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Lehrstuhl fürNumerische Mathematik
Lattice Boltzmann methods: Applications
Porous media flowDissolution process of a sugar ball
Heat transfer (→ programming tutorial 3)
July 23, 2019
TUM, Munich
Lehrstuhl fürNumerische Mathematik
Porous media flow
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 0 Approach to porous media simulation
Meso- to Macro-scale approach to porous media
• Here: No “Lattice Boltzmann DNS”
• We do not resolve individual porous mediaparticle with the mesh
• Look at the porous medium “from above”
• Only consider the effect of the mediumspresence on particle collisions
• Additional force contribution to thecollision operator
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 1 A LBM model of porous media flow
Porous media flow
• Generalized Navier-Stokes model [P. Nihiarasu et al., 1997]
∇ · u = 0
∂tu + (u · ∇)(u
ε
)= −1
ρ∇(εp) + νe∆u + F
• Porosity ε ∈ (0,1]
• Effective viscosity νe = Jν
• Total body force F due topresence of porous medium+ external forces
• Force model for porous medium
F = −ενK
u− εFε√K|u|u + εG︸ ︷︷ ︸
linear drag︸ ︷︷ ︸
nonlinear drag
• External forces G
• Fluid shear viscosity ν
• Permeability K of porousmedium
• Forchheimer term Fε
• Parameter relations
Fε =1.75√150ε3
K =ε3d2p
150(1− ε)2Re =
U · Lν
Da =K
L2
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 1 A LBM model of porous media flow
Force treatment with LBM
• Modified LBM-Update equation [Zhaoli Guo, T.S. Zhao, 2002]
fi(t+ ∆t, x+ ci∆t) = fi(t, x)−1
τ· (fi(t, x)− f eq
i (t, x)) + ∆tFi
Collision now also incorporates the influence of forces.
• Force contribution in porous medium setting
Fi = ρωi
(1− 1
2τ
)(3(F · ci) +
9
ε(u · ci) · (F · ci)−
3
ε(u · F)
)
• Redefinition of velocity
u =1
ρ
∑i
cifi +∆t
2F
• F depends on u itself → nonlinear relation
• F(u) Quadratic equation → solvable!
• Macroscopic velocity can be computed in every stepJJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 2 Numerical examples
Numerical example I - Poiseuille flow
• Geometry and flow setting:
0
d
x2
x1
• Channel of width H
• Completely filled with porousmedium
• Flow is driven by external force Galong channel direction
• No-slip boundary conditions atchannel walls• Analytical solution
u =
(u(y)
0
)satisfies
νeε
∂2u
∂y2+G− ν
Ku− Fε√
Ku2 = 0
For Fε = 0 (no nonlinear drag) thesolution is (r =
√νε/(Kνe)):
u =GK
ν
(1−
cosh [r(y −H/2)]
cosh(rH/2)
)
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 2 Numerical examples
Numerical example I - Poiseuille flow
0 0.2 0.4 0.6 0.8 1 1.2-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.7
0 0.2 0.4 0.6 0.8 1 1.2-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.9
0 0.2 0.4 0.6 0.8 1 1.2-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.95
• Good numerical approximation ofBrinkmann solution
• Forchheimer term Fε hasincreasing influence with higherporosity
• For ε → 1 we get back to thePoiseuille-channel profile
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 2 Numerical examples
Numerical example II - Couette flow
• Geometry and flow setting:x2
0
H
x1
• Channel of width H
• Completely filled with porousmedium
• Flow is driven by prescribed velocityU0 in channel direction at topboundary
• No-slip boundary condition atbottom wall
• Analytical solution
u =
(u(y)
0
)with (in case of Fε = 0): u = U0
sinh(ry)
sinh(rH)
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 2 Numerical examples
Numerical example II - Couette flow
0 0.2 0.4 0.6 0.8 1 1.2-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.96
0 0.2 0.4 0.6 0.8 1 1.2-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.99
• Again good numerical approximation of Brinkmann solution
• Also again the Forchheimer term Fε (nonlinear drag) slows the flow down
• For ε→ 1 we get back the linear Couette profile
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 2 Numerical examples
Numerical example III - Free-flow to Darcy-flow
• Geometry and flow setting:
0
d
x2
x1
• Channel of width H
• Only a fraction r is filled withporous medium
• Flow is again driven by external forceG along channel direction
• No-slip boundary conditions atchannel walls
0
d
x2
x1
• Channel can also contain an obstacle
• For example a building
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 2 Numerical examples
Numerical example III - Free-flow to Darcy-flow
0 2 4 6 8 10 12-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.9
0 0.5 1 1.5 2 2.5 3 3.5-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.95
• Poiseuille-profile in the free flowregion
• Good accordance with analyticalsolution in porous media region
• Transition zone around theinterface
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 2 Numerical examples
Numerical example III - Free-flow to Darcy-flow
0 2 4 6 8 10 12-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.9
0 0.5 1 1.5 2 2.5 3 3.5-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Numerical solution with F =0
Brinkman-solutionNumerical solution with F 0
ε = 0.95
• Poiseuille-profile in the free flowregion
• Good accordance with analyticalsolution in porous media region
• Transition zone around theinterface 0 20 40 60 80 100
10
20
30
40
50
60
70
80
90
100
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 3 Literature
Main Literature
• General LBM
Timm Kruger et al - The Lattice Boltzmannmethod
• Good introduction
• Covers a variety of applications and scenarios
• Code examples included
• About the presented porous media simulations
• Guo, Zhaoli, and T. S. Zhao. ”Lattice Boltzmann model for incompressibleflows through porous media.” Physical Review E 66.3 (2002): 036304.
