0 90 6 0 30 0 15 6. The current entering the differential relay’s transformer primary restraint coils is labeled IrA. The restraint current is the resultant of the delta connection. CTH3= IC). IrB.78@120° + 3. if you add two vectors that have equal magnitudes and are 60º apart.78 Amp 0 0 12 3.78@ 60° (420° − 360°) 6. Add the currents connected to the polarity marks to the currents connected to the non-polarity marks rotated by 180º as shown below. We must add these currents together vectorially to determine the restraint currents.78Amp 21 0 33 3.78@ 0° + (3. and H3 is now labeled with the capital H designation to indicate current measured at the primary bushings. The CT secondary currents have been relabeled with a lower case Ih1.78@240° + 3. Remember. IrB is connected to the polarity of Ih2 and the non-polarity of Ih1. H2. The current entering H1.78@240°) 30 0 15 180 0 3. you can determine the resultant vector by multiplying the magnitude by √3 and split the difference in angles Ih3 = 3.55@330° 90 6 0 0 12 30 0 15 Ih2 + −Ih1 180 0 Ih1 = 3. However.78Amp 21 0 33 0 24 0 0 270 30 Ih3 + −Ih2 90 6 -Ih2 = 3.78@ 0° + −(3.78@120° + 3. Chapter 2: Percent Differential (87) Element Testing because the MVA and voltage ratings haven’t changed. IrC is connected to the polarity of Ih3 and the non-polarity of Ih2.78@ 420° IrA = 6. you will discover that IrA is connected to the polarity of Ih1 and the non-polarity of Ih3.55 Amp 3. and IrC and is NOT the CT secondary current.55@210° 180 0 Ih1 = 3.78@120°) 12 0 90 6 0 30 0 3.78@300° 33 0 24 0 0 270 30 6.78@120° + 180°) 15 180 0 Ih1 = 3. CTH2 = IB. These CTs have been connected in Delta to compensate for the Delta-Wye shift of the transformer.78 Amp Ih1 + −Ih3 3. The H-winding is connected in Wye and the primary winding currents equal the terminal current. 6A. The Ia current magnitude can be calculated one of two ways. The phase- angles and winding currents will change due to the Delta connection.5kV       151A@0° × 115kV  151A@240° × 115kV  151A@120° × 115kV   3  3  3 Ia = Ib = Ic = 34. and Ic 3 3 currents using the transformer ratio.5kV 34.5kV 34. The transformer ratio is the same as the Wye-Wye example which means that the current magnitude flowing through the X-Bushings have not changed.4A . The transformer secondary winding is connected in Delta and. the CTs are connected in Wye to compensate for the phase shift between the primary and secondary windings. (Remember that transformer ratios are listed as phase-phase values. The new Tap for the primary winding is 6.The Relay Testing Handbook The primary winding Tap setting must change because the measured current at the relay is √3 larger due to the delta-connected CTs. Or we can calculate the Ia. I.5kV =Ia 291A@0° =Ib 291A@240° =Ic 291A@120° 46 .5kV 34. therefore.= IX1 503A Ia = = 290.b. We can use logic and the Delta-Wye rule of thumb that states the current inside the delta is √3 smaller than the current leaving the delta.5kV 34.) Ia VAN Ib VBN Ic VCN = = = IA Vab IB Vbc IC Vca IA×VAN IB×VBN IC×VCN Ia = Ib = Ic = Vab Vbc Vca       151A@0° ×  VAB  151A@240° ×  VBC  151A@120° ×  VCA   3  3  3 Ia = Ib = Ic = 34. IX1 is connected to Ia and –Ic. IX2 is connected to Ib and –Ia.19A@330° + 180° IX1=503@330° Ix1=4.19A@30° IX3=Ic + −Ib 1X3 Ix3= + 180° CT Ratio IX3=291@120° + −(291@240°) 503A@90° IX3=291@120° + (291@240° + 180°) Ix3= + 180° 120 IX3=291@120° + 291@ 60° Ix3=4. Ib.78A@240º 200:5 CTH3 600:5 CTX3 H3 X3 @120º @120º 151A 291A IC Ic Ih3=3.78A@0º 200:5 CTH2 600:5 CTX2 H2 X2 @240º @240º 151A 291A IB Ib Ih2=3.78A@120º H0 We can calculate the X-Bushing currents(Ia.19A@90° + 180° IX3=503@ 90° Ix2=4. The current is entering the non-polarity mark of the CTs and the CT secondary current (1x) will be the primary current divided by the CT ratio and rotated by 180º.5kV 200:5 CTH1 600:5 CTX1 H1 Y: X1 151A 291A @0º @0º IA Ia Ih1=3. Transformer Current CT Output 1X1 Ix1= + 180° CT Ratio IX1=Ia + −Ic 503A@330° Ix1= + 180° IX1=291@ 0° + −(291@120°) (600 5) IX1=291@ 0° + (291@120° + 180°) 503A@330° Ix1= + 180° 120 IX1=291@ 0° + 291@300° Ix1=4.19A@210° + 180° IX2=503@210° Ix2=4. and Ic) using the same calculations from the Delta CTs on the high side.19A@150° (510° − 360°) IX2=Ib + −Ia 1X2 Ix2= + 180° CT Ratio IX2=291@240° + −(291@ 0°) 503A@210° IX2=291@240° + (291@ 0° + 180°) Ix2= + 180° 120 IX2=291@240° + 291@180° Ix2=4. and IX3 is connected to Ic and –Ib. Chapter 2: Percent Differential (87) Element Testing 151 A XFMR 1 503 A 30MVA 115kV:34.19A@270° 47 .19A@510° Ix1=4. 99@330° =IrB 0.99@ 90° Ix1 Ix2 Ix3 Ira = Irb = Irc = Tap2 Tap2 Tap2 4.00 0.55A @330° 6.00 Ir3 = 1.00@270° IrA + Ira IrB + Irb IrC + Irc Ir1 = Ir2 = Ir3 = 2 2 2 0.00@150° =Irb 1.19A @270° Ira = Irb = Irc = 4.2A 4.01 48 .99 1. The Relay Testing Handbook XFMR 1 503 A 30MVA 115kV:34.01 Iop2 = 0.01 Iop3 = 0.99 1.60A =IrA 0.99 + 1.99 + 1.99 + 1.5kV 600:5 CTX1 Y: X1 291A @0º Ia 600:5 CTX2 X2 @240º 291A Ib 600:5 CTX3 X3 @120º 291A Ic We can now calculate the restraint currents for each winding and phase and the operate current for each phase.99@330° + 1.00       Iop1 = IrA + Ira Iop2 = IrB + Irb Iop3 = IrC + Irc ° Iop2 0.2A 4.00@30° = Iop1 0.2A =Ira 1.99@210° + 1.55A @210° 6. Ih1 − Ih3 Ih2 − Ih1 Ih3 − Ih2 IrA = IrB = IrC = Tap1 Tap1 Tap1 6.19A @30° 4. Remember that all calculations are in percent of Tap or per-unit.00 Ir2 = 1.55A @ 90° IrA = IrB = IrC = 6.00@270° Iop1 = 0.99@90° + 1.99@210° =IrC 0.00 Ir1 = Ir2 = Ir3 = 2 2 2 1.99 Ir1 = Ir2 = Ir3 = 2 2 2 Ir1 = 1.00@30° =Irc 1.00 0.60A 6.00@150= = Iop3 0.60A 6.19A @150° 4. 00@270º Ir3=1.99@330º Ira=1.19A@150º Ix2=Ib-Ia=4.55A@210º Ih3-Ih2=6.19A@30º Ih1-Ih3=6.01 Tap 4.99@90º Irc=1.19A %Tap = = = 0.00 Iop2 =0. Chapter 2: Percent Differential (87) Element Testing Ix3=Ic-Ib=4.01 IrA Ira IrA=0. 49 . Most calculations assume a balanced 3-phase condition and are usually only performed on A-phase. As you can see.01 IrB Irb IrB=0.99 %Tap = = = 1.99@210º Irb=1.55A Iop3 Actual 4.00 Actual 6.19A@270º Ix1=Ia-Ic=4.2A IrC Irc IrC=0.00 Tap 6.00@150º Ir1=1.00@30º Ir2=1.00 There is no reason to perform all three calculations when dealing with a balanced 3-phase condition as described previously.60A =0. all the phases have the same operate and restraint currents.55A@330º Ih2-Ih1=6.55A@90º Iop1 =0. 01 IrA Ira IrA=0.78A@0º Ix1=4.5kV 200:5 CTH1 600:5 CTX1 H1 Y: X1 151A 291A @0º @0º IA Ia Ih1=3.19A %Tap = Tap = 6.00@180º W1CTC=12 W2CTC=1 Iop2 =0.78A@240º Ix2=4.00@60º W1CTC=12 W2CTC=1 Iop3 =0.01 IrB Irb IrB=0. Figure 43 displays the same transformer described in Figure 42 with a digital relay.78A@120º Ix3=4.19A@150º 200:5 CTH2 600:5 CTX2 H2 X2 @240º @240º 151A 291A IB Ib Ih2=3.01 IrC Irc IrC=0.19A@270º H0 Actual 6.00@300º W1CTC=12 W2CTC=1 Figure 43: Wye-Delta Transformer Differential Protection Using CT Connections 50 .00 =0.60A = 0. 151 A XFMR 1 503 A 30MVA 115kV:34. The CTs are almost always connected in Wye and the Tap setting calculation is very similar to the simple differential protection described previously.55A Actual 4.19A@30º 200:5 CTH3 600:5 CTX3 H3 X3 @120º @120º 151A 291A IC Ic Ih3=3. and the Tap setting is relatively simple.99@120º Irc=1.2A = 1.99@0º Ira=1.The Relay Testing Handbook iii) Transformer Differential Protection with Digital Relays Most digital relays have correction factors and other algorithms to compensate for the phase shift between Delta-Wye transformers.99@240º Irb=1.99 Iop1 %Tap = Tap = 4. The third designation is the phase shift between windings and can be designated by a connection description. or clock position. The end result is the same with operate currents of 0.) The phase compensation angle replaces the different CT configurations required by electromechanical relays. H3 X3 •• “Y”H3is the secondary designation X3 for the Wye- C c C connected winding. phase-angle. The symbol represents the relay’s phase compensation settings which are “W1CTC=12” and “W2CTC=1” for this application. H0 X0 H0 H1 : X1 H1 Y: X1 A a A a H2 X2 H2 X2 B b B b 51 H3 X3 H3 X3 C c C c . All CTs are typically Wye-connected and the relay’s settings are changed to match the application. Both winding CTs are connected in Wye and are connected to the restrained coils (remember all of the coils have been replaced with digital algorithms now) and we use the same actual-current vs. Beckwith Electric and older SEL relays use the connection description as shown in the following figures: H1 Y:Y X1 H1 Y: X1 A a A a Phase-Angle Compensation H2 X2 H2 X2 B b •• “Y” B is the primary designation b for the Wye- connected winding. C) Phase-Angle Compensation Modern relays use formulas to compensate for the phase shifts between windings on transformers and can even compensate for unwanted zero-sequence currents. (Don’t worry.01A. Unfortunately there is no standard between relays or relay manufacturers and most problems in transformer differential applications occur trying to apply the correct compensation settings. and “Y” or “y” for Wye-connected windings. these settings will be explained in the next section. Tap-current to determine the per-unit or percent-of-Tap current. The winding is always designated with a “D” or “d” for Delta-connected windings. c •• This is a "YY" transformer. Chapter 2: Percent Differential (87) Element Testing Protecting this transformer with a digital relay has simplified the connections substantially. All phase-angle compensation settings are based on the transformer windings and their phase relationship using different codes depending on the manufacturer or relay model. 52 . Y X1 H1 Y: X1 a A a •• “D” is the W2 designation. X0 H0 Y : •• This is a "DACY" transformer. •• “AC” is the second designation because the W1 X3 H3 X3 winding is Delta and the X1 bushing is connected X0 c C H0 c to A and C phases. c •• This is a "DACDAC" transformer. X2 X1 H2 H1 Y: X2 X1 ab AB b a •• This is a "YDAC" transformer. •• “Y” is the W2 designation. •• “AC” is the third designation because the X1 bushing is connected to a and C-Phases. X2 c C H2 X2 c b B b •• This is a "YDAB" transformer. Phase=Angle Compensation if X is W1 X3 X2 H3 H2 X3 X2 bc BC cb •• “D” is the W1 designation. •• Both windings are “AC” because the H1 and X1 H3 X3 H3 X3 C c bushing C is connected to A and C phases. H2 X2 •• “D”H2is the secondary designation X2 for the Delta- B b B b connected winding. X2 H2 X2 X1 b B H1 Y: X1 b •• “D” is the W2 designation. •• “Y” is the W2 designation. X1 H1 X1 a A a Phase-Angle Compensation if H is Winding-1 •• “Y” is the W1 designation. •• “AB” is the second designation because the W1 winding is Delta and the X1 bushing is connected H0 to a and b phases. H0 Phase-Angle Compensation if H is Winding-1 •• “Y” is the W1 designation. a A a •• “AB” is the third designation because the X1 X3 H3 X3 bushing is connected to a and b phases. •• This is a "DABY" transformer. Phase-Angle Compensation if X is Winding-1 X3 H3 H0 X3 c C c •• “D” is the W1 designation. H3 X3 H3 X3 C c C c The Relay Testing Handbook H0 X0 H0 Phase-Angle Compensation H1 : X1 H1 Y: X1 A a A •• “D” is the primary designation afor the Delta- connected winding. f2b. The primary winding in transformer differential protection is determined by the CT connections to the relay. 102. Inc. G2. 105. Z05.g. 47. Z03. 49. use a lagging reference and the angles increase in the clockwise direction. the generator (low voltage) side of a generator set-up transformer]. M-3310—51. G3 •• GE T-60—f1a. the winding that is connected to the primary source [e. use a leading phase- angle reference as shown in Figure 44 where the angles increase in the counter-clockwise direction. 106 •• Schweitzer Engineering Laboratories SEL-387—Z01. Chapter 2: Percent Differential (87) Element Testing Some transformer differential relays define the phase relationship between windings in their settings using the actual phase displacement or clock positions as a simple reference. f3a. 46 •• Schweitzer Engineering Laboratories SEL-587—101. G1. f1b. 104. Some relays. Z06 There is an easy and hard way to determine transformer compensation settings that can be gleaned from the transformer nameplate or the single and three line drawings. The CT’s connected to the following terminals in the most common relays are considered to be the primary windings to the relay: •• GE/Multilin SR-745—Terminals H1. I always recommend looking at the nameplate instead of drawings to ensure you have the most accurate information. 12 o'clock 12 o'clock k 1o k 1o loc 0 or 12 'clo loc 0 or 12 'clo o'c 0º 33 0º 1 ck o'c 0º 11 11 º 11 11 º 1 ck 0 or 3 0 30 3 -30 3 º º 60 0 k 30 0 ck 1 loc 2 o 2 -60º 2o 2 30 1 lo o' c o'c 0º 'clo 'clo 0º º 60 10 or ck ck 10 º 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 3 9 3 90º 270º or -90º 270º 90º 0º 4 ock 0º 12 4 k 8o 8 º 8o 8 º c -12 'clo 'cl 'clo 'clo 12 0º 24 4o 4o or ck ck 0 0 7o ck 24 7o ck 'clo 'clo 'clo 'clo ck 5 o 5 -150 º 21 7 k c 5o 5 º 15 7 6 o'clock r 0º 6 o'clock 15 0 0º o 6 10º 6 2 180º 180º or -180º Figure 44: Phase Relationship / Clock Figure 45: Phase Relationship / Clock Position with Leading Angles Position with Lagging Angles Many people assume that the high voltage winding is the primary winding but this is not true. the grid (high voltage) side of a generator step-up transformer]. Other relays. such as SEL.g. 103. 0º is shifted to a vertical line to make the clock reference work. It is extremely important to understand what reference the relay uses for phase-angles. 50. H3. The primary winding could be the winding that is first energized [e. and Figures 44 and 45 display the phase-angle/clock references. 48. H2. f2a. Z02. such as GE Multilin. f3b •• Beckwith Electric Co. Z04. or arbitrarily chosen by the design engineer. 53 . The Relay Testing Handbook 1) The Hard Way The hard way requires a three line drawing of the transformers and a little vector addition. The first example is a Wye-Wye transformer with Wye-connected CTs as shown in the following example. The currents are the same and all three-phase primary currents are plotted in the first phasor diagram of Figure 47. Follow the current flowing through the H1 terminal into the IA winding. CT1 CT2 H1 400:5 X1 1200:5 PHA IA@0º Ia@0º IH1=IA@0º IX1=Ia@0º H2 X2 PHB IB@-120º Ib@-120º IH2@-120º IX2@-120º H3 X3 PHC IH3@120º IX3@120º IC@120º Ic@120º H0 X0 H1 H2 H3 H4 H5 H6 1A 1B 1C 2A 2B 2C G1 G2 G3 G4 G5 G6 GE/MULTILIN-745 Figure 46: Wye-Wye Transformer You should always start with the Wye-connected winding when determining phase relationships and the primary winding current phasors are plotted in the first vector diagram. We’ll review the 6 most common transformer connections and determine the vector relationship between windings. 54 . Using this information. Only the H1 and X1 currents are used when determining the vector relationships and are plotted in the third phasor diagram. “y” for the secondary connection. The secondary current is plotted in the second phasor diagram 180º from the actual current flowing into the relay to make the relationship between windings easier to understand. The current flowing into the H1 bushing has the same phase relationship as the current flowing through the X1 Bushing. 12 o'clock 12 o'clock 12 o'clock k 0 or 12 k 0 or 12 k 0 or 12 loc 1o loc 1o loc 1o o'c 0° 'clo o'c 0° 'clo o'c 0° 'clo 11 11 33 0° 1 ck 11 11 33 0° 1 ck 11 11 33 0° 1 ck ° ° ° 30 or -3 30 or -3 30 or -3 0° 0° 0° IA=IH1 IH1=Y0 10 lock 10 lock 10 lock 2 o 2 00° 2 o 2 00° 2 o 2 00° -6 -6 -6 Ia=IX1 o'c o'c o'c 0° 0° 0° 'clo 'clo 'clo ° ° ° 60 60 60 10 10 10 IX1=y0 or or or ck ck ck 3 3 3 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 3 9 3 9 3 90° -90° or 270° 90° -90° or 270° 90° -90° or 270° X2 Ic= H2 =I IC IX =I Ib =I Y0y0 = Yy0 20 4 ck 4 ck 4 ck ° ° ° 3 8o 8 40 8o 8 40 8o 8 40 IB 'clo 'clo 'clo H3 r2 r2 r2 'clo 'clo 'clo 4o 4o 4o 12 12 12 °o °o °o ck ck ck 0° 0° 0° 20 20 -1 -1 -1 7o ck 7o ck 7o ck 'clo 'clo 'clo 'clo 'clo 'clo 7 ck 5 o 5 210° 7 ck 5 o 5 210° 7 ck 5 o 5 210° 15 r 15 r 15 r 0° 6 o'clock 0 °o 0° 6 o'clock 0 °o 0° 6 o'clock 0 °o 6 5 6 5 6 5 -1 -1 -1 180° 180° 180° Figure 47: Wye-Wye Transformer Phasor Diagram 55 . Chapter 2: Percent Differential (87) Element Testing As you follow the current through the transformer’s AØ primary winding. we can determine the transformer vector relationship is Y0y0. and “0” to show that both windings are at the 0 o’clock position. All examples in this book use the Figure 44 characteristic to better illustrate the difference between connections. “Y” for the primary winding connection. We usually drop the first 0 and the correct notation would be "Yy0". 