(a) Show from first principles that the hydrostatic force acting on one side of

Question

(a) Show from first principles that the hydrostatic force acting on one side of an immersed vertical flat surface due to a fluid at rest is given by R = rho g Ay where R - hydrostatic force rho density of fluid A -area of surface y - depth to centroid of surface (b) A closed tank, rectangular in plan and with vertical sides, is 2.5 m deep and contains water. A 10 kg solid plastic sphere floats on the water surface, with the water level being 1.8 m above the base of the tank. The enclosed space above the water is pressurised with air to a gauge pressure of 30 kN/m^2. The length and width of the tank are 3 m and 1.2 m respectively. A circular opening on one of the vertical sides is sealed with a steel plate. The opening has a diameter of 25 cm and its centre is located 40 cm above the base of the tank. A U-tube is used to monitor the air pressure in the tank. It consists of two connected tubes, A and B, having cross-sectional areas 600 mm^2 and 300 mm^2, respectively. Tube A is vertical with its end open to atmosphere while tube B is connected to a point 10 cm above the water level in the tank. The surface of separation between the manometric fluid and the compressed air from the tank is located in tube B, while that between the manometric fluid and the atmosphere is in Tube A. (i) Compute the hydrostatic forces acting on the circular steel plate. (ii) Compute the hydrostatic forces acting on the base of the tank. (iii) Determine the manometric reading in millimetres of water. (iv) The level on the manometric fluid in Tube A is observed to increase by 4cm. Estimate the resulting percentage increase in the hydrostatic force on the circular steel plate. Relative density of manometric fluid: 1.4

Explanation / Answer

a) consider a vertical plate submerged in a liquid of density rho on one side, and depth of plate centroid be y

then consider depth x inside the fluid on the plate

consider a small area dA at this location

force on this small area, dF = PdA

where P is hydrostatic pressure at this location

but P = Po + rho*g*x

so dF = rho*g*xdA + PodA

for Po = 0

now definitiaion of centroid of an area is

z = {integral}xdA/{integral}dA

so, zA = {integral}xdA

dF = rho*g*x*dA

integrating both sides we get

F = rho*g*{integrate}x*dA

F = rho*g*yA [ where y is depth of centroid]

b) i) Hydrostatic force on the circular plate = rho*g*yA + Po*A + mg

here mg is weight of the ball floating on the water

where Po = 30,000 Pa

A = pi(0.25/2)^2 = 0.04908 m^2

y = 1.8 - 0.4 = 1.4 m

m = 10 kg

so, F = 0.04908(1000*9.81*1.4 + 30,000) + 9.81*10 = 2244.5647 N

ii) hydrostatic force on the base of the tank, F = (Po + rho*g*d)A + mg

A = area of base of tank =3*1.2 = 3.6 m^2

Po = 30,000Pa

d = depth of water in tank = 1.8 m

m = 10 kg

F = (30,000 + 1000*9.81*1.8)3.6 + 10*9.81 = 171666.9 N

iii) assume that the u tube has water inside it

then the force at the bottom of the tube has to be the same

so, let pressure inside the chamber be P ( gauge pressure)

let atmospheric pressure be Pa

(P + rho*g*h1) = rho*g*h2

here h1 is height of the water column in the left pipe ( connected to the tank)

h2 is height of the water column in the right pipe

A1 is area of cross section of left pipe

A2 is area of cross section of the right tube

(30,000 + 1000*9.81*h1) = 1000*9.81*h2

also, h1*A1 = h2*A2 [ from conservation of volume of the fluid assume incompressibel condiditons]

h1 = H2*A2/A1 = h2*(3/6) = h2/2

30 + 9.81*h2/2 = 9.81*h2

30 = 9.81*h2/2

h2 = 6.116 m

h1 = 3.05 m

so manometric reading is height in the right tube that is h1 = 3.05 m = 3050 mm

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