Question

A rectangular channel of 2.5 m width is carrying a discharge of $4m^3/s.$ Considering that acceleration due to gravity as $9.81m/s^2,$ the velocity of flow (in m/s) corresponding to the critical depth (at which the specific energy is minimum) is

• A. 3.814 m/s
• B. 2.504 m/s
• C. 5.637 m/s
• D. 1.504 m/s

Answer 1

The correct option is B 2.504 m/s

$Q=4m^3/s,B=2.5m$


$\Rightarrow q=\frac{4}{2.5}{m}^3/s/m$

$=1.6m^3/s/m$

$y_c^3=\frac{q^2}{g}\Rightarrow y_c=0.639m$

$\text{At critical depth velocity head}=\frac{y_c}{2}$

$\Rightarrow \frac{V^2}{2g}=\frac{0.639}{2}$

$\Rightarrow V=\sqrt{0.639\times g}$

$=2.504m/s$