1 o'clock would be 30º and not -30º as shown because this relay uses the phase relationship in Figure 45. an opposite current flows out the aØ secondary winding and flows through the X1 bushing. It is important to note that if the phasor diagrams were displayed on the actual relay software. F3a. The first phasor diagram demonstrates the IA and -IC addition and the final result “IH1” is found at the 1 o’clock position or -30º. The second phasor diagram demonstrates the Ia and -Ic addition and the final result “IX1” is found at the 1 o’clock position or -30º. F2a. The current flowing through X1 is “Ia and -Ic” because X1 is connected to the polarity of the aØ winding and the non-polarity of the cØ winding. The secondary current is plotted in the second phasor diagram 180º from the actual current flowing into the relay to make the relationship between windings easier to understand. F2b. F1b.The Relay Testing Handbook Our next example uses a Delta-Delta transformer. There is no Wye-connected winding in this transformer configuration and we will start with the primary winding as defined by the CTs connected to F1a. 56 . and F3b. CT1 CT2 400:5 1200:5 H1 IH1=IA-IC@-30º X1 IX1=Ia-Ic@-30º PHA IA@0º Ia@0º H2 X2 PHB IB@-120º Ib@-120º IH2@-150º IX2@-150º H3 X3 PHC IH3@90º IX3@90º IC@120º Ic@120º F1a F2a F3a M1a M2a M3a 1A 1B 1C 2A 2B 2C F1b F2b F3b M1b M2b M3b GENERAL ELECTRIC T-60 Figure 48: Delta-Delta Transformer Connections The current flowing through H1 is the transformer “IA and -IC” currents because H1 is connected to the polarity of the AØ winding and the non-polarity of the CØ winding. - 180° 180° 180° 12 o'clock 11 o'clock k 0 or 12 k 11 12 loc 1o loc o o'c 0° 'clo o'c 30° 0 o 'clo 11 11 33 0° 1 ck 10 10 r ck 30 ° or 60 ° 0° 12 -3 0° IH1=D1 IH 10 lock 9 ck 1= 2 o 2 00° 1 o 1 30° 'clo D1 -6 -3 o'c 0° 0° 'clo 'clo ° ° 9o 60 90 10 IX1=d1 or or ck ck 3 3 IX 1= d1 9 o'clock 3 o'clock 8 o'clock 2 o'clock 9 3 8 2 90° -90° or 270° 120° -60° or 300° D1d1 = Dd0 D1d1 = Dd0 20 4 ck 0° 3 ck ° 0° 8o 8 7o 7 40 'clo 'clo 27 r2 'clo 'clo 4o 3o 12 15 or °o ck ck 0° 0° -9 -1 7o ck 6o ck 'clo 'clo 'clo 'clo 7 ck 5 o 5 210° 6 ck 4 o 4 240° 15 r 18 r 0° 6 o'clock 0 °o 0° 5 o'clock 0 °o 6 5 5 2 -1 -1 180° -150° or 210° Figure 49: Delta-Delta Transformer Phasors 57 .” You could also change the reference instead of the phasors as shown in the fifth phasor diagram. It is important to note that if the phasor diagrams were displayed on the actual relay software. and “1” to show that both windings are at the 1 o’clock position. . The first “0” is usually dropped from the designation and the Delta-Delta connection is displayed as “Dd0. 12 o'clock 12 o'clock 12 o'clock k 0 or 12 k 0 or 12 k 0 or 12 loc 1o loc 1o loc 1o o'c 0° 'clo o'c 0° 'clo o'c 0° 'clo 11 11 33 0° 1 ck 11 11 33 0° 1 ck 11 11 33 0° 1 ck ° ° ° 30 or -3 30 or -3 30 or -3 IH 0° IX 0° 0° 1= 1= IH 10 lock 10 lock 10 lock IA Ia 1= -IC -Ic -Ic 2 o 2 00° 2 o 2 00° 2 o 2 00° D1 -6 -6 -6 o'c o'c o'c 0° 0° 0° 'clo 'clo 'clo ° ° ° 60 60 60 10 10 10 Ia or or or IA ck ck ck 3 3 3 IX 1= -IC d1 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 9 9 IH3 IX3 3 3 3 90° -90° or 270° 90° -90° or 270° 90° -90° or 270° IB IC Ic Ib D1d1 4 ck 4 ck 4 ck ° ° ° 8o 8 8o 8 8o 8 40 40 40 'clo 'clo 'clo r2 r2 r2 'clo 'clo 'clo 4o 4o 4o 12 12 12 2 2 °o °o °o IH IX ck ck ck 0° 0° 0° 20 20 20 -1 -1 -1 7o ck 7o ck 7o ck 'clo 'clo 'clo 'clo 'clo 'clo 7 ck 5 o 5 210° 7 ck 5 o 5 210° 7 ck 5 o 5 210° 15 r 15 r 15 r 0° 6 o'clock °o 0° 6 o'clock °o 0° 6 o'clock °o 6 1 50 6 1 50 6 1 50 . “D” for the primary winding connection. Chapter 2: Percent Differential (87) Element Testing Only the H1 and X1 currents are used when determining the vector relationships and are plotted in the third phasor diagram. we can determine the transformer vector relationship is D1d1. The primary winding should always be at the 0 or 12 o’clock position and we can rotate both phasors as shown in the fourth phasor diagram to display a “D0d0” transformer. All examples in this book use the Figure 44 characteristic to better illustrate the difference between connections. “d” for the secondary connection. 1 o'clock would be 30º and not -30º as shown because this relay uses the phase relationship in Figure 45. Using this information. The Relay Testing Handbook The next transformer is a Wye-Delta transformer with a Wye side primary because the Wye side, H-Bushing CTs are connected to 47, 49, 51, 46, 48, and 50. CT1 CT2 400:5 1200:5 H1 X1 IX1=Ia-Ic@-30º PHA Ia@0º IA@0º IH1=0º H2 X2 PHB Ib@-120º IB@-120º IH2@-120º IX2@-150º H3 X3 PHC IH3@120º IX3@90º Ic@120º IC@120º H0 47 49 51 55 57 59 1A 1B 1C 2A 2B 2C 46 48 50 54 56 58 BECKWITH ELECTRIC M-3310 Figure 50: Wye-Delta Transformer Connections Starting with the Wye currents, draw a phasor diagram of the current flowing through the primary bushings (IH1, IH2, & IH3) which are equal to the phase (IA, IB, & IC) currents as shown in the first phasor diagram. The current flowing through X1 is “Ia and -Ic” because X1 is connected to the polarity winding of the aØ winding and the non-polarity of the cØ winding. The secondary current is plotted in the second phasor diagram 180º from the actual current flowing into the relay to make the relationship between windings easier to understand. The second phasor diagram demonstrates the Ia and -Ic addition and the final result “IX1” which is found at the 1 o’clock position or -30º. The third phasor diagram displays the relationship between the primary and secondary windings which can be translated to “Y0d1” or “Yd1” using correct notation. 12 o'clock 12 o'clock 12 o'clock k 0 or 12 1o k 0 or 12 1o k 0 or 12 1o loc loc loc o'c 0° 'clo c o'c 0° 'clo c o'c 0° 'clo c 11 11 330 ° 1 k 11 11 330 ° 1 k 11 11 330 1 k ° ° ° ° 30 or -30 30 or -30 30 or -30 ° IX1 ° IH1=Y0 ° IA=IH1 =Ia 10 ck 10 ck 10 ck -Ic -Ic 2 o 2 00° 2 o 2 00° 2 lo lo lo -60 -60 -60 o'c IX1 o'c o'c o'c 'clo 'clo ° ° ° ° ° ° loc =d 60 60 60 2 00° 10 10 10 or or or ck ck Ia 1 k 3 3 3 9 o'clock 3 o'clock 9 o'clock 3 o'clock o'clock 3 o'clock IX3 9 3 9 3 9 3 90° -90° or 270° 90° -90° or 270° 90° -90° or 270° Ic Ib H2 IC Y0d1 = Yd1 =I =I 0° 0° 0° ck ck 0° 4 k 8 8o 8 8o 8 IB H3 c 'clo 'clo 'clo 24 24 24 o'c 'clo 'clo 0° 4 0° 4 loc 4o 4o 4o or or or 12 12 12 8 ° 2 ck ck IX k 0 0° 0° -12 -12 -12 7 7 ck 7 o'c ck o'c o'c ck lo 'clo lo 'clo lo 'clo 7 ck 5 o 5 210° 7 ck 5 o 5 210° 7 ck 5 o 5 210° 150 150 150 ° 6 o'clock or ° 6 o'clock or ° 6 o'clock or 0° 0° 0° 6 -15 6 -15 6 -15 180° 180° 180° Figure 51: Wye-Delta Transformer Phasor Diagrams 58 Chapter 2: Percent Differential (87) Element Testing The next example is a Wye-Delta transformer with a different delta configuration. The Wye or high voltage winding is the primary winding because the Wye side CTs are connected to the SEL-587 relay terminals Z01, Z03, Z05, Z02, Z04, and Z06. CT1 CT2 400:5 1200:5 H1 X1 IX1=Ia-Ib@30º IA@0º PHA Ia@0º IH1=0º H2 X2 PHB IB@-120º Ib@-120º IH2@-120º IX2@-90º H3 X3 PHC IH3@120º IX3@150º IC@120º Ic@120º H0 101 103 105 107 109 111 IBW1 IBW2 ICW1 ICW2 IAW1 IAW2 102 104 106 108 110 112 SCHWEITZER ENGINEERING LABORATORIES SEL-587 Figure 52: Wye-Delta Alternate Transformer Connections Starting with the Wye currents, draw a phasor diagram of the current flowing through the primary bushings (IH1, IH2, & IH3) which are equal to the phase (IA, IB, & IC) currents as shown in the first phasor diagram. The second phasor diagram displays the Delta-winding phasors with the phase currents in phase with the Wye-connected winding and the line current equal to “Ia-Ib” because the X1 bushing is connected to the aØ winding and the bØ non-polarity winding. The third phasor shows the phase relationship between windings and can be described as Y0d11 or “Yd11”. 12 o'clock 12 o'clock 12 o'clock k 0 or 12 k 0 or 12 k 0 or 12 loc 1o loc 1o loc 1o o'c 0° 'clo o'c 0° 'clo o'c 0° 'clo 11 11 33 0° 1 ck 11 11 33 0° 1 ck 11 11 33 0° 1 ck ° ° ° 30 or -3 30 or -3 30 or -3 0° 0° 0° IA=IH1 IH1=Y12 10 lock 10 lock 10 lock 2 o 2 00° 2 o 2 00° 2 o 2 00° Ib -6 -6 -6 11 o'c o'c o'c a- d 0° 0° 0° 'clo 'clo 'clo ° ° ° =I -Ib 1= 60 60 60 10 10 10 1 or or or IX IX ck ck ck 3 3 3 Ia 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 3 9 3 9 3 90° -90° or 270° 90° -90° or 270° 90° -90° or 270° Ib Ic H2 Y0d11 = Yd11 IC =I =I ck 4 ck 4 ck 0° ° ° 8o 8 8o 8 8o 8 40 40 IB 'clo 'clo 'clo H3 24 r2 r2 'clo 'clo 'clo 0° 4 4o 4o 4o or 12 12 12 °o °o ck ck ck 0° 0° 0° 20 20 2 -1 -1 -1 7o ck 7o ck 7o ck 'clo 'clo 'clo 'clo 'clo 'clo 7 ck 5 o 5 210° 7 ck 5 o 5 210° 7 ck 5 o 5 210° 15 r 15 r 15 r 0° 6 o'clock °o 0° 6 o'clock °o 0° 6 o'clock °o 6 1 50 6 1 50 6 1 50 - - - 180° 180° 180° Figure 53: Wye-Delta Alternate Transformer Phasor Diagrams 59 The Relay Testing Handbook The example in Figure 50 can easily become a Delta-Wye transformer by switching the protective relay connections as shown in the next example. CT1 CT2 400:5 1200:5 H1 X1 IX1=Ia-Ic@-30º PHA Ia@0º IA@0º IH1=0º H2 X2 PHB Ib@-120º IB@-120º IH2@-120º IX2@-150º H3 X3 PHC IH3@120º IX3@90º Ic@120º IC@120º H0 55 57 59 47 49 51 1A 1B 1C 2A 2B 2C 54 56 58 46 48 50 BECKWITH ELECTRIC M-3310 Figure 54: Delta-Wye Transformer Connections You should always start with the Wye-winding and the phase and line currents are drawn in the second phasor diagram because the Wye-winding is the secondary winding. 60 The third diagram displays the phasor relationship between windings and the transformer is described as D1y0 based on this diagram. 12 o'clock 12 o'clock 12 o'clock k 0 or 12 k 0 or 12 k 0 or 12 loc 1o loc 1o loc 1o o'c 0° 'clo o'c 0° 'clo o'c 0° 'clo 11 11 33 0° 1 ck 11 11 33 0° 1 ck 11 11 33 0° 1 ck ° ° ° 30 or -3 30 or -3 30 or -3 0° 0° 0° IX IX 1= 10 lock 10 lock 10 lock Ia 1= 2 o 2 00° 2 o 2 00° 2 o 2 00° -Ic -Ic D1 -6 -6 -6 IH1=IA IH1=y0 o'c o'c o'c 0° 0° 0° 'clo 'clo 'clo ° ° ° 60 60 60 10 10 10 or or or ck ck ck 3 3 3 Ia 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 3 9 3 9 3 90° -90° or 270° 90° -90° or 270° 90° -90° or 270° IB IH Ib Ic 2= D1y0 3= IH 4 ck 4 ck 4 ck IC ° ° ° 8o 8 8o 8 8o 8 40 40 40 'clo 'clo 'clo r2 r2 r2 'clo 'clo 'clo 4o 4o 4o 12 12 12 °o °o °o ck ck ck 0° 0° 0° 20 20 20 -1 -1 -1 7o ck 7o ck 7o ck 'clo 'clo 'clo 'clo 'clo 'clo 7 ck 5 o 5 210° 7 ck 5 o 5 210° 7 ck 5 o 5 210° 15 r 15 r 15 r 0° 6 o'clock °o 0° 6 o'clock °o 0° 6 o'clock °o 6 50 6 50 6 50 -1 -1 -1 180° 180° 180° 12 o'clock 11 o'clock k 0 or 12 k 11 12 loc 1o loc o o'c 0° 'clo o'c 30° 0 o 'clo 11 11 33 0° 1 ck 10 10 r ck 30 ° or 60 ° 0° 12 -3 0° IX1=D0 IX 10 lock 9 ck 1= 2 o 2 00° 1 o 1 30° D0 'clo -6 -3 1 IH1=y11 o'c y1 0° 0° 'clo 'clo ° ° 9o = 60 90 10 H1 or or I ck ck 3 3 9 o'clock 3 o'clock 8 o'clock 2 o'clock 9 3 8 2 90° -90° or 270° 120° -60° or 300° D0y11= Dy11 D0y11= Dy11 20 4 ck 0° 3 ck ° 0° 8o 8 7o 7 40 'clo 'clo 27 r2 'clo 'clo 4o 3o 12 15 or °o ck ck 0° 0° -9 -1 7o ck 6o ck 'clo 'clo 'clo 'clo 7 ck 5 o 5 210° 6 ck 4 o 4 240° 15 6 o'clock or 18 5 o'clock or 0° 0° 0° 0° 6 5 5 2 -1 -1 180° -150° or 210° Figure 55: Delta-Wye Transformer Connections 61 . The primary winding must always be at 12 o’clock and we can rotate the phasors or the clock to obtain the correct description D0y11 or “Dy11” as shown in the fourth and fifth phasor diagrams. Chapter 2: Percent Differential (87) Element Testing The first diagram follows the steps from Figure 50 for the Delta-winding because it is now the primary winding. & IC) currents as shown in the second phasor diagram. IB.The Relay Testing Handbook The example in Figure 52 can easily become a Delta-Wye transformer by switching the protective relay connections as shown in the next example where the Delta-winding is connected to SEL-387 terminals Z01. The first phasor diagram displays the Delta-winding phasors with the phase currents in phase with the Wye-connected winding and the line current equal to “Ia-Ib” because the X1 bushing is connected to the aØ winding and the bØ non-polarity winding. & IH3) which are equal to the phase (IA. 62 . draw a phasor diagram of the current flowing through the primary bushings (IH1. Z04. Z02. IH2. Z03. Z05. and Z06 CT1 CT2 400:5 1200:5 H1 X1 IX1=Ia-Ib@30º IA@0º PHA Ia@0º IH1=0º H2 X2 PHB IB@-120º Ib@-120º IH2@-120º IX2@-90º H3 X3 PHC IH3@120º IX3@150º IC@120º Ic@120º H0 Z07 Z09 Z11 Z01 Z03 Z05 IBW1 IBW2 ICW1 ICW2 IAW1 IAW2 Z08 Z10 Z12 Z02 Z04 Z06 SCHWEITZER ENGINEERING LABORATORIES SEL-387 Figure 56: Delta-Wye Alternate Transformer Connections Starting with the Wye currents. - 180° 180° 180° 12 o'clock 1 o'clock k 0 or 12 k 1 loc 1o loc 2o o'c 0° 'clo o'c 2 330° or -30° 'clo 11 11 33 0° 1 ck 12 or 1 30 0° 2 ck ° 0 0° 30 or -3 or -6 0° 0° IX1=D0 D0 10 lock 11 lock 1= 2 o 2 00° 3 o 3 90° -6 27 IH IX IH1=y1 o'c o'c 1= 0° 'clo 'clo ° ° 0° 60 30 10 11 y1 or or ck ck 3 - 9 o'clock 3 o'clock 10 o'clock 4 o'clock 9 3 10 4 90° -90° or 270° 60° 240° or -120° D0y1 = Dy1 D0y1 = Dy1 20 4 ck 0° 5 ck ° ° 50 8o 8 40 9o 9 'clo 'clo -1 r2 'clo 'clo 4o 5o 12 90 or °o ck ck 0° ° 21 -1 7o ck 8o ck 'clo 'clo 'clo 'clo 7 ck 5 o 5 210° 8 ck 6o 6 15 r 12 0° 0° 6 o'clock °o 0° 7 o'clock 18 6 50 7 -1 180° 150° Figure 57: Delta-Wye Alternate Transformer Phasor Diagrams 63 . The primary winding must always be at 12 o’clock and we can rotate the phasors or the clock to obtain the correct description D0y1 or “Dy1” as shown in the fourth and fifth phasor diagrams. . 12 o'clock 12 o'clock 12 o'clock k 0 or 12 k 0 or 12 k 0 or 12 loc 1o loc 1o loc 1o o'c 0° 'clo o'c 0° 'clo o'c 0° 'clo 11 11 33 0° 1 ck 11 11 33 0° 1 ck 11 11 33 0° 1 ck ° ° ° 30 or -3 30 or -3 30 or -3 0° 0° 0° -Ib D1 1 10 lock 10 lock 10 lock Ia 1= IA=IH1 1= 2 o 2 00° 2 o 2 00° 2 o 2 00° -6 -6 -6 -Ib IX IH1=y0 'c 'c 'c IX 0° 0° 0° o o o 'clo 'clo 'clo ° ° ° 60 60 60 10 10 10 or or or Ia ck ck ck 3 3 3 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 o'clock 3 o'clock 9 3 9 3 9 3 90° -90° or 270° 90° -90° or 270° 90° -90° or 270° H2 IB =I Ib =I Ic D11y0 IB H3 4 ck 4 ck 4 ck ° ° ° 8o 8 40 8o 8 40 8o 8 40 'clo 'clo 'clo r2 r2 r2 'clo 'clo 'clo 4o 4o 4o 12 12 12 °o °o °o ck ck ck 0° 0° 0° 20 20 20 -1 -1 -1 7o ck 7o ck 7o ck 'clo 'clo 'clo 'clo 'clo 'clo 7 ck 5 o 5 210° 7 ck 5 o 5 210° 7 ck 5 o 5 210° 15 r 15 r 15 r 0° 6 o'clock °o 0° 6 o'clock °o 0° 6 o'clock °o 6 1 50 6 1 50 6 1 50 . Chapter 2: Percent Differential (87) Element Testing The third diagram displays the phasor relationship between windings and the transformer is described as D11y0 based on this diagram. 64 . This symbol represents a Wye-connected winding and symbol represents a Delta-connected winding. You can use any phase at the 12 o’clock position to make it easier to translate as long as the same phase is used for both windings. or 1 o’clock position. The secondary winding symbol’s AØ position will be either in the 11. or the transformer nameplate which should all have a set of symbols like that describe the phase relationship between windings. Redraw the secondary winding symbol to the right of the primary winding with the same phase relationship as shown on the nameplate. These are the only two symbols you need to determine phase relationship after reviewing the CT connections to determine which winding the relay considers to be the primary winding. 