Stirlingengineforum.com

Should have posted this here:

viewtopic.php?p=27419#p27419

I've been fooling around with the ideal gas law and came to a rather astonishing conclusion, but maybe someone else would like to check the results.

If you compress a quantity of gas (1 mole of air for example) isothermally, that is keeping it from heating up by removing the heat of compression as the air is compressed.

By doubling the pressure the volume will be reduced to half.

If now released from pressure back to 1 atmosphere then the temperature will be half of the starting temperature. (On the Kelvin scale)

At least, that is what the ideal gas law calculator suggests.

So if the air started out at 300°k after compressing it isothermally to just about 30 psi then returning it to 1 atm the temperature will be 150°K

Which is -123°C or -190°F

Those are minus (below zero)

That is assuming the gas does not warm up by absorbing any heat from the surroundings.

In other words, you can halve the volume of the gas by either cutting the temperature in half or doubling the pressure

But if the pressure is doubled (isothermally) the volume is halved which volume back at 1 atmosphere would equate to halving the temperature (on the absolute scale).

30 PSI seems like hardly anything; easy hand pump pressure for a bicycle tire.

Anyway, I thought that was interesting, but the end result is way colder than I was expecting, for such low pressure so am I doing something wrong?

Anyway, what I'm thinking is that compressing air isothermally to a modest pressure, using air cooling of the compressor, the compression would be very easy, not requiring much energy.

Then releasing 1/2 the volume of the gas into a tank of water, it would expand, doubling in volume absorbing ambient heat.

30 PSI could inject air down to more than 60 feet below the surface of water. (0.433 psi/ft).

So the way I figure, a simple air cooled compressor could push air 60 ft under water with very little energy needed for isothermal compression.

So, ...

Maybe that is why you often see the doors open and a big fan running in these KPP power plants?


That is on top of the compressor cooling fins and built in flywheel fan.

Flooid uses a heat pump to keep the compressor even colder, making compression easier and puts the heat into the Flooid.

But is the additional power needed to run a heat pump really make that big of a difference?

Well, a refrigerant is phase changing so can absorb, or move larger quantities of heat, so maybe, but also complicates the system.

Now I think you could also bootstrap the compressor by coupling it with an air motor.

In other words, use the compressed air to drive an air motor which air motor would take some of the load off the compressor, that would result in a cold exhaust which could be used to pre-cool the compressed air, and then this exhaust could go to the water tank.

Scott Robertson had gathered literally hundreds of historic accounts of similar compressed air based "perpetual" or "free energy" type inventions on his https://aircaraccess.com/ website.

Unfortunately, he told me in an email that he had become afraid for his life and deleted nearly everything.


https://aircaraccess.com/achf-intro.html

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Tom Booth wrote:Torque by itself does not break anything as long as there are smooth transitions and things keep moving (no sudden hard stops)

Most machinery shafts and gears have "safety" devices. A soft key in the shaft keyway that will sheer. I've replaced many dozens in lawnmowers where people hit a rock or something.

Not true. The maximum torque will be the limit that will break things regardless of how slowly the torque is put on. Safety devices such as soft keys only protect the shaft and parts with higher maximum torques inherent in the design (shafts, sprockets, pulleys, belts, shelves, e.t.c.). I've even seen an inch and a half diameter steel splined rear axle sheared off by selecting a low gear and slowly applying throttle through an automatic transmission. Just shear torque. It was a 1952 GMC 6x6 dump truck, called a Duce and a Half.

And it was far less than 16900 ft•lbs of torque.

https://www.militarytrader.com/military ... -gmc-xm211

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Tom Booth wrote:If now released from pressure back to 1 atmosphere then the temperature will be half of the starting temperature. (On the Kelvin scale)

PV=nRT

P1•V1/T1 = nR = P2•V2/T2 = P0•V0/T0 = a constant

P0 = atm could put in 14.7 psi.
P1 = 2•atm
P2 = atm

T0=300 K
T1=300 K
T2=?

V0=10 m^3 or any starting point.
V1= ?
V2= ?

P0 to P1

Atm•10/300 = 2•atm•5/300 or

V1 = V0/2, or half the starting volume (Keeping T constant.) As you correctly concluded.

Unfortunately going from P1 back to atmospheric or to P2 there isn't enough information to use that equation. Why? Look below as we try:

P1•V1/T1 = P2•V2/T2
2•atm•5/300=atm•V2/T2

We don't know what T2 or V2 is, and V2 is not 10. That would only be true if the gas gained heat and went back to 300 K, and you say it is adiabatic, no heat exchange.

The equation for adiabatic expansion would need to be used.

It looks something like the following:

V2=V1(P2/P1)^(-5/7)

The 5/7 is 1/gamma for a generic diatomic gas, such as air, oxygen, and or nitrogen. Different value for Helium. Better to use Cv/Cp for a specific real gas. Be my guest.

https://en.m.wikipedia.org/wiki/Adiabatic_process

V2=5•(atm/2•atm)^(-5/7)

Or
V2=5•(1/2)^(-5/7)

Equals about 8.2 m^3, or same pressure at a smaller volume because the gas is colder.

Then V2 gets plugged back in to the earlier equation.

