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Theoretically, if we have system like this enter image description here

and we put it inside Cryogenic storage dewar (thermos) that energy does not go outside we can store any amount of energy and we can get it back, when we need.

I think it is true because the adiabatic process is reversible.

Of course, in real life, we will lose something, let say 10%, because we cannot get ideal thermoisolation, we will have friction, but it is still should be very efficient.

Why is it not in use instead of lithium batteries etc? Or my consideration is wrong? we can put small tube and piston with electric generator/motor as AA battery size and that it.

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    $\begingroup$ Extracting energy reliably from that would be quite complicated (and you'd lose a lot of efficiency in basically whatever you do). Also, just as importantly; the energy density of such a system is likely extremely small compared to a battery. I don't think you would want to carry a device like this for your phone (since it would likely weigh 100's of pounds for the same energy storage/use). $\endgroup$ – JMac Jun 13 '17 at 12:08
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    $\begingroup$ It is used: Compressed air energy storage $\endgroup$ – Karsten Kretschmer Jun 13 '17 at 12:14
  • $\begingroup$ Why is it complicated to extract energy? We put motor which pushes the piston when we store energy it is almost 100% efficient (let say 90% as written here en.wikipedia.org/wiki/…). When we need this energy back, we use the same motor as a generator. Piston will push it back and rotate. Efficiency is high as well. $\endgroup$ – Zlelik Jun 13 '17 at 12:59
  • $\begingroup$ but when I calculate energy for monoatomic gas I see that you are right. If I have like half of AA battery gas and compress it 10 times I will get only 2 Joules of energy which is for 1.5v battery is 0.0004 mAh. to get 500 mAh I will need to compress it 10000000000 times, pressure will be 4.6E+16 atm, which is not possible technically of course. Well, it was a stupid question. $\endgroup$ – Zlelik Jun 13 '17 at 12:59
  • $\begingroup$ @KarstenKretschmer good link, Answers all the questions. $\endgroup$ – Zlelik Jun 13 '17 at 13:09

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