Application of Compressible Volume of Fluid Model in Simulating the Impact and Solidification of Hollow Spherical ZrO₂ Droplet on a Surface

Abstract

Applications of hollow spherical particles in thermal spraying process have been developed in recent years, accompanied by attempts in the form of experimental and numerical studies to better understand the process of impact of a hollow droplet on a surface. During such process, volume and density of the trapped gas inside droplet change. The numerical models should be able to simulate such changes and their consequent effects. The aim of this study is to numerically simulate the impact of a hollow ZrO₂ droplet on a flat surface using the volume of fluid technique for compressible flows. An open-source, finite-volume-based CFD code was used to perform the simulations, where appropriate subprograms were added to handle the studied cases. Simulation results were compared with the available experimental data. Results showed that at high impact velocities (U ₀ > 100 m/s), the compression of trapped gas inside droplet played a significant role in the impact dynamics. In such velocities, the droplet splashed explosively. Compressibility effects result in a more porous splat, compared to the corresponding incompressible model. Moreover, the compressible model predicted a higher spread factor than the incompressible model, due to planetary structure of the splat.

Appendix

To obtain the bulk modulus elasticity of mixture, first of all the modulus of elasticity of each phase is written as (Ref 25):

$$ E_{\text{l}} = \frac{{ - {\text{d}}P}}{{\left( {\frac{{{\text{d}}V_{\text{l}} }}{{V_{\text{l}} }}} \right)}}, $$

(21)

$$ E_{\text{g}} = \frac{{ - {\text{d}}P}}{{\left( {\frac{{{\text{d}}V_{\text{g}} }}{{V_{\text{g}} }}} \right)}}. $$

(22)

Then, the bulk modulus of elasticity of mixture is defined as:

$$ E_{\text{mix}} = \frac{{ - {\text{d}}P}}{{\left( {\frac{{{\text{d}}V}}{V}} \right)}}, $$

(23)

where $V = V_{\text{g}} + V_{\text{l}}$ and ${\text{d}}V = {\text{d}}V_{\text{g}} + {\text{d}}V_{\text{l}}$. Using ${\text{d}}V_{\text{l}}$ and ${\text{d}}V_{\text{g}}$ from relations (21) and (22) results in:

$$ {\text{d}}V = {\text{d}}V_{\text{g}} + {\text{d}}V_{\text{l}} = - \left( {\frac{{V_{\text{l}} {\text{d}}P}}{{E_{\text{l}} }} + \frac{{V_{\text{g}} {\text{d}}P}}{{E_{\text{g}} }}} \right) = - \left( {\frac{V}{{E_{\text{l}} }} + \frac{{V_{\text{g}} }}{{E_{\text{g}} }}} \right){\text{d}}P. $$

(24)

Finally, from Eq 23 and 24 we have:

$$ E_{\text{mix}} = \frac{{ - {\text{d}}P}}{{\left( {\frac{{ - \left( {\frac{{V_{\text{l}} }}{{E_{\text{l}} }} + \frac{{V_{\text{g}} }}{{E_{\text{g}} }}} \right){\text{d}}P}}{{V_{\text{g}} + V_{\text{l}} }}} \right)}} = \frac{1}{{\frac{{V_{\text{l}} }}{{V_{\text{g}} + V_{\text{l}} }}\frac{1}{{E_{\text{l}} }} + \frac{{V_{\text{g}} }}{{V_{\text{g}} + V_{\text{l}} }}\frac{1}{{E_{\text{g}} }}}} = \frac{1}{{\frac{1 - p}{{E_{\text{l}} }} + \frac{p}{{E_{\text{g}} }}}}. $$

(25)