Chapter 6, Problem 6.123P

To determine

The value of $z_3$ for which the flow rate in pipe 3 is equal to $0.2m^3/s$.

Answer to Problem 6.123P

$z_3=89.4m$

Explanation of Solution

Given information:

Diameter of all pipes are $28cm$ and roughness $\in =1mm$

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 $z_3=95m$

Guess $h_a=95m$, therefore

${\left(h_f\right)}_1=z_1-h_a=25m-95m=-70m$

The roughness ratio for all pipes will be,

$\frac{\in }{d}=\frac{1\times {10}^{-3}m}{0.28m}=3.571\times {10}^{-3}$

To find Reynolds’s number,

$R_e=\frac{DV}{\nu }$ Try $V_1=10m/s$

Therefore,

${\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(10m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=2.77\times {10}^6$

The friction factor will be equal to,

$\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{2.77\times {10}^6}+{\left(\frac{\text{}3.571\times {10}^{-3}}{3.7}\right)}^{1.11}\right)=0.0276$

Therefore, according to below equation

$\Delta h=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=z-h_j$ $\begin{array}{l}{\left(h_f\right)}_1=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.0276\right)\left(\frac{95m}{0.28m}\right)\left(\frac{V_1{}^2}{2\left(9.81\right)}\right)=70m\\ V_1=12.11m/s\end{array}$

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

${\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(12.11m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=3.36\times {10}^6$

To find the flow rate,

$Q_1=A_1{V}_1=\left[\pi {\left(0.14\right)}^2\left(12.11m/s\right)\right]=0.74m^3/s$

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

Similarly for pipe 2,

${\left(h_f\right)}_2=z_2-h_a=115m-95m=20m$

To find Reynolds’s number,

Try $V_2=10m/s$

Therefore,

${\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(10m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=2.77\times {10}^6$

The friction factor will be same.

$\begin{array}{l}{\left(h_f\right)}_2=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.0276\right)\left(\frac{125m}{0.28m}\right)\left(\frac{V_2{}^2}{2\left(9.81\right)}\right)=20m\\ V_1=5.64m/s\end{array}$

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

${\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(5.64m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=1.56\times {10}^6$

To find the flow rate,

$Q_2=A_2{V}_2=\left[\pi {\left(0.14\right)}^2\left(5.64m/s\right)\right]=0.35m^3/s$

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

For pipe 3,

${\left(h_f\right)}_3=z_3-h_a=95m-95m=0$

$Q_3=0$

Therefore,

$Q=Q_2-Q_1=0.35m^3/s-0.74m^3/s=-0.39m^3/s$

Hence $h_a$ must be lower than $95m$, try $h_a=80m$

${\left(h_f\right)}_1=z_1-h_a=25m-80m=-55m$

Try $V_1=10m/s$

Therefore,

${\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(10m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=2.77\times {10}^6$

The friction factor will be same.

$\begin{array}{l}{\left(h_f\right)}_1=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.0276\right)\left(\frac{95m}{0.28m}\right)\left(\frac{V_1{}^2}{2\left(9.81\right)}\right)=55m\\ V_1=10.73m/s\end{array}$

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

${\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(10.73m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=2.97\times {10}^6$

To find the flow rate,

$Q_1=A_1{V}_1=\left[\pi {\left(0.14\right)}^2\left(10.73m/s\right)\right]=0.66m^3/s$

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

Similarly for pipe 2,

${\left(h_f\right)}_2=z_2-h_a=115m-80m=35m$

To find Reynolds’s number,

$R_e=\frac{DV}{\nu }$ Try $V_2=10m/s$

Therefore,

${\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(10m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=2.77\times {10}^6$

The friction factor will be same.

$\begin{array}{l}{\left(h_f\right)}_2=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.0276\right)\left(\frac{125m}{0.28m}\right)\left(\frac{V_2{}^2}{2\left(9.81\right)}\right)=35m\\ V_1=7.46m/s\end{array}$

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

${\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(7.46m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=2.07\times {10}^6$

To find the flow rate,

$Q_2=A_2{V}_2=\left[\pi {\left(0.14\right)}^2\left(7.46m/s\right)\right]=0.459m^3/s$

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

Similarly for pipe 3,

${\left(h_f\right)}_3=z_3-h_a=95m-80m=15m$

To find Reynolds’s number,

Try $V_3=1m/s$

Therefore,

${\left(R_e\right)}_3=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(1m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=2.77\times {10}^5$

The friction factor will be equal to,

$\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{2.77\times {10}^5}+{\left(\frac{\text{}3.571\times {10}^{-3}}{3.7}\right)}^{1.11}\right)=0.0279$

$\begin{array}{l}{\left(h_f\right)}_3=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)=\left(0.0279\right)\left(\frac{160m}{0.28m}\right)\left(\frac{V_3{}^2}{2\left(9.81\right)}\right)=15m\\ V_3=4.29m/s\end{array}$

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

${\left(R_e\right)}_3=\frac{DV}{\nu }=\frac{\left(0.28m\right)\left(2.48m/s\right)}{1.01\times {10}^{-6}{m}^2/s}=11.893\times {10}^5$

To find the flow rate,

$Q_3=A_3{V}_3=\left[\pi {\left(0.14\right)}^2\left(4.29m/s\right)\right]=0.26m^3/s$

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

Therefore the flow rate will be,

$Q=(Q_2+Q_3)-Q_1=\left(0.26+0.459\right)-0.66=0.059m^3/s$

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

$\begin{array}{l}{Q}_1=0.66m^3/s\\ Q_2=0.459m^3/s\\ Q_3=0.26m^3/s\end{array}$

Calculate the exact value of $z_3$ for the flow rate of $0.02m^3/s$ by interpolation,

We know that,

At $z_3=85m$

$Q_3=0.153m^3/s$

At $z_3=95m$

$Q_3=0.26m^3/s$

By interpolation, the $z_3$ at $Q_3=0.2m^3/s$ is equal to,

$\begin{array}{l}{z}_3=\frac{\left(0.2-0.153\right)\left(95-85\right)}{\left(0.26-0.153\right)}+85\\ =89.4m\end{array}$

Conclusion:

The $z_3$ is equal to $89.4m$ for which the flow rate in pipe 3 is equal to $0.2m^3/s$.