Contents Preface to third edition Acknowledgements Principal notation Introduction viii x xiii xvi 1 Hydrostatics 1 1.1 Fundamentals 2 1.2 Hydrostatic pressure and force 5 1.3 Force on a plane (flat), vertical immersed surface 9 1.4 Location of the resultant force on a vertical surface 11 1.5 Force on a plane, inclined immersed surface 15 1.6 Force on a curved immersed surface 18 1.7 Variation of pressure with direction and buoyancy 22 1.8 The hydrostatic equation 28 1.9 Stratified fluids 28 Summary 32 2 Pressure measurement 35 2.1 Fundamentals 35 2.2 Piezometers 36 2.3 A simple U-tube manometer 38 2.4 A differential U-tube manometer 40 2.5 The inverted U-tube differential manometer 43 2.6 Adjusting the sensitivity of a manometer 48
iv Contents PROOF 2.7 The Bourdon gauge 51 2.8 Surface tension 53 Summary 54 3 Stability of a floating body 57 3.1 Introduction 57 3.2 Factors affecting the stability of a floating body 60 3.3 Calculation of the metacentric height, M 63 3.4 Period of roll 69 Summary 71 4 Fluids in motion 74 4.1 Introduction to the fundamentals 74 4.2 Classifying various types of fluid flow 79 4.3 Visualising fluid flow 83 4.4 The continuity equation 86 4.5 Understanding the momentum equation 88 4.6 Applying the momentum equation 93 4.7 The energy (or Bernoulli) equation 102 4.8 Applying the energy equation 109 4.9 Drag and lift 115 4.10 Free and forced vortices 117 Summary 120 5 Flow measurement 123 5.1 Introduction 123 5.2 The Venturi meter 124 5.3 The Pitot tube 130 5.4 Small and large orifices 132 5.5 Discharge over a sharp crested weir 142 5.6 Calibration of flow measuring devices 152 5.7 Velocity meters 157 Summary 159
Contents v 6 Flow through pipelines 163 6.1 Introduction 164 6.2 Understanding reservoir pipeline flow 166 6.3 Parallel pipelines 175 6.4 Branching pipelines 178 6.5 The development of the pipe friction equations 183 6.6 Head losses at changes of section 201 Summary 206 7 Flow under a varying head time required to empty a reservoir 209 7.1 Introduction 210 7.2 Time to empty a reservoir of uniform cross-section 210 7.3 Time to empty a reservoir of varying cross-section 214 7.4 Flow between two tanks 221 Summary 223 8 Flow in open channels 225 8.1 Fundamentals 226 8.2 Discharge equations for uniform flow 231 8.3 Channel proportions for maximum discharge or velocity 237 8.4 Compound channels and the composite Manning s n 242 8.5 Environmentally acceptable channels 247 8.6 Specific energy and critical depth 247 8.7 Calculation of the critical flow conditions in any channel 257 8.8 Calculation of the critical flow in a trapezoidal channel 262 8.9 Calculation of the critical flow in a rectangular channel 264 8.10 Flow transitions 266 8.11 radually varying non-uniform flow 270 8.12 Surge waves in open channels 292 Summary 297 9 Hydraulic structures 300 9.1 Dams 300 9.2 Sluice gates and other control gates 314
vi Contents PROOF 9.3 Flow around bridge piers and through bridge waterways 319 9.4 Culverts 331 9.5 Broad crested and Crump weirs 341 9.6 Throated flumes 345 Summary 349 10 Dimensional analysis and hydraulic models 352 10.1 Units and dimensions 353 10.2 Dimensional homogeneity 354 10.3 Dimensional analysis using the Rayleigh method 355 10.4 Dimensional analysis using the Buckingham P theorem 359 10.5 Hydraulic models and similarity 364 Summary 374 11 Turbines and pumps 377 11.1 Introduction 377 11.2 Impulse turbines 381 11.3 Reaction turbines 392 11.4 Performance equations and characteristics of turbines 395 11.5 Rotodynamic pumps 398 11.6 Pump performance equations, affinity laws and specific speed 400 11.7 Pump selection for a particular duty 406 11.8 Avoiding problems with cavitation and surge 413 11.9 Introduction to the analysis of unsteady pipe flow 418 11.10 The ram pump 428 Summary 431 12 Introduction to engineering hydrology 435 12.1 The hydrological cycle 436 12.2 Humankind s intervention in the hydrological cycle 439 12.3 Precipitation 445 12.4 Evaporation, transpiration and evapotranspiration 456 12.5 Infiltration and percolation 459
Contents vii 12.6 Surface runoff 464 Summary 475 13 Applications of engineering hydrology 477 13.1 Predicting a catchment s response to rainfall 478 13.2 The unit hydrograph rainfall runoff model 482 13.3 Statistical analysis of riverflow data 492 13.4 Riverine and surface water flood risk management 507 13.5 Surface water sewer design using the modified rational method 519 13.6 Water supply reservoirs 527 13.7 roundwater 532 Summary 547 14 Sustainable Drainage Systems (SUDS) 551 14.1 Introduction 552 14.2 What do SUDS do, and why? 554 14.3 Design of SUDS 558 14.4 Potential problems with SUDS 566 Summary 567 Bibliography and references 569 Appendix 1 Derivation of equations 576 Appendix 2 Solutions to self test questions 592 Appendix 3 raph paper 617 Index 621
