Introduction & motivation
Example discussion
Hands-on experience
Unit conversion in LBM
Timm Krüger
Department of Microstructure Physics and Metal Forming
40237 Düsseldorf, Germany
[email protected]
LBM Workshop (Edmonton, Canada), August 22–26, 2011
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
1 / 31
Introduction & motivation
Example discussion
Hands-on experience
Outline
1
Introduction & motivation
2
Example discussion
3
Hands-on experience
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
2 / 31
Introduction & motivation
Example discussion
Hands-on experience
Outline
1
Introduction & motivation
2
Example discussion
3
Hands-on experience
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
3 / 31
Introduction & motivation
Example discussion
Hands-on experience
Physical observables and units
physical observables
usually dimensional, i.e., a number plus
a dimension
measuring means comparing with a
reference scale (e.g., measuring tape)
physical units
fundamental units, e.g., meter, second,
kilogram (SI)
bethlehem-news.de
uniquely derived units, e.g., newton,
joule, watt
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
4 / 31
Introduction & motivation
Example discussion
Hands-on experience
Computers and physical units
computers can only process binary
numbers
numbers are dimensionless
user has to provide unit conversion
(physical unit ↔ number)
img.ehowcdn.com
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
5 / 31
Introduction & motivation
Example discussion
Hands-on experience
Computers and physical units
computers can only process binary
numbers
numbers are dimensionless
user has to provide unit conversion
(physical unit ↔ number)
img.ehowcdn.com
proper unit conversions are required to
1
set up a computer simulation
2
interpret the results afterwards
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
5 / 31
Introduction & motivation
Example discussion
Hands-on experience
Hydrodynamics and the law of similarity (1)
dimensionless Navier-Stokes equations
ρ
∂u
+ (u · ∇)u
∂t
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
= −∇p + ρν∇2 u
Unit conversion in LBM
6 / 31
Introduction & motivation
Example discussion
Hands-on experience
Hydrodynamics and the law of similarity (1)
dimensionless Navier-Stokes equations
ρ
∂u
+ (u · ∇)u
∂t
= −∇p + ρν∇2 u
∂ ũ
˜ ũ = −∇
˜ p̃ + 1 ∇
˜ 2 ũ
+ (ũ · ∇)
Re
∂ t̃
ũ = u/um ,
2
p̃ = p/(ρum
),
t̃ = tum /H,
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
˜ = ∇H,
∇
Unit conversion in LBM
Re =
um H
ν
6 / 31
Introduction & motivation
Example discussion
Hands-on experience
Hydrodynamics and the law of similarity (1)
dimensionless Navier-Stokes equations
ρ
∂u
+ (u · ∇)u
∂t
= −∇p + ρν∇2 u
∂ ũ
˜ ũ = −∇
˜ p̃ + 1 ∇
˜ 2 ũ
+ (ũ · ∇)
Re
∂ t̃
ũ = u/um ,
2
p̃ = p/(ρum
),
t̃ = tum /H,
˜ = ∇H,
∇
Re =
um H
ν
significance of Reynolds number
only dimensionless number constructible from um , H, and ν
characterizes solutions of Navier-Stokes equations
flows with same Re are equivalent, even if um , H, and ν
are different
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
6 / 31
Introduction & motivation
Example discussion
Hands-on experience
Hydrodynamics and the law of similarity (2)
concept of characteristic dimensionless numbers can be
generalized to more complicated hydrodynamic problems
example: presence of diffusive tracer with diffusivity D
additional dimensionless number (Schmidt number):
Sc = ν/D
flows with same Re and Sc are equivalent
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
7 / 31
Introduction & motivation
Example discussion
Hands-on experience
Hydrodynamics and the law of similarity (2)
concept of characteristic dimensionless numbers can be
generalized to more complicated hydrodynamic problems
example: presence of diffusive tracer with diffusivity D
additional dimensionless number (Schmidt number):
Sc = ν/D
flows with same Re and Sc are equivalent
general recommended approach
always find all independent dimensionless numbers for given
problem before anything else is done
