Question

A rectangular open channel needs to be designed carry a flow of $2.0m^3/s$ under uniform flow conditions. The Manning's roughness coefficient is 0.018. The channel should be such that the flow depth is equal to half the width. and the Froude number is equal to 0.5

The bed slope of the channel to be provided is

• A. 0.0012
• B. 0.0021
• C. 0.0025
• D. 0.0052

Answer 1

The correct option is B 0.0021

Given:
$Q=2m^3/s,n=0.018,B=2y,F=0.5$

we know that

$F=\frac{V}{\sqrt{gD}}$

where V is velocity of flow, D is hydraulic radius

Now $D=\frac{A}{T}=\frac{By}{B}=y$

$\therefore F=\frac{Q}{A\sqrt{gy}}$

$\Rightarrow 0.5=\frac{2}{2y^2\times \sqrt{gy}}$

$\Rightarrow =\frac{1}{y^2\sqrt{gy}}$[Squaring both sides]
$\Rightarrow 0.25=\frac{1}{g{y}^5}$

$y={\left(\frac{1}{0.25\times 9.81}\right)}^{1/5}$

$\Rightarrow y=0.836m$

Using Manning's equation, we get

$Q=\frac{1}{n}A(R{)}^{2/3}\sqrt{S}$

$\Rightarrow 2=\frac{1}{0.018}\times [2\times (0.836{)}^2]\times {\left(\frac{0.836}{2}\right)}^{2/3}\sqrt{S}$

$\Rightarrow \sqrt{S}=0.0461$
$\Rightarrow S=0.0021$