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Design separate sewer system for a town having population of 1 lakh with water supply of 180 lpcd. The permissible sewer slope is 1 in 1000 and n=0.012. Assume DWF is` 1/3` rd of maximum discharge. Also check velocity of flow.
Solution:
The design discharge is,
`Q=(3**0.8**100000**0.18)/86400`
`=0.5\ m^3/s`
Now, Considering the circular sewer running full,
`Q=(pi D^2)/4**1/0.012**(D/4)^(2/3)**(1/1000)^(1/2)`
or, `D=0.83\ m`
Thus, `V=0.5/((pi**0.83^2)/4)=0.924\ m/s`
Check for self Cleansing velocity during dry weather flow:
`q/Q=0.333`
or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`
Solving this, we get,
`theta=156.31^0`
and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.899`
or, `v=0.899**0.924=0.831`
Here 0.60 < 0.831 < 3 m/s. Okay
Check for self cleansing velocity during minimum flow.
Let, `Q_(m i n)=1/2 Q_(DWF)`
Thus,
`q/Q=0.1665`
or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]`
Solving this, we get,
`theta=126.77^0`
and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.74`
or, `v=0.74**0.924=0.683>0.6`
He....
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Quantity of Waste Water
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