**1 Answer**

written 2.4 years ago by | • modified 2.4 years ago |

**Condition for Maximum Discharge for a Given Value of Specific Energy**

The specific energy $(E)$ at any section of a given channel is given by,

$$ \begin{aligned} \qquad E &=h+\frac{V^{2}}{2 g}, \\ &\text { where } V=\frac{Q}{A}=\frac{Q}{b \times h} \\ E &=h+\frac{Q^{2}}{b^{2} \times h^{2}} \times \frac{1}{2 g}\\ &=h+\frac{Q^{2}}{2 g b^{2} h^{2}} \end{aligned} $$

or

$$\begin {aligned} Q^{2} &=(E-h) 2 g b^{2} h^{2} \\ Q &=\sqrt{(E-h) 2 g b^{2} h^{2}}\\ &=b \sqrt{2 g\left(E h^{2}-h^{3}\right)} \end {aligned}$$

Where, E is constant

For maximum discharge, $Q$, the expression $\left(E h^{2}-h^{3}\right)$ should be maximum. Or $$ \begin{aligned} \frac{d}{d h}\left(E h^{2}-h^{3}\right) =0 \end{aligned} $$

$$ \begin{aligned} 2 E h-3 h^{2} &=0 \end{aligned} $$

$$\begin{aligned} 2 E-3 h &=0 \end{aligned} $$

$$ \begin{aligned} E &=\frac{3 h}{2}\end{aligned} ......(i) $$

But specific energy is minimum when it is equal to ${3}/{2}$ times the value of depth of critical flow.

Here in the above equation, the specific energy $(E)$ is equal to ${3}/{2}$ times the depth of flow.

Thus equation (i) represents the minimum specific energy and $h$ is the critical depth.

Hence the condition for maximum discharge for given value of specific energy is that the depth of flow should be critical.