Numerical methods for FCI Part IV Multi-temperature fluid models Hele-Shaw models

Transcription

1 B. Després+ X. Blanc LJLL-Paris VI+CEA Thanks to same collegues as before plus C. Buet, H. Egly and R. Sentis methods for FCI Part IV Multi-temperature fluid s s B. Després+ X. Blanc LJLL-Paris VI+CEA Thanks to same collegues as before plus C. Buet, H. Egly and R. Sentis methods for FCI Part IV Multi-temperature fluid s s p. / 7

2 FCI scenario s During the implosion, pure hydrodynamics is a very strong hypothesis. It is much more relevant to consider multi-temperature s. methods for FCI Part IV Multi-temperature fluid s s p. / 7

3 Plan s One temperature for ions T i and one temperature for electrons T e : T i T e Basic considerations One temperature for the matter, and one temperature for radiatiopn T r for the stability of the ablation front methods for FCI Part IV Multi-temperature fluid s s p. 3 / 7

4 The simplified T i T e s Starting point is (page 9 and page 3 of the notes) D t ρ + ρ u = 0, ρd t u + p = F r, ρd t ε e + p e.u (χ e T e ) + W ei = Q r + S, ρd t ε i + p i u + (χ i T i ) W ei = 0, The unknowns of this system are the density ρ(x, t) R + of the plasma, its velocity u(x, t) R 3 and its pressure p(x, t) R. We also have electronic and ionic values : pressures p e (x, t), p i (x, t) R (with p = p e + p i ), energies ε e (x, t), ε i (x, t) R (with E = E e + E i ), and temperatures T e (x, t), T i (x, t). Here, The terms F r and Q r are the radiative sources, and S is an additional source term ling the laser energy drop. This set of equations is closed by an adapted equation of state (ε e, p e, ε i, p i ) = F(ρ, T e, T i ). We will assume that the fluid is described by a perfect gas EOS p i = (γ i )ρc vi T i = (γ i )ρε i, ε i = C vi T i and the electronic part is described by a perfect gas EOS p e = (γ e )ρc ve T e = (γ e )ρε e, ε e = C ve T e. Since electrons are monoatomic γ e = 5 3. methods for FCI Part IV Multi-temperature fluid s s p. 4 / 7

5 Hydrodynamics of the T i T e s The hydrodynamic part is D t ρ + ρ u = 0, ρd t u + p = 0, ρd t ε e + p e.u = 0, ρd t ε i + p i u = 0, This system is non conservative. For discontinuous functions a and b the product a x b is not defined. What is a shock in such a system? We need to transform it into a conservative system of conservation laws. Universal principles : mass is preserved, momentum is preserved ad total energy is preserved. We get after convenient manipulations the correct conservative formulation (in D) t ρ + x (ρu) = 0, t (ρu) + x ρu + p = 0, t (ρe) + x (ρue + pu) = 0, with p = p i + p e and e = ε i + ε e + u. Question : Is there a fourth conservation laws? methods for FCI Part IV Multi-temperature fluid s s p. 5 / 7

6 Conservation of the electronic entropy s Convenient manipulations show that smooth solutions satisfy t (ρ(αs i + βs e )) + x (ρu(αs i + βs e )) = 0, α, β. For discontinuous solutions, these relations are not equivalent. The correct choice is α = 0 and β =. This is called the Born-Oppenheimer hypothesis. It is related to the fact that me is small. m i Zeldovith-Raizer, Cordier (PhD thesis 96), Degond-Luquin, Massot,... Finally t ρ + x (ρu) = 0, t (ρu) + x ρu + p = 0, t (ρs e ) + x (ρus e ) = 0, t (ρe) + x (ρue + pu) = 0. The mathematical entropy law writes t (ρs i ) + x (ρus i ) 0. methods for FCI Part IV Multi-temperature fluid s s p. 6 / 7

