Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Converstion of radiant flux (watts) to temperature (C)

Tags:
  1. Sep 25, 2014 #1
    Hey all,
    I'm having some unique challenges on an optical system I've created, where I'm arraying several LEDs under 1 optic/reflector. We're beginning to see major degradation of the metallic coating that is applied to the part, and I'm tyring to quanitfy the temperature (converted from radiant flux incident on the reflector surface) at the surface of the reflector. I need to know what temperatures we're seeing inside the optic so I can recommend a coating specifiation for a vendor.

    Seems the internet is flooded with things, or I could rewind back to some physics textbooks, but trying the forum first, in case there are any good suggestions.

    Thanks in advance,
    JD
     
  2. jcsd
  3. Dec 18, 2016 #2
    In the most basic case of only radiative emission I'd say you could simply consider the Stefan-Boltzmann law:

    $$J=\sigma T^4$$

    where J is your incoming flux by unit of surface and ##\sigma## is a known physical constant. Of course this assumes black body behaviour and no other forms of heat dispersion. If you assume a grey body (namely that your optic reflects and absorbs equal percentages of radiation at all wavelengths) nothing really changes because if J is the flux you already know is coming then both sides of the equation get multiplied by the emissivity ##\epsilon##. If instead you add another 'sink' of heat, like some conduction mechanism with thermal resistance R, then you have

    $$\epsilon J = \epsilon \sigma T^4 + \frac{(T-T_{room})}{R}$$

    and that further lowers your equilibrium temperature.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Converstion of radiant flux (watts) to temperature (C)
  1. Watt is it? (Replies: 1)

Loading...