The Carnot efficiency problem

[Tom Booth]

I have build several expansion engines (not exactly Stirling engines), when running on compressed air at perhaps 6 bar, they
easily go down to -50 Celsius on the exhaust, in one stage. At least that's what my (cheap) digital thermometer can go down to.
They also seems to suddenly "get stocked" or freeze up, but I think that because of moisture in the compressed air, as I don´t
do anything special to remove it after compressing.
Btw, this will only happens with load on, because PRESSURE is converted to work during expansion.

We have learnt over and over again that it is HEAT that is converted to work, but there shouldn't be any thing wrong with at gas at 200 bar being at 10 kelvin, and still do "work" on a piston ?
We are also told, that all the energy in a compression process, turns into heat. So when put it in at storage tank to go down to ambient temperature, what energy are you left with ? Still a lot in my opinion.
OK, we then tell our self, that the stored energy comes from the ambient heat energy, and then the equation comes to an equal.

If we imagine an engine/turbine expand a gas at 200 bar/10 kelvin out in freezing/vacuum space will it run at all ? ? ?
If yes, then by what heat ?
Carnot tells us there is still "a lot" energy down to 0 kelvin, so . . .

Thanks for some confirmations, like how cold the exhaust from an "expansion engine" can get running on just 90 PSI or so ("6 bar").

Also: "We have learnt over and over again that it is HEAT that is converted to work" for some reason I often get a lot of push back when I say that, even on science/physics forums.

Anyway, yes, still energy to be had all the way down to absolute zero, theoretically.

My little LTD or any air engine or motor running on air that is not completely "dry" can have issues with H2O freezing and jamming up the works.

And also: "Btw, this will only happens with load on, ..."

I think the heat and/or pressure of the gas is converted to work, but the result is the same, a temperature fall.

It doesn't look like it but this engine is running under a load, because the piston was too tight in the cylinder so I let it run on ice with grinding paste on the piston.

The bottom of the engine kept getting stuck, frozen to the ice underneath it.

This doesn't make sense to me if the engine is supposed to be passing heat through to the "sink" how could the ice re-freeze after already having starting to melt before putting the engine on top. Then re-freeze over and over with this warm engine sitting on top sending heat down into it?


https://youtu.be/2b2dIR8Eql8


This re-freezing of the ice under the engine happened over and over.

[Tom Booth]

BTW, information on the subject seems difficult to find but in reading about air-cycle refrigeration or watching videos, passing mention is sometimes made of the fact(?) that for a gas (air) energy output, such as giving up energy to turn an expansion turbine in an air-cycle system results in a temperature drop only.

That is, if I understand it correctly, no matter what sort of energy a gas might have that it can transfer, the only effect of the transfer of energy will be a temperature drop. Heat, "internal energy", kinetic energy, "spin", "work" whatever.

Maybe that is only theoretical for an "ideal gas" but in general it is what these air-cycle refrigeration systems depend on.

If a gas transfers energy (out) in any form, the temperature falls. So I believe an argument based on the idea the gas is only giving up "internal energy" and not "heat" is kind of missing the point. Either way the temperature goes down.

[matt brown]

If a gas transfers energy (out) in any form, the temperature falls. So I believe an argument based on the idea the gas is only giving up "internal energy" and not "heat" is kind of missing the point. Either way the temperature goes down.

Heat is an ambiguous term, internal energy is not. An isothermal expansion is an xlnt example, and I could care less that a "real one" is impossible. During such, temperature AND internal energy remain constant, yet work can be done. If you need to know nitty-gritty thermo then this is the process to start with. I'm well aware that thermo edu glosses over most stuff, but there isn't enough time to pick everything apart, especially when most students will be engineers, not physicists.

I prefer to consider "heat" engines as pressure differential machines where PV work is achieved via timely volume and temperature differentials (creating a cycle of energy differentials). In this manner, PVT is an order or merit.

[Tom Booth]

If a gas transfers energy (out) in any form, the temperature falls. So I believe an argument based on the idea the gas is only giving up "internal energy" and not "heat" is kind of missing the point. Either way the temperature goes down.

Heat is an ambiguous term, internal energy is not. An isothermal expansion is an xlnt example, and I could care less that a "real one" is impossible. During such, temperature AND internal energy remain constant, yet work can be done. ...

Yes, because for every Joule of work that goes out, a Joule of heat goes in to replace it. But can that actually cause expansion and do real work in the real world?

According to Wikipedia:

"For isothermal expansion, the energy supplied to the system does work on the surroundings"

How are we supposed to interpret that?

1) We supply heat and that heat/energy expands the gas and the expanding gas does external work on the surroundings?????

So,... Now does the gas stay expanded with the heat energy we just supplied to do the work and we need to remove that same heat/energy to get the gas to contract????

So we just violated conservation of energy.

Or

2) we supply heat/energy (Joules) and an equal amount of energy goes out as work (Joules). Like an overflowing bath tub. Equal amounts of energy in as out.