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
7 / 31
Introduction & motivation
Example discussion
Hands-on experience
Conversion principle (1)
for each physical quantity Q, one can write
Q = Q̃ × CQ
Q
Q̃
CQ
[Q]
[Q̃] = 1
[CQ ] = [Q]
physical value (incl. unit)
dimensionless value
conversion factor (incl. unit)
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
8 / 31
Introduction & motivation
Example discussion
Hands-on experience
Conversion principle (1)
for each physical quantity Q, one can write
Q = Q̃ × CQ
Q
Q̃
CQ
[Q]
[Q̃] = 1
[CQ ] = [Q]
physical value (incl. unit)
dimensionless value
conversion factor (incl. unit)
user’s task: find all required conversion factors CQ , but
relevant dimensionless parameters must be correct
simulation parameters must be valid
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
8 / 31
Introduction & motivation
Example discussion
Hands-on experience
Conversion principle (2)
example for velocity conversion
u = ũ × Cu
u = 10
m
s,
ũ = 0.1
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
=⇒
Cu = 100 ms
Unit conversion in LBM
9 / 31
Introduction & motivation
Example discussion
Hands-on experience
Conversion principle (2)
example for velocity conversion
u = ũ × Cu
u = 10
m
s,
ũ = 0.1
=⇒
Cu = 100 ms
conversion of dimensionless numbers
dimensionless numbers should generally be invariant, e.g.,
f
Re = Re
⇐⇒
!
CRe = 1
possible exceptions: e.g., Mach number (next slide).
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
9 / 31
Introduction & motivation
Example discussion
Hands-on experience
LBM and the Mach number
usually, LBM Mach number is larger
than in reality
otherwise, simulations would be too
expensive (too many time steps)
no problem since Ma is not important
www.allmystery.de
!
Ma 1, e.g., Ma < 0.3
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
10 / 31
Introduction & motivation
Example discussion
Hands-on experience
Approaching a hydrodynamic problem
1
identify relevant dimensionless numbers
(e.g., Reynolds or Péclet number)
2
write down their definitions,
e.g., Re = uH
ν
3
make use of their invariance during unit conversion,
f and CRe = 1
e.g., Re = Re
4
from this, it is possible to construct a unique set of unit
conversions
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
11 / 31
Introduction & motivation
Example discussion
Hands-on experience
Approaching a hydrodynamic problem
1
identify relevant dimensionless numbers
(e.g., Reynolds or Péclet number)
2
write down their definitions,
e.g., Re = uH
ν
3
make use of their invariance during unit conversion,
f and CRe = 1
e.g., Re = Re
4
from this, it is possible to construct a unique set of unit
conversions
Explicit examples will be shown later!
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
11 / 31
Introduction & motivation
Example discussion
Hands-on experience
Primary conversion factors
for mechanical problems, all quantities have units mql sqt kgqm
quantity
velocity
force
kinematic viscosity
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
unit
m
s
kg sm2
m2
s
Unit conversion in LBM
ql
1
1
2
qt
−1
−2
−1
qm
0
1
0
12 / 31
Introduction & motivation
Example discussion
Hands-on experience
Primary conversion factors
for mechanical problems, all quantities have units mql sqt kgqm
quantity
velocity
force
kinematic viscosity
unit
m
s
kg sm2
m2
s
ql
1
1
2
qt
−1
−2
−1
qm
0
1
0
three independent primary conversion factors required, e.g., for
length, time, mass
length, velocity, energy
length, time, density
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
12 / 31
Introduction & motivation
Example discussion
Hands-on experience
Primary conversion factors
for mechanical problems, all quantities have units mql sqt kgqm
quantity
velocity
force
kinematic viscosity
unit
m
s
kg sm2
m2
s
ql
1
1
2
qt
−1
−2
−1
qm
0
1
0
three independent primary conversion factors required, e.g., for
length, time, mass
length, velocity, energy
length, time, density
All other (secondary) conversion factors are uniquely derived.
