Chapter 6, Problem 6.113P

To determine

(a)

The flow rate in each pipe.

Answer to Problem 6.113P

$\begin{array}{l}{Q}_1=0.023m^3/s\\ Q_2=0.013m/s\end{array}$

Explanation of Solution

Given information:

The total flow rate is equal to $0.036m^3/s$

The loss of co-efficient is equal to $K=1.5$

The volumetric flow rate Q can be defined as below,

$Q=VA$

In the above equation,

$V$ - Velocity of the flow

$A$ -Area

The energy equation is defined as,

$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f+{\displaystyle \sum h_m}$

Where the head loss $h_f$ is equal to,

$h_f=f\frac{L}{D}\left(\frac{V^2}{2g}\right)$

The friction factor can be determined by,

$\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)$

The Reynolds’s number is defined as,

${\mathrm{Re}}_D=\frac{\left(\rho VD\right)}{\mu }$

$\mu$ - Dynamic viscosity

For parallel pipe flow,

$Q_{Total}=Q_1+Q_2$

Assume, the water at $20°C$ will have,

$\rho =998kg/m^3$

$\mu =0.001kg/ms$

Calculation:

For parallel pipe flow, we can say

$Q_{Total}=Q_1+Q_2$

Substitute for known values,

$\frac{\pi }{4}{\left(0.05m\right)}^2{V}_1+\frac{\pi }{4}{\left(0.04m\right)}^2{V}_2=0.036m^3/s\to \left(1\right)$

For pipe 1, the energy equation can be defined as,

$\Delta z=h_1$

Head loss can be written as,

$h_1=f\frac{L_1}{D_1}\left(\frac{V_1{}^2}{2g}\right)$

Guess $f=0.02$

Substitute for known values,

$h_1=f\frac{L_1}{D_1}\left(\frac{V_1{}^2}{2g}\right)=\left(0.02\right)\frac{60m}{0.05m}\left(\frac{V_1{}^2}{2\left(9.81m/s^2\right)}\right)$

Similarly for pipe 2,

Energy equation can be written as,

$\Delta z=h_2+{\displaystyle \sum h_m}$

Therefore, the head loss is equal to,

$h_2+{\displaystyle \sum h_m}=\frac{V_2{}^2}{2g}\left(f\frac{L_2}{D_2}+K\right)$

Guess $f=0.02$

Substitute for known values,

$h_2+{\displaystyle \sum h_m}=\frac{V_2{}^2}{2g}\left(f\frac{L_2}{D_2}+K\right)=\frac{V_2{}^2}{2\left(9.81m/s^2\right)}\left(\left(0.02\right)\frac{55m}{0.04m}+1.5\right)$

But we know that,

$h_1=h_2+{\displaystyle \sum h_m}$

Therefore,

$\left(0.02\right)\frac{60m}{0.05m}\left(\frac{V_1{}^2}{2\left(9.81m/s^2\right)}\right)=\frac{V_2{}^2}{2\left(9.81m/s^2\right)}\left(\left(0.02\right)\frac{55m}{0.04m}+1.5\right)\to \left(2\right)$

Solve equation 1 & 2 to find relevant velocities,

According to equation 2,

$V_1=1.1V_2$

Substitute in equation 1,

Therefore we get,

$\begin{array}{l}{V}_1=11.59m/s\\ V_2=10.54m/s\end{array}$

Calculate the relevant Reynolds’s numbers,

$\begin{array}{l}{\mathrm{Re}}_{D_1}=\frac{\rho VD}{\mu }=\frac{\left(998kg/m^3\right)\left(11.59m/s\right)\left(0.05m\right)}{0.001kg/ms}=578341\\ {\mathrm{Re}}_{D_2}=\frac{\rho VD}{\mu }=\frac{\left(998kg/m^3\right)\left(10.54m/s\right)\left(0.04m\right)}{0.001kg/ms}=420756.8\end{array}$

Calculate the roughness ratio,

For pipe 1,

$\frac{\epsilon }{d}=\frac{0.15mm}{50mm}=3\times {10}^{-3}$

For pipe 2,

$\frac{\epsilon }{d}=\frac{0.15mm}{40mm}=3.75\times {10}^{-3}$

Calculate the relevant friction factors,

For pipe 1,

$\frac{1}{f_1{}^{1/2}}=-2.0\log \left(\frac{\in /d}{3.7}+\left(\frac{2.51}{{\mathrm{Re}}_D{f}_1{}^{1/2}}\right)\right)=-2.0\log \left(\frac{3\times {10}^{-3}}{3.7}+\left(\frac{2.51}{578341f_1{}^{1/2}}\right)\right)$

By solving,

$f_1=0.0264$

For pipe 2,

$\frac{1}{f_2{}^{1/2}}=-2.0\log \left(\frac{\in /d}{3.7}+\left(\frac{2.51}{{\mathrm{Re}}_D{f}_2{}^{1/2}}\right)\right)=-2.0\log \left(\frac{3.75\times {10}^{-3}}{3.7}+\left(\frac{2.51}{420756.8f_2{}^{1/2}}\right)\right)$

By solving,

$f_2=0.0282$

Calculate the relevant velocities for the above found friction factors,

Substitute in equation 2,

$\left(0.0264\right)\frac{60m}{0.05m}\left(\frac{V_1{}^2}{2\left(9.81m/s^2\right)}\right)=\frac{V_2{}^2}{2\left(9.81m/s^2\right)}\left(\left(0.0282\right)\frac{55m}{0.04m}+1.5\right)\to \left(2\right)$

Solve further,

$V_1=1.127V_2$

Substitute in equation 1,

$\frac{\pi }{4}{\left(0.05m\right)}^2\left(1.127V_2\right)+\frac{\pi }{4}{\left(0.04m\right)}^2{V}_2=0.036m^3/s$

Therefore we get,

$\begin{array}{l}{V}_1=11.69m/s\\ V_2=10.37m/s\end{array}$

Now, find the relevant flow rates,

$Q_1=\frac{\pi }{4}{\left(0.05m\right)}^2\left(11.69m/s\right)=0.023m^3/s$

$Q_2=\frac{\pi }{4}{\left(0.04m\right)}^2\left(10.37\right)=0.013m/s$

Conclusion:

The flow rate in pipe 1 is equal to $0.023m^3/s$

The flow rate in pipe 2 is equal to $0.013m/s$.

To determine

(b)

The overall pressure drop.

Answer to Problem 6.113P

$\Delta P=2160303.36Pa\approx 2.16MPa$

Explanation of Solution

Given information:

The total flow rate is equal to $0.036m^3/s$

The loss of co-efficient is equal to $K=1.5$

The pressure drop is defined as,

$\Delta P=\rho g{h}_f=\frac{\rho V^2}{2}\left(f\frac{L}{D}\right)$

Assume, the water at $20°C$ will have,

$\rho =998kg/m^3$

Calculation:

The pressure drop is same in both pipes, therefore,

For pipe 1,

$\Delta P=\rho g{h}_f=\frac{\rho V_1{}^2}{2}\left(f_1\frac{L_1}{D_1}\right)$

Substitute for known values,

$\Delta P=\frac{\rho V_1{}^2}{2}\left(f_1\frac{L_1}{D_1}\right)=\frac{\left(998kg/m^3\right){\left(11.69m/s\right)}^2}{2}\left(\left(0.0264\right)\frac{60m}{0.05m}\right)=2160303.36Pa\approx 2.16MPa$

Conclusion:

The pressure drop is equal to $2.16MPa$.