Chapter 12, Problem 12.1PP

Figure 12.7 shows a branched system in which the pressure at A is $700kPa$ and the pressure at B is $550kPa$. Each branch is $60m$ long. Neglect losses at the junctions, but consider all elbows. If the system carries oil with a specific weight of $8.80kN/m^3$, calculate the total volume flow rate. The oil has a kinematic viscosity of

   $4.8\times {10}^{m}2/s.$

To determine

Total volume flow rate by using the given information.

Answer to Problem 12.1PP

The total volume flow rate is determined below.

Explanation of Solution

First we have to find the head loss at point A and B on applying the energy equation.

   $\frac{p_A}{\gamma }+z_A+\frac{v^2{}_A}{2g}-h_L=\frac{p_B}{\gamma }+z_B+\frac{v^2{}_B}{2g}$

Here $p_A$ =pressure

   $\gamma$ =specific weight

   $z_A$ =Head at A

   $v^2{}_A$ =velocity at A

And then parallel system head losses for both the parts are same

   ${(h_L)}_a={(h_L)}_b$

   $=\frac{p_A-p_B}{\gamma }$

Then substitute $p_A=700kN/m^2,P_B=550kN/m^2,\gamma =8.80kN/m^3$ values into above equation we get

   ${(h_L)}_a={(h_L)}_b$

   $=\frac{700-500}{8.80}$

   $=17.05m$

Then the roughness of steel pipe is $\epsilon =\left(4.6\times {10}^{-5}\right)m$

Inside diameter of the pipe is $0.1023m$

Find the relative roughness of the elbow

   $\begin{array}{l}\frac{\epsilon }{D}=\frac{4.6\times {10}^{-5}}{0.1023}\\ =4.49\times {10}^{-4}\end{array}$

Corresponding to friction factor value is $f_r=0.017$

Find the total head loss of the upper section

   $\begin{array}{l}\left(h_L\right)=h_{fric}+2h_{elb}\\ =fa\left(\frac{L}{D}\right)\frac{u_a{}^2}{2g}+2fr\left(\frac{L_e}{D}\right)\frac{u_a{}^2}{2g}\end{array}$

For 90⁰ elbows curved, effective length ratio is $\left(\frac{L_e}{D}\right)=30$

Substitute required values we will get

   ${(h_L)}_a=f_a\left(\frac{60}{0.1023}\right)\frac{u_a{}^2}{2g}+2\times 0.017\times (30)\frac{u_a{}^2}{2g}$

   $=\left[587f_a+1.02\right]\frac{u_a{}^2}{2g}$

From the table of text book DN 80 pipes the inside diameter of the pipe is 0.0779m

Find the relative rough ness

   $\begin{array}{l}\frac{\epsilon }{D}=\frac{4.6\times {10}^{-5}}{0.0779}\\ =5.9\times {10}^{-4}\end{array}$

Find the total head loss of the lower branch we will get

   $\begin{array}{l}\left(h_L\right)=h_{fric}+2h_{elb}\\ =fa\left(\frac{L}{D}\right)\frac{u_a{}^2}{2g}+fr{\left(\frac{L_e}{D}\right)}_v\frac{u_a{}^2}{2g}+2fr{\left(\frac{L_e}{D}\right)}_e\frac{u_a{}^2}{2g}\end{array}$

Substitute above values we get

   $\begin{array}{l}\left(h_L\right)=f_b\left(\frac{60}{0.0779}\right)\frac{u_b{}^2}{2g}+2\times 0.018\times (30)\frac{u_b{}^2}{2g}+0.018\times (240)\frac{u_b{}^2}{2g}\\ =\left[770f_b+5.4\right]\frac{u_b{}^2}{2g}\end{array}$

Assume $f_a=f_b=0.02$

Find the flow velocity at upper branch

Substitute above value we get

   $\begin{array}{l}17.05=\left[587\times 0.02+1.02\right]\frac{u_a{}^2}{2g}\\ v_a=5.08m/s\end{array}$

Find the Reynolds number for the flow

   $N_R=\frac{u_a{D}_a}{v}$

   $\begin{array}{l}{N}_R=\frac{5.08\times 0.1023}{4.8\times {10}^{-6}}\\ =1.08\times {10}^5\end{array}$

Find the relative roughness we get

   $\begin{array}{l}\frac{\epsilon }{D}=\frac{4.6\times {10}^{-5}}{0.1023}\\ =4.49\times {10}^{-4}\end{array}$

Then the value of friction factor is $F_a=0.02$

Find the flow velocity at B

Substitute above values we get

   $\begin{array}{l}17.05=[770\times f_b+5.4]\frac{u_b^2}{2\times 9.81}\\ 334.521=\left[770\times 0.02+5.4\right]u_b^2\\ v_b=4.01m/s\end{array}$

Calculate the Reynolds number for the flow

   $N_R=\frac{u_a{D}_a}{v}$

Substitute above values we get

   $\begin{array}{l}{N}_R=\frac{4.01\times 0.0023}{4.8\times {10}^{-6}}\\ =6.51\times {10}^4\end{array}$

Find the relative roughness value

   $\begin{array}{l}\frac{\epsilon }{D}=\frac{4.6\times {10}^{-5}}{0.0779}\\ =5.9\times {10}^{-4}\end{array}$

Then the value of friction factor is $F_a=\mathrm{0.022.}$

Recomputed the velocity of flow using above equation

   $\begin{array}{l}17.05=[770\times f_b+5.1]\frac{u_b^2}{2\times 9.81}\\ 334.521=\left[770\times 0.002+5.1\right]u_b^2\\ v_b=3.87m/s\end{array}$

Find the Reynolds number for the flow

   $N_R=\frac{u_a{D}_a}{v}$

Substitute above values we get

   $\begin{array}{l}{N}_R=\frac{3.87\times 0.0779}{4.8\times {10}^{-6}}\\ =6.51*{10}^4\end{array}$

Find the value of relative roughness

   $\begin{array}{l}\frac{\epsilon }{D}=\frac{4.6\times {10}^{-5}}{0.0779}\\ =5.9\times {10}^{-4}\end{array}$

From Moody's diagram, the value of relative roughness and Reynolds number $F_b=\mathrm{0.022.}$

Find the flow rate at A,

We get

   $Q_a=A_a{u}_a$

   $\begin{array}{l}{Q}_a=(8.213\times {10}^{-3})\times 5.08\\ =4.17\times {10}^{-2}{m}^3/s\end{array}$

Find the flow rate at B

   $\begin{array}{l}Q=A_b{u}_b\\ Q_b=(4.768\times {10}^{-3})\times 3.87\\ =1.85\times {10}^{-2}{m}^3/s\end{array}$

Total volume flow rate is

   $\begin{array}{l}Q=Q_a+Q_b\\ =(4.17\times {10}^{-2})+(1.85\times {10}^{-2})\\ =\boxed{6.02\times {10}^{-2}{m}^3/s}\end{array}$