Chapter 6, Problem 6.129P

To determine

The flow rate in each pipe.

Answer to Problem 6.129P

$\begin{array}{l}{Q}_1=0.0388m^3/s\\ Q_2=-0.0223m^3/s\\ Q_3=0.0241m^3/s\\ Q_4=-0.0365m^3/s\end{array}$

Explanation of Solution

Given information:

All pipes are made of cast iron.

The diameter of all pipes are $8cm$ and the length id equal to $45m$

The fluid is water at $20°C$

The basic rules of pipe networks are,

“The net pressure change in a closed loop must be equal to zero”

“The net flow into a junction must be equal to zero”

“All pressure changes must satisfy moody and minor loss friction co-relation”

The head loss $h_f$ can be defined as,

$h_f=\frac{8Lf{Q}^2}{{\pi }^{2}g{d}^5}=K{Q}^2$

In above equation,

$L$ - Length of pipe

$f$ - Friction factor

$Q$ - Flow rate

$d$ - Diameter

$K$ - Loss co efficient

The total head loss is equal to zero in a closed loop.

The pressure difference $\Delta P$ can be defined as,

$\Delta P=\rho g{h}_f$

Where, $\rho$ - density

Calculation:

To solve this problem,

Assume pressure at junction,

$p_a=500kPa$

But we know that,

$h_f=\frac{\Delta P}{\rho g}=\frac{p_1-p_a}{\rho g}=\frac{\left(950\times {10}^{3}Pa-500\times {10}^{3}Pa\right)}{9790N/m^3}=45.96m$

We can define head loss as,

$h_f=\frac{8Lf{Q}^2}{{\pi }^{2}g{d}^5}$

In above equation,

Assume,

$f_1=0.02$

Therefore,

$\begin{array}{l}45.96m=\frac{8\left(45m\right)\left(0.02\right)Q_1^2}{{\pi }^2\left(9.81m/s^2\right){\left(0.08m\right)}^5}\\ Q_1=0.045m^3/s\end{array}$

Similarly, find the flow rate for all pipes

For pipe 2,

$h_f=\frac{\Delta P}{\rho g}=\frac{p_2-p_a}{\rho g}=\frac{\left(350\times {10}^{3}Pa-500\times {10}^{3}Pa\right)}{9790N/m^3}=-15.32m$

$\begin{array}{l}-15.32m=\frac{8\left(45m\right)\left(0.02\right)Q_2^2}{{\pi }^2\left(9.81m/s^2\right){\left(0.08m\right)}^5}\\ Q_2=-0.0259m^3/s\end{array}$

For pipe 3,

$\begin{array}{l}\frac{\left(675\times {10}^{3}Pa-500\times {10}^{3}Pa\right)}{9790N/m^3}=\frac{8\left(45m\right)\left(0.02\right)Q_3^2}{{\pi }^2\left(9.81m/s^2\right){\left(0.08m\right)}^5}\\ Q_3=0.028m^3/s\\ \end{array}$

For pipe 4,

$\begin{array}{l}\frac{\left(100\times {10}^{3}Pa-500\times {10}^{3}Pa\right)}{9790N/m^3}=\frac{8\left(45m\right)\left(0.02\right)Q_4^2}{{\pi }^2\left(9.81m/s^2\right){\left(0.08m\right)}^5}\\ Q_4=-0.0424m^3/s\end{array}$

To find the exact value of the flow rate,

Calculate the roughness ratio

$\frac{\epsilon }{d}=\frac{0.26mm}{80mm}=3.25\times {10}^{-3}$

Calculate the Reynolds’s number for each pipe

$\mathrm{Re}=\frac{\rho Vd}{\mu }=\frac{4\rho Q}{\pi \mu d}$

For pipe 1,

${\mathrm{Re}}_1=\frac{4\rho Q_1}{\pi \mu d}=\frac{4\left(998kg/m^3\right)\left(0.045m^3/s\right)}{\pi \left(0.001kg/ms\right)\left(0.08m\right)}=714764.85$

