Chapter 8, Problem 148P

To determine

The percentage increase in net power output.

Answer to Problem 148P

The percentage increase in net power output is $14.30\%$.

Explanation of Solution

Given information:

The diameter of the cast iron pipe is $0.35\text{m}$ and the length of the cast iron pipe is $200\text{m}$. The head loss during pump is $140\text{m}$ and the density of water at $20°\text{C}$ is $\text{998}\text{kg}/{\text{m}}^{\text{3}}$. The viscosity of water at $20°\text{C}$ is $\text{0}\text{.001002}\text{kg}/\text{m}\cdot \text{s}$ and the combined turbine efficiency is $80\%$. The diameter of the pipe is tripled in order to reduced the pipe loss.

Write the expression for energy equation for a control volume.

  $\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1+h_{\text{pump}}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+z_2+h_{\text{turbine}}+h_L$  ...... (I)

Here, the pressure at point $1$ is $P_1$, the pressure at point $2$ is $P_2$, the height at $1$ is $z_1$, the height at $2$ is $z_2$, the velocity at point $1$ is $V_1$, the velocity at point $2$ is $V_2$,the head loss due to pump is $h_{\text{pump}}$, the head loss due to turbine is $h_{\text{turbine}}$, the kinetic energy correction factor at point $1$ is ${\alpha }_1$, the kinetic energy correction factor at point $2$, acceleration due to gravity is $g$, the density of fluid is $\rho$ and the head loss in the pipe is $h_L$.

Write the expression for average velocity.

  $V=\frac{4\dot{V}}{\pi {\left(D\right)}^2}$....... (II)

Here, the volume flow rate is $\dot{V}$, the diameter of the pipe is $D$ and the relative velocity is $V$.

Write the expression for Reynolds number for flow in pipe.

  $\mathrm{Re}=\frac{\rho VD}{\mu }$...... (III)

Here, the Reynolds number for flow in pipe is ${\mathrm{Re}}_A$ and the density of fluid is $\rho$ and

Write the expression for Colebrook equation for pipe A.

  $\frac{1}{\sqrt{f}}=-2\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$...... (IV)

Here, the relative velocity through the pipe is $\epsilon /D$ and the friction factor of the pipe $f$.

Write the expression for the head loss in pipe.

  $h_L=\frac{fL{V}^2}{2gD}$..... (V)

Here, the head loss in the pipe is $h_L$, the friction factor in pipe is $f_A$, the length of the pipe is $L_A$ and the velocity ion the pipe is $L_B$.

Write the expression for percentage increase in net power output.

  $\%\text{increase}=\frac{h_{\text{turbine}}-h_{\text{turbine,}\text{e}}}{h_{\text{turbine,}\text{e}}}\times 100$  ...... (VI)

Calculation:

Substitute $P_{atm}$ for $P_1$, $0$ for $V_1$, $P_{atm}$ for $P_2$, $140\text{m}$ for $h_{\text{pump}}$ and $0$ for $V_2$ in Equation (I).

  $\begin{array}{l}\frac{P_E}{\rho g}+{\alpha }_1\frac{\left(0\right)}{2g}+0+140\text{m}=\frac{P_{\text{atm}}}{\rho g}+{\alpha }_2\frac{\left(0\right)}{2g}+0+h_{\text{turbine}}+h_L\\ 140\text{m}=h_{\text{turbine}}+h_L\\ h_{\text{turbine}}+h_L=140\text{m}\end{array}$....... (VII)

Substitute $0.35\text{m}$ for $D$ and $0.55{\text{m}}^3/\text{s}$ for $\dot{V}$ in Equation (II).

  $\begin{array}{c}V=\frac{4\left(0.55{\text{m}}^3/\text{s}\right)}{\pi {\left(0.35\text{m}\right)}^2}\\ =\frac{2.2{\text{m}}^3/\text{s}}{1.099\times 0.35{\text{m}}^2}\\ =5.7194\text{m}/\text{s}\end{array}$

Substitute $\text{998}\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ and $0.35\text{m}$ for $D$ in Equation (III).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(\text{998}\text{kg}/{\text{m}}^{\text{3}}\right)\left(5.7194\text{m}/\text{s}\right)\left(0.35\text{m}\right)}{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{1997.786\text{kg}/\text{m}\cdot \text{s}}{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =1993798.822\end{array}$

Refer the Table-3.1, "Roughness height for certain common Pipe materials" to obtain the value of the roughness height $0.26\times {10}^{-3}\text{m}$ corresponding through cast iron pipe.

Substitute $0.35\text{m}$ for $D$, $1993798.822$ for $\mathrm{Re}$ and $0.26\times {10}^{-3}\text{m}$ for $\epsilon$ in Equation (IV).

