Merge Fluid Mechanics

FLUID MECHANICS Support Materials CENTRE FOR PROFESSIONAL DEVELOPMENT & LIFELONG LEARNING UNIVERSITY OF MAURITIUS CONTRIBUTORS FLUID MECHANICS (Support Materials) was prepared by Associate Professor M. Nowbuth, from the Faculty of Engineering, University of Mauritius. August 2008 All rights reserved. No part of the work may be reproduced in any form, without the written permission from the University of Mauritius, Réduit, Fluid Mechanics ii Aug 2008 TABLE OF CONTENTS ABOUT THE COURSE Unit 1 Unit 2 Unit 3 Unit 4 Unit 5 Unit 6 Unit 7 Unit 8 Unit 9 Unit 10 Unit 11 Properties of Fluids Fluid Pressure Measurement of Fluid Pressure Hydrostatic Forces on Plane Surfaces Hydrostatic Forces on Curved Surfaces Pressure Diagrams Buoyancy Hydrodynamics – Fluid Dynamics (in Motion) Principles Of Conservation Of Mass & Energy Flow Rate Measurements – Orifices & Weirs Flow Rate Measurements – Venturimeters Fluid Mechanics iii Aug 2008 ABOUT THE COURSE Welcome to FLUID MECHANICS 1 (CIVE1104) is the introductory module of a series of Fluid Mechanics modules which comes in levels 2, 3 and 4, namely, CIVE2211, CIVE3104, CIVE4007. The aims of this module are to help you: • • • Appreciate the fundamental principles governing fluid at rest and in motion. Learn about flow measuring devices. Learn about forces exerted by fluids within a system. This module introduces you to the fundamentals behind the approach of analysis fluids at rest or in motion. In general, you tend to find this introductory module of Fluid Mechanics confusing and vague, since you are most probably used to analysing the more concrete solid mechanics behaviour. This manual has thus been structured in such a way that you are gradually introduced to the various concepts, through a set of theoretical notes, diagrams and simple examples. The complexity of examples will grow as you proceed through the contents of the modules stepwise. Your attention is drawn to the fact that all the units forming this module are interlinked, so the units should be studied progressively for the first time. Once you become familiar with the various units, you can then consult each unit on its own. You are strongly advised to ensure that the contents of this module are well assimilated and that you have properly understood the fundamental concepts governing the analysing of fluids. These basic concepts will be used again and again in the more advanced level modules of Fluid Mechanics. A series of tutorials are included in the manual, and unless you actually attempt them on your own, you will fail to get the thorough understanding of the individual units. Fluid Mechanics iv Aug 2008 LEARNING OBJECTIVES FOR THE COURSE By the end of the course, you will be able to do the following: Unit 1 Identify the basic concepts, theories and equations related to the properties of fluids. Unit 2 Derive the equations governing variation of pressure with depth and variation of pressure along a horizontal plane Unit 3 Be conversant with the various apparatus used to measure the pressure exerted by a fluid. Unit 4 Learn how this fluid pressure is expressed in terms of forces, commonly termed hydrostatic forces when the fluid is at rest. Unit 5 Calculate the individual forces acting at different points along solid and curved surfaces. Unit 6 Use the different approaches of calculating the resultant hydrostatic force acting on vertical plane surface only. Unit 7 Learn about equilibrium conditions in liquid medium and governing factors influencing stability of structures in such medium. Unit 8 Discuss about the concepts and approaches used to analyse fluids in motion (hydrodynamics). Unit 9 Define and differentiate between two main principles used to analyse fluids in motion, the principles of Continuity and the principles of Conservation of Energy. Unit 10 Apply the two main principles: the principles of Continuity and the principles of Conservation of Energy. Unit 11 Measure with a venturimeter, the flow rate within a closed conduit. Fluid Mechanics v Aug 2008 HOW TO PROCEED COURSE MATERIALS The manual is self-contained. HOW DO I USE THE COURSE MANUAL? Take a few minutes now to glance through the entire manual to get an idea of its structure. Notice that the format of the different units is fairly consistent throughout the manual. For example, each unit begins with an OVERVIEW, and LEARNING OBJECTIVES sections. The OVERVIEW provides a brief introduction to the unit and provide perquisite skills and knowledge you will have to possess to proceed successfully with the unit. You should then read the LEARNING OBJECTIVES. These objectives identify the knowledge and skills you will have acquired once you have successfully completed the study of a particular unit. They also show the steps that will eventually lead to the successful completion of the course. The learning objectives also provide a useful guide for review. WHERE DO I BEGIN? You should begin by taking a look at the TABLE OF CONTENTS in the MANUAL. The table provides you with a framework for the entire course and outlines the organisation and structure of the material you will be covering. The Course Schedule indicates how you should allocate your workload and what you should be working on in each week to be ready for the respective class. You should stick to the Course Schedule to ensure that you are working at a steady space and that your workload does not pile up. Fluid Mechanics vi Aug 2008 Proposed * Course Schedule (CIVE 1104) Session 01 02 03 04 05 06 07 08 09 10 11 12 13 14 Student’s Workplan Read Unit 1 Read Unit 2 Read Unit 3. Read Unit 4 Read Unit 5. Read Unit 6 Read Unit 7 Read Unit 8. Read Unit 9 Revision CLASS TEST Read Unit 10 Read Unit 11 Revision 15 * Any change will be communicated by your respective tutor in class. Fluid Mechanics vii Aug 2008 NOTE (to b confirmed by lecturer): For this module you are required to submit/present three practical reports which will be assigned to you by your lecturer during the course of the module. ASSESSMENT → COURSE GRADING SCHEME: Continuous Assessment: Examinations: 30 marks 70 marks → FINAL EXAMINATIONS: s Scheduled and administered by the Registrar’s Office s A two-hour paper at the end of the Semester. STUDY TIPS Much of your time in the course will be spent reading. Your comprehension and assessment of what you read are likely to be best if you heed the following tips: 1. Organise your time. It is best to complete each assigned reading in one sitting. The logical progression of thought in a chapter/unit can be lost if it is interrupted. 2. Be an active reader. Use question marks to flag difficult or confusing passages. Put exclamation marks beside passages you find particularly important. Write short comments in the margins as you go. For example, if you disagree with an author’s argument or if you think of examples which counter the position presented, note your opinions in the margin. If you prefer to leave your book pages unmarked, you can make your notations on “postit-notes”. 3. Read critically. You must evaluate, as well as appreciate and understand, what you read. Ask questions. Is the author’s argument logical? Are there alternatives to the author’s Fluid Mechanics viii Aug 2008 explanations or to the conclusions drawn? experience? Does the information fit with your 4. Take notes. If you make notes on an article or chapter right after finishing it, you reap a number of benefits. First, note-taking allows you an immediate review of what you have just read. (You will find that this review helps you recall information). Second, it gives you an opportunity to reassess your flagged or margin comments. Finally, it gives you a second shot at deciphering any confusing passages. 5. Review your scribbling! Whether or not you make separate notes on your readings, review your flags, underlining and marginalia. considered significant or difficult. Study closely those passages you 6. Write down your ideas in a course journal. As you progress through the course, the new information you absorb will stimulate new thoughts, questions, ideas, and insights. These may not be directly related to the subject matter, but may be of great interest to you. Use these ideas to focus your personal involvement in this and other courses. 7. Your ability to explain the subject matter to others is a good test of your true comprehension of the material. Try explaining the material you are learning to others, classmates or friends, without resorting to jargon. Even if some of them are not directly involved with the techniques discussed in this course, many of the concepts may be of interest to them. 8. Activities found in units will not be marked. We strongly recommend that you do not skip any of them. They will help you prepare for the graded assignments. Now, it’s time to get to work. Good luck and enjoy the course! Fluid Mechanics ix Aug 2008 UNIT 1 PROPERTIES OF FLUIDS Unit Structure 1.0 1.1 1.2 1.3 Overview Learning Objectives Introduction States of Matter 1.3.1 1.3.2 Difference Between Solid and Fluid Technical Terms commonly Used to Describe the Properties of Fluids 1.3.2.1 Mass Density or Density, ρ 1.3.2.2 Specific weight, w 1.3.2.3 Relative Density ( R) or Specific gravity, s 1.3.2.4 Specific volume, v 1.3.2.5 Viscosity 1.3.2.6 Coefficient of Dynamic (µ) and Kinematic (γ) Viscosity γ 1.3.3 Real & Ideal fluids 1.3.3.1 Surface Tension 1.3.3.2 Vapour Pressure 1.4 1.5 1.6 1.7 1.8 1.9 Bulk Modulus Cavitation Activities Summary Worked Examples Tutorial 1.0 OVERVIEW This first unit emphasizes the need to be familiar with technical terms related to the description and analysis of fluids in motion and at rest, their exact definitions and their S.I. (Standard International) units. When analysing fluids at rest or in motion, students will need to know how the equations being used for the analysis were derived, what are the assumptions behind the derivation of the Fluid Mechanics 1 Aug 2008 equations and does the equation hold true for this particular situation. The results of an analysis are very much dependent on the validity of the equations used in the analysis, and students will have to be well conversant with the equations used. The implication is that there will be no more learning equations by heart and replacing numbers. Students will need to derive equations before applying them and from there, get a feel about whether the results they obtain have a practical and realistic meaning. So in this unit, you will gradually be introduced to technical terms commonly used in describing properties of fluids, the basic units of length, mass and time, the units of various fluid parameters and the need for assumptions in the analysis of fluid mechanics. This is a very important unit, it is simple, and it should be properly studied and appreciated, for it will come up throughout the entire course on Fluid Mechanics, until the final level of your course, in basically all the different units. 1.1 LEARNING OBJECTIVES At the end of this unit, students should be able to do the following: 1. State the main difference between a solid and a fluid. 2. Differentiate between real & ideal fluid. 3. Identify situations when a real fluid has to be treated as an ideal fluid 4. Elaborate on the different ways of expressing fluid properties such as density, viscosity and compressibility. 5. Define cavitation. 6. Explain how cavitation is taken care of in design of pipelines. 1.2 INTRODUCTION Unit 1 introduces students to the basic concepts, theories and equations related to the properties of fluids. The main properties differentiating behaviour of a solid and a fluid, basic properties used to describe fluids, importance of SI units of each term within an equation and the need to resort to assumptions during analysis of fluids behaviour will be introduced and discussed in the course of the unit. Fluid Mechanics 2 Aug 2008 1.3 STATES OF MATTER Matter can exist in 3 different states - Solid, Liquid and Gas. The state of a substance dictates to a large extent the behaviour of that substance under static or dynamic conditions. Both liquids and gases are considered to be under the category fluids. 1.3.1 Difference between a Solid & a Fluid A solid has a definite shape while a fluid takes the shape of the vessel containing it. In a more technical term, a solid offers resistance to a force while a fluid cannot resist an applied force. Hence, in order to better distinguish between a solid and a fluid medium, we shall examine the response of each substance to an applied shear force: Apply of a force: Solid: Offers resistance to the deforming force Liquid and Gas (Fluid): Deform continuously as long as the force is applied. Other properties of a fluid: It flows under its own weight. It takes the shape of any solid body with which it comes into contact. Definition of a fluid: A fluid (liquid or gas) is a substance which deforms continuously under the action of shearing forces. Fluid Mechanics 3 Aug 2008 1.3.2 Technical Terms Commonly Used to Describe the Properties of Fluids The properties of a fluid are characterised by its density, its viscosity and its degree of compressibility. The fluid property density can be expressed in several ways, mass density also commonly referred to as density, specific weight, specific gravity, relative density or specific volume. Similarly viscosity of a substance can either be expressed as the coefficient of dynamic viscosity or the coefficient of kinematic viscosity. The degree to which a fluid can be compressed is expressed by the term bulk modulus of compressibility. Different symbols are used to differentiate between different terms and each term is associated with a specific SI unit. (Students need to pay particular attention to the definition, symbol and unit for each of these terms. Students would need to learn and remember these basic technical terms, to be able to recognise precisely in which way the density of a substance has been stated in a given question.) 1.3.2.1 Mass Density or Density, ρ The density of a substance is defined as the mass per unit volume, ρ = mass / volume units: kg m-3 Example: Mass density of water at 4oC is 1000 kg m-3 Fluid Mechanics 4 Aug 2008 1.3.2.2.1.1 Specific weight, w Another way of expressing the density of a substance, is by the term specific weight, where specific weight is the weight per unit volume, w = weight / volume units: Mass x Acceleration due to gravity / volume kg m s-2 / m-3 or N / m3 Example: Specific Weight of water at 4oC is 9.81 kN/ m3 1.3.2.3 Relative Density (R) or Specific gravity, s The specific gravity of a substance, also known as the relative density of the substance, is the ratio of its density or specific weight to that of water under standard conditions of temperature and pressure. The standard pressure for water is usually taken as one atmosphere and temperature as 4oC, s = Density of substance / Density of water or Specific weight of substance / Specific weight of water Example: Relative Density or Specific gravity of mercury = 13.6 Fluid Mechanics 5 Aug 2008 1.3.2.4 Specific volume, v The specific volume v, is defined as the volume occupied by a unit mass of the substance, it is the reciprocal of the density, ρ v = 1/ρ Units: m3 per unit weight of substance Example: The specific volume of 1 kg of water = 10-3 m3 per unit weight of water kg 1.3.2.5 Viscosity When a shearing force is applied to a liquid which is initially as rest, the fluid cannot resist the shearing forces. The fluid will flow in such a way, that the fluid in contact with the boundary, will have the same velocity as the boundary, while successive layers of the fluid will move with increasing velocity, away from the boundary (Figure 1.1). Fluid Mechanics 6 Aug 2008 Open channel running completely full A A’ B B’ Applied force, F Stationary base of channel Velocity of fluid at surface, U Velocity of fluid in contact with the base of the channel is 0. D C Figure 1.1: Application of a shearing force to a fluid initially at rest Fluid Mechanics 7 Aug 2008 To explain the behaviour of a fluid when shearing force is exerted, consider an element of a fluid (ABCD), which is initially at rest between two solid surfaces separated by a small distance, h, along the y direction (Figure 1.1). Now, suppose that the upper solid surface is moved in the xdirection, by applying a force, F (Figure 1.1). The elemental fluid will undergo a change in shape as illustrated by element A’B’CD (Figure 1.1). The bottom layer of the fluid immediately in contact with the boundary will have the same velocity of the boundary, i.e, the velocity at y=0, will be equal to 0 (Figure 1.1). The layer of fluid in contact with the moving upper solid surface, will have the same velocity as the velocity of the moving surface, velocity U. Successive fluid layers will have velocity which increases from bottom to the top surface (Figure 1.2). APPLYING A SLIDING FORCE AT SURFACE OF LIQUID A δ A1 B B1 Applied force, F δ δ D C Stationary base of channel Successive fluid layers sliding above each other, since each has a different velocity, Top layer, velocity, V Bottom layer, velocity 0 Shearing stress, τ = F/shearing area = F/AB x thickness of elemental fluid Shear Strain = δθ = δx / δy Shear Stress α Rate of change of shear strain τ α δθ / δt τ α (δx/ δy) / δt = δu/ δy (where δx/ δt = δu) Figure 1.2: Consider a small elemental fluid under the application of a shearing force With reference to Figure 1.2; application of the force F, has caused a shearing action to be exerted within the fluid element, whereby Fluid Mechanics 8 Aug 2008 Shearing force per unit area, shear stress, τ = F / AB x s The consequent, shear strain acting on the fluid element, ϕ , can be represented by the term x/y. The shear strain will continue to increase with time and the fluid will flow. It is found experimentally that the rate of shear strain is directly proportional to the shear stress, for a fluid: Assuming that shear stress is proportional to rate of shear strain, τ τ α δϕ / time α δx / δy per unit time where distance x per unit time, can also be expressed as the velocity, u τ α δu/δy Removing the symbol of proportionality results in the following equation: τ = µ δu/δy ………………………equation A whereby the term µ is known as the coefficient of absolute or dynamic viscosity of the fluid, Units: kg m -1 s-1 The relationship defined by equation A is known as Newton’s Law of Viscosity. 1.3.2.6 Coefficient of Dynamic (µ) and Kinematic (γ) Viscosity γ Viscosity is the property of a fluid which offers resistance to fluid deformation by the application of a tangential, shearing force (Figure 1.2). For liquids, viscosity arises mainly from the cohesive force of molecules. The coefficient of dynamic or absolute viscosity, µ , is a function of temperature, for a specific fluid. When the temperature of a fluid increases, the cohesive forces within fluid body 9 Aug 2008 Fluid Mechanics decreases. This results in a decrease in the shear stress, hence, eventually, in a decrease in the coefficient of viscosity. Another way of defining coefficient of viscosity, is by using the coefficient of kinematic viscosity, which is the coefficient of dynamic viscosity divided by the density of the fluid: γ = µ/ρ where γ = the coefficient of kinematic viscosity, units : m2 s-1 1.3.3 Real and Ideal Fluids Fluids that obey Newton’s equation, (equation A), are classified under the group of Newtonian fluids, e.g air and water. Fluids such as tar and polymers do not obey equation A, the relation between shear stress and the rate of shear strain in these cases is non-linear (Figure 1.3). The concept of an ideal or perfect fluid is based on theoretical considerations because all real fluids exhibit viscous property. When µ = 0, then u/y = 0, hence shear stress, τ , vanishes. Therefore, a real fluid, with coefficient of viscosity very small, and velocity gradient very small can be considered as being frictionless, i.e, no shearing action takes place. An ideal fluid is one which has zero viscosity, and in many problems arising in fluid mechanics, we often have to resort to the assumption that the fluid we are dealing with is an ideal one, (however, only theoretical solutions are obtained in these cases). Fluid Mechanics 10 Aug 2008 Shear Stress, τ Newtonian fluid τ α δv/δy δ µ is constant – LINEAR RELATIONSHIP Non-newtonian fluid δ τ α δv/δy µ is not constant – NON LINEAR RELATIONSHIP Ideal fluid, τ = 0, µ is zero Velocity gradient, δv/δy δ Figure 1.3: Real & Ideal fluids 1.3.3.1 Surface Tension Consider a molecule p, of fluid well inside the body of the fluid (Figure 1.4). Molecule q is at the surface, and unequal forces are acting on B, a resultant downward attractive force is exerted on q Molecule p is at rest, hence equal & balancing forces are acting on p Figure 1.4: Liquid molecules and forces of attraction Fluid Mechanics 11 Aug 2008 Molecule p, is attracted equally in all directions by the surrounding molecules within a small sphere of radius, a. But molecule, q, which is on the surface of the liquid, will experience a resultant pull inward due to the unbalanced cohesive force of attraction. The surface molecules are being pulled inward towards the bulk of the liquid, this effect causes the liquid surface to behave as if it were an elastic membrane under tension. The surface tension, σ , is measured as the force acting across unit length of a line drawn in the surface, and surface tension in a liquid has a tendency to contract to a minimum surface area for a given volume. 