modeling compressible multiphase flow and …...for simplicity, but consistant, with modelling gas...

Multiphase Flow andTransport in

saturated-unsaturatedporous media

Alain Bourgeat,Universite ClaudeBernard Lyon 1Institut Camille

Jordan-UMR 5208Contributors:F.Smaı, IRSN

Fontenay aux Rosesand ICJ-UCBLyon1;

M.Jurak, Univ. Zagreb

Physical assumptions

Unsaturated flowequations

Saturated flowequations

Construction of asaturated/unsaturatedmodel

Three Numerical Tests

Conclusions

Modeling compressible Multiphase

Flow and Transport in

saturated-unsaturated porous media:

Phase appearance-disappearanceApplication to gas migration in underground nuclear waste

repository

Alain Bourgeat, Universite Claude Bernard Lyon 1 InstitutCamille Jordan-UMR 5208

Contributors:F.Smaı, IRSN Fontenay aux Roses and ICJ-UCBLyon1;

RICAM; Linz- Austria; Oct3-7, 2011

Conclusions

IntroductionThe Context

I in a deep geological Nuclear Waste Repository there could behydrogen gas generation due to corrosion of the steelengineered barriers

I the flow could be saturated (' only the liquid phase) insome regionsand unsaturated (a ”liquid + gas” mixture) in other ones

I Problem in simulations, using standard models, when there isa phase appearance/disappearance

– - variable switching or variable substitution– - small artificial non-zero saturation

Conclusions

IntroductionA repository

Figure: A Repository Zone; Waste Packages(containers sets); StorageUnits

Conclusions

IntroductionEngineered Barriers

Figure: LONG TERM RISKS, from pressure buid-up: hydraulic headpressure gradient ⇒RN transport enhancement; mechanical damagingof the host rock and the barrier; perturbation of the seals resaturation;...!

Conclusions

Saturated-Unsaturated two phases flowPhysical assumptions -(i)

For simplicity, but consistant, with modelling gas migration indeep geological Nuclear Waste Repositories :

I 2 phases : liquid (incompressible), denoted l and gas(compressible)denoted g

I 2 components : water,w and and hydrogen,h

I Mass conservation for each component w, h

I Generalized Darcy law for each phase , α ∈ {g, l}

qα = −Kkr,α(Sα)

µα(∇pα − ραg) (1)

I local mechanical equilibrium of phases ≈ Capillary pressurelaw : pg − pl = pc(Sg).

I Ideal gaseous Mixture of Ideal gas,with Dalton’s law of partialpressures, and diluted liquid solution

I Isothermal flow, with the two phases locally at the sametemperature

I No chemical reactions

Conclusions

Saturated-Unsaturated two phases flowPhysical assumptions -(ii)

I Diffusion of component i in phase α ( and infinite dilution )

jhl = −ΦSlD∇ρhl , jwl = −jhl ,

I Local thermodynamical equilibrium liquid solution/ gasmixture

pwg = xwg pg = pwv exp

((pg − pc)− pwvRTρw,∗l /Mw

)xwl (2)

phg = xhgpg = KhH exp

(pl − p0

RT/vh,∞l

)xhl ; (3)

∼Raoult-Kelvin and Henry laws; with:xiα, the i-component molar concentration in theα-phase, pwv the pure water saturated vapor pressure , ρw,∗lthe pure liquid water mass density, Kh

H the Henry’s constant

at p0 (a reference pressure) , vh,∞l the hydrogen molarconcentration at infinite dilution .

Conclusions

Saturated-Unsaturated two phases flowPhysical assumptions -(ii)

I Diffusion of component i in phase α ( and infinite dilution )

jhl = −ΦSlD∇ρhl , jwl = −jhl ,

I Local thermodynamical equilibrium liquid solution/ gasmixture

pwg = xwg pg = pwv exp

((pg − pc)− pwvRTρw,∗l /Mw

)xwl (2)

phg = xhgpg = KhH exp

(pl − p0

RT/vh,∞l

)xhl ; (3)

∼Raoult-Kelvin and Henry laws; with:xiα, the i-component molar concentration in theα-phase, pwv the pure water saturated vapor pressure , ρw,∗lthe pure liquid water mass density, Kh

H the Henry’s constant

at p0 (a reference pressure) , vh,∞l the hydrogen molarconcentration at infinite dilution .

