7-trickling filter f11

7-Trickling Filter_F11.doc

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Attached Growth Reactors (4th

ME 887)

Attached growth processes:

1) Trickling filters

2) Rotating biological contactors (RBC), Submerged RBC (SBC)

3) Combined attached and suspended growth processes

Other attached growth processes:

4) Activated-sludge processes with fixed-film packing

5) Submerged attached growth processes

6) Attached growth denitrification processes

Classification

Classes based on physical settings:

1) Non-submerged attached growth processes

2) Suspended growth processes with fixed-film packing

3) Submerged attached growth aerobic processes

Classes based on oxygen

1) Aerobic Processes

a. Trickling Filters, TF (High Rate, Low Rate, Roughing Filters)

2) Rotating Biological Contactors, RBC

3) Packed-bed Reactors

4) Activated Biofilter, ABF

5) Biological Tower

2) Anoxic Process

1) Fixed-film denitrification

3) Anaerobic Processes

1) Anaerobic filter process

2) Expanded bed

Non-submerged attached growth processes

Principal advantages and disadvantages of aerobic attached growth processes over the

activated sludge process


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Advantages

1) Less energy required

2) Simpler operation with no issues of mixed liquor inventory control and sludge

wasting

3) No problems of bulking sludge in secondary clarifiers

4) Better sludge thickening properties

5) Less equipment maintenance needs

6) Better recovery from shock toxic loads

Disadvantages

For rock filters

1) Poor effluent quality in terms of BOD and TSS concentrations

2) Greater sensitivity to lower temperatures

3) Odor production

4) Uncontrolled solids sloughing events

Actual limitations:

1) make it difficult to accomplish biological nitrogen and phosphorus removal compared

to single-sludge biological nutrient removal suspended growth designs

2) results in an effluent with a higher turbidity than activated-sludge treatment.

Suspended growth processes with fixed-film packing

- the placement of packing material in the aeration tank of the activated-sludge

process

The advantages for the activated sludge process:

1) Increased treatment capacity

2) Greater process stability

3) Reduced sludge production

4) Enhanced sludge settleability

5) Reduced solids loadings on the secondary clarifier

6) No increase in operation and maintenance costs

Submerged attached growth aerobic processes

Theses are upflow and downflow packed-bed reactors and fluidized-bed reactors that do

not use secondary clarification.

To be continued …..


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Trickling Filter (4th

ME 890)

1893 First operation in England (3rd

ME 403)

A. Basic Mechanism

1. Conceptual model - See Diagram

2. Biofilms

a. The biological

slime layer consists of:

1) aerobic,

anaerobic, and

facultative bacteria,

protozoans, fungi, and

higher animals (e.g.,

sludge worms, filter-

fly larvae, rotifers,

snails, etc.)

b. Facultative system

1) Although TF

is classified as an

aerobic treatment

system, primary

organisms are

facultative because

aerobic-anaerobic

occur interchangeably.

2) Facultative

bacteria are

predominating

microorganisms.

c. Thin aerobic film

1) Although TF is classified as an aerobic treatment device, the microbial film on the

filter medium is aerobic to depth of only 0.1-0.2 mm. The zone next to the

medium is anoxic to anaerobic.

NH4+

Nutrient

N, P

Organic carbon

BOD

Mediu

m S

urfa

ce Air

O2 O2

Reoxygenation

: : Wastewater

Metabolic

end products

CO2 CO2

NO3-

Anaerobic Aerobic

0.1 - 0.2 mm

Biofilm

Schematic diagram of biofilm and biological process

on the medium surface in a Trickling Filter


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d. Common organisms

1) Bacteria - Achromobacter, Flavobacterium, Pseudomonas, Alcaligenes

2) Filamentous forms - Sphaerotilus natans, Beggiatoa

3) Nitrifying bacteria - Nitrosomonas, Nitrobacter

4) Fungi - Fuzarium, Mucor, Penicillium, Geotrichum, Sporatichum

5) Algae - Phormidium, Chlorella, Ulothrix

6) Protozoa - predominantly ciliata group, Vorticella, Opercularia, Epistylis.

3. Biological variation with depth

a. Algal uptake at the upper surface - The surface of the bed may support algal growth when temperature and sunlight

conditions are optimum. - Do not take a direct part in waste degradation, but add oxygen

- troublesome because they can cause clogging of the filter surface (which produce

odors).

b. Nitrification at the bottom

- The lower portion of a deep filter frequently supports populations of nitrifying

bacteria.

c) Higher organisms

- Snails are troublesome, they consume nitrifying bacteria.

4. Growth Phases

a. Exponential growth phase near surface of the bed.