CHAPTER 1 Hydrostatics This chapter introduces some of the fundamental quantities involved in hydraulics, such as pressure, weight, force, mass density and relative density. It then considers the variation of pressure intensity with depth below the surface of a static liquid, and shows how the force on a submerged surface or body can be calculated. The principles outlined are used to calculate the hydrostatic forces on dams and lock gates, for example. These same principles are applied in Chapter 2 in connection with pressure measurement using piezometers and manometers, and in Chapter 3 to the analysis of floating bodies. Thus the sort of questions that are answered in this chapter are: What is meant by pressure? What is the difference between force and weight? What is the difference between mass and weight? How and why does pressure intensity vary with depth in a liquid? How can we calculate the pressure intensity at any depth? How can we calculate the force on a flat immersed surface, such as the face of a dam? How can the hydrostatic force be calculated when the immersed surface is curved? Does hydrostatic pressure act equally in all directions, and if it does why? How can the buoyancy force on a body be calculated? What do we do if the liquid is stratified with layers of different density? 1
2 Understanding Hydraulics PROOF 1.1 Fundamentals 1.1.1 Understanding pressure and force Have you ever asked yourself why a trainer will not damage a soft wooden floor, but a stiletto heel will? The answer is because the average pressure, P AV, exerted on the floor is determined by the weight of the person, W, and the area of contact, A, between the sole of the shoe and the floor. Thus: P AV = W A (1.1) So, because a trainer has a flat sole with a large area of contact, it exerts a relatively small pressure on the floor (Fig. 1.1). On the other hand, the sharp point of a stiletto means that much of the weight is transmitted to the floor over a small area, giving a large pressure. Similarly a drawing pin (or a thumb tack in American) creates a large, penetrative pressure by concentrating a small applied force at a sharp point. I understand that, but can you now tell me what is the difference between weight and force? The answer is basically none. Weight is simply one particular type of force, namely that resulting from gravitational attraction. So equation (1.1) can also be written as P AV = F/A, where F is the force. This can be rearranged to give: F = P AV A (1.2) The unit of force is the Newton (N), named after Sir Isaac Newton, so pressure has the units N/m 2. A Newton is defined as the force required to give a mass of 1kg an acceleration of 1m/s 2. Hence: Figure 1.1 Illustration of the pressure exerted on a floor by two types of shoe. The stiletto is the more damaging because the weight is distributed over a small area, so giving a relatively large pressure
Hydrostatics 3 Force = mass acceleration F = Ma (1.3) where M represents mass and a is the acceleration. For weight, W, which is the force caused by the acceleration due to gravity, g, this becomes: Weight = mass gravity W = Mg (1.4) On Earth, gravity, g, is usually taken as 9.81m/s 2. 1.1.2 Understanding the difference between mass and weight OK, so what is the essential difference between mass and weight, and why is it important? It is important to have a clear understanding of the difference between mass and weight, because without it you will make mistakes in your calculations. The essential difference is that mass represents the amount of matter in a body, which is constant, so mass stays the same everywhere in the universe, while weight varies according to the local value of gravity since W = Mg (equation (1.4) and Fig. 1.2). So what is mass density and weight density What is meant by relative density? And how heavy is water? Density, r, is the relationship between the mass, M, of a substance and its volume, V. Thus: r = MV (1.5) Figure 1.2 of gravity The concept of weight, which varies according to the local value
4 Understanding Hydraulics PROOF Box 1.1 Remember It is important to realise that water is heavy! Each cubic metre of water weighs 9.81 10 3 N, that is one tonne. Thus every cubic metre weighs about the same as a large car. Figure 1.3 Illustration of the weight of water The density of fresh water (r) is 1000kg/m 3. This can be thought of as the mass density of the water, since it gives the mass per unit volume. Alternatively, the weight (W) per unit volume may be quoted, which is the weight density, w (also called the specific weight). Using equations (1.4) and (1.5), weight density can be expressed in several ways: w = W V or w = Mg V or w = rg (1.6) Thus the weight density of fresh water is 1000 9.81N/m 3. Another term you may come across is the relative density (or specific gravity) of a liquid, s. This is the ratio of the density of a substance, r S, to the density of fresh water, r. Of course, the same value can be obtained by using the ratio of the weight densities (equation (1.6)), since g is the same for both substances. Thus: s = r r or s = w w S S (1.7) where w S is the weight density of the substance. Since s represents a ratio of the mass or weight of equal volumes of the two substances, it has no dimensions. For example, water has a relative density of 1.0 while mercury has a relative density of 13.6. Box 1.2 Using relative density It is important to remember that s usually has to be multiplied by the density of water before it can be used in your calculations, otherwise the answer you obtain will be wrong, both numerically and dimensionally. For example, the density of mercury (r M ) is 13.6 1000kg/m 3. Quoting the relative density as 13.6 is just a shorter and more convenient way of writing this.
