On a class of numerical schemes. for compressible flows

Transcription

1 On a class of numerical schemes for compressible flows R. Herbin, with T. Gallouët, J.-C. Latché L. Gastaldo, D. Grapsas, W. Kheriji, T.T. N Guyen, N. Therme, C. Zaza. Aix-Marseille Université I.R.S.N. Cadarache C.E.A Cadarache

2 Some key words: Compressible Euler and Navier Stokes Implicit or semi-implicit (or explicit) schemes Internal energy balance formulation (rather than total energy formulation) even in the presence of shocks All Mach scheme Finite volume / non conforming finite element Upwinding with respect to the material velocity (no Riemann solver) Colocated or staggered grids Stability via a kinetic energy inequality Consistency in the Lax-Wendroff sense if a conservative numerical scheme for a hyperbolic system of conservation laws converges, then it converges towards a weak solution.

3 The Euler (NS) equations: total energy vs. internal energy Euler (Navier-Stokes) equations: t ϱ + div(ϱu) = 0, t (ϱu) + div(ϱu u) divτ + p = 0, t (ϱe) + div [ (ϱe + p)u ] = div(τ u), (mass) (mom) (tot.en) p = (γ 1) ϱe, E = 1 2 u 2 + e. For regular functions, (mom) u & (mass) (kin.en): 1 2 t (ϱ u 2 ) div(ϱ u 2 u) + p u = div(τ ) u. (kin.en) Subtracting from (tot.en) yields the internal energy balance: t (ϱe) + div(ϱeu) + p divu = τ : u, (int.en) which implies e 0. Incompressible schemes use the internal energy (or temperature) equation.

4 Internal energy Dealing with the internal energy: positive internal energy convenient for incompressible problems ρe is not a conserved variable conserved variables : ρ, ρu, ρe Formulations based on variables other than the conserved variables (non-conservative variables) fail at shock waves. They give the wrong jump conditions; consequently they give the wrong shock strength, the wrong shock speed and thus the wrong shock position. (Toro, 1999) density Test 5 of [ Toro chapter 4] - Density at t = 0.035, n = 2000 cells, with and without corrective source terms, and analytical solution. 5 0 with without exact x

5 Right and wrong shock speed for Bürgers Bürgers equation: for regular positive solutions (B) : t u + x (u 2 ) = 0 (BS) : t u x u3 = 0. No longer true with irregular solutions: Rankine Hugoniot gives σ = u l + u r and σ = 4 3 (u l + u r ). Weak solutions of (B) weak solutions of (BS). Figure: Explicit upwind Scheme for (B) (left) and (BS) (right) with different mesh sizes, CFL = 1.

6 Bürgers, numerical diffusion Scalar conservation law: t u + x (f (u)) = 0, f 0 Upwind scheme: h δt (un+1 i u n i ) + f (u n i ) f (u n i 1 ) = 0. Upwinding is formarly similar to add a numerical diffusion. t u + x (f (u)) x ( hf (u) δt f 2 (u) x u) = 0 2 The CFL condition states that hf (u) δt f 2 (u) 0 (i.e. δt f (u) h) In the case of the Bürgers equation it gives t u + x (u 2 ) x ((hu 2δt u 2 ) x u) = 0, x R, t R +

7 Bürgers, non conservative numerical diffusion In the case of the equivalent equation t (u 2 ) + (4/3) x (u 3 ) = 0, u > 0 upwinding is formally consistent with Dividing by 2u, this leads to t (u 2 ) x (u3 ) x ((2hu 2 4δt u 3 ) x u) = 0, t u + x (u 2 ) 1 u x ((hu2 2δt u 3 ) x u) = 0, x R, t R + This is a numerical diffusion (thanks to the CFL condition) but not in conservative form. A non conservative diffusion may lead to wrong discontinuities

8 Benefit from a non conservative numerical diffusion? t u + x (u 2 ) = 0 + initial condition (B) t (u 2 ) x (u3 ) = 0 + initial condition (BS) Explicit upwind scheme on (BS) formally equivalent to: t u + x (u 2 ) 1 u x ((hu2 2δt u 3 ) x u) = 0. }{{} non conservative numerical diffusion. Negative result for a non conservative diffusion :-( Non conservative numerical diffusion on (B) yields { wrong shock velocity for (B) correct shock velocity for (BS) Positive result for a non conservative diffusion? :-) Non conservative numerical diffusion on (BS) yields { wrong shock velocity for (BS) correct shock velocity for (B)? How do we choose the non conservative numerical diffusion?

