activated sludge process
TRANSCRIPT
Suspended growth biological processes
Farid khan
Wastewater treatment processesWastewater treatment processes
Stages of Wastewater Treatment• Primary
– Contaminants (60% of solids and 35% of BOD removed)
• Oil & Grease
• Total Suspended Solids (Css or TSS) – 60% Removed
• Pathogens• BOD – 35% removed
– Processes• Screens• Grit Settling• Scum Flotation• Primary Settling
• Secondary– Contaminants
• BOD – 90% Removed• TSS – 90% Removed
– Processes• Trickling Filter – rotating disk• Activated Sludge – Suspended and mixed• Oxidation ponds – lagoons
(promote contact between microbes and contaminants)
Stages of Wastewater Treatment
• Tertiary– Contaminants
• Nutrients• Dissolved solids (e.g., salt, other ions, etc.)
– Processes• Nitrogen removal (Denitrification) – bacteria• Phosphorus removal – precipitation or bacteria• Other chemicals – adsorption and precipitation
Stages of Wastewater Treatment
Secondary TreatmentSecondary Treatment
• This involves treating the liquid part of the wastewater biologicallybiologically. It is carried out after primary treatment (which removes some of the solid material).
• The purpose of this process is to remove the organic matter and the nitrogen from the wastewater.
• A group of microorganisms called bacteria are ‘employed’ to do the job.
Two Types (based on growth condition)Two Types (based on growth condition)
1.1. Suspended GrowthSuspended Growth
Organisms are suspended in the treatment fluid. This fluid is commonly called the “mixed liquor”. Example: Activated sludge.
2.2. Attached growth or Fixed FilmAttached growth or Fixed Film
Organisms attached to some inert media like rocks or plastic. Example: Trickling filter.
Secondary TreatmentSecondary Treatment
Suspended growth vs. Fixed film biological treatment
• Suspended growth:– Biomass well-mixed, in
suspension– Diverse microbiology– High energy aeration systems– Process control follows from
modeling– Shocked more easily– Biomass recycled– High solids production, easy
to settle– High efficiency– DO 2 to 3 mg/L– Reliable N & P removal
• Fixed film:– Biomass layered, attached– More diverse (snails!), stratified micr
obiology– Frequently relies on draft for aeration– Process control is empirically based
on organic and hydraulic loading– Resilient to shock loads– Biomass not typically recycled– Low solids production, hard to settle
(low floc formers)– Lower efficiency– Higher DO required– unreliable N & P removal
Activated Sludge Historical development
• Invented in 1914 by Ardern and Lockett in England. They aerated a batch of sewage, allowing the generated sludge to settle, decanting the supernatant, adding a fresh batch of sewage.
• During aeration sewage is mixed with a large mass of previously grown organisms.
• The solids formed are flocculent and can be removed from the liquid by settling.
• Separate control is exercised over the solid and liquid phases, so that the solids retention time in the process is much longer than the hydraulic retention time.
Characteristics of activated sludge Process
• The mostly widely used biological process for the treatment of municipal and industrial wastewaters.
• Strictly aerobicexcept anoxic variation for denitrification.
• Parts: i) aeration tank, ii) a settling tank, iii) solids recycle, and iv) a sludge wasting line
• activated sludge: microbial aggregates (flocs) in the aeration tank.→Flocs stay in suspension with mixing by aeration.
• Recycle of the activated sludge is crucial to maintaining a high concentration of cells.
Basic Porcess of Activated Sludge
Activated sludge + Wastewater + O2
CO2 , H
2O
NO3, SO
4 , PO
4
New biomass
Raw wastewater or effluent from primary treatment
Aerobic reactor
Secondary
settler
Recycle sludge
Wastage sludge
Effluent
Schematic of Municipal Sewage Treatment
• Aerobic heterotrophic bacteria are main “activated”organisms. →Microbial community is highly diverse and competative.
• Floc formation is really key –individual bacteria do not settle fast enough to be captured in the settling tank. Cells not in flocs are washed out.
• The suspended flocs particles are called “activated”.
Characteristics of activated sludge Process
Why “activated sludge”?
The process involves the production of an activated mass of microorganisms capable of aerobic stabilization of organic material in wastewater.
Activated sludge
Sludge particle
Bulk mixed liquor with free floating microorganisms
Activated sludge Microbiology
Microorganisms in the activated sludge system
• Activated sludge floc– Bacteria: major component– Fungi: low pH, toxicity, N defi
cient waste– Protozoa: gazing on bacteria– Rotifers: multicellular organis
m (help to floc formation)– Organic/ inorganic particle
Process reactor configuration (fig. 8-1)
1. Plug-flow process
Process reactor configuration
2. Complete mix activated sludge (CMAS) or Complete stir tank reactor (CSTR)
Process reactor configuration
3. Sequencing batch reactor (SBR)
Activated Sludge Principles • Wastewater is aerated in a tank • Bacteria are encouraged to grow by providing
• Oxygen • Food (BOD)• Nutrients • Correct temperature • Time
• As bacteria consume BOD, they grow and multiply • Treated wastewater flows into secondary clarifier • Bacterial cells settle, removed from clarifier as sludge • Part of sludge is recycled back to activated sludge tank,
to maintain bacteria population • Remainder of sludge is wasted
Schematic of activated sludge unit
• Assumptions:• Effluent bacteria concentration is 0• Concentration of substrate or BOD in sludge is 0 • Sludge waste flowrate (Qw) is much smaller than Q
Key Characteristics and Terms• Mixed Liquor Suspended Solids (MLSS)
– Indication of microbial population– Usually between 2000 and 5000 mg/L– Maintained by adjusting WAS– Mixed liquor volatile suspended solids (MLVSS) approx. = 0.7-0.8 × MLSS
• Food to Microorganism ratio
– Also termed sludge loading rate (SLR)– F traditionally on BOD basis but now often on COD basis– M biomass fraction under aeration only (MLSS or MLVSS)– F:M typically maintained between 0.1 – 0.4 kgBOD / kgVSS. d
Designed based on loadingloading (the amount of organic matter added relative to the microorganisms available)
Commonly called the food-to-microorganisms ratio, F/M
F measured as BOD. M measured as volatile suspended solids concentration (VSS)
F/M is the amount of BOD/day per amount of MLVSS in the aeration tank
Design of Activated SludgeDesign of Activated Sludge
Design of Activated SludgeDesign of Activated Sludge
Influent organic compounds provide the food for the microorganisms and is called substrate (S)
The substrate is used by the microorganisms for growth, to produce energy and new cell material.
