CIV E 530 - OPEN-CHANNEL HYDRAULICS FALL 2015 - MIDTERM 2 - SOLUTION

PROBLEM 1

• S_o = f F₁²

  F₁² = S_o / f = 0.004 / 0.004 = 1

  F₁ = 1   ANSWER.

• F₁² = v²/(gy₁) = q²/(gy₁³) = 1

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

• v₁ = q / y₁ = 0.5 / 0.294 = 1.7 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 = (1.8 - 0.294) / 0.004

  Δx = 376.5 m.    ANSWER.

  Problem 2

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