1.3.3.2 Vapour Pressure Molecules of liquids that possess sufficient kinetic energy leave the liquid surface and become vapour. If the vapour is confined to a space, an equilibrium condition is obtained when the amount of vaporisation is equal to the amount of condensation. The vapour pressure at this condition is called saturation vapour pressure, which depends on the temperature and increases with its rise. The degree of molecular activity increases with increasing temperature, and therefore, the vapour pressure will also increase (Figure 1.5). Boiling will occur when the vapour pressure is equal to the pressure above the liquid. By reducing the pressure, boiling can be made to occur at temperatures well below the boiling point. Fluid Mechanics 12 Aug 2008 1 Few water molecules possess enough energy & go into vapour state 2 More water molecules possess enough energy to go into vapour state 3 Increasing heat, further increases the number of molecules in vapour state 4 Air space eventually becomes completely saturated & cannot take more vapour molecules & At this point an equilibrium is achieved,number of water Molecules going into vapour state is the same as the number of molecules going back into liquid state Figure 1.5: Saturated vapour conditions 1.4 BULK MODULUS – COMPRESSIBILITY PROPERTIES OF FLUIDS Compressibility is basically the change in volume of a substance when subjected to a compressive force. All matter whether solids, liquids or gases is to some extent compressible. While gases are highly compressible, solids and liquids offer much resistance to compressive forces. Compressive forces causing major changes in volume, also result in changing the density of a substance (density=mass/volume). The degree of compressibility of a substance is expressed by the term Bulk Modulus, K. Consider a substance of original volume V, subjected to a compressive force, and thereby undergoes a change in volume, expressed by term δv. If the force which is applied to compress the fluid is increased from P to P + δP, then the relationship between the change in pressure and the change of volume as defined by the property Bulk Modulus is given by: Fluid Mechanics 13 Aug 2008 Change in volume / Original Volume = Change in Pressure / Bulk Modulus -δV/V = - δP/K δ where K is the Bulk Modulus. K = -V δP/δV δ Bulk Modulus of a fluid can also be expressed in terms of density - Considering unit mass of a substance V=1/ρ; Then K = ρ dp/dρ ρ NOTE: When the change in volume is very small then a liquid can be assumed to be incompressible and this happens to be an assumption often made in fluid mechanics problems. As far as a gas is concerned, the compressibility of a gas is very large, and can rarely be ignored. If the pressure applied is very small, only in such a case, a gas can be assumed to be incompressible. Units: Typical values: same as pressure (N/m2) 2.05 x 109 N/m2 (water) 1.62 x 109 N/m2 (oil) 1.5 CAVITATION Consider a pipeline running full (Figure 1.6). Gases are known to be soluble in water under given temperatures and pressures. Under certain conditions along a pipeline, localised zones of low pressures (at A & B) can occur. If the pressure in such areas falls below the vapour pressure at which certain gases were initially soluble in water, at those points the gases are no longer soluble, and are thus released. Vapour bubbles are formed at A and B, but being in vapour state and thus light, they will move towards the highest points within the pipeline (C). Fluid Mechanics 14 Aug 2008 The highest points within the pipeline may be regions of high pressures, whereby the gas molecules are once again soluble in the water. At this point, (C), the gas molecules will collapse suddenly to go back to the liquid phase. To change from gaseous to liquid state, the gaseous molecules will have to lose excess energy. This is achieved by striking either against each other or by striking against the walls of a container. This phenomenon can cause serious problem, and is known as cavitation. As this process takes place over time, usually years in the case of pipes, at location C, the pipe gradually gets damage, until it breaks. To avoid such damage to pipelines, air valves are normally positioned at highest points along pipelines, thus allowing the gaseous molecules to escape. C Gaseous molecules going back to liquid state A Gas molecules released from water at zones of low pressures B Figure 1.6: Cavitation within a pipeline 1.6 (a) ACTIVITIES True/False Statements 1. Solids and fluids behave in the same way under the action of stress. 2. Liquids and gases behave in the same way under the action of all types of stresses. 3. Liquids are often assumed to be ideal since the fluid property viscosity is often an unknown parameter in analysis of fluids. Fluid Mechanics 15 Aug 2008 4. Density and specific weight are terms describing the same property of a fluid. 5. Temperature affects the property of liquids and gases in the same way. 6. Cavitation is the process of release of gases from a liquid. 7. Air valves are always needed in water pipeline owing to cavitation process. 8. Under all conditions, a liquid can be safely assumed to be incompressible. (b) Main Questions: 1. Derive with the help of sketches, the equation relating shear stress and velocity gradient, for a fluid in motion. 2. Show that the bulk modulus of compressibility can also be expressed in terms of the pressure and density parameters. 3. Describe the process of cavitation with the help of well labelled sketches. 4. Differentiate with the help of curves, Newtonian and non-Newtonian fluids. 1.7 SUMMARY This basic unit introduced the basic technical terms commonly used to describe the properties of fluids. The terms introduced will be used in the coming units. In this unit, we explained the: 1. Importance of getting the right meaning of various technical terms in fluid mechanics. 2. Importance of getting the right units for the various technical terms in fluid mechanics. 3. Meaning of the different ways of expressing the fluid properties, density and viscosity. 4. Derivation of shear stress and velocity gradient relationship. 5. Concept behind the process of cavitation. Fluid Mechanics 16 Aug 2008 6. Importance of the fluid property bulk modulus of compressibility for a liquid and for a gas. The next unit will be concerned with fluid pressure: what is fluid pressure, how it varies over space and how it is measured. Once again you are strongly advised to ensure that the various concepts illustrated in this unit are clear to you before you proceed to UNIT 2. 1.8 WORKED EXAMPLES Example 1 – Shearing force & Viscosity properties Example 1 – Shearing forces & Viscosity properties…1/2 • The space between two large flat and parallel walls 25 mm apart is filled with a liquid of absolute viscosity 0.7 N m-2 s. A thin flat plate 100mm x 100mm, is located mid way between the two large walls. If the thin flat plate is being towed at 75mm/s, determine the force exerted by the liquid on the thin flat plate. 25mm V=75mm/s Fluid Mechanics 17 Aug 2008 Example 1 – Shearing forces & Viscosity properties…2/2 25mm V=75mm/s • Shear stress = coefficient of dynamic viscosity x velocity gradient • τ = µ δV/δy = 0.7 x ((75 x 10-3 – 0) / 25/2 x 10-3} = 4.2 N/m2 • Shearing force acting on the thin plate – Force = Shearing stress x surface area of the plate – Force = 2 x 4.2 x (100 x10-3)2 = 0.084 N • NOTE: – Shearing forces are acting on the thin plate on the top surface as well as on the bottom surface. – The velocity distribution is assumed to be linear in this example, maximum at the centre and zero at the contact of the walls. 1.9 TUTORIAL Question 1 The space between two large flat and parallel walls 25 mm apart is filled with a liquid of absolute viscosity 0.7 N m-2 s. Within this space a thin flat plate, 250 by 250 mm is towed at a velocity of 150 mm s-1 at a distance of 6 mm from one wall, the plate and its movement being parallel to the walls, determine the force exerted by the liquid on the plate. Question 2 The velocity distribution for viscous flow between stationary plates is given as follows: v = dP/dx (By - y2) 2 µ If glycerine (µ = 0.62 N s m-2) is flowing and the pressure gradient dP/dx is 1.6 kN/m3, what is the velocity and shear stress at a distance of 12 mm from the wall if the spacing B is 5 cm? What are the shear stress and velocity at the wall? Fluid Mechanics 18 Aug 2008 Question 3 The velocity profile in laminar flow through a round pipe is expressed as U = 2V (1- r2/ro2) Where V is the average velocity, r is the radial distance from the centre line of the pipe, and ro is the pipe radius. Draw the dimensionless shear stress profile τ/τo against r/ro. What is the value of wall stress when fuel oil having absolute viscosity µ = 4 x 10-2 N s/m2 flows with an average velocity of 4m/s in a pipe of diameter 150mm? Question 4 A bush of 165mm length and 103mm internal diameter slides on a vertical column of 100mm diameter, the clearance space being filled with oil. If a 3.5kg bush mass slides with a velocity of 1m/s, determine the coefficient of dynamic viscosity of the oil. Fluid Mechanics 19 Aug 2008 UNIT 2 Unit Structure FLUID PRESSURE 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 Overview Learning Objectives Introduction Pascal’s Law for pressure at Point Variation of pressure vertically in a fluid under gravity Equality of pressure at the same level in a static fluid Basic general equation for fluid statics (at rest) Activities Summary Worked Examples Tutorial 2.0 OVERVIEW In this unit, student will be introduced to the concept of pressure exerted by a fluid. Student will need to understand and appreciate variation of pressure with depth and along a horizontal plane. These concepts will form the basis for the coming units, measurement of pressure and hydrostatics. Some of the concepts learnt in this unit will also be used in flow measurement in later units. Students are strongly advised to make sure that the concept illustrated in this unit is clear, for this unit forms an introductory unit to many others that will follow later on in this course. Fluid Mechanics 20 Aug 2008 2.1 LEARNING OBJECTIVES At the end of this Unit, students should be able to do the following: 7. Define and prove Pascal’s Law 8. Derive the equation governing variation of pressure with depth 9. Derive the equation governing variation of pressure along a horizontal plane 2.2 INTRODUCTION Fluids exert a force on the walls on the vessel containing it. The magnitude of this force which can also be expressed as pressure (force per unit area) varies with the depth of fluid from the top water surface. In this unit, you will learn how fluid exerts pressure at a point and how this pressure varies both with depth and along horizontal planes. 2.3 PASCAL’S LAW FOR PRESSURE AT A POINT When a fluid is at rest, there are no shearing forces acting on the fluid, and consequently, the fluid is in equilibrium. Under these conditions, the only force that such a fluid can sustain acts normally on a surface within the fluid. The normal force per unit area (acting inwards) is termed the fluid pressure, p. The famous seventeenth century mathematician, B. Pascal, first established that the pressure at any point within a stationary fluid is the same in all directions. Consider equilibrium of a small fluid element in the form of a triangular prism ABCDE surrounding a point in the fluid (Figure 2.1): Fluid Mechanics 21 Aug 2008 B ps A px δy y θ θ δs F C δx E θ py D Figure 2.1: Pressure at a Point If the fluid is at rest, the following pressure forces will be acting over the prism: the pressure acting at right angles to plane ABFE, px, the pressure acting at right angles to plane CDEF, py the pressure acting at right angles to plane ABCD, ps Since the fluid is at rest, the sum of all the forces in any direction must be equal to zero: Solving for resultant forces in the x direction: = px (area ABFE) - ps sin θ (Area ABCD) = px (δ zδy) - ps (δy/δs) δsδz = px δ zδy - ps δyδz………………..equation A For equilibrium conditions, equation A, is equal to 0 Therefore, px = ps Fluid Mechanics 22 Aug 2008 Similarly resolving for forces in the y direction py = ps Hence, pz = py = ps Given that ps, is the pressure on a plane at any angle θ , the x, y and z axes have not been chosen with any particular orientation, and the element is so small that it can be assumed to be a point, therefore indicating that the pressure at a point is the same in all directions. This is known as Pascal’s Law to a fluid at rest. 2.4 VARIATION OF PRESSURE VERTICALLY IN A FLUID UNDER GRAVITY Consider a cylindrical element of fluid of constant cross-sectional area A, totally surrounded with fluid of mass density, ρ: (Figure 2.2) Elevation = Z1 Cross sectional area, A Height = h Liquid density, ρ Elevation = Z2 Pressure = P1 Figure 2.2: Cylindrical elemental fluid The pressure acting on the bottom of the elemental fluid is p1, and on the top of the elemental fluid is p2 Fluid Mechanics 23 Aug 2008 Resolving for vertical forces acting on the elemental fluid: Force due to pressure p1 = p1 * A Force due to pressure p2 = p2 * A Force due to weight of the fluid = Volume x Density x gravity = A (z2 - z1) ρ g Since fluid is at rest, and consequently under equilibrium conditions, the resulting vertical force acting on the fluid is equal to 0: p1 A - p2 A + A (z2 - z1) ρ g = 0 p1 - p2 = - (z2 - z1) ρ g Thus in any fluid under gravitational attraction pressure decreases with increase of height h. 2.5 EQUALITY OF PRESSURE AT THE SAME LEVEL IN A STATIC FLUID Consider a cylindrical element (Figure 2.3): Liquid density, ρ h P Pressure = P1 Q 2 h Pressure = P2 Cross sectional area, A Figure 2.3: Cylindrical elemental fluid If P and Q are two points at the same level in a fluid at rest, a horizontal prism of fluid of constant cross-sectional are A will be in equilibrium: Forces acting on the fluid element in the horizontal direction is P1A and P2A, similar in magnitude but in opposite direction. Since the fluid is at rest, there is no horizontal resultant force acting on the elemental fluid. Fluid Mechanics 24 Aug 2008 For static equilibrium the sum of the horizontal forces must be equal to zero: P 1 A = P2 A Hence P1 = P2 Thus, the pressure at any two points at the same level in a body of fluid at rest will be the same, or fluid pressure is the same along a horizontal plane. 2.6 BASIC GENERAL EQUATION FOR FLUID STATICS (AT REST) Consider an elemental cylindrical fluid, with pressure p acting on one end and pressure p + δp acting on the other end of the fluid (Figure 2.4). The weight of the fluid also acts downwards. Cross sectional area, A δs Pressure = P + δp θ θ Liquid density, ρ Pressure = P Weight of elemental fluid mg Figure 2.4: Cylindrical inclined elemental fluid Since the fluid is at rest, the all forces must be in equilibrium. Fluid Mechanics 25 Aug 2008 Resolving horizontal forces acting on the elemental fluid: pA - (p + δp )A - mg cos θ = 0 pA - (p + δp )A - A δs ρ g cos θ = 0 δp =- δs ρ g cos θ In differential form : dp/ds = - ρ g cos θ In a horizontal plane (x, y direction) When θ =0 then cos θ =0, hence : dp/dx = 0, implying that pressure is constant everywhere in a horizontal plane. In a vertical plane ( z direction ) When θ =90 then cos θ =1, hence : dp/dz = ρ g, implying that in a vertical plane, pressure varies with height, whereby integrating with respect to z, gives, p = z ρ g. Replacing dscos θ by dz, then we end up with the general equation: dp/dz = - ρ g p = z ρ g ……………….equation 1 Hence NOTE: From equation 1, pressure can also be expressed as an equivalent head (z) of liquid with density ρ. Fluid Mechanics 26 Aug 2008 2.7 ACTIVITIES Main Questions: 5. 6. 7. 8. Define Pascal’s Law and illustrate it with the help of sketches Derive Pascal’s Law Show that the pressure exerted by a fluid is dependent of the depth of the liquid Show with the help of sketches that the pressure of a fluid is the same along a horizontal plane. 9. Explain why the pressure of a liquid can either be given in terms of N/m2 or in terms of a head of liquid. 2.8 SUMMARY Unit 2 introduced the student with the concept of fluid pressure, how it varies at a point, with depth and along a horizontal plane. The mathematical expression relating pressure to depth of liquid must be clear to the student before he moves on to the next unit. The next unit will be concerned with the measurement of the pressure exerted by a fluid, using the basic concepts illustrated in unit 2. Unless these concepts are clear, student will not be able to appreciate contents of Unit 3. Fluid Mechanics 27 Aug 2008 2.9 WORKED EXAMPLES Example 1 – Pressure & Force Example 1 – Pressure & Force • A mass of 50kg acts on a piston of area 100cm2. What is the intensity of pressure on the water in contact with the underside of the piston, if the piston is in equilibrium? • Intensity of pressure = Force / Area • P = (50 x 9.81) / (100 x 10-4) = 49.05 kN/m2. 