Chapter 6, Problem 6.125P

To determine

The flow rate in each pipe and the direction of it.

Answer to Problem 6.125P

$\begin{array}{l}{Q}_1=6.78\times {10}^{-3}{m}^3/s\left(towards\right)\\ Q_2=-1.4\times {10}^{-3}{m}^3/s\left(away\right)\\ Q_3=-5.81\times {10}^{-3}{m}^3/s\left(away\right)\end{array}$

Explanation of Solution

Given information:

Pressures in each pipe is equal to,

$\begin{array}{l}{p}_1=200kPa\\ p_2=160kPa\\ p_3=100kPa\end{array}$

The fluid is water at $20°C$

The pipe is made of cast iron.

The pressure drop $\Delta P$ is defined as,

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

In the above equation,

$f$ - Friction factor

$L$ - Length of pipe

$D$ - Diameter of pipe

$\rho$ - Density of the fluid

$V$ - Velocity of the flow

The pressure drop in each pipe can be defined as below,

$\begin{array}{l}{p}_1-p_B=f_1\left(\frac{L_1}{D_1}\right)\left(\frac{\rho V_1{}^2}{2}\right)\\ p_2-p_B=f_2\left(\frac{L_2}{D_2}\right)\left(\frac{\rho V_2{}^2}{2}\right)\\ p_B-p_3=f_1\left(\frac{L_3}{D_3}\right)\left(\frac{\rho V_3{}^2}{2}\right)\end{array}$

Consider flow in pipe (1) is towards B and the flow in pipe (3) is away from B and the flow in pipe (2) cannot be confirmed.

Calculation:

For pipe 1,

Assume,

$V_1=0.5m/s$

Calculate the roughness ratio,

$\frac{\epsilon }{D_1}=\frac{0.26mm}{120mm}=2.16\times {10}^{-3}$

Calculate the Reynolds’s number,

${\mathrm{R}}_e=\frac{\rho V{D}_1}{\mu }=\frac{\left(998kg/m^3\right)\left(0.5m/s\right)\left(0.12m\right)}{0.001kg/ms}=59880$

Find the friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}=-1.8\log \left(\frac{6.9}{{\mathrm{R}}_e}+{\left(\frac{\epsilon /d}{3.7}\right)}^{1.11}\right)\\ =-1.8\log \left(\frac{6.9}{59880}+{\left(\frac{2.16\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ =0.026\end{array}$

Calculate the pressure at point B,

$\begin{array}{l}{p}_1-p_B=f_1\left(\frac{L_1}{D_1}\right)\left(\frac{\rho V_1{}^2}{2}\right)\\ 200\times {10}^{3}Pa-p_B=\left(0.026\right)\left(\frac{800m}{0.12m}\right)\left(\frac{\left(998kg/m^3\right){\left(0.5m/s\right)}^2}{2}\right)\\ p_B=178.37kPa\end{array}$

Calculate the flow rate,

$Q=VA=\left(0.5m/s\right)\left(\pi {\left(0.06m\right)}^2\right)=5.654\times {10}^{-3}$

For pipe 2,

Assume,

$V_2=0.5m/s$

Calculate the roughness ratio,

$\frac{\epsilon }{D_2}=\frac{0.26mm}{80mm}=3.25\times {10}^{-3}$

Calculate the Reynolds’s number,

${\mathrm{R}}_e=\frac{\rho V{D}_2}{\mu }=\frac{\left(998kg/m^3\right)\left(0.5m/s\right)\left(0.08m\right)}{0.001kg/ms}=39920$

Find the friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}=-1.8\log \left(\frac{6.9}{{\mathrm{R}}_e}+{\left(\frac{\epsilon /d}{3.7}\right)}^{1.11}\right)\\ =-1.8\log \left(\frac{6.9}{39920}+{\left(\frac{3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ =0.0294\end{array}$

Calculate the pressure at point B,

$\begin{array}{l}{p}_2-p_B=f_1\left(\frac{L_1}{D_1}\right)\left(\frac{\rho V_1{}^2}{2}\right)\\ 160\times {10}^{3}Pa-p_B=\left(0.0294\right)\left(\frac{600m}{0.08m}\right)\left(\frac{\left(998kg/m^3\right){\left(0.5m/s\right)}^2}{2}\right)\\ p_B=-259kPa\end{array}$

