A open channel of most economical section having form of half hexagonal shape with horizontal bottom is required to give maximum discharge of 20.2 m³/s of water. The slope of channel bottom is 1 in 2500, C=60, determine the dimensions of the cross section.

Solution :

Given :

1. Max discharge, $\quad Q=20.2 \mathrm{~m}^3/ \mathrm{~s} $

2. Bed slope, $\quad i=\frac{1}{2500}$

3. 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

1. Half of the top width = one of the sloping side

2. 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

$$ b=\frac{2 d}{\sqrt{3}}=\frac{2}{\sqrt{3}} \times 2.852=3.293 \mathrm{~m} $$