Heat goes in as say 100,000 Joules the gas expands and an equal amount of energy goes out as work. Another 100,000 joules. And then what???

Does the water in a bathtub "expand" when the tub is overflowing???

Does the gas stay expanded???

So we expanded the gas "isothermally" and did 100,000 Joules of work for FREE and now we can take our previously supplied heat/energy back out to contract the gas???

Let's see, 100,000 Joules in. 100,000 Joules out as work and 100,000 Joules back out as heat to complete the cycle?

So we've just turned 100,000 Joules of heat energy into 200,000 joules??? (100,000 joules of heat AND 100,000 joules of work).

Seems like pure fiction to me. An impossible process that violates conservation of energy.

Maybe you could clarify how this works or what the quoted text is supposed to mean because it looks like pure fantasy world academic Caloric water wheel nonsense to me.

The energy that causes water to fall is gravity. So water can expand a cylinder hydraulically and do work and then the water can all be drained out and the process repeated. That's Caloric theory. In Caloric theory, heat is a fluid and is indestructible and cannot be converted into anything else. That's Caloric theory.

Not so with heat in REALITY.

In reality heat itself is energy. Joules of heat and joules of work are energy. You can convert one into the other but you can't produce work out of thin air, and get back the same heat that you just converted into work.

Isothermal expansion is not just impossible for a REAL engine. If heat is energy then isothermal expansion is impossible in principle

Can you, perhaps, produce a definition of "isothermal expansion" that actually corresponds with observable reality in some way, that does not violate conservation of energy?


You "could care less that a "real one" is impossible".

Well I'm a mechanic. I work on real engines in the real world not purple winged unicorn engines that only ever existed in a young Sadi Carnot's imagination where heat can be used to power a heat engine and then come out the other side unchanged like water through a water wheel. I can't make my engines run on fantasy like the academics apparently can.

That (isothermal expansion aka Caloric theory) was an interesting THEORY for some time, but if I'm not mistaken it's been proven wrong.

When heat is converted into work the heat no longer exists. It cannot pass through to the sink because it is gone, converted.

When the heat goes out as work the expanded gas collapses back in on itself. The heat doesn't need to be removed, it's already gone. The expansion can no longer be sustained.

The Carnot cycle is nonsense. An impossible fiction.

[Tom Booth]

Here is a definition from the Encyclopedia Britannica. That should be authoritative shouldn't it?

an isothermal expansion, involves keeping the gas at a constant temperature. As the gas does work against the restraining force of the piston, it must absorb heat in order to conserve energy. Otherwise, it would cool as it expands (or conversely heat as it is compressed). This is an example of a process in which the heat absorbed is converted entirely into work with 100 percent efficiency. The process does not violate fundamental limitations on efficiency, however, because a single expansion by itself is not a cyclic process.

https://www.britannica.com/science/ther ... -processes

Hmmm.....?????

The heat is used to expand the gas, but the heat is "converted entirely into work with 100 percent efficiency".

Seems suspiciously like a violation of conservation of energy there to me.

[Tom Booth]

This is kind of interesting:

Resize_20230803_015015_5500.jpg (37.57 KiB) Viewed 5253 times

Isothermal Expansion

This shows the expansion of gas at constant temperature against weight of an object's mass (m) on the piston. Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant.

https://chem.libretexts.org/Bookshelves ... _Expansion

Someone help me out here because I don't get it.

In an Isothermal expansion of a gas the "volume is constant" or pressure AND volume is constant?

Well it makes sense the volume should be constant, like the overflowing bath, the water level stays the same.

Equal energy in as out...

No expansion. No change in volume. No work?

So why is it called "expansion" if volume is constant?

I may be stupid but this is absurd.

[matt brown]

Yes, because for every Joule of work that goes out, a Joule of heat goes in to replace it. But can that actually cause expansion and do real work in the real world?

It's a slow process for sure vs adiabatic, but all isothermal processes are not equal. As my deep dive into gammas is revealing, there's a world of difference between alphas and gammas.

According to Wikipedia:

"For isothermal expansion, the energy supplied to the system does work on the surroundings"

So,... Now does the gas stay expanded with the heat energy we just supplied to do the work and we need to remove that same heat/energy to get the gas to contract????

sq cycle.jpg (58.59 KiB) Viewed 5251 times

No !!! Per this PV, isothermal expansion 3-4 converted all heat input into work output, so you don't "remove" any energy to compress the gas back (per previous post, the temperature AND internal energy remains constant, despite PV work output). All you need to do is use the same expansion work output as work input to compress the gas and "sink" the same heat energy out during compression that was supplied as heat input during expansion. This is why I said studying isothermal processes is a good start for nitty gritty thermo where the headbanger for most is that an infinite isothermal expansion STILL has the same temperature AND internal energy, we just can't use it for anything beyond a small scale.

So we just violated conservation of energy.

