In thread about computers in a vacuum, it was pointed out that a vacuum is a very good insulator. So why did the Apollo 13 spacecraft get so cold so fast when they powered it down? Shouldn’t it have held its heat very well? Or was this just in the movie?
While vacuum is a good insulator, it is not a perfect one, so things will still reach thermal equilibrium.
Since the earth’s atmosphere warms the planet about 40º or so through the greenhouse effect, and the average temperature on earth is around 60ºF or so ( 57 sounds right to me)I assume that the equilibrium temperature for an object in earth orbit with no insulating blanket of air would be below freezing. Obviously, the reflectivity of the object would make a big difference, but the Apollo probes looked like they had a high albedo to me, which would make them even colder.
The astronauts and the equipment that were still on were generating heat, and the surface of the ship was losing it faster until it got pretty cold and was in thermal equilibrium with space.
There are four mechanisms for heat transfer.
Conduction – in a solid object or in solid objects in contact, heat flows in proportion to the temperature gragient. (Ok, liquids and gases will conduct too, but since they are prone to movement it is a bit more complex.) A vacuum will not conduct because there is not contact. Yaay for vacuums.
Convection – heat passes between a solid and a fluid (liquid or gas). The fluid moves and transports the heat energy away from the surface. Convection can be further categorised into forced convection (eg, a fan) or natural convection (convection currents are set up due to changes in density and the action of gravity.) No gas in a vacuum, so you are safe from this in space too.
Phase change – a substances change from one state to another they either release or absorb energy. Hold your finger down on the flyspray for a while and see how cold it gets. But this wasn’t happening with Apollo 13 either.
Radiation – heat is transferred by infra red radiation between surfaces. The transfer rate is proportional to the fourth power of the difference in temperature. So, if there is a 2 degree difference in temperature, then heat transfer is proportional to 16. If there is a 10 degree difference, then the heat transfer coefficient is multiplied by 10000 – 625 times as much. Apollo 13 is in space where the ambient temperature is about 3K if I recall correctly. The guys inside were hoping to stay around 293K (20 degrees C or 68 F) There’s a whopping 290 degrees temperature difference. Hence heat loss by radiation is a bit of a problem. It was probably not quite as bad as that because they were in line of sight to the sun for much of the time and were receiving some incoming radiation. But the shadow of the moon would have been a tad chilly. The best solution to reducing this loss is to have highly reflectorised surfaces so that the radiation bounces back. And that’s why they made it shiny.
In laymans terms, think about a red-hot poker. It radiates light as well as heat. Cooler things, like your body, also radiate light(!)(*) and heat, but at lower levels and at frequencies which your eyes cannot see.
- Light used loosely, I really mean electro-magnetic energy
Another factor may have been all those frozen hot dogs they took with 'em.
Just a wild guess here, but wouldn’t the thermal conduction between the crew cabin and the liquid oxygen/hydrogen in the fuel tanks have cooled things down in short order? Or were the tanks sufficiently insulated to prevent this, even with the heaters off?
There was no liquid Hydrogen aboard Apollo and the LOX was for breathing and fuel cell power. The main engine on the Command/Service module (as well as the descent stage of the LM) was hypergolic, i.e. the two chemicals ignited the moment they mixed.
And the LOX storage tanks were highly insulated otherwise the oxygen wouldn’t stay liquid very long!
If the LOX is liquid, how do you get it to stay on the bagel?
That reminds me of one of my science epiphanies. How does the earth not burn to a crisp, seeing as how every day the sun pumps in more light that gets converted to heat? There’s no wind to take away the hot air into space, and we’re not touching anything to pass along heat.
The answer is that night side of earth is always pouring out infrared light. That takes energy to do, and you get a 20-30 degree temperature swing between day and night.
Is there something wrong with this statement? Isn’t a reflective exterior designed to prevent the temperature from RISING too fast from solar radiation? How would it prevent any loss?
