The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

[Tom Booth]

Taking this sample problem:

T1=400°K
T2=300°K
efficiency = 1-300/400 = 1/4 = 25%

https://www.toppr.com/ask/question/the- ... pectively/

That is fine. I agree with it!

The question is 25% of what?

25% of T1, all the heat down to absolute zero. Or 25% of 400 joules using "fools" normalization.

100 joules available to convert to work.
100 joules added to the engine

100% thermal efficiency.

"Carnot efficiency" then is a kind of euphemism for 100% thermal efficiency.

Always.

Not 25% of T1-T2 or the mere 100 joules added above the equilibrium baseline.

T1 is the full volume of "the caloric" without canceling out T2

For a proper energy assessment T2 is zero.
100 joules added, 100 joules converted, zero rejected

[Tom Booth]

If we assume from all this then, that Tesla's proposed "self acting engine" that runs on ambient heat due to the presence of a "perpetual" cold, how would such an engine be optimized?

Compress_20240427_222542_2565.jpg (27.56 KiB) Viewed 2345 times

Well, we need very good insulation to protect the ice from the surrounding ambient heat.

The piston has been moved up away from the "cold hole".

It might also be helpful to make the displacer thicker (taller) to put more distance between the heat expanding the air from above.

Of course, the entire engine including the displacer, as far as material constraints allow, should be of non-heat conducting materials

So how does this work?

Starting with the displacer raised up and the working fluid already cooled by the ice, the displacer is lowered admitting some heat and expanding the working fluid.

As the working fluid begins to expand, the displacer rises back up discontinuing the heat input as the power piston continues moving and the expanding gas does work turning the engine with the result that the working fluid cools back down due to conversion of the supplied heat into "work".

As the engine revolves around towards TDC the displacer again falls admitting some heat that again expands the gas and the process repeats.

Probably making sure the power piston is big enough and it's traveling distance is far enough would help ensure complete expansion and heat utilization.

As soon as possible, as heat is introduced and the piston reaches TDC the "cold hole" is covered by the displacer to ensure that the "heat of compression" does not heat any working fluid in contact with the ice.

In other words, the displacer should make positive contact with the cold plate as heat is being introduced above so as to "protect" the ice from heat generated from compression as the piston reaches TDC and heat is introduced.

As cold air tends to sink and hot air rises, some means of forcing hot ambient air downward to make contact with the engine and transfer heat to the top hot plate should be incorporated, such as fan blades attached to the flywheel to keep the ambient air in rapid circulation above the engine.

In spite of all such efforts, inevitably, in all likelihood, some heat is still going to reach the ice.

Theoretically the engine could gradually store up enough energy from power generation derived from ambient heat, some form of heat pump could be operated intermittently as needed to "recharge the cold battery" if necessary.

Also the engine itself can act as a continuous refrigeration system by slightly extending the pistons travel to expand and.cool the working fluid at the end of the power stroke while the bottom cold plate is still exposed to the expanded and cooled working fluid, in effect "trapping" additional cold in the ice.

Recently Kontax released a new engine design which might be useful for experimenting.

Compress_20240428_200834_4703.jpg (24.72 KiB) Viewed 2345 times

https://stirlingengine.co.uk/products/K ... p593860804

Not fully optimized as described above, but it's a place to start perhaps, with the elevated power piston, heat of friction is kept away from the cold side while running on ice.

[Fool]

Start out with three hundred Homunculus's, little humans, inside an engine pushing out. Add one hundred more. As they push out they get converted to work. After 100 have been converted to work expansion stops.

Pushing the piston back in using the work that came out, causes 100 Homunculus's to reappear using up all the work. The gas is now 100 Homunculus's hot. Cooling it by removing 100 Homunculus's brings it back to the cycles beginning. Zero total heat equals zero total work.

Now again compress the piston instead of transforming work for Homunculus's, cool the engine as work goes in. Lose a Homunculus, push it in. Lose another Homunculus push it in. Still fighting 300 Homunculus the whole way. 75 units of work are needed to return to the cycle starting point.

Why 75. If we expanded it until all were converted to work, 400 Homunculus's would be converted. We would end up after compression with again 300 Homunculus's. That is 75% of the work back in. So 75 Homunculus's, 75%, would reappear for the back work of 100 forward work.

