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Transport in Porous Media

, Volume 126, Issue 3, pp 683–699 | Cite as

A Collection of Analytical Solutions for the Flash Equilibrium Calculation Problem

  • Jiří MikyškaEmail author
Article
  • 61 Downloads

Abstract

We describe an interesting family of closed-form solutions for the flash equilibrium calculation problem. These solutions can be used as benchmark solutions for verification of numerical solvers of the flash equilibrium problem for multicomponent mixtures. To obtain a problem possessing an analytical solution, we consider a special form of the free energy. Although this form of the free energy is artificial, it captures qualitatively several features that are present in the realistic cases too. The procedure is first illustrated on a 1-D two-phase case and is further generalized to multicomponent mixtures in two and more phases, and also to a problem including the capillary pressure effect.

Keywords

Phase equilibrium problem Multicomponent mixtures Flash equilibrium calculation Analytical solution Closed-form solution Capillary pressure effect 

1 Introduction

Phase stability testing and phase equilibrium calculation are important problems in reservoir simulation with many applications including enhanced oil recovery or \(\hbox {CO}_2\) sequestration. Most of the existing codes use a kind of the so-called PTN-flash, i.e., calculation of phase equilibrium of a multicomponent mixture in a system with prescribed pressure, temperature, and overall mole numbers (or mole fractions) (Michelsen 1982a, b; Michelsen and Mollerup 2004; Firoozabadi 1999, 2015). Recently, alternative flash formulations have received some attention including the VTN-formulation (with prescribed total volume, temperature, and mole numbers) (Mikyška and Firoozabadi 2012; Jindrová and Mikyška 2013, 2015a, b; Polívka and Mikyška 2014; Michelsen 1999; Castier 2014) and UVN formulation (prescribed total internal energy, volume, and mole numbers) (Castier 2010; Qiu et al. 2014; Saha and Carroll 1997; Castier 2009). The development of alternative flash formulations can be partially motivated by their relevance for modeling unconventional reservoirs, in which capillary effects are strong and cannot be neglected. For a formulation of the flash equilibrium calculation including the capillary effects in the VTN-settings, we refer the reader to Kou and Sun (2018).

With the development of new formulations of the equilibrium flash problem and their new implementations, we are faced with the problem of how to verify the correct function of the implemented code. In other fields of applied mathematics, say computational fluid dynamics or flow and transport in porous media, the common practice is to test a numerical code on a simple problem with known analytical solution (Buckley and Leverett 1942; McWhorter and Sunada 1990; Chen et al. 1992; McWhorter and Sunada 1992; Fučík et al. 2007, 2008, 2016). The numerical solution obtained by a numerical model is compared to the analytical solution and convergence of the numerical solutions toward the analytical one can be established (Fučík and Mikyška 2011). To the best of author’s knowledge, it seems that in the context of phase equilibrium calculations such a comparison of numerically computed solutions with analytical benchmark solutions has never been done. This is probably caused by the fact that the development of flash equilibrium calculators is directed toward models using cubic or more complex realistic equations of state, while the flash equilibrium cannot be solved analytically even for a two-component system described by the van der Waals equation of state.

The main goal of this paper is to develop a collection of suitable problems with known analytical (or closed-form) solutions which can be used as benchmark solutions when testing convergence of the numerical solutions computed by flash equilibrium calculation codes. The procedure will be based on a unified formulation of the phase equilibrium problem described recently in Smejkal and Mikyška (2018), which can include all the three formulations (PTN, VTN, and UVN) mentioned above. To obtain an analytical solution, we assume that the energy function in this formulation is piecewise quadratic. Although such a form does not correspond to realistic equations of state, it mimics some important features of realistic free energies in the PTN-flash problem, which is discussed in the text. Next, we describe the benchmark problem in the simplest case—1D-flash with two phases. The solution procedure is then generalized to more dimensions and more phases. Other possible generalizations are discussed in the text. Then, we introduce another benchmark solution whose free energy function is a polynomial of order 4. As this function is smooth, this problem can be used for testing VTN-flash solvers. Moreover, the VTN-formulation allows for an easy formulation of the phase equilibrium problem with capillary pressure effect, which may play an important role if the fluid is confined in a nanoporous medium. We present an analytical solution for a simplified VTN-flash problem including capillarity in the last section.