9 Non conservative numerical diffusion on (BS) Viscous Bürgers: Multiplying by 2u: t u + x (u 2 ) ε xx u = 0. (B) ε t (u 2 ) x (u3 ) 2uε xx u = 0. (BS) ε Discretize (BS) ε instead of (BS): t (u 2 ) x (u3 ) uε 0 h xx u = 0, (BS) ε with 2ε = ε 0 h.

10 Non conservative numerical diffusion on (BS) t (u 2 ) x (u3 ) uε 0 h xx u = 0, (BS) ε with 2ε = ε 0 h. Centered finite volume with non conservative diffusion ( (n)) 2 ( (n 1)) 2 4δt u i = u i + u(n 1) + u (n 1) 3 i 1 i 3h 2 + δt [ h 2 ε 0h u (n 1) i u(n 1) i u (n 1) i u (n 1) i+1 2 2u (n 1) i ] + u (n 1). i+1 Centered Scheme for (BS) ε0 h, ] uε 0 h xx u = ε 0 h [ x (u x u) + x ( u2 2 )

11 From Bürgers to Euler For regular solutions, Bürgers: t u + x (u 2 ) = 0 t (u 2 ) x (u3 ) = 0. Euler: t ϱ + div(ϱu) = 0, t (ϱu) + div(ϱu u) + p = 0, t (ϱe) + div [ (ϱe + p)u ] = 0. t ϱ + div(ϱu) = 0, t (ϱu) + div(ϱu u) + p = 0, t (ϱe) + div(ϱeu) + p divu = 0. we had an equation, we now have a system... Idea: add a non conservative corrective term to the internal energy equation. Which term? Inspiration comes from copying the formal derivation of the internal energy equation at the discrete level.

12 Total energy= kinetic energy + internal energy Total energy (Euler): t (ϱe) + div [ (ϱe + p)u ] = 0, E = e u 2 = t (ρe) + div(ρeu) + p divu t (ρ u 2 ) div(ρ u 2 u) + u p = 0. From mass balance, for regular z: t (ρz) + div(ρzu) =( t ϱ + div(ϱu)) }{{} z + ρ t z + ρu z, = 0. = 1 2 t (ρui 2 ) + 1 [ ] 2 div(ρu2 i u) = ρu i t u i + ρu i u u i = u i ρ t u i + ρu u i = [ u i t (ρu i ) + div(ρu i u) ] = u i i p, 1 i 3. From momentum balance: = 1 2 t (ρ u 2 ) div(ρ u 2 u) = u [ t (ρu) + div(ρu u) ] = u p. and t (ρe) + div(ρeu) + p div(u) = 0.

13 Strategy for a numerical scheme for Euler How to get correct weak solutions of Euler equations while solving the internal energy balance? More precisely: General idea: (int.en) d + (kin.en) d (tot.en) d 1- (mom) d and (mass) d discrete kinetic energy (E k ) balance 2- (int.en) d : ð t E k + div d (E k u) + u d p + R = 0 R : non conservative residual term ð t (ρe) + div d (eu) + pdiv d u = R R 0 3- Perform Lax-Wendroff consistency analysis of the scheme: (a) Suppose bounds and convergence for a sequence of discrete solutions control in BV and L, convergence in L p, for p 1. (b) Let ϕ be a regular function, interpolate, test the kinetic energy balance, test the internal energy balance, and pass to the limit in the scheme. The corrective term in the internal energy balance is such that, at the limit, the weak form of the total energy equation is recovered.

14 Required discrete properties Discrete transport property, i.e. discrete equivalent of t (ρz) + div(ρzu) = ρ t z + ρu z, z = u i. = Compatible discretization of mass and momentum balance equation Discrete duality i.e. discrete equivalent of div(pu) = pdivu + u p. Positivity of the residual R 0 in the discrete kinetic energy balance equation (to ensure the positivity of the internal energy). Points to be taken care of when designing the scheme(s).

