The effect of capillary forces onimmiscible two-phase flow inheterogeneous porous media

Report 94-103

C.J. van DuijnJ. MolenaarM.J. de Neef

Technische Universiteit DelftDelft University of Technology

Faculteit der Technische Wiskunde en InformaticaFaculty of Technical Mathematics and Informatics

ISSN 0922-5641

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The e�ect of capillary forces on immiscibletwo-phase ow in heterogeneous porous mediaC.J. van Duijn, J. Molenaar and M.J. de NeefDepartment of Mathematics, Delft University of Technology,P.O. Box 5031, 2600 GA Delft, The NetherlandsAbstractWe consider the one-dimensional two-phase ow including capillarye�ects through a heterogeneous porous medium. The heterogeneity isdue to the spatial variation of the absolute permeability and the poros-ity. Both these quantities are assumed to be piecewise constant. Atinterfaces where the rock properties are discontinuous, we derive, by aregularisation technique, conditions to match the values of the satura-tion on both sides. There are two conditions: a ux condition and anextended pressure condition. Applying these conditions we show thattrapping of the wetting phase may occur near heterogeneities. To il-lustrate the behaviour of the saturation we consider a time-dependentdi�usion problem without convection, a stationary convection-di�usionproblem, and the full time-dependent convection-di�usion problem (nu-merically). In particular the last two problems explicitly show the trap-ping behaviour.Key words: capillary pressure, immiscible ow, heterogeneities, trap-ping.1 IntroductionCapillary forces combined with the spatial variation of rock properties in anoil reservoir, may reduce the recovery factor of the reservoir signi�cantly. Forinstance, large quantities of oil are di�cult to remove from parts of reservoirswith small-scale heterogeneities, such as cross-bedded reservoir zones, or mayeven remain trapped (cf. Kortekaas [11], Weber [16]). Laboratory experimentswith scale models (Dawe [6], Ledeboer [12]) con�rm this trapping behaviour.Likewise it can be di�cult to remove oil spills from heterogeneous soils.To investigate the in uence of heterogeneities on trapping, we consider inthis paper a two-phase wetting-nonwetting uid ow, which is directed per-pendicular to an interface across which the reservoir has an abrupt change1

in permeability and porosity. In a simpli�ed approach this leads to a one-dimensional ow problem for which we will investigate the role of convectionand capillary di�usion in relation to the discontinuous rock properties.In particular we focus on the conditions to be posed at interfaces acrosswhich the permeability or porosity is discontinuous. Since the ow problem isdescribed by a partial di�erential equation that is second-order in space, twointerface conditions are needed. One condition follows from uid conserva-tion across the interface and leads to a continuity requirement for the wettingor nonwetting uid ux. The second condition is an extension of the capil-lary pressure continuity. This condition turns out to depend critically on thequalitative behaviour of the capillary pressure.According to Leverett [14] the capillary pressure in a medium with rockpermeability k and rock porosity � is inversely proportional to (k=�)1=2, andproportional to the J -Leverett function, which is a function of the (reduced)water saturation Sw only (0 � Sw � 1). The quantity (k=�)1=2 may be as-sociated with a mean pore diameter, and the J -Leverett function is typicalfor the lithology of the porous medium. If the permeability and porosity arediscontinuous, then consequently the capillary pressure curves on the left- andright-hand side of the discontinuity are di�erent, as depicted in Figure 1.1.The upper curves in Figure 1.1 correspond to the side where the porous medi-um has a �ne structure (small mean pore diameter), whereas the lower curvescorrespond to the side where it has a coarse structure (larger mean pore di-ameter).The capillary pressure at Sw = 1 may either be zero (Figure 1.1, left) orpositive (Figure 1.1, right). This pressure is called the threshold pressure,bubbling pressure or nonwetting phase entry value (see Bear [2]). It representsthe minimum pressure needed for a nonwetting uid to enter a medium thatis initially saturated by a wetting uid.If the threshold pressure is zero, we typically have capillary pressure curvesas in Figure 1.1 (left). For every saturation on one side of the interface, therecorresponds a saturation on the other side such that the capillary pressure iscontinuous. In this case the second interface condition is just continuity ofcapillary pressure.If, however, the threshold pressure is positive, we see from Figure 1.1 (right)that there is a threshold saturation S� such that continuity of capillary pressurecan only be achieved if the wetting phase saturation on the side correspondingto the lower curve is below or equal to S�, but not above this value. We willshow in this paper, that in the latter case the wetting phase saturation on theside corresponding to the upper curve is equal to one, and that the capillarypressure across the interface is discontinuous. Since this is an extension ofthe capillary pressure continuity condition, we shall refer to it as the extendedpressure condition. An important consequence of the extended pressure con-dition is that the ow problem admits solutions describing entrapment of the2

