Calculate the quantity of wastewater to be carried by separate system with the following data and design a half flowing sanitary sewage with a slope of 1 in 600.Check for self cleaning velocity and limiting velocity for concrete sewer pipe. Area to be served = 500 ha, population density = 100 persons/ha, waters supply rate = 20 lpcd, peak factor =3 and 80% of water supply is converted to sewage. Assume any data if required.

Solution:

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

`Q_(DWF)=(0.8**100**500**0.02)/(86400)`

`=0.00926\ m^3/s`

The peak sanitary discharge is, `=3**0.00926=0.0278\ m^3/s`

Now, For half full circular sewer,

`q=0.0278=(pid^2)/8**(1/0.014)**(d/4)^(2/3)**(1/600)^(1/2)`

or, `d=0.351\ m`

Adopting commercially available size, `d=0.35\ m`

`A=(pid^2)/8=0.048\ m^2`

`V=Q/A=0.1874/0.2513=0.57\ m/s`

Here, `v=0.57 <0.6\ m/s`.

For cement concrete pipes the limiting value i.e, maximum value is about 3.0 and the minimum self cleansing velocity is 0.6m/s

Since the velocity is less than self cleansing velocity, flow does not maintain the self cleansing velocity of `0.6\ m/s`

Solution:

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

`Q_(DWF)=(0.8**100**500**0.02)/(86400)`

`=0.00926\ m^3/s`

The peak sanitary discharge is, `=3**0.00926=0.0278\ m^3/s`

Now, For half full circular sewer,

`q=0.0278=(pid^2)/8**(1/0.014)**(d/4)^(2/3)**(1/600)^(1/2)`

or, `d=0.351\ m`

Adopting commercially available size, `d=0.35\ m`

`A=(pid^2)/8=0.048\ m^2`

`V=Q/A=0.1874/0.2513=0.57\ m/s`

Here, `v=0.57 <0.6\ m/s`.

For cement concrete pipes the limiting value i.e, maximum value is about 3.0 and the minimum self cleansing velocity is 0.6m/s

Since the velocity is less than self cleansing velocity, flow does not maintain the self cleansing velocity of `0.6\ m/s`