15 Time discretization : I - Decoupled (explicit) choice Decoupled 1: natural ordering, bad idea..., ρ u e p 1 δt (ϱn+1 ϱ n ) + div(ϱ n u n) ) = 0, (mass) d ϱ n+1 1 δt (ϱn+1 u n+1 ϱ n u n ) + div(ϱ n u n u n ) + p n = 0, (mom) d u n+1. 1 δt (ϱn+1 e n+1 ϱ n e n ) + div(ϱ n e n u n ) + p n divu n+1 = R n, (int.en) d e n+1 p n+1 = (ϱ n+1, e n+1 ) = (γ 1) ϱ n+1 e n+1, (eos) d p n+1 Decoupled 2: better idea..., ρ e p u 1 δt (ϱn+1 ϱ n ) + div(ϱ n u n) ) = 0, (mass) d ϱ n+1 1 δt (ϱn+1 e n+1 ϱ n e n ) + div(ϱ n e n u n ) + p n divu n = R n, (int.en) d e n+1 p n+1 = (ϱ n+1, e n+1 ) = (γ 1) ϱ n+1 e n+1, (eos) d p n+1 1 δt (ϱn+1 u n+1 ϱ n u n ) + div(ϱ n u n u n ) + p n+1 = 0, (mom) d u n+1.

16 Time discretization : I -Decoupled choice 1000 pressure exact 1000 pressure exact pressure pressure ρ u e p x ρ e p u x [E. Toro, Riemann solvers and numerical methods for fluid dynamics, third edition, test 3 of chapter 4].

17 Decoupled scheme : Why choose ρ e p u? Derivation of the entropy balance at the continuous level. For regular functions, an entropy s satisfies t s + u s = 0 or equivalently t (ρs) + div(ρus) = 0; for Euler perfect gas s = ψ(ρ, e) = ln(ρ) 1 γ 1 lne. t ϱ + u ϱ + ϱ divu = 0 ϱψ(ρ, e) t e + u e + p divu = 0 eψ(ρ, e) ϱ [ t s + u s + ϱ ϱs + p ] ϱ es divu = 0. }{{} =0 We need div(ϱu) and p divu at the same time level to mimick this computation at the discrete level. Choice 1 ρ u e p div(ϱ n u n ) and p n divu n+1 Choice 2 ρ e p u div(ϱ n u n ) and p n divu n

18 Time discretization : II - Implicit or semi-implicit choice Implicit ρ n+1 ρ n + div(ρ n+1 u n+1 ) = 0, δt (mass) d 1 δt (ϱn+1 u n+1 ϱ n u n ) + div(ϱ n+1 u n+1 u n+1 ) + p n+1 = 0, (mom) d 1 δt (ϱn+1 e n+1 ϱ n e n ) + div(ϱ n+1 e n+1 u n+1 ) + p n+1 divu n+1 = R n+1, (int.en) d p n+1 = (ϱ n+1, e n+1 ) = (γ 1) ϱ n+1 e n+1, Semi-implicit ( ρ Pressure gradient scaling step: ( p) n+1 n ) 1/2 = ( p n ρ n 1 ) (eos) d Prediction step Solve for ũ n+1 : 1 ( ρnũ n+1 ρ n 1 u n) + div(ρ n ũ n+1 u n ) + ( p) n+1 = 0, (mom) δt d Correction step Solve for p n+1, e n+1, ρ n+1 and u n+1 : 1 δt ρn (u n+1 ũ n+1 ) + ( p n+1 ) ( p) n+1 = 0, ρ n+1 ρ n + div(ρ n+1 u n+1 ) = 0, (mass) δt d 1 δt (ρn+1 e n+1 ρ n e n ) + div(ρ n+1 e n+1 u n+1 ) + p n+1 (div(u n+1 )) = R n+1, (int.en) d ρ n+1 = ϱ ( e n+1, p n+1). (eos)

19 Meshes K p K, ρ K, e K, u K Colocated Advantage: Easier Data structure Easily refined Pressure correction scheme studied for the Euler equations (C. Zaza s thesis). u σw u σn K p K, ρ K, e K u σs u σe u (2) σ N K u σ (1) W u σ (1) p K, ρ K, e E K Staggered: Crouzeix-Raviart (on simplices) Rannacher-Turek (on quadrangles) full velocities on the edges (faces) MAC: normal velocities on the edges (faces) Advantage: Stable in the incompressible limit u (2) σ S

