Question

Design a regime channel for a discharge of 90 cumecs and silt factor is 1.3 use Lacey’s Theory.

(a) Depth = 2.2 m, Base width = 40.5 m, Slope = 1/4670

(b) Depth = 2.12 m, Base width = 40.42 m, Slope = 1/4675

(c) Depth = 2.135 m, Base width = 40.3 m, Slope = 1/4676

(d) Depth = 2.123 m, Base width = 40.1 m, Slope = 1/4672

The question was posed to me by my school principal while I was bunking the class.

My doubt stems from Stable Channels Design  in India topic in chapter Sediment Transport and Irrigation Channels Design of Irrigation Engineering

Answer 1

Right answer is (c) Depth = 2.135 m, Base width = 40.3 m, Slope = 1/4676

Easiest explanation: Given Q = 90 cumecs, f = 1.3

V = [Qf^2/140]^1/6 = 1.014 m/sec

A = Q/V = 90/1.014 = 88.76 m^2

R = 5/2 x V^2/f = 2 m

P = 4.75√Q = 45.06 m

For a trapezoidal channel with 1/2H: 1V slopes

P = b +√5y and A = (b + y/2) y

45.06 = b + √5y, 88.76 = (b + y/2) y

b = 45.06 – 2.24y, substitute this value in area equation

88.76 = (45.06 – 2.24y + y/2) y

Y = 2.135 m

b = 45.06 – 2.24 x 2.135 = 40.3 m

S = (f^5/3 / 3340Q^1/6) = 1/4676.