Not at all.

Heat goes in as say 100,000 Joules the gas expands and an equal amount of energy goes out as work. Another 100,000 joules. And then what???

Does the gas stay expanded???

Yep, until you throw some more heat or work at it.

So we expanded the gas "isothermally" and did 100,000 Joules of work for FREE and now we can take our previously supplied heat/energy back out to contract the gas???

Not for free, we had to supply 100,000J of heat energy.

Let's see, 100,000 Joules in. 100,000 Joules out as work and 100,000 Joules back out as heat to complete the cycle?

So we've just turned 100,000 Joules of heat energy into 200,000 joules??? (100,000 joules of heat AND 100,000 joules of work).

No real cycle yet, just expansion and compression along same isotherm where heat=work and work=heat. A cycle will require another isotherm or some other process.

[Tom Booth]

Yes, because for every Joule of work that goes out, a Joule of heat goes in to replace it. But can that actually cause expansion and do real work in the real world?

It's a slow process for sure ...

Right. So called "Quasi-static".

Takes "infinite" time to expand without moving.

Simply more illogical impossible nonsense I would implore every rational person to ignore unless they want to go completely insane.

[matt brown]

This shows the expansion of gas at constant temperature against weight of an object's mass (m) on the piston. Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant.

In an Isothermal expansion of a gas the "volume is constant" or pressure AND volume is constant?

Gadz, they botched this (like computer generated). That last part should read

"and the pressure times volume is constant".

BTW, if you want to see an xlnt, short, simple explanation of enthalpy, check this out...

https://www.youtube.com/watch?v=l0PueMWC-tY

Note in this video he uses E for internal energy since he a chem head vs thermo guys use U for internal energy.

[matt brown]

Simply more illogical impossible nonsense I would implore every rational person to ignore unless they want to go completely insane.

This is exactly what ICE guys say about ECE guys...

[Tom Booth]

(...)
"...All you need to do is use the same expansion work output as work input to compress the gas and "sink" the same heat energy out during compression that was supplied as heat input during expansion. "
(...)

Work "output" is, well, ... OUTPUT.

Like, turning a crankshaft that turns a generator that lights a city.

"Reversible" work output is another impossible myth.

Well, an air spring, maybe some stored energy in a flywheel could be returned but any real work output is long gone.

Transferring heat in and out of a regenerator is not work output.

These "ideal" engine cycles are virtually one and all complete nonsense with virtually no correspondence to reality.

[matt brown]

Here is a definition from the Encyclopedia Britannica. That should be authoritative shouldn't it?

an isothermal expansion, involves keeping the gas at a constant temperature. As the gas does work against the restraining force of the piston, it must absorb heat in order to conserve energy. Otherwise, it would cool as it expands (or conversely heat as it is compressed). This is an example of a process in which the heat absorbed is converted entirely into work with 100 percent efficiency. The process does not violate fundamental limitations on efficiency, however, because a single expansion by itself is not a cyclic process.

https://www.britannica.com/science/ther ... -processes

Hmmm.....?????

The heat is used to expand the gas, but the heat is "converted entirely into work with 100 percent efficiency".

Seems suspiciously like a violation of conservation of energy there to me.

Yep, they nailed it (we cross posted). I just wish they'd managed to explain this without "must absorb heat in order to conserve energy" which sounds dorky to newbies.

Among a host of mumbo-jumbo, I also don't worry about the 1st and 2nd laws, I'm only concerned with the cluck:buck ratio (input:output ratio).

[matt brown]

These "ideal" engine cycles are virtually one and all complete nonsense with virtually no correspondence to reality.

Again, exactly what ICE guys have been saying about ECE guys for decades.

Most guys that dabble with ECE have a two year walk thru, then wander off and return to reality. But some will go to their grave convinced that the Holy Grail is waiting for them (like PMD guys).

[Tom Booth]

This shows the expansion of gas at constant temperature against weight of an object's mass (m) on the piston. Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant.

In an Isothermal expansion of a gas the "volume is constant" or pressure AND volume is constant?

Gadz, they botched this (like computer generated). That last part should read

"and the pressure times volume is constant".

(...)

I don't think they botched anything. The idea of "isothermal expansion" as part of an "ideal" (aka Carnot) engine cycle is just inherently full of contradictions.

Logically on the other side of the piston you have atmospheric pressure, which is constant.

The piston expands slowly as the temperature increases so as to remain in balance with that external pressure. Ergo, pressure is constant.

If P*V is constant and pressure is constant then volume must also be constant.

That leads to logical knots and conundrums.

See this lengthy conversation for example:

https://chemistry.stackexchange.com/que ... -expansion

This lunacy can twist your brain into knots to the point where the knots can no longer be undone. At that point you can have a career teaching thermodynamics.

[matt brown]

Poor text leads to poor conclusions, see Boyle's Law

https://en.wikipedia.org/wiki/Boyle%27s_law