Hamill 2011 - Solutions Chapter 01

1HYDROSTATICS 1.1 Define clearly what is meant by the following, and give the appropriate units in each case: (a) pressure; (b) force; (c) weight; (d) gravity; (e) mass; (f) mass density (g) weight density; (h) relative density; (i) hydrostatic pressure; (j) buoyancy force. (a) Average pressure, PAV , is defined by equation 1.1 as: PAV = F/A where F is a force (N) and A is the area (m2) over which the force acts. Thus pressure is the force per unit area with the units N/m2. (b) Force is the product of a mass (M) and an acceleration (a). Equation 1.3 shows this relationship as: F = Ma with M in kg and a in m/s2. Force is measured in Newtons, with a Newton being defined as the force required to give a mass of 1 kg and acceleration of 1 m/s2. F = Ma is Newton’s Second Law of Motion. (c) Weight (W) is a force, namely the force on a body when acted upon by gravity. Thus weight and force have the same units, that is the Newton (N). F = Ma can be written as W = Mg, since weight is a force and g is the acceleration due to gravity. (d) Gravity, g, is the force of attraction that tends to move bodies towards the centre of a celestial body, such as the earth or the moon. Bodies subject to gravity do not fall towards the celestial centre at a constant speed, but at a progessively increasing velocity. In other words they accelerate, so gravity is an acceleration that is measured in m/s2. A familiar definition of acceleration is the time in seconds required for a car to go from 0 to 60 miles per hour. This is the time required for a change in velocity to occur, so another definition of acceleration is the rate of change of velocity. Gravity is not constant. It varies according to the size of the celestial body and the distance from its centre. However, on earth the variations are relatively small so the value of gravity is usually assumed to be constant at 9.81 m/s2. The moon is smaller that the earth, and its gravitational acceleration is about 1.6 m/s2. Jupiter is larger than the earth and has an average gravitational acceleration of about 26.9 m/s2. Because Jupiter is not a perfect sphere the value varies according to location. (e) Mass is a universal constant. It is a physical quantity expressing the amount of matter in a body. It is measured in kg. Unlike weight its value is not affected by gravity and does not change. (f) Mass density, ρ, gives the relationship between the mass (M kg) of a substance and its volume (V m3). Equation 1.5 defines mass density as: ρ = M /V, with the units of mass density being kg/m3. Thus ρ is basically the mass per unit volume. For example, water has a mass density of 1000 kg/m3, so every cubic metre of water has a mass of 1000 kg. (g) Weight density, w, is the relationship between the weight (W N) of a substance and its volume (V m3). Thus w = W/V and has the units N/m3. Hence w is the weight of a cubic metre of a substance. Water has a weight density of 9810 N/m3. This follows from W = Mg, that is if we multiply the mass density by gravity we obtain the weight density. (h) Relative density, s, is a ratio denoting the density of a substance, ρS , relative to the density of water, ρ . Thus equation 1.7 is: s = ρS /ρ . Since this is the ratio of two similar quantities, s is dimensionless an has no units. For example, mercury has a mass density of 13 600 kg/m3 while water has a mass density of 1000 kg/m3, so for mercury s = 13 600/1000 or 13.6. In other words mercury is 13.6 times as dense as water. The same result is obtained if weight densities are used instead of mass densities because both 13 600 kg/m3 and 1000 kg/m3 would have to be multiplied by 9.81 m/s2 to obtain the weight density, so the ratio is unaltered. (i) The pressure at a point which is at a depth, h, in the liquid is given by P = ρgh, where P is often referred to as the hydrostatic pressure (in N/m2). Hydrostatic pressure results from the weight of the liquid acting downwards, so the greater the depth, the greater the weight (N) of liquid acting on any surface and the greater the pressure P. (j) Hydrostatic pressure acts equally in all directions, even upwards, and it acts normally (at 90 degrees) to any surface immersed in it. Therefore, on something like a flat bottomed pontoon the hydrostatic pressure (P) acts vertically upwards over its plan area (A) resulting in a vertical force (F = PA) being exerted on the pontoon. This is the buoyancy force, which can be defined simply as the upward vertical force exerted on an immersed object as a result of hydrostatic pressure acting over the submerged surfaces of the body. Figs 1.28 and 1.33 provide good illustrations of a buoyancy force. Archimedes showed that the buoyancy force exerted on a body is equal to the weight of the liquid it displaces, that is F = ρgV, where V is the volume of water displaced by the body. The units are the same as for any other force, N/m2. 