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

PROBLEM 1

• S_o = f F₁²

  F₁² = S_o / f = 0.0025 / 0.0025 = 1

  F₁ = 1   ANSWER.

• F₁² = v²/(gy₁) = q²/(gy₁³) = 1

  y₁ = (q²/g)^1/3 = (0.4²/9.81)^1/3 = 0.254 m.    ANSWER.

• v₁ = q / y₁ = 0.4 / 0.254 = 1.575 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 = (1.5 - 0.254) / 0.0025

  Δx = 498.4 m.    ANSWER.

  Problem 2

  131023 1500