A Fully Equivalent Global Pressure Formulation for Three-Phases - PowerPoint PPT Presentation

Scaling Up for Modeling of Transport and Flow in Porous Media A conference in Honor of Alain Bourgeat A Fully Equivalent Global Pressure Formulation for Three-Phases Compressible Flows . Guy Chavent * Raphaël DiChiara-Roupert** Gerhard Schäfer** * CEREMADE, Université Paris-Dauphine, and INRIA ** IMF Strasbourg Dubrovnik, October 13-16 2008 – p. 1/19

Summary • The three-phases immiscible compressible equations • Why a global pressure ? • Equivalent global pressure reformulation: TD Condition. • An example of Global Capillary Pressure function • TD-interpolation of two-phase data: compatibility condition • Conclusions Dubrovnik, October 13-16 2008 Summary – p. 2/19

Notations ( 1 = water , 2 = oil , 3 = gas ) • dependant variables :  S j = S j ( x, t ) = reduced saturation, 0 ≤ S j ≤ 1 ,   P j = P j ( x, t ) = pressure,  ϕ j = ϕ j ( x, t ) = volumetric flow vector at reference pressure.  Dubrovnik, October 13-16 2008 Notations – p. 3/19

Notations ( 1 = water , 2 = oil , 3 = gas ) • dependant variables :  S j = S j ( x, t ) = reduced saturation, 0 ≤ S j ≤ 1 ,   P j = P j ( x, t ) = pressure,  ϕ j = ϕ j ( x, t ) = volumetric flow vector at reference pressure.  • fluids and rock data : = ρ j /ρ ref  B j = B j ( p j ) = volume factor, j    d j = d j ( p j ) = B j /µ j = phase mobility,     φ = φ ( x, P pore ) = porosity,     K = K ( x ) = absolute permeability,  kr j = kr j ( s 1 , s 3 ) = phase relative permeability,      g = gravity constant ,     Z = Z ( x ) = depth.  Dubrovnik, October 13-16 2008 Notations – p. 3/19

Three Phases Equations • conservation laws : ∂ � � φ B j ( P j ) S j + ∇ · ϕ j = 0 , j = 1 , 2 , 3 . ∂t Dubrovnik, October 13-16 2008 Three-phase equations – p. 4/19

Three Phases Equations • conservation laws : ∂ � � φ B j ( P j ) S j + ∇ · ϕ j = 0 , j = 1 , 2 , 3 . ∂t • Muskat law : ϕ j = − K d j ( P j ) kr j ( S 1 , S 3 )( ∇ P j − ρ j g ∇ Z ) , j = 1 , 2 , 3 . Dubrovnik, October 13-16 2008 Three-phase equations – p. 4/19

Three Phases Equations • conservation laws : ∂ � � φ B j ( P j ) S j + ∇ · ϕ j = 0 , j = 1 , 2 , 3 . ∂t • Muskat law : ϕ j = − K d j ( P j ) kr j ( S 1 , S 3 )( ∇ P j − ρ j g ∇ Z ) , j = 1 , 2 , 3 . • capillary pressure law : � P 12 P 1 − P 2 = c ( S 1 ) , P 32 P 3 − P 2 = c ( S 3 ) , Dubrovnik, October 13-16 2008 Three-phase equations – p. 4/19

An example of three-phase relative permeabilities Dubrovnik, October 13-16 2008 Three-phase equations – p. 5/19

An example of capillary pressures Dubrovnik, October 13-16 2008 Three-phase equations – p. 6/19

Classical resolution : “pressure” equation 3 ∂ � � � φ B j S j + ∇ · q = 0 , ∂t j =1 where q is the global volumetric flow vector: 3 � ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � q = ϕ j = − Kλ − ρg ∇ Z c c j =1  � 3 λ ( s 1 , s 3 , p 2 ) = j =1 kr j d j = global mobility,   = j th fractional flow , � 3 f j ( s 1 , s 3 , p 2 ) = kr j d j /λ j =1 f j = 1 , � 3  ρ ( s 1 , s 3 , p 2 ) = j =1 f j ρ j = global density.  Dubrovnik, October 13-16 2008 classical resolution – p. 7/19

