4.8 EXERCISES 71 Binary Displacement with Mutual Solubility The discussion of the application of the velocity constraint and entropy condition to eliminate nonphysical solutions follows that of Johns [54, Chapter 3] for the Buckley-Leverett problem Examples of solutions for displacement of C10 by CO2 are given by Pande [95, Chapter 4] The description of the dependence of solutions on initial and injection conditions was given first by Helfferich [32] Effects of Volume Change on Mixing A comparison of binary solutions with and without volume change as components change phase is given for CO2 /C10 displacements by Dindoruk [19, Chapter 6] 4.8 Exercises Characteristic curves Consider the equation ∂C ∂C + C2 = ∂t ∂x (4.8.1) The initial composition is Cinit = 0.1, and the injection composition is Cinj = 0.8 Derive expressions for the characteristic curves Plot the appropriate characteristic curves on a t-x diagram Determine whether shocks would occur in this displacement Gas dissolution Consider a laboratory core which contains initially water that is saturated with CO2 in equilibrium with gas at the critical gas saturation, Sgc = 0.05 The equilibrium volume fraction of CO2 dissolved in the saturated water phase is 0.03, and the volume fraction of water in the gas phase is 0.001 At time τ = 0, injection of pure water into the core begins Assuming that effects of volume change as components change phase can be neglected, calculate the saturation profile at τ = 0.5 pore volumes injected Determine how much pure water would have to be injected to remove all the gas present initially in the core Calculate the saturation profile at τ = 0.5 and 1.0 pore volumes injected for the relative permeability functions of Eqs 4.1.14-4.1.19 with Sgc = 0.1, Sor = 0.3, and M = 10 for a displacement in which gas displaces oil The volume fraction of the light component required to saturate the liquid phase is 0.4, and the volume fraction of light component in the equilibrium vapor phase is 0.95 The initial composition is a single-phase mixture in which the light component has a volume fraction of 0.2 The volume fraction of the light component in the injection gas is 0.98 Also calculate a recovery curve for the heavy component How many pore volumes of gas must be injected to recover all of the oil initially in place? Displacement with two-phase initial and injection mixtures Consider the fluid system of problem Calculate the saturation profiles at the same times and calculate a recovery curve for a displacement in which the core initially contains a two-phase mixture in which the volume fraction of the light component is 0.5, and the injection gas is also a two-phase mixture with a light component volume fraction of 0.9 Effect of volume change on shock speed Consider the situation outlined in problem Determine the shock speed for a situation in which the density of the water does not change as it moves between phases, but CO2 that dissolves in the water phase occupies only half the 72 CHAPTER TWO-COMPONENT GAS/OIL DISPLACEMENT Table 4.4: Equilibrium Phase Compositions and Fluid Properties Fluid xCH4 xC10 Initial Oil Equil Liq Equil Vap Injected Gas 0.3519 0.9964 1 0.6481 0.0036 ρ (gmol/l) 4.881 6.481 4.316 4.298 ρ (g/cm3) 0.6945 0.6342 0.0712 0.0690 µ (cp) 0.262 0.015 - volume of CO2 in the vapor phase Assume that the CO2 in the vapor phase has a component density of 1.1364 x 10−3 gmol/cm3 and that water in the liquid phase has a component density of 5.5555 x 10−2 gmol/cm3 CH4 displacing C10 Consider the fluid property data given in Table 4.4 for the CH4 /C10 system at 160 F and 1600 psia Calculate and plot the composition profile as a function of ξ/τ for two differing assumptions about density behavior: (1) when the volume occupied by each component is the pure component volume no matter what phase the component appears in, and (2) when the equilibrium phases assume the densities given in Table 4.4 Assume that the phase compositions in mole fractions given in Table 4.4 are correct for both cases Calculate and plot recovery curves as a function of pore volumes of methane injected for the two density assumptions Chapter Ternary Gas/Oil Displacements In this chapter, we consider the behavior of displacements in which three components and two phases are present Much of the original work on gas drives was done for ternary systems, which display essential features of displacement behavior but are simple enough to analyze The basic physical mechanisms of gas drives were outlined by Hutchinson and Braun [38], who considered what would happen if a porous medium were represented as a series of mixing cells Figs 5.1 and 5.2 summarize their arguments (For another version of the mixing cell argument, see Lake [62].) Suppose that oil with composition O1 is displaced by gas with composition G1 Mixtures of oil O1 with gas G1 in the first mixing cell lie on the