Design section of combined sewer from following data: Area to be served =150 ha, population of locality = 50000, maximum permissible velocity =3.2 m/sec, time of entry = 5 minutes, time of flow = 20 minutes, rate of water supply =270 lpcd, impermeability factor =0.45, maximum discharge =1.8 times DWF. Assume any or suitable data if needed.

Solution:

Assuming `100%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(1**50000**0.270)/(86400)`

`=0.15625\ m^3/s`

The peak sanitary discharge is, `=1.8**0.15625=0.28125\ m^3/s`

The sewer is designed for maximum discharge.

Now, 

`T_c=T_e+T_f=5+20=25\ m i n`

Where, `T_c` is the time of concentration

`T_e` is the time of entry.

`T_f` is the time of flow.

The quantity of storm water will be maximum when storm duration is equal to the time of concentration.

`Thus, t=T_c=25 mi n`

Now, the intensity of rainfall is,

`i=1020/(t+20)=22.67\ (m m)/(h r)` 

Again, the storm discharge is given by,

`Q_(W W F)=(CiA)/360=(0.45**22.67**150)/360=4.35\ m^3/s`

Thus, the discharge for the combined sewer is,

`Q=4.35+0.281`

`=4.531\ m^3/s`

Now, `Q=AV`

or, `4.531=(pi d^2)/4**3.2`

Solution:

Assuming `100%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(1**50000**0.270)/(86400)`

`=0.15625\ m^3/s`

The peak sanitary discharge is, `=1.8**0.15625=0.28125\ m^3/s`

The sewer is designed for maximum discharge.

Now, 

`T_c=T_e+T_f=5+20=25\ m i n`

Where, `T_c` is the time of concentration

`T_e` is the time of entry.

`T_f` is the time of flow.

The quantity of storm water will be maximum when storm duration is equal to the time of concentration.

`Thus, t=T_c=25 mi n`

Now, the intensity of rainfall is,

`i=1020/(t+20)=22.67\ (m m)/(h r)` 

Again, the storm discharge is given by,

`Q_(W W F)=(CiA)/360=(0.45**22.67**150)/360=4.35\ m^3/s`

Thus, the discharge for the combined sewer is,

`Q=4.35+0.281`

`=4.531\ m^3/s`

Now, `Q=AV`

or, `4.531=(pi d^2)/4**3.2`

or, `d=1.34\ m`

Adopting commercially available size, `d=1.40\ m`

`A=(pid^2)/4=1.54\ m^2`

`V=Q/A=0.4.531/1.54=2.94\ m/s`

Here, `0.6<2.94<3`. Hence okay.

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.15625/4.531=0.034`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=83.25^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.464`

or, `v=0.464**2.94=1.36`

Here 0.60 < 1.36 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.017`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=70.11^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.377`

or, `v=0.377**2.94=1.11><0.60`

Hence the design is  okay.