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Chapter 28 Pressure in Fluids

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Chapter 28 Pressure in Fluids CHAPTER 28 PRESSURE IN FLUIDS EXERCISE 128, Page 281 1. A force of 280 N is applied to a piston of a hydraulic system of cross-sectional area 0.010 m 2 . Determine the pressure produced by the piston in the hydraulic fluid. force 280 N Pressure, p = = 28000Pa = 28 kPa area 0.010m2 That is, the pressure produced by the piston is 28 kPa 2. Find the force on the piston of question 1 to produce a pressure of 450 kPa. Pressure, p = 450 kPa = 450000 Pa force Pressure p = hence, force = pressure × area area = 4540000 × 0.010 = 4500 N = 4.5 kN 3. If the area of the piston in question 1 is halved and the force applied is 280 N, determine the new pressure in the hydraulic fluid. 0.010 New area = 0.005m2 2 force 280 N New pressure, p = = 56000Pa = 56 kPa area 0.005m2 331 © John Bird Published by Taylor and Francis EXERCISE 129, Page 283 1. Determine the pressure acting at the base of a dam, when the surface of the water is 35 m above base level. Take the density of water as 1000 kg/m 3 . Take the gravitational acceleration as 9.8 m/s 2 . Pressure at base of dam, p = gh = 1000 kg/m 3 9.8 m/s2 0.35 m = 343000 Pa = 343 kPa 2. An uncorked bottle is full of sea water of density 1030 kg/m 3 . Calculate, correct to 3 significant figures, the pressures on the side wall of the bottle at depths of (a) 30 mm, and (b) 70 mm below the top of the bottle. Take the gravitational acceleration as 9.8 m/s 2 . Pressure on the side wall of the bottle, p = gh (a) When depth, h = 30 mm = 30 103 m , pressure, p = 1030 kg/m 3 9.8 m/s2 30 103 m = 303 Pa (b) When depth, h = 70 mm = 70 103 m , pressure, p = 1030 kg/m 3 9.8 m/s2 70 103 m = 707 Pa 3. A U-tube manometer is used to determine the pressure at a depth of 500 mm below the free surface of a fluid. If the pressure at this depth is 6.86 kPa, calculate the density of the liquid used in the manometer. Take the gravitational acceleration as 9.8 m/s 2 Pressure, p = gh hence, 6.86 103 Pa = × 9.8 m/s2 ×500 103 m 6.86 103 from which, density of liquid, = = 1400 kg/m 3 9.8 500 103 332 © John Bird Published by Taylor and Francis EXERCISE 130, Page 283 1. The height of a column of mercury in a barometer is 750 mm. Determine the atmospheric pressure, correct to 3 significant figures. Take the gravitational acceleration as 9.8 m/s 2 and the density of mercury as 13600 kg/m. Atmospheric pressure, p = gh = 13600 kg/m 3 9.8 m/s2 750 103 m = 99960 Pa = 100 kPa 2. A U-tube manometer containing mercury gives a height reading of 250 mm of mercury when connected to a gas cylinder. If the barometer reading at the same time is 756 mm of mercury, calculate the absolute pressure of the gas in the cylinder, correct to 3 significant figures. Take the gravitational acceleration as 9.8 m/s 2 and the density of mercury as 13600 kg/m. 3 2 3 Pressure, p1 = gh = 13600 kg/m 9.8 m/s 250 10 m = 33320 Pa = 33.32 kPa 3 2 3 Pressure, p2 = gh = 13600 kg/m 9.8 m/s 756 10 m = 100760 Pa = 100.76 kPa Absolute pressure = atmospheric pressure + gauge pressure = p2 + p1 = 100.76 + 33.32 = 134 kPa 3. A water manometer connected to a condenser shows that the pressure in the condenser is 350 mm below atmospheric pressure. If the barometer is reading 760 mm of mercury, determine the absolute pressure in the condenser, correct to 3 significant figures. Take the gravitational acceleration as 9.8 m/s 2 and the density of water as 100 kg/m. 3 2 3 Pressure, p1 = - 1 gh 1 = - 1000 kg/m 9.8 m/s 350 10 m 333 © John Bird Published by Taylor and Francis = - 3430 Pa = - 3.43 kPa 3 2 3 Pressure, p2 = 2 gh 2 = 13600 kg/m 9.8 m/s 760 10 m = 101293 Pa = 101.3 kPa Absolute pressure = atmospheric pressure + gauge pressure = p2 + p1 = 101.3 - 3.43 = 97.9 kPa 4. A Bourdon pressure gauge shows a