Question

Calculate the central depth of a triangular channel section to carry a discharge of 15 cumecs. Consider the available slope as 1 in 9000. Assume the side slopes of the channel be 1.25:1 and manning’s constant is 0.015 for good brick work in lining.

(a) 2.94 m

(b) 3.14 m

(c) 2.25 m

(d) 2.77 m

I have been asked this question during an interview for a job.

My question is taken from Channel Cross-sections in section Lining of Irrigation Canals and its Economics of Irrigation Engineering

Answer 1

The correct choice is (a) 2.94 m

Explanation: Given data- tanQ = 1/1.25; Q = 0.675 radian

A = y^2 (Q + cotQ) = 1.925y^2 and R = y/2

Q = 1/n.A.R^2/3.S^1/2

15 = 1/0.015 x 1.925y^2 x (0.5y)^2/3x 1/9000^1/2

y = 2.94 m.