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Thickness to insulate a space

  1. Feb 5, 2009 #1
    Hello all,

    Maybe this is a simple question. I've done some calculations like this, but for heating in buildings. Now it's different.

    I have some crystals I must keep cold, at -25ºC. So I will surround them with an insulation material wall. Outside we have atmosphere temperature

    I want to see which material is better (Rohacell, Styrodur...), I know the thermal conductivity, I know the temperatures, but I don't know the thickness of the wall to reach the -25º.

    I send a picture to illustrate the problem. I know that on the faces of the wall there is convection and radiation, I think here is the biggest problem.

    So, my question is how to calculate the thinkness of the wall (it's just one material)

    Thank you!
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2009 #2

    mgb_phys

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    The thickness doesn't set the temperature.
    If there is a temperature difference between the inside and the outside, heat will inevitably flow into the crystals and (unless there is some refrigeration source) they will reach the outside temp. All the insulation can do is slow this process.

    You can work out the power flow and so the time to melt from the insulation factor and the thickness.
     
  4. Feb 5, 2009 #3
    Ok, I understand about the balance of temperatures. (the in part is going to get warmer)

    Let's say we have a cooling machine inside which keeps the -25ºc.

    The question is how does the thickness of the wall affects to difference of temperatures.

    I need to know the global "k" (coefficient global of transmission) with the thermal conductivity of the wall.

    I think i'm quite lost

    Thank you
     
  5. Feb 5, 2009 #4

    mgb_phys

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    The insulation can only affect the power reaching the sample.
    If you have a cooling mechanism then the part is going to stay at -25c as long as the heat flow is small enough that the cooling mechanism can cope with it.

    The insulation material will have a thermal conductivity in W/mK so:
    heat flow = conductivity * temperature difference * area/thickness

    The area you know (the surface area of your container) the temperature difference is fixed so for a given cooling power you can set the thickness.
    Alternatively you can use the same equation to work out what the sample will heat up to until the heat flow balances the cooling power. A lower temperature difference means less heat flow.

    see http://en.wikipedia.org/wiki/Thermal_conductivity

    Its exactly the same theory as the resistivity of an electrical resistor if that's easier to think about.
     
  6. Feb 6, 2009 #5
    I was thinking this night (i take the work home)

    i can live with the heat flow

    but the problem is how to calculate de global K of the wall. i have the K of the material but there is still missing what happens in the faces of the wall
     
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