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Wastewater treatment: Activated sludge

Biological Treatment
Activated Sludge


Activated Sludge
Influent

GC

PST

Aeration Tank
Secondary
settling Tank

BS

The aeration tank contains mixed liquor suspended solids (MLSS), a
combination of influent wastewater and return (recycled) activated sludge
from the settling tank.
The MLSS includes:

• Bacteria (both aerobes and facultative)
• Slime: usually in flocs composed of extracellular polymeric substances,
live bacterial cells, cell debris (dead, lysed)
• Free cells
• Protozoa
• Nondegradable organic matter
• Inert suspended solids


Activated sludge floc with slime

• MLSS rapidly (20-45 min)
absorbs organic matter in
wastewater influent.
• Bacteria then solubilize and
oxidize organic matter.
• The food (organic matter) to
microorganisms ratio (F/M)
dictates character of
bacteria and flocs.


Design Considerations
Activated sludge process is defined by:
1. Aeration period = detention time= 3-30 hrs
2. BOD loading = (BOD entering per day)/ (V=volume of aeration tank)
3. F/M= (Q*BODin)/(V*MLSS)
4. Sludge age = mean cell residence time

t sludge =

MLSS *V
SS e * Qe + SS w * Qw
Aeration Tank

SSe= suspended solids in the effluent
SSw= suspended solids in waste sludge
Qe = effluent flow rate
Qw= sludge flow rate


Qw, SSw

SST

Qe, SSe


Types of Activated Sludge Systems

Conventional aeration

Step feed to aeration

High-purity oxygen

Extended aeration


Loading and operation procedure for different aeration processes
Process

BOD load
(lb BOD/d
per 1000 ft3)

MLSS
mg/l

F/M
lb BOD/d
per lb
MLSS

tsludge
days

Aeration
period
hr

R%

% BOD
removal

Conventional

20-40

1000-3000

0.2-0.5

5-15

4-7.5

20-40

80-90

Step aeration

40-60

1500-3500

0.2-0.5

5-15

4-7

30-50

80-90

Extended
aeration

10-20

2000-8000

0.05-0.2

>20

20-30

50-100

85-95

High-purity
oxygen

>120

4000-8000

0.6-1.5

3-10

1-3

30-50

80-90


Example
Consider the following activated sludge system. Calculate:
1. The aeration period
2. BOD loading
3. F/M ratio
4. SS and BOD removal efficiency
5. tsludge
6. Return activated sludge rate
7. What is the required QR if the MLSS is to be 2200 mg/l
MLSS=2600 mg/l
Q=2.14 mgd
BOD=185 mg/l
SS=212mg/l
QR=0.85mgd

Qw=39,000 gpd
SSw=8600 mg/l

Aeration:
4 units each
40x40ft2
15.5 ft deep

BOD=15 mg/l
SSe=15 mg/l
Settling
tank


Solution
V= 4*40*40*15.5* (7.481/106)= 0.74 million gallon
t=V/Q=(0.74/2.99)*(24)=5.9 hr
BOD loading= [2.14*185*8.34]/[4*40*40*15.5/1000]= 33.3 lb BOD/d/1000 ft3
Note that the BOD loading from re-circulated sludge is
negligible

F /M =

2.14 mgd × 185 mg / l × 8.34
= 0.21 lb BOD / day per lb MLSS
0.74 mil gal × 2600 mg / l × 8.34

SS removal= (212-15)/212= 0.93= 93%
BOD removal= (185-15)/185= 0.92= 92%


t sludge =

0.74 mil gal × 2600 mg / l × 8.34
= 5.2 days
(2.14 mgd ×15 mg / l × 8.34) + (0.039 mgd × 8600 mg / l × 8.34)

Return sludge ratio (R)= 0.85/2.14=0.40 =40%

To find QR for a MLSS =2200, we apply mass balance on suspended solids
at the entrance of the aeration tank.
2.14*212 + QR*8600 = (2.14+ QR)*2200
Solve for QR= 0.66 mgd



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