Hydrostatics 5 1.1.3 An application of what you have learned so far the hydraulic jack You may not realise it, but you now have a sufficient understanding of hydrostatics to understand how a hydraulic jack works. The hydraulic jack uses two cylinders (Fig. 1.4), one with a large cross-sectional area (CSA), A, and one with a small area, a. By using a handle, or something similar, a small force, f, is applied to the piston in the small cylinder. From equation (1.2), it can be seen that this generates a pressure in the liquid of P AV = f /a. Now one of the properties of a liquid is that it transmits pressure equally in all directions (more of this later), so this means that the same pressure P AV acts over the whole cross-sectional area (A) of the large piston. As a result, the force exerted on the large piston is F = P AV A (equation (1.2)). Because A > a, the output force F > f, even though the pressure of the liquid is the same. Thus the jack acts as a kind of hydraulic amplifier. This simple but extremely useful effect can be used to lift weights of many tonnes while applying only a relatively small force to the input end of the jack. 1.2 Hydrostatic pressure and force Now let us try to determine how we can work out the hydrostatic force, F, on a dam, or on a lock gate, or on the flap gate at the end of a sewer. The term hydrostatic means, of course, that the liquid is not moving. Consequently there are no viscous or frictional resistance forces to worry about (see section 4.1). Also, in a stationary liquid there can be no shear forces, since this would imply movement. The water pressure must act at right angles to all surfaces with which the liquid comes into contact. If the pressure acted at any other angle to the surface, then there would Figure 1.4 A hydraulic jack. The hydraulic pressure that results from applying a small force to the small piston is transmitted to the large piston, so enabling a relatively heavy load to be raised
6 Understanding Hydraulics PROOF Figure 1.5 calculated Typical examples of situations where the hydrostatic force may have to be be a component of force along it which would cause the liquid to move. However, this component is zero when the pressure is normal to the surface since cos90 = 0. Hence in a static liquid the pressure acts at right angles to any surface. This fact comes in useful later. OK, so the pressure acts at 90 to the surface. Please can you now explain why a submarine can only dive to a certain depth, as in all those old war movies? The answer is quite simple. The pressure intensity increases with depth. Beyond a certain depth the water pressure would crush the hull of the submarine. But what causes the pressure, and how can you calculate what it is? After all, if you were in the submarine you would want to know, right? The weight of the water above the submarine causes the pressure. Remember, every cubic metre of fresh water equals 1 tonne, which is 9810N (that is rg N with r = 1000kg/m 3 and g = 9.81m/s 2 ). This makes it quite easy to calculate the pressure. Try thinking of it like this. Imagine a large body of fresh water. Then consider a column of the liquid with a plan area of 1m 2 extending from the surface all the way to the bottom, as in Fig. 1.6. Now, suppose we draw horizontal lines at one metre intervals from the surface, so that the column is effectively separated into cubes with a volume of 1m 3. Every cube weighs 9.81 10 3 N. Since the pressure on the base of each of the cubes is equal to the weight of all the cubes above it divided by 1m 2 (P AV = W/A), it can be seen that the pressure increases uniformly with depth. Similarly, if the column of liquid has a total depth, d, then the total weight of all the cubes is 9.81 10 3 dn. Dividing this by 1m 2 to obtain the pressure on the base of the column gives 9.81 10 3 dn/m 2. Therefore, at any depth, h, below the water surface the pressure is: P = rgh Nm 2 (1.8) Equation (1.8) shows that there is a linear relationship between pressure, or pressure intensity, and depth. This pressure depth relationship can be drawn graphically to obtain
Hydrostatics 7 Figure 1.6 depth Variation of pressure with a pressure intensity diagram like that in Fig. 1.7. This diagram shows the pressure intensity on a vertical surface that is immersed in a static liquid and which has the same height, h, as the depth of water. The arrows can be thought of as vectors: they are drawn at 90 to the surface indicating the direction in which the pressure acts, while the length of the arrow indicates the relative magnitude of the pressure intensity. When analysing a problem, a pressure intensity diagram is used to help visualise what is happening, while equation (1.8) provides the means to calculate the pressure intensity. The relationship described by equation (1.8) is very useful; it can be used to calculate the pressure at any known depth, or alternatively, to calculate the depth from a known pressure. The fact there is a precise relationship between pressure and depth forms the basis of many instruments that can be used to measure pressure, such as manometers, which are described in Chapter 2. Now one important point. Figure 1.7 only shows the pressure caused by the weight of the water. This is called the gauge pressure, and is Figure 1.7 A pressure intensity diagram corresponding to Fig. 1.6
8 Understanding Hydraulics PROOF Box 1.3 Visualising the size of units You can easily visualise a metre, because it is just over three feet in length, and, of course, you know how long a second is. You may also be aware that a kilogramme is about 2.2lb, that is about the equivalent of a bag of sugar. But do you know how large or small a Newton is? If you use equation (1.8) to work out the pressure at a depth of 0.3m of fresh water you get P = 1000 9.81 0.3 = 2943N/m 2. So every time you have a bath at home, parts of your body are being subjected to almost 3000N/m 2. It does not cause any discomfort, in fact you do not even notice. So you may deduce that a Newton is a relatively small unit of force. For this reason it is frequently not worthwhile quoting a value to less than a Newton (the exception being if you are dealing with very, very small values where accuracy may be affected by rounding off). the pressure most often used by engineers. For convenience, gauge pressure measures the pressure of the water relative to atmospheric pressure, that is it takes the pressure of the air around us as zero. Now in reality, the atmosphere exerts a pressure of about 101 10 3 N/m 2 on everything at sea level (this is equivalent to the pressure at the bottom of a column of water about 10.3m high, that is a head of 10.3m of water). So if we want to obtain the absolute pressure measured relative to an absolute vacuum, that is the total pressure exerted by both the water and the atmosphere, we have to add atmospheric pressure, P ATM, to the gauge pressure (Fig. 1.8). Thus the absolute pressure, P ABS, is: P = rgh+ P ABS ATM Nm 2 (1.9) A good way to think of this is that you can measure the height of a table top either from the floor, which is the most convenient way, or above sea level (ordnance datum). Similarly, it is more convenient to measure temperature above the freezing point of water than above absolute zero. Consequently in this book we will always use gauge pressures (unless stated otherwise). For future reference, note that under some circumstances, such as in pipelines, a pressure less than atmospheric may occur (Fig. 1.8). This is a negative gauge pressure, rgh, but equation (1.9) is still valid. Note also that if absolute pressure is used then the gauge pressure intensity diagram shown in Fig. 1.7 will have to have P ATM added to it, as shown in Fig. 1.9. Now try Self Test Question 1.1. A short guide solution is given in Appendix 2, if you need it. SELF TEST QUESTION 1.1 Oil with a weight density, w O, of 7850N/m 3 is contained in a vertically sided, rectangular tank which is 2.0m long and 1.0m wide. The depth of oil in the tank is 0.6m. (a) What is the gauge pressure on the bottom of the tank in N/m 2? (b) What is the weight of the oil in the tank?
