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Lattice Boltzmann Units Example

Posted by NanoNano  
Re: Lattice Boltzmann Units Example
March 19, 2010 08:19AM
Yes, I always did it in such a way, that firstly I use only m, s, kg.
But next I check the behaviour, of equation for cm, s, kg:
rho_{LB} = 1+1/c_s^2 p_{Phys}dt_{Phys}^2 / dx_{Phys}^2

and I receive strange things:) which I rescribe above.
Re: Lattice Boltzmann Units Example
March 19, 2010 08:59AM
You cannot sum up a dimensionless number and a number with a dimension. That's the explanation. The conversion factor has also to be applied to the average density of your fluid which is 1 in LB units. In other words, the lattice mean density is 1, but it is 1000 kg/m^3 or 1 g/cm^3. Please take this into account.
Re: Lattice Boltzmann Units Example
March 19, 2010 10:12AM
To sum up, If I understand good, it means that equation:
rho_{LB} = 1+1/c_s^2 * p_{Phys} * dt_{Phys}^2 / dx_{Phys}^2

is valid with such a units ??
[ g/cm^3 ] = [g/cm^3] + [g / (cm * s^2)] * s^2 / cm^2

And also such a form is valid
rho_{LB} = 1000+1/c_s^2 * p_{Phys} * dt_{Phys}^2 / dx_{Phys}^2
but with units
[ kg/m^3 ] = [kg/m^3] + [kg / (m * s^2)] * s^2 / m^2

And now going/coming to LBM simulation, when I use second equation, the density will oscillate about 1000
and this is not good for numerical stability, therefore is better to use first equation ??

I know that physicaly these two equations are the same. But do I understand this translation problem in good manner?
Thanks in advance for answer.
Re: Lattice Boltzmann Units Example
March 19, 2010 10:37AM
No matter which equation you take, you set the lattice density to 1. I mean, it is very clear that the same equation cannot lead to different result if another unit system is used.
Example: lattice density is 1. If you take kg, m, s as units, the density conversion factor for water is delta rho = 1000 kg/m^3. right? If you take g, cm, s instead, the conversion factor is delta rho = 1 g/cm^3. However, both conversion factors are exactly the same since 1 kg = 1000 g and 1 m = 100 cm. The density of water in the one unit system is 1000 kg/m^3 leading to a lattice density of 1. The density of water in the other unit system is 1 g/cm^3 leading also to a lattice density of 1. Clear?
Re: Lattice Boltzmann Units Example
March 19, 2010 12:02PM
Thanks, it is now clear like the sun, we were talking about the same but in slightly different manner.

And one more question but still connected with the same subject:
So when I want to use He&Luo imcompressible model where I need q and q0 and in distribution function instead of 'f' we have 'p', in order to obtain q0 (rh0) I should use the conversion factor 1000/3 [kg/m^3]
(or consequently (1 / 3) [g/cm^2]) ??

Thanks in advance for answer
Re: Lattice Boltzmann Units Example
November 04, 2010 01:53PM
jlatt Wrote:
-------------------------------------------------------
2>
> rho_{LB} = 1+1/c_s^2 p_{Phys}dt_{Phys}^2 /dx_{Phys}^2
>
> where c_s^2 is the constant 1/3, which should be
> used without units when you proceed to a
> dimensional analysis of the above equation.

According to Timm, the conversion factor between Pp and Plb is rhop * dxp2 / dtp2, then
Pp=rhop * dxp2 / dtp2*Plb, so Plb=Pp*dtp2/(rhop*dxp2). Given P=cs2*rho,
so rhop=1/cs^2*[Pp*dtp2/(rhop*dxp2)].
I wonder where is the first term 1 of the RHS function Latt mentioned comes from.



Edited 1 time(s). Last edit at 11/04/2010 01:56PM by ilvucy.
Re: Lattice Boltzmann Units Example
November 11, 2010 12:13PM
Hi.
I'm simulating thermal flow in a 2-d micro-couette using LBM. I have a serious problem with thermal unit conversion in LB.
I want to reproduce the results of the DSMC method in (XIAO-JUN GU et. al. "A high-order moment approach for capturing non-equilibrium phenomena in the transition regime", J. Fluid Mech. (2009), vol. 636, pp. 177–216. c ,Cambridge University Press 2009, doi:10.1017/S002211200900768X) Part 6. Results and discussion, for thermal-micro-couette flow in the transition regime (namely, Knudsen number Kn>0.1). But I don't know how to convert the values of the wall temperature (Tw) and the gas constant (R) to lattice units.
as we know the sound speed in the LBM is Cs=C/SQRT(3) and C=SQRT(3RT). Then what should I do with the value of the R and T in the calculation of the C.
Please help me. I'm working on my thesis and I'm in a bad hurry.
I appreciate any help in advance.
Re: Lattice Boltzmann Units Example
December 07, 2010 10:16PM
I have a question regarding multicomponent LBM. What is the significance of the density values we put into our simulations? Are they just indicators (showing which part of a domain is occupied by a certain fluid)?
Re: Lattice Boltzmann Units Example
December 08, 2010 07:04AM
No, of course not. In reality, two different components or phases may have a quite different density (e.g., water and air much more than water and oil). This is of significant physical relevance since the density ratio can influence the flow dynamics. I believe that, in general, not all properties of a multiphase system can be correctly recovered when the density ratio is wrong.

