vortex in bifurcation, difference between Navier-Stokes and LBM simulations

Hello everybody!!
I implemented simple LBM D2Q9 model with BGK, constant pressure conditions at inputs/outputs (GUO2012 extrapolation method) and half bounce-back at walls.
I tested the code in the simple bifurcations etc, below example geometries:
http://mediatorzypodlascy.org/flow/pic1.jpg
http://mediatorzypodlascy.org/flow/pic2.jpg
http://mediatorzypodlascy.org/flow/pic3.jpg
http://mediatorzypodlascy.org/flow/pic4.jpg

In these pictures you can also find streamlines received based on the Navier-Stokes (N-S) flow simulation. Based on them we see vortexes.
However when I simulated flow in this geometry by LBM with comparable conditions to N-S calculations, I could not received any vortex, everywhere there was a laminar flow or something very similar.

Maybe somebody has any idea what I did bad ?
Maybe the LBM model was to simple ?
Maybe I forgot about something ?
Thanks for any help

Gustaw

Hello,

Is your resolution high enough? You can only see vortices if you can resolve them.
Is the Reynolds number high enough? In or close to the Stokes regime, you usually do not observe vortices. You could try to reduce the viscosity and see what happens.
Have you waited long enough? It may also help to add some fluctuations at the beginning to “help” the flow to become unsteady.

Best,
Timm

Hello
Thank you very much Timm for response.
As you suggested I tried to find problem in many ways, below you can find summary:

  1. I tried to increase resolution but I still was not able to see vortex.

2.What does it mean resolve vortexes? I thought that the model which I adapted in my simulation is enough to simulate steady flow and in the same time steady vortexes. (LBM D2Q9 model with BGK, constant pressure conditions at inputs/outputs (GUO2012 extrapolation method) and half bounce-back at walls.) Am I wrong ?

Currently I don’t want to consider unsteady vortexes.

  1. I finish calculation when the “steady flow condition” is reached, i.e. differences between velocities in neighboring steps are smaller than 10e-6. Is it not enough ?

  2. Here, you can find the solution based on the of Navier-Stokes equations.
    link:
    http://mediatorzypodlascy.org/flow/pic1.jpg

  3. Here, you can find results from LBM in the comparable geometry:
    a) parameters: http://mediatorzypodlascy.org/flow/velocityAmplitudesAndParameters.png
    b) mean flow at inlet 005 cm/s: http://mediatorzypodlascy.org/flow/LBM_meanVelocity005cm_s.bmp
    c) mean flow at inlet 010 cm/s: http://mediatorzypodlascy.org/flow/LBM_meanVelocity010cm_s.bmp
    d) mean flow at inlet 022 cm/s: http://mediatorzypodlascy.org/flow/LBM_meanVelocity022cm_s.bmp
    e) mean flow at inlet 045 cm/s: http://mediatorzypodlascy.org/flow/LBM_meanVelocity045cm_s.bmp
    f) mean flow at inlet 115 cm/s: http://mediatorzypodlascy.org/flow/LBM_meanVelocity115cm_s.bmp
    g) mean flow at inlet 250 cm/s: http://mediatorzypodlascy.org/flow/LBM_meanVelocity250cm_s.bmp

As you can see, still I could not receive any vortex.

Do somebody knows any possible reasons of such a situation?
Maybe boundary conditions are not enough good?

Thanks for continue the topic, and any help.

Hello,

could you please report the Reynolds numbers you have in your simulations?

TImm

inlet diameter=8mm, water kinematic viscosity = 1,00E-06 m2/s
b) mean flow at inlet 005 cm/s: Re = 0.008 * 0.05 / 0.000001 = 400
c) mean flow at inlet 010 cm/s: Re = 0.008 * 0.10 / 0.000001 = 800
d) mean flow at inlet 022 cm/s: Re = 0.008 * 0.25 / 0.000001 = 2000
e) mean flow at inlet 045 cm/s: Re = 0.008 * 0.45 / 0.000001 = 3600
f) mean flow at inlet 115 cm/s: Re = 0.008 * 1.15 / 0.000001 = 9200
g) mean flow at inlet 250 cm/s: Re = 0.008 * 2.50 / 0.000001 = 20000

But what is the Reynolds number based on your simulation parameters (not the physical parameters)?

Hi Gustaw,
What is the relaxation time tau in your simulation? U will get comparable results to NS simulation when tau~1 and Mach number (u/c) ~0.01.
Best
Dongke

One response to Timm and dongke:
###############################################################################
To be sure that I calculated Reynolds number, Mach number etc well, I wrote shortly some of the experiment parameters and used equations:

dX_ph = tubeDiameter / N = 0.008 / 48 = 0.0001667 [m]

dT_lb = 1 (lattice time step)
dX_lb = 1 (lattice time step)
Rho_lb = 1 (lattice density)
c = dx_lb/dt_lb = 1
and
u_lb = 0.1 (boundary velocity)
u_ph = 1.5 [m/s] (boundary velocity)
so
u_ph = u_lb * dX_ph / dt_ph => dt_ph = 0.000011111 [s ]
visc_lb = visc_ph * dT_ph / dX_ph^2 = 0.000001 * dT_ph / dX_ph^2 = 0.0004016

finally
tau = 3* n_lb * dx_lb^2 / dt_lb + 0.5 = 0.5012
Mach = u_lb / cs = 0.1 * sqrt(3) = 0.17

Reynolds_lb = N * u_lb / visc_lb = 48 * 0.1 / 0.0004016 = 11952
############################################################################

I can increase tau, but it is hard. On the other side, Mach is quite good, isn’t it ?
lattice Reynolds number is also oddly high, did I calculate it well, hhmmm ?

