Activated Sludge Process
Suspended growth reactor
In this type of reactor, reaction between organic matter and bacteria takes
place in suspension on the surface of the suspended solids (suspended
returned sludge particles).
The main objective of the activated sludge process is to stabilize organic
matter and make it settleable.
A microscopic photo of the reaction between organic matter and bacteria
which takes place in suspension solids on the surface of the suspended
Types of activated sludge process:
1- Conventional aeration.
2- Contact stabilization.
3- Step aeration.
4- Extended aeration.
5- Oxidation ditch (orbital aeration).
Conventional aeration
Advantages of aeration tanks :
1- It does not need large land areas.
2- It is more efficient than trickling filters.
3- No fly and odor problems occurs around the trickling filter.
Disadvantages of aeration tanks:
1- It can not take shock loads and sudden increase in discharge.
2- Needs a supervisor since it is the most complicated systems.
3- High cost of constructions, operation and maintenance.
99
Q
P.S.T
Q
So
Xo
A.T
X
QR.S , Xr
Q+QR.S
Q - Qw
F.S.T
pump
Effluent
Se
Xe
Set
Excess sludge
Qw
Xr
Flow line diagram of activated sludge process
Diffused air system of aeration tank
100
Surface aerators of aeration tank
Aeration tank
Factors affecting design of aeration tank:
1- Temperature: the biological reaction increase with the increase of
temperature.
2- Mixed Liquor is a mixture of raw or settled wastewater and activated sludge
within an aeration tank in the activated sludge process.
Mixed Liquor Suspended Solids (MLSS) is the concentration of suspended
solids in the mixed liquor, usually expressed in milligrams per litre (mg/l)
101
MLSS: (mixed liqueur suspended solids) 2000 - 4000 mg/L
MLVSS: (mixed liqueur volatile suspended solids)
MLVSS = 0.8 MLSS
If MLSS content is too high
- The process is prone to bulking and the treatment system becomes
overloaded
- This can cause the dissolved oxygen content to drop with the effect
that organic matters are not fully degraded and biological 'die off‘
- Excessive aeration required which wastes electricity
If MLSS content is too low
- The process is not operating efficiently and is wasting energy
- Typical Control band 2,000 to 4,000 mg/l
3- Hydraulic retention time = T = 4 - 8 hrs
4- Sludge return rate (R) = 0.2 – 0.3
R = QR / Qd
Return sludge QR.s = 0.2 - 0.3 Qd
5- Өc = SRT= sludge age (average time biomass stays in the system.
Calculated by total solids in system / total solids being wasted =5 - 15 days
Within the limits, a longer SRT results in:
- More efficient biodegradation
- Smaller reactor size
- Lower cost
If SRT drops below the cell regeneration time, biomass will wash out faster
than it forms new cells.
Types of aeration :
1- Diffused air system
2- Surface aerators
Design criteria :
V
y  c  Q  (S0  Se )
X (1  K d c )
Q = discharge influent to the aeration tank (A.T).
= 1.5 x Qave sewage (Q = 0.8 x 1.5 x pop x qave )
So = dissolved BOD5 influent to the A.T.
Se = dissolved effluent BOD5
Xe = suspended solids concentration (S.S)
BOD5 suspended = S.S x 0.7
BOD5 dissolved = total BOD5 - BOD5 suspended
Se = Set – Xe x 0.7
X: total number of microorganisms responsible for the stabilization of organic
matter.
X = MLVSS
MLVSS = 0.8 MLSS
Өc = sludge age =5 - 15 days
y: cell yield coefficient
y = gm MLVSS/gm BOD5
Y observed = y / ( 1+ kd Өc ) = 0.31
Kd = decay coefficient = 0.05 day-1
102
A=V/d
Depth = 3 - 5 m
n≥2
b = 1.5 – 2 d
L ≤ 50 m
Checks :
F/M 
T
Q( S o  S e )
MLVSS  V
V
Q
F / M  0.2  0.4
T  4  8 hrs
T .O.L
L
S Q
L o
V  1000
V
Allowable organic load (L) = 0.3 - 2 kg BOD5 /m³/ day
QR.S
MLSS

Q
TSSinR.S  MLSS
X V
C 
Qw  X r
R
c 
X V
Qw  X r  X e (Q  Qw )
Qw = sludge withdrawal rate
Xr =MLVSS in return sludge
Xe = effluent S.S
Example:
Determine the volume and dimensions of aeration tank and the required air
flow for an activated sludge process treatment plant. Which is designed to
treat the flow Q = 12000 m3/d and fined overall efficiency, F/M ratio, allowable
organic load and the diffused air system. Given the following data:
- BOD5 of primary effluent (So) = 200 mg/l
- Required effluent BOD5 = 30 mg/l
(Set)
- Required effluent suspended solids (Xe) = 20 mg/l
- MLVSS in aeration tank (X) = 2500 mg/l
- MLSS in return sludge = 10000 mg/l
- Cell residence time (c) = 6 days
- Coefficient of decay (kd) = 0.06 day-1
- Cell yield coefficient (y) = 0.65 gm MLVSS/gm BOD5
- Efficiency of aeration = 8%
103
Solution:
Q=12000
Q
So=200mg/l
Xo
P.S.T
A.T
X=2500
Q+QR.
S
QR.S , Xr
OverallefficiencyE % 