0 1pc Sw rr 0 S� 1pc Sw rrFigure 1.1: Capillary pressure pc as a function of the reduced water saturationSw. The upper curves correspond to a porous medium with a �ne structure,the lower ones to a medium with a coarse structure. The threshold pressureat Sw = 1 is either zero (left) or positive (right).nonwetting phase. Without justi�cation the extended pressure condition isused by Dale et al. [5].Physically the extended pressure condition can be interpreted as follows.The capillary pressure must be continuous across an interface unless the non-wetting phase is immobile (Sw = 1) on the low permeability side. The nonwet-ting phase pressure is unde�ned if the nonwetting phase is immobile. Hencethe capillary pressure, which is the di�erence between the phase pressure of thenonwetting phase and the wetting phase, is no longer physically meaningful,and need not be continuous anymore.The extended pressure condition can be derived mathematically by regular-ising the rock heterogeneity. This is done in Section 2. It leads to a boundarylayer equation from which the extended pressure condition is deduced. Yort-sos and Chang [17] used a similar regularisation, but they considered only asteady-state situation and they did not consider the role of the threshold sat-uration S�: their data and examples were always chosen so that the capillarypressure is continuous.Having derived the appropriate interface conditions, we consider two specif-ic ow problems with a single heterogeneity in Section 3. Both these problemsadmit an analytical treatment. One is a di�usion problem, where redistribu-tion of phases is caused by capillary forces only. For this di�usion problemtoo there are cases in which the capillary pressure is discontinuous across theinterface. As this is a degenerate di�usion problem, free boundaries may oc-3

cur. Free boundaries are surfaces, moving in time, that separate the regionsoccupied by the wetting uid, the mixture, and the nonwetting uid (see VanDuijn & Floris [8] and Gilding [9], [10]). Conditions for the occurrence aregiven, and the behaviour of the solution near the free boundary is discussed.The other problem is a stationary convection-di�usion problem. The steady-state solution is of particular interest because it clearly demonstrates the mech-anism of trapping. In particular we show that trapping of the oil occurs forpositive threshold pressures.In Section 4 we present an algorithm for the numerical solution of the fulltime-dependent convection-di�usion problem. In this numerical algorithm theextended pressure condition is incorporated. Therefore it is able to deal withproblems in which the capillary pressure is discontinuous. The di�usion prob-lem of Section 3 is used as a test problem to validate the numerical algorithm.The numerical method is used in Section 5 to study two characteristic ow problems. The �rst describes the interaction of a moving oil-water frontwith a single heterogeneity. We show that the movement of the front stopstemporarily at the interface. Only after buildup of the oil saturation to abovethe critical oil saturation 1 � S� the front passes the heterogeneity.In the other problem we consider the water ooding through a mediumwith a periodic permeability. If there is a small amount of oil in the medium,for example an oil spill, it is shown that this oil cannot be removed from themedium; it is trapped in the highly permeable regions. We also indicate howthe amount of trapped oil can be diminished. In the �nal Section we summariseour conclusions.2 Mathematical Model2.1 Flow EquationsWe consider the ow of two immiscible and incompressible phases in a satu-rated and heterogeneous porous medium. We assume that the heterogeneityof the porous medium, i.e. porosity � and permeability k, varies in one di-rection only, say the x-direction. Further we assume that the uid ow isone-dimensional in that same direction. We characterise the phases by theirreduced saturations: Sw (saturation of the wetting uid) and Sn (saturationof the nonwetting uid), with 0 � Sw; Sn � 1. Since the porous medium isassumed to be saturated we haveSw + Sn = 1: (2.1)The basic equations governing the ow of the two phases are given by the uid-balance equations �@Si@t + @qi@x = 0; i = w;n; (2.2)4

and Darcy's law, qi = ��i@pi@x ; i = w;n: (2.3)Here qi, �i and pi (i = w;n) denote the speci�c discharge, the mobility and thepressure of the wetting and nonwetting phase. In writing equation (2.3) weassumed that the ow is horizontal, so that gravitational forces in the directionof the ow are absent. The mobility of each phase is given by�i = k(x)kri(Sw)�i ; i = w;n; (2.4)where k is the absolute permeability of the porous medium, and kri and �i therelative permeability and viscosity of phase i.The pressure of the nonwetting uid di�ers from the wetting uid becauseof interfacial tension on the microscopic pore level. This pressure di�erence,which is called the capillary pressure pc, obeys the Leverett-relationshippn � pw = pc(x; Sw) = �vuut�(x)k(x)J(Sw): (2.5)Equations (2.1)-(2.5) can be combined into a convection-di�usion equationfor the saturation of the wetting phase Sw only. When we add the equation of(2.2) for i = w and i = n, and use equation (2.1), we �nd that the total owrate q, de�ned by q = qw + qn; (2.6)is constant in space. Using this observation, setting u = Sw, and rede�ning t,x, q and k according tot := tT ; x := xL; q := qTL; k := kL2 ; (2.7)where T is a characteristic time and L a characteristic length, we obtain�@u@t + @F@x = 0; (2.8)with F denoting the dimensionless ux of the wetting phaseF = qfw(u) + k��(u) @@x 0@s�kJ(u)1A : (2.9)Here fw is the fractional ow function of the wetting phase,fw(u) = krw(u)krw(u) + krn(u)=M ;5

with M = �n=�w, and ��(u) = Nckrn(u)fw(u);with Nc = �T=(�nL). The positive numbersM and Nc are called the mobilityratio and the capillary number.We assume throughout this work that the relative permeabilities krw andkrn are continuously di�erentiable on [0; 1], and that J is continuously di�er-entiable on (0; 1]. Further we assume that they satisfy� krw is strictly increasing such that krw(0) = 0 and krw(1) = 1;� krn is strictly decreasing such that krn(0) = 1 and krn(1) = 0;� J 0 < 0 on (0; 1] and J(1) � 0;� limu#0 J 0(u)krw(u) = 0.The conditions krw(1); krn(0) = 1 can be relaxed to allow for krw(1), krn(0) 6= 1.This would only a�ect the mobility ratio M and the capillary number Nc.In the analysis of equation (2.8) we will assume that the permeability andporosity are piecewise constant.2.2 Interface Conditions at a DiscontinuityAt discontinuities in k and � the equations cease to hold. At those points weneed interface conditions. To derive them we consider a medium with a singlediscontinuity at x = 0, i.e. k and � satisfyk(x) = ( k� x < 0;k+ x > 0; and �(x) = ( �� x < 0;�+ x > 0: (2.10)The �rst condition, the ux condition, is straightforward. Assuming that thetime derivative in the uid-balance equation (2.8) is bounded, we integratethis equation in a small neighbourhood of x = 0. This leads to continuity ofthe ux, limx"0 F (x; t) = limx#0 F (x; t) for all t > 0: (2.11)The second condition, the extended pressure condition, requires more at-tention. Before we proceed we introduce, for convenience, the expressionh(x) = vuutk(x)�(x); (2.12)where h(x) = 8<: h� = (k�=��)1=2 for x < 0;h+ = (k+=�+)1=2 for x > 0; (2.13)6

