The Carnot efficiency problem

[Tom Booth]

The rationale seems to be that this energy carrier/caloric, on an absolute scale is actually carrying 100% heat all the way down to absolute zero.

In my cook pot, I have only elevated the temperature of the water 20% on the Kelvin scale, so I can only utilize that added 20%. My engine can be no more than 20% efficient. The other 80% down to absolute zero was already there in the caloric and will leave with it. I'm not able to get at or utilize this additional heat.

This makes sense, but this is not how this is being taught.

What is being taught as the Carnot limit as valid is that because of this 20:80 ratio, only 20% of my added heat from the boiling water is available to be converted to work.

I can find no rational (or experimental) justification for this.

Let's say it took 1,000,000 joules to bring the water to a boil. How many joules can be converted to work?

200,000 joules.

The other 800,000 joules used to elevate the temperature must be "rejected". (According to the academics teaching the Carnot limit that is)

[Tom Booth]

Here's a typical example of the math as it is being routinely applied:

I'm surprised, given the extraordinary sophistication of many of your questions, that you aren't familiar with the reason for the Carnot limit. As heat Q flows out of the hot reservoir, its entropy S goes down by Q/Th. (That's by definition of T.) According to the second law of thermodynamics, no process decreases net S. So there must be a heat flow to the cold reservoir of at least (Q/Th)Tc. That's energy not available to do work. So of the Q energy drawn from the hot source, what's left to do work is at most Q(1-Tc/Th).

Mike W.

Has any of this ever been tested and verified experimentally?

There should be, according to these calculations, "of the Q energy drawn from the hot source" a veritable flood of heat passing through the engine and out the cold side to the sink (ambient).

At best, experimentally, I've been able to measure a mere trickle, indicating an efficiency near 100%

Just for comparison, Nickola Tesla who disputed Carnot and Kelvin's conclusions wrote:

..in passing from hot to cold. If the process of heat transformation were absolutely perfect, no heat at all would arrive at the low level, since all of it would be converted into other forms of energy.

Tesla is rumored to have been knowledgeable in regard to scientific matters of this sort, so I'm not jumping to conclusions.

[Fool]

It is a real mystery. Caloric theory borders on P.I.U. theory. There is no way of detecting a Pink Invisible Unicorn entering the engine then exiting completely whole and unscathed except for being cooler by 20%. The extra entity doesn't help and can't be disproven. Hence caloric theory is only discarded by the process of Occam's razor.

I prefer to think about pressure. During expansion: 100% of the heat, entering the working gas, results in a pressure that does a work output of 100%. Compression to complete the cycle requires 80% of that gained work if, and only if, the gas is cooled by a 20% temperature difference accomplished by heat rejection.

If the gas is not cooled, compression (Adiabatically) requires 100% of the work and zero heat is converted to cyclic useable work output. If cooled more it requires less. Rejecting heat reduces the loss of work. Not rejecting any heat costs all the work for compression.

It costs 80% or more to compress the gas thus completing the cycle. 80% is only for the best path. One needs to look at the path (PV and TS diagram) of the entire cycle to understand the Carnot limit, not just a single process.

[matt brown]

Borrowing some graphics from Urieli's xlnt site...

Urieli exp.png (117.48 KiB) Viewed 12466 times

Tom, note that all heat becomes work during an isothermal expansion, and

Urieli comp.png (117.64 KiB) Viewed 12466 times

all work becomes heat during an isothermal compression which leads to

Urieli Carnot.png (45.47 KiB) Viewed 12466 times

As I've said before, the Carnot buzz is a cycle conclusion, and akin the backwork ratio. The Carnot 'limit' derives from the fact that the two common 'isothermal' cycles (Stirling and Ericsson) have their working processes (expansion and compression) at the extremes of the cycle temperatures, thus ideal maximum Carnot efficiency. Meanwhile, the two common 'adiabatic' cycles (Otto and Brayton) have their working processes in a temperature gradient where the mean expansion temperature is less than Tmax while the mean compression temperature is more than Tmin, thereby resulting in a lower Carnot efficiency.

For a given Tmin (think 300k), any ideal Stirling or Ericsson will parallel Carnot limit and efficiency by increasing Tmax. The only way to beat Carnot here is by capturing some/all of the sink heat...

[Tom Booth]

...

I prefer to think about pressure. During expansion: 100% of the heat, entering the working gas, results in a pressure that does a work output of 100%.

