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Departamento de Ingeniería del Terreno, Cartográfica y Geofísica
UNIVERSITAT POLITÈCNICA DE CATALUNYA
Instituto de Investigaciones Antisísmicas “Ing Aldo Bruschi”
UNIVERSIDAD NACIONAL DESAN JUAN
Laboratorio de GeotecniaUNIVERSIDAD NACIONAL DE
CÓRDOBA
Geotécnia e Ingeniería Sísmica aplicadas a la MineríaSan Juan, Argentina, 16 de Octubre de 2007
Formulación para problemas THM
1
FORMULATION FOR FORMULATION FOR THERMO HYDRO MECHANICALTHERMO HYDRO MECHANICALPROBLEMS IN POROUS MEDIAPROBLEMS IN POROUS MEDIA
Sebastià OlivellaEscuela Técnica Superior de Ingenieros de Caminos, Canales y Puertos
Departamento de Ingeniería del Terreno (UPC)
Jornada Geotecnia e Ingeniería Sísmica, San Juan, 16 de octubre 2007
Formulación para problemas THM
2
Índice
1. PROBLEMAS ACOPLADOS2. FORMULACION BASICA 3. ALGUNOS EJEMPLOS
Formulación para problemas THM
3
PROBLEMAS ACOPLADOSMechanical problemsin soils and rocks
Formulación para problemas THM
4
Mechanical problemsin soils and rocksOne phase flow
PROBLEMAS ACOPLADOS
Formulación para problemas THM
5
Mechanical problemsin soils and rocksOne phase flowTwo phase flow
PROBLEMAS ACOPLADOS
Formulación para problemas THM
6
Mechanical problemsin soils and rocks
Saturated consolidation/unsaturated consolidation
One phase flowTwo phase flow
PROBLEMAS ACOPLADOS
Formulación para problemas THM
7
Mechanical problemsin soils and rocks
Saturated consolidation/unsaturated consolidation
One phase flow
Evaporation/evapotranspirationWaste disposalsGeothermal fluid extraction
Two phase flow
PROBLEMAS ACOPLADOS
Formulación para problemas THM
8
Mechanical problemsin soils and rocks
Saturated consolidation/unsaturated consolidation
One phase flow
Evaporation/evapotranspirationWaste disposalsGeothermal fluid extraction
Solute transportproblems
Two phase flow
CODE_BRIGHT
PROBLEMAS ACOPLADOS
Formulación para problemas THM
9
2. BASIC FORMULATIONGas phase: dry air + water vapour
Solid phase
Liquid phase: water + dissolved air
The three phases are:• solid phase (s) :mineral• liquid phase (l) :water + air
dissolved • gas phase (g) :mixture of
dry air and water vapour
The three species are: • Solid (-): the mineral is
coincident with solid phase • Water (w):as liquid or
evaporated in the gas phase
• Air (a) :dry air, as gas or dissolved in the liquid phase
Formulación para problemas THM
10
Solid phaseVs / V = 1 - φ
Liquid phaseVl / V = Vl / Vh × Vh / V =
Sl × φ
Unsaturated soil: Porosity and degree of Saturation
Total volumeV = Vs +Vl +Vg = Vs + Vh
Gas phaseVg / V = Vg / Vh × Vh / V
= Sg × φ
Formulación para problemas THM
11
Mass of Solid in the Solid phase
Mass of water in the liquid phase
Unsaturated soil: Mass in phases
wl lω ρ
wg gω ρ
Mass of water in the gas phase
sρ
Mass of water in the liquid phase
Mass of air in the gas phase
ag gω ρ
al lω ρ
Formulación para problemas THM
12
Mass of Solid in the Solid phase
Mass of water in the liquid phase
Unsaturated soil: Mass in porous medium
wl l gSω ρ φ
wg g gSω ρ φ
Mass of water in the gas phase
( )1sρ − φ
Mass of water in the liquid phase
Mass of air in the gas phaseag g gSω ρ φ
al l lSω ρ φ
Formulación para problemas THM
13
Liquid
Gas
Solid
Heat in Heat out
Water in
Vapour flux (diffusion)
Heat flux (conduction and advection)
Waterevaporation
Water condensation
Liquid flux (advection)
Heat flux(conduction and advection)
Heat flux (conduction)
Formulación para problemas THM
14
TemperatureWater content
Water fluxes
Formulación para problemas THM
15
ωgwρg qg
φSgωgwρg du/dt
igw
igwωgwρg qgφSgωg
wρg du/dt
the advective flux caused by solid motion: φSg ωg
wρg du/dt where du/dt is the vector of solid velocities, Sg is the volumetric fraction of pores occupied by the gas phase and φ is porosity.
