The Reversible engine

[Tom Booth]

You have 15 or so paragraphs that to me are a ball of confusion nearly impossible to follow or decipher, I'll read through it again, but in the mean time;

Using simple algebra:

This is my THEORY. I don't necessarily consider it anything more than that. A theory to explain my observations. It could, of course be wrong or require modification, and this is, at any rate a SIMPLIFICATION of a very complex system of interactions. On the other hand I don't want to make it TOO simple, by neglecting important elements, such as potential energy, velocity and momentum, along with displacer(heat input) action.

Along with TDC (Top Dead Center or full compression) and BDC (Bottom Dead Center or full expansion) I'll introduce a new acronym NDC (Neutral Dead Center) to indicate the midway point, which can be thought of as a point of equilibrium (neglecting velocity/momentum)

The forces involved in joules are

J = big J is the given internal energy at the start at NDC (equilibrium).


The working fluid starts out at NDC (Neutral Dead Center). With J joules. (The internal pressure is in balance with external atmospheric pressure so these cancel out)

As a Key:
E internal = J joules (given or starting internal energy)
j1 = joules added or taken in to start the engine (for expansion) including work and/or heat.
jv = joules converted to velocity
js = joules converted to external shaft-work
ja = joules added by atmospheric work input
jd = joules converted to displacement work.

We will assume j1=jv=js=ja=jd for simplicity.

At first the internal energy is J

To start the engine someone pushes the flywheel around by hand to get the engine started while also applying heat. So the combination of applied heat (at a specific time, when the engine revolves enough to lift the displacer near TDC), along with the physical work input results in an increase of internal energy at TDC of j1 joules.

So to start the engine, at startup, E internal = J + j1 joules.

At TDC turning the engine over by hand the j1 joules represents potential energy manifesting as increased pressure at TDC

The engine starts.

At TDC due to the increase in E-internal the pressure is greater than the external atmospheric pressure and the piston moves outward "on its own" (without manual work input to start the engine)

In moving out, some of the internal energy is converted to velocity equivalent to the previous internal energy increase (work goes "out" from the working fluid in joules (jv) as an increase in velocity of the piston) and pressure returns to normal.

E-internal = J + j1 - jv

The piston/engine is now in motion. An object in motion stays in motion (until it meets resistance, Atmospheric pressure at this point is not counted as "resistance" as previously noted the system started out in a state of equilibrium with the surroundings)

At NDC during expansion then:

E-internal = J + (j1 - jv)

If j1 = jv then j1 - jv = 0

So as the engine crosses NDC during expansion:
J + 0 = J therefore
E-internal = J at NDC during expansion. But the engine now has velocity/momentum.

Now at NDC during expansion the displacer covers the hot heat exchanger stopping heat input but the piston continues to move due to momentum and the working fluid expands adiabatically doing displacement work jd while also doing additional shaft-work to drive the engine js.

At BDC the piston slows down, the velocity has converted to shaft-work output.

At BDC
E-internal = J - jd - js

At NDC during expansion the "internal energy" of the working fluid already returned to a balance (but the piston/engine still has velocity/momentum.)

At BDC the gas has expanded from NDC adiabatically (-jd) while also outputting shaft work.(-js).

(jd = displacement work in joules during expansion. js = shaft-work output in joules during expansion)

Now at BDC there is an imbalance and internal pressure is less than atmospheric pressure and the piston stops. The velocity has been converted to "shaft work".(work output external to the "system"). The gas has also expanded doing its own displacement work.

Now due to the low internal energy/pressure the piston begins to move in the opposite direction towards TDC.

At that point (BDC to NDC for about 1/4 cycle) the working fluid is COLD having expanded and lost internal energy to external shaft-work output and so is ready to absorb heat more quickly and easily when heat is made available. (Heat is not yet available, but has been building up in the heat exchanger the hot heat exchanger insulated from the working fluid by the displacer).

Traveling back to NDC atmosphere does "work" on the gas so at NDC the piston has again picked up velocity and:

E-internal = J - jd - js + ja

-jd and +ja cancel leaving -jd so:

E-internal at NDC During "compression" is:

E-internal = J - js

The internal energy is still in deficit at NDC during "compression".

