Question

SITUATIONAL PROBLEM A vertical storage tank with a hemispherical bottom and a cylindrical shell of 2 m. inside diameter and 4 m. high is filled with water. A sharp-edge orifice 150 mm diameter with C = 0.62 is located at the lowest point. 26. Determine the time to empty the cylindrical part of the tank through the orifice. A. 134.04 sec C. 152.11 sec B. 145.6 sec D. 160 sec 27. Determine the time to empty the hemispherical tank only through the orifice. A. 60.41 sec C. 75.11 sec B. 53.61 sec D. 48.11 sec 28. Determine the time to empty the whole tank through the orifice. A. 213.61 sec C. 220.41 sec B. 187.65 sec D. 194.45 sec

Answer 1

Height of Cylinderical portion = 4 m

m FICE

Dia of tank = 2 m

Area, A =$\frac{\Pi }{4} (2)^{2}$= 3.1415 m²

Dia of Orifice = 150 mm = 0.15 m

Area, a = $\frac{\Pi }{4} (0.15)^{2}$ = 0.01767 m²

C_d = 0.62

26. For cylinderical portion, H₁= 4+1 = 5m (from bottomost point to top most point) and H₂= 1m ( from bottom most point to starting of cylinderical portion

Time taken, T₁ = $\frac{2A\left [ \sqrt{H_{1}} -\sqrt{H_{2}} \right ]}{C_{d}\times a\times \sqrt{2g}} = \frac{2\times 3.1415\left [ \sqrt{5} -\sqrt{1} \right ]}{ 0.62\times 0.01767\times \sqrt{2\times 9.81}} =160.04 sec$

T₁ = 160sec option - D

27 For hemispherical portion, H₁= 1m ( from bottom most point to starting of cylinderical portion and H₂ = 0 m (bottom most)

Time taken, T₂=  $\frac{\Pi }{C_{d}\times a\times \sqrt{2g}}\left [ \frac{4}{3}RH_{1}^{3/2} - \frac{2}{5}H_{1}^{5/2} \right ] = \frac{\Pi }{0.62\times 0.01767\times \sqrt{2\times 9.81}}\left [ \frac{4}{3}\times 1\times 1^{3/2} - \frac{2}{5}\times 1^{5/2} \right ]$

T₂ = 60.42 sec and option - A

28. Total Time taken to empty the tank = T = T₁ + T₂ = 160.04 + 60.42 = 220.46 seconds and option - C