A concrete lined circular channel of diameter 3m has a bed slope of 1 in 500. Workout the velocity and flow rate for the conditions of i) Maximum velocity, ii)Maximum discharge.

Solution :

Given :

Dia of channel, $ D=3 {~m}$

Bed slope, $i=\frac{1}{500}$

Value of $C=50$

1) Velocity and discharge for maximum velocity

For maximum velocity,

$ \begin{aligned}\theta =128.75^{\circ}&=128.75 \times \frac{\pi}{180}\\& =2.247 ~radians\end{aligned}$

Wetted perimeter,

$\begin{aligned}P =2 \times R \times \theta &=2 \times 1.5 \times 2.247\\ &=6.741 \mathrm{~m}\end{aligned}$

Area of …