wrd ot activated sludge process 445196 7
DESCRIPTION
waste water processTRANSCRIPT
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ACTIVATED SLUDGEPROCESS
ACTIVATED SLUDGEPROCESS
Prepared byMichigan Department of Environmental Quality
Operator Training and Certification Unit
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ACTIVATED SLUDGEPROCESS
ACTIVATED SLUDGEPROCESS
To “Treat”WastewaterTo “Treat”
Wastewater
Remove (reduce) Or “Stabilize”The Material in Wastewater
Remove (reduce) Or “Stabilize”The Material in Wastewater
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SECONDARY TREATMENTSECONDARY TREATMENT
BiologicalBiological Wastewater Treatment
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Microorganisms consume organic matter from the wastewater, using oxygen for
respiration
FoodFood
FoodFoodFoodFood
OO22OO22
OO22OO22
Millions of aerobic and facultative micro-organisms remove pollutants thru living and growing process
SECONDARY TREATMENTSECONDARY TREATMENT
BiologicalBiological Wastewater Treatment
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Activated SludgeActivated SludgeSuspended Growth, Biological Treatment
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SecondaryClarifier
Activated Sludge SystemActivated Sludge System
Air
Pri.Eff. Sec. Eff.Aeration
Tank
Return Activated Sludge (RAS)
Waste Activated Sludge (WAS)
MLSSBiomass
(suspended)
Biomass(suspended)
→Provides Oxygen and Mixing
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Growth rate produces about 0.7 lbs of biological solids per
lb BOD removed
Activated SludgeActivated Sludge
Need favorable conditions for growth and for separation from the water
Biological solids are used over and over
Suspended Growth, Biological Treatment
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Aeration Tank
Secondary Clarifier
Mixed Liquor(MLSS)
Mixed Liquor(MLSS)
PrimaryEffluent
ReturnSludge
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PARTS OF A GENERALIZED BACTERIAL PARTS OF A GENERALIZED BACTERIAL CELL OF THE BACILLUS TYPECELL OF THE BACILLUS TYPE
Capsule
Cell Membrane
Cell Wall
Flagellum
NuclearMatter
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FoodStorage
Slime Layer
New Cells
Oxygen
Soluble Organics
CellMembrane
NH3
CO2
H2O
AdsorbedParticle
Enzymes
Wastewater
(Absorption)
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Mixed LiquorMixed Liquor
FlocculationA process of contact and adhesion whereby the particles of a dispersionform larger-size clusters.
FlocculationA process of contact and adhesion whereby the particles of a dispersionform larger-size clusters.
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ADSORPTION And
ABSORPTION
Aeration Tank
Secondary Clarifier
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Screening
Grit
Influ
ent
Effluent
Disinfect
PrimaryClarifiers Aeration
TanksSecondaryClarifiers
LandApplication
Typical Flow-Through Activated Sludge Plant
RASWAS
Sludge Processingand Storage
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Biological Wastewater Treatment
Three Steps
Adequate MixingEnough Detention Time
1. Transfer of Food from Wastewater to Cell.
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2. Conversion of Food to New Cells and Byproducts.
Acclimated BiomassUseable Food SupplyAdequate D.O.Proper Nutrient Balance
100 : 5 : 1 C : N : P
Biological Wastewater Treatment
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3. Flocculation and Solids Removal
Biological Wastewater Treatment
Proper MixingProper Growth Environment
Secondary Clarification
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3. Flocculation and Solids Removal
Biological Wastewater Treatment
Proper Growth Environment
Filamentous Bacteria – Form Strings
Mixed Liquor Does Not Compact - Bulking
Must Have ControlsMust Have Controls
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Air
SecondaryClarifier
Control FactorsControl Factors
AerationTank
MLSS
OrganicLoad, F:M
Biomass Quantity and Age
D.O.D.O.
Hydraulic LoadSolids Load
Settleability
Sludge Blanket Depth
Return Activated Sludge
Waste Activated Sludge
Pri.Eff. Sec. Eff.
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SecondaryClarifier
Activated Sludge SystemActivated Sludge System
Organic LoadOrganic Load
Pri.Eff. Sec. Eff.AerationTank
= = PoundsPounds of Organics ( of Organics (BODBOD))Coming into Aeration TankComing into Aeration Tank
MLSS
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Pounds =
CALCULATION OF POUNDSCALCULATION OF POUNDS
ConcentrationOf STUFF
In theWater
Conc. x Flow (or Volume) x 8.34 Lbs/gallon
XQuantityOf Water
The STUFFIs In
WeightOf TheWater
X
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Pounds =
CALCULATION OF POUNDSCALCULATION OF POUNDS
Conc. x Flow (or Volume) x 8.34 Lbs/gallon
Flow (Volume) and Concentration must be expressed in specific units.
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Concentration must be expressed as parts per million parts.
Concentration is usually reported asmilligrams per liter.
This unit is equivalent to ppm.
1 mg = 1 mg = 1 mg = ppm liter 1000 grams 1,000,000 mg
ppm = PartsMil Parts
= Lbs.Mil Lbs.
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Flow or Volume must be expressed as millions of gallons:
= MG
i.e.) A tank contains 1,125,000 gallons of water. How many million gallons are there?
1,125,000 gal = 1.125 MG 1,000,000 gal/MG
1,000,000 gal/MGgallons
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Lbs. =
When Volume is expressed as MG and concentration is in ppm,
the units cancel to leave only Pounds.
Concentration x Volume x 8.34 Lbs/gallon
Lbs.M Lbs. X X M gal Lbs.
gal
= Lbs
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Lbs./Day =
When Flow is expressed as MGD and concentration is in ppm,
the units cancel to leave Pounds/Day.