1.2 (a) Explain what is meant by gauge pressure and absolute pressure. (b) What is the approximate numerical value of atmospheric pressure expressed in N/m2 and as a head of water? (c) Calculate atmospheric pressure expressed as a head of mercury (the relative density of mercury is 13.6). 3rd ed, Understanding Hydraulics 1 © Les Hamill 2001, 2011 (c) The mass density of mercury is 13 600 kg/m3.81 × 19 × (1.0 × 38) m2 since the depth h = 38 m. and (d) the force on the side. Thus PAV = 2452.5 N/m2. below the surface of a body of water is P = ρgh N/m2. 2011 .5 A dam that retains fresh water has a vertical face. Understanding Hydraulics 2 © Les Hamill 2001.7 × 0. In reality the atmosphere does exert a considerable pressure on the water surface (and is not zero as assumed when calculating the gauge pressure).2 and Chapter 1. For example. PATM has the value given in part (b) below.5 × (1. so the height of a column of mercury needed to equal atmospheric pressure is 101 × 103/(13 600 × 9. Thus atmospheric pressure is roughly the same as the pressure that exists at the bottom of a 10. absolute pressure must always be positive because it is impossible to have a pressure lower than a vacuum.3/13. (b) At what depth from the surface does the resultant force act? (a) As the questions get more complex it is best to start with F = ρghGA. F = PAV A = 2452.5 P = 4905 N/m2 (b) With atmospheric pressure.81) or 10. h.3 A rectangular tank is 1. Since P = ρgh then h = P/ρg which gives PATM as the equivalent of 101 × 103/(1000 × 9.3 we know that F = PAV A where F is the force due to hydrostatic pressure and PAV is the average pressure intensity acting over an area A. (a) What is the gauge pressure at the bottom of the tank in N/m2 ? (b) What is the absolute pressure at the bottom of the tank? (a) At the bottom of the tank the gauge pressure is: P = ρgh = 1000 × 9.0 × 38) = 7083 × 103 N 3rd ed. PATM .76 m.8). P N/m2. When working with absolute pressure. using gauge pressure. which is the lowest datum possible. However. at a depth.3 m of water.5) = 1226 N. Over a one metre length of the face at the centre of the valley the water has a depth of 38 m.0 × 0. F = 1000 × 9. PABS .2).8 that it is possible to have a negative gauge pressure.5 × (0. as a guide atmospheric pressure is roughly about 101 000 N/m2. 1. The same result is obtained by dividing the head of water equivalent to atmospheric pressure by the relative density of mercury. in the liquid we must add the gauge pressure to the atmospheric pressure thus: PABS = gauge pressure + atmospheric pressure (1.1 and 6. 1.3. Both approaches are correct. equal to 101 000 N/m2.81) = 0. so if we want to calculate the total or absolute pressure.5) = 1226 N Note that using F = ρghGA gives the same answer. This is the hydrostatic pressure of the liquid measured relative to atmospheric pressure.(a) The gauge pressure.6 = 0.5 N/m2 (b) The average pressure intensity on the side of the tank is the same as on the end.7 m wide end of the tank.5 m. With a one metre length of the dam A = (1. that is taking atmospheric pressure as a datum so that PATM = 0 (see Fig 1.5/2) × (1. For example. It is higher at sea level than at the top of Everest. At any point it also fluctuates according to weather conditions. at the depth. h. the absolute pressure is: PABS = ρgh + PATM = 4905 + 101 000 = 105 905 N/m2 1. On the other hand. (c) the force on the end of the tank. since it depends upon the depth of liquid not the area of the surface.0 × 0. Note from Fig 1. as shown by the rise and fall of a barometer. that is 101 × 103 N/m2.7 and Figs 6.81 × 0. (b) the mean pressure intensity on the 1. then the pressure of the water