Air Lift Turbine Generator

[Tom Booth]

Tom - here's how to calculate adiabatic expansion and compression. I've posted this several times since this is

(1) the easiest way to calculate the change in temperature or pressure based on volume change

(2) where table of values eludes PVT relationship

I did another similar post not long ago where I walk thru each calculator step (for any timid guys who come along) but I couldn't locate it.


adiabatic PVT calcs.png

I did not say adiabatic anything. That was Fools interpretation he pulled out of his ass.

I said isothermal compression to 2 ATM results in a volume reduction to 1/2 the heat of compression being removed.

If the pressure is then reduced back to 1 atmosphere the temperature of that same mole of gas will drop by 1/2. That is BEFORE absorbing any heat allowing the gas to expand.

Half of 300°K (approximate ambient) is 150°K or about -123°C or -190°F which is about 80°F below the temperature of dry ice.

It should then be a relatively easy proposition to run a Stirling engine on dry ice to start, use the Stirling engine to compress air isothermally to a mere 30 psi or more, using only simple air cooling by a radiator, then allow the compressed and cooled air to expand through a turbine to reclaim some mechanical work as well as impart additional cooling to the exhausted air by such work extraction.

That exhausted air should then be well below the -109°F temperature of the dry ice. And can therefore be utilized to keep the dry ice refrigerated. Or, after startup, the dry ice should no longer be needed as the cold exhaust can simply be used directly to cool the cold side of the Stirling engine.

This is nothing new, of course. Peter Lindemann has been advocating for this for a decade or more, but personally I had assumed that such a system would require extremely high pressure.

Perhaps high pressure is not actually necessary at all. A mere 30 psi seems quite sufficient for an "ambient heat engine" of any sort.

[matt brown]

I've been following this KPP nonsense for many years and NEVER seen anyone mention that this basic pitch would involve a relatively MASSIVE amount of low pressure air. So, what's up with all these piston air compressors ? If only ~2 bar air is required then why no rotary compressors ? Maybe all these piston compressors are just more schmooze for the schmucks (their 'mark').

I think the basic con hinges on borrowing a bunch of 'free' money to develop a scam and then selling the scam to a bunch of suckers how also borrowed a bunch of 'free' money to purchase it.

There's soooo many red flags with this scam that I can't believe these bozos have lasted this long.

(1) ascending air doesn't need to expand for buoyancy differential, but this adds mystique
(2) if ascending air expands, it doesn't need additional heating...ambient 'isothermal' is adequate for this slow low dV process, but adding compressor 'waste heat' (or whatever) adds more mystique

On a simple energy balance analysis, ideal isothermal waste heat from the compressor equals ideal isothermal input during expansion. It's bogus logic to transfer this waste heat between compressor and uptube when 'ideally' (keyword) this is the same quantity of energy at the same temperature. When unideal compression exceeds isothermal whereby compressor waste heat temperature exceeds ambient then this waste heat transfer makes sense, but at best, energy balance remains (no net gain from transferring unideal compressor waste heat to uptube).

Now, if you added some solar input during ascension whereby temperature exceeds ambient, then you're getting some extra input. However, within 100k constraint of liquid water, there's still little REAL gain between assumed isothermal input vs potential isobaric input extreme (the devil is in the details).

The difference between all these buoyancy scams and my mile high tube is that my scheme was a thought experiment that pondered whether it was possible to "convert gravity into heat". Yeah, totally off the wall, but my hipshot premise was nix latent heat of vaporization under high pressure prior engine (via gravity) and then recover this "nixed" latent heat under low pressure at condenser similar common Rankine cycle. The mile high dimension is a distraction, but was needed for mercury (xlnt pros and cons). Gaming latent heat may seem enticing, but it's far more complex than single phase gas. Kalina attempted this via his clever absorption cycle, but the heat of solution effectively equals the heat of vaporization when summed via energy balance. In retrospect, this was the defining issue that nixed my steam interest while scheming "steam" decades ago (no free lunch) otherwise absorption cycles would be everywhere, large and small.

