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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

Respuesta :

Answer:

[tex]S = \dfrac{1}{2.5}[/tex]

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 = ?

[tex]R = \dfrac{A}{P}[/tex]

[tex]R = \dfrac{y(b + xy)}{b+2y\sqrt{1+x^2}}[/tex]

[tex]R = \dfrac{5(8+ 5)}{8+2\times 5\sqrt{1+1^2}}[/tex]

R = 4.69 m

using manning's equation

[tex]Q = \dfrac{1}{n}AR^{2/3} S^{1/2}[/tex]

[tex]2312= \dfrac{1}{0.013}\times (5(8+5))\times 4.69^{2/3} S^{1/2}[/tex]

[tex]2312=14009.37\times S^{1/2}[/tex]

S = 0.406

[tex]S = \dfrac{1}{2.5}[/tex]

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