Phase exchange for flow in porous media and complementary problems Jérôme Jaffré with A. Sboui, I. Ben Gharbia, J.-C. Gilbert jerome.jaffre@inria.fr Institut National de Recherche en Informatique et Automatique Scaling Up and Modeling for Transport and Flow in Porous Media Dubrovnik, 13-16 October 2008 Dedicated to Alain Bourgeat Supported by Andra and Momas (http://www.gdrmomas.org) J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 1 / 11
Motivation : Couplex-Gas, an Andra-Momas benchmark • In a deep underground nuclear waste disposal • Production of hydrogen from corrosion of waste packets • Migration of this hydrogen ? Formulations with complementary equations can be found in G. Chavent and J. Jaffré, Mathematical models and finite elements for reservoir simulation, (North Holland, 1986) A. Bourgeat, M. Jurak and F . Smaï, Two phase partially miscible flow and transport modeling in porous media ; application to gas migration in a nuclear waste repository A recent work with a new formulation: A. Abadpour and M. Panfilov, Method of negative saturations for multiple compositional flow with oversaturated zones. J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 2 / 11
Formulation I: Phase equations 2 fluid phases: liquid ( i = ℓ ) and gas ( i = g ) Darcy’s law: q i = − K ( x ) k i ( s i )( ∇ p i − ρ i g ∇ z ) , i = ℓ, g K the absolute permeability q i Darcy’s velocities, s i saturations, p i fluid pressures, k i mobilities Phases occupy the whole pore space: s ℓ + s g = 1. Capillary pressure law: p c ( s ℓ ) = p g − p ℓ ≥ 0 , p c decreasing, p c ( 1 ) = 0. Since liquid phase does not disappear, main unknowns will be s ℓ and p ℓ . J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 3 / 11
Fluid components 2 components: water ( j = w ) and hydrogen ( j = h ). Mass density of phase i : ρ i = ρ i w + ρ i h , i = ℓ, g . h = ρ i w = ρ i w Mass fractions: χ i h χ i ( χ i w + χ i , , i = ℓ, g , h = 1 ) . ρ i ρ i Assume • liquid phase contains both components, • gas phase contains only hydrogen. h = ρ g Then ρ g ρ g = ρ g χ g χ g h w = 0 , = 1 , w = 0 . h , ρ g Assume • ρ g = C g p g and ρ ℓ w constant. χ ℓ h is the third main unknown. J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 4 / 11
Conservation of components Diffusion of hydrogen in the liquid phase: j ℓ h = − φ s ℓ ρ ℓ D ℓ h ∇ χ ℓ h . D ℓ φ porosity, h molecular diffusion coefficient. Mass conservation for each component: φ ∂ ∂ t ( s ℓ ρ ℓ χ ℓ w ) + div ( ρ ℓ χ ℓ Water: w q ℓ ) = Q w φ ∂ ∂ t ( s ℓ ρ ℓ χ ℓ h + s g ρ g ) + div ( ρ ℓ χ ℓ h q ℓ + ρ g q g + j ℓ Hydrogen: h ) = Q h . J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 5 / 11
