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fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Recovering the full Navier Stokes equationswith lattice Boltzmann schemes
Francois Dubois∗†, Benjamin Graille† and Pierre Lallemand‡§
NUMKIN 2016: International Workshopon Numerical Methods for Kinetic Equations
Strasbourg, 20 october 2016
∗ Conservatoire National des Arts et Metiers, Paris, France† Universite Paris-Sud, France‡ Centre National de la Recherche Scientifique, Paris, France§ Beijing Computational Science Research Center, China
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Jean-Louis Loday (1946 - 2012)
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
From Boltzmann equation to lattice Boltzmann schemes
Boltzmann equation ∂t f + v•∇f = Q(f )Linearization Q(f ) ' Q(f eq) + dQ(f eq)•(f − f eq)Boltzmann BGK ∂t f + v•∇f = dQ(f eq)•(f − f eq)Discrete velocities
∂t(Mf ) + M•v•∇f =(M dQ(f eq)M−1
)•(M (f − f eq)
)Moments m = M f , meq = M f eq
Multiple Relaxation Times hypothesis :the matrix M dQ(f eq)M−1 is diagonal and real
Time discretization: alternate directions, or “collide-stream”Zero eigenvalues dmk
dt = 0, 0 ≤ k < N: conserved moments WRelaxation of the nonconserved moments mk , k ≥ Ndmkdt = − 1
τk
(mk −meq
k
), m∗k = mk − ∆t
τk(mk −meq
k (W )), sk = ∆tτk
Particle densities f ∗j (x , t) =(M−1 m∗
)j(x , t)
x ∈ Lattice and x + vj ∆t ∈ Lattice, t = n∆tSolve the advection equation ∂t fj + vj•∇fj = 0 with CFL=1!
fj(x + vj ∆t, t + ∆t) = f ∗j (x , t)Lattice Boltzmann scheme DdQq
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Outline
• Fluid flow with a Multi Relaxation TimesD1Q3 lattice Boltzmann scheme
Advection-diffusion with a Multi Relaxation TimesD1Q3 lattice Boltzmann scheme
• Double distribution for Navier StokesNavier Stokes with mass, momentum and total energyDouble distribution with the Bhatnagar Gross Krook
lattice Boltzmann framework
• Navier Stokes with mass, momentum and volumic entropyDouble distribution for a Multi Relaxation Times
D1Q3Q3 lattice Boltzmann schemeLinearized study with a focus on stability
• Definition of a nonlinear schemeFirst numerical experiments
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Fluid flow with a MRT D1Q3 lattice Boltzmann scheme
f0(x , t) : motionless particles at x during the time step ∆t
f+(x , t) : density of particles going from x to x + ∆x during ∆t
f−(x , t) : particles going from x to x −∆x during the time step ∆t
numerical velocity λ = ∆x∆t fixed
discrete velocities v0 = 0, v+ = +1, v− = −1
Mass density ρ ≡ ρ0
∑j=0,+,−
fj = ρ0 (f+ + f− + f0)
Momentum J ≡ ρ0 λ∑
j=0,+,−vj fj = ρ0 λ (f+ − f−)
Energy e ≡ ρ0 λ2 (f+ + f− − 2 f0) third momentum
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Fluid flow with MRT D1Q3 lattice Boltzmann scheme (ii)
Invertible matrix M between the particles and the moments:ρJe
= MD1Q3
