identificaÇÃo de propriedades fÍsicas em problemas de transferÊncia de massa · 2007. 10....

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IDENTIFICAÇÃO DE PROPRIEDADES FÍSICAS EM PROBLEMAS DE TRANSFERÊNCIA DE MASSA

Renato M. Cotta e Helcio R. B. OrlandeDepartamento/Programa de Engenharia Mecânica, POLI/COPPE

Universidade Federal do Rio de Janeiro, [email protected] e [email protected]

2

AGRADECIMENTOS

• Lucilia B. Dantas, Estimativa de Parâmetros na Secagem de Meios Porosos-Capilares, Tese de Doutorado, PEM/COPPE/UFRJ, 2001• Leonardo F. Saker, Estimativa Simultânea dos Coeficientes de Transferênciade Calor e Massa em Meios Porosos Capilares, Tese de Mestrado, PEM/COPPE/UFRJ, 2002• Udilma da Conceição Nascimento, Transporte de Contaminantes em Colunas de Meios Porosos Saturados, Tese de Doutorado, PEM/COPPE/UFRJ, 2005• Paulo H. da Silva Moreira, Identificação do Coeficiente Aparente de Difusãoem Colunas de Meios Porosos Saturados, Projeto de Fim de Curso, DEM/POLI/UFRJ, 2004

PROF. ROBERTO DE SOUZA PROJETO CNEN/COPPE

3

SUMMARY

1. Luikov’s Equations

2. Diffusion and Dispersion in Saturated Columns

4

2. LUIKOV’S EQUATIONS

• Luikov, A. V., (1935), On Thermal Diffusion of Moisture, Zh. Prikl. Khim., vol.8, pp. 1354.• Luikov, A. V., (1966), Heat and Mass Transfer in Capillary-porous Bodies,Pergamon Press, Oxford. • Luikov, A. V., (1975), Systems of differential equations of heat and mass transfer in capillary-porous bodies (review), International Journal of Heat and Mass Transfer, Volume 18, Issue 1, pp. 1-14.• Kirscher, O., (1942), Der Warme-und Stoffaustausch im Trocknunsgut, VDI-Froschungsheft, 415.• Philip, I. R. and de Vries, D. A. (1957), Trans. Am. Geophys. Union, vol.38, pp. 594.

Mikhailov and Ozisik1: “…Krischer’s system was identical to that of Luikov. The system defined by de Vries was also of the Luikov type…”.

1Mikhailov, M. D. and Özisik, M. N., (1984), Unified Analysis and Solutions of Heat and Mass Diffusion, John Wiley, New York

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•Macroscopic model: the variables of interest are not representative of heat and mass transfer at the pore scale (microscopic level), but at the scale of a small volume containing a sufficient number of pores. This representative elementary volume is the smallest differential volume that results in statistically meaningful local average properties such as porosity, saturation and capillary pressure.•Unsaturated capillary porous media.•Experimentally confirmed: Low initial moisture content, narrow pore size distribution and small moisture content heterogeneities.

Remark 1: For a microscopic model, see, p.ex., F. Plourde and M. Prat, (2003), Pore network simulations of drying of capillary porous media. Influence of thermal gradients, International Journal of Heat and Mass Transfer, Volume 46, pp. 1293-1307.Remark 2: A body is said to be capillary-porous, and the pores to be capillaries, if the capillary potential is significantly greater than the gravity potential. In this case, the effects of the gravity force can be neglected. Remark 3: Large-scale heterogeneities of moisture content during drying have been observed for initially saturated porous media.

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HYPOTHESES• Water, in the liquid and vapor phases (i.e., no ice, T > 0oC), and air, fill the pores of the capillary-porous body;• The mass flow rate in the body is extremely slow, so that the temperatures of the interstitial fluids and of the porous matrix are locally identical, that is, they are in thermodynamic equilibrium;• Phase transitions occur between liquid and vapor only;• Chemical reactions do not take place in the medium;• The mass of air, as well as the mass of vapor, is negligible as compared to the mass of liquid water in the pores;• Changes in the volume and porosity of the body, resulting from changes in the moisture content, are neglected;• Radiation heat transfer is negligible;• The medium is isotropic.

