Boundary Conditions in lattice Boltzmann method
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1 Boundar Conditions in lattice Boltzmann method Goncalo Silva Department of Mechanical Engineering Instituto Superior Técnico (IST) Lisbon, Portugal
2 Outline Introduction 1 Introduction Boundar Value Problems 2 Navier-Stokes Boundar Conditions 3 4
3 Outline Introduction Boundar Value Problems 1 Introduction Boundar Value Problems 2 Navier-Stokes Boundar Conditions 3 4
4 Definitions Introduction Boundar Value Problems Boundar Value Problem { Partial Differential Equation Boundar Condition
5 Definitions Introduction Boundar Value Problems Boundar Value Problem { Partial Differential Equation Boundar Condition e.g. Poisson equation: { 2 ϕ ϕ = f(, ), 2 in ϕ = ϕ b, on
6 Definitions Introduction Boundar Value Problems Tpes of Boundar Conditions
7 Definitions Introduction Boundar Value Problems Tpes of Boundar Conditions Dirichlet Boundar Condition ϕ = ϕ b on
8 Definitions Introduction Boundar Value Problems Tpes of Boundar Conditions Dirichlet Boundar Condition ϕ = ϕ b on Neumann Boundar Condition n ϕ = ϕ n = ϕ b on
9 Definitions Introduction Boundar Value Problems Tpes of Boundar Conditions Dirichlet Boundar Condition ϕ = ϕ b on Neumann Boundar Condition n ϕ = ϕ n = ϕ b on Robin Boundar Condition gϕ + h ϕ n = ϕ b on
10 Outline Introduction Navier-Stokes Boundar Conditions 1 Introduction Boundar Value Problems 2 Navier-Stokes Boundar Conditions 3 4
11 Definitions Introduction Navier-Stokes Boundar Conditions Stead isothermal and incompressible Navier-Stokes equations { (u )u = p + ν u + a in u = 0
12 Definitions Introduction Navier-Stokes Boundar Conditions Stead isothermal and incompressible Navier-Stokes equations { (u )u = p + ν u + a in u = 0 Boundar Condition on solid walls u = u b on
13 Definitions Introduction Navier-Stokes Boundar Conditions Stead isothermal and incompressible Navier-Stokes equations { (u )u = p + ν u + a in u = 0 Boundar Condition on solid walls u = u b on Boundar Condition on fluid boundaries u = u in or { p + ν u n n on = (F n) in ν ut n = (F t) in on
14 Definitions Introduction Navier-Stokes Boundar Conditions Periodic or Cclic Boundar Conditions u(, ) = u( + L, ) in u in = u out on
15 Definitions Introduction Navier-Stokes Boundar Conditions Periodic or Cclic Boundar Conditions u(, ) = u( + L, ) in u in = u out on Smmetr Boundar Conditions u n = 0 and u n = 0 on
16 Outline Introduction 1 Introduction Boundar Value Problems 2 Navier-Stokes Boundar Conditions 3 4
17 Introduction and motivation Lattice Boltzmann method (LBM) f α ( + c α t, t + t) = f α (, t) ω(f α f (eq) α ) (,t) in
18 Introduction and motivation Lattice Boltzmann method (LBM) f α ( + c α t, t + t) = f α (, t) ω(f α f (eq) α ) (,t) in Hdrodnamic Boundar Conditions in LBM
19 Introduction and motivation Lattice Boltzmann method (LBM) f α ( + c α t, t + t) = f α (, t) ω(f α f (eq) α ) (,t) in Hdrodnamic Boundar Conditions in LBM Solution on is specified for f α and NOT for {ρ, u, Π}
20 Introduction and motivation Lattice Boltzmann method (LBM) f α ( + c α t, t + t) = f α (, t) ω(f α f (eq) α ) (,t) in Hdrodnamic Boundar Conditions in LBM Solution on is specified for f α and NOT for {ρ, u, Π} f α set in a higher DoF sstem than {ρ, u, Π}, hence: Trivial: f α {ρ, u, Π} Comple: {ρ, u, Π} f α
21 Introduction and motivation Lattice Boltzmann method (LBM) f α ( + c α t, t + t) = f α (, t) ω(f α f (eq) α ) (,t) in Hdrodnamic Boundar Conditions in LBM Solution on is specified for f α and NOT for {ρ, u, Π} f α set in a higher DoF sstem than {ρ, u, Π}, hence: Trivial: f α {ρ, u, Π} Comple: {ρ, u, Π} f α Incorrect upscaling Unwanted behavior, e.g. Knudsen laers