Concentration x Flow x 8.34 Lbs/gallon
Lbs.M Lbs. X X M gal Lbs.
galDay
= Lbs/Day
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EXAMPLE:
How many pounds of suspended solids leave a facility each day if the flow rate is150,000 gal/day and the concentration of suspended solids is 25 mg/L?
Lbs/day = Conc. (mg/L) x Flow (MGD) x 8.34 Lbs
galLbs/day = 25 mg/L x 150,000 gal/day x 8.34 Lbs 1,000,000 gal/MG gal
= 25 x 0.15 x 8.34
= 31 Lbs/day
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SecondaryClarifier
Activated Sludge SystemActivated Sludge System
Organic LoadOrganic Load
Pri.Eff. Sec. Eff.AerationTank
= = PoundsPounds of Organics ( of Organics (BODBOD))Coming into Aeration TankComing into Aeration Tank
MLSS
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Example ProblemExample ProblemBOD LoadingBOD Loading
An activated sludge plant receives 2.0 MGD from the primary clarifiers at 120 mg/L BOD. Calculate the organic loading (Lbs/D BOD) on the activated sludge process.
Work Calculation on Separate PaperAnswer Given on Next Slide
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Example ProblemExample ProblemBOD LoadingBOD Loading
An activated sludge plant receives 2.0 MGD from the primary clarifiers at 120 mg/L BOD. Calculate the organic loading (Lbs/D BOD) on the activated sludge process.
LbsDay
= 120 mg/L X 2.0 MGD X 8.34 Lbs Gal
= 2001.6 Lbs BOD Day
Lbs/day = Conc. (mg/L) x Flow (MGD) x 8.34 Lbs
gal
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OXYGEN DEMAND
Biochemical Oxygen DemandB.O.D.
The Quantity of Oxygen Used in the Biochemical Oxidation of
Organic Material.
5 Day Test
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OXYGEN DEMAND
Biochemical Oxygen DemandB.O.D.
Best to Use a “Moving Average” to Determine the Average Impact
on a Treatment System.
5 Day Test
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BOD Moving Average
Date Pounds of BOD
9/29 2281
9/30 2777
10/1 1374
10/2 2459
10/3 960
10/4 1598
10/5 2076
10/6 1577
10/7 2351
10/5 2281 2777 1374 2459 960 1598 207613,525
13,525 7
= 1932
10/613,525- 2281+ 157712,821
12,821 7
= 1832
Calculate the 7 day moving average of pounds of BOD for 10/5 and 10/6.
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Need to Balance Organic Load (lbs BOD) Need to Balance Organic Load (lbs BOD) With Number of Active Organisms in With Number of Active Organisms in
Treatment SystemTreatment System
Food to MicroorganismRatio
F:M orFM
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How Much Food ?
How is M (Microorganisms) measured?
Mixed Liquor Volatile Suspended Solids (MLVSS)
Lbs/D BOD = FLOW (MGD) X 8.34 Lbs/Gal X P.E. BOD (mg/L)
M = Pounds MLVSSM = Pounds MLVSS(In Aeration Tank)(In Aeration Tank)
F = Pounds BODF = Pounds BOD(Coming into Aeration Tank)(Coming into Aeration Tank)
Primary Effluent BOD
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Mixed Liquor Suspended Solids (MLSS) and
Mixed Liquor Volatile Suspended Solids (MLVSS)
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Mixed Liquor Suspended Solids (MLSS) and
Mixed Liquor Volatile Suspended Solids (MLVSS)
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Wt. of Solids + Paper, mg Wt. of Paper, mg Wt. of Solids, mg
Determining MLSSDetermining MLSS
Solids
Wt. of Solids, mgVolume of Sample, L MLSS, mg/LMLSS, mg/L
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Determining MLVSSDetermining MLVSS
Wt. of Dish + Solids, mgWt. of Dish + Ash, mg
Wt. of Volatile Solids, mg
Solids
VolatileSolids
550 oC
Wt. of Volatile Solids, mgVolume of Sample, L MLVSS, mg/LMLVSS, mg/L
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How Much Food ?
How is M (Microorganisms) measured?
Mixed Liquor Volatile Suspended Solids (MLVSS)
Lbs/D BOD = FLOW (MGD) X 8.34 Lbs/Gal X P.E. BOD (mg/L)
M = Pounds MLVSSM = Pounds MLVSS(In Aeration Tank)(In Aeration Tank)
F = Pounds BODF = Pounds BOD(Coming into Aeration Tank)(Coming into Aeration Tank)
Primary Effluent BOD
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Analysis Gave Us M (MLVSS) In mg/L
How Do We Get Volume ?
Lbs/D BOD = Volume (MG) X 8.34 Lbs/Gal X MLVSS (mg/L)
Volume Of What ?Where Microorganisms Are
Aeration Tank
How Do We Get To Pounds?
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Aeration Tank Volume (MG)
Aeration Tank Volume (MG)
L (ft) X W (ft) X SWD (ft) = Volume (ftL (ft) X W (ft) X SWD (ft) = Volume (ft33))
ftft3 3 X 7.48 gal/ftX 7.48 gal/ft33 = gallons = gallons
gallons / 1,000,000 = million gallons (MG)gallons / 1,000,000 = million gallons (MG)
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Aeration Tank Volume (MG)
Aeration Tank Volume (MG)
Example Calculation:Calculate the volume in million gallons of an aeration tank that is 120 ft long, 35 ft wide, with a SWD of 15 ft.
A.
V = L X W X D
V = 120 ft X 35 ft X 15 ft = 63,000 ft3
63,000 ft3 X 7.48 gal ft3
= 471,240 gallons
= 0.471 MG471,240 gallons / 1,000,000
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Aeration Tank Volume (MG)
Aeration Tank Volume (MG)
Example Calculation:The average BOD load on this aeration tank is 1954 lbs/day. Calculate the organic loading in lbs/day/1000ft3.
