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

$$ \begin{aligned}Q_{\max } &=\text { Area } \times \text { Velocity }\\ &=(b \times \text { depth of flow }) \times \text { Velocity }\\ &=\left(b \times h_{c}\right) \times V_{c} \end{aligned}$$

$(\because$ At critical depth, Velocity will be critical $)$ where $V_{c}$ is critical velocity and it is given by equation $$ V_{c} =\sqrt{g \times h_{c}}=\sqrt{9.81 \times 2.0}\ =4.43{~m} /{s} \ $$

$$ Q_{max } =\left(b \times h_{c}\right) \times V_{c} $$

$$ =(3 \times 2) \times 4.43$$

$$ =26.58 {~m}^{3}/ {~s} $$