This is generally consistent with the published theory and information. 100% conversion to work during a process is not only possible but the consequence of the first law, conservation of energy. So it is supposedly only on the return stroke where the consequences of the Carnot limit come into play in a cyclic process.

Compression to complete the cycle requires 80% of that gained work if, and only if, the gas is cooled by a 20% temperature difference accomplished by heat rejection.

If the gas is not cooled, compression (Adiabatically) requires 100% of the work and zero heat is converted to cyclic useable work output. If cooled more it requires less. Rejecting heat reduces the loss of work. Not rejecting any heat costs all the work for compression.

It costs 80% or more to compress the gas thus completing the cycle. 80% is only for the best path. One needs to look at the path (PV and TS diagram) of the entire cycle to understand the Carnot limit, not just a single process.

In reading Kelvin previously, (the complete commentary on Carnot's (caloric based) theories, it seems clear that the PV diagram he introduced (not sure if it's original with Kelvin but anyway) originates from or is based on the concept (caloric theory) that work is generated by transferring ALL the heat through the engine from source to sink. In that case there is no accounting for heat converted to work in a PV diagram. Work is just a byproduct of heat transfer.

This appears to be more or less the case in your focus on pressure.

Heat is introduced, increasing pressure

You stated: "During expansion: 100% of the heat, entering the working gas, results in a pressure that does a work output of 100%"

Being cautious at this point, if the above statement is true, the conclusions that followed that are false:

In a Stirling engine the working fluid is contained in a hermetically sealed chamber. (Mostly or practically, some leaks in a model engine are possible, but can be ignored for the current purpose)

First heat is added. Next the pressure of the working gas in the chamber increases. Eventually the inertia of the engine, piston, flywheel, crank and load etc. is overcome and the piston is driven out and as you say: "During expansion: 100% of the heat, entering the working gas, results in a pressure that does a work output of 100%"

Now thinking about this logically based on known principles of rapid gas expansion, at the end of the expansion process is there any of this increased pressure remaining?

Conservation of energy would seem to suggest that if the heat originally supplied, resulting in a pressure that "does a work output of 100%" we now have an engine in motion and the heat that caused the pressure is gone.

Newtons 1st law of motion states that an object in motion remains in motion unless acted on by some other force.

Is it logical to believe that after the added heat results in a "work output of 100%" the resulting increase in pressure that resulted from that heat addition remains??????

The increase in pressure was a consequence of heat input which by the end of the expansion stroke has been 100% converted from heat into the mechanical motion of the engine.

The heat that created the increase in pressure is now gone, converted, so it can be logically assumed that the heat that caused the pressure being gone, the pressure that resulted from the presence of that heat should be gone as well

In other words, the conversion of the heat 100% to mechanical motion means there is no longer any increased pressure to prevent the piston from returning. Infact, due to the fact that the chamber is sealed and the piston has changed position the state of the working fluid can now be considered a vacuum. This vacuum results in the piston being sucked back to its starting position, or preferably we can say it is pushed back by atmospheric pressure.

Or please explain how the pressure that set the engine in motion can continue to exist once the cause of that pressure, the added heat, has been converted to mechanical motion.

Your statement: "If the gas is not cooled, compression (Adiabatically) requires 100% of the work and zero heat is converted to cyclic useable work output." overlooks the several facts just pointed out.

The working fluid is in a sealed chamber. If the expansion work results in 100% conversion of the heat added at the start of the cycle to work (mechanical motion of the engine) the heat, which is the causative force behind the pressure is gone from the working fluid so the pressure which was an effect of the added heat, (now gone, having been converted to work), is also gone and the gas returns to its original state and so, of necessity will contract.

This is the observable reality that can be seen and measured and plotted in real time.

https://youtu.be/SHyke4hUNOs

The scenario you describe, where the heat is converted to work, but somehow the pressure caused by the heat remains to fight against the continued motion of the engine through the compression stroke is a violation of conservation of energy.

[Tom Booth]

Borrowing some graphics from Urieli's xlnt site...

Urieli exp.png (117.48 KiB) Viewed 12458 times

Tom, note that all heat becomes work during an isothermal expansion, and


...

Same response.

If all the heat is converted to work during expansion then effectively, there is no "compression". The piston returns due to a vacuum condition.

[Fool]

If there is gas inside a decreasing volume, there is always compression. Outside conditions don't change that.

Heat doesn't disappear in an isothermal process. T1=T2=Th.

