Question

Given to design the irrigation channel to carry 80 cumecs of discharge. The channel slope is 1 in 3000. The critical velocity ratio is 1.3. Take rugosity coefficient as 0.021.

(a) Depth = 3 m, Base width = 15 m

(b) Depth = 3.5 m, Base width = 15.5 m

(c) Depth = 3.3 m, Base width = 15.2 m

(d) Depth = 3.2 m, Base width = 15.1 m

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This is a very interesting question from Stable Channels Design  in India topic in portion Sediment Transport and Irrigation Channels Design of Irrigation Engineering

Answer 1

Right option is (d) Depth = 3.2 m, Base width = 15.1 m

Explanation: Given Q = 80 cumecs, S = 1/3000, m = 1.3, n = 0.021

Critical velocity (Vo) = 0.55my^0.64 (y = depth of the channel)

= 0.55 x 1.3 x (2)^0.64

= 1.114 m/sec

Area (A) = Q/Vo = 80 / 1.114 = 71.8 m^2

Now assume side slope as (1/2): 1(1/2H: V)

Now, A = y (b + y x 1/2)

71.8 = 2(b + 2 x 1/2)

71.8 = 2(b + 1)

b = 34.9 m

Perimeter (P) = b + 2 x √5/2y

= 34.9 + 4.47

= 39.37 m

Now R = A/P = 71.8 / 39.37 = 1.82 m

V = ((1/n + (23 + 0.00155/s)) / (1 + (23 + 0.00155/s)n/\(\sqrt{RR}\)))\(\sqrt{RSRS}\)

V = 0.78 m/sec

Now calculating for critical velocity,

Vo = 0.55 x 1.3 x y^0.64

= 0.715 x y^0.64

Now assume y = 2.5 m

Vo = 0.715 x (2.5)^0.64 = 1.28 m/sec

A = 80 / 1.28 = 62.5 m^2

62.5 = 2.5 (b + 1/2(2.5))

b = 23.75 m

P = 23.75 + 2 x √5/2(2.5) = 29.35 m

R = A/P = 2.13

√R  = 1.46

V = 75.27/1.4 x (1.46/54.8) = 1.432 m/sec

V > V0

Now assume y = 3.2 m

Vo = 0.715 x (3.2)^0.64 = 1.51 m/sec

A = 80 / 1.5 = 53.33 m^2

53.33 = 3.2 (b + 1.6)

b= 15.1 m

P = 15.1 + √5(3.2)

P = 22.2 m

R = A/P = 53.33/22.2 = 2.4 m

√R = 1.56

V = (75.27/1.4) x (1.56/54.8)

= 1.53 m/sec

Therefore V = Vo

Hence use the depth equal to 3.2 m and base width 15.1 m with slope 1/2:1.