Question

15. A rectangular open concrete (n-0.012) channel is to be designed to carry a flow of 5 m/s. The slope is 0.003 m/m and the bottom width of the channel is 5 meters. Determine the normal depth that will occur in this channel. (A) 0,4 m (B) 0.47 m (C) 0.5 m (D) 0.77 m

Answer 1

We have to determine the normal depth of the given rectangular channel;

Let the normal depth of the channel is 'y':

We have to use Manning's Equation for this;

$Q=v.A=\frac{1}{n}*A*R^{\frac{2}{3}}*S^{\frac{1}{2}}$

where,

Q is rate of flow = 5 m³/s.

v is the velocity of flow

A is the wetted area of the channel = Bottom Width * Normal Depth of the channel = 5 * y.

n is the Manning's Roughness Coefficient = 0.012

S is the slope of the channel = 0.003 m/m

R is the Hydraulic Radius of the channel

$R=\frac{Wetted \; Area}{Wetted\; Perimeter}=\frac{A}{P}$

P = Bottom Width + (2*Normal Depth) = 5 + 2y

Putting all the values in Manning's Equation:

$5=\frac{1}{0.012}*5y*\left ( \frac{5y}{5+2y} \right )^{\frac{2}{3}}*0.003^{\frac{1}{2}}$

From this, we get Normal Depth in the channel, y = 0.4285 m

It is more close to option (A), hence, option (A) is the correct answer.