7


Energy Conservation Equation: =

+−∇=∂∂ 2

10 .

iii Ih

tT

c qρ

Mass Conservation Equation: ii I

tu +−∇=

∂∂

i0 .Jρ i = 1 and 2

Subscript i = 0 refers to the dry body, i = 1 to the water vapor and i = 2 to the liquid water

ui , hi, Ii and Ji are the mass content, enthalpy, mass source/sink and mass flux vector for phase i, respectively, while T is the temperature, q is the heat flux vector, ρ0 is the density of the dry body and c is the reduced specific heat.

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The moisture content u is defined as the water relative concentration, that is,

=

= ==2

10

2

1

ii

ii

um

mu

where ui = mi/m0 and mi is mass of phase i.

Since the mass of vapor is negligible as compared to the mass of liquid water, we have that u = u2.

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The mass source/sink results from the water phase change:

tu

II∂∂=−= 021 ρε

where εεεε is the phase-change criterion ( ).10 ≤≤ ε

• εεεε = 0: no phase-change takes place inside the porous medium; moisture is transported in the medium as liquid and the phase change takes place at the body surface. • εεεε = 1: all the liquid changes phase in the pores and it is then transported through the body as vapor.

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The reduced specific heat is given by:

=

+=2

10

iiiuccc

where ci is the specific heat of phase i.

The source/sink term in the energy equation refers to the phase change in the medium:

tu

rIhhIhIh∂∂=−=+ 01212211 )( ρε

where r is the latent heat of vaporization.

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For capillary-porous media, Luikov has shown that the diffusive mass flux can be written in terms of the moisture content gradient and of the temperature gradient.

Constitutive Equations:

Conduction heat flux: Tk ∇−=q

)(0 Tuam ∇+∇−= δρJ

where δ is the thermogradient coefficient and am is the moisture diffusivity.

where k is the body effective thermal conductivity.

Diffusive mass flux:

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LUIKOV’S SYSTEM OF EQUATIONS

( )tu

rTktT

c∂∂+∇∇=

∂∂

00 . ερρ

( ) ( )Tauatu

mm ∇∇+∇∇=∂∂ δ..

For constant thermophysical properties:

tu

cr

TatT

∂∂+∇=

∂∂ ε2

Tauatu

mm22 ∇+∇=

∂∂ δ

where is the thermal diffusivity.cka 0ρ=

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REMARK: A third equation is needed to model fast and intense drying processes, for which vigorous boiling may happen, in special if the temperatures exceed 100 oC. This additional equation involves the pressure gradient in the medium and is also required when the flow is imposed by a pressure gradient. Also, the energy and mass conservation equations above need to be changed.

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BOUNDARY CONDITION – MASS CONSERVATION

00 =+

∂∂+

∂∂− mm j

Tua

nnδρ

where n∂

∂= normal derivative at the boundary surface

δ = thermogradient coefficientam = moisture diffusivityjm = imposed mass fluxρ0 = density of the dry body

n

jm

∂∂+

∂∂−

nnTu

am δρ0

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For the body interacting with the surrounding air:

0)( * =−+

∂∂+

∂∂− uuh

Tua mm nn

δ

)( *0 uuhj mm −= ρ

n

jm

∂∂+

∂∂−

nnTu

am δρ0

where:• mass transfer coefficient: hm [m/s]• moisture content in equilibrium with the surrounding air: u*

Therefore:

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BOUNDARY CONDITION – ENERGY EQUATION

02 =−+∂∂− mq jrjT

kn

where n∂

∂T= normal derivative of temperature at the boundary surface

k = effective thermal conductivityjq = imposed heat fluxjm2 = liquid water mass fluxr = latent heat of vaporization

n∂∂− T

k

jm2 n

jq

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0)()1()( *0 =−−+−+

∂∂− uuhrTThT

k ms ρεn

n∂∂− T

k

jm2 n

jq

For the body interacting with the surrounding air: )( TThj sq −=

where: h = heat transfer coefficient Ts = temperature of the surrounding air

where: ε = phase-change factor

)()1()1( *002 uuh

Tuaj mmm −−−=

∂∂+

∂∂−−= ρεδρε

nnLiquid mass flux:

Therefore:

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3. PARAMETER ESTIMATION

L

xFlow of dry air Heat and moisture transport

Moist porous body

Hot Plate – Heat flux: q

T = T0 and u = u0, for t = 0

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(((( ))))TT

T - )t,x( T,X ,

os

o −−−−

====ττττθθθθ (((( ))))uu

)t,x(uu,X ,*o

o

−−−−−−−−====ττττφφφφ

)TT(kqL Q ,

os −−−−====

Lat

,2====ττττLxX ,====

kLh

Biq =

m

mm a

LhBi =

= Biot number for heat transfer

= Biot number for mass transfer


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• The Luikov number is a ratio between the mass and heat diffusivities. Thus, it compares how the mass transfer and the heat transfer processes develop. • The Posnov number indicates the change in moisture content resulting from the temperature gradient relative to the total moisture change. • The Kossovitch number indicates how the heat expended in the evaporation of the liquid compares to that expended for heating the wet body.

a

aLu m= *

uuTT

Pno

os

−−= δ

os

o

TTuu

cr

Ko−−=

*


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( ) ( ) ( )τ∂

τφ∂ε−∂

τθ∂=τ∂

τθ∂ ,XKo

X,X,X

2

2

in 0 < X < 1 , for τ > 0 ( ) ( ) ( )

2

2

2

2

X,X

LuPnX

,XLu

,X∂

τθ∂−∂

τφ∂=τ∂

τφ∂

( ) ( ) 00,X00,X , , =φ=θ for τ = 0 in 0 < X < 1

( ) ( ) ( ),0

X,0

PnX,0

Q, X,0 =

∂τθ∂−

∂τφ∂−=

∂τθ∂

( ) ( ) ( ) 0,],1 -[1)KoLuBi-(1],1 -[1BiX,1

mq =τφε+τθ−∂

τθ∂

( ) ( ) ( ) ,0],11[(Bi X,1

PnX,1

- m =τφ−+∂

τθ∂+∂

τφ∂

at X = 0 , for τ > 0

at X = 1 , for τ > 0


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DIRECT PROBLEM

Known:• Boundary and initial conditions

• Lu, Pn, Ko, Biq, Bim and ε

Determine:• Temperature distribution θ(X, τ)

• Moisture content distribution φ(X,τ)

INVERSE PROBLEM

Known:• Boundary and initial conditions

• Temperature measurements Ym(τi)

taken at locations Xm , m=1,…,M and

times τi, i=1,…,I

Estimate:• Lu, Pn, Ko, Biq, Bim and ε


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O vetor [M-E(P)]T é dado por

E(P): vetor de temperaturas e umidades estimadas

M: vetor de temperaturas e umidades medidas

NA FORMA MATRICIAL

)]([)]([)( PYPYP θθθθθθθθ −−= TS

[ ])-(,...,)-(,)-(][ 2211 IIT YYY θθθ

≡− (P)Y θθθθ

( ) [ ]iMiMiiiiii YYYY θθθθ −−−=− , ... , , 2211

Minimize:

where:

Hypotheses:

• The errors are additive, with zero mean and normally distributed.

• The statistical parameters describing the errors are known.

• There are no errors in the independent variables.• There is no prior information about P.• The measurements are uncorrelated.• The standard-deviation of the measurements is constant.


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LEVENBERG-MARQUARDT’SMETHOD

where µk is a positive scalar, Jk is the sensitivity matrix, ΩΩΩΩk is a diagonal matrix and k is the number of iterations

])([)(])[( 11 kTkkkkTkkk PYJJJPP θθθθΩΩΩΩ −++= −+ µ

• The Levenberg-Marquardt Method is related to Tikhonov’s regularization approach.

• Compromise between steepest-descent method and Gauss’method.• Simple, powerful and straightforward iterative procedure.

• Capable of treating complex physical situations.

• Easy to program.• Stable and converges fast.

