In an open channel flow, for best efficiency of a rectangular section channel, ratio of bottom width to depth shall be:

Question

Question

This question was previously asked in

SSC JE CE Previous Year Paper 8 (Held on 29 Oct 2020 Morning)

Answer 1

Concept:

For hydraulic efficient rectangular channel for a given area, perimeter(P) should be minimum.

Calculation:

Width of rectangular channel is b and depth of rectangular channel is y

Than Area, A = by

Perimeter, P = b + 2y

P = (A/y) + 2y

For P to be minimum,

dP/dy = 0

(-A/y²) + 2 = 0

A = 2y²

by = 2y²

b/y = 2

Important Points

Rectangular channel section	Trapezoidal channel section
R = y / 2 A = 2y2 T = B = 2y P = 4y D = y	R = y / 2 \(A = \sqrt 3 {y^2}\) \(T = \frac{{4y}}{{\sqrt 3 }}\) \(P = 2\sqrt 3y\) D = 3y / 4