Question 12.P.8: Derive the condition for maximum velocity of flow in a chann......
Derive the condition for maximum velocity of flow in a channel of circular section.
Step-by-Step
Refer problem Problem 12.7.
A={\frac{R^{2}}{2}}\ (\ 2\ \theta-2\sin\ 2\theta),P=2 R\theta,\,V=C\sqrt{R_{h}S_{b}}\,,R_{h}={\frac{A}{P}}
For V to be maximum, A/P should be maximized,
\frac{d(A/P)}{d\theta}\left(P\,\frac{d A}{d\theta}-A\,\frac{d P}{d\theta}\right)/P^{2}=0\quad∴\quad P\,\frac{d A}{d\theta}=A\,\frac{d P}{d\theta}\qquad\qquad(A){\frac{d A}{d\theta}}={\frac{R^{2}}{2}}~(2-2~\mathrm{cos}~2\theta),~{\frac{d P}{d\theta}}=2R\,, Substituting
2R\theta\ {\frac{R^{2}}{2}}\ (2-2\cos2\theta)={\frac{R^{2}}{2}}\ (2\theta-2\ \mathrm{cos}\ 2\theta)\ 2 R2\theta(1-\cos2\theta)=\ 2\theta-\sin2\theta
-\,2\theta\,\cos\,2\theta+\sin\,2\theta=0\ \ \ \mathrm{or}\ \ \ \ \tan\,2\theta=2\theta∴ Solving = 2.247 radians, or 128.75°
Depth for maximum velocity y = R(1 – cos θ) = 1.626 R or 0.813 D
Note: This is using Chezy equation, In case Manning equation is used, the same result is obtained as in equation A.
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