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It occurred to me that all of the kinetic energy not going exactly in the opposite direction of a spacecraft is wasted! Since heat is essentially kinetic "noise" in objects, I wondered how much more boost one could get from eliminating all heat from the exhaust.

What would the delta-v of say, a falcon 9 look like if its raptor engines were equipped with a variation on Maxwell's Demon, and only allowed particles moving exactly the opposite direction to leave the nozzle? A "Cryorocket" or "Kelvin Drive" if you wanted something that sounded like sci-fi.

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    $\begingroup$ It would look like kaboom! You're going to greatly increase the chamber pressure and the heat, the rocket won't be able to take it. $\endgroup$ – Loren Pechtel Dec 14 '18 at 2:04
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Partial answer to a very cool question! While the pun was not intended, I'll take credit for it anyway.

(see also this answer)

One of the functions of the nozzle is indeed to "remove" as much of the heat as is practical and convert that kinetic energy from random to ordered motion. Same mechanism has been hypothesized for water plumes released to space from subsurface oceans of natural satellites.

Of course simply allowing expansion of the ~100 atm chamber pressure to lower pressure is a big part of generating thrust as well.

In the atmosphere the nozzle has to stop expanding when it reaches roughly an equal pressure to the ambient atmosphere. Beyond that expansion, other problems begin.

In space you could continue to expand (longer nozzle with larger exit diameter) and reduce the temperature even farther, but the nozzle gets really big and at some point there is no place to store it during ascent. And at the same time, the incremental benefits further diminish each time the size increases.

Note that some heat is lost by radiation from the red-hot nozzle. We'd need a thermal infrared analysis to understand how much heat is radiated from the nozzle, and compare that to how much heat is lost by the still-hot exhaust That can't as easily be measured quantitatively, since even if transparent materials can produce a black-body spectrum, that doesn't mean they are effective black-body radiators.

Converting the remaining energy in the exhaust from a well-designed engine and matched nozzle into a little bit more thrust by making the nozzle longer in space will probably not boost the exhaust velocity (and therefore boost the Isp) by a lot, but people work hard for every bit of delta-v, so I think that it's simply the practical limitations of a 5, 20, or even 100 meter diameter nozzle that keep this from being done.

Using an engineered "Maxwell's Demon equivalent" will have to be addressed by other answers.

below: "gas characteristics along a de Laval nozzle, T - absolute temperature; p - pressure; v - speed; M - Mach number" from here.

gas characteristics along a de Laval nozzle

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    $\begingroup$ Does anyone know what temperature the exhaust gasses from some typical rockets actually get to? $\endgroup$ – Steve Linton Dec 14 '18 at 8:08
  • $\begingroup$ @SteveLinton there's a ballpark estimate of 1500ºC in the linked "expansion of the ~100 atm chamber pressure..." answer, but if an answer with more details can't be found that would certainly make for an interesting new question by itself. Some quantitative plots of velocity, temperature, pressure as a function of position from chamber through nozzle and into ambient for a canonical modern engine. $\endgroup$ – uhoh Dec 14 '18 at 10:59
  • $\begingroup$ In fact, that's such a compelling question that it can't wait any longer! Quantitative plots of v, T, p, vs position from chamber through nozzle to ambient for a few canonical modern engines? $\endgroup$ – uhoh Dec 14 '18 at 11:03
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    $\begingroup$ Wouldn't the expansion from 100 atm in the chamber to 1 at the end of the nozzle (assuming a liftoff engine, not a space engine) cool it to below atmospheric temperature? $\endgroup$ – Loren Pechtel Dec 15 '18 at 4:53
  • $\begingroup$ @LorenPechtel you should ask that as a separate question because it's quite an interesting one. The answers will involve a reminder that we can't apply the ideal gas law to a complicated system where combustion is taking place, but it will still have some explaining to do. $\endgroup$ – uhoh Dec 15 '18 at 6:08

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