Heat isn't converted to work in an adiabatic process.S1=S2. Zero heat. Temperature drop indicates only a decrease in internal energy, but that is not heat. Less work will be output for an adiabatic process than for an isothermal process. So it's "100% of work", is smaller. And it is all cancelled out by an adiabatic return compression stroke.

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

When pressure does it's work during expansion, pressure returns during compression with all it's heat and temperature ready to oppose that process just as forcefully as it's outward work. It's on the same curve. Without any cooling it will be equal. Zero work output will be available.

You leave out the fact that the outside air pressure that is helping push the piston in was bucking the piston coming out. Thus, the help and hindrance from the atmosphere cancels. Work = Px∆V-Px∆V=0. It's the same P and ∆V, opposite directions add to zero. Hence why outside pressure and buffer pressure are often ignored during these calculations.

PV diagrams were discovered by measurements. PV=nRT was discovered from the data and diagrams.

Every point on a PV diagram can be associated with a specific temperature, and a specific point on the TS diagram.

Constant entropy, S, lines are useful to show adiabatic process with work. Reversible. It is only possible to move to a different line by the outside addition or rejection of heat energy.

Constant enthalpy, H, lines show adiabatic processes without work. Irreversible.

I apologize for the long post. You asked many questions. I'm only attempting to answer. I hope it helps. Some?

[tibsim]

Do you know the metronome cycle? What do you think about it? Here:
URL removed because of it's "anti-Semitic" views. Do not re-post.

[Tom Booth]

If there is gas inside a decreasing volume, there is always compression. Outside conditions don't change that.

Heat doesn't disappear in an isothermal process. T1=T2=Th.

Heat isn't converted to work in an adiabatic process.S1=S2. Zero heat. Temperature drop indicates only a decrease in internal energy, but that is not heat. Less work will be output for an adiabatic process than for an isothermal process. So it's "100% of work", is smaller. And it is all cancelled out by an adiabatic return compression stroke.

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

When pressure does it's work during expansion, pressure returns during compression with all it's heat and temperature ready to oppose that process just as forcefully as it's outward work. It's on the same curve. Without any cooling it will be equal. Zero work output will be available.

You leave out the fact that the outside air pressure that is helping push the piston in was bucking the piston coming out. Thus, the help and hindrance from the atmosphere cancels. Work = Px∆V-Px∆V=0. It's the same P and ∆V, opposite directions add to zero. Hence why outside pressure and buffer pressure are often ignored during these calculations.

PV diagrams were discovered by measurements. PV=nRT was discovered from the data and diagrams.

Every point on a PV diagram can be associated with a specific temperature, and a specific point on the TS diagram.

Constant entropy, S, lines are useful to show adiabatic process with work. Reversible. It is only possible to move to a different line by the outside addition or rejection of heat energy.

Constant enthalpy, H, lines show adiabatic processes without work. Irreversible.

I apologize for the long post. You asked many questions. I'm only attempting to answer. I hope it helps. Some?

You have detailed many "facts' here. Where you get your information I don't know, but so many of your statements are flat out wrong, based on some misunderstanding or misconception, I do not have time right now to address them all, maybe another day, but I would encourage you to review the definitions of the thermodynamics terminology you use

Thermodynamics can be confusing and seemingly contradictory, especially in regard to "heat" and 'work" as sometimes these are treated as equivalent, the same thing, and sometimes as seperated things.

I'll address one point now.

You mention "adiabatic" and say: "Heat isn't converted to work in an adiabatic process."

The first paragraph from the Wikipedia article on "Adiabatic process": (emphasis added)

In thermodynamics, an adiabatic process (Greek: adiábatos, "impassable") is a type of thermodynamic process that occurs without transferring heat or mass between the thermodynamic system and its environment. Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work

So is your statement true?: : "Heat isn't converted to work in an adiabatic process."

Wikipedia says: , "an adiabatic process transfers energy to the surroundings only as work".

[Fool]

You are getting close. Work is not heat. Heat is thermal energy. Energy can be transferred by work without heat. Force times distance. No heat involved.

Temperature is not heat. "Adiabatic cooling" is confusing. A better statement is "adiabatic temperature drop". Cooling seems to imply heat transfer. Adiabatic means without heat transfer. Cooling/"temperature drop" can happen without heat transfer.

The page you quoted and bolded says energy, it carefully avoids the term heat. "Transfers 'energy'...", Not 'heat'. It is well written. Note: the surroundings can transfer work back into the system thus increasing it's internal energy, but it becomes an even trade.

