A Godunov-type method for the diphasic Baer-Nunziato model

Transcription

1 A Godunov-type method for the diphasic Baer-Nunziato model C. Chalons Joint work with : A. Ambroso, P.-A. Raviart with benefit from discussions with F. Coquel (CNRS, Ecole Polytechnique/CMAP) N. Seguin (University Paris 6/LJLL) Strasbourg, April, /35

2 THE 7-EQUATION MODEL In 1D and dimensionless form, the model reads α k α k + u I t x =Θ(p k p l ), t (α k k)+ x (α k ku k )=0, t (α k ku k )+ x (α k( ku 2 k + p k)) p I α k x =α k kg Λ(u k u l ), t (α k ke k )+ x (α k( ke k + p k )u k ) p I u I α k x =α k kgu k p I Θ(p k p l ) u I Λ(u k u l ) withα 1 +α 2 = 1 u I, p I : interfacial velocity and pressure (to be precised) We note that the system is of the general nonconservative form U t + x F(U)+B(U) U x = S(U) 2/35

3 THE 7-EQUATION MODEL In 1D and dimensionless form, the model reads α k α k + u I t x =Θ(p k p l ), t (α k k)+ x (α k ku k )=0, t (α k ku k )+ x (α k( ku 2 k + p k)) p I α k x =α k kg Λ(u k u l ), t (α k ke k )+ x (α k( ke k + p k )u k ) p I u I α k x =α k ku k g p I Θ(p k p l ) u I Λ(u k u l ) We assume that the drag force and pressure relaxation coefficients are given by Θ= θ(u) ǫ 2 for a small parameterǫ. Then we have Λ= λ(u) ǫ 2 u 1 u 2 p 2 p 1 =O(ǫ 2 ), u 2 u 1 =O(ǫ) 3/35

4 ASYMPTOTIC ANALYSIS Following the Chapman-Enskog method, assume that { pr = p 1 p 2 = 0+ǫp 1 r+o(ǫ 2 ) u r = u 1 u 2 = 0+ǫu 1 r +O(ǫ2 ) and set ρ=α 1 ρ 1 +α 2 ρ 2 ρu=α 1 ρ 1 u 1 +α 2 ρ 2 u 2 ρe=α 1 ρ 1 e 1 +α 2 ρ 2 e 2 ρy=α 2 ρ 2 A first-order approximation w.r.t.ǫ of the 7-equation model is given by the following differential drift-flux model t ρ+ x ρu=0 t ρy+ x (ρyu+ρy(1 Y)u r )=0 t ρu+ x (ρu 2 + p+ρy(1 Y)u 2 r )=ρg t ρe+ x (ρeu+pu+ρy(1 Y)u 2 r u)=ρgu with the following Darcy-like differential relation for the relative velocity u r u r =ǫ 2ρY(ρ ρy) ( 1 1 ) xp λ(u) ρ 1 ρ 2 ρ 4/35

5 ASYMPTOTIC ANALYSIS The solutions of the 7-equation model behave asǫ 0 like the ones of the drift-flux model with u r u r =ǫ 2ρY(ρ ρy) ( 1 1 ) xp λ(u) ρ 1 ρ 2 ρ Note that in the case of permanent flows (i.e. long-time limit solutions in the Lagrangian frame) x p= ρg so that the drift law reads u r u r = ǫ 2ρY(ρ ρy) ( 1 1 )g λ(u) ρ 1 ρ 2 See Ambroso-Chalons-Coquel-Galié-Godlewski-Raviart-Seguin See Dellacherie See Guillard-Murrone 5/35

6 MATHEMATICAL PROPERTIES Eigenvalues of the Jacobian matrix F (U)+B(U) are always real and given by where c k is the sound speed of phase k u I u k u k ± c k k=1, 2 The system is weakly hyperbolic : resonance occurs if u I = u k ± c k Eigenvalues u k ± c k are GNL and eigenvalues u k are LD Eigenvalue u I if LD provided that χα 1 1 u I =βu 1 + (1 β)u 2, β= χα 1 1+ (1 χ)α 2 2 whereχis a constant such thatχ [0, 1], for instanceχ=0, 1, 1 2 See for instance Gallouet-Hérard-Seguin 6/35

