For wastewater treatment

1. Lecture 5. Attached Growth Biological Treatment System
Secondary Treatment
Husam Al-Najar
The Islamic University of Gaza- Civil Engineering Department
Advance wastewater treatment and design (WTEC 9320)

2. Cross-section of an attached growth biomass film
Wastewater
Oxygen (the natural or forced pump)
filter media
Biomass : viscous, jelly-
like substance containing
bacteria
Organic matter + NH4
+

3. Remove Nutrient
Remove dissolved organic solids
Remove suspended organic solids
Remove suspended solids
Attached Growth is a biological treatment process in which microorganisms
responsible for conversion of organic matter or other constituents in wastewater are
attached to some inert material such as: rocks, sand or specially ceramic or plastic
materials. This process is also called fixed film process.
Trickling filters (biological tower )
Rotating biological contactors ( RBC )
Packed bed reactors
Fluidized bed biofilm reactors

4. Trickling Filter (TF)- side view
Wastewater
rotating distributor arms
Packing media
Underdrain
rocks, plastic, or
other material

5. Packing Media- Filter media

6. Flow Diagram for Trickling Filters
Recycle
Primary
clarifier
Trickling
filter
Final
clarifier
Waste
sludge
Final
effluent
Influent
Qr
Q
Importance of recirculation
• recirculation flow dilutes the strength of
raw wastewater & allows untreated
wastewater to be passes through the
filter more than once
• maintain constant wetting rate
• dilute toxic wastes
• increase air flow
• A common range for recirculation
ratio 0.5~3.0

7. Single stage configuration
PC SC
TF
PC SC
TF
PC SC
TF
a.
b.
c.

8. Two stage configuration
PC SC
TF
PC SC
TF
TF
TF SC
PC SC
TF
TF SC

9. Stone media TF design
PC SC
TF2
TF1
VF
w
E
1
1
4432
.
0
1
100


VF
w
E
E
2
1
2
1
4432
.
0
1
100



NRC formula:
E2 = BOD removal efficiency for second-stage filter at 20oC, %
E1 = fraction of BOD removal in the first-stage filter
w1 and w2 = BOD load applied, kg/day
V = volume of filter media, m3
R = recycle ration
F = recirculation factor
2
)
10
/
1
(
1
R
R
F




10. • Organic (BOD) loading rate:
– Expressed as kg/m3/d
– Typically, 0.320-0.640 kg/m3/d for single-stage filters
– Typically, 0.640-0.960 kg/m3/d for two-stage filters
• Hydraulic loading rate:
– m3 wastewater/m2 filter*d
– the rate of total influent flow is applied to the surface of the filter media
– Total influent flow = the raw WW + recirculated flow
– Typically, 9.4 m3/m2/d
– Maximum, 28 m3/m2/d
• The effect of temperature on the BOD removal efficiency
ET = BOD removal efficiency at ToC, %
E20 = BOD removal efficiency at 20oC, %
20
20 )
035
.
1
( 
 T
T E
E

11. Example: Calculate the BOD loading, hydraulic loading, BOD
removal efficiency, and effluent BOD concentration of a single-stage
trickling filter based on the following data:
– Design assumptions:
• Influent flow =1530 m3/d
• Recirculation ratio = 0.5
• Primary effluent BOD = 130 mg/L
• Diameter of filter = 18 m
• Depth of media = 2.1 m
• Water temperature =18oC

12. Solution
(1) BOD loading rate (kg/m3/d)
– BOD load = BOD Conc. x Influent flow
= 130 mg/L x 1530 m3/d =198.9 kg/d
– Volume of filter = surface area of filter x depth
= π (18 m x 18m)/4 X 2.1 m = 533 m3
– BOD loading rate = BOD load / volume of filter
=0.37 kg/m3/d
(2) Hydraulic loading rate (m3/m2/d)
– Total flow to the media = influent + recirculation flow
= 1530 m3/d + (1530 m3/d x 0.5)
– Surface area of filter = π (18 m x 18m)/4 = 254 m3
– Hydraulic loading rate = Total flow to the media / area of filter
= 9.04 m3/m2/d