Hydrostatics 9 Figure 1.8 Relationship between gauge pressure and absolute pressure Figure 1.9 Pressure intensity diagram including atmospheric pressure (c) If the bottom of the tank is resting not flat on the ground but on two pieces of timber running the width of the tank, so that each piece of timber has an area of contact with the tank of 1.0m 0.1m, what is the pressure on the timber? 1.3 Force on a plane (flat), vertical immersed surface How do you work out the force on something as a result of the hydrostatic pressure? Say, something like a rectangular gate at the end of a sewer or culvert? OK, there are two thing to remember. First of all, equation (1.2) tells us that F = P AV A, so a force is a pressure multiplied by an area. However, the second thing we have to remember is that the pressure varies with depth. So, on a vertical surface such as the gate in Fig. 1.10, the pressure at the top of the gate is rgh 1. At the bottom of the gate the pressure is rgh 2. Hence the average pressure on the gate is P AV = (rgh 1 + rgh 2 )/2. Now if we multiply this by the area of the gate in contact with the water, A, we get the force, F: F = rg[ ( h1 + h2) 2] A (1.10) For a rectangle, (h 1 + h 2 )/2 is the depth to the centre of the area, that is the vertical depth to the centroid,, of the immersed surface. This depth is represented by h, so the expression for the resultant hydrostatic force, F, becomes: F = rgh A (1.11) This equation can be applied to surfaces of any shape. For geometrical shapes other than a rectangle, the depth to the centroid can be found from Table 1.1. For the full derivation of equation (1.11), see Proof 1.1 in Appendix 1.
10 Understanding Hydraulics PROOF Figure 1.10 A vertical gate at the end of a sewer which discharges to a river. The gate hangs from a hinge at the top: (a) side view, (b) front view, (c) pressure intensity diagram. Note that only the part of the pressure intensity diagram at the same depth as the gate contributes to the hydrostatic force acting on it Table 1.1 eometrical properties of some simple figures Shape Dimensions Location of the Second moment centroid, of area, I Rectangle breadth L D/2 from base LD 3 /12 height D Triangle base length L D/3 from base LD 3 /36 height D Circle radius R centre of the circle pr 4 /4 Semicircle radius R 4R/3p from base 0.1102R 4 The next paragraph can be helpful in some circumstances, since it reconciles what can appear to be different ways to solve a particular problem. However, you may omit it the first time you read the chapter, or if it confuses you. From equation (1.10), the resultant force, F = average pressure intensity area of the immersed surface (A). For simple, flat surfaces like that in Fig. 1.10, the average pressure intensity is (rgh 1 + rgh 2 )/2. If A = DL, then equation (1.10) can be written as F = rg[(h 1 + h 2 )/2]DL. The same expression can be obtained by calculating the area of the trapezoidal pressure intensity diagram in contact with the gate, rg[(h 1 + h 2 )/2]D and multiplying by the length of the gate, L. This can sometimes provide a useful check that what you are doing is correct, or a means of remembering the equation. However, your best approach initially is usually to go straight to equation (1.11).
Hydrostatics 11 Box 1.4 Remember Whenever you are faced with calculating the horizontal hydrostatic force on a plane, vertical immersed surface, the equation F = rgh A is the one to use. This simple equation can solve a lot of problems. We will also use it later on when we progress to the force on inclined and curved immersed surfaces. Remember that A is the area of the immersed surface in contact with the liquid. 1.4 Location of the resultant force on a vertical surface How do you know where the resultant force, F, acts? I assume that there must be some way of working it out? Yes, there is a way of calculating where the resultant force acts, and normally you would work this out at the same time as the magnitude of the force itself. However, the proof is a bit complicated, so I have put it in Appendix 1 (the second half of Proof 1.1). You can go through it later if you want to. For the time being, though, let us try to deduce something about where the force must act. Consider the dam in Fig. 1.11. In this case the pressure intensity diagram is triangular, since the gauge pressure varies from zero (atmospheric pressure) at the surface to rgh at the bottom. The average pressure intensity on the dam is therefore (0 + rgh)/2 or rgh/2. This pressure occurs at, half way between the water surface and the bottom of the dam. Figure 1.11 Pressure intensity on a dam. is the centroid of the wetted area, P is the centre of pressure where the resultant force acts But where would the resultant force act? At, half way down? Above? Below? Can you deduce where it would be? Think of it this way. The resultant force on the dam is the result of the average pressure intensity acting over the area of the dam face in contact with the water. The longer the arrows of the pressure intensity diagram, the greater the pressure. The larger the area of the pressure intensity diagram, the greater the force.
