Maximizing heat absorption through Radiation

  1. Hi All,

    I am currently investigating a method of absorbing heat from radiation. This is being done to harness heat from skin.

    I was wondering what was the best method to do this?
    Is there a particular material that is best suited to do this? I know that having the emissivity value of the body that the radiation is falling on close to 1 is one thing that can be done.
  2. jcsd
  3. Philip Wood

    Philip Wood 1,088
    Gold Member

    The textbook answer is to use a dull, dark, matte surface. This, though, is a bit superficial (no pun intended) because surfaces which are good absorbers of visible light may not be good absorbers of far infrared (as emitted by skin). I'd guess, though, that you wouldn't go too far wrong with a soot-blackened surface.
  4. Thanks for that.

    I want to conduct the heat away. Does this mean that i could coat the conductive material [eg. silver (ideally) or copper (most likely)] with some form of a matt black paint.
  5. Philip Wood

    Philip Wood 1,088
    Gold Member

    Yes, I would think so. The heat will flow through the metal only if you keep the other 'end' of the metal cooler than its black face. The black face, if I understand your set up, absorbs heat from skin, some distance away. Don't forget that the temperature of the black face won't reach anything like that of the skin (say 35°C).
  6. Many thanks again,

    What would you estimate the range of heat absorption from skin radiation? i.e. How far away could i have the metal away from the skin before the heat absorbed from radiation be negligible?

    I know this has a lot of variable so lets just say that the atmosphere is at 10°C and no wind. The heat radiated from the skin is at 35°C and the temperature that i wish the metal be at is ~25°C.
  7. Philip Wood

    Philip Wood 1,088
    Gold Member

    Can't give you a straight answer on this, but here are one or two points.

    Seen from a point on the absorbing surface the skin needs to occupy a decent fraction (say half) of the forward field of view. So if a large patch of skin (the mind boggles!) is available it needn't be as close as a small patch. A rough calculation suggests that you'd need about 0.1 m2 of skin at a distance of 0.1 m from the black surface (assumed much smaller in area) in order to meet this rule of thumb criterion. Maybe the criterion is too stringent.

    If it's radiative transfer you're interested in, then the black surface needs to be below the skin, (which needs to be in a roughly horizontal plane) in order to avoid transfer by convection.

    What's going to make it very difficult to get the black surface to 25°C is that you're conducting heat away from it via the metal. The rate of heat arrival at the black surface will be too low to sustain anything but a very small temperature gradient through the metal.

    It is possible to make some a estimate the net heat arriving per second at the black surface, using Stefan's law. What value have you in mind for the area of your absorbing surface?
  8. I'm only looking at the radiative transfer.
    The absorbing face will be located approximately between 10 - 25mm away from the skin and in front of the face.

    The area of skin (the face) is quite a bit larger than the absorbing surface and located quite close for my application so going by what you have said, it seems that radiation can play a role in the transfer of heat.

    The absorbing face will be about 0.004m^2
  9. Philip Wood

    Philip Wood 1,088
    Gold Member

    Let's make the very crude (and optimistic) assumptions that the absorbing surface is 100% absorbing at all wavelengths, and that it is effectively totally surrounded by the radiating skin, which can also be treated as a black body. In that case the net heat, P, arriving by radiation on the absorbing surface (area A) per second is given by
    [tex]P=\sigma A (T_S^4 - T_A^4)[/tex].
    Putting [itex]\sigma = 5.67 \times 10^{-8} \text{W m}^{-2} \text{K}^{-4},\ T_S = 308 \text{K}, \ T_A = 293 \text{K}, \ A=0.004 \text{m}^2[/itex], gives an absorbed power of 0.37 W.
    I suspect the real figure will be a lot less, maybe by a factor of 3 or 4.
    Last edited: Nov 12, 2012
  10. This is very helpful, thanks.

    Just wondering though about the absorbing face.

    ''Let's make the very crude (and optimistic) assumptions that the absorbing surface is 100% absorbing at all wavelengths,''

    How would i find out how absorbing the face is across the different wavelengths but more specifically infrared?(as skin emits radiation at this wavelength)
  11. Philip Wood

    Philip Wood 1,088
    Gold Member

    Sorry, can't help much here. With Leslie's cube (copper tank filled with hot water), I found that the dull dirty face and - curiously enough - the white-painted face radiated almost as well as the matt-black face. Good emitters are good absorbers (at a given temp and for a given wavelength), so I'd guess that absorption of far infrared from warm surfaces may not be too much of a problem. Beyond that, I can only suggest the internet...
  12. Thanks for your help. You've given me plenty to look into.
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