the advective flux caused by fluid motion: ωg
wρg qg, where qg is the Darcy's flux,
the nonadvective flux: igw, i.e.
diffusive/ dispersive,
THE TOTAL MASS FLUX OF A SPECIES IN A PHASE (E.G. FLUX OF AIR PRESENT IN GAS PHASE jGW IS, IN GENERAL, THE SUM
OF THREE TERMS
Formulación para problemas THM
16
mass or energy per fluxes of mass unit volume of porous media or energy
sources or sinks ofmass and energy
t⎛ ⎞ ⎛ ⎞∂
+∇• =⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠⎛ ⎞
= ⎜ ⎟⎝ ⎠
The following balance equation will be applied
Mass in Mass out(∆ Mass)
Formulación para problemas THM
17
( )( ) ( )1 0 s st∂
ρ − φ + ∇ ⋅ =∂
j
MASS BALANCE OF SOLID
Mass balance of solid:
( )( ) ( )1 1 0 s sd
t dt∂ ⎛ ⎞ρ − φ + ∇ ⋅ ρ − φ =⎜ ⎟∂ ⎝ ⎠
u
ρs solid density and js is the flux of solid.
( )1 1 (1 ) s s s
s
D D dDt Dt dt
⎡ ⎤φ ρ= −φ + −φ ∇ ⋅⎢ ⎥ρ ⎣ ⎦
u
Porosity variation:
Formulación para problemas THM
18
Volumetric strain: y yx d ddd
dt dt dt dtε εε
∇ ⋅ = + +u
( ) (1 1 1 ) s s s
s
ddt
Dt Dt
DD
⎡ ⎤ρ− φ⎢ ⎥ρ ⎣
φ− φ ∇ ⋅+
⎦=
u
Formulación para problemas THM
19
( ) ( ) w w condensation wl l l lS f f
t∂
ω ρ φ + ∇ ⋅ = +∂
j
MASS BALANCE OF WATER
Water is present in liquid an gas phases. The mass balance of water in the liquid and in the gas phase is expressed as:
( ) ( ) w w evaporation wg g g gS f f
t∂
ω ρ φ + ∇ ⋅ = +∂
j
f w is an external supply of water and:
f condensation + f evaporation = 0Adding the two mass balance equations leads to:
( ) ( ) w w w w wl l l g g g l gS S f
t∂
ω ρ φ + ω ρ φ + ∇ ⋅ + =∂
j j
Formulación para problemas THM
20
Flux referred to a fixed framework: jg
w= igw + ωg
wρg qg + φSgωgwρg du/dt = j'g
w + φSgωgwρg du/dt
Flux referred to the solid skeleton: j'g
w= igw + ωg
wρg qg
( ) ( ) w w w w wl l l g g g l gS S f
t∂
ω ρ φ + ω ρ φ + ∇ ⋅ + =∂
j j
Formulación para problemas THM
21
The use of the material derivative leads to:
( ) ( )
( )( ) ( )' '
w ws l l l g g g w w s
l l l g g g
w w w w wl l l g g g l g
D S S DS SDt Dt
dS S fdt
ω ρ +ω ρ+ ω ρ +ω ρ +
+ ω ρ +ω ρ ∇⋅ +∇ ⋅ +
φφ
=φu j j
Porosity appears in this equation as:•a coefficient in storage terms•in a term involving its variation caused by different processes•hidden in variables that depend on porosity (e.g. intrinsic permeability).
