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Heat dissipation in a box

  1. Feb 2, 2012 #1
    Hi all

    I need an some help on thermodynamics. I want to install a camera outside of my house, but the camera can only operate between 0 and +55 degrees celsius. Now I want to make an isolating box around it, and wanted to calculate how thick the material of the box should be to allow outside temperatures between -10 and 40 degrees celsius. I know that if the camera is on, its power is 50Watt. On wikipedia I found that the heat capacity of air is 20.85 J/(mol.K).

    If I assume the camera is a point source of heat, located in the middle of a rectangular box, how can I calculate the temperature at the interior walls of the box, assuming the outside environment is an infinite reservoir at either -10 degrees C or +40 degrees.

    Many thanks for your help!!
  2. jcsd
  3. Feb 2, 2012 #2
    You need to consider the thermal resistances of each part. In electronics you can have a chain of resistors in series with a voltage difference between the ends of a chain, and current will flow through the chain. At each resistor junction there will be a proportion of the voltage depending on the values of the resistors. There is an analogy here with the heat problem, the equations are the same except voltage is replaced by temperature difference, the resistance replaced by thermal resistance, and the current by heat flow. So V=IR becomes ΔT=QR[itex]_θ{}[/itex].

    In your system you have a source of heat (inside the camera housing) which at equilibrium is going to be at a certain temperature, and an ambient outside temperature. Those are your end points, like the voltage across the resistor chain. Heat will flow from the hot end to the colder end, and in this case the total heat flow is 50W. You also have thermal resistances, namely internal camera heat source to camera case, camera case to box (through the air), and box to external ambient. I'm ignoring the box itself as I'm assuming it's metal with low thermal resistance, but if it's plastic there's one more junction, internal box to external box.

    Your camera should have a thermal resistance specification in °C/W. The box should be calculatable, again these are treated just like heatsinks and are specified in °C/W. The more difficult one is the air, but let's assume this can be found. You should have all you need to work out the various temperatures at each junction, and if necessary alter the thermal resistances by re-designing the enclosure to meet your operating temperature criteria.

    Edit: I think that since air has quite a high thermal resistance, the temperature difference between the camera case and the box is going to be quite high at 50W, and you've only got 15 deg C to work with in total. I think you might have to thermally bond the camera case to the box. Also you have two criteria, it must be greater than 10 deg C if the outside temp is -10, but it must be less than 15 deg C if the outside temp is +40.
    Last edited: Feb 2, 2012
  4. Feb 2, 2012 #3
    Thanks for helping me out!
  5. Feb 3, 2012 #4
    I think it is no thermodynamics but heat engineering most likely. It is a problem of heat transfer under convection (natural convection within the box and forced one- since there may be wind- outside). The problem is very difficult for strict solution, but approximate methods of estimations are commonly used. There are a lot of books on the problem, or you may search by Google using such key words as “heat transfer”, convection, Nusselt. For the beginning you may look at http://en.wikipedia.org/wiki/Heat_transfer_coefficient .
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