dilution line that connects O1 to G1 on the ternary diagram in Fig 5.1 Suppose that after mixing some oil with gas in the first cell, the overall composition is M1 That mixture splits into two phases with compositions V1 and L1 Now assume that the less viscous vapor phase with composition V1 moves to the next downstream cell, where it mixes with fresh oil Those mixtures lie on the dilution line that connects V1 with the oil composition, O1 If the new overall composition is M2 , then the phases that form in the second cell have compositions L2 and V2 But when the vapor V2 moves to the next cell and mixes with fresh oil, the dilution line does not pass through the two-phase region Instead, the mixtures are “miscible” after multiple contacts, even though the original gas and oil not form only one phase when mixed in any proportions This displacement is what is known as a vaporizing gas drive because the crucial transfer of components that leads to miscibility is the vaporization of the intermediate component from the oil into the fast-moving vapor phase Mixture V1 is richer in component than the original injection gas is, and mixture V2 is richer still Oil O1 is rich enough in component that miscibility develops If, however, the oil had had composition L2 (or any mixture on the extension into the single-phase region of the tie line that connects V2 and L2 ), mixture of V2 with fresh oil would have given another mixture on the same tie line In that case, the enrichment of the vapor phase with component ceases to change with further contacts in downstream mixing cells Such a vaporizing gas drive is said to be “immiscible.” In vaporizing gas drives, the mixing cell argument indicates that miscibility develops if the original oil composition does not lie within the region of tie line extensions on the ternary diagram Fig 5.2 summarizes a similar argument for a displacement known as a condensing gas drive in which gas G2 displaces oil with composition O2 Mixtures of original oil with gas in the first mixing cell give composition M1 That mixture splits into phases with compositions V1 and L1 Here again, the vapor phase is assumed to move ahead and contact fresh oil, but this time we focus 73 74 CHAPTER TERNARY GAS/OIL DISPLACEMENTS C1 aG1 a1 V M1 a M2 a L1 C3 a V a2 a L2 a O1 C2 Figure 5.1: Mixing cell representation of a vaporizing gas drive on what happens in the first mixing cell There, liquid phase with composition L1 mixes with new injected gas Those mixtures lie on the dilution line that connects L1 with G2 If the new composition after mixing in the first cell is M2 , then the resulting phase compositions are V2 and L2 When liquid phase L2 mixes with new injection gas, the mixtures are single-phase Here again, multiple contacts of the gas with the oil have created mixtures that are miscible This displacement is called a condensing gas drive because it relies on transfer of the intermediate component from the injected gas phase to the oil Ternary condensing gas drives are multicontact miscible if the injection gas composition lies outside of the region of tie line extensions on a ternary phase diagram If, on the other hand, the injection gas had had some composition on the extension of a tie line (say the L2 -V2 tie line), the enrichment of the oil in the first cell with component condensing from the gas would have stopped when that tie line was reached, because additional mixtures of fresh gas with liquid L2 would fall on the same tie line again The mixing cell argument is necessarily a qualitative one because it assumes that only the vapor phase moves from cell to cell In a displacement in a porous medium, the phases would move according to their fractional flows Our task for this chapter, then, is to put the qualitative argument of Hutchinson and Braun on a firm mathematical footing The Riemann problem we will consider is illustrated in Fig 5.3: for a given pair of initial and injection compositions, find the set of compositions that form between the injection composition at the upstream end of the transition zone and the initial composition at the downstream end When three components are present, the flow is no longer constrained to take place along a single tie line, as it is in binary displacements Thus, for three-component flows, an essential part of the problem is to find the collection of tie lines (and their associated phase compositions) that are traversed during a displacement Three-component flows have been considered for systems that range from alcohol displacements [125] to surfactant flooding [35] to gas/oil systems [134, 22] The ideas developed in this and the next chapter come from many sources reviewed at the end of the chapter The development given here draws heavily from the work of Johns [54, Chapter 3], Dindoruk [19, Chapters 3, 4, and 6], and Wang [128] 5.1 COMPOSITION PATHS 75 C1 a1 V a2 V M1 a M2 O2 