A simple and novel coupling method for CFD–DEM modeling with uniform kernel-based approximation

Abstract

The Eulerian–Lagrangian approach is commonly employed in particulate flows, which can be classified into two subcategories: unresolved computational fluid dynamics-discrete element method (CFD–DEM) and resolved CFD–DEM. When the particle sizes are comparable to the cell sizes, the interphase coupling is not straightforward anymore and both the conventional unresolved CFD–DEM and resolved CFD–DEM are not applicable. In this paper, we propose a simple and novel coupling method for projecting and reconstructing the particle and interphase quantities, which is also called semi-resolved CFD–DEM. The particle quantities are uniformly distributed to the surrounding cells by expanding the fluid domain. The fluid phase quantities at the particle location are also reconstructed from the surrounding cells. Then, the relative velocity and the local void fraction for calculating the drag force are corrected. The expanding factor for the fluid domain is determined by comparing the drag force of the semi-resolved CFD–DEM with the resolved CFD–DEM results. It is found that the expanding factor increases linearly with the autocorrelation length. The developed method is validated by the simulation of a particle sedimentation and a sediment transport process, which proves that the present semi-resolved CFD–DEM fills the gap between the resolved and unresolved CFD–DEM. The difference between the implicit and explicit treatment of momentum exchange term is also discussed, and the explicit treatment shows better performance for large particles.

Appendix A. Contact force model

The expressions of the particle contact model can be found in [30] but are repeated here for the reader. When two particles overlap by a normal distance, $\varvec{\delta }_n$, the normal ($\varvec{f}_{n,ij}$) and tangential ($\varvec{f}_{t,ij}$) contact force are given by

$$\begin{aligned} \varvec{f}_{n,ij}= & {} k_n\varvec{\delta }_n- \eta _n\Delta \textbf{U}_n , \end{aligned}$$

(A1)

$$\begin{aligned} \varvec{f}_{t,ij}= & {} {\left\{ \begin{array}{ll} k_t\varvec{\delta }_t- \eta _t\Delta \textbf{U}_t \qquad |\varvec{f}_{t,ij}|\le -\mu _s|\varvec{f}_{n,ij}|\frac{\Delta \textbf{U}}{\left| \Delta \textbf{U}\right| }\\ -\mu _s\left| \varvec{f}_{n,ij}\right| \frac{\Delta \textbf{U}}{\left| \Delta \textbf{U}\right| } \ \ \ \ |\varvec{f}_{t,ij}|>-\mu _s|\varvec{f}_{n,ij}|\frac{\Delta \textbf{U}}{\left| \Delta \textbf{U}\right| } \end{array}\right. } ,\nonumber \\ \end{aligned}$$

(A2)

where $k_n$ and $k_t$ are the normal and tangential spring constants, respectively. $\eta _n$ and $\eta _t$ are the normal and tangential damping coefficients, respectively. $\mu _s$ is the sliding friction coefficient. $\textbf{U}_n$ and $\textbf{U}_t$ are the relative normal and tangential velocities between the particles. The spring constant and damping coefficient are given by the following equations

$$\begin{aligned} k_n= & {} \frac{4}{3}Y^{*}\sqrt{R^*|\varvec{\delta }_n|} , \end{aligned}$$

(A3)

$$\begin{aligned} \eta _n= & {} 2\sqrt{\frac{5}{6}}\beta \sqrt{S_{n}m^*} , \end{aligned}$$

(A4)

$$\begin{aligned} k_t= & {} 8G^{*}\sqrt{R^*|\varvec{\delta }_n|} , \end{aligned}$$

(A5)

$$\begin{aligned} \eta _t= & {} 2\sqrt{\frac{5}{6}}\beta \sqrt{S_{t}m^*} . \end{aligned}$$

(A6)

$S_n$, $S_t$, and $\beta$ are given by

$$\begin{aligned} S_n= & {} 2Y^*\sqrt{R^*|\varvec{\delta }_n|} , \end{aligned}$$

(A7)

$$\begin{aligned} S_t= & {} 8G^*\sqrt{R^*|\varvec{\delta }_t|} , \end{aligned}$$

(A8)

$$\begin{aligned} \beta= & {} \frac{ln(e)}{\sqrt{ln^2(e)+\pi ^2}} , \end{aligned}$$

(A9)

where e is the coefficient of restitution. The effective Young’s modulus, $Y^*$, effective shear modulus, $G^*$, effective radius, $R^*$, and effective mass, $m^*$ are given by

$$\begin{aligned} \frac{1}{Y^*}= & {} \frac{\left( 1-\nu _i^2\right) }{Y_i}+\frac{\left( 1-\nu _j^2\right) }{Y_j} , \end{aligned}$$

(A10)

$$\begin{aligned} \frac{1}{G^*}= & {} \frac{2(2+\nu _i)(1-\nu _i)}{Y_i}+\frac{2(2+\nu _j)(1-\nu _j)}{Y_j} , \end{aligned}$$

(A11)

$$\begin{aligned} \frac{1}{R^*}= & {} \frac{1}{R_i}+\frac{1}{R_j} , \end{aligned}$$

(A12)

$$\begin{aligned} \frac{1}{m^*}= & {} \frac{1}{m_i}+\frac{1}{m_j} , \end{aligned}$$

(A13)

where $\nu$ is the Poisson’s ratio.