Activated Sludge _kinetic Model

8/2/2019 Activated Sludge _kinetic Model

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ACTIVATED SLUDGE : KineticModel

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.



A,T,

V a, X,Se

Secondary Secondary

ClarifierClarifier

Q, So

Q(1+R)

X, Se

(Q-Qw),

Xe, Se

Q w , Xr, SeRQ, Xr, Se

ACTIVATED SLUDGE KINETIC MODEL



Material balance equation for

biomass across full system.

(dx/dt) v a =( Y(ds/dt)u--kdX)Va --XVa /Θc

In steady state conditions (constant MLSSmaintained)

=>(dx/dt)=0

=>1/ Θc =Y(ds/dt)u -- kd

X

=>(ds/dt)=k.SeX Monod’s equation1st order

k+Se

=>Se = ks(1+kd Θc )

Θc (Yk--kd)--1



(dS/dt) v a =Q So - (ds/dt)u v a - (1+R) QSe

under steady state conditions(ds/dt)u = Q(So-Se)

X Va(1/ Θc+ kd)X = Θc Y Q(So-Se)

Va(1+kd Θc )

Sludge recycled : because

1.Increased MLVSS-increased efficiency of

process.2.Better flocculation.

3.Improved performance- acclimatedbiomass.

Material balance equationfor substrate in A.T



ACTIVATED SLUDGE PROCESS

• Concept of mean cell residence time

Mean cell residence timeΘc is the time forwhich cell remains in the system .

The physiological state of themicroorganisms can be controlled by

simply regulating the rate at which cellsare wasted from the system .

Cells could be wasted either directly fromthe aeration tank viz Θc =VX/QwX =V/Qw

Or from the recycled line viz Θc =VX/QwX r



MERITS/DEMERITS OF SLUDGEWASTING FROM THE A.T.Better process control since microorganism

concentration X,Xr need not bedetermined.

Larger volume of waste sludge hence largersludge handling units.



Concept of mean residencetime .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. If it is desirable toretain cells in the system for a longer timehow could be this be achieved in practice?2. Would the retention time of cells in youropinion have any bearing on the physiologicalstate of the microorganisms?

3. Can the physiological state of the

microorganisms be controlled? How?



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 kg

mass of cell remaining in the system for750/150= 5 days

Question :What conclusion can be drawn from these

two examples?



MINIMUM BIOLOGICAL CELLRETENTION TIME

If cells are removed from the system at a rate fastthan the their generation rate, cell washout willoccur.

minΘ

c = ?If Θc = Θc min ( min, than rate of cells leaving = rat

of cells generated ….no treatment is possible i.e.So =Se 1/ Θc =Y(ds/dt)u -- kd

X1/ Θc min =Y(ds/dt)u -- kd

X



,

So=2

So=1.5

So= 1

Cell conc.curves

Substrate

conc.

M.C.R.T

Se,X

Θcmin



INFERENCES FROM THEKINETIC MODELIndependence between So and Se implies

that :

-As long as Θc is held constant ,any change in

So will result in change in X but not in Se .

-It is not necessary to use same influentconcentration for lab as encountered in field

to determine kinetic coefficient .

Advantages ….unknown concentration beforestart up



ASP OXYGEN REQUIREMENT

Oxygen is the ultimate electron acceptor inan aerobic process.

Low oxygen concentration causes process

failuresMethod for computation:

Total oxygen required =Q(BODuo—BODue)

However , not all substrate is oxidized. Partof it is converted to new cells(synthesis).At steady state cells wasted =cells formed

Therefore substrate synthesized to new’



C5H7NO2+5O25CO2 +2H2O+NH3113 32

(Sykes,1975)

5x32 = 1.42 units of O2 per unit of

biomass 113 synthesized

Actual O2 required per day

=Q(BODuo—BODue) -- 1.42Q(Yocs(So-Se))

1000Q=cum/day BOD=mg/l

ocs– unoxidised cells



II LAWRENCE 1975

O2 requirement = Q[{1-1.42Y}(Se-So)]+1.42K d V aX

1000



EXTENDED AERATION

Major objective :minimization of sludge Primary settling tank omitted, larger Θc

and Θ hence PST not provided. Theoretically ,

Absolute growth rate =0 i.e. dy =0dt

Viz , amount of biomass produced duringorganic removal = amount oxidised to

provide for energy requirements



Y(ds/dt)u Va=K dxVa

YQ(So-Se)=K dxVa

Va=YQ(Se-So) (ds/dt)u =

Q(So-Se)/Va

K aX

• No sludge is produced

• Poor aggregation of biomass• Poor settleability

• High energy costs as larger volumes to bekept mixed and O2 needed for

endogenous respiration is also satisfied.



Tempearture effects on ASPparametersArrhenius relationship

d(ln k) = Ea . 1

dt R T2

K= retention rate constant

Ea=activation energy constant

R=gas constant

T=temp

t=temp as continous independent variableintegrating between limits T1 and T2.



Optimum

maxmin

Rategrowth

temp

Enzyme denaturation---physiological state of bacteriaaffected



ln (k 2/k 1)=Ea .(T2-T1)R T2-T1

ln (k 2/k 1)=const(T2-T1)⇒k 2/k 1= e^const(T2-T1)

⇒k 2/k 1=Θ ^(T2-T1) whereΘ= e^const

⇒Biological processes follow arrheniusrelationship within narrow range.

⇒Θ1.01 to 1.04 for ASP

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Activated Sludge _kinetic Model

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