Design a section of combined circular sewer from following data :

Area to be served = 250 ha
• Population = 60,000
• Maximum permissible velocity = 3.2 m/sec
• Time of entry = 5 minutes
• Time of flow =20 minutes
• Rate of water supply = 280 lpcd
• Impermeability factor =0.5
• Maximum discharge =1.85 * DWF.

Solution:

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

`Q_(DWF)=(0.8**60000**0.280)/(86400)`

`=0.1556\ m^3/s`

The peak sanitary discharge is, `=1.85**0.1556=0.28778\ m^3/s`

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 intensty of rainfall is,

`i=1020/(t+20)`

`=1020/(45)=22.67\ ( m m)/(hr)`

Again, the storm discharge is given by,

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

`=(0.5**22.67**250)/360`

`=7.8715\ m^3/s`

Therefore, the combined discharge is,

`Q=0.28778+7.8715=8.1593\ m^3/s`

Now, 8.1593=(pi d^2)/4**3.2`

or, `d=1.801\ m`

Adopting commercially available size `d=1.85\ m`

Solution:

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

`Q_(DWF)=(0.8**60000**0.280)/(86400)`

`=0.1556\ m^3/s`

The peak sanitary discharge is, `=1.85**0.1556=0.28778\ m^3/s`

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 intensty of rainfall is,

`i=1020/(t+20)`

`=1020/(45)=22.67\ ( m m)/(hr)`

Again, the storm discharge is given by,

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

`=(0.5**22.67**250)/360`

`=7.8715\ m^3/s`

Therefore, the combined discharge is,

`Q=0.28778+7.8715=8.1593\ m^3/s`

Now, 8.1593=(pi d^2)/4**3.2`

or, `d=1.801\ m`

Adopting commercially available size `d=1.85\ m`

`A=(pid^2)/4=2.688\ m^2`

`V=Q/A=3.035/ m/s` which is less than `3.2\ m/s`.

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.1556/8.1593=0.0191`

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

Solving this, we get,

`theta=72.142^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.391`

or, `v=0.391**3.035=1.18`

Here 0.60 < 1.18 < 3.2 m/s. Okay

Check for self cleansing velocity during minimum flow.

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

Thus,

`q/Q=0.00955`

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

Solving this, we get,

`theta=60.94^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.32`

or, `v=0.32**3.035=0.97`

Here 0.60 < 0.97 < 3.2 m/s. Okay

Hence the design is okay.