Four-roll mill flow

Dear all

My question is related to the four-roll mill flow, a simple periodic flow with no-inertial terms. This is flow can be found in the phd thesis of Orestis Malaspinas.

This flow was solver by Orestis Malaspinas for both Newtonian and non-Newtonian fluids. I am exclusively interested in the Newtonian case.

I was already able to implement the flow with LB, which is very simple since boundary conditions are periodic and the eq. distribution function has only linear terms.
Although the numerical solution is qualitatively identical to the analytical one the two are quantitatively distinct by a scale factor. This suggests me that my conversion of units t(from physical to lattice space) is not right.

My problem is that this flow has no inertial terms. A flow with no inertial terms has zero Reynolds number! So I cannot relate the velocity with a spatial length scale and the kinematic viscosity! Is there any non-dimensional parameter that I am missing? This flow must be governed by some non-dimensional number.

What I find strange is that if I apply a standard non-dimensional analysis to the momentum eq. of my problem: gx+1/Re*(d^2(u)/dx^2+d^2(u)/dy^2) and gx+1/Re*(d^2(v)/dx^2+d^2(v)/dy^2)=0, i.e. I find that the diffusive term must be weighted by the inverse of the Reynolds number. But by definition Re=0! Therefore I must be mistaken some definition in the physical parameters! Because when I related this analysis to the analytical solution of the flow I obtain mismatching resulting.

u_analy= sin(x)*cos(y)
v_analy= -cos(x)*sin(y)

gx= 2niusin(x)cos(y)
gy= -2
niu*cos(x)*sin(y)

Any hints or suggestions will be very much welcomed and appreciated :slight_smile:

Thanks in advance.

Goncalo

Dear Goncalo,

this is not exactly correct. You can always define a Reynolds number, even if you do not have inertial terms. In this case, the Reynolds number has a more abstract meaning. You can always define it as $v l / \nu$ where v is the velocity scale (which you have), l is the length scale (which you also have) and \nu is the viscosity (which you also have). In flows without inertial terms, and even in some flows with inertial terms, the overall flow pattern does not depend on the Reynolds number except for scaling factors.

Some examples:

  1. Poiseuille flow: You do not have inertial terms, but you can define the Reynolds number. The amplitude of the flow depends on Re, but not the shape.
  2. Couette flow: No inertial terms, linear velocity profile.
  3. Decaying Taylor-Green vortex flow: Inertial terms, but Re only changes decay rate, not the spatial shape of your solutions.

I always find it a bit confusing writing the flow properties in terms of Re since one may take the “wrong” definition, and everything is garbage. It is clearer to write the hydrodynamic solutions in terms of velocity, length scale, and viscosity. Please always have in mind that there is no unique definition of the Reynolds number. You should repeat your analysis without Re, and you will see that everything is fine.

Good luck,
Timm

Thank you Timm

Once more your explanation was pretty clear.

When I achieve some more progresses I will you informed!

Regards

Goncalo