1 HYDROSTATICS 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 is a force (N) and A is the area 2 (m ) over which the force acts. Thus p ressure 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 for ce 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. T hus 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 2 in m/s . 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 2 2 be constant at 9.81 m/s . The moon is smaller that the earth, and its gravitational acceleration is about 1.6 m/s . Jupiter is larger than the earth and has an average gravitational acceleration of about 26.9 m/s 2. Because Jupiter is not a perfect sphere the value varies according to location. (e) Mass is a universal constant. co nstant. It is a physical quantity expressing the amount of matter in a body. It i s measured in kg. Unlike weight its value is not affected by gravity and does not change. (f) Mass density, ρ density, ρ , gives the relationship between b etween the mass ( M kg) of a substance and its i ts volume (V m m3). M kg) 3 Equation 1.5 defines mass density as: ρ as: ρ = = M / V Thus ρ is is basically V , with the units of mass density being kg/m . Thus ρ 3 the mass per unit volume. For example, water has a mass density of 1000 kg/m , so every cubic metre of water has a mass of 1000 kg. 3 (g) Weight density, w, is the relationship relatio nship between the weight (W N) N) of a substance and its volume (V m m ). Thus w 3 = W / V and has the units N/m . 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, ρ substance, ρ S , relative to the density of water, ρ water, ρ . . Thus equation 1.7 is: s = ρ = ρ S / ρ . . Since this is the ratio of two similar quantities, s is dimensionless an has no units. For 3 3 example, mercury has a mass density of 13 600 kg/m while water has a mass density of 1000 kg/m , 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 = ρ gh gh, where P is often referred to as 2 the hydrostatic pressure (in N/m ). Hydrostatic pressure results from the weight of t he 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 immersed in it. Therefore, on something like a flat bot tomed 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 o f 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 = ρ gV is the volume of water displaced by the body. The gV , where V is 2 units are the same as for any other force, N/m . 1.2 (a) Explain what is meant by gauge pressure and absolute pressure. (b) What is the approximate 2 numerical value of atmospheric pressure expressed in N/m and as a head of water? (c) Calculate atmospheric pressure expressed as a head of mercury (the relative density of mercury is 13.6).
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(a) The gauge pressure, P N/m2, at a depth, h m, below the surface of a body of water is P = ρ gh N/m2. This is the hydrostatic pressure of the liqui d measured relative to atmospheric pressure, that is taking atmospheric pressure as a datum so that PATM = 0 (see Fig 1.8). In reality the atmosphere does exert a considerable pressure on the water surface (and is not zero as assumed when calculating the gauge pressure), so if we want to calculate the total or absolute pressure, PABS , at the depth, h, in the liquid we must add the gauge pressure to the atmospheric pr essure thus: PABS = gauge pressure + atmospheric pressure = ρ gh + PATM (1.9) 2 When written like this PABS is the pressure (in N/m ) at the depth, h, measured relative to a datum corresponding to a vacuum, which is the lowest datum possible. When working with absolute pressure, PATM has the value given in part (b) below. Note from Fig 1.8 that it is possible to have a negative gauge pressure. For example, if a pipeline transporting water goes over the crest of a hill that is higher than the ends of the pipeline, then the pressure of the water may fall below atmospheric pressure so that ρ gh has a negative value (see Example 4.7 and Figs 6.1 and 6.2). On the other hand, absolute pressure must always be positive because it is impossible to have a pressure lower than a vacuum. (b) Atmospheric pressure varies over the surface of the earth. It is higher at sea level than at the top of Everest. At any point it also fluctuates according to weather conditions, as shown by the rise and fall of a barometer. However, as a guide atmospheric pressure is roughly about 101 000 N/m2, that is 101 × 103 N/m2. Since P = ρ gh 3 then h = P / ρ g which gives PATM as the equivalent of 101 × 10 /(1000 × 9.81) or 10.3 m of water. Thus atmospheric pressure is roughly the same as the pressure that exists at the bottom of a 10.3 m high column of water. (c) The mass density of mercury is 13 600 kg/m3, so the height of a column of mercury needed to equal atmospheric pressure is 101 × 103 /(13 600 × 9.81) = 0.76 m. The