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How to handle radiation exchange with the sky 1

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Tunalover

Mechanical
Mar 28, 2002
1,179
Would you mind answering a few simple questions about radiation heat transfer? I’d be very grateful for your opinions!

In its simplest terms, I have a painted horizontal aluminum plate that faces the sky. It is heated from below by internal electronic power dissipation and from above by sea level by the sun at a 12:00 position. The plate temperature will not exceed 100degC.

One source of information I have says that the daytime radiation absorbed by the surface facing the sky is A_plate*(S*alpha_plate + sigma*epsilon_sky*T_sky^4) where S is the solar load and the rest of the terms are described by their subscripts and by Greek letters conventions commonly used in radiation heat transfer
Subscripts are preceded with an underscore "_".
(a forum that'll allow Greek letters, mathematical symbols, subscripts, etc. would make this a lot easier!).

Another source of information says that the daytime radiation absorbed by the plate from the sky is A_plate*alpha_plate*(S + sigma*T_sky^4).

Paying close attention to subscripts, is either correct? If neither, then what’s the right answer? alpha_plate is the solar absorptance of the painted surface= 1-rho_plate where rho_plate is the normal solar reflectance determined by test.

Thanks in advance for your help. If you can’t make sense with the Greek letters and subscripts then please notify me so that I can create and e-mail a separate Word document.

Regards,
Tunalover
 
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Actually, neither is completely correct. The 1st term in each is mostly correct, since the A_plate*alpha_plate*S is the solar heat load modified by the absorption of the plate. Note however, that the broadband absorption should be done with a spectral absorption, particularly since solar load is not spectrally flat.

The second term is partly correct in both cases, although the more correct term should have the difference of the 4th power of sky temperature and plate temperature. As with solar load, absorption and emissivity should likewise be evaluated spectrally.

TTFN
 
IRStuff-
Thanks for replying.
The first formula:

alpha_plate is the normal SOLAR absorptance measured (indirectly) with a solar reflectometer. The normal solar reflectance, rho_plate is measured with the reflectometer over the sun's bandwidth. Then alpha_plate=1-rho_plate since the plate is opaque. Therefore, I'm convinced (as I think you are, as well) that the first term alpha_plate*A_plate*S is good.

Are you saying that it's wrong to use alpha_plate in the alpha_plate*A_plate*sigma*T_sky^4 term because I should be using a different alpha, say alpha_plate,sky? You have a point there. Over what bandwidth should that be evaluated over? In the IR band the paint acts gray where epsilon_plate=alpha_plate but that assumes that the radiation from the sky is within the IR band. Is it? If it is, things would work out swimmingly because I had epsilon_plate measured with an emissometer.

The second formula:
I disagree with your contention that there should be a "difference of the 4th power of sky temperature and plate temperature." The minus sign comes in to accommodate incoming AND outgoing energy terms. I am looking only for the incoming terms.

I hate to raise another question with the second formula, but if the sky emits epsilon_sky*sigma*T_sky^4 then shouldn't this be reduced further by a factor of alpha_plate when it is absorbed by the plate? Did I just solve the problem with
A_plate*alpha_plate*(S + sigma*epsilon_sky*T_sky^4) being the incoming energy? Sounds reasonable but I'm no radiation heat transfer expert! Note that my sources were all PhDMEs in the heat transfer discipline and I'm just a working slob design engineer with a BSME.

Tunalover
 
After further thought, it's clear that both equations are missing the sky radiance term, which is mostly solar scatter with an apparent color temperature of 25,000K. The RCA EO Handbook say's that it about 1/4 of the illuminance of the sun. The alpha_plate would be slightly different than that of the normal solar load.

I forgot to point out that the 2nd term actually should be negative most if not all the time, since the sky is probably colder than the plate and the blackbody emission would be from the plate to the sky. The sky blackbody temperature runs around 250 K.

The 2nd term should have a sky emissivity and an absorption coefficient that's different than that for solar, due to the wavelength differences.

I'm BSEE myself...

TTFN
 
Great dialog IRStuff! As for the sky temperature, my research[1] shows that it is different between summer and winter:
Tsky=Tair-A
where A=20 degC in winter
and A=6 degC in summer.
Almost invariably when people talk about "how much heat something emits" they show the net difference between outgoing and incoming terms, thus the difference of absolute temperatures to the fourth power and the tendency of folks such as yourself to look for difference and negatives.

With your help, and with a few small changes in terminology, I think we're finally there. I believe that the following is what should be used to represent the total heat entering the plate from the daytime sky due to the sun and due to the sky itself:

A_plate*(alpha_sun*S+sigma*alpha_grey*epsilon_sky*T_sky^4)

Where alpha_sun= the plate surface's normal solar absorptance integrated over the sun's spectrum
=1-rho_sun where
rho_sun= the plate surface's normal solar reflectance integrated over the sun's spectrum (measured using a solar reflectometer),
alpha_grey= the plate surface's total hemispherical apsorptance
=epsilon_grey since the plate is grey in the IR band (and measured with an emissometer), and
epsilon_sky=the sky's hemispherical emittance obtained by spectrally and geometrically integrating the ratio of the sky's emissive power to the sky's blackbody emissive power (sorry if this is pretty basic stuff). This term was actually determined empirically in studies of solar energy and the HVAC of buildings. It depends on a number of things that would only clutter the picture here.

Thanks again for all your inputs. I've printed everything out and am saving it all for the future.

References:
1. Duffie, J.A., and W. A. Beckman, "Solar Energy Thermal
Processes", New York: John Wiley & Sons, 1974.

Tunalover
 
A couple of comments

> Our industry has often used 250 K for sky background for detection of airborne targets, so it's not clear what your reference's assumptions are, but I do know from actual experience that the daytime sky, is quite cold as far as infrared is concerned, but will vary as a function of elevation angle.

> As I mentioned earlier, there is also a sky radiance component caused by the scattered sunlight and looks like a 25,000 K color temp source. It's only about 25% of the solar load...

TTFN
 
IRStuff-
Putting aside airborne particles and humidity, you're right when you say that the sky is a black body radiator. It's safe (yet liberal) to say then, that epsilon_sky=1.

I would suspect that sky temperature depends on altitude, getting lower with increasing altitude. Designing radar and such, it would seem that a "sky temperature" used in aerospace (high altitudes) would be substantially lower than a "sky temperature" used in terrestrial applications (low altitudes). I'm going to stick with the data that points to higher sky temperature; this is more conservative anyway. It would be interesting to probe the variation of sky temperature with altitude though. If you run into anything, please let me know.

Tunalover
 
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