Chapter 12, Problem 47P

To determine

The mechanical efficiency of the pump.

Explanation of Solution

Given:

The electric power generated by the pump is ${\dot{W}}_{\text{electric}}=25\text{kW}$.

The density of the oil is $\rho =860\text{kg}/{\text{m}}^3$.

The rate of pumping is $\dot{V}=0.1{\text{m}}^3/\sec$.

The inlet diameter of the pipe is $D_1=8\text{cm}$.

The outlet diameter of the pipe is $D_2=12\text{cm}$.

The pressure rise in the oil is $\Delta P=P_2-P_1=250\text{kPa}$.

The efficiency of the motor is ${\eta }_{\text{motor}}=90\%$.

The kinetic energy correction factor is $\alpha =1.05$.

Calculation:

Consider the point 1 is the inlet of the pump and the point 2 is the outlet of the pump.

The elevation difference at the point 1 and point 2 are same. Therefore, $z_1=z_2$.

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

Find the velocity at point 1 $\left(V_1\right)$ using the equation;

  $\begin{array}{c}{V}_1=\frac{\dot{V}}{A_1}\\ =\frac{4\dot{V}}{\pi D_1^2}\end{array}$

Substitute $\dot{V}=0.1{\text{m}}^3/\text{s}$ and $D_1=8\text{cm}$ in the above equation;

  $\begin{array}{c}{V}_1=\frac{4\times 0.1}{\pi \times {\left(8\text{cm}\times \frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}^2}\\ =19.894\text{m}/\sec \end{array}$

Find the velocity at point 2 $\left(V_2\right)$ using the equation;

  $\begin{array}{c}{V}_2=\frac{\dot{V}}{A_2}\\ =\frac{4\dot{V}}{\pi D_2^2}\end{array}$

Substitute $\dot{V}=0.1{\text{m}}^3/\text{s}$ and $D_2=12\text{cm}$ in the above equation;

  $\begin{array}{c}{V}_2=\frac{4\times 0.1}{\pi \times {\left(12\text{cm}\times \frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}^2}\\ =8.842\text{m}/\sec \end{array}$

Write the Energy equation in terms of the heads between the points 1 and 2 as follows;

  $\begin{array}{c}\frac{P_1}{\rho g}+\alpha \frac{V_1^2}{2g}+z_1+h_{\text{pump},u}=\frac{P_2}{\rho g}+\alpha \frac{V_2^2}{2g}+z_2+h_{\text{turbine},\text{e}}+h_L\\ \frac{P_1}{\rho g}+\alpha \frac{V_1^2}{2g}+z_1+h_{\text{pump},u}=\frac{P_2}{\rho g}+\alpha \frac{V_2^2}{2g}+z_1+0+0\\ h_{\text{pump},u}=\frac{P_2}{\rho g}+\alpha \frac{V_2^2}{2g}-\frac{P_1}{\rho g}-\alpha \frac{V_1^2}{2g}\\ =\frac{P_2-P_1}{\rho g}+\alpha \frac{V_2^2-V_1^2}{2g}\end{array}$

Substitute $P_2-P_1=250\text{kPa}$, $\rho =860\text{kg}/{\text{m}}^3$, $g=9.81\text{m}/{\sec }^2$, $\alpha =1.05$, $V_2=8.842\text{m}/\sec$, and $V_1=19.894\text{m}/\sec$ in the above equation;

  $\begin{array}{c}{h}_{\text{pump},u}=\frac{250\text{kPa}\times \frac{1,000\text{N}/{\text{m}}^2}{1\text{kPa}}\times \frac{1\text{kg}\cdot \text{m}/{\sec }^2}{1\text{N}}}{860\times 9.81}+1.05\times \frac{\left({8.842}^2-{19.894}^2\right)}{2\times 9.81}\\ =12.634\text{m}\end{array}$

Find the useful pumping power $\left({\dot{W}}_{\text{pumps},\text{u}}\right)$ using the relation;

  ${\dot{W}}_{\text{pumps},\text{u}}=\rho \dot{V}g{h}_{\text{pump},\text{u}}$

Substitute $\rho =860\text{kg}/{\text{m}}^3$, $\dot{V}=0.1{\text{m}}^3/\text{s}$, $g=9.81\text{m}/{\sec }^2$, and $h_{\text{pump},u}=12.634\text{m}$ in the above equation;

  $\begin{array}{c}{\dot{W}}_{\text{pumps},\text{u}}=860\times 0.1\times 9.81\times 12.634\\ =10.659\times {10}^3\text{W}\times \frac{1\text{kW}}{1,000\text{W}}\\ =10.659\text{kW}\end{array}$

Find the shaft pumping power $\left({\dot{W}}_{\text{pump},\text{shaft}}\right)$ using the equation;

  ${\dot{W}}_{\text{pump},\text{shaft}}={\eta }_{\text{motor}}{\dot{W}}_{\text{electric}}$

Substitute ${\eta }_{\text{motor}}=90\%$ and ${\dot{W}}_{\text{electric}}=25\text{kW}$ in the above equation;

  $\begin{array}{c}{\dot{W}}_{\text{pump},\text{shaft}}=\frac{90}{100}\times 25\\ =22.5\text{kW}\end{array}$

Find the mechanical efficiency of the pump $\left({\eta }_{\text{pump}}\right)$ using the relation;

  ${\eta }_{\text{pump}}=\frac{{\dot{W}}_{\text{pump},\text{u}}}{{\dot{W}}_{\text{pump},\text{shaft}}}\times 100\%$

Substitute ${\dot{W}}_{\text{pump},\text{u}}=10.6\text{kW}$ and ${\dot{W}}_{\text{pump},\text{shaft}}=22.5\text{kW}$ in the above equation;

  $\begin{array}{c}{\eta }_{\text{pump}}=\frac{10.6}{22.5}\times 100\%\\ =47.1\%\end{array}$

Therefore, the mechanical efficiency of the pump is $\underset{\_}{47.1\%}$.