- Microorganisms near the surface of the bed, where food concentration is high, are in

a rapid growth phase.

b. Declining death phase near the bottom. - The lower zone of a bed is in a state of starvation. c. Endogenous growth - Overall operation of a trickling filter may be considered in the endogenous growth

phase.


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b. Filter media 1) provide:

i) a surface for biological growth and ii) voids for passage of liquid and air. 2) the most common media are: i) crushed rock, ii) slag, or iii) field stone - these materials are durable, insoluble, and resistant to spalling. 3) the preferred range of size for the stone media is: i) 3 - 5 in. in diameter (VH) ii) 1 - 4 in (25 - 100 mm) ME, p 404

depth of rock bed = 3 - 8 ft (0.9 - 2.5 m) = 6 ft (1.8 m avg) ME p 404

Natural

draft air

100 ft (30 m)

Wastewater id distributed

by hydraulic head

Filter media 5-7 ft

Underdrain

Effluent

Influent Underdrain collects treated WW

and supply air (O2)


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4) plastic media (Fig. 12.32, VH., p.479; Fig. 12.33, p. 481) i) depth = 14 - 40 ft (4 - 12 m) ME p. 404 ii) types vertical-flow packing (Fig 8-23); cross-flow packing (Fig 10-33); variety of random packings (Fig 10-33) iii) have advantages of: . light weight . chemical resistance . high specific surface (ft

2/ft

3 or m/m

3) with a large percentage of free space.

c. Underdrain system 1) Carries away the effluent - collecting the treated wastewater and any biological solids that have become

detached from the media - carries away the effluent 2) Circulates air - permits circulation of air through the bed. - the need for free passage of air controls the size of opening in the underdrain. 3) Recycle line - to dilute the strength of the incoming wastewater - to maintain the biological slime layer in a moist condition.


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d. Rotary distributor 1) Provides uniform hydraulic load - provides a uniform hydraulic load on the filter surface. 2) Require pressure head (minimum 24 inches) - it is driven by the reaction of the wastwater flowing out of the distributor

nozzles. - requires a minimum pressure head of 24 in. measured from the center of the

arms. 3. Operation a. Spray - primary effluent is sprayed on a bed of media (e.g., crushed rock or plastic

media) coated with biological films. b. Microbial metabolism - As the wastewater flows over the microbial film, the soluble organics are rapidly

metabolized and the colloidal organics adsorbed onto the surface. c. Reoxygenation - Dissolved oxygen extracted from the liquid layer is replenished by reoxygenation

from the surrounding air. d. Sloughing - losing the slime layer - a function of the organic and hydraulic loading on the filter. 4. Operational problems a. Undesirable anaerobic conditions - Undesirable anaerobic conditions can be created in a trickling filter by inhibiting

aeration of the bed. b. Plugging - Plugging of the air passages with excess microbial growth, as a result of organic

overload, can create anaerobic and foul odors. - organic overload 5. Air movement 1) (summer) If air temperature > waste temperature, Tair > Tw, air moves down through the filter bed 2) (winter) If air temperature < waste temperature,

Tair < Tw, air moves up through the filter bed 3) (spring & fall) If air temperature = waste temperature, Tw = Tair, no ventilation, no air movement * June and august are transition zone, then system become anoxic


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6. Advantages and Disadvantages of TF Advantages: a. Low energy input b. accepts qualitative and quantitative shock load very well c. accepts toxic load well d. good sludge settling at secondary clarifier Disadvantages: a. land requirement - much land needed b. performance 1) fair in summer – mediocre, 30 mg/L BOD in effluent 2) poor in winter - can increase performance by covering . cost for cover . corrosion . condensation 5. Level of Treatment a. Secondary treatment systems 1) Normally, TFs are proceeded by primary treatment 2) the standard stone-filled filter must be preceded by a sedimentation tank equipped

with a scum-collecting device. 3) the system includes a final settling tank to remove biological growths that are

washed off the filter media. 4) the sloughed solids are commonly disposed of through a drain line from the bottom

of the final clarfier to the head end of the plant. 5) this return sludge flow is mixed with the raw wastewater and settled in the primary

clarifier. 6. Classification of TF (ME, p. 615, Table 10-13)

- TFs are classified by: a) hydraulic loading rate

b) organic loading rate Classifications are: 1) Low-rate or standard rate (Fig. 12.26, VH p.466) 2) Intermediate-rate 3) High-rate (Fig 12.27, VH p.467) 4) Super high-rate 5) Roughing filters 6) Two-stage (Fig 12.29, VH p. 468)

or 1) Single-stage 2) Two-stage


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Low-rate TFs (Fig. 12.26, VH p.466) - or Standard Rate 1) rarely used in new designs 2) the wastewater passes through only once 3) the effluent is then settled prior to disposal. 4) the sludge line is operated, generally once or twice a day, to waste the accumulated

settled solids. 5) operate intermittently, dosing and resting. This operation is required bacause of the

low hydraulic load on the filter. - low night time flow may result in intermittent dosing

6) a dosing siphone, or similar alternating flow-control system, must be installed ahead of the filter to provide an adequate flow rate to turn the reaction-type rotary distributor.