2 Unified Formulation of the Phase Equilibrium Problem

Given a function f defined on a convex domain \(D\subset \mathbb {R}^n\) and a point \(\mathbf{x}^*\in D\), we are searching for \(p\in \mathbb {N}\) and affine independent vectors \(\mathbf{x}_1,\dots ,\mathbf{x}_p\in D\) and coefficients \(\alpha _1,\dots ,\alpha _p>0\) such that
$$\begin{aligned} \sum _{i=1}^p \alpha _i f(\mathbf{x}_i) \rightarrow \min , \end{aligned}$$
(1)
while satisfying the following constraints
$$\begin{aligned} \sum _{i=1}^p \alpha _i\mathbf{x}_i&= \mathbf{x}^*, \end{aligned}$$
(2a)
$$\begin{aligned} \sum _{i=1}^p \alpha _i&= 1. \end{aligned}$$
(2b)
According to Smejkal and Mikyška (2018), this general formulation includes the above-mentioned PTN-, VTN-, and UVN-formulations. For example, for the VTN-flash, f is the Helmholtz free energy density A / V written as a function of the molar concentrations \(N_i/V\) of individual components (temperature is assumed to be constant), \(\mathbf{x}_i\) is the vector of molar concentrations of individual components in phase i, n is equal to the number of components, and \(\alpha _i\) is the volume fraction (or saturation) of phase i, and \(\mathbf{x}^*\) is the prescribed vector of overall concentrations of all components in the mixture. For realistic equations of state, such as the Peng–Robinson equation (Peng and Robinson 1976), the function f is a smooth (at least twice continuously differentiable) function which is bounded from below. For the PTN-flash, f is the Gibbs free energy per one mole G / N written as a function of \(n_c-1\) independent mole fractions where \(n_c\) is the number of mixture components (pressure and temperature are assumed to be constant). Obviously, the case \(n_c=1\) is excluded by this formulation. The flash dimension n is thus equal to \(n_c-1\). The coefficient \(\alpha _i\) is now the molar fraction of phase i, and finally \(\mathbf{x}^*\) is the vector of prescribed overall mole fractions of the \(n_c-1\) independent components. Compared to the VTN-flash, in the PTN-formulation, there is a complication caused by the fact that for a given pressure, temperature, and mole fractions, the molar volume of the system is not given uniquely. One has to find all roots of the equation of state, where different roots correspond to different phases and the root with the lowest value of the Gibbs free energy of the system is accepted. Therefore, the definition of the Gibbs free energy has the following form
$$\begin{aligned} f(\mathbf{x}) = \min _c\sum _{i=1}^{n_c} x_i\mu _i(cx_1,\dots ,cx_{n_c}), \end{aligned}$$
(3)
where the minimum is taken over all values of c such that \(P^{(EOS)}(cx_1,\dots ,cx_{n_c})=P^*\). In the equations above, \(\mu _i\) denotes the chemical potential of component i, which can be derived from a given equation of state \(P^{(EOS)}\) (Firoozabadi 1999), \(\mathbf{x}=(x_1,\dots ,x_{n_c-1})^T\), and \(x_{n_c}=1-x_1-\dots -x_{n_c-1}\). Both functions \(P^{(EOS)}\) and \(\mu _i\) additionally depend on temperature, but this dependence is not indicated as the temperature in the PTN-flash is assumed to be constant. In the PTN-flash, the function f is also bounded from below, continuous, but because of the root-selection procedure [minimization in Eq. (3)] only piecewise continuously differentiable. The points in which f is not differentiable are exactly the points at which different branches corresponding to different phases intersect. This form of the Gibbs free energy gave the motivation for the family of benchmark problems described below.

Fortunately, the points of non-differentiability are inside the unstable region. The stable phases occur at the points at which f is differentiable. Therefore, using the unified formulation, the following first-order necessary conditions for optimality can be derived:

Theorem 1

Let \(\alpha _1,\dots ,\alpha _p\in (0,1)\) and \(\mathbf{x}_1,\dots ,\mathbf{x}_p\) be the optimal solution of (1) satisfying the constraints (2). Then,
$$\begin{aligned} \nabla f(\mathbf{x}_1) = \nabla f(\mathbf{x}_2)&= \cdots = \nabla f(\mathbf{x}_p), \end{aligned}$$
(4a)
$$\begin{aligned} \nabla f(\mathbf{x}_1)\cdot \mathbf{x}_1 - f(\mathbf{x}_1) = \nabla f(\mathbf{x}_2)\cdot \mathbf{x}_2 - f(\mathbf{x}_2)&= \cdots = \nabla f(\mathbf{x}_p)\cdot \mathbf{x}_p - f(\mathbf{x}_p). \end{aligned}$$
(4b)
According to theorem of Caratheodory (Locatelli and Schoen 2013) ,1\(p\le n+1\). For the VTN-flash, \(\nabla f(x)\) is the vector of chemical potentials of all components, and equations (4a) state the equality of chemical potentials of each component in all phases. From the definition of the Helmholtz free energy density, it follows that \(\nabla f(\mathbf{x})\cdot \mathbf{x} - f(\mathbf{x})\) is equal to minus pressure, and thus, equation (4b) states the equality of pressures in all phases. For the PTN-flash, it follows from the Gibbs–Duhem relation that
$$\begin{aligned} \frac{\partial f}{\partial x_i}(\mathbf{x}) = \mu _i(\mathbf{x}) - \mu _{n_c}(\mathbf{x}),\qquad i=1,\dots ,n_c-1, \end{aligned}$$
and
$$\begin{aligned} \nabla f(\mathbf{x})\cdot \mathbf{x} - f(\mathbf{x}) = -\mu _{n_c}(\mathbf{x}), \end{aligned}$$
therefore both equations together imply equality of chemical potentials of each component in all phases. The necessary conditions will be used in the derivation of the analytical solution.

