Chapter 6, Problem 6.120P

To determine

(a)

The flow rate in each pipe.

Answer to Problem 6.120P

$\begin{array}{l}{Q}_1=101.52m^3/h\\ Q_2=40.032m^3/h\\ Q_3=58.32m^3/h\end{array}$

Explanation of Solution

Given information:

The fluid is water at $20°C$

The total flow rate is $200m^3/h$

In parallel pipe system the pressure drop is same in each pipe.

$\Delta h_{A\to B}=\Delta h_b=\Delta h_c=\Delta h_d$

Total flow is sum of individual flows,

$Q=Q_1+Q_2+Q_3$

The velocity $V$ of the flow is equal to,

$V=\frac{Q}{A}$ Where $A$ is the cross sectional area

The Reynolds’s number is defined as,

$R_e=\frac{DV}{\nu }$

$D$ - Diameter of the pipe

$V$ - Velocity of the flow

$\nu$ - Kinematic viscosity

Roughness ratio is equal to,

$Roughnessratio=\frac{\in }{D}$

$\in$ - Roughness

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 total head loss ${\text{h}}_f$ can be defined as,

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

$L$ - Length of the pipe

Calculation:

For pipes $1,2\&3$

${\left({\text{h}}_f\right)}_1={\left({\text{h}}_f\right)}_2={\left({\text{h}}_f\right)}_3$

To find velocity of the flow in each pipes,

$\begin{array}{l}{V}_1=\frac{Q_1}{A_1}=\frac{Q_1}{\pi {\left(0.06\right)}^2}=88.419Q_1\\ V_2=\frac{Q_2}{A_2}=\frac{Q_2}{\pi {\left(0.04\right)}^2}=198.945Q_2\\ V_3=\frac{Q_3}{A_3}=\frac{Q_3}{\pi {\left(0.05\right)}^2}=127.324Q_3\end{array}$

By equating the head losses in each pipe,

$f_1\left(\frac{800m}{0.12m}\right)\left(\frac{{\left(88.419Q_1\right)}^2}{2\left(9.81m/s^2\right)}\right)=f_2\left(\frac{600m}{0.08m}\right)\left(\frac{{\left(198.945Q_2\right)}^2}{2\left(9.81m/s^2\right)}\right)=f_3\left(\frac{900m}{0.1m}\right)\left(\frac{{\left(127.324Q_3\right)}^2}{2\left(9.81m/s^2\right)}\right)\to \left(1\right)$

If, $f_1=f_2=f_3$

$\begin{array}{l}{Q}_1=2.387Q_2\\ Q_3=1.426Q_2\end{array}$

Therefore,

$\begin{array}{l}{Q}_1+Q_2+Q_3=2.387Q_2+Q_2+1.426Q_2=0.0556m^3/s\\ Q_2=0.0116m^3/s\end{array}$

By substituting,

$\begin{array}{l}{Q}_1=0.0276m^3/s\\ Q_3=0.0165m^3/s\end{array}$

To find relevant Reynolds’s number,

$\begin{array}{l}{\left(R_e\right)}_1=\frac{DV}{\nu }=\frac{\left(0.12m\right)\left(88.419\right)\left(0.0276m^3/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=2.686\times {10}^4\\ {\left(R_e\right)}_2=\frac{DV}{\nu }=\frac{\left(0.08m\right)\left(198.945\right)\left(0.0116m^3/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=1.694\times {10}^4\\ {\left(R_e\right)}_3=\frac{DV}{\nu }=\frac{\left(0.1m\right)\left(127.324\right)\left(0.0165m^3/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=1.927\times {10}^4\end{array}$

The relevant roughness ratio will be,

$\begin{array}{l}{\left(\frac{\in }{D}\right)}_1=\frac{0.26\times {10}^{-3}m}{0.12m}=2.167\times {10}^{-3}\\ {\left(\frac{\in }{D}\right)}_2=\frac{0.26\times {10}^{-3}m}{0.08}=3.25\times {10}^{-3}\\ {\left(\frac{\in }{D}\right)}_3=\frac{0.26\times {10}^{-3}m}{0.1m}=2.6\times {10}^{-3}\end{array}$

