phase exchange for flow in porous media and complementary
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Phase exchange for flow in porous media and complementary problems - PowerPoint PPT Presentation

Phase exchange for flow in porous media and complementary problems Jrme Jaffr 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


  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  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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