Flows of Two Slightly Miscible Fluids in Porous Media: Two-Scale Numerical Modeling

Abstract

We address the two-scale homogenization of the Navier–Stokes and Cahn–Hilliard equations in the case of a weak miscibility of a two-component fluid. To this end a notion of the miscibility strength is formulated on the basis of a correlation between the upscaling parameter and the surface tension. As a result, a two-scale model is derived. Macro-equations turn out to be a generalization of the Darcy law enjoying cross-coupling permeability tensors. It implies that the Darcy velocity of each phase depends on pressure gradients of both phases. Micro-equations serve for determination both of the permeability tensors and the capillary pressure. An example is constructed by analytical tools to describe capillary displacement of oil by mixture of water with carbon dioxide in a system of hydrophobic parallel channels.

Appendices

Appendix A Evolution of a droplet

All our numerical calculations for the Cahn–Hilliard equations are performed with the use of FreeFem++. To tackle nonlinearity, the Newton-Raphson iteration procedure is employed. As for the temporal discretization, an explicit in time method is applied with the time step equal to $\mathrm 5\cdot\,10^{-3}$. The code was validated by applying it to the problem concerning the evolution of a droplet initially located in the center of the domain and having the form of a quadrat, i.e., $\varphi = 1$ inside the quadrat and $\varphi = 0$ outside it when $t = 0$ Kou and Sun (2018). The boundary of the droplet is slightly “blurred". It is clear that the droplet ultimately should take the spherical form Balashov (2021) due to gradient of chemical potential. In the domain

$$\begin{aligned} Y=\{0<y_1<1, \, 0<y_2<1\}, \end{aligned}$$

we consider the following initial boundary value problem

$$\begin{aligned}{} & {} R \frac{\partial \varphi }{\partial t}=\tilde{B}\Delta \mu , \quad \mu =\tilde{\Gamma }\omega ^\prime (\varphi )-\tilde{\Lambda }\Delta \varphi ,\\{} & {} \frac{\partial \varphi }{\partial n} \Big |_{\partial Y_f}=0,\quad \frac{\partial \mu }{\partial n}\Big |_{\partial Y_f}=0, \quad \varphi |_{t=0} = \varphi _0(y), \end{aligned}$$

where

$$\begin{aligned} \varphi _0(y)=0.5(1+\tanh (b(a-\max \{\vert y_1-0.5 \vert , \vert y_2-0.5 \vert \}))). \end{aligned}$$

We use the following data:

$$\begin{aligned} \tilde{\Lambda }=1, \; \tilde{\Gamma }=10^3, \; R=1.2, \; B=1, \; b=10^6, \; a=0.3. \end{aligned}$$

Calculations reveal that the initial quadrat becomes a circle (Fig. 4).

Fig. 4

Initial quadrat becomes a circle with smeared boundary. The number of steps are equal to 0 a, 20 b and 40 c

Appendix B

Here, we prove that equality (12) holds. Indeed, due to equation (8), one can write the derivative of energy as follows:

$$\begin{aligned} \frac{\Gamma }{\Lambda } \frac{d}{dt}\int \limits _{\Omega _f^\varepsilon }w(\varphi )\,dx= \frac{\Gamma }{\Lambda } \int \limits _{\Omega _f^\varepsilon } w'(\varphi )\varphi _t\,dx= -\frac{\Gamma }{\Lambda } \int \limits _{\Omega _f^\varepsilon } w'(\varphi )(R^{-1}\text {div}\,\textbf{j}+\textbf{v}\cdot \nabla \varphi )dx. \end{aligned}$$

By equation (7), we conclude that

$$\begin{aligned} \frac{\Gamma }{\Lambda } \frac{d}{dt} \int \limits _{\Omega _f^\varepsilon } w(\varphi )\,dx= \frac{\Gamma B}{R\Lambda } \int \limits _{\Omega _f^\varepsilon } w'(\varphi )\Delta \mu \,dx- \frac{\Gamma }{\Lambda } \int \limits _{\partial \Omega _f^\varepsilon } w\textbf{v}\cdot \textbf{n}d\sigma = \frac{\Gamma B}{R\Lambda } \int \limits _{\Omega _f^\varepsilon } w'(\varphi )\Delta \mu \,dx. \end{aligned}$$

(B1)

