Modeling of Pore-Scale Two-Phase Phenomena Using Density Functional Hydrodynamics

Abstract

Predictive modeling of pore-scale multiphase flow is a powerful instrument that enhances understanding of recovery potential of subsurface formations. To endow a pore-scale modeling tool with predictive capabilities, one needs to be sure that this tool is capable, in the first place, of reproducing basic phenomena inherent in multiphase processes. In this paper, we overview numerical simulations performed by means of density functional hydrodynamics of several important multiphase flow mechanisms. In one of the reviewed cases, snap-off in free fluid, we demonstrate one-to-one comparison between numerical simulation and experiment. In another case, geometry-constrained snap-off, we show consistency of our modeling with theoretical criterion. In other more complex cases such as flow in pore doublets and simple system of pores, we demonstrate consistency of our modeling with published data and with existing understanding of the processes in question.

Appendices

Appendix 1: 1D Problem for DFH Equations

Let us begin by writing the full system of governing equations for 1D isothermal compositional flow. From (7) and (8), we have

$$\begin{aligned} \partial _t n_i + \partial _x ( n_i v_x + Q_{ ix} ) = 0, \end{aligned}$$

(33)

$$\begin{aligned} \rho (\partial _t v_x + v_x \partial _x v_x ) = \partial _x p_{xx} , \end{aligned}$$

(34)

where symbol $\partial _x$ denotes partial derivative in respect of the only spatial coordinate x. Boundary conditions for this system are the 1D isothermal form of the boundary conditions (10), (11), and (13):

$$\begin{aligned}&\displaystyle v_x {\big |} _{\partial \mathrm{D}} = 0, \end{aligned}$$

(35)

$$\begin{aligned}&\displaystyle Q_{ ix} {\big |} _{\partial \mathrm{D}} = 0, \end{aligned}$$

(36)

$$\begin{aligned}&\displaystyle \partial _x n_i \left| {_{\partial D} } \right. = 0. \end{aligned}$$

(37)

The symbol $\partial D$ in (35)–(37) denotes two ends of the 1D domain. The boundary condition (37) exhibits the fact of the neutral wetting properties (i.e., no contact angle in 1D geometry) leading to no gradient in molar densities at both ends of the domain.

The constitutive relations, which are 1D isothermal analogs of the Eqs. (16)–(19), (28), and (29), are follows:

$$\begin{aligned}&\displaystyle p_{xx} = \sigma _{xx} + \tau _{xx} , \end{aligned}$$

(38)

$$\begin{aligned}&\displaystyle \sigma _{xx} =f-\frac{1}{2}\nu _{ij} \partial _x n_i \partial _x n_j -\varPhi _i n_i , \end{aligned}$$

(39)

$$\begin{aligned}&\displaystyle \varPhi _i =\kappa _i -\nu _{ij} \partial _{xx} n_j , \end{aligned}$$

(40)

$$\begin{aligned}&\displaystyle Q_{ix} =-D_{ij} \partial _x \varPhi _j , \end{aligned}$$

(41)

$$\begin{aligned}&\displaystyle \tau _{xx} =\left( {\eta _\mathrm{v} +\frac{4}{3}\eta _\mathrm{s} } \right) \partial _x v_x . \end{aligned}$$

(42)

where $\nu _{ij}$ is related to $\alpha _{ij}$ by $\nu _{ij} =T\alpha _{ij}$, and $D_{ij}$ is the part of $\mu _{AB}$ corresponding to $A,B=1,\dots ,M$, namely $D_{ij} =T^{-1}\mu _{ij} , i,j=1,\dots ,M$.

The system (33)–(42) is closed after specifying Helmholtz energy density function f. It can be solved numerically to obtain a 1D compositional flow dynamics.

To present a simple analytical solution, we introduce several assumptions:

• The system is in static equilibrium state

• The fluid is two-phase single component

• Helmholtz energy is taken in a special form

• Boundary condition (37) is satisfied at infinity

Under these assumptions, let us turn to analyzing equilibrium solutions of the system (33)–(42). Firstly, we observe that in case of single-component fluid, there is no diffusion. Secondly, in equilibrium there is no flow, and velocity is zero, and all time derivatives are zeros. Therefore, the Eq. (33) is totally discarded, while the Eq. (34) yields the static equilibrium condition

$$\begin{aligned} \partial _x \sigma _{xx} =0, \end{aligned}$$

(43)

which is equal to (20). With regard to (39) and (40), it is easy to see that

$$\begin{aligned} \partial _x \sigma _{xx} =n \partial _x \varPhi , \end{aligned}$$

(44)

where index i has been dropped as according to the first assumption, we now deal with the only chemical component with molar density n. Because n cannot be zero everywhere in the domain, we have from (43) and (44)

$$\begin{aligned} \varPhi =\varLambda . \end{aligned}$$

(45)

To find the unknown constant $\varLambda$, we need to observe that in equilibrium, the second term in (40), which is the definition for $\varPhi$, vanishes everywhere except the interface between phases. At the same time, within the equilibrium phases A and B, their chemical potentials are equal $\kappa _\mathrm{A} =\kappa _\mathrm{B}$. Therefore, we have everywhere

$$\begin{aligned} \varPhi =\kappa _\mathrm{A} =\kappa _\mathrm{B} . \end{aligned}$$

