Geotécnia e Ingeniería Sísmica aplicadas a la Minería

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


2

Índice

1. PROBLEMAS ACOPLADOS2. FORMULACION BASICA 3. ALGUNOS EJEMPLOS


3

PROBLEMAS ACOPLADOSMechanical problemsin soils and rocks


4

Mechanical problemsin soils and rocksOne phase flow

PROBLEMAS ACOPLADOS


5

Mechanical problemsin soils and rocksOne phase flowTwo phase flow

PROBLEMAS ACOPLADOS


6

Mechanical problemsin soils and rocks

Saturated consolidation/unsaturated consolidation

One phase flowTwo phase flow

PROBLEMAS ACOPLADOS


7



One phase flow

Evaporation/evapotranspirationWaste disposalsGeothermal fluid extraction

Two phase flow

PROBLEMAS ACOPLADOS


8



One phase flow

Evaporation/evapotranspirationWaste disposalsGeothermal fluid extraction

Solute transportproblems

Two phase flow

CODE_BRIGHT

PROBLEMAS ACOPLADOS


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


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


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 air in the gas phase

ag gω ρ

al lω ρ


12

Mass of Solid in the Solid 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 air in the gas phaseag g gSω ρ φ

al l lSω ρ φ


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)


14

TemperatureWater content

Water fluxes


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


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)


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:


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


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


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


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).


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


23

Foundation in a saturated soil

Pore pressure

Displacement


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.


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.


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


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


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


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


30


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 (%

)


31


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

)


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.


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


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


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


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)


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).


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.


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


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


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.


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.


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<


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= −


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)


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.


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


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


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.


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


51

3. Algunos ejemplos

• Earth dams (2 types)• Two phase flow• Rock deformation at great depth• Radioactive Waste disposal


52

HM problem: earth dams

El Infiernillo, Mexico


53


Construction history


54


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


55



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


57

Water level

Water level


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


59

No drenado

Drenado

Deformaciones volumetricas(hardening)

Deformaciones de corte.

(Vol =0)


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


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


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


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


64

a) 60X60 ORIGINAL b) 60X60 UPSCALED c) 12X12 UPSCALED

Gas fluxes and degree of saturation


65

TM: ROCK SALT IN SITU TEST

BAMBUS-I and BAMBUS-II: EC research project


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


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)


68

(THM) Residuos Radioactivos en Roca fracturada

Evaporation and condensation of water (fractured rock)

J. Rutqvist & C.-F. Tsang (2003)


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


Heated drift including canister heater

Borehole 76: gas tests at 76-3 and 76-4


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


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)


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



Heated drift including canister heater



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


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

Geotécnia e Ingeniería Sísmica aplicadas a la Minería

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