OpenLB – Open Source Lattice Boltzmann Code Forums on OpenLB General Topics Lattice and physical units of thermal conductivity on phase change simulation

Viewing 2 posts - 1 through 2 (of 2 total)
• Author
Posts
• #8418
aaron
Participant

Hello!

I was trying to understand the implementation of the example: example/thermal/stefanMelting2d.

Reference: https://doi.org/10.1016/j.jcp.2015.03.064.

The literature has reported that Chapman-Enskog expansion is adopted to recover the Navier-Stokes equations to LBM. Referring to the thermal component, we may obtain the lattice thermal conductivity:

\lambda = cs^2 (\tau – 0.5) dt C_p,

It is quite similar to the recovery of the fluid viscosity:

\nu = cs^2 (\tau – 0.5) dt.

My question is about whether cs is defined as c/3 or 1/3.

Since the expansion is like:

f(x+c*dt, t+dt) – f(x,t) = -(f(x,t) – feq(x,t))/ \tau,

And c=dx/dt due to discretized physical space and time.

My point is maybe c cannot be simplified to 1, even though I understand that
c=1 is convenient to get the zero- or first-moment of feq equivalently (term of c canceled out).

Another question may be whether the physical pressure should be recovered as:

p = \rho /3 or p = \rho * c^2 /3

Thank you very much if you can correct me or leave some comments.

#8427
TimBingert
Participant

Hi Aaron,

in our code, the cs^2 is defined depending on the DESCRIPTOR, so in most cases this is 1/3.
Hence, the pressure should be p = \rho * cs^2.
When you take a closer look at the heat conductivity lambda, it is connected to the thermal diffusivity alpha via alpha = lambda/(\rho*C_p). In your first equation, there is no \rho to be seen since this is all in lattice units where we choose \rho = 1 for simplicity.

Hope this helps.
Kind regards
Tim

Viewing 2 posts - 1 through 2 (of 2 total)
• You must be logged in to reply to this topic.