CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > Software User Forums > OpenFOAM

how to calculate the mean value of a surface

Register Blogs Community New Posts Updated Threads Search

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
Old   December 3, 2010, 09:32
Default a problem about steady state solver
  #1
Member
 
chenkai
Join Date: May 2010
Location: munich
Posts: 44
Rep Power: 15
xck1986 is on a distinguished road
Dear OF Users,

I have three problems when I use rhoSimpleFoam.

1. Is the relaxationFactors only used for steady state solver? And how much I should choose them. e.g. In my case when I choose the relaxationFactors for rho bigger than 0.1, the result will not converge. But why?

2. Another problem about the Courant Number. For the steady state solver, do I need to also consider the courant number?

3. And I want to calculate the mean velocity and density of the inlet and oulet of a channel. Anyone who know how can I calculate it? Directly in ParaView or use other functions?


Thanks a lot!

Last edited by xck1986; December 3, 2010 at 09:51.
xck1986 is offline   Reply With Quote

Old   December 3, 2010, 10:27
Default
  #2
Member
 
Diego Villa
Join Date: Mar 2010
Location: Genova Italy
Posts: 37
Rep Power: 16
DiegoNaval is on a distinguished road
Hi
For the first question, also some unsteady solver use the relaxation factor but in a different way, because normally they use its in the inner iteration but not in the last one (see PimpleFoam). Normally the relaxation factor should be closer to 1 but for a steady solution as the rhoSimpleFoam the value could be smaller to stabilize the solution. The results at the end should be not affected to the value, only the velocity of the convergence. I normally see that if the case blow up could be due to the mesh or the turbulent parameter.
For the second question the Courant Number is valid only for a unsteady solution in that case the solution is steady.
For the third I'm not an expert but I think you can do that in Paraview

I hope to be helpful.
DiegoNaval is offline   Reply With Quote

Old   December 3, 2010, 11:07
Default
  #3
Member
 
chenkai
Join Date: May 2010
Location: munich
Posts: 44
Rep Power: 15
xck1986 is on a distinguished road
Quote:
Originally Posted by DiegoNaval View Post
Hi
For the first question, also some unsteady solver use the relaxation factor but in a different way, because normally they use its in the inner iteration but not in the last one (see PimpleFoam). Normally the relaxation factor should be closer to 1 but for a steady solution as the rhoSimpleFoam the value could be smaller to stabilize the solution. The results at the end should be not affected to the value, only the velocity of the convergence. I normally see that if the case blow up could be due to the mesh or the turbulent parameter.
For the second question the Courant Number is valid only for a unsteady solution in that case the solution is steady.
For the third I'm not an expert but I think you can do that in Paraview

I hope to be helpful.

Hi Villa,
thanks a lot for your explain.
Is that means, not matter which relaxation factor I have chosen, it has no influence on the final results? The only things is that if it can help to reach convergence or not?
And another question: can I use a transient solver to solve a steady state problem?
xck1986 is offline   Reply With Quote

Old   December 3, 2010, 11:17
Default
  #4
Senior Member
 
Bernhard
Join Date: Sep 2009
Location: Delft
Posts: 790
Rep Power: 21
Bernhard is on a distinguished road
Wrt 3.

You can add something like this in your ControlDict. I assume it is selfexplanatory for you.
Code:
functions 
{
    massFlow
    {   
         type patchAverage;
         functionObjectLibs
           (   
             "libsimpleFunctionObjects.so"
           );  
         verbose true;
         patches
           (   
             inlet
             outlet
           );  
         fields
           (   
             U   
           );  
         factor 1.; 
    }   
}
Bernhard is offline   Reply With Quote

Old   December 3, 2010, 15:09
Default
  #5
Member
 
Diego Villa
Join Date: Mar 2010
Location: Genova Italy
Posts: 37
Rep Power: 16
DiegoNaval is on a distinguished road
Yes you can solve all steady problem with a transient solver, you must wait the end of the transition... but normally that needs more computational time, in fact exist some steady solver that use a virtual time marching approach. Like Star..
DiegoNaval is offline   Reply With Quote

Old   December 3, 2010, 16:36
Default
  #6
Member
 
chenkai
Join Date: May 2010
Location: munich
Posts: 44
Rep Power: 15
xck1986 is on a distinguished road
Quote:
Originally Posted by Bernhard View Post
Wrt 3.

You can add something like this in your ControlDict. I assume it is selfexplanatory for you.
Code:
functions 
{
    massFlow
    {   
         type patchAverage;
         functionObjectLibs
           (   
             "libsimpleFunctionObjects.so"
           );  
         verbose true;
         patches
           (   
             inlet
             outlet
           );  
         fields
           (   
             U   
           );  
         factor 1.; 
    }   
}
Hi Bernhard,
Thanks a lot for you useful sugguestion!!
xck1986 is offline   Reply With Quote

Reply


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
[Gmsh] Problem with Gmsh nishant_hull OpenFOAM Meshing & Mesh Conversion 23 August 5, 2015 02:09
About surface weighed average anne OpenFOAM Running, Solving & CFD 1 July 6, 2010 15:06
Recover surface from surface mesh in IcemCFD Jerry Tanner CFX 0 August 20, 2008 12:48
BC for free surface pilar FLUENT 3 January 14, 2004 07:36
free convection heat transfer from a heated horizontal surface through a liquid to a thin cooled fin Kaushik FLUENT 1 May 8, 2000 06:47


All times are GMT -4. The time now is 03:01.