3. EAP 582.4 Waste Water Engineering Treatment Principles and Design_Session3

Dr. Abu Ahmed Mokammel Haque ([email protected])

October 3, 2010

School of Civil Engineering, Engineering Campus, Universiti Sains Malaysia, 14300 Nibong Tebal, P. Pinang, Malaysia

AAMH 1

EAP582/4: Wastewater Engineering

Wastewater Treatment Plant Principles' and Design

Dr. ABU AHMED MOKAMMEL HAQUEDr. ABU AHMED MOKAMMEL HAQUESchool of Civil Engineering,

Engineering Campus,Universiti Sains Malaysia

14300 Nibong Tebal, P. Pinang,Malaysia.

E-mail: [email protected], 2010

3

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Pre-Requisite Knowledge and/or Skills

Basic Principles of Environmental Engineering

Basic Principles of Environmental Fluid Mechanics

Mass Balance Techniques

Basic Organic and Inorganic Chemistry

Understanding of Environmental Engineering unit Processes

Basic Computer Spreadsheet Application


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OutlineOutlineObjectives

Introduction- Plant Classifications (Aerobic, Anaerobic, Fixed Media,

Suspended Culture etc)

Type of Treatments- Primary, Secondary, Tertiary (Management Aspect,

Biological oxidation, Kinetics of BOD etc)- Design Aspects- Physical & Chemical Plant (Screen, Grit

Removal, Comminutor, Skimming & Equalization Tanks, Sedimentation Tank, Coagulation & flocculations)

- Design of biological Plant (Activated Sludge, RBC, Anaerobic Digester etc)

Advance Wastewater Treatment

Reclamation and Reuse

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AAMH Teaching Plan

Revision (Sep 06/10)W5

Wastewater Reclamation & Reuse (Sep 01/10)W4

Wastewater Reclamation & Reuse (Aug 29/10)W4

Advance Wastewater Treatment (Aug 25/10)W3

Type of Wastewater Treatments (Aug 23/10)W3



Introduction – Concept, Source, Objectives, Plant Classification (Aug 11/10)

W1

Introduction – Concept, Sources, Objectives, Plant Classification (Aug 09/10)

W1

TOPICSSECTION


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Primary Treatment Layout Complete

COMPLETE

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Biological Treatment Processes

WastewaterBiological ProcessesTreatment ProcessesApplicationsResearch Activities


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Biological Treatment ProcessesWastewaterWastewater• Domestic Wastewater• Commercial Wastewater• Industrial Wastewater

Present State of Wastewaterover 80 % - untreated in Asian mega cities

major components- COD = 250-1000 mg/LTotal N = 20-90 mg/LTotal P = 4-15 mg/L

effects of discharging into natural receiving bodies

oxygen demand by carbon and nitrogen

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Starch industry wastewatermajor component-

COD = 10,000-20,000 mg/L

Effects of discharging into natural receiving bodies (Rivers, lake, Sea etc)

- 20 m3/ton of starch- high COD - high suspended solids- cyanide exposure


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Starch industry wastewaterFactory with 300 T/d of starchWastewater generation 6000 m3/day COD 14,000 mg/LPopulation equivalent 1000,000

•Present treatment method: Anaerobic ponds •Typical loading rates: around 800-1000kg

COD /ha/d•Area requirement: 100 ha

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Strength of Wastewater• Waste-water originates predominantly

from water usage by residences, commercialand industrial establishments, together with groundwater, surface water and storm water.

• Strength of wastewater depends on: the types and source of wastewater generation;concentration level of pollutant constituents;and their toxicity level on surrounding environment.


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6 – 12 mg/LTotal Phosphorus (TP)

26 – 75 mg/LTotal Nitrogen (TN)

< 1 mg/LNitrate-Nitrogen, NO3-N

4 – 13 mg/LAmmonium-Nitrogen, NH4-N

106 – 108CFU/100mLFecal Coliform Bacteria

108 – 1010CFU/100mLTotal Coliform Bacteria

6 – 9 N/ApH

155 – 286 mg/LBOD5

155 – 330 mg/LTotal Suspended Solids, TSS

Range of Concentration UnitComponents

Raw sewage pollutant constituent’s characteristics

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Biological Treatment ProcessesRaw Leachate pollutant constituent’s characteristics

54.6746.67mg/LPhosphate ((PO43-)

2.671.33mg/LNitrate Nitrogen(NO3–N)

46.736.7mg/LNitrite Nitrogen(NO2–N)

2387.51800mg/LAmmonia Nitrogen (NH4–N)

1771.871520.5mg/LTKN

338184mg/LBOD

2403.31773mg/LCOD

490366mg/LSuspended Solid

41533527.5Pt.Co*Color

7.897.12N/ApH

MaximumMinimumUnitParameter

* Platinum-Cobalt Scale (Pt.Co scale or APHA-Hazen Scale )


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Biological Processes .. .. .. .. Aim: any form of life- ‘ survive & multiply ’

Need for energy & organic molecules as building blocks

Made of C, H, O, N, S, P and trace elements

Cell: derives energy from oxidation of reduced food sources (carbohydrate, protein & fats)

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MicroorganismsMicroorganisms

Classification: Heterotrophic- obtain energy from oxidation of organic matter (Organic Carbon)

Autotrophic- obtain energy from oxidation of inorganic matter (CO2, NH4, H+ )

Phototrophic- obtain energy from sunlight


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Biochemical PathwaysBiochemical PathwaysOxidation of organic molecules inside the cell can occur aerobic or anaerobic manner

Generalized pathways for aerobic & anaerobic fermentation

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Biological Treatment ProcessesGlucose