B.
1954 lbs/day63,000 ft3
X 1,000 = 31.0 lbs/day/1000ft3
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Need to Balance Organic Load (lbs BOD) Need to Balance Organic Load (lbs BOD) With Number of Active Organisms in With Number of Active Organisms in
Treatment SystemTreatment System
Food to Microorganism Ratio
F:M orFM
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How Much Food (FF) ? Pounds BOD
How is MM (Microorganisms) measured?
Mixed Liquor Volatile Suspended Solids (MLVSS)
Lbs/D BOD = FLOW (MGD) X 8.34 Lbs/Gal X Pri. Eff. BOD (mg/L)
M = Pounds MLVSSM = Pounds MLVSS
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Pounds =
CALCULATION OF POUNDSCALCULATION OF POUNDS
ConcentrationOf STUFF
In theWater
Conc. x Flow (or Volume) x 8.34 Lbs/gallon
XQuantityOf Water
The STUFFIs In
WeightOf TheWater
X
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Pounds of Volatile Solidsin the Aeration Tank
Volume Aeration Tank, MG X MLVSS, mg/L X 8.34 Lbs/gal
Example Problem:Calculate the pounds of volatile solids in an aeration tank that has a volume of 0.471 MG and the concentration of volatile suspended solids is 1700 mg/L.
Lbs MLVSS =
Lbs = 0.471 MG X 1700 mg/L X 8.34 lbs/gal
= 6678 lbs MLVSS
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Food to Microorganism RatioFood to Microorganism Ratio
Lbs of BOD
Lbs of MLVSS
Example Problem:The 7-day moving average BOD is 2002 lbs and the mixed liquor volatile suspended solids is 6681 pounds. Calculate the F/M ratio of the process.
FM
= 2002 lbs BOD6681 lbs MLVSS
= 0.30
=
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Conventional Activated SludgeF:M
F:M
Extended Aeration Activated Sludge
0.25 - 0.45
0.05 - 0.15
Typical Range:
The F/M Ratio for Best Treatment Will Vary for Different Facilities
Determined by Regular Monitoring andComparing to Effluent Quality
Often Will Vary Seasonally
Food to Microorganism RatioFood to Microorganism Ratio
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Food to Microorganism RatioFood to Microorganism Ratio
Lbs of BOD
Lbs of MLVSS
Calculate Often to Monitor/Control
=
Monthly (Minimum)Weekly (Better)
Use Moving Average
FM
=
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Food to Microorganism Ratio Calculations
F/M Ratio is Used to Determine the Lbs of MLVSS Needed at a Particular Loading Rate
suppose F/M of 0.30 is desiredand BOD loading is 1200 lbs/day
FM
= 0.30
1200 lbs
0.30= 4000 lbs MLVSS
F
F/M= M (Lbs MLVSS)
Lbs BOD lbs MLVSS
=F/M
F0.30
= M
FOR DAILY USEFOR DAILY USE
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If we Know the Pounds of MLVSS Needed and the Volume of the Aeration Tank We Can Calculate MLVSS, mg/L.
Calculate the MLVSS, mg/L givenan Aeration Tank Volume of 0.20 MG.
4000 lbs = 0.20 MG X 8.34 lbs X ? mg/L gal
2398 mg/L4000 lbs
0.20 MG X 8.34 lbs/gal=
Food to Microorganism Ratio Calculations
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Problem A:How many pounds of MLVSS should be maintained in an aeration tank with a volume of 0.105 MG receiving primary effluent BOD of 630 lbs/d ? The desired F:M is 0.3.
= 630 lbs/d 0.3
= 2100 lbs MLVSS F F/M
= M
F:M Calculations
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Problem B:What will be the MLVSS concentration in mg/L ?
2100 lbs = Conc X 0.105 MG X 8.34 lbs/gal
2100 lbs 0.105 MG X 8.34 lbs/gal
= 2398 mg/L
F:M Calculations
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Food to Microorganism Ratio Calculations
F/M Ratio is Used to Determine the Lbs of MLVSS Needed at a Particular Loading Rate
Can you Calculate the Pounds of MLVSS Needed for a Specific F/M
andWhat Concentration That Would Be in an Aeration Tank?
Prove It !
F
F/M= M (Lbs MLVSS)
Lbs BOD lbs MLVSS
=F/M
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Problem C:How many pounds of MLVSS should be maintained in an aeration tank with a volume of 0.471 MG receiving primary effluent BOD of 2502 lbs/d ? The desired F:M is 0.3.
F:M Calculations II
Problem D:What will be the MLVSS concentration in mg/L ?
Work Calculations on Separate PaperAnswers Given on Next Slides
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Problem C:How many pounds of MLVSS should be maintained in an aeration tank with a volume of 0.471 MG receiving primary effluent BOD of 2502 lbs/d ? The desired F:M is 0.3.
F .F/M
= M = 2502 lbs/d 0.3
= 8340 lbs MLVSS
F:M Calculations II
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Problem D:What will be the MLVSS concentration in mg/L ?
8340 lbs = Conc X 0.471 MG X 8.34 lbs/gal
8340 lbs .0.471 MG X 8.34 lbs/gal
= 2123 mg/L
F:M Calculations II
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SecondaryClarifier
Control Factors
Air
PEFE
AerationTank
Return Activated Sludge
Waste Activated Sludge
MLSS
OrganicLoad, F:M
Biomass Quantity and Age Hydraulic Load
Solids Load
SettleabilitySludge Blanket Depth
D.O.
