Orr, F. M. - Theory of Gas Injection Processes Episode 3 pptx

3.4 FLASH CALCULATION 31 Table 3.2: Peng-Robinson Binary Interaction Parameters [99, 18] Component CH4 C2 C3 C4 C5 C6 C7 C8 C9 C10 C12 C16 C20 CH4 C2 C3 CO2 0.100 0.150 0.134 0.130 0.125 0.119 0.100 0.112 0.100 0.102 0.095 0.105 0.093 0.010 0.020 0.020 0.025 0.025 0.035 0.035 0.035 0.035 0.035 0.035 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 0.010 N2 0.120 0.120 0.120 0.120 0.120 0.120 0.120 0.120 0.120 0.120 0.120 0.120 0.120 in subsequent chapters Interaction parameters for CO2 are those recommended by Deo et al [18] Otherwise Table 3.2 reports values of δij recommended by Peng and Robinson [99] Eqs 3.3.7–3.3.16 provide the description of volumetric behavior needed to complete the calculation of the partial fugacity of a component in a mixture Differentiation of Eq 3.3.7 gives an ∂P expression for ∂ni That expression is then substituted into Eq 3.2.8, and the integration is performed The result, after considerable algebraic manipulation, is ln ˆ fij xij P bi bm ˆ = ln φij = (z − 1) − ln z − bm V am + √ 2bm RT 3.4 bi − bm am nc xkj (aα)ik k=1 √ m + ( + 1) bV √ ln m − ( − 1) bV (3.3.17) Flash Calculation The result of the thermodynamic analysis and the use of an equation of state to describe volumetric ˆ behavior is a set of nonlinear equations (Eqs 3.1.30 in which each value of fij is given by an expression like Eq 3.3.17) that must be solved for the compositions of the phases The following procedure can be used: Estimate composition, xij , of each of the phases present Solve Eq 3.3.7 for the molar volume, V , of each phase ˆ Use Eq 3.3.17 to calculate the partial fugacity, fij , for each component in each phase 32 CHAPTER CALCULATION OF PHASE EQUILIBRIUM Check to see if component partial fugacities are equal in all of the phases (Eq 3.1.30) If not, then adjust the estimates of phase compositions and return to step While this procedure will give the compositions of phases that satisfy the requirement that component partial fugacities (or equivalently, chemical potentials) be equal at equilibrium, it should be noted that it is sometimes possible to find solutions for which the resulting phases are not stable [3] In other words, the phase compositions make component chemical potentials equal but not minimize free energy For such situations, the stability of phases can be tested directly See Baker et al [3] or Michelsen [78, 80] for examples and details Given the nonlinearity of Eq 3.3.17, it is easy to see why flash calculations with van der Waals equation did not catch on in the 1880’s when van der Waals [122] performed the first phase equilibrium calculations for binary systems In fact, equation-of state calculations of phase equilibrium did not come into widespread use until the late 1960’s when availability of computing resources made solution of the set of nonlinear equations reasonable Step requires that some sort of initial guess of phase compositions be made For flash calculations for two-phase systems, information about phase compositions is often expressed in terms of equilibrium ratios (also known as K-values), Ki = xi1 , xi2 (3.4.1) where xi1 and xi2 are the mole fractions of component i in phases and 2, typically vapor and liquid The Wilson equation [136], Ki = xi1 Pci Tci exp 5.37(1 + ωi ) − = xi2 P T , (3.4.2) is frequently used to estimate equilibrium K-values from which phase compositions can be estimated by the following manipulations From the estimated or updated K-values, the phase compositions can be obtained from a material balance on each component Consider one mole of a mixture in which the overall mole fraction of component i is zi A material balance for component i gives zi = xi1 L1 + xi2 (1 − L1 ), i = 1, nc (3.4.3) where L1 is the fraction of the one mole of mixture that is phase Elimination of xi1 from Eq 3.4.3 gives xi2 = zi , + L1 (Ki − 1) i = 1, nc (3.4.4) i = 1, nc (3.4.5) Similar elimination of xi2 using Eq 3.4.2 gives xi1 = Kizi , + L1 (Ki − 1) An equation for L1 is obtained by noting that Eqs 3.4.4 and 3.4.5 each sum to unity, so that nc nc xi1 − i=1 nc xi2 = i=1 i=1 zi (Ki − 1) = + L1 (Ki − 1) (3.4.6) 3.4 FLASH CALCULATION 33 Eq 3.4.6, which is known as the Ratchford-Rice equation [103], can be solved for L1 by a NewtonRaphson iteration [80, p 220] Given the value of L1 , Eqs 3.4.4 and 3.4.5 give the phase compositions consistent with the K-values The equilibrium K-values defined in Eq 3.4.2 are related to the equilibrium partial fugacity coefficients The equilibrium relations, Eqs 3.1.30, can be written using Eq 3.1.28 as ˆ ˆ ˆ ˆ fi1 = φi1 xi1 P = φi2 xi2 P = fi2 , i = 1, nc (3.4.7) Rearrangement of Eq 3.4.7 shows that the equilibrium K-value is just the ratio of