pressure of 1.151 MPa. If the absolute pressure is 1.25 MPa, find the atmospheric pressure in millimetres of mercury. Take the gravitational acceleration as 9.8 m/s 2 and the density of mercury as 13600 kg/m. Atmospheric pressure = absolute pressure - gauge pressure = 1.25 MPa – 1.151 MPa = 0.099 MPa = 0.099 106 Pa Atmospheric pressure, p = gh = 13600 kg/m 3 9.8 m/s2 h i.e. 0.099 106 Pa = 13600 kg/m 3 9.8 m/s2 h 0.099 106 Pa from which, height, h = = 0.743 m 13600kg / m32 9.8m / s 1000mm i.e. atmospheric pressure in millimetres of mercury = 0.743 m = 743 mm 1m 334 © John Bird Published by Taylor and Francis EXERCISE 131, Page 285 1. A body of volume 0.124 m 3 is completely immersed in water of density 1000 kg/m 3 . What is the apparent loss of weight of the body? Take the gravitational acceleration as 9.8 m/s 2 . Mass, m = density, ρ × volume, V = 1000 kg/m 3 × 0.124 m 3 = 124 kg Apparent loss of weight of the body, W = ρ × V × g = 124 kg × 9.8 m/s 2 = 1215 N = 1.215 kN 2. A body of weight 27.4 N and volume 1240 cm 3 is completely immersed in water of specific weight 9.81 kN/m 3 . What is its apparent weight? Take the gravitational acceleration as 9.8 m/s 2 and the density of water as 1000 kg/m 3 . Body weight, W 1 = 27.4 N Apparent weight, W 2 = 27.4 - ρ × V × g = 27.4 – (1000 kg/m 3 × 1240 1063 m × 9.8 m/s 2 ) = 27.4 N – 12.152 N = 15.25 N 3. A body weighs 512.6 N in air and 256.8 N when completely immersed in oil of density 810 kg/m 3 . What is the volume of the body? Take the gravitational acceleration as 9.8 m/s 2 . W = ρ oil × V × g i.e. (512.6 – 256.8) = ρ oil × V × g i.e. 255.8 = 810 × V × 9.8 335 © John Bird Published by Taylor and Francis 255.8 from which, volume, V = = 0.03222 m 3 or 32.22 dm 3 810 9.8 4. A body weighs 243 N in air and 125 N when completely immersed in water. What will it weigh when completely immersed in oil of relative density 0.8? Take the gravitational acceleration as 9.8 m/s 2 and the density of water as 1000 kg/m 3 . W = ρ × V × g i.e. (243 – 125) = ρ water × V × g i.e. 118 = 1000 × V × 9.8 118 from which, volume, V = = 0.012041 m 3 1000 9.8 Weight in oil = 243 - ρ oil × V × g = 243 – (0.8 × 1000) × 0.012041 × 9.8 = 243 – 94.4 = 148.6 N 5. A watertight rectangular box, 1.2 m long and 0.75 m wide, floats with its sides and ends vertical in water of density 1000 kg/m 3 . If the depth of the box in the water is 280 mm, what is its weight? Take the gravitational acceleration as 9.8 m/s 2 . Volume of box, V = 1.2 m × 0.75 m × 280 103 m = 0.252 m 3 Weight of box, W = ρ × V × g = 1000 × 0.252 × 9.8 = 2469.6 N = 2.47 kN 6. A body weighs 18 N in air and 13.7 N when completely immersed in water of density 1000 kg/m 3 . What is the density and relative density of the body? Take the gravitational acceleration as 9.8 m/s 2 . 336 © John Bird Published by Taylor and Francis W = ρ × V × g i.e. (18 – 13.7) = ρ water × V × g i.e. 4.3 = 1000 × V × 9.8 4.3 from which, volume, V = = 4.388 104 m 3 1000 9.8 mass 18 N 1 Density of body, ρ = volume 4.388 1043 m 9.8 N / kg = 4186 kg/ m 3 or 4.186 tonnes/ m 3 density 4186kg / m3 Relative density = = 4.186 densityof water 1000kg / m3 7. A watertight rectangular box is 660 mm long and 320 mm wide. Its weight is 336 N. If it floats with its sides and ends vertical in water of density 1020 kg/m 3 , what will be its depth in the water? Take the gravitational acceleration as 9.8 m/s 2 . Volume of box, V = 660 103 m × 320 103 m × D where D = depth of box Weight, W = ρ × V × g i.e. 336 = 1020 kg/m 3 ×( 660 103 m × 320 103 m × D) × 9.8 336 from which, depth of box, D = = 0.159 m = 159 mm 1020 0.66 0.32 9.8 8. A watertight drum has a volume of 0.165 m 3 and a weight of 115 N. It is completely submerged in water of density 1030 kg/m 3 , held in position by a single vertical chain attached to the underside of the drum.
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