Although a reduction of internal energy by exporting work, causing a reduced temperature, appears as if a gas has cooled, the temperature drop is not by an act of heat.

What happens when the temperature drops during and adiabatic process is that "internal energy" has decreased. Internal energy is not heat.

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

Knowing the difference between internal energy and heat helps in the understanding of thermodynamics and the adiabatic process.

An adiabatic line on a PV or TS diagram denotes zero heat transfer, but internal energy can change. Therefore no heat can be converted to work. It is a heatless process. Internal energy is converted. During a cyclic process that work will be put back in, the internal energy will return and the two processes will cancel. No heat will disappear. No internal energy will disappear.

Heat is "converted" in a cyclic process because it takes less heat out to recycle during the same volume change at a lower temperature.

It is like lowering a bowling ball then making it 20% lighter and raising it back up using 80% of the gained energy. The ball then regains it's weight. You'll get 20% energy out per cycle. Except in thermodynamics you are unable to make the ball 100% lighter (zero Kelvin).

Tesla, and many others, don't realize that it costs work, from the outside inward, to reduce the balls weight more than 20%. (Temperature below 20%, "cold hole"). That too, at best, would be an unobtainable even trade.

It is a very difficult process for me to understand. Entropy is even more difficult for me. I have spent many hours attempting to learn this well enough to explain it. It will probably take many hours of explaining it before I can write it well enough so that anyone else can understand. Sorry for the gibberish.

It goes along with the notion that heat isn't contained, internal energy is, and although the two relate to temperature, neither are temperature. Yet both are energy, (so is work), just not the same kind. We are stuck with the nature of how to convert between those energies.

[Tom Booth]

You are getting close. Work is not heat. Heat is thermal energy. Energy can be transferred by work without heat. Force times distance. No heat involved.

Temperature is not heat. "Adiabatic cooling" is confusing. A better statement is "adiabatic temperature drop". Cooling seems to imply heat transfer. Adiabatic means without heat transfer. Cooling/"temperature drop" can happen without heat transfer.

The page you quoted and bolded says energy, it carefully avoids the term heat. "Transfers 'energy'...", Not 'heat'. It is well written. Note: the surroundings can transfer work back into the system thus increasing it's internal energy, but it becomes an even trade.

Although a reduction of internal energy by exporting work, causing a reduced temperature, appears as if a gas has cooled, the temperature drop is not by an act of heat.

What happens when the temperature drops during and adiabatic process is that "internal energy" has decreased. Internal energy is not heat.

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

Knowing the difference between internal energy and heat helps in the understanding of thermodynamics and the adiabatic process.

An adiabatic line on a PV or TS diagram denotes zero heat transfer, but internal energy can change. Therefore no heat can be converted to work. It is a heatless process. Internal energy is converted. During a cyclic process that work will be put back in, the internal energy will return and the two processes will cancel. No heat will disappear. No internal energy will disappear.

Heat is "converted" in a cyclic process because it takes less heat out to recycle during the same volume change at a lower temperature.

It is like lowering a bowling ball then making it 20% lighter and raising it back up using 80% of the gained energy. The ball then regains it's weight. You'll get 20% energy out per cycle. Except in thermodynamics you are unable to make the ball 100% lighter (zero Kelvin).

Tesla, and many others, don't realize that it costs work, from the outside inward, to reduce the balls weight more than 20%. (Temperature below 20%, "cold hole"). That too, at best, would be an unobtainable even trade.

It is a very difficult process for me to understand. Entropy is even more difficult for me. I have spent many hours attempting to learn this well enough to explain it. It will probably take many hours of explaining it before I can write it well enough so that anyone else can understand. Sorry for the gibberish.

It goes along with the notion that heat isn't contained, internal energy is, and although the two relate to temperature, neither are temperature. Yet both are energy, (so is work), just not the same kind. We are stuck with the nature of how to convert between those energies.

Nothing personal but "gibberish" is a pretty accurate description. You make the same statement over and over repeatedly contradicting your own citation:

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

Did you read it?

The internal energy of a thermodynamic system is the energy contained within it, (....) It includes the thermal energy, i.e., the constituent particles' kinetic energies of motion ...

You stated above "Heat is thermal energy."

Your citation Wikipedia states: "...internal energy ...includes the thermal energy".

Please do not continue taking "many hours of explaining it before I can write it well enough so that anyone else can understand" since you pretty obviously don't understand yourself or even read your own references.

And please, we don't need more semantic hair splitting like " "Adiabatic cooling" is confusing. A better statement is "adiabatic temperature drop".

