Unit conversion in LBMlbmworkshop.com/wp-content/uploads/2011/08/2011-08-22_Edmonton_scaling.pdf · Max-Planck-Institut f¨ur Eisenforschung, D usseldorf, Germany¨ Unit conversion

Introduction & motivation Example discussion Hands-on experience

Unit conversion in LBM

Timm Kruger

Department of Microstructure Physics and Metal Forming40237 Dusseldorf, [email protected]

LBM Workshop (Edmonton, Canada), August 22–26, 2011

Max-Planck-Institut fur Eisenforschung, Dusseldorf, Germany Unit conversion in LBM 1 / 31


Outline

1 Introduction & motivation

2 Example discussion

3 Hands-on experience



Outline






Physical observables and units

physical observablesusually dimensional, i.e., a number plusa dimensionmeasuring means comparing with areference scale (e.g., measuring tape)

physical unitsfundamental units, e.g., meter, second,kilogram (SI)uniquely derived units, e.g., newton,joule, watt

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Computers and physical units

computers can only process binarynumbersnumbers are dimensionlessuser has to provide unit conversion(physical unit↔ number)

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proper unit conversions are required to1 set up a computer simulation2 interpret the results afterwards



Computers and physical units

computers can only process binarynumbersnumbers are dimensionlessuser has to provide unit conversion(physical unit↔ number)

img.ehowcdn.com

proper unit conversions are required to1 set up a computer simulation2 interpret the results afterwards



Hydrodynamics and the law of similarity (1)

dimensionless Navier-Stokes equations

ρ

(∂u∂t

+ (u · ∇)u)

= −∇p + ρν∇2u

∂u∂ t

+ (u · ∇)u = −∇p +1

Re∇2u

u = u/um, p = p/(ρu2m), t = tum/H, ∇ = ∇H, Re =

umHν

significance of Reynolds numberonly dimensionless number constructible from um, H, and νcharacterizes solutions of Navier-Stokes equationsflows with same Re are equivalent, even if um, H, and νare different





ρ

(∂u∂t

+ (u · ∇)u)

= −∇p + ρν∇2u

∂u∂ t

+ (u · ∇)u = −∇p +1

Re∇2u


umHν






ρ

(∂u∂t

+ (u · ∇)u)

= −∇p + ρν∇2u

∂u∂ t

+ (u · ∇)u = −∇p +1

Re∇2u


umHν





concept of characteristic dimensionless numbers can begeneralized to more complicated hydrodynamic problemsexample: presence of diffusive tracer with diffusivity Dadditional dimensionless number (Schmidt number):Sc = ν/Dflows with same Re and Sc are equivalent

general recommended approachalways find all independent dimensionless numbers for given

problem before anything else is done




concept of characteristic dimensionless numbers can begeneralized to more complicated hydrodynamic problemsexample: presence of diffusive tracer with diffusivity Dadditional dimensionless number (Schmidt number):Sc = ν/Dflows with same Re and Sc are equivalent

general recommended approachalways find all independent dimensionless numbers for given

problem before anything else is done



Conversion principle (1)

for each physical quantity Q, one can write

Q = Q × CQ

Q [Q] physical value (incl. unit)Q [Q] = 1 dimensionless valueCQ [CQ] = [Q] conversion factor (incl. unit)

user’s task: find all required conversion factors CQ, but

relevant dimensionless parameters must be correctsimulation parameters must be valid




for each physical quantity Q, one can write

Q = Q × CQ

Q [Q] physical value (incl. unit)Q [Q] = 1 dimensionless valueCQ [CQ] = [Q] conversion factor (incl. unit)

user’s task: find all required conversion factors CQ, but

relevant dimensionless parameters must be correctsimulation parameters must be valid




example for velocity conversion

u = u × Cu

u = 10 ms , u = 0.1 =⇒ Cu = 100 m

s

conversion of dimensionless numbersdimensionless numbers should generally be invariant, e.g.,

Re = Re ⇐⇒ CRe!

= 1

possible exceptions: e.g., Mach number (next slide).




example for velocity conversion

u = u × Cu

u = 10 ms , u = 0.1 =⇒ Cu = 100 m

s

conversion of dimensionless numbersdimensionless numbers should generally be invariant, e.g.,

Re = Re ⇐⇒ CRe!