• Fattahi Evati, Ehsan ”High performance simulation of fluid flow in porousmedia using Lattice Boltzmann method” Dissertation, Technische UniversitatMunchen, 2017
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 4 Dissolution of a candy ball
Dissolution process of acandy ball
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 5 Motivation
Motivation
Do you remember the “jawbreaker” hard candy from the 90s?
Figure 1: Jawbreaker-candy consisting of multiple layers of sugar formedto a ball. Image sources: http://asusvilla.wonecks.net/2013/05/21/history-of-jawbreaker-rock-candy/ and
https://www.tumblr.com/search/jawbreaker%20candy
The question everyone was curious about is:
xxxxxxxxxxxxxxxx“How many licks does it take to get to the center?”
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 5 Motivation
Motivation
Do you remember the “jawbreaker” hard candy from the 90s?
Figure 2: Jawbreaker-candy consisting of multiple layers of sugar formedto a ball. Image sources: http://asusvilla.wonecks.net/2013/05/21/history-of-jawbreaker-rock-candy/ and
https://www.tumblr.com/search/jawbreaker%20candy
The question everyone was curious about is:
xxxxxxxxxxxxxxxx“How many licks does it take to get to the center?”
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 6 Experimental setup
Simplification of the “experiment”
• We cannot simulate e.g. tongue roughness or chemical properties of saliva
• Simplify the situation to a uniform candy ball being suspended into a flow channel
• Simulate dissolution due to convection and diffusion
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 6 Experimental setup
Simplification of the “experiment”
• We cannot simulate e.g. tongue roughness or chemical properties of saliva
• Simplify the situation to a uniform candy ball being suspended into a flow channel
• Simulate dissolution due to convection and diffusion
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 6 Experimental setup
Simplification of the “experiment”
• We cannot simulate e.g. tongue roughness or chemical properties of saliva
• Simplify the situation to a uniform candy ball being suspended into a flow channel
• Simulate dissolution due to convection and diffusion
Figure 3: Fluid flow around the sugar ball is visualized by laser refraction of particles.Video link: https://www.youtube.com/watch?v=OvtZ7jpNCUI
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 7 Mathematical formulation
Mathematical model
• The flow is again described by the Navier-Stokes-system
∇ · u = 0
∂u
∂t+ (u · ∇)u = −1
ρ∇p+ ν ·∆u
with no-slip BC at the sugar balls surface
• The “spreading” of sugar into the fluid is described by a coupled (!) Convection-Diffusion-equation
∂ρh∂t
+ u · ∇ρh = D ·∆ρh
where ρh is the sugar density between 0 (no sugar) and 1 (pure sugar). At the sugarballs surface a BC of ρh = 1 is imposed.
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 7 Mathematical formulation
Mathematical model
• The flow is again described by the Navier-Stokes-system
∇ · u = 0
∂u
∂t+ (u · ∇)u = −1
ρ∇p+ ν ·∆u
with no-slip BC at the sugar balls surface
• The “spreading” of sugar into the fluid is described by a coupled (!) Convection-Diffusion-equation
∂ρh∂t
+ u · ∇ρh = D ·∆ρh
where ρh is the sugar density between 0 (no sugar) and 1 (pure sugar). At the sugarballs surface a BC of ρh = 1 is imposed.