- otherwise, the distributor arm would stop turning during periods of low, such as at night, and wastewater would trickle down only under the stalled arm.

- constant hydraulic loading is maintained by suction level controlled pumps or a dosing siphon (not by recirlulation).

Intermediate and High-rate TFs (Fig 12.27, VH p.467)

1) raw wastewater is diluted with recirculation flow so that it is

passed through the filter more than once.

2) a return line from the final clarifier serves a dual function,

as a sludge return and a recirculation line.

3) the combined flow (Q+QR) through the TF is always sufficiently

great to turn the distributor, and dosing siphone is not needed.

4) a high-rate filter is dosed continuously.

Super High-rate TFs

1) high hydraulic loading and great filter depths (ME, Fig 10-32)

2) plastic media

3) most are in the form of packed towers

Roughing filters

1) Organic loading > 100 lb BOD/1000 ft3. d (1.6 kg/m3.d)

- because of the heavy organic loading on roughing beds, their

design must consider the problem of bed plugging.

2) Hydraulic loading is up to 3.2 gal/ft2.min (187 m3/m2.d)

3) Most use plastic media

- the high percentage of void volume in manufactured media permits their application

and use in roughing filters.

4) Treat wastewater prior to secondary treatment.

- those receiving unsettled wastes, as well as other used for

pretreatment of settled strong wastes prior to subsequent

aeration, are referred to as roughing filters.

Two-stage TFs (Fig 12.29, VH p. 468)

a. Two filters in series, usually with an intermediate settling tank.

b. Super high rate-Roughing filters

c. Two-stage filters are used

1) where a high-quality effluent is required,


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2) for treatment of strong wastewater

3) to compensate for lower bacterial activity in treating cold

wastwater.

4) where nitrification is required.

Recirculation (ME, p. 617, Figure 10-31)

Figure 10-31 (ME, p. 617) shows intermediate-rate and high-rate trickling-filter flow

diagrams with various recirculation patterns:

(a) single-stage filters and (b) two-stage filters.

1) the recirculation ratio is the ratio of recirculated flow to the

quantity of raw wastewater

2) a common range for recirculation ratio values is 0.5-3.0

QR

R = ------ = 0.5 - 3.0

Q

3) Purposes of recirculation

a. dilute very strong wastes

b. dilute toxic waste

c. minimize Q and BOD flactuations to filters

d. Return sludge for collections

e. seeding effect - seed input w/ some of viable microbes

4) Recirculation Control

- recirculation is done

a. only during periods of low wastewater flow

b. at a rate proportional to raw-wastewater flow

c. at a constant rate at all times

d. at two or more constant rates predetermined automatically

or by manual control.

5) Recirculation Patterns (Fig 12.28, VH p.467)

a. recirculation with sludge return

b. direct recirculation around dilter

c. recirculation of plant effluent

d. dual recirculation.


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Loading Standards for Stone-Media Trickling Filters a. Standards for Trickling Filters with Stone, Crushed Rock, or Slag Media (Table 12.2, VH p.469) b. The BOD load on a TF is calculated using the raw BOD in the primary effluent applied to the filter, without regarding to the BOD in the recirculated flow. * Criteria for two-stage filters based on the total volume in both filters. c. BOD loadings are expressed in terms of lbs of BOD applied /day lbs of BOD applied/day ------------------------ or ---------------------- 1000 ft

3 of Volume acre-ft

* The hydraulic load is computed using the raw-wastewater flow plus recirculation flow. d. Hydraulic loadings are expressed in terms of million gallons applied average flow (in gpm) applied ---------------------------- or ----------------------------- acre of surface area per day ft

2 of surface area per day

d. Current values used in design (VH, p. 469, Table 12.2) -------------------------------------------------------------------- Trickling BOD loading Hydraulic loading Recirculation Filter lb/1000 ft

3 g/m

3 gpm/ft2 Ratio

--------------------------------------------------------------------- Low-rate 15 (5-25) 240 0.05 (0.03-0.06) 0 High-rate 35 (25-45) 560 0.25 (0.16-0.48) 1.5 (0.5-3.0) Two-stage 55 (45-65) 880 0.25 (0.16-0.48) 1.5 (0.5-3.0) --------------------------------------------------------------------- In a low-rate trickling filter, the relationship between BOD load and hydraulic load depends on the strength of the applied wastewater. - this relationship exists because there is no recirculation flow and depth of the bed is limited to 5-7 ft. 2. Efficiency equations for stone-media TFs a. National Research Council (NRC) formula

1) Based on data collected from filter plants at military installations in the United

States in the early 1940s.

