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Find the bed slope of trapezoidal channel of bed width 6m,depth of water 3m and side slope 3H:4V, when the discharge through the channel is 30m3/sec, C=70.
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Solution :

Given :

  1. Bed width, b=6m

  2. Depth, d=3m

  3. Slope =3H:4V

  4. Discharge, Q=30m³/s

  5. Chezy's constant, C=70

enter image description here

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+2BE2+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=ACmi30.0=24.75×70×1.833×ii=302345.6i=(302345.6)2=1(2345.630)2=16133

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