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 flow,

$\begin{aligned} A &=R^{2}\left(\theta-\frac{\sin 2 \theta}{2}\right)\\ &=1.5^{2}\left[2.247-\frac{\sin \left(2 \times 128.75^{\circ}\right)}{2}\right]\\ &=2.25[2.247-(-0.488)]\\ &=6.1537 {~m}^{2} \end{aligned}$

Hydraulic mean depth

$\begin {aligned}m^{*}=\frac{A}{P}=\frac{6.1537}{6.741}=0.912\end{aligned}$

Now velocity,

$\begin{aligned}V =C \sqrt{m \times i} &=50 \times \sqrt{\frac{0.912 \times 1}{500}}\\ &=2.135{~m} /{s}\end{aligned}$

discharge,

$\begin{aligned}Q=A \times V &=6.1537 \times 2.135\\ &=13.138{~m}^{3} / {s}\end{aligned}$

2) Velocity and discharge for maximum discharge

For maximum discharge

$ \begin {aligned}\theta=154^{\prime \prime} &=\frac{154 \times \pi}{180} \\ &=2.6878\end{aligned} ~radians $

Wetted perimeter

$\begin{aligned}P =2 \times R \times \theta &=2 \times 1.5 \times 2.6878\\ &= 8.0634 {~m}\end{aligned}$

Area of flow,

$\begin{aligned} A &=R^{2}\left(\theta-\frac{\sin 2 \theta}{2}\right)\\ &=1.5^{2}\left[2.6878-\frac{\sin \left(2 \times 154^{\circ}\right)}{2}\right]\\ &=2.25[2.6878-(-0.394)]\\ &=6.934 {~m}^{2} \end{aligned}$

Hydraulic mean depth

$\begin {aligned}m^{*}=\frac{A}{P}=\frac{6.934}{8.0634}=0.8599\end{aligned}$

Now velocity

$\begin {aligned}V=C \sqrt{m i} &=50 \times \sqrt{0.8599 \times \frac{1}{500}} \\ &=2.0735 {~m} /{s}\end{aligned}$.

discharge,

$\begin {aligned}Q=A \times V &=6.934 \times 2.0735\\ &=14.377 {~m}^{3} /{s} \end{aligned}$.