7 Shock relations As a consequence the ionic entropy increases at shocks while the electronic entropy is constant as shocks s Exercise : prove it. Solution < : S + i > S i, S + e = S e. σ (ρ R ρ L ) + (ρ R u R ρ L u L ) = 0, σ `ρ R S e,r ρ L S e,l + `ρr u R S e,r ρ L u L S e,l = 0, σ `ρ R S i,r ρ L S i,l + `ρr u R S i,r ρ L u L S i,l > 0. So ρ R (u R σ) = ρ L (u L σ). This is the constant mass flux D = ρ R (u R σ) = ρ L (u L σ). Therefore DS e,r = DS e,l and DS i,r > DS i,l. Assume a shock and the mass flux is positive D > 0. Then S e,r = S e,l and S i,r > S i,l. CQFD This behavior is absolutely fundamental : it explains that ions and electrons behave differently at shocks. In summary physical considerations show that is the correct eulerian system of conservation laws to analyze for the two temperature T i T e. methods for FCI Part IV Multi-temperature fluid s s p. 7 / 7

8 Lagrangian T i T e s In Lagrange variable in dimension one, one gets t τ mu = 0, t u + mp = 0, p = p i + p e, t S e = 0, t e + m(pu) = 0. Set ρ c i = p i τ S i, ρ c e = p i τ S e and ρ c = p i τ S i pe τ S e The sound speed of the lagrangian system is ρc where c = c i + c e. The natural Lagrangian scheme is now M j t (τ j L τj n ) u j+ + u j = 0, M j t (ul j uj n ) + p j+ p j = 0, (S e ) L j (S e ) n j = 0, M j t (el j ej n ) + p j+ u j+ p j u j = 0, with the solver u j+ = (un j + uj+ n ) + ρc (pn j pj+ n ) p j+ = (pn j + pj+ n ) + ρc (un j uj+ n ), h i (ρc) j+ = (ρc) n j + (ρc) n j+. A pure Lagrangian scheme is such that f n+ = f L for all f. methods for FCI Part IV Multi-temperature fluid s s p. / 7

9 With source terms s A simplified eulerian T i T e system with source terms writes t ρ + x ρu = 0 t ρu + x (ρu + p i + p e ) = 0 t ρε i + x ρuε i + p i x u = τ ei (T e T i ) t ρε e + x ρuε e + p e x u = τ ei (T i T e ) + x (K e x T e ). The relaxation time is τ ei. The electronic diffusion coefficient is K e. We assume that ε i = C vi T i et ε e = C ve T e. The rigorous way to write this is t ρ + x ρu = 0 t ρu + x (ρu + p i + p e ) = 0 t ρs e + x ρus e = τ ei Te (T i T e ) + Te x (Ke x Te ) t ρe + x (ρue + p i u + p e u) = x (K e x T e ),. where the unknowns are the density ρ, the momentum ρu, the electronic entropy ρs e and the total energy ρe. methods for FCI Part IV Multi-temperature fluid s s p. 9 / 7

10 solution base on a splitting strategy s First stage : solve the hydro. Second stage solve the remaining part We use the linear law t ρ = 0 t ρu = 0 t ρε i = τ ei (T e T i ) t ρε e = τ ei (T i T e ) + x (K e x T e ). ε i = C vi T i and ε e = C ve T e. The numerical solution of the system can be computed with an implicit linear solver in case the gas is described by perfect gas equations of state. ρ n+ = ρ L, u n+ = u L, ρ L (T i ) n+ (T j i ) L j C vi = t τ ((T e ) n+ ei j ρ L (Te ) n+ j (Te ) L j C ve = t τ ((T i ) n+ ei j K e,i+ (Te ) n+ j+ + (T i ) n+ ), j (T e ) n+ j (Te )n+ j ) K e,i (Te ) n+ j (Te ) n+ j x. methods for FCI Part IV Multi-temperature fluid s s p. 0 / 7

