Chapter 6, Problem 6.115P

To determine

(a)

The flow rate from reservoir 1 if valve C is closed.

Answer to Problem 6.115P

$Q=0.0152m^3/s$

Explanation of Solution

Given information:

All pipes are made of cast iron and the diameter is equal to $8cm$

If the valve C is closed, the energy equation for the above system can be written as,

$z_1-z_2=h_{f_A}+{\displaystyle \sum h_{m_A}}+h_{f_B}+{\displaystyle \sum h_{m_B}}$

Since, the flow is series and the diameter is same for all pipes, the flow rate and the velocity will be equal in both pipe A and pipe B.

Calculation:

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

$\rho =998kg/m^3$

$\mu =0.001kg/ms$

The loss co efficient will be equal to,

$\begin{array}{l}{K}_{sharpentrance}=0.5\\ K_{linejunction}=0.9\\ K_{Branchjunction}=1.3\\ K_{Submergedexit}=1.0\end{array}$

According to the explanation, the energy equation will be,

$\begin{array}{l}{z}_1-z_2=h_{f_A}+{\displaystyle \sum h_{m_A}}+h_{f_B}+{\displaystyle \sum h_{m_B}}\\ \Delta Z=\frac{V^2}{2g}\left(f\frac{L_A}{D}+K_{sharpentrance}+K_{linejunction}+f\frac{L_B}{D}+K_{Submergedexit}\right)\\ 25m=\frac{V^2}{2\left(9.81m/s^2\right)}\left(f\frac{100m}{0.08m}+0.5+0.9+f\frac{50m}{0.08m}+1.0\right)\end{array}$

Assume,

$f=0.02$

Therefore,

$\begin{array}{l}25m=\frac{V^2}{2\left(9.81m/s^2\right)}\left(\left(0.02\right)\frac{100m}{0.08m}+0.5+0.9+\left(0.02\right)\frac{50m}{0.08m}+1.0\right)\\ V=3.5m/s\end{array}$

Calculate the Reynolds’s number

${\mathrm{Re}}_d=\frac{\left(\rho Vd\right)}{\mu }=\frac{\left(998kg/m^3\right)\left(3.5m/s\right)\left(0.08m\right)}{0.001kg/ms}=279440$

Calculate the roughness ratio

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

Calculate the exact value of friction factor

$\begin{array}{l}\frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{R_e}+{\left(\frac{\text{}\in /d}{3.7}\right)}^{1.11}\right)\\ \frac{1}{f^{1/2}}\approx -1.8\log \left(\frac{6.9}{279440}+{\left(\frac{\text{}3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f=0.0272\end{array}$

Now, calculate the exact value of the velocity

$\begin{array}{l}{z}_1-z_2=h_{f_A}+{\displaystyle \sum h_{m_A}}+h_{f_B}+{\displaystyle \sum h_{m_B}}\\ \Delta Z=\frac{V^2}{2g}\left(f\frac{L_A}{D}+K_{sharpentrance}+K_{linejunction}+f\frac{L_B}{D}+K_{Submergedexit}\right)\\ 25m=\frac{V^2}{2\left(9.81m/s^2\right)}\left(\left(0.0272\right)\frac{100m}{0.08m}+0.5+0.9+\left(0.0272\right)\frac{50m}{0.08m}+1.0\right)\\ V=3.03m/s\end{array}$

Calculate the flow rate

$Q=VA=\left(3.03m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=0.0152m^3/s$

Conclusion:

The flow rate is equal to $0.0152m^3/s$.

To determine

(b)

The flow arte from reservoir 1 if valve C is open.

Answer to Problem 6.115P

$\begin{array}{l}{Q}_A=0.0174m^3/s\\ Q_B=0.00955m^3/s\\ Q_C=0.00784m^3/s\end{array}$

Explanation of Solution

Given information:

All pipes are made of cast iron and the diameter is equal to $8cm$

If the valve C is opened, the energy equation for the above system can be written as,

$z_1-z_2=h_{f_A}+{\displaystyle \sum h_{m_{AB}}}+h_{f_B}+{\displaystyle \sum h_{m_B}}=h_{f_A}+{\displaystyle \sum h_{m_{AC}}}+h_{f_B}+{\displaystyle \sum h_{m_C}}$

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

$Q_A=Q_B+Q_C$

$V_A=V_B+V_C$

Calculation:

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

$\rho =998kg/m^3$

$\mu =0.001kg/ms$

The loss co-efficient will be equal to,

$\begin{array}{l}{K}_{sharpentrance}=0.5\\ K_{linejunction}=0.9\\ K_{Branchjunction}=1.3\\ K_{Submergedexit}=1.0\end{array}$

According to the explanation, the energy equation will be,

$\begin{array}{l}{z}_1-z_2=h_{f_A}+{\displaystyle \sum h_{m_{AB}}}+h_{f_B}+{\displaystyle \sum h_{m_B}}=h_{f_A}+{\displaystyle \sum h_{m_{AC}}}+h_{f_B}+{\displaystyle \sum h_{m_C}}\\ \Delta Z=\frac{V_A{}^2}{2g}\left(f\frac{L_A}{D}+K_{sharpentrance}+K_{linejunction}\right)+\frac{V_B{}^2}{2g}\left(f\frac{L_B}{D}+K_{Submergedexit}\right)\\ 25m=\frac{V_A{}^2}{2\left(9.81m/s\right)}\left(f\frac{100m}{0.08m}+0.5+0.9\right)+\frac{V_B{}^2}{2\left(9.81m/s\right)}\left(f\frac{50m}{0.08m}+1.0\right)\end{array}$