7 MATHEMATICAL PROPERTIES Regarding the interfacial pressure, we set p I =µp 1 + (1 µ)p 2, µ=µ(u) [0, 1] The choice ofµis related to entropy considerations Note s k = s k ( k,ε k ) the specific entropy and T k the temperature of the phase k Letφ=φ(s) be a decreasing function. We set η=η(u)= 2 α k kφ(s k ), q=q(u)= k=1 The following entropy inequality holds true if η t + q x 0 (1 β)t 2 µ=µ(β)= βt 1 + (1 β)t 2 The pair (η, q) is a mathematical entropy pair ifφis convex 2 α k kφ(s k )u k. k=1 7/35

8 NUMERICAL DIFFICULTIES The Riemann problem is difficult to solve due to the large number of waves (7) and intermediate states (6) the presence of the nonconservative products 1 p I x α k and p I u I x α k the possibly strong nonlinearities of the pressure laws p k t u 1 u I u 2 u k c k u k + c k U L U R x FIG.: General structure of a Riemann problem 1 These are defined thanks to the Riemann invariants 8/35

9 OBJECTIVES Our objective is to propose an Approximate Riemann Solver : with the simplest possible structure (with a predefined waves position, without underlying nonlinear algorithms) with an accurate treatment of the nonconservative products able to deal with any equation of state and any choice (u I, p I ) leading to an asymptotic-preserving Godunov-type scheme (upwinding the sources...) 9/35

10 HOW TO GET THE ASYMPTOTIC-PRESERVING PROPERTY? How to get the asymptotic-preserving (AP) property? In a recent work by Chalons, Coquel, Godlewski, Raviart and Seguin (M3AS, 2010), devoted to the Euler equations with friction and gravity t ρ+ x ρu=0 t ρu+ x (ρu 2 + p)=ρ(g 1 ǫ φ(u)) t ρe+ x (ρeu+pu)=ρ(gu 1 ǫ ψ(u)), we proved that a good stretagy to get the AP property was to consider approximate Riemann solvers that are consistent with the integral form of the full model, i.e. including both the convective part and the sources. 10/35

11 CONSISTENCY OF SIMPLE APPROXIMATE RIEMANN SOLVERS We consider the nonconservative system with sources U t + x F(U)+B(U) U x = S(U) A simple Approximate Riemann Solver has the following form x U 1 = U L, t<σ 1, W ( x t ; U L, U R )= U k, σ k 1 < x t<σ k, k= 2,.., m, x U m+1 = U R, t>σ m. whereσ k =σ k (U L, U R ), 1 k m and the intermediate states may depend on =( x, t) Following Gallice (2002), we remind how to design a consistent and simple Approximate Riemann Solver with initial data { UL if x<0, U(x, 0)=U 0 (x)= U R if x>0 11/35

12 CONSISTENCY IN THE INTEGRAL SENSE Definition. The simple Approximate Riemann solver is said to be consistent with the integral form if there exists a matrix B (U L, U R ) and a vector S (U L, U R ) such that on the first hand lim U L, U R U 0 on the other hand B (U L, U R )=B(U), lim U L, U R U 0 F+B (U L, U R ) U x S (U L, U R )= S (U L, U R )=S(U) m σ k (U k+1 U k ) under the usual CFL condition and where we have set U=U R U L and F=F(U R ) F(U L ). k=1 In the case of a usual system of conservation laws, we recover the classical relation m F= σ k (U k+1 U k ) k=1 12/35

13 SO... Our ojective is to propose an Approximate Riemann Solver for α k α k + u I t x = 0, t (α k k)+ x (α k ku k )=0, t (α k ku k )+ x (α k( ku 2 k + p k)) p I α k x =α k kg Λ(u k u l ), t (α k ke k )+ x (α k( ke k + p k )u k ) p I u I α k x =α k kgu k u I Λ(u k u l ) with the property of consistency with the integral form of the full model 13/35

