A sewer pipe is to be laid at a slope1 in 8100.To carry discharge of 600lit/sec ,when depth of water is 75% of the vertical diameter. Find the diameter of the pipe if value of Manning’s constant is 0.025

Solution :

Given :

Discharge, $Q=600 ~Litre / {s}=0.6{~m}^{3} /{s} $

Bed slope, $ i =\frac{1}{8100} $

Manning's, $N =0.025$

Depth of water, $d=75=0.75 D$

From Figure, we have $O C=C D-O D=0.75 D-0.5 D=0.25 D$

In $\Delta$AOC $$\begin{aligned} cos\alpha &=\frac{OC}{OA}=\frac{0.25D}{0.5D}=0.5\\ \alpha &=cos^{-1}0.5= 60°\\ \theta &=180°-\alpha=180°-60°\\ &=120°=2.0946 ~rad \end{aligned}$$

Wetted perimeter (P)

$$ \begin{aligned} P &=2 R \theta=2 \times 0.5 \mathrm{D} \times 2.0496 \\ &=2.0496 D \end{aligned} $$

Area of flow (A)

$$ \begin{aligned} A &=R^{2}\left(\theta-\frac{\sin 2 \theta}{2}\right) \\ &=(0.5 D)^{2}\left[2.0946-\frac{\sin \left( 240^{\circ}\right)}{2}\right] \\ &=0.25 D^{2}\left[2.0946-\left(\frac{-0.866}{2}\right)\right]\\ &=0.25 D^{2}[2.0946+0.433] \\ &=0.6319 D^{2} \\ m &=\frac{A}{P}=\frac{0.6319 D^{2}}{2.0496 D}=0.308 D \end{aligned} $$

Discharge by Manning's formula is given by

$$ \begin{aligned} Q &=\frac{1}{N} \times A \times m^{2 / 3} \times i^{1 / 2} \\ 0.6 &=\frac{1}{0.025} \times 0.6319 D^{2} \times(0.308 D)^{2 / 3} \times\left(\frac{1}{8100}\right)^{1 / 2} \\ 0.6 &=0.128 \times D^{8 / 3} \\ D^{8 / 3} &=\frac{0.6}{0.128}=4.6875 \\ D &=(4.6875)^{3 / 8}\\ &=1.785 {~m} \end{aligned} $$