Question

A 2.44 wide channel with a bed slope of 0.0004 has a depth of flow of 0.61 m.
(a) Determine the discharge of the channel. The manning coefficient is 0.015.
(b) Determine the normal depth if the channel has a flow rate of 2.85 m3/s.
(c) Determine the cross section of greatest hydraulic efficiency for a trapezoidal channel if the design discharge is 10 m3/s, the channel slope is 0.0052 and manning’s n = 0.025.

Answer 1

Y = 0.610 m O. 015 Area = Bxy - 2.44 x 0.16 Perim etu P = B+2y Ā ?0.3904m Hy draulie Radtiug (R] > - 2-u4 + 2%0.6) = 3.c0 m 0.3904 3.C0 2/, 0.015 Q: 2.85 m/ L[R]"(0.000y2yA 0- 015 2.65 = (-00ng2, 2.85 = 2-1315 = APply povwe 3% 0n bom side RAT2.444 (2.13752 %3D o.8 199 : 312 o.S199 = 2-44+2xy O. 336 = 2.44+2xy O.1899 + 0.69 2y = y%2 -0.011 m

Slope = 0.052 n: 0.025 d=60° Hydraulic efficient Trapcaoidal channel -b most ef ficient 29155 TYapezoided channel ) half 9/13 of a Reguloor helagon" 1. --o Regular hexagus au side are equcu S Sin6o = y 8= 2%5 24 メ2 + hydraulie Radius (R3 = O-025 y's+2 3.177 = 8/3 3-177 = 2x1. 542 B = 1- 78) m 1. 542 0.762 m .