partial fugacity coefficients, Ki = ˆ xi1 φi2 = , ˆ xi2 φi1 i = 1, nc (3.4.8) The form of Eq 3.4.8 suggests a simple successive approximation scheme for updating K-values in step of the flash calculation [98, p 103][80, p 245] If the component partial fugacites are not equal at the kth iteration, then new K-values can be estimated from Kik+1 = Kik ˆk fi2 , ˆ fk i = 1, nc (3.4.9) i1 ˆ ˆ Eq 3.4.9 modifies the K-values in the appropriate direction if fi2 = fi1 While iteration with Eq 3.4.9 usually converges to solutions that satisfy the equilibrium relations even when the guess of initial phase compositions is relatively poor, convergence will be very slow for two-phase mixtures near a critical point (a pressure, temperature and composition for which the two phases become identical) For such situations, more sophisticated iterative schemes can and should be used [79, 75] Negative Flash The flash calculation can be performed whether a mixture forms one or two phases Whitson and Michelsen [135] pointed out that Eq 3.4.6 can be solved for L1 equally well when only one phase forms, a calculation that is known as a negative flash When the iteration for L1 has converged for a single-phase system, the resulting value will be in the range L1 < or L1 > The phase compositions calculated with Eqs 3.4.4 and 3.4.5 will be equilibrium compositions that can be combined to make the single-phase mixture In other words, the single-phase mixture is a linear combination of the phase compositions, which means that the single-phase composition must lie on the extension of the line that connects the equilibrium compositions on a phase diagram We will make repeated use of the negative flash to find that line, known as a tie line, for displacement calculations Whitson and Michelsen showed that their negative flash calculation converges as long as L1 lies in the range 1 < L1 < , − Kmax − Kmin (3.4.10) where Kmax and Kmin are the largest and smallest K-values If the single-phase composition is far enough from the two-phase region that the condition 3.4.10 is not satisfied, a modified negative flash suggested by Wang [128] can be used It is based on the 34 CHAPTER CALCULATION OF PHASE EQUILIBRIUM idea that while L1 can vary over wide ranges, the equilibrium phase compositions, xij , are restricted to lie between zero and one The mole fraction of phase 1, L1 can be eliminated by solving Eq 3.4.4 written for component 1, L1 = z1 − x12 (K1 − 1) x12 (3.4.11) Substitution of Eq 3.4.11 into Eq 3.4.4 gives an expression for the phase compositions in phase in which the only unknown is x12 , xi2 = zi x12 (K1 − 1) z1 (Ki − 1) + x12 (K1 − Ki ) (3.4.12) The revised version of Eq 3.4.6 is nc nc (Ki − 1) x12 = i=1 i=1 zi x12 (Ki − 1) (K1 − 1) = z1 (Ki − 1) + x12 (K1 − Ki) (3.4.13) Eq 3.4.13 can be solved for x12 by a Newton-Raphson iteration 3.5 Phase Diagrams It will be convenient for many of the flow problems considered here to represent the solutions as a collection of compositions on a phase diagram Accordingly, we review briefly the terminology and properties of binary, ternary, and quaternary phase diagrams 3.5.1 Binary Systems Fig 3.1 is a typical phase diagram for a binary mixture at some fixed temperature above the critical temperature of component At pressure, P , liquid phase (phase 2) with mole fraction x12 of component is in equilibrium with vapor phase (phase 1) containing mole fraction x11 of component Those equilibrium compositions are connected by a tie line, along which a tie line material balance like Eq 3.4.3 applies There is one tie line in Fig 3.1 for each pressure A mixture with an overall mole fraction z1 of component between x12 and x11 forms two phases Mixtures with z1 < x12 are all liquid, while those with z1 > x11 form only vapor For a given overall composition, the mole fraction of phase present is easily determined by rearrangement of Eq 3.4.3 to be L1 = z1 − x12 x11 − x12 (3.5.1) Eq 3.5.1 is a lever rule, which states that the mole fraction vapor is proportional to the distance from the overall composition to the liquid composition locus divided by the length of the tie line A similar statement applies to systems with any number of components Eq 3.5.1 indicates that L1 ≤ for mixtures that form only liquid, and L1 ≥ for mixtures that are all vapor At the top of the two-phase region in Fig 3.1 is a critical point, at which the liquid and vapor phases, as well as all phase properties, are identical The critical point can be thought of as a tie line with zero length Because phase compositions are equal at a critical point, Eq 3.4.2 indicates that all K-values must be equal to one for a critical mixture 3.5 PHASE DIAGRAMS 35 2000 a 1800 Critical Point 1600 Pressure(psia) 1400 1200 Liquid