E-internal = J -js joules at NDC but the piston is now in motion (with velocity) moving towards TDC.

From NDC to TDC the velocity of the piston slows down due to increasing pressure and the velocity is converted into heat.

At TDC you again have E internal = J + j1 joules with some heat input from the now exposed hot heat exchanger equivalent to js (previous shaft work output)

But this time the j1 joules are supplied by the momentum of the piston rather than the mechanical input used to start the engine.

And from there the cycle continues.

If the shaft work increases due to the application of an external load this will generally be compensated by an increase in j1 due to the general engine dynamics or the engine will slow down and/or stop is j1 cannot increase to balance the load, but describing the complexities involved in varying the external load and how the engine might respond to that is somewhat beyond this basic description of the engine cycle.

Any engine, if it is running at all is doing some amount of "shaft-work" even if that just amounts to spinning it's own wheels.

To summarize the cycle:

1)NDC (compression): Internal energy = J - js (+ velocity)

(At or rather before startup, of course js would be 0 as the engine hasn't done any work yet and velocity is zero, until the engi e is manually started)

2)TDC: Internal energy = J + j1 (j1 as potential energy/pressure) j1 represents work and heat input to drive 1 expansion.

3)NDC (expansion): Internal energy = J or J (+j1 - jv) pressure drops to normal but now the engine has velocity/momentum.

4)BDC: Internal energy = J - jd - js

During expansion internal energy is lost both in terms of PV displacement work and shaft-work.


I think the problem with the usual "ideal" models, TS diagrams etc. is that no distinction is made between jd ("negligible" displacement work) and js (work output in the form of "external" shaft work)

A typical TS diagram in most textbooks and courses, only accounts for displacement or PV work or erroneously (IMO) assumes displacement work and shaft work to be one and the same which IMO they cannot be.

Calculating shaft work would involve more complex math than represented by a simple PV diagram.

So far, all I've been able to find in relation to calculating shaft work is statements such as, that is beyond the scope of this course. Or that will be covered in (some other) chapter in relation to open systems such as turbines.

There seems to be little recognition that heat engines or a "closed system" like a Stirling engine actually does any external "shaft work" that needs to be accounted for.

Of course, that is understandable as in Caloric theory a water wheel does shaft work by displacement of water from a high level to a low level without any reduction in the volume of the fluid. And heat engine thermodynamics as passed down from Kelvin etc al, is based on Caloric theory.

With heat being energy, however, shaft work does result in a reduction in fluid volume.

When gas expands adiabatically, especially when doing external shaft work, its temperature drops, and so it's pressure drops and so after velocity and momentum are exhausted it's volume will subsequently decrease.

So, for the past 200 years, the unavoidable consequence of heat being energy with an unavoidable subsequent reduction in the volume of the gas or working fluid (in a closed system), with shaft work output has been, apparently, entirely missed or neglected, or gone unrecognized.

As if a gas, upon doing work, does not loose energy or decrease in volume, just like water driving a water wheel passes right through with no apparent loss of energy or volume, but continues on its way down the river.

As if a heat engine only does a negligible amount of displacement work so that the actual shaft work can be ignored.

Thermodynamics started out, entirely steeped in Caloric theory. All the math, diagrams etc and has been, unsuccessfully IMO trying to shore up this caricature and somehow shoehorn new findings and new discoveries into an old worn out obsolete model of heat and heat engine theory, with the unfortunate consequence that it no longer models or represents reality. Not when it comes to Stirling type heat engines anyway.

Where you have mass flow through the system, as in a steam turbine, maybe it has some applicability, but for a Stirling engine, I think some fundamentals, like basic conservation of energy are being overlooked.

How do you add heat as joules, to a closed system, then take out an equivalent amount of energy in joules from the same closed system as work, then imagine it is still necessary to also remove another equivalent amount of heat in joules to return to the initial state?