12. redraw its connection symbol with AØ at the 12 o’clock position as shown in the next example.The Relay Testing Handbook 2) The Easy Way The easy way requires a single line drawing. The following table displays the most common transformer connections and their phase relationships. After the primary winding is determined. three line drawings. Chapter 2: Percent Differential (87) Element Testing Phase Symbol Description A a k 12 o’clock 1o loc 0 or 12 ’clo o’c 33 ck 11 11 0° 0o 1 Aa ° r- 30 30 ° Yy0 C B c b A a A a k 12 o’clock 1o loc 0 or 12 ’clo o’c 33 ck 11 11 0° 0o 1 ° Aa 30 r -3 0° C c Dd0 C B c b B b 12 o’clock A a k 1o loc 0 or 12 ’clo o’c 33 ck 11 11 0° 0o 1 Aa ° r- 30 30 ° c Yd1 C B b A a A 12 o’clock a k 1o loc 0 or 12 ’clo o’c 33 ck 11 11 0° 0o 1 ° aA 30 r -3 0° C b c b C B Dy11 B c a 12 o’clock 1o A k loc 0 or 12 ’clo o’c 33 ck 11 11 0° 0o 1 aA ° r- 30 30 ° b Yd11 C B c A a A a k 12 o’clock 1o loc 0 or 12 ’clo o’c 33 ck 11 11 0° 0o 1 Aa ° r -3 30 0° B c Dy1 c b C B C b Figure 58: Transformer Nameplate Phase Relationships 65 . At least two must be selected with a typical maximum of four windings. this setting could be defined in one step using a descriptor such as “TRCON=DABY” or each winding could be defined by “W1CTC=11” and “W2CTC=12”. Settings The following differential settings are described below to help understand what each setting means. This is the most important setting and MUST be correct. differential protection is disabled. Determine the number of CT inputs used and ensure the correct number of windings is selected. If the element is not used. This setting can also be applied by enabling individual windings such as the SEL-387 “E87n” setting where “n” is the winding number. B) Number of Windings This setting defines the number of windings to be used for differential protection. 66 . Some relays will also have ”Latched” or ‘Unlatched” options. Unlatched indicates that the relay output contacts will open when the trip conditions are no longer present. A) Enable Setting Many relays allow the user to enable or disable settings. Depending on the relay manufacturer. the setting should be disabled or OFF to prevent confusion. If all of the E87W settings are set to “N”. Make sure that the element is ON or the relay may prevent you from entering settings.The Relay Testing Handbook 2. A Latched option indicates that the output contacts will remain closed after a trip until a reset is performed and acts as a lockout relay. C) Phase-Angle Compensation This setting defines the phase-angle compensation between windings as described in the previous section. T-30 W1 Connection=Wye A a AA AA aaaa AAA A a aaaa W1 Angle=0 Yy0 A Y/y0° W2 Connection=Wye C B c b CC CC BB ccc BB c bb b b CCC C BBBcccc B bbb b W2 Angle=0 C B c b A a W1 Connection=Delta AAA aaaaa A a a A A AAA A aaaa W1 Angle=0º Dd0 c A D/d0° C B c W2 Connection=Delta CCC C BBBcccc B bbbb b C CCC B BBB cccc bbb C B c b C C B B c b b W2 Angle=0º AAAA W1 Connection=Wye A A a aaaaa AAA a aaa A A aa W1 Angle=0º Yd1 cccc b Y/d30° b CCC BB BB c B b ccc W2 Connection=Delta C C c bb bb CCC C BBB ccc B b C B bb b W2 Angle=30º b A a A W1 Connection=Delta AA AAA aa a AAA a a ca a A A aaaa W1 Angle=0º Dy11 C B bbb b D/y330° c b b C B bb b b W2 Connection=Wye CC CC BB BBB cc c CCC BBB b C C C B B cc c b cc cc W2 Angle=330º AA a A a W1 Connection=Wye AA A A aaaa a AAA A A aaaa a c b b W1 Angle=0º Yd11 CC BB bb b Y/d330° C Bc bb b bb W2 Connection=Delta CC BB ccc b CC C BB B c CC BB cc CC BB cccc b W2 Angle=330º AAA a A a W1 Connection=Delta AA aaaa AA A A aaaa b cc b W1 Angle=0º Dy1 D/y30° cc C CCC BBB c c c B bb CCC C BBBc c c B W2 Connection=Wye bb C B bb b b W2 Angle=30º A a A a c c C B C B b b 67 . Chapter 2: Percent Differential (87) Element Testing A table of the most common transformer connections and relays can be found in the following table.A a A a C B c Connection b GE/Multilin C SR-745 B c b GE T-60. C DACDACyy B c b W2CTC=0 CTCON=YY C B c b CC BBB ccc aa b bb C B c b CAA A aa A a A AA c c A a A a aa AA cc C C ABB c a Transformer Connection a a W1CTC=12 TRCON=YDAC c Yd1 C B c b b C C B c B b b C 02. DACYyy B b W2CTC=11 C B cc b C B c CTCON=YY b C B c b c b C B C C B B c b C B c ca b A A caa A a A a A a W1CTC=12 A a A Transformer Connection TRCON=YDAB a Yd11 A a b b A aa b b A a b A C AB C B cc bb 03. The Relay Testing Handbook Beckwith Electric Connection A A aa A A aa SEL-387 SEL-587 A a M-3310 A a A a A aa A a C AB C B cc b b TransformerC C B B Connection cc b b W1CTC=12 TRCON=YY Yy0 C B c b C B c b C B c b C B c b 01. YDACyy B c b W2CTC=1 CTCON=YY CC BB cc b C B b C B bb C B c b A aa b A b A A aa A A a a A a A A a b b A Transformer Connection A a W1CTC=0a b b TRCON=DACY a Dy11 C C A B c a B b C B C 05. E) Winding Voltage This setting is used in conjunction with the MVA setting to automatically calculate the Tap values for each winding. Different engineers use different criteria for the MVA setting (minimum MVA. 68 . YDAByy C C B B c W2CTC=11 b b CTCON=YY b C C B B c b C B c b C B c C B C C A B B cc C B cc c A c aa A A a A Transformer Connection aaW1CTC=0 TRCON=DABY Dy1 c A a A A A c a a 04. nominal MVA. Yyyy C B c W2CTC=12 b CTCON=YY C AA B c aa b A a A A aa A a A a AA aa A A a W1CTC=0 C C A B B c c a b b TransformerC Connection B cc b TRCON=DACDAC Dd0 C B c b C B b 10. DABYyyAA cc a W2CTC=1 CTCON=YY a C C A B B c ab a C B c b b C C B B c c b C B c b b b C B CC B Bc c b C B B b b C B c b Figure 59: Common b Phase-Angle Compensation Settings A a A a c c D) MVA C B b C B b This setting is used in conjunction with the Winding Voltage setting to automatically calculate the Tap values for each winding. maximum MVA) but the setting should NEVER exceed the maximum transformer MVA. The slope setting is typically set at 20-30%. the 87-Element will use Slope-2 for its calculation. J) Breakpoint This setting defines whether Slope-1 or Slope-2 will be used to determine if the relay should trip. Verify the correct Tap setting using the following formula. G) Tap The Tap setting defines the normal operate current based on the rated load of the protected equipment. This minimum operate current is used to prevent nuisance trips due to noise or metering errors at low current levels. The Breakpoint is defined as a multiple of Tap and if the restraint current exceeds the Breakpoint setting. This setting should be set around 0. and the CT ratio. 69 . I) Slope-2 The Slope-2 setting sets the ratio of operate current to restraint current that must be exceeded before the 87-Element will operate if the restraint current exceeds a pre-defined or user-defined breakpoint between Slope-1 and Slope-2. K) Time Delay The Time Delay setting sets a time delay (typically in cycles) between an 87-Element pickup and trip. This setting is used as the per-unit operate current of the protected device and most differential settings are based on the Tap setting. The Time Delay is typically set at the minimum possible setting but can be set as high as three cycles for maximum reliability on some relays. CTSEC × Power TAP = P-P Volts × 3 × CTPRI H) Slope-1 The Slope-1 setting sets the ratio of operate current to restraint current that must be exceeded before the 87-Element will operate as described previously. the primary voltage. Chapter 2: Percent Differential (87) Element Testing F) Minimum Pickup (Restrained) The Minimum Pickup setting is used to provide more stability to the differential element by requiring a minimum amount of current to flow before the differential element will operate.3 x the nominal or Tap current of the protected device. This setting is rarely used. make sure the condition is not true when testing. or the differential protection may never work correctly. This setting is typically set for 2nd. CT connections are confusing for some people. or 5th harmonics. Current Transformer Connections Zones of protection are defined by the CT locations which makes the CT connections the most important aspect of transformer differential protection. 3. Current flows through the relay coil and exits the relay from terminal G1. M) Harmonic Inhibit Parameters This setting determines what harmonic the differential protection will use to detect transformer inrush or over-excitation which could cause a nuisance trip. BØ(1B) flows into H2. if applicable. Always verify correct blocking operation by operating the end-device instead of a simulation to ensure the block has been correctly applied. and location must match the manufacturer’s requirements and relay settings. polarity. Repeat this step for the other two phases and ground. If enabled. 70 . Use the following figures to determine the correct connections for our examples. Using this procedure on Figure 60 we discover that the Generator AØ(1A) current flows into H1. The CT ratio.” Current flows into the CT polarity mark and out the CT secondary polarity mark (or H1 & X1). and CØ(1C) flows into H3.The Relay Testing Handbook L) Block The block setting defines a condition that will prevent the differential protection from operating such as a status input from another device. Follow G1 back to the CT to ensure the CT secondary connections are a closed loop. but the connections can be correctly interpreted by following the flow of power. Power flows from the generator to the system in this application so we can trace the flow of power from “To Generator” to “To System. Follow the line from the polarity mark to relay terminal H1. 4th. Figure 60: Zones of Protection Example 71 . The secondary current flows out of the non-polarity mark. Using this procedure on Figure 60 we discover that the System AØ(2a) current is connected to H4. through the two neutral jumpers and into G4. BØ(2b) to H5. 2” to “To System” and flows into the non- polarity mark of the CT. Chapter 2: Percent Differential (87) Element Testing The secondary current flows from “Winding No. The current flows through the relay and out H4 to the CT polarity mark. and CØ(2c) to H6. and H6. Always ensure that the CTs are a closed loop and that only one grounding point is connected for each isolated circuit. H5.The Relay Testing Handbook Another way to quickly determine correct CT polarity is to pick a reference point and compare the manufacturer’s drawings to the application drawings. H2. as shown in Figures 61 and 62. Choose the CT side away from the transformer in the examples and follow them to H1. and H3 for each phase. are not as important as the actual CT connections. Many people get fixated on the polarity marks which. Choose the CT side facing away from the transformer on the secondary side and follow the wiring to H4. CT1 CT2 H1 400:5 X1 1200:5 PHA H2 X2 PHB H3 X3 PHC X0 H1 H2 H3 H4 H5 H6 1A 1B 1C 2A 2B 2C G1 G2 G3 G4 G5 G6 GE/MULTILIN-745 Figure 61: CT Connections Example #1 CT1 CT2 400:5 1200:5 H1 X1 PHA H2 X2 PHB H3 X3 PHC X0 H1 H2 H3 H4 H5 H6 1A 1B 1C 2A 2B 2C G1 G2 G3 G4 G5 G6 GE/MULTILIN-745 Figure 62: CT Connections Example #2 72 . Chapter 2: Percent Differential (87) Element Testing The following figures represent the connection drawings for the most popular differential relays applied today. Figure 63: GE/Multilin SR-745 Transformer Protective Relay Connections Figure 64: GE T-60 Transformer Protective Relay Connections 73 . The Relay Testing Handbook Figure 65: Beckwith Electric M-3310 Transformer Protective Relay Connections Figure 66: Schweitzer Electric SEL-587 Transformer Protective Relay Connections 74 . Chapter 2: Percent Differential (87) Element Testing Figure 67: Schweitzer Electric SEL-387 Transformer Protective Relay Connections 4. refer to the manufacturer’s literature to determine if the relay uses some other nominal setting such as rated secondary current (5A usually). 75 . Because 3-phase transformers typically have correction factors to compensate for transformer phase shifts. Record the Pickup and Tap settings. Multiply the Pickup and Tap settings to determine the Minimum Pickup in amps. Minimum Pickup tests should be performed using 3-phase balanced currents. Repeat for all differential windings. You must determine what the expected result should be before performing any test. 3-Phase Restrained-Differential Pickup Testing Performing differential pickup testing is very similar to the test procedure for overcurrent protection with a few extra phases to test. If the relay does not have a Tap setting. The test is performed by applying and raising current in one winding until the element operates. 6109 The Tap settings appear to be appropriate for the application and are very close to the SEL- 387E generated Tap settings. 000. Winding-1 Winding-2 CTR_ (CT Ratio) 240 1600 MVA (XFMR MVA Ration) 230 W_CTC (Winding Connection) 12 1 VWDG_ (Winding Voltage) 230 18 Tap_ 2. We can determine the Minimum Pickup for each winding with the following formula.3 × 2.The Relay Testing Handbook We will use an SEL-387E relay with the following differential settings for the rest of the tests in this chapter.000 × 3 × 240 18.61 O87P 0.41 4.0 This relay has a Minimum Pickup setting (O87P) and we can determine the three-phase Minimum Pickup using the formula O87P × TAP . Winding-1 Winding-2 CTSEC × Power CTSEC × Power TAP1 = TAP2 = P-P Volts × 3 × CTPRI P-P Volts × 3 × CTPRI 1 × 230. 000 TAP1 = TAP2 = 230.383A 76 .30 SLP1 20 SLP2 60 IRS1 3.40569 TAP2 = 4. 000.3 × 4.723A Minimum Pickup = 1.41 Minimum Pickup = 0.000 × 3 × 1600 TAP1 = 2. 000 1 × 230. Winding-1 Winding-2 Minimum Pickup = O87P × Tap Minimum Pickup = O87P × Tap Minimum Pickup = 0.61 Minimum Pickup = 0. It’s always a good idea to verify the Tap settings as described in the previous chapter using the transformer nameplate as shown below for this transformer. W1 RELAY INPUT W2 RELAY INPUT ICW1Ø ICW2Ø IAW1Ø IAW2Ø RELAY IBW1Ø IBW2Ø RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps IAW1 Amps IAW1 Test Amps 0° Test Hz + + B Phase Amps IBW1 Amps IBW1 Test Amps -120° (240°) Test Hz + + C Phase Amps ICW1 Amps ICW1 Test Amps 120° Test Hz (PS = Phase Shift) WINDING 2 + + A Phase Amps IAW2 Amps IAW2 Test Amps IAW1°+PS Test Hz + + B Phase Amps IBW2 Amps IBW2 Test Amps IBW1°+PS Test Hz + + C Phase Amps ICW2 Amps ICW2 Test Amps ICW1°+PS Test Hz Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 69: Simple 3-Phase 87-Element Test-Set Connections with Six Current Channels 77 . RELAY INPUT CØ AØ RELAY RELAY TEST SET BØ WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps IAW1 Amps IAW1 Test Amps 0° Test Hz + + B Phase Amps IBW1 Amps IBW1 Test Amps -120° (240°) Test Hz + + C Phase Amps ICW1 Amps ICW1 Test Amps 120° Test Hz WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection DC Supply - + Element + + Timer - Element Timer Output Input Output + Input Figure 68: Simple 3-Phase 87-Element Test-Set Connections If your test-set has six available current channels. you can use the following connection diagram. and change the output channel for each test until all pickup values have been tested. After all Winding-1 tests are completed. Chapter 2: Percent Differential (87) Element Testing A) 3-Phase Test-Set Connections Connect all three phases to one winding. move the test leads to Winding-2 and repeat. The Relay Testing Handbook B) 3-Phase Pickup Test Procedure Use the following steps to determine pickup. 1. Determine how you will monitor pickup and set the relay accordingly, if required. (Pickup indication by LED, output contact, front panel display, etc…See previous packages of The Relay Testing Handbook for details.) 2. Group all three currents together so that you can change the magnitude of all three simultaneously while maintaining 120º between phases. Set the test current 5% higher than the pickup setting as shown in the following table. Winding-1 Winding-2 Min Pickup × 105% = Test Amps Min Pickup × 105% = Test Amps 0.723A × 1.05 = Test Amps 1.383A × 1.05 = Test Amps 0.75915A 1.452A AØ Test Amps 0.759A@0º 1.452A@0º BØ Test Amps 0.759A@240º 1.452A @240º CØ Test Amps 0.759A@120º 1.452A @120º 3. Apply 3-Phase test current and make sure pickup indication operates. 4. Slowly lower all three currents simultaneously until the pickup indication is off. Slowly raise the currents until pickup indication is fully on (Chattering contacts or LEDs are not considered pickup). Record the pickup values on your test sheet. The following graph displays the pickup procedure. 