Question

A trapezoidal ditch is designed with a bottom width of 3 space f t and side slopes of m equals 1 on both sides. The channel is made of concrete (n equals 0.013) and has a longitudinal slope of 1 percent sign. If the flow rate on a particular day is 125 space f t cubed divided by s, determine the depth of flow in the channel. Assume steady uniform flow.

Answer 1

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:

$Q = \frac{1.49}{n} AR_{h} ^{2/3} S^{1/2}$...........................(1)

$R_{h} = \frac{Area}{Perimeter} \\R_{h} = \frac{3d}{3 + 2d}$

Substituting all appropriate values into equation (1)

$125 = \frac{1.49}{0.013} *3d*(\frac{3d}{3+2d})^{2/3} * 0.01^{1/2}$

$125 = \frac{71.52d^{5/3} }{(3+2d)^{2/3} } \\1.75(3+2d)^{2/3} = d^{5/3}\\(3+2d)^{2/3} = 0.57d^{5/3}\\3+2d = (0.57d^{5/3})^{3/2} \\3+2d = 0.43d^{5/2}\\(3+2d)^{2} = 0.187d^{5} \\9 + 6d + 4d^{2} = 0.187d^{5}\\ 0.187d^{5}- 4d^{2}-6d -9 = 0\\d^{5}-21.39d^{2}-32.09d -48.13 = 0$

On solving the equation above

d = 3.25 ft