Biological Treatment I- Trickling Filter

AIR

Sprinkler

FEED PIPE

TRICKLING FILTER

0.9-3.5m

Diameter : 30 to 60 m 0.1-0.2 mm

Organic present in wastewater is degraded by a population of microbes attached to the filter media. 1. Organics will be adsorbed onto the

biological slimy layer (0.1-0.2 mm thick)

2. the organics gets degraded by aerobic microorganism s.

3. Thickness of slime layer increases as the organisms grow in number

4. Oxygen will be consumed before it can penetrate full depth, anaerobic situation will grow, and anaerobes grow.

5. Further growth of slimy layer shall limit the penetration of food. Food will get exhausted before it can go to the anaerobes

6. Anaerobes then die and no longer can cling to the media surface. The whole slimy layer breaks out and comes out with the treated wastewater. This is called Sloughing. Sloughing is a function of organic and hydraulic loading of the filter

ATTACHED GROWTH PROCESS

Wastewater trickles through a bed of highly permeable porous media of plastic or rock containing microbes and treated water gets collected at the bottom. It resembles like a filter bed but operationally it is much different.

Recirculation of wastewater is not for the same reason as Activated sludge process. It does not help to increase the efficiency of the system.

Recirculation is done only to ensure a low-strength influent and constant hydraulic loading rate to maintain a thin layer of biofilm on the media and avoiding the periodic sloughing of the microbes. Recirculation helps to maintain a minimum wetting ratio so that they do not run dry at certain points inside the filter at any particular point of time.

Types of SystemsPartial BOD removal

Intermediate- and High-Rate FiltersRecirculation of the filter effluent or final effluent permits higher organic loadings, provides higher dosing rates on the filter to improve the liquid distribution and better control of the slime layer thickness, provides more oxygen in the influent wastewater flow, and returns viable organisms. Recirculation also helps to prevent ponding in the filter and to reduce the nuisance from odors and flies.

Low-Rate Filters Dosing tanks are small, usually with only a 2-min detention time based on twice the average design flow, so that intermittent dosing is minimized. Even so, at small plants, low night-time flows may result in intermittent dosing and recirculation may be necessary to keep the packing moist. If the interval between dosing is longer than 1 or 2 h, the efficiency of the process deteriorates because the character of the biological slime is altered by a lack of moisture.

Distribution Systems Clearance of 150 to 225 mm should be allowed between the bottom of

the distributor arm and the top of the bed. The clearance permits the wastewater streams from the nozzles to spread out and cover the bed uniformly. The maximum diameter is 60 m. Hence the designer should consider this fact while fixing the size of the trickling filter units.

Nozzles are spaced unevenly so that greater flow per unit of length is achieved near the periphery of the filter than at the center. Headloss through the distributor is in the range of 0.6 to 1.5 m.

The underdrain system for a rock filter usually has precast blocks of vitrified clay or fiberglass grating laid on a reinforced-concrete subfloor

Underdrains

The floor and underdrains must have sufficient strength to support the packing, slime growth, and the wastewater and should slope to a central or peripheral drainage channel at a 1 to 5 percent grade. The effluent channels are sized to produce a minimum velocity of 0.6 m/s at the average daily flowrate.

Underdrains may be open at both ends, so that they may be inspected easily and flushed out if they become plugged

Airflow Natural ventilation has historically been

the primary means of providing airflow, but it is not always adequate and forced ventilation using low-pressure fans provides more reliable and controlled airflow.

In the case of natural ventilation, the driving force for airflow is the temperature difference between the ambient air and the air inside the pores..

upward flow means the air enters from the bottom ventilation port and flow upwards. As it flows upwards, the DO shall get consumed and when it reaches at the top it will be much depleted. This situation is undesirable because the wastewater at the top has the highest oxygen demand and there the concentration of oxygen shall be the least.

If the wastewater is colder than the ambient air, the pore air will be cold and the direction of flow will be downward. If the ambient air is colder than the wastewater, the flow will be upward.

