Fast SPH simulation for gaseous fluids

Abstract

This paper presents a fast smoothed particle hydro-dynamics (SPH) simulation approach for gaseous fluids. Unlike previous SPH gas simulators, which solve the transparent air flow in a fixed simulation domain, the proposed approach directly solves the visible gas without involving the transparent air. By compensating the density and force calculation for the visible gas particles, we completely avoid the need of computational cost on ambient air particles in previous approaches. This allows the computational resources to be exclusively focused on the visible gas, leading to significant performance improvement of SPH gas simulation. The proposed approach is at least ten times faster than the standard SPH gas simulation strategy and is able to reduce the total particle number by 25–400 times in large open scenes. The proposed approach also enables fast SPH simulation of complex scenes involving liquid–gas transition, such as boiling and evaporation. A particle splitting and merging scheme is proposed to handle the degraded resolution in liquid–gas phase transition. Various examples are provided to demonstrate the effectiveness and efficiency of the proposed approach.

Appendix A: Derivation of Eqs. (7–8)

This appendix shows the detailed derivations of Eqs. (7) and (8). Since densities in adjacent particles are assumed to change gradually in the gas simulations, we can further consider that for a particle $i$, the densities of its neighboring particles $\rho _{i,j}$ are corrected with the same proportion as itself and in a similar form as Eq. (6):

$$\begin{aligned} \rho _{i,j}=\frac{\bar{\rho }_{i,j}}{1-V W_{ik}} \end{aligned}$$

(27)

Here we want to determine the value of the scalar $V$, which for a given particle $i$ does not vary with adjacent particle $j$. Using the form in Eq. (27), Eq. (4) can be rewritten as

$$\begin{aligned} 0=(1-V W_{ik})\sum _j\frac{m_j}{\bar{\rho }_j}\nabla W_{ij} + \frac{m_k}{\rho _i}\nabla W_{ik} \end{aligned}$$

(28)

Utilizing $V_0$ in Eq. (5), the above equation can be further rewritten as

$$\begin{aligned} m_k=(1-V W_{ik})V_0\rho _i \end{aligned}$$

(29)

Note that it has been assumed in the first place that the final correction form should be in the form of Eq. (27), which is equivalent to requiring $m_k=V\rho _i$. Comparing this with Eq. (29), the following equation can be obtained:

$$\begin{aligned} (1-V W_{ik})V_0 = V \end{aligned}$$

(30)

Solving Eq. (30) yields

$$\begin{aligned} V=\frac{V_0}{1+V_0W_{ik}} \end{aligned}$$

(31)

Substituting Eq. (31) into Eqs. (27) and (29) leads to Eqs. (7–8).