Phase Tie Line 1000 Two Phase Region 800 600 Vapor Phase 400 200 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Mole Fraction Component Figure 3.1: Pressure-composition phase diagram for a two-component mixture A phase diagram like Fig 3.1 can be calculated with an equation of state Fig 3.2 shows such a diagram calulated with the Peng-Robinson equation for a two-component system, CO2 /decane (C10 ) system Also shown in Fig 3.2 are experimental data of Reamer and Sage [104] along with the phase diagram calculated with van der Waals equation of state Critical properties used for CO2 and C10 are given in Table 3.1, and the binary interaction parameter used in the Peng-Robinson flash calculations was δ12 = 0.102 (Table 3.2) Fig 3.2 shows that the Peng-Robinson phase compositions agree much better with the experimental observations than the van der Waals predictions They should, of course, because the value of δ12 was chosen to minimize the disagreement between calculated phase compositions and measured data Even so, with the average value of δij chosen by matching experimental data at several temperatures [18], there is some disagreement between experiment and calculation near the critical point 3.5.2 Ternary Systems A typical vapor/liquid phase diagram for a three-component system is shown in Fig 3.3, which displays phase behavior information at fixed pressure and temperature Because the mole fractions at any composition point in the diagram always sum to one, it is useful to plot equilibrium phase compositions on an equilateral triangle On such a diagram, the three mole (or volume or mass) fractions are read from the perpendicular distances from the composition point to the three sides The corners of the diagram represent 100% of the component with which the corner is labeled, and the opposite side represents the zero fraction The sides of the ternary diagram represent binary mixtures of the two components that lie on that side For gas/oil systems, the component at the top corner of the diagram is usually the lightest component, and the heavies component is usually 36 CHAPTER CALCULATION OF PHASE EQUILIBRIUM 2000 2000 b 1800 b 1800 b 1600 1600 b Pressure(psia) 1400 Experimental Data Van der Vaals Peng-Robinson 1400 b b 1000 b 800 b 600 b 600 b 400 b 400 b 0.1 1200 b 1000 b 800 0.0 b b 1200 200 b b 200 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Mole Fraction CO2 Figure 3.2: Phase behavior of the CO2 /C10 system at 160 F Experimental data shown are from Reamer and Sage [104] placed at the bottom left corner Any mixture of two overall compositions, whether or not either is single phase, must lie on a straight line that connects those compositions For example, if M moles of a mixture with overall compositions, zi1 , are mixed with N moles of mixture with compositions, zi2 , the overall composition of the resulting mixture is given by zi = (M zi1 + N zi2 ) M +N (3.5.2) Eq 3.5.2 is the equation of a straight line (the dilution line) that connects the two overall compositions The region of overall compositions in Fig 3.3 that form two phases is enclosed by the loci of liquid and vapor phase compositions, which taken together are known as the binodal curve Here again, equilibrium liquid and vapor compositions are connected by tie lines In contrast with binary systems, in which there is only a single tie line at fixed temperature and pressure, a ternary system has an infinite number of tie lines A critical point on a ternary phase diagram is often referred to as a plait point An example of a calculated (Peng-Robinson) ternary phase diagram that includes a plait point is shown in Fig 3.4 for mixtures of CO2 , butane (C4 ), and decane (C10 ) at 160 F (71 C) and 1250 psia (85 atm) Mixtures containing no C4 lie on the CO2 /C10 side of the diagram Hence, that side shows one tie line, at 1250 psia, from the pressure composition phase diagram in Fig 3.2 Also shown in Fig 3.4 are experimental data reported by Metcalfe and Yarborough [76] and by Orr and Silva [87] Again, the Peng-Robinson equation of state shows reasonable agreement with 3.5 PHASE DIAGRAMS 37 Comp.1 80 30 us oc rL 20 po Va 10 90 70 40 a Pla 60 it P 50 Liquid Locus oin t 50 60 40 70 Tie Lines 30 80 20 90 10 90 80 70 60 50 40 30 20 Comp.2 10 Comp.3 Ternary Diagram Figure 3.3: Ternary phase diagram the experimental observations Calculated vapor compositions agree quite well with experimental observations, while calculated liquid phase compositions show slightly lower CO2 mole fractions Tie line slopes are represented quite well by the equation of state Tie line slopes turn out to be important in the analysis of flows in systems with three or more components, so it is useful that the equation of state captures that