It all depends upon how your reflector is designed. A one way mirror blocks only a portion of the light travelling in one direction, but all the light traveling in another direction. If your reflector is designed in a similar fashion, then it’ll prevent all the solar radiation from entering into the system, but it won’t block all of it from escaping. The Apollo CSM was designed to loose heat, in order to keep the electronics from frying, as well as block heat from the sun. So when they closed the blinds in the CSM, it blocked heat from getting in, but since the CSM was designed to shed heat as well, what heat was in the capsule leaked out as well, and for reasons of power management, they couldn’t turn the heaters on to maintain an even temperature.
Hail Ants, there had to be liquid hydrogen aboard Apollo, because a fuel cell works by combining LOX and LH to generate electricity. I don’t have time to dig for a cite at the moment, but there most definately was LH aboard the Apollo craft. IAC, here’s an article on fuel cells
Yes, I understand that you can vary the reflector design, More specifically, I was responinding to the statement that the space craft looks shiny (I presumed this meant the outer surface rather than the inner one). What I’m saying is that I imagine that exterior reflectivity has more to do with preventing overheating than preventing cooling.
True, but the Apollo CSM was designed to shed heat, and this couldn’t have been shut down when they had to power down almost everything in order to conserve power. So you have a system which is designed to throw off heat at a certain rate, and then you reduce some of the heat which is being put into the system (shutting down most of the systems would have reduced much of the heat generated by the electronics), but don’t reduce the amount of heat being shed by the system by the same amount, then you’re going to end up with a system that’s going to cool down and stay cool.
I agree entirely with what you’re saying. That appears to put both of us at odds with the statement of** j_sum1** that I quoted. Or am I still missing something?
Well, I can’t speak for j_sum1, so I can’t say for certain what he/she meant. I know that NASA had to deal with not only the sunlit side heating up, but the side in darkness cooling off. One of their solutions was ferrofluids to help equalize the temp, but IIRC, NASA abandoned the idea for Apollo and put the CSM on a slow rotation. Now, when the CSM went around the Moon’s far side it wouldn’t have undergone any heating whatsoever, and would have chilled dramatically. The fact that it was reflective would have blocked any feeble infrared radiation (if any) coming from the far side (which does get exposed to sunlight on a fairly regular basis).
Can we please call this the Moon’s night side? Just to avoid any confusion.
This isn’t quite right. The fourth power of the temperature difference between two objects isn’t particularly interesting. Radiated power does scale with the fourth power of the temperature, but that just means that an isolated object at 30 deg C only looses heat 14% faster that that object at 20 deg C ( (303 K/293 K)[sup]4[/sup] = 1.14 .)
In Apollo’s case, the input heat flux is dominated by the sun. Some numbers:
- At earth, the sun’s radiation amounts to about 1400 W/m[sup]2[/sup].
- A 293 K black body radiates heat with a flux of [symbol]s[/symbol]T[sup]4[/sup] = (57 nW/K[sup]4[/sup]/m[sup]2[/sup])(293 K)[sup]4[/sup] = 420 W/m[sup]2[/sup].
Assuming a spherical blackbody Apollo 13 of emissivity [symbol]e[/symbol]=1 radius r, the net heat flux out would be:
F = (420(4[symbol]p[/symbol] r[sup]2[/sup])) - 1400([symbol]p[/symbol] r[sup]2[/sup]) W/m[sup]2[/sup] = (880 W)(r/m)[sup]2[/sup]
Taking r = 3 m (??), a sunlit spherical black Apollo looses heat at about 8 kW.
I have no idea what the emissivity of the Apollo surface was or what its absorption spectrum looked like or what the conductivity of the outer walls were. All of these factors are terribly important but highly tunable by the designers. I’d guess that the designers chose materials such that the net radiative heat loss matched the internal heat production (electronics, lights, fans, humans…) Given this, the rate of cooling would be equal to the typical power output of the devices that were shutdown (several kW easily). No flux calculation needed.
I see that Tuckerfan has already pointed this out. Shoulda read your posts more closely…
But for half of every month it’s the **day **side.
Far side makes more sense, from our geocentric POV.
And since Apollo missions were scheduled so that there would be daylight at the landing site, they’re equivalent in this situation.
I know Apollo 11 was scheduled so the angle of the sun at the landing site maximized contrast for the landing. Does anyone know if the other missions were scheduled the same way?