Similarly,

$\begin{array}{l}\Delta Z=\frac{V_A{}^2}{2g}\left(f\frac{L_A}{D}+K_{sharpentrance}+K_{branchjunction}\right)+\frac{V_C{}^2}{2g}\left(f\frac{L_C}{D}+K_{Submergedexit}\right)\\ 25m=\frac{V_A{}^2}{2\left(9.81m/s\right)}\left(f\frac{100m}{0.08m}+0.5+1.3\right)+\frac{V_C{}^2}{2\left(9.81m/s\right)}\left(f\frac{70m}{0.08m}+1.0\right)\end{array}$

But, we also know

$V_A=V_B+V_C$

Assume the friction factor is same for all pipe and it is equal to,

$f=0.02$

$\begin{array}{l}25m=\frac{V_A{}^2}{2\left(9.81m/s\right)}\left(\left(0.027\right)\frac{100m}{0.08m}+0.5+0.9\right)+\frac{V_B{}^2}{2\left(9.81m/s\right)}\left(\left(0.027\right)\frac{50m}{0.08m}+1.0\right)\\ 25m=1.791V_A{}^2+0.911V_B{}^2\to \left(1\right)\end{array}$

Similarly,

$\begin{array}{l}25m=\frac{V_A{}^2}{2\left(9.81m/s\right)}\left(\left(0.027\right)\frac{100m}{0.08m}+0.5+1.3\right)+\frac{V_C{}^2}{2\left(9.81m/s\right)}\left(\left(0.027\right)\frac{70m}{0.08m}+1.0\right)\\ 25m=1.811V_A{}^2+1.255V_C{}^2\to \left(2\right)\end{array}$

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

By solving we get,

$\begin{array}{l}{V}_A=3.48m/s\\ V_B=1.91m/s\\ V_C=1.57m/s\end{array}$

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

$\begin{array}{l}{\mathrm{Re}}_A=\frac{\left(\rho Vd\right)}{\mu }=\frac{\left(998kg/m^3\right)\left(3.48m/s\right)\left(0.08m\right)}{0.001kg/ms}=277843.2\\ {\mathrm{Re}}_B=\frac{\left(\rho Vd\right)}{\mu }=\frac{\left(998kg/m^3\right)\left(1.91m/s\right)\left(0.08m\right)}{0.001kg/ms}=152494.4\\ {\mathrm{Re}}_C=\frac{\left(\rho Vd\right)}{\mu }=\frac{\left(998kg/m^3\right)\left(1.57m/s\right)\left(0.08m\right)}{0.001kg/ms}=125348.8\end{array}$

$\begin{array}{l}\frac{1}{f_A{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{277843.2}+{\left(\frac{\text{}3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_A\approx 0.0272\\ \frac{1}{f_B{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{152494.4}+{\left(\frac{\text{}3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_B\approx 0.0275\\ \frac{1}{f_C{}^{1/2}}\approx -1.8\log \left(\frac{6.9}{125348.8}+{\left(\frac{\text{}3.25\times {10}^{-3}}{3.7}\right)}^{1.11}\right)\\ f_C\approx 0.0277\end{array}$

$\begin{array}{l}25m=\frac{V_A{}^2}{2\left(9.81m/s\right)}\left(\left(0.0272\right)\frac{100m}{0.08m}+0.5+0.9\right)+\frac{V_B{}^2}{2\left(9.81m/s\right)}\left(\left(0.0275\right)\frac{50m}{0.08m}+1.0\right)\\ 25m=1.804V_A{}^2+0.927V_B{}^2\to \left(1\right)\end{array}$

$\begin{array}{l}25m=\frac{V_A{}^2}{2\left(9.81m/s\right)}\left(\left(0.0272\right)\frac{100m}{0.08m}+0.5+1.3\right)+\frac{V_C{}^2}{2\left(9.81m/s\right)}\left(\left(0.0277\right)\frac{70m}{0.08m}+1.0\right)\\ 25m=1.824V_A{}^2+1.283V_C{}^2\to \left(2\right)\end{array}$

According to the equations,

$\begin{array}{l}{V}_A=3.46m/s\\ V_B=1.9m/s\\ V_C=1.56m/s\end{array}$

The relevant flow rate,

$\begin{array}{l}{Q}_A=V_{A}A=\left(3.46m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=0.0174m^3/s\\ Q_B=V_{B}A=\left(1.9m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=0.00955m^3/s\\ Q_C=V_{C}A=\left(1.56m/s\right)\left(\pi {\left(0.04m\right)}^2\right)=0.00784m^3/s\end{array}$

Conclusion:

The flow rate in pipe A is equal to $0.0174m^3/s$

The flow arte in pipe B is equal to $0.00955m^3/s$

The flow arte in pipe C is equal to $0.00784m^3/s$.