Chapter 14, Problem 111RQ

To determine

The power produced by the plant and the percentage increase in net power.

Explanation of Solution

Given:

Temperature of water is $20°\text{C}$.

Flow rate of water is $0.6{\text{m}}^3/\text{s}$.

Diameter of the pipe is $3\left(0.35\text{ m}\right)$.

Length of the pipe is $\text{200 m}$.

The difference in elevation is $140\text{ m}$.

Generator efficiency is 80%.

Calculation:

Refer Table A-15 “Properties of saturated water” from Appendix 1 and obtain the following properties of water:

  $\begin{array}{l}\text{Density, }\rho =998\text{kg}/{\text{m}}^3\\ \text{Dynamic viscocity, }\mu =1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\end{array}$

Apply energy equation to point 1 at the free surface of water in the reservoir and point 2 at the turbine exit.

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

Calculate the average velocity.

  $\begin{array}{c}V=\frac{\dot{V}}{A_c}\\ =\frac{\left(0.6{\text{m}}^3/\text{s}\right)}{\left[\frac{\pi }{4}{\left(\text{0}\text{.35 m}\right)}^2\right]}\\ =6.236\text{m}/\text{s}\end{array}$

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{\left(998\text{kg}/{\text{m}}^3\right)\left(6.236\text{m}/\text{s}\right)\left(0.35\text{ m}\right)}{\left(1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =2.174\times {10}^6\end{array}$

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

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2.0\log \left(\frac{0.00026\text{ m}/0.35\text{ m}}{3.7}+\frac{2.51}{2.174\times {10}^6\sqrt{f}}\right)\end{array}$

Solving the above equation, $f=0.01847$.

Calculate the head loss.

  $\begin{array}{c}{h}_L=f\frac{L}{D}\frac{V^2}{2g}\\ =\left(0.01847\right)\frac{200\text{ m}}{\text{0}\text{.35}\text{m}}\frac{{\left(6.236\text{m}/\text{s}\right)}^2}{2\left(9.81{\text{m}/\text{s}}^2\right)}\\ =20.92\text{ m}\end{array}$

Calculate $h_{turbine}$.

  $\begin{array}{c}{h}_{turbine}=z_1-h_L\\ =140\text{ m}-20.92\text{ m}\\ =119.08\text{ m}\end{array}$

Calculate the power produced by the turbine.

  $\begin{array}{c}{\dot{W}}_{turbine}={\eta }_{turbine}\dot{V}\rho g{h}_{turbine}\\ =0.80\left(0.6{\text{m}}^3/\text{s}\right)\left(998\text{kg}/{\text{m}}^3\right)\left(9.81{\text{m}/\text{s}}^2\right)\left(119.08\text{m}\right)\\ =560\text{ kW}\end{array}$

When the pipe diameter is tripled:

Calculate the average velocity.

  $\begin{array}{c}V=\frac{\dot{V}}{A_c}\\ =\frac{\left(0.6{\text{m}}^3/\text{s}\right)}{\left[\frac{\pi }{4}{\left(\text{3}\times \text{0}\text{.35 m}\right)}^2\right]}\\ =0.6929\text{m}/\text{s}\end{array}$

The Reynolds number is,

  $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{\left(998\text{kg}/{\text{m}}^3\right)\left(0.6929\text{m}/\text{s}\right)\left(3\times 0.35\text{ m}\right)}{\left(1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =0.7247\times {10}^6\end{array}$

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

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2.0\log \left(\frac{0.00026\text{ m}/1.05\text{ m}}{3.7}+\frac{2.51}{0.7247\times {10}^6\sqrt{f}}\right)\end{array}$

Solving the above equation, $f=0.01545$.

Calculate the head loss.

  $\begin{array}{c}{h}_L=f\frac{L}{D}\frac{V^2}{2g}\\ =\left(0.01545\right)\frac{200\text{ m}}{\text{1}\text{.05}\text{m}}\frac{{\left(0.6929\text{m}/\text{s}\right)}^2}{2\left(9.81{\text{m}/\text{s}}^2\right)}\\ =0.0720\text{ m}\end{array}$

Calculate $h_{turbine}$.

  $\begin{array}{c}{h}_{turbine}=z_1-h_L\\ =140\text{ m}-0.072\text{ m}\\ =139.928\text{ m}\end{array}$

Calculate the power produced by the turbine.

  $\begin{array}{c}{\dot{W}}_{turbine}={\eta }_{turbine}\dot{V}\rho g{h}_{turbine}\\ =0.80\left(0.6{\text{m}}^3/\text{s}\right)\left(998\text{kg}/{\text{m}}^3\right)\left(9.81{\text{m}/\text{s}}^2\right)\left(139.928\text{m}\right)\\ =657.575\text{ kW}\end{array}$

Thus, the power produced by the plant is $\text{657}\text{.575 kW}$.

Calculate the percent increase in the net power.

  $\begin{array}{c}\%\text{ increase}=\frac{657.575-560}{560}\times 100\%\\ =17.42\%\end{array}$

Thus the increase in the net power output is $17.42\%$.