An Orthorhombic Lattice Boltzmann Model for Pore-Scale Simulation of Fluid Flow in Porous Media

Abstract

One application of the lattice Boltzmann equation (LBE) models is in combination with tomography to simulate pore-scale flow and transport processes in porous media. Most LBE models in the literature are based on cubic lattice, and if the voxels in a tomography image are not cubic or cannot be divided into cubes due to computational limitations, these models will lose most of their advantages. How to deal with such images is, hence, an interest in use of the LBE model to simulate pore-scale processes. In this paper, we present an orthorhombic LBE model based on the single-relaxation time approach with the relaxation parameter varying with lattice directions. The equilibrium distribution functions in the standard LBE model were modified to correct the anisotropy induced by the non-cubic lattice, and the calculations of the fluid density and momentum were also redefined in order to maintain the conservation of mass and momentum during the collision. We tested the model against analytical solution for fluid flow in a tube, and against the standard cubic-based LBE model for fluid flow in a duct with an island inside. The model was then applied to simulate fluid flow in a 3D image in attempts to analyse the errors if the voxels in the image are not cubic but are assumed to be cubic.

Appendix

We prove here that the proposed lattice Boltzmann model recovers the Navier–Stokes equations at the limits using the Chapman–Enskog expansion, and determine the relationship between the grid shape parameters and the six relaxations parameters. Assuming the time step is small and in the order of $\varepsilon$, that is, $\delta t=\varepsilon$, each particle distribution function can then be expanded as follows:

$$\begin{aligned} f_i =f_i^{(0)} +\varepsilon f_i^{(1)} +\varepsilon ^{2}f_i^{(2)} +\varepsilon ^{3}f_i^{(3)} +\cdot \cdot \cdot \end{aligned}$$

(13)

Using the Taylor expansion, the terms on the left-hand side of Eq. (1) can be expanded around $({\varvec{x}},\, t)$ as follows:

$$\begin{aligned} \varepsilon \left( {\partial _t +e_{i_\alpha } \partial x_\alpha } \right) f_i +0.5\varepsilon ^{2}\left( {\partial _t +e_{i_\alpha } \partial x_\alpha } \right) ^{2}f_i =-\tau _i \left[ {f_i -f_i^\mathrm{eq} } \right] +O\left( {\varepsilon ^{3}} \right) . \end{aligned}$$

(14)

After substituting Eqs. (13) into (14), equalling the terms with the same power of $\varepsilon$ on both sides of Eq. (14) gives

For terms in $\varepsilon ^{0}$:

$$\begin{aligned} f_i^{(0)} =f_i^\mathrm{eq}, \end{aligned}$$

(15)

For terms in $\varepsilon$:

$$\begin{aligned} \left( {\partial _t +e_{i\alpha } \partial x_\alpha } \right) f_i^{(0)} =-\tau _i f_i^{(1)}, \end{aligned}$$

(16)

For terms in $\varepsilon ^{2}$:

$$\begin{aligned} \left( {\partial _t +e_{i\alpha } \partial x_\alpha } \right) f_i^{(1)} +0.5\left( {\partial _t +e_{i\alpha } \partial x_\alpha } \right) ^{2}f_i^{(0)} =-\tau _i f_i^{(2)}. \end{aligned}$$

(17)

Substituting Eqs. (16) into (17) yields

$$\begin{aligned} \left( {1-0.5\tau _i } \right) \left( {\partial _t +e_{i\alpha } \partial x_\alpha } \right) f_i^{(1)} =\tau _i f_i^{(2)}. \end{aligned}$$

(18)

Summing (16) over all the 19 directions leads to

$$\begin{aligned} \frac{\partial \rho }{\partial t}+\rho _0 \left( {\frac{\partial u_x }{\partial x}+\frac{\partial u_y }{\partial y}+\frac{\partial u_z }{\partial z}} \right) =0. \end{aligned}$$

(19)

We take the momentum in the $x$ direction as an illustrative example. Multiplying $e_{ix}$ to both sides of Eq. (16) and then summing it over all the 19 directions give, with the help of Eq. (6), the following equation:

$$\begin{aligned} \sum \limits _{i=0}^{18} {e_{ix} \left( {\frac{\partial f_i^\mathrm{eq} }{\partial t}+e_{ix} \frac{\partial f_i^\mathrm{eq} }{\partial x}+e_{iy} \frac{\partial f_i^\mathrm{eq} }{\partial y}+e_{iz} \frac{\partial f_i^\mathrm{eq} }{\partial z}} \right) =0}. \end{aligned}$$

