Modélisation et simulation d écoulements diphasiques avec contrainte granulaire

Transcription

1 Modélisation et simulation d écoulements diphasiques avec contrainte granulaire T. Gallouët, P. Helluy, J.-M. Hérard, J. Nussbaum LATP Marseille IRMA Strasbourg EDF Chatou ISL Saint-Louis

2 Context Flow of weakly compressible grains (powder, sand, etc.) inside a compressible gas; importance of the granular stress when the grains are packed; averaged model (the solid phase is represented by an equivalent continuous medium); two-velocity, two-pressure mathematical model; relaxation of the pressure equilibrium in order to gain stability; stable and useful approximation for real applications.

3 Granular stress R

4 Averaging process

5 Outlines Two-phase granular flow models Unstable one-pressure model More stable two-pressure model Entropy and closure Thermodynamically coherent granular stress Shape of the granular stress Entropy and granular stress Hyperbolicity study Stability of the relaxed system Instability of the one-pressure model Numerical approximation: splitting method Convection step Relaxation step Numerical results Stability study 1D Combustion chamber 3D Combustion chamber Conclusion Bibliography

6 Two-phase granular flow models

7 Unstable one-pressure model A gas phase (G) k = 1, a solid phase (S) k = 2. Seven unknowns: partial densities ρ k (x, t), velocities u k (x, t), internal energies e k (x, t), gas volume fraction α 1 (x, t). Pressure law: p k = p k (ρ k, e k ). Other definitions: m k = α k ρ k, α 2 = 1 α 1, E k = e k + u2 k 2. The whole six-equation PDE system reads [3] m k,t + (m k u k ) x = 0, (m k u k ) t + (m k u 2 k + α k p k ) x p 1 α k,x = 0, (m k E k ) t + ((m k E k + α k p k )u k ) x + p 1 α k,t = 0, (1) with k = 1, 2 and with the algebraic relation p 2 = p 1 + R( ).

8 More stable two-pressure model The previous system is generally not hyperbolic. Therefore, we relax the algebraic pressure relation. The seven-equation PDE system reads [1] m k,t + (m k u k ) x = 0, (m k u k ) t + (m k uk 2 + α k p k ) x p 1 α k,x = 0, (m k E k ) t + ((m k E k + α k p k )u k ) x + p 1 α k,t = 0, α 1,t + u 2 α 1,x = P. (2) And the source term P has to contain the granular modeling.

9 Entropy and closure The gas entropy s 1 satisfies the standard relation (T k is the temperature of phase k) T 1 ds 1 = de 1 p 1 ρ 2 dρ 1. 1 For the entropy s 2 of the solid phase, we take into account the granular stress. T 2 ds }{{} 2 = de }{{} 2 energy variation internal energy variation p 2 ρ 2 dρ 2 2 }{{} pressure work R m 2 dα 2 }{{} granular work After some computations, we find the following entropy dissipation equation ( k m k s k ) t + ( k m k u k s k ) x = P T 2 (p 1 + R p 2 ).

10 A natural choice that ensures positive dissipation is P = 1 τ p α 1 α 2 (p 1 + R p 2 ), τ p 0. The choice of the pressure relaxation parameter τ p is very important. We can also write p 1 + R p 2 = τ p α 1 α 2 (α 1,t + u 2 α 1,x ). for τ p = t eq = 0, we recover the instantaneous six-equation equilibrium model p 2 = p 1 + R; τ p = t eq p ref where t eq is the characteristic time of the pressure equilibrium and p ref a reference pressure; for τ p > 0, the system should be more stable.

11 Thermodynamically coherent granular stress

12 Shape of the granular stress Usually, R = R(α 2 ) [4]

13 Entropy and granular stress How to choose the granular stress R for compressible solid phases? Exemple: for a stiffened gas equation of state p 2 = p 2 (ρ 2, e 2 ) = (γ 2 1)ρ 2 e 2 γ 2 π 2, γ 2 > 1. We suppose that R = R(ρ 2, α 2 ). We express that the entropy is a closed differential form (1/T 2 is an integrating factor of the energy form) ds 2 = 1 ( de 2 p 2 dρ 2 R ) dα 2. T 2 m 2 We find Particular choice ρ 2 2 R(ρ 2, α 2 ) = ρ γ 2 2 r(α 2). r(α 2 ) = λα γ 2 2 R = λm γ 2 2.