Phase equilibrium: Henry’s law In the presence of gas phase Henry’s law reads Hp g = ρ ℓ h . To integrate Henry’s law in a formulation which includes the case with no gas phase, introduce the liquid pressure p ℓ = p g − p c ( s ℓ ) , and • Either gas phase exists: 1 − s ℓ > 0 and H ( p ℓ + p c ( s ℓ )) − ρ ℓ h = 0 • Or gas phase does not exist: s ℓ = 1 , p c ( s ℓ ) = 0 and Hp ℓ − ρ ℓ h ≥ 0 In other words, last inequality means • for a given pressure p ℓ mass fraction is too small for the hydrogen component to be partly gaseous, • for a given mass fraction ρ ℓ h the pressure p ℓ is too large for the hydrogen component to be partly gaseous. Thus we close the system with the complementary constraints � H ( p ℓ + p c ( s ℓ )) − ρ ℓ H ( p ℓ + p c ( s ℓ )) − ρ ℓ � ( 1 − s ℓ ) = 0 , 1 − s ℓ ≥ 0 , h ≥ 0 . h J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 6 / 11
Phase diagram A phase diagram tells also how a component separates into the liquid and gas phases: P Liquid f(P,C)>0 liquid liquid + gas gas 0 = ) C , Liquid + Gaz s ℓ = 1 0 < s ℓ < 1 s ℓ = 0 P ( f f(P,C)<0 p g = p ℓ − p c ( s ℓ ) P p ℓ p g ρ ℓ ρ ℓ h + ρ g C ρ g h Gaz f(P,C)>0 C We now concentrate on the separation one liquid phase – twophase zones. J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 7 / 11
Henry’s law versus phase diagram • In the twophase zone, with Henry’s law HP − C = Hp g − ( ρ ℓ h + ρ g ) = − ρ g < 0. • When moving to the separation between liquid and twophase zones HP − C = − ρ g → 0. • In the liquide zone HP − C = Hp ℓ − ρ ℓ h > 0. P liquid liquid + gas Liquid s ℓ = 1 0 < s ℓ < 1 HP−C>0 P p ℓ p g = p ℓ − p c ( s ℓ ) 0 ρ ℓ ρ ℓ = C h + ρ g C − h P H Liquid + Gaz Thus, with Henry’s law, the curve HP−C<0 separating the liquid and twophase zones is the straight line HP − C = 0. C J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 8 / 11
A nonlinear problem with complementary equations φ ∂ � s ℓ ρ ℓ ( 1 − χ ℓ � � ρ ℓ ( 1 − χ ℓ � h ) + div h ) q ℓ = Q w ∂ t φ ∂ � s ℓ ρ ℓ χ ℓ � h + ( 1 − s ℓ ) C g ( p ℓ + p c ( s ℓ )) + ∂ t ρ ℓ χ ℓ h q ℓ + C g ( p ℓ + p c ( s ℓ )) q g + j ℓ � � div = Q h h q ℓ = − K ( x ) k ℓ ( s ℓ )( ∇ p ℓ − ρ ℓ g ∇ z ) q g = − K ( x ) k g ( 1 − s ℓ )( ∇ ( p ℓ + p c ( s ℓ )) − C g ( p ℓ + p c ( s ℓ )) g ∇ z ) j ℓ h = − φ s ℓ ρ ℓ D ℓ h ∇ χ ℓ h � H ( p ℓ + p c ( s ℓ )) − ρ ℓ χ ℓ � ( 1 − s ℓ ) = 0 , 1 − s ℓ ≥ 0 , h H ( p ℓ + p c ( s ℓ )) − ρ ℓ χ ℓ h ≥ 0 . J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 9 / 11
The discretized problem Discretized with cell-centered finite volumes : N , the number of cells. x ∈ R 3 N , vector of unknowns for s ℓ , p ℓ , χ ℓ h H : R 3 N → R 2 N for discretized conservation equations F : R 3 N → R N for discretized 1 − s ℓ G : R 3 N → R N for discretized H ( p ℓ + p c ( s ℓ )) − ρ ℓ χ ℓ h Problem in compact form H ( x ) = 0 , F ( x ) ⊤ G ( x ) = 0 , F ( x ) ≥ 0 , G ( x ) ≥ 0 . J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 10 / 11
Task H ( x ) = 0 , F ( x ) ⊤ G ( x ) = 0 , F ( x ) ≥ 0 , G ( x ) ≥ 0 . Construction of general purpose fast and robust solvers for nonlinear problems with complementary constraints Additional difficulties: • Unknowns s ℓ and χ ℓ h are bounded • Vertical tangent of p c at s = 1 in the Van Genuchten model Other examples for flow in porous media • Black-oil model • Dissolution-precipitation J. Jaffré (INRIA) Complementary Problems Dubrovnik, October 2008 11 / 11
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