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Test-case: Drying of ceramics

Biq = 2.5Bim = 2.5Lu = 0.2Pn = 0.084Ko = 49ε = 0.2

ρ0 = 2000 kg/m3

kq= 0.34 W/mK , km= 2.4x10-7 kg/msoMcq= 607 J/kgK , cm= 1.8x10-3 kg/kgoMδ = 0.56 oM/K , r = 2.5x106 J/kg. l = 0.05 mT0 = 24 oC , u0/cm = 80 oM Ts = 30 oC , u* /cm = 40 oMhq = 17 W/m2K , hm = 1.2x10-5 kg/m2soM


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Figura V.2.5.a – Temperatura versus tempo

ao longo de todo o domínio (CASO 5)

Figura V.2.5.b – Umidade versus tempo

ao longo de todo o domínio (CASO 5)

( Biq = 2.5, Bim = 2.5, Lu = 0.2, Pn = 0.084, Ko = 49, εεεε = 0.2)TEMPERATURE AND MOISTURE CONTENT

0 2 4 6 8 10 12 14 16 18 20Dimensionless time, τ

-4.0

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

Dim

ensi

onle

ss te

mpe

ratu

re, θ

X = 0.0

X = 0.1

X = 0.2

X = 0.4

X = 0.5

X = 0.6

X = 0.7

X = 0.8

X = 0.9

X = 1.0


0.0

0.2

0.4

0.6

0.8

1.0

1.2

Dim

ensi

onle

ss M

oist

ure

cont

ent,

φ

X = 0.0

X = 0.1

X = 0.2

X = 0.3

X = 0.4

X = 0.5

X = 0.6

X = 0.7

X = 0.8

X = 0.9

X = 1.0


-5-4-3-2-101234

P ∂θ

/∂P

X = 1.0

X = 0.0

X = 0.4

-5-4-3-2-101234

P ∂θ

/∂P



-5-4-3-2-101234

P ∂θ

/∂P

Biq

Bim

Lu

Pn

Ko

ε

(a) (b)

(c)

Nor

mal

ized

Sen

sitiv

ity C

oeffi

cien

tsN

orm

aliz

ed S

ensi

tivity

Coe

ffici

ents

Nor

mal

ized

Sen

sitiv

ity C

oeffi

cien

ts

280 2 4 6 8 10 12 14 16 18 20

Dimensionless time, τ

1E-5

1E-4

1E-3

1E-2

1E-1

1E+0D

eter

min

ant

Sensors at:

0.0

1.0

0.0 and 1.0


29

Initial-guesses: Lu0 = 1.0 , Biq0 = 5.0 , Ko0 = 60.0

Parameters Exact Estimatedσ = 0

Estimatedσ = 0.01 Ymax

Confidence intervals

Biq 2.5 2.5 2.500 (2.500,2.502)Lu 0.2 0.2 0.20 (0.19,0.21)Ko 49.0 49.0 48.99 (48.92,49.05)

Levenberg-Marquardt’s Method


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DIRECT PROBLEM INVERSE PROBLEM2D Formulation

1D Formulation

Simulated Measurements

l R0

Flow of dry airHeat and moisture transport

Insolute

Heat supply rate: q

l R0

Flow of dry airHeat and moisture transport

Heat loss by convection

Heat supply rate: q

Thermally Insulated

0=m

0=m

0=m

0=m


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ANALYSIS OF THE RESIDUALS: imimim YtR θθˆ )( −=

0 20 40 60 80 100Measurement

-20.0

-10.0

0.0

10.0

20.0

Res

idua

l

Independent Correlated

0 20 40 60 80 100Measurement

-20.0

-10.0

0.0

10.0

20.0

Res

idua

l


32

ra=1Biqr=0.1

0 1 2 3 4 5 6Dimensionless time, τ

-0.10

-0.05

0.00

0.05

0.10

0.15

0.20T

empe

ratu

re r

esid

ual

Measurements

Measurements with random errors

Errorless measurementsMeasurements with random errors


33

4. FUNCTION ESTIMATION

L

xFlow of dry air Heat and moisture transport

Moist porous body

Hot Plate – Heat flux: q

T = T0 and u = u0, for t = 0

Biq(τ) and Bim(τ)