5A ELEMENT PICK-UP 4A 3A PICKUP SETTING 2A 1A STEADY-STATE PICKUP TEST Figure 70: Pickup Test Graph 78 Chapter 2: Percent Differential (87) Element Testing 5. Compare the pickup test result to the manufacturer’s specifications and calculate the percent error as shown in the following example. The 87-Element pickup setting is 0.723A and the measured pickup was 0.731A. Looking at the “Differential Element” specification in Figure 71, we see that the acceptable metering error is ±5% ±0.10A. Differential Element Unrestrained Pickup Range: 1–20 in per unit of tap Restrained Pickup Range: 0.1–1.0 in per unit of tap Pickup Accuracy (A secondary) 5 A Model: ±5% ±0.10 A 1 A Model: ±5% ±0.02 A Unrestrained Element Pickup Time (Min/Typ/Max): 0.8/1.0/1.9 cycles Restrained Element (with harmonic Blocking) Pickup Time (Min/Typ/Max): 1.5/1.6/2.2 cycles Restrained Element (with harmonic Restraint) Pickup Time (Min/Typ/Max): 2.62/2.72/2.86 cycles Figure 71: SEL-387E Specifications We can calculate the manufacturer’s allowable percent error for Winding-1. 100 × (5% × Setting) + 0.1A Allowable Percent Error = Setting 100 × (5% × 0.723) + 0.1A Allowable Percent Error = 0.723 18.831% Allowable Percent Error The measured percent error can be calculated using the percent error formula below. Actual Value - Expected Value × 100 = Percent Error Expected Value 0.731A - 0.723A × 100 = Percent Error 0.723A 1.1% Error The Winding-1 test is within the manufacturer’s tolerance and passed the test. 6. Repeat the pickup test for all windings that are part of the differential scheme. 79 The Relay Testing Handbook DIFFERENTIAL TEST RESULTS PICK UP 0.3 TIME DELAY 0 SLOPE 1 20% TAP1 2.41 SLOPE 2 60% TAP2 4.61 MINIMUM PICK UP TESTS (Amps) 3 PHASE MFG % ERROR W1 PICKUP 0.731 0.723 1.11 W2 PICKUP 1.398 1.383 1.08 W3 PICKUP NA NA C) 1-Phase Test-Set Connections The most basic test-set connection uses only one phase of the test-set with a test lead change between every pickup test. After the Winding-1 A-phase pickup test is performed, move the test leads from Winding-1 A-Phase amps to B-Phase amps and perform the test again. Repeat until all enabled phases are tested on all windings. RELAY INPUT RELAY AØ PU RELAY TEST SET WINDING 1 A Phase Input=Pickup Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps AØ Test Amps 0° Test Hz + + B Phase Amps C2 Amps + + C Phase Amps C3 Amps WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection DC Supply - + Element + + Timer - Element Timer Output Input Output + Input Figure 72: Simple 87-Element Test-Set Connections 80 you can use the following connection diagram and change the output channel for each test until all pickup values have been tested. Chapter 2: Percent Differential (87) Element Testing You can also connect all three phases to one winding and change output channels instead of changing leads. After all Winding-1 tests are completed. TEST #1 RELAY INPUT TEST #2 RELAY INPUT TEST #3 RELAY INPUT AØ PU BØ PU CØ PU RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps AØ Test Amps 0° Test Hz + + B Phase Amps C2 Amps BØ Test Amps 0° Test Hz + + C Phase Amps C3 Amps CØ Test Amps 0° Test Hz WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection DC Supply - + Element + + Timer - Element Timer Output Input Output + Input Figure 73: Simple 3-Phase 87-Element Test-Set Connections If your test-set has six available current channels. move the test leads to Winding-2 and repeat. TEST #1 TEST #2 TEST #3 TEST #4 TEST #5 TEST #6 RELAY INPUT RELAY INPUT RELAY INPUT RELAY INPUT RELAY INPUT RELAY INPUT A1Ø PU B1Ø PU C1Ø PU A2Ø PU B2Ø PU C2Ø PU RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps A1Ø Test Amps 0° Test Hz + + B Phase Amps C2 Amps B1Ø Test Amps 0° Test Hz + + C Phase Amps C3 Amps C1Ø Test Amps 0° Test Hz WINDING 2 + + A Phase Amps C4 Amps A2Ø Test Amps 0° Test Hz + + B Phase Amps C5 Amps B2Ø Test Amps 0° Test Hz + + C Phase Amps C6 Amps C2Ø Test Amps 0° Test Hz Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 74: Simple 3-Phase 87-Element Test-Set Connections with Six Current Channels 81 . 4. 4. Make sure pickup indication operates. if required. set the fault current at 0.) 2. Determine the expected pickup: Winding-1 Winding-2 MinimumPickup = O87P × TAP1 × A MinimumPickup = O87P × TAP2 × A MinimumPickup =0. 8.1 = 2.3 × 2. Slowly lower the current until the pickup indication is off. 12 1. 1.395 × 1. 10. 9.The Relay Testing Handbook D) 1-Phase Pickup Test Procedure A single-phase test procedure is slightly more complicated because correction factors may apply. 6.723 × 1. (Pickup indication by LED. Set the fault current 10% higher than the pickup setting.3 × 4. WnCTC Setting A 0 1 Odd: 1.634A ) for Winding-2. 82 . 3.634A (2. Record the pickup values on your test sheet. Use the following steps to determine pickup. 11 √3 Even: 2.723 MinimumPickup = 2. We have found through experimentation that the correction factor is 1 when WnCTC equals 12. The following graph displays the pickup procedure.795A ) for Winding-1 or 2.795A (0. Digital transformer relays use algorithms to compensate for the different transformer phase shifts as described earlier and apply compensation factors to compare windings with different configurations. front panel display. output contact.1 = 0.41 × (W1CTC (12))1 MinimumPickup =0. 5. 7. the table does not seem to be correct when the WnCTC equals 12. Slowly raise current until pickup indication is fully on (Chattering contacts or LEDs are not considered pickup). The compensation factors for the example SEL- 387 can be found in the Testing and Troubleshooting section of the instruction manual as shown in the following table. For example.5 However. etc…See previous packages of The Relay Testing Handbook for details.61 × (W1CTC (1)) 3 MinimumPickup = 0.395 3. Determine how you will monitor pickup and set the relay accordingly. Looking at the “Differential” specification in Figure 76.421A.10A. we see that the acceptable metering error is ±5% ±0. Compare the pickup test result to the manufacturer’s specifications and calculate the percent error as shown in the following example. The 87-Element pickup Winding-2 pickup setting is 2. Figure 76: SEL-387E Specifications 83 . Chapter 2: Percent Differential (87) Element Testing 5A ELEMENT PICK-UP 4A 3A PICKUP SETTING 2A 1A STEADY-STATE PICKUP TEST Figure 75: Pickup Test Graph 5.395A and the measured pickup was 2. 18% and passed the test.21975 100 × Allowable Percent Error = 2.41 W1CTC 12 SLOPE 2 60% TAP2 4.24 W3 PICKUP NA NA NA NA 5.422 2.730 0.18% Allowable Percent Error The measured percent error can be calculated using the percent error formula below.395 0.395 1.97 W2 PICKUP 2.731 0. For example. such as the SEL-387.11 1. Restrained-Differential Timing Test Procedure It is very important to ensure that all phases are assigned to operate the output contact because some relays.3 TIME DELAY 0 SLOPE 1 20% TAP1 2.395A × 100 = Percent Error 2.421 2.395 9. 6. the SEL-387 word bit “87R1” is an A-Phase differential trip and “87R” is a differential trip on any phase. 100 × (5% × Setting) + 0.11 1.1A Allowable Percent Error = Setting 100 × (5% × 2.732 0. allow the designer to assign differential trips for each phase individually.61 W2CTC 1 MINIMUM PICK UP TESTS (Amps) A PHASE B PHASE C PHASE MFG % ERROR W1 PICKUP 0.07 1.1% Error The Winding-2 test is within the manufacturer’s tolerance of 9. Assigning the incorrect word bit to the final output is an easy mistake that has occurred in the field.395) + 0.1A Allowable Percent Error = 2.421A .24 0.Expected Value × 100 = Percent Error Expected Value 2. Actual Value .395A 1.723 1. Repeat the pickup test for all phase currents that are part of the differential scheme DIFFERENTIAL TEST RESULTS PICK UP 0.425 2.2. I strongly recommend performing single-phase timing 84 .The Relay Testing Handbook We can calculate the manufacturer’s allowable percent error for Winding-2. 72/2.86 cycles Output Contacts: Standard: Make: 30 A.9 cycles Restrained Element (with harmonic Blocking) Pickup Time (Min/Typ/Max): 1. Carry: 6 A continuous carry at 70°C.0 in per unit of tap Pickup Accuracy (A secondary) 5 A Model: ±5% ±0. However. Dropout time: Less than 5 ms typical.10 A 1 A Model: ±5% ±0.3 cycles) Total Time 52. The timing test procedure is very straightforward. 4 A continuous at 85°C: 1 s Rating: 50 A: MOV protected: 270 Vac.5/1.02 A Unrestrained Element Pickup Time (Min/Typ/Max): 0. 360 Vdc. Breaking Capacity (10000 operations): Figure 77: SEL-387 Differential and Output Relay Specifications The time delay setting for our example is 0 cycles which means that there is no intentional delay. we first must discover what an acceptable result is.6 cycles.86 cycles (47. Reason Delay Restrained Element (with harmonic restraint) Pickup 2. Apply a single-phase current 10% greater than the Minimum Pickup setting into each input related to the percent differential element and measure the time between the applied current and the relay trip signal. Pickup time: Less than 5 ms. there are software and hardware delays built into the relay that must be accounted for.7 ms) Time (Max) [Software] Output Contacts Operate Time [Hardware] 5 ms (0.7ms or 3. Differential Element Unrestrained Pickup Range: 1–20 in per unit of tap Restrained Pickup Range: 0. Use Figure 77 to determine the acceptable tolerances from the manufacturer’s specifications. 40 J.2 cycles Restrained Element (with harmonic Restraint) Pickup Time (Min/Typ/Max): 2.0/1. Figure 78 indicates that the maximum allowable time is 60ms or 3.6/2.1–1. Chapter 2: Percent Differential (87) Element Testing tests on every current-input connected to the differential element using the final trip output to ensure that the relay trips on all phases and all windings.8/1.16 cycles Figure 78: SEL-387 Differential Minimum Trip Time 85 . As with all tests.62/2. 44 2.48 3.723 × 1.The Relay Testing Handbook Perform a timing test using the following steps: 1. 3-Phase Restrained-Differential Slope Testing Differential slope testing is one of the most complex relay tests that can be performed and requires careful planning. and information from the manufacturer regarding the relay’s characteristics. 3. 2.410 2. Run the test plan on the first phase related to restrained-differential.16 OK OK OK W3 NA NA NA NA NA 6. 5. Record the test results.395 × 1. This section will discuss 3-Phase testing that will require at least six current channels to be performed correctly. The math is also simplified when you apply balanced three-phase conditions. Refer to the “1-Phase Restrained-Differential Slope Testing” section of this chapter if your test equipment has less than six-channels. Connect the test-set input(s) to the relay output(s) that are programmed to operate when the restrained-differential relay operates. mimicking an ideal-world steady state scenario.634A (2. a good understanding of the differential relay’s operating fundamentals.634A ) for Winding-2) 4. (0.16 OK OK OK W2 2.490 2. Configure your test-set to start a timer when current is applied and stop the timer and output channels when the appropriate input(s) operate(s).1 = 2.45 2.634A 2.1 = 0. Six-phase restrained-differential testing is achieved by applying 3-phase balanced currents in each winding. RESTRAINED DIFFERENTIAL TIMING TESTS (cycles) WINDING TEST A PHASE (cy) B PHASE (cy) C PHASE (cy) MFG (cycles) % ERROR W1 0. 86 . Choose a connection diagram from Figures 72-74. Six-channel restrained-differential testing is actually very similar to the simple testing discussed in the previous chapter after the phase-angles between windings have been applied correctly.500 3.795A (0. and then increasing all three currents in one winding simultaneously until the relay operates.723 A 2. Set a single-phase current at least 10% higher than the Minimum Pickup setting using the calculations in the “1-Phase Pickup Test Procedure” section of this chapter.795A ) for Winding-1 or 2. Repeat the test on all phases related to the restrained-differential. DABYyy W2CTC=1 CTCON=YY C Bc b C A B ba AA ab a A c a Transformer Connection W1CTC=12 TRCON=YDAC C A B cc ab CC A BB c ab 02. YDACyy W2CTC=1 CTCON=YY C Bc b C A B ca b C AA B a b a bb Transformer Connection W1CTC=0 TRCON=DACY C AA B aa bb CC A BB ca 05. YDAByy W2CTC=11 CTCON=YY CC A BB c a b c C B cc b b C B c C B Figure 79: Common Phase-Angle Compensation Settings c 87 . DACDACyy W2CTC=0 CTCON=YY C A B c a b C AA B c aa b A c a C A B cc ab Transformer Connection W1CTC=0 TRCON=DABY CC A BB c a C Bc bb 04. Chapter 2: Percent Differential (87) Element Testing We discussed how to interpret the phase-angle-shift settings previously in this chapter which can be summarized by the following figure. A a AA aa Beckwith Electric Connection SEL-387 SEL-587 A a M-3310 C A B c a b CC A BB cc a bb C B c b Transformer Connection W1CTC=12 TRCON=YY C A B c a b 01. Yyyy W2CTC=12 CTCON=YY C A B c a b A a A a C A B c a b CC A BB cc a bb Transformer Connection W1CTC=0 TRCON=DACDAC C B c b 10. DACYyy W2CTC=11 CTCON=YY c b C B c b C B c b CA B c a AA aca Transformer Connection W1CTC=12 TRCON=YDAB A a b C A B a bb 03. The Relay Testing Handbook Let’s figure out the test-settings for balanced full-load conditions for each case. The full load condition for Winding-1 is the Tap1 setting. The other two phases will be 120º apart to create a three-phase balanced condition. The Capital “A” (Winding-1) and lower case “a” (Winding-2) are both at 0º so our A-Phase test conditions will be at 0º. All three-phases for Winding-2 should start at Tap2. All three-phases for Winding-1 should start at Tap1. A-Phase for Winding-2 will start at 180º to be opposite Winding-1 and the other two phases will be 120º apart as shown in the following table. The first case is SEL-387 when both W1CTC and W2CTC=12. Connection SEL-387 W1AØ W1BØ W1CØ W2AØ W2BØ W2CØ A a W1CTC=12 Tap1@ Tap1@ Tap1@ Tap2@ Tap2@ Tap2@ W2CTC=12 0º -120º 120º 180º 60º -60º C B c b A a W2BØ = TAP2@60º C B c b W1CØ = TAP1@120º A a 0 90 60 12 c W2AØ = TAP2@180º W1AØ = TAP1@0º 30 0 15 C B 180 0 b 21 0 33 24 0 0 0 0 270 3 A a W1BØ = TAP1@-120º c C B b W2CØ = TAP2@-60º A a Figure 80: Yy12 or Yy0 3-Phase Differential Restraint Test Connections b C B c A a b C B c 88 . Both settings are the same so there is no phase-angle shift for this connection. A-Phase for Winding-2 will start at 180º to be opposite Winding-1 and the other two phases will be 120º apart as shown in the following A table. Both settings are the same so there is no phase-angle shift for this connection. The other two phases will be 120º apart to create a three-phase balanced condition. Chapter 2: Percent Differential (87) Element Testing The next case is SEL-387 when both W1CTC and W2CTC=0. a C Connection B c b SEL-387 W1AØ W1BØ W1CØ W2AØ W2BØ W2CØ A a W1CTC=0 Tap1@ Tap1@ Tap1@ Tap2@ Tap2@ Tap2@ W2CTC=0 0º -120º 120º 180º 60º -60º C B c b A a c W2BØ = TAP2@60º C B b W1CØ = TAP1@120º A a c 12 0 90 60 C B W2AØ = TAP2@180º W1AØ = TAP1@0º 30 0 b 15 180 0 21 0 33 24 0 0 0 0 270 3 A a b W1BØ = TAP1@-120º C B c W2CØ = TAP2@-60º A a b Figure 81: Dd0 3-Phase Differential Restraint Test Connections C B c 89 . The Capital “A” (Winding-1) and lower case “a” (Winding-2) are both at 0º so our A-Phase test conditions will be at 0º. The Relay Testing Handbook TheA next casea is SEL-387 with W1CTC=0 and W2CTC =1. The Capital “A” (Winding-1) is at 0º so our A-Phase test conditions will be at 0º. The other two phases will be 120º apart to create a Cthree-phase B c balancedb condition. The lower case “a” is at -30º so there is a phase shift between windings. Remember that we apply Winding-2 current 180º out-of-phase so the A-Phase for Winding-2 will start at 150º ( −30° + 180° ) to be opposite Winding-1. The other two phases will A a be 120º apart as shown in the following table. C B c b Connection SEL-387 W1AØ W1BØ W1CØ W2AØ W2BØ W2CØ A a W1CTC=0 Tap1@ Tap1@ Tap1@ Tap2@ Tap2@ Tap2@ c C B W2CTC=1 0º -120º 120º 150º 30º -90º b A aW2AØ = TAP2@150º W1CØ = TAP1@120º W2BØ = TAP2@30º c C B b 12 0 90 60 0 30 15 180 0 W1AØ = TAP1@0º A a 21 0 33 24 0 0 0 0 270 3 b C B c W1BØ = TAP1@-120º A a