Question

A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.0 m below the water level. The rate of flow from the leak is found to be $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{}{\mathit{m}}^{\mathbf{3}}\mathbf{/}\mathit{m}\mathit{i}\mathit{n}$. Determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole.

Short answer

1. The speed of the water leaving is$V=17.7m/s$
2. Diameter of the hole is$D=1.73\times {10}^{-3}m$

Step-by-step solution

Step 1:

Expression of the volume flow rate R in terms of area and speed of the water in the hose:

$R=A_1\times V_1$

Step 2:

Part(a):

Given:

Therateofflow(v),$v=2.5\times {10}^{-3}$

Height(h),$h=16m$

Velocity at which water leaves:

$$\begin{gathered} V=\sqrt{2gh} \\ V=\sqrt{2\times 9.8\times 16} \\ V=17.7m/s \end{gathered}$$

Step 3:

Part(b):

The area of the hole is:

$A=\frac{V}{v\times t}$

$$\begin{gathered} A=\frac{2.5\times {10}^{-3}}{17.7\times 60} \\ A=2.353\times {10}^{-6}{m}^2 \end{gathered}$$

Diameter of the hole is:

$\begin{array}{rcl}D& =& 2\sqrt{\frac{A}{\pi }}\\ D& =& 2\sqrt{\frac{2.33\times {10}^{-6}}{3.14}}\\ D& =& 1.73\times {10}^{-3}m\\ & & \end{array}$