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

[MikeB]

Tom,
If you could build an engine that had ice at the cold end, but that ice never, ever melted, what role would it have within the engine?
My intuition says that if no heat/energy transfers into the ice, then it isn't having any effect on the engine cycle.

[Tom Booth]

Tom,
If you could build an engine that had ice at the cold end, but that ice never, ever melted, what role would it have within the engine?
My intuition says that if no heat/energy transfers into the ice, then it isn't having any effect on the engine cycle.

It puts an upper limit on fools magical Qcz

[Tom Booth]

...
Heat is flowing through an engine. The flow isn't any particle, or caloric, nor does the "flow" cause the work. It is necessary to produce pressure increase and decrease. Heat in, higher pressure. Heat out lower pressure, back work. Calorics have been discarded completely in main stream science. That was accomplished about 100 years ago. Heat flows more the way light flows.

...

OK, let's go with that last statement.

How is light converted into electricity in a photovoltaic panel?

Does the light "flow through" the PV panel?

Does the light, after being converted to electricity need to be "rejected"?

Light is energy. Heat is energy.

[Tom Booth]

Well, right off the bat I see the same conceptual difficulty with:

Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th.

We add 100 joules at a temperature of 400°K

As stated previously "internal energy from heat ADDED to the cycle all the way from zero K Kelvin to Th" counts the baseline "internal energy" of 300 joules as if it is carried along or included in the 100 joules actually supplied as "heat".

Then later this same baseline internal energy is supposedly "rejected" though in actuality it was never added in the first place.

You already conceded previously:

Qcz is not rejected during the running. Nor is it added. It is the base energy that DQh is added to. Earlier definition discarded.

viewtopic.php?p=21686#p21686

This is the fundamental fallacy of the whole Carnot Limit so-called "LAW". It's an accounting error.

The baseline "internal energy" between 0°K and 300°K cannot be included with "heat added".

The only way that statement begins to make sense is in the context of Caloric theory where heat is considered to be a fluid that passes through the engine and temperature is the measure of a quantity of a fluid.

I see that you are challenging the Qcz term. It technically should use the U variable name. I used 'Q' because I'm only modeling the internal thermal energy. Since it can be represented as coming only from heat input, I thought it would be okay to represent it as a name similar to heat. It is just a name. U gets confused with 1/2MV^2 and Mgh and I^2R, E.T.C.... Qhz and Qcz are concerning temperature values, Kelvin scale/absolute.

If the gas is at 300 K, one doesn't need to ask how it got there. We just use it as an absolute quantity of energy.

The comparison to cooling is only to bring in absolute zero K, and have some reference to calculate the absolute energy contained in the engine at rest before even heating it to Th.

It is an absolute accounting of total energy. Sort of like having a pot of money, say 300 dollars. Someone asks to get paid the next amount you put into that pot. You put in $100. And you pull out $100, but the dollars you pull out are a mix of all the dollars you have. So instead of paying the $100 you last put in, there will be a 25~75 mix of bills that you pay out.

The person who you paid has to use the old bills for food, and the new ones for gasoline (silly rule made up for demonstration only). He would have $25 to spend on gasoline. And $75 to reject for food.

My derivation shows how absolute energy gets substituted for Delta energy. The same amount gets used by the engine but it becomes clear as to were it must be spent. 25 J on work, 75 J on back work, and or, heat transfer. The transfer of lower temperature heat, helps get more power out of the complete cycle by saving energy to complete the compression.

Hence it's often called 'the heat of compression'.

The two equations that are equal to 'n', do not imply that the numbers are equal:

Case in point, example:

n= (DQh-DQc)/DQh=(100-25)/100=0.25

n=(Qhz-Qhc)/Qhz=(400-300)/400=0.25

It's the same ratio of heat going out either way. The accounting is the same either way. You put into your bank $100, you take out $100. You have a bank account of $400, you have a balance of $300. The same $100 amount, just more information the second way. When you deposit a check, don't you ask for a printout of your balance? I do. More information, right.

The difference is that I know what Qcz is, so can use it in calculations to find 'n' and then DQc.

This is the fundamental fallacy of the whole Carnot Limit so-called "LAW". It's an accounting error.

Not really. There are other way to derive/prove the Carnot Theorem. I have come up with this way because the other methods require more advanced mathematics. Integral Calculus specifically. I wanted it more understandable by more people.

