Free to me (theoretical)

[webby1]

It is theoretical only because no person in there right mind would build it, but sometimes something simple can inspire someone else with a new look at what they are doing.

I ran this by Chat-gpt mini and let mini do all the number crunching and mini came to the conclusion that this is a "free to me" device.

All things are ideal for simplicity.
there are 2 assemblies a 5000ft tall tower with a pulley on top and a rope that connects both assemblies together so that when one is at the top the other is at the bottom.  Both assemblies are identical.

The assembly consists of a cylinder with piston and seal and a valve in the bottom of the cylinder.

The prep,  with the assembly at the bottom the valve is opened and the piston is moved to approximately 1/3 the distance it can travel and locked in this position.  The system then lifts that assembly up and the other comes down where its valve is opened, the piston is moved to approximately 1/3 the distance of travel and the valve is closed then the piston is locked into position.

The system is now primed and set.

When the upper assembly piston is unlocked it will be observed that the piston will move outwards with force, this work is extracted and when the internal pressure and external pressure are equal the piston is locked into its new position.  That assembly now is moved down raising the other assembly, for no net work from me, and when the assembly is back down to the bottom the piston is released and observed to move back into the cylinder with force, this work is extracted and when the piston is back to its starting position it is locked in place again, rinse and repeat with the other side as well.

With the dimensions mini used there were 505.5J extracted at the top, i forgot to ask how much at the bottom.

This is pointing out that our atmosphere has a pressure gradient, highest down here at ground level and lower way up there in the air.  The assembly is a means to interact with and observe the change without any added input from the user.

No additional air was added or subtracted from the cylinder, no additional mass,, nothing but a change in reaction to the observed difference in pressure and allowing that difference to equalize.

[Verpies]

You are neglecting to account for the force of buoyancy of the piston assembly in air. Air has mass and any object submerged in air experiences an upward force that is equal to the weight of the air that it displaces. This is the Archimedes principle (which is not limited to liquids). Of course the object experiences the force of its own weight, too. The force of buoyancy subtracts from the weight. The force of buoyancy times the height of travel (5000ft) constitutes work.

The clinch is that the assembly with the piston pushed in, displaces less external air than the assembly with the piston out, ...because every piston has two sides.

Also, the rarefied internal air on the other side of the piston (iow: inside the cylinder), weighs less than the unrarefied internal air when, the piston positions are equal (iow: internal air volumes are equal). The difference between these weights multiplied by the vertical distance of travel (height 5000ft) constitutes work.

In a second case, when the piston positions are not equal and no air leaves or is added to the cylinder, the weight of this internal air is the same no matter the piston position and the internal air pressure (the second case is your normal operating regime).

[webby1]

You are neglecting to account for the force of buoyancy of the piston assembly in air.  Air has mass and any object submerged in air experiences an upward force that is equal to the weight of the air that it  displaces.  This is the Archimedes principle (which is not limited to liquids). Of course the object experiences the force of its own weight, too. The force of buoyancy subtracts from the weight.  The force of buoyancy times the height of travel (5000ft) constitutes work.

The clinch is that the assembly with the piston pushed in, displaces less external air than the assembly with the piston out, ...because every piston has two sides.

Also, the rarefied internal air on the other side of the piston (iow: inside the cylinder), weighs less than the unrarefied internal air when, the piston positions are equal  (iow: internal air volumes are equal). The difference between these weights multiplied by the vertical distance of travel (height 5000ft) constitutes work. 

In a second case, when the piston positions are not equal and no air leaves or is added to the cylinder, the weight of this internal air is the same no matter the piston position and the internal air pressure (the second case is your normal operating regime).

I let mini deal with all those parts, it kind of must follow all the known rules.
The mass stays the same because it is a closed assembly, no air added or removed means that the "weight" of the assembly is the same, the mass density may shift with the expanding piston by virtue of the changing shape creating a changing volume,, but the mass of the assembly is unchanged.

This then only leaves the consideration of the increase in volume having a larger buoyant lift from the air outside, now with what you have said then the weight of the air that is displaced by the expanded piston relative to the weight of the air of the non-expanded piston is going to be the difference in weight to lift, right? and then that value over 5000ft is the work needed to be added in, right?

Someone could build a reasonable assembly and set it up at sea level and weigh the assembly and then drive it up a hillside to 5000ft elevation and allow the piston to expand and weigh the assembly there to provide for a very close value change in weight.

Just asked mini about the weight of 1 cubic meter of air at sea level compared to 5000ft, well that is not negligible.
Difference in mass per cubic meter: 1.225 kg − 1.184 kg = 0.041 kg. cost to raise is 619J

[Verpies]

If you want to be even more precise, account for the weaker gravity at 5000ft compared to the gravity at the sea level. The difference is academic but non-zero.