Formulación para problemas THM
22
Porosity from solid balance leads to:
( ) ( )
( )( ) ( )
1
' '
w ws l l l g g g w w
l l l g g gs
w w w w wl l l g g g l
s s
g
D DDt
S SS S
Dt
S St
fdd
ρω ρ +ω ρ −φ+ ω ρ +ω ρ +
ρ
+ ω ρ +ω ρ +∇⋅ + =
φ
∇ ⋅ j ju
For incompressible solid:
( ) ( )
( )( )
' '
w ws l l l g g g w w
l g
w w wl l l g g g
D S SDt
dS S fdt
ω ρ + ω ρφ + ∇ ⋅ + =
= − ω ρ + ω ρ ∇ ⋅ +
j j
u
Formulación para problemas THM
23
Foundation in a saturated soil
Pore pressure
Displacement
Formulación para problemas THM
24
( ) ( ) a a dissolution al l l lS f f
t∂
ω ρ φ + ∇ ⋅ = +∂
j
MASS BALANCE OF AIR
( ) ( ) a a liberation ag g g gS f f
t∂
ω ρ φ + ∇ ⋅ = +∂
j
where f a is an external supply of air and the internal source terms that represent phase change are:
f dissolution + f liberation = 0Adding the two equations leads to:
( ) ( ) a a a a al l l g g g l gS S f
t∂
ω ρ φ+ ω ρ φ +∇ ⋅ + =∂
j j
An internal production term is not present because the total mass balance inside the medium is considered in this equation.
Formulación para problemas THM
25
∇ ⋅ + =σ b 0
MOMENTUM BALANCE FOR THE MEDIUM
Momentum balance = equilibrium of stresses (inertial terms neglected)
where σ is the stress tensor and b is the vector of body forces.
Formulación para problemas THM
26
( )( )1 ( ) Qs s l l l g g g c es el ege e S e S f
t∂
ρ −φ + ρ φ+ ρ φ +∇⋅ + + + =∂
i j j j
INTERNAL ENERGY BALANCE FOR THE MEDIUM
Internal energy balance for the porous medium (es, el, eg):
• ic is energy flux: conduction through the porous medium•(jes, jel, jeg) are advective fluxes of energy caused by mass motions • f Q is an internal/external energy supply.
Advective heat fluxes due to mass movement
Formulación para problemas THM
27
The internal energy in each phase can be calculated as a function of internal energy of each component.
( )w w a a w w a ag g g g g g g g g g g g g
w w a ag g g g g
e e e e e
e e e
ρ = ω ρ + ω ρ = ω + ω ρ
= ω + ω
Following this decomposition, the advective fluxes of energy in each phase are calculated as:
wg g g
wg g g
ag g g
wg
ag
w a
w w a ag g g g g
w w wg g g
a a ag g g
w wg
a a w ag g g g g g g g g
a
gg
g g
g
g
d
e e
e e
e e
e e e e e
Sdt
dSdt
dSdt
ω ρ
ω ρ
= +
⎛ ⎞= + +⎜ ⎟⎝ ⎠⎛ ⎞= + +⎜ ⎟⎝ ⎠
⎛ ⎞= + = +
ω ρ φ
ω ρ
+ ρ +⎜ ⎟
φ
⎝ ⎠φ
q
q
q
j j
i
i
i i
j
j
j
u
u
uj j j
Formulación para problemas THM
28
SOLID
GAS LIQUID
FINAL SET OF BALANCE EQUATIONS
MASS BALANCE OF WATER. Unknown: Pl
( ) ( ) wwg
wlgg
wgll
wl fSS
t=+⋅∇+φρω+φρω
∂∂ jj
MASS BALANCE OF AIR. Unknown: Pg
INTERNAL ENERGY BALANCE FOR THE MEDIUM. Unknown: T
( )( )1 ( ) Qs s l l l g g g c Es El EgE E S E S f
t∂
ρ −φ + ρ φ+ ρ φ +∇⋅ + + + =∂
i j j j
∇⋅ + =σ b 0 MOMENTUM BALANCE FOR THE MEDIUM. Unknown: u
SPECIES MASS BALANCE (REACTIVE TRANSPORT). Unknown: c
RcSt ll =⋅∇+ρφ∂∂ j)(
( ) ( ) aag
algg
agll
al fSS
t=+⋅∇+φρω+φρω
∂∂ jj
Formulación para problemas THM
29
THM problem: Febex mock-up
SECTION A11
SECTION A9SECTION A8
SECTION A5
SECTION A2SECTION A1
SECTION A10
SECTION A7
SECTION A6
SECTION A4
SECTION A3
SECTION AB
Formulación para problemas THM
30
THM problem: Febex mock-up
Heater A Heater B
Zone A Zone B
0 200 400 600 800 1000 1200 1400Time (day)
10
20
30
40
50
60
70
80
90
100
Rel
ativ
e H
umid
ity (%
)
0 200 400 600 800 1000 1200 1400Time (day)
10
20
30
40
50
60
70
80
90
100
Rel
ativ
e H
umid
ity (%
)
Formulación para problemas THM
31
THM problem: Febex mock-up
Heater A Heater B
Zone A Zone B
0 200 400 600 800 1000 1200 1400Time (day)
0
1
2
3
4
5
6
7
8
9
10
11
12
Nor
mal
Str
ess
(MPa
)
0 200 400 600 800 1000 1200 1400Time (day)
0
1
2
3
4
5
6
7
8
9
10
11
12
Nor
mal
Str
ess
(MPa
)
Formulación para problemas THM
32
Boundary conditions
( ) ( ) ( ) ( ) ( )( )0 00 0 0wg
w wg g g
w wg gg g gg g gP Pj j ω γ −= + +β ω − ρ ωω ρ
where the superscript ()0 stands for prescribed values.