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 8 Simulation of the sugar convection / diffusion
Simulation without shape change of the sugar ball
Figure 4: Simulation of ρh for different values of D = 0.1, 0.02, 0.002.
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 9 LBM implementation of Convection-Diffusion-equation
How to solve Convection-Diffusion equation with LBM
• For the flow simulation ones uses populations fi, i = 1, ..., 9, equilibrium populationsf eqi , performs collision and streaming and boundary treatment.
• For the simulation of ρh one uses a second set of populations hi and conducts thesame steps, i.e.
ρh =∑i
hi and v =∑i
cihi
and collision and streaming etc. for hi whenever they are applied to fi.
• Some differences:
1. Different boundary conditions −→ different boundary treatment2. Different relaxation frequency ω → ωh and different material parameters ν → D3. The equilibrium heq
i is formed with u and not v in order to couple the sugardensity with the flow, i.e.
heqi = ωiρh
(1 + 3(ci · u) +
9
2(ci · u)2 − 3
2|u|2
)JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 9 LBM implementation of Convection-Diffusion-equation
How to solve Convection-Diffusion equation with LBM
• For the flow simulation ones uses populations fi, i = 1, ..., 9, equilibrium populationsf eqi , performs collision and streaming and boundary treatment.
• For the simulation of ρh one uses a second set of populations hi and conducts thesame steps, i.e.
ρh =∑i
hi and v =∑i
cihi
and collision and streaming etc. for hi whenever they are applied to fi.
• Some differences:
1. Different boundary conditions −→ different boundary treatment2. Different relaxation frequency ω → ωh and different material parameters ν → D3. The equilibrium heq
i is formed with u and not v in order to couple the sugardensity with the flow, i.e.
heqi = ωiρh
(1 + 3(ci · u) +
9
2(ci · u)2 − 3
2|u|2
)JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 9 LBM implementation of Convection-Diffusion-equation
How to solve Convection-Diffusion equation with LBM
• For the flow simulation ones uses populations fi, i = 1, ..., 9, equilibrium populationsf eqi , performs collision and streaming and boundary treatment.
• For the simulation of ρh one uses a second set of populations hi and conducts thesame steps, i.e.
ρh =∑i
hi and v =∑i
cihi
and collision and streaming etc. for hi whenever they are applied to fi.
• Some differences:
1. Different boundary conditions −→ different boundary treatment2. Different relaxation frequency ω → ωh and different material parameters ν → D3. The equilibrium heq
i is formed with u and not v in order to couple the sugardensity with the flow, i.e.
heqi = ωiρh
(1 + 3(ci · u) +
9
2(ci · u)2 − 3
2|u|2
)JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 10 Shape change of the sugar ball
Shape change of the sugar ball
• The shape of the sugar ball changes due to the sugar dissolving in the fluid.
• The normal velocity by which the surface moves is given by:
vn = −D ∇ρh · n+ k u · n
• Hence we need the quantities n and ∇ρh via a “post-processing” step
• In the discrete setting, we have to think about how to implement e.g. a normalrecession speed of vn = 0.5 geometrically, since we can not cancel a half cell forexample...
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 10 Shape change of the sugar ball
Shape change of the sugar ball
• The shape of the sugar ball changes due to the sugar dissolving in the fluid.
• The normal velocity by which the surface moves is given by:
vn = −D ∇ρh · n+ k u · n
• Hence we need the quantities n and ∇ρh via a “post-processing” step
• In the discrete setting, we have to think about how to implement e.g. a normalrecession speed of vn = 0.5 geometrically, since we can not cancel a half cell forexample...
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 10 Shape change of the sugar ball
Shape change of the sugar ball
• The shape of the sugar ball changes due to the sugar dissolving in the fluid.
• The normal velocity by which the surface moves is given by:
vn = −D ∇ρh · n+ k u · n
• Hence we need the quantities n and ∇ρh via a “post-processing” step
• In the discrete setting, we have to think about how to implement e.g. a normalrecession speed of vn = 0.5 geometrically, since we can not cancel a half cell forexample...
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 10 Shape change of the sugar ball
Shape change of the sugar ball
• The shape of the sugar ball changes due to the sugar dissolving in the fluid.
• The normal velocity by which the surface moves is given by:
vn = −D ∇ρh · n+ k u · n
• Hence we need the quantities n and ∇ρh via a “post-processing” step
• In the discrete setting, we have to think about how to implement e.g. a normalrecession speed of vn = 0.5 geometrically, since we can not cancel a half cell forexample...