Classical resolution : “pressure” equation 3 ∂ � � � φ B j S j + ∇ · q = 0 , ∂t j =1 where q is the global volumetric flow vector: 3 � ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � q = ϕ j = − Kλ − ρg ∇ Z c c j =1  � 3 λ ( s 1 , s 3 , p 2 ) = j =1 kr j d j = global mobility,   = j th fractional flow , � 3 f j ( s 1 , s 3 , p 2 ) = kr j d j /λ j =1 f j = 1 , � 3  ρ ( s 1 , s 3 , p 2 ) = j =1 f j ρ j = global density.  Solve for the oil pressure P 2 ? Dubrovnik, October 13-16 2008 classical resolution – p. 7/19

Behaviour of individual phase pressures oil pressure Case of a two-phase water-oil flow : oil and water pressures water pressure are singular near front space boundary oil saturation 1 water oil 0 space Dubrovnik, October 13-16 2008 Behaviour of phase pressures – p. 8/19

Let us have a dream... e r u s s e r p l o i water pressure space Does there exists a pressure field ( x, t ) ❀ P such that : Dubrovnik, October 13-16 2008 Let us have a dream... – p. 9/19

Let us have a dream... e r u s s e r p l o i water pressure space Does there exists a pressure field ( x, t ) ❀ P such that : P is smooth , P water ≤ P ≤ P oil Dubrovnik, October 13-16 2008 Let us have a dream... – p. 9/19

Let us have a dream... e r u s s e r p l o i ? global pressure ? water pressure space Does there exists a pressure field ( x, t ) ❀ P such that : P is smooth , P water ≤ P ≤ P oil and P governs the global volumetric flow vector q : � � q = − Kd ∇ P − ρg ∇ Z ? ( where d ( s 1 , s 3 , p ) = λ ( s 1 , s 3 , p 2 ) ) Dubrovnik, October 13-16 2008 Let us have a dream... – p. 9/19

Searching for a Global Pressure P ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � • how to replace : q = − Kλ − ρg ∇ Z c c � � by : q = − Kd ∇ P − ρg ∇ Z ? Dubrovnik, October 13-16 2008 searching for a global pressure – p. 10/19

Searching for a Global Pressure P ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � • how to replace : q = − Kλ − ρg ∇ Z c c � � by : q = − Kd ∇ P − ρg ∇ Z ? • define the Global Pressure : P = P 2 + P cg ( S 1 , S 3 , P ) Dubrovnik, October 13-16 2008 searching for a global pressure – p. 10/19

Searching for a Global Pressure P ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � • how to replace : q = − Kλ − ρg ∇ Z c c � � by : q = − Kd ∇ P − ρg ∇ Z ? • define the Global Pressure : P = P 2 + P cg ( S 1 , S 3 , P ) where the global capillary function : ( s 1 , s 3 , p ) ❀ P cg ( s 1 , s 3 , p ) is required to satisfy: Dubrovnik, October 13-16 2008 searching for a global pressure – p. 10/19

Searching for a Global Pressure P ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � • how to replace : q = − Kλ − ρg ∇ Z c c � � by : q = − Kd ∇ P − ρg ∇ Z ? • define the Global Pressure : P = P 2 + P cg ( S 1 , S 3 , P ) where the global capillary function : ( s 1 , s 3 , p ) ❀ P cg ( s 1 , s 3 , p ) is required to satisfy: • For any saturation and pressure fields S 1 ( x, t ) , S 3 ( x, t ) , P ( x, t ) : f 1 ( S 1 , S 3 , P − P cg ( S 1 , S 3 , P )) ∇ P 12 ∇ P cg ( S 1 , S 3 , P ) = c ( S 1 ) + f 3 ( S 1 , S 3 , P − P cg ( S 1 , S 3 , P )) ∇ P 32 c ( S 3 ) + ∂P cg /∂P ( S 1 , S 3 , P ) ∇ P Dubrovnik, October 13-16 2008 searching for a global pressure – p. 10/19

