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

[Fool]

In fact I'll do it again.

Qh heat added from Tc to get to Th.

Uh total energy contained in the gas at Th
Uc total energy contained in the gas at Tc

Qh=Uh-Uc
100=400-300
Heat equals the change in internal energy.

Starting point one:
Uc=MCvTc
300=MCv300

Maximum point two:
Uh=MCvTh
400=MCv400

MCv= Mass times constant chosen so it is one.

If an engine forward stroke starts while containing Uh 400 Joules, the most work it can produce is 400 J. This will be when it gets to absolute zero Kelvin. We want a maximum. The piston returns to the starting point costing zero back work, because it is at zero Kelvin. The buffer pressure has added a total of nothing because the same pressure opposed the forward stroke, as helped the back stroke. P∆V work, out forward stroke, in backward stroke, same P∆V

Now, back at the starting volume and zero K, 300 J must be added to complete the cycle. That 300 Joules must come from the forward work. Why? Because we are only adding 100 J. 300 J more must come from somewhere to get to the 400 J second position and 300 J starting first position. Forward work is the only energy available.

If the atmosphere were expected to add the Joules, it will do it at Tc for the entire return stroke and 300 J of work will be needed to compress the gas.

If the buffer pressure force compresses it, 300 J Uc will be rejected as energy of compression to the cold sink as heat, to keep the reverse work at a minimum.

So, it starts with Uh and returns the amount Uc to complete the cycle. The maximum efficiency to start at Uc, add Qh, expand all the way to zero borrowing from the Uc energy, return the extra gained by over expanding to become a cold hole by adding back the borrowed heat Uc, during the return stroke, to complete the cycle is:

n=(Uh-Uc)/Uh

Maximum efficiency after all ideas have been tried.

Substituting in the temperature energy formulas Ut=MCvTh:

n=(MCvTh-MCvTc)/(MCvTh)

Canceling MCv top and bottom.
n=(Th-Tc)/Th

This raises the question, Why is Qc subject to the same efficiency as Uc? It's the same engine operating from the same temperatures. It should get the same efficiency even if we caculated all the work Uc could supply on the single forward stroke. Again it is the return stroke that gets it.

If it is obfuscated with buffer pressure, then the forward stroke gets it. There is no efficiency win from applying buffer pressure to a full cycle. It cancels too. Or as Matt says, there's no free lunch from buffer pressure in regard to efficiency.

This is Proof that mathematically shows the cancelations, and scientifically describes why it is an absolute maximum when operating in a full cycle from two limiting temperatures.

If you want to beat the following equation:

n=(Th-Tc)/Th

You will need to measure the temperatures and :

n=W-Qsupplied

Qsupplied is the outside energy of your energy source and not necessarily the Qh absorbed by the internal gas. It would be better to know Qh. A good indicator diagram will give that.

Measure Work, and the wattage of the heat source, and elapse time for the run.

And if you want to do good science you will run the experiment many times.

[Fool]

A mathematical proof is mathematical proof. Take it to a PhD physics, engineering, or mathematics professor. If they disagree I would love to know why.

[Fool]

This one too.

From Wikipedia:

Wikipedia

Stirling_cycle_pV.svg.png (10.59 KiB) Viewed 2834 times

Work Energy coming out of the engine is positive.

Heat energy going in is positive.

Vt = Volume top dead center.
Vb = Volume bottom dead center.
M Mass of the working gas.
R gas constant for working gas
ln() = Natural log function.
Th and Tc Temperatures hot an cold
W12 work for each process, respectively. Four total. 2 isothermal. Two constant volume.
Q12, Heat transfered for each of four process respectively.

The following is for the ideal Stirling Cycle as depicted in the above graphic.

W12 = M•R•Th•ln(Vb/Vt) positive because work is coming out.
W23 = 0. Zero volume change.
W34 = M•R•Tc•ln(Vt/Vb) negative because work is going in.
W41 = 0. Zero volume change.

Q12 = W12 = M•R•Th•ln(Vb/Vt) positive because the energy is going in.

Q23 = -M•Cv•(Th-Tc) negative because the energy is coming out.