and where h� > h+ is assumed. The expression for the ux (2.9), rewritten interms of h, then becomesF = qfw(u) + �(x)h2(x)��(u) @@x J(u)h(x)! : (2.14)We want to model the discontinuity in � and h as the mathematical ideali-sation of a physically realistic situation in which a rapid, but smooth, variationin these functions occurs. We do this by regularising the functions � and h.The idea is the following.We approximate for " > 0 the discontinuous � and h by smooth monotonefunctions �" and h" such that �" and h" only di�er from � and h in a smallinterval [�"; "] around the discontinuity. For this regularised problem we wantto obtain a relation between the saturations at the end points of this intervalfor vanishing ". These saturations are identi�ed with the saturation at theleft- and right-hand side of the discontinuity. To achieve this we blow up theinterval [�"; "], by rescaling x, such that it is mapped onto [�1; 1]. Then wedetermine for vanishing " the di�erential equation that should be satis�ed bythe limit function in the blown-up interval. From this di�erential equationa relation can be derived between the saturations at the end points of theinterval. This relation is then the second interface condition.Let " > 0 denote the regularisation parameter. The discontinuous functionsh and � are now approximated by the smooth functions h" and �" such that�"(x) = ��x"� and h"(x) = h�x"� ; (2.15)and �"(x) = �(x), h"(x) = h(x) outside the interval [�"; "]. The functions �and h are smooth and monotone on [�1; 1]. The solution corresponding to theregularised problem is denoted by u", the corresponding ux by F".In the expression for the ux (2.14) we replace � and h by �" and h".Further we �x t (dropping the explicit time dependence), rescale x by puttingy = x=", and write v"(y) = u"(x). In terms of these variables, the ux takesthe form F" = qfw(v") + 1"�(y)h2(y)��(v") ddy J(v")h(y) ! : (2.16)Now, assume that the saturation v", 0 � v" � 1, converges as " # 0 towardsa limit function v, and moreover assume that during this limit process the uxF" remains bounded, uniformly in ". Then the limit function v should satisfy��(v) ddy J(v)h(y)! = 0 for � 1 < y < 1: (2.17)This equation yields a relation between v(�1) and v(+1), which are identi�edwith the value of u at the left- and right-hand side of x = 0, i.e. with u(0�) =7

u� and u(0+) = u+. To obtain the relation we solve the ordinary di�erentialequation (2.17) starting at y = �1 with v(�1) = u�. This yields the value ofv(+1) = u+ corresponding to v(�1) = u�.The di�erential equation in (2.17) is satis�ed byv(y) = 0; v(y) = 1 (2.18)(since ��(0) = ��(1) = 0) and byJ(v)h(y) = constant: (2.19)If u� = 0, the only admitted solution is v(y) � 0 on [�1; 1], implying u+ = 0.If u� = 1, then v(y) � 1 on [�1; 1], since (2.19) only has increasing solutionsfor decreasing h. Consequently u+ = 1.Now consider 0 < u� < 1. Then v(y) satis�es to the right of y = �1J(v(y)) = J(u�)h� h(y): (2.20)Since h(y) is decreasing from h� to h+, this expression is well-de�ned in theinterval �1 � y � 1 only if J(u�)h+h� � J(1); (2.21)i.e. if u� � S�, where S� is determined byJ(S�)h� = J(1)h+ : (2.22)Then, for 0 < u� � S� we �nd thatJ(u�)h� = J(u+)h+ ; (2.23)which expresses continuity of capillary pressure.If, however, S� < u� < 1, then (2.20) is only de�ned in the interval �1 �y � y� < 1, where y� satis�esJ(1) = J(u�)h� h(y�): (2.24)At y = y� we have v = 1. The solution is continued by v = 1 in the remainingpart, y� < y � 1. This way we have constructed an increasing solution v(y)such that v(�1) = u�, v satis�es (2.20) in �1 � y < y�, and v(y) = 1 fory� � y � 1. In particular u+ = 1. 8