EMP Pathway

Pyruvic Acid

ADP ATP

Energy

Lactic Acid TCA Cycle H+ Respiration H2O

CO2 O2http://dwb4.unl.edu/Chem/CHEM869P/CHEM869PLinks/www.bact.wisc.edu/mi

crotextbook/metabolism/Fermentation.html Embden-Meyerhoff-Parnas Pathway (EMP)

TCA = Tri-Carboxylic Acid cycleATP = Adenosine Tri-PhosphateADP = Adenosine Di-PhosphateNAD+ = Nicotinamide adenine dinucleotide


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aerobic pathways contains-EMP pathways, TCA cycle, respiration

anaerobic pathways contains-EMP pathways

released energy stored as ATP molecules

excess food is stored as Glycogen

C6H12O6 + 6O2 +38 ADP + 38 Pi 6 CO2 +38 ATP + 44 H2O

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Biological growthConcept of Microbiology

- BOD exertion

- Exponential growth (batch)

- Monod kinetics

- Haldane kinetics

under toxic conditions


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Biological Oxygen Demand (BOD)

Biological/Biochemical Oxygen DemandBiological/Biochemical Oxygen Demand ((BOD):the amount of oxygen used by microorganism in the oxidation of organic matter (ammonia, nitrite) in water or wastewater.

Any ammonia present in a waste stream may also be oxidized by nitrifying bacteria in a process called nitrification. Nitrification also demands oxygen, which is referred to as nitrogenous BOD (NBOD). A general equation for the overall nitrification process is shown below.

OXIDATIONammonia +oxygen +carbon dioxide +nitrifying bacteria

nitrate + water + new cells + energy

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Biological Oxygen Demand (BOD)Biological/Biochemical Oxygen Demand (BOD)Biological/Biochemical Oxygen Demand (BOD)The Oxygen (O2) demand by microbes during the degradation of Organic matter (OM) to CO2 + H2O under aerobic conditions at a particular temperature and incubation period.

Standard BOD BOD on 5 days at 20oC which indicate pollution strength of wastewater or waste stream.

OM + DO (Dissolve Oxygen) + Heterotrophic Microbes CO2 + H2O + More cells + Energy

BOD Test UsedBOD Test Used• To determine the polluting strength of waste stream.

• To determine the size and efficiency of waste treatment facilities.


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Biological CharacteristicsBOD Test

Provide microbes with initial DO and measured DO consumption after 5 days in a close environmentAlso provide nutrients for microbes growth (Except domestic wastewater)BOD = [(DO)i – (DO)f], Saturation level of normal (DO)i = 9 mg/L @ 20oC

• is conducted in air tight bottles to prevent reaeration of samples.• Due to limited solubility of oxygen in water (about 9 mg/L at

20oC) concentrated waste must be diluted to ensure DO viability throughout the test period.

• The samples should have adequate amount of microorganism; if not then seeding is required

• Presence of nutrients and absence of toxic substance is ensured.• Samples are incubated for 5 days at 20oC.• DO of the samples is measured befor and after incubation to

calculate the BOD.

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Biological CharacteristicsCalculation:Calculation: BOD TestBOD TestFor domestic Wastewater ---- Do, Microorganism, Nutrient presentFor Industrial Wastewater --- Add Seed Microorganism (Domestic

Settle Sewage content 106 – 107 MO/mL)Case 1Case 1(DO)f = 5 mg/L; BOD = (DO)saturation - (DO)f = (9 – 5)*1 = 4 mg/LCase 2Case 2(DO)f = 0 mg/L; BOD = (DO)saturation - (DO)f = (9 – 0)*1 = 9 mg/L;

Invalid Result, BOD cannot be determined, therefore need the dilution of the waste sample --- Test to be valid upto (DO)f should not < 1 mg/L

Case 3Case 3(DO)f = 8.8 mg/L; BOD = (DO)saturation - (DO)f = (9 – 8.8)*1 = 0.2

mg/L; Invalid Result, Test to be valid (DO) consumption ≥ 2.0 mg/L

Conditions Conditions -- ◙◙ Provide DO ( If not Dilute the Sample) ◙◙ MO (If not add seed MO from domestic settle sewage) ◙◙ Nutrient ◙◙ Dilution Dilution (DO) consumed 5 days ≥ 2.0 mg/L and (DO)f ≥ 1 mg/L


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Standard BOD bottles in the lab

When dilution water is not seeded:

When dilution water is seeded;-

Where :

D1 = DO of diluted sample immediately after preparation, mg/L

D2 = DO of dilute sample after 5 day incubation, mg/L

B1 = DO of seed control before incubation mg/LB2 = DO of seed control after incubation mg/Lf = fraction of seeded dilution water volume in sample to volume of seeded dilution in seed control.P= fraction of wastewater sample volume to total combined volume

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Question : An unseeded BOD test was conducted on domestic wastewater 20mL of the sample was diluted to 1L by aerated dilution water. Calculated the BOD of the sample if the initial and 5 days DO were 7.6 and 3.7 mg/L respectively.

EXAMPLE:

Solution

D1 = 7.6 mg/LD2= 3.7mg/L

P = 20 /100

= 0.02

= (7.6- 3.7)/ 0.02= 195 mg/L

Fraction of water sample


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Biological Oxygen Demand (BOD)Question : The following information was obtained from a seeded 5 days BOD test on a wastewater sample. Dilution water was prepared with a seed dilution of 1 in 200 and the BOD bottles were prepared by diluting the wastewater sample to 1 in 200. The initial DO of diluted sample was 8.5 mg/L and the final 5 day BOD was 2.2 mg/L. The corresponding initial and final DO of the seed dilution water was 8.8 and 7.6 mg/L respectively. Calculate the BOD5 of the wastewater sample.