For pipe 2,

${\mathrm{Re}}_2=\frac{4\rho Q_2}{\pi \mu d}=\frac{4\left(998kg/m^3\right)\left(0.0259m^3/s\right)}{\pi \left(0.001kg/ms\right)\left(0.08m\right)}=411386.88$

For pipe 3,

${\mathrm{Re}}_3=\frac{4\rho Q_3}{\pi \mu d}=\frac{4\left(998kg/m^3\right)\left(0.028m^3/s\right)}{\pi \left(0.001kg/ms\right)\left(0.08m\right)}=444742.57$

For pipe 4,

${\mathrm{Re}}_4=\frac{4\rho Q_4}{\pi \mu d}=\frac{4\left(998kg/m^3\right)\left(0.0424m^3/s\right)}{\pi \left(0.001kg/ms\right)\left(0.08m\right)}=673467.32$

Calculate the friction factor for all pipes,

For pipe 1,

$\begin{array}{l}\frac{1}{f_1{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{714764.85}+{\left(\frac{3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_1\approx 0.0269\end{array}$

For pipe 2,

$\begin{array}{l}\frac{1}{f_2{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{411386.88}+{\left(\frac{3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_2\approx 0.027\end{array}$

For pipe 3,

$\begin{array}{l}\frac{1}{f_3{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{444742.57}+{\left(\frac{3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_3\approx 0.027\end{array}$

For pipe 4,

$\begin{array}{l}\frac{1}{f_4{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \approx -1.8\log \left(\frac{6.9}{673467.32}+{\left(\frac{3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_4\approx 0.0269\end{array}$

Friction factors are almost same for all four pipes.

Now, calculate the exact flow rates in each pie,

For pipe 1,

$\begin{array}{l}45.96m=\frac{8\left(45m\right)\left(0.0269\right)Q_1^2}{{\pi }^2\left(9.81m/s^2\right){\left(0.08m\right)}^5}\\ Q_1=0.0388m^3/s\end{array}$

For pipe 2,

$\begin{array}{l}-15.32m=\frac{8\left(45m\right)\left(0.027\right)Q_2^2}{{\pi }^2\left(9.81m/s^2\right){\left(0.08m\right)}^5}\\ Q_2=-0.0223m^3/s\end{array}$

For pipe 3,

$\begin{array}{l}\frac{\left(675\times {10}^{3}Pa-500\times {10}^{3}Pa\right)}{9790N/m^3}=\frac{8\left(45m\right)\left(0.027\right)Q_3^2}{{\pi }^2\left(9.81m/s^2\right){\left(0.08m\right)}^5}\\ Q_3=0.0241m^3/s\end{array}$

For pipe 4,

$\begin{array}{l}\frac{\left(100\times {10}^{3}Pa-500\times {10}^{3}Pa\right)}{9790N/m^3}=\frac{8\left(45m\right)\left(0.0269\right)Q_4^2}{{\pi }^2\left(9.81m/s^2\right){\left(0.08m\right)}^5}\\ Q_4=-0.0365m^3/s\end{array}$

Net junction flow is equal to zero, therefore,

$\left(0.0388m^3/s\right)-\left(0.0223m^3/s\right)+\left(0.0241m^3/s\right)-\left(0.0365m^3/s\right)=+0.0041$

The assumed $p_a$ is little low. By assuming $p_a=515kPa$ we can get more accurate answers.

But for now, we can say that,

$\begin{array}{l}{Q}_1=0.0388m^3/s\\ Q_2=-0.0223m^3/s\\ Q_3=0.0241m^3/s\\ Q_4=-0.0365m^3/s\\ \end{array}$

Conclusion:

The assumed $p_a$ is little low. By assuming $p_a=515kPa$ we can get more accurate answers.

But for now, we can say that,

$\begin{array}{l}{Q}_1=0.0388m^3/s\\ Q_2=-0.0223m^3/s\\ Q_3=0.0241m^3/s\\ Q_4=-0.0365m^3/s\end{array}$.