a a L1 a aG2 a L2 C3 C2 Figure 5.2: Mixing cell representation of a condensing gas drive We begin by formulating the eigenvalue problem that determines wave velocities and allowed composition variations, and we develop the idea of a composition path Next we consider the behavior of shocks, which play important roles in the behavior of solutions In Sections 5.3 and 5.4, example solutions are described that show in detail the patterns of flow behavior associated with vaporizing and condensing gas drives Section 5.5 shows that the key patterns of shocks and rarefactions (continuous composition variations) for ternary systems can be catalogued in a simple way based on the lengths of two key tie lines and whether tie lines intersect on the vapor side or the liquid side of the two-phase region Section 5.6 introduces the important concept of multicontact miscibility Effects on ternary systems of volume change as components transfer between phases and calculation of component recovery are reviewed in the remaining sections 5.1 Composition Paths The conservation equations for a three-component system without volume change are ∂F1 ∂C1 + ∂τ ∂ξ ∂F2 ∂C2 + ∂τ ∂ξ = 0, (5.1.1) = 0, (5.1.2) where Ci = ci1 S1 + ci2 (1 − S1 ), i = 1, 2, (5.1.3) Fi = ci1 f1 + ci2 (1 − f1 ), i = 1, (5.1.4) and 76 CHAPTER TERNARY GAS/OIL DISPLACEMENTS C1 Gas Composition •B • • • • •A Oil Composition C3 C2 (a) Composition Path Injection Gas Composition B Oil Composition A Produced Fluids ξ (b) Initial Condition Figure 5.3: Riemann problem for displacement of a three-component oil by gas The analysis of this chapter will show that much of the behavior of solutions to Eqs 5.1.1 and 5.1.2 is controlled by the properties of tie lines (in particular, two key tie lines that extend through the injection gas and initial oil compositions), and hence it is convenient to parametrize the flow problem based on the equation of a tie line The equation of any tie line can be written C2 = α(η)C1 + φ(η), (5.1.5) where α is the slope of the tie line, φ is the intercept at C1 = 0, and η is a parameter that determines which tie line is being represented For example, η could be chosen to be the volume fraction of component in the vapor phase, c11 , [54], or it could be taken to be the slope of the tie line [52] Any convenient parameter is suitable as long as it uniquely identifies a tie line Eq 5.1.5 is just another version of Eqs 5.1.3, so expressions for α and φ in terms of the phase compositions can be obtained by eliminating S1 from Eqs 5.1.3 written for components and 2, α= c21 − c22 , c11 − c12 φ= c22 c11 − c21 c12 c11 − c12 (5.1.6) 5.1 COMPOSITION PATHS 77 Similarly, elimination of f1 from Eqs 5.1.4 indicates that F2 = α(η)F1 + φ(η) (5.1.7) Substitution of the expressions for C2 and F2 into Eq 5.1.2 gives dα dφ ∂η dα dφ ∂η + + F1 + = (5.1.8) dη dη ∂τ dη dη ∂ξ In the original form of the conservation equations, F1 and F2 are functions of C1 and C2 only Given a value of C1 and C2 , a flash calculation can be performed to find the equilibrium phase compositions, and the phase saturations can then be calculated (Eq 5.1.3) From the phase saturations, relative permeabilities of the phases can be evaluated Phase viscosities can be calculated from phase compositions All the information required to calculate phase fractional flows is then available, and Eqs 5.1.4 can be used to evaluate them Similarly, in the parametrization of the problem based on tie lines, F1 is a function of C1 and η Setting η determines the tie line, and fixing C1 gives the location of the overall composition on the tie line, from which all other properties can be obtained Hence the derivatives of F1 in Eq 5.1.1 can be written in terms of C1 and η, and Eq 5.1.1 becomes C1 ∂F1 ∂C1 ∂F1 ∂η ∂C1 + + = ∂τ ∂C1 ∂ξ ∂η ∂ξ (5.1.9) Eqs 5.1.8 and 5.1.9 can be rearranged into the form, ∂u ∂u + A(u) = 0, ∂τ ∂ξ (5.1.10) where u = (C1 , η)T and A(u) = ∂F1 ∂C1 ∂F1 ∂η F1 +p C1 +p , (5.1.11) with p given by p= dφ dη dα dη = dφ dα (5.1.12) The physical interpretation of p can be seen by considering two adjacent tie lines (labeled A and B) The equations of the tie lines are C2 = αA C1 + φA , (5.1.13) C2 = αB C1 + φB , (5.1.14) and where α and φ are defined by Eqs 5.1.6 If the tie lines are adjacent to each other then we can write αB = αA + dα ∆η, dη (5.1.15) 78 CHAPTER TERNARY GAS/OIL DISPLACEMENTS and dα ∆η, (5.1.16) dη where again, η is a parameter that determines the tie line The point at which the two tie lines x x intersect, (C1 , C2 ) can be found by solving Eqs 5.1.13 and 5.1.14 using 5.1.15 and 5.1.16 The result is φB = φA + e C1 = dφ dη − dα dη = −p (5.1.17) e Thus, C1 is a point on the locus of intersections of infinitesimally separated tie lines That curve is known as an envelope curve [19] Fig 5.4 shows an envelope curve for a ternary system with constant K-values Any tie line can be constructed by drawing a tangent to