CIV E 530 - OPEN-CHANNEL HYDRAULICS SPRING 2016 - MIDTERM 2 - SOLUTION

PROBLEM 1

• S_o = f F₁²

  F₁² = S_o / f = 0.0035 / 0.0035 = 1

  F₁ = 1   ANSWER.

• F₁² = v²/(gy₁) = q²/(gy₁³) = 1

  y₁ = (q²/g)^1/3 = (0.75²/9.81)^1/3 = 0.386 m.    ANSWER.

• v₁ = q / y₁ = 0.75 / 0.386 = 1.94 m/s    ANSWER.

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

• S_o = Δy/Δx

  Δx = Δy / S_o = (y₂ - y₁) / S_o = (2.0 - 0.386) / 0.0035

  Δx = 461.14 m.    ANSWER.

  Problem 2

  141111 1145