Stirlingengineforum.com

Fool wrote: ↑Thu Nov 14, 2024 12:31 am .

No. Energy moves into an engine, some comes out as work, some as heat. Perfect efficiency is impossible. Breaking the Carnot limit is impossible.

Conservation of energy only works one way. It is conserved, all the energy coming out must equal all the energy going in minus the energy stored. Regardless of the forms. We can model an engine as if no energy is stored, not that we have to.

Flywheels store energy. Buffer pressures store energy.

Your opinion on this "is bogus. Illogical and really just plain silly."

.

Fool wrote: ↑Wed Jun 12, 2024 9:56 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 calculated 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.

Fool wrote: ↑Sat Nov 16, 2024 5:14 am .

Fool wrote: ↑Wed Jun 12, 2024 9:56 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 calculated 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.

.

Fool wrote: ↑Sat Nov 16, 2024 11:24 am .

You asked for the truth in this thread. It might as well start and end with it.

.

Fool wrote: ↑Sat Nov 16, 2024 5:14 am .

Fool wrote: ↑Wed Jun 12, 2024 9:56 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.

This continual 300J example is devoid of reality. Yeah, I get your drift Fool, but you're missing actual details...

(1) if 3 legged Lenoir 300-0k cycle then
(a) 300-0k adiabatic expansion
(b) 0-0k isobaric compression
(c) 0-300k isochoric input

(2) if Otto 300-0k cycle then
(a) 300-0k adiabatic expansion
(b) 0-0k isochoric cooling
(c) 0-0k adiabatic compression
(d) 0-300k isochoric heating

(3) if Atkinson 300-0k cycle then similar Otto but with isobaric compression

(4) if Stirling 300-0k cycle then
(a) 300 isothermal expansion
(b) 300-0k isochoric regen (cooling)
(c) 0-0k isothermal compression
(d) 0-300k isochoric regen (heating)

There's a major difference between consuming the internal energy during adiabatic expansion vs preserving it during isothermal expansion. I don't have any problem with zero K as a passing thought experiment, but dwelling on it is lame.

The first 3 cycles above have adiabatic expansion and are inline Tom's continual magic cycle where Wpos occurs during expansion but where Wneg=0 during compression. In effect, these 3 cycles are the same where 100cc 300k gas expands to 2000cc zero K gas deriving Wpos from sucking (lol) all internal energy out of gas. Assuming this 20x expansion reached zero K (and U=0 which it doesn't) then 2000cc gas is "compressed" back to 100cc at zero K (where Wneg=0) before 300k input restores cycle to start state. No doubt, nothing more than a bogus attempt at a gas spring cheat.

Meanwhile, cycle 4 is more realistic, but still suffers from the major flaw of all cold hole schemes where anything short of zero K will tax the cycle twice (1) backwork of compression above zero k (2) cooling required to maintain zero K sink.