# Can steam freeze via adiabatic expansion?

Context:

At the end of animation movie Steamboy, a huge steam engine's pressure vessel breaks, which results in the escaped steam freezing to ice.

I've been thinking about this -- is it really possible to freeze steam through rapid adiabatic expansion to atmospheric pressure? What are the initial steam temperature / pressure conditions allowing for this?

Bonus question: has a catastrophic failure of a steam pressure vessel ever resulted in such an outcome?

• I did very rough calculations, and it seems that for rapid (irreversible) adiabatic expansion, it may not be possible. But I also tried with adiabatic reversible expansion and that might do the trick. I will show my calculations if you want to see them in an answer (but they are very rough, remember). – Shreyansh Darshan May 16 '18 at 16:46
• I'd be glad to! I got the feeling I could get an answer from h/p charts, but I forgot how to read them... related: engineering.stackexchange.com/questions/11588/… – Florian Castellane May 16 '18 at 17:20
• Ah, I got it! I was looking for a T/s chart. in adiabatic expansion, s is constant, so I have to follow the s lines in the chart. Now, to find one for water with low temperatures and labelled phases... – Florian Castellane May 16 '18 at 17:39
• Well, I haven't yet studied about h/p, t/s diagrams. I just did the standard $P^{\gamma-1}T^\gamma = constant$ and some information from wikipedia about pressure (100 atm) inside a high pressure locomotive, and used the temperature at which water would boil at that pressure (600K), for the initial conditions. Plugging in the values seemed to yield a temperature of about 200 K (of course the temperature won't drop down to 200K because there the latent heat to overcome, but it was indicative of formation of ice). – Shreyansh Darshan May 17 '18 at 6:55
• For irreversible adiabatic, I equated work done by steam to the change in it's internal energy, and used ideal gas equation along with that to find the final temperature, which came around 500 K. I didn't post it as an answer because I think the diagrams may be a better approach and these calculations aren't very accurate ($C_p$ and $C_v$ of water keep changing with P and T, which I haven't accounted for). – Shreyansh Darshan May 17 '18 at 7:02