Phase diagram

Conclusions

Unsaturated two-phase flowAn example of Unsaturated liquid+gas mixture flow (no vaporized water)

The components mass conservation reads :

∂t(Slρ

wl ) + div (ρwl ql + jwl ) = Fw, (4)

hl + Sgρg

)+ div

(ρhl ql + ρgqg + jhl

)= Fh; (5)

Usual primary variables: (pl, Sl), (pl, pg), (pl, pc).

Where : ρil, the i-component mass concentration;

ρg ' ρhg = Cvpg( = ideal gas); pg = pc + pl;

ρhl = Chpg( = Henry’s law) .and constants :

Ch =Mh

, Cv =Mh

RT(Mh = Hydrogen molar mass) .

Phase diagram

Conclusions

∂t(Slρ

hl + Sgρg

)+ div

)= Fh; (5)

ρhl = Chpg( = Henry’s law) .

and constants :

Ch =Mh

, Cv =Mh

Phase diagram

Conclusions

∂t(Slρ

hl + Sgρg

)+ div

)= Fh; (5)

ρhl = Chpg( = Henry’s law) .and constants :

Ch =Mh

, Cv =Mh

Phase diagram

Conclusions

Unsaturated/Saturated flowPhase Diagram

C h(p l

+pc(0))

Sg = 0

ρhl ≤ Ch(pl + pc(0))

Sg > 0

ρhl = Chpg

≥ pl + pc(0)

pg = pl + pc(Sg)

Figure: Henry’s law:ρhl = Chpg. Localization of the saturated state,Sg = 0, and the unsaturated state, Sg > 0 .

Conclusions

Saturated flow ( One phaseLiquid saturated flow

I Sl ≡ 1, pg is then indeterminate

I Classical Darcy law for liquid flow (water + dissolvedhydrogen)

I Dissolved hydrogen transported by diffusion and convectionI The

liquid solution flow (water + dissolved hydrogen) is described by :

div(ρwl ql − jhl

)= Fw, Φ

∂ρhl∂t

+ div(ρhl ql + jhl

)= Fh; (6)

ql = −Kλl(1)(∇pl − (ρwl + ρhl )g

), jhl = −ΦD∇ρhl . (7)

I Primary variables are then (pl, ρhl ); the amount of dissolved

hydrogen, ρhl ,is now an independent unknown( no morerelated to the pg); and a gas phase cannot be taken inaccount.

I the couple (Saturation,Phase Pressure) cannot describe theflow in such a liquid saturated region ( one-phase flow).

How to have a unique model for both saturated and unsaturatedflows ?

Conclusions

I Sl ≡ 1, pg is then indeterminateI Classical Darcy law for liquid flow (water + dissolved

hydrogen)

I Dissolved hydrogen transported by diffusion and convectionI The

liquid solution flow (water + dissolved hydrogen) is described by :

div(ρwl ql − jhl

)= Fw, Φ

∂ρhl∂t

+ div(ρhl ql + jhl

)= Fh; (6)

), jhl = −ΦD∇ρhl . (7)

Conclusions

hydrogen)I Dissolved hydrogen transported by diffusion and convection

I Theliquid solution flow (water + dissolved hydrogen) is described by :

div(ρwl ql − jhl

)= Fw, Φ

∂ρhl∂t

+ div(ρhl ql + jhl

)= Fh; (6)

), jhl = −ΦD∇ρhl . (7)

Conclusions

div(ρwl ql − jhl

)= Fw, Φ

∂ρhl∂t

+ div(ρhl ql + jhl

)= Fh; (6)

), jhl = −ΦD∇ρhl . (7)

Conclusions

div(ρwl ql − jhl

)= Fw, Φ

∂ρhl∂t

+ div(ρhl ql + jhl

)= Fh; (6)

), jhl = −ΦD∇ρhl . (7)

Conclusions

div(ρwl ql − jhl

)= Fw, Φ

∂ρhl∂t

+ div(ρhl ql + jhl

)= Fh; (6)

), jhl = −ΦD∇ρhl . (7)

Conclusions

div(ρwl ql − jhl

)= Fw, Φ

∂ρhl∂t

+ div(ρhl ql + jhl

)= Fh; (6)

), jhl = −ΦD∇ρhl . (7)

