Experimentally confirming the conversion of heat energy to work

[Tom Booth]

As soon as you add in WORK it's a whole 'nother ball game.

It seems to me you're the one who cannot get that concept through your head no matter how many times it is stated.

When a gas does 1000 joules or work doing something like driving a piston engine it is the equivalent of removing 1000 joules of heat.

If you are so sure of this, why are you always asking for experimental confirmation of a Carnot engine?

I'm not. It's already recognized there is no such thing as a Carnot engine so no experiment involving a Carnot engine is possible.

The "Carnot Limit" however, is applied to ALL engines, including real engines, so it's applicability to REAL engines is subject to some testing. It takes a little effort but it's possible. Other than myself, however, I can find no references to anyone else making any such efforts.

Transforming heat into work or vice versa HAS been demonstrated experimentally and such experiments are repeatable and the principle is in wide application worldwide on an industrial level.

The Carnot limit has been promulgated.without any evidence, so IMO is open to question.

Can you provide a paper that describes, in detail, the conversion you are so sure of?

What "detail"?

I've already provided a dozen or so references including those I posted ten years ago on the thermodynamics thread which you claim to have read.

You argue the other extreme as I do, so why be so dismissive of another idea?

I'm not.

But you asked for experimental confirmation and you have it.

Go down to Amerigas. They have hundreds of tanks full of liquified bottled gases that were produced by converting heat into work.

You're just being intentionally obtuse it seems to me.(I hope).

[Tom Booth]

Anyway, I could care less about the damn Carnot Limit.

It's basically when other people keep throwing it up as a reason for dismissing my actual experiments and actual first hand observations that anyone can repeat and verify, then I have no choice but to ask where or when was there ever any actual proof of the Carnot Limit?

Why should anyone dismiss actual experimental results on the basis of some offshoot of the obsolete Caloric theory that has never been experimentally or otherwise verified?

[Tom Booth]

You inquire of chat GPT occasionally don't you?

Here is part of what Google's AI has to say about turbo-expanders:

Compress_20240619_214234_4415.jpg (85.04 KiB) Viewed 1683 times

"Cools a gas stream by extracting work from it."

Up to 90% efficiency.

Quite a few references listed lower down on the screen.

PDF: https://files.chartindustries.com/LAT4.pdf

Turboexpanders achieve these temperature targets by extracting relatively large amounts of energy and driving down temperatures accordingly. As such, they can be viewed as very efficient refrigeration machines.

http://gasprocessingnews.com/articles/2 ... operation/

PDF: https://simmsusa.com/wp-content/uploads ... anders.pdf

https://oaktrust.library.tamu.edu/bitst ... t33-15.pdf

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

[VincentG]

In applications that require the refrigeration of process gas, the distinction of a turboexpander is that it expands the gas stream for its own sake, and mechanical work is generated as a byproduct. This is not to say that the side effect of mechanical work is not useful. On the contrary, most turboexpanders likely drive a compressor or generator. In this case, the compressor or generator serves as a loading or braking device—a sink for the expander’s energy. Another common term for this type of machine is “compander,” although this is less common in the natural gas processing industry.

Doesn't seem very definitive to me, this is the first link I clicked on. But I will read more, thank you.

[Tom Booth]

This looks like a good reference I haven't seen before:

http://gasprocessingnews.com/articles/2 ... operation/

Lots of basic information regarding heat into work.

Lots of things here I've already stated previously above.

A turboexpander is a rotating machine with an expansion turbine that converts the energy contained in a gas into mechanical work, ...

In applications that require the refrigeration of process gas, the distinction of a turboexpander is that it expands the gas stream for its own sake, and mechanical work is generated as a byproduct. This is not to say that the side effect of mechanical work is not useful. On the contrary, most turboexpanders likely drive a compressor or generator. In this case, the compressor or generator serves as a loading or braking device—a sink for the expander’s energy. Another common term for this type of machine is “compander,” although this is less common in the natural gas processing industry.

