CIV E 530 - OPEN-CHANNEL HYDRAULICS SPRING 2013 - MIDTERM 2 - SOLUTION

PROBLEM 1

1. S_o = f F₁²

   F₁² = S_o / f = 0.003 / 0.003 = 1

   F₁ = 1   ANSWER.

2. F₁² = v²/(gy₁) = q²/(gy₁³) = 1

   y₁ = (q²/g)^1/3 = (0.5²/9.81)^1/3 = 0.294 m.    ANSWER.

3. v₁ = q / y₁ = 0.5 / 0.294 = 1.699 m/s    ANSWER.

4. Because the u/s flow is critical, the water-surface flow profile type is C₁.    ANSWER.

5. S_o = Δy/Δx

   Δx = Δy / S_o = (y₂ - y₁) / S_o = (2.0 - 0.294) / 0.003

   Δx = 568.67 m.    ANSWER.