Choice of suitablevariables

Choice i Model

Choice ii Model

Capillary PressureCurve, and Inverse

Conclusions

Construction of asaturated(1-phase)/unsaturated(2-phases )model

I According to the phase state , the primary variables are:

I two-phase unsaturated : Liquid pressure pl/Phase SaturationSl; like in (4)-(5)

I one-phase saturated : Liquid pressure pl/Hydrogen massconcentration, ρhl , like in (6) - (7)

I Considering pl and ρhl as main variables in both casessaturated and unsaturated , then eqs. (4)-(7) reduce to aunique couple of equations:

∂t(Slρ

(ρhtot)

+ div(ρhl ql + ρgqg + jhl

)= Fh ; (9)

I with ρhtot = Slρhl + CvpgSg ; Cvp

hg = ρhg (∼ ideal gas)

Choice i Model

Choice ii Model

Conclusions

I According to the phase state , the primary variables are:I two-phase unsaturated : Liquid pressure pl/Phase SaturationSl; like in (4)-(5)

∂t(Slρ

(ρhtot)

)= Fh ; (9)

Choice i Model

Choice ii Model

Conclusions

∂t(Slρ

(ρhtot)

)= Fh ; (9)

Choice i Model

Choice ii Model

Conclusions

∂t(Slρ

(ρhtot)

)= Fh ; (9)

Choice i Model

Choice ii Model

Conclusions

∂t(Slρ

(ρhtot)

)= Fh ; (9)

Choice i Model

Choice ii Model

Conclusions

Construction of asaturated(1-phase)/unsaturated(2-phases)modelChoice of suitable variables

I the Total Hydrogen mass concentration ρhtot ≡ ρhl Sl + ρhgSg ,is definite in both states of flow (according to the flow state):

I two-phases unsaturated state :

ρhtot = a(Sg)(pl + pc(Sg)) ;Sg > 0 (10)

with: a(Sg) = Ch(1− Sg) + CvSg ∈ [Ch, Cv]. (11)

I one-phase saturated :ρhtot = ρhl ; Sg = 0.

I There is now two possible choices for the main variables ineq. (8) and (9):

Choice i : Liquid pressure, pl / Total Hydrogen concentration, ρhtotChoice ii : Liquid pressure, pl / Hydrogen mass concentration, ρhl

Choice i Model

Choice ii Model

Conclusions

Construction of asaturated(1-phase)/unsaturated(2-phases)modelChoice of suitable variables

I the Total Hydrogen mass concentration ρhtot ≡ ρhl Sl + ρhgSg ,is definite in both states of flow (according to the flow state):

I two-phases unsaturated state :

ρhtot = a(Sg)(pl + pc(Sg)) ;Sg > 0 (10)

with: a(Sg) = Ch(1− Sg) + CvSg ∈ [Ch, Cv]. (11)

I one-phase saturated :ρhtot = ρhl ; Sg = 0.

I There is now two possible choices for the main variables ineq. (8) and (9):

Choice i : Liquid pressure, pl / Total Hydrogen concentration, ρhtotChoice ii : Liquid pressure, pl / Hydrogen mass concentration, ρhl

Choice i Model

Choice ii Model

Conclusions

Choice i:saturated(1-phase)/unsaturated(2-phases )modelLiquid pressure, pl / Total Hydrogen concentration, ρhtot

I In system (8)-(9) we compute Sg = Sg(pl, ρhtot), (from

(11)); Sl = 1− Sg, andρhl = ρhl (pl, ρ

htot) = min(Chpg(pl, ρ

htot), ρ

htot), pg(pl, ρ

htot) =

pl + pc(Sg(pl, ρhtot)).

I Noticing a() is > Ch, increasing and pg > pl + pc(0); theState of flow is then characterized by:

unsaturated : ρhtot > Ch(pl + pc(0))

saturated : ρhtot ≤ Ch(pl + pc(0))

ρhtot ≡ ρhl

I Then:I 1st equation is parabolic(unsaturated)/elliptic(saturated) in plI 2nde equation is parabolic in ρhtot

Choice i Model

Choice ii Model

Conclusions

htot), ρ

htot), pg(pl, ρ

htot) =

saturated : ρhtot ≤ Ch(pl + pc(0))

ρhtot ≡ ρhlI Then:

I 1st equation is parabolic(unsaturated)/elliptic(saturated) in plI 2nde equation is parabolic in ρhtot

Choice i Model

Choice ii Model

Conclusions

htot), ρ

htot), pg(pl, ρ

htot) =

saturated : ρhtot ≤ Ch(pl + pc(0))ρhtot ≡ ρhl

Choice i Model

Choice ii Model

Conclusions

htot), ρ

htot), pg(pl, ρ

htot) =

saturated : ρhtot ≤ Ch(pl + pc(0))ρhtot ≡ ρhl

Choice i Model

Choice ii Model

Conclusions

Choice i:saturated(1-phase)/unsaturated(2-phases )modelExistence of solutions

I After an ad hoc variable change, the Alt-Luckhaus theoremapplies, and the existence of a solution could be proved (F.