1 + 1 -1 -1 does not add up to 1

1=1 starting internal energy equals itself

1+1=2 original internal energy plus added heat

1+1-1=1 original internal energy, plus added heat minus work output = original internal energy. (In a closed system at increased volume that would result in "contraction" or a subsequent reduction in volume)

1+1-1-1= -1

You don't have a cycle.

[Tom Booth]

In the above I noticed I made a mistake.

I wrote:

E-internal = J - jd - js + ja

-jd and +ja cancel leaving -jd so:

That, perhaps obviously should have been:

E-internal = J - jd - js + ja

-jd and +ja cancel leaving -js so:

OR, I should have written:

E-internal at NDC during compression = J - js

Which I did state correctly, (or as intended) in the several lines following that.

[Fool]

Tom,

"E-internal = J - jd - js + ja

-jd and +ja cancel leaving -jd so:

E-internal at NDC During "compression" is:

E-internal = J - js

The internal energy is still in deficit at NDC during "compression"."

You appear to have left out the term +jv from that equation.

Shouldn't it be:

E-internal = J - jd - js + ja + jv

Or did the energy input to accelerate the piston in the opposite direction not count? It seems to cancel out js.

"A typical TS diagram in most textbooks and courses, only accounts for displacement or PV work or erroneously (IMO) assumes displacement work and shaft work to be one and the same which IMO they cannot be."

Why? In the PV diagram, the area enclosed is the total amount of work available to operate the engine and output on the shaft. If mechanical losses are zero, your assumption in your theory, the shaft work will equal the gas volume work that is depicted on a PV diagram. If you want to talk real engines it will be much less.

We are discussing the thermodynamics here, not mechanics.

Shaft work comes from P•∆V work minus mechanical losses. It is difficult to add up all the mechanical losses using kinematic theory, but it is easy to measure, just add a brake and a torque meter (spring scale and lever).

In your theory, shaft work appears to be coming from jv energy. That will tend to slow the piston down with greater load, ultimately bringing it to a stop at NDC.

"At TDC you again have E internal = J + j1 joules with some heat input from the now exposed hot heat exchanger equivalent to js (previous shaft work output)."

I find it interesting that the equations, for your theory of a heat engine, lack any variables for heat input. Might I suggest jh and jc. Of course, I assume you will either leave it out or claim it is zero.

Your description appears to have the engine running on inertia without any possible means for heat input. T hot plate equals T gas. Heat doesn't "build" in a constant temperature hot plate. It only comes in if cooled by a cooler working gas volume.

I also see that you have mixed total cycle energies with 1/4 process energies. I don't think you intend to have shaft load only on the NDC to BDC 1/2 stoke. Isn't a shaft load from a brake or generator always applied? Shouldn't js be in every equation for every 1/4 cycle?

Also for the TDC to NDC to BDC you've described the Carnot cycle. Heated expansion followed by adiabatic expansion. Good one.

[Tom Booth]

Tom,

"E-internal = J - jd - js + ja

-jd and +ja cancel leaving -jd so:

E-internal at NDC During "compression" is:

E-internal = J - js

The internal energy is still in deficit at NDC during "compression"."

You appear to have left out the term +jv from that equation.

Shouldn't it be:

E-internal = J - jd - js + ja + jv

(...)

The quantity of interest is the "internal energy" of the gas or "working fluid".

jv is not a form of "internal energy" so is not included in E-internal.

I mention velocity several times as a form of energy that needs to be recognized and accounted for, similar to heat and "work" it can be converted to other forms of energy, but it is not "internal energy".

This gives rise to some seemingly "weird" and nearly inexplicable phenomenon, like the steam injector, where the exhaust steam from a steam boiler is used in a loop, to inject water back into the same boiler against its own pressure.

This is accomplished by converting pressure to velocity creating a low pressure vacuum that picks up water, then by slowing down the stream the velocity is converted back into pressure higher than the boiler pressure.

This seems incredible. Like attaching a hose to an air tank and letting the air go out from the tank through a venturi where the high velocity air stream picks up and entrains additional air, then using the same air hose with a greater volume of air at higher pressure, to put more air back into the same tank the air came out of in the first place.