You, I don't know why, always want to add atmospheric Homunculus's. 300 inside, 300 outside. They only can push. Inside pushes out. Outside pushes in. No energy because no motion or volume change. Forces cancel, energy only flows in or out. Flowing in negative. Flowing out positive.

Heat the gas up by adding Joules, 100 J equals 100 Homunculus's. Now there are 400 Homunculus's pushing out.

Expand it until 100 are converted to work. The work is stored in the inertia MV of the engine.

Compressing it back without cooling causes 1 Homunculus to reappear for 1 Joule. Giving again 400 Homunculus's inside for -100 J of work input. Cooling it to get to the starting point, cost again 100 Homunculus's, (300 inside, 300 outside.) Zero zero again.

Compress it with cooling. One Homunculus's gone, 300 outside 299 inside. Out side compresses inside causing one Homunculus to reappear. It will cost 75 Homunculus's to return to the starting point.

Why 75? Same reason as above. It's cooler for the return stroke. 299/300 Homunculus's the whole way. If it were expanded until all Homunculus's were converted the outside Homunculus's would have to be zero so it could Expand that far. The maxim would be 400 J. 300 J would be used to make 300 Homunculus's reappear to return to zero. This is a 75% use of Joules per Homunculus. Lending, 75% back work to forward work.

It would be more efficient with more Homunculus's on the inside.

If you added only one Homunculus, to the inside, it would expand only one J, one Homunculus. It would cost 299/300 J to return, a short round trip, and a small amount of energy saved for output. 1/300 J maximum, less after friction, this is the reason LTD engines produce so little power on so little temperature difference. Milliwatts.

[Tom Booth]

Start out with three hundred Homunculus's, little humans, inside an engine pushing out. Add one hundred more. As they push out they get converted to work. After 100 have been converted to work expansion stops.

Pushing the piston back in using the work that came out, causes 100 Homunculus's to reappear using up all the work. The gas is now 100 Homunculus's hot. Cooling it by removing 100 Homunculus's brings it back to the cycles beginning. Zero total heat equals zero total work.

...

More like 100 are converted to momentum.

The piston continues moving using up 100 "homunculus power" of momentum which allows another 100 to be converted to work powering an outside load.

This leaves 200 inside and 300 outside.

100 reappear inside during compression restoring the 300<=>300 balance

Another 100 are added to begin another cycle.

There is no need for "Pushing the piston back in using the work that came out".

The "work that came out" (momentum) is used to continue the expansion, which assists turning another 100 to work.

Theoretically that additional work could provide some additional momentum, to produce more work output.

The piston returns easily by atmospheric pressure, which is still 300 as the inside pressure dwindles precipitously.

This PV diagram (real readings) shows the precipitous dive down in pressure at BDC under load (large egg shaped tracing)

Compress_20240428_222416_6993.jpg (22.45 KiB) Viewed 2333 times

Video source: https://youtu.be/dvomod6SsA0

Atmospheric pressure cuts through the center of the egg.

Internal pressure does not return to equilibrium until AT TDC.

As far as I can tell, this engine is not subject to any external below ambient cooling. The pressure drop under load (additional work out), however, is much greater than no load.

[matt brown]

Taking this sample problem:

T1=400°K
T2=300°K
efficiency = 1-300/400 = 1/4 = 25%

https://www.toppr.com/ask/question/the- ... pectively/

That is fine. I agree with it!

The question is 25% of what?

25% of T1, all the heat down to absolute zero. Or 25% of 400 joules using "fools" normalization.

100 joules available to convert to work.
100 joules added to the engine

100% thermal efficiency.

"Carnot efficiency" then is a kind of euphemism for 100% thermal efficiency.

Always.

Not 25% of T1-T2 or the mere 100 joules added above the equilibrium baseline.

T1 is the full volume of "the caloric" without canceling out T2

For a proper energy assessment T2 is zero.
100 joules added, 100 joules converted, zero rejected

I think I figured out a simpler way to describe this...

First, the source input is ALL at 400k, not 300-400k which is regen

The bugaboo is that 'relative' 100J between 300k and 400k, but if we consider that internal energy (U) is linear temperature (Kelvin) we can relate U=T whereby 100k=100U, 200k=200U, 300k=300U, 400k=400U.

