Temperature of satellite

[LSAteam]

Just curious, I'm writing a thermal analysis code to get into the swing of things as it were, and thought I'd use this as a metric. What sort of Absorptivity are you assuming in your model? These can be highly variable in thermal control equipment (0.05-0.9), more variable than Emissivity (0.5-0.9).

Have you just assumed emissivity and absorptivity are equal values?

I made results like yours eventually, by assuming the two were equal and fixing a few unit errors I had made... (cm to m by a factor of 1000? well, even NASA makes unit errors now and then!)

Still indeed this does highlight quite the thermal issue.

[lavalamp]

The extent to which a material emits radiation of a certain wavelength and that which it absorbs radiation at the same wavelength are the same. However, the satellite would be generally be emitting longer wavelengths than that falling upon it from the sun.

Regardless, for the simplified simulation I did, I set a value for the emissivity that was constant across all wavelengths. This is more than good enough for a first approximation I feel, especially since the earth and satellite would emit entirely in the IR range, and the sun emits a very large fraction of it's power as IR too.

For an improved analysis, you could try and obtain data on the emissivity of, for example, aluminised mylar or aluminium foil, across a range of wavelengths from 100 μm down to 100 nm. Additionally the sun and earth could be modelled as black body emitters with some effective temperature and a constant emissivity, and you would have to calculate the total power absorbed by the satellite based on it's emissivity across the range of wavelengths emitted by the sun and earth independently. Also calculating the power radiated by the satellite based on its temperature too.

Honestly though I can't see that such a detailed model would affect the outcome by more than a couple of degrees, and since you only care about keeping it in a window of a hundred degrees or so, all you have to do is get close enough.

Making sure that heat can be dumped from powered components is more important than the overall temperature of the satellite. For that you'll probably looking at designing in heatspreaders, if your components are low power enough though you might get away with relying on the copper in your PCB to do that job for you.

[LSAteam]

The physics there is true, but generally when designing thermal control on a satellite, satellite IR emissions are considered in the IR waveband as emissivity. The sun though tends to considered through the whole band as absorbtivity.They can differ by an extremely large ammount, as well. White paint for instance has a low absorbtivity, but has an emissivity almost equal to that of black paint.

Testing foil and other materials for their values would be an interesting experiment to set up, though. Personally I think i'd consider patch heating for the eclipse side, as it takes suprisingly little power to do so.

The risk in letting temperature vary too freely is battery or solder failures, given the effort to get the satellite there it isn't worth the risk if there are simple fixes to the problem.

[lavalamp]

[Delta_V]

Hello Lavalamp,

How did you derive the model of radiation from the earth's surface? (220 W/m2 on night side, 220 W/m2 +110 W/m2 sin(theta) on day)? Do you have any sources for more accurate models? Otherwise I might just average the diurnal variation of temp at various points on the earth's surface and the oceans, find an average emissivity, and try to find some sort of albedo for the surface and clouds to get a more accurate model. Any ideas?

Delta_V

[lavalamp]

I seem to recall I sat down and had a think about it, then calculated some numbers. As it turns out, they were calculated incorrectly since they are too low, but most heating comes from the sun so it doesn't matter toooooo much.

There are MUCH more accurate models available though. A good one is the Clouds and Earth's Radiant Energy Systems (CERES) Energy Balanced and Filled (EBAF) Top-of-Atmosphere (TOA). Link:
https://climatedataguide.ucar.edu/clima ... and-filled

In this model, the surface of the earth is divided into 1 x 1 degree cells and each is given a flux value for both long wave (essentially infra-red) which you can assume to be constant day and night, and short wave (essentially reflected visible light) which you can vary with some cosine function during the day and set to 0 at night.

If you take some of these cells that are directly below the spacecraft within a 45 degree cone say (the number of visible cells will vary as a function of the altitude of the satellite), and add up their fluxes, you'll probably be pretty close.

If you want to be more accurate, you could open up your cone below the satelite all the way until the sides make a tangent with the edge of the earth, this will now contain every visible cell and you can scale the individual fluxes from these by the viewing angle away from the normal, assuming that each cell is a diffuse emitter. This will be somewhat more computationally intensive of course.

Likely this model will be WAY overkill for simply getting a rough idea of the temperature of a satellite though. Like I said earlier, the sun is by far the largest factor.

[Xan]

1. The averaged temperature of the satellite.
2. Temperature variations.

1.

Accurate calculation of radiation fluxes from the Earth and the Sun is the "looking for fleas".

The most significant influence on the temperature of the satellite have coefficient of absorption of sunlight and coefficient of radiation in infrared.

