Question

3. A hemispherical tank of diameter 4 m contains water upto a height of 1.5 m. An orifice of darmeter 50 mm is provided at the bottom. Find the time required by water (i) to fall from 5 m to 2 m (ii) for completely emptying to tank. Take Ca 0.6

Answer 1

Dia = 4m radius = 2m dia of orifice (d) = 50 mm = 0.05 m area (a) = pi/4 (0.05)^2 = 0.001963 m^2 initial height H_1 = 5m cd = 0.6 i)H_2 = 2m time T = pi/cd times a times squareroot 2g [4/3 R(H_1^3/2 -H_2^3/2]-2/5(H_1^5/2 -H_2^5/2] = pi/0.6 times 0.001913 times squareroot 2 times 9.81 [4/3 times 2 (5^3/2 -2^3/2) -2/3(5^5/2 -2^5/2)] = 602.189[22.271 -20.0979] = 1308.616 seconds = 21.81 min T = 21.81 min ii) H_2 = 0 T = pi/cd times a squareroot 2g [4/3 RH_1^3/2 -2/5 H_1^5/2] = pi/0.6 times 0.001963 times squareroot 2 times 9.81 [4/3 times 2 times 5^3/2 -2/5 times 2^5/2] = 602.189 [29.814 -2.262] T = 16591.51 sec = 276.52 min T = 276 min 52 sec