The Carnot efficiency problem

Discussion on Stirling or "hot air" engines (all types)
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Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

Fool wrote: Fri Jul 28, 2023 5:32 pm ...

I also understand that the points on a PV diagram are 'states', meaning a single temperature can be calculated for any single point. Adding heat or work would move from that point, on a specific path, to a new point determinable by the amount of heat or work being applied. The path is determined by heat and or work.

...
Yes, but, my point is, in a real engine these things happen simultaneously. Heat input and work output.

If I can get from point A to point B by adding 500,000 joules of heat, without doing any external work, what if the engine is simultaneously outputting 450,000 joules of work to an attached load. Turning an electric generator for example. It isn't going to get to point B. Maybe it will get 10% of the way from A to B.

In effect, the 450,000 joules of work has canceled out all but 50,000 joules of heat input, so that 450,000 joules of heat input along with work output is not represented on the PV diagram.

In other words, it appears that the PV diagrams do not account for the real time conversion of heat into work. A PV diagram assumes that any work output is a consequence of the working fluid taking in and transporting "caloric" without any of this "caloric" being lost. Without any "caloric" being converted along the way.

At that rate of heat input and work output, to actually get all the way from A to B will take about an additional 4 million joules of both heat input and work output, all of which will be unaccounted for on the PV diagram.
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Re: The Carnot efficiency problem

Post by Tom Booth »

Here is a screenshot of a real time PV diagram.
Resize_20230729_234917_7771.jpg
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The outer oval is a PV tracing with the engine under load. The smaller inner oval is the PV diagram tracing of the engine running no-load.

The video the screenshot is from is here:

https://m.youtube.com/watch?v=dvomod6SsA0

It appears that the "under load" oval indicates more work as might be expected, but I wonder.

When the load is taken off the engine speeds up so there is less time for heat exchange.

The heat being applied to the engine, according to the video description, is from a cup of hot water, so the "available heat" is basically constant. Gradually decreasing an insignificant amount over the course of the recording.

I think if my theory is correct, if the heat input were increased, increasing the RPM the "under load" tracing would return to looking like the lean "no-load" tracing. The additional heat input and work output would go unrepresented. An engine running with little heat and no load would trace the same path as an engine with a high heat input and a heavy load. More heat and more work but the same "path" on the PV diagram.

That is speculation at this point. As of now, it at least LOOKS LIKE the additional work is being represented by the larger area within the larger oval.

I would like to see what happens when the heat input and work output are varied simultaneously.
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Re: The Carnot efficiency problem

Post by Tom Booth »

kalapacs wrote: Sun Jul 30, 2023 12:19 am
I think the electricity as an analogy is good, because in every energy conversion device, a part of the absorbed power becomes a loss. This must also be taken into account for electrical equipment. If a drilling machine is poorly designed, its efficiency will also be poor,
(...)
That is the crux of the issue, I think. (Highlighted in bold)

If a _______fill in the blank______machine is poorly designed, it's efficiency will also be poor.

We know enough about electricity to design electrical power equipment of very high efficiency. An electrical transformer can be nearly 100% efficient, and I guess we don't want to forget about superconductors.

"Carnot efficiency" however, dictates that the "blank" can never be filled by "Heat Engine".

We supposedly can never learn enough about how a heat engine operates to have anywhere near 100% efficiency.

Designing a more efficient heat engine is simply not possible as this heat engine efficiency limit is a "Law of the universe".

In fact, we can do nothing at all to improve heat engine efficiency so why bother even trying? Don't waste your time!

In my opinion, we already have heat engines that exceed the so-called "Carnot efficiency" by far, but nobody noticed, because nobody expected it or bothered looking for it or actually tested for it.

