Question

If a trapezoidal channel has side slopes of 1:1, hydraulic depth is 5 feet, the bottom width is 8 feet, flow is 2,312 cubic feet per seconds and the Manning's "n" is 0.013, what is the slope of the channel? g

Answer 1

`S = (1)/(2.5)`

Step-by-step explanation:

given,

side slope = 1 : 1

hydraulic depth(y) = 5 ft

bottom width (b)= 8 ft

x = 1

Q = 2,312 ft³/s

n = 0.013

slope of channel = ?

`R = (A)/(P)`

`R = (y(b + xy))/(b+2y√(1+x^2))`

`R = (5(8+ 5))/(8+2* 5√(1+1^2))`

R = 4.69 m

using manning's equation

`Q = (1)/(n)AR^(2/3) S^(1/2)`

`2312= (1)/(0.013)* (5(8+5))* 4.69^(2/3) S^(1/2)`

`2312=14009.37* S^(1/2)`

S = 0.406

`S = (1)/(2.5)`