•The first term is the mass inflow or outflow that takes place when a flow rate of gas (jg0) is prescribed.
•The second term is the mass inflow or outflow that takes place when gas phase pressure (Pg
0) is prescribed at a node. The coefficient γg is a leakage coefficient, i.e., a parameter that allows a boundary condition of the Cauchy type.
•The third term is the mass inflow or outflow that takes place when vapour mass fraction is prescribed at the boundary. This term naturally comes from the nonadvective flux (Fick's law). Mass fraction and density prescribed values are only required when inflow takes place. For outflow the values in the medium are considered.
Formulación para problemas THM
33
T(t), ρ0vap(t), I(t)
Q(t)=0.70 I(t)Q(t)=0.60 I(t)
No horizontal flows and
displacements
jwg= βg (ρ0vap - ρvap )
jwg
jwl= Q if Pl < P atm.
jwl
βg = 0.002
Pl Patm. ≤
0 1 2m
T(t), ρ0vap(t), I(t)
Q(t)=0.70 I(t)Q(t)=0.60 I(t)
No horizontal flows and
displacements
jwg= βg (ρ0vap - ρvap )
jwg
jwl= Q if Pl < P atm.
jwl
βg = 0.002
Pl Patm. ≤
0 1 2m
ROAD PAVEMENT PROBLEMS
Formulación para problemas THM
34
0
10
20
30
40
50
60
70
80
90
100
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35
Time (days/10)
Rai
n (m
m/1
0 da
ys),
Tem
p (º
C)
0,62
0,64
0,66
0,68
0,7
0,72
0,74
0,76
0,78
Rel
ativ
e hu
mid
ity
Rain (mm/(10 days))
Temperature (ºC)
Relative humidity
Formulación para problemas THM
35
Constitutive laws
Darcy ⇒ liquid and gas flow
Fick ⇒ vapor and air diffusion (dispersion)
Fourier ⇒ heat conduction
Retention curve ⇒ saturation versus suction
Water V-P-T ⇒ water density
Ideal gases law ⇒ gas density
Henry’s law ⇒ dissolved air
Psychrometric law ⇒ vapour concentration
Mechanical model ⇒ stress – strain response
Formulación para problemas THM
36
Enthalpy of water
Depending on the pressure of water (p, MPa) and the enthalpy per unit mass (h, J/kg) three regions are distinguished.From 100 oC to 180 oC, vapor pressure changes by one order of magnitude. While atmospheric pressure (0.1 MPa) is the vapor pressure at 100 oC, 1 MPa is the vapor pressure at 180 oC.
0.01
0.1
1
10
100
0.E+00 1.E+06 2.E+06 3.E+06 4.E+06
h (J/Kg)
Pres
sure
(MPa
)
T=100 oC
T=180 oC
12 3
Pressure-enthalpy diagram for pure water. Isotherm lines for 100 oCand 180 oC are represented.
(1) Single phase region (liquid water)(2) two phase region (liquid water+vapor)(3) single phase region (heated vapor)
Formulación para problemas THM
37
SOIL DESATURATION
•Liquid pressure decrease or air pressure increase: two phase flow with nearly immiscible fluids. Pg≈ Pa (moderate T).
