written 3.1 years ago by
RakeshBhuse
• 3.2k
|
•
modified 3.1 years ago
|
Solution :
Given :
Bed width, b=6m
Depth, d=3m
Slope =3H:4V
Discharge, Q=30m³/s
Chezy's constant, C=70

From Fig.
BE=3×34=2.25m
Length of CD,
CD=AB+2×BE=6.0+2×2.25=10.50 m
Wetted perimeter,
P=AD+AB+BC=AB+2BC(∵BC=AD)=AB+2√BE2+CE2=6.0+2√(2.25)2+(3)2=13.5 m
Area of flow, (A)
A= Area of trapezoidal ABCD
=(AB+CD)×CE2=(6+10.50)2×3.0=24.75 m2
Hydraulic mean depth, (m)
m=AP=24.7513.50=1.833
Discharge, (Q)
Q=AC√mi30.0=24.75×70×√1.833×i√i=302345.6i=(302345.6)2=1(2345.630)2=16133