# Is Carnot efficiency valid?

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8 minutes ago, exchemist said:

That's ballocks. It is an ideal theoretical engine cycle, with constant temperature heat input and output. The environment has nothing to do with it.

I thought the equation was based on the hot and cold "reservoirs" The engine sits between.

Is that not correct?

like a water wheel between an elevated body of water and a lower body of water.

the Carnot "efficiency" is a measure of the "height of the fall" not the actual mechanical efficiency of the water mill (or engine) to convert "available" energy. It is the temperature difference, not how well the engine actually utilized that temperature difference. The actual engine has nothing whatsoever to do with it.

Edited by Tom Booth

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20 minutes ago, Ghideon said:

Thanks for your comments! I guess proportions are also a concern; a really large piece of ice on a low power engine allows for other factors to have significant impact? Also freezers are maybe not built to precision, hysteresis may allow for the ice to have different temperatures. A few degrees may make a difference when a small engine melts the ice.

Exactly.

The block can have any temperature distribution compatible with Fourier's Heat Equation. Regular opening and closing of the freezer door might produce a nice decaying harmonic oscillation with depth for example.

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2 minutes ago, Ghideon said:

The logical consequence of your idea seems to be that you are just one simple component short of creating a major breakthrouh in physics. But I do not believe in such machinery therefore my question was rethorical. But I would be happy to be proven wrong.

Great, and I'm more than happy to prove my hypothesis wrong. So IMO it's a fantastic suggestion for an actual experiment.

There is though, I believe, at least one very real example of a "self-running" cooler/heat engine utilizing evaporative cooling.

Could a similar device utilizing a compression expansion cycle for refrigeration instead of evaporative cooling be devised or actually exist?

I'm not entirely convinced that such a thing is impossible, especially since finding out that a rather intelligent guy who pretty much handed us the modern world on a silver platter thought it could work.

I'm not making any claims whatsoever. I just have a hunch that maybe he was just a little smarter than some of those other guys; Carnot, Kelvin et al

12 minutes ago, sethoflagos said:

Exactly.

The block can have any temperature distribution compatible with Fourier's Heat Equation. Regular opening and closing of the freezer door might produce a nice decaying harmonic oscillation with depth for example.

Not that your addressing me I guess but it seems to me that a small bit of ice sandwiched between the cold sides of a couple of Stirling engines is a no-brainer.

Either the ice melts or it doesn't.

I'm willing to concede that if the ice melts that proves my theory wrong conclusively. The engine is not a heat pump pulling heat from the cold side. We can all go home.

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10 minutes ago, Tom Booth said:

I thought the equation was based on the hot and cold "reservoirs" The engine sits between.

Is that not correct?

like a water wheel between an elevated body of water and a lower body of water.

the Carnot "efficiency" is a measure of the "height of the fall" not the actual mechanical efficiency of the water mill (or engine) to convert "available" energy. It is the temperature difference, not how well the engine actually utilized that temperature difference. The actual engine has nothing whatsoever to do with it.

If you express the total potential energy of water in terms of its height above mean sea level, then the amount of it that you can extract from an overshot waterwheel is limited to (1 - ht/hm) where  hm and ht are the heights above msl of the mill race and tail race respectively. Do you see anything familiar here? Can you imagine all that waste potential energy disappearing down the tail race? Why isn't waterwheel efficiency a part of this limit?

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1 hour ago, Tom Booth said:

Arguably, an actual Carnot engine does not meet that criteria of falsifiability, would you agree?

How so? The sentence does not really make sense; how can an engine be falsified? The efficiency claim could be falsified, though. Build an engine that exceeds the efficiency limit, and it would be falsified. That’s how you falsify scientific claims - with an experiment.

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1 hour ago, exchemist said:

Don't be ridiculous. This is a total ranting rhetorical muddle as usual and contains a stupid straw man. There is no such thing as a "Carnot engine". You have made that up.

There is a Carnot cycle, which, as several people have told you several times, is a theoretical optimum heat engine cycle whose thermal efficiency, according to the theory of thermodynamics, no real engine can exceed.

The theory of the Carnot cycle would thus obviously be falsified if someone were to produce a heat engine exceeding Carnot cycle efficiency. So it is - obviously-  a falsifiable theory, in Popper's terms.

Those are excellent points that partially overlap with what I was thinking at this point.