[Tom Booth]

You two always seem to like complicating and confusing what are very simple principles.

A normal shop compressor looses ALL the energy used for compression as "waste heat". That waste heat can be recovered 100% by using it to pre-heat water for some other purpose.

In the mean time, you end up with a tank full of compressed air capable of doing work. ALL the energy used to compress the air was lost, so how can you end up with compressed air to do additional useful work?

The "trick" is the air was simply MOVED.

Put into the tank it is now in a circumstance where if released it can expand using the "free" heat in the surrounding atmosphere to do the expanding, and in the process of expanding, drive some pneumatic shop equipment.

Same principle here.

As far as energy balance there is no gain, but in the process the air was moved into a circumstance where it can be used to do useful work.

Obviously Rosch makes no effort to utilize any waste heat from the compressor. The energy saving comes from economically removing the heat, which standard shop compressors are already equipped for with built in cooling fins and fans.

IMO this setup is not much different from Tripler making liquid air and finding that he only needed 3 gallons of liquid air to power his steam engine that drove his compressor, boiling the liquid air using the freely available ambient heat.

What he says he found was that for every 3 gallons of liquid air he used to power his liquid air machinery he could manufacture 10 more gallons of liquid air. Keep 3 to keep the machine running and sell the other 7 gallons.

He was distributing liquid air all over the country to anyone who wanted it at a nominal cost.

Prof. Dewar produced the first ounce of liquid air at a cost of $3,000, but that now Mr. Tripler claims that he can produce it by his apparatus for five cents a gallon.

https://www.gutenberg.org/cache/epub/36 ... mages.html

[Tom Booth]

...
(1) ascending air doesn't need to expand for buoyancy differential, but this adds mystique
(2) if ascending air expands, it doesn't need additional heating...ambient 'isothermal' is adequate for this slow low dV process, but adding compressor 'waste heat' (or whatever) adds more mystique

On a simple energy balance analysis, ideal isothermal waste heat from the compressor equals ideal isothermal input during expansion. It's bogus logic to transfer this waste heat between compressor and uptube when 'ideally' (keyword) this is the same quantity of energy at the same temperature. When unideal compression exceeds isothermal whereby compressor waste heat temperature exceeds ambient then this waste heat transfer makes sense, but at best, energy balance remains (no net gain from transferring unideal compressor waste heat to uptube).

Now, if you added some solar input during ascension whereby temperature exceeds ambient, then you're getting some extra input. ...

IMO, this is baseless as well as contradictory.

Why should "solar input" heat be any better than waste heat directly from the compressor, which is itself only second hand "solar input" taken from the compressed air.

Using the tower water to cool the compressor also does double duty. Keeps the compressor cool for near isothermal compression while adding that removed "solar input" heat directly to the boyancy tank water, which extra heat you claim doesn't, but then claim does provide some extra input.

But the ambient heat is freely available and inexhaustible and engineering a custom water cooled compressor, in-tank heat exchangers, extra transfer pipe etc. would increase the cost for what might only be a slight improvement in output, so maybe not really worth it.

The various individuals associated with this appear to be high profile. IMO, not likely to be the kind of people who would participate in a scam.

Take John Crowley for example:

https://phgd.group/en/index.php/about/

https://bifrostonline.org/john-crowley/

https://x.com/jpgcrowley/with_replies

https://www.ki-tech.global/managment-team/john_crowley

https://m.youtube.com/watch


I've similarly looked into several others in association with this enterprise. I really have difficulty imagining so many high profile individuals associating themselves with some fly-by-night "free energy" scam.

Any one of these individuals associated with what you call a "scam" are 1000X more reputable than any of you anonymous clowns in this message board.

Each having thousands of followers and contacts on LinkedIn, Twitter/X, etc.

If there are "red flags" suggesting a scam it's with you and "fool", anonymous trolls.

[matt brown]

Why should "solar input" heat be any better than waste heat directly from the compressor, which is itself only second hand "solar input" taken from the compressed air.

