CIV E 444 - APPLIED HYDRAULICS
FALL 2009
HOMEWORK No. 3 SOLUTION

  1. The formula is: So = (f/8) F2

    f / 8 = 0.016 / 8 = 0.002

    So = 1 = 0.002 F2

    F = (1 / 0.002)1/2 = 22.3

  2. Use ONLINE CHANNEL 15 with the given data to calculate:

    Q = 21,618 cfs.

    V = 3.659 fps.

    F = 0.18.

  3. Since Q/Qo is independent of the slope or roughness, use ONLINE CHANNEL 03 for given input set:   D = 3 ft, S = 0.01, n = 0.013.

    Try varying flow depth from y = 2.7 ft to 3.0 ft, in increments of Δy = 0.03 ft.

    Find maximum Q = 71.74 cfs at y = 2.82 ft.

    Therefore, the maximum discharge in a circular culvert occurs when the flow depth is (2.82/3.00) × 100 = 94% of the diameter.

  4. Use ONLINE CHANNEL 01 with Q = 18,000 cfs, b = 380 ft, z = 3, S = 0.0001, n - 0.022 to find:

    Design flow depth y = 12.573 ft.