Answer:
d = 3.25 ft
Explanation:
n = 0.013
side slope, m = 1
Flow rate, Q = 125 ft³/s
Width, w = 3 ft
Longitudinal slope, S = 1% = 0.01
Let the depth = d
Area, A = wd = 3d
The equation for the flow rate can be given as:
[tex]Q = \frac{1.49}{n} AR_{h} ^{2/3} S^{1/2}[/tex]...........................(1)
[tex]R_{h} = \frac{Area}{Perimeter} \\R_{h} = \frac{3d}{3 + 2d}\\[/tex]
Substituting all appropriate values into equation (1)
[tex]125 = \frac{1.49}{0.013} *3d*(\frac{3d}{3+2d})^{2/3} * 0.01^{1/2}[/tex]
[tex]125 = \frac{71.52d^{5/3} }{(3+2d)^{2/3} } \\1.75(3+2d)^{2/3} = d^{5/3}\\(3+2d)^{2/3} = 0.57d^{5/3}\\3+2d = (0.57d^{5/3})^{3/2} \\3+2d = 0.43d^{5/2}\\(3+2d)^{2} = 0.187d^{5} \\9 + 6d + 4d^{2} = 0.187d^{5}\\ 0.187d^{5}- 4d^{2}-6d -9 = 0\\d^{5}-21.39d^{2}-32.09d -48.13 = 0\\[/tex]
On solving the equation above
d = 3.25 ft