Summarising, we have found that u� and u+ are related by the extendedpressure condition which reads8><>: J(u�)h� = J(u+)h+ if u� � S�;u+ = 1 if S� < u� � 1; (2.25)where u� and u+ correspond to the limit value of u at the side of the highestpermeability and the lowest permeability, respectively. If J(1) = 0 we haveS� = 1, and in that case we recover the familiar capillary pressure continuityfor all 0 � u� � 1. Note that the relation between u� and u+ is independentof the chosen regularisation functions h and �.3 Behaviour of Saturation in HeterogeneousMedia: Analytical ObservationsIn this section we discuss the behaviour of the wetting saturation in a porousmedium with a heterogeneity of the idealised form (2.13), where h and � arediscontinuous at x = 0, and constant elsewhere. Throughout this section weassume that J(1) > 0. Two speci�c ow problems are studied. First weconsider, for q = 0, the di�usion problem in which the redistribution is causedby capillary forces only. It is solved by means of a similarity transformation.This problem is of interest because it allows for a precise characterisation ofthe free boundaries, separating the regions where u = 0, 0 < u < 1 and u = 1,and because it shows explicitly how to use the extended pressure condition.Moreover, it serves as a case for which it is possible to verify the convergenceof numerical approximations. This will be done in Section 4 where we discussthe algorithm.Next we consider a stationary convection-di�usion problem. Because thesaturation equation is now exactly solvable, this allows for an explicit investi-gation of the trapping mechanism and the capillary pressure near the hetero-geneity.3.1 Di�usion ProblemWe set q � 0 in (2.14) and consider the di�usion equations@u@t = h� @@x D(u)@u@x! for x < 0; (3.1)@u@t = h+ @@x D(u)@u@x! for x > 0; (3.2)9

where D(u) = ���(u)J 0(u): (3.3)The conditions imposed on krw and J imply D(u) > 0 and bounded on (0; 1)with D(0) = D(1) = 0. We look for a solution of these equations satisfying att = 0 u(x; 0) = ( 1 x < 0;0 x > 0; (3.4)and at x = 0 continuity of ux and the extended pressure condition. Thisproblem is studied in detail by Van Duijn & De Neef [7]. It is solved by meansof a similarity transformation. Setu(x; t) = g(�); with � = xpt: (3.5)Then u satis�es (3.1){(3.4) if g satis�es the equations12�g0 + h� (D(g)g0)0 = 0 for � < 0; (3.6)12�g0 + h+ (D(g)g0)0 = 0 for � > 0; (3.7)and the boundary conditionsg(�1) = 1 and g(1) = 0: (3.8)Here the primes denote di�erentiation with respect to �. At � = 0, the solu-tions are matched by the ux continuity conditionlim�"0 ��h�D(g)g0(�) = lim�#0 �+h+D(g)g0(�); (3.9)and by the extended pressure condition (2.25), with u+ and u� replaced byg+ = lim�#0 g(�) and g� = lim�"0 g(�) if h� > h+, and vice versa, by g� andg+ if h� < h+.It is shown in [7] that this problem has a unique solution for any valueof the parameters h� and ��. This solution is non-increasing for � < 0 andfor � > 0. Moreover, if h� � h+ then the corresponding capillary pressure iscontinuous, and 0 < g+ � g� < 1 with g+ < S�. This is shown in Figure 3.1,left. If h� > h+ then a solution may occur with a discontinuous capillarypressure satisfying g+ = 1 and S� < g� < 1 (see Figure 3.1, right). Whetherthis occurs depends on the actual shape of the di�usion function D(g) and theratio h�=h+.If D(g) vanishes at g = 0 and g = 1 su�ciently fast, then two free bound-aries emerge from the origin x = 0, having a �nite speed of propagation. Theprecise conditions are as follows (e.g. see Van Duijn & Floris [8]). If for some0 < � < 1 Z 11�� D(g)1 � g dg <1; (3.10)10

0 1g+ g�S�pc gp+c p�c yK �r r rr 0 1S� g�pc p�c p+c y1i rr rFigure 3.1: Pressure-saturation path for similarity solution. The arrows indi-cate the direction of increasing similarity variable �. h� < h+ gives a contin-uous capillary pressure (left), and h� > h+ possibly leads to a discontinuouscapillary pressure (right).then there exists a number an < 0 such thatg(�)( = 1 � � an;< 1 � > an; or u(x; t)( = 1 x � anpt;< 1 x > anpt: (3.11)If Z �0 D(g)g dg <1; (3.12)there exists a number aw > 0 such thatg(�)( > 0 � < aw;= 0 � � aw; or u(x; t)( > 0 x < awpt;= 0 x � awpt: (3.13)Here the positions of the free boundaries in time are given by �n(t) = anpt(the nonwetting front) and �w(t) = awpt (the wetting front). In many casesof practical interest both conditions (3.10) and (3.12) are satis�ed.Near the free boundaries we obtain from (3.6) and (3.7)12an = lim�#an �D(g(�))g0(�)�0�g0(�) h� = lim�#an D(g(�))1 � g(�) g0(�)h� (3.14)�12aw = lim�"aw �D(g(�))g0(�)�0g0(�) h+ = lim�"aw D(g(�))g(�) g0(�)h+ (3.15)From these limits we directly obtain the behaviour of the solution near a free11

boundary. From (3.15) we �ndlim�"aw g0(�)8>>><>>>: = 0 if D0(0) =1;2 (0;1) if D0(0) 2 (0;1);=1 if D0(0) = 0: (3.16)Similarly we �nd from (3.14)lim�#an g0(�)8>>><>>>: = 0 if D0(1) = �1;2 (�1; 0) if D0(1) 2 (�1; 0);= �1 if D0(1) = 0: (3.17)Because (3.6){(3.7) are nonlinear, one cannot expect to �nd similarity solu-tions in closed form. Nonetheless, the qualitative observations made here givea fairly complete picture. To obtain quantitative information, one has to usea numerical solution procedure for the ordinary di�erential equations. This isexplained in [7]. Figure 3.2 shows the result of such computation. The dataare taken from Table 4.1. In this speci�c case the di�usion function is givenby D(g) = 12 g1=2(1 � g)2g2 + (1� g)2 :It satis�es the conditions (3.10) and (3.12), and therefore free boundaries occurwhere g tends to 1 or 0. At the free boundaries we have g0(a+n ) = �1 andg0(a�w) = 0, since D0(1) = 0 and D0(0) = 1. The capillary pressure at � = 0is discontinuous in this case (g+ = 1).Note that the di�usion problem discussed here is in many respects similarto the one-dimensional hysteresis problem studied by Philip [15]. In his contextu denotes the water saturation in unsaturated ow, the top pc-curve is a dryingcurve, and the bottom pc-curve is a wetting curve. However, since he considersa drying and a wetting curve that form a closed loop (the hysteresis loop), hissolutions have continuous capillary pressure.3.2 Stationary Convection Di�usion ProblemThis example will clearly demonstrate the e�ect of entrapment of the nonwet-ting phase. We assume here that the total ow rate q is positive and constantin time, and that the saturation is stationary, i.e. @u=@t � 0. In (2.7) wechoose the characteristic time T and the characteristic length L now such thatL=T = q. The capillary number is then given by Nc = �=(�nq), and thedimensionless total ow rate is equal to one.12

