Stirlingengineforum.com

Tom, by "removing rocks" you are reducing the inside pressure. Reduced pressure on water reduces the temperature at which it boils.

That is a delta T. Sorry. That then will be input for the Carnot Theorem.not very large for very little change.

Only during an adiabatic process does Caloric/heat not "fall"/flow,. You have described an adiabatic process. That is why it will generate zero CYCLIC work. That is why adiabatic bounce won't continue forever, zero area enclosed.

Even your description conveys zero work. Rocks are removed, forward stroke. All the same rocks are put back on, return stroke. Equal and opposite "rock" work out and in. Zero cyclic total.

You can't beat Carnot. Lots of proof. Even your own thought example proves it.

This is the thread to hammer this out. Other treads should be used to discuss heat exchangers, materials, projects, and other ideas.

Fool wrote: ↑Tue Apr 16, 2024 10:59 am Tom, by "removing rocks" you are reducing the inside pressure. Reduced pressure on water reduces the temperature at which it boils.

That is a delta T. Sorry. ...

No it isn't.

Reducing the boiling point only means the water boils/changes phase to a gas much more readily at 100°C it does not create a ∆T.

The "infinite" 100°C "reservoir" will maintain the temperature at 100°C in any case.

Fool wrote: ↑Tue Apr 16, 2024 10:59 am ...
Only during an adiabatic process does Caloric/heat not "fall"/flow,. You have described an adiabatic process. That is why it will generate zero CYCLIC work.
...

No I haven't

1st convenient search reference:

Isothermic reactions are characterized as reactions that occur in constant temperature. Recall that phase changes (such as condensation, evaporation, etc.) occur at a constant temperature (melting point, boiling point,e etc.). Consider the following example. Water boils at 100∘C; this means that when liquid water is placed at its boiling point, it will convert rapidly to gas (water vapor). The reaction will happen at a constant temperature (at 100∘C) and, therefore, will be an isothermic process.

https://www.varsitytutors.com/physical_ ... -processes

I think maybe you need to brush up on your thermo or something.

The following quote from the following website :

http://www.iapws.org/faq1/boil.html

In the following table, we list the vapor pressure for water as a function of temperature as taken from the latest IAPWS formulation for general and scientific use.
Temperature
(degrees Celsius) Vapor Pressure
(MPa)
0.01 0.000 612
25 0.003 17
50 0.012 35
75 0.0386
100 0.1014
150 0.4762
200 1.555
250 3.976
300 8.588
350 16.529
373.946 22.064
The first value in the table is for water's triple point, which is the thermodynamic state where vapor, liquid, and solid coexist. The last value is for water's critical point. The critical point is the end of the vapor pressure curve; there the vapor and liquid phases become identical and at higher temperatures there is only a single fluid phase.

Updated April 26, 2000

Fool wrote: ↑Tue Apr 16, 2024 12:30 pm The following quote from the following website :

http://www.iapws.org/faq1/boil.html

In the following table, we list the vapor pressure for water as a function of temperature as taken from the latest IAPWS formulation for general and scientific use.
Temperature
(degrees Celsius) Vapor Pressure
(MPa)
0.01 0.000 612
25 0.003 17
50 0.012 35
75 0.0386
100 0.1014
150 0.4762
200 1.555
250 3.976
300 8.588
350 16.529
373.946 22.064
The first value in the table is for water's triple point, which is the thermodynamic state where vapor, liquid, and solid coexist. The last value is for water's critical point. The critical point is the end of the vapor pressure curve; there the vapor and liquid phases become identical and at higher temperatures there is only a single fluid phase.

Updated April 26, 2000

Right, at 1 atmosphere (0.1014 Mpa) water boils at 100°C at lower pressure it still boils, better and faster or more vigorously at 100°C

At higher pressure it will condense.

What's your point?

Maybe you are forgetting temperature is a measure of the average kinetic energy of a substance?

I've even demonstrated this recently.

https://youtu.be/KBtv3skwfDw

With water held at the boiling/condensation point just a very slight pressure change causes a transition between condensing and vigorous boiling.

Do you really imagine such a rapid pressure change has any significant effect on the temperature of the water/water vapor ?

The liquid, some but very little. The gas, a little more. My guess is the gas follows the adiabatic temperature drop curve. The liquid has more mass so requires more boiling loss temperature drop energy, so won't change much on account of very little work is applied.

