The Carnot efficiency problem

[Tom Booth]

Another clip:

These devices are either centrifuges or axial-flow turbines, instrumental in creating work from pressurized gas to be used to drive a compressor or generator.

The extracted gas heat energy is transformed into the mechanical energy of the shaft in the turbine through the turbine wheel. This mechanical energy is then transformed into thermal energy via an integrated turbine brake, which is then drained out of the turbine by means of a coolant.

Depending on the application, some small cryogenic turbines are made to do "work" against a simple break, rather than driving a compressor, generator or other work.

The break of course, gets hot while the gas driving the turbine gets cold..

This phenomenon seems bizarre, or almost magical as if the heat were teleported out of the gas to the break.

I proposed this arrangement for cooling a Stirling engine years ago:

Self_Cooling_3.jpg (37.76 KiB) Viewed 10988 times

Not sure if anyone on the forum understood the purpose of the little expansion turbine with the brake.

Of course typical cryogenic turbine spins at something like 500,000 RPM runs on pressurized gas at 200 bar and results in temperatures in the low cryogenic range so I'm not so sure any turbine attached to an LTD Stirling "compressor" (or vacuum pump) could produce any measurable cooling at all, but I thought it might be worth trying.

It wouldn't need to go down to 4K to liquify helium or anything, but maybe produce the few degrees cooling required to run a P-90 type Ultra LTD.

[Fool]

https://en.m.wikipedia.org/wiki/Liquefaction_of_gases

Claude's process
Edit
Air can also be liquefied by Claude's process in which the gas is allowed to expand isentropically twice in two chambers. While expanding, the gas has to do work as it is led through an expansion turbine. The gas is not yet liquid, since that would destroy the turbine. Commercial air liquefication plants bypass this problem by expanding the air at supercritical pressures.[1] Final liquefaction takes place by isenthalpic expansion in a thermal expansion valve.

That appears to imply that final conversion to get any liquid uses the irreversible Jouel-Thompson Linde process. Excellent.

[Tom Booth]

https://en.m.wikipedia.org/wiki/Liquefaction_of_gases

Claude's process
Edit
Air can also be liquefied by Claude's process in which the gas is allowed to expand isentropically twice in two chambers. While expanding, the gas has to do work as it is led through an expansion turbine. The gas is not yet liquid, since that would destroy the turbine. Commercial air liquefication plants bypass this problem by expanding the air at supercritical pressures.[1] Final liquefaction takes place by isenthalpic expansion in a thermal expansion valve.

That appears to imply that final conversion to get any liquid uses the irreversible Jouel-Thompson Linde process. Excellent.

What you think a Wikipedia article with no citation "appears to imply" is irrelevant.

The article states: "Commercial air liquefaction plants bypass this problem..."

That is, liquefaction inside the turbine does happen and over time can lead to avoidable problems.

Over time, with high volume continuous operation, cavitation could eventually eat away at the very expensive turbine, so if liquefaction can be delayed until a later stage that can save money as expansion valves are relatively inexpensive, but cavitation can still be a problem.

So your statement: "That appears to imply that final conversion to get any liquid uses the irreversible Jouel-Thompson Linde process. Excellent." Is, as usual, false, and misleading.

[Fool]

Cavitation happens in a liquid from work input. Condensation happens in a gas from work output. Both can erode Impellers.

There isn't anything misleading about my last statement. They appear to use Claude's process to decrease the temperature to just above liquidation temperature. Then expand it further, maybe using a cold finger, to get part of the gas to liquefy. The cold finger, and other further expansions without a turbine , is the Joule-Thompson process.

[Tom Booth]

Cavitation happens in a liquid from work input. Condensation happens in a gas from work output. Both can erode Impellers.

There isn't anything misleading about my last statement. They appear to use Claude's process to decrease the temperature to just above liquidation temperature. Then expand it further, maybe using a cold finger, to get part of the gas to liquefy. The cold finger, and other further expansions without a turbine , is the Joule-Thompson process.

You don't know what your talking about. Please go away and find another "student" to "help"

[Tom Booth]

I suspect you never heard of turboexpanders or the Claude process prior to hearing about it through me here, but already you are an expert.