On the other hand, if we multiply equation (8) by $\Delta \varphi$ and integrate by parts, we get the equality

$$\begin{aligned} \frac{d}{dt} \int \limits _{\Omega _f^\varepsilon }\frac{|\nabla \varphi |^2}{2} dx + \int \limits _{\Omega _f^\varepsilon } \frac{B|\nabla \mu |^2}{R\Lambda }dx= - \frac{\Gamma B}{R\Lambda } \int \limits _{\Omega _f^\varepsilon } w'(\varphi )\Delta \mu \,dx+ \int \limits _{\Omega _f^\varepsilon } (\textbf{v}\cdot \nabla \varphi )\Delta \varphi \,dx. \end{aligned}$$

(B2)

The last integral in the above formula can be written as

$$\begin{aligned} \int \limits _{\Omega _f^\varepsilon } (\textbf{v}\cdot \nabla \varphi )\Delta \varphi \,dx= -\int \limits _{\Omega _f^\varepsilon } \frac{2\eta |\nabla \textbf{v}|^2}{\Lambda } \,dx. \end{aligned}$$

Summation of equalities (B1) and (B2) results in formula (12).

Appendix C

Here, we derive some formulas useful in homogenization. Our calculations are based on the following differentiation formula

$$\begin{aligned} \frac{\partial f(x,x/\varepsilon )}{\partial x_i}= \left( \frac{\partial f(x,y)}{\partial x_i}+ \frac{1}{\varepsilon } \frac{\partial f(x,y)}{\partial y_i}\right) \Big |_{y=x/\varepsilon }. \end{aligned}$$

Calculations reveal that

$$\begin{aligned}{} & {} \frac{\partial \textbf{v}^\varepsilon (x)}{\partial x_j}= \varepsilon ^2\sum _{i=0} \varepsilon ^i \frac{\partial \textbf{v}^i (x,y)}{\partial x_j}+ \varepsilon \sum _{i=0} \varepsilon ^i \frac{\partial \textbf{v}^i (x,y)}{\partial y_j},\\{} & {} \text {div}(\eta ^\varepsilon D^\varepsilon )= \text {div}_y(\eta ^0 D_y^0)+O(\varepsilon ),\quad \eta ^0=\eta (\varphi ^0),\quad 2(D_y^0)_{ij}=\frac{\partial v_i^0}{\partial y_j}+ \frac{\partial v_j^0}{\partial y_i},\\{} & {} \frac{\partial \varphi ^\varepsilon (x)}{\partial x_j} = \sum _{i=0,i\ne 1} \varepsilon ^i\left( \frac{\partial \varphi ^i (x,y)}{\partial x_j}+\frac{1}{\varepsilon }\frac{\partial \varphi ^i (x,y)}{\partial y_j}\right) ,\\{} & {} \frac{\partial ^2 \varphi ^\varepsilon (x)}{\partial x_k\partial x_j} \\{} & {} \quad = \sum _{i=0,i\ne 1} \varepsilon ^i\left( \frac{\partial ^2 \varphi ^i (x,y)}{\partial x_k\partial x_j}+\frac{1}{\varepsilon ^2} \frac{\partial ^2 \varphi ^i (x,y)}{\partial y_k\partial y_j} +\frac{1}{\varepsilon } \frac{\partial ^2 \varphi ^i (x,y)}{\partial x_k\partial y_j}+ \frac{1}{\varepsilon } \frac{\partial ^2 \varphi ^i (x,y)}{\partial y_k\partial x_j} \right) ,\\{} & {} \Delta \varphi ^\varepsilon (x)= \sum _{i=0,i\ne 1} \left( \varepsilon ^i \Delta _x \varphi ^i(x,y) +\varepsilon ^{i-2}\Delta _y \varphi ^i(x,y)+ 2\varepsilon ^{i-1}\sum _k\frac{\partial ^2 \varphi ^i (x,y)}{\partial y_k \partial x_k}\right) , \end{aligned}$$

where $y=x/\varepsilon .$ According to the theory of two-scale homogenization, the variables x and y on the right-hand side of these formulas are considered independent variables when the derivatives are substituted into the equations.