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
12 / 31
Introduction & motivation
Example discussion
Hands-on experience
Finding secondary conversion factors
1
set primary conversion factors,
e.g., for length, time, and density (Cl , Ct , Cρ )
2
express secondary units in terms of primary units,
e.g., for the energy
[E] =
3
[Cρ ][Cl ]5
kg m2
kg m5
[ρ][l]5
=
=
=
s2
m 3 s2
[t]2
[Ct ]2
read off secondary conversion factor, e.g., for the energy
CE =
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Cρ Cl5
Ct2
Unit conversion in LBM
13 / 31
Introduction & motivation
Example discussion
Hands-on experience
Finding secondary conversion factors
1
set primary conversion factors,
e.g., for length, time, and density (Cl , Ct , Cρ )
2
express secondary units in terms of primary units,
e.g., for the energy
[E] =
3
[Cρ ][Cl ]5
kg m2
kg m5
[ρ][l]5
=
=
=
s2
m 3 s2
[t]2
[Ct ]2
read off secondary conversion factor, e.g., for the energy
CE =
Cρ Cl5
Ct2
Only the unit of the secondary quantity has to be known!
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
13 / 31
Introduction & motivation
Example discussion
Hands-on experience
Outline
1
Introduction & motivation
2
Example discussion
3
Hands-on experience
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
14 / 31
Introduction & motivation
Example discussion
Hands-on experience
Gravity-driven planar Poiseuille flow
H
relevant input parameters
channel height (wall distance)
viscosity
density
gravity (force per volume)
H
ν
ρ
f = ρg
g
relevant output parameters
maximum velocity
(Poiseuille law):
um =
fH 2
8ρν
um
Reynolds number:
Re :=
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
um H
gH 3
=
ν
8ν 2
Unit conversion in LBM
ρ, ν
15 / 31
Introduction & motivation
Example discussion
Hands-on experience
Physical parameters
example input parameters
channel height
viscosity
density
gravity
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
H
ν
ρ
g
10−3 m
2
10−6 ms
kg
103 m
3
10 sm2
Unit conversion in LBM
16 / 31
Introduction & motivation
Example discussion
Hands-on experience
Physical parameters
example input parameters
channel height
viscosity
density
gravity
H
ν
ρ
g
10−3 m
2
10−6 ms
kg
103 m
3
10 sm2
resulting output parameters
um = 1.25 ms ,
Re = 1250
These values are defined by the physical problem!
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
16 / 31
Introduction & motivation
Example discussion
Hands-on experience
Choose simulation parameters
resolution
H̃ = 20,
H = 10−3 m
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
=⇒
CH = 5 × 10−5 m
Unit conversion in LBM
17 / 31
Introduction & motivation
Example discussion
Hands-on experience
Choose simulation parameters
resolution
H̃ = 20,
H = 10−3 m
=⇒
CH = 5 × 10−5 m
density
ρ̃ = 1,
ρ = 103
kg
m3
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
=⇒
Cρ = 103
Unit conversion in LBM
kg
m3
17 / 31
Introduction & motivation
Example discussion
Hands-on experience
Choose simulation parameters
resolution
H̃ = 20,
H = 10−3 m
=⇒
CH = 5 × 10−5 m
density
ρ̃ = 1,
ρ = 103
kg
m3
=⇒
Cρ = 103
kg
m3
relaxation time
τ = 0.6
The user may choose any other set of parameters!