f0f+f−
, MD1Q3 = ρ0
1 1 10 λ −λ−2λ2 λ2 λ2
• RelaxationThe density and momentum remain at equilibrium
and do not change: ρ∗ = ρ , J∗ = J
The equilibrium energy is a given function of the two momentsat equilibrium: eeq = eeq(ρ, J)
(the Gaussian distribution for the BGK model)
The energy after relaxation is obtained by a simple evolution:e∗ = e + se (eeq − e) with 0 < se < 2
The particle distribution after relaxation:f ∗0f ∗+f ∗−
=(MD1Q3
)−1
ρJe∗
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
MRT D1Q3 fluid lattice Boltzmann scheme (iii)
• Advectionf0(x , t + ∆t) = f ∗0 (x , t)f+(x , t + ∆t) = f ∗+(x −∆x , t)f−(x , t + ∆t) = f ∗−(x + ∆x , t)
• Analysis with the Taylor expansion method[FD, ESAIM ProcS, 18, 2007]
A numerical Chapman-Enskog expansion withthe small parameter ε ≡ λ
D replaced by ε ≡ ∆xL
Equivalent partial differential equations up to order 2:
∂tρ+ ∂xJ = O(∆x2)
∂tJ + ∂x(
23λ
2 ρ+ 13e
eq)− 1
3 σe ∆t ∂xθe = O(∆x2)
with σe = 1se− 1
2 , θe ≡ ∂teeq + λ2 ∂xJ ' 3λ2ρ ∂xu
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
MRT D1Q3 fluid lattice Boltzmann scheme (iv)
1D “isentropic” Navier Stokes equationsmass ∂tρ+ ∂xJ = 0
momentum ∂tJ + ∂x(ρ u2 + p
)− ∂x
(ρ ν ∂xu
)= 0
D1Q3 fluid lattice Boltzmann schememass ∂tρ+ ∂xJ = O(∆x2)
momentum
∂tJ + ∂x(
23λ
2 ρ+ 13e
eq)− ∂x
(σe ∆x λ ρ ∂xu
)= O(∆x2)
Energy at equilibrium: eeq = 3(ρ u2 + p
)− 2λ2 ρ
Relaxation parameter and kinematic viscosity:1
se− 1
2= σe =
ν
λ∆x
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Some algebraic tools...
Moments m ≡ M f
Momentum-velocity tensor Λk` ≡ λ∑j
Mkj vj (M−1)j`
ΛD1Q3 =
0 1 023λ
2 0 13
0 λ2 0
with the ordering (ρ , J , e) of the moments
Defect of conservation θ` ≡ ∂tmeq` +
∑p
Λ`p ∂xmeqp
Second order partial differential equationsatisfied by the ko conserved moment Wk
∂tWk +∑`
Λk` ∂xmeq` −∆t
∑`
σ` Λk` ∂xθ` = O(∆x2)
Straitforward generalization to two and three dimensions
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Advection-diffusion with a MRT D1Q3 LB scheme
g0(x , t) : motionless particles at x during the time step ∆t
g+(x , t) : density of particles going from x to x + ∆x during ∆t
g−(x , t) : particles going from x to x −∆x during the time step
numerical velocity λ = ∆x∆t fixed
discrete velocities v0 = 0, v+ = +1, v− = −1
Scalar ζ ≡ ζ0
∑j=0,+,−
gj = ζ0 (g+ + g− + g0)
Momentum ψ ≡ ζ0 λ∑
j=0,+,−vj gj = ζ0 λ (g+ − g−)
Associated energy ε ≡ ζ0 λ2 (g+ + g− − 2 g0)
third momentum
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Advection-diffusion with a MRT D1Q3 LB scheme (ii)
Invertible matrix M between the particles and the moments:ζψε
= MD1Q3
g0
g+
g−
, MD1Q3 = ζ0
1 1 10 λ −λ−2λ2 λ2 λ2
• RelaxationThe scalar ζ remains at equilibrium : ζ∗ = ζ
The momentum ψ and associated energy ε at equilibriumare given functions of the scalar ζ :
ψeq = ψeq(ζ) , εeq = εeq(ζ)