20 Space discretization: Finite volume discretization of the mass equation K ρ n+1 ρ n + div(ρ n+1 u n+1 ) = 0, (mass) δt (mass) discretization of the fluxes: K δt K (ρ n+1 K ρ n K ) + ρ n+1 ρ n σ E(K ) δt F n+1 = σ ρn+1 K,σ σ un+1 σ n K,σ, numerical flux through σ. F n+1 K,σ = 0, + (ρ n+1 u n+1 n K ) = 0. K σ F ρ K,σ K u σ K ρ n+1 σ upwind approximation of ρ n+1 at the face σ with respect to u n+1 σ n K,σ. Positive density: ρ n+1 > 0 if (ρ n > 0 and ρ > 0 at inflow boundary) σ

21 Discrete internal energy equation and E.O.S. 1 δt (ρn+1 e n+1 ρ n e n ) + div(ρ n+1 e n+1 u n+1 ) + p n+1 (div(u n+1 )) = R n+1 Discretization by upwind finite volume of the discrete internal energy K δt (ρn+1 e n+1 ρ n K K K en K ) + σ E(K ) eσ n+1 upwind choice positivity of e (if R n+1 0) K K (divu) K = σ u σ n K,σ. σ E(K ) F n+1 K,σ en+1 σ + K p n+1 K (divu n+1 ) K = R n+1 K, discrete E.O.S. p n+1 K = (γ 1)ρ n+1 K e n+1 K. (eos) d

22 Discretization of the momentum equation 1 ( ρnũ n+1 ρ n 1 u n) + div(ρ n ũ n+1 u n ) + ( p) n+1 (mom) n δt (mom) n D σ + ρ n ũ n+1 ρ n 1 u n ρ n ũ n+1 u n n K D σ δt D σ }{{} C(ρ n, u n ) + D σ ( p) n+1 = 0. Space discretization D σ (ρ n D δt σ ũ n+1 σ ρ n 1 u n D σ σ) + Fσ,ɛ n ũn+1 ɛ + D σ ρn D σ ρ n 1 ( p n ) σ = 0, ɛ E(D σ) D σ }{{} C d (ρ n, u n ) Grad-div duality : K p K (divu) K + D σ u σ ( p) σ = 0 K T σ E D σ ( p n ) σ = σ (pl n pn K ) n K,σ for σ = K L. ρ n D σ? F n σ,ɛ? K ρ Dσ F K,σ L u σ D σ σ F σ,ɛ

23 Discretization of the convection operator Choice of ρ n D σ, ρ n 1 D σ and F n σ,ɛ in C d (ρ n, u n, ũ n+1 )? Discretize C d (ρ n, u n, ũ n+1 ) so as to obtain a discrete kinetic energy balance. Copy the continuous kinetic energy balance: (mom) u & (mass) (kin.en) with some formal algebra... using t ρ + div(ρ u) = 0. Do the same at the discrete level? Momentum on dual cells, mass on primal cells... Idea: reconstruct a mass balance on the the dual cells Choose ρ Dσ = 1 ( ) DK,σ ρ K + D L,σ ρ L Dσ F σ,ɛ : linear combination of the primal fluxes (F K,σ ) σ E(K ). so that a discrete mass balance holds on the dual cells D σ: σ E int, D σ δt (ρ n+1 D σ ρ n D σ ) + ɛ E(D σ) F n+1 σ,ɛ = 0. Then take C d (ρ n, u n, ũ n+1 ) = Dσ δt with u n+1 ɛ (ρ n D σ ũ n+1 σ ρd Fσ,ɛ n ũn+1 ɛ n 1 σ u n σ) + ɛ E(D σ) = 1 2 (un+1 σ + u n+1 σ )

24 Discrete kinetic energy balance: computation of R σ Continuous setting: Multiply continuous momentum by u: ( ) t (ρ u) + div(ρ u u) + p = 0 u... with some formal algebra... using t ρ + div(ρ u) = 0, continuous kinetic energy balance: t ( 1 2 ρ u 2 ) + div ( ( 1 2 ρ u 2 ) u ) + p u = 0 (kin.en) Discrete setting: Similarly, multiply discrete momentum by ũ n+1 σ : ( Dσ (ρ n D δt σ ũ n+1 σ ρ n 1 σ u n σ) + ) F σ,ɛũn+1 n ɛ + D σ ( p n+1 ) σ = 0 ũ n+1 σ... with some algebra... using ɛ E(D σ) Mass on dual cell: Dσ (ρ n δt Dσ ρn 1 σ ) + F n σ,ɛ = 0. ɛ E(Dσ ) Scaling of the pressure ρ n 1 ( p) n+1 2 = ρ n p n 2. Correction equation 1 δt ρn u n+1 + p n+1 = 1 δt ρn ũ n+1 + ( p) n+1 discrete kinetic energy balance: 1 2 D [ σ ρ n δt σ u n+1 σ 2 ρ n 1 D σ u n σ 2] D σ ( p n+1 ) σ u n+1 σ ɛ=d σ D σ Fσ,ɛ n ũn+1 σ ũ n+1 σ +Rn+1 σ +Pn+1 σ = 0 with Rσ n+1 0, (kin.en) σ