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
17 / 31
Introduction & motivation
Example discussion
Hands-on experience
Make use of Reynolds number
Reynolds number in physical and dimensionless systems
Re =
um H
,
ν
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
f = ũm H̃
Re
ν̃
Unit conversion in LBM
18 / 31
Introduction & motivation
Example discussion
Hands-on experience
Make use of Reynolds number
Reynolds number in physical and dimensionless systems
Re =
um H
,
ν
f = ũm H̃
Re
ν̃
equality of Reynolds numbers
! f
Re = Re
=⇒
ν
um H
=
ν̃
ũm H̃
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
=⇒
Cν = Cu CH
Unit conversion in LBM
18 / 31
Introduction & motivation
Example discussion
Hands-on experience
Make use of Reynolds number
Reynolds number in physical and dimensionless systems
Re =
um H
,
ν
f = ũm H̃
Re
ν̃
equality of Reynolds numbers
! f
Re = Re
=⇒
ν
um H
=
ν̃
ũm H̃
Cν = Cu CH
=⇒
consistency check
[Cν ] = [Cu ][CH ] =
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
m2
s
X
Unit conversion in LBM
18 / 31
Introduction & motivation
Example discussion
Hands-on experience
Get time conversion factor
lattice spacing and time step
f = 1 and ∆t
f =1
for convenience, choose ∆x
=⇒
∆x = CH ,
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
∆t = Ct
Unit conversion in LBM
19 / 31
Introduction & motivation
Example discussion
Hands-on experience
Get time conversion factor
lattice spacing and time step
f = 1 and ∆t
f =1
for convenience, choose ∆x
=⇒
∆x = CH ,
∆t = Ct
LBM viscosity
ν=
τ−
1
2
cs2 ∆t,
cs2 =
1 ∆x 2
3 ∆t 2
=⇒
ν=
τ − 12 CH2
3 } Ct
| {z
ν̃
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
19 / 31
Introduction & motivation
Example discussion
Hands-on experience
Get time conversion factor
lattice spacing and time step
f = 1 and ∆t
f =1
for convenience, choose ∆x
∆x = CH ,
=⇒
∆t = Ct
LBM viscosity
ν=
τ−
1
2
cs2 ∆t,
cs2 =
1 ∆x 2
3 ∆t 2
=⇒
ν=
τ − 12 CH2
3 } Ct
| {z
ν̃
time conversion factor
Ct =
τ−
3
1
2
CH2
= 8.3̄ × 10−5 s
ν
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
19 / 31
Introduction & motivation
Example discussion
Hands-on experience
Get velocity conversion factor
secondary conversion factor
[u] =
[H]
[t]
=⇒
Cu =
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
CH
Ct
=⇒
Cu = 0.6 ms
Unit conversion in LBM
20 / 31
Introduction & motivation
Example discussion
Hands-on experience
Get velocity conversion factor
secondary conversion factor
[u] =
[H]
[t]
=⇒
Cu =
CH
Ct
=⇒
Cu = 0.6 ms
compute maximum lattice velocity
ũm = um /Cu
=⇒
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
ũm = 2.083̄
Unit conversion in LBM
20 / 31
Introduction & motivation
Example discussion
Hands-on experience
Get velocity conversion factor
secondary conversion factor
[u] =
[H]
[t]
=⇒
Cu =
CH
Ct
=⇒
Cu = 0.6 ms
compute maximum lattice velocity
ũm = um /Cu
=⇒
ũm = 2.083̄
consistency of Reynolds number
Re =
um H ! ũm H̃
=
= 1250
ν
ν̃
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
X
Unit conversion in LBM
20 / 31
Introduction & motivation
Example discussion
Hands-on experience
Correct simulation parameters
problem
Simulation parameters are consistent,
but not valid for LBM simulations (ũm 0.3).
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
21 / 31
Introduction & motivation
Example discussion
Hands-on experience
Correct simulation parameters
problem
Simulation parameters are consistent,
but not valid for LBM simulations (ũm 0.3).
correction approach
Cu =
3
CH
=
Ct
τ−
1
2
ν
CH
decrease CH =⇒ more expensive
decrease τ =⇒ LBM may become unstable
User has to find a consistent and valid set of parameters!
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
21 / 31
Introduction & motivation
Example discussion
Hands-on experience
Example correction
change resolution and relaxation parameter
CH = 5 × 10−5 m → CH∗ = 1 × 10−5 m
τ = 0.6 → τ ∗ = 0.55
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
22 / 31
Introduction & motivation
Example discussion
Hands-on experience
Example correction
change resolution and relaxation parameter
CH = 5 × 10−5 m → CH∗ = 1 × 10−5 m
τ = 0.6 → τ ∗ = 0.55
corrected velocity conversion factor
Cu∗ =
τ∗
3
−
1
2
ν
= 6 ms
CH∗
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
=⇒
∗
ũm
= 0.2083̄ X
Unit conversion in LBM
22 / 31
Introduction & motivation
Example discussion
Hands-on experience
Example correction
change resolution and relaxation parameter
CH = 5 × 10−5 m → CH∗ = 1 × 10−5 m
τ = 0.6 → τ ∗ = 0.55
corrected velocity conversion factor
Cu∗ =
τ∗
3
−
1
2
ν
= 6 ms
CH∗
=⇒
∗
ũm
= 0.2083̄ X
consistency of Reynolds number
Re =
∗ H̃ ∗
um H ! ũm
=
= 1250 X
ν
ν̃ ∗
The user has found a consistent and valid parameter set.