The momentum and energy after relaxation are obtained bysimple evolutions: ψ∗ = ψ + sψ (ψeq − ψ) , 0 < sψ < 2
ε∗ = ε+ sε (εeq − ε) , 0 < sε < 2
Particle distribution g after relaxation:g∗0g∗+g∗−
=(MD1Q3
)−1
ζψ∗
ε∗
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Advection-diffusion with a MRT D1Q3 LB scheme (iii)
• Advectiong0(x , t + ∆t) = g∗0 (x , t)g+(x , t + ∆t) = g∗+(x −∆x , t)g−(x , t + ∆t) = g∗−(x + ∆x , t)
• Analysis with the Taylor expansion method
ΛD1Q3 =
0 1 023λ
2 0 13
0 λ2 0
Equivalent partial differential equation up to order 2:
∂tζ + ∂xψeq − σψ ∆t ∂xθψ = O(∆x2)
with σψ = 1sψ− 1
2 , θψ ≡ ∂tψeq + ∂x(
23λ
2 ζ + 13ε
eq)
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Advection-diffusion with a MRT D1Q3 LB scheme (iv)
Linear advection-diffusion equation
∂tζ + ∂x(u0 ζ
)− ∂x
(κ ∂xζ
)= 0
Thermal D1Q3 lattice Boltzmann scheme∂tζ + ∂xψ
eq − ∂x(σψ ∆t θψ
)= O(∆x2)
Momentum at equilibrium ψeq = u0 ζ
Defect of equilibrium θψ =(
23λ
2 − u20
)∂xζ + 1
3 ∂xεeq + O(∆x)
Energy at equilibrium εeq = α λ2 ζ
Thermal conductivity(
2+α3 λ2 − u2
0
) (1sψ− 1
2
)∆t = κ
Constraint for numerical stability −2 < α < 1
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
1D Navier Stokes equations
Conserved variables
ρ , J ≡ ρ u , ρE ≡ ρ(i +
1
2u2)
State equation: polytropic perfect gasp = (γ − 1) ρ i = ρ r T , i = cv T
Sound velocityc2 = γ p
ρ = γ (γ − 1) i = (γ − 1) cp T , γ =cpcv
Kinematic viscosity, thermal conductivity and Prandtl numberPr =
ρ ν cpκ
Mass conservation∂tρ+ ∂xJ = 0
Momentum conservation∂tJ + ∂x
(ρ u2 + p
)− ∂x
(ρ ν ∂xu
)= 0
Conservation of energy∂t(ρE)
+ ∂x(ρE u + p u
)−∂x
(ρ ν u ∂xu
)− ∂x
(κ ∂xT
)= 0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
A first and natural idea
Two distributions: fj for mass and momentumgj for total energy
Conserved momentsρ = ρ0
∑j
fj , J = ρ0 λ∑j
vj fj , ρE = ρ0 λ2∑j
gj
Non conserved momentse = ρ0 λ
2(f+ + f− − 2 f0
), ψ = ρ0 λ
3∑
j vj gj ,
ε = ρ0 λ4(g+ + g− − 2 g0
)Particle representation
f ≡(f0 , f+ , f− , g0 , g+ , g−
)tMomentum representation
m ≡(ρ , J , ρE , e , ψ , ε
)t, m = M f
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Previous work with the BGK framework
Two particle distributions for the full Navier Stokes equations
∂t fj + vj•∇fj = − 1τf
(fj − f eqj
)∂tgj + vj•∇gj = − 1
τg
(gj − g eq
j
)+ Z
τf
(fj − f eqj
)see Guo, Zheng, Shi, Phys. Rev. E, 75, 2007
Li, He, Wang, Tao Phys. Rev. E, 76, 2007Li, Luo, He, Gao, Tao, Phys. Rev. E, 85, 2012Karlin, Sichau, Chikatamarla Phys. Rev. E, 88, 2013. . .
Adapt this idea for the Multi Relaxation Times approach ?Change the relaxation step !
i conserved moments W ∗k = Wk (k < N)
ii first non-conserved moments m∗k = (1− sk)mk + sk meqk
iii second non-conserved momentsm∗` = m` − s`
(m` −meq
`
)+ K`k
(mk −meq
k
)with mk in the first family of nonconserved moments. . .
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Add a source term !
Two distributions. . .