![Page 60: Wrd Ot Activated Sludge Process 445196 7](https://reader036.vdocuments.us/reader036/viewer/2022081501/563dbb7f550346aa9aadb485/html5/thumbnails/60.jpg)
Time
Gro
wth
Rat
e o
f O
rgan
ism
s
X
X
Graph Showing Growth Phases in a Biological
System
When Food Supply is Introduced into a
Biological Treatment System that is in Start-up
Abundance of Food
Few Organisms
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Time
Gro
wth
Rat
e o
f O
rgan
ism
sLag
Growth
Organisms Begin to Acclimate Producing Needed Enzymes
Food Begins to be Consumed
Organism Population Begins to Increase
Graph Showing Growth Phases in a Biological
System
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Time
Gro
wth
Rat
e o
f O
rgan
ism
sLag
GrowthLog
Growth
FoodRapidlyConsumed Organisms Acclimated
Organism PopulationRapidly Increases
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Time
Gro
wth
Rat
e o
f O
rgan
ism
sLag
GrowthLog
GrowthDecliningGrowth
Food
Organism Population Growth Limited by Food
Supply
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Time
Gro
wth
Rat
e o
f O
rgan
ism
sEndogenous
GrowthLag
GrowthLog
GrowthDecliningGrowth
Food
Food Supply Depleted -Organism Growth Rate
Continues Decline
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Time
Gro
wth
Rat
e o
f O
rgan
ism
sEndogenous
GrowthLag
GrowthLog
GrowthDecliningGrowth
Food
Sludge Production
Stored Food Metabolized -
Organisms Feed on One Another
(Producing Less Sludge)
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Time
Gro
wth
Rat
e o
f O
rgan
ism
sEndogenous
GrowthLag
GrowthLog
GrowthDecliningGrowth
Food
Sludge Production
Summary
Graph Showing Growth Phases in a Biological
System
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This graph illustrates that the activities of
Microorganisms in a biological treatment
system is related to the Average Age of the
Organisms in the System or the “CRT” of the System
Note: The CRT is Controlled in an
Activated Sludge System by Wasting which will be
discussed later.
Graph Showing Growth Phases in a Biological System
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Cell Residence Time, CRTMean Cell Residence Time, MCRTSludge Age, SA
MCRT = Total MLVSS, lbs (Aerator + Clarifier)Total MLVSS Wasted + Effluent TSS, lbs/d
SA, days = Suspended Solids in Aerator, lbsSuspended Solids in PE, lbs/day
The Average Length of Time in Days that an Organism Remains in the Secondary Treatment System
Biomass Age
The SA and MCRT Calculations are Seldom UsedThe Most Common (and Best for Most Processes) Is the
Cell Residence Time
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Cell Residence Time
The Average Length of Time in Days that an Organism Remains in the Secondary Treatment System
CRT, days = Total MLVSS, lbsTotal MLVSS Wasted, lbs/d
Cell Residence Time, CRT
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The Average Length of Time in Days that an Organism Remains in the Secondary Treatment System
CRT, days = Total MLVSS, lbsTotal MLVSS Wasted, lbs/d
Example:
MLVSS = 6681 lbsMLVSS Wasted = 835 lbs/d
Calculate the CRT.
6681 lbs835 lbs/d
CRT =
CRT = 8.0 Days
Cell Residence Time, CRT
Cell Residence Time
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Like The F/M RatioThe CRT for Best
TreatmentWill Vary for Different
Facilities
Determined by Regular Monitoring and
Comparing to Effluent Quality
Often Will Vary Seasonally
Cell Residence Time
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Conventional Activated Sludge
Aerator Detention Time
F:M
CRT
Aerator Detention Time
F:M
CRT
Extended Aeration Activated Sludge
4 - 8 Hrs.
0.25 - 0.45
4 - 6 Days
16 - 24 Hrs.
0.05 - 0.15
15 - 25 Days
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Time
Gro
wth
Rat
e o
f O
rgan
ism
sEndogenous
GrowthLag
GrowthLog
GrowthDecliningGrowth
Food
Sludge Production
ConventionalTreatment
Extended Air
Young Sludge Old Sludge
![Page 74: Wrd Ot Activated Sludge Process 445196 7](https://reader036.vdocuments.us/reader036/viewer/2022081501/563dbb7f550346aa9aadb485/html5/thumbnails/74.jpg)
Young SludgeYoung Sludge
• Start-up or High BOD Load
• Few Established Cells
• Log Growth
• High F:M
• Low CRT
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Young SludgeYoung Sludge
Poor FlocculationPoor SettleabilityTurbid Effluent
WhiteBillowingFoam
High O2
Uptake Rate
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Old SludgeOld Sludge
• Slow Metabolism
• Decreased Food Intake
• Low Cell Production
• Oxidation of Stored Food
• Endogenous Respiration
• Low F:M
• High CRT
• High MLSS
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Old SludgeOld Sludge
Dense, Compact Floc
Fast Settling
Straggler Floc
Slurp
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SecondaryClarifier
Control Factors
Air
PEFE
AerationTank
Return Activated Sludge
Waste Activated Sludge
MLSS
OrganicLoad, F:M
Biomass Quantity and Age Hydraulic Load
Solids Load
SettleabilitySludge Blanket Depth
D.O.
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Cell Residence Time
The Average Length of Time in Days that an Organism Remains in the Secondary Treatment System
CRT, days = Total MLVSS, lbsTotal MLVSS Wasted, lbs/d
Waste Activated Sludge (WAS)
The CRT for Facility is Controlled/Maintained by Wasting the Appropriate Amount of
Excess Biomass
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SecondaryClarifier
Control Factors
Air
PEFE
AerationTank
Return Activated Sludge
Waste Activated Sludge(WAS)
MLSS
OrganicLoad, F:M
Biomass Quantity and Age Hydraulic Load
Solids Load
SettleabilitySludge Blanket Depth
D.O.