12 Understanding Hydraulics PROOF Box 1.5 Note that the centre of pressure, P, is always below the centroid,, of the surface in contact with the water. In many problems it is not obvious where P is located, so this has to be calculated using equation (1.12). However, as the depth of immersion of the surface increases, P moves closer to. This is apparent from equation (1.12): the distance between P and is (h P - h ). If A and I have constant values, then the equation can be rearranged as (h P - h ) = C/h where C represents the value of the constants. Thus (h P - h ) decreases as h increases. Look at the triangular area that forms the top half of the pressure intensity diagram, and compare it with the area of the trapezoidal bottom half. The area of the bottom part of the diagram is much larger, indicating that the resultant force would act below half depth. In fact, the resultant force acts horizontally through the centroid of the pressure intensity diagram. For the triangular pressure intensity diagram in Fig. 1.11, this is located at h/3 from the base (but note that this is only the case when the pressure intensity diagram is triangular). The point, P, at which the resultant force acts is called the centre of pressure (Fig. 1.11). With more complex problems, like that in Fig. 1.10, there is no simple rule to give the location of P, but if h P is the vertical depth to the centre of pressure then this can be calculated from: hp = ( I Ah)+ h (1.12) where the value in the brackets gives the vertical distance of P below the vertical depth to the centroid of the surface, h. The appropriate expression for the second moment of area calculated about an axis through the centroid, I, can be found from Table 1.1. For a rectangle I = LD 3 /12, where L is the length of the body and D its height. A is the surface area of the body. The derivation of equation (1.12) can be found in Appendix 1. Examples 1.1 and 1.2 show how equations (1.11) and (1.12) are used to solve a couple of typical problems, one involving the flap gate at the end of a sewer and the other a lock gate. Study these carefully and then try Self Test Question 1.2 (a short solution is given in Appendix 2). SELF TEST QUESTION 1.2 A rectangular culvert (a large pipe) 1.8m wide by 1.0m high discharges to a river. At the end of the culvert is a rectangular gate which seals off the culvert when the river is in flood (as in Fig. 1.10). The gate hangs vertically from hinges at the top. If the flood level in the river rises to 1.9m above the top of the gate, calculate the magnitude and location of the resultant hydrostatic force on the gate caused by the water in the river. EXAMPLE 1.1 A rectangular gate is 2m wide and 3m high. It hangs vertically with its top edge 1m below the water surface. (a) Calculate the pressure at the bottom of the gate. (b) Calculate the
Hydrostatics 13 resultant hydrostatic force on the gate. (c) Determine the depth at which the resultant force acts. (a) From equation (1.8), P = rgh Therefore P = 1000 9. 81 ( 3 + 1) = 39. 24 10 3 Nm 2 (b) From equation (1.11), F = rgh A Figure 1.12 Now where h = 1+( 3 2) = 2. 50m A = 2 3= 6m Thus F = 1000 9. 81 2. 50 6 3 = 147. 15 10 N (c) From equation (1.12) h = ( I Ah )+ h P = LD 12 = 2 3 12 = 4. 50m 3 3 4 I Aand h are as above so h = ( 4506. 250. )+ 250. P = 280. m 2 EXAMPLE 1.2 A lock on a canal is sealed by a gate that is 3.0m wide. The gate is perpendicular to the sides of the lock. When the lock is used there is water on one side of the gate to a depth of 3.5m, and 2.0m on the other side. (a) What is the hydrostatic force of the two sides of the gate? (b) At what height from the bed do the two forces act? (c) What is the magnitude of the overall resultant hydrostatic force on the gate and at what height does it act? (a) Using F = rgh A F1 = 1000 9. 81 ( 3. 5 2) ( 3. 5 3. 0) 3 = 180. 26 10 N F2 = 1000 9. 81 ( 2. 0 2) ( 2. 0 3. 0) 3 = 58. 86 10 N (b) Since both pressure intensity diagrams are triangular, both forces act at onethird depth from the bed: Y1 = 353. = 117. m Y2 = 203. = 067. m (c) Overall resultant force F R = F 1 - F 2 Figure 1.13 F R = 121.40 10 3 N Taking moments about O to find the height, Y R, of the resultant: 121.40 10 3 Y R = 180.26 10 3 1.17-58.86 10 3 0.67 Y R = 1.41m above the bed.
14 Understanding Hydraulics PROOF The value of Y R obtained in part (c) of the above example may have surprised you. Possibly you expected Y R to be somewhere between 0.67 m and 1.17 m, whereas it is actually 1.41m. This is a situation where the pressure intensity diagrams (which are not really needed to conduct the calculations) can be used to visualise what is happening. In Fig. 1.13 the slope of the two pressure intensity triangles is the same, since the water has the same density on both sides of the gate. Thus if the triangle on the right is subtracted from the triangle on the left, the Figure 1.14 Net pressure intensity result is as in Fig. 1.14. This is the net pressure intensity on the gate. The diagram is diagram for Example 1.2 more rectangular than either of the triangles so, employing a similar argument to that used with Fig. 1.11, this indicates that Y R would be higher above the base than either Y 1 or Y 2. Figure 1.15 The dam on the bottom left of the photograph is holding back a considerable quantity of water. The force exerted by the water on the structure must be calculated before the dam can be designed. Many lay people believe, incorrectly, that the greater the volume of water stored behind the dam, the larger the force on the structure. This is not the case. Equation (1.8) indicates that the pressure on the dam is related to the depth of water, while the force is the product of the average pressure and the area of the dam in contact with the water (equation (1.2))
Hydrostatics 15 1.5 Force on a plane, inclined immersed surface I understand how to work out the force on a flat vertical surface, but how about one that is inclined at an angle to the water surface? Surely this is much more difficult? The answer is no. The calculations are still very simple and almost identical to those above. There are three things that you should remember when analysing these situations: (1) The resultant force acts at right angles to the immersed surface. (2) The hydrostatic pressure on the inclined surface is still caused only by the weight of water above it, so P = rgh. (3) When calculating the location of the resultant force on an inclined surface, always use equation (1.13) (never equation (1.12), see below). To illustrate simply that the resultant force can be calculated in the same way as for a vertical surface, consider this. The pressure at the top of the rectangular, inclined surface in Fig. 1.16a is rgh 1 while that at the bottom is rgh 2. Thus the average pressure intensity on the surface is rg(h 1 + h 2 )/2, or rgh since h = (h 1 + h 2 )/2. The resultant force is the average pressure intensity multiplied by the