Question

A triangular channel (n=0.012) carries water at a flow rate of 10 cumecs. The critical slope of channel is ____$\times {10}^{-4}$

• A. 18.08

Answer 1

The correct option is A 18.08

For critical flow,

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

$\frac{(10{)}^2\times 6d_c}{9.81\times (3d_c^2{)}^3}=1$

$d_c^5=2.265$

$d_c=1.178m$


$V_c=\frac{1}{n}{R}^{2/3}{S}_2^{1/2}$

$\frac{10}{3d_c^2}=\frac{1}{0.012}\times {\left(\frac{3d_c^2}{2\sqrt{10}{d}_c}\right)}^{2/3}{S}_c^{1/2}$

Substituting $d_c=1.178m$

We get, $S_c=1.808\times {10}^{-3}$

$=18.08\times {10}^{-4}$