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 equilib-rium vapor phase is 0.95.. The initial
Trang 1Binary 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 C10by
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
1 Characteristic curves Consider the equation
∂C
∂t + C
2∂C
The initial composition is C init = 0.1, and the injection composition is C inj = 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
2 Gas dissolution Consider a laboratory core which contains initially water that is saturated with CO2 in equilibrium with gas at the critical gas saturation, S gc = 0.05 The equilibrium volume fraction of CO2dissolved 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
3 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 S gc = 0.1, S or = 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 equilib-rium 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?
4 Displacement with two-phase initial and injection mixtures Consider the fluid system of problem 3 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
5 Effect of volume change on shock speed Consider the situation outlined in problem 2 De-termine 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
Trang 2Table 4.4: Equilibrium Phase Compositions and Fluid Properties
(gmol/l) (g/cm3) (cp)
Initial Oil 0 1 4.881 0.6945 -Equil Liq 0.3519 0.6481 6.481 0.6342 0.262 Equil Vap 0.9964 0.0036 4.316 0.0712 0.015 Injected Gas 1 0 4.298 0.0690
-volume of CO2 in the vapor phase Assume that the CO2 in the vapor phase has a
compo-nent 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.
6 CH4displacing C10 Consider the fluid property data given in Table 4.4 for the CH4/C10system
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 compo-nent volume no matter what phase the compocompo-nent 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
Trang 3Ternary 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 O 1 is displaced by gas with composition G 1 Mixtures of
oil O 1 with gas G 1 in the first mixing cell lie on the dilution line that connects O 1 to G 1 on the ternary diagram in Fig 5.1 Suppose that after mixing some oil with gas in the first cell, the
overall composition is M 1 That mixture splits into two phases with compositions V 1 and L 1 Now
assume that the less viscous vapor phase with composition V 1 moves to the next downstream cell,
where it mixes with fresh oil Those mixtures lie on the dilution line that connects V 1 with the oil
composition, O 1 If the new overall composition is M 2, then the phases that form in the second
cell have compositions L 2 and V 2 But when the vapor V 2 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 do 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 V 1 is richer in component 2 than the original injection
gas is, and mixture V 2 is richer still Oil O 1 is rich enough in component 2 that miscibility
develops If, however, the oil had had composition L 2 (or any mixture on the extension into the
single-phase region of the tie line that connects V 2 and L 2 ), mixture of V 2with fresh oil would have given another mixture on the same tie line In that case, the enrichment of the vapor phase with component 2 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 G 2 displaces oil with composition O 2 Mixtures of original oil with gas in the first
mixing cell give composition M 1 That mixture splits into phases with compositions V 1 and L 1 Here again, the vapor phase is assumed to move ahead and contact fresh oil, but this time we focus
73
Trang 4C2
a
a
a
a
a
a
a
a
a
O1
G1
L1 L2
M1
M2
V1
V2
Figure 5.1: Mixing cell representation of a vaporizing gas drive
on what happens in the first mixing cell There, liquid phase with composition L 1 mixes with
new injected gas Those mixtures lie on the dilution line that connects L 1 with G 2 If the new
composition after mixing in the first cell is M 2 , then the resulting phase compositions are V 2 and
L 2 When liquid phase L 2 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 L 2 -V 2 tie line), the enrichment of the oil in the first cell with component 2 condensing from the gas would have stopped when that tie line was reached, because additional
mixtures of fresh gas with liquid L 2 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]
Trang 5C2
C3
a
a
a
L1 L2
M1 M2
V1
V2
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
The conservation equations for a three-component system without volume change are
∂C1
∂τ +
∂F1
∂C2
∂τ +
∂F2
where
C i = c i1S1+ c i2(1− S1), i = 1, 2, (5.1.3) and
F i = c i1f1+ c i2(1− f1), i = 1, 2. (5.1.4)
Trang 6C2
C3
(a) Composition Path
A Oil Composition
Gas Composition B
•
•• •
• •
•
ξ
Oil Composition A Injection Gas
(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
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 1 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 1 and 2,
α = c21− c22
c11− c12, φ =
c22c11− c21c12
Trang 7Similarly, elimination of f1 from Eqs 5.1.4 indicates that
Substitution of the expressions for C2 and F2 into Eq 5.1.2 gives
C1dα dη +dφ
dη
∂η
∂τ +
F1dα dη + dφ
dη
∂η
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 satu-rations, 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
∂τ +
∂F1
∂C1
∂C1
∂ξ +
∂F1
∂η
∂η
Eqs 5.1.8 and 5.1.9 can be rearranged into the form,
∂u
∂τ + A(u)
∂u
where u = (C1, η) T and
A(u) =
∂F 1
∂C1
∂F1
∂η
0 F1+p
C1+p
with p given by
p =
dφ dη dα dη
= dφ
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
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
α B = α A+dα
Trang 8φ B = φ A+dα
where again, η is a parameter that determines the tie line The point at which the two tie lines intersect, (C1x , C2x) can be found by solving Eqs 5.1.13 and 5.1.14 using 5.1.15 and 5.1.16 The
result is
C1e=−
dφ dη dα dη
Thus, C1e 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,
C1e , 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
To solve Eq 5.1.10 we calculate the trajectory of a fixed composition (fixed values of C1and 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
dC1 = ∂C1
∂τ dτ +
∂C1
∂τ dτ +
∂η
Eqs 5.1.18 and 5.1.19 can be rearranged to give
∂C1
∂τ = − ∂C1
∂ξ
dξ
dτ =−λ ∂C1
∂η
∂τ = − ∂η
∂ξ
dξ
dτ =−λ ∂η
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
∂F 1
∂C1 − λ ∂F1
∂η
0 F1+p
C1+p − λ
∂C
1
∂ξ
∂η
∂ξ
Eq 5.1.22 is an eigenvalue problem that has nontrivial solutions if and only if
det
∂F 1
∂C1 − λ ∂F1
∂η
0 F1+p
C1+p − λ
=
∂F1
∂C1 − λ F1+ p
C1+ p − λ = 0. (5.1.23)
Trang 9C2
C3
Envelope Curve
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Figure 5.4: Tie-line construction from an envelope curve for a ternary system with constant
K-values: 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 composi-tion 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
λ t= ∂F1
∂C1, λ nt =
F1+ p
(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 do 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
Trang 10t=
1 0
, nt=
⎛
⎝ λ nt1−λ t
∂F1
∂η
⎞
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, λ t = λ nt = 1, and Eq 5.1.23 is satisfied for any composition variation Therefore, there are no discrete composi-tion direccomposi-tions in the single-phase region, and composicomposi-tion variacomposi-tions in any direccomposi-tion are allowed For compositions in the two-phase region, however, there are two separate composition directions
associated with e t and e nt Next we consider the properties of those eigenvector directions
The existence of discrete directions given by the eigenvectors indicates that arbitrary compo-sition variations cannot satisfy the conservation equations Instead variations in compocompo-sition 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,
∂u
∂τ =−w ξ
and
∂u
∂ξ = w
1
in Eq 5.1.10 gives
A(w) − ξ
τ I
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 > 0 [39], as direct evaluation of the partial derivatives shows.
... discrete composi-tion direccomposi-tions in the single-phase region, and composicomposi-tion variacomposi-tions in any direccomposi-tion are allowed For compositions in the two-phase region, however,... p(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... 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