Download - ASP Kinetic Model
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ACTIVATED SLUDGE : Kinetic
Model
Assumptions:
1.Complete mixing in aeration tank.
2.Influent substrate concentration remains constant.3.No microbial solids in raw water.
4.No microbial activity in clarifier.
5.Good efficiency of separation in clarifier and no sludge accumulates
in it.6.Steady state conditions prevail.
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A,T,
Va, X,Se
Secondary
Clarifier
Q, So
Q(1+R)
X, Se
(Q-Qw),
Xe, Se
Qw, Xr, SeRQ, Xr, Se
ACTIVATED SLUDGE KINETIC MODEL
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Material balance equation for biomass
across full system.
(dx/dt)va =(Y(ds/dt)u--kdX)Va --XVa /cIn steady state conditions (constant MLSS
maintained)
=>(dx/dt)=0=>1/c =Y(ds/dt)u -- kd
X
=>(ds/dt)=k.SeX Monods equation 1st
orderk+Se
=>Se = ks(1+kdc )c (Yk--kd)--1
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(dS/dt) va =Q So - (ds/dt)u va - (1+R) QSeunder steady state conditions
(ds/dt)u = Q(So-Se)X Va(1/ c+ kd)
X = c Y Q(So-Se)
Va(1+kdc )
Sludge recycled : because
1.Increased MLVSS-increased efficiency of process.
2.Better flocculation.3.Improved performance- acclimated biomass.
Material balance equation forsubstrate in A.T
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ACTIVATED SLUDGE PROCESS
Concept of mean cell residence time
Mean cell residence time c is the time for whichcell remains in the system .
The physiological state of the microorganisms can becontrolled by simply regulating the rate at which
cells are wasted from the system .Cells could be wasted either directly from the
aeration tank viz c =VX/QwX =V/Qw
Or from the recycled line viz c =VX/QwX r
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MERITS/DEMERITS OF SLUDGEWASTING FROM THE A.T.Better process control since microorganism
concentration X,Xr need not be determined.
Larger volume of waste sludge hence larger sludgehandling units.
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Concept of mean residence time .
Eg 1#Q=1000cum /day Va =250 cumX=3000 mg/l Qw =50 cum/dayMass of cells leaving the system= 750kg
viz cell remaining in system =750/150 =5 daysQuestions : 1. Ifit is desirable to retain cells in the system for a longertime how could be this be achieved in practice?
2. Would the retention time of cells in your opinion haveany bearing on the physiological state of themicroorganisms?
3. Can the physiological state of the microorganisms becontrolled? How?
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Eg 2#
Q=1000cum /day Va= 250 cumX=3000cum/day Qw=10cum /day
Xr=1500mg/l
Mass of cells leaving the system =150 kg
Mass of cell in the AT=750 kgmass of cell remaining in the system for 750/150= 5
days
Question :What conclusion can be drawn from these two
examples?
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MINIMUM BIOLOGICAL CELL
RETENTION TIME
If cells are removed from the system at a rate faster than thetheir generation rate, cell washout will occur.
min c = ?
If c = c min ( min, than rate of cells leaving = rate of cellsgenerated .no treatment is possible i.e.So =Se1/c =Y(ds/dt)u -- kd
X1/c min =Y(ds/dt)u -- kdX
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,
So=2
So=1.5
So= 1Cell conc.curves
Substrate
conc.
M.C.R.T
Se,X
cmin
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INFERENCES FROM THE KINETIC
MODELIndependence between So and Se implies that :
-As long as c is held constant ,any change in So willresult in change in X but not in Se .
-It is not necessary to use same influent concentrationfor lab as encountered in field to determine kineticcoefficient .
Advantages .unknown concentration before start up
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ASP OXYGEN REQUIREMENT
Oxygen is the ultimate electron acceptor in anaerobic process.
Low oxygen concentration causes process failures
Method for computation:Total oxygen required =Q(BODuoBODue)
However , not all substrate is oxidized. Part of it isconverted to new cells (synthesis).At steady statecells wasted = cells formed
Therefore substrate synthesized to new biomassdoesnt exert any oxygen demand.
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C5H7NO2+5O25CO2 +2H2O+NH3113 32 (Sykes,1975)
5x32 = 1.42 units of O2 per unit of biomass113 synthesized
Actual O2 required per day
=Q(BODuoBODue) -- 1.42Q(Yocs(So-Se))
1000
Q=cum/day BOD=mg/locs unoxidised cells
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II LAWRENCE 1975
O2 requirement = Q[{1-1.42Y}(Se-So)]+1.42KdVaX
1000
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EXTENDED AERATION
Major objective :minimization of sludge
Primary settling tank omitted, larger c and
hence PST not provided.Theoretically ,Absolute growth rate =0 i.e. dy =0
dt
Viz , amount of biomass produced during organicremoval = amount oxidised to provide for energy
requirements
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Y(ds/dt)u Va=KdxVa
YQ(So-Se)=KdxVa
Va=YQ(Se-So) (ds/dt)u = Q(So-Se)/Va
KaX No sludge is produced
Poor aggregation of biomass
Poor settleability High energy costs as larger volumes to be kept
mixed and O2 needed for endogenous respirationis also satisfied.
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Tempearture effects on ASP parameters
Arrhenius relationship
d(ln k) = Ea . 1
dt R T2K= retention rate constant
Ea=activation energy constant
R=gas constantT=tempt=temp as continous independent variable
integrating between limits T1 and T2.
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Optimum
maxmin
Rategrowth
temp
Enzyme denaturation ---physiological state of bacteriaaffected
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ln (k2/k1)=Ea .(T2-T1)R T2-T1
ln (k2/k1)=const(T2-T1)k2/k1= e^const(T2-T1)
k2/k1= ^(T2-T1) where = e^const
Biological processes follow arrhenius relationshipwithin narrow range.
1.01 to 1.04 for ASP