Solution :

Given :

Width of channel, $ b=3 {~m}$

Specific energy, $E=3 {~kg}.{m} /{kg}=3{~m}$

For the given value of specific energy, the discharge will be maximum, when depth of flow is critical.

$$ h_{c}=h=\frac{2}{3}{E}=\frac{2}{3} \times 3=2 {~m} $$

$\therefore$ Maximum discharge, $Q_{\max }$ is given by

$$ \begin{aligned}Q_{\max } …