The population of a town is 80,000 persons with a water supply rate of 145 lpcd. Assuming 80% of water supply contributes for sewage flow, taking Manning's N as 0.013, average slope as 1:400 and peak factor as 3. Determine minimum diameter of sewer required to carry maximum discharge if it runs at 0.75 depths?

Solution:

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

`Q_(DWF)=(0.8**80000**0.145)/(86400)`

`=0.1074\ m^3/s`

The peak sanitary discharge is, `=3**0.1074=0.3222\ m^3/s`

Given, 

`d/D=1/3`

or, `1/2(1-cos (theta/2))=0.75`

or, `theta=240^0`

Now, we know,

`q/Q=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)=0.9118`

Solving this equation, we get,

`0.3222/Q=0.9118\ m^3/s`

or, `Q=0.3533\ m^3/s`

or, `(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/400)^(1/2)=0.4559`

or, `d=0.632\ m`

Solution:

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

`Q_(DWF)=(0.8**80000**0.145)/(86400)`

`=0.1074\ m^3/s`

The peak sanitary discharge is, `=3**0.1074=0.3222\ m^3/s`

Given, 

`d/D=1/3`

or, `1/2(1-cos (theta/2))=0.75`

or, `theta=240^0`

Now, we know,

`q/Q=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)=0.9118`

Solving this equation, we get,

`0.3222/Q=0.9118\ m^3/s`

or, `Q=0.3533\ m^3/s`

or, `(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/400)^(1/2)=0.4559`

or, `d=0.632\ m`