may fall below atmospheric pressure so that ρgh has a negative value (see Example 4.5) = 858 N (d) Similarly.0 m long and 0.0 m long side of the tank. In this case: F = PAV A = 2452. if a pipeline transporting water goes over the crest of a hill that is higher than the ends of the pipeline.4 For the tank in question 1. (b) Atmospheric pressure varies over the surface of the earth.81 × (0. measured relative to a datum corresponding to a vacuum. (a) Calculate the resultant force on this unit length of the face. h m. for part (d): F = 1000 × 9.3 m high column of water.7 m wide and contains fresh water to a depth of 0. (a) Gauge pressure at the liquid surface = 0 Gauge pressure at the bottom of the tank = 4905 N/m2 Therefore the mean or average pressure intensity PAV = (0 + 4905)/2 = 2452. (c) From equation 1. calculate (a) the mean pressure intensity on the 0. Thus hG = 38/2 = 19 m. thus 10.9) = ρgh + PATM When written like this PABS is the pressure (in N/m2) at the depth.76 m. 81 × 0. F =ρghGA = 1000 × 9.33 m.1 + (0.0491/[π × 0.1 × 1.5 + (1.82 × 103 N Again hP = hG + (IG/AhG) where for a circle IG = πR4/4 = (π × 0.693/2) = 0. acts.1) m2.2 m wide.8/2) = 4. F = 1000 × 9. Its vertical height is 0. this being the angle of the banks of a trapezoidal river channel.81 × 4.1 m wide by 1. and during a flood the hinge is 3.0206/[3. In this case IG = (2.21 × 103 N Fig Q1. of the gate is 0.693 m.54/4) = 0. What is the force exerted by the floodwater on the gate.0]) = 4.96 = 4.78 m2 so: hP = 4.(b) The resultant force acts at ⅔ of the water depth when measured from the surface. P.8 m high discharges to a river channel as in Fig 1. that is the point at which the resultant force.78 × 4.1) IG = LD3/12.0 m above the hinge. also hinged at the top. thus: LP = LG + (IG/ALG). and at what depth from the surface does it act? (b) A circular gate. What is the force exerted by the floodwater on the gate.447 × 0. this time hG = 3. P. and the centre of pressure.7 A gate at the end of a sewer measures 0.8 m by 1.52× 4.016 m 1.4 m and A = (1.83)/12 = 1.8 × 2. It is hinged along its top edge and hangs at an angle of 30 degrees to the vertical. hangs vertically at the end of a pipe discharging to the river.4]) = 4. The area.8 m as shown in Fig Q1. hP = 4.96 m2. (a) Calculate the hydrostatic force on the gate and the vertical distance between the centroid of the gate.8 × 2. the gate having the same dimensions as the culvert.0491 m4.461 m hP is often referred to as the vertical depth to the centre of pressure.6 (a) A rectangular culvert 2. hG = 0. G.2 = 0.10. (b) F = ρghGA with hG = 3.81 × 4.8 × cos 30° = 0.52) = 30. During a flood the river rises to 3.0 × (π × 0.5 + 0.12 gives the depth from the water surface. 1.5 m above the hinge.0206 m4 and A = (1.7 Because the gate is inclined to the vertical equation 1.8 × 2.447 m. The vertical depth from the water surface to the centroid.1) = 163.16 × 103 N Equation 1. what is the force and the distance GP now? (c) Has the value of GP increased or decreased.4 × (1. F = 1000 × 9. At the end of the culvert is a vertical flap gate which is hinged along its top edge.0 m. to the resultant force as: hP = hG + (IG/AhG) where for a rectangle (Table 1.5 = 4. The gate has a radius of 0.13 must be used to find the location of P. 2011 . that is: = ⅔ × 38 = 25. G.0 + (0. A. and why has it changed in this manner? (a) The inclined dimension of the gate is 0.8 × 1.1 m above the top of the hinge. 3rd ed. (b) If the river level increases to 2.7. hP. Understanding Hydraulics 3 © Les Hamill 2001.4 + (1.5 m. and at what depth from the surface does it act? (a) Again starting with F = ρghGA.1) = 3. when the river level is 0.5 m below the water surface. F. of the gate. 7.776 – 480. hence P is closer to G.2 × 0. F = 1000 × 9. The distance between G and P is.52 × 2. As H increases P moves closer to G.22 1.115 m. a measure of the amount by which the pressure varies over an immersed surface.515 m.020 × cos 30° = 0. (b) Again F = ρghGA with hG = 2.104 × cos 30° = 0.0 m while the head in the river is 2.10 × 103 N As above. and the distance GP.52) = 46 228.090 m. The increase in pressure over the 1 m increment is always 9810 N/m2. P. so P moves close to G. Now if the same vertical surface of height.104 m The equivalent vertical distance between G and