22 Lattice structure c7 c3 c6 D2Q9 model c4 c8 c1 c5 c2 c9 c 1 = (0, 0) c 2 = (1, 0) c 3 = (0, 1) c 4 = ( 1, 0) c 5 = (0, 1) c 6 = (1, 1) c 7 = ( 1, 1) c 8 = ( 1, 1) c 9 = ( 1, 1)
23 Definitions Introduction is a fluid node if c so that + c t { } is a boundar node if c so that + c t / { } c7 c3 c6 c4 c2 c8 c5 c9 c7 c3 c6
24 Periodic Boundar Conditions Periodicit u in = u out on c6 c6 c2 c2 c9 c9 c7 c7 c4 c4 c8 c8
25 Smmetr Boundar Conditions Smmetr u n = 0 and u n = 0 (also called free-slip boundar) on c 8 c 5 c 9 c c 3 7 c 6
26 Bounceback Boundar Conditions A ver intuitive idea: A hard wall reflects particles back to where the originall came from
27 Bounceback Boundar Conditions A ver intuitive idea: A hard wall reflects particles back to where the originall came from As a result: There is no flu crossing the wall, i.e. the wall is impermeable There is no relative transverse motion between fluid and wall, i.e. no-slip at the wall
28 Bounceback Boundar Conditions REMEMBER: LBM algorithm can be operated in 2 steps: Collision step: f α (, t) = f α (, t) ω(f α f α (eq) ) (,t) Streaming step: f α ( + c α t, t + t) = f α (, t)
29 Bounceback Boundar Conditions The Bounceback method can be implemented following 2 reasonings: Full-wa bounceback: inversion of particle velocit takes place during the collision step Half-wa bounceback: inversion of particle velocit takes place during the streaming step
30 Bounceback Boundar Conditions Full-wa Bounceback: fᾱ( b, t) = f α ( b, t) t- t t t t+ t Streaming Collision Streaming
31 Bounceback Boundar Conditions Full-wa Bounceback: fᾱ( b, t) = f α ( b, t) t- t t t t+ t Streaming Collision Streaming Half-wa Bounceback: fᾱ( f, t + t) = f α ( f, t) t t t+ t/2 t+ t Collision Streaming
32 Bounceback Boundar Conditions Half-wa bounceback in 2D: t c9 c5 c8 t+ t c7 c3 c6
33 Bounceback Boundar Conditions: summar Pros Mass is eactl conserved Stable for ω close to 2 (i.e. for high Re) Local Fleibilit in handling wall, edges, corners both in 2D and 3D Ver simple to implement from a programming viewpoint Cons Velocit accurac ma decrease from 2 nd to 1 st Pressure accurac ma decrease from 1 st to 0 th In SRT model momentum is not eactl conserved (viscosit dependent slip velocit), which is equivalent to sa the boundar location is not eactl defined (viscosit dependent slip length)
34 Bounceback Boundar Conditions: Eercise Question: Use the half-wa bounceback scheme to find the unknown populations at t + t Bottom wall Top wall t c 5 t c 8 c9 c 7 c 3 c 6 t+ t? t+ t?
35 Bounceback Boundar Conditions: Eercise Question: Use the half-wa bounceback scheme to find the unknown populations at t + t Solution: Bottom wall Top wall t c5 t c8 c9 c7 c3 c6 t+ t c7 c3 c6 t+ t c8 c5 c9
36 Momentum echange Force (per unit volume): F (t+ t 2 ) = p t (t+ t 2 ) = 1 t (p(t + t) p(t))
37 Momentum echange Force (per unit volume): F (t+ t 2 ) = p t (t+ t 2 ) = 1 t (p(t + t) p(t)) t t+ t c7 c3 c6 c8 c5 c9
38 Momentum echange Force (per unit volume): F (t+ t 2 ) = p t (t+ t 2 ) = 1 t (p(t + t) p(t)) t t+ t c7 c3 c6 c8 c5 c9 Momentum echange (per unit volume) between the fluid/wall surface: p(, t + t 2 ) = [ (cᾱ)fᾱ(, t + t) (c α ) f ] α (, t) α
39 Momentum echange Momentum echange between the fluid/wall surface: p(, t + t 2 ) = α c α [fᾱ(, t + t) + f ] α (, t)
40 Momentum echange Momentum echange between the fluid/wall surface: p(, t + t 2 ) = α c α [fᾱ(, t + t) + f ] α (, t) Remember: Half-wa Bounceback: fᾱ( f, t + t) = f α ( f, t)
41 Momentum echange Momentum echange between the fluid/wall surface: p(, t + t 2 ) = α c α [fᾱ(, t + t) + f ] α (, t) Remember: Half-wa Bounceback: fᾱ( f, t + t) = f α ( f, t) p(, t + t 2 ) = 2 c α fα (, t) α