>>> P1•V1/T1 = P2•V2/T2
2•atm•5/300=atm•V2/T2 <<< earlier equation.
Plugging it in:

2•atm•5/300 = atm•8.2 / T2

Rearranging the terms:
T2 = atm•8.2 / (2•atm•5/300)

Canceling 'atm' top and bottom:

T2 = 8.2/(2•5/300) = 246 K
or about -27 C. -16.6 F.

Still very cold.

As a check:

2•atm•5/300 = atm•8.2/246

Dividing both sides by atm:
2•5/300 = 8.2/246
Or
0.033333 = 0.033333

Or go use an online, isothermal compression, and an adiabatic expansion calculator.

It's best to calculate the exact numbers than hope for "very little" this or that. Not as easy, but worth it. The devil is in the details.

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Too late to edit it but a previous post should have been formated as below:

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Tom Booth wrote:Torque by itself does not break anything as long as there are smooth transitions and things keep moving (no sudden hard stops)

Most machinery shafts and gears have "safety" devices. A soft key in the shaft keyway that will sheer. I've replaced many dozens in lawnmowers where people hit a rock or something.

Not true. The maximum torque will be the limit that will break things regardless of how slowly the torque is put on. Safety devices such as soft keys only protect the shaft and parts with higher maximum torques inherent in the design (shafts, sprockets, pulleys, belts, shelves, e.t.c.). I've even seen an inch and a half diameter steel splined rear axle sheared off by selecting a low gear and slowly applying throttle through an automatic transmission. Just shear torque. It was a 1952 GMC 6x6 dump truck, called a Duce and a Half.

And it was far less than 16900 ft•lbs of torque.

https://www.militarytrader.com/military ... -gmc-xm211

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I meant to supply this thread with the following link:

https://en.m.wikipedia.org/wiki/Pulser_pump

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Tom Booth wrote:Now I think you could also bootstrap the compressor by coupling it with an air motor.

Why have the buoyancy chain? Just heat the air motor with solar, and cool the air compressor with evaporation?

Use the double M turbine.

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Fool wrote: ↑Fri Dec 06, 2024 7:59 am .

Tom Booth wrote:Now I think you could also bootstrap the compressor by coupling it with an air motor.

Why have the buoyancy chain? Just heat the air motor with solar, and cool the air compressor with evaporation?

Use the double M turbine.

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Are you suggesting some variation of this "perpetual motion" scheme is possible?

Be that as it may, the point here is to determine the feasibility of the KPP concept.

I'm not sure I trust your personal determination that '"Unfortunately going from P1 back to atmospheric or to P2 there isn't enough information to use that equation"

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


At any rate, using your numbers have we still managed to get very cold air into the bottom of the tank with a minimal amount of work using only readily available ambient air cooling during compression?

Looks that way to me.

The removed heat is reabsorbed as the canisters slowly rise somewhat increasing buoyancy force all along the way up.

Tom Booth wrote: ↑Fri Dec 06, 2024 9:54 am .

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
...

Sorry, I meant 150°K not -150°. No such thing as negative Kelvin I guess.

So 150°K or about -123°C or -190°F

If using the tank water for water cooling the compressor, in theory, the "heat of compression" extracted is also recovered and then goes toward power production.

It's near perfect isothermal compression and isothermal expansion. The expanding air can also be used to pre-cool the intake air going to the compressor.

But it appears the Rosch system already functions without any additional tricks. Their compressor looks like it just came from Harbor Freight.

Well, supposing that works, for real, then Scott Robertsons collection of "free range air cars" running on compressed air and somehow powering an on board compressor to roam the countryside indefinitely without fuel doesn't seem so far fetched.

Plenty of air cooling to be had to cool your compressor while driving down the road. You don't even need a fan.

Air cool the air being compressed. Drive car with air motor. Use cold exhaust to pre-cool the air intake. You can even use a scoop/ram to partially compress the air.

All kinds of ways to wring some of the solar energy out of the air.

The buoyancy thing is still kind of interesting, but I don't really think there is any mystery about the working principle.

Like so many other similar "perpetual" engines I've looked into, the "extra energy" is harnessed from ambient heat.

The process always seems to involve isothermal compression as a starting point.

Same basic principle as heat pumps. A heat pump is just a gas compressor that compresses a gas directly into cooling tubes/heat exchanger (or "condenser") immediately, rather than into a holding tank.

Simultaneous cooling of the air during compression or even before compression or at worst immediately after compression results in the ability to compress the air using little energy.

Then the air can be re-expanded to do work using "free" ambient heat as the main power source.

There is a website that appears to be tracking and currently updating developments regarding the KPP system.

https://novam-research.com/rosch-gaia-k ... -plant.php

Still no sensible or credible explanation given for exactly how the system works.

The author or compiler of information IMO seems to have a balanced view.

At least the site is current and pretty thorough and all the links, websites and videos etc. seem to be working.

One thing I wanted to mention is that the air buoyancy canisters are apparently steel or some type of metal.

I think that is significant in terms of my thermal transfer assumptions.

While thermal transfer would still be possible using plastic canisters, it would presumably, be less effective.

That the standard canisters are steel was on a KPP venture in Mexico website, I came across possibly via the Wayback Archive I have not been able to re-locate, or it may have been a PDF I downloaded.

OK, I found this offline website in my downloads folder, which, if I remember right, I had to first run through a website translator.
The photo caption mentions:

"The Kinetic Power Plant uses steel cylinders with holes..."

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.