•Vapour pressure increase: gas pressure is dominated by vapour pressure. Pg≈ Pv (relatively high T).
Formulación para problemas THM
38
Thermal conductivity
Fourier's law to compute conductive heat flux, i.e.:
i c T= −λ∇Where λ is thermal conductivity of the porous medium.
(1 ) (1 ) dry solid gas sat solid liqφ φ φ φ− −λ = λ λ λ = λ λ
2 31 2 3( )solid solid o a T Ta aTλ = λ + + +
The definition of dry and saturated thermal conductivities permits to separate the dependence on porosity and the dependence on degreeof saturation. Finally,
( )1 llsat d
Sy
Sr
−λ = λ λOn the other hand, the dry and saturated thermal conductivities can be calculated as described above or directly determined experimentally.
Formulación para problemas THM
39
Thermal conductivity
Thermal conductivity is used in Fourier's lawto compute conductive heat flux, i.e.:ic T= −λ∇with
( )λ λ λ= −sat
Sdry
Sl l1
(1 )
(1 )
dry solid gas
sat solid liq
−φ φ
−φ φ
λ = λ λ
λ = λ λ
λ λsolid solid o a T a T a T= + + +( ) 1 22
33
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
Degree of saturation
Ther
mal
Con
duct
ivity
(W/m
K)
Experimental data
Preoperational
Operational
BENTONITE THERMAL CONDUCTIVITY
Formulación para problemas THM
40
Numerical modelling of drying induced by temperature gradients
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0 0.02 0.04 0.06 0.08 0.1Distance from hot side (m)
Deg
ree
of s
atur
atio
n
Calculated. Time(d): 0Calculated. Time(d): 0.5Calculated. Time(d): 2Calculated. Time(d): 14Experimental: Time (d):14
t=0
t=2d
t=0.5d
t=14d
t=14d experiment
T=120 oC T=30 oC
Formulación para problemas THM
41
Liquid phase density
As a first approximation, the density of liquid phase can be calculated as:
0 exp( ( ) )ll l lo T cPPρ = ρ β − +α + γ
where a water compressibility (β) and a volumetric expansion coefficient (α) are defined. Reference values of 4.5 10-4 MPa-1 and -3.4 10-4 oC-1 can be used in this expression.
Formulación para problemas THM
42
Gas phase density
Gas density is calculated as the sum of vapor and air density. This can be written as:
avg ρ+ρ=ρ
If ideal gases law is used to calculate vapor and air density, it follows that:
v v a ag v a
P MT T
P MR R
ρ = ρ +ρ = +
where the Mv = 0.018 kg/mol and Ma = 0.028 kg/mol are the molecular masses for vapor and dry-air. R is the ideal gas constant (8.3143 J/molK) and T is temperature. On the other hand, partial pressures principle is assumed to hold in the vapor-air mixture (Pg=Pv+Pa), which implies that:
g
aavvg P
MPMPM +=
i.e. the gas phase has variable molecular mass depending on its composition. If Pg is given, air pressure can be calculated as Pa = Pg - Pv where it is assumed that vapor pressure is also known.
Formulación para problemas THM
43
Vapor partial pressure
Vapor pressure depends on temperature and suction:
( , ) ( ) ( , )5239.7( ) 136075exp
273
( , ) exp(273 )
v v
v
wc
l
g l
P P F
P
MF PR
s P P
T s T s T
TT
sTT
= ×
−⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞−
= ⎜ ⎟+ ρ⎝ ⎠= −
In which Pv(T) is the boundary between (1) and (2) in the phase diagram for pure water and F(s,T) is a reduction term due to capillary effects (psychrometric law).
Relative humidity inside soil pores is assumed to be F
h
vvoP P<
Formulación para problemas THM
44
Retention curve
This law relates capillary pressure and degree of saturation. One of the commonly used models is the van Genuchten model (van Genuchten, 1980) that can be written as:
min
max min
le
g l
1/(1- )-P PS S 1+S
S S P
−λλ⎛ ⎞− ⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟− ⎝ ⎠⎜ ⎟
⎝ ⎠
This equation contains the parameters λ and P. The first (λ) essentially controls the shape of the curve while the second (P) controls its height, so this latter can be interpreted as a measure of the capillary pressure required to start the desaturation of the soil. Since capillary pressure can be scaled with surface tension (Milly, 1982) it appears that P also does. This can be shown if Laplace's law is recalled:
2g lP P
r=−
σ
where σ is surface tension and r is the curvature radius of the meniscus.
