Question

A short reach of a 3 m wide rectangular open channel has its bed level falling in the direction of flow at a slope of 1 in 10000. It carries a discharge of 5 cumecs and its manning's roughness coefficient is 0.012. The flow in the reach is gradually varying. At a certain section in this reach, the depth of flow was measured as 0.6 m. The rate of change of water depth with distance (i.e., $\frac{dy}{dx}$) at this section is $\times {10}^{-3}$.$Takeg=10m{s}^{-2}$

• A. 11.5

Answer 1

The correct option is A 11.5
$S_0=\frac{1}{10,000}$
Q=5 cumecs, n=0.012
y=0.6 m,B=3 m

$\frac{dy}{dx}=\frac{S_0-S_f}{1-F_r^2}$
$F=\frac{V}{\sqrt{gy}}=\frac{Q}{By\sqrt{gy}}=\frac{5}{3\times 0.6\sqrt{10\times 0.6}}=1.134$
$NowQ=\frac{1}{n}\left(By\right)R^{\frac{2}{3}}{S}_f^{\frac{1}{2}}$
$5=\frac{1}{0.012}(3\times 0.6)[\frac{3\times 0.6}{3+(2\times 0.6)}{]}^{\frac{2}{3}}{S}_f^{\frac{1}{2}}$
$S_f=3.44\times {10}^{-3}$
$\frac{dy}{dx}=\frac{\frac{1}{10,000}-3.44\times {10}^{-3}}{1-(1.134{)}^2}=0.011675=11.67\times {10}^{-3}$