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Chapter:
Design a sewer for a population of 100,000 persons with water supply per capita of `120` l/d. It is expected that `80%` of water is converted into sewage. The DWF estimated will be `1/3` rd of maximum discharge in this separate sewer. The permissible slope is `1:1000` and rugosity coefficient is taken as 0.012. For self-cleaning purpose at least `0.75` m/sec velocity need to be developed in drain.
Solution:
Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by
`Q_(DWF)=(0.8**100000**0.12)/(86400)`
`=0.11\ m^3/s`
The peak sanitary discharge is, `=3**0.11=0.33\ m^3/s`
Now,
`Q=(pi d^2)/4**(1/0.012)**(d/4)^(2/3)**(1/1000)^(1/2)`
or, `d=0.713\ m`
Adopting commercially available size, `d=0.75\ m`
Now, velocity is,
`v=0.33/((pi**0.713^2)/4)`
`=0.8346\ m/s`
Check for self Cleansing velocity during dry weather flow:
`q/Q=0.11/0.33=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**0.8346=0.7503`
Here 0.75 < 0.7503 < 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[....
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