behavior well 3.5.3 Quaternary Systems When four components are present, an equilateral tetrahedron can be used to plot phase compositions in much the same way an equilateral triangle is used for ternary systems Fig 3.5 is an example of a quaternary diagram for mixtures of CO2 , methane CH4 , butane (C4 ), and decane (C10 ) at 160 F (71 C) and 1600 psia (109 atm) A ternary diagram like Fig 3.4 is one of the faces of the quaternary diagram At that temperature and pressure, there are plait points in the CO2 /C4 /C10 and CO2 /CH4 /C4 ternary faces, and there is a locus of critical points in the interior of the quaternary diagram that connects those plait points The two-phase region on the CO2 /CH4 /C10 face is a band of tie lines that spans that face The locus of liquid compositions is now a surface in the interior of the diagram, as is the locus of vapor compositions Those surfaces meet at the locus of critical points, on which the compositions of liquid and vapor are identical Each composition point on the surface of vapor compositions is connected to another composition point on the surface of liquid compositions by a tie line Thus, the space enclosed by the binodal surface is densely packed with tie lines The geometry of tie lines in the four-component diagram of Fig 3.5 is worth a bit of study because displacement behavior is intimately connected with the geometry of surfaces of tie lines, as the analysis of subsequent chapters will show 38 CHAPTER CALCULATION OF PHASE EQUILIBRIUM CO2 10 b a a b b a 90 bb b b a 20 80 30 70 40 b a a ba bbb b b bb a b b 60 50 50 60 40 70 30 80 Peng-Robinson binodal curve Peng-Robinson tie lines Metalfe and Yarborough experiment Orr and Silva experiment 90 a b 20 10 90 80 70 60 50 40 30 20 C4 10 C10 Figure 3.4: Ternary phase diagram (in mole fractions) calculated with the Peng-Robinson equation of state for the CO2 /C4 /C10 system at 1250 psia (85 atm) and 160 F (71 C) Measured phase compositions reported by Metcalfe and Yarborough [76] and by Orr and Silva [87] are also shown 3.5.4 Constant K-Values ˆ Eq 3.3.17 shows clearly that φij can depend quite strongly on the composition of phase j through the parameters, am and bm , and it follows, therefore, from Eq 3.4.8 that K-values are also functions of composition For example, equilibrium K-values for a phase diagram that includes a critical point must be strong functions of composition (and, of course, temperature and pressure), because Kvalues are all equal to one at the critical point and differ significantly from one for the remainder of the two-phase region If the pressure is low enough for many gas/oil systems, however, a critical point will not be present, and K-values will depend only weakly on composition For those systems, it is reasonable to assume that K-values are constant The constant K-value limit is a useful one, because the solutions for pure convection simplify considerably, and hence, we will pause for a moment to explore the properties of the phase diagrams that result from the assumption of constant K-values If K-values are constant, then the liquid and vapor portions of the binodal curve are straight lines on a ternary diagram Consider Eq 3.4.6 with L1 = 0, which gives an expression for compositions on the liquid locus, i=1 zi (Ki − 1) = i=1 zi Ki − 3 zi = i=1 zi Ki − = (3.5.3) i=1 Because the Ki are constants, Eq 3.5.3 is the equation of a straight line (It is a bit easier to see that it is a straight line if z3 = − z1 − z2 is eliminated from Eq 3.5.3.) Similarly, when L1 = 1, 3.5 PHASE DIAGRAMS 39 CO2 • • CH4 C4 C10 Figure 3.5: Quaternary phase diagram calculated with the Peng-Robinson equation of state for mixtures of CO2 , methane, butane and decane at 160 F (71 C) and 1600 psia (109 atm) The dashed lines are tie lines that lie on the four ternary faces of the diagram the equation for the vapor phase compositions is i=1 3 c zi (Ki − 1) zi zi = zi − =1− = Ki Ki Ki i=1 i=1 i=1 n (3.5.4) Again, Eq 3.5.4 is the equation of a straight line Hence, for a ternary system with constant Kvalues, the two-phase region must be a band across the diagram that intersects the two sides with one K-value greater than one and the other K-value less than one Two phases cannot form on a side where both K-values are greater than one or both are less than one Fig 3.6, for example, shows a phase diagram for K1 = 1.5, K2 = 2.6, and K3 = 0.01 Those K-values roughly reproduce the behavior of the CO2 /CH4 /C10 face of Fig 3.5, for example The intersection points on the binary sides of the diagram can be easily obtained from Eq 3.5.3 written for two components with K-values above and below one For example, on the component 1/component side of the diagram (CO2 /C10), x12 = − K3 , K1 − K3 x11 = K1 − K3 K1 − K3 (3.5.5) For four–component systems, Eqs 3.5.3 and 3.5.4 indicate that the surfaces of vapor and liquid compositions are planes Fig 3.7 is an example of such a phase diagram for K1 = 2.5, K2 = 1.5, K3 = 0.5, and K4 = 0.05 Those values are roughly equivalent to K-values for the CO2 /CH4 /C4 /C10 system at 160 F (71 C) and 1600 psia (109 atm) in regions of the phase diagram far from the critical locus As Fig 3.7 shows, the liquid and vapor phase planes intersect all four ternary sides of the diagram when two K-values are greater than one and two are less than one If only one 40 CHAPTER CALCULATION OF PHASE EQUILIBRIUM Comp.1 10 90 20 80 30 70 40 60 50 50 60 40 70 30 80 20 90 10 90 80 70 60 50 40 30 20 Comp.2 10 Comp.3 Figure 3.6: Ternary phase diagram for a system with constant K-values, K1 = 1.5, K2 = 2.6, and K3 = 0.01 K-value is less than one (or only one is greater than one), the two–phase region does not intersect the ternary face on which all K-values are greater than (or less than) one 3.6 Additional Reading Thermodynamic Background, State Functions, Chemical Potential and Fugacity van Ness and Abbott’s text, Classical Thermodynamics of Nonelectrolyte Solutions [123], gives concise but readable derivations of the material in Sections 3.1 and 3.2 The various equations of state in widespread use are reviewed in considerable detail by Walas in Phase Equilibria in Chemical Engineering [126] A shorter description of phase equilibrium calculations with an equation of state is given by Lake in Chapter of Enhanced Oil Recovery [62] A description of flash calculations and K-values can also be found there A detailed account of the fundamental thermodynamic structure of phase equilibrium as well as the many computational issues that arise in solving practical problems is given by Michelsen and Mollerup [80] Phase Diagrams Binary and ternary phase diagrams are described in detail by Walas [126, Chapter 5] Properties of ternary diagrams as well as approximate representations for two-phase regions in ternary diagrams are described by Lake [62, Chapter 4] Examples of phase diagrams for systems with constant K-values are reported by Dindoruk [19] 3.7 Exercises Use the Peng-Robinson equation of state to calculate the molar volume and molar density of CH4 and C10 at 160 F and 1600 psia 3.7 EXERCISES 41 2(CO2) 3(C4) 1(CH4) 4(C10 ) Figure 3.7: Quaternary phase diagram for a system with constant K-values, K1 = 2.5, K2 = 1.5, K3 = 0.5, and K4 = 0.05 Solid lines outline the planes of liquid and vapor compositions Dashed lines are tie lines in three of the four faces Consider a binary system in which x1L = 0.6 and x1V = 0.99 The overall composition is z1 = 0.75 The mass density of the liquid phase is ρL = 0.7 g/cm3, and the vapor phase density is ρV = 0.25 g/cm3 The molecular weights of components and are 44 and 142 respectively Calculate the mole fractions of the liquid and vapor phases Also calculate the volume fractions of the liquid and vapor phases Consider a two-phase mixture in a ternary system in which the vapor phase has composition, y1 = 0.95, y2 = 0.04, and y3 = 0.01 The overall composition of the mixture is z1 = 0.8, z2 = 0.1, z3 = 0.1 Calculate the composition of the liquid phase Calculate the phase compositions and volume fractions for a binary system with K-values (based on volume fractions), K1 = 1.5 and K2 = 0.1 The overall volume fraction is 0.6 Use the Peng-Robinson equation of state to calculate the compositions of the equilibrium liquid and vapor phases for a two-phase mixture of CH4 and C10 at 160 F and 1600 psia A ternary mixture has overall composition (mole fractions), z1 = 0.6, z2 = 0.2, z3 = 0.2 The K-values (based on mole fractions) are K1 = 3.0, K2 = 1.5, K3 = 0.1 Calculation the compositions and volume fractions for the equilibrium liquid and vapor phases Derive the fugacity expression for the Redlich-Kwong equation of state: P = am RT − V − bm T 1/2V (V + bm ) (3.7.1) 42 CHAPTER CALCULATION OF PHASE EQUILIBRIUM nc √ am = xi (3.7.2) nc bm = xi bi (3.7.3) 5/2 = 0.42748R2Tc Pc (3.7.4) 0.08664RTc Pc (3.7.5) bi = Chapter Two-Component Gas/Oil Displacement In this chapter, we use a simple two-phase flow problem, displacement of an oil by a gas, to illustrate application of the method of characteristics and to develop many of the themes that recur, with variations, in the more complex multicomponent flows considered subsequently The introduction given here to the method of characteristics is by no means comprehensive For a much more careful development of the ideas behind the mathematical techniques applied here, see Volumes I and II of First Order Partial Differential Equations by Rhee, Aris and Amundson [106, 107] (see Chapters and especially) Their