Chapter 10, Problem 10.81P

To determine

To compute: y_2,V_2,Fr₂, h_f, percentage dissipation and horse power.

Answer to Problem 10.81P

$\begin{array}{l}{y}_2=5.75ft\\ Velocity{V}_2=4.35ft/s\\ F{r}_2=0.32\\ h_f=4.66ft\\ \%dissipation=44\%\\ Powerdissipated=13.2hp/ft\\ Crititcaldepth{y}_c=2.69ft\end{array}$

Explanation of Solution

Given information:

q = 25 ft³ /s-ft

y₁=1 ft

Concept:

$\begin{array}{l}Froudenumber,\\ Fr=\frac{V}{V_c}\\ Fromcontinuityequation,\\ q=y_1{V}_1=y_2{V}_2\\ E=y+\frac{V^2}{2g}\\ $ \frac{2y_2}{y_1}=-1+{(1+8F{r}_1{}^2)}^{1/2}$h_f=\frac{{\left(y_2-y_1\right)}^3}{4y_2{y}_1}\\ \%dissipation=\frac{h_f}{E_1}\\ Powerdissipated=\rho gq{h}_f\\ Criticaldepth,y_c={\left(\frac{q^2}{g}\right)}^{1/3}\end{array}$

$\begin{array}{l}Fromcontinuityequation,\\ q=y_1{V}_1\\ V_1=\frac{q}{y_1}=V_1=\frac{25f{t}^3/ft-s}{1ft}\\ V_1=1\text{}ft/s\\ F{r}_1=\frac{V_1}{V_{1c}}\\ F{r}_1=\frac{25}{\sqrt{g{y}_1}}\Rightarrow \frac{25}{\sqrt{32.2*1}}\\ F{r}_1=4.41\\ $ \frac{2y_2}{y_1}={(1+8F{r}_1{}^2)}^{1/2}-1$$ \frac{2y_2}{1}={(1+8*{4.41}^2)}^{1/2}-1$y_2=5.748ft\\ \\ Fromcontinuityequation,\\ q=y_2{V}_2\\ V_2=\frac{q}{y_2}\\ V_2=\frac{25f{t}^3/ft-s}{5.748ft}\\ V_2=4.35ft/s\\ F{r}_2=\frac{V_2}{V_{2c}}\\ F{r}_2=\frac{4.35}{\sqrt{g{y}_2}}\Rightarrow \frac{4.35}{\sqrt{32.2*5.748}}\\ F{r}_2=0.32\\ h_f=\frac{{\left(y_2-y_1\right)}^3}{4y_2{y}_1}\\ h_f=\frac{{\left(5.85-1\right)}^3}{4*5.85*1}\\ h_f=4.66ft\\ \end{array}$

$\begin{array}{l}\%dissipation=\frac{h_f}{E_1}\\ E_1=y_1+\frac{V_1{}^2}{2g}\\ E_1=1+\frac{{25}^2}{2*32.2}\\ E_1=10.7ft\\ \%dissipation=\frac{4.66}{10.7}*100\%\\ \%dissipation=44\%\\ Powerdissipated=\rho gq{h}_f\\ Powerdissipated=62.4*25*4.66/550\\ Powerdissipated=13.2hp/ft\\ Criticaldepth,y_c={\left(\frac{q^2}{g}\right)}^{1/3}\\ Criticaldepth,y_c={\left(\frac{{25}^2}{32.2}\right)}^{1/3}\\ y_c=2.69ft\end{array}$

Conclusion:

$\begin{array}{l}{y}_2=5.75ft\\ Velocity{V}_2=4.35ft/s\\ F{r}_2=0.32\\ h_f=4.66ft\\ \%dissipation=44\%\\ Powerdissipated=13.2hp/ft\\ Crititcaldepth{y}_c=2.69ft\end{array}$.