This article’s primary focus is cryogenic turboexpanders loaded by compressors, although many of the principles expounded are applicable to other types of expanders, such as an expander-generator.

Turboexpander applications. Turboexpanders were introduced in the mid-1930s when the first machine was designed and installed for air separation. The first turboexpander for a natural gas application was designed and installed in the early 1960s. Today, more than 5,000 units are in operation globally.

Cryogenic turboexpanders¹ find use in many applications. They are standard in the natural gas industry for liquefaction (FIG. 1) and dewpoint control. They are also used in the petrochemical industry for ethylene plants, air separation, refrigeration and power generation.

The two main markets for these machines are hydrocarbon processing and air separation plants. In both cases, there is a desire to change the state of a process gas to a specific pressure and temperature. Turboexpanders achieve these temperature targets by extracting relatively large amounts of energy and driving down temperatures accordingly. As such, they can be viewed as very efficient refrigeration machines.²

How does a turboexpander operate? A refrigeration cycle requires that the gas be greatly expanded to reduce its temperature. This is referred to as a Joule–Thomson (J-T) effect, and it can be accomplished with a valve. The J-T valve (or throttling valve) achieves a constant enthalpy expansion adiabatically, with no work output.

The expander is, in some sense, also a valve because it also accomplishes a sharp pressure drop; however, it accomplishes more than a valve because it also extracts work from the gas expansion via a turbine. By requiring the expanding gas to perform work, the resulting temperature is further reduced and the efficiency of the refrigeration cycle is improved.

The kinetic energy (work) produced by the turbine is absorbed by a “loading” element that is mechanically coupled to the turbine via a spindle or shaft. This can be a dyno (oil brake), an electric generator or a centrifugal compressor stage. For the latter two, turboexpanders afford the opportunity to utilize energy that would otherwise not be available with a J-T valve.

...
Over the years, many technological advances in design and manufacturing have allowed turboexpanders to contribute to improvements in the efficiency of multiple gas processes.

"Doesn't seem very definitive?" It's absolutely 100% definitive and confirms everything I've already said.

[VincentG]

Likely a more accurate description is that a valve lets air expand against the pressure of the atmosphere it's expanding against, largely nullifying the cooling effect. The turbo expander in contrast allows an extra buffer space that allows a more free expansion after the initial restriction.

It's just surprising to me that it's the one area of physics/ science where the entire description of the process is as simple as the one liner "the gas, made to do work, cools further".

Really? No further explanation of the mechanism?

[matt brown]

It's just surprising to me that it's the one area of physics/ science where the entire description of the process is as simple as the one liner "the gas, made to do work, cools further".

Really? No further explanation of the mechanism?

Yeah, it's like someone doesn't want to give up a trade secret for fear everyone will capitalize on it.

The fantasy here is that this bugger is turning 100,000 rpm and gushing out LOX, but any car guy will quickly realize that a turbine has severe limitations. The main drawback against a turbine is that it's not positive displacement, so this 'slip' will increase gas temp and lower efficiency when both issues are paramount as k lowers. The gimmick is to increase gas density as T drops until JT valve is effective, despite turbine decreasing density during primary expansion. The small lab units for recompressing helium use a multi pass scheme.

[Tom Booth]

Likely a more accurate description is that a valve lets air expand against the pressure of the atmosphere it's expanding against, largely nullifying the cooling effect. The turbo expander in contrast allows an extra buffer space that allows a more free expansion after the initial restriction.

It's just surprising to me that it's the one area of physics/ science where the entire description of the process is as simple as the one liner "the gas, made to do work, cools further".

Really? No further explanation of the mechanism?

Oh my oh my, yeah, I'm sure these two guys got it all wrong and vincentG and "Matt Brown" know oh so much better than these two professional engineers from LA Turbine:

Compress_20240620_031418_8307.jpg (72.67 KiB) Viewed 1664 times

You seem to deliberately avoid the clear explanations given, quoted in bold and/or underlined in my previous post that directly address your question and the topic of your post.