With the above calculations at hands, equation (6) becomes

$$\begin{aligned}{} & {} \sum _{i=0}\varepsilon ^i \mu ^i(x,y)= \tilde{\Gamma } \omega '(\xi ) -\\{} & {} \varepsilon ^2 \tilde{\Lambda } \sum _{i=0,i\ne 1} \left( \varepsilon ^i \Delta _x \varphi ^i(x,y) +\varepsilon ^{i-2}\Delta _y \varphi ^i(x,y)+ 2\varepsilon ^{i-1}\sum _k\frac{\partial ^2 \varphi ^i (x,y)}{\partial y_k \partial x_k}\right) , \end{aligned}$$

with

$$\begin{aligned} \xi =\sum _{i=0,i\ne 1}\varepsilon ^i\varphi ^i(x,y). \end{aligned}$$

Clearly, one can write this equation as an equality of two expansions with respect to $\varepsilon$.

Appendix D

Here, we give details of construction of a one-dimensional solution $S(\tau ,x)$, $\Pi _1(\tau ,x)$, $\Pi _2(\tau ,x)$ to the system (42)-(45), where $S=S_2$.

For such a one-dimensional flow, the equation $\text {div}(\textbf{v}^{(1)}+\textbf{v}^{(2)})=0$ becomes

$$\begin{aligned} \frac{\partial }{\partial x}(v^{(1)}_1+v^{(2)}_1)=0, \quad \text {i.e.,}\quad v^{(1)}_1+v^{(2)}_1=V. \end{aligned}$$

Passing to pressures, we obtain that

$$\begin{aligned} -\xi C\frac{\partial \Pi _1}{\partial x}-E\frac{\partial \Pi _2}{\partial x}=V(\tau ). \end{aligned}$$

(D3)

where

$$\begin{aligned} C=C^{(1)}_{11}+C^{(2)}_{11},\quad E=E^{(1)}_{11}+E^{(2)}_{11}. \end{aligned}$$

Due to (70) and (71), equation (D3) is equivalent to

$$\begin{aligned} -A\left( \frac{\partial \Pi _1}{\partial x}+\xi S\frac{\partial \Pi _2}{\partial x}\right) =V. \end{aligned}$$

(D4)

The macro-equation (42) for $S=S_2$ can be formulated as

$$\begin{aligned} \frac{\partial (\xi S)}{\partial \tau } +\frac{\partial }{\partial x} \left( -\xi ^{-1}C^{(2)}_{11}(x,\tau )\frac{\partial \Pi _1}{\partial x}-E^{(2)}_{11}(x,\tau )\frac{\partial \Pi _2}{\partial x} \right) =0. \end{aligned}$$

(D5)

Due to (70) and (71), equation (D5) admits the form

$$\begin{aligned} \frac{\partial (\xi S)}{\partial \tau } -\frac{\partial }{\partial x} \left( A S \left( \frac{\partial \Pi _1}{\partial x}+\xi S \frac{\partial \Pi _2}{\partial x}\right) \right) =0. \end{aligned}$$

(D6)

It follows from (D4) that equation (D6) becomes

$$\begin{aligned} \frac{\partial (\xi S)}{\partial \tau } +V\frac{\partial S }{\partial x}=0. \end{aligned}$$

(D7)

This macro-equation describes displacement with a discontinuity front, when at the initial moment of time on the interval $0<x<1$, the function S takes the value $S_+$, $0\le S_+\le 1$, and the boundary condition $S=S_-$, $S_+\le S_-\le 1$ is kept at the input boundary $x=0$:

$$\begin{aligned} S|_{t=0}=S_+,\quad S|_{x=0}=S_-. \end{aligned}$$

The functions $S,\Pi _1,\Pi _2$ satisfy the following macro-problem

$$\begin{aligned}{} & {} \frac{\partial (\xi S)}{\partial \tau } -\frac{\partial }{\partial x} \left( AS\frac{\partial \Pi _1}{\partial x}+A\xi S^2\frac{\partial \Pi _2}{\partial x} \right) =0, \end{aligned}$$

(D8)

$$\begin{aligned}{} & {} \frac{\partial }{\partial x} \left( \frac{\partial \Pi _1}{\partial x}+\xi S\frac{\partial \Pi _2}{\partial x}\right) =0, \end{aligned}$$

(D9)

$$\begin{aligned}{} & {} \Pi _1+\frac{(\xi S-2c_2(\cdot ))}{1-2S}\Pi _2=\frac{\tilde{p}_c}{1-2S},\quad \tilde{p}_c=p_c(S)+\nu _{2,1}(\cdot ),\, \nu _{2,1}=\nu _2-\nu _1, \end{aligned}$$

