Chapter 6, Problem 6.116P

To determine

The flow rate in $m^3/h$ ?

Answer to Problem 6.116P

$\begin{array}{l}{Q}_A=0.01035m^3/s\\ Q_B=0.0165m^3/s\\ Q_C=0.0269m^3/s\end{array}$

Explanation of Solution

Given information:

The pipes are made of asphalted cast iron.

The diameter of all pipes is equal to $8cm$

The total pressure drop is equal to $750kPa$

The fluid is water at $20°C$

The energy equation for the above system can be written as,

$\begin{array}{l}\frac{\Delta P}{\rho g}=h_{f_A}+h_{f_C}=\frac{V_A{}^2}{2g}\left(f\frac{L_A}{D}\right)+\frac{V_C{}^2}{2g}\left(f\frac{L_C}{D}\right)\\ \frac{\Delta P}{\rho g}=h_{f_B}+h_{f_C}=\frac{V_B{}^2}{2g}\left(f\frac{L_B}{D}\right)+\frac{V_C{}^2}{2g}\left(f\frac{L_C}{D}\right)\end{array}$

Since, the flow is parallel in pipe A and B, and the diameter is same for all pipes,

$Q_C=Q_A+Q_B$

$V_C=V_A+V_B$

Calculation:

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

$\rho =998kg/m^3$

$\mu =0.001kg/ms$

The roughness ratio is equal to,

$\frac{\epsilon }{d}=\frac{0.12mm}{80mm}=1.5\times {10}^{-3}$

According to the explanation, the energy equation will be,

Assume the friction factor as,

$f=0.022$

$\begin{array}{l}\frac{\Delta P}{\rho }=\frac{V_A{}^2}{2}\left(f\frac{L_A}{D}\right)+\frac{V_C{}^2}{2}\left(f\frac{L_C}{D}\right)\\ \frac{750\times {10}^{3}Pa}{\left(998kg/m^3\right)}=\frac{V_A{}^2}{2}\left(\left(0.022\right)\frac{250m}{0.08m}\right)+\frac{V_C{}^2}{2}\left(\left(0.022\right)\frac{150m}{0.08m}\right)\\ 751.5=34.375V_A{}^2+20.625V_C{}^2\to \left(1\right)\end{array}$

Similarly,

$\begin{array}{l}\frac{\Delta P}{\rho g}=\frac{V_B{}^2}{2g}\left(f\frac{L_B}{D}\right)+\frac{V_C{}^2}{2g}\left(f\frac{L_C}{D}\right)\\ \frac{750\times {10}^{3}Pa}{\left(998kg/m^3\right)}=\frac{V_B{}^2}{2}\left(\left(0.022\right)\frac{100m}{0.08m}\right)+\frac{V_C{}^2}{2}\left(\left(0.022\right)\frac{150m}{0.08m}\right)\\ 751.5=13.75V_B{}^2+20.625V_C{}^2\to \left(2\right)\end{array}$

But, we know that,

$V_C=V_A+V_B\to \left(3\right)$

By solving above equations,

$\begin{array}{l}{V}_A=2.1m/s\\ V_B=3.3m/s\\ V_C=5.4m/s\end{array}$

Now find the Reynolds’s number and friction factor separately for each pipe

$\begin{array}{l}{\mathrm{Re}}_A=\frac{\left(\rho Vd\right)}{\mu }=\frac{\left(998kg/m^3\right)\left(2.1m/s\right)\left(0.08m\right)}{0.001kg/ms}=167664\\ {\mathrm{Re}}_B=\frac{\left(\rho Vd\right)}{\mu }=\frac{\left(998kg/m^3\right)\left(3.3m/s\right)\left(0.08m\right)}{0.001kg/ms}=263472\\ {\mathrm{Re}}_C=\frac{\left(\rho Vd\right)}{\mu }=\frac{\left(998kg/m^3\right)\left(5.4m/s\right)\left(0.08m\right)}{0.001kg/ms}=431136\end{array}$

$\begin{array}{l}\frac{1}{f_A{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{167664}+{\left(\frac{1.5\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_A\approx 0.0229\\ \frac{1}{f_B{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{263472}+{\left(\frac{\text{}1.5\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_B\approx 0.0225\\ \frac{1}{f_C{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{431136}+{\left(\frac{\text{}1.5\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_C\approx 0.0222\end{array}$

Replace friction factor in energy equation with above found friction factors,

$\begin{array}{l}\frac{\Delta P}{\rho }=\frac{V_A{}^2}{2}\left(f\frac{L_A}{D}\right)+\frac{V_C{}^2}{2}\left(f\frac{L_C}{D}\right)\\ \frac{750\times {10}^{3}Pa}{\left(998kg/m^3\right)}=\frac{V_A{}^2}{2}\left(\left(0.0229\right)\frac{250m}{0.08m}\right)+\frac{V_C{}^2}{2}\left(\left(0.0222\right)\frac{150m}{0.08m}\right)\\ 751.5=35.781V_A{}^2+20.8125V_C{}^2\to \left(1\right)\end{array}$

Similarly,

$\begin{array}{l}\frac{\Delta P}{\rho g}=\frac{V_B{}^2}{2g}\left(f\frac{L_B}{D}\right)+\frac{V_C{}^2}{2g}\left(f\frac{L_C}{D}\right)\\ \frac{750\times {10}^{3}Pa}{\left(998kg/m^3\right)}=\frac{V_B{}^2}{2}\left(\left(0.0225\right)\frac{100m}{0.08m}\right)+\frac{V_C{}^2}{2}\left(\left(0.0222\right)\frac{150m}{0.08m}\right)\\ 751.5=14.0625V_B{}^2+20.8125V_C{}^2\to \left(2\right)\end{array}$

Therefore, according to above equations,

The new velocities will be,

$\begin{array}{l}{V}_A=2.06m/s\\ V_B=3.29m/s\\ V_C=5.35m/s\end{array}$

Therefore,

Calculate the flow rates,

$\begin{array}{l}{Q}_A=V_{A}A=\left(2.06m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=0.01035m^3/s\\ Q_B=V_{B}A=\left(3.29m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=0.0165m^3/s\\ Q_C=V_{C}A=\left(5.35m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=0.0269m^3/s\end{array}$

Conclusion:

The flow rate in pipe A is equal to $0.01035m^3/s$

The flow arte in pipe B is equal to $0.0165m^3/s$

The flow arte in pipe C is equal to $0.0269m^3/s$.