The Joule-Thomson Effect and Models We Know • Hacker News Board

Re: The Joule-Thomson Effect and Models We Know

#11

Post by hashimotonomora » Fri, Dec 17, 2021, 9:35 PM UTC

bernulli wrote: post #9

Pulcinella wrote: post #2

One of my favorite things about the JT effect is how gases not being “perfect” is actually “better” than if they were. What I mean is, if gases all behaved as some kind of perfect, platonic* ideal of a gas and followed the ideal game law exactly, there would be no temperature change. But because they don’t, the Joules-Thomson effect is what allows for refrigeration. *Helium is probably the closest to some platonic id…

Why wouldn't there be any temperature change in ideal gases? When I compress an ideal gas in a bike pump, it heats up; when I expand it through a nozzle, it cools down. Often I can approximate these processes as isentropic, and then the temperatures are uniquely determined by the expansion as either a function of pressure ratio T2 = T1 (p2/p1)^(g-1/g) or volume ratio; T2 = T1 (V2/V1)^g, where g is gamma, the ratio of…

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.

Re: The Joule-Thomson Effect and Models We Know

#12

Post by BlueTemplar » Fri, Dec 17, 2021, 9:36 PM UTC

avmich wrote: post #5

Good article. Certain tempting things left unexplained though :) . Like, why properties of H, He and Ne are such that they are at this part of their diagram at normal conditions? Or where dispersion bonding stores the kinetic energy - when two atoms bond this way, both momentum and energy should be preserved, so some places to put excess of energy should be present, otherwise the pair should be unstable. Like, requir…

- H and He (and Ne ?) are small.

- The energy is stored in the bond ?

Re: The Joule-Thomson Effect and Models We Know

#13

Post by BlueTemplar » Fri, Dec 17, 2021, 9:39 PM UTC

bernulli wrote: post #9

Pulcinella wrote: post #2

One of my favorite things about the JT effect is how gases not being “perfect” is actually “better” than if they were. What I mean is, if gases all behaved as some kind of perfect, platonic* ideal of a gas and followed the ideal game law exactly, there would be no temperature change. But because they don’t, the Joules-Thomson effect is what allows for refrigeration. *Helium is probably the closest to some platonic id…

Why wouldn't there be any temperature change in ideal gases? When I compress an ideal gas in a bike pump, it heats up; when I expand it through a nozzle, it cools down. Often I can approximate these processes as isentropic, and then the temperatures are uniquely determined by the expansion as either a function of pressure ratio T2 = T1 (p2/p1)^(g-1/g) or volume ratio; T2 = T1 (V2/V1)^g, where g is gamma, the ratio of…

But the isentropic part is what is being violated here... (I was surprised that the article didn't mention that even though it managed to demonstrate the real behaviour in another way.)

Re: The Joule-Thomson Effect and Models We Know

#14

Post by Pulcinella » Fri, Dec 17, 2021, 9:42 PM UTC

bernulli wrote: post #9

Pulcinella wrote: post #2

One of my favorite things about the JT effect is how gases not being “perfect” is actually “better” than if they were. What I mean is, if gases all behaved as some kind of perfect, platonic* ideal of a gas and followed the ideal game law exactly, there would be no temperature change. But because they don’t, the Joules-Thomson effect is what allows for refrigeration. *Helium is probably the closest to some platonic id…

Why wouldn't there be any temperature change in ideal gases? When I compress an ideal gas in a bike pump, it heats up; when I expand it through a nozzle, it cools down. Often I can approximate these processes as isentropic, and then the temperatures are uniquely determined by the expansion as either a function of pressure ratio T2 = T1 (p2/p1)^(g-1/g) or volume ratio; T2 = T1 (V2/V1)^g, where g is gamma, the ratio of…

I didn’t mean that there would be no temperature changes. But that in the example where you allow the gas to expand into a new volume there would be no temperature change. Remember, temperature relates to the speed the gas particles are moving. If the barrier between the two sides of the container was removed and the gas allowed to expand into that new volume, why would the ideal gas particles slow down and decrease the temperature?

Re: The Joule-Thomson Effect and Models We Know

#15

Post by chermi » Fri, Dec 17, 2021, 9:53 PM UTC

hashimotonomora wrote: post #11

bernulli wrote: post #9

Earlier quoted context omitted.

