1. science
2. physics
3. thermodynamics

I saw the comments from part 4 and Mark Euthanasius was indeed correct. I did my integrals
backwards (long day haha – thanks for that). Let’s try again one more time
𝑉2
𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 = − ∫ 𝑃𝑑𝑉
𝑉1
𝑥2
𝑊𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑 = ∫ 𝐹𝑛𝑒𝑡 𝑑𝑥
𝑥1
𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 = 𝜌𝑊 𝑉𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑔
𝐹𝑛𝑒𝑡 = 𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 − 𝐹𝑤𝑒𝑖𝑔ℎ𝑡
𝑃𝑉 = 𝑛𝑅𝑇
However now taking into consideration that the volume of water displaced in the float changes
𝑉𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 =
∴ 𝑉𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 =
𝑛𝑅𝑇
𝑃(𝑥)
𝑛𝑅𝑇
𝑃𝑎𝑡𝑚 + 𝜌𝑔𝑥
Similarly at constant temp for compressed gas, the work required to compress that gas from atm to
P2 (This is work from compressors point of view, so I will keep the negative sign)
To clarify on what is going on here, the compressor is taking a unit volume of air call it V1 and
compressing it to the volume it is injected into the bottom float as V 2. However, it is more convenient
to work in pressures. Therefore V1=nRT/P1 where P1 is Patm, and P2 is the pressure the gas needs to be
to allow it to be injected into the bottom float.
𝑛𝑅𝑇
𝑃2
𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 = − ∫
𝑛𝑅𝑇
𝑃𝑎𝑡𝑚
𝑃𝑑𝑉
𝑛𝑅𝑇 𝑛𝑅𝑇
𝑃𝑎𝑡𝑚
∴ 𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 = 𝑃𝑎𝑡𝑚 (
−
) = 𝑛𝑅𝑇 (1 −
)
𝑃𝑎𝑡𝑚
𝑃2
𝑃2
(Note that the negative sign here is multiplied into the brackets)
But 𝑃2 is pressure of water at bottom of column where gas enters the float
𝑃2 = 𝑃𝑎𝑡𝑚 + 𝜌𝑔𝑥
∴ 𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒 = 𝑛𝑅𝑇 (1 −
𝑃𝑎𝑡𝑚
)
𝑃𝑎𝑡𝑚 + 𝜌𝑊 𝑔𝑥
Also
𝐹𝑛𝑒𝑡 = 𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 − 𝐹𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑔𝑎𝑠 ≅ 𝐹𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦
0
0
𝑛𝑅𝑇
𝑊𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑 = ∫ 𝐹𝑛𝑒𝑡 𝑑𝑥 = ∫ 𝜌𝑊 𝑔 (
) 𝑑𝑥
𝑃𝑎𝑡𝑚 + 𝜌𝑊 𝑔𝑥
𝑥
𝑥
𝑃𝑎𝑡𝑚
= 𝑛𝑅𝑇 ln (
)
𝑃𝑎𝑡𝑚 + 𝜌𝑊 𝑔𝑥
(Integral is now the correct way round: Final point minus Initial point i.e. nRT(ln(Patm) – ln(Patm + rho g
x)))
Note the integral is from x to 0, which is from the bottom of the column at distance 𝑥 meters to the
top of the column.
Define
𝑊𝑒𝑥𝑡𝑟𝑎𝑐𝑡𝑒𝑑
𝜂=|
|
𝑊𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑖𝑠𝑒
𝑃𝑎𝑡𝑚
𝑛𝑅𝑇 ln (
)
𝑃𝑎𝑡𝑚 + 𝜌𝑊 𝑔𝑥
=|
|
𝑃𝑎𝑡𝑚
𝑛𝑅𝑇 (1 −
)
𝑃𝑎𝑡𝑚 + 𝜌𝑊 𝑔𝑥
𝑃𝑎𝑡𝑚
ln (
)
𝑃𝑎𝑡𝑚 + 𝜌𝑊 𝑔𝑥
∴𝜂=|
|
𝑃𝑎𝑡𝑚
(1 −
)
𝑃𝑎𝑡𝑚 + 𝜌𝑊 𝑔𝑥
Setting 𝜌 = 1000 [
𝑘𝑔
𝑚3
units)
Giving
𝑚
] , 𝑔 = 9.8 [ 2 ] , 𝑃𝑎𝑡𝑚 = 101325 [𝑃𝑎]
𝑠
(sorry everyone I work in SI
THERE IS A MASSIVE PROBLEM WITH THIS INCASE
IT ISN’T EVIDENT. THIS IMPLIES FREE ENERGY.
And in the real world there is a factor that I realise I have neglected that is absolutely essential to how
this machine operates.
I neglected to work out the energy required for the displacement of the water in each float as the float
rises. The trapped gas must use energy to expand and push the water out. The only source of energy
for this work is the enthalpy of the gas.
Immersed in the water (which has a high heat capacity and high heat conductivity) the gas in the float
cools as it rises and pushes the water out of the float – BUT THE SURROUNDING WATER HEATS THE
FLOAT AND THE GAS WITHIN IT! If the floats are moving slow enough, there is enough time for heat
to transfer BETWEEN the surrounding water and the expanding gas to keep it at constant temperature.
And that's it.
The buoyancy generator is NOT a Free Energy device. It does not break any conventional laws of
thermodynamics. It is simply a type of Energy Pump. It converts ambient heat energy (that is in the
air) into buoyant work which is then converted into electricity. It operates almost exactly the same as
an air-conditioner or "heat-pump". The energy you put into the system is used to create an entropy
difference to get energy to flow in the direction you want. Like an air-con can do 3kW of heating with
a 1kW electric motor input, these buoyant generators can output more electric energy than it takes
to drive the compressor. All you need is a heat reservoir – atmospheric air is a pretty good one.