Chapter 10, Problem 10.95P

To determine

i.

The velocity V₄ for the section 4 for the given flow.

Answer to Problem 10.95P

The velocity V₄for the given flow is 2.64 m/s.

Explanation of Solution

Given:

y₃ = 0.3 m

h = 0.1 m

Concept Used:

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

Calculation:

$\begin{array}{l}\mathrm{Sec}tion2is\sup ercrititcaland\sec tion4issubcrititcal.Therefore,itissafetoassume\sec tion3is\\ critical.\\ F{r}_3=1\\ F{r}_3=\frac{V_3}{V_{3c}}\\ V_3=F{r}_3*V_{3c}\\ V_3=1*\sqrt{g{y}_3}\\ V_3=1*\sqrt{9.81*0.3}\\ V_3=1.72m/s\\ Foruniformflow,\\ q=y_4{V}_4=y_3{V}_3\\ y_4{V}_4=y_3{V}_3\\ y_4{V}_4=0.3*1.72\\ y_4{V}_4=\mathrm{0.515.....}(1)\\ Fromconservationofenergy,\\ E_4=E_3=E\\ y_4+\frac{V_4{}^2}{2g}=y_3+\frac{V_3{}^2}{2g}+h\\ y_4+\frac{V_4{}^2}{2g}=0.3+\frac{{1.72}^2}{2*9.81}+0.1\\ y_4+\frac{V_4{}^2}{2g}=0.55=E\mathrm{....}(2)\\ Solving(1)and(2),weget,\\ V_4=2.64m/s\end{array}$

Conclusion:

The velocity V₄ for the given flow is 2.64 m/s.

To determine

ii.

The depth y₄ for the given flow.

Answer to Problem 10.95P

The depth y₄for the given flow is 0.195 m

Explanation of Solution

Given:

y₃ = 0.3 m

h = 0.1 m

Concept Used:

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

Calculation:

$\begin{array}{l}\mathrm{Sec}tion2is\sup ercrititcaland\sec tion4issubcrititcal.Therefore,itissafetoassume\sec tion3is\\ critical.\\ F{r}_3=1\\ F{r}_3=\frac{V_3}{V_{3c}}\\ V_3=F{r}_3*V_{3c}\\ V_3=1*\sqrt{g{y}_3}\\ V_3=1*\sqrt{9.81*0.3}\\ V_3=1.72m/s\\ Foruniformflow,\\ q=y_4{V}_4=y_3{V}_3\\ y_4{V}_4=y_3{V}_3\\ y_4{V}_4=0.3*1.72\\ y_4{V}_4=\mathrm{0.515.....}(1)\\ Fromconservationofenergy,\\ E_4=E_3=E\\ y_4+\frac{V_4{}^2}{2g}=y_3+\frac{V_3{}^2}{2g}+h\\ y_4+\frac{V_4{}^2}{2g}=0.3+\frac{{1.72}^2}{2*9.81}+0.1\\ y_4+\frac{V_4{}^2}{2g}=0.55=E\mathrm{....}(2)\\ Solving(1)and(2),weget,\\ y_4=0.195m\end{array}$

Conclusion:

The depth y₄ for the given flow is 0.195 m.

To determine

iv.

The depth y₁ for the given flow.

Answer to Problem 10.95P

The depth y₁for the given flow is 0.165 m

Explanation of Solution

Given:

y₃ = 0.3 m

h = 0.1 m

Concept Used:

$\begin{array}{l}Fr=\frac{V}{V_c}\\ E=cons\tan t=E_1=y_1+\frac{V_1{}^2}{2g}\\ Foruniformflow,\\ q=y_1{V}_1=y_2{V}_2\\ \frac{2y_1}{y_2}=\sqrt{1-8F{r}_2{}^2}-1\end{array}$

Calculation:

$\begin{array}{l}q=y_3{V}_3=y_2{V}_2\\ y_2{V}_2=\mathrm{0.515.....}(3)\\ y_2=\frac{0.515}{V_2}\\ Fromconservationofenergy,\\ E_2=E_3=E\\ y_2+\frac{V_2{}^2}{2g}=\mathrm{0.55......}(4)\\ Solving(1)and(2),weget,\\ 0.051V_2{}^2+\frac{0.515}{V_2}=0.551\\ 0.051V_2{}^3-0.551V_2+0.515=0\\ Onsolvingthecubicequation,weget,\\ V_2=1.04m/s\\ y_2=\frac{0.515}{V_2}\\ y_2=\frac{0.515}{1.04}\\ y_2=0.495m\\ F{r}_2=\frac{V_2}{V_{2c}}\\ F{r}_2=\frac{1.04}{\sqrt{g{y}_2}}\Rightarrow \frac{1.04}{\sqrt{9.81*0.495}}\\ F{r}_2=0.472\\ \frac{2y_1}{y_2}=\sqrt{1-8F{r}_2{}^2}-1\\ \frac{2y_1}{0.495}=\sqrt{1-8*{0.472}^2}-1\\ y_1=0.165m\end{array}$.

Conclusion:

The depth y₁ for the given flow is 0.165 m.

To determine

iii.

The velocity V₁ for the given flow.

Answer to Problem 10.95P

The velocity V₁ for the given flow is 3.12 m/s.

Explanation of Solution

Given:

y₃ = 0.3 m

h = 0.1 m

Concept Used:

$q=y_1{V}_1=y_2{V}_2$

Calculation:

$\begin{array}{l}q=y_1{V}_1=y_2{V}_2\\ V_1=\frac{0.515}{y_1}\\ V_1=\frac{0.515}{0.165}\\ V_1=3.12m/s\end{array}$

Conclusion:

The velocity V₁ for the given flow is 3.12 m/s.