42 Momentum echange Momentum echange between the fluid/wall surface: p(, t + t 2 ) = α c α [fᾱ(, t + t) + f ] α (, t) Remember: Half-wa Bounceback: fᾱ( f, t + t) = f α ( f, t) p(, t + t 2 ) = 2 c α fα (, t) α Force on the fluid due to the wall : F(, t + t 2 ) = 2 t c α fα (, t) b S α
43 Transverse force on the fluid due to the bottom wall F (, t + t 2 ) = 2 t b S (c α ) fα (, t) α
44 Transverse force on the fluid due to the bottom wall F (, t + t 2 ) = 2 t b S (c α ) fα (, t) α t t+ t c7 c3 c6 c8 c5 c9
45 Transverse force on the fluid due to the bottom wall F (, t + t 2 ) = 2 t b S (c α ) fα (, t) α t t+ t c7 c3 c6 c8 c5 c9 Transverse force b bottom wall: F (, t + t 2 ) = 2 t ( f 9 f 8 ) (,t) b S
46 Normal force on the fluid due to the bottom wall F (, t + t 2 ) = 2 t b S (c α ) fα (, t) α
47 Normal force on the fluid due to the bottom wall F (, t + t 2 ) = 2 t b S (c α ) fα (, t) α t t+ t c7 c3 c6 c8 c5 c9
48 Normal force on the fluid due to the bottom wall F (, t + t 2 ) = 2 t b S (c α ) fα (, t) α t t+ t c7 c3 c6 c8 c5 c9 Normal force b bottom wall: F (, t + t 2 ) = 2 t ( f 5 + f 8 + f 9 ) (,t) b S
49 Momentum echange: Eercise Question: Write the formulas of the transverse and normal forces at the top wall Remember: t t+ t c7 c3 c6 c8 c5 c9
50 Momentum echange: Eercise Question: Write the formulas of the transverse and normal forces at the top wall Solution: Transverse force b top wall: F (, t + t 2 ) = 2 t ( f 6 f 7 ) (,t) b S Normal force b bottom wall: F (, t + t 2 ) = 2 t ( f 3 + f 6 + f 7 ) (,t) b S
51 Eercise I Introduction Eercise I: Poiseuille flow with bounceback walls
52 Introduction The solution of the Navier-Stokes not onl requires the no-slip velocit condition on walls but also demands these equations to be valid near the wall
53 Introduction The solution of the Navier-Stokes not onl requires the no-slip velocit condition on walls but also demands these equations to be valid near the wall Taking advantage of the Chapman-Enskog epansion... f = f (0) (ρ, u) + ɛf (1) ( u) + O(ɛ 2 )
54 Introduction The solution of the Navier-Stokes not onl requires the no-slip velocit condition on walls but also demands these equations to be valid near the wall Taking advantage of the Chapman-Enskog epansion... f = f (0) (ρ, u) + ɛf (1) ( u) + O(ɛ 2 )...it can be shown: ( ) f α (0) = w α ρ + cα u + (cαcα c2 c 2 s I) : uu s 2c 4 s f (1) α = w α (c αc α c 2 si) ωc 2 s : ( u + ( u) T )
55 Zou He boundar condition Boundar node and solid node coincide Onl unknown incoming populations are modified Set ρ or u in f (0) α (ρ, u) Construct f (1) α from the smmetr requirement
56 Zou He velocit boundar condition Known: u = 0 f α = (f 1, f 2, f 4, f 5, f 8, f 9 ) c 7 c 3 c 6 c 4 c 2 c 8 c 5 c 9
57 Zou He velocit boundar condition Known: u = 0 f α = (f 1, f 2, f 4, f 5, f 8, f 9 ) c 7 c 3 c 6 c 4 c 2 Unknown (4 variables): ρ c 8 c 5 c 9 f α = (f 3, f 6, f 7 )
58 Zou He velocit boundar condition Known: u = 0 f α = (f 1, f 2, f 4, f 5, f 8, f 9 ) c 7 c 3 c 6 c 4 c 2 Unknown (4 variables): ρ c 8 c 5 c 9 f α = (f 3, f 6, f 7 ) 3 Equations (2 linearl independent): f α = ρ c α f α = u
59 Zou He velocit boundar condition Smmetr of f (1) α (3 equations): c 7 c 3 c 6 c 4 c 2 Bounceback of non-equilibrium populations c 8 c 5 c 9
60 Zou He velocit boundar condition Smmetr of f (1) α (3 equations): c 7 c 3 c 6 c 4 c 2 c 8 c 5 c 9 Bounceback of non-equilibrium populations Introduce etra variable (problem overspecified): Transverse momentum correction N
61 Zou He velocit boundar condition Smmetr of f (1) α (3 equations): c 7 c 3 c 6 c 4 c 2 c 8 c 5 c 9 Bounceback of non-equilibrium populations Introduce etra variable (problem overspecified): Transverse momentum correction N Problem is well specified: 6 eqs. and 6 unknowns
