4 ms·
JT is isoenthalpic not isoentropic. For an ideal gas, h = u + pv = CT + RT = f(T). So an isoenthalpic process is necessarily isothermic for an ideal gas. Expan
by hashimotonomora 5y ago
JT is isoenthalpic not isoentropic. For an ideal gas, h = u + pv = CT + RT = f(T). So an isoenthalpic process is necessarily isothermic for an ideal gas.
Expansion through a nozzle is extremely chaotic and generates entropy. You can’t model JT that way.
When you compress air in an air pump you are doing work against the system and increasing its internal energy u = q - w, which can be explained using ideal gases by knowing that u = u(T). But this is not because the pressure increases but because of your work.
- chermi 5y agoThe emphasis on the entropy generation kind of confuses the point from my POV. What's important about the nozzle setup is that, by construction, it generates the JT throttling process. That is, the procedure is isenthalpic. Focus on the thermodynamic consequences of that. Sorry for butting in. It took me a long time to get comfortable with throttling. Non-equilibrium stat mech stuff can really throw you (well, at least me) off if you come at it too microscopically at first. Edit -- H. Callen's thermo book has a great little section on it. Best book on thermo out there if you're into a real postulate-and-construct approach. One of my favorite books of all time. https://en.m.wikipedia.org/wiki/Thermodynamics_and_an_Introduction_to_Thermostatistics https://en.m.wikipedia.org/wiki/Thermodynamics_and_an_Introd...
- bernulli 5y agoYou can very much model expansion in a nozzle as isentropic, but I did miss the part where the original comment was restricted to JT. Thanks!
- deleted 5y ago[deleted]