25 Choice of R K R σ = Dσ ρ n 1 δt functions. ũ n+1 D σ σ u n σ 2 0 for regular functions, but NOT for discontinuous By definition of ρ Dσ, for σ = K L, R σ = D K,σ ρ n 1 δt ũ n+1 K σ u n σ 2 + D L,σ δt ρ n 1 ũ n+1 L σ u n 2 σ R K = 1 2 D K,σ ρ n 1 ũ n+1 δt K σ σ E(K ) u n 2 σ R K R σ = 0 K T σ E

26 Passage to the limit: total energy recovered Kinetic energy (kin) n σ = Dσ 2δt (ϱn+1 σ un+1 σ 2 ϱ n σ un σ 2 ) + 1 F n 2 σ,ɛũn+1 σ ũn+1 σ + ( p) n+1 σ un+1 σ = R n+1 σ + Pn+1 σ, ɛ=dσ D σ Internal energy (int) n K = K δt (ρn+1 K e n+1 K ρ n K en K ) + σ E(K ) F n K,σ en σ + K pn+1 K (divu n+1 ) K = R n+1 K, ϕ : test function Multiply (kin) σ by interpolate ϕ n σ and (int) K by interpolate ϕ n K n σ E(kin) n σ ϕn σ + (int) n K ϕn K = δtr K ϕ n K δtr σϕ n σ + δtp σϕ n σ. n K T n K T n σ E n σ E }{{}}{{}}{{} T [ ] ρ E t ϕ + (ρ E + p) u ϕ Ω ρ 0 (x) E 0 (x) ϕ(x, 0) Ω In particular, the pressure terms combine themselves to converge to pu ϕ.

27 Numerical tests - I Euler, high Mach ρ p u (1) u (2) Mach 3 facing step (Woodward Collela) MAC space discretization, uniform grid, δt = h/4 = 0.001, (u1 + c = 4 at the inlet boundary). Artificial viscosity µ = 0.001, which roughly corresponds to the numerical viscosity associated to an upwinding of the convection term µupw ' ρ u h/2 divided by 5.

28 Numerical tests - II Flow past cylinder, low Mach Flow past a cylinder, benchmark Schäfer and S. Turek, Mach 0.003, Re 100. Pressure correction scheme, Rannacher-Turek FE. coarse mesh fine mesh Mesh Space unks c d,max c l,max St m m m m Reference range Table: Drag and lift coefficients and Strouhal number.

29 Numerical tests - II Flow past cylinder, high Mach Flow past a cylinder, Mach ' 3, Re ' 100. pext = γ /10 ρ, c = 0.1. mes = impermeability and and perfect slip condition at the upper and lower boundaries u = (1, 0)t. at inlet. δt = Rannacher-Turek FE. ρ p u (1) u (2) t = 5, mesh of 106

30 Summary: main features of the class of schemes Colocated or staggered discretization Upwind choice for ρ, positivity of the density, Compatible discretization of (mass) and (mom) & careful choice of ρ Dσ and fluxes in (mom) to recover a mass conservation on the dual cells discrete kinetic energy inequality Compatible discretization of (mass) and (int.en) & upwind choice for e K under CFL positivity of e Conservation of total mass Existence of a solution to the scheme (topological degree argument) Velocity and pressure are constant through the contact discontinuity: p n and u n constant p n+1 and u n+1 constant. Consistency : under compactness assumptions, the discrete solution tends to a weak solution of the Euler systems. Numerical discontinuous solutions have correct shocks.

31 Recent and on going work Higher order schemes for the explicit MUSCL and additional viscosity. Convergence of the scheme to an entropy weak solution Low Mach limit for barotropic NS (with K. Saleh, M.-H. Vignal). Convergence for barotropic NS (γ 3) and strong-weak error estimates for barotropic NS (with D. Maltese, and A. Novotny). Reactive flows. Analysis of the MAC scheme in primitive variables for incompressible flows (K. Mallem s thesis)