The relevant friction factor will be,

$\begin{array}{l}\frac{1}{f_1{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{2.686\times {10}^4}+{\left(\frac{\text{}2.167\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\approx 0.0285\\ \frac{1}{f_2{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{1.694\times {10}^4}+{\left(\frac{\text{}3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\approx 0.0323\\ \frac{1}{f_d{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{1.927\times {10}^4}+{\left(\frac{2.6\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\approx 0.0307\end{array}$

By substituting these values in equation (1),

$\left(0.0285\right)\left(\frac{800m}{0.12m}\right)\left(\frac{{\left(88.419Q_1\right)}^2}{2\left(9.81m/s^2\right)}\right)=\left(0.0323\right)\left(\frac{600m}{0.08m}\right)\left(\frac{{\left(198.945Q_2\right)}^2}{2\left(9.81m/s^2\right)}\right)=\left(0.0307\right)\left(\frac{900m}{0.1m}\right)\left(\frac{{\left(127.324Q_3\right)}^2}{2\left(9.81m/s^2\right)}\right)$

$\begin{array}{l}{Q}_1=2.54Q_2\\ Q_3=1.462Q_2\end{array}$

Therefore,

$\begin{array}{l}{Q}_1+Q_2+Q_3=2.54Q_2+Q_2+1.462Q_2=0.0556m^3/s\\ Q_2=0.01112m^3/s\end{array}$

By substituting,

$\begin{array}{l}{Q}_1=0.0282m^3/s\\ Q_3=0.0162m^3/s\end{array}$

$\begin{array}{l}{Q}_1=101.52m^3/h\\ Q_2=40.032m^3/h\\ Q_3=58.32m^3/h\end{array}$

Conclusion:

The flow rate in each pipe is as below,

$\begin{array}{l}{Q}_1=101.52m^3/h\\ Q_2=40.032m^3/h\\ Q_3=58.32m^3/h\end{array}$.

To determine

(b)

The pressure drop across the system

Answer to Problem 6.120P

$1.1413\times {10}^{8}N/m^2$

Explanation of Solution

Given information:

The fluid is water at $20°C$

The total flow rate is $200m^3/h$

In parallel pipe system the pressure drop is same in each pipe.

$\Delta h_{A\to B}=\Delta h_b=\Delta h_c=\Delta h_d$

Total flow is sum of individual flows,

$Q=Q_1+Q_2+Q_3$

The velocity $V$ of the flow is equal to,

$V=\frac{Q}{A}$ Where, $A$ is the cross sectional area

The Reynolds’s number is defined as,

$R_e=\frac{DV}{\nu }$

$D$ - Diameter of the pipe

$V$ - Velocity of the flow

$\nu$ - Kinematic viscosity

Roughness ratio is equal to,

$Roughnessratio=\frac{\in }{D}$

$\in$ - Roughness

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 total head loss ${\text{h}}_f$ can be defined as,

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

$L$ - Length of the pipe

Calculation:

To find the pressure drop across the system,

By finding the head loss across any one of the pipe, we can able to find the pressure drop

Therefore, to find the head loss across pipe 2,

$\begin{array}{l}{\left({\text{h}}_f\right)}_2=\left(5.5624\right)\left(\frac{600m}{0.08m}\right)\left(\frac{{\left(198.945\times 0.01177m^3/s\right)}^2}{2\left(9.81m/s^2\right)}\right)\\ =1.1658\times {10}^4\end{array}$

Consider the specific weight of the water as $9790N/m^3$

$\begin{array}{l}{p}_1-p_2=\gamma {\left({\text{h}}_f\right)}_{total}\\ =\left(9790N/m^3\right)\left(1.1658\times {10}^{4}m\right)\\ =1.1413\times {10}^{8}N/m^2\end{array}$

Conclusion:

The pressure drop $p_1-p_2$ is equal to $1.1413\times {10}^{8}N/m^2$.