W2CØ = TAP2@-90º A a b C B c C B c bFigure 82: Dy1 3-Phase Differential Restraint Test Connections The A next casea is SEL-387 with W1CTC=12 and W2CTC =1. The Capital “A” (Winding-1) is at 0º so our A-Phase test conditions will be at 0º. The other two phases will be 120º apart to create a Cthree-phase B c balanced b condition. The lower case “a” is at -30º so there is a phase shift between windings. Remember that we apply Winding-2 current 180º out-of-phase so the A-Phase for A Winding-2 willa start at 150º ( −30° + 180° ) to be opposite Winding-1. The other two phases will be 120º capart as shown in the following table. C B b Connection SEL-387 W1AØ W1BØ W1CØ W2AØ W2BØ W2CØ A a W1CTC=12 Tap1@ Tap1@ Tap1@ Tap2@ Tap2@ Tap2@ c C B W2CTC=1 0º -120º 120º 150º 30º -90º b A a b C B c 90 A a b C B c Chapter 2: Percent Differential (87) Element Testing W2AØ = TAP2@150º W1CØ = TAP1@120º W2BØ = TAP2@30º 0 90 60 12 0 30 15 180 0 W1AØ = TAP1@0º 21 0 33 24 0 0 0 270 30 A a W1BØ = TAP1@-120º C B c b W2CØ = TAP2@-90º A a C B c bFigure 83: Yd1 3-Phase Differential Restraint Test Connections A next caseais SEL-387 with W1CTC=0 and W2CTC =11 The Capital “A” (Winding-1) is at 0º The so our A-Phase c test conditions will be at 0º. The other two phases will be 120º apart to create a C three-phase B balanced condition. The lower case “a” is at 30º so there is a phase shift between b windings. Remember that we apply Winding-2 current 180º out-of-phase so the A-Phase for Winding-2 A will astart at 210º (-150º) (30° + 180° ) to be opposite Winding-1. The other two phases will be 120º apart as shown in the following table. c C B b Connection SEL-387 W1AØ W1BØ W1CØ W2AØ W2BØ W2CØ A a W1CTC=0 Tap1@ Tap1@ Tap1@ Tap2@ Tap2@ Tap2@ b W2CTC=11 0º -120º 120º -150º 90º -30º C B c A a b W2BØ = TAP2@90º C B c W1CØ = TAP1@120º 0 90 60 12 0 30 15 180 0 W1AØ = TAP1@0º 21 0 33 24 0 0 00 270 3 W2AØ = TAP2@-150º W1BØ = TAP1@-120º W2CØ = TAP2@-30º Figure 84: Dy11 3-Phase Differential Restraint Test Connections 91 C B c b A a c TheC Relay Testing B Handbook b TheA next case ais SEL-387 with W1CTC=12 and W2CTC =11 The Capital “A” (Winding-1) is at 0º so ourcA-Phase test conditions will be at 0º. The other two phases will be 120º apart to create Ca three-phase B balanced condition. The lower case “a” is at 30º so there is a phase shift between b windings. Remember that we apply Winding-2 current 180º out-of-phase so the A-Phase for A Winding-2 awill start at 210º (-150º) (30° + 180° ) to be opposite Winding-1. The other two phases will be 120º apartb as shown in the following table. C B c Connection SEL-387 W1AØ W1BØ W1CØ W2AØ W2BØ W2CØ A a W1CTC=12 Tap1@ Tap1@ Tap1@ Tap2@ Tap2@ Tap2@ b C B W2CTC=11 0º -120º 120º -150º 90º -30º c W2BØ = TAP2@90º W1CØ = TAP1@120º 0 90 60 12 0 30 15 180 0 W1AØ = TAP1@0º 21 0 33 24 0 0 00 270 3 W2AØ = TAP2@-150º W1BØ = TAP1@-120º W2CØ = TAP2@-30º Figure 85: Yd11 3-Phase Differential Restraint Test Connections 92 Follow the other side of the CT to terminal Z07 (IAW2) which is where we will connect the second channel current from the test-set. This is where we will connect the first current from our test-set. Keep following the primary buss through the transformer to the Phase A' CT and then follow the secondary to terminal Z08. 93 . Chapter 2: Percent Differential (87) Element Testing A) Test-Set Connections The connection diagram for the SEL-387 is as follows: Figure 86: Schweitzer Electric SEL-387 Transformer Protective Relay Connections Follow the AØ primary buss through the Phase A CTs then follow the CT secondary to terminal Z01(IAW1). This is the neutral of the CTs so we will connect the test-set’s second current-channel-neutral-terminal to Z08. Connect the neutral of the test-set current channel to Z02 by following the other side of the CT to its relay terminal. Follow the other phases to determine the following connections when testing B or C phases. Remember that there is an easier way that will be discussed later in this chapter. This section will discuss this method in detail.The Relay Testing Handbook The test-set connections for a 3-Phase Restrained-Differential Slope test are displayed in the next figure. and what results to expect. 94 . what test points will work best. The first method involves a significant amount of calculations to determine exactly how the element is supposed to operate. SEL-387 RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency Z01 + + IAW1 C1 Amps W1AØ Test Amps 0° Test Hz Z02 Z03 + + IBW1 C2 Amps W1BØ Test Amps -120° (240°) Test Hz Z04 Z05 + + ICW1 C3 Amps W1CØ Test Amps 120° Test Hz Z06 WINDING 2 (PS = Phase Shift) Z07 + + IAW2 C4 Amps W2AØ Test Amps W2AØ Test° Test Hz Z08 Z09 + + IBW2 C5 Amps W2BØ Test Amps W2BØ Test° Test Hz Z10 Z11 + + ICW2 C6 Amps W2CØ Test Amps W2CØ Test° Test Hz Z12 Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 87: 3-Phase Restrained-Differential Slope Test-Set Connections B) Pre-Test Calculations There are several different methods to test the restrained-differential slope. 00 0.50 2.00 2.00 1. Chapter 2: Percent Differential (87) Element Testing The following information details the example test settings and expected characteristic curve.50 0.61 O87P 0.00 3.50 Slope 1 1.41 4.0 Differential Protection 20%/60% Characteristic Curve 4.00 Operate Current Slope 2 (Iop)xTAP 2.00 1.50 4.00 Trip Area 3.00 8.50 3.30 SLP1 20 SLP2 60 IRS1 3.00 6.00 5.00 7. Winding-1 Winding-2 W_CTC 12 1 Tap_ 2.00 Restraint Area 0.00 Restraint Current (Ir) x TAP Figure 88: Percentage Differential Protection Dual Slope Characteristic Curve 95 .00 4. respectively. For example. Figure 90: SEL-387 Definition of IOP and IRT 96 . If SLP2 = OFF. Then we can calculate the transition point between the Minimum Pickup and Slope-1 operation. Different relays have different slope calculations and we can refer to the SEL-387 relay instruction manual to determine the relay’s differential calculation as shown in Figures 89 and 90. Figure 89: SEL-387 Slope-1 Differential Formulas The resulting A-phase.3 × TAP1 ) as we calculated earlier in this chapter. The Winding-1 Minimum Pickup is 0.The Relay Testing Handbook We first need to change the restraint and operate current values to amps instead of multiples of Tap. B-phase. The magnitude of this result is IOP. Method: Decide where you want to cross the differential characteristic by picking a restraint value IRT. 100 O87P • < IRT < IRS1 SLP1 The value of IOP corresponding to the selected IRT equals the following: SLP1 IOP = • IRT 100 Both IRT and IOP are in multiples of tap. In each element a phasor addition sums the winding currents. for a balanced through-load current of 4 amps. and C-phase currents from each winding then go to the differential elements 87R-1. The magnitudes of the winding currents are then summed in a simple scalar addition and divided by two. Because this test is for the SLP1 threshold. IRS1 and SLP2 are not functional. and -3. select a point above the O87P intersection point and below IRS1.723A ( 0. Slope-1 will start operating when the operate current is greater than the minimum pickup. This result is IRT. these calculations produce ideal results of IOP = 0 and IRT = 4. -2. which is a vertical line on the graph. 915A.523I AW 2 = 97 .41) +  AW 2 × 2. Chapter 2: Percent Differential (87) Element Testing Notice that the IRT current is defined as “The magnitudes of the winding currents are then summed in a simple scalar addition and divided by 2.915A I AW 1 0.3 × 5 < IRT I AW 1 I 1. Use the first IOP formula to calculate the transition We should calculate the expected Winding-1 between Minimum Pickup and Slope-1.5 × 4.61   Min IRT for Slope1 = 1.61A  4.723 + 0.41 4. current for a Winding-2 current less than 6.5 < I AW 1 = 0. The IOP current is defined as “In each element a phasor addition sums the winding currents. Remember that both of these formulas are still in per-unit and not actual Amp values.5 × TAP2 I  I AW 1 =( 0.” The magnitude of this result is IOP which can be translated into IW1 − IW 2 for our calculations because we will use balanced 3-phase currents at the theoretical normal operating angles.3 × < IRT 20 I AW 1 I = O87P + AW 2 TAP1 TAP2 0. 100 IOP=O87P = IW1 − IW2 O87P × < IRT SLP1 IW1 O87P + IW2 = 100 0.61 IW 1 + IW 2  I  1.3 × 2.61  Min IRT for Slope1 = 6.” This can be translated into I +I W1 2 because we will use balanced 3-phase currents at the theoretical normal operating W2 angles.3 + AW 2 2.3 + AW 2  × 2.5 < IRT = 0.41  Min IRT for Slope1 = 1.41 2 4. 20 4.945 I AW 1 = I AW 2 10IW 1 − IW 1 = IW 2 + 10IW 2 4.20 × IRT = IW 1 = 1.6388 I AW 2 = 1.222IW 2  I + IW 2   IW 1 − IW 2  = 0.20 ×  IW 1 + IW 2  I AW 1 I 2 ×  IW 1 − IW 2  = 1.565I AW 1 9IW 1 = 11IW 2 9 IW 1 = IW 2 11 IW 2 = 0.818IW 1 98 .222 AW 2 =  IW 1 + IW 2  2.945I AW 2 10IW 1 − 10IW 2 =IW 1 + IW 2 2.222 × I AW 2 10 ×  IW 1 − IW 2  = IW 1 + IW 2 4.61I AW 1 = 2.41 4.The Relay Testing Handbook Remember that these values are in per-unit.61 9IW 1 = 11IW 2 I AW 1 = 0.61 0.20 ×  W 1 2 I AW 1 I = 1.41 × 1. Now we can calculate the Slope-1 expected results: Change to amps with the following formulas: SLP1 IOP = × IRT 100 20 IOP = × IRT 100 IOP 0.222IW 2 0.61I AW 1 =2.639I AW 2 11 IW 1 = IW 2 9 1 I AW 1 = I AW 2 IW 1 = 1.222 AW 2 TAP1 TAP2 2 ×  IW 1 − IW 2  = 0. Chapter 2: Percent Differential (87) Element Testing We also need to determine the maximum restraint current before we accidentally start testing Slope-2 when testing Slope-1.222 IW 2 = 2.99 4.99 I AW 2  IW 1 + IW 2  = 2.99 TAP2 2 I AW 2 1.98 IW 2 = 2.61 1.69 99 .69 × 4. We can use the following formula to calculate the Don’t forget that the calculation so far has been restraint current (Ir) at the transition point between in per-unit.69 = 2.69 = 2.99 I AW 2 = 12. IRT < IRS1 IRT < 3 IRT = 2.222IW 2 + IW 2 =2 × 2.222IW 2 = 5.406 A 2.98 5. The maximum W2 current in amps is Slope-1 and Slope-2 by setting the restraint calculated as follows: slightly below the IRS1 setting.61 2 I= AW 2 2.222IW 2 + IW 2  = 2. 2 IOP 0. lSince this test is for the SLP2 threshold.7IW 1 + 1.The Relay Testing Handbook Now we can define the expected test currents when Slope-2 is required.2 IW 2 = + IW 1 − IW 2 + 1.7 0.2 = 1.0 × −0.3IW 2 1.3IW 2 − 1.3IW 2  60    20 − 60   IOP =  × IRT  + 3.6IRT + −1.2 2 0.2 = 1.3IW 2 =IW 2 0.3IW 1 + 0.2  I + IW 2  IW 2 = IW 1 − IW 2 + 1.2 = 0.2 = 0.3 IW 1 − IW 2 + 1.857IW 2 − 1.6IRT + 3.3 ×  IW 1 + IW 2  1.3IW 2 100 .3IW 2 − 1.0 ×   100    100   0.2 = 1. select a point above the IRS1 setting.6 ×  W 1 1.2 IOP = 0. Figure 91: SEL-387 Slope-2 Differential Formulas  SLP2    SLP1 − SLP2   IOP =  × IRT  + IRS1 ×    100    100  0.6IRT + 3.3 1.7IW 1 + 1.2 = =IW 1 1.538IW 1 + 0.923 IW 1 − 0.6 ×  − 1.3IW 2 + IW 2 0.40 = IW 1 = − 0. The following information is found in the test section of the SEL-387 instruction manual: Method: Decide where you want to cross the differential characteristic by picking a restraint value IRT.7 = IW 1 1.714  IW 1 + IW 2  0.7IW 1 + 1.3 2 0.2 = 0.0 ×  IW 1 =  0.7 IOP 0.7IW 1 1.7IW 1 + 1. IRT > IRS1 The value of IOP that corresponds to the selected IRT is as follows: SLP2  SLP1 − SLP2  IOP = • IRT + IRS1•   100  100  Both IRT and IOP are in multiples of tap. which is a vertical line on the graph.3IW 2  IW 1 − IW 2  =0.3IW 1 + 1.2 = 0.7   100   1.2   −40   1. 714 × 2.223IAW1 + 0.857IW2 − 1.41 IAW2 = (0.714 = 0.00 Operate Current Slope 2 (I W1 in Amps) 15.131)) Differential Protection Characteristic Curve 25.61) =IAW1 0.971 * I AW 2 ) − 4.61) + (0.61 I= AW1 (0.61 IAW1 = (0.255 The actual characteristic curve of this 87-Element using Winding-1 and Winding-2 currents can be plotted where W2 (Restraint Current) is a group of arbitrary numbers and W1 (Operate Current) uses the following formulas: If (I AW 2 < 6.714 = 0.41 4.41 4.029IAW1 + 4.403IAW2 × 2.915A then I AW 1 = 0.714 = 0.131 •• •• The spreadsheet calculation for these equations could be: EXPECTED = IF (I AW 2 < 6.41) − (1.406.714) × 2.41 IAW1 IAW2 = 0.403IAW2 − 1. The following calculations change the per-unit to actual current applied.923 TAP1 TAP2 TAP2 TAP1 IAW1 I IAW2 I = 1.538 AW1 + 0.923 × 4.00 Trip Area 20. Chapter 2: Percent Differential (87) Element Testing Remember that the calculations so far are in per-unit.923 2.639 * I AW 2 .00 Restraint Area O87P 0.131 =IAW2 1.538W1 + 0. 0.857 × AW2 − 1.971IAW2 − 4. 0.923 IAW1 I IAW2 I = 1.41) I= AW2 (0.523I AW 2 •• If (I AW 2 > 6.857 × AW2 − 1.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Restraint Current (I W2 in Amps) Figure 92: Percentage Differential Protection Dual Slope Characteristic Curve in Amps 101 .923 2.223IAW1 + 0.61 4.971IAW2 − 4.00 10.714 =W2 0.(0.639I AW 2 •• If (I AW 2 > 12.523 * I AW 2 ). IF (I AW 2 < 12.923) × 4.723 + 0.403IAW2 − 1.00 Slope 1 5. =IW1 1.538 AW1 + 0.406A) then IAW1 = 0.915.223IAW1 × 4.406A) then I AW 1 = 0.723 + (0.61 2.915A and <12. 9A as shown in Figure 93.905A − 4.131 I AW 1 = 0.44A from the x-axis to the characteristic curve.639I AW 2 =IAW1 0.00 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Restraint Current (I W2 in Amps) Figure 93: Using Graphs to Determine Pickup Settings It is more convenient to use formulas to determine the expected pickup current as shown below. follow the current up until it crosses the line.971IAW2 − 4.00 (I W1 in Amps) 10. follow 18.The Relay Testing Handbook You could use this graph to determine the expected pickup current for a Slope-1 test by choosing a W2 current after the Minimum Pickup and less than Slope-2.00 Slope 2 Operate Current 15.00 Slope 1 Restraint Area 5. Slope-1 Slope-2 =IAW1 0. Differential Protection Characteristic Curve 25. The relay should operate at approximately 5.44A).44A ) − 4.971 (18.639 ( 9.774A 102 .00 0. Find that current on the x-axis.7A as per Figure 93. For example. Test Slope-2 by choosing a current below the Slope-2 portion of the graph (4xTap=18.892A IAW1 = 13.131 I AW 1 = 0. The expected pickup current for the Slope-2 test is 13.22A ) IAW1 17.00 Trip Area 20. we could test Slope-1 by applying 2xTap to both windings and increase the W1 current until the relay operates. and follow it to the y-axis.131 = I AW 1 = 5. then follow the crossover point back to the other axis to determine the expected pickup. Chapter 2: Percent Differential (87) Element Testing The test sheet for this test could be similar to the test sheet below with the “EXPECTED” calculation: EXPECTED = IF (I AW 2 < 6.69 7.915.639 * I AW 2 .723 + (0.29 3 16.971 * I AW 2 ) − 4.17 2 12.0 103 .523 * I AW 2 ).0.61 W2CTC 1 SLOPE TESTS (Amps) TEST IAW2 (A) I_W1 PHASE (A) EXPECTED (A) % ERROR 1 10.0 A 7.30 SLP1 20 SLP2 60 IRS1 3.(0.0 A 6.41 W1CTC 12 SLOPE 2 60% TAP2 4. IF (I AW 2 < 12.368 13.0 A 11.16 C) Post-Test Calculations It is often easier to pick random points and perform a calculation to determine if the test result is correct rather than perform all of the calculations above.421 11.3 TIME DELAY 0 SLOPE 1 20% TAP1 2. We’re using the same information for this test as summarized below: Winding-1 Winding-2 W_CTC 12 1 Tap_ 2.131)) DIFFERENTIAL TEST RESULTS PICK UP 0.390 0.14 4 18.347 0.406.61 O87P 0.0 A 13.668 0.41 4.401 6.0.405 0. 0 A 21.091 104 .0 A 9.” This can be translated into W1 W 2 because we will 2 use balanced 3-phase currents at the theoretical normal operating angles. The magnitude of this result is IOP which can be translated into IW1 − IW 2 for our calculations because we will use balanced 3-phase currents at the theoretical normal operating angles.0 A 5.3 × < IRT 20 20 IOP = × IRT   −40   100 IOP = 0.0 A 3.401 7 12.6IRT − 1.309 12 22.The Relay Testing Handbook IRT is defined as “The magnitudes of the winding currents are then summed in a simple I +I scalar addition and divided by 2.779 3 4.421 10 18.6IRT + 3.3 TIME DELAY 0 SLOPE 1 20% TAP1 2. Remember that both of these formulas are still in per-unit and not actual Amp values.128 6 10.0 × −0.0 A 0. O87P SLP1 SLP2  SLP2    SLP1 − SLP2   IOP =  × IRT  + IRS1 ×    100    100  100 O87P × < IRT SLP1 SLP1 IOP = × IRT  60    20 − 60   100 IOP =  × IRT  + 3.368 11 20.5 < IRT IOP 0.0 A 17.2 = The following test results were recorded: DIFFERENTIAL TEST RESULTS PICK UP 0.40 = IOP 0.0 A 11.825 4 6.0 ×   0.197 14 26.0 A 23.0 A 1.0 A 6.20 × IRT = 1.0 A 19.6IRT + 3.47 9 16.0 A 7.61 W2CTC 1 PICKUP TESTS TEST IAW2 (A) IAW1 (A) IRT IOP EXPECTED ACTUAL E R R O R (% ) 1 0.143 15 28. IOP is defined as “In each element a phasor addition sums the winding currents.69 8 14.0 A 13.3 × 5 < IRT   100   IOP 0.0 A 2.0 ×   100 100    100   0.878 5 8.0 A 15.41 W1CTC 12 SLOPE 2 60% TAP2 4.726 2 2.25 13 24. 