2.10 TUTORIAL Question 1 A gas holder at sea level contains gas under a pressure head equal to 9cm of water. If the mass densities of air and gas are assumed to be constant and equal to 1.28kg/m3 and 0.72 kg/m3 respectively, calculate the pressure head in cm of water in a distribution main 260 m above sea level. Question 2 A pump delivers water against a head of 15 m of water. It also raises the water from a reservoir to the pump against a suction head equal to 250mm of mercury. Convert these heads into N/m2 and find the total head against which the pump works in N/m2 and in metres of water. Question 3 Fluid Mechanics 28 Aug 2008 In a hydraulic jack a force F is applied to the small piston to lift the load on the large piston. If the diameter of the small piston is 15mm and that of the large piston is 180 mm calculate the value of F required to lift 1000kg. Question 4 A mercury manometer is used to measure pressure drop between two points along a horizontal pipe through which water drops. If the manometer shows a reading of 0.8m, what is the corresponding pressure drop? Assume the relative density of mercury is 13.6. Question 5 Two pipes, A and B, are in the same elevation. Water is contained in A and rises to a level of 1.8m above it. Carbon tetrachloride, specific gravity 1.59 is contained in B. The inverted Utube is filled with compressed air at 300kN/m2 and barometer reads 760mm of mercury. Determine (a) the pressure difference in kN/m2 between A and B if the elevation of A is 0.45m, and (b) the absolute pressure in mm of mercury in B. Fluid Mechanics 29 Aug 2008 UNIT 3 MEASUREMENT OF FLUID PRESSURE Unit Structure 3.0 3.1 3.2 3.3 3.4 Overview Learning Objectives Introduction Gauge and Absolute Pressure Piezometer 3.4.1 Advantages and disadvantages of using Piezometer 3.5 U tube Manometers 3.5.1 Pressure difference measurement using U tube manometers 3.6 3.7 3.8 3.9 3.10 3.11 Enlarged ends U tube manometers Inverted U tube manometers Activities Summary Worked Examples Tutorial 3.0 OVERVIEW Unit 3 deals with the various apparatus which are used to measure the pressure exerted by a fluid. Students are introduced from the simplest pressure measuring device to the more complex ones. The advantages and disadvantages related to each of these apparatus are highlighted, so that the students are not only aware of the existence of the various apparatus, but also of the need for the use of a particular apparatus. Students should however note that the concepts already described in Unit 2 form the basis behind pressure measurement, thus, the concepts of Unit 2 will need to be well appreciated and understood before students set out to study Unit 3. Fluid Mechanics 30 Aug 2008 3.1 LEARNING OBJECTIVES At the end of this unit, students should be able to do the following: 1. Differentiate between Gauge Pressure and Absolute Pressure 2. Describe how a Barometer and a Piezometer works 3. State the advantage(s) and disadvantage(s) of using piezometers for pressure measurements 4. Describe how a vertically and inverted U tube manometer works 5. List the advantage(s) and disadvantage(s) of using U tube manometers for pressure measurements 6. Analyse how an enlarged and an inverted U tube manometer measures pressure difference 3.2 INTRODUCTION Fluids can exert either a positive or a negative pressure (suction pressure). There are several devices which are commonly used to measure pressure, but not all of them are able to measure both positive and negative pressures. While some devices can measure positive pressures, they are limited physically by the magnitude of the pressure to be measured, hence the need to appreciate their limitations. So, while various devices can be used to measure pressure, it is very important to appreciate and understand the advantages and disadvantages associated with the use of a particular apparatus. 3.3 GAUGE AND ABSOLUTE PRESSURE In practice, pressure is always measured by the determination of a pressure difference, i.e, relative to a particular datum. (A datum being a reference point or a reference line which forms the basis of a particular analysis, for example in the case of land surface elevation, the mean sea level is taken as datum). If the datum is total vacuum, then the difference between the pressure of the fluid in question and that of a vacuum is known as the ABSOLUTE PRESSURE of the fluid. More commonly, the pressure difference is determined between the pressure of the fluid concerned and the pressure of the surrounding atmosphere, i.e, the atmosphere is taken as the datum. So, when measuring the pressure of a fluid with respect to the atmosphere, this pressure Fluid Mechanics 31 Aug 2008 is referred to as GAUGE PRESSURE. Pressure gauges are mechanical devises which are used to measure the pressure of either liquids or fluids. They are calibrated on gauge pressure, i.e, taking atmosphere as datum. Hence, ABSOLUTE PRESSURE = ATMOSPHERIC PRESSURE + GAUGE PRESSURE Example - GAUGE AND ABSOLUTE PRESSURE Consider a vessel containing liquid, as shown in Figure 3.1, Point A is located at the top water surface, and point B is located at the bottom of the vessel. From what we have learnt in Unit 2, Pressure = z ρ g, where z is the vertical distance from the top water surface to the point under consideration, ρ is the density of the liquid and g, the acceleration due to gravity. IN TERMS OF ABSOLUTE PRESSURE Consider Figure 3.1: A Top water surface hd D hb C B hc Figure 3.1: Vessel containing liquid Figure 3.1 illustrate a vessel containing liquid to a maximum depth, hb. Points A, B, C and D have been positioned on the diagram with the objective of defining the relationship between gauge pressure and absolute pressure. Now, Point A is open to the atmosphere, so pressure at point A is simply Atmospheric pressure. Fluid Mechanics 32 Aug 2008 PA = Atmospheric Pressure Similarly, Point B is located at vertical distance hb from top water surface, so the pressure at point B is Pressure at top water surface + Pressure due to the column of liquid above point B, so PB = Atmospheric Pressure + hb ρ g Therefore, pressures at point C and D: PC = Atmospheric Pressure + hc ρ g PD = Atmospheric Pressure + hd ρ g NOW IN TERMS OF GAUGE PRESSURE As mentioned in earlier, gauge pressure is measured, taking the atmosphere as datum, so PA = 0 P B = hb ρ g P C = hc ρ g P D = hd ρ g NOTE: Gauge pressure refers only to the pressure recorded by the column of liquid above the point at which the pressure is being measured. Refer to relationship: ABSOLUTE PRESSURE = ATMOSPHERIC PRESSURE + GAUGE PRESSURE Fluid Mechanics 33 Aug 2008 3.4 PIEZOMETER A piezometer is simply a small diameter tube open at both ends. It is usually connected to pipelines for measuring the pressure of the fluid in the pipe at different positions (refer also to the Bernoulli’s experiment carried out during the Practical sessions of Fluid Mechanics). Consider a pipeline running full under pressure (Figure 3.2). Imagine now that a piezometer is connected to the top of the pipeline as shown in Figure 3.2. Liquid will rise in the piezometer tube to a height h. Now the relationship between the height of a column of liquid and the pressure at its base is given by the following equation: P=hρg Piezometer h H A Pipeline running full under pressure Figure 3.2: Pressure measurement by a Piezometer tube Hence Gauge Pressure at point A = H ρ g. Fluid Mechanics 34 Aug 2008 If the tube were closed at the top, the space above the liquid surface were a perfect vacuum, then the height of the column of liquid would then correspond to the absolute pressure of the liquid, and Absolute Pressure at A = P (gauge pressure) + Atmospheric pressure. This principle is used in the well-known mercury barometer. Advantages and disadvantages of using PIEZOMETER Piezometers are simple tube and relative inexpensive, so a cheap pressure measuring device. With reference to the equation (P= h ρ g), it can be noted that the higher the pressure the higher will be the value of h. When a piezometer is being used for pressure measurement, it is limited by the density of the fluid being measured, so density cannot be changed. Hence, if a piezometer were to be used to measure very high pressure, we would need a very long tube, and practically this is not possible. So piezometers can only be used to measure relatively small pressure, and besides piezometers cannot be used to measure negative pressure (pressure lower than atmospheric pressure), also commonly known as suction pressures. 3.5 U TUBE MANOMETERS Manometers, usually U shaped (Figure 3.3), are devices in which columns of a suitable liquid (usually mercury) are used to measure the difference in pressure between a certain point and the atmosphere, or between two points neither of which is necessarily at atmospheric pressure. For measuring small gauge pressures of liquids, simple piezometer tubes may be adequate, but for larger pressures some modifications are necessary. A common type of manometer is employing a transparent U - tube set in a vertical plane (Figure 3.3). Fluid Mechanics 35 Aug 2008 mercury Figure 3.3: U tube mercury manometer This mercury U tube manometer is connected to a pipe or other container containing liquid (A) under pressure. The lower part of the U - tube contains a liquid B (mercury in this case) immiscible with A and of greater density (Figure 3.4). h liquid A P1 liquid B mercury DATUM LINE y P Q Figure 3.4: U tube mercury manometer connected to pipe flowing full under pressure with liquid A After equilibrium is achieved, the pressure is the same at any two points in a horizontal plane when equilibrium is achieved. To solve for the value of P1, the first step is to identify a horizontal plane, commonly known as the DATUM LINE. For ease of analysis, this line is taken at the lowest level of mercury level in the U tube manometer. Having drawn the DATUM line, the next step is then to identify two reference points along this line and referring to Figure Fluid Mechanics 36 Aug 2008 3.4, these reference points are P and Q. Therefore P and Q are in the same horizontal plane and hence, the pressure at P and Q are equal and in equilibrium. Let the pressure in the pipe at is centre line be P1. Then provided the fluid is of constant density, the pressure at P is P1 + ρ A g y If the other side of the U tube is open to the atmosphere, the gauge pressure at Q is given by ρB g h From unit 2, we have learnt that pressure at two points lying on the same horizontal plane is similar, therefore, since P and Q are along the same horizontal line: P1 + ρ A g y = ρ B g h Hence, P1 = ρ B g h - ρ A g y Fluid Mechanics 37 Aug 2008 3.5.1 Pressure difference measurement -U tube MANOMETERS Consider the U tube manometer in Figure 3.5, connected in such a way so as to measure the pressure difference between points 1 and 2. P1 P2 x Liquid A Liquid B y P Q Mercury Figure 3.5: U tube manometer measuring pressure difference Pressure at P = P1 + ρA g (y+x) Pressure at Q = P2 + ρhg g y + ρB g x Therefore Thus: P1 + ρA g (y+x) = P2 + ρhg g y + ρB g x P1 - P2 = (ρB g x - ρA g x - ρA g y + ρhg g y) ρ Fluid Mechanics 38 Aug 2008 3.5.2 Advantages and Disadvantages of Using U Tube Manometers U tube manometers can be used to measure both positive and negative pressures and besides by using a manometer liquid of high density, U tube manometers can be used to measure high pressures. 3.6 ENLARGED ENDS U TUBE MANOMETERS Consider a U tube manometer with enlarged ends (Figure 3.6). The right hand side of the U tube manometer contains liquid A and the left hand side contains liquid B. Liquid B is lighter than liquid A, hence it floats on liquid A. Liquid B ha hb 1 2 Liquid A Figure 3.6: Enlarged ends U The first step in the analysis is to define the relationship between ha and hb when this tube is not connected to any system. Drawing a horizontal line at the intersection of the two liquids: P1 = P2 P1 = ha ρa g , while P2 = hb ρB g Fluid Mechanics 39 Aug 2008 Hence, ha ρA g = hb ρB g ha = hb (ρB / ρA)……………………………………..equation 1 ρ The second step is then to apply the pressure difference. Suppose pressure in the right hand side is higher than pressure in the left hand side. Then the pressure will cause the liquid in the right hand side to move downwards. This downward movement in the right limb will be accompanied by an upward movement small limb at the point of contact of the two liquids and another upward movement, this time in the left limb, as shown by the small arrows located next to the side of the tubes, Figure 3.7. The volume of the liquid moving is the same in all three cases. PA PB x x ha 3 new old datum hb 4 y Figure 3.7: Enlarged ends U tube manometer – measuring pressure difference Note: When both enlarged ends of the U tube manometer have similar diameter, then on both sides the liquid moves by the same height, X. The diameter of the tube in which the two liquids meet is smaller, hence the liquid moves by a height h. The third step in this analysis is now to draw a NEW DATUM, which is a line drawn at the new position of the interface of the liquid, as shown in Figure 3.6. Points 3 and 4 are positioned in the right and left limb, and since both lies on the same horizontal line, P3 = P4. Fluid Mechanics 40 Aug 2008 Pressure at Point 3 = PA + (ha – x ) ρA g…………………….equation 2 Pressure at Point A = PB + (hb – y + x) ρB g………………….equation 3 The final step would be to combine equations 1,2 and 3, and simply for the difference in pressure: PA - PB = x ρ A g - y ρ B g + x ρ B g NOTE: Students are advised to follow all the steps while solving similar problems. 3.7 INVERTED U TUBE MANOMETER U tube manometer can also be connected in an inverted position as shown in Figure 3.8. As described in the above sections, the first step is to draw the DATUM line, and in the case of an inverted U tube manometer, this datum is at the highest position of the mercury level. If we refer once again to the relationship between pressure and head of liquid, P = h ρ g, we will find that this relationship implies that as the head of liquid above the point under consideration increases, the pressure also increases. Thus starting from a particular point and moving upward, then will imply a decreasing pressure. Referring to Figure 3.8: Along the Datum Line, at points X and Y, we know that Px = Py since both points X and Y lie on the same horizontal plane. Px = P1- ha ρa g- H ρhg g Py = P 2 - h b ρ b g Hence, equating Px and Py, we have, P1- ha ρa g- H ρhg g = P2- hb ρb g P1 - P2 = ha ρa g + H ρhg g - hb ρb g Fluid Mechanics 41 Aug 2008 Mercury x H y DATUM LINE Liquid B hb ha Liquid A P2 P1 Figure 3.8: Inverted U tube 3.8 ACTIVITIES Main Questions: 10. 11. 12. 13. Differentiate between absolute and gauge pressures. Explain how a piezometer tube works. List the limitations of using piezometers for pressure measurements. Why is the manometer liquid usually denser than the liquid whose pressure is being measured? 14. Explain with the help of sketches how pressure difference is measured using enlarged ends U tube manometers. 15. Explain with the help of sketches how pressure difference is measured using inverted U tube manometers. Fluid Mechanics 42 Aug 2008 3.9 SUMMARY Unit 3 has introduced students to the various apparatus commonly used to measure pressure and pressure difference. Their advantages and limitations has been highlighted, together with detailed explanation of how the pressure is actually measured using these apparatus. The next unit, Unit 4, will now introduce the student to the approach adopted to calculating the force exerted by a fluid on the walls of its container, the hydrostatic force. In Units 2 and 3, you have learnt that pressure varies with depth, so bearing this in mind, the next unit will explain how hydrostatic forces are calculated. 3.10 WORKED EXAMPLES Example 1 – Simple application of pressure head equation, P = h ρ g Example 1 – Pressure = hρg ρ • Calculate the pressure in the ocean at a depth of 2000m assuming that salt water is (a) incompressible with a constant density of 1002kg/m3, (b) compressible with a bulk modulus of 2.05 GN/m2 and a density at the surface of 1002 kg/m3. (a) Assuming constant density Pressure = h ρ g = (2000) x 1002 x 9.81 = 19.66 MN/m2. (b) Considering the salt water as being compressible incomplete part (b) NOTE: Student should pay particular attention to the S. I. unit of each term in the equation of P=hρg, and at the end of the calculation always specify the S. I. unit associated with the answer. Fluid Mechanics 43 Aug 2008 Example 2 – Difference between Absolute and Gauge Pressure Example 2 – Gauge & Absolute Pressure • What will be (a) the gauge pressure and, (b) the absolute pressure of water at a depth of 12m below the free surface. Assume the density of water to be 1000kg/m3 and the atmospheric pressure 101 kN/m2. Absolute Pressure = Atmospheric Pressure + Gauge Pressure Gauge pressure = h ρ g = 12 X 1000 x 9.81 = 117.72 kN/m2 Absolute pressure = 101 + 117. 72 = 218.72 kN/m2 Example 3 – Pressure head as a function of different types of liquid (density dependency) Example 3 – Pressure & density dependency • Determine the pressure in N/m2 at (a) a depth of 6m below the free surface of a body of water and (b) at a depth of 9m below the free surface of a body of oil of specific gravity 0.75. Pressure P = h ρ g Specific gravity of a substance = Density of the substance/ density of water at standard temperature and pressure (1000 kg/m3) Pressure at point A = 6 x 1000 x 9.81 = 58.86 kN/m2 6m water A Pressure at point B = 9 x (1000 x 0.75) x 9.81 = 66.22 kN/m2 9m B Oil Fluid Mechanics 44 Aug 2008 Example 4 – Pressure expressed as a depth of a given liquid Example 4 – Expressing pressure as a depth of liquid • What depth of oil, specific gravity 0.8, will produce a pressure of 120 kN/m2. What would be the corresponding depth of water. Pressure = h ρ g Depth of oil corresponding to pressure of 120kN/m2: 120 x 1000 = hoil x (0.8 x 1000) x 9.81 hoil = (120 x 1000 ) / ( 0.8 x 1000 x9.81) = 15.29 m Corresponding depth of water, h water = (120 x 1000) / (1000 x 9.81) = 12.23m Example 5 – Negative gauge pressure Example 5 – Negative gauge pressure & absolute pressure • A U tube manometer is connected to a pipe in which a fluid is flowing under a negative gauge pressure of 50mm of mercury. What is the absolute pressure in the pipe in N/m2, if the atmospheric pressure is 1 bar? Gauge pressure is negative and its magnitude is 50mm mercury. Atmospheric pressure = 1 bar = 101 kN/m2 Absolute pressure = Atmospheric pressure + Gauge pressure = (101x 1000 ) - (50 x 0.001 x 1000 x 13.6 x 9.81) = 94.32 kN/m2 Fluid Mechanics 45 Aug 2008 Example 6 – Barometric pressure expressed as head of mercury Example 6 – Barometric pressure in terms of equivalent head of mercury…1/2 • What is the gauge pressure and absolute pressure of the air in the figure below, if the barometric pressure is 780mm of mercury and (a) the liquid is water of density 1000kg/m3 and (b) oil of specific weight 7.5 x102 N/m3. Air A 0.5m Pressure at A = Pressure at B (along same horizontal line) B Absolute Pressure at A = atmospheric pressure Absolute Pressure at B = Pair + (Pressure due to 0.5m of liquid) liquid Therefore: Absolute pressure of air = Atmospheric pressure - (Pressure due to 0.5m depth of liquid) Example 6 – Barometric pressure in terms of equivalent head of mercury…2/2 Liquid is water: Absolute Pressure of air = Atmospheric pressure – pressure due to 0.5m depth of water = (780 x 0.001 x 1000 x 13.6 x 9.81) – (0.5 x 1000 x9.81) = 99.16 kN/m2 Gauge pressure of air = Pressure due to 0.5m depth of water = - (0.5 x 1000 x 9.81) = -4.9 kN/m2 Liquid is oil: Absolute Pressure of air = Atmospheric pressure – pressure due to 0.5m depth of oil = (780 x 0.001 x 1000 x 13.6 x 9.81) – (0.5 x 7.5 x 102) = 103.7 kN/m2 Gauge pressure of air = Pressure due to 0.5m depth of oil= - (0.5 x 7.5 x 102) = - 3.75 kN/m2 Fluid Mechanics 46 Aug 2008 Example 7 – U tube manometer E x a m p le 7 – U tu b e m a n o m e te r… 1 /2 • A U tu b e m a n o m e te r a s illu s tra te d in th e fig u re b e lo w is u s e d to m e a s u re th e p re s s u re a b o v e a tm o s p h e ric o f w a te r in a p ip e . If th e m e rc u ry , liq u id Q , is 3 0 c m b e lo w A in th e le ft h a n d lim b a n d 2 0 c m a b o v e A in th e rig h t h a n d lim b , w h a t is th e g a u g e p re s s u r e a t A . S p e c ific g ra v ity o f m e rc u ry is 1 3 .6 . D 20cm A L iq u id Q (m e rc u r y ) 30cm w a te r B C Example 7 – U tube manometer…2/2 • Pressure at B = Pressure at C, since B and C are on the same horizontal line PB = PA + Pressure due to the height of water between A and B PC = Pressure due to the height of mercury between D and C PB = PC PA + (30 x 10-2 x 1000 x 9.81) = (30+20) x 10-2 x 13.6 x 1000 x 9.91 PA = 63.7kN/m2 D A B C Fluid Mechanics 47 Aug 2008 Example 8 – U tube manometer to measure negative pressure Example 8 – Negative gauge pressure (suction pressure) • In the figure below, fluid P is water and fluid Q is mercury. If the specific weight of mercury is 13.6 times that of water and the atmospheric pressure is 101.3kN/m2, what is the absolute pressure at A when h1 is 15cm and h2 is 30cm. NOTE: When the pressure in the pipe is lower than atmospheric pressure, the gauge pressure is negative and is known as a Suction Pressure. A h1 Fluid P B h2 C PB = PC PB = PA + h1ρPg+ h2ρQg PC=0 (gauge pressure) or atmospheric pressure PA + h1ρPg+ h2ρQg =101.3 x 103 PA = (101.3 x 103) – (15x10-2)x1000x9.81(30x10-2)x13.6 x1000x9.81 Absolute pressure, PA = 59.5 kN/m2 Fluid Q Fluid Mechanics 48 Aug 2008 Example 9 – U tube manometer with enlarged ends Example 9 – Enlarged U tube manometer…1/5 • An oil and water manometer consists of a U-tube 4mm diameter with both limbs vertical. The right-hand limb is enlarged at its upper end to 20mm diameter. The enlarged end contains oil with its free surface in the enlarged portion and the surface of separation between water and oil is below the enlarged