3 Benchmark Solution for the 1D Two-Phase Flash

Let us take \(n=1\) and consider a function f in the following form
$$\begin{aligned} f(x) = \min \{f_1(x), f_2(x)\}, \end{aligned}$$
(5)
where
$$\begin{aligned} f_i(x) = a_ix^2 + b_ix + c_i,\qquad i=1,2, \end{aligned}$$
(6)
for some \(a_1, a_2>0\), and \(b_1,b_2,c_1,c_2\in \mathbb {R}\). The motivation for this form of the free energy is that it is sufficiently simple to allow derivation of analytical solutions and the formula (5) mimics the root-selection procedure (3) that is performed in the PTN-flash calculation. As \(a_i>0\), both parabolas described by \(f_i\) are convex for \(i=1,2\). In order to have f non-convex, parameters \(a_i\), \(b_i\), and \(c_i\) have to be selected so that the graphs of \(f_1\) and \(f_2\) intersect each other at some point. The intersection point is given by the following equation
$$\begin{aligned} f_1(x)-f_2(x) = (a_1-a_2)x^2 + (b_1-b_2)x + c_1-c_2 = 0. \end{aligned}$$
(7)
If \(a_1=a_2\) and \(b_1\ne b_2\), then the above equation is linear and has only 1 solution \(P=\frac{c_2-c_1}{b_1-b_2}\). For \(a_1\ne a_2\), the equation is quadratic and may have zero, one, or two solutions depending on the sign of the discriminant
$$\begin{aligned} D(a_1,a_2,b_1,b_2,c_1,c_2):=(b_1-b_2)^2-4(a_1-a_2)(c_1-c_2). \end{aligned}$$
(8)
If \(D\le 0\), then one of the parabolas lies above the other one (or touches at a point if \(D=0\)), and the resulting f is convex. No two-phase region appears in this case. If \(D>0\), then we have two intersection points given by
$$\begin{aligned} P_{1,2} = \frac{b_2-b_1\pm \sqrt{D}}{2(a_1-a_2)}, \end{aligned}$$
(9)
around each of which there is a two-phase region. For each two-phase region, it remains to find the phase properties, i.e., the points \(x_1\ne x_2\in \mathbb {R}\) such that optimality conditions (1) are satisfied for \(p=2\). These conditions are rewritten as
$$\begin{aligned} f_1'(x_1)&= f_2'(x_2), \end{aligned}$$
(10a)
$$\begin{aligned} x_1f_1'(x_1)-f_1(x_1)&= x_2f_2'(x_2)-f_2(x_2), \end{aligned}$$
(10b)
or
$$\begin{aligned} 2a_1x_1 - 2a_2x_2&= b_2-b_1, \end{aligned}$$
(11a)
$$\begin{aligned} a_1x_1^2 - a_2x_2^2&= c_1-c_2. \end{aligned}$$
(11b)
Expressing \(x_1\) from Eq. (11a) and substituting the result into (11b), we derive the following equation
$$\begin{aligned} 4a_2(a_2-a_1)x_2^2 + 4a_2(b_2-b_1)x_2 + (b_2-b_1)^2-4a_1(c_1-c_2) = 0. \end{aligned}$$
(12)
We discuss two cases. First, if \(a_1=a_2\) and \(b_1\ne b_2\), then the last equation is linear and provides a single solution \(x_2\) from which \(x_1\) can be recovered using Eq. (11a). The phase properties (boundary of the two-phase zone) are given by
$$\begin{aligned} x_{1,2} = \frac{c_1-c_2}{b_2-b_1}\pm \frac{b_2-b_1}{4a}, \end{aligned}$$
(13)
where a denotes the common value of \(a_1\) and \(a_2\). For any \(x^*\in (x_1,x_2)\), the phase properties are given by \(x_1\) and \(x_2\). For all other \(x^*\), the system remains in a single phase.
Next we discuss the case when \(a_1\ne a_2\). In this case, Eq. (12) is quadratic and its discriminant is equal to
$$\begin{aligned} \tilde{D} = 16a_1a_2 D, \end{aligned}$$
(14)
where D is given by (8). As \(\tilde{D}>0\), we get two roots \(x_2\) corresponding to two different two-phase regions
$$\begin{aligned} x_2^{(1,2)} = \frac{b_1-b_2}{2(a_2-a_1)}\pm \frac{\sqrt{a_1a_2D}}{2a_2(a_2-a_1)}, \end{aligned}$$
(15)
from which the corresponding \(x_1^{(1,2)}\) can be established using (11a) as
$$\begin{aligned} x_1^{(1,2)} = \frac{(b_1-b_2)(a_1+a_2)}{2a_1(a_2-a_1)}\pm \frac{\sqrt{a_1a_2D}}{2a_1(a_2-a_1)}. \end{aligned}$$
(16)
For any \(x^*\in (\min \{x_1^{(i)},x_2^{(i)}\},\max \{x_1^{(i)},x_2^{(i)}\})\), where \(i\in \{1,2\}\), the phase properties are given by \(x_1^{(i)}\) and \(x_2^{(i)}\). For all other \(x^*\), the system remains in a single phase.
In any case, once the phase properties have been established, the amounts of the phases can be computed readily by solving the system
$$\begin{aligned} \alpha _1 x_1^{(i)}+\alpha _2 x_2^{(i)}&= x^*, \end{aligned}$$
(17)
$$\begin{aligned} \alpha _1 + \alpha _2&= 1. \end{aligned}$$
(18)

