Chapter 10, Problem 10.51P

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To determine

(a)

If the flow of the unfinished concrete duct is critical.

Answer to Problem 10.51P

The flow is super critical.

Explanation of Solution

The concrete duct which is unfinished has a diameter of $1.5m$ that flows at $8.0m^3/s$

Froude number is given as:

$Fr=\frac{V}{\sqrt{gy}}$

$\to (1)$

$\begin{array}{l}\begin{array}{l}Where,\hfill \\ V=flowvelocity\hfill \\ g=specificdensity\hfill \end{array}\\ y=hydraulicdepth\end{array}$

Velocity of the flow

$V=\frac{Q}{A}$

$\to (2)$

$\begin{array}{l}A=area\\ Q=disch\arg e\end{array}$

Calculation:

From equation (2)

$V=\frac{Q}{A}$

$\begin{array}{l}=\frac{Q}{A}\\ =\frac{Q}{\frac{\pi }{4}{(1.5)}^2}\Rightarrow \frac{8}{\frac{3.14}{4}{(1.5)}^2}=\frac{8}{\frac{3.14}{4}(2.25)}\\ =4.527m/s\end{array}$ Angle is calculated for the circular channel which is partially filled:

${\theta }_d={90}^0-{25}^0\Rightarrow {65}^0$

$\begin{array}{l}{\theta }_r=65*\frac{\pi }{180}\Rightarrow 1.13rad\\ \end{array}$

$[\because unitstoradians]$

Area of the flow is given as:

$A={\left(\frac{D}{2}\right)}^2({\theta }_r-\sin {\theta }_d\cos {\theta }_d)$

Substituting the values:

$\begin{array}{l}={\left(\frac{1.5}{2}\right)}^2(1.13-\sin 65\cos 65)\\ ={\left(\frac{2.25}{2}\right)}^2(1.13-\sin 65\cos 65)\\ =1.2656(1.13-\sin 65\cos 65)\\ =0.420m^2\\ \end{array}$

Hydraulic depth is given as:

$y=\frac{A}{widthattop}$

$\begin{array}{l}=\frac{A}{2\left(\frac{D}{2}\right)\sin \theta }\Rightarrow \frac{0.420}{2\left(\frac{1.5}{2}\right)\sin 65}\\ =0.3m\end{array}$

Substituting all equations (1):

$Fr=\frac{V}{\sqrt{gy}}$

$\begin{array}{l}=\frac{V}{\sqrt{gy}}\\ =\frac{4.527}{\sqrt{9.81*0.3}}\Rightarrow 2.685\end{array}$

The flow is determined to be super critical as the Froude’s number is more than 1.

Conclusion:

Thus, the flow of the unfinished concrete duct is determined.

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To determine

(b)

The critical flow rates.

Answer to Problem 10.51P

Critical flow rate is $Q_c=3.07m^3/\sec$

Explanation of Solution

Given Information:

$$

The concrete duct which is unfinished has a diameter of $1.5m$ that flows at $8.0m^3/s$

Concept Used:

$Q_c=\frac{\alpha A{R}_b{}^{2/3}*S_c{}^{1/2}}{n}$

$\to (1)$

$\begin{array}{l}where,\\ A=area\\ S_c=Criticalslope\\ R=hydraulicchannel\\ \alpha =conversionfactorn\\ n=manning’sfactor \end{array}$

Calculation:

Substituting we have,

$\begin{array}{l}8=\frac{1.76\times 1.486\times 0.25\times 1.5\times S_c{}^{1/2}}{0.014}\\ S_c{}^{1/2}=\frac{8\times 0.014}{1.76\times 1.486\times 0.25\times 1.5}\\ S_c{}^{1/2}=\frac{8\times 0.014}{0.98076}\Rightarrow \frac{0.112}{0.98076}\end{array}$

$S_c=0.001$

From equation (1) we have,

$\begin{array}{l}=\frac{1.486\times 1.76\times {(0.25\times 1.5)}^{2/3}\times {0.001}^{1/2}}{0.014}\\ =\frac{2.61536\times {\left(0.375\right)}^{2/3}\times {0.001}^{1/2}}{0.014}\\ =3.07m^3/\sec \end{array}$

Conclusion:

Thus, the critical flow rate is $3.07m^3/\sec$.

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To determine

(c)

The critical slope.

Answer to Problem 10.51P

Critical slope is $S_c=0.0019$

Explanation of Solution

Given Information:

$$

The concrete duct which is unfinished has a diameter of $1.5m$ that flows at $8.0m^3/s$

Concept Used:

$Q_c=\frac{\alpha A{R}_b{}^{2/3}*S_c{}^{1/2}}{n}$

$\to (1)$

$\begin{array}{l}where,\\ A=area\\ S_c=Criticalslope\\ R=hydraulicchannel\\ \alpha =conversionfactorn\\ n=manning’sfactor \end{array}$

Calculation:

Substituting we have:

$\begin{array}{l}\frac{n{Q}_c}{\alpha A{R}_b{}^{2/3}}=S_c{}^{1/2}\\ S_c{}^{1/2}=\frac{3.07\times 0.014}{0.98076}\\ S_c=0.0019\end{array}$

Conclusion:

Thus, the critical slope is determined.

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To determine

(d)

The Froude’s number.

Answer to Problem 10.51P

Froude’s number is $F_r=1.027$

Explanation of Solution

Given Information:

$$

The concrete duct which is unfinished has a diameter of $1.5m$ that flows at $8.0m^3/s$.

Concept Used:

Froude number is given as:

$Fr=\frac{V}{\sqrt{gy}}$

$\to (1)$

$\begin{array}{l}\begin{array}{l}Where,\hfill \\ V=flowvelocity\hfill \\ g=specificdensity\hfill \end{array}\\ y=hydraulicdepth\end{array}$

Velocity of the flow

$V=\frac{Q}{A}$

$\to (2)$

$\begin{array}{l}A=area\\ Q=disch\arg e\end{array}$

Calculation:

Substituting all equation (1)

$Fr=\frac{V}{\sqrt{gy}}$

$\begin{array}{l}=\frac{V}{\sqrt{gy}}\Rightarrow \frac{Q}{A\sqrt{gy}}\\ =\frac{3.07}{1.767}\times \frac{1}{\sqrt{9.81\times 0.3}}\Rightarrow 1.0127\end{array}$

The flow is determined to be critical.

Conclusion:

Thus, the Froude number is determined.

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To determine

(e)

The slope of the duct when the flow is uniform.

Answer to Problem 10.51P

$S_c=0.2698$

Explanation of Solution

Given Information:

$$

The concrete duct which is unfinished has a diameter of $1.5m$ that flows at $8.0m^3/s$ . $$

Concept Used:

$\begin{array}{l}\frac{n{Q}_c}{\alpha A{R}_b{}^{2/3}}=S_c{}^{1/2}\\ where:\\ \\ A=area\\ S_c=Criticalslope\\ R=hydraulicchannel\\ \alpha =conversionfactorn\\ n=manning’sfactor \end{array}$

Calculation:

The slope is equal in the duct with the uniform flow. So the duct slope with the critical flow is given as:

$S_c=0.2698$

Conclusion:

Thus, the slope of the duct when the flow is uniform is determined.