Answer:
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 *[tex]\sqrt{n² +1}[/tex]
(b + 2*2*h)/2 =h *[tex]\sqrt{n² +1}[/tex]
(b+ 4h)/2 = h *[tex]\sqrt{n² +1}[/tex]
(b+ 4h)/2 = h *[tex]\sqrt{2² +1}[/tex] =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