61 TEST4 = IOP − 2 2.385 13 24.392 6 10.0 A 2.309 5.0 + 3.508 105 .41 4.0 A 0.020 0.41 4.0 A 11.0 A 6.105 1.878 1.197 6.759 14 26.0 A 9.588 8 14.41 4.965 2.41 4.413 0.0 + 2.368 18.41 4.413 TEST6 IOP = 0.41 W1CTC 12 SLOPE 2 60% TAP2 4.61 TEST6 IRT = 2.897 0.726 0.0 A 13.41 4.726 1.0 A 23.487 7 12.206 3.0 A 7.3 TIME DELAY 0 SLOPE 1 20% TAP1 2.893 9 16.455 TEST4 IOP = 0.401 10. Chapter 2: Percent Differential (87) Element Testing First we apply the formula for IRT Then we apply the formula for IOP IW1 + IW2 IRT = 2 IOP = IW1 − IW2 IAW1 I + AW2 IRT = TAP1 TAP2 IAW1 I 2 IOP = − AW2 TAP1 TAP2 3.0 13.586 2.455 0.0 A 17.61 TEST6 = IOP − 2 2.151 0.0 A 3.133 15 28.0 A 1.308 6.368 4.61 W2CTC 1 PICKUP TESTS TEST IAW2 (A) IAW1 (A) IRT IOP EXPECTED ACTUAL ERROR (%) 1 0.421 4.932 0.128 1.642 Now our test sheet is shown below: DIFFERENTIAL TEST RESULTS PICK UP 0.401 2.61 TEST4 IRT = 1.301 2 2.586 0.61 TEST10 = IOP − TEST10 IRT = 2.308 5 8.61 2 TEST10 IRT = 4.779 0.143 7.0 TEST4 IRT = 2.642 11 20.0 A 19.487 13.0 A 21.368 18.305 4 6.47 3.828 3.014 12 22.726 TEST10 IOP = 1.825 1.25 5.0 TEST6 IRT = 2.304 3 4.345 2.878 6.0 A 5.091 7.401 10.483 0.268 10 18.0 A 15.0 + 6.69 2.878 6. 00 60.0 A 23.30% 4.18 13 24.642 + 1.61 W2CTC 1 PICKUP TESTS TEST IAW2 (A) IAW1 (A) IRT IOP EXPECTED ACTUAL ERROR (%) 1 0.2 SLP2 IRT = IRT 100 0.21 12 22.151 0.32 1.133 60.825 1.22 10 18.00 60.0 A 3.893 60.13 0.759 60.345 2.2 Test10 SLP2 = 100 × 0.18% IOP + 1.2 2.24 11 20.18 0.413 0.0 A 5.00 20.30 0.0 A 13.0 A 21.301 0.401 2.487 Test6 SLP1=100 × IOP + 1. the expected result formulas are fairly simple.00 60.421 4.00 60.897 0. If IRT <1.30 0.20 14 26.305 0.487 20.726 0.00 20.105 1.00 60.46 8 14.00 20.00 60.30 0.308 0.29 1.41 2 2.30 0.47 3.0 A 2.5 < IRT < 3 = Slope-1 IRT > 3 = Slope-2 SLP1  SLP2    SLP1 − SLP2   IOP = × IRT IOP =  × IRT  + IRS1 ×   100  100    100  IOP SLP1 =  SLP2  IRT 100 IOP=  × IRT  − 1.13 9 16.878 1.304 0.268 60.13 0.13 0.0 A 9.368 4.44 3 4.25 5.726 1.30 1.206 3.455 0. IOP = 0.31 2.21 15 28.2 =  × IRT   100  IOP Test6 SLP1=100 × IOP + 1.11 0.588 IRT Test7 SLP1=100 × 2.508 60.392 20.69 2.3 TIME DELAY 0 SLOPE 1 20% TAP1 2.41 W1CTC 12 SLOPE 2 60% TAP2 4.828 3.13% The final test sheet could look like the following: DIFFERENTIAL TEST RESULTS PICK UP 0.197 6.08 0.0 A 17.15 0.385 60.30 0.30 1.642 60.51 4 6.588 20.483 0.965 2.24 106 .091 7.0 A 7.0 A 0.586 0.2  100  IOP SLP1 = 100 ×  SLP2  IRT IOP + 1.726 Test10 SLP2 = 60.897 1.128 1.The Relay Testing Handbook With IOP and IRT.586 2.0 A 15.14 0.12 0.54 5 8.0 A 1.779 0.014 60.0 A 11.0 A 6.5.143 7.59 6 10.413 SLP2 = 100 × IRT Test6 SLP1=20.2 Test10 SLP2 = 100 × Test7 SLP1=20.00 60.0 A 19.89 7 12.3 1.00 60.020 0.932 0.309 5. SLP2)) •• ACTUAL=IF(IRT<(O87P × 100 / SLP1).3 TIME DELAY 0 SLOPE 1 20% TAP1 2.0 A 13.413 0.41 W1CTC 12 SLOPE 2 60% TAP2 4. Slope-2 will occur if the IRT is greater than 3.401 2.IF(IRT<3.487 7 12.61 W2CTC 1 PICKUP TESTS TEST IAW2 (A) IAW1 (A) IRT IOP EXPECTED ACTUAL ERROR (%) 6 10.100*(IOP/IRT).69 2.309 5.368 4.588 10 18.897 0.642 11 20.IOP.O87P. DIFFERENTIAL TEST RESULTS PICK UP 0.5 and 3. Differential Protection Characteristic Curve Trip Area 20 Slope 2 Operate Current (I W1 in Amps) Rise2 Run2 10 Slope 1 Run1 Rise1 Restraint Area 0 0 10 20 30 Restraint Current (I W2 in Amps) Figure 94: Determine Slope by Rise/Run Calculation 107 .SLP1. Chapter 2: Percent Differential (87) Element Testing The formulas used for this test report are: •• IRT=((IAW2 / TAP2) + (IAW1 / TAP1)) / 2 •• IOP=(IAW1/TAP1)-(IAW2/TAP2) •• EXPECTED=IF(IRT<(O87P × 100 / SLP1).0 A 15.0 A 6. You can tell which test will be Slope-1 if the IRT is between 1.726 1.IF(IRT<3.345 2.0 A 7.100*((IOP+1. Follow the Post-Test Calculation in the previous section to the point that IRT and IOP are calculated and pick two test points from each slope.014 You can use the graphical Rise over Run formula to determine the slope if you know the correct IOP and IRT for any two test results for a given slope as shown in the following example.2)/IRT))) D) Alternate Slope Calculation There is another way to test slope that does not require as much preparation or calculation. 101 0.97 -0. The difficult part of this procedure is determining what starting current to apply and what the expected pickup should be.345 .014 .28 10 18.0 A 6.413 0.IRT10 0. Calculate IRT after each test to determine whether O87P.487 7 12. Slope-1.642 11 20.372 %Slope1 = 100 × %Slope2 = 100 × 0.00 59. 108 .487 2.86 4. and Slope-2 settings are tested.IOP10 %Slope1 = 100 × %Slope2 = 100 × IRT7 .69 2.IOP6 IOP11 .4.0 A 15.09% The final test sheet could look like the following: DIFFERENTIAL TEST RESULTS PICK UP 0.00A restraint current and determine pickup. Slope-1.1.345 2. or Slope-2.0.014 60.309 5.3 TIME DELAY 0 SLOPE 1 20% TAP1 2.642 %Slope1 = 100 × %Slope2 = 100 × 2. Increase the restraint current and determine pickup and repeat until the O87P.0 A 13.00 20.897 0.401 2. and raise all three currents in one winding until the relay operates.726 0.0 A 7.413 5. You can use one of the following test procedures to test the relay slope settings: •• Pre-Test Calculations—Use the calculations to determine the actual restraint and expected operate currents and insert these values into the procedure.726 1. Apply 3-phase balanced current into Winding-1 and Winding-2 using the phase shift described in the previous section of this chapter.484 0.06 E) 3-Phase Differential Slope Test Procedure The slope test procedure seems straightforward.588 .588 20.IRT6 IRT11 . •• Post-Test Calculations—Start at 0.The Relay Testing Handbook Determine slope using the rise-over-run graphical method as shown in Figure 94 and the following formulas: Rise Rise %Slope1 = 100 × %Slope2 = 100 × Run Run IOP7 .86% %Slope2 = 60.61 W2CTC 1 PICKUP TESTS TEST IAW2 (A) IAW1 (A) IRT IOP EXPECTED ACTUAL ERROR (%) 6 10.368 4.897 .619 %Slope1 = 20.2.41 W1CTC 12 SLOPE 2 60% TAP2 4. The pickup indication should be ON because we have applied a a 200% slope as per the calculations earlier in this document.0A into the relay so this test should be completed as quicklyb as possible. The b C B c A a 109 b C B c . C B (Pickup c bindication by LED.23A 5. Weaare applying current greater than 5.23 c C B b Connection SEL-387 W1AØ W1BØ W1CØ W2AØ W2BØ W2CØ A a W1CTC=12 5. The applied W2 current c C B for our example will be: b Connection SEL-387 W1AØ W1BØ W1CØ W2AØ W2BØ W2CØ A a W1CTC=12 0A 0A 0A 10. Apply balanced. 3-phase restraint current through C B c b W2 at the phase-angle described in the “3-Phase Restrained-Differential Slope Testing” A chapter of this chapter. output contact. Determine how you will monitor pickup and set the relay accordingly.41 W1 = 10 × 4.0A@ 10. etc…See previous packages of The Relay Testing Handbook for details. but opposite (accounting for phase shift) 3-Phase current in W1.61 C B c b W1 = 10 × 0. front panel display.0A@ 10.) A a 2.0A@ c C B W2CTC=1 0º -120º 120º 150º 30º -90º b A3. Chapter 2: Percent Differential (87) Element Testing Follow these steps to test the Differential Slope settings using six current channels: A a 1.0A@ 10. Determine the Slope-1 restraint current by selecting a value between the Minimum- Pickup and Slope-2 transition points.0A@ 10. Apply a A a an equal.0A@ c C B W2CTC=1 0º -120º 120º 150º 30º -90º b A a Relay should not operate. You may want to apply the previous step for a moment to ensure C B thec 87-Element operates and setup the next step offline.23A 10. You can use the Tap ratios to determine what the current should be.523 A a W1 = 5. A 4.23A 5. For example: C B c bb C B TAP1 c W1 = W2 × TAP2 A a 2. if required. 401 6. Test Slope-2 by repeating steps 2-6 with a restraint current greater than the Slope-1 to Slope-2 transition level. 8.390 0.10 A 1 A Model: ±5% ±0. We measured the pickup to be 6.0 A 13. Compare the pickup test result to the manufacturer’s specifications and calculate the percent error as shown in the following example.0A. you could also use the pulse or jog method to minimize the amount of current applied to the relay. All of the results in the test sheet are less than 0.5% ± 0.The Relay Testing Handbook 5.390A using the Slope-1 formula when IAW2 = 10.1A .69 7.175A.405 0. Because the applied current is higher than 10.5% so we can consider them acceptable.401A.668 0.421 11.41 W1CTC 12 SLOPE 2 60% TAP2 4. Repeat steps 2-5 with another restraint current between the Minimum-Pickup and Slope-2 transition points until the required number of tests are completed.61 W2CTC 1 SLOPE TESTS (Amps) TEST IAW2 (A) I_W1 PHASE (A) EXPECTED (A) % ERROR 1 10.0 in per unit of tap Pickup Accuracy (A secondary) 5 A Model: ±5% ±0. we see that the acceptable metering error is ±0.02 A 110 . 6. Some organizations want more than one test point to determine slope.29 3 16.347 0.0 A 11.3 TIME DELAY 0 SLOPE 1 20% TAP1 2.) The measured pickup for our example is 6.368 13.17 2 12. DIFFERENTIAL TEST RESULTS PICK UP 0. Looking at the “Differential Element” specification below.1–1. Raise the W1 current until the element operates.0 A 7.0A. 7.16 Slope-1 In the TEST 1 example. we calculated that the 87-Element should operate when IAW1 = 6. Differential Element Unrestrained Pickup Range: 1–20 in per unit of tap Restrained Pickup Range: 0.0 A 6.14 4 18. (Review the Instantaneous Overcurrent (50-Element) pickup procedure in previous packages of The Relay Testing Handbook for details. 1A ” from the specifications above. the allowable percent error is 1.39A 2.401-6.1A × 100 = Allowable Percent Error Expected (0. Chapter 2: Percent Differential (87) Element Testing IAW1-EXPECTED × 100 = Percent Error EXPECTED 6. change all connections and references to W2 to the next winding under test.39A 0.390 × 100 = Percent Error 6.6%.032) + 0.005 × 6.390 0. 9.1A × 100 = Allowable Percent Error 6.1A × 100 = Allowable Percent Error 6. (W2 becomes W3 for W1-to-W3 tests) 111 . Using “ ±0.5% ± 0.1% Allowable Percent Error Rule of Thumb Remember that there are other factors that will affect the test result such as: •• Relay close time. Repeat the pickup test for all windings that are part of the differential scheme. If more than two windings are used. •• Test-set sensing time. Maximum Accuracy Tolerance × 100 = Allowable Percent Error Expected (0. (The longer it takes the contact to close the larger the test result will be if ramping the pickup current.39A ) + 0. Some of these error factors can be significant and a rule-of-thumb 5% error is usually applied to test results.39A (0.) •• Test-set analog output error.132 × 100 = Allowable Percent Error 6.5% × EXPECTED) + 0.17% Allowable Percent Error We can calculate the manufacturer’s allowable percent error. the current in one Wye phase CT is reflected into two phases of the Delta CT as shown in Figure 95. Our example will use the same settings used earlier in this chapter for the SEL-387 which are summarized to: Winding-1 Winding-2 E87W_ Y Y W_CT Y Y W_CTC 12 1 VWDG_ 230 18 Tap_ 2. 1-Phase Restrained-Differential Slope Testing 3-Phase differential slope testing is relatively straight-forward as you simulate balanced 3-phase conditions that the relay would expect to see in an ideal world to make all of the calculations easier.41 4.0 A) Understanding the Test-Set Connections Wye-Wye and Delta-Delta connections are not a problem as there is no phase shift and the current in any phase-winding is reflected in the same phases on the other windings.30 SLP1 20 SLP2 60 IRS1 3. The 1-Phase testing procedure described in this section uses the same calculations and test procedure described in the previous “3-phase Differential Slope Testing” section but the connections are completely different. Then apply the W2 current to the phases 112 . we must apply the W1 current to the phases W2 usually measures.61 O87P 0. Schweitzer Engineering Laboratories does have a test procedure and calculation for single-phase testing but I have found that the procedure described here is easier to apply and calculate. However in a Delta-Wye or Wye-Delta transformer.The Relay Testing Handbook 7. CT1 CT2 400:5 1200:5 H1 X1 IX1=Ia-Ic@-30º PHA Ia@0º IA@0º IH1=0º H2 X2 PHB Ib@-120º IB@-120º IH2@-120º IX2@-150º H3 X3 PHC IH3@120º IX3@90º Ic@120º IC@120º H0 Figure 95: YDac Transformer Connection When testing transformer differential elements on a single-phase basis. W2=30° D/y30° DABYyy DABY 0 1 A-C A-N B-A B-N C-B C-N W1=0. Connect the C-Phase W1 test current to ICW1 and ICW1 and W2 test current to ICW2. W2=330° Y/d330° YDAByy DACY 12 11 A-B A-N B-C B-N C-A C-N 12 0 A-B A-B B-C B-C C-A C-A D/d0° D/d0° DACDACyy DACDAC 0 0 A-N A-N B-N B-N C-N C-N 0 12 A-B A-B B-C B-C C-A C-A W1=0. Most digital relays apply zero- sequence filtering to prevent a transformer differential trip for ground faults on the power system. Zero-sequence is a problem in transformer differential protection because a phase-to-ground fault outside the transformer on the power system can cause zero-sequence current to flow through one winding of the transformer and not appear in the other winding due to a delta or tertiary winding. Zero-sequence filtering is applied on the SEL-387 relay when the W_CTC setting equals 12. Another problem with 1-Phase testing is the unbalance or zero sequence current that is applied to the relay. connect the W1 test current to IAW1 and ICW1 in series and the W2 test current is applied to IAW2 for an A-Phase test. W2=30° Y/d30° YDACyy YDAC 12 1 A-C A-N B-A B-N C-B C-N W1=0. W2=330° D/y330° DACyy DACY 0 11 A-B A-N B-C B-N C-A C-N 1 0 A-N A-C B-N B-A C-N C-B 1 12 A-N A-C B-N B-A C-N C-B 1 1 A-N A-N B-N B-N C-N C-N 11 0 A-N A-B B-N B-C C-N C-A 11 12 A-N A-B B-N B-C C-N C-A 11 11 A-N A-N B-N B-N C-N C-N Figure 96: Transformer Relay Connections for Single-Phase Differential Testing 113 . Connect the B-Phase W1 test current to IBW1 and IAW1 and W2 test current to IBW2. GE T-60. Use the table in Figure 96 to determine the correct 1-Phase test connections. For the example in Figure 95. both windings will use a Phase- Phase connection. When both W_CTC settings are equal. Chapter 2: Percent Differential (87) Element Testing W1 normally measures. A phase-to-phase connection is required to compensate for the zero-sequence compensation. W2=0 Y/y0° Yyyy YY 12 12 A-B A-B B-C B-C C-A C-A W1=0. T-30 SR-745 Beckwith SEL-587 SEL-387 A Phase B Phase C-Phase W1 W2 W1 W2 W1 W2 W1 W2 W1=0. 00A 0° Test Hz Z06 WINDING 2 Z07 + IAW2 Z08 Z09 + IBW2 Z10 Z11 + ICW2 Z12 Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 98: 1-Phase Restrained-Differential Slope Test-Set AØ Yd1 Connections 114 . Using the table in Figure 96. the following test-set connections will be required for each phase: SEL-387 RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency Z01 + + IAW1 C1 Amps W1AØ Test Amps 0° Test Hz Z02 Z03 + + IBW1 C2 Amps W2AØ Test Amps 180° Test Hz Z04 Z05 + + ICW1 C3 Amps 0.The Relay Testing Handbook B) Test-Set Connections The connection diagram for the SEL-387 is as follows: Figure 97: Schweitzer Electric SEL-387 Transformer Protective Relay Connections The winding settings for this transformer are W1CTC=12 and W2CTC=1. Chapter 2: Percent Differential (87) Element Testing SEL-387 RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency Z01 + + IAW1 C1 Amps W1BØ Test Amps 0° Test Hz Z02 Z03 + + IBW1 C2 Amps W2BØ Test Amps 180° Test Hz Z04 Z05 + + ICW1 C3 Amps 0.00A 0° Test Hz Z06 WINDING 2 Z07 + IAW2 Z08 Z09 + IBW2 Z10 Z11 + ICW2 Z12 Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 99: 1-Phase Restrained-Differential Slope Test-Set BØ Yd1 Connections SEL-387 RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency Z01 + + IAW1 C1 Amps W1CØ Test Amps 0° Test Hz Z02 Z03 + + IBW1 C2 Amps W2CØ Test Amps 180° Test Hz Z04 Z05 + + ICW1 C3 Amps 0.00A 0° Test Hz Z06 WINDING 2 Z07 + IAW2 Z08 Z09 + IBW2 Z10 Z11 + ICW2 Z12 Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 100: 1-Phase Restrained-Differential Slope Test-Set CØ Yd1 Connections 115 The Relay Testing Handbook C) 1-Phase Differential Slope Test Procedure Follow these steps to test the Differential Slope settings using the Post-Test Calculation method from the previous “3-Phase Differential Slope Testing” Section: 1. Determine how you will monitor pickup and set the relay accordingly, if required. (Pickup indication by LED, output contact, front panel display, etc…See previous packages of The Relay Testing Handbook for details.) 