CIV E 444 - APPLIED HYDRAULICS
SPRING 2016
LAB No. 13 SOLUTION


  1. The Creager formula is:   Qp = 46 C A0.894A - 0.048

    Qp = 46 C (355)0.894 × (355) - 0.048

    Assume the maximum value of C = 100.

    Qp = 241,361 cfs         [Creager]

    For TR-55, assume T = 10000 yr; CN = 100 (the maximum); and ap = 0 (lowest diffusion).

    Assume Type II storm, because for all other conditions being the same, it gives the highest peak (it is the most intense type storm).

    Qp = 236,075 cfs         [TR-55]

  2. α = [(∑ Ai )2 / (∑ Ki )3] [ ∑ (Ki3/ Ai2) ]

    Conveyance K = (1.486/n) A R 2/3

    The calculations are shown in the following table.

    The wetted perimeter inbank is P = 100 + 2(10 - 2) = 116 ft.

    The wetted perimeter in the flood plain is P = 500 + 2 = 502 ft.

    iSegment BdVnAPRR2/3K K3/ A2
    1Left flood plain 500220.05 10005021.9921.58347,047 104,134,780
    2Inbank channel 1001050.03 10001168.6214.204208,238 9,029,837,855
    3Right flood plain 500220.05 10005021.991.58347,047 104,134,780
     Sum    3000   302,3329,238,107,415

    α = [(∑ Ai )2 / (∑ Ki )3] [ ∑ (Ki3/ Ai2) ] = [(3000)2/(302,332)3] [ 9,238,107,415 ]

    α = 3.008