Question

10 m 10 m road 1.5 m n 0.015 n-0.018 n 0.018 1.5 m n-0.025 n-0.032 3 m Figure 1 a) Determine the composite roughness using Colebatch's formula for the channel (Figure 1) for the flow depth up to the top of the bank (point A). The slope of the channel is 0.0015.*

Answer 1

Ans) We know, according to colebatch equation,

Composite roughness $(N_{eq})$ = [$\sum$$N_i^{3/2}$ Ai / A]^(2/3)

where, Ni = Roughness of $i^{th}$ channel

Ai = Area of $i^{th}$ channel

A = Total Area of channel

Area of 1st rectangular channel from left = 10 x 1.5 = 15 $m^2$

Also, to width of trapezoidal channel = B + 2mD

where, m = side slope (m : 1) , so for 2 : 1 slope, m = 2

D = Depth = 1.5 m

B = Bottom width = 3 m

=> Top width = 3 + 2(2)(1.5) = 9 m  

Area of 2nd trapezoidal channel = (10 x1.5) + 0.5(3 + 9)(1.5) = 24 $m^2$

Area of 3rd rectangular channel at right = 10 x 1.5 = 15 $m^2$

Putting values in Colebatch equation,

  => $N_{eq}$ = [($0.015^{3/2}$ x 15) + ($0.032^{3/2}$ x 24) + ($0.025^{3/2}$ x 15) / (15 + 24 + 15)]^(2/3)

=> $N_{eq}$ = [(0.0275 + 0.1374 + 0.0593) / 54]^(2/3)

=> $N_{eq}$ = (0.00415)^(2/3)

=> $N_{eq}$ = 0.0258

Hence, composite roughness of given channel is 0.0258