2 In a trapezoidal channel having side slope at an angle of

2. In a trapezoidal channel having side slope at an angle of 14 with the vertical and normal depth of flow is 2.5 m. Channel having base width of 12.25 ft. Assume Chezy\'s constant C as 53 and Slope 0.002. Calculate the daily discharge from the channel section. S = 0.001

Solution

Side slope of channel=14^o

normal depth of flow=2.5m

channel base width=12.25 ft=3.73 m

slope of channel =0.002

width of water at top = 3.73+2*2.5*tan14=4.98 m

Area of cross section=0.5*2.5*(4.98+3.73)=10.88 m²

wetted perimeter = 3.73+2*2.5/cos14 = 8.88 m

hydraulic radius = R=10.88/8.88=1.22 m

velocity of flow = C*sqrt(R*i)=53*sqrt(1.22*0.002)=2.62 m/s

Flow rate = 10.88*2.62=28.48 m³/s