4 Extension of the Benchmark Solution to the Multidimensional Case

Let us assume that \(n\in \mathbb {N}\), \(n>1\) and consider the function f to be in the form of (5) with
$$\begin{aligned} f_i(\mathbf{x}) = \mathbf{x}^T{\mathbb {A}}_i\mathbf{x} + \mathbf{b}_i\cdot \mathbf{x} + c_i,\qquad i=1,2, \end{aligned}$$
(19)
where \(\mathbb {A}_1,\mathbb {A}_2\in \mathbb {R}^{n,n}\) are symmetric and positive definite matrices, \(\mathbf{b}_1,\mathbf{b}_2\in \mathbb {R}^n\), and \(c_1,c_2\in \mathbb {R}\). As \(\mathbb {A}_i\) are positive definite, both parabolas described by \(f_i\) are convex for \(i=1,2\). In order to have f non-convex, parameters \(\mathbb {A}_i\), \(\mathbf{b}_i\), and \(c_i\) have to be selected so that the graphs of \(f_1\) and \(f_2\) intersect each other at some point. The intersection points are given by the following equation
$$\begin{aligned} f_1(\mathbf{x})-f_2(\mathbf{x}) = \mathbf{x}^T(\mathbb {A}_1-\mathbb {A}_2)\mathbf{x} + (\mathbf{b}_1-\mathbf{b}_2)\cdot \mathbf{x} + c_1-c_2 = 0. \end{aligned}$$
(20)
In the multidimensional case, we will assume that the parameters of the problem have been selected so that \(f_1(\mathbf{x})-f_2(\mathbf{x})\) changes sign in the feasible region [given, e.g., by Eq. (34)].
For a given value \(\mathbf{x}^*\in \mathbb {R}^n\), we are seeking for \(\mathbf{x}_1,\mathbf{x}_2\in \mathbb {R}^n\) and \(\alpha _1,\alpha _2\in (0,1)\) such that the optimality conditions (1) are satisfied for \(p=2\). As \(\nabla f_i(\mathbf{x}) = 2\mathbb {A}_i\mathbf{x}+\mathbf{b}_i\), these conditions can be rewritten as:
$$\begin{aligned} 2\mathbb {A}_1\mathbf{x}_1 - 2\mathbb {A}_2\mathbf{x}_2&= \mathbf{b}_2-\mathbf{b}_1, \end{aligned}$$
(21a)
$$\begin{aligned} \mathbf{x}_1^T\mathbb {A}_1\mathbf{x}_1 - \mathbf{x}_2^T\mathbb {A}_2\mathbf{x}_2&= c_1-c_2. \end{aligned}$$
(21b)
These equations have to be complemented with the mass balance equations
$$\begin{aligned} \alpha _1 \mathbf{x}_1+\alpha _2 \mathbf{x}_2&= \mathbf{x}^*, \end{aligned}$$
(22a)
$$\begin{aligned} \alpha _1 + \alpha _2&= 1. \end{aligned}$$
(22b)
Unlike in the 1D case, the evaluation of phase amounts \(\alpha _k\) cannot be separated from the evaluation of phase properties \(\mathbf{x}_k\). First, we express \(\mathbf{x}_1\) from Eq. (21a) and \(\alpha _1\) from Eq. (22b) as
$$\begin{aligned} \mathbf{x}_1&= \mathbb {A}_1^{-1}\mathbb {A}_2\mathbf{x}_2 + \frac{1}{2}\mathbb {A}_1^{-1}(\mathbf{b}_2-\mathbf{b}_1), \end{aligned}$$
(23)
$$\begin{aligned} \alpha _1&= 1-\alpha _2, \end{aligned}$$
(24)
Eliminating \(\mathbf{x}_1\) and \(\alpha _1\) from (21b) and (22a), we derive
$$\begin{aligned}&\mathbf{x}_2^T(\mathbb {A}_2\mathbb {A}_1^{-1}\mathbb {A}_2-\mathbb {A}_2)\mathbf{x}_2 + (\mathbf{b}_2-\mathbf{b}_1)^T\mathbb {A}_1^{-1}\mathbb {A}_2\mathbf{x}_2 \nonumber \\&\quad + \frac{1}{4}(\mathbf{b}_2-\mathbf{b}_1)\mathbb {A}_1^{-1}(\mathbf{b}_2-\mathbf{b}_1) - c_1 + c_2 = 0, \end{aligned}$$
(25)
and
$$\begin{aligned} \frac{1-\alpha _2}{2}\mathbb {A}_1^{-1}(\mathbf{b}_2-\mathbf{b}_1)+(1-\alpha _2)\mathbb {A}_1^{-1}\mathbb {A}_2\mathbf{x}_2+\alpha _2\mathbf{x}_2&= \mathbf{x}^*. \end{aligned}$$
(26)
Next, we express \(\mathbf{x}_2\) from Eq. (26) as
$$\begin{aligned} \mathbf{x}_2 = \left[ (1-\alpha _2)\mathbb {A}_2+\alpha _2\mathbb {A}_1\right] ^{-1} \left[ \mathbb {A}_1\mathbf{x}^* - \frac{1-\alpha _2}{2}(\mathbf{b}_2-\mathbf{b}_1)\right] . \end{aligned}$$
(27)
Combining (27) with (25), we derive a single equation for determination of \(\alpha _2\) which reads as
$$\begin{aligned}&\left[ \mathbf{x}^{*T}\mathbb {A}_1 - \frac{1-\alpha _2}{2}(\mathbf{b}_2-\mathbf{b}_1)^T\right] \left[ (1-\alpha _2)\mathbb {A}_2+\alpha _2\mathbb {A}_1\right] ^{-1}\nonumber \\&\quad \times (\mathbb {A}_2\mathbb {A}_1^{-1}\mathbb {A}_2-\mathbb {A}_2)\left[ (1-\alpha _2)\mathbb {A}_2+\alpha _2\mathbb {A}_1\right] ^{-1} \left[ \mathbb {A}_1\mathbf{x}^* - \frac{1-\alpha _2}{2}(\mathbf{b}_2-\mathbf{b}_1)\right] \nonumber \\&\quad + (\mathbf{b}_2-\mathbf{b}_1)^T\mathbb {A}_1^{-1}\mathbb {A}_2\left[ (1-\alpha _2)\mathbb {A}_2+\alpha _2\mathbb {A}_1\right] ^{-1} \left[ \mathbb {A}_1\mathbf{x}^* - \frac{1-\alpha _2}{2}(\mathbf{b}_2-\mathbf{b}_1)\right] \nonumber \\&\quad + \frac{1}{4}(\mathbf{b}_2-\mathbf{b}_1)\mathbb {A}_1^{-1}(\mathbf{b}_2-\mathbf{b}_1) - c_1 + c_2 = 0. \end{aligned}$$
(28)
Solving \(\alpha _2\) from (28) can be a formidable task in general. An easy case occurs if \(\mathbb {A}_1=\mathbb {A}_2\). In this case, the quadratic term in Eq. (28) vanishes and (28) reduces into a linear equation from which \(\alpha _2\) can be determined readily as:
$$\begin{aligned} \alpha _2 = \frac{1}{2}+2\frac{c_1-c_2-(\mathbf{b}_2-\mathbf{b}_1)^T\mathbf{x}^*}{(\mathbf{b}_2-\mathbf{b}_1)^T\mathbb {A}^{-1}(\mathbf{b}_2-\mathbf{b}_1)}, \end{aligned}$$
(29)
where we have denoted by \(\mathbb {A}\) the common value of \(\mathbb {A}_1\) and \(\mathbb {A}_2\). In order to have \(\alpha _2\in (0,1)\), the following condition must hold
$$\begin{aligned} |c_1-c_2-(\mathbf{b}_2-\mathbf{b}_1)^T\mathbf{x}^*| < \frac{1}{4} (\mathbf{b}_2-\mathbf{b}_1)^T{\mathbb {A}}^{-1}(\mathbf{b}_2-\mathbf{b}_1), \end{aligned}$$
(30)
which defines the two-phase region. If \(\mathbf{x}^*\) satisfies (30), then \(\alpha _2\) is given by (29), \(\mathbf{x}_2\) by (27), which reduces to
$$\begin{aligned} \mathbf{x}_2 = \mathbf{x}^* - \frac{1-\alpha _2}{2}\mathbb {A}^{-1}(\mathbf{b}_2-\mathbf{b}_1), \end{aligned}$$
(31)
while \(\alpha _1\) is given by (24) and \(\mathbf{x}_1\) is given by (23), which reduces to
$$\begin{aligned} \mathbf{x}_1&= \mathbf{x}_2 + \frac{1}{2}\mathbb {A}^{-1}(\mathbf{b}_2-\mathbf{b}_1) = \mathbf{x}^* + \frac{\alpha _2}{2}\mathbb {A}^{-1}(\mathbf{b}_2-\mathbf{b}_1). \end{aligned}$$
(32)
Note that along with Eq. (30), there may be additional constraints that refine the two-phase region. For the PTN-flash, the two-phase region is further limited by the following conditions
$$\begin{aligned} \mathbf{x}_i&\in \mathcal {D},\qquad i\in \{1,2\}, \end{aligned}$$
(33)
where
$$\begin{aligned} \mathcal {D} = \left\{ \mathbf{x}\in \mathbb {R}^n : (\forall i\in \widehat{n})(x_i\ge 0) \wedge \sum _{i=1}^nx_i\le 1\right\} . \end{aligned}$$
(34)
For \(\mathbf{x}^*\) not satisfying (30) and (33), the system remains in the single phase.