P is hP – hG = 0. (c) by taking moments about the hinge. Relatively speaking this is a smaller variation than 0 to ρgh. namely 0. and dividing this by the average pressure on the immersed surface gives the pressure variation in the last column. was immersed with its top edge a distance.54) /[4 × π × 0.0] = 0.52) = 15 409.8 A circular gate of 0. Measured above the centroid of the gate. The relative variation of pressure on a 1 m high vertical surface that starts with its top edge at the water surface and then moves down in 1 m increments is shown in Table Q1.6).0 × (π × 0. The gate is at the end of a pipe discharging to a river. h.0104 m (b) F = ρghGA with hG = 2.96 × 2.515]) = 0. (c) The distance between G and P has decreased as the water depth increased. 2011 .119 m (see Box 1. Therefore LG = 0. Assuming that the gate is initially vertical: (a) calculate the force exerted by the water in the pipe on the gate.8/2) = 0.347 /cos 30° = 2. Depth.0312 m (c) Taking moments about the hinge through the centroid of the gate: net turning moment = 46 228. the head in the pipe is 6.5 m radius is hinged so that it rotates about its horizontal diameter.52 × 6.0 m. This is as large as GP can get.67 0. h.017 m. in a sense.96 × 0. Table Q1. The equivalent vertical distance between G and P is hP – hG = 0.1) IG = πR4/4.96 = 22. This is as predicted by Box 1. in a liquid is 0 to ρgh.0 + (0.96 m2.347 × 0.2 = 0. Thus: LP – LG = (0.5 N hP = hG + (IG/AhG) where for a circle (Table 1. LP = LG + (IG /ALG) so the inclined distance between G and P is: LP – LG = (IG/ALG) = (0.5 N hP – hG = (π × 0. namely (h/2 – h/3) or h/6.83 ) /12 = 0.8 × 1.13 results in an incorrect answer.710]) = 0.0 × (π × 0.1/cos 30° = 0.7 Relative variation of pressure with depth. below the surface of the liquid the variation in pressure would be from ρgH to ρg(H + h).1 m vertically above the top of the gate which is equivalent to an inclined distance of 0. that is it rotates about a horizontal line passing through the centroid of the gate.347 m and A = 0.0512 /[0.96 m2. Note that using equation 1. LP – LG = (IG/ALG) with LG = hG /cos 30° = 2.710 m.115 + (0.776 = 0 Nm 3rd ed.5 × 0. Explain your answer.0512 m4.00 0.The water surface is 0.0312 = 480. Understanding Hydraulics 4 © Les Hamill 2001.81 × 2.12 instead of 1. F = 1000 × 9. IG = LD3/12 = (1.0 m.693/2) = 2.0] = 0. It is easy to show that the variation in pressure on a flat.0 m. vertical immersed surface that extends from the water surface to a depth.54) /[4 × π × 0. and the centre of pressure.81 × 2. and the distance GP between the centre of the gate. As before A = 0. This is a relatively large variation resulting in a relatively large distance between G and P. (b) calculate the force exerted by the river water on the gate. hP – hG = (π × 0.81 × 6. This decreases with depth.0104 – 15 409.29 0. h 0m 1 2 3 4 5 P = ρgh 0 N/m2 9 810 19 620 29 430 39 240 49 050 Average pressure -------4 905 N/m2 14 715 24 525 34 335 44 145 Pressure variation ----2.0512 /[0.020 m. using the results from above determine the net turning moment on the gate caused by the two forces acting at their respective centres of pressure on opposite sides of the gate.5.40 0. (a) F = ρghGA with hG = 6. F = 1000 × 9.5 × 0. G. H. sin θ = 35/40 so θ = 61. FR = 103 × (19. V = 513. 2011 .914 m2 Volume of displaced water. Calculate the magnitude and direction of the resultant hydrostatic force on a unit length of the gate.365 m Area CDB = ½ × 19.10 The dam in the diagram has a curved face.0/4. cos θ = 2. If there is no turning moment on the valve.6202 + 48.0) = 19.0 = 0.461 m3 per m length Vertical force on gate. the flow in the pipeline will not open or close the valve. FV = ρgV = 1000 × 9.378 m2 Length AD = 4. being part of a 40 m radius circle.464 m Area ACD = ½ × 3.206 × 103 N/m Fig Q1. Regardless of the depths on the two sides of the gate the net turning moment is always zero.365 × 35 = 338. 1.9 A gate which is a quarter of a circle of radius 4.0 × (2.2062)1/2 = 52.349 m2 Length CD = 40 cos θ = 19.The net turning moment is zero because the first force is exactly three times the other while the lever arm (that is the distance GP) of the first force is exactly one-third of the other.914 = 48. φ = tan–1 (48 206 /19 620) = 67. 