Elmer Discussion Forum

Hello !

I applied 1 Joule to 1 kg cube of water. I set heat flux = 0 to effectively insulate the water cube.

Expected change of temperature ΔT = 0.00024108 K = 0.24108 mK
Q = mcΔT = (1 kg)(4148 J/kg*K)(ΔT) = 1 J ∴ ΔT = 0.00024108 K = 0.24108 mK

I discovered that the external temperature does not matter when Heat Flux = 0.
I discovered that the initial temperature affects the rise of T. Numbers: https://imgur.com/a/F7VId8m
Project Directory shared: https://drive.google.com/drive/folders/ ... sp=sharing

One source of difference is the slew rate of the heat applied in pulses instead of constantly.
Can you help. Do you know why the temperature rise is different at different temperatures? ; I don't see any temperature dependencies in my project.

Hi

Ok, I had to have a look at this. You have several things that work as intended but their combination not so much.

First of all, your temperature rise is very small dT=2.4e-6 K. So when you look for temperature change this small you better check your tolerances. When iteration proceeds the change is compared to the norm of the solution, i.e. dT/T. If you start from T=293 this is ~8e-9.

Everything would have gone even with these setting but you happended to have relaxation factor 0.7. So this meant that if your reference temperature was zero you had to iterate quite a few times. But if the reference temperature was 293 the solution was deemed good after 1st iteration which resulted the increase in temperature being 30% too small.

So there were many things that would have saved you:
1) Using heat source that resulted to significant temperature change compared to the starting temperature
2) Setting the relaxation factor to 1
3) Increasing nonlinear tolerances further

-Peter

Good advice Peter. Had not considered that solution tolerance, magnitude of result (meaning small), and relaxation factor could combine to satisfy the specified convergence tolerance, but the solution is not really converged to the accuracy required. I suppose this could happen to any of the solvers.

Hi

Well, sort of the accuracy is what was asked. The nonlinear tolerance was 1.0e-7 and the model would have required accuracy 1e-9. But this was really quite a rare combination. It could indeed happen to many of the solvers. I would think that it could happen to even some other codes. First time for me.

-Peter

I adjusted the tolerances as suggested and the 273 case now produces the expected 0.24Mk results.

Quick question. Why does the nonlinear convergence tolerance matter here if this model/simulation does not have Radiation = 0 and Heat Flux = 0 for all boundary conditions?
I thought that would make this linear.

Thanks!!!

Hi, Because you used "nonlinear system relaxation factor < 1". -Peter

Thanks!