You are still living in a "Caloric" mindset trying to drive a wedge between "heat" and "work". Your anxiety to keep these two things seperated and distinct is not justified. That is not a clear picture of reality. On an atomic level heat and work are identical, or almost identical.

The difficulty, generally, is that for decades, in the field of thermodynamics, "heat" meant Caloric, (a substance not "the particles kinetic energies of motion".)

"Work" was something completely different from "Caloric"

Later Joule discovered, I think it was recognized by others earlier, but Joule demonstrated it experimentally, that these two things that were formerly considered completely separate were actually identical.

Some people, have still not caught up with this.

Technically heat (internal thermal energy) is not converted into "work", since at an atomic, microscopic level, both are simply "the particles kinetic energies of motion".

As a gas, the particles are moving individually, as "work" the particles are joined together in a solid and moving in unison.

A heat engine causes the usual random motion of gas particles to move in unison by carefully controlling the heat input to set up an oscillation of the gas particles which in turn transfers this unified oscillatory motion to the solid piston. In this way the "heat" (the (gas) particles (unified/collective) kinetic energies of motion) is converted into "work" (the (solid pistons) particles kinetic energies of motion). The microscopic kinetic energy of gas particles is converted to macroscopic kinetic energy or the mechanical motion of the engine.

This is not that difficult to understand.

Thermodynamics is difficult to understand because it is an old outdated obsolete "science" with a confusing and contradictory cultish lingo inherited from a long dead and buried theory of heat.

[Tom Booth]

Suppose we consider "internal energy" our BANK ACCOUNT, or the energy account of the working fluid.

We supply X number of joules as heat (let's say 100 joules per second from a 100 watt heating element) resulting in a rise in temperature. 100 joules added to the account.

The 100 joules, for the engine to continue running must go out or be "withdrawn" as either joules of work (the mechanical revolution of the engine) or joules of "waste heat".

Either way the credits and debits are in Joules. Once the Joules are in the "bank" as "internal energy" joules of heat are indistinguishable from joules of work, both result in a rise in temperature of the gas when "deposited" and a fall in the temperature of the gas when "withdrawn".

The, what I would call, NORMAL efficiency formula follows this simple rule of accounting.

W = Qh - Qc

Work output in Joules "W" is the Heat input in Joules "Qh" minus the waste Heat output in Joules "Qc".

You seem to be wanting to dispute even this universally accepted formula saying:

...no heat can be converted to work. It is a heatless process. Internal energy is converted. During a cyclic process that work will be put back in, the internal energy will return and the two processes will cancel. No heat will disappear.

You allow for ZERO conversion of heat into work, which is pure Caloric theory where ALL the heat must pass through to be "rejected" to the "cold reservoir".

This is the confused state of modern thermodynamics generally. Some concession has been made to recognize the equivalency of heat and work but it really is an attempt to preserve Caloric theory in a somewhat disguised form.

These issues can easily be resolved by a simple experiment. Measure the actual Joules of heat being "withdrawn" to the sink, a simple water bath should do, but is anyone doing this?

No! It's too difficult!!!! That is a very very hard experiment to perform, you need extra extra super duper astronomically expensive equipment. Blah blah blah...

Complete nonsense.

Even if Qc were found to be zero, that would be no "overunity", no violation of conservation of energy.

On top of that we are surrounded by an endless, unlimited supply of ambient heat/energy. That is more difficult to get at, but if some of W the work output were used to reduce the temperature of the sink that would still not be any energy from nothing.

Carnot efficiency limit says, quite arbitrarily and with no evidence that Qc can be no less than 80% of all the Joules supplied at Qh. (More or less depending on the actual ∆T but using boiling water as heat source and ambient as "sink").

The last measurement taken in one of my experiments detected only 0.38% of the joules from the 85 joules per second heat source actually measurable at the cold plate.

True, the cold plate certainly must have been dissipating some heat to the surrounding ambient but we have a long long way to go before reaching 80%

I'll be doing additional experiments when I get the opportunity, with a new set of thermocouples when I can spare the expense, but why is it in 200 years there is no record of anyone ever having done any such simple experiment, and there doesn't seem to be any thermodynamicists physicists/scientists willing to do any such experiment today. All they seem capable of is banning me from the forums for making the suggestion or complaining about how impossible such an experiment would be.

I guess with six figure salaries doing what they do there is no incentive to prove themselves wrong.

[Tom Booth]

I have a little time to begin addressing the issues you brought up, one at a time.