13. (3) Effluent BOD (mg/L)
– BOD removal efficiency for first-stage filter at 20oC, %
%
7
.
75
)
035
.
1
(
2
.
81
)
035
.
1
( 2
20
18
20
18 

 

E
E
VF
w
E
1
1
4432
.
0
1
100

 36
.
1
)
10
/
5
.
0
1
(
5
.
0
1
)
10
/
1
(
1
2
2







R
R
F
%
2
.
81
36
.
1
37
.
0
4432
.
0
1
100
4432
.
0
1
100
1
1 




VF
w
E
100
)
7
.
75
100
(
/
130
)
/
(


 L
mg
L
mg
BOD
Effluent = 31.6

14. Example: A municipal wastewater having a BOD of 200 mg/L is to be
treated by a two-stage trickling filter. The desired effluent quality is 25
mg/L of BOD. If both of the filter depths are to be 1.83 m and the
recirculation ratio is 2:1, find the required filter diameters. Assume the
following design assumptions apply.
– Design assumptions:
• Influent flow =7570 m3/d
• Recirculation ratio = 2
• Depth of media = 1.83 m
• Water temperature =20oC
• BOD removal in primary sedimentation = 35%
• E1=E2 = 0.65

17. A rotating biological contactors{RBC}
 Rotating biological contactors consist of a series of closely spaced circular
disks of polyvinyl chloride (PVC) that are submerged in wastewater and rotated
through it.
 The cylindrical disk are attached to a horizontal shaft and are provided at
standard unit sizes of approximately 3.5 m in diameter and 7.5 m in length.
 The surface area of disks for a standard unit is about 9300 m2, and 13900 m2 for
high density units
 The RBC unit is partially submerged (typically 40%) in a tank containing
wastewater , and the disks rotate slowly at about 1.0 to 1.6 revolutions per
minute.
 As the RBC disks rotate out of the wastewater , aeration is accomplished by
exposure to the atmosphere.

18. RBC components

21. RBC process design considerations:-
The following are the main design parameters needed to design the RBC
System :-
• Staging of the RBC units
• Organic loading rate
• Hydraulic loading rate
RBC staging:-
• The RBC process application typically consists of a number of units operated in
series.
• For this purpose, RPC is divided into stages . Number of stages depends on the
treatment goals . For BOD removal “ 2 ” to “ 4” stages are needed and “ 6” or
more stages for nitrification.

22. 1st Stage 2nd Stage 3rd Stage 4th Stage
To
secondary
clarifier
Influent
RBC unit
NOTE : the number of shafts in each stage depends on the treatment efficiency required.
 The separation between stages is accomplished by using baffles in a single tank or by
a series of separate tanks.
 As the wastewater flow through the system , each subsequent stage receives an
influent with a lower organic matter concentration than the previous stage.
 The RBC units may be arranged parallel or normal to the direction of wastewater flow.

23. d
m
le
so

2
lub
d
m
BOD
g

2
5 )
(
d
m
gN

2
d
m
BOD s

2
5 )
(
Organic loading rate:
The organic loading rate for RBC in typically in the range 4-10 g (BOD)
for BOD removal only.
If both BOD removal and nitrification, the range is 2.5- 4
The maximum 1st stage organic loading is 12-15
Nitrifying bacteria can not develop in RBC until (BOD5) drops to less than 15 mg/ L.
The maximum nitrogen surface removal rate that has been observed to be about 1.5
.

24. Hydraulic loading rate:
The typical hydraulic loading rate of 0.08-0.16 2
3
/ m
m for 5
BOD removal
and 0.03-0.08 2
3
/ m
m for both 5
BOD removal and nitrification.
The hydraulic detention time ( ) is 0.7-1.5 hrs for 5
BOD removal and
1.5-4 hrs for both 5
BOD removal and nitrification.
The volume of RBC tank has been optimized at 0.0049 2
3
/ m
m for one
shaft of 9300 2
m .
A tank volume of 45 3
m is needed. Based on this volume and a hydraulic
loading rate of 0.08 d
m
m .
/ 2
3
the detention time is 1.44 hrs. Atypical side
wall depth is 1.5m to achieve 40% submergence.

