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Heat transfer in deep space

  1. Oct 12, 2012 #1
    So, I know that it has been sort of asked in other forms before, but they mostly deal with human beings and so I've yet to find an answer to the specific part of this quandary: if a mechanical system is exposed to space (with just the skin of the ship itself as the only boundary between heat production and space), how effective would space be as a heat sink? I'm assuming that the only real heat loss would be via radiation (miniscule matter in space), so I'm also assuming an albedo of at least 0.5, if not better (Earth is 0.3, iirc?), to help offset the effect of the heat gained by solar radiation. What sort of cooling effect could be reasonably expected for a high temperature system vs. space?
     
  2. jcsd
  3. Oct 13, 2012 #2

    Physicist50

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    If - for example - the door of a rocket was suddenly opened, you wouldn't expect any cooling effect, just a release in pressure. The heat would keep moving in the same direction and wouldn't discharge. But it couldn't continue heating up obviously, because there's nothing to heat up, just a vacuum. Imagine the sun, when the nuclear fission within the star has completed, and it releases photons, since light includes heat, (infrared radiation) it still warms Earth and all the rest of the solar system as far as it can go. The heat doesn't discharge when released into space. In summary, there's no cooling effect when heat is released into a vacuum.
     
  4. Oct 13, 2012 #3

    Drakkith

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    Physicist50, you explanation is a little off.

    Heat is not something you can simply dump into space like air or water. It is energy. Typically it refers to the average kinetic and vibrational energy of particles in a material. Dumping air into space from your ship wouldn't do anything (ignoring things like evaporation) because the particles in the ships structure and everything else still have the same amount of energy.

    However, heat, also known as thermal energy, causes anything above absolute zero to emit radiation. This radiation carries energy with it so when emitted the material loses a small amount of energy, cooling off. The hotter something is the more energy it will emit as EM radiation. That is the only practical way to cool things off while in space unless you want to use a lot of material to keep something cool by evaporation. Literally letting something boil off into space, taking the thermal energy with it that you don't want. However that requires a lot of extra weight and is expensive. By the way the Sun uses fusion, not fission to produce energy in its core.

    http://en.wikipedia.org/wiki/Thermal_energy
    http://en.wikipedia.org/wiki/Temperature
    http://en.wikipedia.org/wiki/Heat
     
  5. Oct 14, 2012 #4
    So essentially, I'm correct in my assumption that it is plausible, I'd just have to figure out the math to determine exactly when it becomes effective. Cool, thanks guys! Or girls. Whatever, lol
     
  6. Oct 16, 2012 #5
    Empty space acts as a black body at 3K, so if you cool a space ship at 300K, empty space is perfectly cold and takes all heat your craft emits.

    But if you're near Earth (or Moon, Mars...) the craft receives heat from these bodies. Very important for low-orbiting satellites, and the craft's IR emissivity absorbs heat from these bodies as well as it emits because it's the same wavelength, so emissivity isn't a means of action for this part of the balance.

    Both absorptivity and emissivity can be varied a lot. Varying them in opposite directions is more difficult but possible. α can be between 0.02 and 0.99 at least, with 0.2 to 0.9 being rather easy; it depends much on cleanliness. ε can be between 0.01 and 0.99 at least, with 0.05 to 0.95 being rather easy. Corrugations increase both.

    Very hot is more difficult and very cold is more. Clean metals can radiate very little, like 0.01 for gold, while absorbing 0.3 for instance. Glass or thin plastic films can be transparent to light but emit IR; they're backed by a mirror to reject light from the object they cover ("second surface mirrors" or SSM). Thin polyimide films commonly absorb 0.3 to 0.6 depending on the thickness and emit 0.9: they're the gold-colour outer skin of satellites. Other plastics would have absorbed less but degrade at Sunlight. Mirrored glass absorbs <0.05 and radiates >0.8; it's special glass that withstands rays, for instance cerium-doped silica, and costs a lot. Do NOT use paints, theses sh*t don't adhere.

    Beyond materials, designers put the parts of controlled temperature in contact with a part that has the proper temperature.

    This choice results first from the too small heat conductivity of solids. As soon as a craft is 1m wide (and even before), its Sun side is too hot and its shade side too cold, so designers use heat pipes or even active circulation to keep the temperature under control. This permits an active regulation as well.

    Then you can have for instance an always-too-cold radiator and circulate to it the amount of fluid that keeps your ship not too hot.
    Or you can have a sun shader to keep a telescope very cold, reducing the noise of its cameras.

    In many cases a single shader or insulating layer wouldn't suffice, even with very small emissivity, so Multi-Layer Insulation (MLI) was invented, see Wiki for instance.
     
  7. Oct 16, 2012 #6
    Enthalpy gives a good answer, so I won't repeat it. You might Google the Space Shuttle to see how they got rid of heat. They dumped a huge heat load via radiation. The radiators lines the inside of the payload bay doors. That is why they were always in such a hurry to open the doors and point them away from the sun. Remember that heat transfer via radiation is proportional to the forth power of delta T,so it is not a small amount.
     
  8. Oct 16, 2012 #7

    jbriggs444

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    The heat transfer is the delta of the fourth powers, not the fourth power of the delta. Further, just because something is proportional to a fourth power does not mean that it is large. The constant in the proportionality matters.
     
  9. Oct 19, 2012 #8
  10. Oct 19, 2012 #9
    Yes, right page at Wiki. You don't need the complicated formula that depends on the wavelength: the simple one with T^4 is just fine.

    The wavelength is interesting just to understand that 300K heat is emitted around 10µm wavelength, so the emissivity at this wavelength can be adjusted differently from the absorptivity at Sunlight wavelengths.
     
  11. Oct 21, 2012 #10
    Awesome, thanks Enthalpy! I'll see if I can figure it out from here.
     
  12. Oct 28, 2012 #11
    Finally getting some time to look at this. For Stefan's constant, in US units, I'm looking at 0.1714 x 10^-8 BTU hr-1 x ft^-2 x R^-4 ? R seems to be the temperature in this case, so I'm assuming rankine scale. Safe assumption will be to put all temps in rankine, then?

    For the view factor, I'm assuming a pipe that is some portion exposed to space, and some portion enclosed by the body of a vehicle. O) <-- like that, in a manner of speaking. How would I go about finding the view factor for that?

    Sorry if these questions seem stupid.
     
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