2) The NRC formula for a single-stage TF

100 E

E = ------------------------ ==> ---- = E1

w 100

1 + 0.0561 (---)0.5

VF

where E = BOD removal at 20oC, %

w = BOD load applied, lb/day

V = volume of filter media, 103 ft

3 (ft

3 x 10

-3)

F = recirculation factor

w

--- = volumetric BOD loading, lb/1000 ft3/day

V


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1 + R

F = -------------

(1 + 0.1 R)2

recirculation flow QR

where R = recirculation ratio = --------------------- = ---

raw wastewater flow Q

c. The NRC formula for the second stage of a two-stage TF

100

E2 = ------------------------

0.0561 w2

1 + (------)(----)0.5

1-E1 VF

where E2 = BOD removal of the second stage at 20

oC, %

E1 = fraction of BOD removal in the first stage,

w2 = BOD load applied to the second stage, lb/day

w2

--- = BOD loading, lb/1000 ft3/day

V

c. Effect of wasewater temperature on stone-filled TF efficiency

E = E20 (1.035)

T-20

where E = BOD removal efficiency at temperature T, oC

E20 = BOD removal efficiency at 20

oC

Examples 12.4 & 12.6 (VH p. 470)

Design low rate TF (Low Rate Filter = 15 lb/1000 ft3 day)

The raw-wastewater flow from a municiparity is 1.5 mgd with an average BOD

strength 180 mg/L.

Use a design BOD loading of 15 lb/1000 ft3/day, a hydraulic loading of 2-4 mil

gal/acre/day and a depth between 5 and 7 ft.

Assume that 1) overflow rates of 600 gpd/ft2 for the primary clarifier and 800

gpd/ft2 for the final clarifier, 2) the BOD removal in the primary clarifier is 35%, and

(3) the recirculation ratio R = 0.

1. Determine the diameter and depth dimensions for a low-rate trickling

filter secondary.

2. Calculate the 17 oC BOD removal efficiency a low-rate trickling

filter secondary.

(Solution)

Given: Raw wastewater flow = 1.5 MGD

Average BOD in raw sewage = 180 mg/L

Design BOD loading = 15 lb/1000 ft3/day


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Hydraulic loading = 2 - 4 MG/acre/day

Filter depth = 5 - 7 ft

Overflow rate = 600 gpd/ft2 for the primary clarifier

= 800 gpd/ft2 for the final clarifier

BOD removal efficiency

in the primary clarifier = 35%

Recirculation ratio =, r = 0

a. The BOD loading to the primar settling tank

lb/MG

(180 mg/L)(1.5 MGD)(8.34 ----- ) = 2252 lb/day

mg/L

a. The BOD load on the TF

BOD load = (1-0.35)(2252 lb/day = 1464 lb/day

b. The hydraulic load to the TF

Q = 1.5 MGD

c. Volume of filter media required

BOD loading 1464 lb/day

V = ------------------- = ------------------- = 97,600 ft3

Design BOD loading 15 lb/1000 ft3/day

d. Surface area of the filter

i) At 5 ft depth,

Volume 97,600 ft3

A = -------- = ---------- = 19,520 ft2

Depth 5 ft

19520 ft2 1 acre

= (---------)(---------) = 0.45 acre

43560 ft2

Q 1.5 MGD

Hydraulic load = --- = ---------- = 3.4 MGD/acre

A 0.45 acre

(satisfy thecriteria: 2 -4 Mgal/acrea-day)

ii) At 6 ft depth,

Volume 97,600 ft3

A = ------- = ---------- = 16,270 ft2 = 0.37 acre

Depth 6 ft

Hydraulic load

Q 1.5 MGD

--- = ---------- = 4.1 MGD/acre (Little too high)

A 0.37 acre

e. Use two 100-ft-diameter filters

97,600 ft3

V = 3.14 r2 h = (3.14)(50 ft)

2(h) = ---------- = 48,800 ft

3

2

48,800 ft3

h = ------------- = 6.2 ft


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(3.14)(50 ft)