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Sludge Wasting RatesSludge Wasting Rates
CRT(days) = Lbs of MLVSS in aerators Lbs/day WAS VSS
Therefore:
Lbs WAS VSS =day
Lbs of MLVSS in aerators CRT (days)
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Sludge Wasting RatesSludge Wasting Rates
With a known RAS VSS concentration, the WAS Flow in MGD can be calculated:
= WAS (MGD) lbs/day WAS VSS RAS VSS (mg/L) x 8.34 lbs gal
MGD x 1,000,000 = gallons per day
Lbs/ day = mg/L x 8.34 lbs gal
x ? MGD
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Sludge Wasting RatesSludge Wasting Rates
If wasting is to be done over a 24 hr. period:
WAS (gpm) = gallons/day .1440 minutes/day
If wasting is to be done over a shorter period:
WAS (gpm) = gallons/day .min wasting to be done/day
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Sludge Wasting RatesSludge Wasting RatesExample CalculationsExample Calculations
Problem #1:
A cell residence time of 5.8 days is desired. With 5800 pounds of MLVSS in the aeration tanks, calculate the pounds of VSS that must be wasted per day.
lbs/day = 5800 lbs 5.8 days
= 1000 lbs/day
Need to Waste 5800 lbs in 5.8 Days
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Sludge Wasting RatesSludge Wasting RatesProblem #2:Calculate the flow rate in MGD that must be pumped in order to waste the number of pounds calculated in Problem #1 given a Return Sludge concentration of 9000 mg/L VSS.
lbs/day = conc. x 8.34 lbs/gal x MGD
1000 lbs = 9000 mg/L x 8.34 lbs/gal x MGD day
= 0.0133 MGD = 13,300 gal/day
1000 lbs/day9000 mg/L x 8.34 lbs/gal
= MGD Wasted
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Problem #3:
Calculate the wasting rate in gallons per minute if the wasting was done in 24 hours.
Sludge Wasting Rates
13,300 gal day
1 day 24 hrs
x x 1 hour 60 min
= 9.24 gals min
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Sludge Wasting Rates
Problem #4:
Calculate the wasting rate in gallons per minute if the wasting was done in 4 hours.
13,300 gal 4 hr
x 1 hr 60 min
= 55.4 gal/min
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Sludge WastingSludge Wasting
Excess Biological Solids eliminated from the secondary treatment system to control the cell residence time of the biomass
When to Waste:Continuous (Whenever Possible)
Or If Necessary (Piping, Pumping or Valve Limitations)
Intermittent - During Low Load Conditions
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Sludge WastingSludge WastingWhere to:
RASWAS
Solids Handling
Primary Clarifiers
Disadvantage - Are Solids Really Wasted?Advantage - Co-Settling
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Sludge WastingSludge WastingWhere to:
Solids Handling
Disadvantage – Thinner Solids to Solids Process
Advantage – Know Solids are Out of the System
Sludge Processing (Thickening, Stabilization, etc.)
RASWAS
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Sludge WastingSludge Wasting
How Much:
Secondary Sludge Wasting One of the Most Important Controls
Wasting Controls the Most Important Aspect of Treatment, the Biomass
Population
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Sludge WastingSludge Wasting
How Much:
Proper Wasting ControlAnd
Metering is Essential
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SecondaryClarifier
Control Factors
Air
PEFE
AerationTank
Return Activated Sludge
Waste Activated Sludge
MLSS
OrganicLoad, F:M
Biomass Quantity and Age Hydraulic Load
Solids Load
SettleabilitySludge Blanket Depth
D.O.
![Page 94: Wrd Ot Activated Sludge Process 445196 7](https://reader036.vdocuments.us/reader036/viewer/2022081501/563dbb7f550346aa9aadb485/html5/thumbnails/94.jpg)
Time
Gro
wth
Rat
e o
f O
rgan
ism
sEndogenous
GrowthLag
GrowthLog
GrowthDecliningGrowth
Food
Sludge Production
ConventionalTreatment
Extended Air
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Growth rate produces about 0.7 lbs of biological solids per
lb BOD removed
Activated SludgeActivated Sludge
Need favorable conditions for growth and for separation from the water
Biological solids are used over and over
Suspended Growth, Biological Treatment
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Yield Coefficient (Y)
Growth Rate
Y =
Pounds of Biological Solids Produced
Per Pound of BOD Removed
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Yield Coefficient (Y)Growth Rate
Pounds of Biological Solids Produced
Per Pounds of BOD Removed
ExampleExample
Average Concentration Of BOD Entering Aeration125 mg/L
Average Concentration of BOD from Secondary System5 mg/L
Average Plant Flow2.0 MGD
Average RAS Concentration (Wasting from Return)
8000 mg/L
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Yield Coefficient (Y)Growth Rate
Pounds of Biological Solids Produced
Per Pounds of BOD Removed
ExampleExample
BOD Removed = 125 mg/L – 5 mg/L = 120 mg/L
At 2.0 MGDLbs BOD Removed = 2 MGD X 8.34 X 120 mg/L
= 2002 Lbs/Day
At Y= 0.7At Y= 0.7Biomass Produced = 2002 Lbs/Day X 0.7
= 1401 Lbs/Day
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Yield Coefficient (Y)
At 2.0 MGDLbs BOD Removed = 2 MGD X 8.34 X 120 mg/L
= 2002 Lbs/Day
At Y= 0.7At Y= 0.7Biomass Produced = 2002 Lbs/Day X 0.7
= 1401 Lbs/Day
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Yield Coefficient (Y)
At 2.0 MGDLbs BOD Removed = 2 MGD X 8.34 X 120 mg/L