end. The left-hand limb contains water only, its upper end being open to the atmosphere. When the right hand side is connected to a cylinder of gas the surface of separation is observed to fall by 25mm, but the surface of the oil remains in the enlarged end. Calculate the gauge pressure in the cylinder. Assume that the specific gravity of the water is 1.0 and that of the oil is 0.9. • oil Diameter = 20mm water Diameter = 4mm Example 9 – Enlarged U tube manometer…2/5 hoil 1 2 hwater • Step 1: Work out the relationship between hoil and hwater when there is no pressure difference between the two top ends of the U tube. • Working in terms of Gauge pressure: P1 = hoil ρoil g P2 = hwater ρwater g P1 = P2 (points 1 and 2 lie on the same horizontal line within the hoil ρoil g = hwater ρwater g Hence: hoil = hwater ρwater / ρoil = hwater/0.9….equation 1 system) Fluid Mechanics 49 Aug 2008 Example 9 – Enlarged U tube manometer…3/5 • • Step 2: Calculating x When the pressure is applied, the level of oil in the larger end goes down by a value of x, while the level of oil in the smaller diameter pipe, moves down by an amount y, since the change in volume in the two cases is the same. Volume change in smaller diameter pipe = (25x10-3) x Π/4 x (4 x 10-3)2 This volume equals (X) x Π/4 x (20 x 10-3)2 Hence X = 1mm x y hoil 1 y 3 2 4 hwater • • • Example 9 – Enlarged U tube manometer…4/5 Pressurised Gas cylinder Pg x y hoil 1 y 2 hwater 3 4 • Step 3: Working in terms of Gauge pressure: P3 = Pg + (y + hoil - x) ρoil g P4 = (2y + hwater ) ρwater g P3 = P4 Hence: Pg + (y + hoil - x) ρoil g = (2y + hwater ) ρwater g Pg = (2y + hwater ) ρwater g – (y + hoil - x) ρoil g …equation 2 Fluid Mechanics 50 Aug 2008 Example 9 – Enlarged U tube manometer…5/5 • Step 4: Replacing values of X, Y and relationship from equation 1 into equation 2: X=1mm, Y=25mm and hoil=hwater/0.9 Pg = (2y + hwater ) ρwater g – (y + hoil - x) ρoil g …equation 2 Pg = (2 x 0.025 + hwater) x 1000 x 9.81 – (0.025 + hwater/0.9) ) x 0.9 x1000 x 9.81 Pg = 278.6N/m2 x y hoil 1 y 3 2 4 hwater Fluid Mechanics 51 Aug 2008 3.11 TUTORIAL Question 1 (a) For a U tube manometer with enlarged ends containing two manometer liquids, derive the formula for the difference in pressures applied to the two enlarged ends. (b) An oil and water manometer consists of a U tube 7 mm diameter with both limbs vertical. The right-hand limb is enlarged at its upper end to 25 mm diameter. The enlarged end contain oil, of density 900 kg/m3, the free surface of the oil is in the enlarged portion of the limb and oil/water interface is in the smaller diameter tube. The left hand limb contains water only, and its upper end is opened to the atmosphere. A pressure of 350 N/m2 is applied to the right hand side of the manometer. If the surface of the oil remains in the enlarged end and the oil/water interface remains in the smaller diameter tube, calculate the depth by which the surface separation between the oil and the water will move. Question 2 A manometer consists of two tubes A and B open to the atmosphere, with vertical axes and uniform cross-sectional areas 500 mm2 and 800 mm2 respectively, connected to a U tube C of cross-sectional area 70 mm2 throughout. Tube A contains a liquid of relative density 0.8; tube B contains one of relative density 0.9. The surface of separation between the two liquids is in the vertical side of C connected to tube A. Tube B is now close and the space above the liquid is pressurised such that the surface of separation in tube C rises by 60 mm. Determine the pressure applied. Question 3 A manometer consists of a U tube of diameter d, the upper part of each limb being enlarged to diameter D, where the ration of D/d is equal to 5. The small tube contains water and on top the water surface there is a liquid of relative density 0.95 in both limbs. The free surfaces are in the enlarged parts of the U tube, while the surfaces of separation between the two liquids are in the small tube. Initially the surfaces are level. Fluid Mechanics 52 Aug 2008 When a pressure difference is applied to the top of the U tube in the enlarged ends, this causes the interface to move by a depth of 1 cm. Calculate the difference in pressure. Question 4 The sensitivity of a U-tube gauge is increased by enlarging the ends, and one side is filled with water and the other side with oil, (specific gravity, 0.95). If the area A of each enlarged end is 50 times the area of the tube, calculate the pressure difference corresponding to a movement of 25 mm of the surface separation between the oil and water. Question 5 The inclined U tube manometer as indicated in the diagram below, gives zero reading when A and B are set at the same pressure. The cross sectional are of the reservoir is 50 times that of the tube. For an angle of inclination θ of 30o and with a manometric fluid of specific gravity 0.8, find the difference in pressure between A and B in N/m2, when the gauge reads 110mm. B A L θ h Fluid Mechanics 53 Aug 2008 UNIT 4 HYDROSTATIC FORCES ON PLANE SURFACES Unit Structure 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 Overview Learning Objectives Introduction Magnitude of the Hydrostatic Force – Vertical plane Magnitude of the Hydrostatic Force – Inclined plane Centre of Pressure Activities Summary Additional Reading Materials Worked examples Tutorial 4.0 OVERVIEW In units 2 and 3, you have been introduced to the concepts of a fluid exerting pressure on the walls of the vessel in which it is contained. You are now familiar with the relationship between pressure (P) and an equivalent head of fluid (h), from the relationship (P = h ρ g). You have also learnt about the various devices that can be used to calculate pressure or pressure difference. In this unit, you will now learn how this fluid pressure is expressed in terms of forces, commonly termed hydrostatic forces when the fluid is at rest. Since pressure exerted by a fluid is depth dependent, this particular characteristic is taken into consideration when calculating hydrostatic forces. Fluid Mechanics 54 Aug 2008 4.1 LEARNING OBJECTIVES At the end of this unit, you should be able to do the following: 1. Derive the equation governing the resultant force exerted by a fluid on plane surfaces. 2. Clearly define the exact meaning of each term making up the equation for hydrostatic force on plane surface. 3. Analyse how the worked examples were solved. 4.2 INTRODUCTION Fluid exerts pressure on the walls of the vessel in which it is contained, and this pressure can also be expressed in terms of a force, given that Force = Pressure x Cross Sectional Area. Now we also know that the pressure exerted by a fluid varies with depth, increasing as the depth of liquid above the point under consideration, increases. Thus, the hydrostatic force acting on the plane surface also varies from point to point, being higher in magnitude, at the bottom of the vessel, ending up with a series of individual forces. These individual forces can then be combined to get the resultant force acting. Similarly, the point of action of the resultant force can be obtained by taken moments of each individual force, as being described in the next section. 4.3 MAGNITUDE OF THE RESULTANT HYDROSTATIC FORCE –VERTICAL PLANE Consider a vessel containing liquid, of density ρ , to a depth H, Figure 4.1, and consider the surface ABCD. As we move down from AB to CD, we know that the pressure will be increasing from a value of 0 to a value of Hρg. Now imagine that surface ABCD is now divided into very small layers of area δA. Consider now point 1, which is located at depth h1 from top water level. The pressure acting at point 1 is given by h1ρg, and from there, the hydrostatic force acting at point 1, will be given by the Fluid Mechanics 55 Aug 2008 relationship, force= pressure x area. Thus force acting at point 1, will be F1, where F1= h1ρg δA. Similarly, the forces acting at point 2, 3 and 4, can be expressed by the following equations: h2ρg δA, h3ρg δA and h4ρg δA A h1 top water surface H h4 h2 h3 Cross sectional area of each layer, δA B D C Figure 4.1: Magnitude of Hydrostatic Force Fluid Mechanics 56 Aug 2008 These individual forces can be represented as shown in Figure 4.2, as a series of parallel forces acting perpendicularly to the surface ABCD. h1 F1=h1 ρ g δA F2=h2 ρ g δA F3=h3 ρ g δA F4=h4 ρ g δA B h h4 h3 h2 Position of centroid of ABCD C Figure 4.2: Individual hydrostatic forces Similarly, going down to the bottom of the side ABCD, there would be a series of individual forces. When vertical surface of a plane is being analysed, these individual forces are parallel and thus can be linearly combined (refer to section 4.8), to get the final resultant force, which will be: Resultant force acting on ABCD = ∑ hi ρ g δA where i will vary from 1 to the total number of layers into which the solid surface has been divided for analysis purposes. If the area is made extremely small, then the above equation can also be written as ∫ ρg h δA, where ∫ h δA represents the first moment of area about the horizontal surface of the liquid and this can also be written as A h, where A is the total surface area of the body and h, is the distance from the centroid of the body to the horizontal surface of the liquid. This results in equation 1 (I think it is equation 1instead of 2) being written as follows: Fluid Mechanics 57 Aug 2008 FR= ρg A h ……………………..equation 1 ρ = density of the liquid g = acceleration due to gravity A = cross sectional area of the surface in contact with the liquid h = vertical depth from top liquid level to the centroid of the solid surface Note: 1. The centroid of a rectangular body lies at the centre of that body. 2. You need to know how to derive this equation, and also to be clear about the exact meaning of each of the term in the equation. Where 4.4 MAGNITUDE OF THE RESULTANT HYDROSTATIC FORCE – INCLINED PLANE Consider surface XY, which is submerged in the liquid. We will consider here, the hydrostatic force acting on the left hand side of surface XY. Four points 1, 2, 3 and 4 have been identified on that side, and these points are located at depth h1, h2, h3 and h4, respectively from the top water level. Refer to the relationship between depth of liquid and pressure, P = h ρ g, where h is the vertical depth of liquid above point under consideration. Thus applying this relationship, we have pressure acting at point 1, being P1 = h1 ρ g. Similarly at points 2, 3 and 4, the pressures will be h2 ρ g, h3 ρ g and h4 ρ g. Once again imagine that surface XY has been divided into small layers, δA. Hence the corresponding forces acting at point 1, 2, 3 and 4, will be h1 ρ g δA, h2 ρ g δA, h3 ρ g δA and h4 ρ g δA. Note here, that these individual forces are also parallel to each other, and they can thus be combined linearly as was the case for the vertical plane surface. Fluid Mechanics 58 Aug 2008 h h4 h3 h2 h1 1 2 4 3 X Position of centroid of XY Y Figure 4.3: Inclined plane surface So referring to equation 1, the resultant hydrostatic force acting on an inclined plane surface will be given by FR = ρg A h …………………..equation 2 Fluid Mechanics 59 Aug 2008 Where ρ = density of the liquid g = acceleration due to gravity A = cross sectional area of the surface in contact with the liquid h = vertical depth from top liquid level to the centroid of the solid surface 4.5 CENTRE OF PRESSURE The hydrostatic force, as illustrated in previous sections, is given by equation 1 or 2. This force is, as indicated before, the summation of individual forces acting on the surface of an inclined body. Now each individual force acts at a particular point. The next step will be then to find out the location of the resultant hydrostatic force. This location is also known as the point of action of the hydrostatic force, or the Centre of Pressure. Consider an elemental force acting on the surface, whereby δF = ρg δA h (Figure 4.4). Take moments about point O on the surface of the liquid then δM = δF x s. θ h D FR = h ρ g A δF = h ρ g h O X s Position of centroid of XY Y Position of centre of pressure Figure 4.4: Centre of Pressure Fluid Mechanics 60 Aug 2008 Total moments of individual elemental forces acting on the surface of the inclined body: ∑ δF x s = ∑ ρg δA h x s M = = ∑ ρg δA (s sin θ) x s ∫ ρg δA s2 sin θ …… ……………………..equation 3 = The total moments can also be given in terms of the resultant hydrostatic force, at position of centre of pressure, D from point O, given by: M = FR x D sin θ M = ρg A h D sin θ ……………………………….equation 4 Hence, equating 3 and 4 gives: ρg A h D sin θ= ∫ ρg δA s2 sin θ ∫ s2 δA is the second moment of area about the surface of the liquid and about the axis passing through the point O. This can be written (Theorem of Parallel Axes ) as the second moment of area about the centroid of a body + A y2, where y is the distance between the axis passing through the centroid of the body and the axis passing through point O (refer to section 4.8). Therefore, ρg A h D / sin θ= ρg sin θ ( Icg + A (h / sin θ) 2) D = Icg sin2 θ / A h + h Where D Icg = vertical distance from position of centre of pressure to the top water level = second moment of area of solid surface about its centroid 61 Aug 2008 Fluid Mechanics A h θ = cross sectional area of solid surface in contact with the liquid = vertical distance from centroid of body to the top water level = angle of inclination of body with the horizontal plane 4.6 ACTIVITIES 1. Illustrate with the help of sketches the centroid of a rectangle and a triangle. 2. A rectangular body, of width 1m and depth 2m, is inserted in a vessel containing water. If the rectangular body is just submerged in the vessel, what is the value of h in the equation of resultant hydrostatic force? 3. Refer to the case in part 2. If the top of the rectangular body is now 1m below the top water level, what is the value of h in the equation of the resultant hydrostatic force? 4. Refer to the case in part 2. Now the top of the rectangular body is 0.5m above the top water level in the vessel. hydrostatic force? What is the value of h in the equation of the resultant 5. Derive the equation for the hydrostatic force acting on the inclined plane rectangular surface of a dam. 6. Derive the centre of pressure for the hydrostatic force acting on a plane vertical rectangular surface. 7. The resultant hydrostatic force acts as the centroid of the solid body on which the force is being exerted Discuss. 8. How does your equation derived in part 5 change, when the body is now circular in shape, such as a valve in the side of a reservoir? Fluid Mechanics 62 Aug 2008 4.7 SUMMARY Unit 4 has illustrated how to calculate the hydrostatic force exerted by fluids on the surface of solids bodies, and also the position of location of that force. You are advised to go over the derivations of the equations and make sure that you have understood them and can easily reproduce them. Your attention is also drawn to the exact meaning of each of the terms in the equations derived. You will be required to derive the equations governing the resultant force and centre of pressure, but you need not do so when solving problems all the time. You should derive the equations when asked to work from first principles or when directly asked to do so. Unit 5 is a continuation of the concepts described in Unit 4. Unit 5 will illustrate the governing principles for deriving hydrostatic forces and centre of pressure when the force is being exerted on a curved surface, as compared to a plane surface as was illustrated in Unit 4. There are some important differences between the equations derived in Unit 4 and those to be derived in Unit 5. So, students are strongly advised to get the concepts discussed here clear before moving on to the next unit. NOTE: Additional reading materials have been included in section 4.8, to help you better appreciate the following: 1. the meaning of resultant force, 2. the position of action of a resultant force, 3. the meaning of the Theorem of Parallel Axes, 4. the application of the Theorem of Parallel Axes. While section 4.8, is not an integral part of Unit 4, it does help you to get a better understanding of the contents of the unit, and you are advised to go through section 4.8 at least once. Fluid Mechanics 63 Aug 2008 4.8 ADDITIONAL READING MATERIALS 1. The concept of resultant forces: RESULTANT FORCES F1=20kN F2=50kN Solid body, A F3=30kN Consider 3 forces acting on solid body A: In terms of magnitude only, the resultant force acting on A is simply the summation of the forces, taking into consideration that not all the three forces are acting in the same direction: Hence FR = (20 + 50) – 30 = 40 kN 2. The concept of the point of action of Resultant Forces: POINT OF ACTION OF RESULTANT FORCE F1=20kN h1 Solid body, A F3=30kN h3 F2=50kN h2 y Z Consider point Z: Each of the individual forces exert a moment about point Z, the magnitude of the resultant moment of the forces about Z, will either cause the body to overturn or to stay unmoved. Hence, the resultant force, FR, should be located such that it produces the same resultant moment about Z. So taking moments about Z: F1 x h1 + F2 x h2 – F3 x h3 = FR x y, where y is the perpendicular distance form the point of action of FR and point Z. Solve for position of Z. Fluid Mechanics 64 Aug 2008 3. Proof of Theorem of Parallel Axes: Consider a rectangle of width b and depth d. In this exercise, the second moment of area of the rectangle about two axes will be derived. The first axis is indicated by line aa, which is a line located at the bottom of the rectangle. The second axis passes through the centroid of the rectangle, line gg. Proof – Theorem of Parallel Axes…. (part 1) b b Small element, thickness δh d h g d g h a a Second moment of area of small element about line aa: = Area x (distance)2 = b δh x h2 Hence, total second moment of area of rectangle about line aa: = ∫ b δh x h2 with limits from h= 0 to d = bd /3 3 Second moment of area of small element about line gg, line gg passing through the centroid of rectangle = Area x (distance)2 = b δh x h2 Hence, total second moment of area of rectangle about line gg: = ∫ b δh x h2 with limits from h= +d/2 to -d/2 = bd /12 = Icg 3 Fluid Mechanics 65 Aug 2008 Having derived the second moment of area of the rectangle about axes passing through aa and gg, the next step will be to make use of the theory in the Theorem of Parallel Axes, to show that the second moment of area passing through line aa can be obtained directly, if the second moment of area about the centroidal axis is known. Hence this second part sets out to prove the Theorem of Parallel Axes. Proof – Theorem of Parallel Axes…. (part 2) b d g g The Theorem of Parallel Axes states that: The second moment of area of a body about any axis, is the sum of the second moment of area of the body about its centroidal axis Icg and the area of the body x (perpendicular distance from the given axis to the centroidal axis)2 Ixx = Icg + (AY2) where Y is the perpendicular a a Second moment of area of rectangle about line aa = bd3/3 Second moment of area of rectangle about line gg, Icg = bd3/12 Distance from the centroid to the axis xx. Hence, applying the above theorem, to find the second moment of area of rectangle about the axis aa: Iaa = Icg + AY2 Iaa = bd3/12 + (bd) (d/2)2 Iaa = bd3/3 Fluid Mechanics 66 Aug 2008 4.9 WORKED EXAMPLES Example 1 – Hydrostatic force exerted on a plane surface Example 1 – Rectangular plane surface…1/4 • A rectangular plane area, immersed in water, is 1.5m by 1.8m with the 1.5m side horizontal and the 1.8m side vertical. Determine the magnitude of the force on one side and the depth of its centre of pressure if the top edge is: a. in the water surface b. 0.3m below the water surface c. 30m below the water surface (a) (b) (c) 30m 0.3m 1.8m 1.8m 1.8m Width = 1.5m Example 1 – Rectangular plane surface…2/4 (a) Width = 1.5m Where is the density3of the liquid in contact with the plane surface (kg/m ) 2 g is the acceleration due to gravity (m/s ) A is the wetted area of the rectangular surface in contact with 2 the liquid (m ) y is the vertical distance from the top liquid surface to the centre