ࡱ> 6345Kb( /./ 0DTimes New Roman8Cbbv 0b( 0_DMonotype Sorts8Cbbv 0b( 0_ DArialpe Sorts8Cbbv 0b( 0_"0DTahomae Sorts8Cbbv 0b( 0_"@DSymbole Sorts8Cbbv 0b( 0_PDArial Unicode MSbbv 0b( 0_" a .  @n?" dd@  @@`` ((           Y(&"'/5& ) ;     !#$%&'"3 0e0e     A@ A1  8c8c     ?1 d0u0@Ty2 NP'p<'p@A)BCD|E? ȫMMMJG 33@8LM, g4BdBdv 0b ppp@ g4BdBdv 0bXpA pp<4!d!d` 0b$Db<4KdKd` 0b$Db ʚ;2Nʚ;<4dddd4b{ 0 bb:2___PPT9/ 0? $O <I_Nuevos Enfoques sobre Medicin y Escalamientos de Desplazamientos Inmiscibles en Medios Porosos`` H N1ContentsA brief analysis of the origin and uses of Darcys law in Reservoir Engineering. Differences between: Conduction of fluids. Injection of fluids. Production of fluids. Limitations of the Relative Permeability concept. Different Sceneries. Conclusions. HgZAZUZgAUxPDarcys Law (I)RDarcys law describes homogeneous fluid flow through linear porous materials. This law, which relates flow velocity with pressure gradient, has three components. A geometric factor given by length and area of porous medium. A fluid related factor (viscosity). A property of the porous material (Permeability). Q = K . A . DP / ( . L) jCC cccfDarcys Law (II)Permeability is usually defined as  The ability of a porous material to conduct fluids". t&7&C%C$G$G$C$C[Darcys Law (III) Only injection rate or production rate can be measured experimentally. But... during the flow of an uncompressible homogeneous fluid: The rate of conduction is equal to: Injection rate and Production rate. So, measuring one of these rates is enough to obtain the value of the other rates. ZHZSZC G C G&C?c c gc g c gcSCZDarcys Law (IV)For multiphase fluid flow, Darcys law was  corrected using a different factor for each flowing phase. This factor is known as relative permeability curve. Qw = K . Krw . A . DPw / (w . L) .......... [2] Qo = K . Kro . A . DPo / (o . L) .......... [3] The value of relative permeability is different for every fluid saturation. 4fMCckcg3o3kccckckccckcg3o3kccckckcccMCtjyQDarcys Law (V)However ..... A new problem arises during unsteady multiphase fluid flow. The rate of conduction of each phase losses its equivalence with Injection rate, and Production rate. When more than one phase is flowing, it is not possible to measure production rate or injection rate to determine the conduction rate of each phase. And Darcys law applied to each phase is based on conduction rate exclusively.JZfZZJC c gcgc gc g c gcCC G C GC GHC GC\Darcys Law (VI)In order to solve this "inconsistency" two ways were found. One of them during measurement, and the other during calculation. Experimentally a method was developed in order to re-create Darcys requirements (Injection = Conduction = Production). This methodology was known as  steady state . Simultaneous Injection rate is maintained until production rate equals injection rate of each phase. Through calculation, equations were solved during unsteady state displacement in order to obtain values in a dimensionless section. This is known as  unsteady state or Welge methodology. ^~h~h]Darcys Law (VII)lBriefly, to allow the use of Darcys law during multiphase fluid flow, it was necessary to generate a unique saturation at the position of calculation. In this way, once again: Injection = Conduction = Production During  steady-state measurements the whole sample has the same saturation. Using  unsteady-state methodology, all calculations are made in a single point (the outlet face) were the saturation is unique (point saturation).