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Design a sewer to serve a population of 35000 the daily per capita water supply allowance being 160 litre of which 80% finds its way into the sewer .

The slope available for the sewer to be laid is 1 in 600 and the sewer should be designed to carry four times the dry weather flow when running full. What would be the velocity of flow in the sewer when running full?

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Given:

Population=35000

Daily per capita water supply=160 litres

Water supplied=80%

Slope available=1 in 600

Sewer to be designed to carry 4xD.W.F

Find:

1. Average sewage discharge

assume n=0.012 in Mannings formula.

Average water supplied daily =$35000*160=5.6*10^6$litres/day

Average water supplied in cumecs

$\frac{5.6*10^6}{1000*24*60*60}m^2/sec$ or cumecs=0.0648cumecs

average sewage discharge

=80% of water supplied

=$\frac{80}{100}*0.0648$

=0.0518 cumecs

D.W.F=0.0518=0.052cumecs

2. Maximum discharge(Q):-

maximum discharge 'Q' for which server should be designed running full

$Q=4*0.052=0.208cumecs$

3. The diameter of sewer pipe:-

using mannings formula

$Q=\frac{1}{N}*A*R^\frac{2}{3}*s^\frac{1}{2}$

When circular sewer run full

Hydraulic mean depth$(R)=\frac{D}{4}$

$Q=\frac{1}{N}*A*R^\frac{2}{3}*s^\frac{1}{2}$

$0.208=\frac{1}{0.012}*\frac{ π}{4}*D^2*(\frac{D}{4})^\frac{2}{3}*(\frac{1}{600})^\frac{1}{2}$

$=\frac{0.208*0.012*4*4^\frac{2}{3}*600^\frac{1}{2}}{ π }$

$=D^\frac{8}{3}$=0.196

D=$(0.196^\frac{8}{3})$=0.54

D=0.54m use 0.54m diameter sewer pipe

4. Velocity of flow (V) when running full

$V=\frac{Q}{A}$=$\frac{0.208}{\frac{ π }{4}*0.54^2}$

$= \frac{4*0.208}{π *0.54^2}$

$V=0.908m/sec$