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Temperature dependent heat capacity

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Hi dear all,
I have a simple 2 layered heater structure: a phase change material X is in top of TiN. This X material has a time dependent resisitivity, thermal conductivity and heat capacity. At a temperature around 872 K all these parameters changes abruptly. Comsol shows good results when temperature is less than 872 K, but whenever it passes 872 K it runs for days with no results. To find why this happens I assigned constant numbers to 2 of them and applied only 1 temperature dependent parameter. For instance, heat capacity and thermal conductivity are assigned as constant with resisitivity changing with temperature. As a result, I found that due to heat capacity simulation lasts long. So is there any solution for this problem? Or how to solve it? Heat capacity is 202 till 874 K then it jumps to 14000 till 882 K and after this temperature-at 883 K it is 202.

2 Replies Last Post Jun 20, 2016, 11:06 a.m. EDT

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Posted: 8 years ago Jun 20, 2016, 2:44 a.m. EDT
When asking question you should always upload jpeg screenshot of your screen and problems you are facing . I am also working on PCM simulation . it seems your problem is because you are using a very steep function ie 202 < Tm and 140000 >=Tm. you must use of of many functions provided ( step , ramp or analytical ) with comsol to get convergence quickly.
When asking question you should always upload jpeg screenshot of your screen and problems you are facing . I am also working on PCM simulation . it seems your problem is because you are using a very steep function ie 202 < Tm and 140000 >=Tm. you must use of of many functions provided ( step , ramp or analytical ) with comsol to get convergence quickly.

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Posted: 8 years ago Jun 20, 2016, 11:06 a.m. EDT
Hi Tohir,

you could try to extend the area which absorbed the latent heat and use the solution as initial solution for your case. For example, you could define a function with a melting interval for the heat capacity. Assume your dT is 50K (or 100K), then your heat capacity will be 202 for 0<T<874 K, 14000 for 874<T<974 K and again 202 for T>974 K. The best way is to use an auxiliary or a parametric sweep for dT with continuation for the last parameter, so that the solver takes the solution for the last dT as initial for the next one. Hope this informatiom will help you!

Best regards
Antoni Artinov
Hi Tohir, you could try to extend the area which absorbed the latent heat and use the solution as initial solution for your case. For example, you could define a function with a melting interval for the heat capacity. Assume your dT is 50K (or 100K), then your heat capacity will be 202 for 0

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