CIV E 530 - OPEN-CHANNEL HYDRAULICS

SPRING 2015 - MIDTERM 2 - SOLUTION

PROBLEM 1

    • So = f F12

      F12 = So / 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.

    • So = Δy/Δx

      Δx = Δy / So = (y2 - y1) / So = (1.8 - 0.294) / 0.004

      Δx = 376.5 m.    ANSWER.

    Problem 2

     
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