Chapter 10, Problem 10.93P

To determine

i.

The velocity V₁ for the given flow.

Answer to Problem 10.93P

The velocity V₁for the given flow is 1.5 m/s 

Explanation of Solution

Given:

y₁ = 1 m

y₃ = 0.4 m

Concept Used:

$\begin{array}{l}{E}_1=y_1+\frac{V_1{}^2}{2g}\\ Foruniformflow,\\ q=y_1{V}_1=y_2{V}_2\end{array}$

Calculation:

$\begin{array}{l}{E}_1=E_3\\ y_1+\frac{V_1{}^2}{2g}=y_3+\frac{V_3{}^2}{2g}\\ 1+\frac{V_1{}^2}{2*9.81}=0.4+\frac{V_3{}^2}{2*9.81}\mathrm{........}(1)\\ Foruniformflow,\\ q=y_1{V}_1=y_3{V}_3\\ 1*V_1=0.4*V_3\mathrm{.....}(2)\\ Onsolving(1)and(2),weget\\ V_1=1.5m/s\\ V_3=3.74m/s\end{array}$

Conclusion:

The velocity V₁ for the given flow is 1.5 m/s.

To determine

ii.

The velocity V₃ for the given flow.

Answer to Problem 10.93P

The velocity V₁for the given flow is 3.74 m/s

Explanation of Solution

Given:

y₁ = 1 m

y₃ = 0.4 m

Concept Used:

$\begin{array}{l}{E}_1=y_1+\frac{V_1{}^2}{2g}\\ Foruniformflow,\\ q=y_1{V}_1=y_2{V}_2\end{array}$

Calculation:

$\begin{array}{l}{E}_1=E_3\\ y_1+\frac{V_1{}^2}{2g}=y_3+\frac{V_3{}^2}{2g}\\ 1+\frac{V_1{}^2}{2*9.81}=0.4+\frac{V_3{}^2}{2*9.81}\mathrm{........}(1)\\ Foruniformflow,\\ q=y_1{V}_1=y_3{V}_3\\ 1*V_1=0.4*V_3\mathrm{.....}(2)\\ Onsolving(1)and(2),weget\\ V_1=1.5m/s\\ V_3=3.74m/s\end{array}$

Conclusion:

The velocity V₁ for the given flow is 3.74 m/s.

To determine

iii.

The depth y₄ for the given flow.

Answer to Problem 10.93P

The depth y₄for the given flow is 0.89 m

Explanation of Solution

Given:

y₁ = 1 m

y₃ = 0.4 m

Concept Used:

$\begin{array}{l}Fr=\frac{V}{V_c}\\ $ \frac{2y_4}{y_3}=-1+{(1+8F{r}_3{}^2)}^{1/2}$\end{array}$

Calculation:

$\begin{array}{l}F{r}_3=\frac{V_3}{V_{3c}}\\ F{r}_2=\frac{3.74}{\sqrt{g*y_2}}\Rightarrow \frac{3.74}{\sqrt{9.81*0.4}}\\ F{r}_2=1.89\\ Theflowis\sup ercritical\\ $ \frac{2y_4}{y_3}=-1+{(1+8F{r}_3{}^2)}^{1/2}$$ \frac{2y_4}{0.4}=-1+{(1+8*{1.89}^2)}^{1/2}$y_4=0.89m\end{array}$.

Conclusion:

The depth y₄ for the given flow is 0.89 m.

To determine

iv.

The bump height h for the given flow.

Answer to Problem 10.93P

The height of the bump is 0.2m 

Explanation of Solution

Given:

y₁ = 1 m

y₃ = 0.4 m

Concept Used:

$\begin{array}{l}Fr=\frac{V}{V_c}\\ E_1=y_1+\frac{V_1{}^2}{2g}\\ Foruniformflow,\\ q=y_1{V}_1=y_2{V}_2\end{array}$

Calculation:

$\begin{array}{l}F{r}_2=\frac{V_2}{V_{2c}}\\ Forthebumpflow,theflowmustbecrititcal\\ F{r}_2=1\\ V_2=V_{2c}*F{r}_2\\ V_2=\sqrt{g{y}_2}*1\\ V_2=\sqrt{9.81*y_2}\mathrm{.....}\left(3\right)\\ Foruniformflow,\\ q=y_1{V}_1=y_2{V}_2\\ 1*1.5m/s=y_2*V_2\\ Substituteequation(3)\\ \sqrt{9.81}*y_2{}^{3/2}=1.5\\ y_2=0.612m\\ V_2=\sqrt{9.81*y_2}\\ V_2=\sqrt{9.81*0.612}\\ V_2=2.45m/s\\ \\ E_1=E_2\\ y_1+\frac{V_1{}^2}{2g}=y_2+\frac{V_2{}^2}{2g}+h\\ 1+\frac{{1.5}^2}{2*9.81}=0.612+\frac{{2.45}^2}{2*9.81}+h\\ h=0.2m\end{array}$

Conclusion:

The height of the bump is 0.2m .