Next, FOR THE SAME VOLUME RATIO the work capacity varies linear T and U. When 400k=400U then 400Q=400W, but when 300k=300U then 300Q=300W, and you can't have Q>300 at 300k for THIS same volume ratio. In this manner, 400k yields a 400 value series, and 300k yields a 300 value series. Each series yields Q=W (1.0 eff) where the only major difference is that at 400k, more source input yields more work output. In effect, at 400k, everything transpires at a higher 'rate' than at 300k.

That covers the expansion process where Q=W=1.0 eff but to complete the cycle we need a compression process. And since this is a 300-400k cycle, if we rewrite Q and W with a "direction", then during 400k expansion 400Qin=400Wout while during 300k compression 300Win=300Qout. And again we find that Win is 3/4 Wout and that Qout is 3/4 Qin and Carnot wins again.

You seem to be always dismissing the backwork of compression as if the ambient compression of your LTD is a free lunch thanks to Mother Nature. If not for the backwork, than heck yeah, your LTD would be 1.0 eff.

[Tom Booth]

I may be wrong, but given that pressure and temperature are more or less associated, I can't help but wonder how the working fluid falls, and stays, so far BELOW atmospheric pressure, all the way back to TDC.

Resize_20240428_230612_2510.jpg (122.68 KiB) Viewed 2327 times

A drop in pressure down to around atmosphere, can be attributed to the displacer shifting the gas to the ambient/atmosphere(cold) side, but can that account for the pressure dropping far down below atmosphere? Would that not also indicate a more or less parallel temperature drop under load as well, due to the added work load when the sponge is being applied?

The most frequently voiced theory seems to be that under load the engine slows down providing more time for heat exchange.

Why there is such a strong and persistent denial from some that a drop in temperature can be attributed directly to the work output I don't know.

Work output by a gas in an expansion engine is a commonly used industrial method for reaching cryogenic temperatures, for liquefaction of difficult to liquify gases, like helium. If a gas won't liquify by normal gas liquefaction methods, by cooling and high pressure, doing work in an expansion engine will do the job.

Why is the cooling of the gas that results from work output so underestimated and undervalued, often as if it is not a factor at all. Matt and Fool on the forum, and there have been others, consistently po po the suggestion that work output during expansion results in any significant cooling that could explain the power piston returning with such apparent ease and force.

I think probably because old school thermodynamics did not recognize the actual CONVERSION of heat into work.

Yes, expanding gas PRODUCES work, but heat is not itself CONVERTED, so as to "produce" COLD.

The heat is converted. Cold is what is felt in its absence when converted into something else. I can understand it's a hard pill to swallow, that something so tangible to us, that we can feel, something that keeps us warm and alive can simply VANISH.

[Tom Booth]

...
You seem to be always dismissing the backwork of compression as if the ambient compression of your LTD is a free lunch thanks to Mother Nature...

I don't think LTD's are functionally any different than so-called "thermoacoustic" or lamina flow, thermal lag.

The pistons return EASILY, and swiftly with as much force as the power stroke due primarily to loss of internal energy of the gas due to WORK output.

It is the definition of a heat engine that it converts HEAT into WORK.

I think the cooling that results from work output has hardly been recognized at all in thermodynamics, historically, and now in more recent times, though it is begrudgingly admitted SOME heat is converted to work, the quantity of heat transformed so as to "disappear" is still greatly underestimated.

In this experiment I heat the "regenerator" from end to end so it cannot be said that any part of that is a "sink", for heat in the working fluid.

I've replaced the aluminum engine body, that some consider a "sink" for heat, with a wooden engine body I turned on a lathe. Wood is not an effective heat sink I don't think.

I can't see much heat leaving through the glass power cylinder, as glass is a poor conductor of heat, but just to be sure, I wrapped it with Styrofoam.

There is no "cooling" of the working fluid possible other than by WORK; conversion of the heat into mechanical motion.

If it were not 100% this engine could not complete a single cycle. However there is no problem whatsoever with the piston returning:

https://youtu.be/LG09AXAjpio

How forcefully does the piston return?

Inverted it can lift a substantial weight.

https://youtu.be/e-7DFp_B0y4

You don't think the work output, (CONVERSION of heat into work) is sufficient to allow the piston to return?

Well, all I can say is your blinded by your academic/thermodynamics indoctrination or something. Plainly, I've shared an abundance of evidence that these little model Stirling engines continue to run and do not overheat even when high heat is supplied and there is no heat outlet or effective sink.

