Lattice Boltzmann Method for Moving Boundaries

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

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

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

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.

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

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

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

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 )

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

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 )

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)

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

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

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 )

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

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

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

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

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

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

Total Force on Cylinder Method: extrapolation

Total Force on Cylinder Method: equilibrium distribution functions

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

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

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

Transformation Matrix Transformation from phase space to moment space M = 1 1 1 1 1 1 1 1 1 4 1 1 1 1 2 2 2 2 4 2 2 2 2 1 1 1 1 0 1 0 1 0 1 1 1 1 0 2 0 2 0 1 1 1 1 0 0 1 0 1 1 1 1 1 0 0 2 0 2 1 1 1 1 0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1

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

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

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

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

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

References M. Bouzidi, M. Firdaouss and P. Lallemand Momentum Transfer of a Boltzmann-Lattice Fluid with Boundaries 2001. P. Lallemand and L.-S. Luo Theory of the Lattice Boltzmann Method: Dispersion, Dissipation, Isotropy, Galilean Invariance and Stability 2000. P. Lallemand and L.-S. Luo Lattice Boltzmann Method for Moving Boundaries 2003.