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January 4, 2015, 21:27 
how to use UDF with cylinderical coordinate system in fluent

#1 
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Peter Aestas
Join Date: Dec 2013
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Hi my friends,
i am working on plasma distribution problems, and it is more convenient for me to written the set of equations and boundary conditions in cylindrical component form, but as far as i knew, fluent only has Cartesian coordinate system so how should i do to put the cylindrical component form into UDF and use them in fluent? I am desperate for your help~ 

January 5, 2015, 04:59 

#2 
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Join Date: Nov 2010
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Hi,
Can you provide more details on the nature of equations? Meanwhile, have a look at this doc, a nice summary of field variables/functions in Fluent. You might be able to find most of the access functions you need. hope it helps. cheers! 

January 5, 2015, 07:57 

#3  
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Peter Aestas
Join Date: Dec 2013
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Thanks for your replay, my friend.
it is in the paper titled "2D expansion of the lowdensity interelectrode vacuum arc plasma jet in an axial magnetic field",with a 2D MHD model, but the energy equations are not considered.The pdf is too large to upload.~ i wonder whether the equation set 3 can be calculated in fluent. And you can see, the self magnetic field Bθ is more simple to express in cylinder system~ Quote:


January 5, 2015, 09:39 

#4 
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Peter,
Unfortunately, I don't have access to the full paper. But Looking at Eqs. 3.1 to 3.6, you need to convert the Cartesian velocities into axial, radial, and tangential ones. I've googled around a bit and stumbled upon these slides. On slide 34 and 35, you'll find a way to convert the velocities. However, I am not quite sure if you can use the very same approach for other terms in the equations. It might be a bit tricky. cheers 

January 5, 2015, 10:37 

#5 
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Peter Aestas
Join Date: Dec 2013
Posts: 60
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Thanks my friend, i got what you mean by the slides, use Cartesian in fluent while use cylinder system in UDF.
It is a bit tricky to me, but i will try two ways, to take the same approach in the slides, or just convert the basic equation set 2 into Cartesian system, and the tangential component has to be divided. BTW, the original auther of the paper didnot use fluent to calculate, insead they program themselves 

January 5, 2015, 11:00 

#6 
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Peter Aestas
Join Date: Dec 2013
Posts: 60
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Hi, Sun,
I think if it's a 3D model in Cylinder Coordinate, it can't directly be used in fluent, because the momentum components are written as Vx,Vy,Vz in fluent.As a consequence, the source terms in the momentum equation can't directly be used from Cylinder to Cartesian. But my problem is a 2D axisymmetric model, the z and r component in the figure just equals the x and y component in Cartesian, the only difference is the tangential component vs y compenent. I think this may be done by the approach you told me. I would give it a try 

January 5, 2015, 12:49 

#7 
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Alright great and thanks for the update. I guess that method might actually work for the 2D case.
good luck. 

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