Question

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

Answer 1

Answer:

$S = \dfrac{1}{2.5}$

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 = ?

$R = \dfrac{A}{P}$

$R = \dfrac{y(b + xy)}{b+2y\sqrt{1+x^2}}$

$R = \dfrac{5(8+ 5)}{8+2\times 5\sqrt{1+1^2}}$

R = 4.69 m

using manning's equation

$Q = \dfrac{1}{n}AR^{2/3} S^{1/2}$

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

$2312=14009.37\times S^{1/2}$

S = 0.406

$S = \dfrac{1}{2.5}$