It wouldn't be consistent reliable science if there weren't multiple cross checks. That is why one data set won't overthrow a 200 year old theory. It will take many many plus, cross checks and theories. That is why people keep asking for additional tests on your experiment. That's why I ask for power output, input temperatures, theory, vacuum tests, oven tests, upside down tests. If we weren't trying to help we wouldn't be asking.

The best way to look at Qcz is the mathematical definition:

Qcz = M•Cv•Tc

It is the starting point of the engines cycle. And the end point. Starting point and end point must be the same for a complete cycle. It is also the point reached after DQc is releasd, regardless of how much that will be.

Heat is flowing through an engine. The flow isn't any particle, or caloric, nor does the "flow" cause the work. It is necessary to produce pressure increase and decrease. Heat in, higher pressure. Heat out lower pressure, back work. Calorics have been discarded completely in main stream science. That was accomplished about 100 years ago. Heat flows more the way light flows.

The parts of Thermodynamics that are kept are the ones that can be derived from current understandings of Thermodynamics, including but not limited to, Kinetic Theory, Statistical Theory, and Quantum Mechanical Theory. No, I don't know all those, so don't ask for those proofs. Just take comfort that if current theories don't agree, someone is working on it. From what I do know of those theories, Entropy, and Carnot, allegedly, hold up well. Kind of like Relativity Theory, both Einstein and Carnot have been picked to death, and still prevail. No matter how smart we are, we'd have zero chance of changing that. And, so far, I only see evidence of support for Carnot. Not even Matt has anything substantial, claims yes, but no working devices. Improvements perhaps. Throwing Carnot out? Not even close. All your points don't add up. My derivation is consistent with others, I'm surprised no one else has mentioned it. I suppose it is considered non ground breaking.

Let's return to Fool's post above.

First of all, I agree with the conclusion at the end: "My derivation is consistent with others".

Yes! Yes! Yes! It certainly is!!!!

Fool is correct in that regard. He has made no mistake in that he has reproduced exactly the GLARING ERROR that is being perpetuated by this farcical Carnot efficiency limit. method of determining so-called "efficiency".

I don't really know how anyone can outline, step by step, how Carnot efficiency is calculated in such a manner and NOT see it is illegitimate. This is not valid mathematics. Not valid accounting.

[Tom Booth]

I see that you are challenging the Qcz term. It technically should use the U variable name. I used 'Q' because I'm only modeling the internal thermal energy. Since it can be represented as coming only from heat input, I thought it would be okay to represent it as a name similar to heat. It is just a name.

The problem is what I've underlined.

The pre-existing internal energy CANNOT be legitimately "represented as heat input"

It isn't "heat" transfered into the engine.

It's NOT "heat input".

[Tom Booth]

If the gas is at 300 K, one doesn't need to ask how it got there. We just use it as an absolute quantity of energy.

Yes we do!

How the "heat" got into the engine is the whole basis of "efficiency".

You have 300 joules that make up the very pre-existing mass and energy of the "working fluid" in equilibrium with the surroundings.

We supply an additional 100 joules of heat that tips the balance. In determining a balance with a scale you do not include the weight of the scale do you? No! You zero it out. Of course it matters where the energy came from.

[Tom Booth]

The comparison to cooling is only to bring in absolute zero K, and have some reference to calculate the absolute energy contained in the engine at rest before even heating it to Th.

It is an absolute accounting of total energy. Sort of like having a pot of money, say 300 dollars. Someone asks to get paid the next amount you put into that pot. You put in $100. And you pull out $100, but the dollars you pull out are a mix of all the dollars you have. So instead of paying the $100 you last put in, there will be a 25~75 mix of bills that you pay out.

The person who you paid has to use the old bills for food, and the new ones for gasoline (silly rule made up for demonstration only). He would have $25 to spend on gasoline. And $75 to reject for food.

Except the 75 joules rejected cannot be spent or used for "food" or anything. That's where the analogy breaks down.

Well mack, I know you put in ten hours at $10/hour and earned $100 dollars, but the $100 is 25% of the $400 in the payroll account. And 25% of $100 is $25 dollars. Here is your $25 dollars pay.

Wait a minute. What does how much you have in your account have to do with how long I worked or how much I earned or how much I'm owed?

Well, you see. I already had $300 in the payroll account. I transfered $100 into the account to cover your pay for the day. That makes $400 total.

$100 is 25% of $400 so your pay amounts to 25%, understand?

25% of the $100 I transfered into the bank is 25 dollars.