V
pump
S o  S et
100
So
200  30
200
Effluent
Se
Xe=20mg/l
Set=
30mg/l
Excess
sludge
Qw
Xr
Xe
Set
suspended
Q
100  85%
y  c  Q  (S0  Se )
X (1  K d c )
Dissolved
F.S.
T
70% organic
30% inorganic
BOD5 dissolved = BOD5 eff. – S.Seff x 0.7
Se =Set – Xe x 0.7
= 30 -20 x 0.7 = 16 mg/l
V
0.65  6 12000  (200  16)
 2532.71m 3
2500(1  0.06  6)
d= 3 – 5 m
Assume d = 4m
V
d
2532.71
area 
 633.17m 2
4
area 
Number of tanks n =2
Area of one tank =
b = 1.5 – 2 d
L
633.17
 316.6m 2
2
take b = 8m
A 316.6

 39.57 m  50m
b
8
104
Checks :
F /M 
Q( So  S e )
MLVSS  V
F/M 
gmBOD5
12000(200  16)
 0.349
2500  2532.71
gmMLVSS.day
T
F/M = 0.2 - 0.4
V
Q
2532.71
 24  5.07 hrs
12000
T
T = 4 - 8 hrs
So  Q
V 1000
kgBOD5
200 12000
L
 0.948
2532.711000
m3 .day
L
Allowable organic load (L) = 0.3 - 2 kg BOD5 /m³/ day
Allowable organic load (L) = 0.3 - 2 kg BOD5 /m³/ day
The design of the return sludge pipe:
QR . S
MLSS

Q
TSSinR.S  MLSS
2500
(
)
QR . S
0
.
8

 0.45
2500
Q
10000  (
)
0.8
QR.S  0.45  12000  5454.5 m 3 / d
R
MLVSS = 0.8 MLSS
QR . S
 2

v
24  60  60
4
5454.5
 2

1
24  60  60
4
  0.28  0.3m
Determination of excess sludge:
C 
6
X V
Qw  X r
2500  2512.06
Qw  (10000  0.8)
Qw  130.83m 3 / d
c 
6
X V
Qw  X r  X e (Q  Qw )
2500  2512.06
Qw  (10000  0.8)  20  (12000  Qw )
Qw  126.6m 3 / d
105
Design of diffused air system:
yobs  Q  ( S o  S e )
1000
0.3112000  (200  17.5)
Px 
 678.9kg / d
1000
( S  S )  Q 10 3
Theoritica l pure oygen (O2 )  o e
 1.42 Px
BOD 5
(
)
BODU
Px 
(200  17.5) 12000 10 3
Theoritica l pure oygen (O2 ) 
 1.42  678.9  2256.55kgO2 / d
(0.68)
Theoritica l pure oygen (O2 )
S tan dard oxygen required ( SOR ) 
0.65
2256.55
S tan dard oxygen required ( SOR ) 
 3471.61kgO2 / d
0.65
SOR
Theoritica l air demand 
1.2  0.232(O2 ratio )
Theoritica l air demand 
3471.61
 12469.88kg / m 3
1.2  0.232
Actual air required 
theoritical air demand
effeciency of oxygen
12469.88
 155873.54m3 / d
0.08
Blowers capacity  1.5  actual air
Actual air required 
Blowers capacity  1.5 155873.54  233810.3m3 / d
Head of blowers  1.3  depth of waterinA.T
Head of blowers  1.3  4  5.2m
No. of blowers 
blowers capacity
design cpacity of each blower
Design of air pipe in aeration tank:
Blower capacity 
2
4
v
v  15m / s
106
Design of surface aerators:
yobs  Q  ( S o  S e )
1000
0.31 12000  (200  17.5)
Px 
 678.9kg / d
1000
( S o  S e )  Q  10 3
Theoritica l pure oygen (O2 ) 
 1.42 Px
BOD 5
(
)
BODU
Px 
(200  17.5)  12000  10 3
Theoritica l pure oygen (O2 ) 
 1.42  678.9  2256.55kgO2 / d
(0.68)
Theoritica l pure oygen (O2 )
S tan dard oxygen required ( SOR ) 
0.65
2256.55
S tan dard oxygen required ( SOR ) 
 3471.61 kg O2 / d
0.65
Total aerators capacity  2  SOR
Total aerators capacity  2  3471.61  6943.22 kg O2 / d
NO. of aerators 
2  SOR
O2 / aerator
107

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