1
$\begingroup$

A simple air compressor compress carbon dioxide/water saturated helium/oxygen air in a tank to liquefy water from the air by pressure. Abstract the water, then refrigerate the carbon dioxide/helium/oxygen air in the tank by having a coil and tank in space. The $CO_2$ will liquefy at -56 Celsius or so depending on pressure.

How would the carbon dioxide be captured as a liquid and the helium and oxygen gas released back into the cabin?

Would there be any benefit to this method over chemical scrubbing?

|improve this question
$\endgroup$
  • 3
    $\begingroup$ You are essentially describing a cold trap. I think this must be possible, but there is one obvious problem, and a number of better techniques. The obvious problem is water. If the air is not totally dry the cold trap will fill up with ice, which interfere with gas flow, valves, etc. The better techniques use various chemicals which react reversibly with CO2. Typically they absorb it at room temperature and release it when heated. These approaches are cheaper and simpler. $\endgroup$ – Steve Linton Nov 25 '18 at 20:10
  • $\begingroup$ I'm trying to understand the starting atmosphere. Is it He + O2 + CO2 ? Or do you have a particular planet's atmosphere in mind? $\endgroup$ – DrSheldon Nov 25 '18 at 23:43
  • $\begingroup$ @DrSheldon yes that is the atmosphere $\endgroup$ – Muze Dec 29 '18 at 16:05
  • 1
    $\begingroup$ This is a significantly changed description of the process. One issue is that the tank, even if kept in the shade will not cool quickly. It can only lose heat by radiation, and even with large radiator fins it wil take a long time to get cool enough. $\endgroup$ – Steve Linton Dec 29 '18 at 16:28
  • 1
    $\begingroup$ Another issue is that saying $CO_2$ liquefies at -56 Celsius is an oversimplification. At that temperature the vapour pressure of liquid $CO_2$ is about 5 atmospheres, so your liquid would be in equilibrium with about 5 atm of gaseous $CO_2$ together with however much $O_2$ and He. If you want the proportion of $CO_2$ in the gas part to be (say) less than 0.1% (2.5 times Earth;s air) you would need about 5000 atmospheres of oxygen and helium, making the compressor and tank less than simple. 13 atmospheres at -120C might be a more reasonable target $\endgroup$ – Steve Linton Dec 29 '18 at 16:33
2
+50
$\begingroup$

Saying $CO_2$ liquefies at -56 Celsius is an oversimplification. At that temperature the vapour pressure of liquid $CO_2$ is about 5 atmospheres, so your liquid would be in equilibrium with about 5 atm of gaseous $CO_2$ together with however much $O_2$ and He. If you want the proportion of $CO_2$ in the gas part to be (say) less than 0.1% (2.5 times that in Earth's air) you would need about 5000 atmospheres of oxygen and helium, making the compressor and tank less than simple. The same link gives data for other temperatures, for instance at -120C the vapour pressure is 0.013 atm, so compressing your air to 13 atmospheres and cooling to -120C might be a more reasonable target

A further issue is the cooling. Compressing the gas will heat it considerably. That heat could, perhaps be lost by conduction back to the rest of the spaceship (although that really just shifts the problem). After that, if you don't want to put in still more energy for refrigeration, the best you can do is put it in a black container with lots of radiator fins and put it on the shady side of the spaceship to cool by radiation. This is not especially effective, especially as the container gets colder. For instance cooling by radiation at 200K (-73C) assuming no incoming energy at all, loses about 80 $W/m^2$ (using the formulae from here). The specific heat capacity of helium is around 3000J per kilogram Kelvin (at constant volume), so it's going to take a while to cool down much gas.

|improve this answer
$\endgroup$
  • $\begingroup$ Can the carbon dioxide be cooled at the rate of production? $\endgroup$ – Muze Dec 29 '18 at 17:01
  • $\begingroup$ @Muze obviously yes, if you have a big enough reserve of helium and oxygen and big enough cooling tanks, compared to the amount of production. I would guess that the equipment for chemical scrubbing is likely to be simpler, lighter and cheaper though. $\endgroup$ – Steve Linton Dec 29 '18 at 18:36
  • $\begingroup$ Would it be easier to 3D print out compressor parts or to manufacture the chemicals in the long run? $\endgroup$ – Muze Dec 29 '18 at 19:25
  • $\begingroup$ @Muze - some of the CO2 scrubber compounds release CO2 when heated, so adding heat and lowering the pressure is all that is required to get near unlimited life. Tricky bit at the moment is to do something with the resulting CO2. $\endgroup$ – GremlinWranger Dec 29 '18 at 23:32
  • 1
    $\begingroup$ @Muze doesnt change the fundamentals a given area can only radiate so much. If you use find to increase the area you do have to make sure that the fins and the thing you are cooling all get to the same temperature. On Mars you have to worry about the fact that the atmosphere is too warm for the temperatures we were considering. $\endgroup$ – Steve Linton Dec 30 '18 at 10:06
4
$\begingroup$

It's possible, but not simple or cheap. This process is called air separation and is the primary method for producing (liquid) nitrogen and oxygen.

Pure gases can be separated from air by first cooling it until it liquefies, then selectively distilling the components at their various boiling temperatures.

In ambient air, there's only a tiny fraction of CO2 (400 ppm) so you need to process large amounts of air to get a tiny amount of CO2.

|improve this answer
$\endgroup$
  • $\begingroup$ "Pure gases can be separated from air by first cooling it until it liquefies, then selectively distilling the components at their various boiling temperatures." That requires a temperature range where all gases are liquid. But CO2 is solid at the temperature of liquid air, that is why CO2 is separated before liquefing the remaining air. Solid CO2 would disturb the process of liquefaction of air. $\endgroup$ – Uwe Nov 25 '18 at 21:27

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.