written 6.1 years ago by
teamques10
★ 66k

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modified 6.1 years ago

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
$V=\frac{Q}{A}$=$\frac{0.208}{\frac{ π }{4}*0.54^2}$
$= \frac{4*0.208}{π *0.54^2}$
$V=0.908m/sec$