= 2002 Lbs/Day
At Y= 0.7At Y= 0.7Biomass Produced = 2002 Lbs/Day X 0.7
= 1401 Lbs/Day
RAS at 8000 mg/LRAS at 8000 mg/L
1401 lbs/day 1401 lbs/day .8000 mg/L X 8.34 lbs/gal8000 mg/L X 8.34 lbs/gal
= 20,998 gallons WAS= 20,998 gallons WAS
(To Balance Solids Produced)
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Yield Coefficient (Y)
At 2.0 MGDLbs BOD Removed = 2 MGD X 8.34 X 120 mg/L
= 2002 Lbs/Day
At Y= 0.5At Y= 0.5Biomass Produced = 2002 Lbs/Day X 0.5
= 1001 Lbs/Day
RAS at 8000 mg/LRAS at 8000 mg/L
1001 lbs/day 1001 lbs/day .8000 mg/L X 8.34 lbs/gal8000 mg/L X 8.34 lbs/gal
= 15,002 gallons WAS= 15,002 gallons WAS
(To Balance Solids Produced)
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Yield Coefficient (Y)
The Difference:
20,998 gallons15,002 gallons 5,996 gallons
6000 gal/day X 365 day/year = 2,190,000 gallons per year
Per DayPer Day
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0 0.20 0.40 0.60 0.80 1.00 1.20
F:M RatioF:M Ratio
ConventionalExtendedAir
High Rate
300
200
100
Slu
dg
e V
olu
me
Ind
exS
lud
ge
Vo
lum
e In
dex
A Major Advantage of Extended Aeration(Old Sludge Age)
Less Solids Produced
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Yield Coefficient (Y)
Growth RateY =
Pounds of Biological Solids Produced
Per Pound of BOD Removed
How to Determine Y for a Facility?How to Determine Y for a Facility?
Use Monthly Average of Pounds of Solids WastedDivided by
the Monthly Average of Pounds of BOD Removed
Should be Monitored Regularly (Monthly)
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Returned from Secondary Clarifier
Activated SludgeActivated Sludge
Need favorable conditions for growth and for separation from the water
Biological solids are used over and over
Suspended Growth, Biological Treatment
![Page 106: Wrd Ot Activated Sludge Process 445196 7](https://reader036.vdocuments.us/reader036/viewer/2022081501/563dbb7f550346aa9aadb485/html5/thumbnails/106.jpg)
Aeration Tank
Secondary Clarifier
Mixed LiquorMixed Liquor
PrimaryEffluent
ReturnSludge
![Page 107: Wrd Ot Activated Sludge Process 445196 7](https://reader036.vdocuments.us/reader036/viewer/2022081501/563dbb7f550346aa9aadb485/html5/thumbnails/107.jpg)
SecondaryClarifier
Control Factors
Air
PEFE
AerationTank
Return Activated Sludge
Waste Activated Sludge
MLSS
OrganicLoad, F:M
Biomass Quantity and Age Hydraulic Load
Solids Load
SettleabilitySludge Blanket Depth
D.O.
![Page 108: Wrd Ot Activated Sludge Process 445196 7](https://reader036.vdocuments.us/reader036/viewer/2022081501/563dbb7f550346aa9aadb485/html5/thumbnails/108.jpg)
Return Activated SludgeReturn Activated Sludge
Biological Solids (Mixed Liquor Solids) which have settled in the secondary clarifier, continuously returned to the aeration system.
Why:
• Control sludge blanket in clarifier
• Maintain a sufficient population of
active organisms in service
It’s Not the FoodIt’s the Bugs
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Return Activated SludgeReturn Activated Sludge
Biological Solids (Mixed Liquor Solids) which have settled in the secondary clarifier, continuously returned to the aeration system.
Why:
• Control sludge blanket in clarifier
• Maintain a sufficient population of
active organisms in service
Not a Means of Controlling MLSS
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Return Activated SludgeReturn Activated Sludge
Biological Solids (Mixed Liquor Solids) which have settled in the secondary clarifier, continuously returned to the aeration system.
Why:
• Control sludge blanket in clarifier
• Maintain a sufficient population of
active organisms in service
Controls Solids Depth in
Seconday Clarifier
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• 1 – 3 Feet Depth
• Too Much – Solids Over Weir
• Too Little – Thin RAS Concentration (More Volume When Wasting)
RAS Control:
Return Activated SludgeReturn Activated Sludge
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• Consistent Flow Rate
• % Influent Flow
• RAS Metering
RAS Control:
Setting the RAS Rate
How Much is Enough (or Too Much)
Sludge Blanket Depth
Sludge Judging
Return Activated SludgeReturn Activated Sludge
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QRQ
Q + RQ
RAS
MLSS
(RQ + Q) X 8.34 lbs/gal X MLSS (mg/L)
Lbs of Material Into Clarifier
RQ X 8.34 lbs/gal X RAS (mg/L)
Lbs of Material Out of Clarifier
RAS Mass BalanceRAS Mass Balance
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Lbs Into Clarifier Lbs Into Clarifier = Lbs Out of ClarifierLbs Out of Clarifier
Q
RQ
Q + RQ
RAS SS
MLSS
(Q + RQ) X 8.34X MLSS = RQ X 8.34 X RAS
(Q + RQ) X MLSS = RQ X RAS SS
(RQ X MLSS) + (Q X MLSS) = RQ X RAS SS
Q X MLSS = RQ X RAS SS - RQ X MLSS
Q X MLSS = RQ X (RAS - MLSS)
Q X MLSS RAS SS - MLSS
= RQ
RAS Mass BalanceRAS Mass Balance
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Q X MLSS RAS SS - MLSS
RQ =
Most People Forget the Derivation of the Formula and Just Memorize the Formula
Return Activated SludgeReturn Activated Sludge
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Q X MLSS RAS SS - MLSS
RQ =
Units for RQ will Match Units for Q
100 X MLSS RAS - MLSS
% RQ =
Return Activated SludgeReturn Activated Sludge
To Express RQ as %% of Influent Flow:
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Return Rates - Example Calculations