11 An example relevant for ICF in direct drive o The Piston velocity is W p > 0. The domain is Ω(t) = nx L = 0 < x < x R (t) = xr 0 twp. t s x L x(t) R x (0) R Radiation push The ionic temperature is the discontinuous curve. The electronic temperature is the continuous curve. The initial temperature is the blue curve (5 000 K). 9e+06 e+06 7e+06 6e+06 Kelvin 5e+06 4e+06 3e+06 e cm The ionic part of the gas is violently heated by the shock. The electronic temperature is continuous everywhere. The temperature relaxation is visible behind the shock. In front of the shock a prehating phenomenon is visible. This calculation shows the great importance of shocks for ICF flows in the context of direct drive. methods for FCI Part IV Multi-temperature fluid s s p. / 7

12 The Rankine-Hugoniot relation for the T i T e s Start from The Rankine-Hugoniot relations are t ρ + x ρu = 0 t ρu + x (ρu + p i + p e ) = 0 t ρs e + x ρus e = τ ei Te (T i T e ) + Te x (Ke x Te ) t ρe + x (ρue + p i u + p e u) = x (K e x T e ), σ[ρ] + [ρu] = 0, σ[ρu] + [ρu + p i + p e ] = 0, [T e ] = 0, σ[ρs e ] + [ρus e ] = Te [Ke x Te ], σ[ρe] + [ρue + p i u + p e u] = [K e x T e ].. Notice that the continuity of T e is provided by diffusion operator. Problem Prove that the continuity of S e is recovered in the limit K e 0 +. All numerical results support the conjecture. Works by Lefloch, Coquel and coworks (Chalon, Berthon,...) on a similar problem. methods for FCI Part IV Multi-temperature fluid s s p. / 7

13 Discretization of boundary conditions Assume an additional no-heat flux physical conditions. The boundary conditions are x = 0 = x L, u 0,t = 0, x T e = 0 s and x = x R (t) = x R (0) tw p, u xr (t),t = Wp, x Te = 0 One has boundary conditions and 4 equations #(bc ) =, #(eq ) = 4. Question : how can we do? There is no problem for the second stage of the algorithm t ρ = 0 t ρu = 0 t ρε i = τ ei (T e T i ) t ρε e = τ ei (T i T e ) + x (K e x T e ). plus homogeneous Neumann conditions x T e = 0 at boundaries. So the real problem is for the Lagrangian stage of the algorithme. We are left with #(bc ) =, #(eq ) = 4. But it works : this is the miracle of Lagrangian scheme+splitting for this problem!! methods for FCI Part IV Multi-temperature fluid s s p. 3 / 7

14 Solution Structure of the discrete problem s M j t (τ j L τj n ) u j+ + u j = 0, M j t (ul j uj n ) + p j+ p j = 0, (S e ) L j (S e ) n j = 0, M j t (el j ej n ) + p j+ u j+ p j u j = 0, #(bc ) =, #(what is needed ) =. Recall the rule : the equations of the discrete Riemann solver are more important than its solution j pb p L + (ρc) L (u B u L ) = 0, u P = W p. j pb = p L + (ρc) L (u B u L ), u P = W p. So we use in the last j = J max < : p j+ = p j + (ρc) j ( W p u j ), u j+ = W p. that we plug into the scheme. Finally : boundary condition is enough for the hydro!! methods for FCI Part IV Multi-temperature fluid s s p. 4 / 7

15 A simple grey non equilibrium s page 33 of the notes, one group Z E r = E ν dν 0 plus hydrodynamics. A rigorous justification is possible, but with an almost physical scaling. Consider the simplified for +matter (ρ) +.(ρu) = 0, t (ρu) +.(ρu u) + (p + pr ) = 0, t (ρe + Er ) +.((ρe + Er )u + (p + pr )u) =.( t 3σt T r 4 ), Er +.(uer ) + pr.u =.( t 3σt T r 4 ) + σa(t 4 Tr 4 ), with the grey hypothesis σ t = σ a + σ s and p r = Er. Fundamental is 3 E r = at 4 r, a = π5 k 4 See Buet+D. for a rigorous justification. 5c 3 h 3 = Stefan-Boltzmann constant. Trick : Define ε r E r = ρε r. The equation rewrites ρεr +.(uρεr ) + pr.u =.( T 4 r t 3σ ) + σa(t 4 T 4 r ) t and the radiative pressure rewrites p r = (γ r )ρε r with γ r = 4 3. methods for FCI Part IV Multi-temperature fluid s s p. 5 / 7