The molecules going into gas will be at the temperature of the boiling point, which is reduced by lower pressure. It is why lowering the pressure allows non boiling to suddenly boil. Reduce the boiling point temperature and suddenly those molecules that were below the temperature are now at the new lower point , so start boiling. Look at the following video for a demo of liquid temperature drop from reduced pressure and boiling. Boiling removes energy from the liquid and adds it to the gas, zero heat. All done with work input.

https://m.youtube.com/watch?v=glLPMXq6yc0

You still are, mechanically getting zero cyclic work out. That is also part of the Carnot theorem. Work out equals Th-Tc. You've described nothing more than an adiabatic steam spring.

Zero delta heat-in-out equals zero cyclic work. The only way to get work out in a full cycle is to add heat and reject heat so that there is an enclosed area inside the indicator digram's path. Invented in 1796, long before Carnot or Kelvin. Pressure change without heat added or rejected is an adiabatic process. I already pointed out the temperature drop. Yes slightly during phase change because, isotherms are very close to isentropes. Your depiction hasn't mentioned heat addition or rejection, just internal energy usage, also known as adiabatic.

https://en.m.wikipedia.org/wiki/Indicat ... %20engines.

An indicator diagram is completely separate from theory, and is applied to working running engines experimentally. They are fairly easy to make and calibrate. Rudimentary ones could be made with a rubber band, or steel spring, a syringe or graphite piston, and a piece of card paper put on a spinning part of the engine and pen or pencil.

Fool wrote: ↑Tue Apr 16, 2024 9:11 pm ...Boiling removes energy from the liquid and adds it to the gas, zero heat. All done with work input.

...

The transfer of energy: "Boiling removes energy from the liquid and adds it to the gas" by definition, IS heat.

https://m.youtube.com/watch?v=glLPMXq6yc0

You still are, mechanically getting zero cyclic work out. That is also part of the Carnot theorem. Work out equals Th-Tc. You've described nothing more than an adiabatic steam spring.

No

My "ideal engine" is in contact with a "reservoir" at 100°C so heat/energy is readily available for transfer.as needed to boil the water at 1 atm.

Your video demonstration takes place in a vacuum. A vacuum is insulation, even from ambient heat. The water in the vacuum boils drawing on the "internal (thermal) energy" of the water itself. That is not "work".

No offense, but, did you really spend five years earning a BSME degree in Mechanical Engineering? Your comprehension of common thermo terminology and processes is hopelessly confused and muddled.

Zero delta heat-in-out equals zero cyclic work. The only way to get work out in a full cycle is to add heat and reject heat so that there is an enclosed area inside the indicator digram's path.

You are confusing heat and temperature. Not the same thing. This is the fundamental flaw in the Carnot Limit nonsense. It conflates heat and temperature making them equivalent, just as you are doing here.

η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Q is a specific quantity of heat/energy in joules
T is a measure of the "average" kinetic energy of an undefined quantity of a substance.

I think it is obvious Q and T are not necessarily equivalent. Your whole dissertation from the beginning has been a strenuous exercise in trying to MAKE them equivalent. Force an equivalence which simply does not exist in fact or in reality. Your mathematical contortions are invalid, just as the supposed equivalence represented by η = 1 – (Qc / Qh) = 1 – (Tc / Th) is invalid.

A drop of boiling water at 100°C does not "contain" or have the capacity to transfer out the equivalent number of Joules of heat as a 100 gallon tank full of boiling hot water at the same 100°C temperature.

Temperature is not a measure of Joules of heat available from some substance or "reservoir".

A Joule is a specific well defined quantity. Temperature is not a measure of any quantity. That is a false, obsolete, Caloric theory mindset where it was imagined that temperature measured the depth of some universal "fluid".

So no, my ideal engine is not "adiabatic".

If the water in your vacuum chamber were in contact with a "reservoir" of heat at 100°C it would absorb heat from that reservoir as it boiled (changes phase) rather than drawing on its own internal energy. There would be isothermal heat transfer between the reservoir and the boiling water and there would be no drop in temperature.

Boiling water in a vacuum is adiabatic.

Boiling water in contact with a 100°C reservoir at 1atm DOES involve heat transfer. It is not adiabatic, it would be isothermal.

For a guy with a BSME you don't even understand basic thermo concepts and terminology.

Fool wrote: ↑Tue Apr 16, 2024 9:11 pm ....

You still are, mechanically getting zero cyclic work out. That is also part of the Carnot theorem. Work out equals Th-Tc. You've described nothing more than an adiabatic steam spring.

...

Actually, work out is Qh-Qc.

That represents joules of heat transfered in less joules of heat transfered out.

Yes, the Carnot theorem naively conflates temperature T with heat Q.

Sorry to say Carnot was a naive young man with his head in the clouds and a lot of airy ideas. A Steam engine does not operate on the same principle as a water wheel.

How so much illogical convoluted nonsense continues to be perpetuated after 200 years without anyone doing some kind of reality check, (experiment) is beyond me.

