Presentation on theme: "Multiphase shock relations and extra physics"— Presentation transcript:
1 Multiphase shock relations and extra physics Modelling of interfaces separating compressible fluids and mixtures of materialsMultiphase shock relationsand extra physicsErwin FRANQUET and Richard SAURELPolytech Marseille, UMR CNRS IUSTI
2 A Multiphase model with 7 equations For solving interfaces problems and shocks into mixtures Baer & Nunziato (1986)Saurel & al. (2003)Chinnayya & al (2004)
3 Reduction to a 5 equations model When dealing with interfaces and mixtures with stiff mechanical relaxation the 7 equations model can be reduced to a 5 equations modelInfinite drag coefficientInfinite pressure relaxation parameterKapila & al (2001)Not conservative
4 To deal with realistic applications shock relations are mandatory 7 unknowns : α1, Y1, ρ, u, P, e, σ4 conservation laws :Mixture EOS :One of the variable behind the shock is given (often σ or P)An extra relation is needed : jump of volume fraction or any other thermodynamic variableHow to determine it ?
5 Informations from the resolution of the 7 equations model Impact of an epoxy-spinel mixture by a piston at several velocitiesFully dispersed wavesWhy are the waves dispersed in the mixture ?
8 Consequences The two-phase shock is smooth shock = succession of equilibrium states (P1=P2 and u1=u2)We can use the 7 equations model in the limit :That is easier to integrate between pre and post shock states.In that case, the energy equations reduce to :And can be integrated as :andAs P1=P2 at each point, we have :The mixture energy jump is known without ambiguity :To fulfill the mixture energy jump each phase must obey :Saurel & al (2005)
9 Some properties Identifies with the Hugoniot adiabat of each phase Symmetric and conservative formulationEntropy inequality is fulfilledSingle phase limit is recoveredValidated for weak and strong shocks for more than 100 experimental dataThe mixture Hugoniot curve is tangent to mixture isentrope
11 The reduced model (with 5 equations) is now closed + Mixture EOS+ Rankine-Hugoniot relationsConsequences :A Riemann solver can be builtThis one can be used to solve numerically the 5 equations modelThe second difficulty comes from the average of the volume fraction inside computational cells :It is not a conservative variableIt necessitates the building of a new numerical method
12 A new projection method Saurel & al (2005) xxi-1/2xi+1/2t n+1u*i-1/2u*i+1/2S+i-1/2S-i+1/2Euler equations contextVolume fractions definitionu1, P1, e1u2, P2, e2u3, P3, e3V1V2V3
13 Relaxation systemConservation and entropy inequality are preserved if :This ODE system is solved in each computational cell so as to reach the mechanical equilibrium asymptotic state ( )It can be written as an algebraic system solved with the Newton method.
14 Comparison with conventional methods Conventional Godunov average supposes a single pressure, velocity and temperature in the cell. In the new method, we assume only mechanical equilibrium and not temperature equilibrium.It guarantees conservation and volume fraction positivityThe method does not use any flux and is valid for non conservative equationsIn the case of the ideal gas and the stiffened gas EOS with the Euler equations both methods are equivalent. Results are different for more complicated EOS (Mie-Grüneisen for example)The new method gives a cure to anomalous computation of some basic problems:- Sliding lines- Propagation of a density discontinuity in an uniform flow with Mie Grüneisen EOS.It can be used in Lagrange or Lagrange + remap codes.
15 Propagation of a density discontinuity in an uniform flow with JWL EOS P = PCJ = Pau = 1000 m/sρ = ρCJ = 2182 kg/m3ρ = 100 kg/m30,51
16 Shock tube problem in extreme conditions Euler equations and JWL EOS P = PCJ = Paρ = ρCJ = 2182 kg/m3P = Paρ = 100 kg/m30,51The Godunov method fails in these conditions
17 Shock tube problem with almost pure fluids Liquid-Gas interface with the 5 equations model 0,81P = 105 Paαair =P = 109 Paαwater =ρwater = 1000 kg/m3ρair = 50 kg/m3Stiffened Gas EOS
18 Shock tube problem with mixtures of materials Epoxy-Spinel Mixture 0,61P = 105 Paαepoxy = 0,5954P = 1010 Paρepoxy = 1185 kg/m3ρspinel = 3622 kg/m3Stiffened Gas EOS
19 Shock tube problem with mixtures of materials (2)
20 Mixture Hugoniot tests Comparison with experiments and the 7 equations model New method7 equations modelP = 105 Paαepoxy = 0,5954ρepoxy = 1185 kg/m3ρspinel = 3622 kg/m3UpPistonEpoxy-Spinel Mixture
21 2D impact of a piston on a solid stucture RDX(Mie-Grüneisen EOS)TNT(JWL EOS)Copper (Stiffened Gas EOS)U = 5000 m/sAir (Ideal Gas EOS)