CHAPTER 8 PRESSURE IN FLUIDS EXERCISE 18, Page 81 1. A force of 80 N is applied to a piston of a hydraulic system of cross-sectional area 0.010 m. Determine the pressure produced by the piston in the hydraulic fluid. Pressure, p = force area = 80 N 0.010 m 8000 Pa = 8 kpa That is, the pressure produced by the piston is 8 kpa. Find the force on the piston of question 1 to produce a pressure of 450 kpa. Pressure, p = 450 kpa = 450000 Pa Pressure p = force area hence, force = pressure area = 4540000 0.010 = 4500 N = 4.5 kn. If the area of the piston in question 1 is halved and the force applied is 80 N, determine the new pressure in the hydraulic fluid. New area = 0.010 0.005 m New pressure, p = force area = 80 N 56000Pa = 56 kpa 0.005m 1
EXERCISE 19, Page 8 1. Determine the pressure acting at the base of a dam, when the surface of the water is 5 m above base level. Take the density of water as 1000 kg/m. Take the gravitational acceleration as 9.8 m/s. Pressure at base of dam, p = gh = 1000 kg/m 9.8 0.5 m = 4000 Pa = 4 kpa. An uncorked bottle is full of sea water of density 100 kg/m. Calculate, correct to significant figures, the pressures on the side wall of the bottle at depths of (a) 0 mm, and (b) 70 mm below the top of the bottle. Take the gravitational acceleration as 9.8 m/s. Pressure on the side wall of the bottle, p = gh (a) When depth, h = 0 mm = 0, pressure, p = 100 kg/m 9.8 (b) When depth, h = 70 mm = 70, pressure, p = 100 kg/m 9.8 0 = 0 Pa 70 = 707 Pa. 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 Pressure, p = gh hence, 6.86 10 Pa = 9.8 m/s 500 from which, density of liquid, = 6.86 10 9.8 500 10 = 1400 kg/m
EXERCISE 10, Page 8 1. The height of a column of mercury in a barometer is 750 mm. Determine the atmospheric pressure, correct to significant figures. Take the gravitational acceleration as 9.8 m/s and the density of mercury as 1600 kg/m. Atmospheric pressure, p = gh = 1600 kg/m 9.8 = 99960 Pa = 100 kpa 750. A U-tube manometer containing mercury gives a height reading of 50 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 significant figures. Take the gravitational acceleration as 9.8 m/s and the density of mercury as 1600 kg/m. Pressure, p 1 = gh = 1600 kg/m 9.8 = 0 Pa =. kpa Pressure, p = gh = 1600 kg/m 9.8 = 100760 Pa = 100.76 kpa 50 756 Absolute pressure = atmospheric pressure + gauge pressure = p + p 1 = 100.76 +. = 14 kpa. A water manometer connected to a condenser shows that the pressure in the condenser is 50 mm below atmospheric pressure. If the barometer is reading 760 mm of mercury, determine the absolute pressure in the condenser, correct to significant figures. Take the gravitational acceleration as 9.8 m/s and the density of water as 100 kg/m. Pressure, p 1 = - 1 gh 1 = - 1000 kg/m 9.8 50
= - 40 Pa = -.4 kpa Pressure, p = gh = 1600 kg/m 9.8 = 1019 Pa = 101. kpa 760 Absolute pressure = atmospheric pressure + gauge pressure = p + p 1 = 101. -.4 = 97.9 kpa 4. A Bourdon pressure gauge shows a pressure of 1.151 MPa. If the absolute pressure is 1.5 MPa, find the atmospheric pressure in millimetres of mercury. Take the gravitational acceleration as 9.8 m/s and the density of mercury as 1600 kg/m. Atmospheric pressure = absolute pressure - gauge pressure = 1.5 MPa 1.151 MPa = 0.099 MPa = Atmospheric pressure, p = gh = 1600 kg/m 9.8 h 6 0.099 10 Pa i.e. 6 0.099 10 Pa = 1600 kg/m 9.8 h from which, height, h = 6 0.099 10 Pa 1600kg / m 9.8m / s = 0.74 m i.e. atmospheric pressure in millimetres of mercury = 0.74 m 1000mm = 74 mm 1m 4
