CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > Main CFD Forum

Periodic BC + FVM

Register Blogs Members List Search Today's Posts Mark Forums Read

Reply
 
LinkBack Thread Tools Display Modes
Old   March 3, 2005, 04:28
Default Periodic BC + FVM
  #1
Kilda
Guest
 
Posts: n/a
Hi all,

I am trying to implement, Periodic BC to the Unsteady finite volume code, that I have, which solves the solution of equations by line by line TDMA.

So, no longer I can use this Line by line TDMA, for periodic BC, as I will be having some more enteries in the Matrix that I am solving, so I need to go into Cyclic Tridiagonal systems, where in, there are two additional enteries that are positoned at the end of the First row and at the begining of the last row.

I am not sure how these values at the end of the first row, and the begining of the last row are generated, as per my reading, the last entry in the first row is the East co-efficient of the last node (considering 1D) and the first entry in the last row, is West co-efficient of the first node in the grid, but these co-efficients are ZERO, which means that should I consider the co-efficeints at the 2nd and the (n-1) th node?????????

Any help is HIGHLY appreciated

kilda
  Reply With Quote

Old   March 4, 2005, 08:44
Default Re: Periodic BC + FVM
  #2
andy
Guest
 
Posts: n/a
Is you "cyclic tridiagonal system" a function/subroutine you have got from somewhere and do not understand the arguments or do you not understand how to solve a periodic tridiagonal system using code for a non-periodic system?

The coefficients in the corners are the coefficients linking your first point to the last point and the last point to the first point. If they are zero then you do not have a periodic problem.
  Reply With Quote

Old   March 4, 2005, 20:17
Default Re: Periodic BC + FVM
  #3
Kilda
Guest
 
Posts: n/a
Hi Andy,

Thanks for the reply, I am trying to figure out what the co-efficients will be at the end of first row, and the begining of the N th row, in the cyclic system?????

As per my understanding so far, (you have to correct me), the last entry (alpha) in the first row will be the west co-efficient of the Nth node(last node), and

the first entry in the last row(beta), will be the east co-efficient of the first Node (starting node).

The matrix will luk something like this

b1 c1 0 0 0 alpha

a2 b2 c2 0 0 0

0 a3 b3 c3 0 0

0 0 a4 b4 c4 0

0 0 0 a5 b5 c5

beta 0 0 0 a6 b6

thanks,

kilda

  Reply With Quote

Old   March 6, 2005, 09:39
Default Re: Periodic BC + FVM
  #4
andy
Guest
 
Posts: n/a
Assuming west is i-1 and east i+1 then alpha is the west coefficient of the first point and beta is the east coefficient of the last point. This seems an odd question and I suspect it may not be where your problem/misunderstanding lies.
  Reply With Quote

Old   March 9, 2005, 05:02
Default Re: Periodic BC + FVM
  #5
Kilda
Guest
 
Posts: n/a
Hi andy,

Thanks for the reply.

The fact that I am applying PBC, is that there will be no i-1 for the 1st point(boundary point,1st row in the matrix) and no i+1 to the Nth (last boundary point, last row in Matrix).

Where as for the interior nodes there will be no alpha and beta...(which u will know!!!)

Kilda.
  Reply With Quote

Old   March 15, 2005, 21:28
Default Re: Periodic BC + FVM
  #6
Wen Long
Guest
 
Posts: n/a
The point is, since you are imposing periodic bc, that means the first node value u(1) = the point west to the last node value u(N+1), so that (because derivatives at 1 and N are also periodic) u(2) = u(N+2), u(0) = u(N),,,

Hence the west point of your first node u(1) is u(0), and you should substitute u(0) by u(N);;

Also, the left point of your last node u(N) is u(N+1), and you should substitute it by u(1),

Then, you will have a cyclic equation.

sat your discretization for any point i=1,,,N is:

a*u(i-1) +b*u(i) + c*u(i+1), i=1,,,N

then you have(N=6):

c-----------

b c 0 0 0 a u(1)

a b c 0 0 0 u(2)

0 a b c 0 0 * u(3) = (R1,R2,...,RN)^T

0 0 a b c 0 u(4)

0 0 0 a b c ..

c 0 0 0 a b u(N) c-----------

Google Morison-Sherman formula for solving the upper equation.

Wen
  Reply With Quote

Old   March 15, 2005, 21:30
Default Re: Periodic BC + FVM
  #7
Wen Long
Guest
 
Posts: n/a
The point is, since you are imposing periodic bc, that means the first node value u(1) = the point east to the last node value u(N+1), so that (because derivatives at 1 and N are also periodic) u(2) = u(N+2), u(0) = u(N),,,

Hence the west point of your first node u(1) is u(0), and you should substitute u(0) by u(N);;

Also, the East point of your last node u(N) is u(N+1), and you should substitute it by u(1),

Then, you will have a cyclic equation.

say your discretization for any point i=1,,,N is:

a*u(i-1) +b*u(i) + c*u(i+1), i=1,,,N

then you have(N=6):

c-----------

b c 0 0 0 a u(1)

a b c 0 0 0 u(2)

0 a b c 0 0 * u(3) = (R1,R2,...,RN)^T

0 0 a b c 0 u(4)

0 0 0 a b c ..

c 0 0 0 a b u(N) c-----------

Google Morison-Sherman formula for solving the upper equation.

Wen
  Reply With Quote

Reply

Thread Tools
Display Modes

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 On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Periodic boundary condition Arif FLUENT 2 January 17, 2014 06:48
Symmetry and Periodic boundary conditions CFD_Fluent_User FLUENT 0 September 20, 2011 07:58
problem about periodic boundary condition in Fluent winnawinna FLUENT 0 December 28, 2010 23:32
[ICEM] Specifying Periodic Vertices Causes Mesh Overlap Josh ANSYS Meshing & Geometry 10 July 8, 2010 02:39
comments on FDM, FEM, FVM, SM, SEM, DSEM, BEM kenn Main CFD Forum 2 July 18, 2004 18:28


All times are GMT -4. The time now is 06:48.