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 10 Shape change of the sugar ball
Dealing with the recession rate
• For us ∆t = 1, hence if (always for one certain cell) in timestep t we have vn = 0.7,in timestep t+ 1, we have vn = 1.5, etc. ...
• We sum up all those values (for the given cell) and let the cell “break away” as awhole as soon as a certain “breakaway value” B is reached.
• In this way we can still use our discrete lattice while still incorporating the differentspeed by which certain parts of the sugar ball vanish.
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 10 Shape change of the sugar ball
Dealing with the recession rate
• For us ∆t = 1, hence if (always for one certain cell) in timestep t we have vn = 0.7,in timestep t+ 1, we have vn = 1.5, etc. ...
• We sum up all those values (for the given cell) and let the cell “break away” as awhole as soon as a certain “breakaway value” B is reached.
• In this way we can still use our discrete lattice while still incorporating the differentspeed by which certain parts of the sugar ball vanish.
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 10 Shape change of the sugar ball
Dealing with the recession rate
• For us ∆t = 1, hence if (always for one certain cell) in timestep t we have vn = 0.7,in timestep t+ 1, we have vn = 1.5, etc. ...
• We sum up all those values (for the given cell) and let the cell “break away” as awhole as soon as a certain “breakaway value” B is reached.
• In this way we can still use our discrete lattice while still incorporating the differentspeed by which certain parts of the sugar ball vanish.
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 10 Shape change of the sugar ball
Final results
Figure 5: Shape of the sugar ball after: 6751, 8851, and 11401 timesteps, compared toexperimental results.
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 11 Thermal LBM
Thermal LBM simulation:Rayleigh-Benard
convection
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 12 Theory: Boussinesq Approximation
Thermo-coupled Navier-Stokes equations
Want to couple a temperature field to the still incompressible Navier-Stokes equations.Do not want to take into account the energy equation or compressibility terms.
• Assume a linear temperature (T ) dependency of the density ρ
ρ = ρ0 − βρ0(T − T0), ρ0, T0 are reference values
• Variations in density lead to a flow (hot air streams up) → One way coupling
• Consider the body force due to gravity on the rhs of Navier-Stokes: F = ρg, where
g = (0,−9.81)>
F
ρ0= g︸︷︷︸
Gravity pulling downoften neglected again
−Heat gradient pulling up (due to− in front)︷ ︸︸ ︷
β(T − T0)g
• Temperature enters the Navier Stokes equation as a force term on the rhs
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 12 Theory: Boussinesq Approximation
Thermo-coupled Navier-Stokes equations
Need an evolution equation for the temperature field
• Heat is transported by the flow → Backwards coupling
• Use a convection-diffusion equation / convection enhanced heat equation
∂T
∂t+ u · ∇T − κ∆T = 0
where u is the velocity field from the Navies-Stokes equation.
• Those assumptions are known as Boussinesq approximation.
• They take into account effects caused by buoyancy
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 12 Theory: Boussinesq Approximation
Navier-Stokes equations with Boussinesqapproximation
• Choosing the reference temperature T0 = 0 for simplicity
• Neglect the downwards pulling gravity term g, since we are only interested in thebuoyant force
• One arrives at the following system
∂u
∂t+ (u · ∇)u+
1
ρ0∇p− ν∆u = g − βT g
∇ · u = 0
∂T
∂t+ u · ∇T − κ∆T = 0
• With suitable boundary and initial conditions
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 13 LBM Adaption of buoyancy terms and convection/diffusion
Characteristic number for thermal flow configurations
In order to characterize a thermal flow, the different processes that participate have tobe set into relation.
• Reynolds number:
Re =UL
ν∼ inertial forces
viscous forces
• Prandtl number:
Pr =ν
κ∼ viscous diffusivity
heat diffusivity
• Rayleigh number:
Ra = Re2 Pr ∼ thermal transport via diffusion
thermal transport via convection
where U is a characteristic velocity scale, L a characteristic length scale, ν the kinematicviscosity and κ the heat conductivity of the material.
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 13 LBM Adaption of buoyancy terms and convection/diffusion
Treatment via Lattice Boltzmann method
Have to incorporate the temperature field and the buoyancy force term on the rhs ofNavier-Stokes.