1.888 = 513.0 m of fresh water as shown in the diagram.5 so θ = 60° Area sector ABC = (60°/360°) × π × 4. allowing for rounding errors.0 sin 60° = 3.02 = 8.914 m3 per m length Vertical force on gate. it means that a valve designed on this principle will be able to operate smoothy and will remain in a set position.045° Area sector CBE = (61.464 = 4.620 × 103 N/m Resultant force on gate.0 V = 513.9° to horizontal passing upward through C. FH = ρghGA = 1000 × 9.81 × 1. Calculate the magnitude and direction of the resultant hydrostatic force per metre length.0 m holds back 2.9 Horizontal force on gate.05 × 103 N/m 3rd ed.914 × 1.888 m2 Area BDE = area CBE – area CDB = 852.461 × 1.461 = 5037.349 – 338. for example.464 × 2. V = 4. This is a useful result because.0 × 1.81 × 513. The dam holds back water to a depth of 35 m. Understanding Hydraulics 5 © Les Hamill 2001.0 V = 4.05 × 103 N/m Angle of resultant.464 m2 Area ABD = area ABC – area ACD = 8.461 m2 Volume of displaced water.0 = 3. FV = ρgV = 1000 × 9.378 – 3.045°/360°) × π × 402 = 852.81 × 4. 81 × hS = 69. 1.632 + 5037. The side walls are 8 m high. The weight of the lock is 7500 × 103 × 9. The side wall are 8 m high.63 × 103 N/m Resultant force on gate.5 m. of water. then: 1000 × 9. 1. (c) The volume of sand required = W /ρS g = 22 352 × 103/(2600 × 9. FH = ρghGA = 1000 × 9. that is they are immiscible. If the transducer indicates a gauge pressure of 69. If hS is the unknown thickness of the saline layer.93 m. so if the lock floats it will displace 7500 × 103 kg of sea water. The buoyancy force is equal to the weight of the water displaced. Obviously. with fresh water (1000 kg/m3 ) overlying saline water (1025 kg/m3 ). Thus the blanket thickness should be greater than 0. Understanding Hydraulics 6 © Les Hamill 2001.30 m 3rd ed.10 Horizontal force on gate.3 m.6 × 103 N/m Angle of resultant. Assuming that the internal dimensions are the same as the external (that is ignoring the thickness of the walls) then the minimum depth of sand required = 876 /(60 × 30) = 0.81 × 9540 = 95 927 × 103 N. φ = tan–1 (5037.3 m then the volume of water displaced = 5. how thick is the layer of saline water? (a) A stratified fluid is one that consists of layers of fluid of different densities lying on top of each other. Therefore the sum of the pressures resulting from the layers of fresh water and saline water must equal this. The water in the estuary is stratified at the point where the measurement is taken.25 × hS = 69 730 hS = 4.81 × (35/2) × (35 × 1.052)1/2 = 7840.73 × 103 26 487 + 10 055.0) = 6008. M. The freeboard is (8 – 4. (b) If the depth of immersion of the lock structure is 5.7 + 1025 × 9. The lock is 60 m long by 30 m wide in plan and is shaped like an open shoe box. Water sampling shows that the fresh water extends from the water surface to a depth of 2. (b) The transducer indicates a pressure = 69. which is now = 1025 × 9.05 /6008.3 × 60 × 30 = 9540 m3.73 × 103 N/m2.49 m. assuming the structure is watertight? (c) If the additional weight is to be provided by a blanket of sand (density 2600 kg/m3).81) = 876 m3. V = M /ρSW the volume displaced is: V = 7500 × 103/1025 = 7317 m3 The depth of immersion = V / plan area of the lock = 7317 /(60 × 30) = 4.12 (a) Explain what is meant by a stratified fluid. with the lighter fluids above. how thick must the layer of sand be? (1 tonne = 1000kg) (a) A floating body displaces its own mass.11 A 7500 tonne reinforced concrete lock structure has been constructed in a dry dock.07 m This is the draft of the lock. The difference between the buoyancy force and lock weight = (95 927 – 73 575) × 103 = 22 352 × 103 N 3 Therefore a weight in excess of 22 352 × 10 N is required to sink the lock onto the sea bed.07) = 3.81 × 2. 2011 . FR = 103 × (6008.63) = 40° to horizontal passing through C. so the lock will float. This can only occur if the fluids do not mix with each other. (a) Will the lock structure float in sea water of density 1025 kg/m3. the densest fluid would be at the bottom.Fig Q1.7 m. Since volume.81 = 73 575 × 103 N. and if so. what is its draught and freeboard? (b) What additional weight will be required to sink the structure onto the sea bed if the depth of water is 5.73 x 103 N. (b) A pressure transducer is used to measure the hydrostatic pressure on the sea bed in a tidal estuary.