asp kinetic model

8/4/2019 ASP Kinetic Model

1/19

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.


2/19

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


3/19

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


4/19

(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


5/19

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


6/19

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.


7/19

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?


8/19

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?


9/19

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


10/19

,

So=2

So=1.5

So= 1Cell conc.curves

Substrate

conc.

M.C.R.T

Se,X

cmin


11/19

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


12/19

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.


13/19

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


14/19

II LAWRENCE 1975

O2 requirement = Q[{1-1.42Y}(Se-So)]+1.42KdVaX

1000


15/19

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


16/19

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.


17/19

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.


18/19

Optimum

maxmin

Rategrowth

temp

Enzyme denaturation ---physiological state of bacteriaaffected


19/19

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

asp kinetic model

Documents