same result is obtained by dividing the head of water equivalent to atmospheric pressure by the relative density of mercury, thus 10.3/13.6 = 0.76 m. 1.3 A rectangular tank is 1.0 m long and 0.7 m wide and contains fresh water to a depth of 0.5 m. 2 (a) What is the gauge pressure at the bottom of the tank in N/m ? (b) What is the absolute pressure at the bottom of the tank? (a) At the bottom of the tank the gauge pressure is: P = ρ gh = 1000 × 9.81 × 0.5 2 P = 4905 N/m (b) With atmospheric pressure, PATM , equal to 101 000 N/m2, the absolute pressure is: PABS = ρ gh + PATM = 4905 + 101 000 = 105 905 N/m2 1.4 For the tank in question 1.3, using gauge pressure, calculate (a) the mean pressure intensity on the 0.7 m wide end of the tank; (b) the mean pressure intensity on the 1.0 m long side of the tank; (c) the force on the end of the tank, and (d) the force on the side. (a) Gauge pressure at the liquid surface = 0 2 Gauge pressure at the bottom of the tank = 4905 N/m Therefore the mean or average pressure intensity PAV = (0 + 4905)/2 = 2452.5 N/m2 (b) The average pressure intensity on the side of the tank is the same as on the end, since it depends upon the depth of liquid not the area of the surface. Thus PAV = 2452.5 N/m2. (c) From equation 1.2 and Chapter 1.3 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. In this case: F = PAV A = 2452.5 × (0.7 × 0.5) = 858 N (d) Similarly, F = PAV A = 2452.5 × (1.0 × 0.5) = 1226 N Note that using F = ρ ghG A gives the same answer. For example, for part (d): F = 1000 × 9.81 × (0.5/2) × (1.0 × 0.5) = 1226 N. Both approaches are correct. 1.5 A dam that retains fresh water has a vertical face. Over a one metre length of the face at the centre of the valley the water has a depth of 38 m. (a) Calculate the resultant force on this unit length of the face. (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 = ρ ghG A. With a one metre length of the dam 2 A = (1.0 × 38) m since the depth h = 38 m. Thus hG = 38/2 = 19 m. 3 F = 1000 × 9.81 × 19 × (1.0 × 38) = 7083 × 10 N
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(b) The resultant force acts at = ⅔ × 38 = 25.33 m.
⅔ of
the water depth when measured from the surface, that is:
1.6 (a) A rectangular culvert 2.1 m wide by 1.8 m high discharges to a river channel as in Fig 1.10. At the end of the culvert is a vertical flap gate which is hinged along its top edge, the gate having the same dimensions as the culvert. During a flood the river rises to 3.5 m above the hinge. What is the force exerted by the floodwater on the gate, and at what depth from the surface does it act? (b) A circular gate, also hinged at the top, hangs vertically at the end of a pipe discharging to the river. The gate has a radius of 0.5 m, and during a flood the hinge is 3.5 m below the water surface. What is the force exerted by the floodwater on the gate, and at what depth from the surface does it act? 2
(a) Again starting with F = ρ ghG A, this time hG = 3.5 + (1.8/2) = 4.4 m and A = (1.8 × 2.1) m . 3 F = 1000 × 9.81 × 4.4 × (1.8 × 2.1) = 163.16 × 10 N Equation 1.12 gives the depth from the water surface, hP, to the resultant force as: 3 hP = hG + ( I G / AhG) where for a rectangle (Table 1.1) I G = LD /12. 3 4 2 In this case I G = (2.1 × 1.8 )/12 = 1.0206 m and A = (1.8 × 2.1) = 3.78 m so: hP = 4.4 + (1.0206/[3.78 × 4.4]) = 4.461 m hP is often referred to as the vertical depth to the centre of pressure, P, that is the point at which the resultant force, F , acts. (b) F = ρ ghG A with hG = 3.5 + 0.5 = 4.0 m. 2 F = 1000 × 9.81 × 4.0 × (π × 0.5 ) 3 = 30.82 × 10 N 4 4 4 Again hP = hG + ( I G / AhG) where for a circle I G = π R /4 = (π × 0.5 /4) = 0.0491 m . 2 hP = 4.0 + (0.0491/[π × 0.5 × 4.0]) = 4.016 m 1.7 A gate at the end of a sewer measures 0.8 m by 1.2 m wide. It is hinged along its top edge and hangs at an angle of 30 degrees to the vertical, this being the angle of the banks of a trapezoidal river channel. (a) Calculate the hydrostatic force on the gate and the vertical distance between the centroid of the gate, G, and the centre of pressure, P, when the river level is 0.1 m above the top of the hinge. (b) If the river level increases to 2.0 m above the hinge, what is the force and the distance GP now? (c) Has the value of GP increased or decreased, and why has it changed in this manner? (a) The inclined dimension of the gate is 0.8 m as shown in Fig Q1.7. Its vertical height is 0.8 × cos 30° = 0.693 m. The vertical depth from the water surface to the centroid, G, of the gate, hG = 0.1 + (0.693/2) = 0.447 m. The area, A, of the gate is 0.8 × 1.2 = 0.96 m2. F = ρ ghG A = 1000 × 9.81 × 0.447 × 0.96 = 4.21 × 103 N
Fig Q1.7
Because the gate is inclined to the vertical equation 1.13 must be used to find the location of P, thus: LP = LG + ( I G / ALG).
