Chapter 8, Problem 186P

To determine

The equation for unknown flow rates or diameters for each pipe section in the pipe networks and branching pipes.

The analogy between the electric current in electric circuits and fluid flow in pipe networks.

Answer to Problem 186P

The total discharge in the branched pipe is $\frac{\pi }{4}\left(D_2^2\left(\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}\right)+\left(D_1^2\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}\right)\right)$.

The analogy of network pipe is $-\frac{\left(\frac{8f_1{L}_1{V}_1{}^2}{D_1{}^5}\left(Q_1^2\right)+\frac{8f_2{L}_2{V}_2{}^2}{D_2{}^5}\left(Q_2^2\right)+\frac{8f_3{L}_3{V}_3{}^2}{D_3{}^5}\left(Q_3^2\right)\right)}{2\left(\frac{8f_1{L}_1{V}_1{}^2}{g{D}_1{}^5{\pi }^2}\left(Q_1\right)+\left(\frac{8f_2{L}_2{V}_2{}^2}{g{D}_2{}^5{\pi }^2}{Q}_2\right)+\left(\frac{8f_3{L}_3{V}_3{}^2}{g{D}_3{}^5{\pi }^2}{Q}_3\right)\right)}$.

Explanation of Solution

The following figure represents the branched pipes.

Figure-(1)

Write the expression for the area of pipe.

  $A=\frac{\pi D}{4}$........... (I)

Here, the diameter of pipe is $D$ and the area of the pipe is $A$.

Write the expression for the discharge rate in the pipe.

  $Q=AV$........... (II)

Here, the discharge in the pipe is $Q$ and the velocity in the pipe is

  $V$.

Write the expression for head loss in the pipe.

  $h_f=\frac{fL{V}^2}{2gD}$........... (III)

Here, the loss head is $h_f$, the friction factor is $f$, the acceleration due to gravity is $g$ and the length of the pipe is $L$.

Write the expression for total discharge in branched pipes.

  $Q=Q_1+Q_2$........... (IV)

Here, the total discharge in pipe is $Q$, the discharge of the pipe 1 is $Q_1$ and the discharge of the pipe 2 is $Q_2$.

Write the expression for head loss in the pipe 1.

  $\begin{array}{l}{h}_{f_1}=\frac{f_1{L}_1{V}_1^2}{2g{D}_1}\\ V_1^2=\frac{2g{D}_1{h}_{f_1}}{2g{D}_1}\\ V_1=\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}\end{array}$........... (V)

Here, the velocity in pipe 1 is $V_1$, the diameter of the pipe 1 is $D_1$, the head loss at pipe 1 is $h_{f_1}$, the friction factor of pipe 1 is $f_1$, the acceleration due to gravity is $g$ and the length of the pipe 1 is $L_1$.

Write the expression for head loss in the pipe 2.

  $\begin{array}{l}{h}_{f_2}=\frac{f_2{L}_2{V}_2^2}{2g{D}_2}\\ V_2^2=\frac{2g{D}_2{h}_{f_2}}{2g{D}_2}\\ V_2=\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}\end{array}$........... (VI)

Here, the velocity in pipe 2 is $V_2$, the diameter of the pipe 2 is $D_2$, the head loss at pipe 2 is $h_{f_2}$, the friction factor of pipe 2 is $f_2$ and the length of the pipe 2 is $L_2$.

The following figure represents the T-pipe network.

Figure (2)

Write the expressions for head loss in pipe 1.

  $h_{f_1}=\frac{8f_1{L}_1{V}_1{}^2}{g{D}_1{}^5{\pi }^2}\left(Q_1^2\right)$........... (VII)

Write the expressions for head loss in pipe 2.

  $h_{f_2}=\frac{8f_2{L}_2{V}_2{}^2}{g{D}_2{}^5{\pi }^2}\left(Q_2^2\right)$........... (VIII)

Write the expressions for head loss in pipe 3.

  $h_{f_3}=\frac{8f_3{L}_3{V}_3{}^2}{g{D}_3{}^5{\pi }^2}\left(Q_3^2\right)$........... (IX)

Here, the discharge in pipe 3 is $Q_3$, the velocity in pipe 3 is $V_3$, the diameter of the pipe 3 is $D_3$, the head loss at pipe 3 is $h_{f_3}$, the friction factor of pipe 3 is $f_3$ and the length of the pipe 3 is $L_3$.

Write the expression for loop method.

  $\delta Q=-\frac{h_{f_1}+h_{f_2}+h_{f_3}}{2\left(\frac{h_{f_1}}{Q_1}+\frac{h_{{}_{2}f}}{Q_2}+\frac{h_{f_3}}{Q_3}\right)}$........... (X)

Here, the analogy of network pipe is $\delta Q$.

