Calculation+for+Activated+Sludge+Process

COMPUTATION FOR ACTIVATED SLUDGE PROCESS Wastewater Flow rate : So

=

600

θc

=

5

Y

=

0.6

kd

=

0.06

MLVSS/MLSS F/M

Mean Cell Residence Time (set, usually 5 - 15 days) Yield Coefficient (assume) Decay Constant; assume, Range = 0.04 - 0.075 d-1, Typical = 0.06

0.25

MLVSS =

3 1000 m /day

=

BOD5 influent

=

=

m3/hr

50

0.75 (Assume Ratio) Food to Microorganism Ratio, range = 0.05 to 1.0

4000 mg/L

Computation of Effluent BOD5 : Assume Biological Solids in Effluent

:

30 mg/L 65% is biodegradable

Biodegradable Portion of Effluent Biological Solids: 19.5 mg/L BOD L of cells = 1.42 (Biodegradable Portion of Effluent Biological Soild) =

27.69

mg/L

Degradable Fraction of Volatile Suspended Solids (VSS) BOD 5 of VSS in Effluent

Substrate Concentration, S

U =

(F/M)E 100

Therefore, Total Effluent BOD5

=

19.383

=

1 UY

=

0.225

0.7

mg/L (1/θ c + k d )

S =

1.925926 mg/L

=

21.30893 mg/L

Computation of Aerator Tank Volume

:

U = Substrate Utilization Rate

E = Process Efficiency, 90-95 percent

V =

θcQY (So-S) X(1+kdθc)

=

345.0427 m3

Compute for Hydraulic Detention Time: θ =

V Q

= 0.345043 day

=

8.28102564 hours

Quantity Of Sludge That Must Be Wasted Each Day Px = YobsQ(So-S) P x : net waste activated sludge produced each day, kg VSS/day Y obs : observed yield, g/g =

Y 1+k d θ c

Px =

=

0.4615385 mg/mg

276.03 kg VSS/day

based on Total Suspended Solids: Px (SS) =

368.05 kg/day

Compute Sludge Wasting Rate: a. if wasting is accomplished from the aeration tank: Qwa =

=

V θc 3

69.00855 m /day

b. if wasting is done from the sludge return line: VX Qwr = θcXr =

3 86.2747 m /day

Oxygen Requirement kg O2/day =

Q(So - S) f

x

1 kg 1000 g

- 1.42 Px

= F/M =

462.423 kg O2/day 0.434729

Volume of Air Required, assume air contains, 23.2% O2 by weight Theo. Air Rate =

3 1659.619 m /day

Determine Actual Air Requirement assuming 8% oxygen transfer efficiency: 3

=

20745.24 m /day

=

14.40642 m /min

Design Air Requirement =

28.81283 m /min

3

3

=

34.6042106 kg/min

Power required to meet oxygen requirement requirement, assume transfer rate is 0.8 kg O 2/kW h =

24.08453 kW

75 d-1, Typical = 0.06

Utilization Rate

ficiency, 90-95 percent

e is 0.8 kg O 2/kW h