01-0.7 0 1an awg �{ S�Figure 3.2: Similarity solution g as function of � (di�usion problem, q = 0).The capillary pressure is discontinuous.For a stationary saturation we �nd from equation (2.8) that the ux isconstant in space, leading to the �rst-order equation (see (2.14)):fw(u) + �(x)h2(x)��(u) ddx J(u)h(x)! = constant: (3.18)This equation was studied by Yortsos & Chang [17] for discontinuous as well asfor regularised h and �. They considered solutions with continuous capillarypressure. In our case the capillary pressure is also allowed to be discontinuous.We investigate the case of water injection when the heterogeneity is such thath� > h+. For water injection we haveu(�1) = u(+1) = 1: (3.19)These conditions imply that the constant in (3.18) is one, givingfw(u) + �(x)h2(x)��(u) ddx J(u)h(x)! = 1; (3.20)for �1 < x <1. From this equation we see that u = 1 is a solution. Whenu(x) < 1 we �nd, subtracting fw(u) from both sides of (3.20) and dividing by(1� fw(u)), that u satis�es�(x)h2(x)NcMkrw(u) ddx J(u)h(x)! = 1; (3.21)13

or, since � and h satisfy (2.10) and (2.13), respectively,��h�krw(u)J 0(u)dudx = 1NcM for x < 0; (3.22)�+h+krw(u)J 0(u)dudx = 1NcM for x > 0: (3.23)Equation (3.23) only has decreasing solutions when u < 1. Hence the onlysolution that satis�es the boundary condition at x =1 is (see also Yortsos &Chang [17]) u(x) = 1 for all x > 0: (3.24)Equation (3.22) and the boundary condition at x = �1 imply thatdudx < 0 for all x < 0; when u(x) < 1: (3.25)Thus for a non-trivial solution we are left with equation (3.22), subject tou(0�) = u� < 1; (3.26)where u� is related to u+ by the extended pressure condition. Therefore wemust have u� � S�; (3.27)where u� = S� gives a solution with continuous capillary pressure, while u� >S� leads to a discontinuous pressure. The solution u(x) becomes equal to oneat the �nite value x1 < 0, which is found by integrating equation (3.22) from0 to x1, x1 = ��h�NcM Z 1u� J 0(s)krw(s) ds� ��h�NcM�J(1)� J(u�)� > �1: (3.28)The right-hand derivative of u at x = x1 is given byu0(x1) = 1NcMJ 0(1)��h� > �1:In Figure 3.3 some solutions are shown with data taken from Table 4.1.We conclude that the stationary problem allows for a family of non-trivialsolutions, corresponding to each S� � u� < 1. All these solutions describe en-trapment of the nonwetting uid. The amount of trapped uid is proportionalto the product of the capillary number and the mobility ratio, NcM . Thisfollows directly from (3.22). If x is scaled by NcM��h� we see that (3.22)becomes independent of these parameters. Thus, in the scaled x-coordinatethe solution for x < 0 only depends on u�. Hence, also the volume of the14

01-0.5 0 0.5u x{ S�Figure 3.3: Steady-state solutions u(x) which display entrapment of the non-wetting phase; u(�1) = u(1) = 1 and q = 1. The lowest curve satis�esu� = S� = 1=10 and has a continuous capillary pressure. All other curveshave discontinuous capillary pressure.trapped uid only depends on u�. In the original x-coordinate therefore, thevolume of the trapped uid is proportional to NcM��h�; the proportionalityfactor is a decreasing function of u�.Later, in Section 5, we will return to these stationary solutions and identifythem as asymptotic pro�les for t!1 of the saturation distribution.4 Numerical MethodIn standard numerical simulations it is assumed that the capillary pressure iscontinuous at interfaces where rock heterogeneities occur. The contribution ofcapillary di�usion to the ux is calculated by approximating the derivative ofthe capillary pressure. However, the preceding analysis shows that the capillarypressure may be discontinuous, in which case the derivative of the capillarypressure is formally unde�ned. We propose here a numerical algorithm inwhich the extended pressure condition is incorporated. It enables us to dealwith problems in which the capillary pressure is discontinuous.The di�usion problem discussed in Section 3.1 is used as a test problemto validate the numerical method for problems with a discontinuous capillarypressure. 15

