Chapter 10, Problem 10.77P

To determine

To estimate:

The flow rate in unit gal/min with $y_2$ having the range $0.5ft<y_2<5ft$.

Answer to Problem 10.77P

$y_2$	Q
1.5	30173.6984
2.25	30173.6984
3	27526.92998
3.75	22233.2585
4.5	17469.0305

Explanation of Solution

The given figure is:

The discharge for sluice gate is having a free outflow have dissipation when the outflow is drowned. Q drops sharply, and given H=9 in.

Using the figure, the ratio is calculated as:

$\begin{array}{l}\frac{y_1}{H}=\frac{6}{\left(\frac{9}{12}\right)}\\ \frac{y_1}{H}=\frac{6}{\left(\frac{3}{4}\right)}\\ \frac{y_1}{H}=\frac{24}{3}\Rightarrow 8\end{array}$

The limits given for the tail water:

$0.5ft<y_2<5ft$

$\frac{0.5}{H}<\frac{y_2}{H}<\frac{5}{H}$

$\frac{0.5}{\left(\frac{9}{12}\right)}<\frac{y_2}{H}<\frac{5}{\left(\frac{9}{12}\right)}\Rightarrow \frac{2}{3}<\frac{y_2}{H}<\frac{20}{3}$

$[\because H=\frac{9}{12}]$

From the given figure with the intersection lines:

$\frac{y_2}{H}=2,\frac{y_1}{H}=8,c_d=0.57$

Flow rate equation:

$Q=c_{d}Hb\sqrt{2g{y}_1}$

$\to (1)$

$\begin{array}{l}where,Q=c_{d}Hb\sqrt{2g{y}_1}\\ Q=Flowrate\\ b=gapwidth\\ g=acceleration\end{array}$

$\begin{array}{l}=0.57\frac{9}{12}\times 8\sqrt{2\times 32.2\times 6}\\ =3.42\sqrt{386.4}\Rightarrow 67.227f{t}^3/s\end{array}$

$=67.227f{t}^3/s\times \frac{60}{0.13368}\Rightarrow 30173.6984gal/\min$

Height of the tail water is calculated:

$\frac{y_2}{H}=2\Rightarrow y_2=2H$

$y_2=2\times \frac{9}{12}\Rightarrow \frac{3}{2}=1.5ft$

From equation (1):

$Q=c_{d}Hb\sqrt{2g{y}_1}$

$\begin{array}{l}=0.57\frac{9}{12}\times 8\sqrt{2\times 32.2\times 6}\\ =3.42\sqrt{386.4}\Rightarrow 67.227f{t}^3/s\end{array}$

$=67.227f{t}^3/s\times \frac{60}{0.13368}\Rightarrow 30173.6984gal/\min$

Height of the tail water:

$\begin{array}{l}\frac{y_2}{H}=3\Rightarrow y_2=3H\\ y_2=3\times \frac{9}{12}\Rightarrow 2.25ft\end{array}$

At intersection lines the flow rate equation, $\frac{y_2}{H}=4,\frac{y_1}{H}=8,c_d=0.52$ :

$Q=c_{d}Hb\sqrt{2g{y}_1}$

$\to (1)$

$\begin{array}{l}where,Q=c_{d}Hb\sqrt{2g{y}_1}\\ Q=Flowrate\\ b=gapwidth\\ g=acceleration\end{array}$

$\begin{array}{l}=0.52\frac{9}{12}\times 8\sqrt{2\times 32.2\times 6}\\ =3.12\sqrt{386.4}\Rightarrow 61.33f{t}^3/s\\ =61.33\times \frac{60}{0.13368}\Rightarrow 27526.92998gal/\min \end{array}$

$\begin{array}{l}Heightofthetailwater,\\ \frac{y_2}{H}=4\Rightarrow y_2=4H\\ y_2=4\times \frac{9}{12}\Rightarrow 3ft\end{array}$

From equation (1), with the intersection lines $\frac{y_2}{H}=5,\frac{y_1}{H}=8,c_d=0.42$ :

$Q=c_{d}Hb\sqrt{2g{y}_1}$

$\begin{array}{l}=0.42\frac{9}{12}\times 8\sqrt{2\times 32.2\times 6}\\ =2.52\sqrt{386.4}\Rightarrow 49.5357f{t}^3/s\\ =49.5357\times \frac{60}{0.13368}\Rightarrow 22233.2585gal/\min \end{array}$

Height of the tail water:

$\begin{array}{l}\frac{y_2}{H}=5\Rightarrow y_2=5H\\ y_2=5\times \frac{9}{12}\Rightarrow 3.75ft\end{array}$

At intersection lines the flow rate equation $\frac{y_2}{H}=6,\frac{y_1}{H}=8,c_d=0.33$ :

$Q=c_{d}Hb\sqrt{2g{y}_1}$

$\to (1)$

$\begin{array}{l}where,Q=c_{d}Hb\sqrt{2g{y}_1}\\ Q=Flowrate\\ b=gapwidth\\ g=acceleration\end{array}$

$\begin{array}{l}=0.33\frac{9}{12}\times 8\sqrt{2\times 32.2\times 6}\\ =1.98\sqrt{386.4}\Rightarrow 38.921f{t}^3/s\\ =38.921\times \frac{60}{0.13368}\Rightarrow 17469.0305gal/\min \end{array}$

$\begin{array}{l}Heightofthetailwater:\\ \\ \frac{y_2}{H}=6\Rightarrow y_2=6H\\ y_2=6\times \frac{9}{12}\Rightarrow 4.5ft\end{array}$ Conclusion:

Thus, the flow rate in unit gal/min with $y_2$ having the range $0.5ft<y_2<5ft$ .is calculated.