(D10)

where the notation $f(\cdot )$ implies that the function f depends not only on the macro-variables $(x,\tau )$, but on the micro-equations (55)-(59) as well. The fast time $t'$ in equations (D8)-(D10) is set to be equal to the stabilization time $t'_*(x)$. Due to the properties (60), the macro-equations admit a solution given by constants:

$$\begin{aligned} S=1,\quad \Pi _1=-p_c(1),\quad \Pi _2=\text {const}, \end{aligned}$$

and

$$\begin{aligned} S=0,\quad \Pi _1=p_c(0),\quad \Pi _2=\text {const}. \end{aligned}$$

We look for a piecewise continuous solution of equations (D8)-(D10) with the discontinuity front $x=\sigma (t)$. Equation (D9) is equivalent to the system of equations (D4) and (D7). We introduce the jump $[S]_{\sigma }$ of the discontinuous function S(x) at the line of discontinuity $x=\sigma (\tau )$:

$$\begin{aligned}{}[S]_{\sigma }=\lim \limits _{\delta \rightarrow 0}S(\sigma (\tau )+\delta ,\tau )-\lim \limits _{\delta \rightarrow 0}S(\sigma (\tau )-\delta ,\tau ). \end{aligned}$$

By the Hugoniot condition Shelukhin and Perepelitsa (1999)

$$\begin{aligned} \xi [S]_{\sigma }\dot{\sigma }=V[S]_{\sigma }, \end{aligned}$$

it results from the scalar conservation law (D7) that the front satisfies the equation

$$\begin{aligned} \xi \dot{\sigma }=V,\quad \dot{\sigma }\equiv \frac{d \sigma }{d \tau }. \end{aligned}$$

Following the method of Shelukhin and Perepelitsa (1999) we introduce the self-similar variable $\zeta =x/\sigma (\tau )$ and look for a solution in the form of a centered wave $S(x,\tau )=S(\zeta )$ behind the front. In the domain $0<x<\sigma (\tau )$, equation (D7) becomes

$$\begin{aligned} \left( -\zeta \dot{\sigma }+V\right) S'=0,\quad S'\equiv \frac{d S}{d \zeta },\quad 0<\zeta <1. \end{aligned}$$

The equality $S=\text {const}=S_-$ holds behind the front. Before the front, i.e., in the domain $x>\sigma (\tau )$, we assume that $S=\text {const}=S_+$.

To simplify calculations, the data $\Pi _i^0, \Pi _i^1$ ($i=1,2$) in the boundary conditions

$$\begin{aligned} \Pi _i\big |_{x=0}= \Pi _i^0,\quad \Pi _i\big |_{x=1}= \Pi _i^1,\quad i=1,2, \end{aligned}$$

are assumed to satisfy the special restrictions

$$\begin{aligned}{} & {} \Pi _1^0+\xi \Pi _2^0 S\big |_{x=0}=0,\quad \Pi _1^0+\Pi _2^0 \frac{\xi S-2c_2}{1-2S}\Big |_{x=0}=\frac{\tilde{p}_c}{1-2S}\Big |_{x=0}, \end{aligned}$$

(D11)

$$\begin{aligned}{} & {} \Pi _1^1+\xi \Pi _2^1 S\big |_{x=1}=0,\quad \Pi _1^1+\Pi _2^1 \frac{\xi S-2c_2}{1-2S}\Big |_{x=1}=\frac{\tilde{p}_c}{1-2S}\Big |_{x=1}. \end{aligned}$$

(D12)

We integrate equations (D4) over the intervals $0<x<\sigma$ and $\sigma<x<1$ paying attention that the reduced pressures $\Pi _1$ and $\Pi _2$ are continuous. As a result, we obtain the equalities

$$\begin{aligned} \Pi _1^\sigma +\xi S_- \Pi _2^\sigma =-\frac{V\sigma }{A},\quad \Pi _1^\sigma +\xi S_+ \Pi _2^\sigma =\frac{V(1-\sigma )}{A}, \end{aligned}$$