the envelope curve that intersects the liquid and vapor loci Thus, we have shown that −p is the overall volume fraction, e C1 , on the envelope curve The relationship between p and the envelope curve suggests that the geometry of tie lines will play an important role in the behavior of solutions The remainder of this chapter will show that statement to be true 5.1.1 Eigenvalues and Eigenvectors To solve Eq 5.1.10 we calculate the trajectory of a fixed composition (fixed values of C1 and η), just as we did for binary displacements As before, we ask how constant values of C1 and η propagate, and hence we set dC1 and dη to zero in ∂C1 dτ + ∂τ ∂η dτ + ∂τ dC1 = dη = ∂C1 dξ = 0, ∂ξ ∂η dξ = ∂ξ (5.1.18) (5.1.19) Eqs 5.1.18 and 5.1.19 can be rearranged to give ∂C1 ∂τ ∂η ∂τ ∂C1 dξ ∂C1 = −λ , ∂ξ dτ ∂ξ ∂η dξ ∂η = − = −λ , ∂ξ dτ ∂ξ = − (5.1.20) (5.1.21) where λ = dξ/dτ is the wave velocity of some fixed composition defined by the values of C1 and η Substitution of Eqs 5.1.20 and 5.1.21 into Eq 5.1.10 gives −λ ∂F1 ∂C1 ∂F1 ∂η F1 +p C1 +p − λ ∂C1 ∂ξ ∂η ∂ξ = (5.1.22) Eq 5.1.22 is an eigenvalue problem that has nontrivial solutions if and only if det −λ ∂F1 ∂C1 ∂F1 ∂η F1 +p C1 +p − λ = ∂F1 −λ ∂C1 F1 + p − λ = C1 + p (5.1.23) 5.1 COMPOSITION PATHS 79 C1 • • • • • • • • C3 • • • • • C2 Envelope Curve • • Figure 5.4: Tie-line construction from an envelope curve for a ternary system with constant Kvalues: K1 = 2.5, K2 = 1.5, K3 = 0.05 In Eq 5.1.23, values of λ, the eigenvalues, give the wave velocities at which a given overall composition propagates, and the associated eigenvectors, (∂C1 /∂ξ , ∂η/∂ξ)T are directions in composition space along which compositions can vary while also satisfying the material balance equations Expansion of the determinant in Eq 5.1.23 gives a quadratic equation, which has the obvious solutions λt = ∂F1 , ∂C1 λnt = F1 + p C1 + p (5.1.24) (The simple form of the eigenvalue problem , Eq 5.1.23, is the reason for the parametrization of the problem in terms of tie line slope and intercept [52, 54].) The fact that each composition can have two possible wave velocities means that we will have to select which velocity applies at a given composition To that, we must consider the eigenvectors associated with each of the eigenvalues Substitution of the eigenvalues into Eq 5.1.22 shows that the associated eigenvectors are 80 CHAPTER TERNARY GAS/OIL DISPLACEMENTS ⎛ et = ent = ⎝ , λnt −λt ∂F1 ∂η ⎞ ⎠ (5.1.25) The first entry in each eigenvector corresponds to changes in C1 , and the second to changes in η The magnitude of an eigenvector is arbitrary – it indicates only a direction in composition space Hence, the unit value in the first position of each eigenvector is selected for convenience For compositions in the single-phase region, F1 = C1 everywhere As a result, λt = λnt = 1, and Eq 5.1.23 is satisfied for any composition variation Therefore, there are no discrete composition directions in the single-phase region, and composition variations in any direction are allowed For compositions in the two-phase region, however, there are two separate composition directions associated with et and ent Next we consider the properties of those eigenvector directions The existence of discrete directions given by the eigenvectors indicates that arbitrary composition variations cannot satisfy the conservation equations Instead variations in composition that are consistent with the material balance equations must lie on curves in composition space that are obtained by integrating along the eigenvector directions Those curves are known as composition paths [31] In other words, the expressions for the eigenvectors are ordinary differential equations for the allowed composition paths Self-Similarity and Coherence The idea of a composition path is quite important, because a solution to a Riemann problem must lie on a sequence of such paths if the conservation equations are to be satisfied To see why that statement is true, consider again Eq 5.1.10, and assume for the moment that a solution to it is a function of ξ/τ only, u(ξ, τ ) = w(ξ/τ ) Such solutions are said to be self-similar (the papers of Lax [67, 68] and Isaacson [39] show that problems of this sort are indeed self-similar) Substitution of the derivatives, ξ ∂u = −w , ∂τ τ (5.1.26) ∂u =w , ∂ξ τ (5.1.27) and in Eq 5.1.10 gives A(w) − ξ I w = τ (5.1.28) As long as w = 0, ξ/τ must be an eigenvalue of A and w must be an eigenvector At any value of ξ/τ , therefore, the solution u must lie on a curve in composition space that is tangent to an eigenvector The idea that a given overall composition, u = (C1 , η)T , translates with a wave velocity, λ = ξ/τ is what some investigators refer to as coherence [31] To show that the solution, u(ξ, τ ), depends only on ξ/τ , and hence that λ = ξ/τ , we note that if u(ξ, τ ) is a solution to Eq 5.1.10, then u(aξ, aτ ) is also a solution for any a > [39], as