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rrakos1-Visitor
1-Visitor
October 10, 2018
Solved

Thermal initialization

  • October 10, 2018
  • 1 reply
  • 3581 views

Hello,

 

my task is simple. I have 2 metal part with 1 interface between them. The first body temperature is T1, the second body temp. is T2. Goal is to see the modified temperature after certain time. My question is, how can I initialize the different temperatures? Based on creo simulate:

1.) I can define 1 temperature in full field, this solution of course is not ok

2.) I can define temp. distribution from previous steady simulation result but it seems, the result is total wrong with this solution.

Any idea? I can't believe that it is a complicated task for simulate...

 

Thanks

Roland

Best answer by SweetPeasHub

I had a little struggle with this too.  It is actually a formulation issue. 

The boundary surface between the parts is not two surfaces, it is one shared surface.  Logically that surface cannot have two temperatures at the same time.

So what is the workaround?  I will try to explain, but I am going completely on memory because our thermal license is being used...

 

  1. Intermediate volume.
  2. Steady state followed by transient.

 

Make a small volume region for an intermediate region. (yellow area in image)

 

 

Define starting temperature all the same.

Define boundary temperature constraint on all blue as the cold temp.

Define boundary temperature constraint on all red as the hot temp.

Be sure to also select the internal shared surface with the middle volume, on both sides of the middle volume, respectively.

 

Solve for steady state temperature.  The intermediate volume will have a thermal gradient.

Use this analysis as the input for transient, but do not use the boundary temperature constraints, of course.

 

Note: The smaller the intermediate volume the closer you get to your theoretical time=0 condition.  How small depends on your specific situation, materials, etc...

 

Also, the other theoretical method for this would be a "moving - contact" transient analysis where the two parts are initially not touching.  As far as I know Creo does not have this type of coupled structure-thermal capability.

 

 

Thermal2.JPG

 

 

1 reply

S
SweetPeasHub17-PeridotAnswer
17-Peridot
October 10, 2018

I had a little struggle with this too.  It is actually a formulation issue. 

The boundary surface between the parts is not two surfaces, it is one shared surface.  Logically that surface cannot have two temperatures at the same time.

So what is the workaround?  I will try to explain, but I am going completely on memory because our thermal license is being used...

 

  1. Intermediate volume.
  2. Steady state followed by transient.

 

Make a small volume region for an intermediate region. (yellow area in image)

 

 

Define starting temperature all the same.

Define boundary temperature constraint on all blue as the cold temp.

Define boundary temperature constraint on all red as the hot temp.

Be sure to also select the internal shared surface with the middle volume, on both sides of the middle volume, respectively.

 

Solve for steady state temperature.  The intermediate volume will have a thermal gradient.

Use this analysis as the input for transient, but do not use the boundary temperature constraints, of course.

 

Note: The smaller the intermediate volume the closer you get to your theoretical time=0 condition.  How small depends on your specific situation, materials, etc...

 

Also, the other theoretical method for this would be a "moving - contact" transient analysis where the two parts are initially not touching.  As far as I know Creo does not have this type of coupled structure-thermal capability.

 

 

Thermal2.JPG

 

 

R
rrakos1-VisitorAuthor
1-Visitor
October 11, 2018

Thanks for your help, this solution is perfect! But you have to be careful that in the first (static) simulation the center layer has to include very low thermal conductivity. 

 

My problem that it was a simple test case. My real geometry is more difficult. Separation of center layer is more complicate in this model....but I have no more choice Smiley Very Happy

 

Thanks again!

Roland

    S
    SweetPeasHub17-Peridot
    17-Peridot
    October 11, 2018

    I just edited the answer with what I thought was your tweak, but now I that I re-read your comments, do you mean high thermal conductivity, low thermal resistance, so the small layer does not act as an insulator?

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