• Introduce second set of populations gi, i = 1, ..., 9
• Define the macroscopic temperature density
T =
9∑i=1
gi compare to ρ =
9∑i=1
fi
• A “temperature momentum” is not needed, since the temperature is transported bythe flow u→ use the same velocity as for u (coupling).
• Define the temperature equilibrium analogously to the flow equilibrium using thesame velocity u
geqi = ωiT
(1 + 3(ci · u) +
9
2(ci · u)2 − 3
2|u|2)
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 13 LBM Adaption of buoyancy terms and convection/diffusion
Treatment via Lattice Boltzmann method
Have to incorporate the temperature field and the buoyancy force term on the rhs ofNavier-Stokes.
• Remember the update equation for the i-th population
fi(x+ ∆t ci, t+ ∆t) = fi(x, t)−∆t
τν(fi(x, t)− f eq
i (x, t)) + Ji
• Choose the buoyancy force term as follows [1]:
Ji = 3ωiρ(x, t) βT (x, t)(−g · ci
)• The populations for the temperature have their own update equation:
gi(x+ ∆t ci, t+ ∆t) = gi(x, t)−∆t
τκ(gi(x, t)− geq
i (x, t))
• Dependency of the relaxation parameters on viscosity and heat conductivity (∆t = 1normalization) by Chapman-Enskog: τν = 3ν + 1
2, τκ = 3κ+ 12
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 13 LBM Adaption of buoyancy terms and convection/diffusion
Simulation of a Rayleigh-Benard convection cell• Initial flow at rest and isothermal
• Heating at the bottom of the domain
• Cooling at the top of the domain
• Bottom and top are solid walls
• Periodic flow and temperature BCleft and right
� = �bottom
� = �top
periodic
u = �
u = �
periodic
• Boundary conditions for u = 0: Bounce back scheme
• Boundary conditions for T = Tgiven: Anti bounce back scheme [2]
gopp(i)(x, t+ ∆t) = −g∗i (x, t) + 2ωiTgiven
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Lehrstuhl fürNumerische Mathematik 14 Numerical results
Formation of a Benard cell
Velocity and temperature fields for Ra = 50000, Nx = 100,Ny = 50t = 3000
t = 20000
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Lehrstuhl fürNumerische Mathematik 14 Numerical results
Influence of the Rayleigh number
• Analyze the behaviour for different values of the Rayleigh number
• Small Rayleigh numbers → diffusion dominated heat transfer
• Larger Rayleigh numbers → convection dominated heat transfer
Stable temperature fields after t = 20000 time stepsRa = 1000 Ra = 5000
Ra = 50000 Ra = 100000
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 14 Numerical results
Turbulent behavior for high Rayleigh numbers
Turbulent behavior at Ra = 1 · 106
t = 4000 t = 5000
t = 6000 t = 7000
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Lehrstuhl fürNumerische Mathematik 14 Numerical results
Karman vortex street with heated object
An other example is the Karman vortex street:
• Flow around an object within a channel (see last tutorial)
• New: The object is heated/cooled
• Inspect heat transport in the flow around the object
Ny = 100, Nx = 400, Re = 120t = 30000
JJ J I II 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Lehrstuhl fürNumerische Mathematik 14 Numerical results
Karman vortex street with heated object
Temperature field of a hot object at: t = 0, 1000, 2000
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Lehrstuhl fürNumerische Mathematik 14 Numerical results
Karman vortex street with heated object
Temperature field of a cold object at: t = 0, 1000, 2000
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Lehrstuhl fürNumerische Mathematik 15 Conclusion
Conclusion
• Temperature appears in a convection diffusion equation coupled to Navier-Stokes
• The Boussinesq approximation considers buoyancy effects via a force term on the rhsof Navier-Stokes
• In LBM the quantity of a convection diffusion equation is represented by a secondset of populations
• Coupling via u appearing in the equilibrium of the second set of populations
• Incorporation of force terms via modification of the collision operator
• Behavior of heat flows depending on characteristic Rayleigh number
Overall astonishingly easy implementation of coupled PDE system.
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Lehrstuhl fürNumerische Mathematik 16 References
References
References
[1] KAO, P.-H.; YANG, R.-J. Simulating oscillatory flows inRayleigh-Benard convection using the lattice Boltzmann method.International Journal of Heat and Mass Transfer, 2007, 50. Jg.,Nr. 17-18, S. 3315-3328.
[2] KRUGER, Timm, et al. The lattice Boltzmann method. SpringerInternational Publishing, 2017, 10. Jg., S. 978-3.
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