Question

5. A cylindrical tank is having a hemispherical base. The height of cylindrical portion is 5 m and diameter is 4 m. At the bottom of this tank an orifice of diameter 200 mm is fitted. Find the time required to completely emptying the tank. Take Cd 0.6 5 m 4.0 m- 2.0m ORIFICE

Answer 1

C_d = 0.6 Diameter of cylindrical portion = 4 m height of water above the spherical base = 4 m area of orifice(d) = pi/4 times 0.2 = 0.157 m^2 Let us split the tank into two portion i.e, cylindrical and hemispherical and then find out the time required to empty both of them independently. We know that the initial height of water above the orifice H_1 = 5 + 2 H_1 = 7m and the surface area of tank A = pi/4 times D^2 = pi/4 times 4^2 = 12.566 m^2 \r\n Time required to empty the cylindrical portion T_1 = 2A(squareroot H_1 - squareroot H_2)/c_d times a^2 times squareroot 2 g = 2 times 12.566 times (squareroot 7 - squareroot 2)/0.6 times 0.157 times squareroot 2 times 9.81 = 74.17 almostequalto 75s and the time required to empty the hemispherical base, T_2 = 14 pi R^5/2/15 c_d a squareroot 2 g = 14 times pi times (2)^5/2

Answer 2

569.3973 seconds.