Assuming you’d have an oxygen mask on before hand, and you depressurized the aircraft at FL340, would the inside and outside temperatures quickly become the same?

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    $\begingroup$ It depends on your definition of "quickly", and whether or not you continue pumping hot air, (albeit at significantly lower pressure) into the cabin. $\endgroup$ – Michael Hall Jan 17 at 17:52
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    $\begingroup$ For those who didn't know what FL340 was, it seems to mean an altitude of 34,000 feet (measured by air pressure): en.wikipedia.org/wiki/Flight_level $\endgroup$ – Ben Crowell Jan 17 at 23:07
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    $\begingroup$ @BenCrowell Yes, that's exactly what it means. It's a pressure altitude of 34,000 feet. $\endgroup$ – reirab Jan 17 at 23:14
  • $\begingroup$ Are you thinking about the worst greek air disaster? en.wikipedia.org/wiki/Helios_Airways_Flight_522 $\endgroup$ – EarlGrey Jan 18 at 0:51

This depends entirely on how you depressurized it.

Depressurizing the aircraft by fully opening the outflow valve will lower the temperature somewhat, but will not bring it to that outside.

The expansion process will be closest to free (Joule) expansion, which causes only small decrease in temperature, not an adiabatic one that corresponds to the atmospheric lapse rate. And while some energy will be lost through the air's initial expansion as it escapes, it will quickly be replaced by new warm air coming in through the packs (cabin pressurization and air conditioning system).

That air supply is quite reliable, as it's bled from engines at about $200\ \mathrm{^\circ C}$ – it then needs to be air-conditioned so as not to roast the passengers. The packs will automatically maintain set temperature, putting less effort into cooling the air if the cabin gets too cold.

If you also disable the engines, so no new air can come in, the airplane will lose some heat. Still, in all other regards, the fuselage will remain a well-insulated vessel with some heat sources on the inside. Also, the less-dense air is much less effective at heat transfer, and has much less heat capacity, so it will get chilly, but the passengers aren't going to freeze to death or even get a frostbite.

The heat capacity of air at ambient pressure will be about $280\ \frac{\mathrm{J}}{\mathrm{m}^3\cdot\mathrm{K}}$. A 737 encloses about $200\ \mathrm{m}^3$, giving a total of $56\ \frac{\mathrm{kJ}}{\mathrm{K}}$. For comparison, the plane's passengers have a specific heat capacity of $3.5\ \frac{\mathrm{kJ}}{\mathrm{kg}\cdot\mathrm{K}}$, giving, for 120–160 of them, about $40\ \frac{\mathrm{MJ}}{\mathrm{K}}$. So a $\sim70\ \mathrm{^\circ C}$ one-time drop in air temperature can technically match the outside, but it can only cool the passengers by at most $0.1\ \mathrm{^\circ C}$.

In practice, the cooling won't be instant, and it has to compete with heat exchange with the still-warm interior. Passenger bodies producing $12{-}18\ \mathrm{kW}$ of waste heat (at rest) will also restore cabin air temperature by $\sim1\ \mathrm{^\circ C}$ every 4 seconds. The larger the hole, the lower the temperature minimum that will be reached. With a broken window, the effect of this one-time expansion heat loss will have been mostly undone before anyone notices it, amidst the chaos.

Tearing the top of the fuselage off, on the other hand, will most definitely let the passengers feel the cold outside. But it will be the powerful wind blowing in, not the air leaving the airplane, that will chill its interior.

Breaking a single window or making a lot of bullet holes (really a lot - it will take a crate of ammo to overcome the packs' air inflow) is about the same as fully opening the outflow valve. It will produce some cooling effect, but it will definitely not bring the interior to the outside temperature, as the cooling effect is one-time, and will be rapidly compensated for.

This kind of event has happened a few times in aviation history. It has caused some fatalities – a person sitting right next to the broken window has been killed by the mechanical force of the escaping air. No one has been frozen to death, not even in the Aloha Flight 243, which represents about the worst possible case for depressurization (which still leaves survivors and not just a debris field).