A Carnot cycle is an extreme idealisation. When I said before it's just based on

(1) Conservation of energy

(2) Existence, to a good approximation, of heat reservoirs

I wasn't quite thorough. You need the gas doing the work to be ideal. You were quite right when you said,

On 1/31/2023 at 7:27 PM, exchemist said:

So if the gas laws are true, the Carnot efficiency formula is true.

If the "working agent" is an ideal gas, and all the intermediate states are of equilibrium, its energy exchanges with the rest of the universe can be expressed as proportional to its temperature. Thereby Carnot's limit as a function of both temperatures. It's not --as OP has been repeatedly suggesting-- because the basic concept of efficiency is based on temperatures. It's not. It's based on energy exchanges.

Anything, repeat anything, that deviates from this behaviour, would result in further "leaking out of entropy" to the rest of the universe and will make things worse in terms of efficiency. This is intuitively clear and only properly understood once the concept of entropy is introduced.

It would be an interesting exercise --which I'm not going to do-- to replace the gas for a real gas, with an equation of state more similar to Van Der Waals, a virial expansion, etc, to see that things would only be made worse with real gases --never mind mechanical elements that introduce irreversible work and consequently extra dissipation.

Another interesting approach would be a treatment based on statistical mechanics, which allows you in principle to make the isotherms not exactly isothermal, and introduce fluctuations in temperature.

Edited by joigus
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15 minutes ago, sethoflagos said:

If you express the total potential energy of water in terms of its height above mean sea level, then the amount of it that you can extract from an overshot waterwheel is limited to (1 - ht/hm) where  hm and ht are the heights above msl of the mill race and tail race respectively. Do you see anything familiar here? Can you imagine all that waste potential energy disappearing down the tail race? Why isn't waterwheel efficiency a part of this limit?

Well, "the potential energy of water" is different from the potential energy of a given quantity of heat.

What does "the potential energy of water" (your usage of the phrase) actually mean?

You are really talking about gravity, we are not decomposing water into hydrogen and oxygen gas molecules or some such thing along those lines using the water itself as fuel.

Heat entering a heat engine is not a substance or liquid like water subject to a gravitational force that is somehow pulling the heat through the engine. The heat, as sensible, measurable "heat" that can be detected with a thermometer does not go through the engine, it only goes INTO the engine to be converted to mechanical work. The "work" goes out, the "heat" disappears and never comes out the other side as heat. At least not all of it, only some percentage of what went in.

Taking the water analogy, the water would disappear while passing over the mill wheel.

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1 hour ago, Tom Booth said:

My apologies if that end paragraph was posted before I responded to the beginning of your lengthy post.

Unfortunately almost everything you have said in this thread and previously leads to the inevitable conclusion

2 hours ago, Tom Booth said:

As I understand it, there absolutely cannot be < 0 (less than zero).

But you don't understand it. That is the problem.

When myself and others tell you that you are misquoting an equation relating to Carnot efficiency you  respond quoting dead philosophers instead of Mathematics.

You clearly have a facility with mechanics and the maths is actually very elementary so you should have no trouble understanding a properly phrased explanation.

NO it is impossible to buld a material physical 'Carnot Engine'.  It is purely a theoretical concept to explore the theory of Thermodynamics.

It is impossible in theory as well as practice for reasons that Seth and Joigus have hinted at and for reasons exchemist and swansont have tried to explain to do with your misunderstanding of various definitions in thermodynamic theory.

You tell me that we do not need to explore what a 'reservoir' is defined as yet you do not seem to understand that the basic (Th-Tc)/Th  efficiency formula is invalid for any real world heat engine.

Considering the amount of time and effort you put into this, I really cannot understand why you do not want to explore the finer points of this issue.

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25 minutes ago, joigus said:

If the "working agent" is an ideal gas, and all the intermediate states are of equilibrium, its energy exchanges with the rest of the universe can be expressed as proportional to its temperature. Thereby Carnot's limit as a function of both temperatures. It's not --as OP has been repeatedly suggesting-- because the basic concept of efficiency is based on temperatures. It's not. It's based on energy exchanges.

Therefore:

20 minutes ago, Tom Booth said:

Well, "the potential energy of water" is different from the potential energy of a given quantity of heat.

is answered by 'energy is energy'. And 'energy exchanges are energy exchanges'.

24 minutes ago, Tom Booth said:

Taking the water analogy, the water would disappear while passing over the mill wheel.

No. Some of the water's potential energy is transformed into shaftwork just as some proportion of heat (thermal energy) can be converted into shaftwork.

