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Thermal Resistances

  1. Mar 16, 2012 #1
    I searched for something relevant and came up short so forgive me if there is already a thread started on this.

    I am taking a heat transfer class right now and am trying to figure out a few things.

    My question arose when going over thermal resistances and how it relates to the general equations for conduction and convection.

    I know the general equation for heat transfer due to conduction is q = kA(ΔT/L)
    Also the heat transfer due to convection is q = hA(Ts-T)
    From my understanding these are the general relations, does this mean that we are neglecting the thermal resistances when we use these relations as if we were neglecting friction in a dynamics problem?

    Are we then able to find the values for the resistances for a material based on its geometry and heat transfer coefficient and all that good stuff and then use a relationship like resistors in series?

    Hopefully I'm making sense in all of this. Basically if I could sum it all up, are the general relations for heat transfer due to convection and conduction neglecting thermal resistance?

    Small aside, can the thermal resistance of certain materials be understood sort of like the emissivity when finding heat transfer due to radiation in that the smaller the emissivity (the less the material acts like a blackbody) the less heat is transferred due to radiation?

    EDIT: I just realized that my wording is off... the value of k for conduction is similar to resistance based on the material properties of the material you are dealing with, this allows less heat to transfer through plastic then say metal, this leads me to believe k can be thought of more like emissivity than thermal resistance since k is also a material property.

    I am still however a little confused as to what thermal resistances are since k is also sort of like a thermal resistance.
     
    Last edited: Mar 16, 2012
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  3. Mar 16, 2012 #2

    AlephZero

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    They are two different ways of describing the same thing. The equations for conduction, convection, etc describe the physical process by which the heat is transferred. The idea of thermal resistance ignores that level of detail, and for a particular physical object it says "with this much temperature difference, there is thiis much heat flow".

    From an engineering point of view, if you want to use something like a computer heat sink, or an insulating material, you don't really care about how it does what it does. You just want to know what it does, i.e how much or how little heat is going to flow through it for a given temperature difference. For example if I'm designing an electronic circuit and I have a transistor that is dissipating 50W of heat and I want to keep its temperature below some limit, I can easily work out that I need a heat sink with a thermal resistance less than X degrees K / watt. Then I can look in a heat sink supplier's catalog and select a suitable part. In principle I could work out what each heat sink design in the catalog would do, based on its geometry, material properties, etc, but life's too short for that, and it wouldn't make much sense if every customer had to do that work, compared with the supplier telling them "the answer", i.e. the thermal resistance.

    Yes. But for complcated "real world" devices it might be easier to measure its thermal resistance, rather than calculate it from the physics equations.
     
  4. Mar 16, 2012 #3
    So essentially thermal resistance simplifies problems so that for problems such as your example you would only need to know the power dissipated by the transistor, the free stream temperature, and the surface temperature you wish to not exceed? Then it's just referencing tabulated values for thermal resistances of heat sinks to find the one you need?

    I have a few questions concerning this.

    Would you then just look for a heat sink that covers the area of the transistor and also has the required thermal resistance?

    How is the thermal resistance of the heat sink calculated? I'm guessing another heat transfer problem involving fins, but how does the convection coefficient come into play?

    I hope I'm making sense, thanks a ton for your help so far it has really been beneficial.
     
  5. Mar 16, 2012 #4
    Thermal resistances simplify problems wherein there is no heat generation within the system. If there is a KNOWN heat generation entering the system, like if you have a sliding block which is experiencing friction on one face, you can find the the heat flux on that side. But if it's like a wire with current passing through it, you can not use thermal resistances.

    There's also a thing we call "shape factor" which helps us define thermal resistances for things like complicated geometries in certain configurations, so we don't have to do like a finite element analysis. It just simplifies things, and these are generally from experiments.

    Heat sinks usually have fins, and if you're in an introductory heat transfer class, you should cover fins as a special isolated topic. These basically have a high thermal conductivity so they can disperse heat well and they have a high surface area, so they can transfer heat well to the environment convectively. In addition, heat sinks are usually right next to fans where they can have a boosted convection coefficient.

    If you want to know more about it, look up fins and I'd also suggest reading through the chapters on convection in any intro heat transfer book.
     
  6. Mar 16, 2012 #5

    AlephZero

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    That's right.

    Usually the transistor is bolted onto a larger metal plate with then has cooling fins etc. The temperature difference caussed by the heat conduction through the plate is much smaller than the difference between the surface of the fins and the ambient temperature. There are some pictures here: http://www.johnhearfield.com/Eng/Heatsinks.htm

    The usual way to do the math is to use Newton's law of cooling, but with a separate formula to give the heat transfer coefficient as a function of the size of the heat sink, the spacing of the fins, the orientation (horizontal or vertical), etc. There are are many different formulas that have been derived from experiments rather than theory.

    In a first course on heat transfer, you will probably just be given the heat transfer coefficient you need to solve a problem, without any explanation where the number came from.

    Calculating heat transfer coefficients "from first principles" is hard, because you have to calculate what is the air flow pattern over the object as the air heats up and its density changes. For example the heat transfer coefficient of the same rectangular flat plate will be different if it is horizontal, or vertical with the longest or shortest side pointing upwards. Adding cooling fins makes the calculations even harder. This requires knowing about fluid dynamics and computer methods to model fluid flow, not just knowing about heat transfer.
     
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