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Design-of-sewers-Tutorial
Design a sanitary sewer to carry dry wear flow of `0.26\ m^3` /sec laid at a gradient of 1 in 500.Assume manningâ??s `n=0.012` and sewer runs at `3/5` th depth during peak flow condition.
Take peak factor =3.
Solution:
`q=a/n (r)^(2/3) s^(1/2)`
where, `q=0.26\ m^3/s`
`s=1/500`
`n=0.012`
`d/D=3/5`
or, `1/2 (1-cos (theta/2))=3/5`
or, `theta=203.07^0`
Thus, the area is,
`a/A=theta/360-(sin theta)/(2 pi)`
`=0.6265`
`r/R=[1-(360 sin theta)/(2 pi theta)]`
`=1.1106`
Thus,Â
`q/Q=(a/A)(r/R)^(2/3)`
`=0.6265**(1.1106)^(2/3)`
`=0.6718`
or, `(0.26**3)/Q=0.6719`
or, `Q=1.1609`
or, `1/0.012**(pi D^2)/4**(D/4)^(2/3)**(1/500)^(1/2)=1.1609`
or, `D=1\ m`
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