CIV E 530 - OPEN-CHANNEL HYDRAULICS

FALL 2008 - MIDTERM 2 - SOLUTION

PROBLEM 1

    • So = Sc F12

      F12 = So / Sc = 0.0035 / 0.0035 = 1

      F1 = 1   ANSWER.

    • F12 = v2/(gy1) = q2/(gy13) = 1

      y1= (q2/g)1/3 = (2.52/9.81)1/3 = 0.86 m.    ANSWER.

    • v1 = q/y1 = 2.5 / 0.86 = 2.91 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 = (4 - 0.86) / 0.0035 = 897.14 m.    ANSWER.

Problem 2

 
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