A main combined sewer is to be designed to serve an area of 12 sq.km with a population density of 250 persons/ha. The average rate of sewage flow is 250 lpcd. The maximum flow is 100% in excess of average together with rainfall equivalent of 15 mm in 24 hours, all of which are runoff. Determine the capacity of sewer. Taking the maximum velocity of flow as 3 m/sec. Determine the size of sewer.

Solution:

The dry weather flow is given by,

`Q_(DWF)=((250**1200)**(250)*1**1.5)/(1000**86400)`

`=1.302\ m^3/s`

Here, `i=15/24=0.625 (m m)/(h r)`

The storm flow is given by,

`Q_(W W F)=(CiA)/(360)`

`=1**0.625**1200)/360`

`=2.083\ m^3/s`

Thus, the total discharge of the combined sewer is,

`Q=1.302+2.083`

`=3.385\ m^3/s`

or, `3.385=(pi D^2)/4**3`

or, `D=1.2\ m`

Solution:

The dry weather flow is given by,

`Q_(DWF)=((250**1200)**(250)*1**1.5)/(1000**86400)`

`=1.302\ m^3/s`

Here, `i=15/24=0.625 (m m)/(h r)`

The storm flow is given by,

`Q_(W W F)=(CiA)/(360)`

`=1**0.625**1200)/360`

`=2.083\ m^3/s`

Thus, the total discharge of the combined sewer is,

`Q=1.302+2.083`

`=3.385\ m^3/s`

or, `3.385=(pi D^2)/4**3`

or, `D=1.2\ m`