Given: MLSS = 2400 mg/LRAS SS = 6500 mg/LFlow = 2.0 MGD
1. Calculate the Return Sludge Rate in MGD needed to keep the solids in the process in balance.
RAS, MGD = 2.0 MGD X 2400 mg/L6500 mg/L - 2400 mg/L
= 48004100
= 1.17 MGD
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Given:MLSS = 2400 mg/LRAS SS = 6500 mg/L
2. Calculate the Return Sludge Rate in % of plant influent flow needed to keep the solids in the process in balance.
% RAS = 100 X 2400 mg/L 6500 mg/L - 2400 mg/L
% RAS = 240000 4100
= 58.5 %
Return Rates - Example Calculations
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Return Rates - Practice Calculations
Given: MLSS = 2700 mg/LRAS SS = 8200 mg/LFlow = 2.5 MGD
1. Calculate the Return Sludge Rate in MGD needed to keep the solids in the process in balance.
2. Calculate the Return Sludge Rate in % of plant influent flow needed to keep the solids in the process in balance.
Work Calculations on Separate PaperAnswers Given on Next Slides
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Return Rates - Practice Calculations
Given: MLSS = 2700 mg/LRAS SS = 8200 mg/LFlow = 2.5 MGD
Calculate the Return Sludge Rate in MGD needed to keep the solids in the process in balance.
RAS, MGD = 2.5 MGD X 2700 mg/L8200 mg/L - 2700 mg/L
= 67505500
= 1.23 MGD
1.
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Given:MLSS = 2700 mg/LRAS SS = 8200 mg/L
2. Calculate the Return Sludge Rate in % of plant influent flow needed to keep the solids in the process in balance.
% RAS = 100 X 2700 mg/L 8200 mg/L - 2700 mg/L
% RAS = 270,000 5500
= 49.1 %
Return Rates - Example Calculations
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Return Activated SludgeReturn Activated Sludge
Q X MLSS RAS SS - MLSS
RQ =
Units for RQ will Match Units for Q
100 X MLSS RAS - MLSS
% RQ =
To Express RQ as %% of Influent Flow:
In Summary
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SecondaryClarifier
Control Factors
Air
PEFE
AerationTank
Return Activated Sludge
Waste Activated Sludge
MLSS
OrganicLoad, F:M
Biomass Quantity and Age Hydraulic Load
Solids Load
Settleability
D.O.
Sludge Blanket Depth
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Biological Wastewater TreatmentThree Steps
1. Transfer of Food from Wastewater to Cell.
2. Conversion of Food to New Cells and Byproducts.
3. Flocculation and Solids Removal
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Biological Wastewater TreatmentThree Steps
3. Flocculation and Solids Removal
Even if the First Two Steps are Effective,If Settling and Separation is Poor
RAS Will be Thin and/or Solids May Be Lost in the Effluent
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Settleometer TestSettleometer Test
What Determines the Volume of Settled Sludge?What Determines the Volume of Settled Sludge?
Time
Mass of Solids Compaction
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Determination of the Settling Properties(Compaction) of MLSS
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Although a 1000mL Graduated Cylinder
May be UsedA Settleometer
Designed for this Test is Best
The Wider Container More Approximates a Clarifier
Settleometer TestSettleometer Test
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Although a 1000mL Graduated Cylinder
May be UsedA Settleometer
Designed for this Test is Best
Settleometer TestSettleometer Test
A Settleometer has a Capacity of 2000 mL
Graduated in mL/Liter
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Settleometer TestSettleometer Test
Collect SampleBelow Scum Line
Set up Settling Test Immediately
Also Determine MLSS, mg/L on a Portion of Same Sample
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MixGently
Settleometer TestSettleometer TestFill Settleometer to
1000 Graduation
Start Timer
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Settleometer TestSettleometer Test
While Settling Observe:
Color of ML and SupernatantSupernatant TurbidityStraggler Floc
Record Settled Sludge VolumeEvery 5 Minutes for 30 Minutes
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Settled Sludge
Volume
Sludge Blanket
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Minutes
Set
tled
Slu
dg
e V
olu
me,
m
Ls
1000
900
800
700
600
500
400
300
200
100
5 10 15 20 25 30 35 40 45 50 55 60
Good
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Minutes
Set
tled
Slu
dg
e V
olu
me,
m
Ls
1000
900
800
700
600
500
400
300
200
100
5 10 15 20 25 30 35 40 45 50 55 60
Not Good(Settling Too Fast)
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Indication of “Old” Sludge
Leaves Straggler Flocin Effluent
Settleometer TestSettleometer Test
Too Fast
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Minutes
Set
tled
Slu
dg
e V
olu
me,
m
Ls
1000
900
800
700
600
500
400
300
200
100
5 10 15 20 25 30 35 40 45 50 55 60
Not Good(Settling Too Slow)
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Settleometer TestSettleometer Test
Not Compacting (Bulking)
Solids Washed Outin High Flows
Too Slow
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Watch for Indications of Denitrification
Gas Bubbles in Settled Sludge
Rising Sludge
Solids SeparationSolids Separation
RateRate
CharacteristicsCharacteristics
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Sludge Volume Index (SVI)
The volume in milliliters occupied by one gram of activated sludge which has settled for 30 min.
mLs Settled in 30 minSVI = MLSS Conc, grams/L
=mLs Settled
MLSS, mg/L1000
The volume compared to weight.