area of the surface, and since the pressure acts at right angles to the inclined surface the actual area, A, should be used. Thus F = rgh A, as in equation (1.11). Note that the inclination of the surface is automatically taken into account by the value of h. For example, if h 1 in Fig. 1.16a is fixed, and the surface rotated upwards about its top edge, then h will decrease so that h = h 1 when it is horizontal. Similarly, the maximum possible value of h would be obtained when the surface is vertical. One other important point, the resultant force on the inclined surface, F, has components in both the vertical and horizontal directions. These can be calculated separately, as in section 1.6 and Example 1.4, but the procedure outlined above is quicker for flat (plane) surfaces. To calculate the location of the resultant force, the following equation should be used: Figure 1.16 (a) Force on an inclined surface. (b) When the surface is inclined always use the dimensions L and L P with equation (1.13) (never the vertical dimensions h and h P with equation (1.12))
16 Understanding Hydraulics PROOF Box 1.6 Using equations (1.12) and (1.13) Remember that when you have an inclined surface, always use equation (1.13) to find the location of the resultant force. You can then calculate the vertical depth of the centre of pressure, P, below the surface from L P if you want to (see Example 1.3). Never try to do this by using equation (1.12) instead of equation (1.13). The reason for this is that I is calculated in the plane of the surface. For example, with a rectangular inclined surface, I is still taken as LD 3 /12 where D is the actual inclined dimension of the surface, so the remainder of the terms in equation (1.13) must have the same orientation for consistency (see the derivation of the equation in Appendix 1). The same argument applies to vertical surfaces and equation (1.12). LP = ( I AL)+ L (1.13) This is similar to equation (1.12), but the inclined lengths, L P and L, are used to denote the location of the centre of pressure and centroid of surface (Fig. 1.16b), not the vertical depths. EXAMPLE 1.3 A sewer discharges to a river. At the end of the sewer is a circular gate with a diameter (D) of 0.6m. The gate is inclined at an angle of 45 to the water surface. The top edge of the gate is 1.0m below the surface. Calculate (a) the resultant force on the gate caused by the water in the river, (b) the vertical depth from the water surface to the centre of pressure. (a) Vertical height of gate = 0.6sin45 = 0.424m Vertical depth to = h = 1.000 + 0.424/2 = 1.212m Area of gate, A = pd 2 /4 = p0.6 2 /4 = 0.283m 2 F = rgh A = 1000 9.81 1.212 0.283 = 3365N Figure 1.17 An inclined, circular gate at the end of a sewer
Hydrostatics 17 (b) Slope length to, L = 1.212/sin45 = 1.714m For a circle (Table 1.1) I = pr 4 /4 = p(0.3) 4 /4 = 0.0064m 4 L = ( I AL )+ L = ( 0. 0064 0. 283 1. 714)+ 1. 714 = 1. 727 P m Vertical depth to P, h P = L P sin45 = 1.727sin45 = 1.221m Figure 1.18 This vertical lift gate on the Old Bedford River provides another example of where the engineer may be required to calculate the resultant hydrostatic force. If the horizontal force is large it may be difficult for a vertical lift gate to slide up and down, the gate being pushed hard against the guide channels. In the fens of East Anglia much of the drainage is controlled by using pumps and sluice gates like the one above
18 Understanding Hydraulics PROOF 1.6 Force on a curved immersed surface I suppose that you are now going to tell me that working out the force on a curved surface is just as easy as calculating the force on a flat or inclined surface? Well, the calculations are perhaps a little longer, but no more difficult. Let me clarify this by breaking the analysis of the force on an immersed curved surface down into steps. (1) The resultant force (F) acts at right angles to the curved surface. This force can be thought of as having both a horizontal (F H ) and a vertical (F V ) component (Fig. 1.19). (2) To calculate the horizontal component of the resultant force (F H ), project the curved surface onto a vertical plane, as in Fig. 1.20. This effectively is what you would see if you looked at the curved surface from the front. Calculate the force on this projected vertical surface as you would any other vertical surface using F H = rgh A, where A is the area of the projected vertical surface (not the area of the actual curved surface). (3) Calculate the vertical component of the resultant force (Fig. 1.21) by evaluating the weight of the volume (V) of water above the curved surface, that is: F V = rgv (4) The resultant force, F, is given by: ( H V ) 2 2 1 2 F = F + F (5) The direction of the resultant force (Fig. 1.22) can be found from: (1.14) (1.15) tanf = F V F H (1.16) This gives the angle, f, of the resultant to the horizontal. Remember, the resultant also acts at 90 to the curved surface, so it passes through the centre of curvature (for example, the centre of the circle of which the surface is a part). Figure 1.19 Pressure intensity on a curved surface. F passes through the centre of curvature, C Figure 1.20 Projection of the curved surface onto a vertical plane
Hydrostatics 19 Figure 1.21 The vertical component of force, F V, caused by the weight of water above the surface Figure 1.22 The direction of the resultant force, F, which must also pass through C (6) The above steps enable the resultant force on the upper side of the surface to be calculated. Always remember that there is an equal and opposite force acting on the other side of the surface. This fact comes in useful later, because it is always easier to calculate the force on the upper surface, even if this is not the surface in contact with the water. EXAMPLE 1.4 Step 1 Step 2 Step 3 A surface consists of a quarter of a circle of radius 2.0m (Fig. 1.23). It is located with its top edge 1.5 m below the water surface. Calculate the magnitude and direction of the resultant force on the upper surface. Project the curved surface onto a vertical plane and calculate F H F H = rgh A where A is the area of the projected vertical surface. Since the length of the gate is not given, calculate the force per metre length with L = 1.0m. Thus A = 2 1.0 = 2.0m 2 per metre length The value of h is that for the projected vertical surface: h = 1.5 + (2.0/2) = 2.5m. F H Calculate F V from the weight of water above the surface F V = rgv where V is the volume of water above the curved surface. Again using a 1m length: 2 3 V = ( 14 p 20. 10. )+( 20. 15. 10. ) = 614. m per metre length F V = 1000 9. 81 2. 5 2. 0 = 49. 05 10 3 Nm 3 = 1000 9. 81 6. 14 = 60. 23 10 Nm. Calculate the magnitude and direction of the resultant force ( H V ) = + F = F + F 10 ( 49. 05 60. 23 ) = 77. 68 10. 2 2 12 3 2 2 12 3 Nm = - 1( F F ) = - 1 f tan tan ( 60. 23 49. 05) = 50. 8. V H The resultant passes through the centre of curvature, C, at an angle of 50.8.