It's one of the best brief overviews I've come across for some time. What do you expect? Your going to get an eight year (or however many) degree in engineering from a one page blurb on a website?

The information provided is quite thorough, clear and concise, authoritative and backed up by a century of hard experimental science since Joule's experiments along with decades of practical industrial application.

But of course you two clowns trolling the forum know better.

Give me a break.

If you want more detailed in depth information do some research. You act like a baby that needs to be spoon fed every morsel and spits most of it out on your bib.

[VincentG]

The following is from this paper on entropy.
https://www.sfu.ca/~mbahrami/ENSC%20388 ... ntropy.pdf

Entropy Change 
The entropy balance is easier to apply that energy balance, since unlike energy (which has  many forms such as heat and work) entropy has only one form. The entropy change for a  system during a process is: 
 Entropy change = Entropy at final state ‐ Entropy at initial state  system  S final
S  S initial
  
Therefore, the entropy change of a system is zero if the state of the system does not  change during the process. For example entropy change of steady flow devices such as  nozzles,  compressors,  turbines,  pumps,  and  heat  exchangers  is  zero  during  steady  operation. 
Mechanisms of Entropy Transfer 
Entropy can be transferred to or from a system in two forms: heat transfer and mass flow.  Thus, the entropy transfer for an adiabatic closed system is zero. 
Heat Transfer: heat is a form of disorganized energy and some disorganization (entropy)  will flow with heat. Heat rejection is the only way that the entropy of a fixed mass can be  decreased. The ratio of the heat transfer Q/ T (absolute temperature) at a location is  called entropy flow or entropy transfer 
Sheat  Q Entropy tr  
T
ansfer with heat (T  const.)
Since T (in Kelvin) is always positive, the direction of entropy transfer is the same of the  direction of heat transfer. 
When two systems are in contact, the entropy transfer from warmer system is equal to  the  entropy  transfer to the colder system  since the  boundary has no  thickness  and  occupies no volume. 
Note that work is entropy‐free, and no entropy is transferred with work.  
Mass Flow: mass contains entropy as well as energy, both entropy and energy contents of  a system are proportional to the mass. When a mass in the amount of m enters or leaves  a system, entropy in the amount of ms (s is the specific entropy) accompanies it. 
Entropy Balance for a Closed System 
A closed system includes no mass flow across its boundaries, and the entropy change is  simply the difference between the initial and final entropies of the system.  
The entropy change of a closed system is due to the entropy transfer accompanying heat  transfer and the entropy generation within the system boundaries: 

[Fool]

This entire thread seems to confuse 'work out of a volume of gas' verses 'work out of a system'. Not to mention the lack of considering the full cycle.

Expanding gas means work out of the volume of gas. The work out helps the expansion.

A system made up of internal and external gas pressures. Only has work out while the internal pressure is higher than the external pressure. Any further expansion is work into the system. It trades work for temperature/internal energy drop, no net gain, energy balance.

The reverse direction is also true. Higher external pressure during compression is work out of the system only until the pressure inside becomes higher. When the inside is higher than outside compression becomes work into the system. Any further compression is work into the system. It trades work for temperature/internal energy rise, no net gain, energy balance.

Try to visualize a double ended piston with a chamber on each end. Piston dead centered, equal pressure, temperature, and volume both sides equal. Disturb the system by putting in work moving the piston towards one side. One side increases in pressure, second/two side decreases. Let go piston returns to the center, it may oscillate some before stopping at dead center. If frictionless, and adiabatic, it will oscillate forever, unless the initial work is taken back out. The work out will equal the work put in, or less. First law of thermodynamics.

