A modular, partitioned, discrete element framework for industrial grain distribution systems with rotating machinery

Abstract

A modular discrete element framework is presented for large-scale simulations of industrial grain-handling systems. Our framework enables us to simulate a markedly larger number of particles than previous studies, thereby allowing for efficient and more realistic process simulations. This is achieved by partitioning the particle dynamics into distinct regimes based on their contact interactions, and integrating them using different time-steps, while exchanging phase-space data between them. The framework is illustrated using numerical experiments based on fertilizer spreader applications. The model predictions show very good qualitative and quantitative agreement with available experimental data. Valuable insights are developed regarding the role of lift vs drag forces on the particle trajectories in-flight, and on the role of geometric discretization errors for surface meshing in governing the emergent behavior of a system of particles.

Appendices

Appendix 1: The case of a conical disk with no vanes

Let us consider a conical disk of radius R spinning around its vertical axis with angular velocity $\omega$. Let the Cartesian coordinate system $\left\{ \mathbf {O},\mathbf {i},\mathbf {j},\mathbf {k} \right\}$ be centered at the cone’s vertex. The forces acting on the particle are

$$\begin{aligned} \mathbf {f}_g = -g\mathbf {k} ,\quad \mathbf {f}_R = f_R \mathbf {n},\quad \mathbf {f}_f = \mu f_R \mathbf {n}_{\mathbf {v}}. \end{aligned}$$

(53)

The three forces above correspond to the weight, $\mathbf {f}_g$, the reaction against the cone, $\mathbf {f}_R$, and the friction resistance, $\mathbf {f}_f$. Furthermore, $\mathbf {n}$ is the inner normal to the cone at the particle’s position, $\mathbf {n}_{\mathbf {v}}$ the normalized (when non-zero) relative velocity of the cone with respect to the particle and $f_R$ is the magnitude of the normal reaction to the particle’s forces against the cone. Note that $f_R$ is a priori unknown but must be positive (no attractive forces). Consequently, the Second Law of Newton applied to the particle yields the following system:

$$\begin{aligned} \begin{aligned} m\ddot{x}&= f_R (\mu \mathbf {n}_{\mathbf {v}}\cdot \mathbf {i} - \sin \alpha \cos \theta )\\ m\ddot{y}&= f_R (\mu \mathbf {n}_{\mathbf {v}}\cdot \mathbf {j} - \sin \alpha \sin \theta )\\ m\ddot{z}&= f_R (\cos \alpha + \mu \mathbf {n}_{\mathbf {v}}\cdot \mathbf {k}) - mg, \end{aligned} \end{aligned}$$

(54)

where $\theta$ is the standard polar angular coordinate and $\alpha$ the cone angle ($\alpha = 0$ for a flat disk). Since the particle is forced to move on the surface of the cone, the three coordinates x, y and z are linked by

$$\begin{aligned} z = \tan \alpha \sqrt{x^2 + y^2} \end{aligned}$$

(55)

which can readily be derived yielding

$$\begin{aligned} \ddot{z} = \tan \alpha \frac{(xy^2 + x^3)\ddot{x} + (x^2y + y^3)\ddot{y} + (y\dot{x}-x\dot{y})^2}{\left( x^2 + y^2 \right) ^{3/2}} \end{aligned}$$

(56)

In order to be solved, this system of equations must be provided with initial conditions. Let us consider the following:

$$\begin{aligned} \mathbf {x}(0)=\begin{pmatrix} r_0\\ 0\\ 0 \end{pmatrix},\quad \dot{\mathbf {x}}(0)=\begin{pmatrix} 0\\ v_0\\ 0 \end{pmatrix}, \end{aligned}$$

(57)

where $r_0 \in (0, R)$, with R the radius of the disk.