It seems impossible and defies logic, but apparently it works, with steam at least.

Some have claimed it is possible to do the same thing with air.

https://www.aircaraccess.com/achf-intro ... lf-filling

https://youtu.be/s4MNHyoQr3Q?si=_pB-G5Qf4Bt7Iw-x

[Tom Booth]

(...)

In your theory, shaft work appears to be coming from jv energy. That will tend to slow the piston down with greater load, ultimately bringing it to a stop at NDC.

A load on an engine could potentially slow down or stop any kind of engine. So what?

If the heat input is not adequate you either increase the heat or lighten up on the load if possible

"At TDC you again have E internal = J + j1 joules with some heat input from the now exposed hot heat exchanger equivalent to js (previous shaft work output)."

I find it interesting that the equations, for your theory of a heat engine, lack any variables for heat input. Might I suggest jh and jc. Of course, I assume you will either leave it out or claim it is zero.

It states in my "Key": "j1 = joules added or taken in to start the engine (for expansion) including work and/or heat."

You just quoted in the last paragraph as well: ""At TDC you again have E internal = J + j1 joules with some heat input from the now exposed hot heat exchanger equivalent to js (previous shaft work output)."

Sorry if that is not clear.

jh might suffice but the way I see it, as illustrated in my other thread "Aligning heat vectors":

https://www.stirlingengineforum.com/vie ... f=1&t=5556

"Heat" input at TDC arrives from multiple sources.

1) heat of compression from ambient pressure, which is kind of a recycling of displacement work That is, during the power stroke atmosphere is displaced, on the return stroke atmosphere does "work" pushing the piston back, which culminates in "heat of compression" generated in the gas at TDC when velocity (of the piston and flywheel if present etc. primarily) is converted into heat.

2) heat is applied to maintain the "heat" in the heat exchanger, as you quoted: ""At TDC you again have E internal = J + j1 joules with some heat input from the now exposed hot heat exchanger..."

Maybe the wording is not clear, but j1 includes " some heat input from the now exposed hot heat exchanger".

The combination of "heat vectors" results in a temperature at TDC which is hotter than any one "heat vector" by itself. The gas itself, as I've stated previously is COLD during "compression" and so is able to take in heat from the hot heat input side to some degree just before and just after TDC, but at TDC the working fluid is likely too hot to take in heat from the heat exchanger and likely, if anything heat is going the other way at exactly TDC.

The gas volume is also reduced at TDC which concentrates the "internal energy"

So T1 is a little ambiguous. It includes all the heat AND work that contribute to the "explosive" concentration of heat and energy at TDC.

As far as a jc, that is a contradiction in terms.

Joules-cold.

I assume you are trying to weasel "heat rejection" into the formula.

In that case you are correct: "I assume you will either leave it out or claim it is zero."

There may, of course be incidental loses to vibration, friction, radiation etc. but there is no provision for intentionally "rejecting" or "wasting" heat by deliberately transferring it to a cold sink ON PURPOSE, no.

I consider the whole Carnot idea that a heat engine operates by simply escorting heat through the engine to the sink contrary to reason as well as the first law of thermodynamics; conservation of energy.

You only get heat from a heat engine by fighting tooth and nail to prevent the heat from passing through the engine to a cold "sink". You want to convert the heat to work output not throw it away.

[Tom Booth]

That last paragraph should have been:

You only get WORK from a heat engine by fighting tooth and nail to prevent the heat from passing through the engine to a cold "sink". You want to convert the heat to work output not throw it away.

I know, that "goes against 200 years of thermodynamics theory"

Oh well.

[Tom Booth]

This is, I think, my favorite post from the science forum:

...you are not going to overturn 150 years of engineering experience and thermodynamic theory with some badly done Mickey Mouse experiments in your garage.

https://www.scienceforums.net/topic/128 ... nt=1229667


LOL...

It's kind of fitting as that discussion also touched on "reversability".

A heat engine is not a heat pump "running backwards".

A heat pump only MOVES heat.

A heat engine CONVERTS heat.

Completely different functions.

