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Design-of-sewers-Tutorial

Design a sewer for separate system to carry a maximum flow of `0.5` `m^3`/s at a slope of 1 in 1000. Sewer should run 0.7 times depth at maximum flow. Assume n= 0.012.

Solution:

Here, `d/D=0.7`

or, `1/2(1-cos (theta/2))=0.7`

or, `theta=227.15^0`

Now,

`q/Q=theta/360(1-(360 sin theta)/(2pi theta))^(5/3)=0.5/Q`

or, `0.5/Q=0.8372`

or, `Q=0.5972`

or, `(pid^2)/4**(1/0.012)(d/4)^(2/3)(1/1000)^(1/2)`

or, `d=0.8873\ m`

Now,

`v=1/(0.012)**0.2218^(2/3)**(1/1000)^(1/2)`

`=0.96 >0.6\ m/s` okay

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Design a sewer for separate system to carry a maximum flow of `0.5` `m^3`/s at a slope of 1 in 1000. Sewer should run 0.7 times depth at maximum flow. Assume n= 0.012.

Design of sewers Tutorial

Sanitary-Engineering