of gravity of the wetted surface (m). Hydrostatic force acting on one side of rectangular surface = F = ρ g A y b=1.8m Icg of a rectangular body: Icg =bd3/12 Icg = 1.5 x (1.8)3/12 = 0.729 m4 Hydrostatic force acting on one side of rectangular body = 1000 x 9.81 x (1.8 x 1.5 ) x 1.8/2 = 23.8 kN Position of centre of pressure D = Icg/Ay + y = (0.729/(1.5x1.8X0.9) + 0.9 = 1.2 m Fluid Mechanics 67 Aug 2008 Example 1 – Rectangular plane surface…3/4 Since the top edge of the rectangular body is below the water surface, there is a need to calculate the new value of y, i.e the vertical distance from the top water surface to the centre of gravity of the rectangular body: y= 0.3 + (1.8/2) = 1.2 m 0.3m 1.8m (b) Hydrostatic force F = ρ g A y Hydrostatic force = 1000 x 9.81 x (1.8 x 1.5 ) x 1.2 = 31.8 kN Icg of a rectangular body: Icg =bd3/12 Icg = 1.5 x (1.8)3/12 = 0.729 m4 Position of centre of pressure D = Icg/Ay + y = (0.729/(1.5x1.8X1.2) + 1.2 = 1.42 m Example 1 – Rectangular plane surface…4/4 (c) 30m New value of y in this case: y = 30 + (1.8/2) = 30.9 m 1.8m Hydrostatic force F = ρ g A y Hydrostatic force = 1000 x 9.81 x (1.8 x 1.5 ) x 30.9 = 818 kN Icg of a rectangular body: Icg =bd3/12 Icg = 1.5 x (1.8)3/12 = 0.729 m4 Position of centre of pressure D = Icg/Ay + y = (0.729/(1.5x1.8X30.9) + 30.9 = 30.91 m Example 2 – Hydrostatic force and its dependency on density of the liquid exerting it Fluid Mechanics 68 Aug 2008 Example 2 – Hydrostatic force & density of liquid • One end of a rectangular tank is 1.5m wide by 2m deep. The tank is completely filled with oil of specific weight 9kN/m2. Find the resultant pressure on this vertical end and the depth of the centre of pressure from the top. Vertical depth from top liquid surface to centre of gravity of plane vertical surface y = 2/2 = 1m A 2m B width = 1.5m Icg = = 1.5 x 23 = 1 m4 bd3/12 Hydrostatic force F = ρ g A y = γ A y F = 9 x 1000 x (2 x 1.5) x 2/2 = 27 kN Centre of Pressure D = Icg/Ay + y = 1/ (2 x 1.5)x1 + 1 = 1.33 m Side under study NOTE: Specific weight = Density x acceleration due to gravity γ = ρg Example 3 – Hydrostatic force exerted on a circular opening Example 3 – Hydrostatic force on a circular opening..1/2 • A culvert draws off water from the base of a reservoir. The entrance to the culvert is closed by a circular gate 1.25m in diameter, which an be rotated about its horizontal diameter. Show that the turning moment on the gate is independent of the depth of water if the gate is completely immersed and find the value of this moment. Circular opening Diameter of circular gate = 1.25m Hinged about the horizontal diameter Fluid Mechanics 69 Aug 2008 Example 3 – Hydrostatic force on a circular opening..2/2 Icg for circular body = Π r4/4 = 22/7 x (1.25/2)4 / 4 = 0.12m4 Hydrostatic force F = ρ g A y = 1000 x 9.81 x (Π x 1.252/4) x y = 12043.5 y kN Centre of pressure D = Icg/Ay + y = 0.12/(Πx1.252/4)x y + y = (0.1/y)+y The hydrostatic force acts at D and since the opening is hinged along its horizontal diameter, the moment of the hydrostatic force is given by the hydrostatic force multiply by the vertical distance from D to the hinge. Taking moment about the horizontal diameter of the circular opening, M = F (D-y) M = 12043.5 y x { (0.1/y)+y – y } = 1177.2 y x 0.1/y Note: y is eliminated in the equation for M M = (12043.5 x 0.1) = 1204 Nm, thus independent of y. D y D-y Example 4 – Turning moment exerted by hydrostatic force about a hinge Example 4 – Hydrostatic force and turning moments..1/4 • A rectangular sluice door is hinged at the top at A and kept closed by a weight fixed to the door. The door is 120cm wide and 90cm long and the centre of gravity of the complete door and weight is at G, the combined weight being 9810N. Find the height of the water h on the inside of the door which will just cause the door to open. 30cm A W G B 600 h Fluid Mechanics 70 Aug 2008 Example 4 – Hydrostatic force and turning moments..2/4 30cm A 0.9sin60 = 0.78m W G 600 h B • Hydrostatic force acting on door AB = (1000 x 9.81 x 1.2 x 0.9) x (h-0.78/2) = 10595(h-0.39) N Centre of pressure D = (1.2 x 0.93/12)/(1.2x0.9)(h-0.39) + (h-0.39) m =0 .07/(h-0.39) + (h-0.39) m Under equilibrium conditions: Clockwise moment about hinge A = Anticlockwise moments about hinge A • • Example 4 – Hydrostatic force and turning moments..3/4 • Under equilibrium conditions: Clockwise moment about hinge A = Anticlockwise moments about hinge A Hydrostatic force will create a clockwise moment trying to open the gate, while the weight of the door and the gate will induce an anticlockwise moment trying to close the gate: W 30cm h-0.78 h • D F Clockwise moment about A = {10595(h-0.39) } x (D-h+0.78)/sin60o = {10595(h-0.39) } x { [0 .07/(h-0.39) + (h-0.39) ] – h + 0.78} /sin60o = 10595(h-0.39) x { 0.07(h-0.39) + 0.39 }/sin60o Anticlockwise moment about A = (W x 0.3) = 9810 x 0.3 = 2943 Nm Equating these two moments and solving for h, yields h=0.88m Fluid Mechanics 71 Aug 2008 Example 4 – Hydrostatic force and turning moments..4/4 Mathematical solution: Replacing (h-0.39) by X Clockwise moments: 10595(h-0.39) x { 0.07(h-0.39) + 0.39 }/sin60o = (10595X) x (0.07X+0.39)/sin60o = (7416.5+10595X2) /sin60o Nm = 8564.6X2 + 4771.3X Anticlockwise moments: 2943 Nm Equating moments: 8564.6X2 + 4771.3X- 2943 =0 X = -4771.3 E e ( 4771/32-4x8564.5x-2943) = h-0.39 = , therefore h = m Example 5 – Circular opening and moment exerted by hydrostatic force Example 5 – Circular openings & moment of hydrostatic force…1/2 • A sluice gate closes a circular opening 0.30m diameter and is hinged 1m below the surface of the water which acts on its face. If the centre of the opening lies at a depth of 1.25m find the force on the gate due to the fluid pressure. Find also the minimum force that must be applied by a clamp which lies 0.5m below the hinge, in order to keep the gate closed. Sluice gate 1m 1.25m 0.3m 0.5m clamp Circular opening, 03m diameter Fluid Mechanics 72 Aug 2008 Example 5 – Circular openings & moment of hydrostatic force…2/2 1m 1.25m 0.3m 1m 0.5m clamp Hydrostatic D-1 force, F Fclamp • • • Hydrostatic force, F = ρ g A y = 1000 X 9.81 x (π x 0.32/4) x 1.25 = 867N Centre of pressure, D = Icg/Ay + y = (π x 0.154/4)/[(π x 0.32/4)x1.25] + 1.25 = 1.255m Taking moments about the hinge: 867 x (1.255-1) = Fclamp x 0.5 Therefore Fclamp = 442 N 4.10 TUTORIAL Question 1 A square aperture in a vertical side of a tank has one diagonal vertical and is completely covered by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the aperture are 2m long and the tank contains water to a height of 1.5 m above the centre of the aperture. Calculate the force exerted on the plate by the water, the moment of this force about the hinge, and the position of the centre of pressure. How will the above values change if, instead of water, the tank contained a liquid of relative density 1.25? Question 2 The figure below shows a rectangular gate AB hinged at the top A and kept closed by a weight fixed to the door. The door is 120 cm wide and 90 cm long and the combined centre of gravity of the complete door and the weight is at G, the combined weight being 1000N. Find the height of water h on the inside of the door that will just cause the door to open. Fluid Mechanics 73 Aug 2008 Question 3 (a) A closed channel full of water has a cross-section in the form of an equilateral triangle of sides 2.5 m and lies horizontally on one of its sides. Its end is closed by a triangular vertical gate having similar side lengths, i.e, 2.5m. The gate is supported by a bolt at each corner. Calculate the magnitude and position of the force acting on the gate. Find also the force acting on each bolt. Icg = bh3/36 Question 4 (a) A vertical dock gate is 5.5 m wide and has water to a depth of 7.3 m on one side and to a depth of 3 m on the other side. Find the resultant horizontal force acting on the dock gate and the position of its line of action. (b) To what position does this line tend as the depth of water on the shallow side rises to 7.3 m ? Fluid Mechanics 74 Aug 2008 Question 5 (a) Derive from first principles, an expression for the force acting on one side of a plane surface, cross sectional area A submerged in a liquid of density ρ. (b) A square shaped gate, of side length 0.3 m, closes an opening. The gate is hinged at the top, (Figure 1), below the surface of the water which acts on its face. If the centre of the opening lies at a depth of 1.25 m below top water level, calculate the magnitude and position of the force acting on the gate. Find also the force that must be applied by a clamp which lies 0.5 m below the hinge in order to keep the gate closed. Square opening 0.3m side length Fluid Mechanics 75 Aug 2008 UNIT 5 HYDROSTATIC FORCES ON CURVED SURFACES Unit Structure 5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 Overview Learning Objectives Introduction Magnitude of the Horizontal Component of the Hydrostatic Force – Curved Surface Magnitude of the Vertical Component of the Hydrostatic Force Resultant Hydrostatic Force Acting on Curved Surface Activities Summary Worked Examples Tutorial 5.0 OVERVIEW In Unit 4, the concept of fluid exerting a pressure has been converted to the fluid exerting a resultant hydrostatic force on the walls of the vessel in which it is contained. At this point, it is advisable that you once again go back to the contents of Unit 4, to appreciate the methodology guiding the approach to deriving the resultant hydrostatic force and the position of action of this force, the centre of pressure. You must probably have noted that the individual forces acting on the solid plane surface were all parallel, and thus could be linearly combined. In the case of a curved surface, there will once again be individual forces acting at different points along the solid surface, but this time they will not be parallel forces. So your attention is drawn to this particular difference. 5.1 LEARNING OBJECTIVES At the end of this unit, you should be able to do the following: 1. Derive the equation governing the resultant force exerted by a fluid on curved surfaces. 76 Aug 2008 Fluid Mechanics 2. Clearly define the exact meaning of each term making up the equation for hydrostatic force on plane surface. Analyse how the worked examples were solved. Differentiate between solution of problems between plane and curved surface. 3. 4. 5.2 INTRODUCTION Referring to what has been explained in Unit 4, a fluid exerts pressure on the surface of the vessel in which it is contained. This pressure varies with depth, and in terms of gauge pressure, this fluid pressure will be a minimum (zero) at the top liquid surface and a maximum at the bottom liquid level. Force is equal to pressure x cross sectional area, and since fluid pressure varies with depth, this implies that the force exerted by the fluid also varies with depth. The force exerted by a fluid, also known as the hydrostatic force, can be illustrated as a series of individual forces acting on the surface of the solid. When the surface area is plane, as shown in Unit 4, these forces can be linearly combined. However, when the surface is curved, a slightly different approach is adopted to combine these forces and to eventually find their point of action of the resultant force. 5.3 MAGNITUDE OF THE RESULTANT HORIZONTAL COMPONENT OF THE HYDROSTATIC FORCE – CURVED SURFACE Consider a vessel containing liquid, of density ρ , to a depth H, Figure 5.1, and consider the curved surface ABCD. As we move down from AB to CD, we know that the pressure will be increasing from a value of 0 to a value of Hρg. Now, as was the case with the plane surface in Unit 4, imagine that the curved surface ABCD is now divided into very small layers of area δA. Consider now layer 1, which is located at depth h1 from top water level. The pressure acting at 1 is given by h1ρg, and from there, the hydrostatic force acting at point 1, will be given by the relationship, force= pressure x area. Thus, force acting at point 1, will be F1, where F1= h1ρg δA. Similarly, the forces acting at Fluid Mechanics 77 Aug 2008 point 2, 3 and 4, can be expressed by the following equations: h2ρg δA, h3ρg δA and h4ρg δA (same situation as described for plane surface in Unit 4). top water surface A H h4 h3 h2 h1 Width of AB = b B D δA C Figure 5.1: Hydrostatic force on a Curved Surface These individual forces can be represented as shown in Figure 5.2, as a series of individual forces acting perpendicularly at their point of contact with the curved surface ABCD. Note in the case of a curved surface, that these individual hydrostatic forces are NOT parallel to each other. Fluid Mechanics 78 Aug 2008 B h1 h2 F2=h2 ρ g δA h3 F1=h1 ρ g δA H h4 F3=h3 ρ g δA F4=h4 ρ g δA C Figure 5.2: Individual hydrostatic forces Similarly, going down up to the bottom of the side ABCD, there would be a series of individual forces acting perpendicularly on the curved surface, but these forces are not parallel to each other, and therefore cannot be combined linearly, as was the case for the plane surface. However each inclined individual force can be considered in a different way. An inclined force has a vertical and a horizontal component, Figure 5.3. Fluid Mechanics 79 Aug 2008 H Fi sin θ Fi cos θθ F C Figure 5.3: Horizontal and Vertical Components of Individual forces Consider a general inclined individual force Fi, making an angle θ with the horizontal. The corresponding horizontal force is Fi cos θ and the corresponding vertical force is Fi sin θ, as illustrated in Figure 5.3. Now consider all the individual inclined forces acting on the curved surface, each will have a vertical and a horizontal component, Figure 5.4. Fluid Mechanics 80 Aug 2008 B ’ B H C Figure 5.4: Horizontal and vertical components of individual forces Now consider only all the horizontal component of these individual forces. From Figure 5.5, it can be noted that when all the horizontal components of the individual forces are being considered, they can be considered as if they are acting on a vertical plane surface, given by B’C x width of curved surface (b), i.e, the projected area of curved surface ABCD. Fluid Mechanics 81 Aug 2008 B’ B H C’ Figure 5.5: Resultant horizontal force acts on Projected Area = B’C x width of curved surface BC Thus the horizontal components of the individual forces therefore act on the projected area of the curved surface. Since here we are once again dealing with a vertical plane surface (the projected area), we can resort back to the methodology adopted for plane vertical surface in Unit 4, i.e, combine the individual forces linearly. Therefore, the Resultant Horizontal component of the hydrostatic force acting on AB’CD = ∑ hi ρ g δA where i will vary from 1 to the total number of layers into which the solid surface has been divided for analysis purposes, and which eventually yields, FH= ρg A h Fluid Mechanics 82 Aug 2008 Where ρ = density of the liquid g = acceleration due to gravity A = cross sectional area of the projected surface in contact with the liquid h = vertical depth from top liquid level to the centroid of the projected surface; and this force acts at the centre of pressure related to the projected vertical surface. 5.4 MAGNITUDE OF THE RESULTANT VERTICAL COMPONENT OF THE HYDROSTATIC FORCE Having obtained the horizontal component of the resultant force, we need now to work out the vertical component of the vertical force. B’ B H C’ Figure 5.6: Vertical component of resultant forces = weight of liquid above or below curved surface Consider Figure 5.6, the vertical component of the hydrostatic force is exerted by the weight of the liquid. Hence the vertical component of the resultant hydrostatic force acting on surface BC, will simply be the weight of the fluid above the curved surface or opposing the curved surface. Fluid Mechanics 83 Aug 2008 Weight of liquid = shaded area x width of curved surface x density x acceleration due to gravity Weight of liquid = A b ρ g = Vρ g , and this weight acts at the centroid of the body of liquid, and is also the vertical component of the resultant hydrostatic force. Hence, FV= V ρ g Where V is the volume of liquid above or against the curved surface ρ is the density of the liquid g is the acceleration due to gravity 5.5 RESULTANT HYDROSTATIC FORCE ON CURVED SURFACE From section 5.3, it was shown that the horizontal component FH acts at the centre of pressure related to the projected vertical surface B’C. Similarly, in section 5.4, it has been shown that the vertical component of the hydrostatic force FV acts at the centroid of the liquid above the curved surface, and these are illustrated in Figure 5.7. Fluid Mechanics 84 Aug 2008 B’ B D H FV FH Position of the centroid of the liquid above curved surface C Figure 5.7: Resultant force acting on Curved Surface Thus the resultant hydrostatic force acting on the curved surface, is reported as follows: FH θ FV FR FR = { (FH)2 + (FV)2 } ½ Acting at an angle, θ, where θ is given by tan-1 (FV/FH). Fluid Mechanics 85 Aug 2008 5.6 ACTIVITIES 1. Differentiate between the concepts behind the analysis of the resultant hydrostatic force acting on curved surface as compared to those used in the analysis of a plane surface. 2. What do you understand by the term projected area? What is the shape of the projected surface of a sphere? 3. Given the equation of the curved surface under study, explain, based upon the principles of first moment of area, how you would locate the position of centroid of the liquid above the curved surface? 4. Derive the equation governing the resultant hydrostatic force acting on a curved surface. 