Calculation:

Substitute $D_1$ for $D$ and $A_1$ for $A$ in Equation (I).

  $A_1=\frac{\pi D_1^2}{4}$........... (XI)

Here, the area of pipe 1 is $A_1$.

Substitute $D_2$ for $D$ and $A_2$ for $A$ in Equation (I).

  $A_2=\frac{\pi D_2^2}{4}$........... (XII)

Here, the area of pipe 2 is $A_2$.

Substitute $V_2$ for $V$, $Q_2$ for $Q$ and $A_2$ for $A$ in Equation (II).

  $Q_2=A_2{V}_2$........... (XIII)

Substitute $V_1$ for $V$, $Q_1$ for $Q$ and $A_1$ for $A$ in Equation (II).

  $Q_1=A_1{V}_1$........... (XIV)

Substitute $\frac{\pi D_1^2}{4}$ for $A_1$ and $\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}$ for $V_1$ in Equation (X).

  $Q_1=\left(\frac{\pi D_1^2}{4}\right)\left(\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}\right)$........... (XV)

Substitute $\frac{\pi D_2^2}{4}$ for $A_2$ and $\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}$ for $V_2$ in Equation (IX).

  $Q_2=\left(\frac{\pi D_2^2}{4}\right)\left(\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}\right)$........... (XVI)

Substitute $\left(\frac{\pi D_2^2}{4}\right)\left(\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}\right)$ for $Q_2$ and $\left(\frac{\pi D_1^2}{4}\right)\left(\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}\right)$ for $Q_1$ in Equation (IV).

  $\begin{array}{c}Q=\left(\frac{\pi D_2^2}{4}\right)\left(\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}\right)+\left(\frac{\pi D_1^2}{4}\right)\left(\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}\right)\\ =\frac{\pi }{4}\left(D_2^2\left(\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}\right)+\left(D_1^2\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}\right)\right)\end{array}$

The total discharge in the branched pipe is $\frac{\pi }{4}\left(D_2^2\left(\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}\right)+\left(D_1^2\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}\right)\right)$.

Substitute $\frac{8f_3{L}_3{V}_3{}^2}{g{D}_3{}^5{\pi }^2}\left(Q_3^2\right)$ for $h_{f_3}$, $\frac{8f_2{L}_2{V}_2{}^2}{g{D}_2{}^5{\pi }^2}\left(Q_2^2\right)$ for $h_{f_2}$ and $\frac{8f_1{L}_1{V}_1{}^2}{g{D}_1{}^5{\pi }^2}\left(Q_1^2\right)$ for $h_{f_1}$ in Equation (X).

The analogy of network pipe is $-\frac{\left(\frac{8f_1{L}_1{V}_1{}^2}{D_1{}^5}\left(Q_1^2\right)+\frac{8f_2{L}_2{V}_2{}^2}{D_2{}^5}\left(Q_2^2\right)+\frac{8f_3{L}_3{V}_3{}^2}{D_3{}^5}\left(Q_3^2\right)\right)}{2\left(\frac{8f_1{L}_1{V}_1{}^2}{g{D}_1{}^5{\pi }^2}\left(Q_1\right)+\left(\frac{8f_2{L}_2{V}_2{}^2}{g{D}_2{}^5{\pi }^2}{Q}_2\right)+\left(\frac{8f_3{L}_3{V}_3{}^2}{g{D}_3{}^5{\pi }^2}{Q}_3\right)\right)}$.

Conclusion:

The analogy of network pipe is $-\frac{\left(\frac{8f_1{L}_1{V}_1{}^2}{D_1{}^5}\left(Q_1^2\right)+\frac{8f_2{L}_2{V}_2{}^2}{D_2{}^5}\left(Q_2^2\right)+\frac{8f_3{L}_3{V}_3{}^2}{D_3{}^5}\left(Q_3^2\right)\right)}{2\left(\frac{8f_1{L}_1{V}_1{}^2}{g{D}_1{}^5{\pi }^2}\left(Q_1\right)+\left(\frac{8f_2{L}_2{V}_2{}^2}{g{D}_2{}^5{\pi }^2}{Q}_2\right)+\left(\frac{8f_3{L}_3{V}_3{}^2}{g{D}_3{}^5{\pi }^2}{Q}_3\right)\right)}$.

The total discharge in the branched pipe is $\frac{\pi }{4}\left(D_2^2\left(\sqrt{\frac{2g{D}_2{h}_{f_2}}{f_2{L}_2}}\right)+\left(D_1^2\sqrt{\frac{2g{D}_1{h}_{f_1}}{f_1{L}_1}}\right)\right)$.