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in 0 < X < 1 and τ > 0

at X = 0 , for τ > 0

at X= 1 , for τ > 0

for τ = 0 ,in 0 < X < 1

2

2

2

2

XX ∂

∂−∂

∂=∂∂ φβθα

τθ

2

2

2

2

XPnLu

XLu

∂

∂−∂

∂=∂∂ θφ

τφ

QX

−=∂∂θ

QPnX

−=∂∂φ

)1()()1()1()( φτεθτθ −−−−=∂∂

mBiLuKoqBiX

)1()( φτθφ −+∂∂=

∂∂

mBiX

PnX

0)0,()0,( == XX φθ

PnLuKoεα += 1

LuKoεβ =


35

DIRECT PROBLEM

Known:

• Boundary and initial conditions

• Thermophysical properties

Determine:• Temperature distribution θ(X, τ)

• Moisture distribution φ(X, τ)

INVERSE PROBLEM

Known:

• Initial condition

• Boundary condition at X=0

• Thermophysical properties

• Temperature measurements

• Moisture content measurements

• Total moisture measurements

Estimate:

Bim(τ) and Biq(τ)


36

Minimization of the following functional:

== =

= =

−+

−+

+

−=

=

ff

f

dwPBiBidwCBiBiX

dwMBiBiX

BiBiS

qm

N

nnqmn

I

iiqmi

qm

τ

τΓ

τ

τφ

τ

τθ

τττΓτττφ

τττθ

ττ

0

2

0 1

2*

0 1

2

)](),;([)](),;,([

)](),;,([

)](),([

2max

,2max

,2max P

Cw

C

Cw

M

Cw ΓΓ

φφθθ ===

=Γ 10

,, CCC φθ


37

Direction of Descent:

Conjugation Coefficients:

Gradient Directions:and

Search steps size: β1k and β2

k Sensitivity Problemsand

Adjoint Problem

Iterative Procedure:

Polak-Ribiere’s version of the conjugate gradient method

)()()( 111 τβττ kkk

mk

m dBiBi +=+

)()()( 221 τβττ kkk

qk

q dBiBi +=+

)()()]([)( 11

11 τψτγττ qkkkkm

k ddBiSd ++−∇= −

)()()]([)( 21

22 τψτγττ qkkkkq

k ddBiSd ++−∇= −

)](i[ m τkBS∇)](i[ q τkBS∇


38

Lu = 0.2Pn = 0.084

Ko = 49ε = 0.2Q = 0.9τf = 0.2

Test-case: Ceramicsk = 0.34 W/mK

km=2.4x10-7kg/msoMc = 607 J/kgK

r = 2.5x106J/kgT0 = 24oCu0 = 80oM

δ = 0.56oM/K tfinal = 1785 s

L = 0.05m


Bim(τ) = Biq(τ) = 2.5

0 0.04 0.08 0.12 0.16 0.2tempo

0

0.4

0.8

1.2

∆θ1(Y

,τ)

PosiçãoY = 1.0Y = 0.75Y = 0.5Y = 0.25Y = 0.0

X = 1.0X = 0.75X = 0.5X = 0.25X = 0.00

Dimensionless Time

∆θ1

0 0.04 0.08 0.12 0.16 0.2tempo

0

0.4

0.8

1.2

∆θ1(Y

,τ)

PosiçãoY = 1.0Y = 0.75Y = 0.5Y = 0.25Y = 0.0

X = 1.0X = 0.75X = 0.5X = 0.25X = 0.00

Dimensionless Time

∆θ1

Temperature sensitivity function resulting from variations in Bim(τ)

Temperature sensitivity function resulting from variations in Biq(τ)


Bim(τ) = Biq(τ) = 2.5

Moisture content sensitivity functionresulting from variations in Bim(τ)

Moisture content sensitivity function resulting from variations in Biq(τ)


Bim(τ) = Biq(τ) = 2.5

Moisture content total weight sensitivity function

0 0.04 0.08 0.12 0.16 0.2tempo

0

2

4

6

∆φ'(τ

)

Sensibilidade em relação a:

BimBiq

Dimensionless Time

∆Γ

0 0.04 0.08 0.12 0.16 0.2tempo

0

2

4

6

∆φ'(τ

)

Sensibilidade em relação a:

BimBiq

Dimensionless Time

∆Γ


0 0.04 0.08 0.12 0.16 0.2tempo

2

3

4

5

6

Bi q

Posição dos SensoresExataY = 0.975Y = 0.650

Dimensionless Time

ExactTest-case 1Test-case 3

0 0.04 0.08 0.12 0.16 0.2tempo

2

3

4

5

6

Bi q


Dimensionless Time


0 0.04 0.08 0.12 0.16 0.2tempo

2

3

4

5

6

Bi m


Dimensionless Time


0 0.04 0.08 0.12 0.16 0.2tempo

2

3

4

5

6

Bi m


Dimensionless Time


Estimation of Biq(τ) by using only temperature measurements

Estimation of Bim(τ) by usingonly temperature measurements

- X = 0.975- X = 0.65

- X = 0.85- X = 0.65


Estimation of Bim(τ) by using only moisture content measurements

Estimation of Bim(τ) by using only total moisture weight measurements

0 0.04 0.08 0.12 0.16 0.2tempo

2

3

4

5

6

Bi m


Dimensionless Time


0 0.04 0.08 0.12 0.16 0.2tempo

2

3

4

5

6

Bi m


Dimensionless Time


0 0.04 0.08 0.12 0.16 0.2tempo

2

3

4

5

6

Bi m

Posição dos SensoresExataSem ErroCom Erro

Dimensionless Time

ExactWithout random errorWith random error

0 0.04 0.08 0.12 0.16 0.2tempo

2

3

4

5

6

Bi m

Posição dos SensoresExataSem ErroCom Erro

Dimensionless Time

ExactWithout random errorWith random error

- X = 0.85- X = 0.65


ExataEstimada

Dimensionless Time

ExactEstimated

ExataEstimada

ExataEstimada

Dimensionless Time

ExactEstimated

ExataEstimada

Dimensionless Time

Dimensionless Time

Simultaneous estimation of Bim(τ) and Biq(τ) by using temperature and moisture content measurements

X = 0.975


h

L

y

x

Flow of dry air Heat and moisture transport

Moist porous sheet

Hot Plate – Heat supply rate: q

Biq(X) and Bim(X)


46

0 2 4 6 8 10X

2

3

4

5

6

Bi q

ExactConfiguration 1Configuration 2Configuration 3

1.11 x 10-10.850133

4.58 x 10-20.850332

4.38 x 10-20.925 331

RMSError

Y Location

Number of

Sensors

Config.

TEMPERATURE SENSORS

Table . Sensor configurations used for the estimation of Biq(X) with known Bim(X).


47

Results obtained for Bim(X) with known Biq(X) by using temperature measurements of 17 sensors

equally spaced

0 2 4 6 8 10X

2

3

4

5

6

Bi m

EstimatedExact

0 2 4 6 8 10X

2

3

4

5

6

Bi m

EstimatedExact

Results obtained for Bim(X) with known Biq(X) by using moisture content measurements of 17 sensors

equally spaced


48

0 2 4 6 8 10X

2

3

4

5

6

Bi q

EstimatedExact

0 2 4 6 8 10X

2

3

4

5

6

Bi m

EstimatedExact

Simultaneous estimation of Biq(X) and Bim(X):14 temperature sensors and 15 moisture content sensors


h

L

y

x

Flow of dry air Heat and moisture transport

Moist porous sheet

Hot Plate – Heat supply rate: q

Biq(X,τ) and Bim(X,τ)


50

Exact

Estimated

X τ B

i q

Exact

Estimated

X τ B

i q

Results obtained for Biq(X,τ) with known Bim(X,τ) by using temperature measurements

X

τ Bi m

Exact

Estimated

X

τ Bi m

Exact

Estimated

Results obtained for Bim(X,τ) with known Biq(X,τ) by using temperature measurements