5.7 SUMMARY Unit 5 was a continuation of Unit 4, whereby in this unit you have has been explained how to work out the resultant force acting on a curved surface. The next unit will also be concerned with the calculation of hydrostatic force on plane surface based upon a slightly different approach, the Pressure Diagrams. You need to learn both methodologies. However, before moving to Unit 6, you are strongly advised to go back once more to Units 4 and 5, to make sure the concepts described there are clear, else you might get confused while going through Unit 6. Fluid Mechanics 86 Aug 2008 5.8 WORKED EXAMPLES Example 1 – Hydrostatic force on curved surface in the shape of the quadrant of a circle Example 1 – Quadrant of circle …1/4 • A sluice gate as shown below, consists of a quadrant of a circle of radius 1.5m pivoted at its centre O. Its centre of gravity is at G as shown. When the water is level with the pivot O, calculate the magnitude and direction of the resultant force on the gate due to the water and the turning moment required to open the gate. The width of the gate is 3m and it has a mass of 6000kg. 1.5m radius O W G 0.6m 0.6m water 1.5m Fluid Mechanics 87 Aug 2008 Example 1 – Quadrant of circle..2/4 • Horizontal component of force acting on Projected area, FH = ρ g A y Where is the density of the liquid exerting the hydrostatic force (kg/m3) g the acceleration due to gravity (m/s2) A is the projected area of the curved surface (m2) y is the vertical distance from the top liquid surface to the centre of gravity of the projected area (m) • • • Projected area of curved surface is a rectangular surface, width 3m, and depth 1.5m. y is given by 1.5/2 = 0.75m, measured from the top liquid surface. Hence, horizontal component of force acting on Projected area, FH = ρ g A y = 1000 x 9.81 (1.5 x 3) x (1.5/2) = 33.1 kN Example 1 – Quadrant of circle..3/4 • • Vertical component of the hydrostatic force is given by the weight of the liquid either above the curved surface or opposing the curved surface. In this example, the vertical component of the force will be given by the weight of the water which is opposing the presence of the curved surface. Weight of the water acting against the curved surface • Weight = volume of quadrant x density x g Weight = (πr2)/4x3 x 1000 x 9.81 Weight = Vertical component of the hydrostatic force = (πr2)/4x3 x 1000 x 9.81 FV = (πx 1.52)/4x3 x 1000 x 9.81 = 52 kN Fluid Mechanics 88 Aug 2008 Example 1 – Quadrant of circle..4/4 • Resultant hydrostatic force acting on curved surface, FR FR = ? (FH2 + FV2) FR = ? {33.12 +522} FR = ? {33.12 +522} = 61.6 kN Point of action of resultant hydrostatic force, θ = tan -1 (52/33.1) θ = 52.5o • Turning moment required to open the sluice gate: Taking moment about the centre of the quadrant, W x 0.6 = 6000 x 0.6 x 9.81 = 35.3 kNm Note: Since the line of action of the resultant hydrostatic force passes through the centre of the quadrant, the hydrostatic force exerts no moment about the pivot O. Example 2 – Hydrostatic force acting on a curved surface, the surface of a dam Example 2 – Force acting on a Dam..1/9 • The face of a dam is curved according to the relation Y=X2/2.4 where y and x are in metres. The height of the free surface above the horizontal plane through A is 15.25m. Calculate the resultant force F due to the fresh water acting on unit breadth of the dam, and determine the position of the point B at which the line of action of this force cuts the horizontal plane through A. H =15.25m Y A X Fluid Mechanics 89 Aug 2008 Example 2 – Force acting on a Dam..2/9 15.25m Y A X • Horizontal component of force acting on Projected area, FH = ρ g A y = 1000 x 9.81 (15.25 x 1) x (15.25/2) = 1143.7 kN per m width of dam Example 2 – Force acting on a Dam..3/9 A • Vertical component of the hydrostatic force: – Weight of the water above the curved surface = Shaded area x width of the dam x density of water x g • • • • Equation of dam: Y=X2/2.4, and when Y = 15.25, X = 6.05 Shaded area = ∫ (2.4 Y)1/2 dy from y=0 to y=12.25 gives Shaded area = {2/3 x 2.4 x Y} 3/2 from y=0 to y=15.25 Shaded area = 120.5 m2 Fluid Mechanics 90 Aug 2008 Example 2 – Force acting on a Dam..4/9 A • Vertical component of the hydrostatic force: – Weight of the water above the curved surface FV = Shaded area x width of the dam x density of water x g = 120.5 x 1 x 1000 x 9.81 = 1.18 MN per m width of the dam Resultant hydrostatic force acting on dam: • FR = ? (FH2 + FV2) FR = ? {1143.72 +11802} = 1643 kN per m width of dam Example 2 – Force acting on a Dam..5/9 D centre of pressure for force FH FH FV A Z FR B H θ (H-D) tanθ • • • Distance AB = Distance AZ + ZB AZ = horizontal distance from point A to the line of action of the vertical component of the force, FV ZB = (H-D) tanθ Fluid Mechanics 91 Aug 2008 Example 2 – Force acting on a Dam..6/9 FH A FV Z H θ (H-D) tanθ FR B • Calculating distance D: D = Icg/Ay +y= (bd3/12)/(bd)xd/2 + d/2 D = (1x15.253)/{12x(1x15.25)x15.25/2} + 15.25/2 D=10.17m (distance ZB) Example 2 – Force acting on a Dam..7/9 • A’ Small element thickness dX H-Y H Calculating position of centroid of the curved surface: Consider a small element of thickness dX, within the curve surface: First moment of area of this small element about line AA’: A X Y=X2/2.4 B ? {(H-Y)dX x X} = area above curved surface x distance of centroid of curved surface from line AA’. Fluid Mechanics 92 Aug 2008 Example 2 – Force acting on a Dam..8/9 A’ • ? {(H-Y)dX x X} = area above curved surface x distance of centroid of curved surface from line AA’. H-Y H Small element thickness dX • ? {(H-X2/2.4)dX x X} , Integrating from X =0 to X =6.05 (see slide no. 3) A X Y=X2/2.4 B • Total area above curved surface = 120.5 (see slide no. 3) Example 2 – Force acting on a Dam..9/9 • A’ ?{(H-X2/2.4)dX x X} , Integrating from X =0 to X =6.05 (see slide no. 3) Small element thickness dX H-Y H Integrating equation results in { HX – X3/(3x2.4)} and from X=0 to X=6.05, gives 61.51. • Total area above curved surface = 120.5 (see slide no. 3) Therefore X = (120.5/61.51) = = 1. 96m (distance AZ) A X Y=X2/2.4 B • Distance AB = AZ + ZB = 1.96 + 10.17 = 12.13m Fluid Mechanics 93 Aug 2008 5.9 TUTORIAL Question 1 The curved surface of a dam retaining water is shaped according to the relationship y=x2/4, where x and y are measured in metres from the origin. The origin is being defined as the point of intersection of the base of the dam to the horizontal. Calculate (a) the resultant force acting on the dam, in terms of magnitude and direction. (b) the point of action of the resultant force, measured from the origin. Question 2 2.5 (a) The profile of the water face of a dam is given by the equation 44.75 y = x , where the coordinates of x and y in metres are measured from an origin set at the point of intersection of the flat floor and the curved face of the dam. The depth of the water is 4 m. Calculate the magnitude and direction of the resultant hydrostatic force per meter width of the dam. (b) Calculate also the horizontal distance at which the resultant force cuts a horizontal line at floor level. Question 3 (a) A sluice gate consists of a quadrant of a circle of radius 1.5m pivoted at its centre at O. Its centre of gravity is at G as shown. When the water is level with the pivot O, calculate the magnitude and direction of the resultant force on the gate due to the water and the turning moment required to open the gate. The width of the gate is 3 m and it has a mass of 6000 kg. Fluid Mechanics 94 Aug 2008 Question 4 The face of a dam is curved according to the relation y=x2/2.4, where y and x are in metres. The height of the free surface above the horizontal plane through A is 15.25 m. Calculate the resultant force F due to the fresh water acting on unit breadth of the dam, and determine the position of the point B at which the line of action of this force cuts the horizontal plane through A. Fluid Mechanics 95 Aug 2008 Fluid Mechanics 96 Aug 2008 UNIT 6 Unit Structure PRESSURE DIAGARAMS 6.0 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 Overview Learning Objectives Introduction Pressure diagrams Magnitude of the Vertical component of the Hydrostatic Force Position of Centre of Pressure Activities Summary Worked Examples Tutorial 6.0 OVERVIEW In Units 3, 4 and 5, the student was introduced to the concept of pressure exerted by fluids and the conversion of this fluid pressure into a resultant hydrostatic force on a plane and a curved surface. In Units 4 and 5, one particular approached to calculating resultant hydrostatic force was presented. In Unit 6, the student will now be introduced to a second approach of calculating the resultant hydrostatic force acting on vertical plane surface only. Fluid Mechanics 97 Aug 2008 6.1 LEARNING OBJECTIVES At the end of this unit, students should be able to do the following: 1. Understand the meaning of pressure diagrams 2. Understand why the method of pressure diagrams can be applied only to plane surface 3. How to derive the equation for calculating resultant hydrostatic force on vertical plane surface using the pressure diagram method. 4. How to derive the equation for calculating the centre of pressure for the resultant hydrostatic force on a vertical plane surface using the pressure diagram method. 6.2 INTRODUCTION As discussed in Unit 4, a fluid at rest exerts pressure and this pressure can be converted to a force, the hydrostatic force exerted on the walls of the vessel in which it is contained. One particular method for calculating this pressure was illustrated in Unit 4. A second, more simplified way of calculating the resultant hydrostatic force and its centre of pressure will be illustrated in this unit. 6.3 PRESSURE DIAGRAMS The method of Pressure Diagram is a graphical method for calculating hydrostatic forces on solid surfaces and the centre of pressure of these forces. Pressure diagrams are convenient for plane vertical surfaces only. Pressure diagram is simply the graph showing the distribution of pressure from one extremity of the plane vertical surface to the other. For example, from Figure 6.1, we need to calculate the hydrostatic force acting on vertical plane surface AB. The two extremities of the plane vertical surface is A and B, where A is located at the top water level and B is located at the bottom edge. The fluid pressure at A in terms of Gauge Pressure (Unit 3), is zero, and the fluid pressure at B is (Hρg). Thus plotting these two points yields a triangular shaped graph as shown in Figure 6.1: Fluid Mechanics 98 Aug 2008 depth A PA=0 H Liquid density, H B Pressure head PB=H ρ g Figure 6.1: Pressure Diagram – vertical plane surface AB Similarly, consider a vertical plane surface, submerged in a body of fluid, Figure 6.2. The extremities of this plane vertical surface is A and B, where the fluid pressure at A is (Yρg) and the fluid pressure at B is [(H+Y)ρg], yielding a trapezoidal shaped pressure diagram. depth Y Liquid density, ρ A H B PA=Y ρ g Pressure head PB=(H+Y) ρ g Figure 6.2: Pressure Diagram – a submerged vertical surface AB Fluid Mechanics 99 Aug 2008 Note: The plane surface AB should either be rectangular or square, else the width of the plane surface will vary and so will the shape of the pressure diagram. 6.4 MAGNITUDE OF THE HYDROSTATIC FORCE RESULTANT VERTICAL COMPONENT OF THE The magnitude of the resultant hydrostatic force on a vertical plane surface using the pressure diagram is simply, the AREA of the pressure diagram multiply by the width of the plane surface. Referring to Figure 6.1, the hydrostatic force acting on vertical plane surface AB is the area of the triangular pressure diagram multiply by the width of AB: Resultant hydrostatic force = (AREA of Pressure diagram) x Width of surface AB = { ½ Hρg x H } x B = ½ H2ρ g B Referring to Figure 6.2, the hydrostatic force acting on vertical plane surface AB, will therefore be given by Resultant force acting on AB (Figure 6.2) = Area of trapezium x Width of AB = { ½ (Yρg + (H+Y)ρg} H x B = ½ H B (H+2Y)ρg ρ NOTE: Student should know about how to make use of the method of pressure diagram to calculate the resultant hydrostatic force acting on a plane surface. The key point in this method is to have a clear diagram and draw your pressure diagram accurately. The rest is just calculating the area of simple diagrams, triangles or trapezium. Fluid Mechanics 100 Aug 2008 6.5 POSITION OF CENTRE OF PRESSURE The position of action of the resultant hydrostatic force, i.e, the centre of pressure, is simply the CENTROID of the pressure diagram. Referring to Figure 6.1, the pressure diagram in this case is in the form of a triangle. The CENTROID of a triangle is located at position 1/3 the height of the triangle. Hence in Figure 6.1, the centroid will be H/3 from point B (Figure 6.3): dept PA=0 H Position of centroid of triangle H/ Pressure PB=H ρ g Figure 6.3: Pressure Diagram – CENTROID OF PRESSURE DIAGRAM Similarly, the position of centre for the resultant force in Figure 6.2, will be the position of centroid of the trapezium. The position of centroid of a trapezium can be calculated by dividing the trapezium into 2 simple shapes, a rectangle and a triangle, the position of centroid of which are known. Next by taking moments about the base of the trapezium, the position of centroid of the trapezium can be calculated. Fluid Mechanics 101 Aug 2008 depth H H/2 x H/ x Pressure head Take moments about line XX to find the position of centroid of the trapezium Figure 6.4: Pressure Diagram – centroid of Trapezium Calculation of the centroid of a trapezium is best illustrated with a numerical example. Fluid Mechanics 102 Aug 2008 6.6 ACTIVITIES 5. Draw pressure diagrams acting on the plane vertical surface AB for the following cases: (a) A 1.5m B 1m 1.5m A B (b) (c) A (d) Pressurised air A 5kN/m2 1.5m 0.5m B B Width of AB = 2.5m in all cases & fluid is vessel is water 6. Calculate the position of the centroid of the following diagrams in part 1. 7. Calculate the resultant and centre of pressure of the various cases illustrated in part 1. 8. A closed tank rectangular in plan with vertical sides is 1.8m deep and contains water to a depth of 1.2m: (a) Calculate the hydrostatic force acting on one side of the tank if its width is 3m and its centre of pressure, and (b) If the air space above the water is filled with pressurised air, 35 kN/m2. Calculate the hydrostatic force and its centre of pressure. 4. A rectangular plane area, immersed in water is 1.5m by 1.8 m with the 1.5 m side horizontal and the 1.8 m side vertical. Determine the magnitude of the force on one side and the depth of its centre of pressure (measured from the base), if the top edge is 30 m below the water surface. Fluid Mechanics 103 Aug 2008 6.7 SUMMARY In this unit you have been introduced to a different method for calculating resultant hydrostatic force and centre of pressure of pressure, for vertical plane surface. Your attention is drawn to the fact that the plane vertical surface needs to be either rectangular or square for making use of the method of pressure diagrams. Unit 7 will be a continuation of hydrostatics, i.e, fluids at rest, but bears no direct relation to what has been discussed in units 3 to 6. 6.8 WORKED EXAMPLES Example 1 – Calculation of hydrostatic force with the top surface under atmospheric pressure Example 1 – Simple application…1/4 • A tank, rectangular in plan with vertical sides, is 1.8m deep and contains water to a depth of 1.2m. If the length of one wall of the tank is 3m, calculate the resultant force on this wall and the height of the centre of pressure measured from above the base. A 1.8m B Width of wall (AB) = 3m Fluid Mechanics 104 Aug 2008 Example 1 – Simple application…2/4 A 1.8m B H=1.8 Gauge pressure = 1.8ρg ρ height H=0, Gauge pressure=0 Pressure • At position A, the gauge pressure = 0 and at position B, the gauge pressure is given by the height of the liquid exerting the pressure, hρg. Hydrostatic force per m width of the wall is given by the area of the pressure diagram = Area of the shaded triangle Area of shaded triangle = {½ x base x height } = • • {½ x 1.8ρg x 1.8 } ρ Example 1 – Simple application…3/4 A 1.8m B H=1.8 Gauge pressure = 1.8ρg ρ height H=0, Gauge pressure=0 Pressure • Area of shaded triangle = {½ x base x height } = {½ x 1.8ρg x 1.8 } ρ = 1.62 ρg • Total hydrostatic force acting on AB = shaded area x width of AB = 1.62 ρg x 3 = 47.7 kN Fluid Mechanics 105 Aug 2008 Example 1 – Simple application…4/4 A 1.8m b/3 height H=0, Gauge pressure=0 Centroid B H=1.8 Gauge pressure = 1.8ρg ρ Pressure h/3 • Position of action of the hydrostatic force, i.e, Centre of pressure, is given by the position of the centroid of the pressure diagram. Position of Centroid of a triangle, is located as shown in the diagram above: – Hence measured from the base, the position of centroid, hence Centre of Pressure of the hydrostatic force acting on AB is 1.8/3 = 0.6m measured from the base. • Example 2 – Calculation of hydrostatic force with the top surface under pressure Example 2 – Slightly more complex example…1/4 • A closed tank, rectangular in plan with vertical sides, is 2.4m deep and contains water to a depth of 1.8m. Air is pumped into the space above the water until the air pressure is 35kN/m2, if the length of one wall of the tank is 3m, calculate the resultant force on this wall and the height of the centre of pressure measured from above the base. Pressurised Air A 2.4m Width of wall (AB) = 3m 1.8m B Fluid Mechanics 106 Aug 2008 Example 2 – Slightly more complex example…2/4 H=0, Gauge pressure= Pressure of Air = 35kN/m2 Pressurised Air A height 1.8m B Pressure H=2.4m, Gauge pressure = 1.8ρg + Pressure of Air ρ • At A the pressure at the water surface, is equal to the same as the pressure of the air, i.e, 35kN/m2. • At B, the pressure is equal to the pressure exerted by the pressurised air and the pressure exerted by the 1.8m column of water. Example 2 – Slightly more complex example…3/4 Pressurised Air A height H=0, Gauge pressure= Pressure of Air 1.8m B Pressure H=2.4m Gauge pressure = 1.8ρg + Pressure of Air ρ • Hydrostatic force acting on AB is given by the product of the shaded area and the width of the wall (3m). – FAB = { (35x1000 x 2.4) + (½ x 1.8ρg x 1.8)} x 3 = 299.7 kN Fluid Mechanics 107 Aug 2008 Example 2 – Slightly more complex example…4/4 Shaded Area = (35x1000x12.4) Shaded Area = (½ x 1.8ρg x 1.8) 2.4/2=1.2m 1.8/3=0.6m BASE • • The position of action of the hydrostatic force acting on AB, is the position of centroid of the pressure diagram. The pressure diagram is therefore divided into two simple diagrams, a rectangle and a triangle, with known position of centroid, and the centroid of the pressure diagram is calculated by taking the first moment of area about the base: { (35x1000x2.4)x1.2+ (½x1.8ρgx1.8) x 0.6} = {(35x1000x2.4) + (½x1.8ρgx1.8) } Y Therefore, Y = 1.11m 6.9 TUTORIAL Question 1 (a) A gate 3 m wide and 2 m deep divides a storage tank. On one side of the gate there is petrol of specific gravity 0.78 stored at a depth of 1.8m, while on the other side there is oil of specific gravity 0.88 stored to a depth of 0.9m. Using the pressure diagram method, determine the resulting hydrostatic force acting on the gate, and the position of action of this force. (b) 'The centre of pressure is always located below the centroid of the wetted surface', Discuss. Fluid Mechanics 108 Aug 2008 Question 2 A lock gate is 3m wide. Find the initial turning moment necessary to force the gate to open when the water levels across the gate are 3m and 4m respectively above the sill, using the pressure diagram method. Question 3 Find the resultant pressure and the center of pressure on a vertical square plate 1.8m side, given the center of the plate is 1.2m below the surface of the water, using the pressure diagram method. Question 4 a. A gate 3 m wide and 2 m deep divides a storage tank. On one side of the gate there is petrol of specific gravity 0.78 stored at a depth of 1.8m, while on the other side there is oil of specific gravity 0.88 stored to a depth of 0.9m. Using the pressure diagram method, determine the resulting hydrostatic force acting on the gate, and the position of action of this force. b. 'The centre of pressure is always located below the centroid of the wetted surface', Discuss. Fluid Mechanics 109 Aug 2008 UNIT 7 BUOYANCY Unit Structure 7.0 Overview Learning Objectives Introduction Resultant Force Acting on a Completely Submerged Body Definition –Buoyancy Sink or Float? Stability of Submerged Bodies Metacentre Summary 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.0 OVERVIEW Unit 7 deals with the forces exerted by a fluid on a solid body in contact with the fluid. This unit just like units 1 to 6 is concerned with fluids at rest. 7.1 LEARNING OBJECTIVES At the end of this unit, students should be able to do the following: 1. Define what you understand by the term Buoyancy & Archimedes Principle. 2. Differentiate with the help of sketches between, stable equilibrium, unstable equilibrium and neutral equilibrium. 3. Explain what is meant by the term metacenter. 4. To relate the position of metacentre, centre of gravity of a body to the stability of the body in the fluid. Fluid Mechanics 110 Aug 2008 7.2 INTRODUCTION Buoyancy is the term used to describe the force of resistance exerted by a fluid on a body which is either partially or completely submerged into it. Buoyancy is a term which is very much relevant in boat construction. 