The Flooid scheme adds solar to the mix to expand the ascending gas beyond ambient isothermal constraint which equals the most ideal compressor waste heat. The juggernaut here is that this moves the temperature of ascending gas beyond ambient isothermal and towards hotter isobaric, but there's little further gain between these two processes before hotter isobaric potential taps out due to water's boiling point.

Using the tower water to cool the compressor also does double duty. Keeps the compressor cool for near isothermal compression while adding that removed "solar input" heat directly to the buoyancy tank water, which extra heat you claim doesn't, but then claim does provide some extra input.

This is a buoyancy scheme, so (per Stanford report) if 200L barrels are half filled with 100L 2 bar air at bottom then air expands to 200L 1 bar at top (assuming 10m tower per report). However, each barrel could remain 100L (no expansion) and scheme would still has a massive positive buoyancy, but expanding air to 200L increases this buoyancy, and thereby increases the work output potential.

But the ambient heat is freely available and inexhaustible and engineering a custom water cooled compressor, in-tank heat exchangers, extra transfer pipe etc. would increase the cost for what might only be a slight improvement in output, so maybe not really worth it.

I'd pass on everything except water cooling the compressor.

The various individuals associated with this appear to be high profile. IMO, not likely to be the kind of people who would participate in a scam.

I've similarly looked into several others in association with this enterprise. I really have difficulty imagining so many high profile individuals associating themselves with some fly-by-night "free energy" scam.

Finance and energy seem to be the major scams, probably due to gaming numbers. Go ahead and buy some BitCoin, I'll stay with the "yellow stuff"...

[matt brown]

Overall, these buoyancy schemes reminder me of this...

air lift scheme.png (11.27 KiB) Viewed 1996 times

The air slugs are not to scale (nothing is) and the black balls are regulated valves in an air line between the blue down tube and red up tube. Assume both tubes are identical and that Maxwell's Demon is sitting atop down tube and placing magic wafers above air bubbles periodically whereby down tube air slugs will coincide up tube air slugs. In this manner, down tube compression equals up tube expansion but nothing happens until we get shit moving. I hate to overload our Demon, but he will also need to open and close the shown air valves when necessary (and 2 unshown fluid valves below each air valve). Per A, down tube air valve is always 5 bar while up tube air valve is always 4 bar. So, our trusty Demon operates these valves and advances the up tube fluid above air valve until the 5 bar air slug in down tube is exhausted per B. Then our Demon closes both air valves and opens both fluid valves and fluid advances thru the engine/generator until the next down tube air slug reaches it's air valve.

Almost believable except for

(1) equal pressure on both sides of engine
(2) magic wafers atop down tube which must isolate air slugs from fluid

[Tom Booth]

Why should "solar input" heat be any better than waste heat directly from the compressor, which is itself only second hand "solar input" taken from the compressed air.

The Flooid scheme adds solar to the mix to expand the ascending gas beyond ambient isothermal constraint which equals the most ideal compressor waste heat. The juggernaut here is that this moves the temperature of ascending gas beyond ambient isothermal and towards hotter isobaric, but there's little further gain between these two processes before hotter isobaric potential taps out due to water's boiling point.

None of that addresses the question of why solar heated water should be a useful improvement but compressor waste heat heated water is not.

Either way, cold air injected into hotter water will expand both isothermally as well as isobarically.

If you read the Flooid articles their idea of "solar" is the ambient heat "stored" in the air, and released by their air compressor.

Using the tower water to cool the compressor also does double duty. Keeps the compressor cool for near isothermal compression while adding that removed "solar input" heat directly to the buoyancy tank water, which extra heat you claim doesn't, but then claim does provide some extra input.

This is a buoyancy scheme, so (per Stanford report) if 200L barrels are half filled with 100L 2 bar air at bottom then air expands to 200L 1 bar at top (assuming 10m tower per report). However, each barrel could remain 100L (no expansion) and scheme would still has a massive positive buoyancy, but expanding air to 200L increases this buoyancy, and thereby increases the work output potential.

True in either case regardless how the water is heated.

Compressed ambient air is going to go higher than ambient temperature. The 100L injected air at 30 psi is going to expand to 200L cooling the water. The water will take in ambient heat, but extra heat input from the compressor would help to at least maintain the water temperature.