25. Bar
screen
Grit
removal
Primary
Sedimentation
RBC
units
Final
clarifier
effluent
1st stage 2nd
stage
RBC configuration

26. Design equation of RBC:
The following empirical equation developed by Optaken( US EPA,1985 ):
)
(
0195
.
0
)
(
039
.
0
1
1 1
Q
A
S
Q
A
S
s
n
s
n





Where n
S = soluble 5
BOD concentration in stage(n), (mg/L)
s
A = disk surface area on stage(n), 2
m
Q= flow rate, d
m /
3
Rate of nitrification is related to the soluble BOD load/m2
 
d
m
N
g
BOD
rn


 2
5 )
(
1
.
0
1
5
.
1

27. RBC design Example: Design a rotating biological contractor to treat an
influent soluble 5
BOD of 90 mg 5
BOD /L. The flow(Q)= 4000 3
m /d
Solution:
 Assume 1st
stage ( 5
BOD ) organic loading= 15g/ 2
m .d
5
BOD (loading)= ( 5
BOD ) concentration in
Q

=
d
g
mg
g
m
L
d
m
L
mg
000
,
360
10
10
4000
90 3
3
3
3




Disk area= 2
2
24000
.
15
000
,
360
m
d
m
g
d
g
 (first area stage)
Use
shaft
m2
9300 so number of shafts needed for the first stage:
N= 6
.
2
9300
24000
2
2

shaft
m
m
say 3 shafts

28.  Calculate S1, the BOD concentration after the first stage:
Sn=
)
(
0195
.
0
)
(
039
.
0
1
1 1
Q
As
S
Q
As
n



For the first stage n=1, Sn=S1, Sn-1 = S0
S0 = 90 mg/L, AS =
d
m
Q
m
3
2
4000
,
27900
9300
3 


m
d
d
m
m
Q
AS
98
.
6
/
4000
27900
3
2


S1=
L
mg
L
mg
14
75
.
29
98
.
6
0195
.
0
90
98
.
6
039
.
0
1
1








So we need one more stage.

29.  Add another stage and calculate S2:
Assume two shafts in the second stage:
AS= 2
2
18600
9300
2 m
m 

d
m
Q
AS
65
.
4
4000
18600


Sn=S2, Sn-1=S2-1=S1=29.75mg/L
S2=
L
mg
L
mg
14
86
.
16
65
.
4
0195
.
0
75
.
29
65
.
4
039
.
0
1
1








So we need one more stage.
 Add another stage and calculate S3:
Assume one shaft in the third stage:
AS= 2
2
9300
9300
1 m
m 

m
d
d
m
m
Q
AS
33
.
2
4000
9300
3
2



30. Follow Example RBC design example:
Sn=S3,Sn-1=S3-1=S2=16.86mg/L
S3=
L
mg
L
mg
14
13
33
.
2
0195
.
0
86
.
16
33
.
2
039
.
0
1
1








OK
So three stages are enough.
 Check for the hydraulic loading:
HLR=
shaft
each
of
Area
shafts
of
number
total
Q

Nshafts=3+2+1= 6 shafts
HLR=
d
m
m
d
m


 2
3
3
072
.
0
9300
6
4000
, typical range(0.08-0.16), which is a
little bit lower than the range.

