CIV E 444 - APPLIED HYDRAULICS FALL 2009 HOMEWORK No. 12 SOLUTION 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] With CN = 80, run ONLINE TR-55 with Type I storm (Coastal Southern California), to determine Qp = 3,338 cfs. Target peak discharge = 0.8 × 3,338 = 2,670 cfs. Run ONLINE TR-55 again, trying smaller values of CN, say, 76 and 75. For CN= 76, Qp = 2,756 cfs       [Too high!] For CN= 75, Qp = 2,621 < 2,670 The target CN is 75. The watershed will be improved to reduce CN from 80 to 75. 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. i Segment B d V n A P R R2/3 K K3/ A2 1 Left flood plain 500 2 2 0.05 1000 502 1.992 1.583 47,047 104,134,780 2 Inbank channel 100 10 5 0.03 1000 116 8.621 4.204 208,238 9,029,837,855 3 Right flood plain 500 2 2 0.05 1000 502 1.99 1.583 47,047 104,134,780   Sum         3000       302,332 9,238,107,415 α = [(∑ Ai )2 / (∑ Ki )3] [ ∑ (Ki3/ Ai2) ] = [(3000)2/(302,332)3] [ 9,238,107,415 ] α = 3.008 Use ONLINE CHANNEL 04 to find Sc = 0.0025. The Darcy-Weisbach f:   f = 8 Sc = 0.02       [Roberson Chapter 4-1 Notes].