written 2.6 years ago by
RakeshBhuse
• 3.2k
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modified 2.6 years ago
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Solution :
Given :
Max discharge, $\quad Q=20.2 \mathrm{~m}^3/ \mathrm{~s} $
Bed slope, $\quad i=\frac{1}{2500}$
Chezy's constant, $\quad C=60 $
As the channel is the form of a half hexagon as mentioned in question.
That means the angle between the sloping side with horizontal will be $60^{\circ}$.
$$
\begin{aligned}
\therefore \tan \theta &=\tan 60^{\circ}=\sqrt{3}=\frac{1}{n} \\
\therefore \quad n &=\frac{1}{\sqrt{3}} \\
\end{aligned}
$$
Let, $b=$ width of the channel, $d=$ depth of the flow.
As the channel given is of most economical section, hence the condition given below should be satisfied
Half of the top width = one of the sloping side
hydraulic mean depth =half of depth of flow
Consider
$$\begin {aligned}
\frac{b+2 n d}{2} &=d \sqrt{n^{2}+1} \\
For \space n=\frac{1}{\sqrt{3}}, \\
\frac{b+2 \times \frac{d}{\sqrt{3}}}{2} &=d \sqrt{\left(\frac{1}{\sqrt{3}}\right)^{2}+1} \\
\frac{\sqrt{3} b+2 d}{2 \sqrt{3}} &=\frac{2 d}{\sqrt{3}} \\
\frac{\sqrt{3} b+2 d}{2} &=2 d \\
b &=\frac{2 \times 2 d-2 d}{\sqrt{3}}=\frac{2 d}{\sqrt{3}}.......(i)
\end{aligned}$$
Area of flow, (A)
$$\begin{aligned}
A &=(b+n d) d \\
&=\left(\frac{2}{\sqrt{3}} d+\frac{d}{\sqrt{3}}\right) d \\
&\left(\because n =\frac{1}{\sqrt{3}} ,b=\frac{2 d}{\sqrt{3}}\right)\\
&=\frac{3}{\sqrt{3}} d^{2}\\
&=\sqrt{3} d^{2}
\end{aligned}$$
We have $m=\frac{d}{2}$
For discharge $Q =A C \sqrt{m i} $
$\begin{aligned}
\therefore 20.2 &=\sqrt{3} d^{2} \times 60 \times \sqrt{\frac{d}{2} \times \frac{1}{2500}}\end{aligned}$
$\begin{aligned}\therefore 20.2 &=1.4694 d^{5/2} \end{aligned}$
$\begin{aligned} \therefore d^{5/2} &=\frac{20.2}{1.4696}=13.745 \end{aligned}$
$\begin{aligned}\quad \therefore d &=(13.745)^{2 / 5}=2.852 \mathrm{~m}
\end{aligned}$
Substituting this value in equation (i), we get
$$
b=\frac{2 d}{\sqrt{3}}=\frac{2}{\sqrt{3}} \times 2.852=3.293 \mathrm{~m}
$$