[/quote]

But the ambient heat is freely available and inexhaustible and engineering a custom water cooled compressor, in-tank heat exchangers, extra transfer pipe etc. would increase the cost for what might only be a slight improvement in output, so maybe not really worth it.

I'd pass on everything except water cooling the compressor.

The various individuals associated with this appear to be high profile. IMO, not likely to be the kind of people who would participate in a scam.

I've similarly looked into several others in association with this enterprise. I really have difficulty imagining so many high profile individuals associating themselves with some fly-by-night "free energy" scam.

Finance and energy seem to be the major scams, probably due to gaming numbers. Go ahead and buy some BitCoin, I'll stay with the "yellow stuff"...
[/quote]
[/quote]

That does not address the credibility issue. Neither the apparent high credibility of the Ki-Tach associates or your and "fools" complete lack thereof.

[Fool]

.

Tom you are so far away from understanding thermodynamics that it is impossible to communicate the principles with you. Get a calculus book. Ask for help learning it. You may as well add a trigonometry book and arithmetic book too.

Then get a good thermodynamics book. Considering you use such terms as, adiabatic, ideal gas, buoyancy, force, pressure, area, volume, bounce, kinetics, as if you understand their definitions, you still, obviously, have zero concept of their ramifications.

Seemed easy enough using the online calculator.

1 mole was about 0.8 something gallons of gas

At 2 ATM the volume reduced to 0.4 (about, as I recall. (At 300°K)

Return to 1 ATM at 0.4 (before expansion) temperature now equals -150°K

P1•V1/T1 = P2•V2/T2
2atm•0.4/300=1atm•0.4/T2

Solve for T2

You have just described a constant volume process. V1=V2=0.4, zero change, ∆V=0.

The ramifications of that are that an environmental temperature must be at or lower than 150 K so heat can be rejected until the temperature of the gas gets to that temperature. There is zero expansion to reduce the temperature.

Considering all you have to work with is water at standard atmospheric temperature, or above if used to cool the compressor, to get that low a temperature you will need to put work into a heat/refrigerator pump to get it.

If the heat is put into the water column, it's temperature will rise, and the isothermal compression will now be at a higher temperature, making the work of compression higher. The devil is in the details.

A link to John Crowley:

https://independent.academia.edu/JohnCrowley45

You'd think that he'd at least mention PKK or Flooid Power if he were connected to a free energy device that actually worked.

My guess is the power companies that list his name, do so without any real connection. Like Titan Submarine's claims of NASA and Boeing.

Your links yielded zero information to enlighten us regarding that alleged connection. And Real Men don't Twitter or 'X'.

.

[Fool]

.

The difference between solar heat and compressor rejected heat, isn't the heat. It is the source. One is sourced from the sun, and is used as Qh.

The other is sourced from either external work, or internal work. Either way the ramifications are that the work output will be reduced by the amount input.

.

[Tom Booth]

Seemed easy enough using the online calculator.

1 mole was about 0.8 something gallons of gas

At 2 ATM the volume reduced to 0.4 (about, as I recall. (At 300°K)

Return to 1 ATM at 0.4 (before expansion) temperature now equals -150°K (edit: corrected to 150°K)

P1•V1/T1 = P2•V2/T2
2atm•0.4/300=1atm•0.4/T2

Solve for T2

You have just described a constant volume process. V1=V2=0.4, zero change, ∆V=0.

Yes. So what?

The ramifications of that are that an environmental temperature must be at or lower than 150 K so heat can be rejected until the temperature of the gas gets to that temperature.

No, according to the ideal gas law, when the pressure is reduced from 2 ATM to 1 ATM at constant volume, the temperature of the gas drops by 50% (absolute/Kelvin).

Or please explain why the ideal gas law is wrong in this instance.

There is zero expansion to reduce the temperature.