∂t fj + vj•∇fj + 1τf
(fj − f eqj
)= 0
∂tgj + vj•∇gj + 1τg
(gj − g eq
j
)= Z
τf
(fj − f eqj
)Try to approach with a lattice Boltzmann scheme
the full Navier Stokes equations asa system of conservative partial differential equationswith a source term
∂tW + ∂xF (W )− ∂x(Φ(W ,∇W )
)= S
Present choice W = (ρ , J , ζ)ζ ≡ ρ s the volumic entropy
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Navier Stokes with mass, momentum and volumic entropy
“Conserved” variablesρ , J ≡ ρ u , ζ ≡ ρ s
State equation: polytropic perfect gasp = (γ − 1) ρ i = ρ r T = p0
( ρρ0
)γexp
(γ (s−s0)cp
)Sound velocity
c2 = γ pρ = γ (γ − 1) i = (γ − 1) cp T , γ =
cpcv
Kinematic viscosity, thermal conductivity and Prandtl numberPr =
ρ ν cpκ
Mass conservation∂tρ+ ∂xJ = 0 , J ≡ ρ u
Momentum conservation∂tJ + ∂x
(ρ u2 + p
)− ∂x
(ρ ν ∂xu
)= 0
Production of entropy∂tζ + ∂x
(ζ u)− ∂x
(κT ∂xT
)= ρ ν
T
(∂xu)2
+ κT 2
(∂xT
)2
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
D1Q3Q3 for the (ρ, J , ζ) formulation
Two distributions: fj for mass and momentumgj for volumic entropy
Conserved momentsρ = ρ0
∑j
fj , J = ρ0 λ∑j
vj fj , ζ = ρ0 cp∑j
gj
Non conserved momentse = ρ0 λ
2(f+ + f− − 2 f0
), ψ = ρ0 cp λ
(g+ − g−
)ε = ρ0 cp λ
2(g+ + g− − 2 g0
)Particle representation
f ≡(f0 , f+ , f− , g0 , g+ , g−
)tMomentum representation
m ≡(ρ , J , ζ , e , ψ , ε
)t, m = M f
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
D1Q3Q3 for the (ρ, J , ζ) formulation (ii)
Matrix MD1Q3Q3 between particles and moments
MD1Q3Q3 = ρ0
1 1 1 0 0 00 λ −λ 0 0 00 0 0 cp cp cp−2λ2 λ2 λ2 0 0 0
0 0 0 0 cp λ −cp λ0 0 0 −2 cp λ
2 cp λ2 cpλ
2
Equilibrium of non-conserved momentaeeq = eeq
(ρ, J, ζ
), ψeq = ψeq
(ρ, J, ζ
), εeq = εeq
(ρ, J, ζ
)Relaxation of non-conserved momenta
e∗ = e + se(eeq − e
), ψ∗ = ψ + sψ
(ψeq − ψ
),
ε∗ = ε+ sε(εeq − ε
)Time iteration of the D1Q3Q3 scheme
f0(x , t + ∆t) = f ∗0 (x , t), f+(x , t + ∆t) = f ∗+(x −∆x , t)f−(x , t + ∆t) = f ∗−(x + ∆x , t), g0(x , t + ∆t) = g∗0 (x , t)
g+(x , t + ∆t) = g∗+(x −∆x , t), g−(x , t + ∆t) = g∗−(x + ∆x , t) .
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Linearization
Reference state W0 = (ρ0 , ρ u0 , ρ s0) , c20 = γ p0
ρ0
Mass conservation∂tρ+ ∂xJ = 0
Momentum conservation∂tJ +
(c2
0 − u20 −
s0 c20
cp
)∂xρ+ 2 u0 ∂xJ +
c20cp∂xζ
−ν0 ∂2xρ+ ν0 u0 ∂
2xJ = 0
Production of entropy∂tζ − u0 s0 ∂xρ+ s0 ∂xJ + u0 ∂xζ
− ν0Pr
((γ − 1) cp − γ s0
)∂2xρ− γ ν0
Pr ∂2x ζ = 0
• Decoupled simplified system
Momentum conservation∂tJ + (c2
0 − u20) ∂xρ+ 2 u0 ∂xJ − ν0 ∂
2xρ+ ν0 u0 ∂
2xJ = 0
Production of entropy∂tζ + u0 ∂xζ − γ ν0
Pr ∂2x ζ = 0
two systems analogous to the two first simple ones !