2. Apply a restraint current into the first test phase of Winding-2. The pickup indication should be on. 3. Apply an equal, but opposite current in W1 using the Tap ratios to determine what the current magnitude should be. For example: TAP1 × A1 W1 = W2 × TAP2 × A2 2.41 × 1 W1= 5 × 4.61 × 1.732 W1= 5 × 0.302 W1 = 1.510 4. The pickup indication should now be off. Raise the Winding-1 current until the relay operates. Record the value on your test sheet. (Remember that you can also use the pulse or jog method to minimize the amount of current applied to the relay. Review the Instantaneous Overcurrent (50-Element) pickup procedure in previous packages of The Relay Testing Handbook for details.) 5. Calculate the IRT, IOP, and Slope using the formulas in the Post-Test Calculation section of the “3-Phase Differential Slope Testing” in this document. Use a 1.732 correction factor for windings with odd W_CTC settings and use the IRT calculation to determine if the test was O87P, Slope-1, or Slope-2. Raise the W2 current and repeat 2-5 until all elements of the differential protection have been tested. ¾¾ ( ) ( IRT=  IAW2 / ( TAP2 × A2 ) + IAW1 / ( TAP1 × A1)  / 2 ) IOP= IAW1/ ( TAP1 × A1)  - IAW2/ ( TAP2 × A2 )  ¾¾ ¾¾ EXPECTED=IF(IRT<(O87P × 100 / SLP1),O87P,IF(IRT<3,SLP1,SLP2)) ¾¾ ACTUAL=IF(IRT<(O87P × 100 / SLP1),IOP,IF(IRT<3,100*(IOP/IRT),100*((IOP+1.2)/IRT))) 116 Chapter 2: Percent Differential (87) Element Testing 6. Compare the pickup test result to the manufacturer’s specifications and calculate the percent error as shown in the following example. DIFFERENTIAL TEST RESULTS PICK UP 0.3 TIME DELAY 0 CORR SLOPE 1 20% TAP1 2.41 W1CTC 12 1 SLOPE 2 60% TAP2 4.61 W2CTC 1 1.732 PICKUP TESTS TEST IAW2 (A) IAW1 (A) IRT IOP EXPECTED ACTUAL ERROR (%) 1 0.0 A 0.736 0.153 0.305 0.30 0.31 1.80 2 5.0 A 2.239 0.778 0.303 0.30 0.30 0.94 3 15.0 A 5.54 2.089 0.420 20.00 20.11 0.57 4 20.0 A 7.394 2.786 0.563 20.00 20.21 1.06 5 25.0 A 9.923 3.624 0.986 60.00 60.33 0.54 6 30.0 A 12.726 4.519 1.523 60.00 60.26 0.44 TEST IAW2 (A) IBW2 (A) IRT IOP EXPECTED ACTUAL ERROR (%) 7 0.0 A 0.732 0.152 0.304 0.30 0.30 1.24 8 5.0 A 2.24 0.778 0.303 0.30 0.30 1.08 9 15.0 A 5.544 2.090 0.422 20.00 20.19 0.93 10 20.0 A 7.382 2.784 0.558 20.00 20.05 0.26 11 25.0 A 9.902 3.620 0.978 60.00 60.16 0.26 12 30.0 A 12.711 4.516 1.517 60.00 60.17 0.28 TEST IAW2 (A) IBW2 (A) IRT IOP EXPECTED ACTUAL ERROR (%) 13 0.0 A 0.732 0.152 0.304 0.30 0.30 1.24 14 6.0 A 2.544 0.904 0.304 0.30 0.30 1.38 15 15.0 A 5.535 2.088 0.418 20.00 20.02 0.12 16 21.0 A 7.753 2.924 0.587 20.00 20.08 0.38 17 24.0 A 9.342 3.441 0.871 60.00 60.17 0.28 18 27.0 A 11.035 3.980 1.197 60.00 60.23 0.38 7. Repeat steps 2-6 for all three phases. 8. Repeat the pickup test for all windings that are part of the differential scheme. If more than two windings are used, change all connections and references to W2 to the next winding under test. (W2 becomes W3 for W1 to W3 tests) 117 As you can see. Figure 101: Transformer Inrush Waveform 118 . Also the waveform is extremely distorted and doesn’t display the nice round peaks we normally see. This current only occurs in the first- energized winding and can be up to 10x the transformer’s nominal current that. and there are several options available to prevent transformer inrush from tripping the differential relay. We do not want the transformer differential to operate every time a transformer is energized. would be the definition of transformer differential and the differential relay should operate. The most common technique used to prevent differential operation during inrush conditions is called Harmonic Restraint or Harmonic Blocking. Under normal operating conditions. There is a significant DC offset in the first few cycles where the center point between the positive and negative peaks is not the x-axis. but this would prevent the relay from operating during high-impedance faults. in any other circumstance. this is not your typical sine wave. You could increase the pickup setting. but an internal transformer fault would not be isolated until the time delay had passed that will cause additional damage for the extended time caused by this workaround.The Relay Testing Handbook 8. Figure 101 displays a typical transformer inrush waveform. Harmonic Restraint Testing Transformers are inductive machines that require a magnetic field to operate. A large inrush of current is required in the first few cycles after energization to create the magnetic field necessary for transformer operation. It is possible to block the 87-Element for a preset time after a circuit breaker is closed. there are very small losses inside the transformer that are typically less than 3%. These harmonic detectors will prevent differential operation if the harmonics exceed a pre-defined setpoint.4th and/or 5th harmonics in a waveform. If your test equipment allows you to directly adjust the applied harmonic for an output current channel. Harmonic detectors were built into transformer differential relays which measure the percent of 2nd. SEL-387 RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency Z01 + + IAW1 C1 Amps W1AØ Test Amps 0° Test Hz Z02 Z03 + + IBW1 C2 Amps W1BØ Test Amps 0° Test Hz Z04 Z05 + + ICW1 C3 Amps W1CØ Test Amps 0° Test Hz Z06 WINDING 2 Z07 + IAW2 Z08 Z09 + IBW2 Z10 Z11 + ICW2 Z12 Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 102: Simple 3-Phase 87-Element Test-Set Connections 119 . you can use the standard timing test connections diagram shown in Figure 102. A) Harmonic Restraint Test Connections There are two possible connections for testing Harmonic Restraint depending on your test equipment. Chapter 2: Percent Differential (87) Element Testing Protective relay designers analyzed these waveforms and realized that the distortion was caused by even harmonic content that typically only occurred during transformer energization and not during transformer faults. The harmonic percentage is the ratio of harmonic current to fundamental current. Other relays may require the harmonic to start below the setting and ramp up. SEL-387 RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency Z01 + + IAW1 C1 Amps W1AØ Test Amps 0° Test Hz Z02 Z03 + + IBW1 C2 Amps W1AØ Harmonic Amps 0° Harmonic Hz Z04 Z05 + + ICW1 C3 Amps 0. We will modify our timing test procedure to apply the pulse method which should work on most relays. Apply the connection diagram from Figure 102. Use the following procedure to test the Harmonic Restraint when a harmonic setting is available on your test-set. (Typically the 2nd .00A 0° Test Hz Z06 WINDING 2 Z07 + IAW2 Z08 Z09 + IBW2 Z10 Z11 + ICW2 Z12 Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 103: Simple.) 2.The Relay Testing Handbook Otherwise you will need two channels to test the Harmonic Restraint connected in parallel as shown in Figure 103. 4th. 1. Some relays will only restrain when the current is pulsed to look like the sudden inrush of a transformer. 120 . Turn the test-set channel harmonic setting on and choose the correct harmonic. Connect the test-set input(s) to the relay output(s) that are programmed to operate when the restrained-differential relay operates. Some other relays may require the harmonic to be above the setpoint and ramp down so you actually measure the dropout instead of the pickup. or 5th Harmonic. This test configuration includes one channel at the fundamental frequency and the second channel set at the harmonic frequency. Higher Current 3-Phase 87-Element Test-Set Connections B) Harmonic Restraint Test Procedure with Harmonics Harmonic Restraint application can be very different depending on the relay manufacturer and model. (25% for a 20% setting) 6. Set a single-phase current at least 200% higher than the Minimum Pickup setting using the calculations in the “1-Phase Pickup Test Procedure” section of this chapter. This is the Harmonic Restraint pickup result. Return to the last test harmonic when the element operated and increase the harmonic in 0.6 121 . Apply the current and make sure the relay differential element does not operate.3 15.3 TIME DELAY 0 PCT2 15 SLOPE 1 20% TAP1 2.7 2.6 15. Apply the current and make sure the relay differential element operates.6 35.8 35.3 1.2 15.446A ) for Winding-1 or 4.) 4. DIFFERENTIAL TEST RESULTS PICK UP 0.1% increments and apply until the relay does not operate.41 PCT4 15 SLOPE 2 60% TAP2 4.0 1.3 0. 5. 10. 7. (1. Set the harmonic percentage 5% higher than the pickup setting.1 15.61 PCT5 35 HARMONIC RESTRAINT TESTS (Amps) TEST HARMONIC IAW1 (%) IBW1 (%) ICW1 (%) EXPECTED (%) % ERROR 1 2nd 14.0 3 5th 34.0 4.9 35.0 = 1.0 -0.8 15. Increase the harmonic in 1% increments and apply until the relay does not operate.2 15. Repeat steps 3-10 for all phases. Chapter 2: Percent Differential (87) Element Testing 3.446A (0.395 × 2.6 1. 11. 9. Record on your test sheet. Apply the test current and ensure the relay operates.79A (2. Reduce the harmonic percentage to 5% below the pickup setting (15% for a 20% setting) 8.79A ) for Winding-2.0 -1.7 2 4th 15.723 × 2.3 2.0 =4. Multiply the harmonic pickup setting (PCT2=15%) by the applied current from Step 2 (1.446A (0.0 = 1. This is the Harmonic Restraint pickup result. Increase the harmonic current in 1% increments and apply until the relay does not operate. 122 . different relays have different operating characteristics when applying harmonic restraint and we’ll use the pulse method with the following procedure. 1. Reduce the harmonic percentage to 5% below the expected current (0.446A ) for Winding-1 or 4. 6. 8.10 = 0. and increase the harmonic in 0. Connect the test-set input(s) to the relay output(s) that are programmed to operate when the restrained-differential relay operates. As described in the previous section.446 × 0. Set the current 2 Channel frequency to the full harmonic frequency.15 =0. 3.95 = 0. 5th = 300Hz) We’ll call this “harmonic current” from now on.79A ) for Winding-2) Apply the current and make sure the relay differential element operates. Repeat steps 2-8 for all phases.217 × 1. 2. Set your test-set’s harmonic current output 10% higher than the expected harmonic current (0.The Relay Testing Handbook C) Harmonic Restraint Test Procedure with Current Testing without a harmonic setting on your test-set requires two current channels connected in parallel with one current channel set at the fundamental frequency (60 Hz in our example) and the second current at the harmonic frequency.217A ) . Harmonics are described in multiples of the fundamental frequency so a 2nd harmonic would be 120 Hz in our example or 100 Hz in a system with a 50Hz nominal frequency.239A ) . Apply the fundamental and harmonic currents simultaneously and the relay should not operate.0 = 4.1% increments and apply until the relay does not operate.79A (2. 4th = 240Hz. 5. Return to the last test harmonic when the element operated.723 × 2. 7. Apply the connection diagram from Figure 103. This is your expected current. (1. Record on your test sheet. Set the current Channel 1 at the fundamental frequency (60Hz) with a magnitude at least 200% higher than the Minimum Pickup setting using the calculations in the “1-Phase Pickup Test Procedure” section of this chapter. (2nd = 120 Hz.395 × 2. The relay should operate.217 x 0. 4. 9.206A) and apply the fundamental and harmonic currents simultaneously. 0% × EXPECTED) + 0.446 A 0.446 A 0.515 0.217 1.503 0.1A × 100 = Allowable Percent Error 0.1A × 100 = Allowable Percent Error 0.218 0.3 0. Chapter 2: Percent Differential (87) Element Testing DIFFERENTIAL TEST RESULTS PICK UP 0.506 -0.3 TIME DELAY 0 PCT2 15 SLOPE 1 20% TAP1 2.011) + 0.111 × 100 = Allowable Percent Error 0.446 A 0.221 0.3 2.1A × 100 = Allowable Percent Error EXPECTED (0.05 × 0.226 0.61 PCT5 35 HARMONIC RESTRAINT TESTS (Amps) HARM FUND IAW1 (A) IBW1 (A) ICW1 (A) EXPECTED (A) % ERROR PCT2 1.217A 51% Allowable Percent Error 123 .0 PCT5 1.6 D) Evaluate Results Use the harmonic specifications below to determine if the test results are acceptable.6 PCT4 1.220 0.0 4.4 1.519 0.220 0.217A (0.217 -1.6 1.7 2. All of the example test results are less than 5% so we do not need to proceed further We could calculate the exact allowable percent error with the following calculation which is a huge number because of the very low expected current: Maximum Accuracy Tolerance × 100 = Allowable Percent Error EXPECTED (5.41 PCT4 15 SLOPE 2 60% TAP2 4.217A 0.217) + 0.214 0. apply current at a lower value and review the relay’s measured values to make sure your test-set is actually producing an output and your connections are correct. Try a test point further from the break point.5. or 0. the phase shift will look backwards.577 multiple. •• Are you applying the currents into the same phase relationships? •• Did you calculate the correct Tap? •• Don’t forget to use the complete operate and restrained calculations. •• Are the 2 winding currents 180º apart? •• If the pickup tests are off by a 3 . If a Delta-winding is set as the reference. •• Slope-2 tests near the break point can yield seemingly erratic results. the relay probably is applying correction factors. Tips and Tricks to Overcome Common Obstacles The following tips or tricks may help you overcome the most common obstacles: •• Before you start. 1.The Relay Testing Handbook 9. 124 . and you should review the manufacturer’s literature. •• Always use a Wye-connected winding’s phase A as the reference when setting angles. 0 A Figure 104: Simple Differential Protection with Worst Case CT Error and External Fault With pickup settings in the 8-10x Tap range.0 A I1 = 45. unrestrained-differential protection becomes nearly identical to the Instantaneous Overcurrent (50-Element) test procedures in previous packages of The Relay Testing Handbook. Chapter 3: Unrestrained Differential Testing Chapter 3 Unrestrained-Differential Testing Electrical engineers are always concerned about close-onto-fault conditions. Additional protection can be applied which is set higher than the expected inrush current to detect faults on the first-energized winding of the transformer. 125 . There are no restraint coils to prevent mis-operations at high current levels and.0 A I2 = 55. 1000 A 100:5 CT 100:5 CT G FAULT Iop = 10. in fact. especially with transformer protection. The unrestrained-differential element operation is very much like the early differential relays described in the “Application” section of this chapter as shown in Figure 104. the element is typically set at 8-10x Tap. What would happen if a transformer circuit breaker closed onto a fault in the first-energized winding? It is conceivable that the inrush condition could mask the fault for several cycles and cause additional damage or system disturbances. the setting should be disabled or OFF to prevent confusion. the primary voltage. 