Design the separate sewer system for a town having population of 1 lakh with water supply of 180 lpcd. The permissible sewer slope is 1 in 1000 and n=0.012. Assume DWF is` 1/3` rd of maximum discharge. Also check velocity of flow.

Solution:

The design discharge is,

`Q=(3**0.8**100000**0.18)/86400`

`=0.5\ m^3/s`

Now, Considering the circular sewer running full,

`Q=(pi D^2)/4**1/0.012**(D/4)^(2/3)**(1/1000)^(1/2)`

or, `D=0.83\ m`

Thus, `V=0.5/((pi**0.83^2)/4)=0.924\ 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**0.924=0.831`

Here 0.60 < 0.831 < 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**0.924=0.683>0.6`

Hence the d....

Solution:

The design discharge is,

`Q=(3**0.8**100000**0.18)/86400`

`=0.5\ m^3/s`

Now, Considering the circular sewer running full,

`Q=(pi D^2)/4**1/0.012**(D/4)^(2/3)**(1/1000)^(1/2)`

or, `D=0.83\ m`

Thus, `V=0.5/((pi**0.83^2)/4)=0.924\ 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**0.924=0.831`

Here 0.60 < 0.831 < 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**0.924=0.683>0.6`

Hence the design is  okay.