14 Hyperbolicity study

15 Stability of the relaxed system (τ p > 0) Let Y = (α 1, ρ 1, u 1, s 1, ρ 2, u 2, s 2 ) T. In this set of variables the system becomes B(Y ) = Y t + B(Y )Y x = (P, 0, 0, 0, 0, 0, 0) T u 2 ρ 1(u 1 u 2) α 1 u 1 ρ 1 c 2 p 1 ρ 1 u 1,s1 1 ρ 1 u 1 u 2 ρ 2 p 1 p 2 m 2 c 2 2 ρ 2 u 2 p 2,s2 ρ 2 u 2, c k = γ k (p k + π k ) ρ k. det(b(y ) λi ) = (u 2 λ) 2 (u 1 λ)(u 1 c 1 λ)(u 1 +c 1 λ)(u 2 c 2 λ)(u 2 +c 2 λ). The system is hyperbolic [7, 2].

16 Instability of the one-pressure model (τ p = 0) When τ p 0, formally, we end up with a standard single pressure model p 2 = p 1 + R. We can remove an equation (for example the volume fraction evolution) and we find a six-equation system Z = (ρ 1, u 1, s 1, ρ 2, u 2, s 2 ) T. Z t + C(Z)Z x = 0. In the simple case where R = λm γ 2 2, let = α 1 α 2 + δ(α 1 ρ 2 a α 2 ρ 1 c 2 1 ), and δ = α1 1/γ 2 2 λγ 2 ρ γ. 2 2

17 Then we find The eigenvalues can be computed only numerically. We observe that when R 0, the system is generally not hyperbolic. We observe also that when R, we recover hyperbolicity.

18 Numerical approximation: splitting method The two-pressure model is hyperbolic and more stable. The one-pressure model is commonly used in applications on coarse meshes (helped by the high damping of the numerical viscosity...). We can approximate the one-pressure model by the more general two-pressure model. A time step is made of a convection stage and a relaxation stage. In the relaxation stage, we apply the exact or relaxed pressure equilibrium.

19 Convection step Let w = (α 1, m 1, m 1 u 1, m 1 E 1, m 2, m 2 u 2, m 2 E 2 ) T, The system can be written w t + f (w) x + G(w)w x = (P, 0, 0, 0, 0, 0, 0) T In the first half step the source term is omitted. We use a standard Rusanov scheme w i wi n t f n i+1/2 = f (w n i, w n f (a, b) = + f i+1/2 n f i 1/2 n + G(wi n ) w i+1 n w i 1 n = 0, x 2 x i+1) numerical conservative flux, f (a) + f (b) λ (b a). 2 2 For λ large enough, the scheme is entropy dissipative. Typically, we take λ = max [ ρ ( f (a) + G(a) ), ρ ( f (b) + G(b) )].

20 Relaxation step In the second half step, we have formally to solve α 1,t = P, m k,t = u k,t = 0, (m k e k ) t + p 1 α k,t = 0. Because of mass and momentum conservation we have m k = m k and u k = u k. In each cell we have to compute (α 1, p 1, p 2 ) from the previous state w p 2 = p 1 + λm γ τ α 2 α2 p α 1 α 2 t, m 1 e 1 + m 2 e 2 = m1e 1 + m2e 2, (m 1 e 1 m 1e 1) + p 1 (α 1 α 1) = 0.

21 After some manipulations, we have to solve H(α 2 ) = (π 2 π 1 )(α 1 + (γ 1 1)(α 1 α1))(α 2 + (γ 2 1)(α 2 α2)) +(α 2 γ 2 2 m γ 2 2 θ(α τ p α 2 α2 2) + A 2 )(α 1 + (γ 1 1)(α 1 α (1 α 2 ) t 1)) +A 1 (α 2 + (γ 2 1)(α 2 α 2)) = 0, with A k = α k(p k + π k ) > 0. Theorem: for stiffened gas laws and the simple granular law, H is concave, H(0) < 0 and H(β 2 ) > 0 with 0 < β 2 = (1 α 2 ) + α 2 γ 1 γ 1 < 1. There is a unique α 2 in ]0, β 2 [ ]0, 1[ such that H(α 2 ) = 0.

22 Numerical application

23 Stability study We have constructed an entropy dissipative approximation of a non-hyperbolic system! What happens numerically? Consider a simple Riemann problem in the interval [ 1/2, 1/2]. γ 1 = and γ 2 = We compute the solution at time t = The CFL number is 0.9. Data: (L) (R) ρ u p ρ u p α

24 τ p = 0, R = 0 Figure: Void fraction, 50 cells, τ p = 0, R = 0.

25 τ p = 0, R = 0 Figure: Void fraction, 1, 000 cells, τ p = 0, R = 0.

26 τ p = 0, R = 0 Figure: Void fraction, 10, 000 cells, τ p = 0, R = 0.

27 τ p = 0, R = 0 Figure: Void fraction, 100, 000 cells, τ p = 0, R = 0. Linearly unstable but non-linearly stable...

