Find a discharge through a circular pipe of diameter 3 m, and pipe is laid at a slope of 1 in 1000 .Take value of Chezy's constant 70.

1. If the depth of the water in the pipe is 1 m.

2. If the depth of the water in the pipe is 2.5 m

Solution :

Given :

Diameter of pipe, $\quad D=3.0$

Radius, $\quad R=\frac{D}{2}=\frac{3.0}{2}=1.50 {~m}$

Bed slope, $\quad i=\frac{1}{1000}$

Chezy's constant, $\quad C=70$

1. Depth of water in pipe, d=1.0 m

From Figure, we have

$ \begin{aligned} O C &=O D-C D=R-1.0 \\ O C &=O D-C D=R-1.0 \\ &=1.5-1.0=0.5 {~m} \\ A O &=R=1.5 {~m} \end{aligned} $

Also

$\begin{aligned}\cos \theta &=\frac{O C}{A O}=\frac{0.5}{1.5}=\frac{1}{3}\\ \therefore \quad \theta &=70.53^{\circ}=70.53 \times \frac{\pi}{180}\\ \end{aligned}$

$\quad\quad\quad =1.23 \mathrm {~radians} $ $$\quad\left(\because 180^{\circ} =\pi\right) \mathrm {~radians} $$

Wetted perimeter is given by following equation,

$\begin{aligned} P &=2 R \theta=2 \times 1.5 \times 1.23 \\ &=3.69 \mathrm{~m}\end{aligned}$

Wetted Area (A),

$\begin{aligned} A &=R^{2}\left(\theta-\frac{\sin 2 \theta}{2}\right) \\ & =1.5^{2}\left(1.23-\frac{\sin \left(2 \times 70.53^{\circ}\right)}{2}\right) \\ &=2.25\left(1.23-\frac{\sin \left (141.08^{\circ} \right)}{2}\right) \\ &=2.25 \left(1.23-\frac {\sin \left (180 ^{\circ} - 141.08^{\circ}\right)}{2}\right) \\ \quad &=2.25 \left(1.23-\frac{\sin 38.94^{\circ}}{2}\right)\\ &=2.06 {~m}^{2} \end{aligned}$

Hydraulic mean depth (m),

$\begin{aligned} m =\frac{A}{P}=\frac{2.06}{3.69} =0.5582{~ m} \end{aligned}$

Discharge (Q),

$\begin {aligned} Q &=AC \sqrt{mi}\\ &=2.06×70 \sqrt{0.5582 \times \frac{1}{1000}} \\ &=3.407 {~m}^{3} /{s} \end{aligned}$

2. Depth of water in pipe, d = 2.5 m

From above Figure

$ \begin{aligned} O C &=C D-O D=2.5-R \\ &=2.5-1.5=1.0 \mathrm{~m}\\O A &=R=1.5 \mathrm{~m} \\ \cos \alpha &=\frac{O C}{OA}=\frac{1.0}{1.5}=0.667\\ \alpha &=48.16^{\circ} \\ \theta &=180^{\circ}-\alpha \\ &=180 ^{\circ} -48.16^ {\circ} \\ &=131.84^{\circ} \\ &=131.84 \times \frac{\pi}{180}=2.30 \mathrm{~radians} \end{aligned} $

Wetted perimeter is given by following equation,

$\begin{aligned} P &=2 R \theta=2 \times 1.5 \times 2.3 \\ &=6.9 \mathrm{~m}\end{aligned}$

Wetted Area (A),

$ \begin{aligned} A &=R^{2}\left(\theta-\frac{\sin 2 \theta}{2}\right)\\ &=1.5^{2}\left(2.30-\frac{\sin \left(2 \times 131.84^{\circ}\right)}{2}\right) \\ &=2.25\left(2.30-\frac{\sin 263.68^{\circ}}{2}\right) \\ &=2.25\left(2.30-\frac{\sin \left(180^{\circ}+83.68^{\circ}\right)}{2}\right) \\ &=2.25\left[2.30-\frac{\left(-\sin 83.58^{\circ}\right)}{2}\right] \\ &=2.25\left(2.30+\frac{\sin 83.68^{\circ}}{2}\right)\\ &=6.293 \mathrm{~m}^{2} \\ \end{aligned} $

Hydraulic mean depth,

$ \begin {aligned} m=\frac{A}{P}=\frac{6.293}{6.90}=0.912 \mathrm{~m} \end{aligned}$

Discharge (Q),

$\begin {aligned} Q &=AC \sqrt{mi}\\ &=6.239×70 \sqrt{0.912 \times \frac{1}{1000}} \\ &=13.303 {~m}^{3} /{s} \end{aligned}$

Final Answers

1. When d=1.0 m then $Q=3.407 {~m}^{3} /{s}$

2. When d=2.5 m then $Q =13.303 {~m}^{3} /{s}$