# derivative in CEL expressions

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 October 13, 2011, 05:21 #41 New Member   Join Date: Sep 2011 Posts: 8 Rep Power: 8 Sorry Marjan, that's about the limit of my knowledge on the subject. Let us know if you find a solution!

March 15, 2012, 03:58
#42
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Quote:
 Originally Posted by singer1812 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]

April 5, 2012, 03:08
#43
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Quote:
 Originally Posted by happy does any one help me understand how the Update Loop = TRANS_LOOP] added to additional variable commands?? need more explaintions please, ASAP

 April 5, 2012, 08:51 #44 Senior Member   Edmund Singer P.E. Join Date: Aug 2010 Location: Minneapolis, MN Posts: 512 Rep Power: 14 What do you want to know?

April 6, 2012, 01:27
how I can add update loop into my AV?&How we apply derivative at B.Cs
#45
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Quote:
 Originally Posted by happy 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 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?

 April 6, 2012, 09:11 #46 Senior Member   Edmund Singer P.E. Join Date: Aug 2010 Location: Minneapolis, MN Posts: 512 Rep Power: 14 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.

April 8, 2012, 18:51
yes as you said
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Quote:
 Originally Posted by singer1812 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

April 9, 2012, 00:13
misunderstand me
#48
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Quote:
 Originally Posted by singer1812 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
Regards

 April 9, 2012, 08:57 #49 Senior Member   Edmund Singer P.E. Join Date: Aug 2010 Location: Minneapolis, MN Posts: 512 Rep Power: 14 Dont need transloop. Just setup a CEL expression for the derivative and put it into the BC under flow direction.

 April 9, 2012, 09:02 #50 Senior Member   Edmund Singer P.E. Join Date: Aug 2010 Location: Minneapolis, MN Posts: 512 Rep Power: 14 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?

April 9, 2012, 21:10
my model
#51
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Quote:
 Originally Posted by singer1812 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

April 9, 2012, 21:17
how I input my neumann BC through flow dirrection
#52
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Quote:
 Originally Posted by singer1812 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?
Regards

 April 10, 2012, 09:50 #53 Senior Member   Edmund Singer P.E. Join Date: Aug 2010 Location: Minneapolis, MN Posts: 512 Rep Power: 14 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. happy likes this.

April 11, 2012, 22:20
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Quote:
 Originally Posted by singer1812 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.

April 12, 2012, 04:19
#55
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Quote:
 Originally Posted by singer1812 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.

 July 14, 2014, 18:06 TRANS_LOOP operations #56 New Member   Join Date: Feb 2013 Posts: 16 Rep Power: 6 Hello, I am tring to use the TRANS_LOOP operation to solve the following equation in a CFX Domain: XX=1-(1/((1-XXold)^-a+B*exp(-C/(D*T[K]))*F*n))^a I am doing the following steps: · Create an additional variable XX which is defined by the expression : XX value: 1-(1/((1-XXold)^-a+B*exp(-C/(D*T[K]))*F*n))^a The variable is function of temperature. · Create another additional variable XXold which defined by the expression: XXold value: ave(XXold)@domain+XX()@domain*Time Step Size. · Initially, I run a simulation with initial values in Additional Variable Models XX=0 and XXold=0. But when I set for each variable its CEL expression I am getting this following message: "Bad expression value 'XXold val' detected in parameter 'Additional Variable Value' in object '/FLOW:Flow Analysis 1/DOMAIN:domain/SOLID MODELS/ADDITIONAL VARIABLE:XXold'. CEL error: The following unrecognised name was referenced: XX." Is there anyone who can help me to overcome this problem please? Thank you in advance for your advices and suggestions

 July 15, 2014, 17:48 #58 New Member   Join Date: Sep 2011 Posts: 8 Rep Power: 8 Hi emotep, I could be wrong, but on first glance, you have ave(XXold) and XX(). I think the issue amy be here.......

 July 15, 2014, 21:17 #59 New Member   Join Date: Feb 2013 Posts: 16 Rep Power: 6 what do you suggest me. I am not using a variable such MASSFLOW or velocity for the variable I want to calculate. It is in fact independent. To calculate this I have already set an additional variable and express a CEL. Thank you for your answer

 July 16, 2014, 23:31 #60 New Member   Join Date: Oct 2012 Posts: 8 Rep Power: 7 this is covered here: http://www.edr.no/blogg/ansys_blogge...rations_in_cfx

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