= 1

possible exceptions: e.g., Mach number (next slide).



LBM and the Mach number

usually, LBM Mach number is largerthan in realityotherwise, simulations would be tooexpensive (too many time steps)no problem since Ma is not important

Ma!� 1, e.g., Ma < 0.3

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Approaching a hydrodynamic problem

1 identify relevant dimensionless numbers(e.g., Reynolds or Peclet number)

2 write down their definitions,e.g., Re = uH

ν

3 make use of their invariance during unit conversion,e.g., Re = Re and CRe = 1

4 from this, it is possible to construct a unique set of unitconversions

Explicit examples will be shown later!



Approaching a hydrodynamic problem

1 identify relevant dimensionless numbers(e.g., Reynolds or Peclet number)

2 write down their definitions,e.g., Re = uH

ν

3 make use of their invariance during unit conversion,e.g., Re = Re and CRe = 1

4 from this, it is possible to construct a unique set of unitconversions

Explicit examples will be shown later!



Primary conversion factors

for mechanical problems, all quantities have units mql sqt kgqm

quantity unit ql qt qmvelocity m

s 1 −1 0force kg m

s2 1 −2 1kinematic viscosity m2

s 2 −1 0

three independent primary conversion factors required, e.g., forlength, time, masslength, velocity, energylength, time, density

All other (secondary) conversion factors are uniquely derived.








s 2 −1 0










s 2 −1 0





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 ] =kg m2

s2 =kgm3

m5

s2 =[ρ][l]5

[t ]2=

[Cρ][Cl ]5

[Ct ]2

3 read off secondary conversion factor, e.g., for the energy

CE =CρC5

l

C2t

Only the unit of the secondary quantity has to be known!



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 ] =kg m2

s2 =kgm3

m5

s2 =[ρ][l]5

[t ]2=

[Cρ][Cl ]5

[Ct ]2

3 read off secondary conversion factor, e.g., for the energy

CE =CρC5

l

C2t

Only the unit of the secondary quantity has to be known!



Outline






Gravity-driven planar Poiseuille flow

relevant input parameters

channel height (wall distance) Hviscosity νdensity ρgravity (force per volume) f = ρg

relevant output parametersmaximum velocity(Poiseuille law): Reynolds number:

um =fH2

8ρν Re :=umHν

=gH3

8ν2

H

ρ, ν

gum



Physical parameters

example input parameters

channel height H 10−3 mviscosity ν 10−6 m2

sdensity ρ 103 kg

m3

gravity g 10 ms2

resulting output parameters

um = 1.25 ms , Re = 1250

These values are defined by the physical problem!



Physical parameters

example input parameters

channel height H 10−3 mviscosity ν 10−6 m2

sdensity ρ 103 kg

m3

gravity g 10 ms2

resulting output parameters

um = 1.25 ms , Re = 1250

These values are defined by the physical problem!



Choose simulation parameters

resolution

H = 20, H = 10−3 m =⇒ CH = 5× 10−5 m

density

ρ = 1, ρ = 103 kgm3 =⇒ Cρ = 103 kg

m3

relaxation time

τ = 0.6

The user may choose any other set of parameters!