A Carnot engine is a "winged unicorn". There is no such thing. It is a fiction. It's mode of operation is a fiction. It's so-called "efficiency limit" is a fiction. It's alleged superiority above any other engine is pure mythology. Complete nonsense, the flight of a young man's imagination who later thought better of it as he matured in reason and discarded the whole theory. Others however, took up the error with a fanatical zeal and so we've had this idea inflicted on us down through history and now I find myself saddled with the chore of dismantling it bit by bit, A mostly thankless job, while under constant attack from the "Carnot efficiency" zealots and "second law" fanatics.
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Re: The Carnot efficiency problem

Post by Tom Booth »

matt brown wrote: Sat Jul 29, 2023 12:48 pm
Tom Booth wrote: Sat Jul 29, 2023 6:57 am
Well, to keep things simple (LOL, maybe not so simple, but anyway)
(...)

Tom, according to this, an adiabatic expansion would have a lower temperature after expansion when driving a load vs no load.

...
And?

You point this out because....?

My impression based on your statement and the context is that you think because: "according to this, an adiabatic expansion would have a lower temperature after expansion when driving a load vs no load" is a basis for an offhand dismissal. (???)

Do you have some evidence or citation to the contrary?

I've been saying this same thing in here for the past decade or more.

Quoting myself:
In other words, by making a model (or any other) Stirling Engine do some work rather than just freewheeling, more heat would be converted into work and the temperature differential would increase proportionately to the amount of work being done.
From the beginning of the Stirling Engine Thermodynamics thread;

viewtopic.php?f=1&t=478#p1225

That is not, or was not MY idea, it was what I was reading in the literature about how it was found that many "impossible" to liquify gases were finally made cold enough to liquify.

The method that was found capable of liquifying these difficult or impossible to liquify gases was, and still is, adiabatic expansion through an engine (or turbine) that has a load applied.

This is the basis for air cycle refrigeration. Adiabatic expansion driving an engine that is driving a load.

This is the basis for the Claude method of gas liquidation, etc.
How does air cycle work?

...Air is compressed... this air is then expanded to a lower temperature than before it was
compressed. Work must be taken out of the air during the expansion,... Work is taken out of the air by an expansion turbine, which removes energy as the blades are driven round by the expanding air. This work can be usefully employed to run other devices, such as generators or fans.
https://www.grimsby.ac.uk/documents/frp ... search.pdf

The gist of all this is that cooling or refrigeration (to cryogenically cold temperatures) is accomplished by exactly that. Expanding the gas adiabatically in a cylinder to drive an engine, the load on the engine removes "work" and this method produces temperatures cold enough to liquify even helium.

One of my references from back in 2010 on the thermodynamics thread:
"Another way to cool a gas is to have it do work adiabatically" (without heat transfer) "against a piston in an engine, and this has no temperature boundary like the Joule-Kelvin effect. Gas liquefaction using such engines is now common not only for helium, but also for air. These liquefaction plants are small and easily handled, so that every lab can have its own source of liquid air"

Helium -

(Updated link on the internet archive:)

https://web.archive.org/web/20100615230 ... helium.htm
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

That should be "The Claude method of gas liquefaction" not 'liquidation'.

Autospell.

This (adiabatic expansion of a gas through an engine with a load applied) is the basis for many low temperature industrial processes.
Resize_20230730_134812_2273.jpg
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https://www.airliquide.com/group/air-liquide-brief
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Tom Booth wrote: Sun Jul 30, 2023 10:16 am
matt brown wrote: Sat Jul 29, 2023 12:48 pm

Tom, according to this, an adiabatic expansion would have a lower temperature after expansion when driving a load vs no load.

...
And?

You point this out because....?

My impression based on your statement and the context is that you think because: "according to this, an adiabatic expansion would have a lower temperature after expansion when driving a load vs no load" is a basis for an offhand dismissal. (???)

Do you have some evidence or citation to the contrary?