Q34 = W34 = M•R•Th•ln(Vt/Vb) negative because the energy is coming out. Vt smaller than Vb, ln() function negative for values less than one.

Q41 = -M•Cv•(Tc-Th) positive because the energy is going in. Tc smaller than Th, two negatives combined to become a positive.

Q23 and Q41 cancel each other, being an equal and opposite regenerator processes. Ideally.


n = Wout/Qin = (W12 + W34) / (Q12)

Substituting:
n={(M•R•Th•ln(Vb/Vt))+(M•R•Tc•ln(Vt/Vb)}/{M•R•Th•ln(Vb/Vt)}

Using the log identity ln(x) = -ln(1/x), for ln(Vt/Vb) = -ln(Vb/Vt), the equation becomes:
n={(M•R•Th•ln(Vb/Vt))-(M•R•Tc•ln(Vb/Vt)}/{M•R•Th•ln(Vb/Vt)}

Canceling M•R•ln(Vb/Vt) top and bottom:
n=(Th-Tc)/Th
Ideally and a maximum for the temperatures given.

It is direct observable science from the observed relationship between temperature, pressure, volume, energy, mass and the linear coefficients of heat Cv, and ideal gas constant R for the real gas, Rn for Nitrogen, or whatever.

Again this doesn't equate Q to T. The differences just cancel out in the equations of n=W/Q, PV=MRT, and Q=MCvT, for a full cycle. Doesn't work for single strokes. Real engines will be worse.

[Tom Booth]

... As you've said lots of people come up with that idea and discuss it.

My guess is the reason you don't hear about it is that they have all failed. Of course, YouTube is full of magic shows like that. But failures are reported, if you know how to look for them.
...

Assuming you are talking about a Stirling engine operating a heat pump as a "fuel pump" to supply its own high grade heat, I'd love to see any such failure report.

Since you are proclaiming such reports exist "if you know how to look for them" I would assume you do and have.

If not more empty words you cannot back up, please provide one or more such report.

Both Stirling engines and heat pumps are not well understood by many. To build a DIY version of either one is a challenge, to build a machine that combines the features of both with no funding and facing endless smear campaigns even if you DO produce a working example is pretty poor odds.

Historically there are many examples. Charles E. Tripler with his liquid air machine seemed like a completely honest and straight up guy.

He was the victim of a smear campaign, just like you Mr "fool" are conducting against me.

[Tom Booth]

This one too.

From Wikipedia:

Stirling_cycle_pV.svg.png

Work Energy coming out of the engine is positive.

Heat energy going in is positive.

Vt = Volume top dead center.
Vb = Volume bottom dead center.
M Mass of the working gas.
R gas constant for working gas
ln() = Natural log function.
Th and Tc Temperatures hot an cold
W12 work for each process, respectively. Four total. 2 isothermal. Two constant volume.
Q12, Heat transfered for each of four process respectively.

The following is for the ideal Stirling Cycle as depicted in the above graphic.

W12 = M•R•Th•ln(Vb/Vt) positive because work is coming out.
W23 = 0. Zero volume change.
W34 = M•R•Tc•ln(Vt/Vb) negative because work is going in.
W41 = 0. Zero volume change.

Q12 = W12 = M•R•Th•ln(Vb/Vt) positive because the energy is going in.

Q23 = -M•Cv•(Th-Tc) negative because the energy is coming out.

Q34 = W34 = M•R•Th•ln(Vt/Vb) negative because the energy is coming out. Vt smaller than Vb, ln() function negative for values less than one.

Q41 = -M•Cv•(Tc-Th) positive because the energy is going in. Tc smaller than Th, two negatives combined to become a positive.

Q23 and Q41 cancel each other, being an equal and opposite regenerator processes. Ideally.


n = Wout/Qin = (W12 + W34) / (Q12)

Substituting:
n={(M•R•Th•ln(Vb/Vt))+(M•R•Tc•ln(Vt/Vb)}/{M•R•Th•ln(Vb/Vt)}

Using the log identity ln(x) = -ln(1/x), for ln(Vt/Vb) = -ln(Vb/Vt), the equation becomes:
n={(M•R•Th•ln(Vb/Vt))-(M•R•Tc•ln(Vb/Vt)}/{M•R•Th•ln(Vb/Vt)}

Canceling M•R•ln(Vb/Vt) top and bottom:
n=(Th-Tc)/Th
Ideally and a maximum for the temperatures given.