If heat is added to one side. Regardless of how much work is removed, no more than the amount put in, the piston will stop at a new position closer to side two. Side one will have a larger volume. The pressures will be equal, but higher. This is one stroke, heat in energy out. I think I proved in the let's beat up Carnot thread that maximum work out will be quite a bit less. Now becomes the question of how to return the system to dead center. Side one is hotter than side two, so side one must be cooled. Heat must be rejected. To run this as a cycle, heat must be added, then removed. Expansion reduces the amount of heat needed to be rejected, by the amount of work the system is able to perform. There is no way of making the system perform more work that ∆P•∆V. All that can be done is a collection of that work energy that would otherwise go into the oscillation of the piston. That collection, oscillation dampening, is called a load. Adding the load, won't cool it any more than the increase in oscillation it would otherwise gain. This is the second law of thermodynamics and is for a full cycle with a load. Try to calculate the numbers for such a system using PV=nRT rather than denial of science by opinion.

You can make the gas do more work and cool more, but it will cost you the same amount of work to get that cooling.

Unless you can show us your magic path hypothesizing how it's done, your opinion is just blowing in the wind. To deny any PV diagram is to deny science.

But of course you two clowns trolling the forum know better.

Give me a break.

Ignoring that logical fallacy, pure speculation, based on your opinion, Goodwin would shudder, that comment is very much four fingers pointing back at yourself. Quit quoting people that would most likely agree with standard thermodynamic theory, rather than Tom Booth opinion. Sheesh you are rude. Yes I'm a fool to even point this out.

Please try to use scientific principles, instead of cherry picked historic character quotes. And quit putting your opinions into their accounts. Opinions are like trolls, and they all stink. Learn more science. Use only science. Analogies are not science, they are used for the entertainment of the layman.

[VincentG]

Unless you can show us your magic path hypothesizing how it's done, your opinion is just blowing in the wind. To deny any PV diagram is to deny science.

If this is for me, I haven't got to that point, but I don't deny PV diagrams.

[Tom Booth]

(...)
Try to visualize a double ended piston with a chamber on each end. Piston dead centered, equal pressure, temperature, and volume both sides equal. Disturb the system by putting in work moving the piston towards one side. One side increases in pressure, second/two side decreases. Let go piston returns to the center, it may oscillate some before stopping at dead center. If frictionless, and adiabatic, it will oscillate forever, unless the initial work is taken back out. The work out will equal the work put in, or less. First law of thermodynamics.

If heat is added to one side. Regardless of how much work is removed, no more than the amount put in, the piston will stop at a new position closer to side two. Side one will have a larger volume. The pressures will be equal, but higher. This is one stroke, heat in energy out. (...)

Your problem here is that you do not recognize, or apparently do not properly or fully understand, the long established fact of the equivalence of heat and work.

The first paragraph above could be rewritten:

Try to visualize a double ended piston with a chamber on each end. Piston dead centered, equal pressure, temperature, and volume both sides equal. Disturb the system by putting in HEAT moving the piston towards one side. One side increases in pressure, second/two side decreases. Let go piston returns to the center, it may oscillate some before stopping at dead center. If frictionless, and adiabatic, it will oscillate forever, unless the initial HEAT is taken back out. The HEAT out will equal the HEAT put in, or less. First law of thermodynamics.

Substitute heat for work and the resulting effect would be exactly the same. Your second paragraph, implying that the response would be different for heat is absolutely wrong.

Further, the paragraph could be rewritten:

Try to visualize a double ended piston with a chamber on each end. Piston dead centered, equal pressure, temperature, and volume both sides equal. Disturb the system by putting in HEAT moving the piston towards one side. One side increases in pressure, second/two side decreases. Let go piston returns to the center, it may oscillate some before stopping at dead center. If frictionless, and adiabatic, it will oscillate forever, unless the initial heat OR WORK is taken back out. The work out will equal the heat put in, or less. First law of thermodynamics.

That is how heat is converted into work.

Put heat in, adding energy to start the piston oscillating, take work out which takes out energy and stops the oscillation.

The work out will equal the heat put in, or less. First law of thermodynamics.

When adding heat or work, in one way or another that addition must be periodic at the right moment, and at the same frequency as the piston's natural frequency of oscillation, (or at whatever semi- controlled frequency if the oscillations are restricted by a crank and flywheel)

Other than a bit of extra heat in to compensate for friction, for every Joule of heat added to increase the oscillation, a Joule of work can be taken out reducing the oscillation.