I guess nobody really cared or had enough interest.

Fool wrote: ↑Tue Apr 16, 2024 10:59 am Tom, by "removing rocks" you are reducing the inside pressure. ....

...That is why it will generate zero CYCLIC work. ...

Even your description conveys zero work. Rocks are removed, forward stroke. All the same rocks are put back on, return stroke. Equal and opposite "rock" work out and in. Zero cyclic total.

You can't beat Carnot. Lots of proof. Even your own thought example proves it.
...

Perhaps you make a valid argument. The same could be said, however, of Kahn's Carnot engine.


https://youtu.be/aAfBSJObd6Y


If not, why not?


He removes rocks to expand the gas then replaced all the rocks to compress the gas. If my "ideal" engine is bogus then the Carnot engine also does zero work and is just as bogus.

Tom Booth wrote: ↑Wed Apr 17, 2024 6:18 am
Perhaps you make a valid argument. The same could be said, however, of Kahn's Carnot engine.

https://youtu.be/aAfBSJObd6Y

If not, why not?

He removes rocks to expand the gas then replaced all the rocks to compress the gas. If my "ideal" engine is bogus then the Carnot engine also does zero work and is just as bogus.

Many of Khan's thermo videos are lacking. To simplify stuff, let's consider Khan's video as a crude Stirling cycle with 600k source, 300k sink, 12 equal rocks, an inverted piston, and a cylinder where temperature can be regulated.

(1) starting at BDC with 12 rocks and 600k cylinder, remove 6 rocks to get piston to TDC
(2) remove all rocks, and regulate cylinder to 300k
(3) put rocks back on, one at a time
(4) piston will move from TDC after 3 rocks
(5) piston will reach BDC with 6 rocks
(6) compare rock count and piston motion vs temperature
(7) news flash...rock count during 300k compression was 1/2 rock count during 600k expansion
(8) whereby Wpos/Wneg = .5 but any Wnet is in the other 6 rocks we removed/discarded

Carnot wins again, this time with rocks...

matt brown wrote: ↑Wed Apr 17, 2024 8:56 pm

Tom Booth wrote: ↑Wed Apr 17, 2024 6:18 am
Perhaps you make a valid argument. The same could be said, however, of Kahn's Carnot engine.

https://youtu.be/aAfBSJObd6Y

If not, why not?

He removes rocks to expand the gas then replaced all the rocks to compress the gas. If my "ideal" engine is bogus then the Carnot engine also does zero work and is just as bogus.

Many of Khan's thermo videos are lacking. To simplify stuff, let's consider Khan's video as a crude Stirling cycle with 600k source, 300k sink, 12 equal rocks, an inverted piston, and a cylinder where temperature can be regulated.

(1) starting at BDC with 12 rocks and 600k cylinder, remove 6 rocks to get piston to TDC
(2) remove all rocks, and regulate cylinder to 300k
(3) put rocks back on, one at a time
(4) piston will move from TDC after 3 rocks
(5) piston will reach BDC with 6 rocks
(7) compare rock count and piston motion vs temperature
(8) news flash...rock count during 300k compression was 1/2 rock count during 600k expansion
(9) whereby Wpos/Wneg = .5 but any Wnet is in the 6 rocks we removed/discarded

Carnot wins again, this time with rocks...

LOL

Of course, with a mythical engine you can make up any nonsense you like. Including an engine that has full compression at BDC.

The guy putting on and removing rocks is still doing all the work.

Mythical negative output Carnot engine looses to any engine that actually exists.

Since in reality a Carnot engine passes ALL the heat through to the "cold reservoir' Q2 = Q1

Work output = big fat zero

Which is why it won't run at all without outside assistance.

Gadz Tom, for a guy who fancies himself as a world authority on engines, you must have missed that our common vertical piston configuration was called "inverted" in early ICE days, so "BDC" here in Vmin and "TDC" is Vmax.

If you (and anyone else) can't grasp source/sink issue (Carnot limit) from Khan's video, then you're wasting your time chasing the Holy Grail.

And as for any cold boiling nonsense, just clickbait. The video link that Fool supplied has the real story...very little water will flash to steam with pressure drop, surely nothing near vigorous boil. The latent heat of vaporization for water STP is 540 cal/g, and few molecules will have enough 'heat' to vaporize despite ANY pressure drop.

matt brown wrote: ↑Wed Apr 17, 2024 8:56 pm
(8) whereby Wpos/Wneg = .5 but any Wnet is in the other 6 rocks we removed/discarded

oops, that should be Wpos/Wneg = 2 or (Wpos-Wneg)/Wpos = .5