Chapter 14, Problem 76P

To determine

The increase or decrease in the pumping power required.

Explanation of Solution

Given:

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

Internal diameter of the pipe is $70\text{cm}$.

Length of the pipe is $1500\text{ m}$.

Surface roughness is 3 mm.

Thickness of the lining is 2 cm.

Surface roughness of the petroleum based lining is 0.04 mm.

Internal diameter of the pipe when lining is applied is $66\text{cm}$.

Density of water is $1000\text{kg}/{\text{m}}^3$.

Viscosity of water is $1\times {10}^{-6}{\text{m}}^2/\text{s}$.

Calculation:

Case 1: The tank is not lined with petroleum based lining.

Calculate the velocity of flow.

  $\begin{array}{c}V=\frac{\dot{V}}{A_c}\\ =\frac{\left(1.5{\text{m}}^3/\text{s}\right)}{\frac{\pi }{4}{\left(0.7\text{ m}\right)}^2}\\ =3.898\text{m}/\text{s}\end{array}$

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{VD}{\upsilon }\\ =\frac{\left(3.898\text{m}/\text{s}\right)\left(0.7\text{ m}\right)}{\left(1\times {10}^{-6}{\text{m}}^2/\text{s}\right)}\\ =2.728\times {10}^6\end{array}$

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

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2.0\log \left(\frac{0.003\text{ m}/0.7\text{ m}}{3.7}+\frac{2.51}{2.728\times {10}^6\sqrt{f}}\right)\end{array}$

Solving the above equation, $f=0.02904$.

Calculate the head loss.

  $\begin{array}{c}{h}_L=f\frac{L}{D}\frac{V^2}{2g}\\ =\left[\left(0.02904\right)\frac{\text{1500 m}}{0.7\text{ m}}\right]\frac{{\left(3.898\text{m}/\text{s}\right)}^2}{2\left(9.81{\text{m}/\text{s}}^2\right)}\\ =48.19\text{ m}\end{array}$

Calculate the power required by the pump.

  $\begin{array}{c}{\dot{W}}_{pump}=\dot{V}\rho g{h}_L\\ =\left(1.5{\text{m}}^3/\text{s}\right)\left(1000\text{kg}/{\text{m}}^3\right)\left(9.81{\text{m}/\text{s}}^2\right)\left(48.19\text{ m}\right)\\ =709.1\text{ kW}\end{array}$

Case 2: The tank is lined with petroleum based lining.

Calculate the velocity of flow.

  $\begin{array}{c}V=\frac{\dot{V}}{A_c}\\ =\frac{\left(1.5{\text{m}}^3/\text{s}\right)}{\frac{\pi }{4}{\left(0.66\text{ m}\right)}^2}\\ =4.384\text{m}/\text{s}\end{array}$

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{VD}{\upsilon }\\ =\frac{\left(4.384\text{m}/\text{s}\right)\left(0.66\text{ m}\right)}{\left(1\times {10}^{-6}{\text{m}}^2/\text{s}\right)}\\ =2.893\times {10}^6\end{array}$

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

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2.0\log \left(\frac{0.00004\text{ m}/0.66\text{ m}}{3.7}+\frac{2.51}{2.893\times {10}^6\sqrt{f}}\right)\end{array}$

Solving the above equation, $f=0.01175$.

Calculate the head loss.

  $\begin{array}{c}{h}_L=f\frac{L}{D}\frac{V^2}{2g}\\ =\left[\left(0.01175\right)\frac{\text{1500 m}}{0.66\text{ m}}\right]\frac{{\left(4.384\text{m}/\text{s}\right)}^2}{2\left(9.81{\text{m}/\text{s}}^2\right)}\\ =26.16\text{ m}\end{array}$

Calculate the power required by the pump.

  $\begin{array}{c}{\dot{W}}_{pump}=\dot{V}\rho g{h}_L\\ =\left(1.5{\text{m}}^3/\text{s}\right)\left(1000\text{kg}/{\text{m}}^3\right)\left(9.81{\text{m}/\text{s}}^2\right)\left(26.16\text{ m}\right)\\ =384.9\text{ kW}\end{array}$

The change in the pump power requirement is,

  $\frac{709.1-384.9}{709.1}=0.457$

Thus, by lining the concrete tank with petroleum based lining decreases the power requirement of the pump by $45.7\%$.