Atrapezoidal channel is designed to carry 25 m3 /s on a slope of 0.0015 m/m. the channel is unlined, and in order to prevent erosion, the maximum allowable velocity is 1.5 m/s. the side slope must be no steeper than 2h on 1v and the manning n value is 0.03. in order to meet these requirements, what flow depth and bottom width should be used

the flow depth = 2.59m and bottom width =1.24m

Explanation:

A trapezoidal channel

Q discharge rate= 25 m3 /s

slope of 0.0015 m/m.

V, maximum allowable velocity is 1.5 m/s.

The side slope must be no steeper than 2H on 1V and the

Manning n value is 0.03.

In order to meet these requirements, what flow depth and bottom width should be used?

solution

Q = V * A

A = Q/V = 25/1.5 =16.666m²

the area of a trapezoidal channel is given by =

0.5 (a+b)* h

a= top width

b= bottom width

h= height

Note: The side slope must be no steeper than 2H on 1V

side slope, n = horizontal/vertical = 2H/1V = 2/1 = 2

bed slope of 0.0015

Area of trapezodal = (b+nh)*h

using the eqn

(b + 2*n*h)/2 =  h *

(b + 2*2*h)/2 =h *

(b+ 4h)/2 = h *

(b+ 4h)/2 = h * =h* 2.24

b+4h = 2(2.24h) = 4.48h

b= 4.48h-4h = 0.48h

Area of trapezodal =[b + (b + 2*n*h)]/2 *h =(b + b + 2*2*h)/2 =

(2b+4h)/2 *h

(2b+4h)/2 *h = (0.96h+4h)/2 *h = 2.48h*h = 2.48h²

A = 2.48h²

therefore, A=16.666m² = 2.48h²

h² = 6.72

h = 2.59m

b= 0.48h =0.48*2.59=1.24m

top width =b+2nh = 1.24 + (2*2*2.59) = 11.6m

the flow depth = 2.59m and bottom width =1.24m

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.