Searching for a Global Pressure P ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � • how to replace : q = − Kλ − ρg ∇ Z c c � � by : q = − Kd ∇ P − ρg ∇ Z ? • define the Global Pressure : P = P 2 + P cg ( S 1 , S 3 , P ) where the global capillary function : ( s 1 , s 3 , p ) ❀ P cg ( s 1 , s 3 , p ) is required to satisfy: • For any saturation and pressure fields S 1 ( x, t ) , S 3 ( x, t ) , P ( x, t ) : f 1 ( S 1 , S 3 , P − P cg ( S 1 , S 3 , P )) ∇ P 12 ∇ P cg ( S 1 , S 3 , P ) = c ( S 1 ) + f 3 ( S 1 , S 3 , P − P cg ( S 1 , S 3 , P )) ∇ P 32 c ( S 3 ) + ∂P cg /∂P ( S 1 , S 3 , P ) ∇ P • P min ≤ P 1 ≤ P ≤ P 3 ≤ P max Dubrovnik, October 13-16 2008 searching for a global pressure – p. 10/19

Searching for a Global Pressure P ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � • how to replace : q = − Kλ − ρg ∇ Z c c � � by : q = − Kd ∇ P − ρg ∇ Z ? • define the Global Pressure : P = P 2 + P cg ( S 1 , S 3 , P ) where the global capillary function : ( s 1 , s 3 , p ) ❀ P cg ( s 1 , s 3 , p ) is required to satisfy: • For any saturation and pressure fields S 1 ( x, t ) , S 3 ( x, t ) , P ( x, t ) : f 1 ( S 1 , S 3 , P − P cg ( S 1 , S 3 , P )) ∇ P 12 ∇ P cg ( S 1 , S 3 , P ) = c ( S 1 ) + f 3 ( S 1 , S 3 , P − P cg ( S 1 , S 3 , P )) ∇ P 32 c ( S 3 ) + ∂P cg /∂P ( S 1 , S 3 , P ) ∇ P • P min ≤ P 1 ≤ P ≤ P 3 ≤ P max • ∂P cg /∂P ( s 1 , s 3 , p ) ≤ k < 1 Dubrovnik, October 13-16 2008 searching for a global pressure – p. 10/19

Searching for a Global Pressure P ∇ P 2 + f 1 ∇ P 12 + f 3 ∇ P 13 � � • how to replace : q = − Kλ − ρg ∇ Z c c � � by : q = − Kd (1 − ∂P cg /∂P ) ∇ P − ρg ∇ Z • define the Global Pressure : P = P 2 + P cg ( S 1 , S 3 , P ) where the global capillary function : ( s 1 , s 3 , p ) ❀ P cg ( s 1 , s 3 , p ) is required to satisfy: • For any saturation and pressure fields S 1 ( x, t ) , S 3 ( x, t ) , P ( x, t ) : f 1 ( S 1 , S 3 , P − P cg ( S 1 , S 3 , P )) ∇ P 12 ∇ P cg ( S 1 , S 3 , P ) = c ( S 1 ) + f 3 ( S 1 , S 3 , P − P cg ( S 1 , S 3 , P )) ∇ P 32 c ( S 3 ) + ∂P cg /∂P ( S 1 , S 3 , P ) ∇ P • P min ≤ P 1 ≤ P ≤ P 3 ≤ P max • ∂P cg /∂P ( s 1 , s 3 , p ) ≤ k < 1 Dubrovnik, October 13-16 2008 searching for a global pressure – p. 10/19