13 sensors at Y = 0.85


51

Exact

Estimated

Bi m

X

τ

Exact

Estimated

Bi m

X

τ

Results obtained for Bim(X,τ), with known Biq(X, τ), by using moisture content measurements

13 sensors at Y = 0.85


52

Exact

Estimated

Bi m

X τ

X

Exact

Estimated

Bi m

X τ

X

Bi q

X

τ

Exact

Estimated

Bi q

X

τ

Exact

Estimated

Results using temperature and moisture content measurements -simultaneous estimation of Bim(X,t) and Biq(X,t)

13 temperature sensors and 13 moisture content sensors


550 20 40 60 80 100

Iteração

400

800

1200

S[B

i m,B

i q]

Powell-BealePolak-RibiereFlecher-Reeves


56

Method # Iterations

Final Functional

RMS Error (Bim)

RMS Error (Biq)

Powell - Beale

63 151.936 4.379 x 10-1 6.892 x 10-1

Polak - Ribiere

75 197.790 5.117 x 10-1 7.087 x 10-1

Fletcher - Reeves

95 430.310 5.748 x 10-1 8.378 x 10-1


Estimation of Diffusion Coefficient

CENTER FOR ANALYSIS AND SIMULATION

IN ENVIRONMENTAL ENGINEERING - CASEE

Estimation of Diffusion Coefficient



2

2*

z

CD

tC

∂∂

∂∂ = for t > 0 and z > 0

0=zC

∂∂

at z = 0 , for t > 0

>=<<=

==azt

aztCzfC

and0for0

0and0for)( 0

−+

+==tD

zaerf

tD

zaerf

CtzC

tzC**0

*

2221),(

),(



][][)( C(P)YC(P)YP −−= TSMinimization of:

Levenberg-Marquardt’s method: )]([)( 11 kTkkTkk PCYJJJPP −Ωµ++= −+

Inverse Problem: Estimate P = [ D* , a ] by using concentration measurements.

• Mass diffusion coefficient of KBr in sand• 4 conductivity cells located at z = 2.5 , 22.5, 32.5 and 42.5 mm

• Measurements were taken until 90 hours after the beginning of the experiment

• Frequency of 1 measurement per cell every 20 minutes.



0 20 40 60 80 100Time (hours)

-0.2

-0.1

0

0.1

0.2R

esid

uals

D* = 0.83 x 10-5 cm2/sCell 1Cell 2Cell 3Cell 4

D* = 8.29 x 10 –6 cm2/s 8.06 x 10 –6 cm2/s < D* < 8.53 x 10 –6 cm2/s



Estimation of Dispersion Coefficient

Valve

Electrical ConductivityCell

Tracer

Condutivimeter

Computer

Timer

WaterReservoir

Water constant level

Valve

Electrical ConductivityCell

Tracer

Condutivimeter

Computer

Timer

WaterReservoir

Water constant level

zs

a

r2

r1



2

2

( , ) ( , ) ( , )c z t c z t c z tR D V

t z z∂ ∂ ∂= −

∂ ∂ ∂; for 0 z L< < and 0t >

0

, 0

( ,0) ,

,

b s

s s

b s

C em z z

c z C em z z a z

C em z a z L

< <= < < + + < <

for 0t =

(0, ) bc t C= in 00 >= tandz

( , )( , )m m b

c L tD h c L t h C

z∂ + =

∂ in z L= and 0t >

where D is the dispersion coefficient, R is the retardation factor, V is the porous velocity and L is the column length

Inverse Problem: Estimate P = [ D , R , hm ] by using outflow concentration measurements.



Time (s)

Nor

mal

ized

Con

entr

atio

n(C

/Co)

0 2000 4000 6000 8000 10000 120000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Technique ITechnique II

V = 5.19x10-3 cm/s



Time (s)

Nor

mal

ized

Con

cent

ratio

n(C

/Co)

0 2000 4000 6000 8000 10000 120000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

EstimatedMeasured

V = 5.19x10-3 cm/sDe = 1.21x10-3 cm2/sRe = 1.032hme = 1.49x10-3 cm/s

identificaÇÃo de propriedades fÍsicas em problemas de transferÊncia de massa · 2007. 10....

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