Thanks for continue the topic, and any help.

I usually caculate the parameters as the following steps:

  1. define the space step delta x.
  2. define the time step delta t, and the relaxation time tau is determined accordingly.
  3. check the Mach number is less than 0.1 or not; and tau is in the zone [1.0/1.99~2.0].

If Ma is higher than 0.1 and tau~0.5 (1/tau<=1.99 for SRT LBM), We can run LBM very well but can’t get good results as NS solver.

I think the Mach and 1/tau (1.995) is too high. I suggest making Ma ~ 0.01 and 1/tau ~ 1.85 as I have suggested earlier.
make:
(1) u_lb=0.001
and then:
(2) dt_ph = 0.00000011111 s.
(3) Ma = 0.0017.
(4) tau = 0.62 (1/tau = 1.61).

Maybe it will work. :slight_smile:

dongke Wrote:

I usually caculate the parameters as the following
steps:

  1. define the space step delta x.
  2. define the time step delta t, and the
    relaxation time tau is determined accordingly.
  3. check the Mach number is less than 0.1 or not;
    and tau is in the zone [1.0/1.99~2.0].

If Ma is higher than 0.1 and tau~0.5 (1/tau<=1.99
for SRT LBM), We can run LBM very well but can’t
get good results as NS solver.

I think the Mach and 1/tau (1.995) is too high. I
suggest making Ma ~ 0.01 and 1/tau ~ 1.85 as I
have suggested earlier.
make:
(1) u_lb=0.001
and then:
(2) dt_ph = 0.00000011111 s.
(3) Ma = 0.0017.
(4) tau = 0.62 (1/tau = 1.61).

Maybe it will work. :slight_smile:
Hello dongke
I tried to do calculations as You proposed.
However, I do not know how You received
(4) tau = 0.62 (1/tau = 1.61).
for water kinematic viscosity equals about 0.000001 m^2/s
for your calculation kinematic visc = 0.001 m^/s

Gustaw
Maybe I made a mistake yesterday. Today, I will test the 2D problem by my code and show my results later.
Dongke

I have tested the problem by simulating the 2D cylinder flow.
The parameters (Real World) are :
1.0 ! density (density per node)
1.0E-6 ! dynamic viscocity
0.008 ! lengthx: physical length in x direction
500 ! Nx: total meshes in x direction
1.5 ! u_in: inlet velocity

And the LB parameters (LBM World) are:
u_in(LB) = 0.0100251
omega = 1.995
tau = 1.0/1.995
delta_x = 1.6e-005
delta_t = 1.06934e-007

Boundary conditions
topwall: outlet partial u_y / partial y =0
left and right walls: solid wall
bottom wall: inlet (u_x, u_y) = (0, u_in)

The MRT-LBM simulation results is in:

http://www.calcflows.com/wp-content/uploads/2012/07/cylinder-MRT-LBM-vortex_vectors.avi
http://www.calcflows.com/wp-content/uploads/2012/07/cylinder-MRT-LBM-vortex_streamlines.avi

The SRT-LBM simulation can not be implemented for omega>1.99 (omega= 1.995, it is high enough).

If we want to implement the SRT-LBM code, we’d better make a very small delta x and make omega small enough (omega<1.99).

Have a good day. :slight_smile:

dongke Wrote:

Gustaw
Maybe I made a mistake yesterday. Today, I will
test the 2D problem by my code and show my results
later.
Dongke

I have tested the problem by simulating the 2D
cylinder flow.
The parameters (Real World) are :
1.0 ! density (density per node)
1.0E-6 ! dynamic viscocity
0.008 ! lengthx: physical length in x direction
500 ! Nx: total meshes in x direction
1.5 ! u_in: inlet velocity

And the LB parameters (LBM World) are:
u_in(LB) = 0.0100251
omega = 1.995
tau = 1.0/1.995
delta_x = 1.6e-005
delta_t = 1.06934e-007

Boundary conditions
topwall: outlet partial u_y / partial y =0
left and right walls: solid wall
bottom wall: inlet (u_x, u_y) = (0, u_in)

The MRT-LBM simulation results is in:

http://www.calcflows.com/wp-content/uploads/2012/0
7/cylinder-MRT-LBM-vortex_vectors.avi
http://www.calcflows.com/wp-content/uploads/2012/0
7/cylinder-MRT-LBM-vortex_streamlines.avi

The SRT-LBM simulation can not be implemented for
omega>1.99 (omega= 1.995, it is high enough).

If we want to implement the SRT-LBM code, we’d
better make a very small delta x and make omega
small enough (omega<1.99).

Have a good day. :slight_smile:
By the way, what software did You use to visualize flow ?

Hi Gustaw
I usually do data-visualization by Paraview, Gnuplot, and Tecplot. I used Tecplot to create the animate as shown.
Best
Dongke