5 Examples of Benchmark Solutions

In this section, we provide an example of a benchmark solution for each case discussed above.

Example 1

In the first example, we consider a one-dimensional flash problem with parameters \(a_1=a_2=10\), \(b_1=-10\), \(b_2=0\), \(c_1=5\), and \(c_2=0.5\). For this settings, Eq. (13) provides \(x_1=0.2\), \(x_2=0.7\), and for \(x^*=0.4\) we get \(\alpha _1=0.6\) and \(\alpha _2=0.4\). This solution is shown in Fig. 1.

Fig. 1

Example 1: one-dimensional benchmark solution with a single two-phase zone

Example 2

Next, we consider a one-dimensional flash problem with parameters \(a_1=10\), \(a_2=80\), \(b_1=-10\), \(b_2=-70\), \(c_1=5\), and \(c_2=17\). For this settings, Eq. (15) and (16) provide boundaries of two two-phase zones at the following positions:
$$\begin{aligned} x^{(1)}_1&=0.115587109997048, \qquad x^{(1)}_2=0.389448388749631, \end{aligned}$$
(35)
$$\begin{aligned} x_1^{(2)}&=0.741555747145809, \qquad x^{(2)}_2=0.467694468393226. \end{aligned}$$
(36)
This solution is shown in Fig. 2
Fig. 2

Example 2: one-dimensional benchmark solution with two two-phase zones

Example 3

Finally, we consider a two-dimensional flash problem with \(n=2\), \(\mathbb {A}_1=\mathbb {A}_2=\mathbb {A}=10\mathbb {I}_2\), \(\mathbf{b}_1=3\mathbf{e}\), \(c_1=1\), \(\mathbf{b}_2=5\mathbf{e}\), \(c_2=0\), where \(\mathbb {I}_2\) is the identity matrix on dimension 2, and \(\mathbf{e}=(1,1)^T\). This may represent the PTN-flash problem with 3 components and two phases. In Fig. 3, we present the resulting two-phase region, taking into account Eq. (30) and the non-negativity of the mole fractions. In Fig. 3, we also present resulting phase splits for different values of \(\mathbf{x}^*\). The values of \(\mathbf{x}^*\) are indicated by the black points in the two-phase zone, while the resulting phase properties are denoted by the colored points. The arrows indicate the tie-lines along which we obtain the same split phases.

Fig. 3

Example 3: two-dimensional flash problem with a single two-phase zone. Values of \(\mathbf{x}^*\) (black points) and split phases for 5 different flashes (colored points) corresponding to these values of \(\mathbf{x}^*\) are presented

6 Further Generalizations

The solutions described above can be easily generalized to the case when
$$\begin{aligned} f(\mathbf{x}) = \min \{f_i(\mathbf{x}):i\in \widehat{p}\}, \end{aligned}$$
(37)
where \(\widehat{p}=\{1,2,\dots ,p\}\) and \(p\in \mathbb {N}\), \(p\ge 2\) is an arbitrary natural number. We assume that each \(f_i\) is of the form (6) for the 1D-case or (19) for the multidimensional case. Such generalizations can serve two purposes that will be illustrated on the examples below.

Example 4

Let us consider a one-dimensional flash problem with \(n=1\), \(p=3\), \(a_1=a_2=a_3=10\), \(b_1=-10\), \(c_1 = 5\), \(b_2 = 0\), \(c_2 = 0.5\), \(b_3 = -5\), and \(c_3 = 2.4\). This may represent the PTN-flash problem with 2 components and two phases. In Fig. 4, we present the resulting phase properties.

Fig. 4

Example 4: one-dimensional flash problem with several (false) local solutions that do not correspond to the global minimum of the objective functions. The true two-phase splitting is indicated by full green line, while the false local solutions are indicated by the dashed green lines

For \(x^*=0.425\), the two-phase flash problem can be solved using Eq. (13), in which the indices 1 and 2 are replaced by any pair of indices i and j, where \(i\ne j\in \widehat{3}\). The true solution corresponds to the pair of functions \(f_i\) and \(f_j\) with the minimum value of the objective function. As can be seen in Fig. 4, the optimal phase split properties are
$$\begin{aligned} x_1^{(12)} = 0.2,\qquad x_2^{(12)} = 0.7. \end{aligned}$$
(38)
Note that there are also other solutions corresponding to the points
$$\begin{aligned} x_1^{(23)} = 0.505,\qquad x_2^{(23)} = 0.255, \end{aligned}$$
(39)
and
$$\begin{aligned} x_1^{(13)} = 0.395,\qquad x_2^{(13)} = 0.645, \end{aligned}$$
(40)
which are corresponding to local minima of the objective function, but not the global one. This example is thus possessing false solutions that are local but not global and can be used as a test of convergence of an iterative solver toward the global minimum of the respective thermodynamic potential.