The only place a heat pump is a heat engine "running backwards" is in the "ideal" dream world of Carnot/Caloric theory and, unfortunately it has also been preserved in modern thermodynamics theory as well.

[Tom Booth]

To try to make this clear.

Suppose you have a wood or coal fire or any kind of furnace on a cart.

The work done by a heat pump, refrigerator, air conditioner etc. Is like the work needed to roll the cart down the road a distance. The work involved in that is relatively easy.

The work output of a heat engine comes from actually utilizing or CONVERTING heat into work. Using up the heat of the fire to produce work. A heat engine doesn't just move the heat from point A to point B, it consumes heat as "fuel".

A Stirling engine is not MOVING heat, like a "heat pump running in reverse."

The cold produced by a refrigerator results from moving heat which leaves it cold in the area from which heat was taken, but the heat ends up somewhere else.

The cold produced by a Stirling engine results from the CONVERSION of heat into some other form of energy, like electricity, if the engine is driving a generator. Heat conversion also leaves it cold, or at a lower temperature, in the area from which the heat was taken or supplied, but it is not cold because the heat was simply moved.

The heat used up by a heat engine is gone altogether not just moved through the engine from one side to the other.

[Fool]

Tom,

"A heat pump only MOVES heat.

A heat engine CONVERTS heat."

Last I checked, a heat engine input heat from a hot plate converting some of it to work and output on the shaft.

A heat pump inputs work into the shaft converting of it to heat as well as picking up a lot of heat from the cold plate and outputting it to the hot plate.

One, guess which, moves heat from hot to cold, the other, moves heat from cold to hot, a direction of heat travel that is the reverse of the first. One also requires work input, guess which one, the other gives work output.

So a heat pump is a heat engine run in, reverse heat travel and work travel and direction of rotation. The PV diagram runs counter as well.

You have to stop your partial/half cycle and missing parameters analysis to get a good understanding of thermodynamics and heat engines.

You also need to document, instrument and measure your experiment better if you intend to learn from them, or challenge mainstream science. Classical thermodynamics has too many intertwined logical ties to discard because of any one well done experiment by any well respected scientist. Someone from the general public, as we both are, won't have a chance unless very pain stakingly instrumented, measured, and documented.

Linus Pauling ran into that wall, and was eventually disproven on his theory of vitamin C. He still has a large group of duped followers, their cognitive dissonance has their fraudulent belief prevailing over obvious science.

[Fool]

Tom,
"The cold produced by a Stirling engine results from the CONVERSION of heat into some other form of energy, like electricity, if the engine is driving a generator."

It seems to me that, an expanding volume of gas is pushing on a piston mass, or air mass (laminar flow), and does work accelerating that mass. Work is converted to kinetic energy, E=1/2mV^2, momentum increases. An external generator is not needed. External forces, a generator and propeller are examples, keep the piston from over speeding, but the volume of gas doesn't know it. It just pushes on the piston blindly as volume increases, which it does acknowledge. Increasing volume is depicted on the PV diagram.

Momentum, mV, mass times Velocity, is a big factor. As you have repetitively pointed out, but seem to only include it where it is beneficial.

Momentum requires work to increase. It gives back work when decreasing.

[Tom Booth]

Tom,

"A heat pump only MOVES heat.

A heat engine CONVERTS heat."

Last I checked, a heat engine input heat from a hot plate converting some of it to work and output on the shaft.

A heat pump inputs work into the shaft converting of it to heat as well as picking up a lot of heat from the cold plate and outputting it to the hot plate.

One, guess which, moves heat from hot to cold, the other, moves heat from cold to hot, a direction of heat travel that is the reverse of the first. One also requires work input, guess which one, the other gives work output.

So a heat pump is a heat engine run in, reverse heat travel and work travel and direction of rotation. The PV diagram runs counter as well.

You have to stop your partial/half cycle and missing parameters analysis to get a good understanding of thermodynamics and heat engines.

(...)

