A town has a population of 80000 persons with a per capita water supply of 180 liter per day. Design a sewer running 0.7 times full at maximum discharge. Take a constant value of n=0.013 at all depths of flow. The sewer is to be laid at a slope of 1 in 500.Take peak factor of 3.

Solution:

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

`Q_(DWF)=(0.8**80000**0.18)/(86400)`

`=0.133\ m^3/s`

The peak sanitary discharge is, `=3**0.133=0.4`

By question,

`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)/(2 pi theta))^(5/3)=0.8372`

 or, `0.4/Q=0.8372\ m^3/s`

or, `Q=0.4778\ m^3/s`

or, `(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/500)^(1/2)=0.4559`

or, `d=0.738\ m`

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

`v=1/0.013**0.1875^(2/3)**(1/500)^(1/2)`

or, `v=1.127\ m/s`

Check for self Cleansing velocity during dry weather flow:

`q/Q=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**1.127=1.01`

Solution:

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

`Q_(DWF)=(0.8**80000**0.18)/(86400)`

`=0.133\ m^3/s`

The peak sanitary discharge is, `=3**0.133=0.4`

By question,

`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)/(2 pi theta))^(5/3)=0.8372`

 or, `0.4/Q=0.8372\ m^3/s`

or, `Q=0.4778\ m^3/s`

or, `(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/500)^(1/2)=0.4559`

or, `d=0.738\ m`

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

`v=1/0.013**0.1875^(2/3)**(1/500)^(1/2)`

or, `v=1.127\ m/s`

Check for self Cleansing velocity during dry weather flow:

`q/Q=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**1.127=1.01`

Here 0.60 < 1.01 < 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[1-(360 sin theta)/(2 pi theta)]`

Solving this, we get,

`theta=126.77^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.74`

or, `v=0.74**1.127=0.83`

Here 0.60 < 0.83 < 3 m/s. Okay

Hence the design is okay.