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)

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+

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

Trickling Filter (TF)- side view

Wastewater

rotating distributor arms

Packing media

Underdrain

rocks, plastic, or

other material

Packing Media- Filter media

Flow Diagram for Trickling Filters

Recycle

Primary

clarifierTrickling

filter

Final

clarifier

Waste

sludge

Final

effluentInfluent

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

Single stage configuration

PC SC TF

PC SC TF

PC SC TF

a.

b.

c.

Two stage configuration

PC SC TF

PC SC TF

TF

TF SC

PC SC TF TF SC

Stone media TF design

PC SC TF2 TF1

VF

wE

1

1

4432.01

100

VF

w

E

E2

1

2

1

4432.01

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

RF

• 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 EE

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

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

(3) Effluent BOD (mg/L)

– BOD removal efficiency for first-stage filter at 20oC, %

%7.75)035.1(2.81)035.1( 22018

2018 EE

VF

wE

1

1

4432.01

100

36.1)10/5.01(

5.01

)10/1(

122

R

RF

%2.81

36.1

37.04432.01

100

4432.01

100

1

1

VF

wE

100

)7.75100(/130)/(

LmgLmgBODEffluent = 31.6

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

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.

RBC components

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.

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.

dm

leso

2

lub

dm

BODg

2

5 )(

dm

gN

2

dm

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 .

Hydraulic loading rate:

The typical hydraulic loading rate of 0.08-0.16 23 / mm for 5BOD removal

and 0.03-0.08 23 / mm for both 5BOD removal and nitrification.

The hydraulic detention time ( ) is 0.7-1.5 hrs for 5BOD removal and

1.5-4 hrs for both 5BOD removal and nitrification.

The volume of RBC tank has been optimized at 0.004923 / mm for one

shaft of 93002m .

A tank volume of 45 3m is needed. Based on this volume and a hydraulic

loading rate of 0.08 dmm ./ 23 the detention time is 1.44 hrs. Atypical side

wall depth is 1.5m to achieve 40% submergence.

Bar

screen

Grit

removal

Primary

Sedimentation

RBC

units Final

clarifier

effluent

1st stage

2nd

stage

RBC configuration

Design equation of RBC: The following empirical equation developed by Optaken( US EPA,1985 ):

)(0195.0

)(039.011 1

Q

A

SQ

A

Ss

ns

n

Where nS = soluble 5BOD concentration in stage(n), (mg/L)

sA = disk surface area on stage(n), 2m

Q= flow rate, dm /3

Rate of nitrification is related to the soluble BOD load/m2

dm

NgBODrn

25 )(1.015.1

RBC design Example: Design a rotating biological contractor to treat an

influent soluble 5BOD of 90 mg 5BOD /L. The flow(Q)= 4000 3m /d

Solution:

Assume 1st stage ( 5BOD ) organic loading= 15g/ 2m .d

5BOD (loading)= ( 5BOD ) concentration inQ

=d

g

mg

g

m

L

d

m

L

mg000,360

10

10400090

33

33

Disk area= 2

2

24000

.15

000,360m

dmg

dg

(first area stage)

Use shaft

m2

9300 so number of shafts needed for the first stage:

N= 6.29300

240002

2

shaftm

m say 3 shafts

Calculate S1, the BOD concentration after the first stage:

Sn=)(0195.0

)(039.011 1

QAs

SQ

Asn

For the first stage n=1, Sn=S1, Sn-1 = S0

S0 = 90 mg/L, AS = dm

Qm3

2 4000,2790093003

md

dm

m

Q

AS 98.6/4000

279003

2

S1=L

mg

L

mg1475.29

98.60195.0

9098.6039.011

So we need one more stage.

Add another stage and calculate S2:

Assume two shafts in the second stage:

AS=22 1860093002 mm

d

m

Q

AS 65.44000

18600

Sn=S2, Sn-1=S2-1=S1=29.75mg/L

S2=L

mg

L

mg1486.16

65.40195.0

75.2965.4039.011

So we need one more stage.

Add another stage and calculate S3:

Assume one shaft in the third stage:

AS=22 930093001 mm

m

d

d

m

m

Q

AS 33.2

4000

93003

2

Follow Example RBC design example:

Sn=S3,Sn-1=S3-1=S2=16.86mg/L

S3=L

mg

L

mg1413

33.20195.0

86.1633.2039.011

OK

So three stages are enough.

Check for the hydraulic loading:

HLR=shafteachofAreashaftsofnumbertotal

Q

Nshafts=3+2+1= 6 shafts

HLR=dm

md

m

2

3

3

072.093006

4000

, typical range(0.08-0.16), which is a

little bit lower than the range.

Is nitrification possible in any of the three stages?:

*Nitrification is only possible when soluble 5BOD loading is less than

10g dm

BOD

2

1st stage =

dm

gBOD

m

g

d

m

2

5

3

3

9.1293003

1904000 <10(no

nitrification)

2nd

stage = dm

gBOD

m

g

d

m

2

5

3

3

4.693002

175.294000

(nitrification occurs)

3rd

stage = dm

gBOD

m

g

d

m

2

5

3

3

25.793001

186.164000

(nitrification occurs)

*Rate of nitrification is related to the soluble 5BOD loading by the

following equation:

dm

NgBODrn

25 )(1.015.1

* So for 2nd

stage→ dm

Ngrn

254.04.61.015.1

* And for 3rd

stage→ dm

Ngrn

2413.025.71.015.1

*If the ammonia concentration in the influent to the 2nd

stage is 30

mgN/L, find the effluent ammonia concentration.

dm

Ngrn

254.0 for 2

nd stage

Nitrogen removal = d

gm

dm

Ng10044)93002(54.0 2

2

Concentration = L

mg

m

d

d

g51.2

4000

100443

So→ N2 = L

mg5.2751.230

→ rn = 0.413 dm

Ng

2 for 3

rd stage,

Nitrogen removal = d

gm

dm

gN3841)93001(413.0 2

2

Concentration = L

mg

m

d

d

g96.0

40003841

3

So→ N3 = 27.5 L

mgN

L

mgN5.2696.0

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.

Appendix

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

nq

CDt

• hydraulic loading (q)

Where:

Q= influent flow rate L/min

A=filter cross section area m2

A

Qq

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).

)/( nqkD

o

e eS

S

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

Solution

– Diameter of tower trickling filter, m

a. Correct k for temperature effect

187.0)035.1(23.0)035.1( 201420

20 T

T kk

b. Determine the hydraulic loading rate

c. Determine the tower area

d. Determine the tower diameter

)/( nqkD

o

e eS

S

filtertwoforeachmDiameter

mmtowerofNoArea

17

1.2262/2.452./ 22

,A

Qq

smLqqforsolvee q 2)/1.6187.0( /3875.0"."125

25 5.0

2

22

3

2.452/3875.0

/2.175

/3875.0

/140,15m

smL

sL

smL

dm

q

QA