6.1 Three-Phase Flash Problem

Finally, let us consider a general n-dimensional flash problem with 3 functions \(f_i\) of the form
$$\begin{aligned} f(\mathbf{x}) = \min \{f_i(\mathbf{x}):i\in \widehat{3}\}, \end{aligned}$$
(41)
where
$$\begin{aligned} f_i(\mathbf{x}) = \mathbf{x}^T\mathbb {A}{} \mathbf{x} + \mathbf{b}_i\mathbf{x}+c_i, \end{aligned}$$
(42)
\(\mathbb {A}\) is a symmetric positive definite matrix, \(\mathbf{b}_i\in \mathbb {R}^n\), and \(c_i\in \mathbb {R}\) for \(i\in \widehat{3}\). As \(\nabla f_i(\mathbf{x})=2\mathbb {A}{} \mathbf{x}+\mathbf{b}_i\), and \(\mathbf{x}^T\nabla f_i(\mathbf{x})-f_i(\mathbf{x})=\mathbf{x}^T\mathbb {A}{} \mathbf{x}-c_i\), the optimality conditions (1) can be rewritten for \(p=3\) as
$$\begin{aligned} \mathbb {A}{} \mathbf{x}_1+\mathbf{b}_1&= 2\mathbb {A}{} \mathbf{x}_2+\mathbf{b}_2 = 2\mathbb {A}{} \mathbf{x}_3+\mathbf{b}_3, \end{aligned}$$
(43a)
$$\begin{aligned} \mathbf{x}_1^T\mathbb {A}{} \mathbf{x}_1-c_1&= \mathbf{x}_2^T\mathbb {A}\mathbf{x}_2-c_2 = \mathbf{x}_3^T\mathbb {A}{} \mathbf{x}_3-c_3. \end{aligned}$$
(43b)
From (43a), we express \(\mathbf{x}_1\) and \(\mathbf{x}_2\) in terms of \(\mathbf{x}_3\) as
$$\begin{aligned} \mathbf{x}_1 = \mathbf{x}_3+\frac{1}{2}\mathbb {A}^{-1}(\mathbf{b}_3-\mathbf{b}_1),\qquad \mathbf{x}_2 = \mathbf{x}_3+\frac{1}{2}\mathbb {A}^{-1}(\mathbf{b}_3-\mathbf{b}_2). \end{aligned}$$
(44)
Substituting these expressions to (43b), we derive upon some rearrangement
$$\begin{aligned} (\mathbf{b}_3-\mathbf{b}_i)^T\mathbf{x}_3 = c_i-c_3-\frac{1}{4}(\mathbf{b}_3-\mathbf{b}_i)^T\mathbb {A}^{-1}(\mathbf{b}_3-\mathbf{b}_i), i\in \widehat{2}, \end{aligned}$$
(45)
which is a set of two linear equations for \(n\ge 2\) unknowns components of vector \(\mathbf{x}_3\). If the vectors \((\mathbf{b}_3-\mathbf{b}_1)\) and \((\mathbf{b}_3-\mathbf{b}_2)\) are linearly independent, then this system of equations possesses a solution \(\mathbf{x}_3\). Once this solution is established, (44) can be used to evaluate \(\mathbf{x}_1\) and \(\mathbf{x}_2\). For \(\mathbf{x}^*\) from the convex hull of \(\mathbf{x}_1\), \(\mathbf{x}_2\), and \(\mathbf{x}_3\), the coefficients \(\alpha _1\), \(\alpha _2\), and \(\alpha _3\) can be found uniquely such that the constraints equations (2) are satisfied for \(p=3\). The convex hull of \(\mathbf{x}_1\), \(\mathbf{x}_2\), and \(\mathbf{x}_3\) thus defines the three-phase zone (possibly with some additional constraints). For different values of \(\mathbf{x}^*\) within the three-phase zone, the phase split properties are the same, only the amounts of the phases differ. For \(\mathbf{x}^*\) outside the three-phase zone, \(\mathbf{x}^*\) can be either single phase or two phase. This has to be tested by solving the two-phase flashes with any pair of functions \(f_i\) for \(i\in \widehat{3}\) and comparing the values of the objective functions (the minimum value of F is preferred).

Example 5

In the last example, we investigate the two-dimensional flash problem with 3 functions \(f_i\) of the form as described in the previous section with \(n=2\), \(p=3\), \(\mathbb {A} = 16\mathbb {I}_2\), \(\mathbf{b}_1=(-16,-8)^T\), \(\mathbf{b}_2=(-8,-16)^T\), \(\mathbf{b}_3=(-8,-8)^T\), \(c_1 = 5\), \(c_2=5\), \(c_3=2\). This may represent a PTN-flash problem with three components splitting up to three phases. In Fig. 5, we present the resulting three-phase region (the cyan region in the middle) and connected two-phase regions (yellow regions). In Fig. 5, we also present resulting phase splits for different values of \(\mathbf{x}^*\). The values of \(\mathbf{x}^*\) are indicated by the black points in the two-phase zone, while the resulting phase properties are denoted by the colored points. The arrows indicate the tie-lines along which we obtain the same split phases. Note that in the 3-phase region, the phase split properties are constant.

Fig. 5

Example 5: two-dimensional flash problem with a three-phase region (cyan region) and several two-phase regions (yellow). Values of \(\mathbf{x}^*\) (black points) and split phases for several different flashes (colored points) corresponding to these values of \(\mathbf{x}^*\) are presented