( )( )o
o
TP PT
σ=
σ( )(1 )1/ 1g l e-P P P S
−λ− λ= −
Formulación para problemas THM
450.1
1.0
10.0
100.0
1000.0
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00Degree of Saturation
Suct
ion
(MPa
)
> 1.73 [1.73 , 1.67] [1.67 , 1.62][1.62 , 1.57] [1.57 , 1.52] < 1.52
BC Drying BC WettingGRIMSEL MOCK_UP
RETENTION CURVES FOR THE BENTONITETest Data
Dry Density (gr/cm3)
Retention Curves
Experimental retention curve
Van Genuchten law
Different densities
Swelling not permitted (except in low density samples)
Formulación para problemas THM
46
Darcy’s law
Advective fluxes of liquid and gas are calculated using darcy’s law in the generalized form:
( )rk Pααα α
α
= − ∇ −ρµ
kq g
Where α = l, g if the law is used for liquid or gas phase. Intrinsic permeability (k) relative permeability (krα) and viscosity (µα) should be calculated.
Intrinsic permeability
This parameter depends primarily on the porous medium structure.
2 3 3
3 2 2
(1 )(1 ) (1 )
oo
o
⎡ ⎤− φ= =⎢ ⎥φ − −⎣ ⎦
φ φφ φ
k k a
where φo is a reference porosity and ko is the intrinsic permeability at the reference porosity.
Formulación para problemas THM
47
Intrinsic permeability
For a continuum medium (Kozeny’s model)
oo
o
o
oo
φφ
φφ
φφ
matrix for ty permeabili intrinsic :porosity reference :
)1()1( 3
2
2
3
k
kk −−
=
which is used in Darcy’s law:
( )qk
gαα
αα αµ
ρ= − ∇ −k
Pr
0.00E+00
5.00E-14
1.00E-13
1.50E-13
2.00E-13
2.50E-13
3.00E-13
3.50E-13
4.00E-13
0.30 0.35 0.40 0.45 0.50 0.55Porosity (n)
K (m
/s)
Experimental Data
Kozeny's Model (Preoperational)
Kozeny's Model (Operational)
Exponential Model
BENTONITE. INTRINSIC PERMEABILITY
Formulación para problemas THM
48
Liquid viscosity
Water viscosity is required in Darcy’s law.
exp273.15l
BT
A ⎛ ⎞µ = ⎜ ⎟+⎝ ⎠Where A = 2.1 10-2 MPa s and B = 1808.5 K-1.
Gas viscosity
273 1 0 ( : intrinsic permeability)11
273
g kk
g
A b C Dk kbBT
PT
+µ = = − ≥
⎛ ⎞ ++⎜ ⎟+⎝ ⎠Where A = 1.48 10-12 MPa s, B = 119.4 K-1, C = 0.14 and D = 1.2 1015. A reduction coefficient takes into account the possibility of Knudsen diffusion (Klinkenberg effect). As the mean free path of the gas particles becomes comparable to the pore size (low permeability material) slip effects (non laminar flow) may result in an apparent lower viscosity (apparent higher gas conductivity).
( ) ( ) ( )1r k r r ko o og g g g g
k b k k bP P PP P
α α αα α α α α α α
⎛ ⎞ ⎛ ⎞= − + ∇ −ρ = − ∇ −ρ − ∇ −ρ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟µ µ µ⎝ ⎠ ⎝ ⎠
k k kq g g g
Formulación para problemas THM
49
Fick’s Law. Molecular Diffusion
To calculate vapor and dissolved air migration:
( )mii iS Dα α αα = φ− ρ τ ∇ωi I
where α=l, g and i=a, w depending if diffusion takes place in the liquid (dissolved air) or in the gas phase (vapor).
( )273 nvap
gm
TP
D D⎛ ⎞+
= ⎜ ⎟⎜ ⎟⎝ ⎠
where τ is a tortuosity reduction coefficient, D = 5.9 10-6 m2/s/K-nPa and n=2.3.
