Solution :

Given :

1. Bed width, $b=6m$

2. Depth, $ d=3m$

3. Slope $=3H:4V$

4. Discharge, $Q=30 m³/s$

5. Chezy's constant, $C=70$

From Fig.

$BE = 3 × \frac{3}{4}= 2.25 m$

Length of CD,

$\begin {aligned}C D &=A B+2 \times B E\\ &=6.0+2 \times 2.25 \\ &=10.50 \mathrm{~m} \end {aligned}$

Wetted perimeter,

$\begin {aligned} \quad P &=A D+A B+B C\\ &=A B+2 B C \quad(\because B C=A D)\\ &=A B+2 \sqrt{B E^{2}+C E^{2}}\\ &=6.0+2 \sqrt{(2.25)^{2}+(3)^{2}}\\ &=13.5 \mathrm{~m}\end{aligned}$

Area of flow, $(A)$

$\quad A=$ Area of trapezoidal $A B C D$

$\begin{aligned} \space \space \quad &=\frac{(A B+C D) \times C E}{2}\\ &=\frac{(6+10.50)}{2} \times 3.0\\ &=24.75 \mathrm{~m}^{2}\end{aligned}$

Hydraulic mean depth, $(m)$

$\begin{aligned} \space\space m=\frac{A}{P}=\frac{24.75}{13.50}=1.833\end {aligned}$

Discharge, $(Q)$

$\begin{aligned}Q &=A C \sqrt{m i}\\ 30.0 &=24.75 \times 70 \times \sqrt{1.833 \times i}\\ \sqrt i &=\frac{30}{2345.6}\\ i &=\left(\frac{30}{2345.6}\right)^{2}=\frac{1}{\left(\frac{2345.6}{30}\right)^{2}}\\ &=\frac{1}{6133} \end{aligned}$