# Lattice Boltzmann Units Example

Posted by Nano
 Re: Lattice Boltzmann Units Example March 19, 2010 08:19AM Registered: 10 years ago Posts: 36
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 Registered: 11 years ago Posts: 464
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 Registered: 10 years ago Posts: 36
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?
 Re: Lattice Boltzmann Units Example March 19, 2010 10:37AM Registered: 11 years ago Posts: 464
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 Registered: 10 years ago Posts: 36
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]) ??

 Re: Lattice Boltzmann Units Example November 04, 2010 01:53PM Registered: 10 years ago Posts: 1
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 Registered: 8 years ago Posts: 3
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.
I appreciate any help in advance.
 Re: Lattice Boltzmann Units Example December 07, 2010 10:16PM Registered: 8 years ago Posts: 13
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 Registered: 11 years ago Posts: 464
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 Registered: 8 years ago Posts: 13
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!!!

Sincerely,
 Re: Lattice Boltzmann Units Example December 14, 2010 07:09PM Registered: 11 years ago Posts: 464
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 Registered: 8 years ago Posts: 13

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 Registered: 11 years ago Posts: 464
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 Registered: 8 years ago Posts: 13
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 Registered: 8 years ago Posts: 13
You were right about F, it is proportional to density.

Thanks
 INSTALLATION OF PALABOS March 13, 2011 08:04AM Registered: 8 years ago Posts: 1
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 Registered: 1 year ago Posts: 4
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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