Stirlingengineforum.com

Fool wrote: ↑Sun May 26, 2024 6:14 am 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 wrote: ↑Wed Jun 05, 2024 12:42 am From Wikipedia:

Wikipedia

Stirling_cycle_pV.svg.png (10.59 KiB) Viewed 2836 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.

Fool wrote: ↑Wed Jun 12, 2024 8:46 am
... 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.
...

Fool wrote: ↑Wed Jun 12, 2024 10:09 am This one too.

Fool wrote: ↑Wed Jun 05, 2024 12:42 am 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.

Fool wrote: ↑Wed Jun 12, 2024 10:06 am 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.

VincentG wrote: ↑Wed Jun 12, 2024 12:03 pm 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.

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,

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

Tom Booth wrote:Your "ideal" Stirling engine cycle is pure imagination.

Your "ideal" Stirling engine cycle is pure imagination.

VincentG wrote: ↑Wed Jun 12, 2024 1:47 pm

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 wrote: ↑Mon Mar 25, 2024 9:48 am 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.

Fool wrote: ↑Tue Oct 29, 2024 12:13 pm

Tom Booth wrote: ↑Mon Mar 25, 2024 9:48 am 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.

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.

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.

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.

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.