Question 12.P.7: Derive the expression for depth of flow in a channel of circ......

Refer figure.

Let the flow depth be y, and Let the angle subtended be θ.

Wetted perimeter P = 2 R θ

Flow area               =(2\ \theta/2\ \pi)\ \pi R^{2}-R\ \sin\theta\ R\ \mathrm{cos}\ \theta=R^{2}\theta-(R^{2}/2)\ \sin\ 2\theta

=(R^{2}/2)\ (2\theta -\sin\ 2\theta)

Using Chezy’s equation,

{Q}=A V=A C~\sqrt{\frac{A}{P}~S_{b}}\;=C\left[\frac{A^{3}}{P}~S_{b}\right]^{0.5}

Q will be maximum if A³/P is maximum. The condition is determined, using

\left[\frac{d{(A^{3}/P)}}{d\theta}\right]=0\quad i.e.,\quad\frac{P\times3\times A^{2}\ \frac{d A}{d\theta}-A^{3}\ \frac{d P}{d\theta}}{P^{2}}=0

or              3P\ {\cfrac{d A}{d\theta}}=A\,{\cfrac{d P}{d\theta}}

P=2\;R\;\theta,\;\;\;\;∴\;\;\;\frac{d P}{d\theta}=2\;R, \\ A=\left(R^{2}/2\right)\left[2\theta-\sin\,2\theta\right] \\ {\frac{d A}{d\theta}}={\frac{R^{2}}{2}}\ (2-2\cos2\theta), Substituting

3\theta(1-\cos2\theta)=[\theta-(\sin2\theta/2)]

This is a transcendental equation to be solved by trial, θ = 2.69 radian or about 154°. (check using radian mode in the calculator)

The area for flow      =R^{2}\,\theta-\frac{R^{2}}{2}\,\sin\,2\theta=3.0818\,R^{2}

Perimeter                   =2 R\theta=\;5.3756\;R\;\;\;\;∴\;\;\;R_{h}=0.5733\;R

Depth of flow             =R-R\cos\theta={\mathrm{R}}({\mathrm{1}}-{\mathrm{cos~}}\theta)=1.8988\ R{\mathrm{~or~}}0.9494\,D

Note: This uses Chezy equation:

In case Manning equation is used, then the flow,

Q=\frac{A}{N}\left(\frac{A}{P}\right)^{3/2}\;S_{b}^{\mathrm{~}1/2}=\frac{1}{N}\,\frac{A^{5/2}}{P^{3/2}}\;S_{b}^{\mathrm{~}1/2}

\frac{A^{5/2}}{P^{3/2}}\,~\mathrm{or}\,~~\frac{A^{5}}{P^{3}} should be maximized with θ as independent variable

P^{3}\times5\times A^{4}\,{\frac{d A}{d\theta}}-A^{5}\,3P^{2}\,{\frac{d P}{d\theta}}=0, Substituting,

5P\,{\frac{d A}{d\theta}}-3A\,{\frac{d P}{d\theta}}=0 and Substituting for \frac{d A}{d\theta}\mathrm{~and~}\frac{d P}{d\theta}

5\times\ 2R\theta\ {\frac{R^{2}}{2}}\ (2-2\cos2\theta)=3\ {\frac{R^{2}}{2}}\ (2\theta-\sin2\theta)\times2R \\ 5\theta(2-2\cos2\theta)-3(2\theta-\sin2\theta)=0,\quad{\mathrm{or}}\quad4\theta-10\theta\cos2\theta+3\sin2\theta=0

Solving,                  θ = 2.5007    radian       or          143.28°

In this case depth   R-R\ \cos\,\theta=R(1-\cos\,\theta)=1.802\ R\quad{\mathrm{or}}\quad0.901\,D