The rate of new cell production as a result of the use of substrate may be written mathematically as:
dX/dt = - Y dS/dt
Y is called the yield and is the mass of cells produced per mass of substrate used (g SS/g BOD)
Activated sludge process modeling • Biochemical reaction
• The concentration of biomass, X (mg/L), increases as a function of time due to conversion of food to biomass:
Where is the specific growth rate constant (d-1). This represents the mass of cells produced/mass of cells per unit of time.
(Chap.7-6)
• Biomass production
• Where kd represents the endogenous decay rate (d-1) (i.e., microorganism death rate).
– Substituting the growth rate constant:
• Substrate utilization
Where Y is the yield (mg of biomass produced/mg of food consumed)
• Y range:– Aerobic: 0.4 - 0.8 mg/mg
• 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
• Since the hydraulic retention time (HRT),
= V/Qo, then
XM
F
oS
The higher the waste rate, the higher the ratio.
0.2-0.5 lb/BOD5/day/lb MLSS is normal
A low ratio means that the microbes are starving.
Food/Microbe Ratio
Typical range of F/M ratio in activated sludge processes
Treatment Process F/MKg BOD5/Kg MLSS/day
Extended aeration 0.03 - 0.8
Conventional 0.8 - 2.0
High rate > 2.0
Mean Cell Residence Time(θc) or Solids Retention Time (SRT) or Sludge Age
Mean cell residence time (MCRT, θc) is the mass of cells in the system divided by the mass of cells wasted per day.
Consider the system:
θc = VX/QX = V/Q
At SS the amount of solids wasted per day must equal the amount produced per day:
θc = XV / [Y(dS/dt)V] = X / Y(dS/dt)
Mean Cell Residence Time(θc) or Solids Retention Time (SRT)
• θc = 1/μ = 1 / μmax S/(KS +S)
minimal θc = 1/μmax
• SRT typically 4-20 days, HRT usually 6-24 hours
Now consider a CSTR with cell recycle:
• Influent biomass + biomass production = effluent biomass + sludge wasted
• Substitute biomass production equation
• Assume that influent and effluent biomass concentrations are negligible and solve
Mass balance of biomass production
Activated sludge process modeling
Mass balance of food substrate • Influent substrate + substrate consumed = effluent
susbtrate + sludge wasted substrate•
• Substitute substrate removal equation•
• Assume that no biochemical action takes place in clarifier. Therefore the substrate concentration in the aeration basin is equal to the substrate concentrations in the effluent and the waste activated sludge. Solve:
•
Overall equations – Combine the mass balance equations for food and biomass:
• The cell residence time is:
• and the hydraulic retention time is,= V/Qo
Substitute and rearrange:
• Compute the F/M ratio
Other important Operating Parameters
• Organic loading rate
• Oxygen supply
• Control and operation of the final settling tank
Final settling tank
Functions:ClarificationThickening
Sludge settleability is determinedby sludge volume index (SVI)sludge
SVI (ml/g) = ___________
MLSS
V x 1000 where V is volume of settled sludge after 30 min
SVI One-liter graduated cylinder,30 minute settling
period SVI = (mL/L)/(g/L) = mL/g, i.e., volume occupied by
one gram of settled solids
A high SVI (>150 ml/g) indicates bulking
1-L
mL
Settling Problem in Activated Sludge Processes
Settling well
Settling problem
Definition of BODDefinition of BODMicroorganisms (e.g., bacteria) are responsible for decomposing organic waste. When organic matter such as dead plants, leaves, grass clippings, manure, sewage, or even food waste is present in a water supply, the bacteria will begin the process of breaking down this waste. When this happens, much of the available dissolved oxygen (DO) is consumed by aerobic bacteria, robbing other aquatic organisms of the oxygen they need to live. Biological Oxygen Demand (BOD) is a measure of the oxygen used by microorganisms to decompose this waste. If there is a large quantity of organic waste in the water supply, there will also be a lot of bacteria present working to decompose this waste. In this case, the demand for oxygen will be high (due to all the bacteria) so the BOD level will be high. As the waste is consumed or dispersed through the water, BOD levels will begin to decline.
BOD bottles
BOD as afunction oftime
5-day BOD test
BOD5 (mg/l) = __________D0 – D5
P
P is volume fraction of 1 liter used in test
D is dissolved oxygenconcentration at Time=0 and Time = 5 days
Sample calculation
Determine the 5-day BOD for a 15 ml sample that is diluted with dilution water to a total volume of 300 ml when the initialDO concentration is 8 mg/l and after 5 days, has been reduced to 2 mg/l.
D0 = 8D5 = 2P = 15 ml/300ml = 0.05
BOD (mg/l) = _______ = 120 0.05
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