2. Calculate the 17oC BOD removal efficiency

a. Use the NRC formula for a single-stage trickling filter

100 100

E = -------------------- = ---------------------- = 82.2%

1 + 0.0561 (w/VF)0.5 1 + 0.0561 (15/1)

0.5

where E = BOD removal efficiency at 20oC, %

w/V = BOD loading, lb/1000 ft3/day = 15


1 + R 1 + 0

= ------------ = ---------- = 1.0

(1 + 0.1R)2 [1+0.1(0)

2

QR

R = recirculation ratio = ---- = 0

Q

b. Effect of temperature

E = E20 (1.035)T-20

= (82.2%)(1.035)17-20

= 74%

c. The plant efficiency (overall efficiency)

ET = 100 - 100 [(1 - 1o efficiency)(1 - 2

o efficiency)]

= 100 - 100 [(1 - 0.35)(1-0.74)]

= 83 %


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Loading to Trickling Filter

Q So Q NoVolumetric BOD Loading Volumetric TKN Loading

V V= =

Q So Q NoSpecific BOD Loading Specific TKN Loading

A A= =

Specific surface area, m2/m

3 Table 9-2 (4

th ME 898)

Total surface area, m2 = (Specific surface area, m

2/m

3) (Total volume m

3)

Example 9-2 (4th

ME 911) Trickling Filter Loading

A 10-m diameter single-stage trickling filter contains conventional cross-flow plastic

packing at a depth of 6.1 m. Primary effluent with the characteristics given below is

applied to the filter.

Primary effluent wastewater characteristics

------------------------------------------

Parameter Unit Value

------------------------------------------

Flow rate m3/d 4000

BOD g/m3 120

TSS g/m3 180

TKN g/m3 25

------------------------------------------

Note: g/m3 = mg/L

1) What is the volumetric BOD and TKN loading?

2) What is the specific TKN loading?

3) What is the approximate BOD removal efficiency at 20°C?

4) Can nitrification be expected?

(Solutions)

Given:

a. Filter depth, D = 6.1 m

b. Filter diameter = 10 m


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1. Determine the volumetric BOD and TKN loading

Volumetric BOD LoadingQSo

VVolumetric TKN Loading

QNo

V= =

a) Determine the volume of the trickling filter packing material.

A = π r2 = π (10/2 m)

2 = 78.5 m

2

V = A D = π r2 D = π (10/2 m)

2 (6.1 m) = 478.85 m

3 = 479 m

3

b) Determine the BOD loading

3

3 3

3 3

4000 120 1

10 1.00

479 .

m g kg

d m gQ So kgBOD loading rate

V m m d

= = =

2. Determine the TKN loading rate.

TKNQ So

V

m

d

g

m

kg

g

m

kg TKN

m d= =

=

4000 25 1

10

479

0 21

3

3 3

3 3

.

.

3. Estimate the approximate BOD removal efficiency.

From Fig. 9-8 (4th

ME 910), at a loading of 1.0 kg BOD/m3.d, the BOD removal

efficiency is about 82 %.

4. What is the specific TKN loading?

From Table 9-2 (4th

ME 898), the specific surface area of the packing (plastic-

conventional) is ~ 90 m2/m

3

Total surface area, AT = (90 m2/m

3)(479 m

3) = 43,110 m

2

Q No (4000 m3/d) (25 g/m

3)

Specific TKN loading = ----------- = -----------------------------

AT 43,110 m2

= 2.3 g TKN/m2. d


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5. Can nitrification be expected?

Based on the data given in Table 9-5 (4th

ME 909),

• the BOD loading (1.0 kg/m3-d) for combined BOD removal and nitrification is

too high.

• the BOD loading ranging from 0.1 to 0.3 kg BOD/m3. d is appropriate.

• The specific TKN loading (2.3 g TKN/m2. d) for combined BOD removal and

nitrification is too high.

• TKN loading ranging from 0.2 to 1.0 g TKN/m2. d is appropriate.

Thus, no nitrification can be expected

At the higher BOD loading, the heterotrophic bacteria outcompete the nitrifying bacteria

for sites on the packing surface and thus control the slime biomass population.