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The water surface is 0.1 m vertically above the top of the gate which is equivalent to an inclined distance of 0.1/cos 30° = 0.115 m. Therefore LG = 0.115 + (0.8/2) = 0.515 m. 3 3 4 I G = LD /12 = (1.2 × 0.8 ) /12 = 0.0512 m . 2 As before A = 0.96 m . LP = LG + ( I G / ALG) so the inclined distance between G and P is: LP – LG = ( I G / ALG) = (0.0512 /[0.96 × 0.515]) = 0.104 m The equivalent vertical distance between G and P is hP – hG = 0.104 × cos 30° = 0.090 m. Note that using equation 1.12 instead of 1.13 results in an incorrect answer, namely 0.119 m (see Box 1.6). 2 (b) Again F = ρ ghG A with hG = 2.0 + (0.693/2) = 2.347 m and A = 0.8 × 1.2 = 0.96 m . 3 F = 1000 × 9.81 × 2.347 × 0.96 = 22.10 × 10 N As above, LP – LG = ( I G / ALG) with LG = hG /cos 30 ° = 2.347 /cos 30° = 2.710 m. Thus: LP – LG = (0.0512 /[0.96 × 2.710]) = 0.020 m. The equivalent vertical distance between G and P is hP – hG = 0.020 × cos 30° = 0.017 m. (c) The distance between G and P has decreased as the water depth increased. This is as predicted by Box 1.5. The distance between G and P is, in a sense, a measure of the amount by which the pressure varies over an immersed surface. It is easy to show that the variation in pressure on a flat, vertical immersed surface that extends from the water surface to a depth, h, in a liquid is 0 to ρ gh. This is a relatively large variation resulting in a relatively large distance between G and P , namely (h /2 – h /3) or h /6. This is as large as GP can get. Now if the same vertical surface of height, h, was immersed with its top edge a distance, H , below the surface of the liquid the variation in pressure would be from ρ gH to ρ g( H + h). Relatively speaking this is a smaller variation than 0 to ρ gh, hence P is closer to G. As H increases P moves closer to G. 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 Q 1.7. The increase in pressure over the 1 m increment is always 9810 N/m2, and dividing this by the average pressure on the immersed surface gives the pressure variation in the last column. This decreases with depth, so P moves close to G. Table Q1.7 Relative variation of pressure with depth. Depth, h 0m 1 2 3 4 5
P = ρ gh 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 ----2.00 0.67 0.40 0.29 0.22
1.8 A circular gate of 0.5 m radius is hinged so that it rotates about its horizontal diameter, that is it rotates about a horizontal line passing through the centroid of the gate. The gate is at the end of a pipe discharging to a river. Measured above the centroid of the gate, the head in the pipe is 6.0 m while the head in the river is 2.0 m. Assuming that the gate is initially vertical: (a) calculate the force exerted by the water in the pipe on the gate, and the distance GP between the centre of the gate, G, and the centre of pressure, P; (b) calculate the force exerted by the river water on the gate, and the distance GP; (c) by taking moments about the hinge, using the results from abov e 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. Explain your answer. (a) F = ρ ghG A with hG = 6.0 m. 2 F = 1000 × 9.81 × 6.0 × (π × 0.5 ) = 46 228.5 N 4 hP = hG + ( I G / AhG) where for a circle (Table 1.1) I G = π R /4. 4 2 hP – hG = (π × 0.5 ) /[4 × π × 0.5 × 6.0] = 0.0104 m (b) F = ρ ghG A with hG = 2.0 m. 2 F = 1000 × 9.81 × 2.0 × (π × 0.5 ) = 15 409.5 N 4 2 hP – hG = (π × 0.5 ) /[4 × π × 0.5 × 2.0] = 0.0312 m (c) Taking moments about the hinge through the centroid of the gate: net turning moment = 46 228.5 × 0.0104 – 15 409.5 × 0.0312 = 480.776 – 480.776 = 0 Nm
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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 o ther, allowing for rounding errors. Regardless o f the depths on the two sides of the gate the net turning moment is always zero. This is a useful result because, for example, it means that a valve designed on this principle will be able to operate smoothy and will remain in a set position. If there is no turning moment on the valve, the flow in the pip eline will not open or close the valve. 1.9 A gate which is a quarter of a circle of radius 4.0 m holds back 2.0 m of fresh water as shown in the diagram. Calculate the magnitude and direction of the resultant hydrostatic force on a unit length of the gate. cos θ = 2.0/4.0 = 0.5 so θ = 60° Area sector ABC = (60 ° /360°) × π × 4.02 = 8.378 m2 Length AD = 4.0 sin 60° = 3.464 m Area ACD = ½ × 3.464 × 2.0 = 3.464 m2 Area ABD = area ABC – area ACD = 8.378 – 3.464 = 4.914 m2 Volume of displaced water, V = 4.914 × 1.0 3 V = 4.914 m per m length 3 Vertical force on gate, F V = ρ gV = 1000 × 9.81 × 4.914 = 48.206 × 10 N/m