Well what happened to my other 75 dollars?

Entropy.

Sorry, it's just a law of the universe. Can't be helped, now run along.

[Tom Booth]

My derivation shows how absolute energy gets substituted for Delta energy. The same amount gets used by the engine but it becomes clear as to were it must be spent. 25 J on work, 75 J on back work, and or, heat transfer. The transfer of lower temperature heat, helps get more power out of the complete cycle by saving energy to complete the compression.

This is getting even more convoluted with sloppy accounting: "75 J on back work, and or, heat transfer." ?

Carnot efficiency theory, or whatever you want to call it dictates that the 75 joules go into the "cold reservoir" period. No option for using it for "back work". It is rejected as heat, because supposedly it is "impossible" for it to be converted to work positive, negative, "back" or otherwise. It must, supposedly, go into the "cold reservoir".

You might try and say the expanding gas is "working against" atmospheric pressure. The outside atmospheric pressure is why you need to include the 300 joules.

Again, your trying to include the weight of the scale.

Before adding heat, the working fluid and outside pressure were in equilibrium. These "zero out".

[matt brown]

Carnot efficiency theory, or whatever you want to call it dictates that the 75 joules go into the "cold reservoir" period. No option for using it for "back work". It is rejected as heat, because supposedly it is "impossible" for it to be converted to work positive, negative, "back" or otherwise. It must, supposedly, go into the "cold reservoir".

I can't believe anyone can botch Carnot so bad...

In your 300-400k cycle example, assuming 100J is the Cv heat between 300k and 400k, this J always remains inside engine. IOW, this J = heat of regeneration. If we consider the following notation:

Rin = regen in
Rout = regen out
Qin = heat from source
Qout = heat to sink
Win = work in
Wout = work out

then (ideally) Rin = Rout = 100J

Finite Qin and Qout values will depend upon actual volume ratio,
but when isothermal

Qin = Wout aka expansion
Qout = Win aka compression

So, let's consider this "100J" simply as OLD heat that is a cycle constant.
Meanwhile, let's consider Qin and Qout as NEW heat/s that are cycle variables.

During this 300-400k cycle, Rin = R out = 100J OLD heat goes round and round WITHIN engine.
Meanwhile, Qin = xxJ NEW heat from source during expansion
and Qout = yyJ NEW heat to sink during compression

Qout is a consequence of compression (Win) adding NEW heat to gas that must go to sink (Qout) during isothermal compression while 100J OLD heat remains constant !!!

And since Q = W here, for this 300-400k cycle, work of compression (Wneg) will be 3/4 work of expansion (Wpos) thanks to U linear T. Thus, Carnot wins (again) with .25 efficiency via yyJ/xxJ = Qout/Qin = 3/4

Maybe cheat Carnot by manipulating buffer pressure ??? dream on...buffer pressure won't change compression values, only where the compression force is coming from.

[matt brown]

During this 300-400k cycle, Rin = R out = 100J OLD heat goes round and round WITHIN engine.
Meanwhile, Qin = xxJ NEW heat from source during expansion
and Qout = yyJ NEW heat to sink during compression

In retrospect, better wording would be...

During this 300-400k cycle, Rin = R out = 100J OLD heat goes round and round WITHIN engine.
Meanwhile, Qin = xxJ NEW heat to gas during expansion
and Qout = yyJ NEW heat to gas during compression

[Tom Booth]

Carnot efficiency theory, or whatever you want to call it dictates that the 75 joules go into the "cold reservoir" period. No option for using it for "back work". It is rejected as heat, because supposedly it is "impossible" for it to be converted to work positive, negative, "back" or otherwise. It must, supposedly, go into the "cold reservoir".

I can't believe anyone can botch Carnot so bad...

In your 300-400k cycle example, assuming 100J is the Cv heat between 300k and 400k, this J always remains inside engine. IOW, this J = heat of regeneration. If we consider the following notation:

Rin = regen in
Rout = regen out
Qin = heat from source
Qout = heat to sink
Win = work in
Wout = work out

then (ideally) Rin = Rout = 100J

Finite Qin and Qout values will depend upon actual volume ratio,
but when isothermal

Qin = Wout aka expansion
Qout = Win aka compression

So, let's consider this "100J" simply as OLD heat that is a cycle constant.
Meanwhile, let's consider Qin and Qout as NEW heat/s that are cycle variables.