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
22 / 31
Introduction & motivation
Example discussion
Hands-on experience
Get force conversion factor
secondary conversion factor
[f ] =
[ρ][H]
[t]2
=⇒
Cf =
Cρ CH
Ct2
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
=⇒
Cf = 3.6 × 109
Unit conversion in LBM
N
m3
23 / 31
Introduction & motivation
Example discussion
Hands-on experience
Get force conversion factor
secondary conversion factor
[f ] =
[ρ][H]
[t]2
=⇒
Cf =
Cρ CH
Ct2
=⇒
Cf = 3.6 × 109
N
m3
compute lattice force density
f̃ = f /Cf
=⇒
f̃ = 2.7̄ × 10−6
Everything for a successful simulation is prepared!
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
23 / 31
Introduction & motivation
Example discussion
Hands-on experience
Alternative routes
start with H̃ and ũm
1
choose H̃ and ũm =⇒ CH , Cu
2
find viscosity conversion factor Cν = CH2 /Ct =⇒ ν̃
3
identify relaxation time τ from ν̃ = (τ − 12 )/3
4
find remaining conversion factors & check validity
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
24 / 31
Introduction & motivation
Example discussion
Hands-on experience
Alternative routes
start with H̃ and ũm
1
choose H̃ and ũm =⇒ CH , Cu
2
find viscosity conversion factor Cν = CH2 /Ct =⇒ ν̃
3
identify relaxation time τ from ν̃ = (τ − 12 )/3
4
find remaining conversion factors & check validity
start with ũm and τ
1
choose ũm and τ =⇒ Cu , ν̃
2
find viscosity conversion factor Cν from ν̃
3
find length conversion factor CH = Cν /Cu =⇒ H̃
4
find remaining conversion factors & check validity
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
24 / 31
Introduction & motivation
Example discussion
Hands-on experience
Alternative routes
start with H̃ and ũm
1
choose H̃ and ũm =⇒ CH , Cu
2
find viscosity conversion factor Cν = CH2 /Ct =⇒ ν̃
3
identify relaxation time τ from ν̃ = (τ − 12 )/3
4
find remaining conversion factors & check validity
start with ũm and τ
1
choose ũm and τ =⇒ Cu , ν̃
2
find viscosity conversion factor Cν from ν̃
3
find length conversion factor CH = Cν /Cu =⇒ H̃
4
find remaining conversion factors & check validity
and so on. . .
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
24 / 31
Introduction & motivation
Example discussion
Hands-on experience
Outline
1
Introduction & motivation
2
Example discussion
3
Hands-on experience
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
25 / 31
Introduction & motivation
Example discussion
Hands-on experience
Gravity-driven Poiseuille flow with diffusive tracer
H
relevant input parameters
channel height (wall distance)
viscosity
tracer diffusivity
density
gravity (force per volume)
H
ν
D
ρ
f = ρg
g
um
relevant dimensionless parameters
Reynolds number:
um H
gH 3
Re :=
=
ν
8ν 2
Schmidt number:
ν
Sc :=
D
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
ρ, ν, D
26 / 31
Introduction & motivation
Example discussion
Hands-on experience
Formulary
dimensionless parameters
Re =
g̃ H̃ 3
ũm H̃
=
,
ν̃
8ν̃ 2
Sc =
ν̃
D̃
dimensionless viscosity and diffusivity
ν̃ =
τν − 12
,
3
D̃ =
τD −
3
1
2
gravity
f̃ = ρ̃g̃
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
27 / 31
Introduction & motivation
Example discussion
Hands-on experience
Physical parameters and numerical restrictions
physical parameters
Re = 100,
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Sc = 3
Unit conversion in LBM
28 / 31
Introduction & motivation
Example discussion
Hands-on experience
Physical parameters and numerical restrictions
physical parameters
Re = 100,
Sc = 3
numerical restrictions
relaxation times τν and τD shall both be at least 0.55
(stability)
center velocity shall be ũm ≤ 0.05 (compressibility)
system height shall be H̃ ≤ 150 (efficiency)
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
28 / 31
Introduction & motivation
Example discussion
Hands-on experience
Physical parameters and numerical restrictions
physical parameters
Re = 100,
Sc = 3
numerical restrictions
relaxation times τν and τD shall both be at least 0.55
(stability)
center velocity shall be ũm ≤ 0.05 (compressibility)
system height shall be H̃ ≤ 150 (efficiency)
task
Find a conclusive and valid set of simulation parameters!