You ignored my questions relating to the limit (1 - ht/hm):

1 hour ago, sethoflagos said:

Do you see anything familiar here? Can you imagine all that waste potential energy disappearing down the tail race? Why isn't waterwheel efficiency a part of this limit?

The ratio is reject energy / input energy just as TC/TH is in the Carnot limit.

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1 hour ago, Tom Booth said:

I thought the equation was based on the hot and cold "reservoirs" The engine sits between.

Is that not correct?

like a water wheel between an elevated body of water and a lower body of water.

the Carnot "efficiency" is a measure of the "height of the fall" not the actual mechanical efficiency of the water mill (or engine) to convert "available" energy. It is the temperature difference, not how well the engine actually utilized that temperature difference. The actual engine has nothing whatsoever to do with it.

Are you trying to be obtuse? These reservoirs are just concepts denoting a heat source and a heat sink, whose temperatures remains constant irrespective of whether heat is drawn from or given to them They are nothing to do with any "environment". Just a heat input at one temperature and a heat output at another. That's it. Doesn't matter from where, or to where.

The concept of a cycle is specifically intended to represent the working processes of a theoretical, idealised heat engine, containing a fixed amount of an ideal gas, which goes through a cycle of expansion and contraction and does work in the process, all according to the gas laws.

So it's not a real engine (of course) but a distillation, to its essentials, of the simplest imaginable heat engine: a gas, alternately expanded and contracted and pushing a piston, and the heat flows involved in that.

As it turns out, the efficiency formula that results is beautifully simple, as you have realised, and only depends on the input and output temperatures. But it is a formula for a heat engine: the most efficient one that is possible to imagine.

Edited by exchemist
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32 minutes ago, Tom Booth said:

Well, "the potential energy of water" is different from the potential energy of a given quantity of heat.

What does "the potential energy of water" (your usage of the phrase) actually mean?

You are really talking about gravity, we are not decomposing water into hydrogen and oxygen gas molecules or some such thing along those lines using the water itself as fuel.

Heat entering a heat engine is not a substance or liquid like water subject to a gravitational force that is somehow pulling the heat through the engine. The heat, as sensible, measurable "heat" that can be detected with a thermometer does not go through the engine, it only goes INTO the engine to be converted to mechanical work. The "work" goes out, the "heat" disappears and never comes out the other side as heat. At least not all of it, only some percentage of what went in.

Taking the water analogy, the water would disappear while passing over the mill wheel.

You misunderstand the analogy.

The heat flow is analogous to the energy being converted from potential energy and extracting work. Your focus on water being a fluid is misplaced. It’s the energy of the fluid, not the fluid itself. The potential energy is reduced. Nothing has to happen to the water.

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47 minutes ago, sethoflagos said:

If you express the total potential energy of water in terms of its height above mean sea level, then the amount of it that you can extract from an overshot waterwheel is limited to (1 - ht/hm) where  hm and ht are the heights above msl of the mill race and tail race respectively. Do you see anything familiar here? Can you imagine all that waste potential energy disappearing down the tail race? Why isn't waterwheel efficiency a part of this limit?

I can see your point to some extent. It makes sense and I think I understand your point. But...

Let's visit the two engines together using the same piece of ice between them as the lower cold reservoir.

The two engines first exist at equilibrium with the environment. There is no hotter or colder "reservoir" correct?

So we create a mock cold reservoir using our refrigerator to make a thin  sliver of cold. Ice, cold metal plate, whatever.

Now we have ambient heat entering the engines from both hot sides, but this heat cannot get to the cold reservoir any other way except through one or the other or both engines.

Once the heat reaches the gas inside the engine(s) the gas expands and drives the piston(s) performing work and the heat "goes out" as mechanical shaft work. The engines are running.

From my reading, I've come to the understanding that actually it is theoretically possible to convert 100% of the heat used in the expansion of the gas into work output. I've read this statement in many places. Is this statement not correct, that ALL of the heat can be converted to work on the outward "power stroke" ?

So, consider this for a moment.

If I push a car up a hill against gravity is not that comparable to gas expanding and pushing a piston out against atmospheric pressure?

What if I use up all my strength pushing the car up the hill and collapse?

Haven't I added potential energy to the car?

Hasn't the expanding gas added potential energy to the piston?

As  a result of the expansion work the gas that did the work falls in temperature, like I fell down exhausted.

What happens to the car now?

What happens to the piston once the energy from the expanding gas has been exhausted?

The car rolls back down hill. Does it not? There is a return to equilibrium.

