MEASURING OXYGEN TRANSFERIN AN ACTIVATED SLUDGE TANK

IN OPERATION

MEASURING OXYGEN TRANSFERIN AN ACTIVATED SLUDGE TANK

IN OPERATION

UNIVERSITY OF AALBORG, SEPTEMBER 2009“ERASMUS TEACHING STAFF VISIT” - SEMINAR B

Salvatore Nicosia, Assoc. Prof. in Environmental EngineeringDIIAA, Dipartimento di Ingegneria Idraulicaed Applicazioni Ambientali, Università di Palermo Viale delle Scienze, 90128 Palermo, ITALY(nicosia@idra.unipa.it)

2

THE PHENOMENA AND REACTIONS GENERALLY MAKING UP THE UNIT OPERATION OF OXYGEN TRANSFER (PICTORIAL)

Waste water in with its DO

Mixed liquor out with the same DO as in the tank

O2 from atmosphere

O2 supplied

Recycle sludge with its DO

Non-utilized O2

O2 transferred to mixed liquor and used up

ACTIVATED SLUDGE TANKACTIVATED SLUDGE TANK

VML

3

THE FLUXES OF OXYGEN MASS TRANSFER AND USE (QUANTITATIVE; WO as kgO2/h)

ACTIVATED SLUDGE TANKACTIVATED SLUDGE TANK

Win,atm = kLaatm (Ceq – CML) VML

Wtransf = kLaML

(Ceq – CML) VMLWin,rec = Qrec Crec

Wout,ML = (Q + Qrec) CML

Wout,atm

VML

Win,aer

Win,ww = Q Cin

4

THE BALANCE EQUATION FOR O2 (kgO2/h) - 1

VOURCQQVCCakCQCQdt

dCV MLrinMLeqMLLrrinin

ML

Mass of oxygen entering Mass of oxygen exiting + used

Now we must state the system conditions during the oxygen transfer measure ()

O2 intake from atmosphere is usually neglected; this simplification loses significance for surface aeration systems, ad is relevant only if bubble aeration is made.

5

THE BALANCE EQUATION FOR O2 - 2

AERATION MODE

WATER FLOW PATTERN

CYCLIC CONTINUOUS

BATCH without internal use (physical transfer)

with internal use (chem. or biol.)

CONTINUOUS with internal use (chem. or biol.), possibly + convective loss/gain.

with internal use (chem. or biol.), possibly + convective loss/gain.

Possible and practical conditions during an oxygen transfer measure

6

ADVANTAGES AND DRAWBACKS OF THE OPERATIONAL CONDITIONS

AERATION MODE

WATER FLOW PATTERN

CYCLIC CONTINUOUS

BATCH • Physical transfer requires reagents to de-oxygenate water• O2 transfer along with biological uptake is inexpensive and realistic

CONTINUOUS • Aeration must be repeatedly interrupted • Extra instruments required • More complicated mathematics

• No perturbation of steady state • Only one value of transfer coefficient is obtained; instable measures can make it erratic

7

HOW CAN WE SOLVE THE EQUATION? (1)

A- The simplest system: physical batch re-aeration

Wtransf = kLaML

(Ceq – CML) VML

Wout,atm

VML

Win,aer

ACTIVATED SLUDGE TANKACTIVATED SLUDGE TANK

Win,atm = kLaatm (Ceq – CML) VML

8

HOW CAN WE SOLVE THE EQUATION? (2)

Physical batch re-aeration (cont)

VCCakdt

dCV MLeqMLL

ML VCCak

dt

dCV MLeqMLL

ML

MLeqMLLMLeq CCak -

dt

CCd

MLeqMLLMLeq CCak -

dt

CCd

) t-(t a(-k exp CC CC 0L0eqeq ) t-(t a(-k exp CC CC 0L0eqeq

0MLL

ML,0eq

MLeq ttak - CC

CC

0MLLML,0eq

MLeq ttak - CC

CC

By changing the variable CML into (Csat – CML), where Csat is the actual saturation concentration in that basin, not foreseeable theoretically we get:

a 1st order differential equation, linear, homogeneous, which yields

or, as well

9

RATIOS OF DO DIFFERENCES (NAT. LOG)

-2

-1,5

-1

-0,5

0

0 0,2 0,4 0,6 0,8 1t, h

VIEWING IT GRAPHICALLY… (1)

• Graphically, kLa is the slope of the straight line that interpolates the DO concentrations.• In order to draw the line we must assign Csat a value, which is an asymptotic one we can estimate but not measure it.• Ceq is taken as the value that makes best adhere the experimental points to the interpolating line.• It is advisable to make Ceq range between the highest DO value measured during the experiment and the concentration in clean water.