~$   ^A Question (I)bBoth methodologies ( steady-state and  unsteady-state ) gives the same result when applied to homogeneous media. However ..... Is this a good answer for Reservoir Calculations?gA Question (II)In other words: The solution for the particular case where Injection = Conduction = Production is useful for unsteady reservoir conditions?. h<   /_ Answer (I) A very simple example will give a satisfactory answer for this question. Let us suppose that we have a thin, horizontal porous system. If the system is filled with gas it is obvious that its ability to conduct water is null. Using Darcy law we can say that: Kw = 0 when Sw = 0 However..... Although water cannot be  conducted there is no special impediment to inject water in the porous system. An empty rock allows water injection. Although water can be injected, water cannot be produced until it reaches de outlet end. 0 1  $8 !, ` Answer (II) Using this model we may ask: Which is the system ability to conduct water when Sw = 50%? Half of the system with Sw = 100% and the other half with Sw = 0% There are two  well defined capacities to conduct water at this moment: The capacity "X" (obtained through Darcys calculations) where Sw = 100% A null capacity where Sw = 0 So.... Which is the overall value for the water conduction ability?~YZBZJZgZDZYBJgDbNIa Answer (III) Option 1: Kw = X ? Option 2: Kw = 0 ? Option 3: Kw = X/2 ? Option 4: ...... While there is not a well defined Conduction ability, always exists a well defined Injection ability and also a well defined Production ability. pp ' !  > bReal Scenaries In Reservoir Simulation through Darcys calculations only Conduction ability is used. For this reason Darcys equation is unable to reproduce Injection or Production rates in unsteady systems. And ..... All reservoirs are Unsteay Systems during production.j: J  \e8Reference Frame in Natural Reservoirs (The actual world)9& 8In Reservoir Engineering frequently: Natural porous media are heterogeneous. Multiphase flow is the result of an equilibrium between viscous, capillary and gravity forces. This equilibrium varies with time and with physical location. Reservoir calculations are founded on average phase saturation. In a single cell (Reservoir Simulation) The whole reservoir (Material Balance). The properties of interest are Production and Injection abilities.&Z ZRZCZ& R   c Solutions (I) Every Scenery has its own solution. Typical situations are: Homogeneous systems with dominant viscous forces. Very few reservoirs fall in this category, but it is the typical laboratory scenery.. Heterogeneous systems under viscous forces predominance. Stratified reservoirs without connected layers. Heterogeneous systems under viscous and capillary forces predominance. Stratified reservoirs with connected layers. A cross-flow happens as a consequence of imbibition phenomena.2<ZZ<  dSolutions (II)Other typical situations. Heterogeneous systems with heavy oil where imbibition phenomena could be dominant. Gravity dominated systems. Mainly in high permeability rocks with remarkable thickness and density differences and low viscosities (expanding gas cap, basal aquifers, ...). Reservoirs dominated by capillary forces. Mainly in "tight sands reservoirs..OOD F-Conclusions (I)QLaboratory measurements are adequate to describe the ability to conduct fluids in porous media with homogeneous fluids saturation. In Reservoir Engineering, the ability to inject or to produce fluids in non-homogeneous fluids saturation is