Chapter 4, Problem 4.2.1P

To determine

The drawing of the hydraulic gradient line and the energy grade line of the given system.

Explanation of Solution

Given:

$D=12in.$

Formula used:

$\begin{array}{l}{h}_{Lf}=\frac{fL{V}^2}{2gD}\\ where,h_{Lf}isheadloss\\ fisfrictionfactor\\ Visvelocity\\ Lislengthofpipe\\ gisaccelerationduetogravity\\ Disdiameterofpipe\end{array}$

Calculation:

The given figure is shown below:

The flow between reservoir 1 and reservoir 2 is given by

$\begin{array}{l}\frac{P_1}{\gamma }+\frac{V_1^2}{2g}+Z_1=\frac{P_2}{\gamma }+\frac{V_2^2}{2g}+Z_2+h_{Lf}+{\displaystyle \sum h_{L_m}}\\ 0+0+1000=0+0+950+h_{Lf}+{\displaystyle \sum h_{L_m}}\\ h_{Lf}+{\displaystyle \sum h_{L_m}}=50\end{array}$

The head loss is given by

$\begin{array}{l}{h}_{Lf}=\frac{fL{V}^2}{2gD}\\ h_{Lf}=\frac{f\left(600\right)V^2}{2\left(32.2\right)\left(\frac{12}{12}\right)}\\ h_{Lf}=9.32f{V}^2\end{array}$

The total minor head loss is given by

$\begin{array}{l}{\displaystyle \sum h_{L_m}}=h_{L_{entrance}}+h_{L_{globevalve}}+h_{L_{exit}}\\ {\displaystyle \sum h_{L_m}}=k_e\frac{V^2}{2g}+k_{valve}\frac{V^2}{2g}+\frac{V^2}{2g}\\ {\displaystyle \sum h_{L_m}}=\left(k_e+k_{valve}+1\right)\frac{V^2}{2g}\\ 50=\left(0.5+1.5+1\right)\frac{V^2}{2\left(32.2\right)}+9.32f{V}^2\\ 50=0.0466V^2+9.32f{V}^2\\ 50=V^2\left(0.0466+9.32f\right)\\ V^2=\frac{50}{0.0466+9.32f}\\ V=\sqrt{\frac{50}{0.0466+9.32f}}\cdots \cdots \cdots \cdots \left(1\right)\end{array}$

The relative roughness of pipe is given by

$\begin{array}{l}\mathrm{Re}lativeroughness=\frac{K_s}{D}\\ \frac{K_s}{D}=\frac{0.0005}{\frac{12}{12}}=0.0005\end{array}$

The value of the friction factor is 0.0165 from Moody’s diagram.

Now, substituting the value of the friction factor in the equation (1)

$\begin{array}{l}V=\sqrt{\frac{50}{0.0466+9.32f}}\\ V=\sqrt{\frac{50}{0.0466+9.32\left(0.0165\right)}}\\ V=15.8ft/s\end{array}$

The discharge is given by

$\begin{array}{l}Q=AV\\ Q=\left(\frac{\pi }{4}{D}^2\right)V\\ Q=\left[\frac{\pi }{4}\times {\left(\frac{12}{12}\right)}^2\right]\left(15.8\right)\\ Q=12.4f{t}^3/s\end{array}$

The velocity head is given by

$\begin{array}{l}\frac{V^2}{2g}=\frac{{15.8}^2}{2\left(32.2\right)}\\ \frac{V^2}{2g}=3.876\end{array}$

The total head loss in the flow from A to B is given by

$\begin{array}{l}{\displaystyle \sum h_L}=k_e\frac{V^2}{2g}+\frac{fL{V}^2}{2gD}\\ {\displaystyle \sum h_L}=0.5\left(\frac{{15.8}^2}{2\times 32.2}\right)+\frac{0.0165\left(200\right){15.8}^2}{2\left(32.2\right)\left(\frac{12}{12}\right)}\\ {\displaystyle \sum h_L}=1.938+12.792\\ {\displaystyle \sum h_L}=14.73ft\end{array}$

The hydraulic gradient line at B is given by

$\begin{array}{l}HG{L}_B=1000-{\displaystyle \sum h_L}-\frac{V^2}{2g}\\ HG{L}_B=1000-14.73-\frac{{15.8}^2}{2\left(32.2\right)}\\ HG{L}_B=1000-14.73-3.876\\ HG{L}_B=981.394ft\end{array}$

The energy gradient line at B is given by

$\begin{array}{l}EG{L}_B=HG{L}_B+\frac{V^2}{2g}\\ EG{L}_B=981.394+\frac{{15.8}^2}{2\left(32.2\right)}\\ EG{L}_B=985.27ft\end{array}$

Keep the hydraulic gradient line at point C the same as the hydraulic gradient line at point B.

The hydraulic gradient line at point C is given by

$\begin{array}{l}HG{L}_C=HG{L}_B\\ HG{L}_C=981.394ft\end{array}$

The energy gradient line at C is given by

$\begin{array}{l}EG{L}_C=HG{L}_C+\frac{V^2}{2g}\\ EG{L}_C=981.394+\frac{{15.8}^2}{2\left(32.2\right)}\\ EG{L}_C=985.27ft\end{array}$

The hydraulic gradient line at D is given by

$\begin{array}{l}HG{L}_D=EG{L}_C-\frac{fL{V}^2}{2gD}-\frac{V^2}{2g}\\ HG{L}_D=985.27-\frac{0.0165\left(380\right)\left({15.8}^2\right)}{2\left(32.2\right)\left(1\right)}-\frac{{15.8}^2}{2\left(32.2\right)}\\ HG{L}_D=985.27-24.305-3.876\\ HG{L}_D=957.089ft\end{array}$

The energy gradient line at D is given by

$\begin{array}{l}EG{L}_D=HG{L}_D+\frac{V^2}{2g}\\ EG{L}_D=957.089+\frac{{15.8}^2}{2\left(32.2\right)}\\ EG{L}_D=960.965ft\end{array}$

The hydraulic gradient line at E is given by

$\begin{array}{l}HG{L}_E=EG{L}_D-k_{valve}\frac{V^2}{2g}-\frac{V^2}{2g}\\ HG{L}_E=960.965-1.5\left(\frac{{15.8}^2}{2\times 32.2}\right)-\frac{{15.8}^2}{2\left(32.2\right)}\\ HG{L}_E=960.965-5.814-3.876\\ HG{L}_E=951.275ft\end{array}$

The energy gradient line at E is given by

$\begin{array}{l}EG{L}_E=HG{L}_E+\frac{V^2}{2g}\\ EG{L}_E=951.275+\frac{{15.8}^2}{2\left(32.2\right)}\\ EG{L}_E=955.151ft\end{array}$

The hydraulic gradient line at F is given by

$\begin{array}{l}HG{L}_F=EG{L}_E-\frac{fL{V}^2}{2gD}-\frac{V^2}{2g}\\ HG{L}_F=955.151-\frac{0.0165\left(10\right)\left({15.8}^2\right)}{2\left(32.2\right)\left(1\right)}-\frac{{15.8}^2}{2\left(32.2\right)}\\ HG{L}_F=955.151-0.6396-3.876\\ HG{L}_F=950.635ft\end{array}$

The energy gradient line at F is given by

$\begin{array}{l}EG{L}_F=HG{L}_F+\frac{V^2}{2g}\\ EG{L}_F=950.635+\frac{{15.8}^2}{2\left(32.2\right)}\\ EG{L}_F=954.511ft\end{array}$