Question

2. Design a trapezoidal vegetative waterway that has a side slope of 4:1, a slope of 4%, flow equal to Q= 75 cfs, an erosion resistant soil, and the grass mixture grown to 2 inches in height

Answer 1

Ans) We know, discharge (Q),

Q = (1.47A/N) (R^2/3) (S^1/2)

where, A = cross sctional area of channel

N = Manning roughness coefficient = 0.035 for high grass

R = Hydraulic depth

S = Bed Slope = 4% or 0.04

For trapezoidal channel, A = BD + mD^2

where, B = Bottom width

D = Flow depth = 1.50 m

m = Side slope (m:1) = (4 :1) so m = 4

=> A = BD + 4 D^2

We know, for most economical trapezoidal channel ,top width will be twice of side wall length

Top width = B + 2mD = B + 8D

Side wall length = D = 4.123 D

Hence, condition for most economical section :

=> B + 8D = 2(4.123D)

=> B + 8D = 8.246 D

=> B = 0.246 D

Also, R = Area/Wetted perimeter

Area = BD + 4 D^2 = (0.246)D^2 + 4D^2 = 4.246 D^2

For trapezoidal channel, wetted perimeter = B + 2D = 0.246D + 2D = 8.492D

=> R = 4.246D^2 / 8.492D = 0.5 D

Putting values ,

=> 75 = [1.47(4.246D^2) / 0.035] (0.5D)^0.67 (0.04^1/2)

=> 75 = 178.33 D^2 x (0.5D)^0.67 x 0.2

=> 2.1 = (0.50)^0.67 x D^2.67

=> 3.341 = D^2.67

=> D = 1.57 ft

Hence, Bottom width, B = 0.246(1.57) = 0.386 ft

Top width = B + 2mD = 0.386 + 2(4)(1.57) = 12.94 ft  

Depth = 1.57 ft