Right. There is only a reduction in pressure, from 2 atmospheres to 1 atmosphere. That results in a drop in temperature, before any expansion takes place. (according to the ideal gas law)

Considering all you have to work with is water at standard atmospheric temperature, or above if used to cool the compressor, to get that low a temperature you will need to put work into a heat/refrigerator pump to get it.

Of course. Work was used to compress the gas isothermally from 1 to 2 atmospheres. Compressing a gas from 1 to 2 atmospheres isothermally with an air compressor requires some work.

If the heat is put into the water column, it's temperature will rise, and the isothermal compression will now be at a higher temperature, making the work of compression higher. The devil is in the details.

The great volume of water in the tank will rise in temperature as I imagine it, at the top of the tank, but that heat transfered to the water would be reabsorbed by the cold gas as it rises and expands in the tank.

A link to John Crowley:

https://independent.academia.edu/JohnCrowley45
...
....Real Men don't Twitter or 'X'.

As an anonymous Troll, your moronic opinions about what constitutes a "real man" are irrelevant.

Regarding compressed air, I would consider any compression that can be achieved by a common shop compressor relatively low pressure.

The gas leaving this shop compressor from at most, about 150psi has hardly had opportunity to expand, but is nevertheless absorbing heat from the ambient air as it leaves the pressurized tank.

https://youtu.be/2hYQtB4QkEY

Still, it is creating frost/ice as it hits the wall. It is perfectly reasonable to conclude that the compressed air injected into the KPP buoyancy cylinders does likewise when released into the water tank.

Initially, injected into the bottom of the tank it would still be kept partially pressurized due to the depth/water pressure, expanding and absorbing heat as it slowly rises and the pressure is gradually reduced.

I think the best arrangement would probably be a cross current heat exchanger.

That is, copper tubing with the hot compressed air inside starting at the top of the water tank transferring heat of compression to the water as the compressed air spirals down through the tubing.

That would result in most of the heat of compression transfered to the water at the top of the tank.

By the time the air is released at the bottom it would have transfered all the heat of compression to the water, which it would then reabsorb on the way back up inside the buoyancy containers.

Of course, none of this is absolutely necessary.

[matt brown]

You two always seem to like complicating and confusing what are very simple principles.

A normal shop compressor looses ALL the energy used for compression as "waste heat". That waste heat can be recovered 100% by using it to pre-heat water for some other purpose.

In the mean time, you end up with a tank full of compressed air capable of doing work. ALL the energy used to compress the air was lost, so how can you end up with compressed air to do additional useful work?

The "trick" is the air was simply MOVED.

Put into the tank it is now in a circumstance where if released it can expand using the "free" heat in the surrounding atmosphere to do the expanding, and in the process of expanding, drive some pneumatic shop equipment.

Same principle here.

As far as energy balance there is no gain, but in the process the air was moved into a circumstance where it can be used to do useful work.

I’ll do this in baby steps…

(1) Consider a common piston/cylinder surrounded in air. If we compress and expand a full (or partial) cyl of air with the same process (isothermal, isobaric or adiabatic) then Wneg=Wpos and Wnet=0. In effect a gas spring (duh)

(2) Now, consider PKK scheme where each barrel is replaced with pis/cyl and prior operation (lol) each cyl is ‘filled’ with air whereby inner cyl pressure equals outer water pressure. Since this will be under water, let’s assume that compression and expansion is ideal isothermal and that the air prefill is full cyl of 1 bar air at top, 1/2 full cyl of 2 bar at bottom, and all other cyls somewhere in between. Ready, set, go…still nothing since the nasty remains Wneg=Wpos and Wnet=0 (just another gas spring)

(3) Same as (2) but no air in down side cyls and up side cyls are 1/2 filled with 2 bar air at bottom by addon compressor. Let's assume that the compressor has ideal isothermal compression similar up side cyls with ideal isothermal expansion. Ready, set, go...viola it works...until compressor drains down. Simply connecting an output generator to the compressor will not produce Wnet, since the Wpos of buoyancy equals the Wneg of compressor !!!

Transforming the previous mgh Wneg of downside air filled cyls into the PV Wneg of an ‘isolated’ compressor is the main schmooze. Twisting this into some type of heat engine is lame since up side ‘expansion’ is minor aspect.