(46)

To move further, we must specify particular form of Helmholtz energy density f. We take it to be as follows

$$\begin{aligned} f=A(n-n_\mathrm{A} )^{2}(n-n_\mathrm{B} )^{2}, \end{aligned}$$

(47)

where $n_\mathrm{A} ,n_\mathrm{B}$ are known values of molar density corresponding to the equilibrium phases A and B, and A is positive model coefficient, which can be fixed to fit compressibility data. This is done by noticing that bulk modulus K is related to Helmholtz energy using the thermodynamic relations $K=\frac{\partial p}{\partial n}n$ and $p=n\kappa -f$, where p is pressure. The expression in (47) has only one free parameter, which means that bulk modulus for only one of the two phases can be fit, and the second one appears determined by the first.

It is easy to see that the simple expression (47) allows for description of two-phase single-component equilibrium: The function is nonnegative and has two minimums corresponding to two equilibrium phases (Fig. 16).

Fig. 16

Schematic representation of the model double-well Helmholtz energy function

According to the definition (47), $\kappa _A =\kappa _B =0$ in equilibrium phases corresponding to molar densities $n_\mathrm{A} ,n_\mathrm{B}$. Now with regard to (40) and (46) and remembering that $\kappa =\frac{\partial f}{\partial n}$, we come to the following equation describing equilibrium solution:

$$\begin{aligned} \frac{\partial f}{\partial n}-\nu \partial _x^2 n=0. \end{aligned}$$

(48)

Multiplying this equation by $\partial _x n$, we arrive to $\partial _x f-\frac{1}{2}\nu \partial _x (\partial _x n)^{2}=0$, and then to

$$\begin{aligned} f=\frac{1}{2}\nu (\partial _x n)^{2}, \end{aligned}$$

(49)

where we took notice of the fact that $\partial _x n$ is zero at infinity. Rearranging terms in (49) and using (47), we have

$$\begin{aligned} \partial _x n=\sqrt{\frac{2A}{\nu }}(n-n_\mathrm{A} )(n_\mathrm{B} -n). \end{aligned}$$

(50)

This equation is integrated analytically and yields

$$\begin{aligned} n(x)=\frac{e^{a x}n_\mathrm{B} +n_\mathrm{A} }{e^{ax}+1}=\frac{n_\mathrm{B} +n_\mathrm{A} }{2}+\frac{n_\mathrm{B} -n_\mathrm{A} }{2}th\frac{ax}{2}, \end{aligned}$$

(51)

where $a=\sqrt{\frac{2A}{\nu }}$ and integration constant was fixed from the condition $n(0)=\frac{n_\mathrm{B} +n_\mathrm{A} }{2}$.

A schematic representation of the solution (51) is given in Fig. 17.

Fig. 17

Structure of the interfacial zone in 1D static equilibrium solution

It is instructive to note that the shape of the interface is determined by both Helmholtz energy (i.e., through coefficient A) and coefficient $\nu$. This is the manifestation of the fact that interface is the finite thickness zone, whose properties must necessarily be dependent on thermodynamics.

To conclude overview of this 1D problem, we calculate interfacial tension using the Eq. (21), which in our case reads

$$\begin{aligned} \gamma =\int \limits _{-\infty }^{+\infty } {\nu (\partial _x n)^{2}\mathrm{d}x} . \end{aligned}$$

(52)

Comparing (52) with (49), we have

$$\begin{aligned} \gamma =2\int \limits _{-\infty }^{+\infty } {f\mathrm{d}x} =2\int \limits _{n_A }^{n_B } {\frac{fdn}{\partial _x n}} =\sqrt{2\nu }\int \limits _{n_A }^{n_B } {\sqrt{f}\mathrm{d}n} , \end{aligned}$$

(53)

which is integrable given the convenient Helmholtz energy expression in (47) and yields

$$\begin{aligned} \gamma =\sqrt{2\nu A}\int \limits _{n_A }^{n_B } {(n-n_\mathrm{A} )(n_\mathrm{B} -n)\mathrm{d}n} =\frac{1}{3}(n_{\mathrm{B}}-n_{\mathrm{A}})^{3}\sqrt{\frac{\nu A}{2}}. \end{aligned}$$