7 Benchmark Solution for the VTN-Flash Equilibrium Calculation

In this section, we will present another benchmark solution whose free energy function is smooth. Such a solution can be used for the VTN-based flashes. Unlike previously, this solution is available for \(n=1\) only. We will consider f to be a polynomial of order 4. Using a suitable rescaling, the coefficient at the highest order term can be set to one. Therefore, we have
$$\begin{aligned} f(x) = x^4+ax^3+bx^2+cx+d,\qquad a,b,c,d\in \mathbb {R} \end{aligned}$$
(46)
and
$$\begin{aligned} f'(x) = 4x^3+3ax^2+2bx+c. \end{aligned}$$
(47)
We are seeking for \(x_1,x_2\in \mathbb {R}\), \(x_1\ne x_2\) satisfying conditions (10), which can be rewritten as
$$\begin{aligned} 4(x_1^3-x_2^3)+3a(x_1^2-x_2^2)+2b(x_1-x_2)&= 0, \\ 3(x_1^4-x_2^4)+2a(x_1^3-x_2^3)+b(x_1^2-x_2^2)&= 0. \end{aligned}$$
We note that the solution \(x_1\) and \(x_2\) does not depend on the values of c and d, which can be, without loss of generality, chosen arbitrarily, e.g., \(c=d=0\). As we are seeking the non-trivial solutions of this system, we can divide by the nonzero factor \((x_1-x_2)\) to yield
$$\begin{aligned}&4(x_1^2+x_1x_2+x_2^2)+3a(x_1+x_2)+2b = 0, \\&3(x_1+x_2)(x_1^2+x_2^2)+2a(x_1^2+x_1x_2+x_2^2)+b(x_1+x_2) = 0. \end{aligned}$$
Let us first solve this system for \(a=0\). For \(a=0\), the system reduces significantly into
$$\begin{aligned}&4(x_1^2+x_1x_2+x_2^2)+2b = 0, \end{aligned}$$
(48)
$$\begin{aligned}&3(x_1+x_2)(x_1^2+x_2^2)+b(x_1+x_2)= 0. \end{aligned}$$
(49)
Equation (49) can be fulfilled only if \(x_1=-x_2\) or \(3(x_1^2+x_2^2)+b=0\). In the former case, we substitute \(x_1=-x_2\) into (48). The resulting equation \(4x_1^2+2b=0\) has a solution in the following form
$$\begin{aligned} x_1 = \pm \sqrt{\frac{-b}{2}}, \qquad x_2=-x_1, \end{aligned}$$
(50)
provided that \(b\le 0\); otherwise, it has no solution. If \(x_1\ne -x_2\), then the latter condition (\(3(x_1^2+x_2^2)+b=0\)) can be combined with (48) to get a false solution \(x_1=\pm \sqrt{\frac{-b}{6}}\) and \(x_2=-x_1\), but this solution is to be disregarded as it contradicts the assumption \(x_1\ne - x_2\) that has been used during its derivation. The only valid solution of the problem for \(a=0\) is thus given by Eq. (50). Note that when MATLAB’s symbolic package is used, it indicates the false solution as a valid solution. This is to emphasize the necessity of checking the results obtained by the symbolic packages.
To derive the benchmark solution for the case of \(a\ne 0\), we note that it is possible to rewrite Eq. (46) as
$$\begin{aligned} f(x)&= x^4+ax^3+bx^2+cx+d \nonumber \\&= \left( x+\frac{a}{4}\right) ^4 + \left( b-\frac{3}{8}a^2\right) \left( x+\frac{a}{4}\right) ^2 + \text {the lower order terms}, \end{aligned}$$
(51)
which indicates that we can solve the flash problem for the transformed variable \(y=x+\frac{a}{4}\) with parameters \(\tilde{a}=0\) and \(\tilde{b}=b-\frac{3}{8}a^2\). As \(y_{1,2}=\pm \sqrt{\frac{-\tilde{b}}{2}}\), we have
$$\begin{aligned} x_{1,2} = \pm \frac{1}{4}\sqrt{3a^2-8b} -\frac{a}{4}, \end{aligned}$$
(52)
provided that \(3a^2-8b\ge 0\).

Example 6

We consider the one-dimensional VTN-flash problem with the free energy of the form (46) and parameters \(a=-2000\), \(b=1.18\times 10^6\), \(c=0\), and \(d=0\). For this settings, Eq. (52) provides \(x_1=100\), \(x_2=900\), and for \(x^*=220\) we get \(\alpha _1=0.85\) and \(\alpha _2=0.15\). This solution is shown in Fig. 6.

Fig. 6

Example 6: one-dimensional benchmark solution of the VTN-flash problem with parameters \(a=-2000\), \(b=1.18\times 10^6\), \(c=0\), and \(d=0\)

