Modélisation de l'écoulement diphasique dans les injecteurs Diesel

CAV2001:sessionB6.005 3It can be clearly seen from figure 1 that the sound speed of the homogeneous mixture decreasesdramatically as soon as the fluid is not composed of a single phase. This can be explained by themultiple reflexions of the waves between the mixture’s components interfaces. As a matter of fact,10 4 Void fraction (−)10 3Sound velocity (m/s)10 210 110 010 −10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Fig. 1 – Sound speed, a, versus void fraction, α.we can almost consider that, as soon as cavitation appears in a numerical cell, the flow becomeslocally supersonic. We will see in sections 2.4.2 and 2.4.4 that this conclusion is very importantwith regards to boundary conditions.As we consider the flow as isentropic, we can assume that :a 2 = dPdρ . (7)Considering the acoustic speed as constant in pure vapour and pure liquid, we can calculate theequation of state for the two-phase region H by integrating the expression 7 between the saturationpressure P sat l and the indicated pressure P (see figure 2) :⎧⎪⎨⎪⎩P = ρa 2 g if ρ < ρ gP = P sat l + ρ ga 2 gρ l a 2 l (ρ g − ρ l )ρ g2 a 2 g − ρ l2 a 2 lP = ρa 2 lif ρ > ρ l .log[ ]ρg a 2 g(ρ l + α(ρ g + ρ l ))ρ l (ρ g a 2 g − ρ l a 2 l ) if ρ g < ρ < ρ lThis equation is used in the code to evaluate analytically P from ρ calculated by the continuityequation.2.3 Numerical schemeThe numerical scheme for advection consists in a quasi-second order upwind scheme (QSOU)as reported in Amsden et al. 1989, which assures the monotonicity of the solution. That is to saythat no negative values (by oscillations) can be obtained, despite pressure and density ratios aretremendous. This is very important with regards to calculation robustness. Furthermore, as wehave to consider the transient behaviour of cavitation, the time advancement is made thanks to athird-order Runge-Kutta scheme.(∂ ρ∂ t) k ( ) kThe procedure is as follows : defined at cell centers and ∂ ρU∂ t defined at vertices arecalculated from equations 1 and 2 thanks to a finite-volume method. Then, according to Runge-(8)