Question

Urgent help please

QUESTION 15 A trapezoidal channel (n=0.023) has a 2-foot wide base, slide slopes of 1:1, bed slope of 0.45% and conveys 300 CFS. What is the normal depth? 2.3.72 feet 6.2.38 feet 5.38 feet d. 0.92 feet

Answer 1

Let y be the normal depth

Q= AV and V = (1.49/n)*(R^2/3)*(S^1/2)

R= A/P where P is wetted perimeter . Thus for normal depth y and side slope 1:1 and bed width 2' , P = 2+2*1.414*y = 2+2.828y and A = 0.5(2+2+2y)y = (2+y)y

Thus R = (2+y)y/(2+2.828y)

Given S = 0.45% = 0.0045 and n = 0.023 thus 300=(1.49/0.023)*R^2/3 * (0.0045^1/2) *A

Thus A^(5/3) / P^(2/3) = 69.03 and on cubing both sides we get A^5/ P^2 = 328937.6763

Thus ( 2y +y^2)^5 / (2+2.828y)^2 = 328937.6763

Substituting y = 3.72 we get LHS = 27827.55 < RHS

Substituting y = 5.38 we get LHS = 332964.013 > RHS

Substituting y = 2.38 we get LHS = 1614.96

Substituting y = 0.92 we get LHS = 6.67

Thus option C is most correct option.