What about the piston, if 100% of the heat used to expand the gas has been converted to work and the temperature of the gas, and therefore also the pressure falls, the engine is left out of balance. Will there not be a return to equilibrium?

The outside atmospheric pressure that the expanding gas was working against will push the piston back and equilibrium will be restored.

What quantity of heat that went into the engine originally had not been converted? How much has gone into the "cold reservoir", our sliver of ice or whatever, between the two engines?

None so far. But aha!

The work done by atmosphere will heat up the gas and that heat will need to be removed. Right?

Well, what temperature is the gas at the end of the power stroke?

Hasn't it fallen to the temperature of the ice, or perhaps below?

What happens to the car when it reaches the bottom of the hill we pushed it up adding potential energy? Does it stop rolling at the bottom of the hill?

Atmosphere pushes the piston in restoring equilibrium to the system.

Is it then possible to add another dose of ambient heat to drive the piston out again?

Why not?

My basic point is that there is no actual fluidic substance that was raised up flowing out of the engine carrying it's stored potential energy with it. The engine is hermetically sealed.

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What happens to the best of compression when atmospheric pressure pushes the piston back.

Well, the gas is in the process of collapsing. The gas absorbs the heat raising the temperature back up to the temperature of the hot side by the time it is, maybe 2/3 of the way back. At this time the displacer moves off the "heat source" side. But where did the heat actually come from?

Atmospheric pressure?

At the last moment the input heat and the "heat of compression" collide, but the piston has also acquired momentum and keeps going compressing the gas further to a temperature even greater than the "heat source".

The bat meets the ball so to speak.

Grand slam.

The piston goes out with increased velocity.

Has any heat reached the "cold reservoir" yet?

Edited by Tom Booth
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3 hours ago, Tom Booth said:

Great, and I'm more than happy to prove my hypothesis wrong. So IMO it's a fantastic suggestion for an actual experiment.

Let's outline the logical next step by using one addition. Assume the setup works; We have the Stirling engine running and cold plate insulated with an adjacent Stirling engine or good enough insulation. Place the running engine(s) in a well-insulated box.

All the work performed by the engine(s) will be friction loss in the engine or movement of air inside the box. These losses will heat the air inside the insulated box. Conservation of energy means no energy enters or exits the box and no energy is permanently lost in the box, just changing between heat and mechanical work. The running engine(s) keep the cold side cool, and the losses are heating the air. In principle the box is a sealed system, and the engine will never stop since a permanent difference in temperature is established. This goes against my understaning of physics, but it seems to follow logically from your ideas and your conclusion of experiments?

3 hours ago, Tom Booth said:

There is though, I believe, at least one very real example of a "self-running" cooler/heat engine utilizing evaporative cooling.

Can you provide a reference? It could add value to the discussion.

(As far as I know it is ok per forum rules to link to further reading)

Edited by Ghideon
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Today I've been removing trash from the basement that is in the way of my running wires and need to hurry and get to the landfill before it closes soon.

There has been a flood of posts while I was working I'll have to read later and respond to.

Not ignoring anyone, just another busy day.

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2 hours ago, Tom Booth said:

The two engines first exist at equilibrium with the environment. There is no hotter or colder "reservoir" correct?

There is no heat flow into the engines because they at they same temperature as the hot reservoir.

2 hours ago, Tom Booth said:

So we create a mock cold reservoir using our refrigerator to make a thin  sliver of cold. Ice, cold metal plate, whatever.

There is no heat flow into the engines because they at they same temperature as the hot reservoir.

2 hours ago, Tom Booth said:

Now we have ambient heat entering the engines from both hot sides, but this heat cannot get to the cold reservoir any other way except through one or the other or both engines.

There is no heat flow into the engines because they at they same temperature as the hot reservoir.

However, the air inside the engines has started to cool as they melt the ice sliver. The air contracts perhaps enough to pull the power piston down from TDC to BDC and move the displacer away from the hot side heat exchanger.

2 hours ago, Tom Booth said:

Once the heat reaches the gas inside the engine(s) the gas expands and drives the piston(s) performing work and the heat "goes out" as mechanical shaft work. The engines are running.

Now at last heat can start to flow into the engine. We reheat the air back to the hot reservoir temperature and if no nett work is extracted and there air no friction losses we may have just enough energy to get the piston back to TDC.

Then we can melt a bit more ice.

Edited by sethoflagos
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16 minutes ago, sethoflagos said:

There is no heat flow into the engines because they at they same temperature as the hot reservoir.

There is no heat flow into the engines because they at they same temperature as the hot reservoir.