Carnot doesn't "Win".

Carnot is obsolete.

If you want to live in the past and cling to caloric theory, feel free. I say it's complete nonsense. I don't know how anyone could view so many of the videos of such experiments that I've posted in here and still be blind to what is so obvious.

Just as Tesla stated back in 1900, a heat engine doesn't require a "sink" for heat. It converts heat.

Even Carnot, in his unpublished lab notes wrote that if heat is motion, or a form of energy, he did not understand why a "cold body" should be needed.

Kelvin and others took thermodynamics backward not forward.

[Tom Booth]

Just to be clear, atmospheric pressure is at the horizontal line.

Compress_20240429_015658_8970.jpg (34.97 KiB) Viewed 2314 times

The piston returns 99% of the way by mother nature's "free lunch" as you call it.

Actually all the way. The piston has momentum. That too comes from atmospheric pressure.

You have head of Newtons laws of motion I assume. Objects once set in motion stay in motion.

I call it the result of heat conversion into work.

Heat is Input of energy. No "free lunch".

Your refrigerator operates by plugging it in. An input of energy. You don't think heat, as a form of energy can result in a refrigerating or cooling effect?

You've never seen a Servel?

Not much work output from a Servel gas refrigerator. Converting heat into work is a more effective cooling process than absorption refrigeration. Ask the people who liquefy helium.

[matt brown]

Why is the cooling of the gas that results from work output so underestimated and undervalued, often as if it is not a factor at all.

Because a heat engine runs on a pressure differential, it ain't a thermo couple.

To borrow Vincent's planet Bob analogy...

On planet Bob there's 400k days and 300k nights where a bunch of Fools generate meager power only once a day via compressing 'air' at night when it's 300k and expanding it the next day when it's 400k and pressure is above ambient. Meanwhile, on planet Tom they generate endless power day and night (albeit a tad less at night) via a clever cold hole trick that some aliens taught them.

I think probably because old school thermodynamics did not recognize the actual CONVERSION of heat into work.

Obviously, your thermo history is lacking...the kinetic theory was the longest running controversy in scientific history. Most think it was the atomic theory and are wrong...the atomic theory wasn't considered valid until mid 19th century during the kinetic theory controversy. Joule found the heat/work conversion mid 19th century, but decades after John Herapath's work which flies under the radar.

Yes, expanding gas PRODUCES work, but heat is not itself CONVERTED, so as to "produce" COLD.

The heat is converted. Cold is what is felt in its absence when converted into something else. I can understand it's a hard pill to swallow, that something so tangible to us, that we can feel, something that keeps us warm and alive can simply VANISH.

I think this is kinda where Vincent was going with his engine in a room lifting a weight. Generally, we're told that 'heat' is not created or destroyed when more properly it's that energy is not created or destroyed. But in Vincent's scenario, if his engine only ran thru one expansion process to lift the weight, then all (OK some) of the kinetic energy of the gas would transform into the potential energy of the weight. Assuming 100% eff than there's no heat loss except heat destroyed to provide for the expansion process. Oddly, this is the exact textbook example where kinetic energy is transformed into potential energy, but they always appear to ignore something disappeared. The entropy fanboys are always quick to point out their heat vs work stuff then harp about the heat death of the universe. And the textbooks are always keen to point out that any change in potential energy is considered part of the 'system' (part of the engine in this case) since no 'work' has passed to the surroundings.

BTW when the doctor takes your temperature, he's measuring internal energy all the way down to zero k.

[matt brown]

Just as Tesla stated back in 1900, a heat engine doesn't require a "sink" for heat. It converts heat.

Where does all this Tesla stuff come from, is there some patent or paper floating around ?

[Stroller]

If heat is being converted to work then that should be easily measurable as a torque at the flywheel. But for all the Watts of heat thrown at the hot side of the engine, we see tiny numbers of fairy lights being illuminated by weeny flywheel driven dynamos.

Conclusion: getting heat from blowtorch into working fluid is hard.

[Fool]

Very good points Stroller, and Matt.

Tom, work comes out of the gas during expansion, dropping the temperature of the gas.

Work goes in during compression, raising the temperature of the gas.

The graphic you've provided displays a higher temperature during expansion, and a lower temperature during compression.

That is the opposite of your theory. It can only be explained by heat going into the gas during and slightly before expansion, and heat being rejected during and slightly before compression.