Maybe you should focus on something simpler...

bucket brigade.png (8.36 KiB) Viewed 1835 times

[Tom Booth]

This is interesting.

Must be a deep tank, I counted 43 floats. 11 more than the large KKP tower.

https://youtube.com/shorts/fuyRC2OwKHQ

Apparently another DIY project.

[Tom Booth]

You two always seem to like complicating and confusing what are very simple principles.

A normal shop compressor looses ALL the energy used for compression as "waste heat". That waste heat can be recovered 100% by using it to pre-heat water for some other purpose.

In the mean time, you end up with a tank full of compressed air capable of doing work. ALL the energy used to compress the air was lost, so how can you end up with compressed air to do additional useful work?

The "trick" is the air was simply MOVED.

Put into the tank it is now in a circumstance where if released it can expand using the "free" heat in the surrounding atmosphere to do the expanding, and in the process of expanding, drive some pneumatic shop equipment.

Same principle here.

As far as energy balance there is no gain, but in the process the air was moved into a circumstance where it can be used to do useful work.

I’ll do this in baby steps…

(1) Consider a common piston/cylinder surrounded in air. If we compress and expand a full (or partial) cyl of air with the same process (isothermal, isobaric or adiabatic) then Wneg=Wpos and Wnet=0. In effect a gas spring (duh)

If you think isothermal compression followed by isothermal expansion is the same as an adiabatic "air spring" I think you're an idiot and don't know what you're talking about.

(2) Now, consider PKK scheme where each barrel is replaced with pis/cyl and prior operation (lol) each cyl is ‘filled’ with air whereby inner cyl pressure equals outer water pressure. Since this will be under water, let’s assume that compression and expansion is ideal isothermal and that the air prefill is full cyl of 1 bar air at top, 1/2 full cyl of 2 bar at bottom, and all other cyls somewhere in between. Ready, set, go…still nothing since the nasty remains Wneg=Wpos and Wnet=0 (just another gas spring)

(3) Same as (2) but no air in down side cyls and up side cyls are 1/2 filled with 2 bar air at bottom by addon compressor. Let's assume that the compressor has ideal isothermal compression similar up side cyls with ideal isothermal expansion. Ready, set, go...viola it works...until compressor drains down. Simply connecting an output generator to the compressor will not produce Wnet, since the Wpos of buoyancy equals the Wneg of compressor !!!

No. The work of compression is greatly reduced by simultaneously removing heat. You ass holes are now contradicting what you have been telling me is the reason a Stirling engine requires heat removal (isothermal compression).to reduce the work of compression. Now you are contradicting yourself by saying isothermal compression (compression with heat removal) does not reduce the work load.

Adding heat during expansion increases the work output. Isothermal expansion results in 100% conversion of heat into work. You have also said that yourself, Matt, so again you are now contradicting your own previous statements here on the forum.

Regardless, the expansion of the gas on the way up is possibly only a slight assist. You already stated a few posts back boyancy does not depend on the gas expanding.

The main thing is getting the air compressed and into the buoyancy cylinders at the bottom of the tank with minimal power input at the compressor.

Isothermal compression with simple air cooling can accomplish that, but I think water cooling could work better and the heat put into the tank could increase the buoyancy, as you have also stated, though again contradicting yourself and common sense by saying solar heat is somehow better than compressor "waste heat". Heat is heat.

Transforming the previous mgh Wneg of downside air filled cyls into the PV Wneg of an ‘isolated’ compressor is the main schmooze. Twisting this into some type of heat engine is lame since up side ‘expansion’ is minor aspect.

Maybe you should focus on something simpler...

Maybe you should focus on trying not to flip flop and habitually contradict yourself, spewing a bunch of BS, as usual.

[Fool]

.

Matt Tom is incapable of learning, or being an adult. He is not worth our time of day.

No, according to the ideal gas law, when the pressure is reduced from 2 ATM to 1 ATM at constant volume, the temperature of the gas drops by 50% (absolute/Kelvin).

Or please explain why the ideal gas law is wrong in this instance.