4.1 The AlgorithmFor the discretisation of (2.8), (2.14) we decompose the domain in N cellsi, i 2 1; � � � ; N , and assume that the discontinuities in h coincide with cellboundaries. For simplicity we restrict ourselves to grids with uniform meshsize �x, and assume that h(x) is a piecewise constant function.On this mesh we de�ne an explicit di�erence scheme that can be writtenin conservative form un+1i = uni � �t�i�x(F ni+1=2 � F ni�1=2); (4.1)with uni the approximation of u(x; t) in i at t = tn, �i the porosity in i andF ni+1=2 the ux at the edge between the cells i and i+1.We �rst consider the calculation of the ux F ni+1=2 at a cell edge that is notan interface, i.e. h(x) is continuous. The ux F ni+1=2 is split in a convective anda di�usive part. The convective part of the ux is discretised by the MUSCL-method [13], a higher order Godunov scheme. In this approach an accurateinterpolation operator is used to obtain approximations uL and uR at the left-and right-hand side of a cell edge. In this interpolation step we use the VanAlbada slope limiter [1] to avoid spurious oscillations. These approximationsare then used to determine the Godunov ux FG(uL; uR). For the discretisationof the di�usive part of the ux we use central di�erencing. The nonlineardi�usion coe�cient D(u) is approximated using a linear interpolation for thevalues of u (cf. [4]). Putting the convective and di�usive part together, the ux F ni+1=2 is approximated byF ni+1=2 = FG(uL; uR)� �ihiD((uni + uni+1)=2)uni+1 � uni�x : (4.2)Next we consider the calculation of the ux F ni+1=2 at an interface where his discontinuous, i.e. where hi 6= hi+1. As u may be discontinuous at this edgewe again introduce variables uL and uR, that are approximations of the leftand right limit values of u (see Chavent & Ja�re [3]). Fully analogous to (4.2)we de�ne approximations FL and FR of F ni+1=2 in i and i+1, respectively:FL = FG(uni ; uL)� �ihiD((uni + uL)=2)uL � uni�x=2 ; (4.3)FR = FG(uR; uni+1)� �i+1hi+1D((uR + uni+1)=2)uni+1 � uR�x=2 : (4.4)The dummy variables uL and uR are eliminated by using continuity of ux(FL = FR) and the extended pressure condition (2.25). This system of twononlinear equations is solved by Newton's method. Thus we �nd uL and uR,and the ux F ni+1=2 = FL = FR at the interface.16

Parameter Value FunctionsNc 1 J(u) = u�1=2M 1 krw(u) = u2k� 0.1 krn(u) = (1� u)2k+ 1�� 1�+ 1Table 4.1: Data set of parameters and functions.After having determined the uxes F ni+1=2 at t = tn we calculate un+1iby the explicit time integration (4.1). The explicit time integration is onlyconditionally stable, therefore we use time steps in our numerical calculationsthat are su�ciently small to avoid numerical instability.4.2 ValidationTo test the convergence of this algorithm we consider the nonlinear di�usionproblem discussed in Section 3.1. In this test problem the data of Table 4.1are used. Unless stated otherwise these data are used in all our numericalcalculations. An accurate numerical approximation of the similarity solutionthat arises in this problem is given by a multiple shooting method. Thisshooting method is based on the integration of (3.6), (3.7), continuity of uxand the extended pressure condition (see Van Duijn & De Neef [7]). Thisnumerical solution is used as the reference solution.Next we calculate the solution of the di�usion problem on the interval(�1; 1) at t = 1 by the discretisation scheme presented in the previous section.The error EN in this discrete solution is measured in the grid `2-norm that isde�ned by E2N = 1N NXi=1(uni � u(xi; 1))2: (4.5)In Table 4.2 the error EN is shown for di�erent discretisation grids. We observea O(N�1) convergence behaviour for EN , although the space discretisation isformally O(N�2) accurate away from the interfaces. This loss of accuracy isdue to the fact that the solution u(x; 1) has a vertical tangent near x = �n (cf.(3.17)).Next we consider the movement in time of the free boundaries. Given thenumerical solution ui at a certain time, the position of the free boundaries is17

N EN20 0.0027140 0.0016980 0.00097160 0.00051Table 4.2: Discretisation error EN for nonlinear di�usion problem on di�erentgrids.approximated by extrapolation. We assume that the solution in the neighbour-hood of the free boundaries satis�es a power law relationship. The exponentsof these power laws are the unique numbers that yield a �nite and non-zerospeed of propagation for the free boundaries in (3.14) and (3.15). For example,using the data of Table 4.1, this means that we assume that the solution in aleft neighbourhood of the right free boundary is of the form u = c(�w�x)2. Forthe approximation of the position �w of this free boundary we determine therightmost point xi such that ui > ", with " = 0:01, and then calculate �w fromui�1, ui and the power law. The other free boundary is treated analogously.In Figure 4.1 this approximation of the free boundaries at di�erent timelevels is indicated by dots. In the calculation a grid of 80 points is used. Thesolid line gives the movement of the free boundaries as determined by thesimilarity solution. We observe a good agreement between the numerical ap-proximation and the reference solution. These results show that our algorithmdeals in a satisfactory way with problems in which the capillary pressure isdiscontinuous.5 Numerical Results and DiscussionWith the numerical algorithm we can study problems that cannot be dealtwith by analytical means. To understand the interaction of a free boundarywith a heterogeneity, we �rst consider the movement of a capillary dispersedoil-water front across an interface where the permeability is discontinuous.Next we consider the case of oil trapping in heterogeneous structures duringwater ooding.18

