Lattice Boltzmann Method for Moving Boundaries

Transcription

1 Lattice Boltzmann Method for Moving Boundaries Hans Groot March 18, 2009

2 Outline 1 Introduction 2 Moving Boundary Conditions 3 Cylinder in Transient Couette Flow 4 Collision-Advection Process for Moving Boundaries

3 Lattice Boltzmann (LB) Equation Define: nodes x 1,..., x n, discrete velocities e 0,..., e m f(x j + e k, t + 1) f(x j, t) }{{} advection = C f(x j, t) =: f c (x j, t) }{{} collision f = (f0,... f m ) T : distribution fk (x, t) = f (x, e k, t): probability that particle is in state (x, e k ) at time t C: collision operator

4 D2Q9 Model δx = 1, δt = 1 Discrete velocities: ( ) 0, 0, k = 0, ( e k = cos ( 1 (k 1)π), sin ( 1 (k 1)π)), k = 1, 2, 3, 4, 2 2 ( cos ( 1 (2k 9)π), sin ( 1 (2k 9)π)) 4 4 2, k = 5, 6, 7, 8.

5 Previous seminar Boundary Conditions in LB Methods: General formulation of boundary conditions Treatment of boundaries: periodic (infinite domain) no-slip free-slip frictional slip, sliding walls open (e.g. fluid inlet) complex geometry

6 No-Slip: bounce-back q = 1/2 wall E A B e i e i dx = 1 f i (x j, t + 1) = fi c (x j, t) e i = e i

7 No-Slip E A B e i q < 1/2 e i wall

8 No-Slip: upwind quadratic interpolation 1-2q q < 1/2 wall E D A B e i e i f i (x A, t + 1) = q(2q + 1)f c i (x A, t) + (1 2q)(1 + 2q)f c i (x A e i, t) q(1 2q)f c i (x A 2e i, t) (q < 1 2 )

9 No-Slip E A B e i q > 1/2 e i wall

10 No-Slip: downwind quadratic interpolation q > 1/2 wall E A D B e i e i 2q - 1 f i (x A, t + 1) = 1 q(2q + 1) f c i (x A, t) + 2q 1 q f c i (x A, t) 1 2q 1 + 2q f c i (x A e i, t) (q 1 2 )

11 Moving Boundary Conditions δf i (x j, t + 1) = f i (x j, t + 1) ( q(2q + 1)f c i (x j, t) + (1 2q)(1 + 2q)fi c (x j e i, t) ) q(1 2q)fi c (x j 2e i, t) (q < 1 2 )(1) ( 1 δf i (x j, t + 1) = f i (x j, t + 1) q(2q + 1) f i c (x j, t) + 2q 1 fi c q (x j, t) 1 2q ) 1 + 2q f i c (x j e i, t) (q 1 2 ) (2)

12 Equilibrium Distribution at Moving Boundary first-order equilibrium distribution at boundary: f (eq) i = fi 0 + α i e i e w e w : velocity of boundary weight coefficients: fi 0 = { 4 9, 1 9, 1 9, 1 9, 1 9, 1 36, 1 36, 1 36, 1 36 } α i = {0, 1 3, 1 3, 1 3, 1 3, 1 12, 1 12, 1 12, 1 12 } conservation (ρ 0 = 1): mass: i f (eq) i = 1 momentum: i e if (eq) i = e w

13 Equilibrium Distribution at Moving Boundary first-order equilibrium distribution at boundary: f (eq) i = fi 0 + α i e i e w e w : velocity of boundary weight coefficients: fi 0 = { 4 9, 1 9, 1 9, 1 9, 1 9, 1 36, 1 36, 1 36, 1 36 } α i = {0, 1 3, 1 3, 1 3, 1 3, 1 12, 1 12, 1 12, 1 12 } conservation (ρ 0 = 1): mass: i f (eq) i = 1 momentum: i e if (eq) i = e w

14 Equilibrium Distribution at Moving Boundary first-order equilibrium distribution at boundary: f (eq) i = fi 0 + α i e i e w e w : velocity of boundary weight coefficients: fi 0 = { 4 9, 1 9, 1 9, 1 9, 1 9, 1 36, 1 36, 1 36, 1 36 } α i = {0, 1 3, 1 3, 1 3, 1 3, 1 12, 1 12, 1 12, 1 12 } conservation (ρ 0 = 1): mass: i f (eq) i = 1 momentum: i e if (eq) i = e w