(20)

This gives

$$\begin{aligned} \frac{\partial \rho _0 u_x }{\partial t}+\frac{\partial \left( {\rho _0 u_x u_x } \right) }{\partial x}+\frac{\partial \left( {\rho _0 u_y u_x } \right) }{\partial y}+\frac{\partial \left( {\rho _0 u_z u_x } \right) }{\partial z}=-\frac{\partial p}{\partial x}. \end{aligned}$$

(21)

Similarly, in the y and z directions, we have

$$\begin{aligned} \begin{aligned}&\frac{\partial \rho _0 u_y }{\partial t}+\frac{\partial \left( {\rho _0 u_y u_y } \right) }{\partial y}+\frac{\partial \left( {\rho _0 u_y u_x } \right) }{\partial x}+\frac{\partial \left( {\rho _0 u_z u_y } \right) }{\partial z}=-\frac{\partial p}{\partial y}, \\&\frac{\partial \rho _0 u_z }{\partial t}+\frac{\partial \left( {\rho _0 u_x u_z } \right) }{\partial x}+\frac{\partial \left( {\rho _0 u_y u_z } \right) }{\partial y}\frac{\partial \left( {\rho _0 u_z u_z } \right) }{\partial z}=-\frac{\partial p}{\partial z}. \end{aligned} \end{aligned}$$

(22)

Multiplying $e_{ix}$ to both sides of Eq. (18) gives

$$\begin{aligned} \sum \limits _{i=0}^{18} {\omega _i \left[ {e_{ix} e_{ix_\alpha } \frac{\partial }{\partial x_\alpha }\left( {\frac{\partial f_i^\mathrm{eq} }{\partial t}+e_{ix_\beta } \frac{\partial f_i^\mathrm{eq} }{\partial x_\beta }} \right) } \right] } =0, \end{aligned}$$

(23)

where $\omega _i =(1/\tau _i -0.5)$, and the subscripts $\alpha$ and $\beta$ are the Cartesian coordinate indexes. Like the traditional lattice Boltzmann model, we assume that the Mach number is small. Hence, the terms involving time derivatives in Eq. (23) can be neglected. This gives

$$\begin{aligned} \begin{aligned}&\frac{1}{3}\frac{\partial }{\partial x}\left[ {\frac{\left( {a_x \omega _x +b_x \omega _{xy} +b_x \omega _{xz} } \right) \delta x^{2}}{\delta t^{2}}\frac{\partial \rho _0 u_x }{\partial x}+\frac{b_y \omega _{xy} \delta x^{2}}{\delta t^{2}}\frac{\partial \rho _0 u_y }{\partial y}+\frac{\omega _{xz} \delta x^{2}}{\delta t^{2}}\frac{\partial \rho _0 u_z }{\partial z}} \right] \\&\quad +\frac{1}{3}\frac{\partial }{\partial y}\left[ {\frac{b_y \omega _{xy} \delta x^{2}}{\delta t^{2}}\frac{\partial \rho _0 u_y }{\partial x}+\frac{b_x \omega _{xy} \delta y^{2}}{\delta t^{2}}\frac{\partial \rho _0 u_x }{\partial y}} \right] \\&\quad +\frac{1}{3}\frac{\partial }{\partial z}\left[ {\frac{\omega _{xz} \delta x^{2}}{\delta t^{2}}\frac{\partial \rho _0 u_z }{\partial x}+\frac{b_x \omega _{xz} \delta z^{2}}{\delta t^{2}}\frac{\partial \rho _0 u_x }{\partial z}} \right] =0. \end{aligned} \end{aligned}$$

(24)

With the help of the continuity equation, rearranging terms in Eq. (12) gives

$$\begin{aligned} \frac{\left[ {a_x \omega _x \!+\! b_x \left( {\omega _{xy} \!+\!\omega _{xz} } \right) -2\omega _{xz} } \right] \delta x^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_x }{\partial x^{2}}\!+\!\frac{b_x \omega _{xy} \delta y^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_x }{\partial y^{2}}\!+\!\frac{b_x \omega _{xz} \delta z^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_x }{\partial z^{2}}=0.\nonumber \\ \end{aligned}$$

(25)

The isotropic viscosity requires

$$\begin{aligned} \begin{aligned}&\left( {a_x \omega _x +b_x \omega _{xy} +b_x \omega _{xz} -2\omega _{xz} } \right) \delta x^{2}=b_x \omega _{xy} \delta y^{2}, \\&\omega _{xy} \delta y^{2}=\omega _{xz} \delta z^{2}. \\ \end{aligned} \end{aligned}$$