I've been saying this same thing in here for the past decade or more.
Yikes, I really wonked on that wording !!! (wish I could edit that post)

An adiabatic expansion into a vacuum has no temperature change for an ideal gas, but real gasses do (not nearly as extreme as 'working' adiabatic expansions). The thermostat in modern ICE is for correcting this load factor issue and rpm sink. Early ICE were big, slow, low compression (read low heat) and low powered, so this variation was largely obscured (and before auto mania took over, most ICE were stationary beasts designed for narrow output range). Modern ICE are far smaller and with a relatively massive power range where load vs no load issues show themselves. You could easily test this via removing thermostat and measuring coolant temperature load vs no load while at same rpm with a few corrective factors thrown into the mix (fuel consumption, work output). This would be an interesting test and likely already on youtube (probably even diesel vs gasoline). I'd bet that the temperature difference (load vs no load) is no where what you're thinking, since there's a built in load on ICE even idling. I saw a video recently of a guy firing up an old 6-71 that had been sitting for years and it ran great for age. This guy ran it at idle for quite a while without a radiator and ran around the engine several times with his digital thermometer showing various low readings (kinda made me wonder about oil temp).

Your idea of stair stepping 'down' an expansion adiabat then compressing up the same adiabat to start state is creative. But this deceptive down motion will move up to higher and higher adiabats. PV plots have P and V linear and everything else is curves such that isotherms slowly converge and adiabats slowly converge, but each process group never terminates (no common point). This is the challenge of scheming any gas cycle, since a cycle requires a return to start state. The notorious cheat is the open cycle which gurus often relate as open referring to PV plot vs open referring to ambient/reservoir.

I tend to stick with conventional PV plots since I'm after work. I really could care less about TS plots, entropy, enthalpy, Gibbs and a bunch of other mumbo-jumbo. Coming from the steam camp, what struck me odd about single-phase gas was the lame output per cycle (rpm) but only those guys who've been around steam will know what I mean. Anyways, I hung in single-phase gas as a sidebar only due to an extensive knowledge of ICE history...how the low powered monsters of past became out modern commonplace ICE. Nevertheless, single-phase gas has little power per stroke until you increase charge pressure and/or the cycle rate (rpm).

A reefer can use COP to their advantage, not as an efficiency model, but as an achievement means whereby a massive mechanical input can achieve a meager reefer effect. However, an engine has no such advantage and often struggles to achieve meager output from massive heat input. Beale invented the FPSE in 1964 and he died in 2016 whereupon his heirs quickly sold off his rights. Many here have heard of this guy and are familiar with his free piston scheme/s, but NASA still has no credible FPSE engine despite dumping mega $$$ into such. The whole free piston scheme appears ok for a reefer, but not an engine.

I still think that something like your Hot Potato is the best chase with an open cycle to ambient or reservoir.
Fool
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Re: The Carnot efficiency problem

Post by Fool »

I like to know, for the engines discussed here, how many of them could have expansion without a load.

If there is no outside restrictions, braking, electric generation, oar in water or air, wouldn't the work just be put fully into accelerating the mass(free piston, crank flywheel, or just a mass of air)?

Faster: the work goes into piston speed. Slower: the work goes into piston speed and friction of the brake.

Work is F•∆X. Or P•∆V. Won't that manifest as acceleration. Which becomes 1/2mV^2 joules? Work? The m varies from engine to engine.

Wouldn't that then mean all heat engines, here, have expansion with work? External load or not?

It is interesting to look at the real LTD Stirling Engine's indicator (PV) diagram and see how; as it is put under load the net work per cycle increases significantly, as does heat input(path), pressure, temperature, and work output(integral of path or area enclosed). It even shows the work per stroke both in and out (integral of a half cycle, the area under each stroke). Tom thanks for finding and displaying that diagram here.

What it doesn't appear to show is power(work per time). No visualization of RPM's.
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Re: The Carnot efficiency problem

Post by Tom Booth »

Matt, I don't know what's wrong with your wording.

You also don't seem to respond directly to the question or issue.

In gas liquefaction and air cycle refrigeration:

1) a gas is compressed to very high pressure and the heat of compression removed. (The compressed gas is cooled to ambient or lower) then the cooled compressed air is

2) released to drive an engine. The expanding gas powers the engine.

