A town has population of 1 lakh with per capita water supply of 200 lpcd. Design a sewer taking n=0.013; slope =1 in 600 and peak factor =2.25.Assume 80% of the water supply is converted to sewage.

Solution:

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

`Q_(DWF)=(0.8**100000**0.2)/(86400)`

`=0.1852\ m^3/s`

The peak sanitary discharge is, `=2.25**0.1852=0.4167\ m^3/s`

The sewer is designed for maximum discharge.

Now,

`Q=(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/600)^(1/2)`

Solving this, we get,

`d=0.73\ m`

 Adopting commercially available size, `d=0.75\ m`

`A=(pid^2)/4=0.442\ m^2`

`V=Q/A=0.4167/0.442=0.94\ m/s`

Here, `0.6....

Solution:

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

`Q_(DWF)=(0.8**100000**0.2)/(86400)`

`=0.1852\ m^3/s`

The peak sanitary discharge is, `=2.25**0.1852=0.4167\ m^3/s`

The sewer is designed for maximum discharge.

Now,

`Q=(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/600)^(1/2)`

Solving this, we get,

`d=0.73\ m`

 Adopting commercially available size, `d=0.75\ m`

`A=(pid^2)/4=0.442\ m^2`

`V=Q/A=0.4167/0.442=0.94\ m/s`

Here, `0.6<0.94<3`. Hence okay.

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.1852/0.4167=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**0.94=0.90`

Here 0.60 < 0.90 < 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=137.84^0`

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

or, `v=0.804**0.94=0.755 >0.60`

Hence the design is  okay.