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Convergence problem with flow facing a backstep

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Old   January 19, 2013, 18:42
Default Convergence problem with flow facing a backstep (case files now attached)
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David
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I am starting with OpenFoam so this can be a silly question. I have experience with other CFD programs but i decide to start learning OpenFoam.

Problem data: Incompressible airfoil facing a backstep, Steady, I am using the icoFoam solver.

I have convergence problems as the Courant goes up as the residuals.

Feel free to ask for further information and thanks in advance.
Attached Files
File Type: zip backstep.zip (10.9 KB, 11 views)

Last edited by GM_XIII; January 20, 2013 at 16:17. Reason: Attach case files
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Old   January 22, 2013, 00:23
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Eric R
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Can we see your output file?
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Old   January 22, 2013, 06:10
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Hi,

icoFoam is a transient solver. You can identify that under:

<OPENFOAM_DIR>/applications/solvers/incompressible/icoFoam/icoFoam.C line 58 (OF-2.1.x)

fvm::ddt(U) -> transient

So you must not set steadyState ddtSchemes in system/fvSchemes. For example: use full implicit Euler scheme.

Search the following folder for all available ddtSchemes:

<OPENFOAM_DIR>/src/finiteVolume/finiteVolume/ddtSchemes

To identify their name for fvSchemes, have a look into their header files (*.H). Search for the string in TypeName. That's what you can put into fvSchemes.

Example:

EulerDdtScheme.H

->

line 71 (OF-2.1.x):

TypeName("Euler");

->

ddtSchemes
{
default Euler;
}
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Old   January 22, 2013, 09:02
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Thank you, I thought that a transient solver could manage steady cases with that fvShemes, my fault of couse. I would like to do it Steady becouse if i do it transient i need to set a very low time step in order to keep the coutant number low. So, what's the best solver to solve this case as Steady?
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Old   January 22, 2013, 10:29
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Use simpleFoam. It has the following properties:

- steady state
  • no fvm::ddt in UEqn.H, which means
  • ddt == 0
- incompressible
  • rho == constant, not solving for rho
  • pressure p is not the real pressure in [N/m2], but it is p / rho [m2/s2]
- time step
  • important: it does not matter which deltaT you select; as best you select dT = 1s
  • convergence is reached with relaxation factors in fvSolution
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Old   January 22, 2013, 11:24
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Hi Maxmeinicke,
I have a small silly question,

[ pressure p is not the real pressure in [N/m2], but it is p / rho [m2/s2] ]

according your post, to get the real pressure in [N/m2], what should we do?

how to convert the simpleFoam simulation pressure in to real pressure????

I am bit confused by the unit of (p/rho ---> m2/s2), can you please explain about it?


Thanks,
Aadhavan
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Old   January 22, 2013, 11:45
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Thank you mate, I finally did it I haven't realised that you could use a turbulent solver and then specify that the case is laminar. Now it is done, I will upload the case files in case someone could be interested.
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Old   January 22, 2013, 14:22
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Congratulations to you

unit calculation:

compressible:

p = F / A = N / m2 = kg * m / s2 / m2 = kg / m / s2

OpenFOAM notation: 'dimensions [1 -1 -2 0 0 0 0];'

incompressible:

p / rho = F / A / rho = N / m2 / (kg / m3) = kg * m / s2 / m2 / (kg / m3) = m2 / s2

OpenFOAM notation: 'dimensions [0 2 -2 0 0 0 0];'

Have a look into the pressure boundary files in time folder 0 for incompressible and compressible tutorial cases.

unit explanation:

In OpenFOAM basic solvers, 'incompressible' means constant density and 'compressible' means the density varies. So in compressible solvers a density transport equation is solved (rhoEqn). If the density is constant, you can divide all terms in the transport equations by the density. You will find out that the density is removed from all terms except the pressure term and the viscosity term. That is why with compressible and incompressible solvers one must be careful with units.

compressible uses:
  • pressure p [Pa]
  • dynamic viscosity mu (greek letter) [kg / m / s]
incompressible uses:
  • pressure p which is actually p / rho [m2 / s2]
  • kinematic viscosity nu (greek letter) [m2 / s]
Connection between units:

p_compr = p_incompr * rho
mu_compr = nu_incompr * rho

As you can see, the units scale with rho. So if you calculate water, mu is 1000 times larger than nu.

To calculate the real pressure for an incompressible solver, multiply with the density. If I would like to write the real pressure to the timefolders, I would derive a new solver and add a new volScalarField p_real.

If someone knows, how to manage writing the real pressure for incompressible solvers to time folders using functions and without creating a new solver, please let us know.

If you have more questions, please ask.

which OF version do you use?

Last edited by maxmeinicke; January 23, 2013 at 05:07. Reason: corrections
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Old   January 23, 2013, 03:47
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Hi,
Thanks for your long reply, I am using OF-2.0.1.

small correction your explanation,

pressure p which is actually p / rho [m2 / m2],

this should be (m2/s2).


Thanks,
Aadhavan
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Old   January 23, 2013, 06:52
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I've created a new solver simpleFoam2.

It is derived from simpleFoam and outputs the real pressure pReal.

just extract the file and read the README...
Attached Files
File Type: gz simpleFoam2.git.tar.gz (48.3 KB, 3 views)
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Old   January 28, 2013, 22:54
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Hi, this paper is useful

http://www.cimec.org.ar/ojs/index.ph...File/3263/3186

Regards.
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Old   January 29, 2013, 02:43
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Hi Santiag,
Can you please say something about it. I tried to look at it but I am not able.
It seems it is in French or some other language.
is there any English version.


Thanks,
Aadhavan
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Old   January 29, 2013, 04:16
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I opened Santiago's link and got an English PDF.
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Old   March 26, 2014, 03:55
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Quote:
Originally Posted by GM_XIII View Post
Thank you mate, I finally did it I haven't realised that you could use a turbulent solver and then specify that the case is laminar. Now it is done, I will upload the case files in case someone could be interested.
can you tel me the boundary conditions for this problem.
I also would like to see your result, may be velocity contort is quiet enough. can you post here?
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Old   March 27, 2014, 04:56
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Quote:
Originally Posted by CFDUser_ View Post
can you tel me the boundary conditions for this problem.
I also would like to see your result, may be velocity contort is quiet enough. can you post here?
It is my mot have uploaded it when i finished it, now it would be hard to find it but if i do i will upload it. By the way, i think i might not change the BC of my initial case, which is uploaded. I. Think i use velocity inlet+ presure outlet (0 pressure).

Regards
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