needed. In unsteady systems the ability to conduct fluids losses its meaning and becomes useless.R @e[0YConclusions (II)In order to fill the gap, experimental measurements must be made honoring real mechanisms occurring at reservoir scale. During scaling-up stage, the following points must be considered: The dominant displacement mechanism. Laboratory  end point measurements. System geometry. Heterogeneity. The scaling-up process must be made by a multidisciplinary team. Relative permeability curves must be  constructed for every case. This operation must be based on available data and accepted reservoir models.N0Z j0Z 0Z jh W W W Darcy  Thank you for your attention ! !rKExamplesQThe Relationship between production and fluid saturation will be analyzed using two very schematic models. A: Water-Oil Displacement in a rectangular horizontal geometry under gravity forces dominance (Segregated flow). B: Water-Oil Displacement in a rectangular non- horizontal geometry under gravity forces dominance (Segregated flow).2kZZkb>3Segregated flow (A). Production at different faces.&4$3 d@3Segregated flow (A). Production at different faces.&4$ eA3Segregated flow (A). Production at different faces.&4$ fB3Segregated flow (A). Production at different faces.&4$ hD3Segregated flow (A). Production at different faces.&4$ tL First RemarkvThe Relationship between production rate and phases saturation may be different for different points in the same cell.jE3Segregated flow (B). Production at different faces.&4$ kF3Segregated flow (B). Production at different faces.&4$ lG3Segregated flow (B). Production at different faces.&4$ mH3Segregated flow (B). Production at different faces.&4$ uM Second RemarkIf capillary forces take part of displacement mechanisms, the relationship between flow rate and phase saturation may depend on: The selected production point. The overall geometry and orientation of the system.&SS/T<^s Pb ` 3` v3fff` ___>?" dd@,v?Kd@  @A@` n?" dd@   @@``PR    @ ` ` p>> zh(  hT ;x\ h ";x\~ h N2?x~ h N2?Dx\&N ;9 h ;9> h  BACDE8F@@?saO@OaNs @;96 h  BCDE4F<@? ;];h]h;@6 h  BCDE4F<@? 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P ,$D0l     0 ,$D0B    `D1??     f1??   ;Swirr 2z      0,$D0B   `D1??     f1??   ;Swirr 22  # l1 ??{ + ,$D  02  # l1 ?? ++,$D  02  # l31 ?? + ,$D  02  # l31 ??@%p,$D  0B   `DԔ??p  ,$D 0B   f.MDԔ?? ,$D 0,   f|1?? ,$D0 bSw avg = Swirr = 25% 2cl 0 0: %0 0:,$D0t  P `0 0 0,$D0  C x1 ?? P `0 B   # l1 ??@P @0 !  f1??   7Q 2 "  f1??@: JSw 2 ql 0  $0 ,$D0(  P `0  0 ,$D0   C x1 ?? P `0 B    # l1 ??@P @0    f1??  7Q 2 #  f1??   JSw 2 H  0޽h ? 3~  .&@!!(    c $$& h<$D 0      f1 ?? P ,$D0z      0 ,$D0B   `D1??     f1??   ;Swirr 2z      0,$D0B   `D1??      f1??   ;Swirr 22   # l1 ??{ + ,$D  02   # l1 ?? ++,$D  02   # l31 ?? + ,$D  02   # l31 ??@%p,$D  0B   `DԔ??p  ,$D 0B   f.MDԔ?? ,$D 0$   f1?? ,$D0 Z Sw avg = 35% 2 z 0 0:  0 0:,$D0(  P `0  0 0,$D0  C xĐ1 ?? P `0 B   # l1 ??@P @0   f1??   7Q 2   f1??@: JSw 2 z 0   0 ,$D0(  P `0  0 ,$D0  C x1 ?? P `0 B   # l1 ??@P @0   f41??  7Q 2   f 1??   JSw 2 2  # l1 ??  ,$D 02  # l31 ??  ,$D 02  # l31 ??@p,$D 02   # l1 ??}  ,$D 0 !  f1 ??P ,$D 0H  0޽h ? 3  800&&(    c $X h<$D 0      f1 ?? P ,$D0z      0 ,$D0B   `D1??     f[1??   ;Swirr 2z      0,$D0B   `D1??      f1??   ;Swirr 22   # l1 ??{ + ,$D  02   # l1 ?? ++,$D  02   # l31 ?? + ,$D  02   # l31 ??@%p,$D  0B   `DԔ??p  ,$D 0B   f.MDԔ?? ,$D 0$   fd1?? ,$D0 Z Sw avg = 45% 2 z 0 0:  0 0:,$D0(  P `0  0 0,$D0  C xa1 ?? P `0 B   # l1 ??@P @0   fd1??   7Q 2   fg1??