I don't claim to be any expert but I've been studying these various gas and air liquefaction processes for many years and have seen mention of the gas liquefying while still inside the turbine itself many times.

The Claude process has been changed and modified and combined with other processes over the past century or more that it has been in use. (Originally a reciprocating engine was used, so not much chance of degradation of the turbine blades)

https://youtube.com/clip/UgkxVagEYQwzl- ... WIqmri5ycJ

[Goofy]

Cavitation happens in a liquid from work input. Condensation happens in a gas from work output. Both can erode Impellers.

There isn't anything misleading about my last statement. They appear to use Claude's process to decrease the temperature to just above liquidation temperature. Then expand it further, maybe using a cold finger, to get part of the gas to liquefy. The cold finger, and other further expansions without a turbine , is the Joule-Thompson process.

@Fool : "No dark sarcasm in the classroom
Teachers, leave them kids alone

Hey, Teacher ! - leave us kids alone" . . . .

[Fool]

Arggg! You fools! I'm one of the students here. I posted for my benefit and others. it was after midnight in the early morning.after a long day. My car had a flat. The phone stopped working. The taxi was late. Ewoks went on strike. Please forgive me. I was trying to come up with a word for, followers, reader's, posters, people, etc but didn't like any of those words. Then I thought, erroneously, that we are all students here learning and describing thermodynamics. It is a complement, and ment as one. I guess you, offended people, know all the answers. Sorry, my error.

Good one Goofy. Have enjoyed Pink Floyd for a number of years now. LOL

Tom,

I studied turbo expanders ever since first hearing of them in the late 1970's. Expert? Thanks for the compliment, but no, and neither are you, from studying what you write here. Do I have a viewpoint that is interesting? Yes. You are striving, reaching, really hard just to oppose what I say without slowing down to contemplate. Hint, you have a viewpoint that is interesting too.

Why does a boat propeller cavitate? Why is it important to know the difference between cavitation and condensation in turbo machinery? Cavitation is from the cold vacuum boiling of the liquid. Condensation is from the cooling and expansion of a pressure stream. Both are detrimental to the rotor blades.

Cavitation requires reduction of power input, i.e., boat motor.

Condensation requires higher initial temperature or less expansion. Less expansion requires a brake or load on it to keep the rotor from over speeding.

Although it is possible to cool a fluid to condensation, liquefaction, in an expansion engine, or rotor, the link was given as a comment to see the logic in cooling something to the point of condensation, but to use the Linde process to do the final cooling. Thus saving the rotor.

That leads to the thought that all modern liquefaction plants are both Linde and Claude process. Do I know that? No. But it, to me, is an interesting viewpoint. I thought maybe others might find it interesting as well. Please have a nice day.

[Tom Booth]

.... You are striving, reaching, really hard just to oppose what I say without slowing down to contemplate. ....

Sorry but I give your comments considerable thought and consideration.

You, however, do virtually nothing other than go from post to post, nit picking, contradicting and ridiculing whatever triviality you see fit to latch onto to try and contradict and criticize so don't play so innocent.

Case in point:

Why does a boat propeller cavitate? Why is it important to know the difference between cavitation and condensation in turbo machinery?

Nobody ever said cavitation and condensation were the same thing that I know of. On and on trying to "correct" some non-issue.

[Tom Booth]

(...)
Even Tesla was unable to conceptualize a machine that would use all the heat. He failed to recognize that it requires a drop in temperature to 0 Kelvin, and how thermodynamics behaves at that temperature, let alone real gases.

(...)

I've come across this ridiculous statement, or the equivalent, so many times in lectures, articles, websites, textbooks, whatever, so many times by what IMO must be some rather arrogant self-inflated egos to think they are so much smarter than Tesla.

Maybe you could explain to me Mr. "Fool" how or why, if I put some heat, let's say 10,000 joules, into a Stirling engine it then "requires a drop in temperature to 0 Kelvin" to have "a machine that would use all the heat".

"Heat" as you delight in pointing out whenever someone uses the word, is the "transfer" of thermal energy.

So starting at ambient, at around 300k, we add or transfer 10,000 joules. Maybe that brings the temperature of the hot side up to 375K.