126 . Some relays will also have “Latched” or ‘Unlatched” options. E) Block The Block setting defines a condition that will prevent the differential protection from operating such as a status input from another device. Unlatched indicates that the relay output contacts will open when the trip conditions are no longer present. make sure the condition is not true when testing. Settings A) Enable Setting Many relays allow the user to enable or disable settings. Make sure that the element is ON or the relay may prevent you from entering settings. B) Minimum Pickup The Minimum Pickup setting is usually set in multiples of Tap and is the minimum amount of current required for the relay to operate. and the CT ratio. Verify the correct Tap setting using the following formula: CTSEC × Power TAP = P-P Volts × 3 × CTPRI D) Time Delay The Time Delay setting sets a time delay between an 87-Element pickup and trip. A Latched option indicates that the output contacts will remain closed after a trip until a reset is performed and acts as a lockout relay. This setting is used as the per-unit operate current of the protected device and most differential settings are based on the Tap setting. If the element is not used. C) Tap The Tap setting defines the normal operate current based on the rated load of the protected equipment. If enabled.The Relay Testing Handbook 1. Always verify correct blocking operation by operating the end-device instead of a simulation to ensure the block has been correctly applied. This setting is rarely used. This setting is typically set at the minimum possible setting but can be set as high as 3 cycles for maximum reliability on some relays. Chapter 3: Unrestrained Differential Testing 2. Repeat until all enabled phases are tested. After the Winding-1 A-phase pickup test is performed. Test-Set Connections The most basic test-set connection uses only one phase of the test-set with a test lead change between every pickup test. RELAY INPUT RELAY AØ PU RELAY TEST SET WINDING 1 A Phase Input=Pickup Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps AØ Test Amps 0° Test Hz + + B Phase Amps C2 Amps + + C Phase Amps C3 Amps WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection DC Supply - + Element + + Timer - Element Timer Output Input Output + Input Figure 105: Simple 87U-Element Test-Set Connections 127 . move the test leads from Winding-1 A-Phase amps to B-Phase amps and perform the test again. W1 RELAY INPUT W2 RELAY INPUT W2 W1 RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps W1 Test Amps 0° Test Hz + + B Phase Amps C2 Amps 0. This connection also checks two windings simultaneously and reduces test times. we can minimize the impact on any relay input by splitting the test current into the analog inputs that operate the unrestrained-differential element as shown in Figure 107.00A 0° Test Hz + + C Phase Amps C3 Amps W2 Test Amps 0° Test Hz WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection + + DC Supply - + Element Timer - Output Input Element Timer Output + Input Figure 107: Parallel 87U-Element Test-Set Connections with Equal W_CTC Settings 128 .The Relay Testing Handbook This element will require a lot of current so the connection in Figure 106 will probably be required if you treat the element as a 50-Element. RELAY INPUT C1 Amps C2 Amps C3 Amps RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency + + A Phase Amps C1 Amps Test Amps / 3 0° Test Hz + + B Phase Amps C2 Amps Test Amps / 3 0° Test Hz + + C Phase Amps C3 Amps Test Amps / 3 0° Test Hz WINDING 2 + A Phase Amps + B Phase Amps + C Phase Amps Alternate Timer Connection + + DC Supply - + Element Timer - Output Input Element Timer Output + Input Figure 106: Parallel 87U-Element Test-Set Connections If we realize that two different relay inputs are used to create the operate current. If the W_CTC settings are not identical. Use the chart to determine connections for B and C-Phase testing. Using the table in Figure 108.00A 0° Test Hz Z06 WINDING 2 Z07 + IAW2 Z08 Z09 + IBW2 Z10 Z11 + ICW2 Z12 Alternate Timer Connection DC Supply - + Element + + Timer Element - Timer Output Input Output + Input Figure 109: 1-Phase Differential Test-Set AØ Yd1 Connections 129 . we need to apply the chart in Figure 108 to apply a connection diagram such as shown in the following connection drawings: GE T-60. W2=330° Y/d330° YDAByy DACY 12 11 A-B A-N B-C B-N C-A C-N 12 0 A-B A-B B-C B-C C-A C-A D/d0° D/d0° DACDACyy DACDAC 0 0 A-N A-N B-N B-N C-N C-N 0 12 A-B A-B B-C B-C C-A C-A W1=0. the following test-set connections will be required for A-Phase. T-30 SR-745 Beckwith SEL-587 SEL-387 A Phase B Phase C-Phase W1 W2 W1 W2 W1 W2 W1 W2 W1=0. W2=30° D/y30° DABYyy DABY 0 1 A-C A-N B-A B-N C-B C-N W1=0. W2=0 Y/y0° Yyyy YY 12 12 A-B A-B B-C B-C C-A C-A W1=0. W2=30° Y/d30° YDACyy YDAC 12 1 A-C A-N B-A B-N C-B C-N W1=0. W2=330° D/y330° DACyy DACY 0 11 A-B A-N B-C B-N C-A C-N 1 0 A-N A-C B-N B-A C-N C-B 1 12 A-N A-C B-N B-A C-N C-B 1 1 A-N A-N B-N B-N C-N C-N 11 0 A-N A-B B-N B-C C-N C-A 11 12 A-N A-B B-N B-C C-N C-A 11 11 A-N A-N B-N B-N C-N C-N Figure 108: Transformer Relay Connections for Single-Phase Differential Testing The winding settings for our example transformer are W1CTC=12 and W2CTC=1. Chapter 3: Unrestrained Differential Testing The connection diagram in Figure 107 only works correctly if both W_CTC settings are identical. SEL-387 RELAY RELAY TEST SET WINDING 1 Magnitude Phase Angle Frequency Z01 + + IAW1 C1 Amps W1AØ Test Amps 0° Test Hz Z02 Z03 + + IBW1 C2 Amps W2AØ Test Amps 0° Test Hz Z04 Z05 + + ICW1 C3 Amps 0. •• Raise current until pickup indication operates. •• Apply a current lower than pickup setting and make sure pickup indication does NOT operate. Determine how you will monitor pickup and set the relay accordingly (if required). 4. •• Tap1 = 2. the table does not seem to be correct when the WnCTC equals 12. The following settings will be used for our SEL-387 example.41 •• Tap2 = 4. •• Apply a current into one-phase higher than the pickup setting and make sure pickup indication operates. •• Repeat for all phases.5 However. (Pickup indication by LED. A single-phase test procedure is slightly more complicated because correction factors may apply. 8. 3. Digital transformer relays use algorithms to compensate for the different transformer phase shifts as described earlier and apply compensation factors to compare windings with different configurations. 7. front panel display.) 130 . 12 1. etc…See previous packages of The Relay Testing Handbook for details. output contact.The Relay Testing Handbook 3. 6. 5. 9. 10. The compensation factors for the example SEL-387 can be found in the Testing and Troubleshooting section of the instruction manual as shown in the following table: WnCTC Setting A 0 1 Odd: 1. We have found through experimentation that the correction factor is actually 1. 11 √3 Even: 2.61 •• U87P = 10 (xTap) Use the following steps to determine pickup: 1. Simple Pickup Test Procedure The simple unrestrained-differential pickup test procedure is identical to an Instantaneous Overcurrent test procedure. 1A and the measured pickup was 24. Chapter 3: Unrestrained Differential Testing 2.0/1. set the fault current at 25. Compare the pickup test result to the manufacturer’s specifications and calculate the percent error as shown in the following example.10 A 1 A Model: ±5% ±0. Make sure pickup indication operates.9 cycles Restrained Element (with harmonic Blocking) Pickup Time (Min/Typ/Max): 1.845 × 1.6/2.61 × (1 = 3) 3 Minimum Pickup = 24.0 in per unit of tap Pickup Accuracy (A secondary) 5 A Model: ±5% ±0. Return to the last test current when the element did not operate and increase the fault current in 0.5/1.2 cycles Restrained Element (with harmonic Restraint) Pickup Time (Min/Typ/Max): 2. Determine the expected pickup.10A.41 × (12 =1)1 Minimum Pickup = 10 × 4. 8. For example.1A Minimum Pickup = 79. 5. Increase the fault current in 1% increments and apply until the relay does operate.305A (24. 6. Differential Element Unrestrained Pickup Range: 1–20 in per unit of tap Restrained Pickup Range: 0. Winding-1 Winding-2 Minimum Pickup = U87P × TAP1 × A Minimum Pickup = U87P × TAP2 × A Minimum Pickup =10 × 2.21A.8/1. 7. The 87U-Element pickup Winding-1 pickup setting is 24. Reduce the fault current 5% below the expected current and apply the fault current momentarily.02 A Unrestrained Element Pickup Time (Min/Typ/Max): 0. Looking at the “Differential” specification in Figure 110.837A ) for Winding-2. Set the fault current 5% higher than the pickup setting.305A ) for Winding-1 or 83.62/2.1% increments and apply until the relay operates.837A (79. The relay should not operate.05 =83.1 × 1.86 cycles Figure 110: SEL-387E Specifications 131 . This is the pickup result. we see that the acceptable metering error is ±5% ±0.05 = 25.1–1. 4. Repeat steps 2-6 for all phases.845A 3.72/2. Record on your test sheet. Expected Value × 100 = Percent Error Expected Value 24.1A Allowable Percent Error = 24.305 100 × Allowable Percent Error = 24.46% Error The Winding-1 test is within the manufacturer’s tolerance of 5.1 5. 9.0 -1.61 UNRESTRAINED PICKUP TESTS (Amps) IAW_ (A) IBW_ (A) ICW_ (A) EXPECTED (A) % ERROR W1 24.1) + 0.1A Allowable Percent Error = Setting 100 × (5% × 24.5 0.21A .130 24.0 132 .1A 0.2 0.1 1.41% Allowable Percent Error The measured percent error can be calculated using the percent error formula below: Actual Value .1 W2 79.850 78.2 0.41 TAP2 4. 100 × (5% × Setting) + 0.1A × 100 = Percent Error 24.210 24.24. Repeat the pickup test for all phase currents that are part of the differential scheme.150 24.The Relay Testing Handbook We can calculate the manufacturer’s allowable percent error for Winding-1. DIFFERENTIAL TEST RESULTS 87U PICK UP 10 TIME DELAY 0 TAP1 2.100 0.840 79.860 79.845 0.41% and passed the test. Determine how you will monitor pickup and set the relay accordingly.61 •• U87P = 10 (xTap) 1. output contact. front panel display. Adding a 1. 2. high-current applications. if required. etc…See previous packages of The Relay Testing Handbook for details. We can use the testing philosophy from the “1-Phase Restrained-Differential Slope Testing” section of the previous chapter to apply equal currents in two windings simultaneously while in-phase to minimize the required current to test the Unrestrained-Differential element. Chapter 3: Unrestrained Differential Testing 4. The following settings will be used for our SEL-387 example. •• W1CTC = 12 •• W1CTC = 1 •• Tap1 = 2.) 133 . Alternate Pickup Test Procedure Unrestrained-differential elements are. Connect the test-set as per Figures 108 and 109 and the descriptions in the “1-Phase Restrained-Differential Slope Testing” section of the previous chapter.41 •• Tap2 = 4. by definition.73 correction factor for single-phase testing can often put the expected pickup above the maximum output of your test-set. (Pickup indication by LED. 1 applied test-set current channels and ensure the relay operates. 12 1. 6.The Relay Testing Handbook 3.52 4. 9. 8. Test Amps Test Amps + Test@Tap = TAP1 × A1 TAP2 × A2  Test Amps Test Amps  TAP1 × A1 ×  + =  Test@Tap × TAP1 × A1  TAP1 × A1 TAP2 × A2   TAP1 × A1 × Test Amps   Test Amps + =  Test@Tap × TAP1 × A1  TAP2   TAP1 × Test Amps  TAP2 × A2 ×  Test Amps + =  Test@Tap × TAP1 × A1 × TAP2 × A2  TAP2  ( TAP2 × A2 × Test Amps ) + ( TAP1 × A1 × Test Amps = ) Test@Tap × TAP1 × A1 × TAP2 × A2 ( TAP2 × A2+TAP1 × A1) Test = Amps Test@Tap × TAP1 × A1 × TAP2 × A2 Test@Tap × TAP1 × A1 × TAP2 × A2 Test Amps = ( TAP1 × A1) + ( TAP2 × A2 )  Test@Tap × TAP1 × A1 × TAP2 × A2 Test Amps = ( TAP1 × A1) + ( TAP2 × A2 )  10 × 2.732A ) from all = 18.43 Test Amps = 10. 4. 5. Momentarily apply 10% more current ( Pickup × 1. 134 . 7. We have found through experimentation that the correction factor is.61 × 1.5 However. 10. Use the following formula to calculate equal currents for any multiple of Tap using the following table for correction factors. 11 √3 Even: 2.1 = 20.732 Test Amps = (2.41 × 1 × 4. WnCTC Setting A 0 1 Odd: 1.732 )  192. 1.41 × 1) + ( 4.61 × 1.39 Test Amps = 18. 3. in fact.52A × 1. the table does not seem to be correct when the WnCTC equals 12. 1% steps until the pickup operates.594A ) and momentarily apply.2 0. The pickup indication should remain OFF. Figure 111: SEL-387 Differential Element Specifications DIFFERENTIAL TEST RESULTS 87U PICK UP 10 TIME DELAY 0 TAP1 2.41 TAP2 4.540 18. 10.61 UNRESTRAINED PICKUP TESTS (Amps) 87U1 87U2 87U3 EXPECTED (A) % ERROR 18. This is the 87U pickup. Change all connections for C phase and repeat steps 4-7. Lower the currents to 95% of the pickup setting ( Pickup × 0.ICW2&IBW2 135 .52A × 0.95= 17.IAW2&ICW2 IBW1.IBW2&IAW2 ICW1. Evaluate the results using “Pickup Accuracy (A Secondary): 5 A Model” from the SEL-387 specification in Figure 111. 6. Change all connections for B phase and repeat steps 4-7. Raise the currents in 1% increments and momentarily apply until the pickup indication turns ON. Subtract 1% and increase in 0.530 18.1 0.550 18. 7.1 IAW1. Chapter 3: Unrestrained Differential Testing 5.95= 18.520 0. Use the following formula to determine the pickup in per-unit: Test Amps Test Amps Per Unit = Pickup + TAP1 × A1 TAP2 × A2 8. 9. Apply the connections from the Simple or Alternate Pickup Test Procedure from the previous sections in this chapter. 7. Pickup time: Less than 5 ms. Set the test current 110% of the expected current using the same test procedure in Step 2. Change the connections for the next phase and repeat. 4 A continuous at 85°C: 1 s Rating: 50 A: MOV protected: 270 Vac. The timer and current output should stop when the relay output operates. 8. Dropout time: Less than 5 ms typical. 3.67ms + 4. 40 J. Configure a test that applies the test current and starts the fault timer simultaneously. Apply the test and record the results. 5.20cycles 136 . 6. 360 Vdc. Figure 112: Preferred SEL-387 Output Contact Specifications Maximum Operating Time Unrestrained Element Pickup Time(Max) + Output Contact Pickup Time = Maximum Operating = Time 1. Carry: 6 A continuous carry at 70°C. The Relay Testing Handbook 5. Change the connections for the next phase and repeat the test. Evaluate the results using “Unrestrained Element Pickup Time” from the SEL-387 specification in Figure 111 and the Output Contact “Pickup Time” specifications in Figure 112.9 Maximum Operating = Time cycles + 4. Timing Test Procedure 1. 4.99ms 60 Maximum Operating = Time 31. Output Contacts: Standard: Make: 30 A.0367ms or 2. Connect the relay output contact to the test-set input.99ms Maximum Operating Time = 0.9cycles+ < 5ms 1. 2. 577 multiple. or 0. •• Are the 2 winding currents 0º apart? •• If the pickup tests are off by a 3 . •• Before you start. 1. the relay probably is applying correction factors and you should review the manufacturer’s literature. Tips and Tricks to Overcome Common Obstacles The following tips or tricks may help you overcome the most common obstacles.5. apply current at a lower value and review the relay’s measured values to make sure your test-set is actually producing an output and your connections are correct. Chapter 3: Unrestrained Differential Testing 6. •• Are you applying the currents into the same phase relationships? •• Did you calculate the correct Tap? 137 . The Relay Testing Handbook 138 . WA. www.mantatest. www. www. Second Edition New York. WA.com 139 . Kenneth.selinc. WA.com Schweitzer Engineering Laboratories (20010606) SEL-587-0. Elmore. Marcel Dekker.com Blackburn.selinc. www. Lewis. -1 Current Differential Relay Overcurrent Relay Instruction Manual Pullman. (September 9. Walter A. Third Edition GEC Alstom T&D Schweitzer Engineering Laboratories (20011003) SEL-300G Multifunction Generator Relay Overcurrent Relay Instruction Manual Pullman. Jeff (AG2000-01) Application Guide Volume IV Determining the Correct TRCON Setting in the SEL-587 Relay When Applied to Delta-Wye Power Transformers Pullman. Walter A. (October 17.selinc. WA. Fault Locator. www. Inc. WA. J. www. David and Gregory. Directional Overcurrent Relay.com Schweitzer Engineering Laboratories (20010808) SEL-351A Distribution Protection System.com Schweitzer Engineering Laboratories (20010625) SEL-311C Protection and Automation System Instruction Manual Pullman.com Tang. Elmore. 2003) Protective Relaying: Theory and Applications.. Schweitzer Engineering Laboratories.. Kenneth. 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