SolutionD1 =8.5mg/L B1= 8.8mg/LD2 = 2.2 mg/L B2= 7.6 mg/L

P = 1 / 200 = 5 x 10‐3

f = ( 200‐ 1)/ 200 = 0.995

1021 mg/L

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Wastewater Type

BOD (mg/L) (1)

Wastewater Volume

(L/unit) (2)

BOD Load (g/unit) (3)

Cow 450-4330 45 - 136 182

Pig 1275 - 13260 11 - 23 114

Chicken & Duck 8500 - 40000 0.2 7 - 15

Sewage 250 - 400 227 68

Wastewater concentration differs depending on its source

BOD Load = (3) = (1) * (2)


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Wastewater type BOD load per day (g/day) (1)

TKN load per day (g/day) (2)

Sewage 55 -

Cow 498 222

Buffalo 347 107

Sheep 102 22

Pig 137 23

BOD load composition and TKN load per day for different wastewater

(1) = BOD (g/L) * Flow rate (L/unit.day)

(2) = TKN (g/L) * Flow rate (L/unit.day)

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Biological Oxygen Demand (BOD)E.g. 1.1 :

Calculate BOD value for a wastewater from 1000 persons with total flow rate of 225 m3/day. Assume BOD rate is 55 gram/person.day.

Solution :A person produces 55 gram BOD/person.day, therefore for 1000 persons, the total BOD produced is

= (1000 persons) (55 gram/person.day)= 55 000 gram/day

Flow rate, Q = 225 m3/day, therefore by dividing 55 000 g/day with flow rate Q, we can get BOD value as;

BOD = 55 000 gram/day225 m3/day

= 244.44 mg/L= 244 mg/L (BOD is written in round number)

What will be the load -------- Population Equivalent???


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Biological Oxygen Demand (BOD)Question:

If the BOD5 for wastewater is 300 mg/L and the total water usage is 200 liter/unit. How much the BOD5 load for this wastewater?

Solution :

BOD5 load = BOD5 × unit usage = 0.3 g/L × 200 L/unit= 60 g/unit

If total population are 10,000 then what will be the load -------- Population Equivalent???

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1st Order Biological Oxygen Demand (BOD)

Modeling of BOD Reaction:The rate of BOD oxidation (“exertion”) is modeled based on the assumption that the amount of organic material remaining at any time “t” is governed by a FIRST-ORDER Kinetic function as given below:

dBODr/dt = - k1BODr

Integrating betn the limits of UBOD and BODt and t =0 and t= t yields [BODr = UBOD(e-k

1t)]

WhereBODr = Amount of waste remaining at time t (days) expressed in oxygen equivalents, mg/LK1 = First-order reaction rate constant, 1/dUBOD = total or ultimate carbonaceous BOD, mg/Lt = time, d


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1st Order of BODModeling of BOD Reaction:

Therefore, BOD exerted upto t is given byBODt = UBOD – BODr = UBOD – UBOD(e-k

1t) = UBOD (1 – e-k

1t)

Calculation of BODDetermine the 1-day BOD and Ultimate first-stage BOD for a

wastewater whose 5-day 20oC BOD is 200 mg/L. The reaction constant k(base e) = 0.23/day. What would have been the 5 day BOD if the test had been conducted at 25oC.

Solve?????

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1st Order Kinetics of BODCalculation:Calculation: Total OM in the systemTotal OM in the system OROR Kinetic of BODKinetic of BODMicrobes consume OM as a function of time (t), No instantaneous consumption of DOBOD excretion OM consumption followed FIRST ORDER EQUATION dL/dt α L ; Where L is the O2 ≅ OM ; Therefore, dL/dt = -kt (-ve sign, OM decreases with time)

tko eLL −=

BOD Rate Constant which explained the property of MO present in the system

k –

BOD EXERTED

Lo – is ultimate O2 demand (UBOD), constant value for a given OM & Heterogeneous microbial Seedk – Speed of Reaction

∫∫ −= oo t

t

L

Lkdt

LdL

tkLL

o

−=⎟⎟⎠

⎞⎜⎜⎝

⎛ln

If Three samples have the same Lo value but different k value k1> k2 > k3

)1( tko

tkoo

o

eLY

eLLY

LLY

−

−

−=

−=

−=

Ultimate O2 demand , O2 for the oxidation of the initial OMLo


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Biological CharacteristicsUltimate BOD

BOD exerted from time 0 to t

BOD remaining at time, t

Lo

0

Yt, Lt Yt

Lt

Time, t

Lo- Yt

K = 0.1-0.5 d-1 ; depends on temperature.KT = Rate constant at ToCK20= Rate constant at 20oCӨ = constant, generally taken as 1.047

BOD curve

KT= K 20 Ө T-20 Van Hoff Arrhenius Equation

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Biological Characteristics

Why 5 days incubation70 – 80% BOD – Lo is exerted in 5 daysNitrification occurs after 5 days

O2NH3 + O2 --- NO2

- --- NO3-

One N atom ≅ 1.5 O2

OM -- CO2 + H2ONow a day BOD on 3 days at 28oC Oxidation take place faster and experimentation completed within short time

•S1 is more easily biodegradable than S2 and S3

• S1 is simple organic matter [Carbohydrate -(C·H2O)n]

• S2 may be Protein (polypeptides – amino acids)

• S3 may be lipid (very resistant for oxidation)

• k depends on Temperature

• kT = k20 (θ)T – 20; θ = 1.047; Constant for Temp. range (20 – 30oC); Micro-organism works high temperature