8 Benchmark Solution for the VTN-Flash Problem with Capillarity

In this section, we present an analytical benchmark solution for the VTN-flash problem with capillarity. For a fluid confined in nanopores, the capillary pressure effect is dominant and may influence the equilibrium between the liquid and vapor phases. We will take \(n=1\) and assume that the Helmholtz free energy density is again of the form (46). To account for the capillary pressure effect, the VTN-flash equations have to be modified. We are seeking for \(x_1,x_2\in \mathbb {R}\), \(x_1\ne x_2\) satisfying
$$\begin{aligned} 4(x_1^3-x_2^3)+3a(x_1^2-x_2^2)+2b(x_1-x_2)&= 0, \\ 3(x_1^4-x_2^4)+2a(x_1^3-x_2^3)+b(x_1^2-x_2^2)&= p_c. \end{aligned}$$
where \(p_c\) denotes the capillary pressure. We assume that \(x_1<x_2\), i.e., 1 corresponds to the vapor phase and 2 corresponds to the liquid phase. Further, we assume that the liquid is the wetting phase, while the gas is the non-wetting phase. Therefore, \(p_c = p_1-p_2 = x_1f'(x_1)-f(x_1)-x_2f'(x_2)+f(x_2)\). For a curved interface in a nanopore of radius r and wetting angle \(\theta \), the capillary pressure is given by the Young-Laplace equation of capillarity which reads as:
$$\begin{aligned} p_c = \frac{2\sigma \cos \theta }{r}, \end{aligned}$$
(53)
where \(\sigma \) denotes the surface tension. In this work, we use the Weinaug–Katz correlation (Weinaug and Katz 1943) to model the surface tension between the two phases. For a single component, the formula looks like
$$\begin{aligned} \sigma = P^4(x_2-x_1)^4, \end{aligned}$$
(54)
where P is the parachor of the component. Combining the last three equations, we end up with the following system of equations describing the VTN-equilibrium including the capillary pressure effect:
$$\begin{aligned} 4(x_1^3-x_2^3)+3a(x_1^2-x_2^2)+2b(x_1-x_2)&= 0, \\ 3(x_1^4-x_2^4)+2a(x_1^3-x_2^3)+b(x_1^2-x_2^2)&= C(x_2-x_1)^4, \end{aligned}$$
where \(C=\frac{2P^4\cos \theta }{r}>0\) is a constant containing information about the pore radius, wetting angle, and parachor. As we are seeking for non-trivial solutions \(x_1\ne x_2\), we factor out the common factor \(x_1-x_2\) from the equations and obtain
$$\begin{aligned}&4(x_1^2+x_1x_2+x_2^2)+3a(x_1+x_2)+2b = 0, \\&3(x_1+x_2)(x_1^2+x_2^2)+2a(x_1^2+x_1x_2+x_2^2)+b(x_1+x_2)+C(x_2-x_1)^3=0. \end{aligned}$$
For \(a=0\), the system reduces into
$$\begin{aligned}&4(x_1^2+x_1x_2+x_2^2)+2b = 0, \\&3(x_1+x_2)(x_1^2+x_2^2)+b(x_1+x_2)+C(x_2-x_1)^3=0. \end{aligned}$$
This system has two solutions that have been found using MATLAB’s symbolic package of the following form
$$\begin{aligned} x_1 = \sqrt{\frac{-b}{2}}\frac{C+1}{\sqrt{3C^2+1}},\qquad x_2 = \sqrt{\frac{-b}{2}}\frac{C-1}{\sqrt{3C^2+1}}, \end{aligned}$$
(55)
and
$$\begin{aligned} x_1 = -\sqrt{\frac{-b}{2}}\frac{C+1}{\sqrt{3C^2+1}},\qquad x_2 = -\sqrt{\frac{-b}{2}}\frac{C-1}{\sqrt{3C^2+1}}, \end{aligned}$$
(56)
from which the one corresponding to the prescribed wettability condition must be chosen.
To obtain solution of the problem with \(a\ne 0\), we use the fact that the free energy can be rewritten as in (51). Using the transformation \(y=x+\frac{a}{4}\), we can solve the problem above with parameters \(\tilde{a}=0\) and \(\tilde{b}=b-\frac{3}{8}a^2\) for \(y_{1,2}\). Transforming back to the x-variables, we derive the following pair of solutions
$$\begin{aligned} x_1 = \frac{1}{4}\sqrt{3a^2-8b}\frac{C+1}{\sqrt{3C^2+1}}-\frac{a}{4},\qquad x_2 = \frac{1}{4}\sqrt{3a^2-8b}\frac{C-1}{\sqrt{3C^2+1}}-\frac{a}{4}, \end{aligned}$$
(57)
and
$$\begin{aligned} x_1 = -\frac{1}{4}\sqrt{3a^2-8b}\frac{C+1}{\sqrt{3C^2+1}}-\frac{a}{4},\qquad x_2 = -\frac{1}{4}\sqrt{3a^2-8b}\frac{C-1}{\sqrt{3C^2+1}}-\frac{a}{4}, \end{aligned}$$
(58)
from which the one corresponding to the prescribed wettability condition must be chosen.

Example 7

We consider the one-dimensional VTN-flash problem with the free energy of the form (46) with parameters \(a=-2000\), \(b=1.18\times 10^6\), \(c=0\) and \(d=0\), and various capillary pressure parameters C. For \(C=0\), we get the same result as in the previous example. Next, we evaluate solutions for parameters \(C\in \{0.2,0.4,0.6,0.8\}\). These solutions are displayed in Fig. 7. As we can see from Fig. 7, for \(C\in (0,\frac{1}{3})\), the phase split consists from a stable vapor phase and the metastable liquid. For \(C\in (\frac{1}{3},1)\), we get equilibrium between the stable vapor and an unstable phase (note that the point \(x_2\) lies in the concave part of the free energy). Such a splitting is not physically relevant. If we increase the value C above one, we would get equilibrium between the metastable vapor and an unstable phase. This is not physically realistic either. Therefore, to get physically relevant examples, the values of C should be kept within the interval \((0,\frac{1}{3})\).

Fig. 7

Example 7: one-dimensional benchmark solutions of the VTN-flash problem including the capillary pressure effect. Parameters of the free energy are \(a=-2000\), \(b=1.18\times 10^6\), \(c=0\), and \(d=0\). The numbers indicate the values of the capillary pressure parameter C

9 Summary and Conclusions

In this paper, a family of simple flash problems has been introduced for which the properties of the split phases are solvable exactly. Although the concept and the involved algebra are very simple, the flash problems can be derived which have desirable features for testing purposes such as multiple two-phase zones, three-phase zone, or existence of local solutions that are not global. The concept introduced here can be generalized readily to multiphase flash problems with any number of components and phases. An alternative formulation of the flash problem has been introduced which mimics properties of the Helmholtz free energy density in the VTN-flash problem. The corresponding benchmark solution can be used for testing the VTN-flash solvers. The VTN-based formulation can be extended to treat problems in a confined medium for which capillarity plays a major role. An open question remains whether the simplicity of such solutions allows for using them for derivation of an analytical solution of a transport problem in a porous medium with multiple phases and components, albeit in a very simplified physical settings. Such a solution would be very useful for testing compositional models, for which no analytical solutions with multiple components in two phases are, to the best of author’s knowledge, available. This may be a topic of further research.

Footnotes

  1. 1.

    If X is an arbitrary subset of \(\mathbb {R}^n\), then any point \(\mathbf x\) from the convex hull of X can be written as a convex combination of at most \(n+1\) points in X, i.e., there exist \(\mathbf{x}_1,\dots ,\mathbf{x}_{n+1}\in X\) and \(\alpha _1,\dots ,\alpha _{n+1}\ge 0\) such that \(\mathbf{x}=\sum \nolimits _{i=1}^{n+1}\alpha _i\mathbf{x}_i\) and \(\sum \nolimits _{i=1}^{n+1}\alpha _i=1\). The proof can be found in Rockafellar (1970).

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© Springer Nature B.V. 2018

Authors and Affiliations

  1. 1.Department of Mathematics, Faculty of Nuclear Sciences and Physical EngineeringCzech Technical University in PraguePrague 2Czech Republic

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