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7-Bedrock

## Thermal initialization

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

1 ACCEPTED SOLUTION

Accepted Solutions
17-Peridot
(To:rrakos)

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.

5 REPLIES 5
17-Peridot
(To:rrakos)

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.

7-Bedrock
(To:SweetPeasHub)

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

Thanks again!

Roland

17-Peridot
(To:rrakos)

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?

7-Bedrock
(To:SweetPeasHub)

Yes, you are right. The low thermal conductivity actually means an insulator but it has to be used for the initial steady analysis only. In the case of next transient simulation, the thermal conductivity is modified to a normal value.

If normal thermal conductivity would be used for steady case then you will be see this:

Altough the outside temperature distribution is ok, but the inside is not OK.

With low thermal conductivity it can be seen this:

Temperature is OK everywhere in full investigated field.

That is why you have to modify a "fictive thermal conductivity" temporarily.

17-Peridot
(To:rrakos)

Ahh... you forgot to include the prescribed temperatures at the interface surfaces on either side of the small volume but you get the right result, so this is a second way to get the initial condition.

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