@: JSw 2 z 0   0 ,$D0(  P `0  0 ,$D0  C xl1 ?? P `0 B   # l1 ??@P @0   fto1??  7Q 2   fr1??   JSw 2 2  # l1 ??  ,$D 02  # l31 ??  ,$D 02  # l31 ??@p,$D 02   # l1 ??}  ,$D 0 !  f1 ??P ,$D 0 "  f1 ??@P ,$D 02 # # l1 ?? :jP ,$D 02 $ # l31 ?? 9i ,$D 02 % # l31 ??@Brp,$D 02 & # l1 ??} Br ,$D 0H  0޽h ? 3!  B!:! +. (    c $ h<$D 0      f1 ?? P ,$D0z      0 ,$D0B   `D1??     f1??   ;Swirr 2z      0,$D0B   `D1??      fh1??   ;Swirr 22   # l1 ??{ + ,$D  02   # l1 ?? ++,$D  02   # l31 ?? + ,$D  02   # l31 ??@%p,$D  0B   `DԔ??p  ,$D 0B   f.MDԔ?? ,$D 0$   f01?? ,$D0 Z Sw avg = 55% 2 z 0 0:  0 0:,$D0(  P `0  0 0,$D0  C xl51 ?? P `0 B   # l1 ??@P @0   f881??   7Q 2   fd;1??@: JSw 2 z 0   0 ,$D0(  P `0  0 ,$D0  C x|@1 ?? P `0 B   # l1 ??@P @0   fB1??  7Q 2   fF1??   JSw 2 2  # l1 ??  ,$D 02  # l31 ??  ,$D 02  # l31 ??@p,$D 02   # l1 ??}  ,$D 0 !  f1 ??P ,$D 0 "  f1 ??@P ,$D 02 # # l1 ?? :jP ,$D 02 $ # l31 ?? 9i ,$D 02 % # l31 ??@Brp,$D 02 & # l1 ??} Br ,$D 0 '  f1 ?? P @,$D 02 ( # l1 ??  ,$D 02 ) # l31 ??f  ,$D 02 * # l31 ??@p,$D 02 + # l1 ??}  ,$D 0H  0޽h ? 3+  **04\*(    c $D h<$D 0      f1 ?? P ,$D0z      0 ,$D0B   `D1??     fI1??   ;Swirr 2z      0,$D0B   `D1??      f+1??   ;Swirr 22   # l1 ??{ + ,$D  02   # l1 ?? ++,$D  02   # l31 ?? + ,$D  02   # l31 ??@%p,$D  0B   `DԔ??p  ,$D 0B   f.MDԔ?? ,$D 0$   f1?? ,$D0 Z Sw avg = 75% 2 z 0 0:  0 0:,$D0(  P `0  0 0,$D0  C x1 ?? P `0 B   # l1 ??@P @0   f01??   7Q 2   f1??@: JSw 2 z 0   0 ,$D0(  P `0  0 ,$D0  C x<1 ?? P `0 B   # l1 ??@P @0   f1??  7Q 2   f4 1??   JSw 2 2  # l1 ??  ,$D 02  # l31 ??  ,$D 02  # l31 ??@p,$D 02   # l1 ??}  ,$D 0 !  f1 ??P ,$D 0 "  f1 ??@P ,$D 02 # # l1 ?? :jP ,$D 02 $ # l31 ?? 9i ,$D 02 % # l31 ??@Brp,$D 02 & # l1 ??} Br ,$D 0 '  f1 ?? P @,$D 02 ( # l1 ??  ,$D 02 ) # l31 ??f  ,$D 02 * # l31 ??@p,$D 02 + # l1 ??}  ,$D 0 ,  f1 ?? P ,$D 0B .  fD3o?? ,$D0B /  fDo?? ,$D0 3 # 20e0e    BC DE F A@3 o 8c8c     ?1 d0u0@Ty2 NP'p<'p@A)BCD|E? @^~ ,$D0 4 # 20e0e    BCDE F A@ o 8c8c     ?1 d0u0@Ty2 NP'p<'p@A)BCD|E?@"" ,$D0H  0޽h ? 3  $(  r  S L h   r  S  h  H  0޽h ? 3  2*(    c $t h<$D 0     # l1 ?? ~ 6,$D0z      0 ,$D0B   `D1??     f1??   ;Swirr 2z      0,$D0B   `D1??      f1??   ;Swirr 22   # l1 ??{ + ,$D  02   # l1 ?? ++,$D  02   # l31 ?? + ,$D  02   # l31 ??@%p,$D  0B   `DԔ??  ,$D 0B   f.MDԔ??" ,$D 0,   f1?? ,$D0 bSw avg = Swirr = 25% 2z 0 0:  0 0:,$D0(  P `0  0 0,$D0  C x1 ?? P `0 B   # l1 ??@P @0   f1??   7Q 2   f1??@: JSw 2 z 0   0 ,$D0(  P `0  0 ,$D0  C xP1 ?? P `0 B   # l1 ??@P @0   f1??  7Q 2   f1??   JSw 2 H  0޽h ? 3  !%(    c $Ȟ h<$D 0     # l1 ?? ~ 6,$D0z      0 ,$D0B   `D1??     f1??   ;Swirr 2z      0,$D0B   `D1??      f1??   ;Swirr 22   # l1 ??{ + ,$D  02   # l1 ?? ++,$D  02   # l31 ?? + ,$D  02   # l31 ??@%p,$D  0B   `DԔ??  ,$D 0B   f.MDԔ??" ,$D 0$   f1?? ,$D0 Z Sw avg = 35% 2 z 0 0:  0 0:,$D0(  P `0  0 0,$D0  C x1 ?? P `0 B   # l1 ??@P @0   f1??   7Q 2   fP1??@: JSw 2 z 0   0 ,$D0(  P `0  0 ,$D0  C xd1 ?? P `0 B   # l1 ??@P @0   f1??  7Q 2   fh1??   JSw 2 2  # l1 ??{  ,$D 02  # l31 ??@p,$D 02   # l1 ?? +,$D 02 ! # l31 ??  ,$D 0 %  0e0e    BCpDEF  A1 8c8c     ?1 d0u0@Ty2 NP'p<'p@A)BCD|E? p @`w  ,$D 0H  0޽h ? 3L"  !!&(!(    c $p h<$D 0     # l1 ?? ~ 6,$D0z      0 ,$D0B   `D1??     fr1??   ;Swirr 2z      0,$D0B   `D1??      fv1??   ;Swirr 22   # l1 ??{ + ,$D  02   # l1 ?? ++,$D  02   # l31 ?? + ,$D  02   # l31 ??@%p,$D  0B   `DԔ??  ,$D 0B   f.MDԔ??" ,$D 0$   fH|1?? ,$D0 Z Sw avg = 50% 2 z 0 0:  0 0:,$D0(  P `0  0 0,$D0  C xX1 ?? 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