16 Hydrodynamic analysis The first task is to discretize the non equilibrium asymptotic set of equations. Setting σ a = 0 and σ s = + then we obtain the simplified hyperbolic set of equations in D t (ρ) + x (ρv) = 0, t (ρv) + x (ρv + p + p r ) = 0, p r = Er 3, t (ρe + Er ) + x (ρev + pv + pr v) = 0, Er = T r 4, t Sr + x (Sr v) = 0, Sr = T r 3. s The solver is compatible with the D Rankine-Hugoniot relations σ[ρ] + [ρv] = 0, σ[ρv] + [ρv + p + p r ] = 0, σ[ρe + E r ] + [ρve + pv + p r v] = 0, σ[s r ] + [vs r ] = 0. All this is compatible with the fact that S r = T 3 r is the number of photons if the is Planckian. methods for FCI Part IV Multi-temperature fluid s s p. 6 / 7

17 Explanation s The group of photons in direction Ω S and with frequency ν > 0 has intensity I (ν, Ω) The energy of is E r = R ν,ω IdνdΩ. v photon = cω, e ν = hν, I (ν, Ω) = n photons e ν. The number of photons is N r = R ν,ω hν I dνdω. The entropy of is S r = k R c 3 ν,ω ν (n log n (n + ) log(n + )) dνdω, n = I ν 3. Asumme a Planckian distribution with radiative temperature T r I = h 3 c ν. hν e ktr Then dimension analysis shows that : E r = αt 4 r, Nr = βt 3 r and S r = γt 3 r. Therefore S r is the number of photons. Read the facinating book by Steven Weinberg : The first three minutes (of universe). methods for FCI Part IV Multi-temperature fluid s s p. 7 / 7

18 Some numerical results The first test problem is a radiative Riemann problem. On the left we plot density, velocity and total pressure versus the position x at t = 0. with 000 cells : CFL= Density Velocity Total pressure 4 Sr/rho s On the right. Radiative entropy s r = T r 3 ρ = Sr versus x at t = 0. with 000 cells : CFL=0.5. One notices ρ the exact preservation of s r across the shock and in the rarefaction fan methods for FCI Part IV Multi-temperature fluid s s p. / 7

19 comparison with a moment Full system. The D solver is implicit. s Tm diffusion Tr diffusion Tm M Tr M cells, T = Right :T and T r for diffusion and the variable Eddington factor. Left : E r, F r and f = Fr Er. One notices that the non-equilibirum diffusion overpredicts the propagation of. It justifies (numerically) higher order s. methods for FCI Part IV Multi-temperature fluid s s p. 9 / 7

20 s : starting point s Chapter 4 of the notes. With H. Egly and R. Sentis. Use a cold coupled with one group for + hydrodynamics + T = T r and ρe + at 4 = ρc v T + at 4 ρc v T. The starting point of our analysis is the compressible Euler with non linear heat flux < : t ρ +.(ρu) = 0 t ρu +.(ρu u) + p = 0 t (ρe) + (ρue + pu κnt n T ) = 0. The Spitzer non linear coefficient is n [ 5, 7 ]. The heat flux boundary condition on the exterior boundary Γ r is non linear κnt n nt Γe = b given. methods for FCI Part IV Multi-temperature fluid s s p. 0 / 7