01-1 0 1t xrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrFigure 4.1: Movement of the free boundaries occurring in the nonlinear dif-fusion problem: numerical approximation (dots) and similarity solution (solidline).5.1 Interaction of oil-water front with a heterogeneityLet us �rst consider the interaction of an oil-water front with a single hetero-geneity at the origin. We use the data of Table 4.1, withk(x) = ( 1; x < 0;0:1; x > 0and the initial distributionu(x; 0) = ( 0; x < �0:25;1; x > �0:25: (5.1)This distribution is shown in Figure 5.1 (top left) together with numericalapproximations of the solution at later times. Two free boundaries emerge;their movement is shown in Figure 5.2.The left free boundary, the wetting front, separates regions without anywater (u = 0) from regions with some water (u > 0). It moves towards theleft, that is upstream. Let �w(t) denote its position. Analogous to the speedof shocks in �rst-order hyperbolic conservation laws, its speed � 0w(t) away fromheterogeneities is given by (see Gilding [9])� 0w(t) = limx#�w(t) F (x; t)�u(x; t) = qf 0w(0)� � limx#�w(t) hD(u)u @u@x: (5.2)For our choice of functional dependencies we have f 0w(0) = 0, and as @u@x(�+w ; t) �19

01-1 0 0.5u x { S� 01-1 0 0.5u x { S�01-1 0 0.5u x { S� 01-1 0 0.5u x { S�Figure 5.1: Interaction of a free boundary with a heterogeneity: solution u(x; t)at t = 0:00; 0:09; 0:13; 0:25.0, it follows that � 0w(t) � 0. This implies that the left free boundary cannotmove to the right, and also that u(x; t) > 0 for any t > t0 if u(x; t0) > 0.The other free boundary, the nonwetting front, moves to the right. Similarto (5.2) its speed � 0n(t) away from heterogeneities is given by� 0n(t) = qf 0w(1)� + limx"�n(t) hD(u)1 � u @u@x: (5.3)For our problem we have f 0w(1) = 0, so the speed � 0n(t) is nonnegative. Whenthis free boundary reaches the interface at x = 0 it stops (see Figure 5.1, topright and bottom left). This can be understood as follows. Suppose there is apositive oil ux qn (cf. (2.6),(2.14),(3.3)),qn = (1 � fw(u))q + �hD(u)@u@x; (5.4)20

00.25-1 -0.5 0t xFigure 5.2: Interaction of a free boundary with a heterogeneity: position inxt-plane.at the right-hand side of the interface and that u(0+; t) = 1, then @u@x(0+; t) >0, which is impossible. So if there is a positive oil ux at x = 0 we musthave u(0+; t) < 1. The extended pressure condition (2.25) then implies thatu(0�; t) < S� with S� = 0:1. As long as the left saturation u(0�; t) is above thethreshold saturation S� there is no oil ux through the heterogeneity at x = 0.Therefore the right free boundary stops at x = 0, and only when u(0�; t) < S�it continues its movement (Figure 5.1, bottom right).In Figure 5.2 the numerical approximation of the position of the free bound-aries is shown. Clearly, the right free boundary stops at the heterogeneity.After a certain waiting time in which the oil saturation builds up, we haveu(0�; t) < S� and the free boundary starts moving again.Finally some remarks about the large-time behaviour for this problem.Consider the stationary convection-di�usion equation discussed in Section 3.2,but now with boundary conditionsu(�1) = u(+1) = 0: (5.5)By using the same argument as in Section 3.2 we �nd that this problem has aunique steady-state solution u � 0. Therefore we havelimt!1 u(0�; t) = limt!1u(0+; t) = 0; (5.6)so there is no trapping of oil. As the speed � 0w(t) is nonnegative this steady-state solution is not attained in �nite time.21

5.2 TrappingNext we consider a case in which trapping of oil does occur. Let us assumethat the medium has a periodic permeability,k(x) = ( 0:1 x 2 (2i; 2i+ 1]; i = 0; 1; � � � ;1 otherwise, (5.7)and a capillary number Nc = 10. Initially there is a single blob of oil present(see Figure 5.3, top left):u(x; 0) = ( 0:1; 1 < x < 2;1; otherwise. (5.8)The saturation between x = 1 and x = 2 is precisely at the threshold saturationS� for the interfaces at x = 1 and x = 2. At t > 0 oil moves to the right due toconvection (see Figure 5.3 (top right)). Consequently the saturation u(2�; t)drops below S�, whereas u(1+; t) > S�.Suppose that oil ows to the left at x = 1. Then (5.4) and the extendedpressure condition (2.25) imply that u(1+; t) < S�. Figure 5.3 shows thatu(1+; t) > S� for t > 0, so there is never any oil upstream of x = 1.The nonwetting front at x = 2 moves to the right immediately. Whenit reaches the heterogeneity at x = 3 it does not stop. If it would stop oilaccumulates at x = 3, giving u(3�; t) < 1. The extended pressure conditionthen implies that u(3+; t) < 1, which is a contradiction. Therefore the frontpasses through the heterogeneity without delay. However, when the frontreaches the heterogeneity at x = 4 it stops (see Figure 5.3, bottom left). Onlywhen u(4�; t) < S� the oil can move further, as discussed in Section 5.1. Theoil saturation now builds up at x = 4�. However there is simply not enoughoil to exceed the oil threshold saturation 1�S�. Therefore the oil-water frontstops de�nitely at x = 4, and the oil remains trapped in the medium.The steady-state solution for this problem consists of pieces of the steady-state solutions discussed in Section 3.2. For 1 < x < 2 there is a piece ofthe steady-state solution with u(2�) = S�. All oil is displaced from the lowpermeable region between x = 2 and x = 3, and for 3 < x < 4 we have a pieceof a steady-state solution with u(4�) > S�. Notice that di�erent amounts ofoil remain trapped in the intervals 1 < x < 2 and 3 < x < 4. The amount ofoil in the former interval corresponds to the maximum amount of oil that canremain trapped in a high permeability region, because u(2�) = S�.Here we have shown the case of the movement of a single blob of oil in amedium with periodic permeability. The same large-time behaviour is foundfor the water ooding of a reservoir with periodic heterogeneity (e.g. a reser-voir with crossbedding) that is initially completely �lled with oil. All oil isdisplaced from the low permeability regions, and oil remains trapped in the22