description of the mathematical structure of many closely related problems of chromatography and packed bed reactors weaves a rich tapestry of mathematics and engineering science that should be high on the reading list of students of first order equations Much useful material can also be found in Courant and Hilbert’s Methods of Mathematical Physics [14] and in Jeffrey’s Quasilinear Hyperbolic Systems and Waves [40] It is with those sources as companions that this version of the theory is offered In the flow considered in this chapter, two components are present Consider a one-dimensional flow in which a gas, composed primarily of a light component such as methane (CH4 ) or CO2 , displaces an oil containing some liquid hydrocarbon, say decane (C10 ) The components are assumed to have limited mutual solubility: some of the C10 vaporizes into the gas phase, and some of the CH4 or CO2 dissolves in the liquid phase This problem is a modest generalization [32] of the well-known problem of displacement of oil by water (solved first by Buckley and Leverett [10], see Lake [62, pp 130-142] for a complete description of the problem in a form similar to that used here, or see Dake [17, pp 356-372] for the conventional description) The two-component problem is relatively straightforward because the entire displacement takes place on a single tie line (Recall that the pressure at which the phase equilibrium is evaluated is taken to be constant – see Fig 3.1.) The solution to the two-component flow reappears as a portion of the solution in multicomponent problems, and hence, it is outlined in some detail here We begin by solving Eq 2.4.4 for flow of two mutually soluble components with no volume change on mixing The basic solution is derived in Section 4.1 The properties of shocks, discontinuities in composition and saturation that arise in the solutions, are examined in Section 4.2 Section 4.3 describes how the behavior of the solution depends on the initial and injection conditions Finally, Section 4.4 examines how displacement behavior changes when components change volume as they transfer between phases 43 44 4.1 CHAPTER TWO-COMPONENT GAS/OIL DISPLACEMENT Solution by the Method of Characteristics For the two-component problem, Eqs 2.4.4 reduce to a single quasilinear equation, ∂C1 ∂F1 + = 0, ∂τ ∂ξ (4.1.1) where the overall composition, C1 , in the two-phase region is given by Eq 2.4.5, C1 = c11 S1 + c12 (1 − S1 ), (4.1.2) and the overall fractional flow of component in the two-phase region is Eq 2.4.6, F1 = c11 f1 + c12 (1 − f1 ) (4.1.3) The equilibrium phase compositions, c11 and c12 are known and fixed by the assumption of local chemical equilibrium For any mixture that forms only a single phase (C1 < c12 or C1 > c11 ), F1 = C1 (4.1.4) Rearrangement of Eq 4.1.2, which is really just another tie-line material balance like Eq 3.4.3 or Eq 3.5.1, shows that S1 = C1 − c12 c11 − c12 (4.1.5) Thus, the saturation is a function only of the overall composition, C1 The overall fractional flow, F1 , is also a function of C1 only For horizontal flow, for example, Eq 2.2.6 reduces to f1 = kr1 /µ1 kr1 /µ1 + kr2 /µ2 (4.1.6) The phase viscosities are fixed, because the phase compositions are fixed, and the phase relative permeabilities are generally assumed to be functions of saturation only Therefore, the value of C1 determines the value of F1 , because F1 depends only on f1 , which depends only on S1 , which in turn depends only on C1 Because F1 is a function of C1 only, Eq 4.1.1 can be written dF1 ∂C1 ∂C1 + = 0, ∂τ dC1 ∂ξ (4.1.7) Because C1 is a function of ξ and τ , we can also write an expression for the ordinary derivative of C1 , ∂C1 dτ ∂C1 dξ dC1 = + dη ∂τ dη ∂ξ dη (4.1.8) where η is a parameter Term by term comparison of Eq 4.1.8 with Eq 4.1.7 indicates that 4.1 SOLUTION BY THE METHOD OF CHARACTERISTICS dC1 dη dτ dη dξ dη 45 = 0, (4.1.9) = 1, (4.1.10) = dF1 dC1 (4.1.11) Eqs 4.1.9–4.1.11 are known as the characteristic equations Integration of Eqs 4.1.9–4.1.11 gives expressions for C1 , τ , and ξ in terms of the parameter η In other words, τ (η) and ξ(η) are characteristic curves along which the derivative of C1 is dC1 /dη Eq 4.1.9 indicates that the derivative is zero, so C1 does not change as η changes along the curves, and therefore, the characteristic curves are curves along which C1 is constant Elimination of η from the expressions for τ and ξ obtained by integrating Eqs 4.1.10 and 4.1.11 gives the trajectory in time and space for some constant value of C1 If the value of C1 is fixed, then the value of dF1 /dC1 is also fixed Hence, Eqs 4.1.10 and 4.1.11 can be integrated easily, and η can be eliminated to obtain a relationship between τ and ξ, ξ= dF1 τ + ξ0 , dC1 (4.1.12) dF where ξ0 is