Chapter 6, Problem 6.118P

To determine

Overall pressure drop $p_1-p_2$ in $lbf/i{n}^2$.

Answer to Problem 6.118P

$p_1-p_2=149.327lbf/i{n}^2$

Explanation of Solution

Given information:

All pipes are concrete, and roughness is equal to $0.04in$

The fluid is water at $20°C$

$Q=20f{t}^3/s$

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 pipe a,

Find the velocity of the flow in pie a,

$\begin{array}{l}V=\frac{Q}{A}\\ =\frac{20f{t}^3/s}{\pi {\left(\frac{6}{12}ft\right)}^2}\\ =25.464ft/s\end{array}$

Determine the Reynolds’s number,

Assume the kinematic viscosity of water at $1atm$ and $20°C$ as

$\begin{array}{l}{R}_e=\frac{DV}{\nu }\\ =\frac{\left(\frac{12}{12}ft\right)\left(25.464ft/s\right)}{1.09\times {10}^{-5}f{t}^2/s}\\ =2.34\times {10}^4\end{array}$

The roughness ratio is equal to,

$\begin{array}{l}Roughnessratio=\frac{\in }{D}\\ =\frac{0.04}{12}\\ =3.33\times {10}^{-3}\end{array}$

To determine the friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\sqrt{b^2-4ac}\\ \approx -1.8\log \left(\frac{6.9}{2.34\times {10}^4}+{\left(\frac{\text{}3.33\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f\approx 0.03114\end{array}$

To determine the head loss in pipe a,

$\begin{array}{l}{\text{h}}_f=f\left(\frac{L}{D}\right)\left(\frac{V^2}{2g}\right)\\ {\left({\text{h}}_f\right)}_a=\left(0.03114\right)\left(\frac{1000ft}{\frac{12}{12}ft}\right)\left(\frac{{\left(25.464ft/s\right)}^2}{2\left(32.2ft/s^2\right)}\right)\\ =313.534ft\end{array}$

For pipes $b,c\&d$

${\left({\text{h}}_f\right)}_b={\left({\text{h}}_f\right)}_c={\left({\text{h}}_f\right)}_d$

To find velocity of the flow in each pipes,

$\begin{array}{l}{V}_b=\frac{Q_b}{A_b}=\frac{Q_b}{\pi {\left(\frac{4}{12}ft\right)}^2}=2.865Q_b\\ V_c=\frac{Q_c}{A_c}=\frac{Q_c}{\pi {\left(\frac{6}{12}ft\right)}^2}=1.273Q_c\\ V_d=\frac{Q_d}{A_d}=\frac{Q_d}{\pi {\left(\frac{7.5}{12}ft\right)}^2}=0.815Q_d\end{array}$

By equating the head losses in each pipe,

$f_b\left(\frac{1500ft}{\frac{8}{12}ft}\right)\left(\frac{{\left(2.865Q_b\right)}^2}{2\left(32.2ft/s^2\right)}\right)=f_c\left(\frac{800ft}{\frac{12}{12}ft}\right)\left(\frac{{\left(1.273Q_c\right)}^2}{2\left(32.2ft/s^2\right)}\right)=f_d\left(\frac{1200ft}{\frac{15}{12}ft}\right)\left(\frac{{\left(0.815Q_d\right)}^2}{2\left(32.2ft/s^2\right)}\right)\to \left(1\right)$

If, $f_b=f_c=f_d$

$\begin{array}{l}{Q}_b=0.264Q_c\\ Q_d=1.425Q_c\end{array}$

Therefore,

$\begin{array}{l}{Q}_b+Q_c+Q_d=0.264Q_c+Q_c+1.425Q_c=20f{t}^3/s\\ Q_c=7.438f{t}^3/s\end{array}$

By substituting,

$\begin{array}{l}{Q}_b=1.964f{t}^3/s\\ Q_d=10.599f{t}^3/s\end{array}$

To find relevant Reynolds’s number,

$\begin{array}{l}{\left(R_e\right)}_b=\frac{DV}{\nu }=\frac{\left(\frac{8}{12}ft\right)\left(2.865\right)\left(1.964f{t}^3/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=3.442\times {10}^5\\ {\left(R_e\right)}_c=\frac{DV}{\nu }=\frac{\left(\frac{12}{12}ft\right)\left(1.273\right)\left(7.438f{t}^3/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=8.687\times {10}^5\\ {\left(R_e\right)}_d=\frac{DV}{\nu }=\frac{\left(\frac{15}{12}ft\right)\left(0.815\right)\left(10.599f{t}^3/s\right)}{1.09\times {10}^{-5}f{t}^2/s}=9.906\times {10}^5\end{array}$

The relevant roughness ratio will be,

$\begin{array}{l}{\left(\frac{\in }{D}\right)}_b=\frac{0.04}{8}=5\times {10}^{-3}\\ {\left(\frac{\in }{D}\right)}_c=\frac{0.04}{12}=3.33\times {10}^{-3}\\ {\left(\frac{\in }{D}\right)}_d=\frac{0.04}{15}=2.667\times {10}^{-3}\end{array}$

The relevant friction factor will be,

$\begin{array}{l}\frac{1}{f_b{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{3.442\times {10}^5}+{\left(\frac{\text{}5\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\approx 0.03068\\ \frac{1}{f_c{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{8.687\times {10}^5}+{\left(\frac{\text{}3.33\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\approx 0.03114\\ \frac{1}{f_d{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\approx -1.8\log \left(\frac{6.9}{9.906\times {10}^5}+{\left(\frac{\text{}2.667\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\approx 0.0255\end{array}$

By substituting these values in equation (1),

$\begin{array}{l}{Q}_b=0.267Q_c\\ Q_d=1.574Q_c\end{array}$

Therefore,

$\begin{array}{l}{Q}_b+Q_c+Q_d=0.267Q_c+Q_c+1.574Q_c=20f{t}^3/s\\ Q_c=7.0398f{t}^3/s\end{array}$

To find the head loss across pipe c,

$\begin{array}{l}{\left({\text{h}}_f\right)}_c=\left(0.03114\right)\left(\frac{800ft}{\frac{12}{12}ft}\right)\left(\frac{{\left(1.273\times 7.0398f{t}^3/s\right)}^2}{2\left(32.2ft/s^2\right)}\right)\\ =31.067ft\end{array}$

Therefore the total head loss will be,

$\begin{array}{l}{\left({\text{h}}_f\right)}_{total}={\left({\text{h}}_f\right)}_a+{\left({\text{h}}_f\right)}_c\\ =313.534ft+31.067ft\\ =344.601ft\end{array}$

To find pressure drop between point 1 and point 2

Consider the specific weight of the water as $62.4lbf/f{t}^3$

$\begin{array}{l}{p}_1-p_2=\gamma {\left({\text{h}}_f\right)}_{total}\\ =\left(62.4lbf/f{t}^3\right)\left(344.601f\right)\\ =21503.102lbf/f{t}^2\\ =149.327lbf/i{n}^2\end{array}$

Conclusion:

The pressure drop $p_1-p_2$ is equal to $149.327lbf/i{n}^2$.