Why wouldn't there be any temperature change in ideal gases? When I compress an ideal gas in a bike pump, it heats up; when I expand it through a nozzle, it cools down. Often I can approximate these processes as isentropic, and then the temperatures are uniquely determined by the expansion as either a function of pressure ratio T2 = T1 (p2/p1)^(g-1/g) or volume ratio; T2 = T1 (V2/V1)^g, where g is gamma, the ratio of…

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 k…

The 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_Introd...

Re: The Joule-Thomson Effect and Models We Know

#16

Post by [deleted] » Fri, Dec 17, 2021, 9:54 PM UTC

hashimotonomora wrote: post #11

bernulli wrote: post #9

Earlier quoted context omitted.

Why wouldn't there be any temperature change in ideal gases? When I compress an ideal gas in a bike pump, it heats up; when I expand it through a nozzle, it cools down. Often I can approximate these processes as isentropic, and then the temperatures are uniquely determined by the expansion as either a function of pressure ratio T2 = T1 (p2/p1)^(g-1/g) or volume ratio; T2 = T1 (V2/V1)^g, where g is gamma, the ratio of…

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 k…

[deleted]

Re: The Joule-Thomson Effect and Models We Know

#17

Post by [deleted] » Fri, Dec 17, 2021, 9:55 PM UTC

Pulcinella wrote: post #14

bernulli wrote: post #9

Earlier quoted context omitted.

Why wouldn't there be any temperature change in ideal gases? When I compress an ideal gas in a bike pump, it heats up; when I expand it through a nozzle, it cools down. Often I can approximate these processes as isentropic, and then the temperatures are uniquely determined by the expansion as either a function of pressure ratio T2 = T1 (p2/p1)^(g-1/g) or volume ratio; T2 = T1 (V2/V1)^g, where g is gamma, the ratio of…

I didn’t mean that there would be no temperature changes. But that in the example where you allow the gas to expand into a new volume there would be no temperature change. Remember, temperature relates to the speed the gas particles are moving. If the barrier between the two sides of the container was removed and the gas allowed to expand into that new volume, why would the ideal gas particles slow down and decrease…

[deleted]

Re: The Joule-Thomson Effect and Models We Know

#18

Post by bernulli » Fri, Dec 17, 2021, 9:57 PM UTC

Pulcinella wrote: post #14

bernulli wrote: post #9

Earlier quoted context omitted.

Why wouldn't there be any temperature change in ideal gases? When I compress an ideal gas in a bike pump, it heats up; when I expand it through a nozzle, it cools down. Often I can approximate these processes as isentropic, and then the temperatures are uniquely determined by the expansion as either a function of pressure ratio T2 = T1 (p2/p1)^(g-1/g) or volume ratio; T2 = T1 (V2/V1)^g, where g is gamma, the ratio of…

I didn’t mean that there would be no temperature changes. But that in the example where you allow the gas to expand into a new volume there would be no temperature change. Remember, temperature relates to the speed the gas particles are moving. If the barrier between the two sides of the container was removed and the gas allowed to expand into that new volume, why would the ideal gas particles slow down and decrease…

I did miss the part where the original comment was restricted to JT. Thanks!

Re: The Joule-Thomson Effect and Models We Know

#19

Post by bernulli » Fri, Dec 17, 2021, 9:58 PM UTC

chermi wrote: post #15

hashimotonomora wrote: post #11

Earlier quoted context omitted.

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 k…

The 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 lea…

You 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!

Re: The Joule-Thomson Effect and Models We Know

#20

Post by mattb314 » Fri, Dec 17, 2021, 10:20 PM UTC

Pulcinella wrote: post #2

One of my favorite things about the JT effect is how gases not being “perfect” is actually “better” than if they were. What I mean is, if gases all behaved as some kind of perfect, platonic* ideal of a gas and followed the ideal game law exactly, there would be no temperature change. But because they don’t, the Joules-Thomson effect is what allows for refrigeration. *Helium is probably the closest to some platonic id…

> Joules-Thomson effect is what allows for refrigeration

I don't think this is true. The "simple" model of refrigeration taught in highschool is just a carnot cycle running backwards, and this can be modeled with an ideal gas. The author of the post covers this the section on "the Thermodynamics 101 Answer"[1], where all you need to drop the temperature of a gas is to let it do work on the piston.

That's not to say that JT is not useful, just that we can explain a theoretical refrigerator without it.

[1] https://mattferraro.dev/posts/joule-thomson#the-thermodynami...