resolution

H = 20, H = 10−3 m =⇒ CH = 5× 10−5 m

density

ρ = 1, ρ = 103 kgm3 =⇒ Cρ = 103 kg

m3

relaxation time

τ = 0.6





resolution

H = 20, H = 10−3 m =⇒ CH = 5× 10−5 m

density

ρ = 1, ρ = 103 kgm3 =⇒ Cρ = 103 kg

m3

relaxation time

τ = 0.6




Make use of Reynolds number

Reynolds number in physical and dimensionless systems

Re =umHν

, Re =umHν

equality of Reynolds numbers

Re != Re =⇒ ν

ν=

um

um

HH

=⇒ Cν = CuCH

consistency check

[Cν ] = [Cu][CH ] =m2

sX





Re =umHν

, Re =umHν


Re != Re =⇒ ν

ν=

um

um

HH

=⇒ Cν = CuCH

consistency check

[Cν ] = [Cu][CH ] =m2

sX





Re =umHν

, Re =umHν


Re != Re =⇒ ν

ν=

um

um

HH

=⇒ Cν = CuCH

consistency check

[Cν ] = [Cu][CH ] =m2

sX



Get time conversion factor

lattice spacing and time step

for convenience, choose ∆x = 1 and ∆t = 1

=⇒ ∆x = CH , ∆t = Ct

LBM viscosity

ν =

(τ − 1

2

)c2

s ∆t , c2s =

13

∆x2

∆t2 =⇒ ν =τ − 1

23︸︷︷︸ν

C2H

Ct

time conversion factor

Ct =τ − 1

23

C2Hν

= 8.3× 10−5 s






=⇒ ∆x = CH , ∆t = Ct

LBM viscosity

ν =

(τ − 1

2

)c2

s ∆t , c2s =

13

∆x2

∆t2 =⇒ ν =τ − 1

23︸︷︷︸ν

C2H

Ct


Ct =τ − 1

23

C2Hν

= 8.3× 10−5 s






=⇒ ∆x = CH , ∆t = Ct

LBM viscosity

ν =

(τ − 1

2

)c2

s ∆t , c2s =

13

∆x2

∆t2 =⇒ ν =τ − 1

23︸︷︷︸ν

C2H

Ct


Ct =τ − 1

23

C2Hν

= 8.3× 10−5 s



Get velocity conversion factor

secondary conversion factor

[u] =[H]

[t ]=⇒ Cu =

CH

Ct=⇒ Cu = 0.6 m

s

compute maximum lattice velocity

um = um/Cu =⇒ um = 2.083

consistency of Reynolds number

Re =umHν

!=

umHν

= 1250 X





[u] =[H]

[t ]=⇒ Cu =

CH

Ct=⇒ Cu = 0.6 m

s


um = um/Cu =⇒ um = 2.083


Re =umHν

!=

umHν

= 1250 X





[u] =[H]

[t ]=⇒ Cu =

CH

Ct=⇒ Cu = 0.6 m

s


um = um/Cu =⇒ um = 2.083


Re =umHν

!=

umHν

= 1250 X



Correct simulation parameters

problemSimulation parameters are consistent,

but not valid for LBM simulations (um � 0.3).

correction approach

Cu =CH

Ct=

3τ − 1

2

ν

CH

decrease CH =⇒ more expensivedecrease τ =⇒ LBM may become unstable

User has to find a consistent and valid set of parameters!



Correct simulation parameters

problemSimulation parameters are consistent,

but not valid for LBM simulations (um � 0.3).

correction approach

Cu =CH

Ct=

3τ − 1

2

ν

CH

decrease CH =⇒ more expensivedecrease τ =⇒ LBM may become unstable

User has to find a consistent and valid set of parameters!



Example correction

change resolution and relaxation parameter

CH = 5× 10−5 m→ C∗H = 1× 10−5 mτ = 0.6→ τ∗ = 0.55

corrected velocity conversion factor

C∗u =3

τ∗ − 12

ν

C∗H= 6 m

s =⇒ u∗m = 0.2083 X


Re =umHν

!=

u∗mH∗

ν∗= 1250 X

The user has found a consistent and valid parameter set.



Example correction


CH = 5× 10−5 m→ C∗H = 1× 10−5 mτ = 0.6→ τ∗ = 0.55


C∗u =3

τ∗ − 12

ν

C∗H= 6 m

s =⇒ u∗m = 0.2083 X


Re =umHν

!=

u∗mH∗

ν∗= 1250 X




Example correction


CH = 5× 10−5 m→ C∗H = 1× 10−5 mτ = 0.6→ τ∗ = 0.55


C∗u =3

τ∗ − 12

ν

C∗H= 6 m

s =⇒ u∗m = 0.2083 X


Re =umHν

!=

u∗mH∗

ν∗= 1250 X




Get force conversion factor


[f ] =[ρ][H]

[t ]2=⇒ Cf =

CρCH

C2t

=⇒ Cf = 3.6× 109 Nm3

compute lattice force density

f = f/Cf =⇒ f = 2.7× 10−6

Everything for a successful simulation is prepared!