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Exponential growth

XdtdxX

dtdx µα =>>>

Log

No.

of C

ells

TimeLa

g ph

ase

Log

gro

wth

pha

se

Stat

iona

ry p

hase

Dea

th p

hase

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Monod kinetics

SKS

S +=

.maxµµ

Substrate Concentration (S)

Spec

ific

grow

th ra

te (

µ)

Max. rateµm

µm/2

ks


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Haldane kinetics(under toxic conditions)

is K

iSSKS

.max++

= µµ

Substrate Concentration (S)

Spec

ific

grow

th ra

te (

µ) i

http://www.graphpad.com/curvefit/introduction63.htm

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Application1. Carbonaceous removal - aerobic

- anaerobic

2. Nitrogen removal - nitrification- denitrification

3. Sulfide removal - anaerobic SO4 reduction- aerobic HS- oxidation


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Biological Carbonaceous Removal

Aerobic- oxidation

bacteriaCHONS + O2 + Nutrients CO2 + NH3 + C5H7NO2

(organic matter) (new bacterial cells)

+ other end products

- Endogenous respirationbacteria

C5H7NO2 + 5O2 5CO2 + 2H2O + NH3 + energy(cells)

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Biological Carbonaceous RemovalIn BOD 5 days incubation70 – 80% BOD – Lo is exerted in 5 days. Nitrification occurs after 5 days

O2NH3 + O2 --- NO2- --- NO3-One N atom ≅ 1.5 O2

OM -- CO2 + H2ONow a day BOD on 3 days at 28oC Oxidation take place faster and experimentation completed within short time


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Anaerobic

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Hydrolysis

Acidogenesis

Methenogenesis

Complex Organics

Intermediates Propionate

H2Acetate

CH4

20% 5%

60% 15%

35% 10% 13%17%

15%

72% 28%

100%

Anaerobic

Schematic of the Anaerobic ProcessReturn 50


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Biological Nitrogen Removal

Nitrification-Energy

NitrosomonasNH4

+ + 1.5 O2 NO2- + H2O + 2 H+ + (240-350 kJ) (1)

NitrobacterNO2

- + 0.5 O2 NO3- + (65-90 kJ) (2)

-AssimilationNitrosomonas

15 CO2 + 13 NH4+ 10 NO2

- + 3 C5H7NO2 + 23 H+ +4 H2O (3)Nitrobacter

5 CO2 + NH4+ +10 NO2

- +2 H2O 10 NO3- + C5H7NO2 + H+ (4)

- Overall reaction

NH4+ +1.83 O2 + 1.98 H CO3

- 0.021 C5H7NO2 + 0.98 NO3- + 1.04 1H2O + 1.88H2CO3

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Factors Affecting Nitrification* temperature * substrate concentration

* dissolved oxygen * pH

* toxic and inhibitory substances

( )[ ])2.7(83.01)15(095.0

4

4 pHeDOK

DONNHK

NNH T

ONm −−⎥

⎦

⎤⎢⎣

⎡+

⋅⎥⎦

⎤⎢⎣

⎡−+

−= −µµ

Factors Affecting Denitrification* temperature * dissolved oxygen

* pH


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Denitrification* Assimilatory denitrification- reduction of nitrate to ammonium by microorganism for protein synthesis

* Dissimilatory denitrification- reduction of nitrate to gaseous nitrogen by microorganism- nitrate is used instead of oxygen as terminal electron acceptor- considered an anoxic process occurring in the presence of nitrate and the absence of molecular oxygen- the process proceeds through a series of four steps

NO NO NO N O N 3-

2-

2 2⎯→⎯ ⎯→⎯ ⎯→⎯ ⎯→⎯

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Denitrification

* Heterotrophic denitrification

- denitrifiers require reduced carbon source for energy and cell synthesis

- denitrifiers can use variety of organic carbon source -methanol, ethanol and acetic acid

NO + 1.08CH OH + H 0.065C H O N 0.47N 0.76CO 2.44H O3-

3+

5 7 2 2 2 2⎯→⎯ + + +


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Biological Sulfate Removal

* Sulfate Removal Cycle

anaerobic

SO4-- HS - S 0 (O2 deficient)

(O2 excess)

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Pond treatment

Activated Sludge Process

Bio-film Process


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BTP – Pond Treatment

Anaerobic Ponds- no biomass recirculation- high HRT- high land area- O2 transfer limitations- inadequate mixing- excess loading

(anaerobic condition-H2S generation)

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- no biomass recirculation- high HRT (MWW 1-3 d; IWW >20 d)- high land area- O2 transfer limitations- inadequate mixing- excess loading

(anaerobic condition-H2S generation)- Temp. ((hydrolysis, acidogenesis, acetogenesis and methanogenesis) [S-42]- TSS >> 50-70%- BOD >>High Temp >>30 – 75%


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BTP – Pond TreatmentMechanism

• After hydrolysis of particulate organic matter, fermenting bacteria convert the readily biodegradable organic substrate into volatile fatty acids (VFAs).

• Higher VFAs are further decomposed, mainly into acetic acid and H2, the typical substrate for the strict anaerobic methanogens.