direct evaluation of the partial derivatives shows 5.1 COMPOSITION PATHS 81 We are free to choose a = 1/τ , which gives u = u(ξ/τ, 1) Thus, u is a function of ξ/τ only, and coherence follows from the differential equations The manipulations of this section have shown that the solution to Eq 5.1.10 can be obtained as a set of compositions, u(ξ, τ ) = (C1 , η)T , that lie on composition paths The next step, therefore, is to examine in more detail the behavior of those paths 5.1.2 Tie-Line Paths The change in η for composition variations along the direction of the eigenvector, et , is zero (see Eq 5.1.25) Because η is a parameter that determines on which tie line the composition point lies, zero variation in η means that the eigenvector direction is that of the tie line Furthermore, it is easy to show that for fixed η, the eigenvalue, λt = ∂F1 /∂C1 reduces to df1 /dS1 In fact the manipulations are identical to those performed in the derivation of Eq 4.1.23 for a binary system, as they must be because the composition variation occurs along a single tie line For the tie-line eigenvector , the path integration is simple Stepwise integration in the direction, et , gives additional compositions that lie on the same tie line Thus, integration of the tie line eigenvector gives a composition path that is just a straight line that coincides with the tie line So far, we have shown that a tie line is a composition path, and the wave speed, λt , for variation along a tie line composition path is simply the familiar Buckley-Leverett wave velocity, df1 /dS1 Those statements hold no matter how many components are present The fact that a tie line is a composition path means that it is possible to have solutions in which compositions vary along a single tie line Indeed, such variations must be possible if threecomponent solutions are to reduce smoothly to binary solutions Consider, for example, the Riemann problem illustrated in Fig 5.5 where initial and injection compositions lie on extensions of the same tie line If so, the entire solution remains on that tie line Such a displacement is really the equivalent of a binary displacement, even though three components are present [32] For such pseudobinary systems, the theory of Chapter applies 5.1.3 Nontie-Line Paths If the initial and injection conditions lie on different tie lines (or tie-line extensions), then composition variations that move between the two tie lines are required, so we consider now the properties of composition variations associated with the second eigenvalue and eigenvector direction To simplify the arguments and the notation, we choose η to be the vapor phase composition, η = c11 = y1 , and we define the liquid phase composition to be x1 = c12 = y1 /K1 , where K1 is the equilibrium K-value for component (see Section 3.4) It is easy to evaluate λnt and ent at three points on any tie line, when the overall composition lies on either the saturated liquid or vapor loci or when F1 = C1 inside the two-phase region We now find λnt and ent for each of those situations When the overall composition lies on the vapor phase portion of the binodal curve, C1 = y1 , and F1 = C1 Eq 5.1.24 shows that λnt = According to Eq 5.1.22 with λnt = 1, 82 CHAPTER TERNARY GAS/OIL DISPLACEMENTS C1 Injection • Gas • • • Initial Oil • C3 C2 Figure 5.5: Riemann problem for displacement of a three-component oil by gas lying on the same tie-line extension (λt − 1) dC1 ∂F1 + = dy1 ∂y1 (5.1.29) When the overall composition is the saturated vapor phase composition, f1 = 1, and hence, F1 = y1 f1 + x1 (1 − f1 ) = y1 (5.1.30) Therefore, ∂F1 /∂y1 = Also, λt = ∂F1 /∂C1 = df1 /dS1 = 0, and hence, substitution of those values into Eq 5.1.29 gives dC1 = dy1 (5.1.31) Eq 5.1.31 indicates that y1 = C1 on the nontie-line composition path that passes through the saturated vapor composition Hence, the composition path is the vapor locus of the binodal curve Similarly, for compositions on the liquid portion of the binodal curve, λnt = The overall composition is C1 = x1 , λt = 0, and differentiation of Eq 5.1.30 indicates that ∂F1 /∂y1 = dx1 /dy1 Eq 5.1.29 then reduces to dx1 dC1 = dy1 dy1 (5.1.32) Integration of Eq 5.1.32 indicates that C1 = x1 Therefore, the liquid portion of the binodal curve is also a composition path If F1 = C1 in the two-phase region, then λnt = (Eq 5.1.24) Comparison of Eqs 5.1.3 and 5.1.4 shows that F1 = C1 when f1 = S1 Points at which f1 = S1 lie on what is known as the EV equivelocity curve (C1 = C1 , because at such points the flow velocities of the two phases are equal There is one equivelocity point on every tie line, and the equivelocity curve, the locus of equivelocity points on the full set of tie lines, must pass through the critical point if one exists 5.1 COMPOSITION PATHS 83 To show that the equivelocity curve (f1 = S1 ) is also a path, we evaluate the derivatives ∂F1 /∂y1 and dC1 /dy1 and show that Eq 5.1.29 is satisfied First, we write F1 and C1 in terms of y1 , F1 = (y1 − x1 )f1 + x1 = y1 (1 − 1 )f1 + , K1 K1 (5.1.33) C1 = (y1 − x1 )S1 + x1 = y1 (1 − 1 )S1 + K1 K1 (5.1.34) For this exercise, we assume that phase viscosities are independent of composition (The manipulations are similar but algebraically more complex if the viscosities in f1 depend on composition, for example) Differentiation of Eq 5.1.33 gives 1 ∂F1 = (1 − )f1 + ∂y1 K1 K1 + y1 (1 − ∂f1 dK1 ) + (f1 − 1) K1 ∂y1 K1 dy1 (5.1.35) Because f1 is a function of S1 only, df1 ∂S1 ∂f1 = , ∂y1 dS1 ∂y1 (5.1.36) and ∂S1 /∂y1 can be obtained by differentiating Eq 5.1.34 (with C1 held constant), which gives y1 − K1 1 y1 dK1 ∂S1 =− 1− S1 − + (1 − S1 ) ∂y1 K1 K1 K1 dy1 (5.1.37) Finally, differentiation of Eq 5.1.34 gives an expression for dC1 /dy1 along the curve f1 = S1 (note that S1 is constant along that curve), 1 y1 dK1 dC1 = 1− S1 + + (f1 − 1) dy1 K1 K1 K1 dy1 (5.1.38) Substitution of Eqs 5.1.35, 5.1.36, 5.1.37, and 5.1.38 into Eq 5.1.29 shows that it is, indeed, satisfied Hence the equivelocity curve is a path Fig 5.6 is a plot of F1 vs C1 for the middle tie line in Fig 5.4 It has the familiar S shape, though the range of C1 is now restricted to remain within the ternary diagram Fig 5.7 shows the behavior of the two eigenvalues on the same tie line If the overall composition lies in the single-phase region, F1 = C1 , and both eigenvalues are one The S shape of the overall fractional flow curve requires that its slope be greater than one E E in the neighborhood of C1 Therefore, λt > 1, and hence, λt > λnt for compositions near C1 , where λnt = If the relative permeability functions (see Eqs 4.1.14-4.1.19) vary smoothly in the neighborhood of the binodal curve, and the relative permeability of a phase with zero saturation is also zero, both physically reasonable assumptions, then λt = df1 /dS1 will go smoothly to zero as either phase boundary is approached Therefore, λt < λnt in the neighborhood of either phase boundary As a result, for smoothly varying functions F1 , λt = λnt at two points on each tie line At those points, called equal eigenvalue points, Eq 5.1.25 indicates that et = ent Thus, integral curves of ent are tangent to tie lines at the equal eigenvalue points, a fact that will play an important role when rarefactions between tie lines are constructed Thus, for the nontie-line paths, we have shown that 84 CHAPTER TERNARY GAS/OIL DISPLACEMENTS Overall Fractional Flow of Component 1, F1 1.0 0.5 FE • CE 0.0 0.0 0.2 0.4 0.6 0.8 1.0 Overall Volume Fraction of Component 1, C1 Figure 5.6: Fractional flow function for a fixed tie line the liquid and vapor branches of the binodal curve are composition paths associated with λnt = 1, the equivelocity curve is also a composition path along which λnt = 1, and nontie-line eigenvectors are tangent to tie line paths at two distinct equal eigenvalue points on each tie line The remaining portions of the nontie-line paths must be obtained by integration of the nontieline eigenvector The complexity of that integration depends on the form of the equation of state For the Peng-Robinson equation, finite differences can be used to evaluate the derivative, ∂F1 /∂C1 and ∂F1 /∂η, that appear in the nontie-line eigenvector Nontie-line Paths for Constant K-Values If the phase behavior is simpler than that represented by the Peng-Robinson equation, it is sometimes possible to perform the integration analytically If equilibrium K-values are constant and phase viscosities are independent of composition, an explicit expression for the nontie-line paths can be obtained [128] The first step is to derive a differential equation for the nontie-line path To that it is convenient to write the definitions of the overall composition, Ci , and the overall fractional flow, Fi , in terms of the liquid phase composition, xi , 5.1 COMPOSITION PATHS 85 Tie-line and Nontie-line Eigenvalues, λt and λnt 3.0 λt 2.5 2.0 1.5 λnt 1.0 0.5 0.0 0.0 0.4 0.2 0.6 0.8 1.0 Overall Volume Fraction of Component 1, C1 Figure 5.7: Eigenvalue variation along a tie line Ci = xi {1 + (Ki − 1)S1 } , (5.1.39) Fi = xi {1 + (Ki − 1)f1 } (5.1.40) An equation for the liquid portion of the binodal curve is obtained from the fact that the liquid locus satisfies x1 + x2 + x3 = 1, (5.1.41) y1 + y2 + y3 = K1 x1 + K2 x2 + K3 x3 = (5.1.42) and the vapor locus Elimination of x3 from Eq 5.1.42 gives an expression for the liquid locus, x2 = − K3 K1 − K3 − x1 K2 − K3 K2 − K3 Eqs 5.1.39 and 5.1.40 can then be written as functions of S1 and x1 , (5.1.43) 86 CHAPTER TERNARY GAS/OIL DISPLACEMENTS C1 = C1 = x1 {1 + (K1 − 1)S1 } , K1 − K3 − K3 C2 = − x1 {1 + (K2 − 1)S1 } , K2 − K3 K2 − K3 F1 = x1 {1 + (K1 − 1)f1 } , − K3 K1 − K3 − x1 {1 + (K2 − 1)f1 } F2 = K2 − K3 K2 − K3 (5.1.44) (5.1.45) (5.1.46) (5.1.47) Substitution of these expressions into Eq 5.1.22 with choice