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

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:

Most efficient channel: A channel is said to be efficient if it carries the maximum discharge for the given cross-section which is achieved when the wetted perimeter is kept a minimum.

1. For a triangular channel:

 The channel with a side slope of 1:1 is the most efficient channel.

2. Trapezoidal channel: The side slope angle should be 60°.