Calculating the effective permeability of sandstone with multiscale lattice Boltzmann/finite element simulations

Abstract

The lattice Boltzmann (LB) method is an efficient technique for simulating fluid flow through individual pores of complex porous media. The ease with which the LB method handles complex boundary conditions, combined with the algorithm’s inherent parallelism, makes it an elegant approach to solving flow problems at the sub-continuum scale. However, the realities of current computational resources can limit the size and resolution of these simulations. A major research focus is developing methodologies for upscaling microscale techniques for use in macroscale problems of engineering interest. In this paper, we propose a hybrid, multiscale framework for simulating diffusion through porous media. We use the finite element (FE) method to solve the continuum boundary-value problem at the macroscale. Each finite element is treated as a sub-cell and assigned permeabilities calculated from subcontinuum simulations using the LB method. This framework allows us to efficiently find a macroscale solution while still maintaining information about microscale heterogeneities. As input to these simulations, we use synchrotron-computed 3D microtomographic images of a sandstone, with sample resolution of 3.34 μm. We discuss the predictive ability of these simulations, as well as implementation issues. We also quantify the lower limit of the continuum (Darcy) scale, as well as identify the optimal representative elementary volume for the hybrid LB–FE simulations.

Appendix: Pressure boundary condition

In the LB method, pressure and density are related through the equation of state

$$p = c^2 \rho,$$

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where c is the speed of sound (for a perfect gas) on the given lattice. To impose a pressure boundary condition, we therefore seek to impose a specified density. Recall that the density ρ and macroscopic flow velocity u are defined in terms of the discrete particle distributions f _i and their associated velocities e _i ,

$$\rho = \sum_{i=0}^{18} f_i ;\quad \rho {\varvec{u}} = \sum_{i=1}^{18} f_i {\varvec{e}}_i.$$

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In our 3D simulations, we have used a D3Q19 lattice consisting of 18 neighbor links and one rest state. Figure 11 illustrates the basic lattice unit and the numbering scheme we have employed.

Fig. 11

a Typical unit cell for a D3Q19 lattice. Rest node is gray, and 18 neighbor nodes are black. Link vectors are omitted for clarity. b Two orthogonal views of the D3Q19 lattice, illustrating the present numbering scheme. Out-of-plane links are noted with parentheses

Consider a typical lattice node located on the inlet boundary. At this node we assign a density ρ_in, and further specify u _y = u _z = 0. Consider now the density distribution after the streaming step in a typical LB iteration. Any particle population coming from an interior node becomes a known quantity. On the other hand, f ₁, f ₇, f ₈, f ₁₁, and f ₁₂ remain undetermined since these populations stream from points outside the domain. We therefore seek to determine these five unknowns such that they satisfy the specified density and velocity conditions.

To shorten notation let us group the particle populations into three sets according to whether e _x = 1, 0, or −1.

$$X^+ = \{ 1,7,8,11,12 \}.$$

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$$X^o = \{ 0,3,4,5,6,15,16,17,18 \}.$$

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$$X^- = \{ 2,9,10,13,14 \} .$$

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In this notation, any f _i ∈ X ⁺ is an unknown. Our first task is to determine the inlet velocity u _x in terms of the known populations. From the density and momentum equations

$$\rho_{\rm in} = \sum_{X^-} f_i + \sum_{X^o} f_i + \sum_{X^+} f_i ,$$

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$$\rho_{\rm in} u_x = \sum_{X^+} f_i - \sum_{X^-} f_i ,$$

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which implies

$$u_x = 1 - {\frac{1}{\rho_{\rm in}}} \left[ \sum_{X^o} f_i + 2 \sum_{X^-} f_i \right] .$$

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Now that the complete vector u is determined, our goal is to find a particle distribution that satisfies Eqs. 5. These relations, however, are insufficient to determine the unknown f _i uniquely. In light of this fact, we have adopted an approach like that of Zou and He [32]. We note that the equilibrium particle distribution also satisfies the momentum equations,

$$\sum_{i=1}^{18} f^{\rm eq}_i {\varvec{e}}_i = \rho {\varvec{u}}.$$

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This implies that the unknown f _i must satisfy the following in order to maintain the appropriate x-momentum:

$$\sum_{X^+} f_i - \sum_{X^-} f_i = \sum_{X^+} f_i^{\rm eq} - \sum_{X^-} f_i^{\rm eq} .$$