Calculate the flow rate,

$Q=VA=\left(0.5m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=2.513\times {10}^{-3}$

For pipe 3,

Assume,

$V_3=0.5m/s$

Calculate the roughness ratio,

$\frac{\epsilon }{D_3}=\frac{0.26mm}{100mm}=2.6\times {10}^{-3}$

Calculate the Reynolds’s number,

${\mathrm{R}}_e=\frac{\rho V{D}_3}{\mu }=\frac{\left(998kg/m^3\right)\left(0.5m/s\right)\left(0.1m\right)}{0.001kg/ms}=49900$

Find the friction factor,

$\begin{array}{l}\frac{1}{f^{1/2}}=-1.8\log \left(\frac{6.9}{{\mathrm{R}}_e}+{\left(\frac{\epsilon /d}{3.7}\right)}^{1.11}\right)\\ =-1.8\log \left(\frac{6.9}{49900}+{\left(\frac{2.6\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ =0.0276\end{array}$

Calculate the pressure at point B,

$\begin{array}{l}{p}_2-p_B=f_1\left(\frac{L_1}{D_1}\right)\left(\frac{\rho V_1{}^2}{2}\right)\\ 100\times {10}^{3}Pa-p_B=\left(0.0276\right)\left(\frac{900m}{0.1m}\right)\left(\frac{\left(998kg/m^3\right){\left(0.5m/s\right)}^2}{2}\right)\\ p_B=69kPa\end{array}$

Calculate the flow rate,

$Q=VA=\left(0.5m/s\right)\left(\pi {\left(0.05m\right)}^2\right)=3.92\times {10}^{-3}$

Calculate the average value of $p_B$

$p_B=\frac{\left(178.37kPa+259kPa+69kPa\right)}{3}=168.79$

The friction factor will not differ a lot; we can use the same friction factor to find relevant velocities,

For pipe (1)

$\begin{array}{l}200\times {10}^{3}Pa-168.79\times {10}^{3}Pa=\left(0.026\right)\left(\frac{800m}{0.12m}\right)\left(\frac{\left(998kg/m^3\right){\left(V_1\right)}^2}{2}\right)\\ V_1=0.6m/s\end{array}$

The flow rate is equal to,

$Q_1=V_{1}A=\left(0.6m/s\right)\left(\pi {\left(0.06m\right)}^2\right)=6.78\times {10}^{-3}{m}^3/s$

The flow in pipe (1) is considered as a flow towards junction B.

For pipe (2)

$\begin{array}{l}160\times {10}^{3}Pa-168.79\times {10}^{3}Pa=\left(0.0294\right)\left(\frac{600m}{0.08m}\right)\left(\frac{\left(998kg/m^3\right){\left(V_2\right)}^2}{2}\right)\\ V_2=-0.28m/s\end{array}$

The flow rate is equal to,

$Q_2=V_{2}A=\left(-0.28m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=-1.4\times {10}^{-3}{m}^3/s$

The flow in pipe (2) is considered as a flow away from junction B.

For pipe (3)

$\begin{array}{l}100\times {10}^{3}Pa-168.79\times {10}^{3}Pa=\left(0.0276\right)\left(\frac{900m}{0.1m}\right)\left(\frac{\left(998kg/m^3\right){\left(V_3\right)}^2}{2}\right)\\ V_2=-0.74m/s\end{array}$

The flow rate is equal to,

$Q_3=V_{3}A=\left(-0.74m/s\right)\left(\pi {\left(0.05m\right)}^2\right)=-5.81\times {10}^{-3}{m}^3/s$

The flow in pipe (3) is considered as a flow away from junction B.

$\begin{array}{l}Q=\left(Q_1-\left(Q_2+Q_3\right)\right)=\left(6.78\times {10}^{-3}{m}^3/s-\left(1.4\times {10}^{-3}{m}^3/s+5.81\times {10}^{-3}{m}^3/s\right)\right)\\ =-0.00043\\ \end{array}$

By further iteration, we an get more accurate answers.

Conclusion:

The flow rates in each pipe is equal to,

$\begin{array}{l}{Q}_1=6.78\times {10}^{-3}{m}^3/s\left(towards\right)\\ Q_2=-1.4\times {10}^{-3}{m}^3/s\left(away\right)\\ Q_3=-5.81\times {10}^{-3}{m}^3/s\left(away\right)\end{array}$.