EXERCISE 11, Page 85 1. A body of volume 0.14 m is completely immersed in water of density 1000 kg/m. What is the apparent loss of weight of the body? Take the gravitational acceleration as 9.8 m/s. Mass, m = density, ρ volume, V = 1000 kg/m 0.14 m = 14 kg Apparent loss of weight of the body, W = ρ V g = 14 kg 9.8 m/s = 115 N = 1.15 kn. A body of weight 7.4 N and volume 140 cm is completely immersed in water of specific weight 9.81 kn/m. What is its apparent weight? Take the gravitational acceleration as 9.8 m/s and the density of water as 1000 kg/m. Body weight, W 1 = 7.4 N Apparent weight, W = 7.4 - ρ V g = 7.4 (1000 kg/m 6 140 9.8 m/s ) = 7.4 N 1.15 N = 15.5 N. A body weighs 51.6 N in air and 56.8 N when completely immersed in oil of density 810 kg/m. What is the volume of the body? Take the gravitational acceleration as 9.8 m/s. W = ρ oil V g i.e. (51.6 56.8) = ρ oil V g i.e. 55.8 = 810 V 9.8 5
from which, volume, V = 55.8 810 9.8 = 0.0 m or. dm 4. A body weighs 4 N in air and 15 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 and the density of water as 1000 kg/m. W = ρ V g i.e. (4 15) = ρ water V g i.e. 118 = 1000 V 9.8 from which, volume, V = 118 1000 9.8 = 0.01041 m Weight in oil = 4 - ρ oil V g = 4 (0.8 1000) 0.01041 9.8 = 4 94.4 = 148.6 N 5. A watertight rectangular box, 1. m long and 0.75 m wide, floats with its sides and ends vertical in water of density 1000 kg/m. If the depth of the box in the water is 80 mm, what is its weight? Take the gravitational acceleration as 9.8 m/s. Volume of box, V = 1. m 0.75 m 80 = 0.5 m Weight of box, W = ρ V g = 1000 0.5 9.8 = 469.6 N =.47 kn 6. A body weighs 18 N in air and 1.7 N when completely immersed in water of density 1000 kg/m. What is the density and relative density of the body? Take the gravitational acceleration as 9.8 m/s. 6
W = ρ V g i.e. (18 1.7) = ρ water V g i.e. 4. = 1000 V 9.8 from which, volume, V = 4. 1000 9.8 = 4.88 10 4 m mass 18 N 1 Density of body, ρ = 4 volume 4.88 10 m 9.8 N / kg = 4186 kg/ m or 4.186 tonnes/ m Relative density = density 4186kg / m = 4.186 densityof water 1000kg / m 7. A watertight rectangular box is 660 mm long and 0 mm wide. Its weight is 6 N. If it floats with its sides and ends vertical in water of density 100 kg/m, what will be its depth in the water? Take the gravitational acceleration as 9.8 m/s. Volume of box, V = 660 0 D where D = depth of box Weight, W = ρ V g i.e. 6 = 100 kg/m ( 660 0 D) 9.8 from which, depth of box, D = 6 100 0.66 0. 9.8 = 0.159 m = 159 mm 8. A watertight drum has a volume of 0.165 m and a weight of 115 N. It is completely submerged in water of density 100 kg/m, held in position by a single vertical chain attached to the underside of the drum. What is the force in the chain? Take the gravitational acceleration as 9.8 m/s. Weight of drum, W = 115 N 7
Upthrust = ρ V g = 100 kg/m 0.165 m 9.8 m/s = 1665.5 N Hence, the force in the chain = 1665.5 115 = 1551 N = 1.551 kn EXERCISE 1, Page 91 Answers found from within the text of the chapter, pages 80 to 91. EXERCISE 1, Page 91 1. (b). (d). (a) 4. (a) 5. (c) 6. (d) 7. (b) 8. (c) 9. (c) 10. (d) 11. (d) 1. (d) 1. (c) 14. (b) 15. (c) 16. (a) 17. (b) 18. (f) 19. (a) 0. (b) 1. (c) 8