21 D cut of an ablation front The isobar regime is p = (γ )ρc v T C. left=cold ρ c right=hot T h s T c Ablation front velocity Fluid velocity ρ h x=0 Two other important ingredients are x=x (t) f x r ε =! n Tc, and u(t = 0, x) = u c (t), x the cold region. T h methods for FCI Part IV Multi-temperature fluid s s p. / 7

22 Quasi-isobar s Let us define the acceleration g = u c (t) and reset the velocity u u uc. After rescaling one gets where D t ρ + ρ.u = 0 ρ D t u + M p = ρ g F r Dt p + γ γ γ p u (HnT n T ) = 0. The Mach number M = ρ u p. The Froude number F r = u g l = u t g is a measure of the acceleration of the particles coming from the left boundary Γ l in the cold region. H = κn(t ) n+ p u = κn(t ) n+ ρ u 3 measures the velocity of the particles in the hot region. methods for FCI Part IV Multi-temperature fluid s s p. / 7

23 Slow Mach number expansion s We perform an asymptotic expansion of all variables with respect to the square of the Mach number, ρ = ρ (0) + M ρ () +,... One gets the quasi-isobar t ρ +.(ρu) = 0 t (ρu) +.(ρu u) + p = ρg (u nt n T ) = 0, ρt =. or also > < t T + u vort. T T T n = 0, t u + u. u + T p = g, u = nt n T + u vort = u therm + u vort,.u vort = 0. methods for FCI Part IV Multi-temperature fluid s s p. 3 / 7

24 : u vort = 0 Conjecture : For well prepared data, the solution of t T T T n = 0 is approximated by max(ε, θ) n where θ T n is a solution of the θ = 0, x Ω(t), nθ = b, x Γ equation : h, θ = 0, x Γ f (t), t Γ f (t) = θ γ(t), x Γ f (t). s A numerical example is Proof in D! Set Θ = T n. Then t Θ n + xx Θ = 0. Progressive waves are defined by Θ = Θ(x + vt). The generating Kull s function is the progressive wave with v = and Θ( ) = K (x) = K n «n + n (x), normalisation K(0) = e. n + x x0 +vt Set T = εk n. Then (T n )( ) = ε n, (T n ) (+ ) = v and ε n ε t T T xx T n = 0. methods for FCI Part IV Multi-temperature fluid s s p. 4 / 7

25 The classical problem (9) s < : p = 0, p = 0, nx = p, x Ω in (t) = blue region, x Ω in (t), x Ω in (t). The Web page of Howison (Ociam, Oxford) for some historical references about (and also with fresh science) : Mr (inv. of the variable pitch propeller) worked on the propeller of the cruisers of her gracious majesty methods for FCI Part IV Multi-temperature fluid s s p. 5 / 7

26 Ablative in D We use a Finite Element Method for the Poisson equation, and markers for the front. The markers move accordingly to the. Solution numerique Solution analytique 0.4 solution numerique solution analytique t=0 t=0.06 R(t) s temps On the left Γ int (t). On the middle t r(t). On the right : The smoothing effect of the equation for convergent front is visible ; In this regime, ablation fronts are stable. The full ablative : u vort 0 The writes Θ = 0, x Ω Ω in, nθ = v, x Ω = Γ ext, Θ = 0, x Ω in = Γ in (t), x (t) = Θ + u vort, x(t) Γ in (t), t V +. (u therm V) = S, x Ω Ω in, ϕ = T V, x Ω Ω in, u vort = ϕ, x Ω Ω in. The vorticity source is S. The thermic velocity if u therm = n+ n T n+ = Θ n Θ. The vorticity is V = ω T. methods for FCI Part IV Multi-temperature fluid s s p. 6 / 7

27 More results Passive vorticity : S 0 but u vort = 0 The initial data is a front Γ int discretize with 00 markers and with a mode 9. We plot the vorticity. s Active vorticity : S 0 and u vort In this regime, ablation fronts may be unstable. Open problem : design a two-temperature. methods for FCI Part IV Multi-temperature fluid s s p. 7 / 7