Chapter 8, Problem 76EP

To determine

The elevation of the higher reservoir for the flow rate of $10{\text{ft}}^{\text{3}}/\text{min}$.

Answer to Problem 76EP

The elevation of the higher reservoir for the flow rate of $10{\text{ft}}^{\text{3}}/\text{min}$ is $12.655\text{ft}$.

Explanation of Solution

Given information:

The temperature of the water is $70°\text{F}$, density of the water is $62.3\text{lbm}/{\text{ft}}^{\text{3}}$, dynamic viscosity of the water is $6.556\times {10}^{-4}\text{lbm}/\text{ft}\cdot \text{s}$, loss coefficient at entrance is $0.03$, loss coefficient for elbow is $0.3$, loss coefficient for valve is $0.2$, loss coefficient at exit is $1.0$, length of the pipe is $60\text{ft}$, diameter of the pipe is $2\text{in}$, roughness of the cast iron pipe surfaces is $8.5\times {10}^{-4}\text{ft,}$ and volume flow rate of the water is $10{\text{ft}}^{\text{3}}/\text{min}$.

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{\rho vD}{\mu }$........... (I)

Here, Reynolds number is $\mathrm{Re}$, density of the water is $\rho$, velocity of the water is $v$, diameter of the pipe is $D,$ and dynamic viscosity of the water is $\mu$.

Write the expression for the volume flow rate of the water.

  $\dot{Q}=\frac{\pi }{4}{D}^{2}v$........... (II)

Here, the volume flow rate of the water is $\dot{Q}$.

Write the expression for the friction factor for turbulent flow.

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

Here, friction factor for turbulent flow is $f$ and roughness of the copper pipe surfaces is $\epsilon$.

Write the expression for the major head loss through the pipe.

  $h_{f,\text{major}}=\frac{fL{v}^2}{2gD}$........... (IV)

Here, length of the pipe is $L$, gravitational acceleration is $g,$ and the major head loss through the pipe is $h_{f,\text{major}}$.

Write the expression for the minor head loss through the pipe.

  $h_{f,\text{minor}}={\displaystyle \sum K_L\left(\frac{v^2}{2g}\right)}$........... (V)

Here, minor head loss through the pipe is $h_{f,\text{minor}}$ and sum of all loss coefficients is ${\displaystyle \sum K_L}$.

Write the expression for the sum of all loss coefficients.

  ${\displaystyle \sum K_L}=K_{L,\text{entrance}}+4K_{L,\text{elbow}}+K_{L,\text{valve}}+K_{L,\text{exit}}$........... (VI)

Here, loss coefficient at entrance is $K_{L,\text{entrance}}$, loss coefficient for elbow is $K_{L,\text{elbow}}$, loss coefficient for valve is $K_{L,\text{valve}}$, and loss coefficient at exit is $K_{L,\text{exit}}$.

Write the expression for the total head loss in the pipe.

  $h_f=h_{f,\text{major}}+h_{f,\text{minor}}$........... (VII)

Here, total head loss in the pipe is $h_f$.

Substitute ${\displaystyle \sum K_L\left(\frac{v^2}{2g}\right)}$ for $h_{f,\text{minor}}$ and $\frac{fL{v}^2}{2gD}$ for $h_{f,\text{major}}$ in Equation (VI).

  $\begin{array}{c}{h}_f=\frac{fL{v}^2}{2gD}+{\displaystyle \sum K_L\left(\frac{v^2}{2g}\right)}\\ =\left(\frac{fL}{D}+{\displaystyle \sum K_L}\right)\frac{v^2}{2g}\end{array}$........... (VIII)

Write the expression for the Bernoulli Equation.

  $\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+h_f$........... (IX)

Here, pressure at point $1$ is $P_1$, velocity at point $1$ in $v_1$, density of the fluid is $\rho$, gravitational acceleration is $g$, total head loss in pipe is $h_f$, elevation of point $1$ is $Z_1$, pressure at point $2$ is $P_2$, velocity at point $2$ in $v_2,$ and elevation of point $2$ is $Z_2$.

Calculation:

Substitute $10{\text{ft}}^{\text{3}}/\text{min}$ for $\dot{Q}$ and $2\text{in}$ for $D$ in Equation (II).