Let us normalize this system of ODE’s to obtain a non-dimensional analog of it . By repeated application of direct substitution and the chain rule to the derivatives, one may obtain the following equivalent non-dimensional system:

$$\begin{aligned} \begin{aligned} \ddot{\xi }_i&= \phi _R (\mu \mathbf {n}_{\mathbf {v}}\cdot \mathbf {i} - \sin \alpha \cos \theta )\\ \ddot{\xi }_j&= \phi _R (\mu \mathbf {n}_{\mathbf {v}}\cdot \mathbf {j} - \sin \alpha \sin \theta )\\ \ddot{\xi }_k&= \phi _R (\cos \alpha + \mu \mathbf {n}_{\mathbf {v}}\cdot \mathbf {k}) - \gamma \end{aligned} \end{aligned}$$

(58)

with

$$\begin{aligned} \begin{aligned} \ddot{\xi _k} = \tan \alpha \frac{(\xi _i \xi ^2_j + \xi ^3_i)\ddot{\xi _i} + (\xi ^2_i \xi _j + \xi _j^3)\ddot{\xi _j} + (\xi _j\dot{\xi _i}-\xi _i\dot{\xi _j})^2}{\left( \xi ^2_i + \xi ^2_j \right) ^{3/2}} \end{aligned} \end{aligned}$$

(59)

and

$$\begin{aligned} \varvec{\xi }(0)=\begin{pmatrix} \rho _0\\ 0\\ 0 \end{pmatrix},\quad \dot{\varvec{\xi }}(0)=\begin{pmatrix} 0\\ \nu _0\\ 0 \end{pmatrix},\, \end{aligned}$$

(60)

where the dot notation has been abused to refer to the derivatives with respect to the non-dimensional time

$$\begin{aligned} \tau = t\omega \end{aligned}$$

(61)

or the units of time needed to complete a 1 radian turn by the spinning disk. The rest of the new non-dimensional parameters are

$$\begin{aligned} \pmb {\xi }= & {} \frac{1}{R}\begin{pmatrix} x\\ y\\ z \end{pmatrix},\quad \phi _R=\frac{f_R}{m\omega ^2R},\quad \gamma =\frac{g}{\omega ^2R},\quad \rho _0=\frac{r_0}{R},\nonumber \\ \nu _0= & {} \frac{v_0}{\omega R}. \end{aligned}$$

(62)

Thus distances have been normalized by the disk’s radius and the forces by the modulus of the centripetal force felt by a particle stuck to the disk’s rim edge. The variable $\gamma$ can be interpreted as a Froude’s number for the system. From this form of the system it becomes apparent, for example, that the solution does not depend on the mass of the particle, since none of the known variables are defined in terms of it.

The system above can be solved numerically. In this work a simple forward Euler scheme has been employed, in which $\phi _R$ is taken from the old time-step in Eqs. 58 and subsequently updated using 59. The rest of the parameter values are fixed to the quantities given in Table 3.

Table 3 Values of the problem parameters used for the cone with no vanes benchmark

Appendix 2: Derivation of Eq. 19

The incremental expression provided by Thornton reads:

$$\begin{aligned} F_{t,\mathrm{el}}^n = \left\{ \begin{array}{ll} F_{t,\mathrm{el}}^{n-1} + k_t^n{{\Delta }\delta _t} &{} \quad if {\Delta } F_n \ge 0\\ F_{t,\mathrm{el}}^{n-1} \left( \frac{k_t^n}{k_t^{n-1}} \right) + k_t^n{{\Delta }\delta _t}&{} \quad if {\Delta } F_n < 0, \end{array}\right. \end{aligned}$$

(63)

where the super-index n indicates the nth time-step and the operator ${\Delta }$ denotes the forward finite difference operator in time. It is direct to see that the first branch of 63 leads to the first branch of Eq. 19 by taking the limit when the time increments tend to 0. Similarly, the second branch becomes, after dividing it through by $k_t^n$

$$\begin{aligned} {\Delta } \left( \frac{F_{t,\mathrm{el}}}{k_t} \right) = {\Delta }\delta _t \end{aligned}$$

(64)

which yields the expression of the second branch in 19 after dividing both sides by the time increment and taking the limit when it tends to 0.