7.3 RESULTANT FORCE ACTING ON A COMPLETELY SUBMERGED BODY Buoyancy is a force exerted by a fluid on a solid body completely or partially immersed in the fluid and buoyancy always pushes upwards against the pull of gravity. How buoyancy works: a) Buoyancy is based on Newton's Third Law of Motion, which states: "For every action, there is an equal and opposite reaction." Here is the process: (i) objects placed in a fluid push particles of the fluid away (displace particles) (ii) following Newton's Third Law, the particles push back on the object (iii) the force with which the particles push back is the buoyant force measured in newtons (N) Referring to Unit 5, it is to be noted here that the method use for calculating forces on curved surfaces applies to any shaped body. The approach is to work out the resultant horizontal component and the resultant vertical component, and combine them to obtain the resultant hydrostatic force acting on the curved surface. Consider the following completely submerged body: Fluid Mechanics 111 Aug 2008 Figure 7.1: Completely submerged body Calculating the resultant HORIZONTAL COMPONENT To obtain the resultant horizontal force acting, we divide the body into two and then project the curved surface, which is a circle in this case, as indicated in Figure 7.1. The result is that there is an equal and opposite horizontal component of hydrostatic force acting on the body, hence no resultant horizontal force will be acting on the completely submerged body. Calculating the resultant VERTICAL COMPONENT For the resultant vertical component, we once again divide the body into two parts from a horizontal plane. Consider equilibrium of vertical forces on the upper surface and on the lower surface of the body: Vertical force acting on the upper surface of the body = weight of fluid displaced by this curved surface Similarly, the vertical force acting on the lower surface = weight of fluid displaced by this curved surface Therefore the resultant vertical force acting on the curved body, also called the upthrust on the body = The weight of the fluid displaced by the body (Archimedes’ Principle). This vertical force will act through the centroid of the volume of fluid displaced by the body, known as the centre of buoyancy. Fluid Mechanics 112 Aug 2008 7.4 DEFINITION - BUOYANCY Any body in a fluid, whether floating or completely submerged, is buoyed up by a force equal to the weight of the fluid that has been displaced – Archimedes Principle. The buoyancy force is given by the product of the volume of fluid displaced, the density of the liquid and the acceleration due to gravity, V ρ (liquid) g. The buoyancy force acts at the centroid of the liquid which has been displaced (Figure 7.2) in the case of a rectangular body ABCD floating in a liquid. In the case of a completely submerged spherical body, the centroid of the liquid which will have been displaced, is the centre of the sphere. A B E cgb cgd F D C cgb – centre of gravity of solid body ABCD cgd – centroid of the displaced fluid, space EFCD Figure 7.2: Centroidal position of displaced liquid NOTE: For a completely submerged body, the centroid of the displaced fluid is located at the same position as the center of gravity of the body. Fluid Mechanics 113 Aug 2008 7.5 SINK OR FLOAT If the downward force exerted by the weight of the body is higher than the vertical upwards buoyancy force, the body will sink, else it floats. The larger the surface area of a solid body in contact with the fluid, the greater the amount of fluid displaced, and hence the higher the vertical upthrust force. 7.6 STABILITY OF SUBMERGED BODY A submerged or partially immersed body can be in stable, unstable or neutral equilibrium position. The body is in stable equilibrium if when titled it regains its original position. If body acquires a different position after having been tilted, then it is said to be under unstable equilibrium. Consider the body ABCD floating in liquid (Figure 7.3): the weight W=mg acts through the centre of gravity of the body cgb, and the upthrust R acts through the centre of buoyancy cgd. A W=mg E cgb cgd B F R D C Figure 7.3: Position of action of Weight and Upthrust acting on ABCD When the body ABCD is titled, this is accompanied by a change in the volume of fluid displaced by the tilted body, and consequently a change in the position of centroid of the dispersed fluid. Fluid Mechanics 114 Aug 2008 The position of cgd changes, but the position of the center of gravity cgb of the body which is independent of the fluid displaced, stays the same, as indicated by Figure 7.4. A B B’ cgd mg A’ D cgb R C Figure 7.4: Tilted object – Resulting restoring moment (anticlockwise) Figure 7.4: Titled object – Resulting restoring moment (anticlockwise) The two forces, the weight of the body ABCD acting at the center of gravity cgb, and the upthrust R acting at the new position of centroid of the displaced fluid (A’B’CD), cgd, exert a moment as indicated by the arrows in Figure 7.4 & Figure 7.5. When this moment is such that it makes the body regain its original position, this is knows as the STABLE condition. If the however the moment is such that the body comes to equilibrium in another position, then we have UNSTABLE condition (Figure 7.6). Fluid Mechanics 115 Aug 2008 B A A’ cgb cgd mg B’ C R D Figure 7.5: Tilted object – Resulting Restoring moment (clockwise) Figure 7.5: Titled Object – Resulting Restoring moment (clockwise) mg cgb cgd R Figure 7.6: Tilted object – Resulting Overturning moment (clockwise) Figure 7.6: Titled object – Resulting Overturning moment (clockwise) Fluid Mechanics 116 Aug 2008 7.7 METACENTRE The metacentre is defined as the theoretical point at which an imaginary vertical line through the centre of buoyancy (cgd) intersects another imaginary vertical line through a new centre of buoyancy created when the body is displaced, or tipped, in the water, however little. Basically the centre of buoyancy of a floating body is the point about which all the body parts exactly buoy. A body is under stable equilibrium if its center of gravity cgb, always lies below the position of metacentre M, as indicated by figures 7.7-7.9. M cgb cgd Note: The center of gravity cgb of the solid body lies below the metacentre point M, hence body is under stable equilibrium Figure 7.7: Position of Metacentre Figure 7.7: Position of Metacentre Fluid Mechanics 117 Aug 2008 M cg b cg d N o te: T he center of gravity cg b of the solid body lies below the m etacentre point M , hence body is under stab le eq u ilib riu m Figure 7.8: P osition of M etacentre Figure 7.8: Position of Metacentre Fluid Mechanics 118 Aug 2008 cgb cgd M Note: The center of gravity cgb of the solid body lies above the metacentre point M, hence body is under unstable equilibrium Figure 7.9: Position of Metacentre Figure 7.9: Position of Metacentre Fluid Mechanics 119 Aug 2008 Figure 7.6: Titled object – Resulting Overturning moment (clockwise) mg cgb cgd R Figure 7.6: Tilted object – Resulting Overturning moment (clockwise) Fluid Mechanics 120 Aug 2008 7.8 SUMMARY In this unit you have learnt how a fluid exerts a vertical force on a solid body. You will also have learnt about how the stability of a body is related to the density of the fluid in which the body is being immersed and the relative position of the metacentre and the centre of gravity of the solid body. Unit 8 will be the first chapter dealing with fluids in motion. Some of the concepts discussed in units 1 to 8 will be referred to in the coming units, and students will need to make sure the concepts behind the analysis of fluids at rest are well understood before they move on to the next part of the module, i.e. fluids in motion. 7.9 TUTORIAL Question 1 A rectangular pontoon 10m by 4m in plan weighs 280kN. A steel tube weighing 34kN is placed longitudinally on the deck. When the tube is in a central position, the centre of gravity for the combined weight lies on the vertical axis of symmetry 250mm above the water surface. Find (a) the metacentric height, (b) the maximum distance the tube may be rolled laterally across the deck if the angle of heel is not to exceed 5o. Question 2 A rectangular tank 90cm long and 60cm wide is mounted on bearings so that it is free to turn on a longitudinal axis. The tank has a mass of 68kg and its centre of gravity is 15cm above the bottom. When the tank is slowly filled with water it hangs in stable equilibrium until the depth of water is 45cm after which it becomes unstable. How far is the axis of the bearings above the bottom of the tank? Question 3 A cylindrical buoy 1.35m in diameter and 1.8m high has a mass of 770kg. Show that is will not float with its axis vertical in sea water of density 1025kg/m3. If one end of a vertical chain is Fluid Mechanics 121 Aug 2008 fastened to the base find the pull required to keep the buoy vertical. The centre of gravity of the buoy is 0.9m from its base. Fluid Mechanics 122 Aug 2008 UNIT 8 Unit Structure 8.0 HYDRODYNAMICS – FLUID DYNAMICS (IN MOTION) Overview Learning Objectives Introduction Types of Flow Uniform & non-uniform Flow Steady & unsteady flow Reynolds’ Number Pathlines & Streamlines Streamlines & Streamtubes Rotational & Irrotational Fluid Activities Summary 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.0 OVERVIEW This unit illustrates the approach and the various concepts commonly used when analyzing a fluid in motion (hydrodynamics). In hydrodynamics assumptions often have to be made while carrying out analysis. In many cases, except when dealing with very complex fluid flows, the motion of fluid can be analysed with high level of accuracy. The concepts used in the analysis of hydrodynamics are very much similar to those used in the motion of solids, with a few minor changes induced by the behaviour of the fluids under stress conditions. Fluid Mechanics 123 Aug 2008 8.1 LEARNING OBJECTIVES At the end of this Unit, you should be able to do the following: 1. 2. Learnt about the concepts and approaches used to analyse fluids in motion. Differentiate between laminar and turbulent fluid flow, steady and unsteady fluid motion, and uniform and non-uniform fluid motion. 3. 4. Differentiate between streamlines and stream tubes. Illustrate with the help of sketches streamlines in a pipeline, in a river, over an obstacle 5. 6. Differentiate between a pathline and a streamline. Differentiate between rotational and irrotational fluids. 8.2 INTRODUCTION Before dealing the mathematical concepts behind the analysis of fluids in motion, it is first of all important to understand the characteristics of different types of flows. The approach used in the analysis of fluids in motion is very much dependent upon the prevailing flow conditions. Some assumptions may not always hold true is all circumstances. Fluid Mechanics 124 Aug 2008 8.3 TYPES OF FLOW Consider the flow in a river (Figure 8.1): flow direction 3 1 4 2 5 Figure 8.1: River flow Figure 8.1: River flow Points 1 to 5 have been highlighted in Figure 8.1, so as to make an important point about the conditions affecting flows in a river. Consider point 1 and 2, both of which lie more or less in the middle of the river. The velocity of flow at these two points should have been the same, however most likely they will not be. The reason being that slightly downstream of point 1 the river meanders slightly and this feature will induce disturbance in flow pattern upstream. Fluid Mechanics 125 Aug 2008 Similarly compare points 3 and 5, both of which are located close to the river bank. The velocity of flow at both points will be influenced by the presence of the solid boundary of the river bank. However point 3 lies within the zone where the river meanders, so the velocity at 3 and 5 will not be similar. Point 4 was highlighted so as to emphasise on the influence of boundary (river bank) on the velocity of flow, the closer the point to the boundary (in this case 3), the higher the impact of boundary. Hence though both points 4 and 5 lie within the meandering zone, the velocity of flow at point 4 will be higher than that at 3. 8.4 UNIFORM & NON-UNIFORM FLOW In section 8.3, it has clearly been illustrated that the flow both along and across a river channel varies from point to point, and such a situation is also technically termed – NON-UNIFORM FLOW. Thus, had the velocity been similar at any point along and across the river channel, the flow would have been termed – UNIFORM FLOW. variation of velocity with distance. Uniform or non-uniform flow is the A simplified illustration of uniform and non-uniform flow is given in Figure 8.2: Fluid Mechanics 126 Aug 2008 Uniform flow – same velocity at any point Non-uniform flow – varying velocity at different points Note: The direction and magnitude of the velocity is illustrated by the arrow at the various points under consideration. Figure 8.2 - Uniform and non-uniform flow Figure 8.2 – Uniform and non-uniform flow 8.5 STEADY & UNSTEADY FLOW Steady or unsteady flow is the variation of velocity with respect to time. A flow may either be uniform or non-uniform, but if it stays the same all the time, then the flow is considered as being steady. Figure 8.3 and Figure 8.4, are simplified illustrations of steady and unsteady flow. Fluid Mechanics 127 Aug 2008 Velocity pattern at time t=0 Velocity pattern at time t=t1 Steady uniform flow Velocity pattern at time t=0 Velocity pattern at time t=t1 Steady non-uniform flow Note: The direction and magnitude of the velocity is illustrated by the arrow at the various points under consideration. Figure 8.3 - Steady flow Figure 8.3: Steady flow Velocity pattern at time t=0 Velocity pattern at time t=t1 Unsteady uniform flow Velocity pattern at time t=0 Velocity pattern at time t=t1 Unsteady non-uniform flow Note: The direction and magnitude of the velocity is illustrated by the arrow at the various points under consideration. Figure 8.4 - Unsteady flow Figure 8.4: Unsteady flow When a fluid flow is uniform and steady it is commonly termed as being LAMINAR FLOW. Unsteady and non-uniform flow is known as TURBULENT FLOW. Fluid Mechanics 128 Aug 2008 8.6 REYNOLDS’ NUMBER The difference behind laminar and turbulent flows has been best illustrated by Reynolds’s Experiment, Figure 8.5: Dye d Laminar flow Dye v ρ Transition Dye Turbulent flow Figure 8.5: Reynolds’ Experiment Figure 8.5: Reynolds’ Experiment In this experiment a coloured dye in injected in a pipeline, diameter d, running full, with a velocity V and liquid of density, ρ . So long as the flow in the pipeline is laminar, the dye appears as a thin coloured thread or streamline within the pipeline. Gradually the flow in the pipeline is varied, hence V varies, laminar flow changes first to a transition flow, as illustrated in Figure 8.5. Finally when the flow rate is even higher, the entire flow in the pipeline becomes turbulent, causing the entire liquid in the pipe to be coloured. Fluid Mechanics 129 Aug 2008 Apart from the flow rate, Reynolds’ also varied the diameter of the pipe and the type of liquid within the pipeline to investigate how these three parameters, V, ρ and d, were influencing type of flow within a pipe. Finally he came up with the relationship: Re = ρ v d / µ where µ is the dynamic viscosity of the liquid Laminar flow conditions will prevail (provided the entire system is undisturbed), if Re≤ 2000. ≤ 8.7 PATHLINES & STREAMLINES A fluid flow (both liquid and air) may be described in two different ways, with respect to distance and with respect to time, commonly referred to as (1) the Lagrangian approach (named after the famous French mathematician Joseph Louis Lagrange), and (2) the Eulerian approach (named after Leonhard Euler, a famous Swiss mathematician), respectively. In the Lagrangian approach, one particle is chosen and is followed as it moves through space with time, i.e, the variation of the particle flow characteristics with time. The line traced out by that one particle is called a particle pathline or a streakline, Figure 8.6. PATHLINE – Consider the movement of a boat every hour Time=t1 Time=t2 Time=t3 Time=t4 Time=t5 Time=t6 Time=t7 Time=t9 Time=t8 Figure 8.6 – Pathline described by moving boat Figure 8.6: Pathline described by moving boat Fluid Mechanics 130 Aug 2008 A Eulerian approach is used to obtain a clearer idea of the airflow at one particular instant. One can imagine taking the picture of the flow of, for instance, surface ocean currents at a particular fixed time. The entire flow field is easily visualized. Just like drawing contour lines, here we join particle having similar velocity together and we produce what is known as streamlines, Figure 8.7. Similar to the properties of contour lines, streamline also does not cross each other. In Figure 8.7, fluid is flowing over a solid body. At different points, the fluids will have different velocity. Lines are used to join points of equal velocity and these produces the streamlines shown below. No flow takes place across steamlines. Streamline – Snapshot of flow at a point in time solid body Figure 8.7 – Flow over a solid body Figure 8.7: Flow over a solid body Thus, a pathline refers to the trace of a single particle in time and space whereas a streamline represents the line of motion of many particles at a fixed time. Refer to section 8.4, Figures 8.4 Fluid Mechanics 131 Aug 2008 to 8.6, it can be seen that under steady conditions, pathlines and streamlines will always be the same. 8.8 STREAMLINES & STREAMTUBES As mentioned in section 8.7, streamlines are formed when fluid particles of similar velocity are joined together. Bearing in mind that the velocity of flow is not constant all throughout, streamlines formed in a pipeline and in a river will appear as shown in Figures 8.8 and 8.9. Fluid Mechanics 132 Aug 2008 a V1 V2 V3 V4 V5 V6 b c a b c streamlines NOTE: Consider a pipeline running full. Across the pipe, at each point the fluid has a different velocity (boundary effects). The same pattern will repeat itself at each section (aa, bb, cc). By joining points of equal velocity, we get streamlines. Figure 8.8 – Streamlines pattern in a pipeline running full Figure 8.8: Streamlines pattern in a pipeline running full flow direction Figure 8.9 – Streamlines in a river Fluid Mechanics 133 Aug 2008 To draw a streamtube, we refer here to the flow through a pipeline running full. Consider a small cylindrical element inside the pipeline, as indicated in Figure 8.10. From this circle, we can draw a series of streamlines from one end of the cylindrical element to the other. This process eventually forms a Streamtube, Figure 8.10. Streamtubes Consider a small circle inside a pipeline running full, based upon the concept of streamlines, we can imagine a series of streamlines stemming from the circle. This gives rise to a Streamtube. streamtube Figure 8.10: Streamtube in a pipeline running full Figure 8.10: Streamtube in a pipeline running full Since a streamtube is formed by a series of streamlines, there cannot be any flow inside a streamtube. 8.9 ROTATIONAL & IRROTATIONAL FLUID As well as steady or unsteady, fluid flow can be rotational or irrotational. If the elements of fluid at each point in the flow have no net angular (spin) velocity about the points, the fluid flow is said to be irrotational. One can imagine a small paddle wheel immersed in a moving fluid. If the wheel translates (or moves) without rotating, the motion is irrotational. If the wheel rotates in a flow, the flow is rotational. Fluid Mechanics 134 Aug 2008 8.10 1. ACTIVITIES Explain with the help of sketches what how boundaries such as river banks influence the flow in a river. 2. 3. 4. 5. 6. 7. Differentiate between steady and uniform flow. Illustrate an unsteady but uniform flow. Does unsteady uniform flow exist in practice? Describe the Reynolds’ Experiment and define the purpose of the Reynolds’ number. Differentiate between a pathline and a streamline. Explain why there cannot be flow across a streamtube. 8.11 SUMMARY This unit has introduced the student to the various terms used to describe the characteristics of fluids in motion. The student need to be familiar with the terms and their meanings, for these will eventually guide the assumptions that can be made during analysis of fluids in motion. The next unit, will now introduce the student to the first two principles used to mathematically analyse fluids in motion, the Principle of Continuity and the Conservation of Energy. Fluid Mechanics 135 Aug 2008 UNIT 9 PRINCIPLES OF CONSERVATION OF MASS & ENERGY Unit Structure 9.0 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 Overview Learning Objectives Introduction Continuity Mass Flow Rate Principles of Continuity Discharge and Mean Velocity Conservation of Energy Bernoulli’s Equation – Pipelines Hydraulic Grade Line Frictional Losses Activities Summary 9.0 OVERVIEW This unit introduces the student to two main principles used to analyse fluids in motion, the principles of Continuity and the principles of conservation of Energy. The third principle is the conservation of Momentum, which will be introduced at a later stage, in the second level of the course. These basic governing principles will always be used in the analysis of the simple or complex cases of fluids in motion, and hence, the need to understand and learn them. 9.1 LEARNING OBJECTIVES At the end of this unit, you should be able to do the following: 1. 