• [CP_CALCULATED_FIELDS id=”1″]

    Graphite requires special attention in thermal modeling as the thin layers of adhesive and coating reduce the spreading performance. Modeling only the graphite layer would produce enhanced results. However, modeling the individual layers will increase mesh-density as the typical thickness for the adhesive and coating can vary from 5 to 25μm. Alternatively, one can lump all three layers (coating + graphite + adhesive) in one single composite “block” with an effective conductivity value. This will relax the mesh-density and produce more accurate results. This calculator estimates the effective in-plane and through-plane conducitivity for a multi-layered system consisting of coating, graphite and adhesive. Click the calculate bottom to see the effect of adding adhesive and coating layers.

    The spreading performance of graphite can be significantly affected (for example from 400 to 200 Wm-¹K-¹) by including the adhesive and coating layers but will produce more accurate results.


     

    Equations used to calculate the effective in-plane and through conductivities:

        \[ \kappa_{in-plane}=\dfrac{\sum_{i=1}^N\kappa_i\tau_i}{\sum_{i=1}^N\tau_i} \]

        \[ \kappa_{through}=\dfrac{\sum_{i=1}^N\tau_i}{\sum_{i=1}^N\dfrac{\tau_i}{\kappa_i}} \]



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5 Responsesso far.

  1. Sen says:

    Can you please check the Thru-plane calculation?
    Thanks

  2. Sen says:

    Can you please check your ‘through-plane’ calculation?
    Thanks

  3. FTD Admin says:

    Hi Sen
    It is fixed now. Thanks!

  4. Sen says:

    Hello

    I actually don’t see the correction. Am I missing something?
    As a quick check, this is what I am doing…
    Say I put Coating and Adhesive thicknesses as 0.
    So its just Graphite.
    Hence when we calculate, the in-plane and thru-plane conductivity that the calculator spits out should be that of Graphite… right? But its not!

    Please let me know if I am missing something?

    Thanks

  5. Farhad says:

    in plane resistance is equivalent of parallel resistances and through plane equivalent resistance is calculated using parallel 1D resistance assumption.
    I don’t understand your question, Sen!

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