It appears well explained by displacer movement and regenerator effect.

More like 100 are converted to momentum.

The piston continues moving using up 100 "homunculus power" of momentum which allows another 100 to be converted to work powering an outside load.

This leaves 200 inside and 300 outside.

100 reappear inside during compression restoring the 300<=>300 balance

Another 100 are added to begin another cycle.

There is no need for "Pushing the piston back in using the work that came out".

The "work that came out" (momentum) is used to continue the expansion, which assists turning another 100 to work.

Theoretically that additional work could provide some additional momentum, to produce more work output.

The piston returns easily by atmospheric pressure, which is still 300 as the inside pressure dwindles precipitously.

You constantly leave so much out and fail to have energy direction correct.

"More like 100 are converted to momentum."
Because 400 are pushing out, 300 in, for 100 J.

"The piston continues moving using up 100 "homunculus power" of momentum which allows another 100 to be converted to work powering an outside load."
200 pushing out, 300 pushing in. 100 disappear from work input used for cooling. No work output. 100 J input to system and stored, like a spring. And 100 homunculus's disappear because of expansion with work input, not output. Absorbed by the atmosphere.

This took two half strokes. Both expanding 100 away.

"This leaves 200 inside and 300 outside."
Yes, and zero net work.

"100 reappear inside during compression restoring the 300<=>300 balance"
Yes and no. because it is not yet at the starting volume. This is only 1/2 backstroke and used the 100 J stored in the system like a spring expanding. Input from atmosphere. Finishing the cycle requires and additional half stroke. Also 100 J were converted to backstroke inward inertia, MV.

Compressing further, back to initial volume requires 100 J to be input from the inertia and 100 more Homunculus's appear. Returning to the original Volume, 100 must now be removed/rejected to restore the 300 300 balance. Finishing the cycle. For net work and heat of zero zero.

Instead if it is cooled for one half of the back stroke, 25 J of work will be saved at the end of the cycle. 25 J maximum available for output.

The most frequently voiced theory seems to be that under load the engine slows down providing more time for heat exchange.

Because the graphic you have supplied shows high temperature during expansion. And low temperature during compression. Opposite of adiabatic processes, hence heat transfer. Obvious from being, hot on top forward stroke, and cold on the bottom return stroke. Slowing it down amplifies that process. Same Delta volume. Larger heat input/output. Larger temperature difference. Larger pressure difference.

The graphic you supplied shows perfect adherence to Carnot's Theorem. It is just a less efficient cycle than the ideals. Look in the "Super-Carnot Cycle graphic proof" thread, where I disprove the proof. Matt was testing me, LOL.

viewtopic.php?t=5646

[Tom Booth]

If heat is being converted to work then that should be easily measurable as a torque at the flywheel. But for all the Watts of heat thrown at the hot side of the engine, we see tiny numbers of fairy lights being illuminated by weeny flywheel driven dynamos.

Conclusion: getting heat from blowtorch into working fluid is hard.

True, a few thimbles full of air has a severely limited heat capacity. Yet this little engine apparently pushed more milliamps than my 110v wall charger.

https://youtu.be/X825itoQn_I

Driving a very cheap, inefficient generator.

[Tom Booth]

Very good points Stroller, and Matt.

Tom, work comes out of the gas during expansion, dropping the temperature of the gas.

You pay lip service to the FACT, which is, of course, undeniable. But you don't seem to really believe it.

Work goes in during compression, raising the temperature of the gas.

The graphic you've provided displays a higher temperature during expansion, and a lower temperature during compression.

That is the opposite of your theory.

No it isn't

It can only be explained by heat going into the gas during and slightly before expansion, and heat being rejected during and slightly before compression.

So what happened to "work comes out of the gas during expansion, dropping the temperature of the gas"

Aren't you contradicting yourself ?

It appears well explained by displacer movement and regenerator effect.
...

Not fully, as I've shown through numerous experiments.

[Tom Booth]

Why is the cooling of the gas that results from work output so underestimated and undervalued, often as if it is not a factor at all.

Because a heat engine runs on a pressure differential, it ain't a thermo couple.
...

Then why is it called a "heat engine"?

Anyway, a drop in temperature is what causes the drop in pressure, just as the increase in temperature causes the increase in pressure.

At least your not denying the premise. "Because...." is a confession of guilt.