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Find the bed slope of trapezoidal channel of bed width 4m, depth of water 3m and side slope of 2 horizontal to 3 vertical
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Find the bed slope of trapezoidal channel of bed width 4m, depth of water 3m and side slope of 2 horizontal to 3 vertical, when the discharge through the channel is 20 $m^3/s$

Taking manning's N = 0.03, in manning's formula $c = \frac{1}{N} m^{1/6}$

Given:

Bed width , b = 4cm

Depth of flow , d = 3m

Side slope = 2 hor to 3 verd

Discharge Q = 20.0 20 $m^3/s$

Manning's N = 0.03

Distance , $BE = d \times \frac{2}{3}$ = 2m

$\therefore$ Top width , Q = AB + 2B

= 4 + 2 x 2

= 8.0 m

$\therefore$ Area of flow,

$A = \frac{(AB + CD)}{2} \times d$

$= \frac{4 + 8}{2} \times 3$

$[ A = 18m^2 ]$

wetted perimeter,

P = AD + AB + BC

= AB + 2BC

$= 4 + 2 \sqrt{BE^2 + EC^2}$

$= 4 + 2 \sqrt{2^2 + 3^2}$

$= 4 + 2 \sqrt{13}$

[ P = 11.21 m ]

$\therefore$ Hydraulic mean depth,

$m = \frac{A}{P}$

$= \frac{18}{11.21}$

[ m = 1.6057]

using mannings formula,

$c = \frac{1}{N} m^1/6$

$\frac{1}{0.03} \times (1.6057)^1/6$

[ c = 36.07]

Now, Q = $AC \sqrt{mi}$

$20 = 18 \times 36.07 \times \sqrt{1.6057 \times i}$

$20 = 822.71 \sqrt i$

$\therefore$ $i = (\frac{20}{822.71})^2$

= 0.0005909

$i = \frac{1}{1692}$