Solution :

Given :

Discharge, $ Q=400$ lps $=0.4 \mathrm{~m}^{3} / \mathrm{s}$

Bed slope, $ i=\frac{1}{2000}$

Chezy's constant, $C=50$

(i) Hydraulic mean depth,

$m=\frac{d}{2}$

(ii) Area of flow,(A)

$\quad A=b \times d=2 d \times d=2 d^{2}$

{iii) Discharge,(Q)

$$ Q=A C \sqrt{m i} $$

$$\begin {aligned} 0.4 &=2 d^{2} \times 50 \times \sqrt{\frac{d}{2} \times \frac{1}{2000}}\\ 0.4 &=2 \times 50 \times \sqrt{\frac{1}{2 \times 2000}} d^{5 / 2}\\ 0.4 &=1.581 d^{5 / 2} \\ d^{5 / 2} &=\frac{0.4}{1.581}=0.253 \\ d &=(0.253)^{2 / 5}=0.577 \mathrm{~m} \end{aligned} $$

(iv) Width,

$b=2 d$

$b =2 d=2 \times .577=1.154 \mathrm{~m} $

For the rectangular channel to be most economical dimensions are,

$b=1.154 \mathrm{~m} $ and $d =0.577 \mathrm{~m}$