(26)

Using this relationship and adding Eq. (22)+ $\varepsilon$ Eq. (25) yields

$$\begin{aligned} \frac{\partial u_x }{\partial t}\!+\!\frac{\partial \left( {u_x u_x } \right) }{\partial x}\!+\!\frac{\partial \left( {u_y u_x } \right) }{\partial y}\!+\!\frac{\partial \left( {u_z u_x } \right) }{\partial z}=-\frac{\partial p}{\rho _0 \partial x}+\frac{\omega _{xz} \delta z^{2}}{3\delta t}\left( {\frac{\partial ^{2}u_x }{\partial x^{2}}+\frac{\partial ^{2}u_x }{\partial y^{2}}+\frac{\partial ^{2}u_x }{\partial z^{2}}} \right) .\nonumber \\ \end{aligned}$$

(27)

Similarly, for the momentums in the $y$ direction, we have the following equation:

$$\begin{aligned} \frac{\left[ {a_y \omega _y \!+\! b_y \left( {\omega _{xy} \!+\!\omega _{yz} } \right) \!-\!2\omega _{yz} } \right] \delta y^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_y }{\partial y^{2}}+\frac{b_y \omega _{xy} \delta x^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_y }{\partial x^{2}}+\frac{b_y \omega _{yz} \delta z^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_y }{\partial z^{2}}\!=\!0,\nonumber \\ \end{aligned}$$

(28)

and for the momentums in the $z$ direction, we have

$$\begin{aligned} \frac{\left( {\omega _z +\omega _{xz} +\omega _{yz} -2b_y \omega _{yz} } \right) \delta z^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_z }{\partial z^{2}}+\frac{\omega _{xz} \delta x^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_z }{\partial x^{2}}+\frac{\omega _{yz} \delta y^{2}}{3\delta t^{2}}\frac{\partial \rho _0 ^{2}u_z }{\partial y^{2}}=0.\qquad \end{aligned}$$

(29)

Since the viscosity must be isotropic, it requires the grid shape parameters, $a_{x},\, b_{x},\, a_{y}$ and $b_{y}$ in the above equations to satisfy

$$\begin{aligned} \begin{aligned}&b_x =\delta x^{2}/\delta z^{2}, \\&b_y =\delta y^{2}/\delta z^{2}. \\ \end{aligned} \end{aligned}$$

(30)

Hence, the relationship between the relaxation parameters in the diagonal directions is obtained as

$$\begin{aligned} \begin{aligned}&b_y \omega _{xy} =\omega _{xz}, \\&b_x \omega _{xy} =\omega _{yz}, \\&b_x \omega _{xz} =b_y \omega _{yz}. \end{aligned} \end{aligned}$$

If the fluid viscosity is $\mu$, we can first calculate the relaxation parameter $\omega _{xz}$ from

$$\begin{aligned} \mu =\omega _{xz} \delta x^{2}/6\delta t. \end{aligned}$$

(31)

The other five relaxation parameters can then be calculated from

$$\begin{aligned} \begin{aligned}&\omega _{yz} \delta y^{2}=\omega _{xz} \delta x^{2}, \\&\omega _{xy} b_y =\omega _{xz}, \\&\left[ {a_x \omega _x +b_x \left( {\omega _{xy} +\omega _{xz} } \right) -2\omega _{xz} } \right] =\omega _{xz}, \\&\left[ {a_y \omega _y +b_y \left( {\omega _{xy} +\omega _{yz} } \right) -2\omega _{yz} } \right] \delta y^{2}=b_y \omega _{yz} \delta z^{2}, \\&\left[ {\omega _z +\left( {\omega _{xz} +\omega _{yz} } \right) -2b_x \omega _{xz} } \right] \delta z^{2}=\omega _{xz} \delta x^{2}. \end{aligned} \end{aligned}$$

(32)

Under these conditions, the model recovers an isotropic viscosity and a symmetrical stress tensor. It is easy to prove that for cubic lattice,$\tau _x =\tau _y =\tau _z =\tau _{xy} =\tau _{xz} =\tau _{yz} =\tau$, the viscosity is given by $\mu =\left( {1/\tau -0.5} \right) \delta l^{2}/\delta t$, where $\delta l$ is the side length of the cubes. However, the proposed model does not converge to the scheme of Qian et al. (1992) when the cuboids reduce to cubes.