chapter 8 trickling filters
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Chapter 8 Trickling FiltersTRANSCRIPT
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TRICKLING FILTERS
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“ATTACHED GROWTH
BIOLOGICAL REACTORS”
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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
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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.
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ALL TRICKLING FILTERS CONSIST OF 3 PARTS:1. Filter medium2. Rotary distributor3. Underdrain system
Filter medium
Rotary distributor
Underdrain system
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Filter medium(i) Provide a surface for biological growth
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Filter medium(ii) Provide voids for passage of liquid and air
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Filter medium
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Rotary distributorProvides uniform hydraulic load on the filter surface
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Underdrain systemCarry away the treated wastewater and the sloughed biomass
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PR
OB
LEM
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PR
OB
LEM
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Disadvantages- Not as efficient as newer treatment
methods - limited future- Large land requirement- Variations in effluent quality- Odour problems, filter flies
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• 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
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• Innovation from trickling filters (essentially deep trickling filters)
• Synthetic / Plastic medium- effectively prevents
blockage- ensures an adequate
flow of air- small area
BIO-TOWERS
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Traditional trickling filters
Biotower
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• Effectively prevents blockage• Ensures an adequate flow of air
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DesignHydraulic loading, Qv
(10.1)
whereQ = flow rate without recirculation,
m3/minA = area
min./ 23 mmA
QQv
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BOD loading
W = li Q (10.2)
where
li = influent substrate concentration, BOD5 mg/L
Q = wastewater flowrate
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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(
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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
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Correction for other temperature can be made by adjusting the treatability factor:
(10.4)
20
20)035.1( T
CT oKK
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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
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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
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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
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The treatability constant must be adjusted for temperature
= 0.055 (1.035)5 = 0.065 min-1
20252025 )035.1( KK
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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
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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:
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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
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Qv0.44 = 0.789
Qv = 0.584 m3/m2.min
44.0)6(065.0494.0vQ
m
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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
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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
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= 22.19 mg/L
le = 22.19 mg/L > 10 mg/L
- not satisfied
ee ll 228.013.17