Find size of circular sanitary sewer to serve population of `50000` with sewage flow is `180` lpcd. The natural slope of ground is 1 in 1250.Take peak low 2.25 of DWF and n=0.012.Sewage should run half full during peak flow condition.

Solution:

Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(0.8**50000**0.18)/(86400)`

`=0.0833\ m^3/s`

The peak sanitary discharge is, `=2.25**1.388**10^-3=0.1874\ m^3/s`

Now, For half full circular sewer,

`q=0.1875=(pid^2)/8**(1/0.012)**(d/4)^(2/3)**(1/1250)^(1/2)`

or, `d=0.778\ m`

Adopting commercially available size, `d=0.80\ m`

`A=(pid^2)/8=0.2513\ m^2`

`V=Q/A=0.1874/0.2513=0.74\ m/s`

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.0833/0.1874=0.444`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=169.9^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.96`

or, `v=0.96**0.74=0.71`

Here 0.60 < 0.71 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q....

Solution:

Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(0.8**50000**0.18)/(86400)`

`=0.0833\ m^3/s`

The peak sanitary discharge is, `=2.25**1.388**10^-3=0.1874\ m^3/s`

Now, For half full circular sewer,

`q=0.1875=(pid^2)/8**(1/0.012)**(d/4)^(2/3)**(1/1250)^(1/2)`

or, `d=0.778\ m`

Adopting commercially available size, `d=0.80\ m`

`A=(pid^2)/8=0.2513\ m^2`

`V=Q/A=0.1874/0.2513=0.74\ m/s`

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.0833/0.1874=0.444`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=169.9^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.96`

or, `v=0.96**0.74=0.71`

Here 0.60 < 0.71 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.222`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=137.8^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.804`

or, `v=0.804**0.74=0.59 <0.60`

Hence the design is not okay.

Let us now take `d=0.70\ m`

`A=(pid^2)/8=0.0.1924\ m^2`

`V=Q/A=0.1874/0.1924=0.97\ m/s`

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.0833/0.1874=0.444`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=169.9^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.96`

or, `v=0.96**0.97=0.93`

Here, `0.6<0.93<3` okay.

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.222`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=137.84^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))=0.804`

or, `v=0.804**0.97=0.77 >0.60`

Hence the design is okay.

Thus, `d=0.7\ m` and `v=0.97\ m/s`