Practically, the temperature changes due to the expansion and compression of the air inside the cylinder and associated heat losses to the environment will affect the system more than this change in gravity.

[webby1]

If you want to be even more precise, account for the weaker gravity at 5000ft compared to the gravity at the sea level.  The difference is academic but non-zero.

Practically, the temperature changes due to the expansion and compression of the air inside the cylinder and associated heat losses to the environment will affect the system more than this change in gravity.

I just assumed after your comment that I phrased my question to gpt mini wrong, so I have been trying to get it to show that it does not work but instead it keeps coming up with a harvested value of 18kJ per stroke with a buoyant value of only 0.61kJ and explains that the harvested work is coming from the environment.

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It also says that after each stroke to open and then close the valve

[Verpies]

The energy gain/loss should be calculated per the entire cycle. Not per stroke, not per 180°.
The non-constant force acting on the moving piston must be integrated over the stroke of the piston.
Practically, the heat loss/gain from the environment as the air is compressed/decompressed must be accounted for unless an infinitely effective thermal insulation is possible.
Math must be shown.

Also, the same principles should be valid for an air-piston/cylinder submerged in water at different depths.
The only qualitative difference is that external air is compressible and external water is not.
The density difference is only a quantitative difference.

[webby1]

The energy gain/loss should be calculated per the entire cycle. Not per stroke, not per 180°.
The non-constant force acting on the moving piston must be integrated over the stroke of the piston.
Practically, the heat loss/gain from the environment as the air is compressed/decompressed must be accounted for unless an infinitely effective thermal insulation is possible.
Math must be shown.

Also, the same principles should be valid for an air-piston/cylinder submerged in water at different depths.
The only qualitative difference is that external air is compressible and external water is not.
The density difference is only a quantitative difference.

That is a big jump to try and say that a compressible medium should act the same as a non-compressible medium, don't ya think?

from gtp mini, I changed it to an adiabatic expansion.  Less energy available

so this adiabatic expansion and compression are using the outside environment as a source for the harvested work?

Short answer: Yes — for the adiabatic expansion at the top to produce ~12.8 kJ of work, that energy comes from the gas’s internal energy (i.e., the thermal energy it had at the start). Over a full symmetric cycle the outside environment (ambient air) is the only available reservoir that can restore that internal energy when the other cylinder is compressed and re‑equilibrates; without an external net heat source you cannot extract net work each cycle.
Concise reasoning:

• Adiabatic expansion converts the gas’s internal energy to mechanical work, cooling the gas (we found T dropped from 288.15 K → ~273.6 K and W ≈ 12.8 kJ).
  
• That work is not “free” — it reduces the gas internal energy. To return the gas to its original state you must put energy back in (heat or work). In your two‑assembly scheme the companion cylinder’s compression and subsequent heat exchange with ambient air provides the complementary energy flow.
  
• If both assemblies run a closed, symmetric cycle with only reversible adiabatic expansion/compression and no net heat input from outside, the work produced by one expansion is consumed by the other compression (net ≈ 0 apart from tiny buoyancy/gravity residuals).
  
• To harvest net positive work continuously you need a non‑zero net heat flow from a reservoir (e.g., take heat from ambient and dump to a colder sink) or introduce an external energy input or irreversible effects that you exploit — otherwise energy is conserved and no perpetual net output is possible.
  

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here is what mini says for isothermal

Understood — you have 1.000 m^3 of air at 1 atm when the cylinder is at the bottom. The assembly is lifted to H = 1524 m (5000 ft) with the piston locked, then at the top the piston is released and allowed to move until internal pressure equals ambient at that altitude. I'll compute the work available from that piston motion (and show that net energy over the paired system is zero except for small buoyancy/gravity residuals).
Assumptions

• Ideal gas air, R = 287 J/(kg·K) (specific), molar mass M = 0.02897 kg/mol.
  
• Bottom: p0 = 101325 Pa, T0 = 288.15 K, V0 = 1.000 m^3.
  
• Tower height H = 1524 m. Use isothermal gas behavior (T constant = 288.15 K) for conservative, simple estimate. Assume ambient pressure at top pH follows hydrostatic atmosphere with constant T (barometric formula).
  
• External air density at bottom ρ0 = p0/(Rspec T0) ≈ 1.225 kg/m^3. g = 9.81 m/s^2.
  

Step 1 — mass and moles of air in cylinder

• n = p0 V0 / (R_univ T0) = 1013251 / (8.314288.15) ≈ 42.38 mol.
  