It is direct observable science from the observed relationship between temperature, pressure, volume, energy, mass and the linear coefficients of heat Cv, and ideal gas constant R for the real gas, Rn for Nitrogen, or whatever.

Again this doesn't equate Q to T. The differences just cancel out in the equations of n=W/Q, PV=MRT, and Q=MCvT, for a full cycle. Doesn't work for single strokes. Real engines will be worse.

If that is supposed to be a Wikipedia article please provide the link.

Other than the generic pv diagram it appears to be something you just made up.

I know search engines seem to be getting worse instead of better at finding things but I've tried to locate your article many times in many different ways and this is all I ever get:

Compress_20240612_133029_9530.jpg (16.57 KiB) Viewed 2829 times

So if this isn't another one of your fake "derivations" please provide the link.

Not that some anonymous Wikipedia post is any kind of authority but it would be something

[Tom Booth]

A mathematical proof is mathematical proof. Take it to a PhD physics, engineering, or mathematics professor. If they disagree I would love to know why.

Copout.

So much for: "the mathematical proofs have been pored over, and over again."

If that were that case you should be able to find one you didn't have to make up yourself.

[VincentG]

Well Tom I've come to the conclusion that you are not defending the "Stirling cycle", but instead the "Booth cycle". The Stirling cycle is isothermal by definition.

Maybe it would be a good start to draw a pv plot and write out the processes of your cycle so they can be better understood.

[Tom Booth]

Well Tom I've come to the conclusion that you are not defending the "Stirling cycle", but instead the "Booth cycle". The Stirling cycle is isothermal by definition.

Maybe it would be a good start to draw a pv plot and write out the processes of your cycle so they can be better understood.

Some theoretician started alleging the "ideal" Stirling cycle would be or should be isothermal, but that is far from reality.

This is an actual PV diagram with the approximate isotherms plotted based on the ideal gas law, which I don't consider valid since the ideal gas law assumes "normal" conditions that don't account for rapid compression and expansion in an engine, Nevertheless any REAL PV readings of a Stirling engine pretty obviously cuts across the isotherms like a knife through a layer cake.

Resize_20230522_031849_9086.jpg (175.66 KiB) Viewed 2820 times

Your "ideal" Stirling engine cycle is pure imagination.

Personally I'm interested in real data. There is no "Booth cycle". I base my conclusions on the observable and available real data. Actual readings and measurements.

It’s important to keep in mind that all these numbers presented are for the ideal Stirling cycle, which will never exist in real life,

https://www.google.com/amp/s/blog.mide. ... s_amp=true

The first attempt at an analysis of the cycle was published in 1871 by Gustav Schmidt. Much as the Otto cycle has become the classic Air standard cycle to describe the spark ignition engine, the cycle described by Schmidt has become the classic ideal Stirling cycle. This is unfortunately mainly because the Schmidt analysis yields a closed form solution rather than its ability to predict the real cycle

https://people.ohio.edu/urieli/stirling ... ermal.html

[Tom Booth]

So, VincentG, there was never any such thing as the "ideal Stirling cycle" for 55 years after Stirlings invention, so IMO it is not at all accurate to say the Stirling cycle is isothermal "by definition".

The Schmidt model is just one theoretical analysis that bears no real resemblance to the actual Stirling cycle and could never exist in reality.

[Fool]

Your "ideal" Stirling engine cycle is pure imagination.

The real cycle is half the efficiency of the ideal Stirling.

It is the: MCv(Th-Tc) / (Area on the PV diagram) = n

Not very much area or heat-in.

Th-Tc = 199.65-181.9 about 17.75° ∆F x5/9 = 9.86 ∆K

I'll let you calculate the rest. Although I don't trust your numbers on that graph. Guessing they are °F. You have access to where they came from.

[VincentG]

Your "ideal" Stirling engine cycle is pure imagination.

I disagree and think it is darn near possible.

Either way, you do idealize adiabats, so can you list the steps to the cycle you propose?

[Tom Booth]

Your "ideal" Stirling engine cycle is pure imagination.