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
D1Q3Q3 for the decoupled simplified system
• Fluid systemequilibrium moment
23λ
2 ρ+ 13e
eq = (c20 − u2
0) ρ+ 2 u0 J
defect of conservationθe = 3 (λ2 − 3 u2
0 − c20 ) ∂xJ − 6 u0 (c2
0 − u20) ∂xρ ' 3λ2∂xJ
coefficient of relaxation se ν0 = σe λ∆x
• Scalar “thermal” equationequilibrium moment ψeq = u0 ζ
defect of conservation θψ =(
23λ
2 − u20
)∂xζ + 1
3 ∂xεeq
compatibility at second order[(23λ
2 − u20
)ζ + 1
3 εeq]σψ ∆t = γ ν0
Pr ζ
then εeq = 3( γPr
σeσψ− 2
3 +u2
0λ2
)λ2 ζ ≡ αλ2 ζ
• Proposed link between two relaxations: σψ = 32γPr σe
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
D1Q3Q3 for the linearized Navier Stokes equations
Coupled linearized Navier Stokes equations∂tρ+ ∂xJ = 0
∂tJ +(c2
0 − u20 −
s0 c20
cp
)∂xρ+ 2 u0 ∂xJ +
c20cp∂xζ
−ν0 ∂2xρ+ ν0 u0 ∂
2xJ = 0
∂tζ − u0 s0 ∂xρ+ s0 ∂xJ + u0 ∂xζ− ν0
Pr
((γ − 1) cp − γ s0
)∂2xρ− γ ν0
Pr ∂2x ζ = 0
Maintain the previous relations for relaxation coefficientsν0 = σe λ∆x , σψ = 3
2γPr σe , σε = 1.5
Linear equilibria eeq = c41 ρ+ c42 J + c43 ζψeq = c51 ρ+ c52 J + c53 ζεeq = c61 ρ+ c62 J + c63 ζ
c41 = 3(1− s0
cp
)c2
0 − 3 u20 − 2λ2 , c42 = 6 u0 , c43 = 3
u20
cpc51 = −u0 s0 , c52 = s0 , c53 = u0 ,c61 = −3 (s0 c0)2 − 6 s0 u
20 + 3 s0 c
20 + 2− 2 s0 − 2
γ
c62 = 6 u0 s0 , c63 = 3 cp u20 + 3 s0 c
20
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
D1Q3Q3 for the linearized Navier Stokes equations (ii)
simulation of a simple linear wave: γ = 1.4 , Pr = 1 , c0 = λ2
u0 = 0 , s0 = 0 , ∆x = 140 , se = 1.9 , ν = 6.579 10−4
numerical stability γ = 1.4 , Pr = 1 , c0 = λ2
u0 = 0.15λ , s0 = 0.2 cp , ∆x = 140 , se = 1.9 , Tf = 120 ∆t
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
D1Q3Q3 for the (ρ, J , ζ) full Navier Stokes equations
Matrix MD1Q3Q3 between particles and moments
MD1Q3Q3 = ρ0
1 1 1 0 0 00 λ −λ 0 0 00 0 0 cp cp cp−2λ2 λ2 λ2 0 0 0
0 0 0 0 cp λ −cp λ0 0 0 −2 cp λ
2 cp λ2 cpλ
2
Equilibrium of non-conserved momentaeeq = eeq
(ρ, J, ζ
), ψeq = ψeq
(ρ, J, ζ
), εeq = εeq
(ρ, J, ζ
)Relaxation of non-conserved momenta
e∗ = e + se(eeq − e
), ψ∗ = ψ + sψ
(ψeq − ψ
),
ε∗ = ε+ sε(εeq − ε
)Time iteration of the D1Q3Q3 scheme
f0(x , t + ∆t) = f ∗0 (x , t), f+(x , t + ∆t) = f ∗+(x −∆x , t)f−(x , t + ∆t) = f ∗−(x + ∆x , t), g0(x , t + ∆t) = g∗0 (x , t)
g+(x , t + ∆t) = g∗+(x −∆x , t), g−(x , t + ∆t) = g∗−(x + ∆x , t) .
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First order analysis
Taylor expansion method: partial differential equationsfor the conserved variables W =
(ρ, J, ζ ≡ ρ s
)∂tWk + Λk` ∂xm
eq` = O(∆t)
with Λk` = λ∑j
Mkj vj M−1j`
Momentum-velocity tensor
Λ =
0 1 0 0 0 02λ2
30 0
1
30 0
0 0 0 0 1 00 λ2 0 0 0 0
0 02λ2
30 0
1
30 0 0 0 λ2 0
ρ J ζ e ψ ε
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First order analysis (ii)
D1Q3Q3 lattice Boltzmann scheme
Equivalent equations at first order∂tρ+ ∂xJ = O(∆t)
∂tJ + ∂x
(2λ2
3 ρ+ 13 e
eq)
= O(∆t)
∂tζ + ∂xψeq = O(∆t)
Navier-Stokes equations of gas dynamics∂tρ+ ∂xJ = 0∂tJ + ∂x
(ρ u2 + p
)= ∂x
(ρ ν ∂xu
)∂tζ + ∂x
(ζ u)− ∂x
(κT ∂xT
)= ρ ν
T
(∂xu)2
+ κT 2
(∂xT
)2
By identificationeeq = 3
(ρ u2 + p
)− 2λ2 ρ
ψeq = ζ u.
