The discharge of water through a rectangular channel of width 8m is 15 m3/sec when depth of flow of water is 1.2m Find i) Specific energy of flowing water, ii) Critical depth, iii) Critical velocity , iv) Value of minimum Specific energy.

Solution :

Given :

Discharge,$\quad Q=15 \mathrm{~m}^{3} / \mathrm{s} $

Width of channel,$\quad b=8 \mathrm{~m} $

Depth, $\quad h=1.2 \mathrm{~m}$

Discharge per unit width,

$\begin{aligned}q=\frac{Q}{b}=\frac{15}{8}=1.875 \mathrm{~m}^{2} / \mathrm{s} \end{aligned}$

Velocity of flow,

$\begin{aligned} V &=\frac{Q}{\text { Area }}\\ &=\frac{15}{b \times h}\\ &=\frac{15.0}{8 \times 1.2}\\ &=1.5625 \mathrm{~m} / \mathrm{s} \end{aligned}$

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