Question

 A vertical cylindrical tank of 1 m diameter is filled with water up to a height of 5 m from its bottom. Top surface of water is exposed to atmosphere. A hole of 5 mm\(^2\) area forms at the bottom of the tank. Considering the coefficient of discharge of the hole to be unity and the acceleration due to gravity to be 10 m/s\(^2\), the rate of leakage of water (in litre/min) through the hole from the tank to the atmosphere, under the given conditions, is _______.

Answer 1

Concept:

The actual discharge through Hole is given by:

\(Q = {C_d} × A × \sqrt {2gh} \)

where,

Cd is the coefficient of discharge, A is the cross-sectional area of the hole, h is the differential head, g = acceleration due to gravity.

Calculation:

Given:

Cd = 1, A = 5 mm² = 5 × 10^-6, h = 5 m, g = 10 m/s².

Now, the actual discharge through hole is given by:

\(Q = {C_d} × A × \sqrt {2gh} \)

\(Q = {1} × 5×10^{-6} × \sqrt {2× 10 × 5} \)

\(Q = 5× 10^{-5} \;m^3/sec\)

\(Q = 3.0 \;L/min\)      [∵ 1 m³ = 1000 L and 1 min = 60 sec]