TRICKLING FILTERS

“ATTACHED GROWTH

BIOLOGICAL REACTORS”

4

Trickling filters : a reactor in which randomly packed solid forms provide surface area for biofilm growth.

A biofilm is a complex aggregation of microorganisms growing on a solid substrate

Over 90% of all bacteria live in biofilms.

Biofilms can form on just about any imaginable surface: metals, plastics, natural materials (such as rocks), medical implants, kitchen counters, contact lenses, teeth, the walls of a hot tub or swimming pool.

ALL TRICKLING FILTERS CONSIST OF 3 PARTS:1. Filter medium2. Rotary distributor3. Underdrain system

Filter medium

Rotary distributor

Underdrain system

Filter medium(i) Provide a surface for biological growth

Filter medium(ii) Provide voids for passage of liquid and air

9

Filter medium

Rotary distributorProvides uniform hydraulic load on the filter surface

Underdrain systemCarry away the treated wastewater and the sloughed biomass

PR

OB

LEM

PR

OB

LEM

16

Disadvantages- Not as efficient as newer treatment

methods - limited future- Large land requirement- Variations in effluent quality- Odour problems, filter flies

• Often operated in series to increase the effective contact time with the biofilm.

• Generally followed by a secondary clarifier to remove sloughed solids.

• CONTACT TIME and ventilation can be improved by recirculation

19

• Innovation from trickling filters (essentially deep trickling filters)

• Synthetic / Plastic medium- effectively prevents

blockage- ensures an adequate

flow of air- small area

BIO-TOWERS

Traditional trickling filters

Biotower

• Effectively prevents blockage• Ensures an adequate flow of air

22

DesignHydraulic loading, Qv

(10.1)

whereQ = flow rate without recirculation,

m3/minA = area

min./ 23 mmA

QQv

23

BOD loading

W = li Q (10.2)

where

li = influent substrate concentration, BOD5 mg/L

Q = wastewater flowrate

24

Eckenfelder formula (with recirculation)

(10.3)

Wherele = effluent substrate concentration, BOD5 mg/LD = depth of the medium, mk = treatability constant relating to the wastewater and the

medium characteristics, min-1 = 0.06 min-1 n = coefficient relating to the medium characteristics = 0.5

nv

nv

QkD

QkD

a

e

R

e

l

l

Re)1(

25

R = ratio of the recycled flow to the influent flow = Qr /Q

Qr = recirculation flow rate

la = BOD5 of the mixture of raw and recycled mixture

=

R

Rll ei

1

Correction for other temperature can be made by adjusting the treatability factor:

(10.4)

20

20)035.1( T

CT oKK

ExampleAn existing 6 m deep tower trickling filter plant cannot meet the effluent standard of less than 10 mg/L. The plant influent BOD is 220 mg/L, the primary effluent BOD is 150 mg/L, the plant final effluent is 30 mg/L and the recirculation flow is 1.5 of the wastewater flow entering the plant. One proposal is to change the recirculation ratio, to reduce the effluent BOD from 30 to 10 mg/L. The constants for the random plastic media, n of 0.44 and K20 of 0.055 min-

1. Design temperature is 25oC.i. Calculate the hydraulic loadingii. By using the recirculation ratio = 2.0, check whether the plant effluent meet the standard

Flow diagram

Bio-tower(trickling filters)

Secondary Clarifier

Plan

t infl

uent

BO

D5:

220

mg/

L

D : 6mn: 0.44

25°C

Primary Clarifier li :

150 mg/L

la :?

Recirculation rate: 1.5

le :30

mg/L

Target le : < 10 mg/L

Bio-tower(trickling filters)

Secondary Clarifier

Plan

t infl

uent

BO

D5:

220

mg/

L

D : 6mn: 0.44

25°C

Primary Clarifier li :

150 mg/L

la :?

Recirculation rate: 1.5

le :30

mg/L

Target le : < 10 mg/L

Find la (BOD5 of the mixture of raw and recycled mixture)

R

Rlll ei

a

1

5.11

)30(5.1150la

= 78 mg/L

30

The treatability constant must be adjusted for temperature

= 0.055 (1.035)5 = 0.065 min-1

20252025 )035.1( KK

31

Find the hydraulic loading, Qv - using Eq. 10.3

nv

nv

QkD

QkD

a

e

R

e

l

l

Re)1(

Where;le = 30 mg/Lla = 78 mg/LD = 6mk = 0.065 min-1 n = 0.44

32

44.0

44.0

)6(065.0

)6(065.0

Re)1(78

30

v

v

Q

Q

R

e

44.0)6(065.0vQ

m Make it simple, if

m

m

Re)R1(

e

78

30

So the Eq. become:

m

m

e

e

5.15.239.0

then,

1.585e-m = 0.975 e-m = 0.61 m = � ln 0.61

and, m = 0.494

34

Qv0.44 = 0.789

Qv = 0.584 m3/m2.min

44.0)6(065.0494.0vQ

m

By using the recirculation ratio, R = 2.0, check whether the plant effluent meet the standard (less than 10 mg/L BOD5)

44.0

44.0

584.0)6(065.0

584.0)6(065.0

2)21(

e

e

l

l

a

e

343.0610.023

610.0

x

343.0

32150

e

e

a

e

ll

l

l

R1

Rll ei

R = 2 ; then,

343.0

32150

e

e

a

e

ll

l

l

37

= 22.19 mg/L

le = 22.19 mg/L > 10 mg/L

- not satisfied

ee ll 228.013.17