Trickling-filter sizing using NRC equation (4th

ME 913)

1

1

1

1

1 0.4432

EW

V F

=

+

(9-11)

where

E1 = BOD removal efficiency for first-stage filter at 20°C, including recirculation,

fraction

W1 = BOD loading to filter, kg/d

V1 = volume of filter packing, m3


The recirculation factor is calculated using Eq. (9-12)

( )F

R

R=

+

+

1

1 012

. (9-12)

where F = recirculation factor

R = recycle ratio, unit less (generally 0 – 2.0)

E

E

W

VF

2

1

2

1

10 4432

1

=

+−

. (9-13)

where E2 = BOD removal efficiency for the second-stage filter at 20C

E1 = fraction of BOD removal in the first-stage filter

W2 = BOD loading applied to the second-stage filter, kg/d


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The effect of wastewater temperature

E ETT

=−

20 1035 20( . ) (9-14)

where ET = BOD removal efficiency at temperature T in °C

E20 = BOD removal efficiency at 20°C

Hydraulic loadingQ

A

T=

where QT = total flow, m3/d

A = surface area, m2

Example 9-3 Trickling Filter Sizing Using NRC Equations (4th

ME 913)

A municipal wastewater having a BOD of 200 g/m3 is to be treated by a two-stage

trickling filter. The desired effluent quality is 25 g/m3 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 diameter.

Assume the following design assumptions apply. (Note: g/m3 = mg/L)

1) Flow rate = 7,570 m3/d

2) filter depth, h = 1.83 m

3) Wastewater temperature = 20°C

4) E1 = E2

(Solution)

Given: So = 200 g/m3; S = 25 g/m

3; Depth, D = 1.83 m; R = 2/1 = 2

1. Compute E1 and E2

So – S (200 – 25) g/m3

Overall efficiency, ET = ----------- (100) = -------------------- (100) = 87.5%

So (200 g/m3)

ET = E1 + E2 (1 - E1) = 0.875

Since E1 = E2, E1 + E1 (1 - E1) = 0.875

Then, solve for E1, E1 = E2 = 0.646

2. Compute the recirculation factor using Eq. (9-12)

( ) ( )F

R

R=

+

+=

+

+=

1

1 01

1 2

1 01 22 08

2 2. . ( )

.


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First Filter

3. Compute the BOD loading for the first filter, W1.

W1 = Q So = (7570 m3/d)( 200 g/m

3)(1 kg/10

3 g) = 1,514 kg BOD/d

4. Compute the volume for the first stage [using Eq. (9-11), ME 913].

EW

VF

1

1

1

1 0 4432

=

+ .

( )

0 6461

1 0 44321514

2 08

.

..

=

+V

Solve for V1, V1 = 476 m3

5. Compute the diameter of the first filter.

3

211

476260

1.83

V mA m

h m= = =

Second Filter

6. Compute the BOD loading for the second-stage filter, W2.

W2 = (1 – E1) W1 = (1 – 0.646)(1,514 kg BOD/d) = 536 kg BOD/d

7. Compute the volume of the second-stage filter using Eq. (9-13).

2

2

1 2

1

0.44321

1

EW

E V F

=

+−

2

10.646

0.4432 5361

1 0.646 (2.08)V

=

+−

Solve for V2, V2 = 1,345 m3


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8. Compute the diameter of the second filter.

3

2

2

1345735

1.83

V mA m

h m= = =

A = π r2 r = (A/ π)

1/2 = (735 m

2/ 3.14)

1/2 = 15.3 m

Diameter, 2r = 30.6 m

9. Compute the BOD loading to each filter.

a. Filter-stage filter:

W1 (1514 kg/d)

BOD loading = ------ = --------------- = 3.18 kg/m3. d

V1 476 m3

b. Second-stage filter:

W2 (536 kg/d)

BOD loading = ------ = --------------- = 0.40 kg/ m3. d

V2 1345 m3

10. Compute the hydraulic loading to each filter.

a. First-stage filter:

( )( )( )( )

3

3 2

21

1 2 7570 /0.061 / .min

260 1440 min/

Tm dQ

Hydraulic loading m mA m d

+= = =

= =0 061 1440 87 843

2

3

2

.

.min

min .

.

m

m d

m

m d (Roughing filter)

b. Second-stage filter:

( )( )( )( )

Hydraulic loadingQ

A

m d

m dm mT= =

+=

2

3

2

3 21 2 7570

735 14400 022

/

min/. / .min

= =0 022 1440 31683

2

3

2

.

.min

min .

.

m

m d

m

m d (High rate filter)


21

See Table 9-1 for classification of trickling filters.

Note:

So – S2

ET = ------------- (1)

So

So – S1

E1 = ------------ (2)

So

S1 – S2

E2 = ------------ (3)

S1

Solve (2) for S1, then substitute into (3); then solve (3) for S2, then substitute (3) into (1).