FILTER

Primary Clarifier Secondary Clarifier

ONE-STAGE WITH RECIRCULATION

FILTER STAGE 1

Primary Clarifier Secondary

ClarifierSecondary Clarifier

TWO-STAGE WITH RECIRCULATION

FILTER STAGE 2

Qr

Q, Si Q, S

Process Design For Trickling Filter

For Trickling Filters with Rock Media: (NRC Equation): Single Stage

VFW

E

44.01

100

E= efficiency of BOD removal in the filterW= BOD loading to the Filter, kg/dayV= volume of the filter media, m3

F= Recirculation FactorR= Recirculation ratio

2)10/1(

1

R

RF

Q

Qr

For Trickling Filters with Rock Media: (NRC Equation): Two Stage

2121

21

21

44.01

100

FVW

E

E21= efficiency of BOD removal in the first filter of two stage processW21= BOD loading to the Filter1 of 2 stage process, kg/dayV21= volume of the filter media, m3

F= Recirculation FactorR21= Recirculation ratio in filter 1 of 2 stage filtration 2222

22

21

22

144.0

1

100

FVW

E

E

For Trickling Filter Made of Plastic Media

Eckenfelder’s Equation

])(exp[ nv

ma

i

QDKSS

S

Germain’s Equation

])(exp[ ,20n

vDi

QDkS

S

S= BOD5 of the settled effluent, mg/LS0= Total BOD5 of wastewater influent to the filter, mg/LD = Depth of the filter. MQv = Q/AA= surface area of the filterSa = Specific surface area of the filter, = surface area/ volume.K= observed reaction rate constant, m/dk20,D= treatability constant for a filter with a specific depth Dn= experimental constant, usually 0.5x= 0.5 for vertical and rock media filters =0.3 for cross-flow plastic media filters

Temperature correction for k is same as that of BOD, but θ= 1.035 (not 1.047)

x

DD D

Dkk

2

11,202,20

Design a trickling filter using rock media to treat wastewater with flowrate 4.5 million liters per day. BOD of the raw sewage 250 mg/L. BOD removed in the primary clarifier is 25%. The treated wastewater needs to discharged to a surface water body where the regulatory limit for BOD5 is 30 mg/L. The fluctuation in the wastewater flowrate can be controlled by keeping a recirculation ratio of 1.4.

Influent BOD5 = 75% of 250 mg/L =187.5 mg/LEffluent BOD5= 30 mg/L

Efficiency, E = %84100*5.187

305.187

VF

WE

44.01

100

2)10/4.11(

4.11

F

846.1W= BOD loading to the Filter, kg/day =4.5*106*187.5/106

=843.75 kg/day

846.1*75.843

44.01

10084

V

3m 96.2438V

Assume a depth of 1.5 m m 5.455.1

96.243844

HV

Dia

It is proposed to use a two stage plant instead of the single stage plant in the previous example. The total volume of filter media remains the same and gets equally divided, (i.e each filter is to contain half of the filter media as earlier). Use the same recirculation ratio. Find out the effluent BOD5.

332221 m 5.12192/m 96.2438 VV

22221 )10/4.11(

4.11

FF

846.1

2121

21

21

44.01

100

FV

WE

W21= BOD5 loading to the first Filter, kg/day =4.5*106*187.5/106

=843.75 kg/day

%77.78

846.1*5.121975.843

44.01

10021

E

W22=BOD loading of the 2nd filter =BOD of wastewater in the effluent of first stage

%11.63

846.1*5.121905.179

144.0

1

100

21

22

E

E

Effluent BOD5 = (1-0.631)*39.80 = 14.68 mg/L

Effluent BOD5 after 1st stage = (1-0.7877)*187.5 = 39.80 mg/L

=(1-0.7877)*843.75= 179.05 kg/day

Design a two stage trickling filter process for wastewater with influent BOD5=250 mg/L. The treated effluent should have a BOD5 of 30 mg/L. For both the filters use same depth of 2 m and recirculation ratio 2. The quantity of wastewater is 8000 cum/day. Assume that efficiencies of the two filters are the same.

Overall efficiency = E = %88100*250

30250

21 EE

FILTER STAGE 1

FILTER STAGE 2

Si

E1Se1 Se

E2

Overall efficiency =E

i

ei

S

SSE

ie SES )1(

ie SES )1( 11,

iee SEESES )1)(1()1( 121,2

ii SEESE )1)(1()1( 12

65.021 EE

Overall Efficiency

After First Stage,

After second Stage,

Equating Se from above

Stage 122221 )10/21(

21

FF

083.2

W21= BOD5 loading to the first Filter, kg/day =8000*103*250/106

=2000 kg/day

2121

21

21

44.01

100

FV

WE

083.2*2000

44.01

10065

21V

321 m 02.641V

m 2.202

02.64144 2121

H

VD

Stage 2 W22= BOD5 loading to the second Filter, kg/day =8000*103*250(1-0.65)/106

=700 kg/day

2222

22

21

22

144.0

1

100

FVW

E

E

083.2*700

65.0144.0

1

10065

22V

322 m 76.1831V

H21 = 2 m

m 14.342

76.183144

22

2222

HV

D H22 = 2 m

1. Compute BOD5 loading to each of the filters2. Compute hydraulic loading onto each of the filters