• Effective anaerobic pond management has to avoid VFA accumulation and the associated drop in pH as methanogens are very sensitive to pH values less than 4-5. [S-42]

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The overall anaerobic decomposition of organic matter can be expressed by the following equation :

Preliminary Treatment- Coarse Screening (large pieces of wood, plastics, can, bottles): to avoid clog pipes/channels- Grit Chamber/Channel – to prevent accumulation of grit to pond and reduce the active pond volume and reduce desludging frequency

3242 83

48283

48243

24dNHCOdbanCHdbanOHdbanNOHC dban +⎟

⎠⎞

⎜⎝⎛ +−−+⎟

⎠⎞

⎜⎝⎛ +−+→⎟

⎠⎞

⎜⎝⎛ +−−+


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Design Parameters:

whereVan = Pond Volume (m3)A = Surface Area (m2)D = Aver. Pond Depth (m)BODin = Influent BOD Concentration (kgBOD/m3)Q = Flow Rate (m3/day) OR Population Equivalentλv = Volumetric Organic Loading Rate (kgBOD/m3/day)

v

inan

QBODDAV

λ*

* ==

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•Design Volumetric Organic Loading Rates for Anaerobic Ponds as a Function of the monthly average Temperature (Mara & Pearson, 1986)

.)( averv TF=λ

300.30>20

2*Temp + 200.02*Temp. – 0.1010 – 20

400.1< 10

BOD Removal (%)Volumetric Organic Loading

(λv = kgBOD/m3/day

Min. Month-Aver. Air Temp. (oC)


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Operational Period of an Anaerobic Pond until Desludging is required:

Where,Van = Pond Volume (m3)n = Operational period betn Desludging (years)PE = no. Population EquivalentSAR = Sludge accumulation Rate, Typically 0.04 m3/PE/year[For a BOD production of 40 g/PE/day and at 20oC >> requires

0.13 m3 pond volume per PE, then n = 1.1 year]

SARPEV

n an

*1*

3=

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Facultative PondsFurther Removal of BOD, nutrients & Pathogen.Depth: usually 1.5 – 2.5 mHRT : varies betn 5 – 30 daysMost effective for MWWTFiltered Effluent BOD = 20 – 60 mg/L and TSS

level = 30 – 150 mg/LAnaerobic Ponds >> Facultative Ponds >>

Maturation Ponds


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•Design based on Algae-Oxygen Production –[Theoretical]

•Design based on First-order BOD degradation constants and ideal flow conditions (CSTR -Continuous Stirred-Tank Reactor or PFR –Plug Flow Reactor) [Semi-Empirical]

•Design based on the dispersed-flow model [Semi-Empirical]

•Design based on surface BOD load [Empirical]

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(1) Design Mechanism

1. Oxidation>> OM>> Aerobic & Facultative Bacteria2. Balance O2 input by Photosynthetic algae & surface

reaeration to the O2 demand exerted by OM.3. Symbiosis of Algae – Bacteria 4. Relation of Algae growth – O2 production

Aerobic Zone

Facultative Zone

Anaerobic Zone

Algae

Light

O2

BacteriaOM

CO2, NH4+, PO4

+

New Cell

Symbiosis

2164518010634322 2

3091690106 OPNOHCLightPONOOHCO +→++++ −−

C106H180O45N16P Represent Algae Biomass. To sustain algae growth & Photosynthesis need supply of macro-nutrients (N, P, K). Required BOD/N/P ratio = 100/5/1 is generally recommended to safe the basic need


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(2) Design Mechanism[accumulation] = [in] - [out] + [generation] - [consumption]

BOD accumulation =BOD influent – BOD effluent – BOD degraded

Si, Se = Ultimate soluble influent and effluent BOD respectively, mg/LQ = Flow Rate (m3/day)V =Volume of the Pond (m3)KT = First order reaction co-efficient of BOD HRT = V/Q = Hydraulic retention time in the Pond (day)Pond with high L/W ratio (>10) behave as Plug-Flow reactor.First-order BOD degradation the effluent BOD (soluble is given by):

increase due to logarithmic increase in PFR.

VSKQSQS eTei −= 1.1

+=≥+=

HRTKS

StKSS

T

ieT

e

i

KT= K 20 Ө T-20

HRTKie

TeSS −=

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(3) Design Mechanism: Dispersed-Flow model

Where n is the model parameter (numbers of mixers in series), If one pond equals one mixer, n> 3 or 4 no BOD removal .

(4) Design Mechanism: EmpiricalWehner-Wilhem model

KT = First-order Reaction coefficient (day-1)

n

T

ie

nHRTK

SS

⎟⎠⎞

⎜⎝⎛ +

=

1

( ) )2/(2)2/(2

21

)1(14

δδ

δ

aai

e

eaeaae

SS

−−−+= δHRTKa T41+=


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BTP – Pond TreatmentSubstrate removal according to the dispersed flow model of

Wehner-Wilhem (Thirumurthi, 1969)

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Various empirical design equationsdeveloped from full-scale performance of facultative ponds (Ellis and Rodrigues, 1995)


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BOD removal rates λr (kg BOD/ha/day) of facultative ponds as function of the applied organic surface loading rates for various regions (Ellis and Rodrigues, 1995)

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BTP – Pond TreatmentOxygen balance in Facultative Ponds

( ) ( ) [ ] [ ] [ ]{ }BODinBODBinNNNdodresphotosat

in CCQaCCQarrrCCKACCQdtdCV −−−−−−+−+−= ..

Where, V = pond volume, (m3); C = DO concentration (mg/L); Cin = DO concentration in Influent (mg/L); t = Time (day); Q = Flow Rate (m3/day); A = pond surface area (m2); K = reaeration mass transfer co-efficient (m/day) CSat = saturation DO concentration (mg/L); rphoto = rate of DO generation by algae photosynthesis (g/m2/day); rres = rate of DO consumption by respiration (g/m2/day); rdod = rate of DO consumption by plant decomposition (g/m2/day); aN = stoichiometric co-efficient for NH4-N oxygen demand; aB = stoichiometric co-efficient for BOD; CN = Concentration ammonium nitrogen (mg/L); CBOD = concentration of BOD (mg/L)


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BTP – Pond Treatment• In the above O2 Balance equation some parameters are not

available Like as aN and aB parameters were estimated for a surface flow wetland to be 4.5 and 1.5 respectively (Kadlec and Knight, 1996).