During this 300-400k cycle, Rin = R out = 100J OLD heat goes round and round WITHIN engine.
Meanwhile, Qin = xxJ NEW heat from source during expansion
and Qout = yyJ NEW heat to sink during compression

Qout is a consequence of compression (Win) adding NEW heat to gas that must go to sink (Qout) during isothermal compression while 100J OLD heat remains constant !!!

And since Q = W here, for this 300-400k cycle, work of compression (Wneg) will be 3/4 work of expansion (Wpos) thanks to U linear T. Thus, Carnot wins (again) with .25 efficiency via yyJ/xxJ = Qout/Qin = 3/4

Maybe cheat Carnot by manipulating buffer pressure ??? dream on...buffer pressure won't change compression values, only where the compression force is coming from.

Looks like a complete hatchet job just chopping up Fool's analysis and making up out of thin air some new null input value Qin == xxJ ?

Then in your last post:

During this 300-400k cycle, Rin = R out = 100J OLD heat goes round and round WITHIN engine.
Meanwhile, Qin = xxJ NEW heat from source during expansion
and Qout = yyJ NEW heat to sink during compression

In retrospect, better wording would be...

During this 300-400k cycle, Rin = R out = 100J OLD heat goes round and round WITHIN engine.
Meanwhile, Qin = xxJ NEW heat to gas during expansion
and Qout = yyJ NEW heat to gas during compression

you take a hatchet to your own new made up nonsense redefining Qout "to sink" as Qout "heat to gas"

Considering Qout is scattered all through your original hatchet job who could know what on earth your driving at.

But somehow in some way you think this adds up to "Carnot wins (again) with .25 efficiency"

Yes, Carnot always wins so long as his devotees are capable of spewing such meaningless irrational mumbo jumbo with mysterious valueless variable that represent who knows what "heat" from who knows where, made up on the fly.

Any "regen" has little or nothing to do with Carnot efficiency which is based on heat input from the "hot reservoir" work output and "waste heat" to the cold reservoir

Any "OLD heat goes round and round WITHIN engine" has no bearing on what crosses the system boundary. So introducing regeneration is just muddying the water and confusing issues.

[Fool]

MikeB, very good point. Could start the engine on ice, then remove the ice and it will, according to Tom, run forever. LOL. Good one. Brilliant.

Tom, freezing a block of ice with a power hungry device, so that the ice can be used by a inefficient heat engine putting out an almost negligible amount of energy while the ice melts doesn't sound to me like responsible use of energy, but hey, have at it.

I can't believe anyone can botch Carnot so bad...

Matt, you are correct about disintegrating the Carnot science. In other words, a Carnot and or a Stirling Engine, has four processes.

Carnot:
1. At Qcz and Tc, constant entropy/adiabatic compression temperature rise, zero Q heat, work in W1*, arrive at V2=Vmin, P2=Pmax>P1, T2=Th.

2. Constant T expansion, heat added, DQh=100 J, work out W2=100 J, V3>V1>V2, P3=P1, T3=T2=Th.

3. Constant Entropy/Adiabatic expansion temperature drop, Saved internal energy, zero Q heat, work out W3=-W1*, V4=Vmax, P4<P1, T4=T1=Tc.

4. Constant T compression, work in W4=75 J, heat out DQc=75 J, V5=V1, P5=P1, T5=T1=Tc, back to the condition of Qcz position one. This is the same rule as for process #2 but in reverse and at 75% of the temperature, so it is less.

*Note: Work-in #1, W1, is equal and opposite to Work-out #3, W3. They cancel each other. Rule of adiabatic cycles.

Stirling:

1. At Qcz, Tc. Constant volume regeneration displacement, 100 J heat from displacer, arriving at V2=V1, P2=Pmax, T2=The.

2. Constant T expansion, DQh added 100 J, work out 100 J, V3=Vmax, P3=P1<P2, T3=Th=T2.

3. Constant volume regeneration displacement, 100 J of heat to the displacer, zero work, V4=V3=Vmax, P4=Pmin, T4=T1=Tc.

4. Constant T compression, work in 75 J, heat out 75 J, V5=V1, P5=P1, T5=T1=TC, back to Qcz starting position. This is the same process as #2 but in reverse and at 75% of the temperature.

Those two are maximum amounts of work and completely idealized. Real cycles will be worse.

At this point I respectfully request that, would Tom please give us an image of a PV diagram with his proposed cycle drawn. Tom is very good at providing images and graphs.

Maybe cheat Carnot by manipulating buffer pressure ??? dream on...buffer pressure won't change compression values, only where the compression force is coming from.

