Chapter 6, Problem 6.124P

To determine

The steady flow rates in each pipe.

Answer to Problem 6.124P

$\begin{array}{l}{Q}_1=2.084f{t}^3/s\\ Q_2=1.605f{t}^3/s\\ Q_3=0.520f{t}^3/s\end{array}$

Explanation of Solution

Given information:

The fluid is water at $20°C$

At three reservoir junction, if all flows are considered positive towards the junction, then

$Q_1+Q_2+Q_3=0\to \left(1\right)$

Assuming the gauge pressure $p_1=p_2=p_3=0$, the head loss $\Delta h$ across each pipe is equal to,

$\Delta h=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=z-h_j$

$f$ - Friction factor

$L$ - Length of the pipe

$D$ - Diameter of the pipe

$V$ -Velocity of the flow

$z$ - Elevation

To solve this type of problems, we should guess the position of $h_j$ and solve the above mentioned equation to find the relevant velocities and flow rates iterating, until the flow rates satisfies the equation 1.

If the $h_j$ guessed was too high, then the sum of equation (1) will be negative, therefore we have to reduce $h_j$ and vice versa.

Calculation:

First of all,

Guess $h_a=50ft$, therefore

${\left(h_f\right)}_1=z_1-h_a=20ft-50ft=-30ft$

The roughness ratio for pipe 1 will be,

$\frac{\in }{d}=\frac{0.0005ft}{\frac{8}{12}ft}=7.5\times {10}^{-4}$

To find Reynolds’s number,

$R_e=\frac{DV}{\nu }$ Try $V_1=5ft/s$

Therefore,

${\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(\frac{8}{12}\right)\left(5ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.058\times {10}^5$

The friction factor will be equal to,

$\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{3.058\times {10}^5}+{\left(\frac{\text{}7.5\times {10}^{-4}}{3.7}\right)}^{1.11}\right)=0.0194$

Therefore, according to the below equation

$\Delta h=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=z-h_j$ $\begin{array}{l}{\left(h_f\right)}_1=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.0194\right)\left(\frac{1800ft}{\frac{8}{12}ft}\right)\left(\frac{V_1{}^2}{2\left(32.2ft/s^2\right)}\right)=30ft\\ V_1=6.073ft/s\end{array}$

According to the above result, the Reynolds’s number will be

${\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(\frac{8}{12}ft\right)\left(6.073ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.714\times {10}^5$

To find the flow rate,

$Q_1=A_1{V}_1=\left[\pi {\left(\frac{4}{12}ft\right)}^2\left(6.073ft/s\right)\right]=2.1198f{t}^3/s$

The flow in pipe 1 is considered as the flow away from the junction

Similarly for pipe 2,

${\left(h_f\right)}_2=z_2-h_a=100ft-50ft=50ft$

To find the Reynolds’s number,

The roughness ratio for pipe 2 will be,

$\frac{\in }{d}=\frac{0.0005ft}{\frac{6}{12}ft}=1\times {10}^{-3}$

To find Reynolds’s number,

Try $V_2=7ft/s$

Therefore,

${\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(\frac{6}{12}\right)\left(7ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.211\times {10}^5$

The friction factor will be equal to,

$\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{3.211\times {10}^5}+{\left(\frac{\text{}1\times {10}^{-3}}{3.7}\right)}^{1.11}\right)=0.02047$

Therefore, according to the below equation

$\Delta h=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=z-h_j$ $\begin{array}{l}{\left(h_f\right)}_2=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.02047\right)\left(\frac{1200ft}{\frac{6}{12}ft}\right)\left(\frac{V_2{}^2}{2\left(32.2ft/s^2\right)}\right)=50ft\\ V_2=8.095ft/s\end{array}$

According to above result, the Reynolds’s number will be

${\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(\frac{6}{12}ft\right)\left(8.095ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.713\times {10}^5$

To find the flow rate,

$Q_2=A_2{V}_2=\left[\pi {\left(\frac{3}{12}ft\right)}^2\left(8.095ft/s\right)\right]=1.589f{t}^3/s$

The flow in pipe 2 is considered as the flow towards the junction.

