Heat Solver

PrateekshaPrateeksha Member Posts: 3

Hi,

I am facing some trouble with Heat solver. Kindly reply to my queries:

  1. What is the minimum value required of "uniform heat source" for creating significant impact?
  2. How to check "Boundary" results? I can see so many results in boundary and unable to figure out the exact match with input power, how to do that?
  3.  When applying uniform heat source to a volume, the volume itself will also be included in the simulation (unlike if thermal boundary condition is used); meaning the thermal conductance of the material should also be relevant.-I gave uniform heat source to a volume (i.e. Graphene) but if I change material to Si or Silica, then also thermal profile doesn't change. It's not considering the thermal properties of material. I cross verified with the Lumerical example also https://support.lumerical.com/hc/en-us/articles/360042833673-Thermally-tuned-waveguide-FDE-. Is it correct?
  4. I used "power boundary condition" in place of "uniform heat source" in the Lumerical example https://support.lumerical.com/hc/en-us/articles/360042833673-Thermally-tuned-waveguide-FDE- and it gives same temperature profile. But it doesn't show same behavior for my structure. In my structure heater is smaller than waveguide and placed just above the waveguide but in the given lumerical example, it's farther and larger than waveguide, please check the attached files. What is the difference in "power boundary condition" and "uniform heat source"?
  5. How the results are dependent on "norm length" in 2D simulations? It's varying with "norm length".

I attached ppt as well as lumerical file. Kindly check it, it will be great help.


Looking forward to your reply!!


Thanks

Prateeksha

Comments

  • kghaffarikghaffari Posts: 45Ansys Employee

    Hi Prateeksha,

    1. This would be different for each project and dependent on specifics such as size, material properties, and design. I suggest keep increasing the power until you do see an increase in your temperature. You can also rely on the Q results to monitor the generated heat in your project and examining the behavior of the system.
    2. All the generated power will be dissipated through the boundaries. If you are using uniform heat source, it will not be listed (since it's not a boundary) but you can rely on the sum of all dissipated power to obtain the generated power. In the list, items with A correspond to area and P correspond to power; you can further discern the boundaries by their labels (e.g. convection, temperature, etc.).
    3. The material is accounted for when using uniform heat source. If you are not seeing any difference when changing materials it might be that: a) lock mesh option in the solver is enabled, so the changes you make are not applied to the simulation b) again, power might be too small for the system so the change in peak temperature is not noticeable.
    4. The power boundary will be applied at the specified surface and unlike the uniform heat source the material within is not accounted for in the simulation. Like you mentioned this could mean different final results depending on the specifics of your design.
    5. When using uniform heat source, the total power specified is attributed to the volume object which depends on the norm length. You will notice that norm length is one of the settings for the uniform heat source. So if the norm length is changed without the total power being updated, I’d expect the results to be different.

    Kind regards,

    Khashayar

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