• The physical reaeration of a pond through the open water surface is the combined effect of molecular diffusion and the vertical mixing of the pond by wind. In addition rainfall increases mixing and rainwater carries DO. The reaeration mass-transfer co-efficient K for situations without wind was estimated by O’Connor and Dobbins as:

Where, D = molecular diffusivity of oxygen in water (m2/day)h = water depth (m)U = Water Speed /(Travel Velocity) (m/day)

hDUK =

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BTP – Pond Treatment• In Facultative/Maturation Ponds the solar radiation is

absorbed by algae present in the water column and the energy is used for photosynthesis in which process Carbon-dioxide (CO2) is consumed and oxygen (O2) is produced. Simultaneously algae consume O2 by respiration. Arceivala (1986) predicted the Net Oxygen production (rphoto – rres) by algae photosynthesis as:

Where, Y = Net Algal Bio-mass yield (mg/cm2/day)S = Average visible radiation (cal/cm2/day)η = Light conversion efficiency (0.06)h = Specific chemical energy of Algae biomass (cal/mg)

hSY η3.1=


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Maturation Ponds (MP)- Shallow ponds, algal biomass is maintained- During daytime large amount of oxygen are produced- Aerobic in nature, depth 1 – 1.5 m- F. Coliform and virus die-off rates very high (probably

reached at 3 to 4 log units)- Cysts and Ova of intestinal parasites are densities

increase and settle on pond bottom and eventually die-off.

- BOD removal is slower, also degradable material is less and already consumed in Facultative ponds.

- Experimental results showed that there was no correlation betn BOD removal in Maturation Ponds with Temperature or retention time (Mara et al., 1990 and 1992).

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- High amount of algal biomass in the effluent quality represents a high Suspended Matter (SM) concentration, may be exceed Final effluent Guidelines.

- Normally O2 demand exerted by these SM is around 0.5 – 0.6 mg BOD5/mg Algal TSS (Arceivala, 1986).

- If MP are designed to optimize algal protein, commonly called HRAR (High Rate Algal Ponds

- Major application of MP is to polish or upgrade facultative pond effluents and achieve substantial reductions to allow safe use of the effluents in Agriculture or Aquaculture.


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Removal of Pathogenic Micro-Organism-Pathogen removal occurs in anaerobic, facultative and maturation ponds, However, Only MP are designed on the basis of required removal rates of pathogens.

105

104

103

103

103

102

104

103

103

1011110

1/L1/L1/L1/L1/L1/L1/L

Enteric virusesSalmonellaeCholera vibrioHookworm ovaAscaris ovaSchistosome ovaEntamoeba histolyica cysts

Pathogens

2 × 108

103105

11/100 mL1/L

Faecal ColiformsHelminth Eggs

Indicators

MaxMin

RangeUnitsIndicators/Pathogen

Common concentration of

pathogens in municipal sewage

Pathogen present in municipal

sewage

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BTP – Pond TreatmentRemoval of Helminth eggs and ProtozoaBoth are removed by sedimentation. Therefore, their removal are mostly affected by Hydraulic Retention Time (HRT) as following formula:

Where, R = % Removal

[ ]20085.0*49.0*41.01*100 HRTHRTeR +−−=


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Light attenuation in algae pond water

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Photo-Oxidation Process (Curtis et al., 1992)


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Modeling of faecal coliform (FC) decayModeling of FC decay is aimed at predicting removal efficiencies in

existing pond systems or at the design of new systems. The model usually comprises a hydraulic flow pattern model and an equation to predict the FC decay coefficient for first order decay:

Where, FC = Count (no./mL); t = Time (days); Kd = First order decay coefficient (day-1)

For a complete Mixed Pond, Above Equation changes into:

Where Ni and Ne are the number of faecal coliform in influent and effluent, respectively and θ is the total retention time.

daymLnotKdtdN

d /100/.*−=

θdi

e

KNN

+=

11

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Design of Maturation Ponds- Calculation depends on # of ponds and Hydraulic

Retention Time per pond.- Assume Depth (1 -1.5 m), - Length: width varies 3 to 10.- Considered as completely mixed pond like as CSTR

Reactor- For n numbers of MP in series with completely mixed

ponds with total retention time θ:

n

do

t

nK

NN

⎟⎠⎞

⎜⎝⎛ +

=θ1

1


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The Kd for faecal coliforms is taken as 2.6/day at 20oC. The temperature dependence of Kdover 5 – 30oC has been determined by Marais (1974). The value of Kd can be corrected for other prevailing Sewage temperature according to:

Where Kd(T) is the die off rate constant at ToC . In practice Maturation Ponds are usually designed to have a total retention time of approximately 10 -20 days.

( ) ( ) 2019.16.2 −= TTdK

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Additional design guidelines for MP:

• The hydraulic retention times in all maturation ponds are equal,this gives the most effective removal at a certain total hydraulic retention time (Marais, 1974).

• The minimum hydraulic retention time per pond to avoid short-circuiting is 3 days (Marais, 1974).

• The organic surface load to the first maturation pond should not exceed 75% of the organic surface load of the preceding facultative pond (Mara et al., 1992).

• The hydraulic retention time in a maturation pond should not exceed the hydraulic retention time in the preceding facultativepond (Mara et al., 1992).