If there is gas inside a decreasing volume, there is always compression. Outside conditions don't change that.
...

One thing I think it is important to understand about an engine is that at BDC (bottom dead canter) the piston is mechanically restrained from moving because the piston, connecting rod and crank angle are in a straight line. If the temperature in the cylinder is dropping and the pressure is dropping, which real time PV readings show they are at that point, a partial vacuum is created. The piston cannot return until the crank rounds BDC and the connecting rod is off on enough of an angle.

For the same reason a "free piston" engine using a diaphragm requires the diaphragm to be weighted. The weight adds momentum which momentarily prevents the piston from returning while a sufficient partial vacuum is created.

When a gas expands into a vacuum it does no "work" and there is no temperature change. The piston cannot "compress" against a vacuum. The outside atmosphere cannot "compress" a vacuum. It can move into a vacuum with virtually no resistance.

At some point, yes, the piston on the return stroke will begin compressing the gas and the temperature will begin to rise, but generally the temperature apparently follows an isotherm on much of the return stroke.

If the compression returned all of the "internal energy" used for expansion, reciprocating engines would not exist.

The picture you paint here:

An adiabatic line on a PV or TS diagram denotes zero heat transfer, but internal energy can change. Therefore no heat can be converted to work. It is a heatless process. Internal energy is converted. During a cyclic process that work will be put back in, the internal energy will return and the two processes will cancel. No heat will disappear. No internal energy will disappear.

Heat is "converted" in a cyclic process because it takes less heat out to recycle during the same volume change at a lower temperature.

Is, as I said, Caloric theory, or effectively no different. The heat is let in to expand the working fluid to push the piston out, then ALL the heat is let out again so the working fluid can contract or be "compressed'.

No heat will disappear. No internal energy will disappear.

That is the production of work from nothing with no expenditure of energy. If no heat "disappears" and no internal energy "disappears" where do you suppose that the energy to drive the load on a heat engine comes from?

You describe the production of horsepower with no energy consumption.

If that were true then all the heat being "rejected" at the sink could be recycled or used over again to drive another engine, and another and another after that and you could have all these engine running perpetually in a circle producing infinite amounts of energy, one engine "rejecting" all its heat into the next.

That is not the case. Heat "disappears", internal energy "disappears", really the heat/internal energy is converted into mechanical motion, there is a transfer of kinetic energy from the working gas to the piston and as a result of this loss of energy the temperature of the working fluid falls and the result is cooling and a partial vacuum.

[Tom Booth]

...

Heat doesn't disappear in an isothermal process. T1=T2=Th.

....

Heat also goes into the working gas of the engine in a so-called isothermal expansion, (we can ignore for the moment that a truly isothermal process in a heat engine is actually impossible and therefore largely irrelevant).

Internal (thermal or otherwise) energy also goes out as work which would certainly result in a drop in temperature except that supposedly this fall in temperature is exactly compensated for by additional heat flow into the engine. As work goes out, an equivalent amount of heat goes in so that the "account balance" on average, remains the same.

If you have $500 in savings and deposit a check for another $500 each Friday while the wife is at the store shopping and spends $500, that does not mean no money is being spent just because the bank balance maintains a $500 balance.

I think it is perfectly legitimate to say that the heat "disappears" and the work "appears" or the heat input is converted to work output so that the temperature remains constant.

If 500 joules goes into the engine as heat and 500 joules simultaneously goes out as work and the temperature, on average remains constant, then it seems logical to assume, or perfectly legitimate to say that the heat input is converted into work output

[Fool]

Well you got me there. That sentence should have read 'does'.

>>Heat does disappear in an isothermal process. T1=T2=Th.<<

100% of the heat is converted to work during an isothermal expansion.

Isothermal is an ideal mathematical construct that is used to show maximum conversion percentage. Like a mathematical description of a circle that shows perfect roundness. Other paths will convert heat to less work.

An adiabatic process is the lowest path and it's reversible. Zero heat is converted to work.

One quantity of internal energy is thermal energy. That doesn't mean heat is the only way to change internal energy. Work, change in pressure, change in volume, are also ways of effecting internal energy and temperature.

Adiabatic merely means, any changes to internal energy are a result of something other than heat.

Adiabatic bounce involves zero heat transfer or conversion. All heat is retained by the system. Internal energy is a state value. Heat and work are a path process. The path taken, isothermal adiabatic etc., Determines the quantity transfered. Path analysis is an important tool of thermodynamics.