For a hydraulically efficient rectangular channel of bed width 5

For a hydraulically efficient rectangular channel of bed width 5
| For a hydraulically efficient rectangular channel of bed width 5 m, the hydraulic radius is

A. 0.5 m

B. 1.25 m

C. 2.75 m

D. 4.25 m

Please scroll down to see the correct answer and solution guide.

Right Answer is: B

SOLUTION

For hydraulically efficient rectangular channel, depth of flow (y) is half of the width (B).

Y = B/2

\({\rm{Hydraulic\;Radius}},{\rm{\;R}} = \frac{{{\rm{Area\;}}\left( {\rm{A}} \right)}}{{{\rm{Wetted\;Perimeter\;}}\left( {\rm{P}} \right)}}\)

Area, A = B × y = By

Wetted Perimeter, P = B + 2y = 4y {∵ B = 2y}

\({\rm{Hydraulic\;Radius}},{\rm{\;R}} = \frac{{{\rm{By}}}}{{4{\rm{y}}}}\)

\(R = \frac{{\bf{B}}}{4} = \frac{5}{4} = 1.25\;{\bf{m}}\)   {∵ B = 5 m, given}

Note:

Conditions of hydraulic efficient channel for other shapes:

1. Triangular channel:  Vertex angel should be 90°.

2. Trapezoidal channel:  Side slope is 1:1 i.e. side slope angle should be 45°.

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