The work output of the air engine is what results in an extreme temperature drop of the air used to drive the engine (or turbine) Enough cooling to liquify helium and other gases which cannot be liquified by any other method.

This type of cooling has no temperature limitation.

That is, potentially it can cool almost down to absolute zero, (inside the power cylinder) and that is by WORK output. That is, by converting heat into work using the compressed gas to drive an engine.

Somewhat remarkably, in practice, the load used to remove "work" to effect cooling is the compressor compressing the gas coupled to the same crankshaft as the engine.

This does not work without an external load on the air engine. (Or the load can be the compressor)

I was a bit shocked when first reading about this as this all seemed very counter intuitive.

Now, aside from the pre-cooling, which is needed to reach extremely cold temperatures, it occured to me that a Stirling engine follows a similar 1 - 2 compression followed by expansion with work output.

If an adiabatic expansion of a compressed gas can result in such mind boggling cold, near absolute zero temperatures, (potentially), if a similar process takes place in a Stirling engine, even just a little bit, compared with the high compression used to liquify Helium, shouldn't there be a similar potentially extreme cooling effect?

So when someone says something like: "Not rejecting any heat costs all the work for compression" I'm like....

I think that potentially, adiabatic expansion with work output might be plenty of cooling all by itself.

If very high compression followed by adiabatic expansion with work output can result in temperatures near absolute zero, the modest compression and adiabatic expansion with work output, in a Stirling engine should at least result in some substantial cooling effect.
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Re: The Carnot efficiency problem

Post by Tom Booth »

I first encountered this phenomenon while being interviewed for a mechanic engine repair job.

The guy I was replacing got his fingers too close to the exhaust port on one of the air tools he was using and the cold exhaust froze his finger which broke off.

His friggin finger froze and broke off.

This is an ordinary tank of compressed air.

https://youtu.be/2hYQtB4QkEY

That is without powering any air wrench. What you see in that video is just normal cooling by ordinary expansion.

Expansion of a compressed gas through an air tool removing additional energy as "work" froze a guys finger so it snapped off.

That shop had a very large compressor and it was out in the Arizona desert where it can get very cold at night so the compressed air was likely pretty cold sitting over night in the tank. So, it partially mimicked the Claude method for making liquid air with just an ordinary shop compressor and impact wrench as an "expansion engine".

When my future employer told me the story about why I was getting the job my mouth must have dropped.

He told me not to worry, I wasn't going to be using that air tool, not that there was anything special about it. The guy had just been using it to remove a lot of big head bolts from a big engine. I would just be working on lawn mowers and chain saws.
Fool
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Re: The Carnot efficiency problem

Post by Fool »

When a piston expands a gas, the gas does work on the piston. This is adiabatic temperature drop with work. It doesn't need the piston to drive a load.

When a piston compresses a gas the piston does work to the gas. This means there is a force, or momentum, driving the piston in. This is adiabatic temperature rise with work. The energy/work, driving the piston in, is absorbed by the gas.

If both strokes are identical and adiabatic, the temperature volume and pressure will be the same as the initial point. Zero net work will be available to output. Area enclosed will be zero. One adiabatic path/line. Work gained minus work applied equals zero.
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Re: The Carnot efficiency problem

Post by Tom Booth »

This again is just compression of a gas in a tank (at room temperature) followed by release (expansion) without any mechanical work output to make dry ice.

Imagine if the CO2 were expanded to drive an engine or turbine. The cold that could potentially result.

https://youtu.be/unodzMmH3dk
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Re: The Carnot efficiency problem

Post by Tom Booth »

Fool wrote: Sun Jul 30, 2023 7:26 pm When a piston expands a gas, the gas does work on the piston. This is adiabatic temperature drop with work. It doesn't need the piston to drive a load.
All the literature I've studied on the subject, some of which I've already posted, states explicitly that to reach the extremely cold temperatures necessary for liquefaction of gases (or air) a load on the expansion engine is necessary.

The work output to the load removes energy from the gas as "work" which results in much more effective cooling than expansion without a load, without work output.


