Chapter 10, Problem 10.80P

To determine

(a)

The downstream Froude number.

Answer to Problem 10.80P

The downstream Froude number is 0.395

Explanation of Solution

Given:

y₁ = 0.25m

y₂ = 1 m

Concept Used:

$\begin{array}{l}{V}_1{}^2=\frac{g{y}_1\eta (\eta +1)}{2}\\ Where,\\ \eta =\frac{y_2}{y_1}\\ Fromcontinuityequation,\\ q=V_1{y}_1=V_2{y}_2\end{array}$

Calculation:

$\begin{array}{l}{V}_1{}^2=\frac{g{y}_1\eta (\eta +1)}{2}\\ Where,\\ \eta =\frac{y_2}{y_1}\Rightarrow \frac{1}{0.25}\\ \eta =4\\ V_1{}^2=\frac{g{y}_1\eta (\eta +1)}{2}\\ V_1{}^2=\frac{9.81*0.25*4*(4+1)}{2}\\ V_1=4.95m/s\\ Now,Fromcontinuityequation,\\ q=V_1{y}_1=V_2{y}_2\\ V_2=\frac{V_1{y}_1}{y_2}\Rightarrow \frac{4.95*0.25}{1}\\ V_2=1.2375m/s\\ F{r}_2=\frac{V_2}{V_{2c}}\Rightarrow \frac{1.2375}{\sqrt{g*y_2}}\Rightarrow \frac{1.2375}{\sqrt{9.81*1}}\\ F{r}_2=0.395\end{array}$

Conclusion:

The downstream Froude number is 0.395.

To determine

(b)

The flow rate per unit width.

Answer to Problem 10.80P

The flow rate per unit width is 1.2375 m³ /m-s

Explanation of Solution

Given:

y₁ = 0.25m

y₂ = 1 m

Concept Used:

$\begin{array}{l}Fromcontinuityequation,\\ q=V_1{y}_1=V_2{y}_2\end{array}$

Calculation:

$\begin{array}{l}Fromcontinuityequation,\\ q=V_2{y}_2\\ q=1.2375*1\\ q=1.2375m^3/m-s\end{array}$.

Conclusion:

The flow rate per unit width is 1.2375 m³ /m-s.

To determine

(c)

The critical depth of water.

Answer to Problem 10.80P

The critical depth of water is 0.538 m

Explanation of Solution

Given:

y₁ = 0.25m

y₂ = 1 m

Concept Used:

$\begin{array}{l}Thecriticaldepthofwater,\\ y_c={\left(\frac{q^2}{g}\right)}^{1/3}\end{array}$

Calculation:

$\begin{array}{l}Thecriticaldepthofwater,\\ y_c={\left(\frac{q^2}{g}\right)}^{1/3}\\ y_c={\left(\frac{{1.2375}^2}{9.81}\right)}^{1/3}\\ y_c=0.538m\end{array}$.

Conclusion:

The critical depth of water is 0.538 m.

To determine

(d)

The percentage of heat dissipation.

Answer to Problem 10.80P

The heat dissipation is 39%

Explanation of Solution

Given:

y₁ = 0.25m

y₂ = 1 m

Concept Used:

$\begin{array}{l}Heatdissipation,\\ heatdissipation=\frac{h_f}{E_2}*100\\ \\ HeadLoss(h_f)=\frac{{(y_2-y_1)}^3}{4y_1{y}_2}\end{array}$

Calculation:

$\begin{array}{l}E=y_2+\frac{V_2{}^2}{2g}\\ E=1+\frac{{1.2375}^2}{2*9.81}\\ E=1.078m\\ HeadLoss(h_f)=\frac{{(y_2-y_1)}^3}{4y_1{y}_2}\\ HeadLoss(h_f)=\frac{{(1-0.25)}^3}{4*1*0.25}\\ HeadLoss(h_f)=0.421\\ \\ Heatdissipation,\\ heatdissipation=\frac{h_f}{E}*100\\ heatdissipation=\frac{0.421}{1.078}*100\\ heatdissipation=39\%\end{array}$.

Conclusion:

The heat dissipation is 39%.