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.