# how symmetry bc is applied for a tensor ?

 User Name Remember Me Password
 Register Blogs Members List Search Today's Posts Mark Forums Read

 LinkBack Thread Tools Display Modes
 January 12, 2012, 22:55 how symmetry bc is applied for a tensor ? #1 New Member   Join Date: Jun 2009 Posts: 27 Rep Power: 9 Hello, I want to understand how the symmetry boundary condition is applied for a tensors of rank 1 and above in OF. Say at a wall with normal n and a velocity vector V at adjacent cell, I think it would be like Code: ``` Vsymm = V - dot(V,n); Vsymm = mag(Vsymm) / mag(V) * V;``` How do you transform a 2 rank and above tensor (f.i 3x3 stress tensor ) at a symmetric wall boundary in a similar way? thanks Last edited by dshawul; January 12, 2012 at 23:56.

 January 19, 2012, 17:15 #2 New Member   Join Date: Jun 2009 Posts: 27 Rep Power: 9 Ok I found how it is applied in OF Code: ```tmp nHat = this->patch().nf(); Field::operator= ( ( this->patchInternalField() + transform(I - 2.0*sqr(nHat), this->patchInternalField()) )/2.0 ); transformFvPatchField::evaluate();``` This averages the reflection of the tensor and itself to get a vector parallel to the wall. But then it doesn't adjust its magnitude to equal the original vector, like I tried to do in my post above. Is this correct ?

 Thread Tools Display Modes Linear Mode

 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 OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post lost.identity CFX 41 May 22, 2013 07:21 [ICEM] Blocking and Symmetry BrolY ANSYS Meshing & Geometry 32 August 24, 2012 03:13 Cirion0000 CFX 0 July 6, 2009 14:26 Matt CFX 2 February 25, 2009 06:41 ZHANG Main CFD Forum 0 June 18, 2007 12:51

All times are GMT -4. The time now is 18:52.

 Contact Us - CFD Online - Top