In the case of dissolved air, the following expression is used for molecular diffusivity:
( )exp
273m TQD D
R⎛ ⎞−
= ⎜ ⎟⎜ ⎟+⎝ ⎠
Where τ is a tortuosity reduction coefficient, D = 1.1x10-4 m2/s and Q =24530 J/mol.
Formulación para problemas THM
50
Mechanical constitutive model
Stress-strain relationship: irreversibleSuction induces swelling and/or collapseStrength depends on suctionEffective stress, suction, temperatureNonlinear laws
γd=20 kN/m3
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0 5 10 15 20
Vertical stress (MPa)
∆e/
e
MeassuredComputed
Swelling pressure depence on dry density
0
5
10
15
20
1.4 1.5 1.6 1.7 1.8Dry density (gr/cm3)
Swel
ling
Pres
sure
(MPa
) Experimental data (FEBEX)
Model
Formulación para problemas THM
51
3. Algunos ejemplos
• Earth dams (2 types)• Two phase flow• Rock deformation at great depth• Radioactive Waste disposal
Formulación para problemas THM
52
HM problem: earth dams
El Infiernillo, Mexico
Formulación para problemas THM
53
HM problem: earth dams
Construction history
Formulación para problemas THM
54
HM problem: earth dams
0
1
2
3
4
5
6
0.0 0.5 1.0 1.5 2.0
tensió vertical (MPa)
def v
ertic
al (%
)
model
mesures reals
Compressionand collapse
Formulación para problemas THM
55
HM problem: earth dams
Formulación para problemas THM
56
HM problem: dam construction
40
60
80
100
120
140
160
180
0 20 40 60 80 100 120 140 160 180 200
Assentaments (cm)
Ele
vaci
ó(m
)
Assent mesurats en el camp
Assent model amb pluja
Assent model en sec
dry
wetmeasured
D1 D2
Formulación para problemas THM
57
Water level
Water level
Formulación para problemas THM
58
00.020.040.060.080.1
0.120.140.160.180.2
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7Tensión media (p)
Tens
ión
de c
orte
(q)
Relleno con capas horizontales
M y HM: Escombrera de residuos salinos sobre lodos mineros
Cu = 30 kPa
Formulación para problemas THM
59
No drenado
Drenado
Deformaciones volumetricas(hardening)
Deformaciones de corte.
(Vol =0)
Formulación para problemas THM
60
Desplazamientos para los casos analizados con 210 días (rápido), 210 semanas (lento) y 210 meses
(muy lento) al final de la colocación de la 2ª etapa.
210 days
210 weeks
210 months
Formulación para problemas THM
610
0.1
0.2
0.3
0.4
0.5
0 1000 2000 3000 4000 5000 6000Time (days)
Max
imum
pre
ssur
e
210 days210 weeks210 months
210 days
210 weeks
210 months
Formulación para problemas THM
62
0
0.1
0.2
0.3
0.4
0.5
0 1000 2000 3000 4000 5000 6000Time (days)
Max
imum
pre
ssur
e
210 days210 weeks210 months
00.020.040.060.080.1
0.120.140.160.180.2
0 0.05 0.1 0.15 0.2 0.25 0.3Tensión media efectiva (MPa)
Tens
ión
de c
orte
(MPa
)
Estadoinicial
final 2a etapa
final 1a etapa
00.020.040.060.08
0.10.120.140.160.18
0.2
0 0.05 0.1 0.15 0.2 0.25 0.3
Tensión media efectiva (MPa)
Tens
ión
de c
orte
(MPa
)
Estadoinicial
final 2a etapa
final 1a etapa
00.020.040.060.080.1
0.120.140.160.180.2
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35
Tensión media efectiva (MPa)
Tens
ión
de c
orte
(MPa
)
Estadoinicial
final 2a etapa
final 1a etapa
210 days 210 weeks
210 months
Formulación para problemas THM
63
5X5
1.2 m 0.1 m
0.005
0.010
0.015
0.020
0.025
0.030
0.035
0 0.2 0.4 0.6 0.8 1
Liquid Saturation
Cap
illar
y pr
essu
re (M
Pa)
BLOCK_1BLOCK_2BLOCK_3BLOCK_4BLOCK_5
0.005
0.010
0.015
0.020
