Convection Temperature Dependence - Not Interpolating Data Table Correctly

[abellX29CJ]

My Nonlinear Transient Heat Transfer analysis is yielding unreasonable results. To troubleshoot, I have created a simple model and ran a Nonlinear Steady State Heat Transfer analysis. The problem appears to be a failure to interpolate the convection coefficient from a temperature-convection data table. I'd like someone to review my findings and help me achieve the desired behaviour. I'm using Autodesk Inventor Nastran Version 2023.3.0.67.

The simple model is a solid block with near perfect thermal conductivity. I have a temperature load on one face and a convection load on the opposite face. This way, I can effectively choose the uniform temperature of the convection surface, which allows me to verify that the correct convection value was interpolated from the table.

The temperature in the simulation is ~392 K, which should result in a convection coefficient of 15 671.1 W/(m^2 K). With an Ambient Temperature of 313 K, I should get a flux of (392 K - 313 K) * (15 671.1 W/(m^2 K)) = 1.238E+6 W/m^2.

However, the simulation results show 2.607E+5 W/m^2. That would require a convection coefficient of 3300 W/(m^2 K). Something is not right.

Now, if I copy and paste the convection coefficient for 392 K from the table into the load definition as a static value and re-run the simulation, I get a heat flux of (392 K - 313 K) * (15 671.1 W/(m^2 K)) = 1.238E+6 W/m^2. That's correct!

 

I don't understand why Nastran cannot interpolate/select the correct value from the table. Thoughts?

-Adam

[John_Holtz · Accepted solution]

Hi @abellX29CJ . Welcome to the Inventor Nastran forum.

Sorry I do not have time this year to investigate your question more fully, but I suspect the problem is your input requires a very, very small tolerance in order to converge to an accurate answer. The default tolerance is not very small. The analysis is trying to add and multiply very large numbers and very small numbers. That makes it difficult to do the calculation accurately. And when the convergence criteria is larger than the calculated temperature difference, it is easy to converge to an inaccurate answer.

I understand your approach to make the calculations easy for you, but in the process you have made it very difficult for the solver. Please try an analysis with realistic values.

• Thermal conductivity: 1E6 W/(m K). Nothing is even close to this.
• Calculated temperature difference: 0.007 degrees (out of 392.) That is very small. 
• Convection coefficient: on the order of 16000 W/(m^2 K). Perhaps something that large occurs in the nozzle of a rocket engine, but not in normal engineering. (Typical values are 5 to 500 depending on the ambient condition.)
• Slope of convection coefficient curve: (15671.1-11153.19 W/(m^2 K))/(1 deg) = 4517.91 W/(m^2 K) per degree. The heat conducted through the material must balance the heat loss to convection. A small "error" in either value can lead to a large difference in the results.

Also, you need to investigate what temperature is used to interpolate the table. Does it use the surface temperature? Does it use the bulk temperature (= 0.5*(surface temperature + ambient temperature))? Does it use something else? Based on this forum post, it is based on the bulk temperature: see Solved: Setting up a temperature dependent convection coefficient - Autodesk Community. What temperature did you use in your calculation? Also see the documentation for the PCONV card.

If you do have somewhat extreme conditions (not the above values!), the other thing you can do is edit the Nastran file and change the convergence criteria. See the documentation for the NLPARM card and How to edit the input file in Inventor Nastran. Instead of input like this:

you need input like this (or with the 3 convergence tolerances on the order of 1E-6):

Good luck.

John

[abellX29CJ]

Thanks for the reply @John_Holtz.

I found part of the solution in the PCONV documentation:

By default, Nastran takes the average temperature of the surface nodes and ambient temperature nodes. It uses that temperature to interpolate the data table. In my case, I do not want that. I want FORM=2, where only the surface node temperatures are used to interpolate. I am simulating heat transfer to water in the nucleate boiling regime, which is why the heat transfer coefficients are unusually high. That heat transfer is an exponential function of the wall temperature, but it is minimally influenced by the ambient (or bulk) water temperature. 

This is the default PCONV entry in my model:

I don't know why FORM is set to zero, and I think column 5 should be blank, rather than "0.0". Maybe something to investigate.

When I edit the card and change FORM to "2", I get much closer to the desired result:

The manual calculation is (392.9 K - 313 K) * H = 1.232E+6 W/m^2

Solving for H gives 15 420 W/(m^2 K). My tabulated value for H at 393 K is 21 300 W/(m^2 K). Could be better.

The second part of the solution was in the convergence tolerances in NLPARM. Here's the default NLPARM card that was generated:

And here is my modified NLPARM:

Combined with the FORM=2 change to the PCONV card, here are the results:

Manual check: (392.0 K - 313 K) * H = 1.229E+6 W/m^2

H = 15 557 W/(m^2 K)

Tabulated value for H at 392 K = 15 700 W/(m^2 K)

That's slightly less than 1% deviation. I think that's good.

Thanks again @John_Holtz , you were right to point me to PCONV and NLPARM.

I'll take what I have learned here and apply it to my complex model. Hopefully I'll get reasonable results. The exponential T vs H relationship makes convergence difficult! I plan to use the temperature results to perform a fatigue analysis. I'm trying to understand some part failures in the field and make some design improvements.

[John_Holtz]

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I plan to use the temperature results to perform a fatigue analysis.

Just a heads up. You will need to do the fatigue calculation by hand since Inventor Nastran does not support temperature loads in a fatigue analysis. See Thermal load input for a multi-axial fatigue analysis in Nastran In-CAD and Inventor Nastran. The PDF attached to this article is a good reference if you need a refresher on calculating the fatigue life: How to understand Fatigue analysis in Nastran.

John