Question

Given that the bed slope of a channel is 1 in 2000 and the discharge is 60 cumecs. The depth of the channel is fixed and is given as 2 m. The tractive force needed for the movement of bedload is? Take critical velocity ratio as 1.1.

(a) 8.89 N/m^3

(b) 8.87 N/m^3

(c) 8.85 N/m^3

(d) 8.83 N/m^3

This question was addressed to me at a job interview.

This interesting question is from Sediment Transport topic in section Sediment Transport and Irrigation Channels Design of Irrigation Engineering

Answer 1

The correct answer is (d) 8.83 N/m^3

The best I can explain: Critical velocity (Vo) = 0.55my^0.64

= 0.55 x 1.1 x (2)^0.64

= 0.943 m/sec

Area = Q / Vo = 60 / 0.943 = 63.63 m^2 (Given Q = 60 cumecs)

A = y (b + y (1/2)) (y = depth, b = base width)

63.63 = 2(b + 1) (for side slope as (1/2:1/2H:V))

b = 30.82 m

Perimeter (P) = b + √5y = 30.82 + 2 x √5

P = 35.3 m

R = A/P = 63.63 / 35.3 = 1.8 m (R = hydraulic mean depth)

Now, tractive force (τo) = γwRS = 9.81 x 10^3 x 1.8 x (1/2000) (γw = 9.81 x 10^3 N/m^3)

= 8.83 N/m^3.