I'm working on modeling the heat transfer of a small payload (approx
10" cube) covered in extruded polystyrene at high altitude (80k-100k ft).
One thing I'm having trouble with is approximating the radiation flux reflected back from the earth (around mid-day) to the payload, which I'm approximating as fully incident on the bottom, and 50% on the sides. I'm also assuming the polystyrene has a emissivity of 0.9.
With the exception of sunlight (fully incident on the top of the payload), can I model the rest of the "sky" at 0K?
On a side note, I've looked at putting reflective materials inside the payload, but I'm trying to prevent possible signal degradation with a GPS reciever inside.
Use this reference for plausible ambient temperatures at the altitude of interest.
GELMAN, MELVYN E., MILLER, ALVIN J., WOOLF, HAROLD M. Regression Technique for Determining Temperature Profiles in the Upper Stratosphere From Satellite-Measured Radiances Monthly Weather Review 1972 100: 542-547
Use 1400 watts /m^2 for insolation. Use a ground temperature under the region of interest for radiation balance - say 10 to 40 degC
Ok, so if I use a range for the albedo of earth, I should get a range of values. Anyone care to double-check my heat flux equations? (please?) :) They are for the top, sides, and bottom surfaces of the payload.
Where ε is the emissivity of the polystyrene (≈0.9, and approximated to be the same as absorbtivity), a is the albedo of the earth, and σ is the S-B constant. The sun is assumed to be directly overhead, and the bottom of the payload is assumed to be perfectly parallel to the surface of the earth. The 1/2 in the Qsides equation is the shape factor between the earth and the sides.
Ok, so some of those symbols in my last post weren't too HTML friendly apparently. Lemme try again:
Qtop = e * Qsun ? e * o * Ttop^4
Qsides = 1/2 * e * a * Qsun ? e * o * Tsides^4
Qbottom = e * a * Qsun ? e * o * Tbottom^4
Where e is the emissivity of the polystyrene (about 0.9, and approximated to be the same as absorbtivity), a is the albedo of the earth, and o is the S-B constant. The sun is assumed to be directly overhead, and the bottom of the payload is assumed to be perfectly parallel to the surface of the earth. The 1/2 in the Qsides equation is the shape factor between the earth and the sides.
You are assuming space is at 0K... your choice. I'd pick 2.7K, for 5th decimal place accuracy. ;>)
The sides are transmitting approximately half to 0K and half to Earth average. This (Earth term) is missing.
The contribution of the sun to the bottom is 0, and I see no contribution of the Earth
I'd have to crack my heat transfer book again, but I seem to recall absorptivity + reflectivity + emissivity = 1 as a simple energy balance. If absorptivity = emissivity = 0.9, that leaves reflectivity as negative... This error could easily be just my memory.
Absorptivity and emissivity can both be equal to 0.9. I see those values all the time in orbital analyses. Both the absorptivity and emissivity are actually "absorptivities".
Absorptivity is the absorptivity of a surface in the solar wavelengths, and emissivity is the absorptivity of a surface in the IR.
I defined the earth term as (a * Qsun), since that's the total flux coming BACK from the earth. You could just as easily make the equation:
Qsides = 1/2 * e * Qearth, where Qearth = a * Qsun
Again, Qearth = a * Qsun. Hope that clarifies.
I think you mean absorbtivity + reflectivity + TRANSMITIVITY = 1, or
100%. If you have 1 unit of light hitting an object, it either has to be absorbed, transmitted through (like glass at some wavelengths), or reflected, therefore accounting for all the light that intersects the object. Sometimes black bodies and gray bodies are assumed to have equal absorbtivity and emissivity values because as a general rule of thumb they're pretty close, even though they're wavelength dependant and not exactly equal.
Thanks for the help. Having clarified, are there any glaring errors that stick out?
I don't think this is correct. It should be energy radiated to-from the Earth, at Earth's surface temperature. Unless you are going to do this experiment at noon, over a *polar icecap* (high reflectance), Qsun is not a good number to use.
contribution
You are exactly correct. I SHOULD have opened my textbook. ;>P
As stated above, the Earth is not usually considered to be a mirror to the Sun, but a thermal sink with its own inherent temperature. I think you'd get more meaningful answers that way.
"Energy radiated to-from Earth"... Wouldn't the energy radiated away from the earth (e*o*T^4) be synonymous with to a*Qsun? I thought albedo was a measure of all radiation emitted by earth in sunlight. (a is usually estimated at 0.3 for earth).
Or does albedo only take into account reflected solar radiation, and does not account for thermal energy emitted by the ground, oceans, etc? If it does not, would the amount of emitted radiation really be significant compared to the reflectance due to the albedo? The emitted radiation alone would be approximately what you'd see on the night side of the earth, which I imagine is significantly lower than the total radiation flux coming from the dayside of earth.
The radiative heat transfer of a surface is descibed by its surface temperature and emissivity, and not "double accounting" for the way the energy arrived at the surface under consideration. You are making this waaaaaaay too complex. The "control volume" of your formulation should be drawn around your cube, and not around your cube *and* the surface of the Earth. Because, there is some non-trivial amount of heat coming from the core of the Earth too (probably not important for noon calculations). ;>)
First off, the only radiation exchange you'll have externally will be with Space. If you've got 6 surfaces, you'll have 6 radiation terms to space.
Of course, you'll have radiation conduction inside the box, that's where the fun comes in, calculating those view factors.
You have 3 energy sources: Solar, Albedo, and Planetary IR. These are not radiation terms, these are Heat Sources, Q's.
Qsol typically has values between 1321 W/m^2 and 1423 W/m^2, depending on what you're looking for, a hot or a cold solution.
The Albedo is the percentage of Solar energy reflected back by the earth, it varies depending on orbital inclination, anamoly angle, and other factors, but it's usually assumed to have values between 0.2 and
0.4, again depending on what you're looking for - a hot or a cold solution.
Qalb = 0.4Qsol (assuming 0.4).
Earth IR's values range from 206 W/m^2 to 286 W/m^2.
Remember to input these items as Energy sources.
The last Q in the energy equation (excluding internal sources) is the radiation terms.
Qrad = sigma*Radk*(T^4 - T^4) sigma = Stephan-Boltzman Radk = emiss*A*VF, in this case the view factor, VF is 1, so only the Area matters.
The following assumes that the box is non-rotating, with the top side normal to the sun.
As far as how much to apply to what surface, the top, space side surface, will only have Qsol. The other 5 sides will only have Albedo and EarthIR acting on them. This is assuming that the box isn't an orbital object, i.e., it's location is fixed wrt a point on the earth.
The bottom side will see full Albedo and Earth IR over it's entire surface area and no solar. The sides will see ~1/2 of the Albedo and Earth IR as the bottom, and no solar. Again, if you're looking to bound the problem, for the cold case, assume Qsol and Qalb = 0 (flying in the night sky) and use the lower QearthIR, and use the highest Q's for the hot case. This way introducing a little bit of error in the amount of energy incident on the sides shouldn't matter.
Sweet. That clarifies alot. I appreciate the help from everyone! Now all I have to do is model it in Matlab and see what the results are. I'll post them once I've put it together.
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