ET = E1 + E2 (1-E1)


22

Tower Trickling Filter with Plastic Packing (4th

ME 916)

S

See

o

kD

q n

=

−

(9-19; 4th

ME 917)

where Se = BOD concentration of settled filter effluent, mg/L (g/m3)

So = influent BOD concentration to the filter, mg/L (g/m3)

D = packing depth, m

q = hydraulic application rate of primary effluent, excluding recirculation, L/m2.s

n = constant characteristic of packing used (normally assumed to be 0.50)

k = wastewater treatability and packing coefficient, (L/s)0.5

/m2 based on n = 0.5

k kD

D

S

S2 1

1

2

0 5

1

2

0 5

=

. .

(9-22; 4th

ME 918)

where k2 = normalized value of k for the site-specific packing depth and influent BOD

concentration

k1 = k value at depth of 6.1 m (20 ft) and influent BOD of 150 mg/L (g/m3)

See Table 9-6 Normalized Germain equation k1 values from pilot-plant

studies for different wastewaters (4th

ME 918)

e.g., k1 = 0.210 (L/s)0.5

/m2 for domestic wastewater

S1 = 150 g BOD/m3 (from pilot-plant studies)

S2 = site-specific influent BOD concentration, g BOD/m3

D1 = 6.1 m (20 ft) packing depth, m (from pilot-plant studies)

D2 = site-specific packing depth, m

k kTT

=−

20201035( . ) (9-20; 4

thME 917)

Rotational speed, n

nR q mm

A DR h=

+( )( )( / min)

( )( )( min/ )

1 10

60

3

(9-1; 4th

ME 899)

where n = rotational speed, rev/min

q = influent applied hydraulic loading rate (q = Q/A), m3/m

2-h

R = recycle ratio

A = number of arms in rotary distributor assembly

DR = dosing rate of distributor arm, mm/pass


23

See Table 9-3. A guideline for trickling filter dosing rate as a function of

BOD loading (4th

ME 899)

Example 9-4. Design of Trickling Filter with Plastic Packing (4th

ME 906)

Given the following design flow rates and primary effluent wastewater characteristics,

determine the following design parameters for a trickling filter design assuming 2 towers

at 6.1 m depth, cross-flow plastic packing with a specific surface area of 90 m2/m

3, a

packing coefficient ‘n’ value of 0.5 and a 2-arm distributor system. The required

minimum wetting rate of 0.5 L/m2.s. Assume a secondary clarifier depth of 4.2 m.

Note: Minimum wetting rate = q + qr where qr = recirculation rate (L/m2.s)

Total pumping rate = (q + qr) (A) where A = filter surface area

Design Conditions:

Item Unit Primary effluent Target effluent

Flow m3/d 15,140

BOD g/m3 125 20

TSS g/m3 65 20

Minimum Temp. °C 14

-------------------------------------------------------------------------------------------

g/m3 = mg/L Flow, Q = 630.8 m

3/hr

Using the above information, determine:

1. Diameter of tower trickling filter, m

2. Volume of packing required, m3

3. Recirculation rate required, if any

4. Total pumping rate, m3/h

5. Flushing and normal dose rate, mm/pass

6. Clarifier diameter, m (assume the ratio of the peak to average flow rate is 1.5)

(Solution)

Given:

Number of towers = 2

Depth of tower = 6.1 m

Surface area (cross-flow plastic packing) = 90 m2/m

3

Packing coefficient ‘n’ value = 0.5

Number of distributor arms = 2

Required minimum wetting rate = 0.5 L/m2.s

Secondary clarifier depth = 4.2 m


24

1. Determine k20 for the design conditions using Eq. (9-22).

k kD

D

S

S2 1

1

2

0 5

1

2

0 5

=

. .

a) From Table 9-6, obtain k1 =0.210 (L/s)0.5

/m2 for domestic wastewater

b) Trickling filter depth D2 = 6.1 m (given)

c) S2 = 125 g/m3 (given)

Calculate k2

( )( )( )

k kD

D

S

S

L s mm

m

g m

g m

L s m

2 11

2

0 5

1

2

0 5

0 5 2

0 5 3

2

0 5

0 5 2

0 21061

61

150

125

0 230

=

=

=

. .

.

..

.

. ( / ) /.

.

/

/

. ( / ) /

a. Correct k2 for temperature effect using Eq. (9-20).

At T = 14°C,

k kTT

= = =− −

2020 14 201035 0 230 1035 0187( . ) . ( . ) . (L/s)

0.5 /m

2

2. Determine the hydraulic application rate and the filter area, volume, and diameter.

a. Using Eq. (9-19), determine the hydraulic application rate, q. (Note that q = Q/A)

S

See

o

kD

qn

=

−

( )( )q

kD

S

S

L m so

e

n

=

=

=

ln

. .

ln

. / .

// .

11 0 5

20187 61

125

20

0 3875


25

b. Determine the tower area.

