CIV E 530 - OPEN-CHANNEL HYDRAULICS

FALL 2013 - MIDTERM 2 - SOLUTION

PROBLEM 1

    • So = f F12

      F12 = So / 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.

    • So = Δy/Δx

      Δx = Δy / So = (y2 - y1) / So = (1.5 - 0.254) / 0.0025

      Δx = 498.4 m.    ANSWER.

    Problem 2

     
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