14 HOWTODEALWITHGENERALEQUATIONSOFSTATE? How to deal with general equations of state? Like in a HLLC-approach, we linearize the system replacingα k p k withπ k : α k α k + u I t x = 0 α k k t + α k ku k x = 0 α k ku k t α k ke k t + x (α k ku 2 k +Π k) p I α k x =α k kg K(u k u l ) + x ((α k ke k +Π k )u k ) p I u I α k x =α k kgu k u I K(u k u l ) Note that if we setπ k,l =α k,l p k,l andπ k,r =α k,r p k,r, the consistency relations F+B (U L, U R ) U x S (U L, U R )= m σ k (U k+1 U k ) k=1 are identical the system under consideration being linearized or not 14/35

15 HOW TO GET THE SIMPLEST POSSIBLE STRUCTURE? How to get the simplest possible structure? We focus on the following structure which consists in : "averaging" the waves u 1, u 2 and u I into a single coupling-wave ũ I approximating the characteristic speeds u k ± c k by u k ± a k /ρ k t u I u k,l a k,l /ρ k,l u k,r + a k,r /ρ k,r U L U R x FIG.: General structure of an approximate Riemann solution 15/35

16 THEN... We determine the intermediate states by : imposing Rankine-Hugoniot jump conditions of the linearized system across the discontinuities of the Approximate Riemann Solver imposing the validity of the consistency relations We then get : explicit formulas for the intermediate states an accurate treatment of the non conservative products (the stationary coupling waves are exactly computed) the propertiesα k (0, 1) and (α k ρ k )>0 the conservativity of (α k ρ k ), (α 1 ρ 1 u 1 +α 2 ρ 2 u 2 ) and (α 1 ρ 1 E 1 +α 2 ρ 2 E 2 ) 16/35

17 NUMERICAL RESULTS We considered several 1D test cases to assess the strategy : Riemann problems : isolated coupling waves general Riemann problems including comparisons with other schemes available in the literature (HLL-scheme proposed by Saurel-Abgrall, the scheme proposed by Andrianov-Saurel-Warnacke, VFRoe-scheme by Gallouët-Hérard-Seguin, Godunov-scheme by Schwendeman-Wahle-Kapila) More typical two-phase flows : Ransom faucet test case sedimentation desequilibrium in velocities bubbly column and asymptotic behavior 17/35

18 TEST 1A Test 1a : isolated coupling wave (Gallouët-Hérard-Seguin) We choose and u I = α 1ρ 1 u 1 +α 2 ρ 2 u 2, χ=0.5 α 1 ρ 1 +α 2 ρ 2 α 1,L = 0.9, (ρ 1, u 1, p 1 ) L = (1, 100, 10 5 ) (ρ 2, u 2, p 2 ) L = (1, 100, 10 5 ) α 1,R = 0.5, (ρ 1, u 1, p 1 ) R = (0.125, 100, 10 5 ) (ρ 2, u 2, p 2 ) R = (0.125, 100, 10 5 ) 18/35

19 TEST 1A 1 Godunov-type VFRoe exact Godunov-type VFRoe exact 1 Godunov-type VFRoe exact FIG.: x versusα 1,ρ 1,ρ 2 19/35

20 TEST 1A 110 Godunov-type VFRoe exact 110 Godunov-type VFRoe exact Godunov-type VFRoe exact Godunov-type VFRoe exact FIG.: x versus u 1, u 2, p 1, p 2 20/35

21 TEST 1B Test 1b : isolated coupling wave (Andrianov-Saurel-Warnacke) We chooseχ=1so that and u I = u 1 α 1,L = 0.8, (ρ 1, u 1, p 1 ) L = (2, 0.3, 5) (ρ 2, u 2, p 2 ) L = (1, 2, 1) α 1,R = 0.3, (ρ 1, u 1, p 1 ) R = (2, 0.3, ) (ρ 2, u 2, p 2 ) R = (0.1941, , 0.1) 21/35

22 TEST 1B pts exact pts exact pts exact FIG.: x versusα 1,ρ 1, u 1 22/35

23 TEST 2 Test 2 : a more general Riemann problem taken from the paper by Schwendeman-Wahle-Kapila We chooseχ=1so that and u I = u 1 α 1,L = 0.8, (ρ 1, u 1, p 1 ) L = (1, 0, 1) (ρ 2, u 2, p 2 ) L = (0.2, 0, 0.3) α 1,R = 0.3, (ρ 1, u 1, p 1 ) R = (1, 0, 1) (ρ 2, u 2, p 2 ) L = (1, 0, 1). We compare our results with the ones given by the schemes proposed by Saurel-Abgrall, Andrianov-Saurel-Warnacke, Schwendeman-Wahle-Kapila 23/35