There is no heat flow into the engines because they at they same temperature as the hot reservoir.

However, the air inside the engines has started to cool as they melt the ice sliver. The air contracts perhaps enough to pull the power piston down from TDC to BDC and move the displacer away from the hot side heat exchanger.

Now at last heat can start to flow into the engine. We reheat the air back to the hot reservoir temperature and if no nett work is extracted and there air no friction losses we may have just enough energy to get the piston back to TDC.

Then we can melt a bit more ice.

That's why we begin by using our big freezer plugged into the utility grid as a start point.

I neglected to mention the ice or whatever is colder than need be to chill the engine on the cold side and the working fluid before moving the displacer to let in heat.

Your scenario positions everything in the most inopportune way.

The action of the displacer is rather essential to the operation acting as a kind of heat valve, to let heat in only at the critical moment. Like a batter waiting for the pitch.

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6 minutes ago, Tom Booth said:

Your scenario positions everything in the most inopportune way.

No. I take care to line up my initial conditions in a consistent way so that I don't get caught in a circular sequence of robbing Peter to pay Paul (which your posts tend to be full of btw).

Eventually the message sinks in that there are no free lunches and life's a bitch.

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54 minutes ago, sethoflagos said:

No. I take care to line up my initial conditions in a consistent way so that I don't get caught in a circular sequence of robbing Peter to pay Paul (which your posts tend to be full of btw).

Eventually the message sinks in that there are no free lunches and life's a bitch.

Sorry if I'm a bit of an optimist and don't entirely share that outlook on life.

Anyway, I spent \$60 and I think the experiments will be fun to watch however it turns out.

I think it (the twin engines) will run for a good long time.

In my experiments so far, the engines running on a cup of hot water keep running for about 3 hours tops.

On a cup of ice, 33 hours. So far, with imperfect insulation.

I know, heat capacity and all that.

Truck is loaded so I'm on my way to the landfill.

Still half a dozen other posts to respond to.

Anyway you haven't really addressed my points, only pulled off the ignition wires, drained the gas tank and smashed the valve lifters.

"See, it can't possibly run."

Well don't worry, I'm a mechanic. Maybe I can fix it up.

BTW, if you look at one of these engines, all the friction producing components are on one side. Running on ice, that's the hot side.

So...

Friction loses?

Heat lost to the hot side is what? A bit of additional heat that either goes right back in to power the engine or returns to the heat source it came from.

We haven't even talked about a regenerator, so far.

Edited by Tom Booth
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So let us return to the OP and explore what Carnot efficiency.

The equation under discussion is

efficiency = eta  =       $\eta = \frac{{{T_h} - {T_c}}}{{{T_h}}}$

So how does this stack up against what we know ?

Well it tells us that

1. @Tom Booth is correct when he says that eta is 1 (or 100% if you will) when Tc is zero.

2. It also precludes the possibility of overunity since to get eta greater than 1 would require Tc to be negative, which it cannot be by definition.

3. It tells us that eta is zero when Tc - Th.
This is effectively the zeroth law of thermodynamics which describes thermal equilibrium.
It is saying that reservoirs Tc and Th are in equilibrium so no heat will pass from one to the other.
Since not heat passes none is converted into heat.

4. It also tells us that Tc and Th must be different with Th > Tc.
This is another way of saying that there must be 2 different reservoirs. One will not do as they are effectively the same, which statement leads to the second Law.

So this shows why consideration of the reservoirs is vitally important.
But can we probe further ?

We can ask what are these reservoirs and how is heat exchanged with them and this brings us to the suprising conclusion.

If you add to or subtract heat from any real physical material object you must change its temperature by the equation

Temp change = mass times specific heat.

Yet we have defined a reservoir as a body that does not change its temperature when heat is added or subtracted !

The only valid conclusion we can draw is that such a reservoir is not a real physical material body.
It can only ever be an unattainable ideal.

A further complication is the question of how is heat exhanged ?

Again we run straight into the zeroth law, but this time it seems to be against us for it is quite adament that heat does not flow between two bodies in themal equilibrium.
So the temperature of the working fluid must be greater than Tc to add energy to it and less than Th to receive heat from it.

This leads to the conclusion that eta itself must be an unattainable 'limit' in some sense
and also to the conclusion that a real material Carnot engine cannot exist.

Chemists and Chemical Engineers will be familiar with a similar situation in regards of concentration where we can regard constant concentration as one that is so strong that it is almost unchanged by the addition or subtraction of solute.