When a piston compresses a gas the piston does work to the gas. This means there is a force, or momentum, driving the piston in. This is adiabatic temperature rise with work. The energy/work, driving the piston in, is absorbed by the gas.

If both strokes are identical and adiabatic, the temperature volume and pressure will be the same as the initial point. Zero net work will be available to output. Area enclosed will be zero. One adiabatic path/line. Work gained minus work applied equals zero.
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Re: The Carnot efficiency problem

Post by Fool »

I'm just guessing, but makes sense, that the load is there just to prevent over revving.
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Fool wrote: Sun Jul 30, 2023 6:16 pm I like to know, for the engines discussed here, how many of them could have expansion without a load.
All of them.
Fool wrote: Sun Jul 30, 2023 6:16 pm If there is no outside restrictions, braking, electric generation, oar in water or air, wouldn't the work just be put fully into accelerating the mass(free piston, crank flywheel, or just a mass of air)?

Faster: the work goes into piston speed. Slower: the work goes into piston speed and friction of the brake.

Work is F•∆X. Or P•∆V. Won't that manifest as acceleration. Which becomes 1/2mV^2 joules? Work? The m varies from engine to engine.

Wouldn't that then mean all heat engines, here, have expansion with work? External load or not?
The average output per stroke is quite low due to backwork ratio (back pressure, friction). Early steam engines were quite similar (low output ratio) but as the output ratio increased with higher cylinder pressures, the famous ball governor became standard or SHTF. All heat engines do some expansive work whether external load or not. Per previous comm, I'll look for a youtube video on load vs no load values.
Fool wrote: Sun Jul 30, 2023 6:16 pm It is interesting to look at the real LTD Stirling Engine's indicator (PV) diagram and see how; as it is put under load the net work per cycle increases significantly, as does heat input(path), pressure, temperature, and work output(integral of path or area enclosed). It even shows the work per stroke both in and out (integral of a half cycle, the area under each stroke). Tom thanks for finding and displaying that diagram here.

What it doesn't appear to show is power(work per time). No visualization of RPM's.
Be cautious with this PV plot. One SE paper I have outlines how typical beta/gamma PV plots are a composite of TWO separate PV plots. Geez, these guys should do a study on how this evolved and slips past everyone. I'm still doing a deep dive on gammas, and baffled by some glaring omissions, but my favorite is their PV plots...simply try to draw a real PV plot for any beta/gamma and you'll see what I mean. This is the only engine cycle I've ever seen that escapes an actual PV plot.
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Re: The Carnot efficiency problem

Post by matt brown »

Tom Booth wrote: Sun Jul 30, 2023 6:54 pm
In gas liquefaction and air cycle refrigeration:

1) a gas is compressed to very high pressure and the heat of compression removed. (The compressed gas is cooled to ambient or lower) then the cooled compressed air is

2) released to drive an engine. The expanding gas powers the engine.

The work output of the air engine is what results in an extreme temperature drop of the air used to drive the engine (or turbine) Enough cooling to liquify helium and other gases which cannot be liquified by any other method.

This type of cooling has no temperature limitation.

That is, potentially it can cool almost down to absolute zero, (inside the power cylinder) and that is by WORK output. That is, by converting heat into work using the compressed gas to drive an engine.
I imagine this would favor a turbine due to minimal contact area. I'm no reefer guy, but got the basics, and attempting compression cycle this low is a no go since there's no cooling possible, not to mention typical compression cycle will lean towards heat addition regardless of design.

I didn't know they still used this process, but likely the most proven and cheapest. I thought that they'd moved on to vacuum chambers. The obscure downside to this expansion process is just how crazy an expansion is required, must require many stages.
Tom Booth wrote: Sun Jul 30, 2023 6:54 pm
Somewhat remarkably, in practice, the load used to remove "work" to effect cooling is the compressor compressing the gas coupled to the same crankshaft as the engine.
This is sweet, since minimal work is required for reefer expansion at these low temperatures while more work is gained from engine expansion further up the scale.
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