# stream function & vorticity boundary condition

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 December 4, 2011, 12:45 stream function & vorticity boundary condition #1 New Member   sara Join Date: Mar 2011 Posts: 7 Rep Power: 0 HI everyone. I've wrote a code to simulate velocity field over a square cylinder. I've used vorticity-stream function solver at 0 angle of attack.(not immersed) but now I don't know how to specify outflow boundary condition for different angles of attack. I used a rectangular as my domain. please help me. thank you for reading my thread.

 December 5, 2011, 10:10 #2 Senior Member   Andrew Join Date: Mar 2009 Location: Washington, DC Posts: 194 Rep Power: 8 I did this for a finite difference method (except I did flow over a step). My exit boundary condition was @x=L, Streamfunction(@x=L) = Streamfunction(@x=L-1). I did this based on assuming that the flow after the step was fully developed and not changing. My vorticity BC was the same. The results were good.

December 6, 2011, 11:53
#3
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sara
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Quote:
 Originally Posted by mettler I did this for a finite difference method (except I did flow over a step). My exit boundary condition was @x=L, Streamfunction(@x=L) = Streamfunction(@x=L-1). I did this based on assuming that the flow after the step was fully developed and not changing. My vorticity BC was the same. The results were good.
but in my case I have vortex shedding which makes it more complicated.
By the way, did free stream velocity have an angle of attack in your case? or it was simply horizontal?

 December 6, 2011, 12:08 #4 Senior Member   Andrew Join Date: Mar 2009 Location: Washington, DC Posts: 194 Rep Power: 8 it had no angle of attack, but why can't you extend the rectangular grid long enough to where the vortex shedding has dissipated and the flow is fully developed?

December 7, 2011, 18:42
#5
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sara
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Quote:
 Originally Posted by mettler it had no angle of attack, but why can't you extend the rectangular grid long enough to where the vortex shedding has dissipated and the flow is fully developed?
Dear Mettler.
Well, I did so.
but still there are lot's of problems.
for example, even if the flow behind the cylinder is fully developed, it still has its angle of attack, so I can't impose the boundary condition you used for outflow.
there are other difficulties in dealing with far field boundaries.