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Second order analysis
Henon’s coefficients
σe =1
se− 1
2, σψ =
1
sψ− 1
2, σε =
1
sε− 1
2
Defects of conservation : θ` ≡ ∂tmeq` + Λ`p ∂xm
eqp
Second order equivalent partial differential equations
∂tWk + Λk` ∂xmeq` = ∆t Λk` σ` ∂xθ` + O(∆t2)
For the D1Q3Q3 lattice Boltzmann scheme:∂tρ+ ∂xJ = O(∆t2)∂tJ + ∂x
(ρ u2 + p
)= ∆t
3 σe ∂xθe + O(∆t2)∂tζ + ∂x
(ζ u)
= ∆t σψ ∂xθψ + O(∆x2)
with θe = ∂teeq + λ2 ∂xJ ' 3λ2 ρ ∂xu Then ν = σe λ∆x
θψ = ∂tψeq + 2
3 λ2 ∂xζ + 1
3 ∂xεeq
= p ∂xs + ∂x[
13
(2λ2 ζ + εeq
)−((ρ u2 + p) s
)]' ∂x
[13
(2λ2 ζ + εeq
)−((ρ u2 + p) s
)]
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Second order analysis (ii)
Entropy equation at second order∂tζ + ∂x
(ζ u)− ∂x
(κT ∂xT
)= ρ ν
T
(∂xu)2
+ κT 2
(∂xT
)2
to compare with
∂tζ + ∂x(ζ u)
−∂2x
{σψ ∆t
[13
(2λ2 ζ + εeq
)−((ρ u2 + p) s
)]}= O(∆x2)
then σψ ∆t[
13
(2λ2 ζ + εeq
)−((ρ u2 + p) s
)]= κ log
(TT0
)Prandtl number Pr =
ρ ν cpκ , ν = σe λ∆x
Relation between the relaxation coefficients σψ = 32γPr σe
Equilibrium value εeq of the third momentum for entropy
εeq =2 ρ cp λ2
γ log(TT0
)+ eeq s a non-trivial relation !
Perfect gas p =ρ0 c2
0γ
( ρρ0
)γexp
(γ s−s0
cp
)=(1− 1
γ
)cp ρT
εeq = 2λ2[ρ (s − s0) +
(1− 1
γ
)cp ρ log
( ρρ0
)]+ eeq s
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.001 ρ0
progressive linear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.001 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.001 ρ0 (ii)
progressive linear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.001 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.001 ρ0 (iii)
progressive linear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.001 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.01 ρ0
progressive nonlinear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.01 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.01 ρ0 (ii)
progressive nonlinear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.01 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.01 ρ0 (iii)
progressive nonlinear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.01 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.1 ρ0
strong nonlinear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.1 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.1 ρ0 (ii)
strong nonlinear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.1 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
First numerical experiments, δρ = 0.1 ρ0 (iii)
strong nonlinear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ
2 , u0 = 0 , s0 = 0 , Tmax = 3∆x = 1
40 ,1
80 ,1
160 , ν = 6.579 10−4 , δρ = 0.1 ρ0
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Conclusion
• Construction of a Multi Relaxation Timeslattice Boltzmann scheme
with two particle distributionsfor the Navier Stokes equations formulated withconservation of mass and momentumand dissipation of volumic entropy
• Linear stability taken into considerationfor coupling two dissipation MRT coefficients
• Interesting first numerical resultsfor strong nonlinear acoustic waves
• Extend the previous work to two an three space dimensionsD2Q9-D2Q5, D3Q19-D3Q7. . .
fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion
Thanks for your attention !
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