This is only the first step of deeper enquiry, however.

When we look at the thermodynamic variables we run into the question of what is meant by Tc and Th.
But that is a story for another post.

Edited by studiot
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15 hours ago, studiot said:

So let us return to the OP and explore what Carnot efficiency.

The machine is constructed to operate within a fixed volume range VC =< V =< VH, and it contains a set amount of working fluid. So if the cycle is in heat engine mode, fixing the low temperature of the working fluid also fixes the minimum pressure on the cold side via PC = nRTC/VC. This in turn sets the maximum pressure available from compression via P= PC (VC/VH)^k which now fixes the minimum temperature from which it can absorb heat by TH = PHVH/nR.

In other words the geometry of the machine has determined the ideal power output it can achieve for any given cold reservoir. There are very few degrees of freedom. Increasing the power output by using a higher temperature heat reservoir is certainly possible but at reduced efficiency due to significantly irreversible heating. In any event this higher reservoir temperature now determines power output per cycle.

So what happens if we don't extract the power produced?

It depends on the machine type, but for the engines we've been discussing:

• The piston accelerates reducing the cycle time and increasing dissipation losses
• The working fluid fails to equilibriate at either reservoir temperature due to shortened contact time.
• Pressure (a little) and temperature fail to equilibriate following compression and expansion strokes.
• The envelope of the PV cycle shrinks until either work output is accommodated by dissipation to the surroundings , or the machine disintegrates.

This is where the image posted previously makes sense.

On 1/31/2023 at 7:48 PM, Tom Booth said:

There is this PV diagrams that was generated in real time taking real measurements using an LTD Stirling engine very much like the one I used I'm my experiment.

The machine can't do much about dV, but it can reduce dT and hence dP by here about 70% to get W as small as necessary.

And what can be said for heat engines puts similar constraints on the loading characteristics of heat pumps and refrigeration cycles.

Their geometry greatly restricts flexibility in their service conditions. And deviations from those service conditions are greeted invariably with reduced thermodynamic efficiency.

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1 hour ago, sethoflagos said:

To add to @studiot's good summary, I'll add just a couple of comments about running heat engines under variable loading conditions.The machine is constructed to operate within a fixed volume range VC =< V =< VH, and it contains a set amount of working fluid. (...)

I think perhaps Studiot intended to continue his outline and I did not wish to interrupt.

One important point he made though, that I would like to highlight:

Regarding your "fixed volume range" I'm not sure about the assumption behind you equation:

Quote

fixed volume range VC =< V =< VH,

It might help, well it would help me anyway, in following your reasoning if you could clearly define your terms.

Vc I assume represents the volume of the gas at is lowest temperature? V cold?

VH the volume of the gas at its highest temperature? V hot? Or is that Enthalpy.

And V, what does this represent? The mean?

So the volume is greatest when the temperature is highest?

The volume is least when the temperature is lowest?

What kind of engine are you referring to?

You do realize, generally speaking: compression decreases volume and increases temperature. Expansion increases volume and decreased temperature. So in an engine Vh > Vc.

If your initial assumption is incorrect, there is no need to consider what follows from that.

Edited by Tom Booth
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15 minutes ago, Tom Booth said:

Regarding your "fixed volume range" I'm not sure about the assumption behind you equation:

Quote

fixed volume range VC =< V =< VH,

VC was originally fully compressed ie piston at BDC. Vvolume fully expanded at TDC. They don't tie to hot and cold as such. Both are clearly defined by the geometry which is the salient point.

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9 minutes ago, sethoflagos said:

VC was originally fully compressed ie piston at BDC. Vvolume fully expanded at TDC. They don't tie to hot and cold as such. Both are clearly defined by the geometry which is the salient point.

So c as in Vc stands for Volume compressed? c=compressed (not cold)

And the H in VH? (Or Vh lower case. Both appear, so this too is confusing.) Is your H the same thing as your h?

Edited by Tom Booth
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1 minute ago, Tom Booth said:

So c as in Vc stands for Volume compressed? c=compressed (not cold)

And the H in VH?

For an ideal Stirling engine maximum compression occurs at the end of the cooling cycle with the working fluid at cold reservoir temperature TC. Conversely, maximum expansion occurs at the end of the heating cycle with the working fluid at hot reservoir temperature TH.

This is the opposite case to most heat engine designs. I can see I wrote minimum pressure when I meant maximum so I've likely got a pair of subscripts reversed somewhere. I'll comb through it later tonight to check.

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