(D13)

where $\Pi _i^\sigma$ is the value of $\Pi _i$ at $x=\sigma$. It follows from (D13) that

$$\begin{aligned} \Pi _1^\sigma =-\frac{V(S_-+\sigma [S]_\sigma )}{A[S]_\sigma },\quad \xi \Pi _2^\sigma =\frac{V}{A [S]_\sigma }. \end{aligned}$$

(D14)

Let us calculate the jumps of the capillary curve (D10):

$$\begin{aligned} \Pi _2^\sigma \left[ \frac{\xi S-2c_2}{1-2S}\right] _\sigma =\left[ \frac{\tilde{p}_c}{1-2S}\right] _\sigma . \end{aligned}$$

Taking into account formula (D14), we arrive at the following equation for the front velocity:

$$\begin{aligned} V\left[ \frac{\xi S-2c_2}{1-2S}\right] _\sigma = \xi A [S]_\sigma \left[ \frac{\tilde{p}_c}{1-2S}\right] _\sigma ,\quad A=-\frac{\xi ^2}{12}. \end{aligned}$$

(D15)

This formula coincides with (77).

The assumption that the phase velocities behind the front and ahead of the front do not depend on the spatial coordinate makes it possible to determine these velocities. Indeed, applying (70) and (71) we conclude that

$$\begin{aligned} v_1^{(1)}= & {} -AS_1 \frac{\partial \Pi _{1}}{\partial x}-A\xi S_1 S_2 \frac{\partial \Pi _{2}}{\partial x}, \end{aligned}$$

(D16)

$$\begin{aligned} v_1^{(2)}= & {} -AS_2 \frac{\partial \Pi _{1}}{\partial x}-A\xi S_2^2 \frac{\partial \Pi _{2}}{\partial x}. \end{aligned}$$

(D17)

We integrate equation (D16) for $v_1^{(1)}$ over the intervals $0<x<\sigma$ and $\sigma<x<1$, keeping in mind formula (D14). As a result, we obtain

$$\begin{aligned} \sigma v_1^{(1)}=-A (1-S_-)\Pi _1^{\sigma }-A\xi (1-S_-)S_-\Pi _2^{\sigma }=\sigma V (1-S_-), \\ (1-\sigma ) v_1^{(1)}=A (1-S_+)\Pi _1^{\sigma }+A\xi (1-S_+)S_+\Pi _2^{\sigma }=(1-\sigma ) V (1-S_+). \end{aligned}$$

Hence,

$$\begin{aligned} v_1^{(1)}=\left\{ \begin{array}{ll} V(1-S_-) &{} \text{ if }\,x<\sigma (\tau )\\ V(1-S_+) &{} \text{ if } \,x>\sigma (\tau ), \end{array} \right. \quad v_1^{(2)}=\left\{ \begin{array}{ll} V S_- &{} \text{ if }\,x<\sigma (\tau )\\ VS_+ &{} \text{ if } \,x>\sigma (\tau ). \end{array} \right. \end{aligned}$$

(D18)

Because of (D4), one can derive from (D16) and (D17) the equality $S_2 v_1^{(1)}=$ $S_1 v_1^{(2)}$. It implies that

$$\begin{aligned} v_1^{(i)}=S_i V,\quad 0<x<1. \end{aligned}$$

(D19)

Formulas (D18) and (D19) are equivalent.

The saturation is given by the formula

$$\begin{aligned} S(x,\tau )=\left\{ \begin{array}{ll} S_- &{} \text{ if }\,x<\sigma (\tau ),\\ S_+ &{} \text{ if } \,x>\sigma (\tau ). \end{array} \right. \end{aligned}$$

Let us determine the reduced pressures. Integration of equation (D4) over the interval $0<x<\sigma$, taking into account the boundary conditions (D11) leads to equalities

$$\begin{aligned} \Pi _1+\xi S \Pi _2=-\frac{Vx}{A},\quad \Pi _1+\Pi _2 \frac{\xi S-2c_2}{1-2S}=\frac{\tilde{p}_c}{1-2S}, \end{aligned}$$

which allow to find $\Pi _i$ on the interval $0<x<\sigma$. As for the interval $\sigma<x<1$, these pressures solve the system

$$\begin{aligned} \Pi _1+\xi S \Pi _2=\frac{V(1-x)}{A},\quad \Pi _1+\Pi _2 \frac{\xi S-2c_2}{1-2S}=\frac{\tilde{p}_c}{1-2S}. \end{aligned}$$