the initial position of the composition C1 for which dC1 is evaluated Eq 4.1.12 is the equation of a straight line, and hence, the characteristic curves are straight lines The velocity at which the composition C1 propagates along a characteristic curve is dF1 dξ = , dτ dC1 (4.1.13) and that velocity remains constant for a given value of C1 The velocity at which a given composition propagates is often called the wave velocity of that composition When two phases with different compositions are flowing simultaneously, the wave velocity of the overall composition of the twophase mixture differs from the physical flow velocity of any of the phases The wave velocity indicates how fast that overall composition moves, not how fast the individual phases, which have compositions quite different from the overall composition, are moving Eqs 4.1.12 and 4.1.13 give the impression that the construction of a solution for a twocomponent gas displacement problem is quite simple For each initial composition C1 , the propadF gation velocity dC1 is evaluated, and all subsequent spatial positions of that composition are given by Eq 4.1.12 So far, however, we have not considered how the discontinuity between the initial and injection compositions, present initially at the inlet, propagates through the porous medium To make the illustration more concrete, we consider the following specific problem: pure cominj ponent (C1 = 1) displaces a single-phase mixture of components and containing volume init percent component (C1 = 0.05) The equilibrium vapor phase contains 95 percent component (c11 = 0.95), and the equilibrium liquid phase contains 20 percent component (c12 = 0.20) We also assume that the relative permeability functions are given by the following simple relations: 46 CHAPTER TWO-COMPONENT GAS/OIL DISPLACEMENT S1 − Sgc − Sgc − Sor kr1 = , kr1 = 0, (4.1.14) S1 < Sgc , − S1 − Sor − Sgc − Sor , (4.1.15) S1 > − Sor , kr1 = 1, kr2 = Sgc < S1 < − Sor , (4.1.16) Sgc < S1 < − Sor , (4.1.17) kr2 = 0, − S1 < Sor , (4.1.18) kr2 = 1, − S1 > − Sgc (4.1.19) Relative permeability functions that are functions of phase saturations raised to a power are known as Corey type curves [13] In Eqs 4.1.14–4.1.19, Sgc represents a critical gas saturation, below which the vapor phase has zero relative permeability Similarly, Sor is a residual oil saturation When the liquid saturation is less than Sor , the liquid phase relative permeability is zero For this example, we assume that Sor = 0.10 and Sgc = 0.05 Substitution of Eqs 4.1.14-4.1.19 into Eq 4.1.6 gives f1 = 0, f1 = f1 (S1 − Sgc = 1, S1 < Sgc , (S1 − Sgc , + (1 − S1 − Sor )2 /M )2 (4.1.20) Sgc < S1 < − Sor , (4.1.21) S1 > − Sor , )2 (4.1.22) where M is the viscosity ratio, µ2 /µ1 For this example, M = Fig 4.1 is a plot of F1 as a function of C1 (see Fig 4.12 for a detailed description of the key points on this plot) Within the two-phase region in Fig 4.1, the fractional flow function has the typical S-shape often seen in measured fractional flow curves Differentiation of Eqs 4.1.3 and 4.1.5 with respect to C1 shows that df1 df1 dS1 df1 dF1 = (c11 − c12 ) = (c11 − c12 ) = (4.1.23) dC1 dC1 dS1 dC1 dS1 Eqs 4.1.20-4.1.22 can be differentiated easily to obtain expressions for df1 /dS1 The resulting values of dF1 /dC1 , plotted in Fig 4.2, are dF1 = 1, dC1 dF1 = 0, dC1 df1 dF1 = , dC1 dS1 dF1 = 0, dC1 dF1 = 1, dC1 < C1 < c12 , S1 = 0, (4.1.24) c12 < C1 < Sgc (c11 − c12 ) + c12 , < S1 < Sgc , (4.1.25) Sgc (c11 − c12 ) + c12 < C1 < Sor (c12 − c11 ) + c11 , Sgc < S1 < − Sor , (4.1.26) Sor (c12 − c11 ) + c11 < C1 < c11 , − Sor < S1 < 1, (4.1.27) c11 < C1 < 1, S1 = (4.1.28) 4.1 SOLUTION BY THE METHOD OF CHARACTERISTICS 47 Overall Fractional Flow of Component 1, F1 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 Overall Volume Fraction of Component 1, C1 Figure 4.1: Overall fractional flow function, F1 (C1 ) The wave velocity is one in the single-phase regions at high and low C1 concentration, is zero in the small regions at either end of the two-phase region where the relative permeability of one of the phases is zero, and is a nonlinear function of composition within the two-phase region Fig 4.2 indicates that more than one composition can have a given wave velocity, a fact that will have important implications for the construction of solutions The behavior of characteristics is commonly plotted on t-x (or in this case, τ -ξ) diagrams Fig 4.3 shows the characteristics associated with the initial composition (C1 = 0.05) They emanate from the τ = axis for ξ > 0, and they all have the same unit slope, because the initial composition is a constant value in the single-phase region For compositions in the single-phase region, differentiation of Eq 4.1.4 according to Eq 