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

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