Flow, Q is

Qm

d

L

m

d

sL s=

=

15140 10

86 4001752

3 3

3

,

,. /

Filter area, A is

AQ

q

L

s

L

m s

m= =

=

1752

0 3875452 2

2

2

.

.

.

.

c. Determine the packing volume.

Packing volume, V = A h = (452.2 m2)(6.1 m) = 2758 m

3

d. Determine the tower diameter

Area/each tower = (452.2 m2) / 2 = 226.1 m

2

A = π r2

rA

=

=

=

π π

1 2 1 22261

8 49

/ /.

. m

Diameter, D = 2r = 2(8.49 m) = 17 m each for two filters

3. Determine the recirculation rate and the recirculation ratio.

a. Determine the recirculation rate.

The required minimum wetting rate = q + qr = 0.5 L/m2.s (given)

where qr = recirculation rate (L/m2.s)

0.3875 L/m2.s + qr = 0.5 L/m

2.s

qr = 0.5 - 0.3875 = 0.1125 = 0.11 L/m2.s


26

b. Determine the recirculation ratio, R.

R = qr / q = 0.11 / 0.39 = 0.28

4. Determine the pumping rate (m3/hr).

Total pumping rate = (q + qr) (A) = (0.5 L/m2.s)( 452.2 m

2) = 226 L/s

226 L m3 3600 s

= --------- -------- --------- = 814 m3/hr

s 103 L h

5. Determine flushing and normal dose rate using the data given in Table 9-3.

a. Determine BOD loading (volumetric).

Q So (15,140 m3/d) (125 g/m

3)(1 kg/10

3 g)

BOD loading = -------- = --------------------------------------------- = 0.69 kg/m3.d

(volumetric) V 2758 m3

b. Determine the dosing rate (from Table 9-3), 4th

ME 899.

The estimated flushing and operation dose rates are:

Flushing dose = 300 mm/pass

Operation dose, DR = 50 mm/pass

6. Determine the distributor speed using Eq. (9-1).

nR q mm

A DR h=

+( )( )( / min)

( )( )( min/ )

1 10

60

3

where n = rotational speed, rev/min

R = recycle ratio = 0.28 (from 3b)

A = number of arms in rotary distributor assembly = 2 (given)

q = influent applied hydraulic loading rate, m3/m

2-h

Q (15,140 m3/d) (d/24 hr) 1.4 m

3

q = ----- = ------------------------------- = ------------

A (452.2 m2) m

2 .hr

a. For flushing

DR = dosing rate, mm/pass of distributor arm = 300 mm/pass (Flushing)


27

( )( )( )

( )( )( )n

R q mm

A DR h

rev i e rev

=+

=+

=

( )( )( / min)

( )( )( min/ )

. .

. / min ( . ., min/ )

1 10

60

1 0 28 14 1000

2 300 60

0 0498 20

3

b. For normal operation

DR = dosing rate, mm/pass of distributor arm = 50 mm/pass (Operation)

( )( )( )

( )( )( )n

R q mm

A DR h

rev i e rev

=+

=+

=

( )( )( / min)

( )( )( min/ )

. .

. / min ( . ., . min/ )

1 10

60

1 0 28 14 1000

2 50 60

0 30 333

3

Note: Because of the different speed requirements for normal and flushing operation, a

distributor drive with variable speed capability should be used.

----------

7. Determine clarifier diameter.

a. Clarifier depth = 4.2 m (given)

b. From Fig. 9-7 (4th

ME 905),

i. the recommended overflow rate (Q/A) for peak = 1.1 m/h

ii. the recommended overflow rate (Q/A) for average flow rate = 2.4 m/h

Because the ratio of the peak to average flow rate = 1.5 (given), the average overflow rate

controls the design.

Flowrate, Q = (15,140 m3/d)(24 h/d) = 630.8 m

3/h

Clarifier area, A = Q / (Q/A) = (630.8 m3) / (1.1 m/h) = 573.5 m

2 for peak flow

Use 2 clarifiers.

Area of each clarifier = 573.5 m2

/2 = 286.7 m2

Diameter of each clarifier

A = π r2

rA

=

=

=

π π

1 2 1 2286 7

9 55

/ /.

.

Diameter = 2 r = 2(9.55) = 19 m each for two clarifiers


28


29


30

Typical low-rate trickling-filter plant

Typical high-rate trickling-filter plant

Single-stage high-rate trickling filters


31

Typical recirculation patterns: a) Recirculation with sludge return; b) Direct recirculation

around filter

Typical recirculation patterns: c) Recirculation of plant effluent; d) Dual recirculation Two-stage Trickling filters

Typical flow diagram

7-trickling filter f11

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