ρ̃, τν , τD , H̃, ũm , f̃
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
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Introduction & motivation
Example discussion
Hands-on experience
Example solution, part 1
identify minimum relaxation parameters
Sc =
ν̃
D̃
=⇒
Sc =
τν −
τD −
1
2
1
2
τD,min = 0.55, Sc = 3
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
=⇒
=⇒
τν = Sc(τD − 0.5) + 0.5
τν,min = 0.65
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Introduction & motivation
Example discussion
Hands-on experience
Example solution, part 1
identify minimum relaxation parameters
Sc =
ν̃
D̃
=⇒
Sc =
τν −
τD −
1
2
1
2
τD,min = 0.55, Sc = 3
=⇒
=⇒
τν = Sc(τD − 0.5) + 0.5
τν,min = 0.65
identify minimum resolution
Re =
ũm H̃
ν̃
=⇒
H̃ = Re
τν,min = 0.65, ũm,max = 0.05, Re = 100
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
τν − 0.5
3ũm
=⇒
Unit conversion in LBM
H̃min = 100
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Introduction & motivation
Example discussion
Hands-on experience
Example solution, part 2
chosen set of simulation parameters
τν = 0.65,
τD = 0.55,
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
H̃ = 100,
ũm = 0.05
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Introduction & motivation
Example discussion
Hands-on experience
Example solution, part 2
chosen set of simulation parameters
τν = 0.65,
τD = 0.55,
H̃ = 100,
ũm = 0.05
validity and consistency already assured
Re =
ũm H̃
= 100,
ν̃
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Sc =
ν̃
D̃
=3
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X
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Introduction & motivation
Example discussion
Hands-on experience
Example solution, part 2
chosen set of simulation parameters
τν = 0.65,
τD = 0.55,
H̃ = 100,
ũm = 0.05
validity and consistency already assured
Re =
ũm H̃
= 100,
ν̃
Sc =
ν̃
D̃
=3
X
set density
ρ̃ = 1
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
(arbitrary)
Unit conversion in LBM
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Introduction & motivation
Example discussion
Hands-on experience
Example solution, part 2
chosen set of simulation parameters
τν = 0.65,
τD = 0.55,
H̃ = 100,
ũm = 0.05
validity and consistency already assured
Re =
ũm H̃
= 100,
ν̃
Sc =
ν̃
D̃
=3
X
set density
ρ̃ = 1
(arbitrary)
set gravity
Re =
g̃ H̃ 3
8ν̃ 2
=⇒
f̃ = ρ̃g̃ =
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
8ρ̃ν̃ 2
H̃ 3
Re = 2 × 10−6
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Introduction & motivation
Example discussion
Hands-on experience
Further comments
systems with identical Re and Sc are equivalent
=⇒ no explicit conversion factors required at this point!
scale can be introduced afterwards, e.g.,
H = 10−3 m
=⇒
CH = 10−5 m
all other conversion factors are then obtained as in the
example discussion
Max-Planck-Institut für Eisenforschung, Düsseldorf, Germany
Unit conversion in LBM
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