Absolutely correct. Misunderstanding that keeps beating us up. Buffer/atmospheric pressure adds nothing to the power, work, or energy, output from an engine. Simple mathematics follows:

Work = Force x distance.

Multiplying the right side by area/area.
W = F•d•(A/A) = (F/A)•(d•A)

F/A = Pressure
d•A = Volume

Substituting into Work equation:
W = (F/A)•(d•A) = P•V

Buffer/Atmospheric pressure is constant, the volume changes for the two strokes for a complete cycle. DeltaV expansion equals DeltaV compression so total work out will be the sum of the two. Since they oppose each other, one must be negative. Negative and equal.

W = P•DV - P•DV = 0.0

Work out equals Work in. So in regard to efficiency, and work out. What happens outside the working gas is nullified.



Matt the big problem with those points is that Tom doesn't believe in 'constant T volume change', neither expansion nor compression. He believes in adiabatic. And he thinks adding an electric generator will change that process. Granted, real processes are somewhere between adiabatic and constant T, unfortunately adding heat during a Stirling Expansion is the only way it will work effectively. The pressure of expansion must be higher than the pressure of compression, and the only way to accomplish that is by adding heat and removing heat. It needs to be for a full cycle. He seems to always narrow the discussion to single processes ignoring the rest of them. Puts buffer pressure on for the return stroke, leaving it off for the forward stroke, etc...

I may start a tread on heat transfer and why there is no such thing as Adiabatic.

[Tom Booth]

Tom,
If you could build an engine that had ice at the cold end, but that ice never, ever melted, what role would it have within the engine?
My intuition says that if no heat/energy transfers into the ice, then it isn't having any effect on the engine cycle.

It's not too difficult to conceptualize if you can get over the idea the heat is a fluid that flows from hot to cold like water flowing downhill and instead view it as what it actually is, at least according to kinetic theory, which at least seems to be a closer approximation.

For example, pushing someone on a swing.

The energy input comes "in" from one side only.

The swing goes out. Reaches a mid way point and continues, reaches an extremity of motion and then returns, a little short of where it started perhaps, but within reach to give it another push and add more kinetic energy.

By timing each "push" just right, the velocity and distance of travel of the swing can be gradually increased. Some mechanical power could be extracted.

Such kinetic energy from the push on a swing is not "flowing through" to any "cold reservoir" anywhere.

The expansion force generated by supplying heat at just the right moment when the piston reaches TDC is an equivalent situation to giving a swing an extra boost each time it returns.

The whole concept of a "flow" of heat through the engine is complete nonsense.

Nevertheless, the swing needs to be "pulled back" to get the process started. The cold working fluid, in order to expand needs to be made cold as a starting position to expand from.

The ice serves the same purpose as pulling back on a swing to get the swinging motion started. After that it is no longer needed.

[Tom Booth]

....

At this point I respectfully request that, would Tom please give us an image of a PV diagram with his proposed cycle drawn. Tom is very good at providing images and graphs.

....

This is a real time tracing of an LTD Stirling:

stirling_pv_real.jpg (287.07 KiB) Viewed 132 times

https://youtu.be/ck8thAoi_rA?si=Khb77JAOWyneplvR


As I've said many times before, I'm not proposing any new engine, just an alternative theory to explain how Stirling engines work based on observations and experiment.

As can be seen, the engine spends nearly all of the "compression" portion of the cycle with the working fluid well below atmospheric pressure, which is indicated by tbe central horizontal line.

The engines internal pressure oscillates above and below atmospheric pressure similar to the way a swing or pendulum moves forward and backward from a center point of equilibrium.

[Stroller]

Morning all, I'm new here, mech eng background, tinker with classic bikes etc.
Looking at Tom's graphs, I think there are two reasons the internal pressure oscillates above and below atmospheric.
One is that the volume of the system is still being reduced by the power piston as the working fluid starts to heat up. The other is the inertia of accelerating moving parts from TDC resisting the expansion of the working fluid as its temperature continues to rise, thus raising pressure. At the other end of the cycle, the opposite effects occur, reducing pressure.

These inertial diabatic effects on the pressure of the working fluid causing the pressure oscillation are small compared to the diabatic effect of the heat source transfering energy to the working fluid, which has to be in equilibrium with the rate at which energy is lost from the cold side, plus the frictional losses in the power train, plus whatever work is being extracted from the motion of the system.