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I'm trying to model the temperature of a large spacecraft for a space colony simulation game I'm working on. In another question, I checked my calculations for the steady-state black-body temperature of an object, considering only insolation and radiation, and it appears I'm on the right track.

My understanding is that this black-body temperature formula works only for passive bodies with no active heating or cooling. Now I want to add active heating and cooling elements. But how?

For cooling, I think I can model radiators as simply increasing the surface area of the craft, with no significant change to insolation (since radiators are placed edge-on to the sun). Please correct me if I'm wrong on that.

For heating, I'm stumped. I can increase the amount of energy dumped into the system, by presuming a nuclear reactor or beamed power or some such, but when I try that, the effect is much smaller than I would expect. I end up having to dump many MW of power into a large craft just to raise it up to room temperature.

So I'm wondering: does it matter how the extra energy is used within the system? Is a kW poured into a big electrical space heater going to get things hotter than a kW spent twirling a beanie, and if so, how?

As a possibly related question, it's claimed that the greenhouse effect significantly raises the temperature of a planet -- for example, Venus's black-body temperature would be 330 K, but due to atmospheric warming, its actual surface temperature is 740 K (*). How is this possible? Isn't it Q_out = Q_in, no matter what? And however this works for Venus, can we do the same thing to warm our spacecraft?

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  • $\begingroup$ A Watt is a Watt is a Watt. In case of the atmosphere you are simply under a blanket that is not a black body radiator and you have to ray-trace the energy balance from atmospheric layer to layer until your photons finally escape into space. There are publicly available codes for that, but they are of no importance to spacecraft design, where radiators are really just a grey body problem. $\endgroup$ – CuriousOne Sep 16 '14 at 21:47
  • $\begingroup$ @CuriousOne - There is a direct analogy to an atmosphere on many spacecraft. It's the Multi-Layer Insulation, that gold-looking stuff that covers many spacecraft. The many layers of reflective material, each separated by vacuum, that comprise MLI blanket act to keep heat trapped within. (It also keeps unwanted heat from getting in.) $\endgroup$ – David Hammen Sep 16 '14 at 22:25
  • $\begingroup$ @DavidHammen: Yes, there is, but one does not usually model that with a radiation code. If the internal temperature of a spacecraft module is important, it is either structurally tied to a radiator or it gets a custom built heat pipe. Those modules usually also have electrical heaters attached. In some cases the problem is solved with a thermally actuated louver. Thermal design of any spacecraft is done very early in the design, when the thermal and electrical budgets are allocated, and it's either done with a spreadsheet or custom software. $\endgroup$ – CuriousOne Sep 16 '14 at 22:36
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Interesting and complicated question. The things to consider:

"Black body radiation" assumes perfect absorption / radiation at all wavelengths. The greenhouse effect comes about from having absorption in the IR: the hot (short wavelength) radiation from the sun can penetrate the atmosphere, but the cooler earth radiates at a lower temperature - longer wavelength. And that longer wavelength light is reflected by the atmosphere (water, carbon dioxide, methane, etc). Think of the story of Winnie the Pooh visiting Rabbit's burrow. He goes in through the hole, has a "little smackerel" of honey (read: the whole pot), and then is too fat to get out again - commemorated on a postage stamp:

enter image description here

That's your photon. It had no difficulty penetrating the atmosphere as a short wavelength photon - but as a long wavelength photon it gets stuck as it tries to leave earth...

If you worry about your spaceship getting too cold (how big is it?) you should probably consider lowering its reflectivity - this directly scales with the heat loss. Notice how space ships are often "shiny metal". This isn't just because paint is expensive to lift into orbit (it is), but also to keep the emitted power down - keep the people inside protected from too much heat loss when not in the sun, and too much heat gain when they are. If you want to simulate the "Venus" effect you would want to create your own greenhouse effect - add a film that is transparent in the visible and opaque in the near IR.

Either way, your black body model needs to take account of the reflectivity as a function of wavelength - and instead of using the simple Stefan-Boltzmann law (which deals with total power per unit area), use the wavelength formulation (Planck law):

$$S_\lambda=\frac{2\pi hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}$$

But yes - the amount of heat that a large object loses through radiation is substantial, even when it is at room temperature. There's a handy calculation on wolframalpha.com - it shows that the heat loss for an emissivity of 0.1 is still over $40 W/m^2$ at 298 K. The best thing you can do to insulate yourself is not let the outer shell get so hot in the first place - if you used a double shell, with the outer being thermally insulated from the inner, then you can see how this will lower the power emitted at the outer shell comes to equilibrium at some temperature $T_o$.

Assuming that the outer shell reflects half its power back to the inner shell, and half to the universe (which is so close to absolute zero that we ignore the difference), you can write

$$\epsilon \sigma T_i^4 = 2 \epsilon \sigma T_o^4$$

since the outer shell loses heat from both surfaces; thus if the inner shell is at $298 K$ the outer shell temperature will be at 250 K, but the inner shield is now losing heat at

$$\epsilon \sigma (T_i^4-T_o^4) = \frac12 \epsilon \sigma T_i^2$$

In other words - you halved it. If you add additional skins, the heat loss will be further reduced.

I must admit that I did that last bit of analysis "by the seat of my pants". It makes sense, intuitively, that the heat loss is reduced by a radiation shield; I have never attempted to come up with a number before, nor do I remember seeing this analysis. There could be a blooper in here - in which case I would be happy to have someone point it out.

I did find an online book that seemed to follow a similar approach but had a cylindrical geometry and uses different reflectivity on the inner and outer faces, which complicated matter further. But they show that multiple layers of shielding can significantly reduce these heat loads - which was really what I tried to say.