(Weight [in grams] of the solids that occupy the Volume.)
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The volume in milliliters occupied by one gram of activated sludge which has settled for 30 min.
SVI Practice Problem:
30 minute settling 260 mL MLSS Conc. 2400 mg/L
mLs Settled
MLSS, mg/L1000
mLs Settled in 30 minSVI = MLSS Conc, grams/L
=
Work Calculations on Separate PaperAnswer Given on Next Slide
Sludge Volume Index (SVI)
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The volume in milliliters occupied by one gram of activated sludge which has settled for 30 min.
SVI Practice Problem:
30 minute settling 260 mL MLSS Conc. 2400 mg/L
260 2.4
= 108SVI =
260 mL 2400 mg/L 1000
SVI =
mLs Settled
MLSS, mg/L1000
mLs Settled in 30 minSVI = MLSS Conc, grams/L
=
Sludge Volume Index (SVI)
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The volume in milliliters occupied by one gram of activated sludge which has settled for 30 min.
mLs Settled
MLSS, mg/L1000
mLs Settled in 30 minSVI = MLSS Conc, grams/L
=
Typical Range for Good Settling 80 - 120The higher the number, the less compact the sludge
Sludge Volume Index (SVI)
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The grams of activated sludge which occupies avolume of 100 mL after 30 min. of settling.
SDI = grams/L of MLSS
mLs settled in 30 min.100
The weight compared to volume.
Sludge Density Index (SDI)
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The grams of activated sludge which occupies avolume of 100 mL after 30 min. of settling
30 min. Settling / 100SDI =
MLSS / 1000
The weight compared to volume.
Sludge Density Index (SDI)
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The grams of activated sludge which occupies avolume of 100 ml after 30 min. of settling
SDI Practice Problem:
30 minute settling 260 mL MLSS Conc. 2400 mg/L
mLs settled in 30 min.SDI =
grams/L of MLSS
100
Work Calculations on Separate PaperAnswer Given on Next Slide
Sludge Density Index (SDI)
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The grams of activated sludge which occupies avolume of 100 ml after 30 min. of settling
2.4 2.6
SDI = = 0.92
SDI Practice Problem:
30 minute settling 260 mL MLSS Conc. 2400 mg/L
mLs settled in 30 min.SDI =
grams/L of MLSS
100
2400 mg/L / 1000 260 mL / 100
SDI =
Sludge Density Index (SDI)
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The grams of activated sludge which occupies avolume of 100 ml after 30 min. of settling
mLs settled in 30 min.SDI =
grams/L of MLSS
100
Typical Range for Good Settling 0.8 - 1.2The lower the number, the less compact the sludge
Sludge Density Index (SDI)
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SVI - SDI Relationship
SDI = 100SVI
SVI = 100SDI
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SVI - SDI RelationshipSVI - SDI Relationship
SVI = 100
SDISDI =
100
SVI
Practice Problems:
a) What is the SDI if the SVI is 133?
b) What is the SVI if the SDI is 0.6?
Work Calculations on Separate PaperAnswers Given on Next Slide
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SVI - SDI RelationshipSVI - SDI Relationship
SVI = 100
SDISDI =
100
SVI
Practice Problems:
a) What is the SDI if the SVI is 133?
b) What is the SVI if the SDI is 0.6?
100/133 = 0.75
100/0.6 = 167
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Return Sludge Concentration Return Sludge Concentration and SDIand SDI
With the Clarifier Solids in Balance, the Settled
Sludge Concentration in the Settleometer
Will Approximate the RAS SS Concentration
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Return Sludge Concentration and SDI
SDI = mLs settled in 30 minutesMLSS, G/L
100
SDI 1.0 =100 mLs settled
1.0 G
1 G 100 mL
1 G 100 G
= 1 %=
1 G 100 mL
= 1000 mg 100 mL
= 10,000 mg 1,000 mL
10,000 mg L
=
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With Clarifier Solids in Balance :
SDI = RAS SS Conc. in Percent
SDI of 0.8
RAS SS = 0.8 % Solids
SDI X 10,000 = RAS SS in mg/L
SDI = 0.8
RAS SS = 8,000 mg/L
Return Sludge Concentration and SDI
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Sludge Volume Index
The volume in milliliters occupied by one gram of activated sludge which has settled for 30 min.
The volume compared to weight.
Sludge Density Index
The grams of activated sludge which occupies avolume of 100 mL after 30 min. of settling.
The weight compared to volume.
In SummaryIn Summary
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Sludge Volume IndexSludge Volume Index
Sludge Density IndexSludge Density Index
mLs settled in 30 min.SDI =
grams/L of MLSS
100
mLs Settled
MLSS, mg/L1000
SVI =
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SVI - SDI Relationship
SDI = 100SVI
SVI = 100SDI
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SVI - SDISVI - SDI
Typical SDI Range for Good Settling 0.8 - 1.2
Typical SVI Range for Good Settling 80 - 120
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This graph illustrates the Relationship Between The
F:M of a System to the Ability of the Biomass to
Settle in Clarifier
Relationship of F:M to Settleability System
It Shows that there are Three Areas of Operation
where the Biomass Normally Settles Well
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These Areas as Defined by F:M Ratio Are
Note: The High rate Mode is Seldom Used
Except when Followed by Additional Treatment
Relationship of F:M to Settleability System
High RateF:M 0.9 to 1.2
ConventionalF:M 0.25 to 0.45
Extended AirF:M Less than 0.2
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The Graph Also Shows the Potential Consequences of Operation with an F:M Out
Of these Ranges
Relationship of F:M to Settleability System
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ACTIVATED SLUDGEPROCESS
ACTIVATED SLUDGEPROCESS
Prepared byMichigan Department of Environmental Quality
Operator Training and Certification Unit