  • $\begingroup$ @Bianfable The plane was turned into a landaulet, with winds free to ravage the interior, and getting scooped up by the remaining part of the fuselage. It's as bad as it gets while still allowing the plane to land (thanks to the cockpit staying enclosed). $\endgroup$ – COVID-20 Jan 17 at 19:33
  • $\begingroup$ What's a "pack(s)"? And can those, or any other part of a plane's HVAC system be run on the APU? And can the system provide heat with the engines off? $\endgroup$ – Mazura Jan 18 at 0:22
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    $\begingroup$ @Mazura Integrated or packaged aircon system. And from the diagrams here, it looks like yes, the APU can supply them, though likely not enough for full performance. $\endgroup$ – SomeoneSomewhereSupportsMonica Jan 18 at 4:46
  • $\begingroup$ @SomeoneSomewhereSupportsMonica, if the APU is running, it can feed the packs, but most APUs won't run at cruise altitude. $\endgroup$ – Jan Hudec Jan 21 at 12:15

Assuming the cabin is pressurized to a pressure equivalent of that at 8000ft, the initial pressure is on board is $P_1$ = 75262.4 Pa, and the temperature is T1 = 20ºC = 293.15K

At FL340, the outside pressure is $P_2$ = 24999 Pa.

Assuming we are talking about rapid decompression, there will be no significant heat transfer into or out of the gas during this process. We can thus approximate the temperature by using the laws of adiabatic cooling.

$P^{(1-\gamma)}T^\gamma = $ constant


$T_2 = \sqrt[\gamma]{\frac{P_1^{(1-\gamma)}T^\gamma}{P_2^{(1-\gamma)}}}$

With $\gamma$ = 1.4 for air.

Filling in the numbers will give $T_2$ = 213.96 K = -59.2ºC

The normal temperature at FL340 is about 222.8 K or -52.4ºC

After a rapid decompression, the temperature on board will thus be slightly below the outside temperature.

This sudden loss of temperature will cause moisture in the air to condensate into fog.

Since all the interior mass of the cabin is still at 20 degrees, the temperature will quickly rise again and the fog will disappear.

Here is a video of a rapid decompression in a pressure camber. Note the fog, and note how quickly the person passes out.

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    $\begingroup$ Excellent explanation that caters to all levels of knowledge! $\endgroup$ – Michael Hall Jan 17 at 20:05
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    $\begingroup$ Keep in mind that this is free expansion, as gas is not doing work against a piston. In free expansion, temperature change is minimal. A smaller hole is not completely free expansion, but much closer to it than it is to a heat engine cycle. $\endgroup$ – COVID-20 Jan 17 at 20:05
  • $\begingroup$ Note that the fog dissipates almost as quickly as it appeared though... $\endgroup$ – Jan Hudec Jan 17 at 20:38
  • $\begingroup$ @JanHudec I guess that is because the fog is very good at catching IR radiation. Since the walls don"t cool down, there is plenty of heat transfer into the fog and since there is very little mass in the air, it warms up very quickly again. $\endgroup$ – DeltaLima Jan 17 at 21:27
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    $\begingroup$ The part I noticed in that video was how much difficulty the guy had getting his mask on. He knew it was coming. He had the calm reassurance of someone standing right there already fitted with O2 who he knew would step in to help him and he couldn't get the strap over his head. It took the helper several tries, too, while the test subject started to lose consciousness. Pretty scary! People, put your own mask on before trying to help anyone else, lest you join the dead! $\endgroup$ – FreeMan Jan 20 at 20:37

More or less...

For gamma = 1,4, an adiabatic expansion from 1013 hPa to 226 hPa will cause an initial temperature of 20ºC = 273K to drop to 191 K.

The Standard Atmosphere temperature for that altitude is approx. that...

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    $\begingroup$ @MichaelHall Pretty sure he is assuming the cabin temperature to be 20°C but forgot to add the 273.15K to make it 293.15K in the cabin at the start of the expansion. $\endgroup$ – AEhere supports Monica Jan 17 at 18:49
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    $\begingroup$ @pericynthion the OP does not specify, so lacking further nuance it is the only meaningful assumption, since the opposite possibility from a thermodynamic perspective would be an isothermal process, would it not? $\endgroup$ – AEhere supports Monica Jan 17 at 18:54
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    $\begingroup$ @MichaelHall I read that as 191K, which is indeed close to the ISA temperature at 12km. $\endgroup$ – AEhere supports Monica Jan 17 at 18:56
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    $\begingroup$ The expansion will most definitely be isochoric, not adiabatic. Adiabatic expansion happens in a sealed cylinder, when the gas is producing work by acting against resistance. Air expansion from a tear in a container is going to produce Joule expansion, which results in no cooling with ideal gases and only a bit of cooling with real gases. $\endgroup$ – COVID-20 Jan 17 at 19:32
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    $\begingroup$ The air in the cabin is not 1013 hPa, it is only 750 hPa. $\endgroup$ – Jan Hudec Jan 17 at 20:19

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