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PHYSICAL REOXYGENATION (kLa = 2 1/h; Ceq = 4)

0,00

0,50

1,00

1,50

2,00

2,50

3,00

3,50

4,00

0 0,2 0,4 0,6 0,8 1

t,h

C,

(g/c

m)

VIEWING IT GRAPHICALLY… (2)

The value of kLa just calculated enables us to draw the curve of DO concentrations versus time as they would have been in a perfect run.

) t-(t a(-k exp CC CC 0L0eqeq ) t-(t a(-k exp CC CC 0L0eqeq

11

HOW CAN WE SOLVE THE EQUATION? (3)

B- Complete system; cyclic aeration mode; differential equation

B CA - dt

dC V ML

ML B CA - dt

dC V ML

ML

1st order, linear, non homogeneous; which can be solved in a well known way.

The oxygen transfer coefficient, kLa, is calculated at last.

We can skilfully group together the variables that contain CML in order to finally get at the differential equation

12

HOW CAN WE SOLVE THE EQUATION? (4)

By substituting this in the general equation and cancelling out common terms, we can finally write

reqrieqieqMLMLL CCQCCQVCCaKVOUR reqrieqieqMLMLL CCQCCQVCCaKVOUR

CCV

Q

V

Qak

dt

dCMLeq

riMLL

ML

CC

V

Q

V

Qak

dt

dCMLeq

riMLL

ML

C- Complete system; cyclic aeration mode; mathematical artifice

If there were equilibrium between oxygen transfer and use, the derivative term would go to zero and we should get

which is easily integrated, linearized etc. as in (A).

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THE EQUATION SOLVED

0ri

MLL0eq

eq ttV

Q

V

QaK

CC

CCln

EXPERIMENTAL RE-OXYGENATION

y = 0,0026x

R2 = 0,1879

0

0,2

0,4

0,6

0,8

1

1,2

1,4

0 60 120 180 240 300 360Time elapsed, s

ln (

(Ceq

- C

) /

(Ceq

- C

0))

EXPERIMENTAL RE-OXYGENATION

y = 0,0026x

R2 = 0,1879

0

0,2

0,4

0,6

0,8

1

1,2

1,4

0 60 120 180 240 300 360Time elapsed, s

ln (

(Ceq

- C

) /

(Ceq

- C

0))

14

THE OPERATION PLAN, 1- DO MEASURE POINTS

V = 410 m3; (F/M) ratio = 0.04; unit power input = 0,058 kW/m3

Bottom turbines, 2 x 7,45 kW 1 Surface

turbine, 9 kW

M.P. A

M.P. B

15

THE OPERATION PLAN, 2- SCHEDULED RUNS

Steady State

S.S. S.S.

Quite ideally…

Aerators off Aerators off

t

DO

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THE FIGURES OF OUR CASE-STUDY

• The slope of the straight interpolar line is 0,0026 s-1 = 9,36 h-1

• The ratio (Qi + Qr)/V = (8 + 8) / 410 = 0,04 h-1

• Therefore, the coefficient of oxygen transfer of the whole of those aeration machines in that basin is 9,32 (gO2transf/m3) / ((gO2deficit/m3)h).

• As the rated coefficient in clean water of the 2 bottom turbines together is 20;

even neglecting as a first approximation the effect of the small surface turbine;

the efficiency of the aeration system is lower than 50%.

17

A POSSIBLE EXPLANATION FOR THIS RESULT

• The bottom aerators are too close to the walls• SS concentration in ML is a bit high (3,87 kg SST/

m3)

Rated influence circle for the bottom aerators

Actual influence circle

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ACCESSORY MEASURE: BIOMASS OUR

y = -0,2311x - 0,0092

R2 = 0,9934

-1,2

-1,0

-0,8

-0,6

-0,4

-0,2

0,0

0,00 0,50 1,00 1,50 2,00 2,50 3,00 3,50 4,00 4,50

(t - t0), min

(DO

- D

O0)

, mg

/l

• A sample of MLSS is put into a stirred bottle, fitted with a DO probe

• Assumed respiration kinetics: 0 – order :d DO / dt = const

(DO – DO0) = kres (t – t0)

kres = 0.23 mg / (l min) = 0.09 (g/gTSS d)

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APPENDIX

20

L’ÉVOLUTION THÉORIQUE DES CONCENTRATIONS D’O.D. AU COURS DU TEMPSDe l’équation intégrée on peut calculer  :

On voit que la concentration en oxygène à saturation dans les conditions de travail, C* n’est rattrapée que après un temps théoriquement infini.

On obtient aussi la vitesse instantanée de réoxygénation de l’eau :

0L0L tt ak exp C *C. ak dt

dC

et le flux unitaire d’oxygène transféré qui est :

)) t-(t a(-k exp C *C. ak)V/ (W 0L0LRO2

où VR est le volume du bassin d‘aération.On obtient ainsi le diagramme suivant qui reconstruit l’essai expérimental.

) t-(t a(-k exp CC CC 0L0eqeq ) t-(t a(-k exp CC CC 0L0eqeq