1
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
5
2. LUIKOV’S EQUATIONS
•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.
6
2. LUIKOV’S EQUATIONS
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
2. LUIKOV’S EQUATIONS
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.
8
2. LUIKOV’S EQUATIONS
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.
9
2. LUIKOV’S EQUATIONS
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.
10
2. LUIKOV’S EQUATIONS
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.
11
2. LUIKOV’S EQUATIONS
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:
12
2. LUIKOV’S EQUATIONS
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=
13
2. LUIKOV’S EQUATIONS
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.
14
2. LUIKOV’S EQUATIONS
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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2. LUIKOV’S EQUATIONS
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:
16
2. LUIKOV’S EQUATIONS
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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2. LUIKOV’S EQUATIONS
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
3. PARAMETER ESTIMATION
20
• 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−−=
*
3. PARAMETER ESTIMATION
21
( ) ( ) ( )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
3. PARAMETER ESTIMATION
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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
3. PARAMETER ESTIMATION
23
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.
3. PARAMETER ESTIMATION
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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
3. PARAMETER ESTIMATION
26
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 2 4 6 8 10 12 14 16 18 20Dimensionless time, τ
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
0 2 4 6 8 10 12 14 16 18 20Dimensionless time, τ
-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
0 2 4 6 8 10 12 14 16 18 20Dimensionless time, τ
0 2 4 6 8 10 12 14 16 18 20Dimensionless time, τ
-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
3. PARAMETER ESTIMATION
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
3. PARAMETER ESTIMATION
30
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
3. PARAMETER ESTIMATION
31
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
3. PARAMETER ESTIMATION
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
3. PARAMETER ESTIMATION
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)
34
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 =
4. FUNCTION ESTIMATION
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)
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
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∇
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
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)
4. FUNCTION ESTIMATION
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)
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
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
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 m
Posição dos SensoresExataY = 0.850Y = 0.650
Dimensionless Time
ExactTest-case 10Test-case 11
0 0.04 0.08 0.12 0.16 0.2tempo
2
3
4
5
6
Bi m
Posição dos SensoresExataY = 0.850Y = 0.650
Dimensionless Time
ExactTest-case 10Test-case 11
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
4. FUNCTION ESTIMATION
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
Posição dos SensoresExataY = 0.850Y = 0.650
Dimensionless Time
ExactTest-case 14Test-case 15
0 0.04 0.08 0.12 0.16 0.2tempo
2
3
4
5
6
Bi m
Posição dos SensoresExataY = 0.850Y = 0.650
Dimensionless Time
ExactTest-case 14Test-case 15
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
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
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)
4. FUNCTION ESTIMATION
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).
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
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)
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
550 20 40 60 80 100
Iteração
400
800
1200
S[B
i m,B
i q]
Powell-BealePolak-RibiereFlecher-Reeves
4. FUNCTION ESTIMATION
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
4. FUNCTION ESTIMATION
Estimation of Diffusion Coefficient
CENTER FOR ANALYSIS AND SIMULATION
IN ENVIRONMENTAL ENGINEERING - CASEE
Estimation of Diffusion Coefficient
CENTER FOR ANALYSIS AND SIMULATION
IN ENVIRONMENTAL ENGINEERING - CASEE
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),(
),(
CENTER FOR ANALYSIS AND SIMULATION
IN ENVIRONMENTAL ENGINEERING - CASEE
][][)( 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.
CENTER FOR ANALYSIS AND SIMULATION
IN ENVIRONMENTAL ENGINEERING - CASEE
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
CENTER FOR ANALYSIS AND SIMULATION
IN ENVIRONMENTAL ENGINEERING - CASEE
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
CENTER FOR ANALYSIS AND SIMULATION
IN ENVIRONMENTAL ENGINEERING - CASEE
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.
CENTER FOR ANALYSIS AND SIMULATION
IN ENVIRONMENTAL ENGINEERING - CASEE
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
CENTER FOR ANALYSIS AND SIMULATION
IN ENVIRONMENTAL ENGINEERING - CASEE
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