Chapter 6, Problem 6.127P

To determine

The flow rate and direction in each pipe?

If $p_A=120lbf/i{n}^2$, determine the pressures at point $B,C,D$ ?

Answer to Problem 6.127P

$\begin{array}{l}{Q}_{AB}=1.253f{t}^3/s\\ Q_{AC}=0.747f{t}^3/s\\ Q_{BC}=1.011f{t}^3/s\\ Q_{CD}=1.876f{t}^3/s\\ Q_{BD}=0.124f{t}^3/s\end{array}$

$p_B=107lbf/i{n}^2$

$p_C=105lbf/i{n}^2$

$p_D=94lbf/i{n}^2$

Explanation of Solution

Given information:

For all pipes, friction factor is equal to $0.025$

The fluid is water at $20°C$

Inlet and exit flow rate is equal to $2f{t}^3/s$

The basic rules of pipe networks are,

“The net pressure change in a closed loop must be equal to zero”

“The net flow into a junction must be equal to zero”

“All pressure changes must satisfy moody and minor loss friction co-relation”

The head loss $h_f$ can be defined as,

$h_f=\frac{8Lf{Q}^2}{{\pi }^{2}g{d}^5}=K{Q}^2$

In above equation,

$L$ - Length of pipe

$f$ - Friction factor

$Q$ - Flow rate

$d$ - Diameter

$K$ - Loss co efficient

The total head loss is equal to zero in a closed loop.

The pressure difference $\Delta P$ can be defined as,

$\Delta P=\rho g{h}_f$

Where, $\rho$ - density

Calculation:

According to the explanation given,

Equate head loss to find loss co efficient in each pipe.

For pipe AC,

$K_{AC}=\frac{8Lf}{{\pi }^{2}g{d}^5}=\frac{8\left(3000ft\right)\left(0.025\right)}{{\pi }^2\left(32.2ft/s^2\right){\left(\frac{6}{12}ft\right)}^5}=60.415$

Similarly, for each pipe

$\begin{array}{l}{K}_{AB}=\frac{8Lf}{{\pi }^{2}g{d}^5}=\frac{8\left(4000ft\right)\left(0.025\right)}{{\pi }^2\left(32.2ft/s^2\right){\left(\frac{8}{12}ft\right)}^5}=19.116\\ K_{BC}=\frac{8Lf}{{\pi }^{2}g{d}^5}=\frac{8\left(5000ft\right)\left(0.025\right)}{{\pi }^2\left(32.2ft/s^2\right){\left(\frac{9}{12}ft\right)}^5}=13.260\\ K_{CD}=\frac{8Lf}{{\pi }^{2}g{d}^5}=\frac{8\left(4000ft\right)\left(0.025\right)}{{\pi }^2\left(32.2ft/s^2\right){\left(\frac{8}{12}ft\right)}^5}=19.116\\ K_{BD}=\frac{8Lf}{{\pi }^{2}g{d}^5}=\frac{8\left(3000ft\right)\left(0.025\right)}{{\pi }^2\left(32.2ft/s^2\right){\left(\frac{3}{12}ft\right)}^5}=1933.284\end{array}$

According to the definitions, the total head loss in a closed loop is equal to zero,

Therefore,

Assume the directions of the flow as in the figure above,

For closed loop ABC,

$19.116Q_{AB}^2+13.260Q_{BC}^2-60.415Q_{AC}^2=0\to \left(1\right)$

For closed loop BCD,

$13.260Q_{BC}^2+19.116Q_{CD}^2-1933.284Q_{BD}^2=0\to \left(2\right)$

The net flow rate is equal to zero at all junctions, therefore,

$\begin{array}{l}{Q}_{AB}+Q_{AC}=2f{t}^3/s\to \left(3\right)\\ Q_{CD}+Q_{BD}=2f{t}^3/s\to \left(4\right)\\ Q_{BD}+Q_{BC}=Q_{AB}\to \left(5\right)\\ Q_{AC}+Q_{BC}=Q_{CD}\to \left(6\right)\end{array}$

By solving above equations,

We get,

$\begin{array}{l}{Q}_{AB}=1.253f{t}^3/s\\ Q_{AC}=0.747f{t}^3/s\\ Q_{BC}=1.011f{t}^3/s\\ Q_{CD}=1.876f{t}^3/s\\ Q_{BD}=0.124f{t}^3/s\end{array}$

To find the pressure at point B,

We know that,

$p_A=120lbf/i{n}^2$

Therefore,

$\begin{array}{l}{p}_B=p_A-\rho g{h}_{f_{AB}}\\ =\left(120lbf/i{n}^2\right)\left(144i{n}^2/f{t}^2\right)-\left(62.4lbf/f{t}^3\right)\left(19.116\right){\left(1.253f{t}^3/s\right)}^2\\ =107lbf/i{n}^2\end{array}$

Similarly,

$\begin{array}{l}{p}_C=p_A-\rho g{h}_{f_{AC}}\\ =\left(120lbf/i{n}^2\right)\left(144i{n}^2/f{t}^2\right)-\left(62.4lbf/f{t}^3\right)\left(60.415\right){\left(0.747f{t}^3/s\right)}^2\\ =105lbf/f{t}^2\end{array}$

$\begin{array}{l}{p}_D=p_B-\rho g{h}_{f_{BD}}\\ =\left(107lbf/i{n}^2\right)\left(144i{n}^2/f{t}^2\right)-\left(62.4lbf/f{t}^3\right)\left(1933.284\right){\left(0.124f{t}^3/s\right)}^2\\ =94lbf/i{n}^2\end{array}$

Conclusion:

The flow rates in each pipe is equal to,

$\begin{array}{l}{Q}_{AB}=1.253f{t}^3/s\\ Q_{AC}=0.747f{t}^3/s\\ Q_{BC}=1.011f{t}^3/s\\ Q_{CD}=1.876f{t}^3/s\\ Q_{BD}=0.124f{t}^3/s\end{array}$

The pressure at point B, C and D are,

$p_B=107lbf/i{n}^2$

$p_C=105lbf/i{n}^2$

$p_D=94lbf/i{n}^2$.