15 Moving Boundary Conditions substitution f i f 0 i + α i e i e w, f i f 0 i α i e i e w (1) = δf i = 2α i e i e w 2 (2) = δf i = q(2q + 1) α ie i e w

16 Moving Boundary Conditions f i (x j, t + 1) = q(2q + 1)f c i (x j, t) + (1 2q)(1 + 2q)f c i (x j e i, t) q(1 2q)f c i (x j 2e i, t) + 2α i e i e w, (q < 1 2 ) f i (x j, t + 1) = 1 q(2q + 1) f c i (x j, t) + 2q 1 q 1 2q 1 + 2q f i c (x j e i, t) + f c i (x j, t) 2 q(2q + 1) α ie i e w (q 1 2 )

17 Moving Wall t fluid solid fluid solid n x e w t+1 Solid nodes become fluid nodes

18 Moving Wall t fluid solid fluid solid n x e w t+1 Solid nodes become fluid nodes

19 Moving Wall Unknown distribution in solid fluid nodes Different methods: 1 extrapolation: f(x, t) = 3f(x, t) 3f(x, t) + f(x + e k, t) e k maximises n e k

20 Moving Wall Unknown distribution in solid fluid nodes Different methods: 1 extrapolation: f(x, t) = 3f(x, t) 3f(x, t) + f(x + e k, t) e k maximises n e k

21 Moving Wall x fluid solid fluid solid x e k n x e w

22 Moving Wall Unknown distribution in solid fluid nodes Different methods: 1 extrapolation: f(x, t) = 3f(x, t) 3f(x, t) + f(x + e k, t) e k maximises n e k 2 equilibrium distribution function: f (eq) i = fi 0 + α i e i e w 3 systematically update distribution functions in non-fluid nodes with velocity e w. Methods produce similar results

23 Moving Wall Unknown distribution in solid fluid nodes Different methods: 1 extrapolation: f(x, t) = 3f(x, t) 3f(x, t) + f(x + e k, t) e k maximises n e k 2 equilibrium distribution function: f (eq) i = fi 0 + α i e i e w 3 systematically update distribution functions in non-fluid nodes with velocity e w. Methods produce similar results

24 Moving Wall Unknown distribution in solid fluid nodes Different methods: 1 extrapolation: f(x, t) = 3f(x, t) 3f(x, t) + f(x + e k, t) e k maximises n e k 2 equilibrium distribution function: f (eq) i = fi 0 + α i e i e w 3 systematically update distribution functions in non-fluid nodes with velocity e w. Methods produce similar results

25 Moving Wall Unknown distribution in solid fluid nodes Different methods: 1 extrapolation: f(x, t) = 3f(x, t) 3f(x, t) + f(x + e k, t) e k maximises n e k 2 equilibrium distribution function: f (eq) i = fi 0 + α i e i e w 3 systematically update distribution functions in non-fluid nodes with velocity e w. Methods produce similar results

26 Cylinder in Transient Couette Flow U 0 -U 0 Typically: d/w = 0.25, U 0 = 0.1, Re = 11.36

27 Two Reference Frames Reference frame at rest cylinder moving with speed U c w.r.t. mesh Moving bounday Reference frame moving with speed U c cylinder fixed w.r.t. mesh

28 Total Force on Cylinder Method: extrapolation

29 Total Force on Cylinder Method: extrapolation

30 Total Force on Cylinder Method: equilibrium distribution functions

31 Discussion Results for moving and fixed boundary in agreement with each other Spatial fluctuation in force number of fluid nodes, hence volume, not conserved error introduced by computing distribution in solid fluid nodes number of lattice lines (edges) varies

32 Collision-Advection Process Step 1: Compute Moments m(x j, t) = Mf(x j, t) Step 2: Relaxation m c (x j, t) = m(x j, t) S(m(x j, t) m (eq) (x j, t)) Step 3: Compute post-collision distributions Cf(x j, t) = M 1 m c (x j, t) Step 4: Advection f(x j + e k, t + 1) = f(x j, t) + Cf(x j, t) Step 5: Moving boundary

33 Collision-Advection Process Step 1: Compute Moments m(x j, t) = Mf(x j, t) Step 2: Relaxation m c (x j, t) = m(x j, t) S(m(x j, t) m (eq) (x j, t)) Step 3: Compute post-collision distributions Cf(x j, t) = M 1 m c (x j, t) Step 4: Advection f(x j + e k, t + 1) = f(x j, t) + Cf(x j, t) Step 5: Moving boundary