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BTP – Activated Sludge Process

- aerobic- suspended-growth - Design equations

PST AT SST

RAS

SW

SW

Influent biomass + biomass production = effluent biomass + sludge wasted

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Substrate utilization rate

XSKS

Ydtds

su

...1 max

+=⎟

⎠⎞

⎜⎝⎛ µ

( ) Xqdtds

u.= ( )SFq =

SKSq

qS +

=.max

[‘q’ is not a constant] >>>>

[qmax → Maximum substrate utilization rate]


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Food to microorganism ratio (F/M) Represents the daily mass of food supplied to the microbial biomass, X, in the mixed liquor suspended solids, MLSS Units are Kg BOD5/Kg MLSS/day

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Mass balance of biomass productionAccumulation= Inflow – Outflow

Where,X = mixed-liquor suspended solids, mg/LQ = Secondary influent flowrate, m3/sQR = return sludge flowrate, m3/sXR = Return activated sludge suspended solids, mg/LQw = Waste activated sludge flowrate, m3/sQe = Effluent flowrate, m3/sXe = Effluent suspended solids, mg/L

( ) eeRWRRR XQXQXQQQX −−−+=0


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(a) Definition Sketch for suspended solids mass balances for return sludge control: Secondary Clarifier mass balance

Aeration Tank

Q Q +QR

XSecondary Clarifier

Qe

Xe

Qw

QR

XR XR

System boundary

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BTP – Activated Sludge ProcessAssuming Xe is negligible and that QwXR is related to the

SRT, solving

Recycling Ratio (QR/Q = R) is then

The required RAS pumping rate can also be estimated by performing a mass balance around the aeration tank.

( ) Rwew XQXQQVXSRT

+−=

[ ]XXSRTXVXQQ

RR −

−=

/

( )( ) 1/

/1−

−=

XXSRTR

R

τ


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(b) Definition Sketch for suspended solids mass balances for return sludge control: Aeration

Tank Mass Balance

Aeration Tank

Q Q +QR

XSecondary Clarifier

Qe

Xe

QwXR

QR

XR

System boundary

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New cell growth can be considered negligible. If the influent Solids are negligible compared to the MLSS, the mass balance around the aeration tank results:

Accumulation= Inflow – Outflow ( )RRR QQXXQ +−=0

XXXR

QQ

R

R

−==


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Sludge Retention Time (SRT): To maintain a given SRT, the excess activated sludge produced each day must be wasted.

V = Volume of the reactor, m3

X = Aeration tank mass concentration, mg/LQw = Waste sludge flowrate from return sludge line, m3/dayXR = Concentration of sludge in the return sludge line, mg/LQe = Effluent flowrate from secondary clarifier, m3/dayXe = Effluent TSS concentration, mg/L

If it is assumed that the concentration of solids in the effluent from the settling tank is low, then

( )eeRw XQXQVXSRT+

=

Rw XQVXSRT =

( )SRTXVXQR

w =

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BTP – Activated Sludge ProcessTo determine the waste flowrate, the solids

concentration in both the aeration tank and the return line must measured. If wasting is done from aeration tank and the solids in the settled efflent are again neglected, then the rate of pumping can estimated by using the following relationships:

Where, Qw = Waste sludge flowrate from the aeration tank, m3/day

Thus, the process may be controlled by daily wasting of a quantity of flow equal to the volume of the aeration tank divided by the SRT

wQVSRT =

SRTVQw =


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BTP: Bio-Film Treatment Process• Biofilm is an aggregate of microorganisms in which cells adhere

to each other and/or to a surface. These adherent cells are frequently embedded within a self-produced matrix of extracellular polymeric substance (EPS).

• Biofilm EPS, which is also referred to as slime (although not everything described as slime is a biofilm), is a polymeric conglomeration generally composed of extracellular DNA, proteins, and polysaccharides in various configurations. Biofilmsmay form on living or non-living surfaces, and represent a prevalent mode of microbial life in natural, industrial and hospital settings.

• The microbial cells growing in a biofilm are physiologically distinct from planktonic cells of the same organism, which, by contrast, are single-cells that may float or swim in a liquid medium.

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BTP: Bio-Film Treatment Process


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Constraints and Opportunities In natural environments. Microbes can

negatively impact environments on a global level including producing and consuming atmospheric gases that affect climate; mobilizing toxic elements such as mercury, arsenic and selenium; and producing toxic algal blooms and creating oxygen depletion zones in lakes, rivers and coastal environments (Eutrophication). Furthermore, the incidence of microbial diseases such as plague, cholera, Lyme disease, and West Nile Virus are linked to global change.

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In industrial environments. Biofouling, biocorrosion, equipment damage and product contamination are constant and expensive problems in industry. Biofilm contamination and fouling occur in nearly every industrial water-based process, including water treatment and distribution, pulp and paper manufacturing and the operation of cooling towers.

In human health. The human body is heavily colonized by microbes that have found it a great place to live. chronic, low-grade infections are related to the biofilm mode of growth.


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BTP: Bio-Film Treatment ProcessBiofilms also offer huge potential for bio-remediating hazardous waste sites, bio-filtering municipal and industrial water and wastewater, and forming biobarriers to protect soil and groundwater from contamination.

The role of microbes in biofuels production e.g., methane, ethanol, hydrogen

The role of microbes in cleaning up pollutants (bioremediation)Biological treatment of pollution or reduction of pollution from current processes

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Bioremediation of existed polluted areaCleanup of Superfund sites, oil spills

Environmental Monitoring: Indicator, Ames test, Microtox, Biosensors, ELISA (Enzyme-linked immunosorbent assay)

Prevention of pollution: (clean technology, Green technology) microbial removal of S compounds from coal, fungal pretreatment of logs before pulp and paper production, biodegradable plastic, biofuels


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A layer of microbes attached and proliferated on the surface of an object.