01 1 2 3 4u x{ S� 01 1 2 3 4u x{ S�01 1 2 3 4u x{ S� 01 1 2 3 4u x{ S�Figure 5.3: Movement of oil blob through periodic medium: solution u(x; t) att = 0; 0:5; 5; 100.high permeability regions. The amount of trapped oil in the high permeabilityregions is determined by the steady-state solution of the convection-di�usionequation that is discussed in Section 3.2. It depends on the factor NcM , so on�=q�w. The amount of trapped oil can be reduced by decreasing this factor.There are several options for achieving this.The total ow rate q can by changed by increasing the injection rate. How-ever it is quite impractical to obtain large changes in NcM in this way. Theinterfacial tension � between water and oil can be lowered by the addition ofsurfactants, which are quite expensive. Finally the viscosity �w of the oodwater can be increased by polymer addition. This last option seems to be mostappealing: already a small polymer concentration leads to an increase of waterviscosity by an order of magnitude. 23

6 ConclusionIn this paper we derived interface conditions at discontinuities in permeabilityand porosity. One condition expresses continuity of the phase uxes, the otheris related to the capillary pressure. In particular this pressure condition hasinteresting consequences: solutions of the ow problem can have a discontinu-ous capillary pressure at a heterogeneity, and non-trivial stationary solutionsare admitted which represent trapped oil. A numerical algorithm has beendeveloped which incorporates these interface conditions. It is used to studytwo characteristic examples.In one example an oil-water front moves through a discontinuity where(k=�)1=2 jumps downward. The front stops for a short time, the waiting time,until enough oil saturation has been built up to move on.In the other example an oil blob, initially present in a medium with peri-odic permeability, is not produced at all during water ooding, but is trappedin the high permeability regions. This explains why considerable amounts ofoil remain trapped in high permeability zones during water ooding of hetero-geneous petrol reservoirs. The essential parameter that determines the amountof trapped oil is �=q�w. This number can be decreased by increasing the wa-ter injection rate q, by adding surfactants to lower the interfacial tension �between oil and water, or by adding polymer to the water to raise its viscosity�w. The last option seems most attractive from the economical point of view.References[1] G. D. van Albada, B. van Leer, and W. W. Roberts. A comparative studyof computational methods in cosmic gas dynamics. Astron. Astrophys.,108:76{84, 1982.[2] J. Bear. Dynamics of Fluids in Porous Media. Dover Publications, 1972.[3] G. Chavent and J. Ja�re. Mathematical models and Finite Elements forReservoir Simulation, volume 17 of Studies in Mathematics and Applica-tions. Elsevier Science Publishers, 1986.[4] C.N.Dawson. Godunov-mixed methods for immiscible displacement. Int.J. Num. Meth. in Fluids, 11:835{847, 1990.[5] M. Dale, S. Ekrann, J. Mykkelveit, and G. Virnovsky. E�ective relativepermeabilities and capillary pressure for 1D heterogeneous porous media.In ECMOR IV, topic C, 1994. 24

[6] R.A. Dawe, M.R. Wheat, and M.S. Bidner. Experimental investigation ofcapillary pressure e�ects on immiscible displacement in lensed and layeredporous media. Transport in Porous Media, 7:83{101, 1992.[7] C.J. van Duijn and M. de Neef. Self-similar pro�les for capillary di�usiondriven ow in heterogeneous porous media. Preprint Delft University ofTechnology, 1994.[8] C.J. van Duijn and F.J.T. Floris. Mathematical analysis of the in uenceof power-law uid rheology on a capillary di�usion zone. J. Pet. Sc. Eng.,7:215{237, 1992.[9] B.H. Gilding. The occurrence of interfaces in nonlinear di�usion-advectionprocesses. Archive for Rational Mechanics and Analysis, 100:243{263,1988.[10] B.H. Gilding. Qualitative mathematical analysis of the Richards equation.Transport in Porous Media, 5:651{666, 1991.[11] T.F.M. Kortekaas. Water/oil displacement characteristics in crossbeddedreservoir zones. Soc. Pet. Eng. J., pages 917{926, december 1985.[12] R.C. Ledeboer. Capillary entrapment in small scale heterogeneous porousmedia. Technical Report 1992-6, Faculty of Mining and Petroleum Engi-neering, Delft University of Technology, 1992.[13] B. van Leer. Towards the ultimate conservative di�erence scheme V. Asecond order sequel to Godunov's method. J. Comput. Phys., 32:101{136,1979.[14] M.C. Leverett. Capillary behavior in porous solids. J. Pet. Tech., pages1{17, august 1940.[15] J.R. Philip. Horizontal redistribution with capillary hysteresis. WaterResour. Res., 27(7):1459{1469, July 1991.[16] K.J. Weber. How heterogeneity a�ects oil recovery. In L.W. Lake andH.B.Carrol, editors, Reservoir Characterization, pages 487{544, Orlando,1986. Academic Press.[17] Y.C. Yortsos and Jincai Chang. Capillary e�ects in steady-state ow inheterogeneous cores. Transport in Porous Media, 5:399{420, 1990.25