34 Collision-Advection Process Step 1: Compute Moments m(x j, t) = Mf(x j, t) Step 2: Relaxation m c (x j, t) = m(x j, t) S(m(x j, t) m (eq) (x j, t)) Step 3: Compute post-collision distributions Cf(x j, t) = M 1 m c (x j, t) Step 4: Advection f(x j + e k, t + 1) = f(x j, t) + Cf(x j, t) Step 5: Moving boundary

35 Collision-Advection Process Step 1: Compute Moments m(x j, t) = Mf(x j, t) Step 2: Relaxation m c (x j, t) = m(x j, t) S(m(x j, t) m (eq) (x j, t)) Step 3: Compute post-collision distributions Cf(x j, t) = M 1 m c (x j, t) Step 4: Advection f(x j + e k, t + 1) = f(x j, t) + Cf(x j, t) Step 5: Moving boundary

36 Collision-Advection Process Step 1: Compute Moments m(x j, t) = Mf(x j, t) Step 2: Relaxation m c (x j, t) = m(x j, t) S(m(x j, t) m (eq) (x j, t)) Step 3: Compute post-collision distributions Cf(x j, t) = M 1 m c (x j, t) Step 4: Advection f(x j + e k, t + 1) = f(x j, t) + Cf(x j, t) Step 5: Moving boundary

37 Step 1: Compute Moments Linear transformation: m = Mf, m = (ρ, e, ɛ, j x, q x, j y, q y, p xx, p xy ) T ρ: density e: related to kinetic energy ɛ: related to kinetic energy square j x, j y : components of momentum density q x, q y : components of energy flux p xx, p xy : components of stress tensor

38 Transformation Matrix Transformation from phase space to moment space M =

39 Step 2: Relaxation Relaxation equation m c = m S(m m (eq) ) m c : post-collision state m (eq) : equilibrium state S = diag ( ) s 1,..., s 9 : diagonal relaxation matrix ρ, j x, j y conserved s 1, s 4, s 6 = 0 e, ɛ, q x, q y, p xx, p yy non-conserved Lattice BGK model: s k = 1 τ, k = 2, 3, 5, 7, 8, 9 τ = relaxation time

40 Step 2: Relaxation Relaxation equation m c = m S(m m (eq) ) m c : post-collision state m (eq) : equilibrium state S = diag ( ) s 1,..., s 9 : diagonal relaxation matrix ρ, j x, j y conserved s 1, s 4, s 6 = 0 e, ɛ, q x, q y, p xx, p yy non-conserved Lattice BGK model: s k = 1 τ, k = 2, 3, 5, 7, 8, 9 τ = relaxation time

41 Step 2: Relaxation Relaxation equation m c = m S(m m (eq) ) m c : post-collision state m (eq) : equilibrium state S = diag ( ) s 1,..., s 9 : diagonal relaxation matrix ρ, j x, j y conserved s 1, s 4, s 6 = 0 e, ɛ, q x, q y, p xx, p yy non-conserved Lattice BGK model: s k = 1 τ, k = 2, 3, 5, 7, 8, 9 τ = relaxation time

42 Equilibrium Distribution Functions Depend only on conserved moments ρ, j x, j y Kinetic theory for Maxwell molecules: e (eq) = 2ρ + 3 ( ) j 2 ρ x + jy 2 ɛ (eq) = ρ 3 ρ ( j 2 x + j 2 y q (eq) x = j x, q (eq) y p (eq) xx = 1 ρ ( j 2 x j 2 y ) = j y ), p (eq) xy = 1 ρ j xj y P. Lallemand and L.-S. Luo

43 Step 3: Compute Post-Collision Distributions Post-collision distributions Cf(x j, t) = M 1 m c (x j, t)

44 Step 4: Advection Advected distributions f(x j + e k, t + 1) = f(x j, t) + Cf(x j, t)

45 Step 5: Moving Boundary t fluid solid fluid solid n x e w t+1

46 References M. Bouzidi, M. Firdaouss and P. Lallemand Momentum Transfer of a Boltzmann-Lattice Fluid with Boundaries P. Lallemand and L.-S. Luo Theory of the Lattice Boltzmann Method: Dispersion, Dissipation, Isotropy, Galilean Invariance and Stability P. Lallemand and L.-S. Luo Lattice Boltzmann Method for Moving Boundaries 2003.