Your analysis or "observation" of how a heat engine operates in "reverse" of a heat pump is superficial as well as erroneous. (IMO of course)

From a close examination of what is actually going on in each of these machines, structurally, there is no similarity between a typical heat pump and Stirling type heat engine. (I'm specifying "Stirling type" to exclude mass flow or "open" systems like turbines or IC engines).

A heat pump compressor has valves with a fluid being pumped through it. A Stirling engine is a closed system. Heat pumps have comparatively HUGE heat exchanger surfaces with yards and yards of tubing looping back and forth many times for heat exchange with the environment on both the heat input and heat output sides, in accordance with its function, to rapidly take in heat from the environment and then rapidly release that heat to the environment in equal quantities. No (or comparatively little) "work" is being converted into heat to make cold in a refrigerator. Some "work" is involved in compressing the refrigerant, adding some amount of heat to the gas being compressed, but the bulk of the heat driven off in a heat pump comes from the compression of the gas which picked up that heat from the environment. not from conversion of work into heat.

A Stirling engine has a comparatively small area where the heat input is focused and concentrated and can carry out it's function of converting heat into work without any enormously long system of tubing with enormous surface area for "heat rejection" back to the environment. Arguably, as demonstrated in my experiments, surface area for heat rejection is not at all necessary and can be eliminated entirely as it is not a heat engines function to return heat to the environment, or simply move heat. In a Stirling engine the heat is converted to a different form of energy which goes out as shaft work, instantaneously. No large radiators with fans blowing air through them are needed for any "heat rejection".

So your statement: "One,(...)moves heat from hot to cold, the other, moves heat from cold to hot, a direction of heat travel that is the reverse of the first" is fallacious.

That would apply to a heat pump used for space heating as opposed to a heat pump used for air conditioning. The central air system is "reversible" in reality.

A heat engine is not an air conditioner running in reverse "pumping" heat in the opposite direction, it is a heat converter. It is not taking in heat and simply transporting it to be released at the opposite side. The heat is converted not simply moved.

In a heat pump no (or very little, comparatively) WORK is converted to heat. The work is USED to move heat, not converted into heat.

It has been found in attempts at application, that Stirling engines are completely useless for moving heat. The US military attempted to use Stirling engines for heating army living quarters. That failed. Though the engines produced power, the "waste heat" for heating the buildings wasn't there. Law of conservation of energy. The math that promised power output as well as "waste heat" did not pan out in practice.

Likewise attempts to use Stirling engines as air conditioners for internet data centers did not work. Stirling engines do not move heat, they prevent the movement of heat. They do not draw heat AWAY from the hot side, they concentrate heat AT the hot side making it even hotter so that this highly concentrated heat can be turned into an explosive force to drive the engine and produce "work".

[Tom Booth]

Tom,
"The cold produced by a Stirling engine results from the CONVERSION of heat into some other form of energy, like electricity, if the engine is driving a generator."

It seems to me that, an expanding volume of gas is pushing on a piston mass, or air mass (laminar flow), and does work accelerating that mass. Work is converted to kinetic energy, E=1/2mV^2, momentum increases. An external generator is not needed. External forces, a generator and propeller are examples, keep the piston from over speeding, but the volume of gas doesn't know it. It just pushes on the piston blindly as volume increases, which it does acknowledge. Increasing volume is depicted on the PV diagram.

(...)

LOL...

Sorry, but this strikes me as just silly, but typical of arm chair academic thinking.

You stated my point for me exactly. A PV diagram does not take any account of the load or external REAL work output that engines are actually designed for. The whole purpose of an engine is to produce external work output and you are treating it as incidental and insignificant.

You have pretty much confirmed everything I have been saying with these words;

"Free expansion" of a gas with a negligible "work" output is not the same as the rapid expansion of a gas doing work with any significant external load.

Even this little LTD.

Imagine if you can, the ENORMOUS amount of work the infinitesimally small, barely if at all visible under the most powerful electron microscope, little air molecules must do using all their collective strength to rotate a comparatively gargantuan, enormous, virtually infinitely big brass flywheel, not only pushing against the piston and all this mechanical apparatus but overcoming air resistance impinging on the flywheel.

