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derivative in CEL expressions

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Old   October 13, 2011, 05:21
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Sorry Marjan, that's about the limit of my knowledge on the subject. Let us know if you find a solution!
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Old   March 15, 2012, 03:58
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Quote:
Originally Posted by singer1812 View Post
No need to use user Fortran. As an example, edit the additional variable in your LIBRARY with the update loop as follows:

ADDITIONAL VARIABLE: OldVel
Option = Definition
Tensor Type = SCALAR
Units = [m s^-1]
Variable Type = Specific
Update Loop = TRANS_LOOP
END

You can then use this to create your derivative.
does any one help me understand how the
Update Loop = TRANS_LOOP]

added to additional variable commands??
need more explaintions please, ASAP
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Old   April 5, 2012, 03:08
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Quote:
Originally Posted by happy View Post
does any one help me understand how the
Update Loop = TRANS_LOOP]

added to additional variable commands??
need more explaintions please, ASAP
No one can help you happy!!!
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Old   April 5, 2012, 08:51
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What do you want to know?
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Old   April 6, 2012, 01:27
Default how I can add update loop into my AV?&How we apply derivative at B.Cs
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Quote:
Originally Posted by happy View Post
does any one help me understand how the
Update Loop = TRANS_LOOP]

added to additional variable commands??
need more explaintions please, ASAP
first, how I can add update loop into my AV. I've tried using comand editor, but when I open it again, I found it is gone!!!. After you show me how it is done. How I can use AV to create derivative - is that by cel expression?

Quote:
Originally Posted by singer1812 View Post
What do you want to know?
Hi singer 1812
Actually, I would like to apply derivative w. r. t. distance at my b.Cs for u, v,w,E, and K. So does I can apply that using your advices into this thread? if so, how we can handle this without using user fortran into CFX solver?

Thanks in advance
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Old   April 6, 2012, 09:11
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You are trying to get d/dx not d/dt?

You dont need this method (update loop) to do that, that data is in already in solution.
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Old   April 8, 2012, 18:51
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Quote:
Originally Posted by singer1812 View Post
You are trying to get d/dx not d/dt?
yes, I need d/dx and not d/dt so which way I should follow, plz?
Regards
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Old   April 9, 2012, 00:13
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Quote:
Originally Posted by singer1812 View Post
You dont need this method (update loop) to do that, that data is in already in solution.
Hi again
Actually U misunderstand me because you talked about d /dx,y,z into post stage but what I'm looking for is set d../dx, d../dy &d../dz=0 at the boundary consitions (outlets) into set up stage.
any input can help
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Old   April 9, 2012, 08:57
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Dont need transloop. Just setup a CEL expression for the derivative and put it into the BC under flow direction.
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Old   April 9, 2012, 09:02
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Ahh, I didnt read your message carefully enough. You need it on the outlet.

To what purpose are you forcing the derivatives on the outlet?
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Old   April 9, 2012, 21:10
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Quote:
Originally Posted by singer1812 View Post
Ahh, I didnt read your message carefully enough. You need it on the outlet.

To what purpose are you forcing the derivatives on the outlet?
I would like to represent the forced plume that its source elevated on the ground into large surroundinf fluid domain. I tried using many BCs but I found 3 journal article used Neumann conditions at the top domain. Actually all the 3 articles did not use ready software solver like CFX or Fluin but they had used their own code. So I want to apply neumann condition for my problem and does not matter if the boundary is outlet or openning. so is this possible through CFX.
Any input can help.

regards
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Old   April 9, 2012, 21:17
Question how I input my neumann BC through flow dirrection
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Quote:
Originally Posted by singer1812 View Post
Dont need transloop. Just setup a CEL expression for the derivative and put it into the BC under flow direction.
the flow direction is available only with opening boundary condition. However, whan I choose the cartesian components that means provide distance (m) and not dv/dx (s^-1). Is it correct?
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Old   April 10, 2012, 09:50
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I dont understand why you are trying to force an outlet to have a certain gradient. Is this purely to match something in liturature?

The opening will only apply those conditions to the flow coming into the domain, not the flow leaving it. The conditions of the flow leaving the domain are a function of what takes place inside.

If you need the gradient to be zero wrt the boundary, use a symmetry BC. If it is non zero, I think you will need to use User Fortran to enforce that.
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Old   April 11, 2012, 22:20
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Quote:
Originally Posted by singer1812 View Post
I dont understand why you are trying to force an outlet to have a certain gradient. Is this purely to match something in liturature?

The opening will only apply those conditions to the flow coming into the domain, not the flow leaving it. The conditions of the flow leaving the domain are a function of what takes place inside.

If you need the gradient to be zero wrt the boundary, use a symmetry BC. If it is non zero, I think you will need to use User Fortran to enforce that.
Thanks singer1812 for your input
If I required any further recomments I will reask you
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Old   April 12, 2012, 04:19
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Quote:
Originally Posted by singer1812 View Post
You are getting it from a fortran routine that you are calling while CFX is running (User Fortran?)

Or are you getting it from a pre-ran fortran routine, and delta_velocity is already set for the CFX run.
This method is right, frist you must run the fortan in your computer.
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