62 Zou He velocit boundar condition 1) Computing ρ... Population velocit set at boundar node: C + = {c 3, c 6, c 7 } c 7 c 3 c 6 c 4 c 2 c 8 c 5 c 9 C 0 = {c 1, c 2, c 4 } C = {c 5, c 8, c 9 } Use the two velocit moments: f α = ρ (c α ) f α = u
63 Zou He velocit boundar condition 1) Computing ρ... c 7 c 3 c 6 c 4 c 2 Relate the two velocit moments: { ρ = ρ+ + ρ 0 + ρ u = ρ + ρ c 8 c 5 c 9
64 Zou He velocit boundar condition 1) Computing ρ... c 7 c 3 c 6 c 4 c 2 Relate the two velocit moments: { ρ = ρ+ + ρ 0 + ρ u = ρ + ρ c 8 c 5 c 9 Solution: ρ = u + ρ 0 + 2ρ i.e. ρ = u + (f 1 + f 2 + f 4 ) + 2(f 5 + f 8 + f 9 )
65 Zou He velocit boundar condition 2) Computing {f 3, f 6, f 7 }... c 7 c 3 c 6 c 4 c 2 c 8 c 5 c 9 Non-equilibrium bounceback with transverse momentum correction: f 3 f (0) 3 = f 5 f (0) 5 f 6 f (0) 6 = f 8 f (0) 8 +N f 7 f (0) 7 = f 9 f (0) 9 N
66 Zou He velocit boundar condition 2) Computing {f 3, f 6, f 7 }... Solution for the unknown incoming populations: f 3 = f u f 6 = f (f 4 f 2 ) u u f 7 = f (f 4 f 2 ) u 1 2 u
67 Zou He boundar condition Pros Local Velocit is 2 nd order accurate Pressure accurac is (at worst) 1 st order accurate In SRT model momentum is conserved (up to 2 nd order) Cons Unstable when ω 0 Mass is not eactl conserved (2 nd order accurate) Fleibilit in handling wall, edges and corners or 2D and 3D domains are being modeled Not so simple to implement (compared to bounceback)
68 Zou He boundar condition: Eercise Question: Use the Zou He procedure to find ρ and the unknown populations at the top wall c 7 c 3 c 6 c 4 c 2 c 8 c 5 c 9
69 Zou He boundar condition: Eercise Question: Use the Zou He procedure to find ρ and the unknown populations at the top wall Solution: ρ = u + (f 1 + f 2 + f 4 ) + 2(f 3 + f 6 + f 7 ) f 5 = f u f 8 = f (f 2 f 4 ) 1 6 u 1 2 u f 9 = f (f 2 f 4 ) 1 6 u u
70 Eercise II Introduction Eercise II: Poiseuille flow Zou He walls
71 Outline Introduction 1 Introduction Boundar Value Problems 2 Navier-Stokes Boundar Conditions 3 4
72 Bounceback vs. Zou He BC scheme Bounceback Zou He
73 Bounceback vs. Zou He BC scheme Bounceback Zou He Boundar location Halfwa On node
74 Bounceback vs. Zou He BC scheme Bounceback Zou He Boundar location Halfwa On node Stable Yes Not as stable
75 Bounceback vs. Zou He BC scheme Bounceback Zou He Boundar location Halfwa On node Stable Yes Not as stable Velocit accurac 2 nd to 1 st 2 nd
76 Bounceback vs. Zou He BC scheme Bounceback Zou He Boundar location Halfwa On node Stable Yes Not as stable Velocit accurac 2 nd to 1 st 2 nd Pressure accurac 1 st to 0 th 1 st
77 Bounceback vs. Zou He BC scheme Bounceback Zou He Boundar location Halfwa On node Stable Yes Not as stable Velocit accurac 2 nd to 1 st 2 nd Pressure accurac 1 st to 0 th 1 st Mass conservative Yes Onl 2 nd
78 Bounceback vs. Zou He BC scheme Bounceback Zou He Boundar location Halfwa On node Stable Yes Not as stable Velocit accurac 2 nd to 1 st 2 nd Pressure accurac 1 st to 0 th 1 st Mass conservative Yes Onl 2 nd Viscosit independent Not in SRT Yes
79 Bounceback vs. Zou He BC scheme Bounceback Zou He Boundar location Halfwa On node Stable Yes Not as stable Velocit accurac 2 nd to 1 st 2 nd Pressure accurac 1 st to 0 th 1 st Mass conservative Yes Onl 2 nd Viscosit independent Not in SRT Yes Fleibilit Yes Not as fleible
80 Bounceback vs. Zou He BC scheme Bounceback Zou He Boundar location Halfwa On node Stable Yes Not as stable Velocit accurac 2 nd to 1 st 2 nd Pressure accurac 1 st to 0 th 1 st Mass conservative Yes Onl 2 nd Viscosit independent Not in SRT Yes Fleibilit Yes Not as fleible Coding simplicit Yes Not as simple
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