Metals with a smooth and clean surface have a low coefficient of radiation in infrared.
Example:
Aluminium 0.03 ... 0.06
Steel 0.07
But steel with coarse finish surface 0.24

Other materials (mineral and organic) have a large coefficient of radiation in infrared.
examples:
rusty iron 0.7
fireclay bricks 0.85
paper 0.8 ... 0.9
varnish and paint 0.85 ... 0.95.
Exception — aluminum paint 0.28.

Coefficient of absorption of sunlight is clearly visible to the naked eye.

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The thermal radiation of the earth = 239 W/m2 at zero altitude.
If the height of the satellite's orbit = 350 km, the Earth takes up 42 percent of the area of the sky, not 50%.
Then the flow from the Earth would be = 200 W/m2.

If a satellite has a spherical shape, then:
its surface, which absorbs sunlight is equal to pi * R^2 (cross section);
its surface, which emits infrared is equal to 4 * pi * R^2 (full surface);
its surface, which absorbs infrared of the Earth, is equal to 42% * 4 * pi * R^2 (part of surface).

I do not consider the flow of the visible light reflected from the Earth (nealy 100 W/m2). It can simply be added to the solar constant, which is equal to 1400 W/m2.

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If the satellite is constantly illuminated by the sun and the absolute black (k_visible = 1, k_infrared = 1), the balance of power would be:

k_visible * 1400 * pi * R^2 + k_infrared * 239 * 0.42 * 4 * pi * R^2 = k_infrared * 4 * pi * R^2 * 0.0000000574 * T^4

where 0.0000000574 - Boltzmann constant.

T^4 = (k_visible * 1400 * pi * R ^ 2 + k_infrared * 239 * 0.42 * 4 * pi * R ^ 2) / (k_infrared * 4 * pi * R^2 * 0.0000000574)

T = sqrt4(k_visible * 1400 + k_infrared * 239 * 0.42 * 4) / (k_infrared * 4 * 0.0000000574) = sqrt4((1400 + 400) / 0.0000002296) = 298 K = 25 C

25 C — it is room temperature.


If a satellite was constantly in the shade, its temperature would be:

T = sqrt4(0 + k_infrared * 239 * 0.42 * 4) / (k_infrared * 4 * 0.0000000574) = sqrt4((400) / 0.0000002296) = 204 K = —69 C

But "constantly in the shade" is not real. Maximum for real orbit is 40% in the shade.
So, for 60% illuminated by the sun:

T == sqrt4((0.6 * 1400 + 400) / 0.0000002296) = 271 K = —2 C

Also quite a decent temperature for electronics.

The same result would be for the satellite, colored by dark paint, because ratio k_visible / k_infrared = ~ 0.8 / 0.8 = ~ 1.

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If the satellite is the sphere from aluminium (k_visible = 0.18, k_infrared = 0.04), the temperature would be:

T = sqrt4(0.18 * 1.0 * 1400 + 0.04 * 239 * 0.42 * 4) / (0.04 * 4 * 0.0000000574) = 413 K = 140 C (100% sun)

T = sqrt4(0.18 * 0.6 * 1400 + 0.04 * 239 * 0.42 * 4) / (0.04 * 4 * 0.0000000574) = 367 K = 94 C (60% sun)


White paint (k_visible = 0.25, k_infrared = 0.85):

T = sqrt4(0.25 * 1.0 * 1400 + 0.85 * 239 * 0.42 * 4) / (0.85 * 4 * 0.0000000574) = 244 K = —29 C (100% sun)

T = sqrt4(0.25 * 0.6 * 1400 + 0.85 * 239 * 0.42 * 4) / (0.85 * 4 * 0.0000000574) = 231 K = —42 C (60% sun)

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So, there is a strong dependence of the temperature from the ratio k_visible / k_infrared.
The best result is obtained at about ratio = 1.
And easy to get the desired temperature, regulate ratio of these coefficients.

-----

2.

To be continued... ?

[Xan]

2.
For aluminum ball with a diameter 2.4 cm (weight 19.6 grams, the heat capacity of 16.7 joules per degree), painted with black paint, the temperature will range from minus 64 to minus 36 degrees.

For an aluminum sphere with a diameter of 5 cm (the same weight), which has 13% of the surface is painted with black paint, the temperature will range from +1 to +53 degrees.

If the paint 9% of the white paint, the temperature will range from +3 to +39 degrees.
If the paint 7% of the white paint, the temperature will range from +12 to +48 degrees.

Conclusion.
I think everything is clear!

It makes sense to do most of the surface of the metal (aluminum), and to adjust the temperature of a small portion of the surface of the paint with white paint.

PS
In fact, to reduce temperature fluctuations, it makes sense to do a double casing.

[Delta_V]

Thanks you guys, all very, very useful.

Delta_V