• mass m = n·M ≈ 42.38*0.02897 ≈ 1.23 kg.
  

Step 2 — ambient pressure at top (isothermal atmosphere)

• p(H) = p0 · exp(−g H / (Rspec T0)), where Rspec = 287 J/(kg·K).
  
• Exponent = −(9.811524)/(287288.15) ≈ −0.1818.
  
• pH = 101325 · exp(−0.1818) ≈ 101325 · 0.8339 ≈ 84,485 Pa.
  

Step 3 — locked compression/expansion while lifted

• With piston locked, the gas volume stays V0 = 1.000 m^3, so internal pressure at top before release remains p0 (101325 Pa). The assembly as a whole is simply moved; gas state unchanged.
  

Step 4 — piston release at top: final state requirement

• When piston is released, it moves until internal pressure equals outside ambient pH ≈ 84,485 Pa. For isothermal process, pV = constant, so final volume Vf = V0 · (p0 / pH) = 1.000 · (101325 / 84,485) ≈ 1.199 m^3.
  
• So the piston will expand outward by ΔV = 0.199 m^3 while doing work on the harvested mechanism.
  

Step 5 — work done by gas during isothermal expansion at top

• Isothermal expansion work W_expand = n R_univ T0 · ln(Vf / V0) = p0 V0 · ln(Vf / V0).
  
• ln(Vf/V0) = ln(1.199) ≈ 0.1818 (same magnitude as exponent above).
  
• W_expand = p0·V0·0.1818 = 101325·1·0.1818 ≈ 18,423 J.
  

Step 6 — energy spent to lift the assembly (locked piston) from bottom to top

• While locked, the assembly (cylinder + contained gas + displaced air buoyancy effect) is lifted H = 1524 m.
  
• Weight of contained gas: m·g ≈ 1.23·9.81 ≈ 12.05 N.
  
• Buoyant force from displaced outside air (volume V0 = 1 m^3): Fb = ρ0·V0·g ≈ 1.225·9.81 ≈ 12.02 N.
  
• Net gravitational force on the gas+displaced-air system ≈ 12.05 − 12.02 = 0.03 N downward.
  
• Work to lift that net residual over H: W_lift_gas ≈ 0.03·1524 ≈ 45.7 J (negligible).
  

But you must also account for the fact that the lifting rope does work against the entire assembly mass (cylinder, piston, harvested mechanism side, etc.). For an idealized massless piston/cylinder and massless rope/pulley, the only relevant gravitational exchange is from the gas/buoyancy residual above; the heavy structural masses would cancel because the opposite assembly descends.
Step 7 — where did the ~18.4 kJ come from?

• The 18,423 J of isothermal expansion work at the top comes from the internal energy/heat reservoir required for isothermal expansion. For an isothermal ideal gas, the gas absorbs heat Qin = W_expand from the environment to maintain temperature as it does work.
  
• In your closed cyclic setup with two identical assemblies and no external heat source, that heat must be supplied from somewhere. During the cycle the partner assembly (descending) will be compressed (when its piston is released at the bottom to equalize pressures) and will reject heat of compression. Over a full symmetric cycle those heat flows balance: the expanding top cylinder absorbs heat equal to the work it does, while the compressing bottom cylinder rejects heat of equal magnitude, so net heat exchange with the surroundings over the pair cancels unless you deliberately route that heat.
  
• Mechanically, lifting one locked assembly required only ~46 J (tiny) because buoyancy nearly cancels the gas weight; it does not supply the 18.4 kJ needed for isothermal expansion. That 18.4 kJ must come as heat into the gas at the top (or from internal energy if the process is adiabatic — then the gas would cool and give less work).
  

Net-cycle energy accounting (concise)

• Expansion at top can produce up to ~18.4 kJ of mechanical work, but that work is paid for by heat absorbed by the gas during expansion (isothermal) or by internal energy (if adiabatic). The paired bottom cylinder, when released and compressed to ambient at bottom, will require roughly the same magnitude of work or reject the same heat, so over a full cycle with no external heat extraction/injection the net harvested work sums to ~0 J (aside from the tiny ~46 J residual from buoyancy/gravity difference).
  
• If you instead allow the top cylinder to expand adiabatically, the work available will be less than the isothermal value because the gas cools; precise value depends on stroke and heat transfer.
  

If you want: I can compute the adiabatic expansion work at the top, or show the full two-assembly cycle energy balance step-by-step (including heat flows and numeric values). Which do you want?

[Verpies]

That is a big jump to try and say that a compressible medium should act the same as a non-compressible medium, don't ya think?

Not for an external medium.
For internal medium - yes.