4.1.13 leads immediately to the conclusion that single phase compositions propagate with unit velocity The characteristics associated with the injected composition are shown in Fig 4.4 Again, that composition is in the single-phase region, and hence, the characteristic lines are parallel, and the compositions on them all have unit wave velocity The stage is now set for evaluating the effect of the step change in composition that occurs at the inlet, and we ask next how the compositions associated with that discontinuity propagate through the porous medium The characteristics associated with the discontinuity initially present at the origin are shown in Fig 4.5 That figure is drawn based on the assumption that all the compositions that lie between the upstream and downstream compositions at the discontinuity 48 CHAPTER TWO-COMPONENT GAS/OIL DISPLACEMENT 2.5 Wave Velocity, dF1/dC1 2.0 1.5 1.0 0.5 0.0 0.0 0.2 0.4 0.6 0.8 1.0 Overall Volume Fraction of Component 1, C1 Figure 4.2: Derivative of the overall fractional flow function, dF1 /dC1 propagate, each with the wave velocity appropriate to the specific intermediate composition In other words, the discontinuity can be viewed as the limit as → of a change in composition from the upstream value, C1 = 1, to the downstream value, C1 = 0.05 over a distance In the limit the characteristics for values of C1 between 0.05 and 1.0 all emanate from the origin Construction of the solution is made more difficult by two facts illustrated in Figs 4.2 and 4.5 First, all but one of the characteristics associated with the discontinuity have wave velocities different from one, and hence the characteristics associated with the initial discontinuity intersect those of the initial and injection compositions Second, there are at least two compositions that have any given wave velocity between zero and the maximum value, as Figs 4.2 and 4.5 illustrate Both conditions seem to imply that more than one composition will exist at a single location, a physical impossibility The inconsistency is resolved by the propagation of discontinuities known as shocks, the subject of the next section 4.2 Shocks When characteristics cross on a t-x diagram, a shock must form The next task then is to find the compositions that form on either side of the shock and to determine how the shock propagates The trajectory of a shock is found from a material balance equation similar to Eq 4.1.1, except that the balance is an integral one, because the differential version of the material balance is no 4.2 SHOCKS 49 1.0 0.8 0 C 05 1= C 05 1= 05 C 1= 0.4 C 05 1= τ C 05 1= 05 0.6 C 1= C 05 1= 0.2 0.0 0.0 0.4 0.2 0.6 0.8 1.0 ξ Figure 4.3: Characteristics for the initial composition 1= C 1= C 1= C 0.8 1= C 1= 1.0 1= C τ 1= C 1= C 0.6 C 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 ξ Figure 4.4: Characteristics for the injection composition 50 CHAPTER TWO-COMPONENT GAS/OIL DISPLACEMENT 845 =0 37 7,0 82 326, C 1= τ 44 5, 82 C1 0.6 C1 =0 0.8 C1=0.282,0 862 1.0 0.4 628 C 1= 0.2 0.0 0.0 0.4 0.2 0.6 0.8 1.0 ξ Figure 4.5: Characteristics for the discontinuity present initially at the inlet longer appropriate because the spatial derivatives that appear in it are not defined Consider the flow situation illustrated in Fig 4.6 At time, τ , the shock is located at distance, ξ, and it moves during a period, ∆τ , to a new position, ξ + ∆ξ The composition of the fluid on the upstream side of the shock is Ci = CiII , and the composition on the downstream side is Ci = CiI Now we write a material balance for component i for a control volume that includes the porous medium between ξ and ξ + ∆ξ Because there is no accumulation of material at the shock, the change in the amount of component i present in the control volume (per unit area of porous medium) is exactly balanced by the net inflow of component i, ∆ξ(CiII − CiI ) = ∆τ (FiII − FiI ) (4.2.1) In the limit as ∆ξ and ∆τ go to zero, the velocity of the shock is obtained by rearrangement of Eq 4.2.1, Λ= F II − FiI dξ = iII dτ Ci − CiI (4.2.2) Eq 4.2.2, called a jump condition or a Rankine-Hugoniot relation, is an integral version of the ... component 1/component side of the diagram (CO2 /C10), x12 = − K3 , K1 − K3 x11 = K1 − K3 K1 − K3 (3. 5.5) For four–component systems, Eqs 3. 5 .3 and 3. 5.4 indicate that the surfaces of vapor and liquid... that Eqs 3. 4.4 and 3. 4.5 each sum to unity, so that nc nc xi1 − i=1 nc xi2 = i=1 i=1 zi (Ki − 1) = + L1 (Ki − 1) (3. 4.6) 3. 4 FLASH CALCULATION 33 Eq 3. 4.6, which is known as the Ratchford-Rice equation... nc (3. 4.7) Rearrangement of Eq 3. 4.7 shows that the equilibrium K-value is just the ratio of partial fugacity coefficients, Ki = ˆ xi1 φi2 = , ˆ xi2 φi1 i = 1, nc (3. 4.8) The form of Eq 3. 4.8