For pipe 3,

${\left(h_f\right)}_3=z_3-h_a=50ft-50ft=0$

$Q_3=0$

Therefore,

$Q=Q_2-Q_1=1.589f{t}^3/s-2.1198f{t}^3/s=-0.5308f{t}^3/s$

Hence $h_a$ must be lower than, try $h_a=40ft$

${\left(h_f\right)}_1=z_1-h_a=20ft-49ft=-29ft$

Try $V_1=5ft/s$

Therefore,

${\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(\frac{8}{12}\right)\left(5ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.058\times {10}^5$

The friction factor will be the same.

$\begin{array}{l}{\left(h_f\right)}_1=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.0194\right)\left(\frac{1800ft}{\frac{8}{12}ft}\right)\left(\frac{V_1{}^2}{2\left(32.2ft/s^2\right)}\right)=29ft\\ V_1=5.971ft/s\end{array}$

According to the above result, the Reynolds’s number will be

${\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(\frac{8}{12}ft\right)\left(5.971ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.65\times {10}^5$

To find the flow rate,

$Q_1=A_1{V}_1=\left[\pi {\left(\frac{4}{12}ft\right)}^2\left(5.971ft/s\right)\right]=2.0842f{t}^3/s$

The flow in pipe 1 is considered as flow away from the junction

Similarly for pipe 2,

${\left(h_f\right)}_2=z_2-h_a=100ft-49ft=51ft$

To find Reynolds’s number,

$R_e=\frac{DV}{\nu }$ Try $V_2=7ft/s$

Therefore,

${\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(\frac{6}{12}\right)\left(7ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.211\times {10}^5$

Friction factor will be the same.

Therefore, according to the below equation

$\Delta h=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=z-h_j$ $\begin{array}{l}{\left(h_f\right)}_2=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.02047\right)\left(\frac{1200ft}{\frac{6}{12}ft}\right)\left(\frac{V_2{}^2}{2\left(32.2ft/s^2\right)}\right)=51ft\\ V_2=8.176ft/s\end{array}$

According to the above result, the Reynolds’s number will be

${\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(\frac{6}{12}ft\right)\left(8.176ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.75\times {10}^5$

To find the flow rate,

$Q_2=A_2{V}_2=\left[\pi {\left(\frac{3}{12}ft\right)}^2\left(8.176ft/s\right)\right]=1.605f{t}^3/s$

The flow in pipe 2 is considered as the flow towards the junction.

Similarly for pipe 3,

${\left(h_f\right)}_3=z_3-h_a=50ft-49ft=1ft$

To find Reynolds’s number,

Try $V_3=1ft/s$

Therefore,

${\left(R_e\right)}_3=\frac{DV}{\nu }=\frac{\left(\frac{9}{12}\right)\left(1ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=6.88\times {10}^4$

The roughness ratio for pipe 3 will be,

$\frac{\in }{d}=\frac{0.0005ft}{\frac{9}{12}ft}=6.667\times {10}^{-4}$

The friction factor will be equal to,

$\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{6.88\times {10}^4}+{\left(\frac{\text{}6.667\times {10}^{-4}}{3.7}\right)}^{1.11}\right)=0.0217$

$\begin{array}{l}{\left(h_f\right)}_3=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.0217\right)\left(\frac{1600ft}{\frac{9}{12}ft}\right)\left(\frac{V_3{}^2}{2\left(32.2ft/s^2\right)}\right)=1ft\\ V_3=1.179ft/s\end{array}$

According to the above result, the Reynolds’s number will be

${\left(R_e\right)}_3=\frac{DV}{\nu }=\frac{\left(\frac{9}{12}ft\right)\left(1.179ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=0.811\times {10}^5$

To find the flow rate,

$Q_3=A_3{V}_3=\left[\pi {\left(\frac{4.5}{12}ft\right)}^2\left(1.179ft/s\right)\right]=0.52f{t}^3/s$

The flow in pipe 3 is considered as the flow towards the junction.

Therefore the flow rate will be,

$Q=(Q_2+Q_3)-Q_1=\left(1.605+0.520\right)-2.084=0.041f{t}^3/s$

By further iteration, we can get this close to zero, but for now, according to the obtained results, we can say,

$\begin{array}{l}{Q}_1=2.084f{t}^3/s\\ Q_2=1.605f{t}^3/s\\ Q_3=0.520f{t}^3/s\end{array}$

Conclusion:

The flow rates in each pipe is equal to,

$\begin{array}{l}{Q}_1=2.084f{t}^3/s\\ Q_2=1.605f{t}^3/s\\ Q_3=0.520f{t}^3/s\end{array}$.