Smaı, PhD Thesis) .

Suppose rmin ≤ ρhtot ≤ rmax and pl ≥ 0 and assume thatinitial and Dirichlet conditions are enough regular.Then there is a weak solution to the simplified formulation.

I Could also certainly be investigated using ”Entropy weaksolutions”, defined by Kruzkov (hyperbolic) and extended byCarillo (parabolic).Remarks:

I no need of capillary pressure for this formulation(h-component eq. in (6) Parabolic −→ Hyperbolic)

I Coefficients in the div operators and in the ∂∂t

operatorscould become discontinuous

Choice i Model

Choice ii Model

Conclusions

Choice i Model

Choice ii Model

Conclusions

Choice i Model

Choice ii Model

Conclusions

Choice ii:saturated(1-phase)/unsaturated(2-phases )Liquid pressure, pl / Hydrogen mass concentration, ρhl

I Introducing in system (8)-(9), from (pc)−1, the extended

gas-phase Pressure p∗g = π + pl; and the extended saturation

S∗g = f(ρhlCh− pl

pc(Sg)

π =ρhlCh

− pl

Figure: pc = pg − pl; p∗g =ρhlCh

I leads to a system with the main variables pl and ρhl

Choice i Model

Choice ii Model

Conclusions

Choice ii:saturated(1-phase)/unsaturated(2-phases )Liquid pressure, pl / Hydrogen mass concentration, ρhl

I Introducing in system (8)-(9), from (pc)−1, the extended

gas-phase Pressure p∗g = π + pl; and the extended saturation

S∗g = f(ρhlCh− pl

pc(Sg)

π =ρhlCh

− pl

Figure: pc = pg − pl; p∗g =ρhlCh

I leads to a system with the main variables pl and ρhl

Choice i Model

Choice ii Model

Conclusions

Choice ii:saturated(1-phase)/unsaturated(2-phases)modelLiquid pressure, pl / Hydrogen mass concentration, ρhl

I 1st equation is parabolic (unsaturated)/elliptic(saturated) inpl, non uniformly ( coefficient of ∇pl → 0 as sg → 1)2nde equation is strictly parabolic in ρhl

I Remarks:I capillary pressure is absolutely necessary for this formulationI pl and ρhl are continuous no matter the discontinuity of the

Saturations ( porous media highly heterogeneous)I The coefficients in all the div and ∂

∂toperators are continuous

Choice i Model

Choice ii Model

Conclusions

Choice ii:saturated(1-phase)/unsaturated(2-phases)modelLiquid pressure, pl / Hydrogen mass concentration, ρhl

I 1st equation is parabolic (unsaturated)/elliptic(saturated) inpl, non uniformly ( coefficient of ∇pl → 0 as sg → 1)2nde equation is strictly parabolic in ρhl

I Remarks:I capillary pressure is absolutely necessary for this formulationI pl and ρhl are continuous no matter the discontinuity of the

Saturations ( porous media highly heterogeneous)I The coefficients in all the div and ∂

∂toperators are continuous

Setting

Simulations

Implementation inthe IRSN code:DIPHPOM

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

Analysis and simulation; Numerical TestsAdvertising

Ongoing benchmark on:”Modelling Multiphase Flows”

http://www.gdrmomas.org/Benchmark/multiphase/

multiphasique.html

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

Analysis and simulation; Quasi-1D scale fieldnumerical simulationsPb 1, Pb 2, Pb 3 in Numerical Test Data Base

Total Hydrogen concentration, ρhtot is denoted X in the following

I Boundary conditions :I Injection of pure gas on left sideI Impervious condition on top and bottom sideI Pure water (Xout = 0) (Test 1) or