There is nothing wrong with the ideal gas law, if it is used correctly and interpreted correctly. Something you seem to miss.

The following:
"when the pressure is reduced from 2 ATM to 1 ATM at constant volume, the temperature of the gas drops by 50% (absolute/Kelvin)."

Should be:
>>> when the Temperature is reduced from 300 K to 150 K at constant volume, the pressure of the gas drops by 50% (absolute). <<<

It is impossible to drop the pressure unless one or more of the following is applied:

a) Volume increases. Otherwise known as adiabatic temperature drop. This is with work.

b) Temperature is reduced by conducting heat out to a colder sink.

C) The mass of the gas is reduced by letting some gas escape, or by escaping, into a lower pressure. This is equivalent to volume increase temperature drop. Otherwise known as adiabatic temperature drop. It is without work.

Don't even bother trying to explain how your idea is different from those three. You are too stupid to defend yourself.

I’ll do this in baby steps…

LOL. That might work for a baby, but it is far above Tom, whom can't even get the first principal of kinetic theory correct. Gasses always push. LOL pass him a pacifier, he's too stupid to suck and walk. He's nothing better than a self craping machine. Thanks for trying. Stupid is what stupid does. Denial of obvious data is the sign of a narcissist. Miss understanding how to interpret an equation hammers it down. Cursing cements it over. Lack of any courtesy posts the epitaph. Cursing proves it.

Sorry Tom unless you wise up there is no hope for you. We've been more than patient and informative. I will now treat you as evil.

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[Tom Booth]

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Matt Tom is incapable of learning, or being an adult. He is not worth our time of day.

No, according to the ideal gas law, when the pressure is reduced from 2 ATM to 1 ATM at constant volume, the temperature of the gas drops by 50% (absolute/Kelvin).

Or please explain why the ideal gas law is wrong in this instance.

There is nothing wrong with the ideal gas law, if it is used correctly and interpreted correctly. Something you seem to miss.

The following:
"when the pressure is reduced from 2 ATM to 1 ATM at constant volume, the temperature of the gas drops by 50% (absolute/Kelvin)."

Should be:
>>> when the Temperature is reduced from 300 K to 150 K at constant volume, the pressure of the gas drops by 50% (absolute). <<<

It is impossible to drop the pressure unless one or more of the following is applied:

a) Volume increases. Otherwise known as adiabatic temperature drop. This is with work.

b) Temperature is reduced by conducting heat out to a colder sink.

C) The mass of the gas is reduced by letting some gas escape, or by escaping, into a lower pressure. This is equivalent to volume increase temperature drop. Otherwise known as adiabatic temperature drop. It is without work.

Don't even bother trying to explain how your idea is different from those three. You are too stupid to defend yourself.

...

Your a numbskull.

Who's the one always claiming "math is science". Here is the math. Don't believe me, use the calculator:

Step 1: find the volume of 1 mole of gas at ambient (300°K) at atmospheric pressure

step_1.jpg (148.83 KiB) Viewed 1798 times

Result: 0.869349 cubic feet

Step 2: isothermal (constant temperature) compression to 2 atmospheres. Note the reduction in volume by exactly 1/2 to 0.434674 cubic feet

step_2.jpg (148.7 KiB) Viewed 1798 times

Step 3: reduce pressure back to 1 atmosphere, Note that with the return to 1 ATM the temperature is reduced by 1/2 from 300°K to 150°K (-123°C or -190°F)

Sstep_3.jpg (143.76 KiB) Viewed 1798 times

Trust the math.

That will be the temperature of the gas upon transitioning from 2 atmospheres pressure back to 1 atmosphere pressure.

During isothermal compression heat/energy had to be removed to maintain the temperature at 300°K

What do you expect?

When the pressure is gone the gas is now colder having had half its heat removed during isothermal compression.

That is, of course the mathematically "ideal" case, but it does give a very realistic idea of just how cold the gas has become due to isothermal compression, or how much heat/energy was removed and the amount of heat that can be re-absorbed by the gas on its way up the buoyancy tube.