Look at how tiny the surface area of the piston is that these infinitely small, invisible molecules have to push against.

https://youtu.be/Cu0IdJbyUfY?si=EeHs_dQqi7xpJA9t

Imagine how much work is required from these tiny air molecules when a real external load of some sort is applied.

This engine, driving a generator and with a drag on the flywheel has what exactly, for a cold heat exchanger?


https://youtu.be/D6F_cDjrEEU?si=dxWt0K1bl9oUkPcn


I'm applying heat from a propane torch, (about 3500°F)

Where is the enormous radiator and fans required to dissipate all that heat after it is "moved" to the "sink".

It doesn't require any radiator because all the heat is "dissipated" as work.

[Tom Booth]

(...)

Momentum, mV, mass times Velocity, is a big factor. As you have repetitively pointed out, but seem to only include it where it is beneficial.

Momentum requires work to increase. It gives back work when decreasing.

Where is it not beneficial?

On expansion velocity is converted to work when the piston slows down at BDC.

With compression velocity is converted to heat when the piston slows down at TDC

The energy conversion to velocity and back is completely internal to the system so no, or very little of that energy is lost.

What is the down side?

[Tom Booth]

(...)

You also need to document, instrument and measure your experiment better if you intend to learn from them, or challenge mainstream science. Classical thermodynamics has too many intertwined logical ties to discard because of any one well done experiment by any well respected scientist. Someone from the general public, as we both are, won't have a chance unless very pain stakingly instrumented, measured, and documented.

(...)

I have no interest in trying to "challenge mainstream science".

Yes, I agree, "Classical thermodynamics has too many intertwined" (il-)"logical ties to discard because of any one well done experiment by any well respected scientist"

So I wouldn't bother. It's become quite obvious over the years that the erroneous thinking surrounding thermodynamics is too deeply entrenched to change.

I have my own reasons for wanting to know the actual truth and have no interest in "proving" anyone or anything right or wrong.

I document as much as I'm able, given my limited resources for anyone that might be interested, but it is certainly clear by this time that the scientific community at large is not interested in upsetting the applecart.

I don't have to prove anything to anybody other than myself.

My impression though is that you, on the other hand, seem to have a strong vested interest in defending the status quo.

[Fool]

Playing the conspiracy card is fruitless in this case. I am merely here as a friend and am only attempting to help.

In electrical engineering one learns that to power something as a motor, and to create electrical power as a generator, one can use the same exact device. In other word a motor and generator are the same thing.

Phillips has a video demonstrating that a single machine can run as a motor, generator, Stirling Heat Engine, Stirling Cold Engine, Stirling Heat Pump, and, Stirling Cryo-Cooler. I appreciate that the fact is amazing. It also proves that there's no difference between a heat pump or engine and that the direction of rotation changes it to an AC/Cryo-Cooler and or Cold Engine. It doesn't matter what it is called, Stirling Machines are capable of all four modes.

The colloquial term "heat pump" describes a home unit used to bring in heat from a colder outside and release it to a hotter inside. Units are also available for both that and air conditioning. Same unit, just has extra hardware to reverse the direction of heat travel. Typically those units use the irreversible Linde Joule-Thompson process, so can't run as an engine.

Using a Stirling Engine to heat a building or cool computers is about as stupid as it sounds. Engines require high delta-T. Heat pumps require low delta-T. Trying to run an engine from computer heat while trying to cool the computer, would lead to the conundrum of the engine cooling it's own hot plate making it less efficient. It also makes it less powerful, but nobody talks about that. People have a fixation on efficiency.

If the temperature rises to help the engine the computers get fried. It would be an evil balance at best. The same is true for heating a space with the cold side. It's the cold side, dummies.

Now using a Stirling Machine as a cryo-Cooler to cool computers would work great. Throw out the irreversible Linde AC. Yes. Stirling Cryo-Coolers are already a profitable business. A Stirling redesigned to work for residential heat pumping and cooling at a much more efficient smaller delta-T would potentially be viable. They could possibly be more efficient and robust, last longer. They probably would be much more expensive. Especially the first ones.