31.  Is nitrification possible in any of the three stages?:
*Nitrification is only possible when soluble 5
BOD loading is less than
10g
d
m
BOD

2
 1st
stage =
d
m
gBOD
m
g
d
m




 2
5
3
3
9
.
12
9300
3
1
90
4000 <10(no
nitrification)
 2nd
stage =
d
m
gBOD
m
g
d
m




 2
5
3
3
4
.
6
9300
2
1
75
.
29
4000
(nitrification occurs)
 3rd
stage =
d
m
gBOD
m
g
d
m




 2
5
3
3
25
.
7
9300
1
1
86
.
16
4000
(nitrification occurs)

32. *Rate of nitrification is related to the soluble 5
BOD loading by the
following equation:
 
d
m
N
g
BOD
rn


 2
5 )
(
1
.
0
1
5
.
1
* So for 2nd
stage→  
d
m
N
g
rn




 2
54
.
0
4
.
6
1
.
0
1
5
.
1
* And for 3rd
stage→  
d
m
N
g
rn




 2
413
.
0
25
.
7
1
.
0
1
5
.
1
*If the ammonia concentration in the influent to the 2nd
stage is 30
mgN/L, find the effluent ammonia concentration.

33. d
m
N
g
rn

 2
54
.
0 for 2nd
stage
Nitrogen removal =
d
g
m
d
m
N
g
10044
)
9300
2
(
54
.
0 2
2




Concentration =
L
mg
m
d
d
g
51
.
2
4000
10044
3


So→ N2 =
L
mg
5
.
27
51
.
2
30 

→ rn = 0.413
d
m
N
g

2
for 3rd
stage,
Nitrogen removal =
d
g
m
d
m
gN
3841
)
9300
1
(
413
.
0 2
2




Concentration =
L
mg
m
d
d
g
96
.
0
4000
3841 3


So→ N3 = 27.5
L
mgN
L
mgN
5
.
26
96
.
0 


34. If complete nitrification is needed a separate nitrification stage should be added
after these stages.
Qin = 4000 m3/d
So = 90 mg/L
N = 30 mgN/L
Shaft 1
Shaft 2
Shaft 3
Shaft 1
Shaft 2
Shaft 1
Se = 13 mg BOD5 /L
Ne = 26 .50 mg N/L
A = 9300m2
A = 9300m2
Proposed design
Note: each shaft has a tank volume of 45m3.

35. Appendix

36. Plastic media TF design
Schulze formula
• The liquid contact time (t) of applied wastewater
Where:
t = liquid contact time, min
D= depth of media (m)
q = hydraulic loading, (m3/m2/h)
C, n = constants related to specific surface & configuration of media
n
q
CD
t 

37. • hydraulic loading (q)
Where:
Q= influent flow rate L/min
A=filter cross section area m2
A
Q
q 

38. Schulze formula for Plastic Media Trickling Filter Design
Where:
Se= BOD concentration of settled filter effluent, mg/L
So= influent BOD concentration to the filter, mg/L
k=wastewater treatability and packing coefficient, (L/s)0.5/m2
D=packing depth, m
q= hydraulic application rate of primary effluent, excluding recirculation, L/m2*s
n=constant characteristic of packing used (assumed to be 0.5).
)
/
( n
q
kD
o
e
e
S
S 


39. Example
Given the following design flow rates and primary effluent wastewater
characteristics, determine the following design parameters for a trickling
filter design assuming 2 reactors at 6.1 m depth, cross-flow plastic packing
with a specific surface area of 90 m2/m3, a packing coefficient n value of
0.5, & a 2-arm distributor system. The required minimum wetting
rate=0.5L/m2*s. Assume a secondary clarifier depth of 4.2m and k value of
0.23.
– Design conditions

40. Solution
– Diameter of tower trickling filter, m
a. Correct k for temperature effect
187
.
0
)
035
.
1
(
23
.
0
)
035
.
1
( 20
14
20
20 

 

T
T k
k
b. Determine the hydraulic loading rate
c. Determine the tower area
d. Determine the tower diameter

  )
/
( n
q
kD
o
e
e
S
S
filter
two
for
each
m
Diameter
m
m
tower
of
No
Area
17
1
.
226
2
/
2
.
452
.
/ 2
2




 ,
A
Q
q
s
m
L
q
q
for
solve
e q



 
 2
)
/
1
.
6
187
.
0
(
/
3875
.
0
".
"
125
25 5
.
0
2
2
2
3
2
.
452
/
3875
.
0
/
2
.
175
/
3875
.
0
/
140
,
15
m
s
m
L
s
L
s
m
L
d
m
q
Q
A 