I disagree and think it is darn near possible.

Either way, you do idealize adiabats, so can you list the steps to the cycle you propose?

I don't "propose" anything.

As I've said many many times and will now repeat. I am only analyzing data from real, already existing Stirling engines.

They should IMO probably best be analyzed using graphs similar to those used for driven oscillators.

PV diagrams originated with the steam engine and IMO are not really applicable to Stirling engines.

[Tom Booth]

I think VincentG, a "isothermal" cycle might be approximated by an in phase oscillation.

That is like the instructor in that video walking very slowly with the pendulum so it doesn't swing.

You could get the piston and displacer to move or work "in phase".

IMO, however, that would not be a practical engine for most common purposes.

An engine "in resonance" seems more likely to bear fruit IMO.

[Fool]

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.

Qcz=MCvTc. It is the internal energy at the start of the cycle top dead center acquired by the room temperature and pressure. 300K and 15 psi. It is before Qh 100J is added. It is 300J.

It is like having 300 dollars in the bank.

There is talk that ambient air is full of energy. This is that energy and contained by the engine's working space, TDC.

Our goal is to determine the efficiency of the engine for a full cycle in reference to thermodynamic variables. First look at the efficiency if no heat is added.

300J, 300K, 1 to 1 temp energy correlation, determined by gas mass and type Cv. If a stroke expands that gas, the gas does work. That work will be equivalent to the temperature reduction. 100K = 100J. If the expansion goes until T becomes zero Kelvin, the engine will put out a maximum of 300J. Ideally.

A zero Kelvin all real gasses would have frozen into a solid chunk. Luck of the draw, the chunk is all located on the head. The piston can now be returned with zero work used. Back at top dead center with no compression or back work needed. 100% of the initial internal energy is converted to work. But wait...

The gas is a frozen chunk, end of cycle not reached. To get there, heat must be put back into the frozen gas until it is 300K again. That will take 300J, or the same amount as the work gained. Leaving a zero energy in zero energy out. So including the base energy has no effect on power in or out of a cycle.

Now let us add 100J every cycle. Start at 300J add 100J, expand to zero Kelvin return free stroke. Return 300J from the 400J work obtained. Grand total per cycle 100J in and out, or 100% efficiency. As predicted by the Carnot formula if you go all the way to zero Kelvin.

So now we want to see how efficient a cycle is if we don't go all the way to zero. Start at 300J add 100J for a total of 400J. Expand it getting 100J of work out.

Shazam! Pushing the piston back in now requires work. It comes from the momentum obtained by the expansion. Why? Because the gas is at a minimum pressure. It is at or close to atmospheric pressure 15 psi, but not zero, infact high, near 75% of the highest value.

I've already shown that going from 400K to 300K and back to TDC staying at 300K, is going from 400J to 300J and getting 100J of work, and now minus back work.

n=work/total Joules available.
n=(400J-300J)/400J = 0.25 or 25%.
Or normalized with temperature:
n=(Th-Tc)/Th=(400K-300K)/400K=0.25 or 25%.

So how much work does it take to compress it. Well it is the work that the efficiency takes away.

100J•(1-0.25) or 75J. That work must be rejected as 'the heat of compression' or the temperature will rise above Tc=300K, and require more back work for an even lower efficiency.

This is true for any complete cycle operating between two temperatures. No specific cycle was specified, therefore it applies to all cycles, all engines.

[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.

Qcz=MCvTc. It is the internal energy at the start of the cycle top dead center acquired by the room temperature and pressure. 300K and 15 psi. It is before Qh 100J is added. It is 300J.

It is like having 300 dollars in the bank.

There is talk that ambient air is full of energy. This is that energy and contained by the engine's working space, TDC.

Our goal is to determine the efficiency of the engine for a full cycle in reference to thermodynamic variables. First look at the efficiency if no heat is added.

300J, 300K, 1 to 1 temp energy correlation, determined by gas mass and type Cv. If a stroke expands that gas, the gas does work. That work will be equivalent to the temperature reduction. 100K = 100J. If the expansion goes until T becomes zero Kelvin, the engine will put out a maximum of 300J. Ideally.