2. Define Conservation of Mass and Conservation of Energy. Differentiate between the governing equation for conservation of mass for a compressible and an incompressible liquid. Fluid Mechanics 136 Aug 2008 3. 4. 5. 6. 7. 8. 9. Differentiate between mass flow rate and discharge or volume flow rate. Appreciate the application of Continuity equation in a branched pipeline. Derive Bernoulli’s equation. Explain the term Energy Head. Differentiate between Total Energy Head and Hydraulic Grade Line Modify Bernoulli’s equation to include frictional losses. Appreciate the application of Bernoulli’s equation. 9.2 INTRODUCTION Analysis of fluids in motion is no very much different from that of solids in motion. Because we are here dealing with a liquid there is some modifications which have to be considered during the analysis. The concepts described below are simple and easy to understand. The principle of conservation of mass is one concept which is used in most if not all the analysis of fluids in motion. The principle of conservation of Energy is used a lot in the analysis of fluids in motion. The most important step in this unit is to learn how to derive the Continuity and Bernoulli’s equation, to understand the meaning of each and every term in these equations and to get used to the units of each term. Fluid Mechanics 137 Aug 2008 9.3 CONTINUITY The concept of Continuity is best illustrated using the anology of a road junction (Figure 9.1). 8 3 B C 4 D A 15 Figure 9.1: Continuum analogy – cars If 15 cars pass by location A, and no car is allowed to stop along the road, then the sum of the total number of cars passing location B, C and D, will be 15. Similarly, if a given volume of liquid flows through a main pipeline (A) in unit time, and this pipe branches out, the sum of the volume of liquid flowing through the two branched pipes will be the same as that which was flowing through the main pipe (Figure 9.2). Fluid Mechanics 138 Aug 2008 7m /s 3m /s 3 3 A 10m /s 3 Figure 9.2: Continuum – flow through a branched pipeline 9.4 MASS FLOW RATE A simple way of measuring the flow of water through a pipe is to allow the water to collect in a bucket at the end of the pipe. The volume of water (V) collected in a given time (t) is noted. The volume of water collected in a given period of time can be converted into mass of water by multiplying the volume by density (V x ρ ). Hence Mass flow rate = m = (V ρ )/ t , and its units are kg/s. Fluid Mechanics 139 Aug 2008 9.5 PRINCIPLES OF CONTINUITY Consider a pipeline with varying diameter along its length (figure 9.3). The velocity of flow at section 1, is V1, where V1, can also be represented by the distance L1 per unit time. Applying the principle of conservation of mass, gives, mass flow per unit time at section 1 is equal to the mass flow per unit time at section 2. Mass flow per unit time at section 1 = (volume x density )/ time, or (cross sectional area of flow x horizontal distance moved my fluid in unit time x density) = A1 L1 ρ1 / t . Similarly the mass flow rate at section 2, will be given by A2 L2 ρ2 / t. Since in general liquids have very low degree of compressibility, they can be safely assumed to be incompressible, hence ρ1 = ρ2 = ρ. Thus, the mass flow rate per unit time, given by A1 L1 ρ1 / t, can be simplified to m = A1 V1 ρ. For continuity of flow, then A2 V2 ρ = A2 V2 ρ, which simplifies to: A1 V1 = A2 V2 = Q, where Q is referred to as the Volume Flow Rate (units are m3/s). Hence, Q = AV, volume flow rate equation, also known as the Discharge. 9.6 DISCHARGE & MEAN VELOCITY Consider the flow of water through a pipeline. The velocity of flow in any system is very much influenced by the surface in which it is in contact. The closer the liquid to the boundary the more significant the impact of the boundary of the liquid. Hence in a pipeline under laminar flow conditions, the maximum velocity is located at the centre. Fluid Mechanics 140 Aug 2008 Velocity flow profile V=0 V Pipeline running full Vmax = V V=0 Figure 9.4: Boundary effects on flow velocity in a pipeline Thus if the flow rate (Q) through the pipeline is known (collect water in a given time period), and the cross sectional area (A) of the pipe is known, then the mean velocity (Vmean) through the pipe is calculated using the discharge equation, Q. Hence, Vmean = Q / A. 9.7 CONSERVATION OF ENERGY In solid mechanics you have learnt about the meaning of Potential energy and Kinetic energy. When a body is located at a given height above a datum, it possesses potential energy by virtue of its position. When a body is moving with a given velocity, V, then this body will possess kinetic energy by virtue of its movement. The same concepts apply for a fluid. Consider Figure 9.5, which shows a small element of a fluid, characterised by pressure P, velocity V, cross sectional area, a, and mass M. Fluid Mechanics 141 Aug 2008 cross sectional area of flow, a P, V A A’ B flow direction Z weight, mg B’ A small fluid element from the main fluid body: Figure 9.5 – Derivation of Bernoulli’s equation 1. Potential energy: The elemental fluid is located at an elevation Z above the datum, hence the potential energy possessed by the elemental fluid is (mass x elevation x acceleration due to gravity) = m g z. 2. Kinetic energy: The elemental fluid is moving with a velocity V. Hence, kinetic energy possessed by the elemental fluid will be, ½ m V2. 3. Pressure energy: The fluid is also under pressure P. This pressure can be converted to a force, by multiplying it by the cross sectional area of flow, which gives, P a. The elemental fluid is moving with a velocity V. In unit time the force will have moved a distance L, which can also be given by the (volume of shaded element/cross sectional area of flow), or L = (mass/density)/a. L = m/ρ a Pressure Energy = Work done by this force = force x distance moved = ( P a) x m/ρ a Therefore, the Pressure energy = P m/ρ Fluid Mechanics 142 Aug 2008 Total Energy Contained by the small fluid element – Bernoulli’s equation: E = P m/ρ + ½ m V2 + m g z ρ Units: Joules Total energy contained by the small fluid element per unit mass, H: H = P/ρ g + V2/2g + Z ρ Units: m Owing to its units, this equation is also commonly known as the total energy Head. 9.8 BERNOULLI’S EQUATION – IN PIPELINES The meaning of Bernoulli’s equation is best illustrated in a system, and a pipeline running full is being considered in Figure 9.6. Assuming that the energy losses from section 1 to section 2 are negligible, this would mean that the total energy head at section 1 is equal to the total energy head at section 2. Usually this equation is applied at two different sections within a system, to work out missing information. Principle of Continuity tends to be used in most of the problems involving application of Energy principles. P2 V2 P1 z2 V1 z1 DATUM A fluid possesses energy by virtue of its position as well as by virtue of its movement. P1/ρg + V1 /2g + Z1 = P2/ρg + V2 /2g + Z2 2 2 Figure 9.6: Principle of conservation of Energy – Bernoulli’s Equation Fluid Mechanics 143 Aug 2008 9.9 HYDRAULIC GRADE LINE Hydraulic grade line is simply the sum of the Pressure head and the elevation head, and this is being illustrated in Figure 9.7. Figure 9.7 shows a reservoir located at a higher elevation Z1 feeding another one located below, at elevation Z2, both reservoirs being connected by a pipeline. At any point along the pipeline, if a piezometer is connected, water will rise into the piezometer tube, to a height equivalent to the pressure at that point (Unit 3 – section 3.4), points A and B in Figure 9.7. A line can then be drawn at points A and B, that join the sum of their pressure head and elevation head at respective points, this line is known as the hydraulic grade line. The Hydraulic Grade Line is always below the Total Energy Head line, and the difference is the velocity head. Total energy head line h1 A Velocity head Z1 Pressure head + Elevation head line commonly known as the Hydraulic Grade Line B h2 Z2 Datum Fluid Mechanics 144 Aug 2008 9.10 FRICTIONAL LOSSES In practice as liquid flows frictional forces have to be overcome, so some energy is lost. The Bernoulli’s equation has then to include the term for frictional losses, hf. P1/ρ g + V12/2g + Z1 = P2/ρ g + V22/2g + Z2 + hf ρ ρ In a pipeline connection, each time there is a feature which disturbs the regular flow pattern, frictional losses are high. Such features may be in the form of sudden change in diameter, bends, reducers, valves and connections to branched pipes. 9.11 8. 9. 10. ACTIVITIES Derive Continuity equation Derive Bernoulli’s equation Explain under what circumstances frictional losses cannot be ignored when applying Bernoulli’s equation. 11. How does Principle of conservation of energy different when it is being used to analyse solid mechanics and fluid mechanics. 9.12 SUMMARY This unit has introduced you to the basic principles which govern the analysis of fluids in motion. The next unit will show you how these principles are being used in practice in flow measurement devices to measure flow of liquid in both closed and open systems. Students are strongly advised to ensure that the concepts described in this unit are clear, for these will crop up in most of the remaining units of the subject, in all levels of the course. Fluid Mechanics 145 Aug 2008 UNIT 10 FLOW RATE MEASUREMENTS – ORIFICES & WEIRS Unit Structure 10.0 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 Overview Learning Objectives Introduction Small Orifice Vena Contracta Flow Measurement by an Orifice Coefficient of Contraction and Velocity Discharge Through a Large Orifice Velocity of Approach (V1) Flow Measurement Through a Rectangular Weir 10.10 Flow Measurement Through a Triangular weir 10.11 Flow Measurement Through a Trapezoidal Weir 10.12 Activities 10.13 Summary 10.14 Worked Examples 10.15 Tutorial 10.0 OVERVIEW In the previous unit, two of the main principles which form the basis of the analysis of fluids in motion were introduced. In this unit, the student will learn how to apply these principles to obtain further information about a system and also how to these principles are used in estimating flow rates of fluids in motion. Fluid Mechanics 146 Aug 2008 10.1 LEARNING OBJECTIVES At the end of this unit, you should be able to do the following: 10. 11. 12. Differentiate between a sharp edged orifice and a streamline office. To applying Bernoulli’s equation to derive the velocity of flow through a small orifice. To justify any assumption made in the derivation of the equation of flow velocity through an orifice. 13. 14. 15. To understand why a larger orifice is analysed in a different approach. To derive the equation governing flow of a fluid over a rectangular and a triangular weir To derive the equation from first principles for flow rate estimation using an irregular trapezoidal weir. 10.2 INTRODUCTION One of the many important characteristics of a fluid in motion is the rate at which it flows. In fluid mechanics practicals, the students will have learnt that to get an estimate of the flow rate of a fluid, fluid can be collected in a bucket for a given time and the flow rate worked out. However in a long complex pipe network this is not possible along the pipes. In such cases a different approach has to be adopted, and the applications of both Continuity equations and Energy equations, have proved to be useful tools for such estimates. Similarly when the flow rate of a river is needed, here also there are different approaches are available. A floating object can be timed over a distance and knowing the cross sectional area of the river, the flow rate of the river can be estimated. This is however a rough approach. Weirs are devises which have proved to be effective in such measurements. In this units, the student will be introduced to several devices which are used to measure flow rate in different situations, and with different accuracies. Fluid Mechanics 147 Aug 2008 10.3 SMALL ORIFICE A small orifice is simply a small circular opening most of the time located on the side of a reservoir or a container. A small orifice can also be located on the bottom of the container. The opening can either be sharp edge or streamline edged (Figure 10.1). The sharp edged opening gives rise to significant disturbances as compared to the streamline opening. Sharp edged opening Streamline opening A Small orifice B Figure 10.1: Small orifice The approach in estimating the flow through a small orifice is to apply the Continuity equation and Bernoulli’s equation, at two points where maximum information is present. This helps in the simplifying the general equations (Figure 10.2). Fluid Mechanics 148 Aug 2008 Enlargement – at orifice Figure 10.2 – Flow pattern through small orifice From Figure 10.2, it can be noted that if the pathway of a fluid particle is highlighted, it would be straight vertical line from the surface of the liquid and curved as it reached the opening. Thus a large flow area is suddenly constricted into a small opening. This sudden change in path gives rise to turbulence and hence heavy losses of energy. Though an orifice is a simple device to measure the flow rate, it does not offer much accuracy owing to the structure of the system. Fluid Mechanics 149 Aug 2008 10.4 VENA CONTRACTA The basic principle behind the application of Bernoulli’s Equation is to identify two points at which maximum information is known. In the case of an orifice, the first such point is at the surface of the liquid. At the surface we know that the pressure is zero gauge pressure or atmospheric pressure and that the velocity of flow is so small that it can safely be considered as being negligible. (Imagine Mare Aux Vacoas reservoir with a small circular opening on its side.) A second such point is just outside the opening of the orifice, since just outside the flow lines becomes straight and parallel as compared to the curved flow lines at the opening of the orifice. This point is also known as the Vena Contracta. The Vena Contracta is smaller in size than that of the orifice and it lies just outside the orifice. At the vena contracta, the pressure of the fluid is once again zero gauge pressure or atmospheric pressure (Figure 3). Enlargement – at orifice 1 vena contracta 2 Figure 10.3: Vena Contracta Fluid Mechanics 150 Aug 2008 10.5 FLOW MEASUREMENT BY AN ORIFICE Having identified the two reference points within the system, the next step is to apply Bernoulli’s equation and Continuity equation (Figure 10.4). 1 H 2 Datum P1/ρg + V12/2g + Z1 = P2/ρg + V22/2g + Z2 Where P1 = 0 gauge pressure, V1≅ 0, Z1= H P2= 0 gauge pressure, Z2= 0 being at datum Figure 10.4– Reference points for application of Bernoulli’s equation As mentioned in the section above, the pressure at points 1 and 2 are both zero gauge pressure, the velocity of flow at point 1 is zero and the elevation at point 2 is zero since in this analysis the datum has been drawn at point 2 itself. Similarly the elevation at point 1 is the height of liquid from the datum to the top water surface, which is equal to H. Fluid Mechanics 151 Aug 2008 1 H 2 Datum P1/ρg + V12/2g + Z1 = P2/ρg + V22/2g + Z2 ρ ρ 0 + 0 + H = 0 + V22/2g + 0 V2= (2gH)½ Figure 10.5– Governing equation for flow through an orifice Simplifying Bernoulli’s equation , yields the governing equation for the velocity of flow through an orifice, as being equal to √2gH (Figure 10.5). The flow rate through the orifice, will simply be the given by Continuity equation, where Q = A V (Figure 10.6). Note, since the reference point 2, was taken at the Vena Contracta position, then the cross sectional area of flow that should be considered, is the area of flow at the vena contracta. However, in practical situation, it is difficult to locate the vena contracta, let alone measure its cross section. Thus assuming that the cross sectional flow area at the vena contracta is not very different from the cross sectional area of flow at the orifice the cross sectional area of the orifice is used in the continuity equation. We end up with a Theoretical flow rate through the orifice. Fluid Mechanics 152 Aug 2008 1 H 2 Datum V2= (2gH)½ Applying Continuity Equation: Q = A V Q = A(vena contracta) x V2 Q = A(vena contracta) x (2gH)½ Q = Aorifice x (2gH)½ x Cd Where Cd = coefficient of discharge Figure 10.6– Flow rate through a small orifice Two assumptions have been made in the derivation of this theoretical velocity, first the velocity of flow at point 1 is negligible and secondly the cross sectional area of flow at the vena contracta is similar to that at the orifice. To cater for the errors likely to be induced by these assumptions, the theoretical discharge is multiplied by a factor of safety known as the Coefficient of Discharge, Cd. Cd is either equal to 1, if any assumptions made is true, or 0 is they are completely wrong. In the case of small orifice, Cd can lie between 0.5 to 0.6. Fluid Mechanics 153 Aug 2008 10.6 COEFFICIENTS OF CONTRACTION & VELOCITY In the case of an orifice, the coefficient of discharge can further be subdivided into two coefficients, the coefficient of velocity and the coefficient of contraction. Discharge through an orifice: Q = Aorifice x (2gH)½ x Cd Cc = coefficient of contraction = Area of orifice/area of vena contracta Cv= coefficient of velocity = velocity of flow at orifice/velocity of flow at vena contracta Cd= coefficient of Discharge = Cc x Cv Figure 10.7: Coefficients of contraction, velocity and discharge The coefficient of velocity as illustrated in Figure 10.7, simply relates the velocity at the orifice to the velocity of flow at the vena contracta. Similarly the coefficient of discharge relates the cross sectional area of flow at the orifice and that at the vena contracta. The product of Cc and Cv yields the coefficient of discharge, Cd. Fluid Mechanics 154 Aug 2008 10.7 DISCHARGE THROUGH A LARGE ORIFICE When the opening is small then the height of liquid causing flow above the centre of the orifice can be safely assumed to be H. However, the opening is large, this assumption may not necessarily hold true. Large orifice 1 2 Figure 10.8– Discharge through a large orifice The flow is more turbulent a large orifice as illustrated by the curved flow lines at the opening (Figure 10.8). Fluid Mechanics 155 Aug 2008 Velocity of flow through an orifice: V = (2gh) ½ 1 Since the orifice this time is large, it is not accurate enough to assume that h varies from 0 to H2. In this case the variable h in the velocity equation varies from h1 to h2. h1 h2 dh h dQ= A V dQ = B dh √2gh Q = ∫ B √2g h dh where the limits of h varies from h2 to h1. B Figure 10.9: Discharge through a large rectangular orifice In the case of a large orifice, then the opening is considered as being made up of a series of small orifices, through which the fluid will flow. Consider the case of a large rectangular opening (Figure 10.9). The opening will be assumed to be made up of a series of small openings, through which the velocity of flow will be equal to √2gh. The discharge is obtained by multiplying this flow velocity by the cross sectional area of flow of the small element, given by B dh. Finally to obtain the total discharge, this equation is integrated with limits of h varying from h2 to h1. 10.8 VELOCITY OF APPROACH (V1) As discussed in section 10.4, the velocity of flow at the surface of the container or reservoir, V1, is often safely assumed to be negligible and hence equal to zero in the analysis. This velocity is commonly known as the velocity of approach. As illustrated in Figure 10.10, when the cross sectional area of flow at the surface is much larger than the cross sectional area of the orifice, then the velocity of approach will be much smaller than the velocity at the orifice, in which case the velocity of approach can be safely assumed as being negligible. Fluid Mechanics 156 Aug 2008 1 P1/ρg + V12/2g + Z1 = P2/ρg + V22/2g + Z2 ρ ρ H Where P1 = 0 gauge pressure, 2 Datum V1≅ 0, Z1= H P2= 0 gauge pressure, Z2= 0 being at datum V1 – Velocity of approach, often safe to assume being so small that it is equal to zero. A 1V 1 = A 2V 2 A1>>>A2, hence V1