20 Understanding Hydraulics PROOF Figure 1.23 SELF TEST QUESTION 1.3 An open tank which is 4.0m wide at the top contains oil to a depth of 3.4m as shown in Fig. 1.24. The bottom part of the tank has curved sides which have to be bolted on. To enable the force on the bolts to be determined, calculate the magnitude of the resultant hydrostatic force (per metre length) on the curved surfaces and its angle to the horizontal. The curved sections are a quarter of a circle of 1.5m radius, and the oil has a relative density of 0.8. Figure 1.24 Tank for Self Test Question 1.3 I understand Example 1.4, but when you described the steps used to analyse the force on a curved surface, in point 6 you said something about always analysing the upper side of the surface. You said that we should do this even if the upper side of the surface was not in contact with the water. How can this be right? No water, no hydrostatic force I would have thought. I suppose this is one of the tricks you have to learn to make hydraulics easy. Think of it like this. The curved surface in Fig. 1.25 is an imaginary one, drawn in a large body of static liquid. Now it is possible to calculate the force on the upper side of this imaginary surface
Hydrostatics 21 using the same procedure as in Example 1.4. However, the surface is only imaginary, so what resists this force? Something must because the liquid is static, that is not moving. The answer is that there is an equal and opposite force acting on the underside of the imaginary surface, so that this balances the force on the top. It does not matter which force you calculate, because they are numerically equal, but it is easier to calculate that on the upper surface. The same is true with real surfaces. Remember this when you encounter Figure 1.25 Equal and opposite forces on a surface problems like Example 1.5 with air on the upper surface and water underneath. Something to note from Example 1.5 is that the vertical component of the resultant force acts upwards, which means that it is a buoyancy force. Sometimes there is a tendency to think of a buoyancy force as being different from the hydrostatic force, but in fact they are the same thing. The buoyancy force on a body, such as a ship, is the result of the hydrostatic pressure acting on the body. This will be explored in more detail in section 1.7. EXAMPLE 1.5 Step 1 A radial gate whose face is part of a circle of radius 5.0m holds back water as shown in Fig. 1.26. The sector of the circle represented by the gate has an angle of 30 at its centre. Water stands to a depth of 2.0m above the top of the upstream face of the gate. The other side of the gate is open to the atmosphere. Determine the magnitude and direction of the resultant hydrostatic force. The gate is 3.5m long. Project the curved surface onto a vertical plane and calculate F H Vertical height of projection = BC = 5.0cos60 = 2.5m. Figure 1.26
22 Understanding Hydraulics PROOF h = 20. +( 252. ) = 325. m, and A = 25. 35. = 875. m 2. F H = rgh A = 1000 9.81 3.25 8.75 = 278.97 10 3 N. Step 2 Step 3 Calculate the vertical component, F V, from the weight of water above the surface In this case calculate the weight of water that would be above the gate if it was not there, that is the weight of the water displaced by the gate. This is shown in the diagram as AEFH. The width of this area, DE, can be calculated as follows: AB = 5.0sin60 = 4.33m, so DE = 5.00-4.33 = 0.67m. The area of ADE (and subsequently AEFH) can be found using geometry, as follows. Area sector ACE = (30/360)ths of a 5.0m radius circle = (30/360)p5.0 2 = 6.54m 2. Area triangle ACD = (1/2) 4.33 2.5 = 5.41m 2. Area ADE = 6.54-5.41 = 1.13m 2. Therefore, the total area AEFH = 1.13 + (0.67 2.00) = 2.47m 2. Volume of water displaced, V = 2.47 3.5 = 8.65m 3. F V = rgv = 1000 9.81 8.65 = 84.86 10 3 N. Calculate the magnitude and direction of the resultant force ( ) = + F = F + F 10 ( 278. 97 84. 86 ) = 291. 59 10 2 2 12 H V 3 2 2 12 3 N. -1 - f = tan ( FV FH) = tan 1 ( 84. 86 278. 97) = 16. 9 Resultant acts at 16.9 to the horizontal passing upwards through the centre of curvature, C. 1.7 Variation of pressure with direction and buoyancy We have already discussed the fact that hydrostatic pressure acts at right angles to any surface immersed in it, so it follows that on the underside of a horizontal surface the resultant force is acting vertically upwards. This is a buoyancy force, and it is caused simply by the hydrostatic pressure on the surface. Try thinking it through like this. Imagine a sphere some distance below the water surface as in Fig. 1.27. The hydrostatic pressure acts at 90 to the surface of the sphere. Looking at this two-dimensionally, as in the diagram, then the smallest pressure intensity is rgh 1 at the top, and the largest is rgh 2 at the bottom. Now, consider what would happen if the diameter of the sphere gradually decreased so that the difference between h 1 and h 2 decreased. This h 3 would cause the two pressures rgh 1 and rgh 2 to become closer numerically. If the diameter of the sphere continued to decrease until it Figure 1.27 Pressure on a sphere
Hydrostatics 23 became infinitesimally small then the difference between h 1 and h 2 would be negligible so that rgh 1 = rgh 2. By the same argument, the pressure intensity in any other direction, such as rgh 3 acting horizontally, would also have the same value (see Proof 1.2, Appendix 1). Thus the pressure at a point in a static liquid acts equally in all directions, up, down, sideways or whatever. That s all very interesting, but does it have any practical purpose, and how can I work out the value of the buoyancy force? Yes it has a practical purpose, and working out the value of the buoyancy force is quite easy. In fact you can do so using what you have already learnt. Let me illustrate by using a similar situation to the sphere in Fig. 1.27, but this time we will make the body a cube because it simplifies the calculations. The cube is shown in Fig. 1.28. The pressure intensities on the vertical sides cancel each other out, so only the pressure acting on the top and bottom faces need be considered. Let the area of each face of the cube be A. Then: Assuming the top and bottom faces are in a horizontal plane, then the pressure is constant over the face so: Pressure on the top face = rgh 1 Pressure on the bottom face = rgh 2 The force on the face is equal to the pressure multiplied by the area of the face, A. So: Force on the top face = rgh 1 A Force on the bottom face = rgh 2 A Since h 2 > h 1 there will be a net force acting vertically upwards, F. This is: F = rgh A - rgh A = rg( h -h ) A 2 1 2 1 Now (h 2 - h 1 )A is the volume of the cube, V, so: F = rgv (1.14) Figure 1.28 Buoyancy force, F Box 1.7 Remember The buoyancy force, F, acts vertically upwards through the centre of gravity of the displaced liquid (such as the centre of the cube). The point at which F acts is called the centre of buoyancy, B. The force, F, is equal to the weight of the volume of liquid displaced by the body, that is rgv. This is known as Archimedes Principle. Now go back and look at Step 2 of Example 1.5. You should be able to see that a buoyancy force is just the vertical force caused by hydrostatic pressure. See also Chapter 3 and Box 3.1.
