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Radiative cooling

  1. Aug 1, 2011 #1
    I'm hoping i'm in the right forum for this, but here's my question. Optical cable can guide radiation to a point, which is the basic of my thoughts. Say you had an insulated box. On the inside of this box, is a liter of water in the shape of a cube, giving it a surface area of .06m^2. From the Stephan-Boltzman equation, I find that the power lost is around 25 watts at room temperature, and 17 watts at freezing. Averaging the two together(which I know won't give me an exact power loss due to radiation, but considering this is just a thought experiment at the moment, the error is negligible) I get a loss of approximately 21 watts. Ignoring the enthalpy of fusion required to actually turn the water to ice, it would take (4.184joule/gC*1000g*25C(C being degrees Celsius) 104600 joules to bring the water to freezing temperature. With one Watt Second being equal to one Joule, it follows that it would take(104600/21) 4980 seconds for the water to radiate the necessary heat, or about 1.4 hours. If one were to line the inside of the insulated box with bundles of optical cable,spliced the cables into one after they were away from the box some(to minimize the surface area available for radiation to reenter the cable, and feed the cables into an infrared solar cell(or any other apparatus to convert the energy into anything but heat, which would simply return to the box though radiation), would the water in the box freeze in a timely manner, or eventually for that matter? Thanks for any thoughts in advance.
     
  2. jcsd
  3. Aug 1, 2011 #2
    The one problem I see with your thought experiment is that the other end of the cables will also be radiating energy, as a function of their temperature. So if your math is correct and the other end of the optical cable is kept at 0 degrees kelvin, then the water will radiate away it's thermal energy in the time frame you derived.

    The laws of thermodynamics, however, still apply. It is not possible to transfer energy up a thermal gradient in bulk. If the other end of the cables are at the same temperature as the water, then they will radiate the same energy as the water, and the net energy flow rate will be zero.
     
  4. Aug 1, 2011 #3
    Since radiation depends on surface area, I assumed that the cables would not radiate inward, is this incorrect?
     
  5. Aug 1, 2011 #4
    You are correct there, and at first glance it looks like we have a contradiction. However, the area of the cable ends inside will still be the same as the area of the cable ends outside. Also, I believe if you have an infinite plane radiating energy, the energy per unit time absorbed depends only on the area of the collecting surface, thus I think the area term can be pulled out, and you get kAn(Tin4 - Tout4)
     
  6. Aug 1, 2011 #5
    I thought I had considered that when I mentioned splicing the cables together. By having each cable feed into one main cable of the same diameter, you end up with a smaller surface area where the radiation exits the cable. Wouldn't that decrease the radiation that entered the box?
     
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