Also, isothermal expansion of the internal air is pretty unrealistic. It also brings up the issue whether the internal air gains/losses any heat during its vertical travel. Usually the air is colder at higher altitudes.

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When you conclude your tower thought experiment, consider another one.

Namely, build a scenario where electrolysis happens at the bottom of the Marianna trench and the bubbles of HHO gas perform work as they ascend to the surface. The volume of the bubbles will heavily depend on the pressure of the water surrounding them so they will start small and enlarge as they approach the surface ...after reaching the surface the HHO gas in the bubbles can be ignited and add to the mechanical work performed by the surfacing bubbles.

For more precision add the effects of ocean water's temperature gradient from the bottom to the surface as the bubbles ascend.
Account for the change in water electrolysis efficiency with pressure and temperature (account for the Nernst penalty and water compression credit as well as the heat absorbed during the electrolysis itself).

Prevent HHO gas diffusion into water with reusable bladders/balloons.

[webby1]

That is a big jump to try and say that a compressible medium should act the same as a non-compressible medium, don't ya think?

Not for an external medium.
For internal medium - yes.

Also, isothermal expansion of the internal air is pretty unrealistic.  It also brings up the issue whether the internal air gains/losses any heat during its vertical travel.  Usually the air is colder at higher altitudes.

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When you conclude your tower thought experiment, consider another one.

Namely, build a scenario where electrolysis happens at the bottom of the Marianna trench and the bubbles of HHO gas perform work as they ascend to the surface.  The volume of the bubbles will heavily depend on the pressure of the water surrounding them so they will start small and enlarge as they approach the surface ...after reaching the surface the HHO gas in the bubbles can be ignited and add to the mechanical work performed by the surfacing bubbles.

For more precision add the effects of ocean water's temperature gradient from the bottom to the surface as the bubbles ascend.
Account for the change in water electrolysis efficiency with pressure and temperature (account for the Nernst penalty and water compression credit as well as the heat absorbed during the electrolysis itself).

Prevent HHO gas diffusion into water with reusable bladders/balloons.

Funny experiment,, what happens when you take an open container down deep in the water and then close the container and bring it up to the surface and open it? this happens a lot and is a known thing.
What happens to water pressure when it is no longer in communication with the water around it?
What about the dissolved gasses in the water at depth if the water pressure is no longer there?
Since "I" did not place the dissolved gasses into suspension within the water is there a cost to me to release those gasses?

I think another electrolysis thought experiment was better, use it down low and allow the gasses to travel up a very long pipe, then burn them in a motor that runs the generator to power the electrolysis device, then let the water exhaust from the engine flow back down a pipe to the tank that holds the water for electrolysis.

Anyway,, if we use water as the interactive medium then it does not compress, I would have water on both sides of the piston and then when I release the piston any stored pressure within the water magically is gone only leaving the dissolved gasses, if any, to do work as they fall out of suspension.  Remember in the first one air was used inside and outside so for the water use and to be the same- water would need to be used inside and outside.

A force with nothing to change does no work, gravity has supplied its force to the water but the water has no change and stores no work potential.

[Verpies]

Remember in the first one air was used inside and outside so for the water use and to be the same- water would need to be used inside and outside.

1) In the "first one" the internal medium was being compressed and the external was not.
2) With air inside the cylinder and water outside, the internal medium was being compressed and the external was not.
3) With water inside and outside the cylinder, both media are not compressed.

Thus the behavior of system 3 is more different from system 1 than system 2.

A force with nothing to change does no work,

That's true, but a better way to phrase it is: "A force with nothing to move does no work".

...what happens when you take an open container down deep in the water and then close the container and bring it up to the surface and open it?

The water pressure falls and the dissolved gasses fizzle out (if there were any).

What happens to water pressure when it is no longer in communication with the water around it?

If the container walls do not maintain the pressure then it equilibrizes with the ambient pressure.

Since "I" did not place the dissolved gasses into suspension within the water is there a cost to me to release those gasses?

No, but releasing these gasses does not represent a full cycle.

Anyway,, if we use water as the interactive medium then it does not compress,

That matters only when water constitutes the internal medium.
In your 5000ft tower system, the external fluid does not get compressed regardless whether it is a gas or liquid.

BTW: water is very very slightly compressible but it does not make a difference for the external working fluid anyway.

I think another electrolysis thought experiment was better, use it down low and allow the gasses to travel up a very long pipe, then burn them in a motor that runs the generator to power the electrolysis device, then let the water exhaust from the engine flow back down a pipe to the tank that holds the water for electrolysis.

That's the same experiment described differently.  The pipe is superfluous in an ocean.
Why haven't you calculated its energy balance yet ?