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Let us denote by ${\bar{f}_i}$ the particle population traveling in the opposite direction to f _i ; that is, e _i and ${\bar{\varvec{e}}_i}$ point in opposite directions. Zou and He noted that by choosing the unknown f _i such that

$$f_i - f_i^{\rm eq} = \bar{f}_i - \bar{f}_i^{\rm eq} , $$

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Eq. 13 is automatically satisfied. They described this approach as a “bounceback rule" for the non-equilibrium distribution, in analogy to the commonly used bounceback rule for dealing with solid interfaces. In the D3Q19 model, the equilibrium distributions are given by

$$f_i^{\rm eq} = \rho \omega_i \left[ 1 + 3 {\varvec{u}} \cdot {\varvec{e}}_i + {\frac{9}{2}} ( {\varvec{u}} \cdot {\varvec{e}}_i )^2 - {\frac{3} {2}} {\varvec{u}} \cdot {\varvec{u}} \right] , $$

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with weights ω _i = 1/3, 1/18, and 1/36 for the rest particle, nearest neighbor links, and diagonal links, respectively.

With this simple heuristic in hand, a scheme emerges for determining the unknown particle populations. First, we fix the unknown component normal to the inlet plane,

$$f_1 = f_2 + \left( f_1^{\rm eq} - f_2^{\rm eq} \right) = f_2 + {\frac{1}{3}} \rho_{\rm in} u_x . $$

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For the four remaining unknowns, we apply a similar bounceback rule to ensure that the x-momentum equation is satisfied. We also note, however, that we can adjust the y and z-momenta without altering the x-momentum by using the populations tangential to the inlet plane, f _i ∈ X ^o. In particular, we choose

$$f_7 = f_{10} + {\frac{1} {2}}\left(f_4+f_{17}+f_{18}\right)-{\frac{1} {2}}\left(f_3+f_{15}+f_{16}\right) + {\frac{1}{6}} \rho_{\rm in} u_x ,$$

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$$f_8 = f_{9} - {\frac{1} {2}}\left(f_4+f_{17}+f_{18}\right)+{\frac{1} {2}}\left(f_3+f_{15}+f_{16}\right) + {\frac{1} {6}} \rho_{\rm in} u_x ,$$

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$$f_{11} = f_{14} + {\frac{1} {2}}\left(f_6+f_{16}+f_{18}\right)-{\frac{1} {2}}\left(f_5+f_{15}+f_{17}\right) + {\frac{1} {6}} \rho_{\rm in} u_x ,$$

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$$f_{12} = f_{13} - {\frac{1} {2}}\left(f_6+f_{16}+f_{18}\right)+{\frac{1} {2}}\left(f_5+f_{15}+f_{17}\right) + {\frac{1}{6}} \rho_{\rm in} u_x.$$

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These algebraic expressions determine the unknown f _i such that ρ = ρ_in and u _y = u _z = 0. An identical approach is applied to outlet nodes, except the unknowns become f _i ∈ X ⁻.

$$u_x = {\frac{1}{\rho_{\rm out}}} \left[ \sum_{X^o} f_i + 2 \sum_{X^-} f_i \right] -1 .$$

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$$f_2 = f_1 - {\frac{1}{3}} \rho_{\rm out} u_x .$$

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$$f_9 = f_{8} + {\frac{1} {2}}\left(f_4+f_{17}+f_{18}\right)-{\frac{1} {2}}\left(f_3+f_{15}+f_{16}\right) - {\frac{1} {6}} \rho_{\rm out} u_x .$$

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$$f_{10} = f_{7} - {\frac{1}{2}}\left(f_4+f_{17}+f_{18}\right)+{\frac{1} {2}}\left(f_3+f_{15}+f_{16}\right) - {\frac{1}{6}} \rho_{\rm out} u_x .$$

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$$f_{13} = f_{12} +{\frac{1}{2}}\left(f_6+f_{16}+f_{18}\right)-{\frac{1} {2}}\left(f_5+f_{15}+f_{17}\right) - {\frac{1}{6}} \rho_{\rm out} u_x .$$

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$$f_{14} = f_{11} - {\frac{1}{2}}\left(f_6+f_{16}+f_{18}\right)+{\frac{1}{2}}\left(f_5+f_{15 }+f_{17}\right) - {\frac{1}{6}} \rho_{\rm out} u_x .$$

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