Fig Q1.9
Horizontal force on gate, F H = ρ ghG A = 1000 × 9.81 × 1.0 × (2.0 × 1.0) = 19.620 × 103 N/m Resultant force on gate, F R = 103 × (19.6202 + 48.2062)1/2 = 52.05 × 103 N/m Angle of resultant, φ = tan–1 (48 206 /19 620) = 67.9° to horizontal passing upward through C. 1.10 The dam in the diagram has a curved face, being part of a 40 m radius circle. The dam holds back water to a depth of 35 m. Calculate the magnitude and direction of the resultant hydrostatic force per metre length. sin θ = 35/40 so θ = 61.045° 2 Area sector CBE = (61.045 ° /360°) × π × 40 = 852.349 m2 Length CD = 40 cos θ = 19.365 m Area CDB = ½ × 19.365 × 35 = 338.888 m2 Area BDE = area CBE – area CDB = 852.349 – 3 38.888 2 = 513.461 m Volume of displaced water, V = 513.461 × 1.0 3 V = 513.461 m per m length Vertical force on gate, F V = ρ gV = 1000 × 9.81 × 513.461 = 5037.05 × 103 N/m
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Fig Q1.10
3
Horizontal force on gate, F H = ρ ghG A = 1000 × 9.81 × (35/2) × (35 × 1.0) = 6008.63 × 10 N/m 3 2 2 1/2 3 Resultant force on gate, F R = 10 × (6008.63 + 5037.05 ) = 7840.6 × 10 N/m –1 Angle of resultant, φ = tan (5037.05 /6008.63) = 40° to horizontal passing through C. 1.11 A 7500 tonne reinforced concrete lock structure has been constructed in a dry dock. The lock is 60 m long by 30 m wide in plan and is shaped like an open shoe box. The side walls are 8 m high. (a) Will the 3 lock structure float in sea water of density 1025 kg/m , 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 5.3 m, assuming the structure is watertight? (c) If the additional weight is to be provided by a blanket of 3 sand (density 2600 kg/m ), how thick must the layer of sand be? (1 tonne = 1000kg) (a) A floating body displaces its own mass, M , of water, so if the lock floats it will displace 3 7500 × 10 kg of sea water. Since volume, V = M / ρ SW the volume displaced is: 3 3 V = 7500 × 10 /1025 = 7317 m The depth of immersion = V / plan area of the lock = 7317 /(60 × 30) = 4.07 m This is the draft of the lock. The side wall are 8 m high, so the lock will float. The freeboard is (8 – 4.07) = 3.93 m. (b) If the depth of immersion of the lock structure is 5.3 m then the volume of water displaced 3 = 5.3 × 60 × 30 = 9540 m . The buoyancy force is equal to the weight of the water displaced, which is now = 1025 × 9.81 × 9540 = 95 927 × 103 N. The weight of the lock is 7500 × 103 × 9.81 = 73 575 × 103 N. The difference between the buoyancy force and lock weight = (95 927 – 73 575) × 103 = 22 352 × 103 N Therefore a weight in excess of 22 352 × 103 N is required to sink the lock onto the sea bed. (c) The volume of sand required = W / ρ S g = 22 352 × 103 /(2600 × 9.81) = 876 m3. 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) = 0.49 m. Thus the blanket thickness should be greater than 0.5 m. 1.12 (a) Explain what is meant by a stratified fluid. (b) A pressure transducer is used to measure the hydrostatic pressure on the sea bed in a tidal estuary. The water in the estuary is stratified at the point 3 3 where the measurement is taken, with fresh water (1000 kg/m ) overlying saline water (1025 kg/m ). Water sampling shows that the fresh water extends from the water surface to a depth of 2.7 m. If the 3 2 transducer indicates a gauge pressure of 69.73 × 10 N/m , 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. This can only occur if the fluids do not mix with each other, that is they are i mmiscible. Obviously, the densest fluid would be at the bottom, with the lighter fluids above. (b) The transducer indicates a pressure = 69.73 x 10 3 N. Therefore the sum of the pressures resulting from the layers of fresh water and saline water must equal this. If hS is the unknown thickness of the saline layer, then: 3 1000 × 9.81 × 2.7 + 1025 × 9.81 × hS = 69.73 × 10 26 487 + 10 055.25 × hS = 69 730 hS = 4.30 m
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