24 Understanding Hydraulics PROOF When we analysed the buoyancy force on the cube in Fig. 1.28 we only considered the hydrostatic forces acting vertically on it. The weight of the cube was irrelevant. However, if we wanted to know whether or not the completely immersed cube would float or sink, we would have to compare the weight of the cube (W ) with the buoyancy force (F), remembering that weight is a force. W = weight density of cube material volume = r S gv N acting vertically downwards. F = weight of liquid displaced by the cube = rgv N acting vertically upwards. Since g and V are the same, it follows that if the density of the substance, r S, that forms the cube is greater than the density of the liquid, r, then the cube would sink (W > F). Conversely, if r S < r, then the cube would float (F > W). If r S = r then the cube has neutral buoyancy and would neither float nor sink, but would stay at whatever depth it was located (F = W). The analysis above explains why a concrete or steel cube would sink, and a cork or polystyrene cube would float. However, this assumes that the cube is solid. If the cube was hollow, its average density would have to be used in the calculations, not the density of the material from which it was made. Submarines provide an interesting example, because they must be able to sink and, more importantly, rise to the surface again. This can be achieved by adjusting the average density of the submarine, by changing its weight by admitting or expelling water from tanks on the outside of the hull. Floating bodies, such as ships and the pontoon in Example 1.7, are quite easy to analyse. If the depth of immersion is constant, then obviously W and F in Fig. 1.29 are exactly equal (otherwise the body would move up or down). Hence the starting point for many calculations involving floating bodies is: Figure 1.29 Floating body or W = F W = rgv (1.17) Thus a floating body of weight W displaces a volume of water (V) that has a weight (rgv) equal to its own. Since W = Mg this can also be written as: or Mg = rgv M = rv (1.18) (1.19) Therefore it is also true to say that a floating body of mass M displaces a volume of water (V) that has a mass (rv) equal to its own. Of course, equation (1.19) is a rearrangement of equation (1.5). Remember to use W with the weight density (rg) and M with the mass density ( r). Typically the body s weight or mass is known, so the relationships above allow the volume of water (V) displaced by a floating body to be calculated. Then for pontoons which are rectangular in plan and cross-section like those in Fig. 3.1: depth of imersion = V/plan area (1.20) By now it should be apparent that a solid steel cube sinks, but a ship made from steel plates floats because it is hollow and can displace a much larger volume of water (V) that has a
Hydrostatics 25 Figure 1.30 Lock gates provide another example of where it may be necessary to calculate hydrostatic forces. The buoyancy, depth of immersion and freeboard (the distance from the deck to the waterline) of the barge may also be the subject of an engineer s calculations mass (or weight) equal to that of the ship. This is why we say that a ship has a displacement of 10000 tonnes, for instance. When W = F the depth of immersion is constant, but if W is increased by adding cargo the ship settles deeper in the water, increasing its displacement and consequently F, until W = F again. From 1876 onwards, British ships have had a Plimsoll line painted on their hull to indicate the maximum safe loading limit. Since the density of water changes according to temperature and salt content, the Plimsoll line includes marks for sea or fresh water, winter or summer, in tropical or northern waters.
26 Understanding Hydraulics PROOF Box 1.8 Try this amaze your friends et an empty fizzy drink bottle, fill it completely with water and put a sachet of ketchup in it (Fig. 1.31a). You need one that just floats, so you may have to try a few different types until you find one that works. Now challenge your friends to concentrate their minds and use the power of thought to make the sachet sink. Unless they really do have telekinetic powers, they won t be able to do it of course. Now here s the trick. When it is your turn, make sure you have your hands around the bottle, and gently squeeze it. Try to disguise the fact you are doing this. If you squeeze hard enough the sachet will sink, and you can claim to have a better brain than all of your friends combined. The reason the sachet sinks is as follows. A body in water has two forces acting on it: its weight (W) acting vertically down and the buoyancy force (F = rgv ) acting vertically up. The weight of the sachet cannot change, so W is constant. However, F depends upon the volume (V ) of water displaced by the sachet. When you squeeze the bottle you are exerting pressure on the water inside. The water is incompressible, but the air in the sachet can be compressed. So by compressing the air, V is reduced and so is F. When W > F the sachet sinks. When you stop squeezing F > W so the sachet rises. Human divers can control their buoyancy and move up and down like this, either by inflating or deflating their dry suits or by controlling the amount of air in their lungs. Usually a Cartesian diver consists of a small length of open ended glass tubing with a bubble at one end (Fig. 1.31b). It can be used instead of the sachet and works in the same way. W Air (a) Figure 1.31 (a) Alternative Cartesian diver using a sachet of sauce. Squeezing and releasing the bottle makes the diver sink and then rise. (b) Conventional glass diver F (b)