Naturally there needs to be sufficient energy "stored" in the oscillating system to act as a buffer to maintain the oscillation but basically if the work out in Joules matches the heat in the system will be in a steady state or "load balanced", where the work out (+ friction or other loses) is equivalent to the heat in.

Your second paragraph above is both theoretically and observably incorrect. Energy input to the system at the right time and the right frequency will have an identical result regardless if that energy input is in the form of work or in the form of heat.

The same is true in regard to the output.

Once an oscillation is fully established, if any of the output is in the form of heat, to that degree, you reduce the oscillation without doing any work.

An important point to keep in mind, I think, is that any changes made to an oscillating system need to be made gradually giving the system time to respond, so if an additional load is applied it needs to occur gradually and the heat input needs to be simultaneously increased gradually to compensate.

[Tom Booth]

Needless to say, contrary to "fools" blathering, this is not "Tom Booth's" opinion it is the well established and recognized 1st law of thermodynamics.

Take this paragraph for example:

The University of British Columbia
PDF Stirling Engine
Hiroko Nakahara – Phys 420

Then the first law of thermodynamics can be introduced by examining two situations. In the first situation, consider adding heat to a system of fixed volume (that is no work is done). Then the internal energy increases by an amount equal to heat added, ∆U = Q. The second case to consider is a gas expanding and pushing a piston. In this case, if no heat is added to or removed from the system, the internal energy of the system decreases by an amount equal to the work done by the system: ∆U = –W. When both cases are considered simultaneously, the internal energy is written as ∆U = Q – W. This equation is the first law of thermodynamics. It should be emphasized that this law is based on careful experimental observation. From my undergraduate education I found that it can be easy to forget which concepts are based on observation and which are derived consequences. Next, three illustrations can be made to show processes for positive, negative, and zero change to the internal energy. Now, since the heat can be written as Q = ∆U + W the internal energy of the system is related its temperature, I can
restate that heat can be transferred into or out of the system both by a temperature gradient and by work done by the system [1, 3].

https://cmps-people.ok.ubc.ca/jbobowsk/ ... oposal.pdf

[Tom Booth]

Just for the sake of clarity, I have no problem with the second law to the effect that the NET work output cannot be 100% of the heat input.

Naturally there will be loses to friction. Though obviously the engine must already be doing work to create friction, therefore the emphasis on "net" work output.

Heat conversion to work is 100% before loses are deducted due to friction, "back work" or whatever.

The "Carnot Limit" however is a radical and unsubstantiated concept that goes far above and beyond the second law insisting that above and beyond friction and other known and somewhat controllable loses an arbitrary quantity of heat must be "rejected" by a percentage based only on the temperature difference. This is derived from the obsolete Caloric theory and nothing else and has no place in modern empirically based science.

Experimentally such huge loses as predicted by the Carnot limit formula, going to the mythical "cold reservoir" cannot be substantiated

Losses to friction is one thing, the arbitrary sequestering of some 90% of the input heat to be reserved for "rejection" to the "cold reservoir" is something entirely different.

[Fool]

The first paragraph above could be rewritten:
Try to visualize a double ended piston with a chamber on each end. Piston dead centered, equal pressure, temperature, and volume both sides equal. Disturb the system by putting in HEAT moving the piston towards one side. One side increases in pressure, second/two side decreases. Let go piston returns to the center, it may oscillate some before stopping at dead center. If frictionless, and adiabatic, it will oscillate forever, unless the initial HEAT is taken back out. The HEAT out will equal the HEAT put in, or less. First law of thermodynamics.
Substitute heat for work and the resulting effect would be exactly the same. Your second paragraph, implying that the response would be different for heat is absolutely wrong.

The piston will be offset from center towards the cooler section, unless pv=nRT, and Q=MCvT are wrong.

VincentG has been demonstrating this by his manual displacer operations.

Adding heat to one side displaces the piston towards the cooler side. It will return only after the heat is removed.