Question

Design an activated sludge treatment unit with the following data for a town of population of 80,000:​

B.O.D. of raw sewage = 200 mg/l.

Per capita Average sewage flow is 180 liters.

B.O.D. removal in primary treatment = 60%,

Raw sludge retained in ASP = 25%

Effluent water should have NO BOD at all to satisfy National Standards.

Answer 1

Sol :- Information provided in the problem :-

Population of town = 80,000:

B.O.D. of raw sewage = 200 mg/l.

Per capita Average sewage flow is = 180 liters.

B.O.D. removal in primary treatment = 60%,

Raw sludge retained in ASP = 25%

Now to design activated sludge treatment unit with the following data for a town can be done in following steps:-

Step 1 - Daily sewage flow Q = (180x80,000 L/day )/1000 =14400 m^3/day

BOD of sewage comming to aeration ( Yo ) = 40% X 200 mg/l. = 80 mg/l (as 60% B.O.D. removal in primary treatment )

BOD Left in effluent (YE ) = 0% X 200 mg/l = 0 mg/l

So BOD removed in activated plant = 80 - 0 = 80 mg/l

Then Efficiency required in Activated plant = 80/80 = 100%

This is the most ideal condition which is next to impossible in practical scinerio so we design fo 85 - 95% efficiency,

Step 2 :- We use F/M ratio as 0.4 to 0.3 and MLSS beween 1500 to 3000 for conventional activated plant

So let us adopt F/M =0.31

Similarly adopt MLSS = 1510 mg/l

Using equation we have :-

F/M = (QxYo) /(VxMLSS)

0.31 = (14400 x 80)/(Vx 1510)

So Volume of Aeration tank (V) = 2461.012 m^3

Step 3 :- Check for Aeration Period (t) :-

t = (V/Q) x 24 h = (2461.012/14400) x 24

t = 4.101 hour ( i.e with in 4 to 6 hour) OK

Step 4 : Check for Sluge retention time S.R.T ()

We know the equation:

V.MLSS = {. Q ( Yo  - YE ) .}/{1+ K.}

Here   = yield coefficient = 1.0 (indian Standard value)

K = Endogeneous respirayion rate constant = 0.06 d^(-1) ....(indian Standard value)

then   2461.012x1510 = {1x14400(80 - 0) .}/{1+ 0.06}

{1+ 0.06} = 0.310

= 4.0 days ( approximately acceptible)

Hence Sluge retention time S.R.T () = 4 days

Step 4 : Check for Return Slidge ratio :(Qr / Q)

Using equation we have :

Qr / Q = {MLSS}/{ (10^6 / SVI) - MLSS)

let us adopt SVI (Sludge volume index) = 100 ml/gram

Qr / Q = {1510}/{ (10^6 / 100) - 1510} = 0.18

So Return Slidge ratio (Qr / Q ) = 0.18

But for Conservation purpose, however provide 0.33 return sludge

Note : The Sluge Pumps for bringing recirculated sluge from Secondary sedimentation tank will thus have a capacity = 33% x Q = 33% x 14400 = 4752 m^3/day

Step 4 :- Tank Dimensions:

Adopt aeration tank od depth 4 meter and width 6 meter . then length of aeration Channel required will be:

Length = (Total volume required)/ (BxD)

Length = (2461.012)/(4x6) = 102.542 meter  

lets take Length = 105 meter

Now provide a continous channel. with 3 aeration chamber, each of 35 m length. Total width of the unit, including 2 baffles each of 0.25 m thickness = 3x6 + (2x0.45) = 18.9 or 19 meter . Total depth provided including free board of 0.6 meter will be 6 + 0.6 = 6.6 meter

Hence overall dimension of Aeration tank will be = 35 mx 19 mx 6.6 m