3
$\begingroup$

Related to Are there any promising next-generation space station cooling technologies?, I'm wondering specifically about a technology that hasn't been mentioned there: Zeeman slowers. Since these are a) used to cool atom beams from at least 290 K to 5-10 K (if not lower) and b) operate without convection or conduction by triggering increased thermal radiation from cooled atoms, it seems that they would be not only suitable for general-purpose cooling in space, but if anything, a good deal more efficient there compared to expansion-compression cooling systems than on Earth, since they skip the whole "pumping heat to enormous radiators" step. And they've been in use for something like three decades in lab setups. What am I missing?

$\endgroup$
  • 1
    $\begingroup$ In what sense are you asking this? If you're asking about thermal management of specific parts of the system then yes Zeeman slowers could work. If you're asking about reducing entropy of the system as a whole, then no, laws of thermodynamics prohibit that. Zeeman slower will by necessity produce more heat than it will remove it, and you end up with even more heat that needs to be radiated or advected (transported) away. $\endgroup$ – TildalWave Nov 8 '15 at 2:22
  • 1
    $\begingroup$ @TildalWave: Waste heat is created by any heat rejection system more complicated than the strictly passive. That's not an inherent thermodynamic obstacle to rejecting heat more efficiently than pumped radiators either in terms of waste heat generated, or in terms of mass used. So yes, a self-cooling Zeeman slower setup would be forced to radiate more heat than the actual system load, but that's not really a theoretical obstacle, any more than it's a problem that the pumps that circulate ammonia in the ISS generate heat. $\endgroup$ – Nathan Tuggy Nov 8 '15 at 2:27
  • $\begingroup$ Zeeman slowers are currently in use to cool down tiny numbers of atoms (less than a nanogram, IIRC) to very low temperatures. I'd be surprised if it could be scaled up enough to satisfy the cooling needs of a space station. $\endgroup$ – Hobbes Nov 8 '15 at 8:06
  • $\begingroup$ @Hobbes: That's a fair point, so I decided to split this up a bit to focus on the two different aspects, so this one can stick with my original (perhaps more dubious) question, and the other can focus on cryogenic setups. $\endgroup$ – Nathan Tuggy Nov 10 '15 at 1:11
2
$\begingroup$

To carry the answer from the other thread over here as well:

No, Zeeman slowers cannot be used as space station coolers

The reason is that the heat flow they provide is in the nanoWatt range.

So no, they are not "close" to being used as general cooling systems in space, for the same reason that placing a home-made water wheel under your kitchen faucet is not "close" to providing you free energy for your home. Sure, you might get a LED to shine faintly but that's about it.

EDIT: Question was asked "Why can't this be scaled up"?

Because there are very narrow tolerances regarding velocity (speed and direction) as the atoms enter the tube. If you have a low density beam then this is fairly easy to maintain, because the atoms are not interacting. But once you scale it up to have a mass flow that is significant enough to provide the amount of cooling that is needed, then we no longer have a particle beam of atoms but a flow of gas or liquid. This means the atoms are bumping into each other and Browninan motion destroys any chance of meeting the narrow tolerances regarding the entry velocity.

$\endgroup$
  • $\begingroup$ The reason a waterwheel under a faucet does not provide free energy is because you're paying for the energy in the water. It has nothing at all to do with the scale of the energy you can pull from it. $\endgroup$ – Nathan Tuggy Mar 31 '16 at 20:32
  • 1
    $\begingroup$ While I pay for the hot water, I get the cold water for free. And this argument is totally beside the point. You still cannot run your house on that energy. 1 mW at 0 EUR does not scale up to 1 kW at 0 EUR. And such is the case with Zeeman slowers. $\endgroup$ – MichaelK Mar 31 '16 at 21:06
  • 2
    $\begingroup$ It would seem the question is why can't they be scaled. $\endgroup$ – kim holder Mar 31 '16 at 21:11
1
$\begingroup$

Too cool down the station, we must transport energy away from the station. Period. That can be done through radiation, or by letting some substance boil of.

The Zeeman slower does neither of that, and is then strictly not usable for "general cooling". In fact, it produces more heat as it uses electricity.

However:

Changing what parts of the spacecraft are hot or not might still have an effect. For instance, if the radiators are hotter, they radiate more energy. The Zeeman slower only makes the surroundings a little hotter, so traditional expansion-compression systems do that part a lot more efficient. The main advantage of the Zeeman slower is to cool a tiny amount of atoms to a really low temperature (just a few Kelvin), so it can be used to cool components that needs to stay really cool. (like super conductors or space-telescope optics). Note that the output of cool atoms from the beam is very small, so any components cooled by it have to be extremely well thermally isolated.

$\endgroup$
  • $\begingroup$ That's a decent analysis, although as far as I'm aware, the slower actually does radiate heat away, and not necessarily to any specific component of the system. It's this observation that made me consider this in the first place. I'll see if I can clarify the question. $\endgroup$ – Nathan Tuggy Apr 1 '16 at 0:27
  • $\begingroup$ @NathanTuggy Of course the slower can actually stick out of the side of the spacecraft, and thus act as radiator. However, that is not any more than a very expensive piece of added surface area. $\endgroup$ – Hohmannfan Apr 1 '16 at 0:34
  • $\begingroup$ It's the cooled atoms that are radiating primarily; directing that radiation out would then be more efficient than a normal radiator for the same temperature of coolant. Of course if the slower structure gets in the way of the initial radiation that benefit is lost. $\endgroup$ – Nathan Tuggy Apr 1 '16 at 0:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.