A zero Kelvin all real gasses would have frozen into a solid chunk. Luck of the draw, the chunk is all located on the head. The piston can now be returned with zero work used. Back at top dead center with no compression or back work needed. 100% of the initial internal energy is converted to work. But wait...

The gas is a frozen chunk, end of cycle not reached. To get there, heat must be put back into the frozen gas until it is 300K again. That will take 300J, or the same amount as the work gained. Leaving a zero energy in zero energy out. So including the base energy has no effect on power in or out of a cycle.

Now let us add 100J every cycle. Start at 300J add 100J, expand to zero Kelvin return free stroke. Return 300J from the 400J work obtained. Grand total per cycle 100J in and out, or 100% efficiency. As predicted by the Carnot formula if you go all the way to zero Kelvin.

So now we want to see how efficient a cycle is if we don't go all the way to zero. Start at 300J add 100J for a total of 400J. Expand it getting 100J of work out.

Shazam! Pushing the piston back in now requires work. It comes from the momentum obtained by the expansion. Why? Because the gas is at a minimum pressure. It is at or close to atmospheric pressure 15 psi, but not zero, infact high, near 75% of the highest value.

I've already shown that going from 400K to 300K and back to TDC staying at 300K, is going from 400J to 300J and getting 100J of work, and now minus back work.

n=work/total Joules available.
n=(400J-300J)/400J = 0.25 or 25%.
Or normalized with temperature:
n=(Th-Tc)/Th=(400K-300K)/400K=0.25 or 25%.

So how much work does it take to compress it. Well it is the work that the efficiency takes away.

100J•(1-0.25) or 75J. That work must be rejected as 'the heat of compression' or the temperature will rise above Tc=300K, and require more back work for an even lower efficiency.

This is true for any complete cycle operating between two temperatures. No specific cycle was specified, therefore it applies to all cycles, all engines.

I think you are forgetting something.

You wrote:

Now let us add 100J every cycle. Start at 300J add 100J, expand to zero Kelvin return free stroke. Return 300J from the 400J work obtained. Grand total per cycle 100J in and out, or 100% efficiency. As predicted by the Carnot formula if you go all the way to zero Kelvin.

It will not be necessary to take 300 Joules from the 400 Joules of work obtained.

You wrote:

Qcz=MCvTc. It is the internal energy at the start of the cycle top dead center acquired by the room temperature and pressure. 300K and 15 psi. It is before Qh 100J is added. It is 300J.

In that scenario, the 300 Joules can just be let in from the environment. So that should be:

Now let us add 100J every cycle. Start at 300J add 100J, expand to zero Kelvin return free stroke. ...

...Return 300J (of heat from the ambient environment) (cut-'from the 400J work obtained'). Grand total per cycle 100J in and (500 joules of work) out, or (400%) (cut-'100%') efficiency. As (not) predicted by the Carnot formula if you go all the way to zero Kelvin.

Cleaning that up:

Now let us add 100J every cycle. Start at 300J add 100J, expand to zero Kelvin return free stroke. Return 300J of heat from the ambient environment Grand total per cycle 100J in and 500 joules of work out, or 400% efficiency. Much higher efficiency than predicted by the Carnot formula if you go all the way to zero Kelvin.

How you imagine starting at 300°K and adding 100 joules of heat will expand the gas to 0°K has me wondering, but it's your scenario. Whatever it takes doesn't really matter though.

However, if that were possible, add 100 joules, expand gas to 0° K, and do 100 joules work, then get 400 joules of work from atmospheric pressure on the return stroke, then add 300 joules environmental heat to get an additional 300 joules of expansion work. Add 100 joules again... Get another 100 joules work.

Hmmm... I think I messed up my calculations.

100 joules in per cycle

100 joules work expansion stroke
100 + 400 joules work "free return stroke'
100 + 400 + 300 free ambient heat expansion work. Back to start of cycle.

So 100 joules heat supplied and
100 + 400 + 300 = 800 joules work out.

So that would be 800% efficiency?

Anyway, you seem to be forgetting that the 300° K or 300 joules ambient energy can be obtained for free from the environment. Or as you stated: "acquired by the room temperature ...300K ... It is before Qh 100J is added. It is 300J"