Chapter 14, Problem 73P

(a)

To determine

The initial velocity from the tank.

Explanation of Solution

Given:

Diameter of the tank is $2.4\text{ m}$.

Level of water in the tank is $4\text{ m}$.

Diameter of the sharp edged orifice is $10\text{cm}$.

Calculation:

Applying energy equation at points 1 at the free surface of water in the tank and point 2 at the exit of the orifice.

  $\begin{array}{l}\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1{}^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2{}^2}{2g}+z_2+h_{turbine}+h_L\\ z_1={\alpha }_2\frac{V_2{}^2}{2g}+h_L\\ z_1={\alpha }_2\frac{V_2{}^2}{2g}+K_L\frac{V_2{}^2}{2g}\\ 2g{z}_1=\left({\alpha }_2+K_L\right)V_2{}^2\\ V_2=\sqrt{\frac{2g{z}_1}{{\alpha }_2+K_L}}\\ V_2=\sqrt{\frac{2\left(9.81\text{m}/{\text{s}}^2\right)\left(4\text{ m}\right)}{1+0.5}}\\ V_2=7.23\text{m}/\text{s}\end{array}$

Thus, the initial velocity from the tank is $7.23\text{m}/\text{s}$.

(b)

To determine

The time required to empty the tank.

Explanation of Solution

Calculation:

The flow rate is given by,

  $\begin{array}{c}\dot{V}=A_{orificce}{V}_2\\ =\frac{\pi D^2}{4}\sqrt{\frac{2gz}{1+K_L}}\end{array}$

The quantity of water flowing through the orifice in time dt is,

  $\begin{array}{l}dV=\dot{V}dt\\ A_{tank}\left(-dz\right)=\frac{\pi D^2}{4}\sqrt{\frac{2gz}{1+K_L}}dt\\ -\frac{\pi D_0{}^2}{4}dz=\frac{\pi D^2}{4}\sqrt{\frac{2gz}{1+K_L}}dt\\ dt=-{\left(\frac{D_0}{D}\right)}^2\sqrt{\frac{1+K_L}{2gz}}dz\end{array}$

Integrating the above equation with respect to limits,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{t_f}{\int }}dt}={\displaystyle \underset{z_1}{\overset{0}{\int }}-{\left(\frac{D_0}{D}\right)}^2\sqrt{\frac{1+K_L}{2gz}}dz}\\ t_f=-{\left(\frac{D_0}{D}\right)}^2\sqrt{\frac{1+K_L}{2g}}{\left[\frac{\sqrt{z}}{\frac{1}{2}}\right]}_{z_1}^0\\ t_f=-{\left(\frac{D_0}{D}\right)}^2\sqrt{\frac{1+K_L}{2g}}\left[0-2\sqrt{z_1}\right]\end{array}$

  $\begin{array}{l}{t}_f=2{\left(\frac{D_0}{D}\right)}^2\sqrt{\frac{z_1\left(1+K_L\right)}{2g}}\\ t_f=2{\left(\frac{2.4\text{ m}}{0.1\text{ m}}\right)}^2\sqrt{\frac{\left(4\text{ m}\right)\left(1+0.5\right)}{2\left(9.81\text{ m}/{\text{s}}^2\right)}}\\ t_f=637\text{ s}\\ t_f=10.6\text{ min}\end{array}$

Thus, the time required to empty the tank is $10.6\text{ min}$.

Neglecting the loss coefficient,

  $\begin{array}{l}{t}_f=2{\left(\frac{D_0}{D}\right)}^2\sqrt{\frac{z_1\left(1+K_L\right)}{2g}}\\ t_f=2{\left(\frac{2.4\text{ m}}{0.1\text{ m}}\right)}^2\sqrt{\frac{\left(4\text{ m}\right)\left(1+0\right)}{2\left(9.81\text{ m}/{\text{s}}^2\right)}}\\ t_f=520\text{ s}\\ t_f=8.7\text{ min}\end{array}$

Thus, the loss coefficient causes the draining time to increase.