Two-phases(Xout 6= 0)(Test 2) and a fixed pressure, on theright side

I Initial conditions :stationary state without injection (Qhin = 0)

Setting

Simulations

Numerical Test I

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Conclusions

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Simulations

Numerical Test I

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Conclusions

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

Analysis and simulationQuasi-1D scale field numerical simulations

I Van Genuchten-Mualem model for capillary pressure andrelative permeabilities

I Fixed temperature, T = 303 K

Porous medium parameters Fluid characteristicsParameter Value Parameter Value

k 5 10−20 m2 Dhl 3 10−9 m2/s

Φ 0.15 (−) µl 1 10−3 Pa.s

Pr 2 106 Pa µg 9 10−6 Pa.s

n 1.49 (−) H(T = 303K) 7.65 10−6 mol/Pa/m3

Slr 0.4 (−) Ml 10−2 kg/mol

Sgr 0 (−) Mg 2 10−3 kg/mol

ρstdl 103 kg/m3

ρstdg 8 10−2 kg/m3

Parameter ValueLx 200 mLy 20 m

Qh 1.5 10−5 m/yearpl,out 106 PaTsimul 5 105 years

For more, see :

http://sources.univ-lyon1.fr/cas test.html

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

Analysis and simulationNumerical test : Implementation

I Fully implicit time discretization of the space/time p.d.e.system

I Newton iteration to solve nonlinearities of the space pdesystem

I Spatial discretization of the pde with a standard linear F.E.from the C++ LIBMESH Library

I GMRES/LU methods (PETSC)

Setting

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Conclusions

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Numerical Test I

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Conclusions

Analysis and simulationTest 1: Gas injection in a fully water saturated host rock

First, only solved Hydrogen density increases; second, the gas phase(Sg)appears ⇒ ∇pg,∇pc(Sg) and ∇pl > 0; finally higher∇pc(Sg)⇒ ∇pl < 0; and no water injection slow down the pl

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Analysis and simulationTest 2: Gas injection in a unsaturated host rock

Injection of Hydrogen increases ρhl , pg; and ( pc(Sg)smallenough)⇒ pl increases up to ”meet” Sg = 0 : a plug of liquidappears and is pushed by the injected gas to the R.H.S. ( withunsaturated Dirichlet cond.); bringing back to Test 1.

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Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 2000

abscissa (m)

0 40 80 120 160 200

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

Analysis and simulationTest 3: Non-equilibrium initial conditions

Starting from Non equilibrium initial condition(discontinuity of thegas-phase pressure at the starting time).

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 10

abscissa (m)

0 0.2 0.4 0.6 0.8 1

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

t = 10 s

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

t = 10

liquid and gas Pressure profiles

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

t = 10

Gas Saturation profiles

t = 5000 s

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

t = 5000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

t = 5000

t = 100000 s

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

t = 100000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

t = 100000

x (m) s

Setting

Simulations

Numerical Test I

Numerical Test II

Numerical Test III

Conclusions

t = 200000 s

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

t = 200000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

t = 200000

t = 1000000 s

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

t = 1000000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

t = 1000000

Conclusions

I Construction of a unique model for both, saturated andunsaturated, flows; handling phase appearance anddisappearance

I Ongoing implementation:I 2-phases, N + 1 components (1 solvent and N solutes)

I Thermal flows (Energy equation)

I Modelling in progress:

I Chemical reactions (... CO2 sequestration)

Conclusions

I Ongoing implementation:I 2-phases, N + 1 components (1 solvent and N solutes)I Thermal flows (Energy equation)

I Modelling in progress:

I Chemical reactions (... CO2 sequestration)

Conclusions

I Ongoing implementation:I 2-phases, N + 1 components (1 solvent and N solutes)I Thermal flows (Energy equation)

I Modelling in progress:I Chemical reactions (... CO2 sequestration)

Conclusions

References

I Bourgeat, A. and Jurak, M. and Smaı, F. Two-phase, partiallymiscible flow and transport modeling in porous media; application to gasmigration in a nuclear waste repository. Computational Geosciences,Volume13, Number 1, mars 2009 .

I Bourgeat, A. and Jurak, M. and Smaı, F. Modelling and NumericalSimulation of Gas Migration in a Nuclear Waste Repository .http://arxiv.org/abs/1006.2914, June 2010

modeling compressible multiphase flow and …...for simplicity, but consistant, with modelling gas...

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