Design combined circular sewer from following available data and draw neat sketch of sewer.

• Population of locality `= 75000`
• Rate of water supply =275 lpcd
• Area to be served = 175 hectares
• Maximum permissible velocity = 3 m/sec
• Time of entry = 5 minutes
• Time of flow = 20 minutes
• Average permeability factor =0.45

Solution:

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

`Q_(DWF)=(0.8**75000**0.275)/(86400)`

`=0.1909\ m^3/s`

The peak sanitary discharge is, `=2**0.1909=0.3819\ 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`

`=(0.45**22.67**175)/360`

`=4.959\ m^3/s`

Therefore, the combined discharge is,

`Q=0.3819+4.959=5.3409\ m^3/s`

Now, 5.3409=(pi d^2)/4**3`

or, `d=1.505\ m`

Adopting commercially available size `d=1.6\ m`

`A=(pid^2)/4=2....

Solution:

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

`Q_(DWF)=(0.8**75000**0.275)/(86400)`

`=0.1909\ m^3/s`

The peak sanitary discharge is, `=2**0.1909=0.3819\ 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`

`=(0.45**22.67**175)/360`

`=4.959\ m^3/s`

Therefore, the combined discharge is,

`Q=0.3819+4.959=5.3409\ m^3/s`

Now, 5.3409=(pi d^2)/4**3`

or, `d=1.505\ m`

Adopting commercially available size `d=1.6\ m`

`A=(pid^2)/4=2.0106\ m^2`

`V=Q/A=2.66\ m/s` which is less than `3\ m/s`.

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.1909/5.3409=0.03574`

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

Solving this, we get,

`theta=84.303^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.471`

or, `v=0.471**2.66=1.25`

Here 0.60 < 1.25 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

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

Thus,

`q/Q=0.01787`

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

Solving this, we get,

`theta=70.97^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.38`

or, `v=0.38**2.66=1.01`

Here 0.60 < 1.01 < 3 m/s. Okay

Hence the design is okay.