of S1 and x1 as the primary dependent variables gives a convenient form of the eigenvalue problem, ⎛ ⎝ df1 dS1 ⎞ dp (f1 −S1 ) dx1 C1 +p F1 +p C1 +p −λ dS1 dx1 ⎠ −λ = 0, (5.1.48) where p= K1 − K2 K1 − K3 x2 x = 1, K2 − 1 − K3 γ (5.1.49) − K3 K2 − K1 − K2 K1 − K3 (5.1.50) where γ= The differential equation for the nontie-line path is obtained by evaluating Eq 5.1.48 for λnt = (F1 + p)/(C1 + p), which yields dx1 + P (S1 )x1 = Q(S1 ), dS1 (5.1.51) where P (S1 ) = df1 dS1 −1 f1 − S1 , (5.1.52) γ {1 + (K1 − 1)S1 } (1 − Q(S1 ) = γ(K1 − 1) + f1 − S1 df1 dS1 ) (5.1.53) Eq 5.1.51 has the solution x1 = e− P (S1 )dS1 P (S1 )dS1 Q(S1 )e dS1 + c0 , (5.1.54) where c0 is a constant of integration Evaluation of Eq 5.1.54 gives x1 = x0 0 f1 − S1 +γ f1 − S1 + 0 γ(K1 − 1) 0 f1 − S1 −1 + (S1 f1 − S1 f1 ) f1 − S1 f1 − S1 2γ(K1 − 1) f1 − S1 S1 S1 f1 dS1 , (5.1.55) 5.1 COMPOSITION PATHS 87 0 where x0 , f1 , and S1 refer to some point that lies on the nontie-line path in question Eq 5.1.55 indicates that if the fractional flow expression is simple enough, the integration can be performed and an explicit solution obtained If we use the relative permeability functions stated in Eqs 4.1.14 - 4.1.19, with phase viscosities independent of composition and Sgc = Sor = 0, the resulting expression for the integral in Eq 5.1.55 is S1 S1 S1 − S1 f1 dS1 = + 1+ M M M M 1+ ⎛ − arctan ⎝ + M M +1 ⎧ ⎨ ln S1 M ⎧ ⎨ −1 M M ⎩ ⎛ arctan ⎝ +1 − M M − M S1 + 1+ S1 M +1 − M M ⎞ ⎠ ⎞⎫ ⎬ ⎠ ⎭ M ⎫ S1 ⎬ ⎩ − S + + (S )2 ⎭ M M M (5.1.56) If Sgc = or Sor = 0, similar but slightly more complex expressions result (see Problem 23) Fig 5.8 shows the geometry of composition paths that result for constant K-values, K1 = 2.5, K2 = 1.5, and K3 = 0.05 While only a few tie lines are shown, the two-phase region is actually filled completely with the two sets of paths, the linear tie-line paths, and the curved nontie-line paths On each tie line there are two equal-eigenvalue points, indicated in Fig 5.8 by dots, at which the nontie-line path is tangent to the tie-line path To construct a solution for given initial and injection compositions, a sequence of composition paths that connect the initial and injection compositions must be selected from that doubly infinite set To that, we must consider what happens at a switch from one path to another 5.1.4 Switching Paths To construct a solution to the Riemann problem illustrated in Fig 5.3, we must find a set of composition paths that connects the initial and injection compositions If the initial and injection compositions lie on the extensions of different tie lines, then the solution may require a switch from a tie-line path to a nontie-line path and vice versa Next, we consider when path switches are permitted A path switch is allowed if and only if it satisfies the velocity constraint It will so if the wave velocity at the switch point decreases (or stays constant) as the combination of paths is traced from downstream to upstream compositions Fig 5.9 shows examples of three possible path switches from a tie-line path to a nontie-line path Point b in Fig 5.9 is the equal eigenvalue point, and Points a and c are additional switch points to be considered Fig 5.10 shows the corresponding eigenvalues for the initial tie line Fig 5.11 shows sketches of the composition profiles that would result for switches at the three points Suppose that the tie-line path is being traced from some downstream composition upward along the initial tie line Suppose also that a path switch occurs at point a, and the nontie-line path is then traversed toward the injection tie line Fig 5.10 shows that at point a, the wave velocity drops as the path is traced through the switch point, and Fig 5.11 shows the zone of constant 88 CHAPTER TERNARY GAS/OIL DISPLACEMENTS C1 Nontie-Line Path a a a a a C3 Tie-Line Path a a C2 Equivelocity Curve Figure 5.8: Tie-line and nontie-line paths for a ternary system with constant K-values: K1 = 2.5, K2 = 1.5, K3 = 0.05 The dots indicate the locations of equal-eigenvalue points where a nontie-line path is tangent to a tie line path state that results (Note that the profiles in Fig 5.11 are plotted against wave velocity, ξ/τ To obtain a location, ξ, of some composition at time, τ , simply multiply the wave velocity by the dimensionless time, τ ) At point a, a single composition has two wave velocities, a situation that is physically acceptable as long as the resulting composition variation is single-valued In other words, two spatial positions for the same composition are allowed, though two compositions at the same position are not The top panel of Fig 5.11 shows that the composition profile for point a is single-valued At point b, the two eigenvalues are equal, so that point is certainly an acceptable switch point, as Fig 5.11 shows A switch at point c, however, would not satisfy the velocity constraint The wave velocity would increase at