Chapter 12, Problem 63P

a)

To determine

The maximum flow of water.

Explanation of Solution

Given:

The efficiency of the pump $\left({\eta }_{pump}\right)$ is 78%.

The input power $\left({\dot{W}}_{pump.in}\right)$ is $\text{5}\text{kW}$.

Height of the pool from the underwater surface $\left(h\right)$ is 30 m.

The diameter of the intake pipe $\left(d_1\right)$ is 0.07 m.

The diameter of the discharge side pipe $\left(d_2\right)$ is 0.05 m.

Calculation:

Draw the free body diagram of the system as in Figure 1.

Consider the acceleration due to gravity (g) as $9.81{\text{m/s}}^2$.

Calculate the work done by the pump $\left({\dot{W}}_{pump,out}\right)$ using the relation;

  $\begin{array}{c}{\eta }_{pump}=\frac{{\dot{W}}_{pump,out}}{{\dot{W}}_{pump,in}}\\ {\dot{W}}_{pump,out}={\eta }_{pump}{\dot{W}}_{pump,in}\\ =0.78\left(5\right)\\ =3.9\text{kW}\end{array}$

Write the Bernoulli’s equation between the points 1 and 2 as follows;

  $\dot{m}\left(\frac{P_1}{\rho }+{\alpha }_1\frac{V_1^2}{2}+g{z}_1\right)+{\dot{W}}_{pump,out}=\dot{m}\left(\frac{P_2}{\rho }+{\alpha }_2\frac{V_2^2}{2}+g{z}_2\right)+{\dot{W}}_{turbine,e}+E_{mech}{}_{,loss}$

    $\begin{array}{c}\dot{m}\left(\frac{P_1}{\rho }+{\alpha }_1\frac{V_1^2}{2}+g{z}_1\right)+{\dot{W}}_{pump,out}=\dot{m}\left(\frac{P_2}{\rho }+{\alpha }_2\frac{V_2^2}{2}+g{z}_2\right)+0+E_{mech}{}_{,loss}\\ \dot{m}\left(\frac{P_{atm}}{\rho }+0+0\right)+{\dot{W}}_{pump,out}=\dot{m}\left(\frac{P_{atm}}{\rho }+0+9.81\left(30\right)\right)+E_{mech}{}_{,loss}\\ {\dot{W}}_{pump,out}=9.81\left(30\right)\dot{m}+E_{mech}{}_{,loss}\end{array}$

Substitute $\dot{m}g{h}_L$ for $E_{mech}{}_{,loss}$.

  $\begin{array}{c}{\dot{W}}_{pump,out}=9.81\left(30\right)\dot{m}+\dot{m}g{h}_L\\ 3.9\times {10}^3=\dot{m}\left(9.81\left(30\right)+9.81\left(4\right)\right)\\ \dot{m}=\frac{3.9\times {10}^3}{9.81\left(30\right)+9.81\left(4\right)}\\ =11.69\text{kg/s}\end{array}$

Calculate the maximum flow of water through the pump $\left(Q\right)$ using the relation;

  $\begin{array}{c}Q=\frac{\dot{m}}{\rho }\\ =\frac{11.69}{1000}\\ =0.01169{\text{m}}^{\text{3}}\text{/s}\end{array}$

Thus, the maximum flow of water through the pump is $\underset{\_}{0.01169{\text{m}}^{\text{3}}\text{/s}}$.

b)

To determine

The Pressure difference across the pump.

Explanation of Solution

Given:

The efficiency of the pump $\left({\eta }_{pump}\right)$ is 78%.

The input power $\left({\dot{W}}_{pump.in}\right)$ is $\text{5}\text{kW}$.

The datum head at point 2 $\left(z_2\right)$ is 30 m.

The diameter of the intake pipe $\left(d_1\right)$ is 7 cm.

The diameter of the discharge side pipe $\left(d_2\right)$ is 5 cm.

Calculation:

Draw the free body diagram of the system as in Figure 1.

Calculate the area of intake pipe using the formula;

  $\begin{array}{c}{A}_1=\frac{\pi d_1{}^2}{4}\\ =\frac{\pi {\left(0.07\right)}^2}{4}\\ =3.848\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Calculate the discharge of water through the pipe at the intake side using the expression;

  $\begin{array}{c}{V}_1=\frac{Q}{A_1}\\ =\frac{0.01169}{3.848\times {10}^{-3}}\\ =3.038\text{m/s}\end{array}$

Calculate the area of intake pipe using the formula;

  $\begin{array}{c}{A}_2=\frac{\pi d_2{}^2}{4}\\ =\frac{\pi {\left(0.05\right)}^2}{4}\\ =1.963\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Calculate the discharge of water through the pipe at the intake side using the expression;

  $\begin{array}{c}{V}_2=\frac{Q}{A_2}\\ =\frac{0.01169}{1.963\times {10}^{-3}}\\ =5.955\text{m/s}\end{array}$

Write the Bernoulli’s equation between the intake side and discharge side as follows;

  $\begin{array}{c}\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1+h_{pump,out}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+z_2+h_{turbine,e}+h_L\\ \frac{P_1}{\rho g}+\left(1\right)\frac{V_1^2}{2g}+0+h_{pump,out}=\frac{P_2}{\rho g}+\left(1\right)\frac{V_2^2}{2g}+0+0+0\\ \frac{P_1-P_2}{\rho g}=\frac{V_2^2-V_1^2}{2g}-h_{pump,out}\\ \frac{P_1-P_2}{\rho g}=\frac{V_2^2-V_1^2}{2g}-\frac{{\dot{W}}_{pump,out}}{\dot{m}g}\end{array}$

  $\begin{array}{c}\frac{P_1-P_2}{\rho g}=\frac{1}{g}\left(\frac{V_2^2-V_1^2}{2}-\frac{{\dot{W}}_{pump,out}}{\dot{m}}\right)\\ P_1-P_2=\rho \left(\frac{V_2^2-V_1^2}{2}-\frac{{\dot{W}}_{pump,out}}{\dot{m}}\right)\\ P_2-P_1=\rho \left(\frac{{\dot{W}}_{pump,out}}{\dot{m}}-\frac{V_2^2-V_1^2}{2}\right)\end{array}$

Substitute 11.69 kg/s for $\dot{m}$, 3.9 kW for ${\dot{W}}_{pump,out}$, 5.955 m/s for $V_2$, 3.038 m/s for $V_1$ and $\text{1000}{\text{kg/m}}^{\text{3}}$ for $\rho$.

  $\begin{array}{c}{P}_2-P_1=1000\left(\frac{3.9\times {10}^3}{11.69}-\frac{{5.955}^2-{3.038}^2}{2}\right)\\ =1000\left(333.62-13.12\right)\\ \text{=320503}\text{.7 Pa}\\ \text{=320}\text{.5}\text{kPa}\end{array}$

Thus, the Pressure difference across the pump is $\underset{\_}{\text{320}\text{.5}\text{kPa}}$.