Question

A constant discharge of $1.2m^3/sec$ is flowing through a triangular channel as shown in the figure:


The critical depth of flow (in m, correct upto two decimal place) will be __________.

• A. 0.975

Answer 1

The correct option is A 0.975

$Q=1.2m^3/sec$.


Area of flow, $A=y_o^2\tan 30{}^{\circ }$

Top width, $T=2y_c\tan 30{}^{\circ }$

For critical flow,

$\frac{Q^{2}T}{g{A}^3}=1$

$\frac{{\left(1.2\right)}^2\times 2y_c\tan 30{}^{\circ }}{9.81\times {\left(y_c^2\tan 30{}^{\circ }\right)}^3}=1$

$y_c={\left[\frac{2\times {\left(1.2\right)}^2}{9.81\times {\left(\tan 30{}^{\circ }\right)}^2}\right]}^{1/5}$

= 0.975 m