28 τ p > 0, R = 0 Figure: Pressure, 50, 000 cells, τ p = 0, 2, 10, R = 0. The relaxation clearly stabilizes the results...

29 τ p = 0, R > 0 Figure: Pressures p 1 and p 2, 10, 000 cells, τ p = 0, R = 500m γ2 2.

30 τ p = 0, R > 0 6 Figure: Eigenvalues: imaginary part I = Im (λ i ) 2, 1, 000 cells, R = 0 or R = 500m γ2 2. i=1 The granular stress slightly improves the stability...

31 1D Combustion chamber We consider now a simplified gun [3, 5]. The right boundary of the computational domain is moving. We activate the granular stress and other source terms (chemical reaction and drag), which are all entropy dissipative. The instabilities would occur on much finer grids... Other source terms are added. m k,t + (m k u k ) x = ±M, (m k u k ) t + (m k u 2 k + α k p k ) x p 1 α k,x = ±Q, (m k E k ) t + ((m k E k + α k p k )u k ) x + p 1 α k,t = S k, α k,t + v 2 α k,x = ±P.

32 The mass transfer term is defined by the simplified relations M = α 2 ρ 2 3ṙ r ṙ = m/s (combustion velocity of the grains) (3) r = 10 3 m (radius of the grains) The momentum source term is given by Q = Mu 2 D D = Cα 1 α 2 ρ 2 (u 1 u 2 ) u 1 u 2 (drag force) C = 3 4r (simplified shape factor) (4) The energy source terms are S 1 = u 2 D + MQ ex S 2 = u 2 D Q ex = J/kg (chemical combustion energy) (5)

33 Figure: Pressure evolution at the breech and the shot base during time. Comparison between the Gough and the relaxation model.

34 Figure: Porosity at the final time. Relaxation model with granular stress.

35 Figure: Velocities at the final time. Relaxation model with granular stress.

36 Figure: Pressures at the final time. Relaxation model with granular stress.

37 Figure: Density of the solid phase at the final time. Relaxation model with granular stress.

38 3D Combustion chamber We consider now a 3D virtual gun [6] of length 434 mm. The section is varying from 42 to 40 mm. The igniter has a length of 50 mm and a radius of 2.5 mm. The 3x4 holes of the igniter are at mm, 63.5 mm and mm from the breech. On each row they are distributed at 0, 90, 180 and 270.

39 3D Gun Figure: Gun

40 Mesh Figure: Mesh of the combustion chamber

41 Temperature Figure: t=0.048 ms

42 Temperature Figure: t=0.100 ms

43 Temperature Figure: t=0.372 ms

44 Temperature Figure: t=0.668 ms

45 Temperature Figure: t=0.943 ms

46 Temperature Figure: t=1.204 ms

47 Temperature Figure: t=1.454 ms

48 Temperature Figure: t=1.697 ms

49 Temperature Figure: t=1.933 ms

50 Temperature Figure: t=2.000 ms

51 Temperature Figure: t=0.100 ms

52 Granular stress Figure:

53 Granular stress Figure:

54 Granular stress Figure:

55 Granular stress Figure:

56 Granular stress Figure:

57 Conclusion Good generalization of the one pressure models; Extension of the relaxation approach to flows with granular stress; Rigorous entropy dissipation and maximum principle on the volume fraction; Stability for a finite relaxation time; The instability is (fortunately) preserved by the scheme for fast pressure equilibrium; The model can be used in practical configurations (the solid phase remains almost incompressible).

58 Bibliography M. R. Baer and J. W. Nunziato. A two phase mixture theory for the deflagration to detonation transition (ddt) in reactive granular materials. Int. J. for Multiphase Flow, 16(6): , Thierry Gallouët, Jean-Marc Hérard, and Nicolas Seguin. Numerical modeling of two-phase flows using the two-fluid two-pressure approach. Math. Models Methods Appl. Sci., 14(5): , P. S. Gough. Modeling of two-phase flows in guns. AIAA, 66: , A. K. Kapila, R. Menikoff, J. B. Bdzil, S. F. Son, and D. S. Stewart. Two-phase modeling of deflagration-to-detonation transition in granular materials: reduced equations. Physics of Fluids, 13(10): , Julien Nussbaum, Philippe Helluy, Jean-Marc Hérard, and Alain Carrière. Numerical simulations of gas-particle flows with combustion. Flow, Turbulence and Combustion, 76(4): , J. Nussbaum. Modélisation et simulation numérique d un écoulement diphasique de la balistique intérieure. PhD thesis, Université de Strasbourg, December R. Saurel and R. Abgrall. A multiphase Godunov method for compressible multifluid and multiphase flows. Journal of Computational Physics, 150(2): , 1999.

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