Chapter 14, Problem 78P

To determine

The power input to the pump.

Explanation of Solution

Given:

Temperature of water is $70°\text{F}$.

Velocity of water is $6\text{ft}/\text{min}$.

Diameter of the pipe is $\text{5 in}$.

Length of the pipe is $\text{125 ft}$.

The elevation difference is 12 ft.

Flow rate of water is $1.5{\text{ft}}^3/\text{s}$.

Pump efficiency is 70%.

Calculation:

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

  $\begin{array}{l}\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1{}^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2{}^2}{2g}+z_2+h_{turbine}+h_L\\ {\alpha }_1\frac{V_1{}^2}{2g}+h_{pump}=z_2+h_L\\ h_{pump}=z_2+h_L-{\alpha }_1\frac{V_1{}^2}{2g}\end{array}$        (I)

Refer Table A-15E “Properties of saturated water” from Appendix 2 and obtain the following properties of water:

  $\begin{array}{l}\text{Density, }\rho =62.3\text{lbm}/{\text{ft}}^3\\ \text{Dynamic viscocity, }\mu =6.556\times {10}^{-4}\text{lbm}/\text{ft}\cdot \text{s}\end{array}$

Calculate the velocity of flow.

  $\begin{array}{c}V=\frac{\dot{V}}{A_c}\\ =\frac{\left(1.5{\text{ft}}^3/\text{s}\right)}{\frac{\pi }{4}{\left(\frac{5}{12}\text{ ft}\right)}^2}\\ =11.0\text{ft}/\text{s}\end{array}$

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{\left(62.3\text{lbm}/{\text{ft}}^3\right)\left(11.0\text{ft}/\text{s}\right)\left(\frac{5}{12}\text{ ft}\right)}{\left(6.556\times {10}^{-4}\text{lbm}/\text{ft}\cdot \text{s}\right)}\\ =435500\end{array}$

Since the Reynolds number is greater than 4000, the flow is turbulent. Hence, the friction factor is determined from the Moody chart as follows:

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2.0\log \left(\frac{0.0005\text{ ft}/\left(5/12\text{ft}\right)}{3.7}+\frac{2.51}{435500\sqrt{f}}\right)\end{array}$

Solving the above equation, $f=0.0211$.

Calculate the total minor loss coefficient.

  $\begin{array}{c}{\displaystyle \sum K_L}=K_{L,entrance}+3K_{L,elbow}+K_{L,exit}\\ =0+3\left(0.3\right)+1.0\\ =1.9\end{array}$

Calculate the head loss.

  $\begin{array}{c}{h}_L=\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\\ =\left[\left(0.0211\right)\frac{\text{125 ft}}{\left(5/12\right)\text{ ft}}+1.9\right]\frac{{\left(11\text{ft}/\text{s}\right)}^2}{2\left(32.2{\text{ft}/\text{s}}^2\right)}\\ =15.5\text{ft}\end{array}$

Calculate $h_{pump}$ form Equation (I).

  $\begin{array}{c}{h}_{pump}=z_2+h_L-{\alpha }_1\frac{V_1{}^2}{2g}\\ =12\text{ ft}+15.5\text{ft}-\frac{{\left(6\text{ft}/\text{s}\right)}^2}{2\left(32.2\text{ft}/{\text{s}}^2\right)}\\ =26.9\text{ft}\end{array}$

Calculate the power required by the pump.

  $\begin{array}{c}{\dot{W}}_{pump}=\frac{\dot{V}\rho g{h}_{pump}}{{\eta }_{pump}}\\ =\frac{\left(1.5{\text{ft}}^3/\text{s}\right)\left(62.3\text{lbm}/{\text{ft}}^3\right)\left(32.2{\text{ft}/\text{s}}^2\right)\left(26.9\text{ ft}\right)}{0.70}\left(\frac{1\text{ lbf}}{32.2\text{ lbm}\cdot \text{ft}/{\text{s}}^2}\right)\left(\frac{1\text{ kW}}{737\text{lbf}\cdot \text{ft}/\text{s}}\right)\\ =4.87\text{ kW}\end{array}$

Thus, the power input to the pump is $4.87\text{ kW}$.