0.025
0.030
0.035
0 0.2 0.4 0.6 0.8 1
Liquid Saturation
Cap
illary
pre
ssur
e (M
Pa)
BLOCK_6BLOCK_7BLOCK_8BLOCK_9BLOCK_10
H: GAS AND WATER H: GAS AND WATER FLOW IN A PLANAR FLOW IN A PLANAR
FRACTURE. THE FRACTURE. THE UPSCALING PROBLEMUPSCALING PROBLEM
Formulación para problemas THM
64
a) 60X60 ORIGINAL b) 60X60 UPSCALED c) 12X12 UPSCALED
Gas fluxes and degree of saturation
Formulación para problemas THM
65
TM: ROCK SALT IN SITU TEST
BAMBUS-I and BAMBUS-II: EC research project
Formulación para problemas THM
66
Features: Tunnel + Observation drifts, σconf= 12 MPa, Axial length: 25 m, hexaedral elements, regular mesh, 65,954 nodal points x 4 dof = 263,816 dof
Backfilled Drift with heaters
Observation driftLevel 750 m
Observation driftLevel 800 m
Formulación para problemas THM
67
1.5 years of heating
Temperature evolution (whole model)
9 years of heating
end process
7. DRIFT CLOSSURE IN THE HEATED AREA (BAMBUS II - Project : 3-DIM modelling studies)
050
100150200250300350400450500
sep-90
sep-91
sep-92
sep-93
sep-94
sep-95
sep-96
sep-97
sep-98
sep-99Date
Drif
t clo
sure
[mm
]horitzontal drift closure (3D-model)horitzontal ( TSDE experiment)vertical drift closure (3D-model)vertical (TSDE experiment)
Formulación para problemas THM
68
(THM) Residuos Radioactivos en Roca fracturada
Evaporation and condensation of water (fractured rock)
J. Rutqvist & C.-F. Tsang (2003)
Formulación para problemas THM
69-10
-7.5
-5
-2.5
0
2.5
5
7.5
10
12.5
15
17.5
20
-20 -17.5 -15 -12.5 -10 -7.5 -5 -2.5 0 2.5 5 7.5 10 12.5 15 17.5 20
Borehole 160: temperature measurements
Borehole 158: temperature measurements
Heated drift including canister heater
Borehole 76: gas tests at 76-3 and 76-4
Borehole 74: gas tests at 74-3 and 74-4
Inner wing heaters
Outer wing heaters
74-4
74-3
76-3
76-4
20
60
100
140
180
220
0
365
730
1095
1460
D ays from start up o f heating
Tem
pera
ture
[ºC
]
158-5
158-10
158-20
158-45
158-55
Temperaturas
Formulación para problemas THM
70
Grado de saturación en la roca y en las fracturas
Degree of saturation (after 4 years)
Matrix(initial degree of saturation = 0.92)
Fracture(initial degree of saturation = 0.05)
Formulación para problemas THM
71
Case 2 analysis (thermomechanical and mechanical coupling)
Gas permeability variation near drift
-10
-7.5
-5
-2.5
0
2.5
5
7.5
10
12.5
15
17.5
20
-20 -17.5 -15 -12.5 -10 -7.5 -5 -2.5 0 2.5 5 7.5 10 12.5 15 17.5 20
Borehole 160: temperature measurements
Borehole 158: temperature measurements
Heated drift including canister heater
Borehole 76: gas tests at 76-3 and 76-4
Borehole 74: gas tests at 74-3 and 74-4
Inner wing heaters
Outer wing heaters
74-4
74-3
76-3
76-4
0
0.5
1
1.5
219
/10/
1997
19/1
0/19
98
19/1
0/19
99
18/1
0/20
00
18/1
0/20
01
Date
k/kp
re-h
eatin
g
59-376-3Base CaseCase 2
0
0.5
1
1.5
2
19/1
0/19
97
19/1
0/19
98
19/1
0/19
99
18/1
0/20
00
18/1
0/20
01
Date
k/kp
re-h
eatin
g59-476-4Base CaseCase 2
Formulación para problemas THM
72
Comentarios finales
• Es conveniente que se puedan elegir las ecuaciones a resolver y fenómenos a incorporar
• Las condiciones de solución numérica cambian mucho según el problema a resolver
• Todos los acoplamientos conocidos hay que incorporarlos en la formulación
• Fenómenos acoplados versus procesos acoplados
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