24 TEST pts exact pts exact pts exact pts exact FIG.:α 1 (x), p 1 (α 1 ),ρ 1 (x),ρ 2 (x) 24/35

25 TEST (a) 1.1 (b) α p x α 1.06 (c) 1 (d) ρ ρ x x FIG.:α 1 (x), p 1 (α 1 ),ρ 1 (x),ρ 2 (x) (by courtesy of Schwendeman-Wahle-Kapila) caption : G HLL (blue), G ASW (green) et G 1 (red) 25/35

26 TEST 2 Riemann invariants associated with u I are K 0 = u I = u 1 K 1 = (1 α 1 )ρ 2 (u 2 u I ) K 2 = (1 α 1 )ρ 2 (u 2 u I ) 2 +α 1 p 1 + (1 α 1 )p 2 K 3 = γp 2 (γ 1)ρ (u 2 u I ) 2 K 4 = p 2 ρ γ 2 Forα 1 varying fromα 1,L toα 1,R, these equations provide a parametrization of the coupling wave u I. This is compared with the "boundary layer" induced by the diffusion of the numerical solution. Constants K i, i=0,..., 4 are calculated using the exact value of the right state associated withα 1 =α 1,R similarly as in the paper Schwendeman-Wahle-Kapila. We observe a very good agreement 26/35

27 TEST 3 : INFLUENCE OF χ chi=0 chi=0.5 chi= chi=0 chi=0.5 chi=1 1.2 chi=0 chi=0.5 chi= FIG.: x versusα 1,ρ 1,ρ 2 27/35

28 TEST 3 : INFLUENCE OF χ chi=0 chi=0.5 chi=1 0 chi=0 chi=0.5 chi= chi=0 chi=0.5 chi=1 1.1 chi=0 chi=0.5 chi= FIG.: x versus u 1, u 2, p 1, p 2 28/35

29 INFLUENCE OF THE SOURCES External forces : g=9.81 Pressure relaxation : Drag force : θ(u)= 1 τ p α 1 (1 α 1 ) p 1 + p 2 (p 1 p 2 ) λ(u)= 1 8 C Da int ρ 1 u 1 u 2 with C D = 0.5, a int = 3(1 α 1), r b = r b 29/35

30 TEST 4 : RANSOM FAUCET TEST CASE VFRoe Godunov-type 15 VFRoe Godunov-type VFRoe Godunov-type VFRoe Godunov-type FIG.: Leftα 2, right u 1. Top 2000 points, bottom points 30/35

31 TEST 5 :SEDIMENTATION t=0.2 t=0.4 t=0.6 t=0.8 t=1.0 t=1.2 t= FIG.:α 1 withτ p = s and 400 cells 31/35

32 TEST 5 :SEDIMENTATION t=0.2 t=0.4 t=0.6 t=0.8 t=1.0 t=1.2 t= /35

33 TEST 7 :BUBBLYCOLUMN Evolution of a two-phase flow in a vartical bubbly column of 1m-length Boundary conditions : inlet, we impose u 1 = 5m/s, u 2 = 15m/s,α 1 = 0.97,ρ 1 = 10 3 kg/m 3,ρ 2 = 1kg/m 3 and the mixture pressure p= Pa outlet, we impose the mixture pressure p= Pa Asymptotic behavior : x p= ρg, p=α 1 p 1 +α 2 p 2, ρ=α 1 ρ 1 +α 2 ρ 2, et u 2 r= α 1(1 ρ 2 ρ 1 )g 33/35

34 TEST 7 :BUBBLYCOLUMN 30 u_r^2 asymptotic relative velocity 0.2 u_r^2 asymptotic relative velocity e+07 p ^7 - rho g (x-1) e e e e e e e e FIG.: Stationary solutions of Test 7. x versus u 2 r, u 2 r (close-up), p 34/35

35 OPEN PROBLEMS Open problems are : the positivity of the internal energies the AP property (seems to be ok from the numerical experiments) an entropy inequality (related to the numerical choice of p I?) 35/35