I would think, in order to use "all the heat" transfered to the engine, the engine would have to use all of the 10,000 joules that had been transfered, bringing the temperature back down to where it started at ambient, or 300k.

So why is it suddenly necessary to use, probably 100 million, billion joules so as to somehow get all the way down to absolute zero just to use up a measly 10,000 joules?

Here's your big chance, Mr smarty pants who knows so much more than Tesla, because I just don't really understand the logic behind this "absolute zero" assertion myself either.

It sounds like a bunch of BS to me, but I will now "slow down and contemplate" and listen carefully as you explain in detail how that works exactly, please.

[matt brown]

Maybe you could explain to me Mr. "Fool" how or why, if I put some heat, let's say 10,000 joules, into a Stirling engine it then "requires a drop in temperature to 0 Kelvin" to have "a machine that would use all the heat".

"Heat" as you delight in pointing out whenever someone uses the word, is the "transfer" of thermal energy.

So starting at ambient, at around 300k, we add or transfer 10,000 joules. Maybe that brings the temperature of the hot side up to 375K.

I would think, in order to use "all the heat" transfered to the engine, the engine would have to use all of the 10,000 joules that had been transfered, bringing the temperature back down to where it started at ambient, or 300k.

So why is it suddenly necessary to use, probably 100 million, billion joules so as to somehow get all the way down to absolute zero just to use up a measly 10,000 joules?

Here's your big chance, Mr smarty pants who knows so much more than Tesla, because I just don't really understand the logic behind this "absolute zero" assertion myself either.

It sounds like a bunch of BS to me, but I will now "slow down and contemplate" and listen carefully as you explain in detail how that works exactly, please.

Here's some HS physics you obviously missed...

Consider a Stirling cycle without any engine application, just the strict cycle itself, and consider everything an ideal 300-600k cycle. A Stirling cycle has isothermal expansion and compression with regen between these processes, and forget your typical harp of no such thing as isothermal in real engines (we all know that). During the isothermal expansion, ALL the source heat becomes work, and the internal energy remains unchanged (aka constant) DESPITE volume and pressure changes. So, to borrow your example somewhat, if the internal energy at 600k is xJ PRIOR expansion, adding 10kJ source heat will produce 10kJ work during expansion, but the internal energy AFTER expansion is still xJ. Great, all heat became work aka 100% efficiency.

The problem is that this is only a single process and NOT A CYCLE...which requires returning the gas to its start state. Now, to compress the gas back to initial volume at 600k will require the same 10kJ work of compression. And since a Stirling cycle has isothermal compression ALL this 10kJ of work will "pass thru" the gas as 10kJ sink heat WHILE the internal energy remains xJ. So, your sink theory is drastically flawed...the sink heat is not coming from internal energy, but from the external work of compression passing THRU THE GAS.

So far, in this example, 10kJ source heat at 600k produced Wpos=10kJ, but compressing the gas back to original volume at 600k will require Wneg=10kJ and produce 10kJ sink heat, whereby Wpos=Wneg and Wnet=0. The only way to achieve Wpos>Wneg (with a Stirling cycle) is to have a temperature differential across the cycle with a higher internal energy during expansion than compression.

Moving to a 300-600k cycle, the internal energy at 600k is 2x the internal energy at 300k. This allows 10kJ of source heat to produce 10kJ of expansion work, but the same compression at 300k will only require 5kJ of compression work (vs 10kJ at 600k). Remember, if the internal energy at 600k=xJ, the internal energy at 300k=.5xJ and this is where Carnot comes into play. The Carnot equation sets temperature extremes as the maximum values for Carnot eff, but only the 3 "regular" isothermal cycles have this potential (Carnot proper, Stirling, and Ericsson). Meanwhile, the "regular" adiabatic cycles (Otto and Brayton) have a Carnot efficiency derived from their temperature extremes of their adiabatic processes. Turning this Carnot limit into a mystery is ridiculous, since it's only a simple mathematical derivation of how PV=mRT plays out.

As for Tesla and the cold hole camp, the reason why 0k would be required for 100% efficiency is simply that to compress any gas back to its original volume IN A CYCLE would require Wneg=0 aka zero internal energy to resist the work of compression.

[Tom Booth]

Honestly, I find this explanation laughable Matt, no offense, but it makes no sense.