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  • $\begingroup$ @ChrisWhite - thanks for your inputs. It's possible that you need the emissivity to be different but I didn't think so. I was calculating that the (colder) outer shield radiates half its power back to the space ship and the other half out (it has twice the surface area, basically). That's where the factor two in my equation came from; that is why the outer shield is colder, and since that's the only one that is facing space, it must be resulting in less power lost. Just found en.wikipedia.org/wiki/Multi-layer_insulation - it agrees with my analysis (see "Function and design" section). $\endgroup$ – Floris Sep 16 '14 at 22:52
  • $\begingroup$ Yes - it should have been $T_i$ thanks for catching that. Fixed it... $\endgroup$ – Floris Sep 16 '14 at 23:32
  • $\begingroup$ In your second line, are you sure it's "perfect absorption/reflection"? I think it should be "perfect absorption/radiation". $\endgroup$ – surelyourejoking Sep 17 '14 at 0:04
  • $\begingroup$ Thanks, this really helps. I see now how insulation (whether it's an atmosphere or something else) breaks the assumption of black-body radiation. I'm still confused about why (or whether) E_in <> E_out in the steady state, though, but I'll mull it over more and see if it clicks. $\endgroup$ – Joe Strout Sep 17 '14 at 23:19
  • $\begingroup$ OK, the Wikipedia link on multi-layer insulation was enlightening. (And fascinating -- I always thought vacuum was the best possible insulator, but now I see that's not true!) Now, I'm just puzzled as to how to apply this to my problem (physics.stackexchange.com/questions/135810). I'm calculating the surface temperature with T_surf = (T_space^4 + Q_in/(e s))^(1/4). Now what? Can I just plug this into the formula you give above to find the internal temperature? I suspect it's more complex than that, but I don't quite see how to put it all together. $\endgroup$ – Joe Strout Sep 17 '14 at 23:38
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You can combat that the same way we avoid getting too cold: apply insulation. The outer surface of the spacecraft may be very cold, but that doesn't mean the internal temperature is that cold. That is what the greenhouse effect does-it insulates the surface from space.

Power dissipated inside the body all becomes heat. If you have solar arrays making power, all the power they make except for what you dump overboard (for example as radio broadcast or electrical thrusters) goes to heat the spacecraft.

How large a craft takes MW to keep warm? Today's communication satellites are running a few to 20 kW and staying room temp or so inside.

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  • $\begingroup$ I'm modeling very large spacecraft -- hundreds of meters in radius. So, yeah, it's a lot of power. But I haven't modeled any effect of insulation, because I don't know how. I see how insulation makes it take longer for an object to reach equilibrium. But in the long run, isn't it going to end up a constant temperature through and through? Or maybe that's true only once no more energy is being put in... I can imagine a temperature gradient being maintained, without violating E_in = E_out. Not sure how to model it, but I can picture it, and that's a start! $\endgroup$ – Joe Strout Sep 17 '14 at 23:22
  • $\begingroup$ If you dissipate heat inside the body, you can maintain a temperature differential, just like your oven at home. You can model it with a conductivity between the various layers (or nodes in a 3D model). Conductivity is measured in W/K (if you are working in layers) or W/m^2/K (if you consider the area). In any case, a low conductivity can maintain a temperature differential as long as heat is being input into the center (say from solar arrays) $\endgroup$ – Ross Millikan Sep 18 '14 at 0:34
  • $\begingroup$ I feel like I'm having dinner in France, and oh-so-close to being able to understand what the waiter is saying! Here's what I know: (1) the energy absorbed at the surface from insolation; (2) the total energy dissipated inside the body; (3) the surface area (and volume, if it matters) of the body. I can probably get (4) some value for the conductivity of the body shell (I know the thickness and material). Using these, I ought to be able to calculate the internal temperature... but how? $\endgroup$ – Joe Strout Sep 18 '14 at 17:26
  • $\begingroup$ Perhaps I can simplify it a bit by calculating the steady-state temperature of the outer surface, ONLY from insolation. I know how to do that. Then we consider energy dissipated inside. This is where I'm lost... how do I calculate the temperature inside, given the outside temperature, inside energy, and thermal conductivity in between? $\endgroup$ – Joe Strout Sep 18 '14 at 17:28
  • $\begingroup$ The outer surface temp considers the dissipation inside as well. You have two inputs-solar and dissipaton. Now find the surface temp that balances that with radiation. Then find the temperature difference from the inside to the surface by dividing the heat flow (the dissipation) by the conductivity. Add that to the outside temp and you have the inside temp. To really do it, you should have a model with lots of nodes, because the outer surface will not be all at the same temp. Then a conductivity between every close pair of nodes and a big model to solve. $\endgroup$ – Ross Millikan Sep 18 '14 at 20:26
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OK, I think I've got it, thanks to your comments above as well as this link, which shows how to calculate the temperature of a solar oven. (My situation is very similar to a solar oven, except that the power dumped inside the craft is electrical -- but watts are watts, right?)

So, I believe that what I need to do is:

  1. Calculate the steady-state temperature of the outside of the craft, as described here, but considering only insolation (no internal energy dissipation).
  2. To calculate internal temperature, observe that P_out = P_in in the steady state, and then apply this critical formula: P_out = U A (T_in - T_out), which describes the power leaving as a function of U (the combined heat transfer coefficient of the walls), A (the area of the walls), and the temperature difference. Using P_out = P_in and solving for T_in, I get T_in = T_out + P_in / (U A).
  3. Now I need only plug in the power dissipated inside the craft, and use the skin temperature found in step 1 for T_out, and I can find T_in.

This all makes sense to me, but I'm obviously no physicist. If anybody sees a mistake here, please let me know!

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