CIV E 530 - OPEN-CHANNEL HYDRAULICS

FALL 2003 - MIDTERM 2 - SOLUTION

PROBLEM 2

(a)

    The water-surface profile upstream of the weir is:

    S1.   ANSWER.

(b)
    q = 2 m3/s/m

    The uniform flow equation, applied upstream, is:

    So = Sc F12

    F12 = So / Sc = 0.01 / 0.002 = 5

    The upstream Froude number is:

    F12 = v2/(gy1) = q2/(gy13)

    Therefore, the upstream flow depth is:

    y1= [q2/(gF12)]1/3 = [22/(9.81 × 5]1/3 = 0.434 m.    ANSWER.

(c)

    The hydraulic jump formula is:

    y2/y1 = (1/2) [(1 + 8F12)1/2 - 1] = 2.7

    Therefore:

    y2 = 2.7 × 0.434 = 1.17 m.    ANSWER.

(d)

    So = Δy/Δx

    Δx = Δy / So = (5.0 - y2) / So = (5.0 - 1.17) / 0.01 = 383 m.    ANSWER.

Problem 3

 
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