All you said, in regard to my actual question is:

"the reason why 0k would be required for 100% efficiency is simply that to compress any gas back to its original volume IN A CYCLE would require Wneg=0 aka zero internal energy to resist the work of compression."

That's an affirmation of an opinion not an explanation.

To put this in perspective, liken atmospheric pressure to gravity.

If I start out on the road at an elevation of 1000 feet and I drive my car up a hill, my car is doing work against gravity.

If the car runs out of gas half way up the hill at 1100 feet and I want to get back down the hill, I can just put the car in neutral and drive the car back down the hill in reverse using the same force of gravity that the engine had to work against to get to 1100 feet in the first place.

That is a readily observable and verifiable fact.

What you are saying is basically equivalent to saying that if I run out of gas at 1100 feet, in order to get back down to 1000 feet someone would have to dig a hole all the way down to zero elevation ABSOLUTE. That is, all the way to the center core of the planet.

The assertion has no rational basis.

Maybe you could explain it another way because just saying "IN A CYCLE" proves nothing at all.

I drove my car up a hill and let gravity pull it back down "IN A CYCLE". I don't need to go 4000 miles to the center of the earth to get my car to roll back down the hill a few thousand feet. So why should a Stirling engine have to go to absolute zero to get the piston to return one cycle, when quite obviously and observably the piston returns to TDC by the same atmospheric pressure it had to work against to get to BDC in the real world all the time, quite easily.

[Tom Booth]

BTW, where did this idea that an "Ideal" Stirling cycle is ALL isothermal expansion (on the expansion stroke) ?

The term "isothermal", I believe, originated with Clausius about 30 years or so after Stirling got his patent. Carnot was skeptical that hot air engines could even exist so not likely he ever said such a thing.

[Tom Booth]

(...)
Turning this Carnot limit into a mystery is ridiculous, since it's only a simple mathematical derivation of how PV=mRT plays out.

(..)

How so?

Who says it's a mystery?

Mathematically the carnot supposed limit is nothing more than a ratio derived from a temperature difference.

[matt brown]

Honestly, I find this explanation laughable Matt, no offense, but it makes no sense.

All you said, in regard to my actual question is:

"the reason why 0k would be required for 100% efficiency is simply that to compress any gas back to its original volume IN A CYCLE would require Wneg=0 aka zero internal energy to resist the work of compression."

That's an affirmation of an opinion not an explanation.

The assertion has no rational basis.

Maybe you could explain it another way because just saying "IN A CYCLE" proves nothing at all.

So why should a Stirling engine have to go to absolute zero to get the piston to return one cycle, when quite obviously and observably the piston returns to TDC by the same atmospheric pressure it had to work against to get to BDC in the real world all the time, quite easily.

In a cycle, everything must return to the start state, so any expansion would require a similar compression. The work of compression aka Wneg must come from somewhere whether it's from partial expansion energy stored in a flywheel or buffer/ambient pressure that was Wneg during expansion. Either way, this Wneg is proportional Wpos via PV=mRT where Wneg/Wpos=eff. The only way to have compression Wneg=0 is at 0k. Your static analogies are fine for lifting and dropping rocks, but don't cut it in thermo since there's kinetic energy at play (except at 0k).

You act as if LTDs are magical...who cares if some toy can run without a flywheeel or whatever. As I've said before, Senft didn't do us any favors with this 'educational' tinker-toy, he just ushered in endless distraction. I suspect that these LTD toys are nothing near a "Stirling", but approximate the three legged Lenoir cycle...yeah, no Wneg or physical heat sink req'd. They simply suck in a little air, mix it with some other air, semi-isolate during heating, then let this air slightly expand and leak a little to ambient, then repeat.

The Lenoir cycle is non compression with Cv input, adiabatic expansion, and isobaric 'cooling' (ambient exhaust). Yep, the same as Otto-Langden and even without Wneg of compression they suck compared to common compression cycles. Why ??? Fool supplied the 'magic' answer a while back - isothermal input is the most efficient, and processes are path dependent. Different paths between the same points (PV plots) do NOT have the same values.

If so concerned with real this and that, why worry about tinker-toys, I hadn't noticed they're real.