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-2\log \left(\frac{0.26\times {10}^{-3}\text{m}/0.35\text{m}}{3.7}+\frac{2.51}{1993798.822\sqrt{f}}\right)\\ =-2\log \left(\frac{7.422857\times {10}^{-4}}{3.7}+\frac{1.258\times {10}^{-6}}{\sqrt{f}}\right)\\ =-2\log \left(2.006\times {10}^{-4}+\frac{1.258\times {10}^{-6}}{\sqrt{f}}\right)\\ =0.0185\end{array}$

Substitute $5.71\text{m}/\text{s}$ for $D$, $0.35\text{m}$ for $D$, $200\text{m}$ for $L$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $0.0185$ for $f$ in Equation (V).

  $\begin{array}{c}{h}_L=\frac{\left(0.0185\right)\left(200\text{m}\right){\left(5.71\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)\left(0.35\text{m}\right)}\\ =\frac{120.635{\text{m}}^3/\text{s}}{6.865{\text{m}}^2/\text{s}}\\ =17.5724\text{m}\end{array}$

Substitute $17.5724\text{m}$ for $h_L$ in Equation (VII).

  $\begin{array}{l}{h}_{\text{turbine,}\text{e}}+17.5724\text{m}=140\text{m}\\ h_{\text{turbine,}\text{e}}=140\text{m-17}\text{.5724}\text{m}\\ h_{\text{turbine,}\text{e}}\text{=122}\text{.427}\text{m}\end{array}$

Substitute $\left(0.35\times 3\right)\text{m}$ for $D$ and $0.55{\text{m}}^3/\text{s}$ for $\dot{V}$ in Equation (II).

  $\begin{array}{c}V=\frac{4\left(0.55{\text{m}}^3/\text{s}\right)}{\pi {\left(3\times 0.35\text{m}\right)}^2}\\ =\frac{2.2{\text{m}}^3/\text{s}}{0.38484{\text{m}}^2}\\ =0.6354\text{m}/\text{s}\end{array}$

Substitute $\text{998}\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ and $\left(0.35\times 3\right)\text{m}$ for $D$ in Equation (III).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(\text{998}\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.6354\text{m}/\text{s}\right)\left(0.35\text{m}\times \text{3}\right)}{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{665.835\text{kg}/\text{m}\cdot \text{s}}{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =664506.646\end{array}$

Substitute $\left(0.35\times 3\right)\text{m}$ for $D$, $664506.646$ for $\mathrm{Re}$ and $0.26\times {10}^{-3}\text{m}$ for $\epsilon$ in Equation (IV).

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-2\log \left(\frac{0.26\times {10}^{-3}\text{m}/\left(0.35\times 3\right)\text{m}}{3.7}+\frac{2.51}{664506.646\times \sqrt{f}}\right)\\ =-2\log \left(\frac{2.4722857\times {10}^{-4}}{3.7}+\frac{3.777\times {10}^{-6}}{\sqrt{f}}\right)\\ =-2\log \left(6.6924\times {10}^{-5}+\frac{3.777\times {10}^{-6}}{\sqrt{f}}\right)\\ =0.0155\end{array}$

Substitute $0.6354\text{m}/\text{s}$ for $V$, $\left(0.35\times 3\right)\text{m}$ for $D$, $200\text{m}$ for $L$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $0.0155$ for $f$ in Equation (V).

  $\begin{array}{c}{h}_L=\frac{\left(0.0155\right)\left(200\text{m}\right){\left(0.6354\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)\left(0.35\times 3\text{m}\right)}\\ =\frac{1.2515{\text{m}}^3/\text{s}}{20.601{\text{m}}^2/\text{s}}\\ =0.06074\text{m}\end{array}$

Substitute $0.06074\text{m}$ for $h_L$ in Equation (VII).

  $\begin{array}{l}{h}_{\text{turbine}}+0.06074\text{m}=140\text{m}\\ h_{\text{turbine}}=140\text{m}-0.06074\text{m}\\ h_{\text{turbine}}=\text{139}\text{.9392}\text{m}\end{array}$

Substitute $\text{122}\text{.427}\text{m}$ for $h_{\text{turbine,}\text{e}}$ and $\text{139}\text{.9392}\text{m}$ for $h_{\text{turbine}}$ in Equation (VI).

  $\begin{array}{c}\%\text{increase}=\frac{\text{139}\text{.9392}\text{m}-\text{122}\text{.427}\text{m}}{\text{122}\text{.427}\text{m}}\times 100\\ =\frac{17.5122\text{m}}{\text{122}\text{.427}\text{m}}\times 100\\ =14.30\%\end{array}$

Conclusion:

The percentage increase in net power output is $14.30\%$.