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An open channel has a trapezoidal cross section with a bottom width of 5 m and side slopes of 2 : 1 (H : V). If the depth of flow is 2 m and the average velocity in the channel is 1 m/s, calculate the discharge in the channel.

Answer :

Manetho

Answer:

Q = 18 m^3/sec

Explanation:

In trapezoidal channel

Area  [tex]A = \frac{1}{2}(a+b)h[/tex]

where a, b are parallel side and h is distance b/w them.

Also, as per question

[tex]tan\theta = \frac{1}{2}\\\Rightarrow \theta = 26.56^{\circ}[/tex]

a = 5 m , b =5+2(2/tanθ) = 13 m

Therefore, A = 0.5(5+13)2 = 18 m^2

Discharge Q = AV  = 18×1 = 18 m^3/sec

V= velocity = 1 m/s (given)

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