Timm
Re: Lattice Boltzmann Units Example
December 14, 2010 05:37PM
Thanks Timm.

I am simulating Co-current immiscible two-phase flow in a 2D channel, the wetting phase flows along the upper
and lower plate while non-wetting phase flows in the center region using SC model.
My objective is to verify my model with the results of:

" Shan-and-Chen-type multiphase lattice Boltzmann study of viscous coupling effects for two-phase flow in porous media"
by Haibo Huang?, Zhitao Li, Shuaishuai Liu and Xi-yun Lu

I am using the same LB kinematic viscosity \nu_{lb} for both fluids (the value is 1/6) and the density of wetting and nonwetting fluids are 1 and 0.1, respectively. After the end of my simulation, I see a single parabolic velocity profile for velocity in the direction of flow {x-direction} which does not account for phase change (density change). In other words, it is just like plotting velocity profile for flow of SINGLE Phase through a channel!!!

Could you please help me on that?

Sincerely,
Re: Lattice Boltzmann Units Example
December 14, 2010 07:09PM
I think that your result is quite logical. The Navier-Stokes equations for Poiseuille flow simplify to
\rho \nu \Delta u = \rho g
where \rho is density (as function of position), \nu is kinematic viscosity (constant throughout your channel), \Delta is the Laplacian operator, u is velocity, and g is gravitational acceleration or any other homogeneous acceleration (constant throughout your channel). You see that the density cancels and that the result MUST be the same as for a single phase fluid. Only if the kinematic viscosity differs, you would expect a modified velocity profile. You should choose different kinematic viscosities.

Best,
Timm
Re: Lattice Boltzmann Units Example
December 14, 2010 09:36PM
Thanks Timm for your quick answer.

I am using a constant driving force for both phases (F) and I think applying the Chapman-Enskog expansion procedure to the lattice-Boltzmann equation for multiphase (SC Model), one obtains the following mass and momentum equations
for the fluid mixture treated as a single fluid



(\sum_a F_a )/(\rho) = - \nu \Delta u          a=1, ... ,9      for 2D
where
\sum_a F_a = F_ext + F_{f-f} + F_{f-s}
where F_{ext}, F_{f-f} , and F_{f-s} are momenta contributed by external force, interaction between phases, and interaction of ¯fluid with solid, respectively. [Z.L. Yang et al. 2001]

In this case, density is not canceled out and fluids with different density should have different velocity profiles!

Furthermore, I see some peaks in velocity perpendicular to the direction of flow at the interface between fluid phases ( from Navier-Stokes, we know that it should be zero everywhere in the doamin ). I think this accounts for the existence of F_{f-f} at the interface. But it changes the velocity magnitude at points around that!
Do you know any solution for this problem?
Re: Lattice Boltzmann Units Example
December 15, 2010 06:01AM
I still think that the density cancels because F is momentum, and momentum is proportional to density. Please carefully recheck your equations and make sure that the difference between force and force density is clear.
Again: For two fluids with the same kinematic viscosity the velocity profile must always be a parabola in the presence of a homogeneous acceleration field. But acceleration is not force and also not force density!

I find it a bit strange that you have a velocity perpendicular to the main flow. Due to the symmetry, this would eventually lead to a compressibility of the fluid. Is your interface flat? It is a 2D simulation, right? Is your system symmetric with respect to the main velocity axis?
Re: Lattice Boltzmann Units Example
December 15, 2010 06:04PM
Thanks again Timm. You are really helpful.

I will check the equations (I am using Palabos for my simulations).

Yes, my interface is flat and 2D and symmetric with respect to the main velocity axis.
Re: Lattice Boltzmann Units Example
December 16, 2010 04:10AM
You were right about F, it is proportional to density.

Thanks
INSTALLATION OF PALABOS
March 13, 2011 08:04AM
Hello


I have difficulties in the installation of Palabos on my computer, I download it and I have the c++ compiler.

is there somebody who can help me?
Re: Lattice Boltzmann Units Example
May 29, 2018 10:32AM
Hello, could you please tell me where did you get this formula?
nu_{Phys}=dt_{Phys}/(dx_{Phys}*dx_{Phys})*nu_{LB}
Even the dimensions do not match and that’s weird to me !
We have “length^2” vs. “second” (for dt) to be canceled and nu(Phys) dimension becomes equals to that of nu(LB).
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