Buoyancy wheel driven by HHO electrolysis.

[sm0ky2]

The point here is that AFTER we extract energy from the phase change
(Liquid to Gas)
We can ALSO extract energy from the stoichiometric chemical mixture
(H+H+O)
Be it in a combustion engine, open flame (10kF), or fuel cell, etc.


We can take this a step further, if your concern is thermal dissipation
Wrap it in peltier/seebeck converters to salvage your heat losses.


The entire system can stay at 1ATM if you choose, just dont put a lid on it




At the very least we improve the efficiency of the paired fuel cells by the buoyant factor


If the system is efficient enough, using both forms of energy is mathematically coherent to OU design.
You can have a (nearly) efficient fuel cell system by itself
You can have a (nearly) efficient phase change buoyancy system by itself


Either of these, by themselves, can stay within the bounds of the thermodynamic perspective.
But NOT both in the same system.
Using both, driven by the same input (it is equivalent in either case)
You double the output at the same cost.
Which allows (mathematically) for a maximum C.O.P. Of 2 (minus losses)
This is for buoyancy in AIR! Greater efficiency can be obtained by using liquid buoyancy.

Electrolysis of water into HHO allows for both situations
you have phase change as well as a fuel cell gas mixture


There is NO law of physics that states we lose energy by letting our hydrogen rise to a higher altitude before converting it back into electricity.


Pressure effects the RATE, not the efficiency so much.
When you bring the oxygen supply with you, altitude doesnt matter.
Look at the numbers for fuel cell efficiency at different altitudes.
Then look at the change in height
And the change in mass density (gas to water)
AT THOSE HEIGHTS (E=mgh)


The math is clear

[sm0ky2]

Low Q]

If electrolysis when conducted at some higher pressure (under water for example)
becomes less efficient in terms of gas quantity produced, than is the gas quantity
produced by electrolysis conducted at some lower pressure (sea level air pressure for
example), it is likely because more heat is produced in place of gas.

This is false


It has more to do with fuel cell design than separation of the molecules under pressure
cells designed to operate at higher pressures can still meet efficiency standards.
Cells designed for lower pressures perform poorly at higher pressures.


And also, most freelance experimenters fail to account for the pressure of the HHO obtained.
In the higher pressure case, the gas output is at a higher pressure/density.
The observed numbers should be Molar, not volumetric. This is a common mistake, even at respectable institutions.


That aside: the situation we are talking about (a reasonable tank of water)
Pressure at any manageable height for a device, is still ~ 1 ATM
the same can be said for any (reasonable) altitude in air.




Then we have the reconversion: fuel cells operate faster under higher pressures.
Which was why high pressure cells were developed
(mainly for automobiles, which require higher current)
not an efficiency issue, but it's important to understand what type of fuel cell you are testing
and to keep those cells under ideal operation conditions.


Often we find that cooling the system (even with additional cost of cooling) can increase overall efficiency of the cell in some cases.


Temperature has a greater impact than pressure


Electrolysis performs better under heat
It takes less electrical energy to separate the molecules through electrolysis when the water is heated. In this scenario, heat is not considered a loss, entirely. Part of the heat is considered a gain. (see Ideal Gas Law)






This is not hard to test and make sure that our laws pf physics are still maintained.
There is only 1 of two outcomes.
1) our physics are correct and this system is (max) COP=2
or
2) out physics is faulty and we need to correct our equations.


Simple experiment:


2 fuel cells, 1 baloon, 1 roll of kite string, 1 generator


Instructions:


1: attach the roll of kite string to the generator shaft


2: Turn on fuel cell 1, and fill baloon enough to lift fuel cell 2


3: let generator produce energy until your kite string runs out


4: turn on fuel cell 2 and reclaim your electricity


5: get out of the way, because theres going to be
        a crater where it impacted the earth and broke thermodynamics.

[Willy]

Electrolysis is some times done with the production vessel sealed. 
This is can be a more efficient way to pressurize the gas, rather than
pressurizing it later by means of for example, electric motors and compressors,
which have their own inefficiencies / losses.

Read- first IF below

"If electrolysis when conducted at some higher pressure (under water for example)
becomes less efficient in terms of gas quantity (MASS IN A GAS STATE)  produced, than is the gas quantity (MASS IN A GAS STATE) produced by electrolysis conducted at some lower pressure (sea level air pressure for example), it is likely because more heat is produced in place of gas.

"In either situation, the equality of the input to output, 1:1 or unity, remains the same (all outputs considered). More gas but less heat or more heat but less gas."

                                Unity, before buoyancy gains.
                                Over Unity after buoyancy

[sm0ky2]

In any case,
The input and output are whatever they are.
60%, 63%?


Add buoyancy energy here


Then add gravity energy on the way down


Thermodynamics is destroyed

[Cloxxki]

Assuming you're trying to extract energy from buoyancy in water, have you looked up at the realistically achieved efficiency of that system?
If you allow the gas to rise at the speed it likes, it's going to rise slowly, a lot of drag. Compare a football released under water. Only when it breaches the surface, does it gain speed. It's like a fancy sports car, driving through a block of syrup. All you do is stir the syrup, or in this case water.
So then, you decide to allow only very slow rising of the gasses. So very high force, very low speed. Which generator would be suggested for that? Or would a weight be raised, later to be dropped? Then you need an efficient way to extact most of the kinetic energy from a falling weight.
How will the gas come back down, collection of water from the combustion, then falling in bulk?
All the efficiencies multiplied, and minding the actual efficiency of producing HHO at the needed pressure, I'm curious whether you can reach 100%, let alone more.
I'd love a realistic schematic based on previously achieved efficiencies for the kinds of production and extraction proposed.