Calculation+for+activated+sludge+process 5i5e3w
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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
=
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
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