Get force conversion factor


[f ] =[ρ][H]

[t ]2=⇒ Cf =

CρCH

C2t

=⇒ Cf = 3.6× 109 Nm3

compute lattice force density

f = f/Cf =⇒ f = 2.7× 10−6

Everything for a successful simulation is prepared!



Alternative routes

start with H and um

1 choose H and um =⇒ CH , Cu

2 find viscosity conversion factor Cν = C2H/Ct =⇒ ν

3 identify relaxation time τ from ν = (τ − 12 )/3

4 find remaining conversion factors & check validity

start with um and τ

1 choose um and τ =⇒ Cu , ν2 find viscosity conversion factor Cν from ν

3 find length conversion factor CH = Cν/Cu =⇒ H4 find remaining conversion factors & check validity

and so on. . .



Alternative routes

start with H and um








and so on. . .



Alternative routes

start with H and um








and so on. . .



Outline






Gravity-driven Poiseuille flow with diffusive tracer

relevant input parameters

channel height (wall distance) Hviscosity νtracer diffusivity Ddensity ρgravity (force per volume) f = ρg

relevant dimensionless parameters

Reynolds number: Schmidt number:

Re :=umHν

=gH3

8ν2 Sc :=ν

D

H

ρ, ν, D

gum



Formulary

dimensionless parameters

Re =umHν

=gH3

8ν2 , Sc =ν

D

dimensionless viscosity and diffusivity

ν =τν − 1

23

, D =τD − 1

23

gravity

f = ρg



Physical parameters and numerical restrictions

physical parametersRe = 100, Sc = 3

numerical restrictionsrelaxation times τν and τD shall both be at least 0.55(stability)center velocity shall be um ≤ 0.05 (compressibility)system height shall be H ≤ 150 (efficiency)

taskFind a conclusive and valid set of simulation parameters!

ρ, τν , τD, H, um, f

















Example solution, part 1

identify minimum relaxation parameters

Sc =ν

D=⇒ Sc =

τν − 12

τD − 12

=⇒ τν = Sc(τD − 0.5) + 0.5

τD,min = 0.55, Sc = 3 =⇒ τν,min = 0.65

identify minimum resolution

Re =umHν

=⇒ H = Reτν − 0.5

3um

τν,min = 0.65, um,max = 0.05, Re = 100 =⇒ Hmin = 100




identify minimum relaxation parameters

Sc =ν

D=⇒ Sc =

τν − 12

τD − 12

=⇒ τν = Sc(τD − 0.5) + 0.5

τD,min = 0.55, Sc = 3 =⇒ τν,min = 0.65

identify minimum resolution

Re =umHν

=⇒ H = Reτν − 0.5

3um

τν,min = 0.65, um,max = 0.05, Re = 100 =⇒ Hmin = 100




chosen set of simulation parameters

τν = 0.65, τD = 0.55, H = 100, um = 0.05

validity and consistency already assured

Re =umHν

= 100, Sc =ν

D= 3 X

set densityρ = 1 (arbitrary)

set gravity

Re =gH3

8ν2 =⇒ f = ρg =8ρν2

H3Re = 2× 10−6





τν = 0.65, τD = 0.55, H = 100, um = 0.05


Re =umHν

= 100, Sc =ν

D= 3 X


set gravity

Re =gH3

8ν2 =⇒ f = ρg =8ρν2

H3Re = 2× 10−6





τν = 0.65, τD = 0.55, H = 100, um = 0.05


Re =umHν

= 100, Sc =ν

D= 3 X


set gravity

Re =gH3

8ν2 =⇒ f = ρg =8ρν2

H3Re = 2× 10−6





τν = 0.65, τD = 0.55, H = 100, um = 0.05


Re =umHν

= 100, Sc =ν

D= 3 X


set gravity

Re =gH3

8ν2 =⇒ f = ρg =8ρν2

H3Re = 2× 10−6



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 theexample discussion


Documents

Unit conversion in LBMlbmworkshop.com/wp-content/uploads/2011/08/2011-08-22_Edmonton_scaling.pdf · Max-Planck-Institut f¨ur Eisenforschung, D usseldorf, Germany¨ Unit conversion