  $\begin{array}{l}\left(10{\text{ft}}^{\text{3}}/\text{min}\right)=\frac{\pi }{4}{\left(2\text{in}\right)}^{2}v\\ \left(10{\text{ft}}^{\text{3}}/\text{min}\right)\left[\frac{1\min }{60\text{s}}\right]=\frac{\pi }{4}{\left(\left(2\right)\left[\frac{1\text{ft}}{12\text{in}}\right]\right)}^{2}v\\ v=7.64\text{ft}/\text{s}\end{array}$

Substitute $62.3\text{lbm}/{\text{ft}}^{\text{3}}$ for $\rho$, $7.64\text{ft}/\text{s}$ for $v$, $6.556\times {10}^{-4}\text{lbm}/\text{ft}\cdot \text{s}$ for $\mu$ and $2\text{in}$ for $D$ in Equation (I).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(62.3\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(7.64\text{ft}/\text{s}\right)\left(2\text{in}\right)}{\left(6.556\times {10}^{-4}\text{lbm}/\text{ft}\cdot \text{s}\right)}\\ =\frac{\left(62.3\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(7.64\text{ft}/\text{s}\right)\left(2\text{in}\right)\left[\frac{1\text{ft}}{12\text{in}}\right]}{\left(6.556\times {10}^{-4}\text{lbm}/\text{ft}\cdot \text{s}\right)}\\ =60700\end{array}$

Since value of the Reynolds number is greater than $4000$, therefore the flow is turbulent.

Substitute $60700$ for $\mathrm{Re}$, $8.5\times {10}^{-4}\text{ft}$ for $\epsilon$ and $2\text{in}$ for $D$ in Equation (III).

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2\log \left(\frac{\frac{\left(8.5\times {10}^{-4}\text{ft}\right)}{\left(2\text{in}\right)}}{3.7}+\frac{2.51}{\left(60700\right)\sqrt{f}}\right)\\ \frac{-1}{2\sqrt{f}}=\log \left(\frac{\frac{\left(8.5\times {10}^{-4}\text{ftt}\right)}{\left(2\text{in}\right)\left[\frac{1\text{ft}}{12\text{in}}\right]}}{3.7}+\frac{2.51}{\left(60700\right)\sqrt{f}}\right)\\ \frac{-1}{2\sqrt{f}}=\log \left(\frac{\frac{\left(8.5\times {10}^{-4}\text{ft}\right)}{\left(0.166\text{ft}\right)}}{3.7}+\frac{2.51}{\left(60700\right)\sqrt{f}}\right)\\ f=0.0320\end{array}$

Substitute $0.03$ for $K_{L,\text{entrance}}$, $0.3$ for $K_{L,\text{elbow}}$, $0.2$ for $K_{L,\text{valve}},$ and $1.0$ for $K_{L,\text{exit}}$ in Equation (VI).

  $\begin{array}{c}{\displaystyle \sum K_L}=0.03+4\times 0.3+0.2+1.0\\ =0.03+1.2+0.2+1.0\\ =2.43\end{array}$

Substitute $P_{\text{atm}}$ for $P_1$, $P_{\text{atm}}$ for $P_2$, $0$ for $v_1$, $0$ for $v_2$ and $0$ for $Z_2$ in Equation (IX).

  $\begin{array}{l}\frac{P_{\text{atm}}}{\rho g}+\frac{{\left(0\right)}^2}{2g}+Z_1=\frac{P_{\text{atm}}}{\rho g}+\frac{{\left(0\right)}^2}{2g}+0+h_f\\ Z_1=h_f\end{array}$........... (X)

Substitute $0.032$ for $f$, $2.43$ for ${\displaystyle \sum K_L}$, $60\text{ft}$ for $L$, $7.64\text{ft}/\text{s}$ for $v$, $32.17\text{ft}/{\text{s}}^{\text{2}}$ for $g,$ and $2\text{in}$ for $D$ in Equation (VIII).

  $\begin{array}{c}{h}_f=\left(\frac{\left(0.032\right)\left(60\text{ft}\right)}{\left(2\text{in}\right)}+2.43\right)\frac{{\left(7.64\text{ft}/\text{s}\right)}^2}{2\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)}\\ =\left(\frac{\left(0.032\right)\left(60\text{ft}\right)}{\left(2\text{in}\right)\left[\frac{1\text{ft}}{12\text{in}}\right]}+2.43\right)\frac{{\left(7.64\text{ft}/\text{s}\right)}^2}{2\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)}\\ =\left(11.52+2.43\right)\frac{{\left(7.64\text{ft}/\text{s}\right)}^2}{2\left(32.17\text{ft}/{\text{s}}^{\text{2}}\right)}\\ =12.655\text{ft}\end{array}$

Substitute $12.655\text{ft}$ for $h_f$ in Equation (X).

  $Z_1=12.655\text{ft}$

Conclusion:

The elevation of the higher reservoir for the flow rate of $10{\text{ft}}^{\text{3}}/\text{min}$ is $12.655\text{ft}$.