the switch point (see Fig 5.10), which would cause the multivalued composition profile sketched in Fig 5.11 Thus, of the three points considered, only points a and b would satisfy the velocity constraint Next we consider whether legal path switches could occur at the intersection of a given nontieline path with the injection gas tie line Points d, e, f, and g in Fig 5.9 are the points to be considered Because point c is not allowed as a switch point, the nontie-line path from c to g cannot be traced, so point g cannot be reached, and it need not be considered further If the nontie-line path from point a, a legal switch point, to point d were traced upstream, a switch at point d would be allowed if λd > λd Fig 5.10 shows, however, that the reverse is true Point d nt t lies below the equal-eigenvalue point on the injection gas tie line, and therefore λd < λd A path nt t switch at point d would create a multivalued solution, which is not permitted Thus, the nontie-line 5.1 COMPOSITION PATHS 89 C1 • Injection Gas g a f a e a d a • a C3 • •b •c C2 Initial Oil Figure 5.9: Possible path switch points path from a to d is not an acceptable solution route because the path switch at d is prohibited by the velocity constraint, even though the path switch at point a is allowed A path switch at point b is allowed, so the nontie-line path could be traced from b to e However, a path switch at point e can be eliminated from further consideration by exactly the same argument used for point d As Fig 5.10 shows, λe < λe , and hence the path segment from nt t b to e can be ruled out as well The only remaining possibility is the segment from b to f Point f lies above the equal-eigenvalue point, and as Fig 5.10 shows, λf > λf Therefore, point f is a switch point that does satisfy the nt t velocity constraint Thus, if a nontie-line path is traced from the initial oil tie line to the injection gas tie line in a displacement with tie lines as shown in Fig 5.9, the upper branch of the nontieline path through point b must be used Similar arguments show that if the positions of the injection gas and initial oil tie lines were exchanged, the lower branch of the nontie-line path with an equal-eigenvalue point on the injection tie line would be the only acceptible solution route In this section we have examined the properties of composition paths They are essential to the construction of a solution to a Riemann problem like that shown in Fig 5.3 because composition paths are traced whenever continuous variations occur However, rarefactions along paths are sometimes prohibited by the velocity constraint, and when that happens, shocks/indexshock are required 90 CHAPTER TERNARY GAS/OIL DISPLACEMENTS Tie-line and Nontie-line Eigenvalues, λt and λnt 3.0 λt 2.5 ad aa ae 2.0 1.5 a aa 1.0 • b f ac a a a 0.5 0.0 0.0 λnt 0.2 0.4 0.6 0.8 1.0 Saturation, S1 Figure 5.10: Wave velocities at the switch points 5.2 Shocks Two kinds of shocks occur in ternary systems: (1) shocks that have a single-phase mixture on one side of the shock and a two-phase mixture on the other, and (2) shocks with two phases present on both sides of the shock The first type arises when the solution path enters or leaves the two-phase region, and the second appears when compositions on different tie lines are connected by a shock In this section we consider the two types in turn 5.2.1 Phase-Change Shocks Reasoning similar to that for binary displacements indicates that the solution route can enter or leave the two-phase region only via a shock Consider, for example, a composition point somewhere on the liquid portion of the binodal curve Such a composition might form if there were a continuous variation in composition from the initial oil composition at downstream locations to the injection gas composition upstream To enter the two-phase region from that point, the solution route would have to follow the tieline path through that composition (The nontie-line path at that point is just the binodal curve, and following it would give other compositions just at the edge of but not inside the two-phase region.) However, the tie-line wave velocity, λt, is zero at the binodal curve, as Fig 5.7 indicates Thus, a continuous composition variation across the binodal curve would cause a discontinuous ... K1 K1 dy1 (5. 1.38) Substitution of Eqs 5. 1. 35, 5. 1.36, 5. 1.37, and 5. 1.38 into Eq 5. 1.29 shows that it is, indeed, satisfied Hence the equivelocity curve is a path Fig 5. 6 is a plot of F1 vs C1... to Eq 5. 1.22 with λnt = 1, 82 CHAPTER TERNARY GAS/ OIL DISPLACEMENTS C1 Injection • Gas • • • Initial Oil • C3 C2 Figure 5. 5: Riemann problem for displacement of a three-component oil by gas lying... K3 K2 − K3 (5. 1.44) (5. 1. 45) (5. 1.46) (5. 1.47) Substitution of these expressions into Eq 5. 1.22 with choice of S1 and x1 as the primary dependent variables gives a convenient form of the eigenvalue