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1) Free-floating, or planktonic, bacteria encounter a submerged surface and within minutes can become attached. They begin to produce slimy extracellular polymeric substances (EPS) and to colonize the surface.

2) EPS production allows the emerging biofilm community to develop a complex, three-dimensional structure that is influenced by a variety of environmental factors. Biofilm communities can develop within hours.

3) Biofilms can propagate through detachment of small or large clumps of cells, or by a type of "seeding dispersal" that releases individual cells. Either type of detachment allows bacteria to attach to a surface or to a biofilm downstream of the original community.

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1) Initial attachment, 2) Irreversible attachment,3) Maturation I, 4) maturation II, 5) dispersion


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BTP: Bio-Film Treatment Process1) Free-floating, or planktonic, bacteria encounter a

submerged surface and within minutes can become attached. They begin to produce slimy extracellular polymeric substances (EPS) and to colonize the surface.

2) EPS production allows the emerging biofilmcommunity to develop a complex, three-dimensional structure that is influenced by a variety of environmental factors. Biofilm communities can develop within hours.

3) Biofilms can propagate through detachment of small or large clumps of cells, or by a type of "seeding dispersal" that releases individual cells. Either type of detachment allows bacteria to attach to a surface or to a biofilm downstream of the original community.

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Bio-film Growth Process


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BiodegradationThe natural process in which microorganisms (bacteria, fungi) are able to completely or partially break down organic compounds to CO2 + H2O or other simple organic molecules that are inert or are readily metabolized by other organisms.

Mineralization

The degradation process is carried to the extreme, in which organic compounds are biodegraded to inorganic compounds.

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Advantages of biofilm processes:

- higher process productivity (loading rates)

- higher biomass holdup- higher mean cell residence time- no need for biomass recirculation- creates suitable environment for each type of bacteria

- sustains toxic loads


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BTP: Bio-Film Treatment ProcessTypes of biofilms: aerobic, anaerobic, anoxic

Process of biofilm formation

- formation of diffuse electrical double layer due to electrostatic forces and thermal motion

- transfer of microorganism to surface

- microbial adhesion

- biofilm formation

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Bio-film Operation

X

Y

BiofilmLiquidFilm

Bulk Liquid

Supp

ort M

ater

ial

(a) Physical concept

Fully Penetrated

Partially Penetrated

SS

Sb

Subs

trate

Con

cent

ratio

n

X

Y

(b) Substrate concentration profile


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BTP: Bio-Film Treatment ProcessBio-film Operation•Diffusion resistance•Inadequate supply of nutrients to inner portions of Biofilm•Limitations on product out diffusion•Attrition of reaction conditions

As biofilm thickness increases effectiveness factor (η) decreases

average rate of substrate consumptionEffectiveness factor η = ----------------------------------------------

substrate consumption at biofilm surface

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Research: Bio-Film

Landfill Leachate Treatment by Swim-bed Biofringe Technology


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Design for Secondary Sedimentation TankThis tank is placed after the biological sedimentation tankThe purpose is to remove the sludge from the biological plant due to the synthesis and microorganism oxidation processTherefore it should be properly designed in order to ensure the discharge of effluent is according the appropriate standardAlso known as humus tankOnly the circular tank is designed specializing for this type of sedimentation tank.

Design of Secondary Sedimentation Tank

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It is very important for treatment unit such as activated sludge plant and aerated lagoonWithout this tank, sludge could be settled. Therefore, final effluent will contain high suspended solid.Design for this tank similar is to the primary tank with a few differences due to the consideration of mixed liquor suspended solid (MLSS) in the secondary tank.Surface Overflow Rate (SOR) also has to be taken into account, it means that how much solid load should be settled for certain area of the plant



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QpuncakFlow rate

< 150Organic load at Qpeak (kg/m2.day)

3Minimum depth of tank (m)

150 – 180Flow rate of weir loading at Qpeak(m3/m.day)

< 30Surface loading rate at Qpeak (m3/m2. day)

2Minimum detention time at Qpeak (hour)

Value of PE < 5000Parameters

Design parameters for secondary sedimentation tank

133,

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Design Parameter for Secondary clarifierDesign of Secondary Sedimentation Tank


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The area of activated sludge tank is determined by:

Where:A = area of sedimentation tank (m2) Q = inlet flow rate to the tank (m3/day) QR = return flow rate of sludge (m3/day) xa = concentration of MLSS (suspended

solid in the biological tank, mg/L) Gminimum = optimum sedimentation (kg/m2.day) because

of two factors:i) Gravitational force in the sedimentation tankii) Force from the pump to return the sludge from sedimentation tank to the aeration tank.


A = (Q + QR) xaGminimum

Eq 9

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Activated sludge process

Figure show how return sludge occur Design parameters are given in Table [Slide 130]



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• It can be seen that, some of the sludge should be returned to the aeration tank (biological tank).

• Purpose: To maintain MLSS in the aeration tank

• In the activated sludge process, MLSS value is important to maintain at certain point in order to ensure good biodegradations process.

• If all the sludge are removed, SS are decreasing.


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Rotary Biological Contactor (RBC)Rotary Biological Contactor (RBC)

inflow outlet

primary sludge

sludge return

primarysedimentation clarifier

Gujer2000

An aerobic, attached-growth wastewater treatment processes


October 3, 2010


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Rotary Biological Contactor (RBC)Rotary Biological Contactor (RBC)

outlet

primary sludge

sludge return

primarysedimentation clarifier

Gujer2000

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Appropriate Treatment Steps

3. EAP 582.4 Waste Water Engineering Treatment Principles and Design_Session3

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