The described solution has a demonstrational value. In majority of practical applications, Helmholtz energy model is much more complex than (47), and obtaining analytical solution even for simple static equilibrium problems requires numerical simulation. Therefore, let us briefly touch an important aspect related to obtaining solutions like (51) numerically. In numerical simulation of partial derivative equations using Eulerian point of view, it is conventional to introduce some type of finite approximations, e.g., finite-differences, finite-volumes, and finite-elements. Regardless of what type of approximation is used, the mere fact of approximation puts constraints on accuracy of the numerical solution, which are determined by numerical grid resolution; these constrains are removed as grid step converges to zero. In case of the solution in (51), there is a constraint on width of the interface, which is quite similar to that described by Kim (2012) with regard to phase-field methods. Indeed, there should be enough numerical cells across the interface for one to be able to calculate finite-difference high-order spatial derivatives of molar density fields. On the other hand, if there are too many cells, the interface profile becomes undesirably diffused. Our experience gives 5–8 as an optimal estimation for the number of cells across the interface. In numerical modeling, we conventionally define the interface as a region where molar density is in the range $n_\mathrm{A} +\frac{n_\mathrm{B} -n_\mathrm{A} }{20}<n<n_\mathrm{B} -\frac{n_\mathrm{B} -n_\mathrm{A} }{20}$. Using this definition in (51), it is easy to obtain the condition

$$\begin{aligned} L=x_\mathrm{B} -x_\mathrm{A} =\frac{4}{a}Arth\frac{9}{10}=m h, \end{aligned}$$

where L is width of the interface, $x_\mathrm{A}$ is coordinate where $n=n_\mathrm{A} +\frac{n_\mathrm{B} -n_\mathrm{A} }{20}$, $x_\mathrm{B}$ is coordinate where $n=n_\mathrm{B} -\frac{n_\mathrm{B} -n_\mathrm{A} }{20}$, h is the cell size, and m is the number of cells. Therefore, given the Helmholtz energy parameter A and coefficient $\nu$, one has the constraint on the numerical grid resolution to adequately represent the solution in (51)

$$\begin{aligned} h=\sqrt{\frac{8\nu }{Am^{2}}}Arth\frac{9}{10}. \end{aligned}$$

(54)

The result in (54) is similar to reported by Kim (2012) for the case of phase-field models. Let us stress again that the constraint like (54) is the result of finite approximation; this constraint does not exist in the original equations.

Appendix 2: Helmholtz Energy Model

This appendix contains description of the model Helmholtz energy used in numerical simulation of examples present in this paper. We followed our previous experience in numerical modeling by DFH (Demyanov and Dinariev 2004a, b; Dinariev and Evseev 2005).

For the homogeneous liquid (Phase A) near thermodynamic equilibrium, Taylor series expansion for its Helmholtz energy can be used in the form

$$\begin{aligned} f_\mathrm{A} (n_i )=f_{\mathrm{A}0} +f_{\mathrm{A}i} (n_i -n_{i\mathrm{A}} )+\frac{1}{2}f_{\mathrm{A}ij} (n_i -n_{i\mathrm{A}} )(n_j -n_{j\mathrm{A}} ), \end{aligned}$$

(55)

where $n_{i\mathrm{A}} ,i=1,2$ is composition of Phase A in molar densities. Expansion coefficients $f_{\mathrm{A0}}$ and $f_{\mathrm{A}i}$ vanish from hydrodynamic equations when Helmholtz energy in the form (55) is used; therefore, only symmetric matrix $f_{\mathrm{A}ij}$ is relevant for modeling. Coefficients $f_{\mathrm{A}ij}$ are constrained by the relation $K_\mathrm{A} =f_{\mathrm{A}ij} n_{i\mathrm{A}} n_{j\mathrm{A}}$, where $K_\mathrm{A}$ is bulk modulus of Phase A. The Helmholtz energy model for the second phase, Phase B, is similar to (55).

For two-phase flow case, it is necessary to calculate Helmholtz energy $f(n_i )$ at arbitrary point $n_i ,i=1,2$. In order to do this, we use interpolation with $f_\mathrm{A} (n_i )$ and $f_\mathrm{B} (n_i )$

$$\begin{aligned} f=\frac{f_\mathrm{A} f_\mathrm{B} }{f_\mathrm{A} +f_\mathrm{B} }. \end{aligned}$$

(56)

Given the symmetry of the matrices $f_{\mathrm{A}ij}$ and $f_{\mathrm{B}ij}$ and the constraint related to bulk moduli, there are two free parameters left in each matrix. One of these parameters (in each matrix) is used to fix the interface thickness as explained in “Appendix 1.” However, in case of two components, the interface should not be necessarily symmetric as it is in Fig. 17; this is why free parameter from each of the matrices is required to fix the shape. The residual free parameter in each matrix is not needed in the problems modeled for this paper. Therefore, we factor it out by assuming additional symmetry of the matrices. This passes analogy with calibration in field theories.

For the surface Helmoltz energy, we use the model

$$\begin{aligned} f_*=\xi _{1i} n_i , \end{aligned}$$

(57)

where parameters $\xi _{1i}$ are found from the system of two linear algebraic equations

$$\begin{aligned} f_{*\mathrm{A}} =\xi _{1i} n_{i\mathrm{A}} , \quad f_{*\mathrm{B}} =\xi _{1i} n_{i\mathrm{B}} , \end{aligned}$$

(58)

where $f_{*\mathrm{A}} ,f_{*\mathrm{B}}$ are known values of surface energy for Phases A and B. The surface energies are related with the contact angle $\theta$ by the Young equation $\cos \theta =\frac{f_{*\mathrm{B}} -f_{*\mathrm{A}} }{\gamma _{\mathrm{AB}} }$.