QUESTION
For the IR heat absorbed from a terrestrial environment (but NOT including sunlight), how can I prove/show that the emittance and absorptance of a particular painted or unpainted metal surface are equal for that IR spectrum? I know it involves Kirchoff’s Law but Kirchoff’s Law is much more than just α=ε (true only for graybodies).
REPLIES
cswilson
Kirchhoff’s law of thermal radiation states that absorptivity α is equal to emissivity ε for a given wavelength λ (and to be strictly correct, for a given angle of incidence). This is widely misinterpreted, so it is very important to understand this in detail.
Absorptivity is the fraction of radiation at that wavelength that is absorbed. Emissivity is the fraction of the theoretical maximum thermal radiation (i.e. the “blackbody” radiation) at a given temperature that the body emits.
So if a body absorbs 90% of 15-micron radiation that strikes it, it will emit 90% of what an ideal blackbody would emit at 15-microns at any given temperature.
What it does NOT state is that the intensity of radiation emitted at a given wavelength is equal to the intensity of radiation absorbed at that wavelength. (This is a very common misconception.) So absorptance does not necessarily equal emittance, even for a given wavelength.
As noted, emissivity/absorptivity can vary with wavelength, so you must be very careful in the general case integrating over wavelengths.
However, for a wide variety of materials, the absorptivity/emissivity over the wavelengths of “thermal infrared” radiation relevant for earth ambient temperatures is nearly constant (often ~0.95), so the graybody approximation works well, meaning that you can get good answers without formal integration, and using α=ε overall. Unpainted polished metals are a significant exception to this generalization.
As to a particular example, an upward facing high-emissivity graybody surface at night is likely to emit more thermal IR than it absorbs, particularly on a clear night. One of my heat transfer profs told us to use as a rule of thumb that a clear night sky radiated downward as if it were at blackbody at about -20C (253K).
Such a surface will reach steady-state temperature below the air temperature immediately above it, with the incoming downward radiation plus the conduction from the warmer air balancing the larger emitted upward radiation. This is why you can get frost on these surfaces even when the air temperature stays several degrees above 0C.
SOURCE
https://www.eng-tips.com/viewthread.cfm?qid=448026
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