[are] my findings
My Problem with Buoyancy Generators Dear S. Allan. I have a problem with all proposed buoyancy generators in that I do not see how it is that they are able to operate in a state of over-unity. I donβt think I am wrong and I welcome criticism, but at this point in time I cannot fathom how it is possible for them to operate. Allow me to explain: In order for the buoyancy generator to work, the amount of energy used to compress the air must be less than the energy extracted through buoyancy. In order to pump air into containers (I like to call them balloons) at the bottom of the swing, the pressure of air must be greater than that of the water at that depth. For instance, say the depth of water used in the generator is π₯ meters, and that there are an arbitrary number of steel balloons that hold the compressed air to become buoyant on both sides of the swing. Here are a few equations. (Note g = 9.8 m/s2 and ππ = 1000 kg/m3) π2 πππππ π π’πππ π = β β« πππ π1 π₯2 πππ₯π‘ππππ‘ππ = β« πΉπππ‘ ππ₯ π₯1 πΉπ΅π’ππ¦ππππ¦ = ππ ππ πΉπππ‘ = πΉπ΅π’ππ¦ππππ¦ β πΉπ€πππβπ‘ π1 π2 π2 π2 = ππ π1 = ππ‘ ππππ π‘πππ‘ π π2 π1 π1 ππ = ππ π ππ πππππ π π’πππ π is the isentropic work required to compress a gas (that is reversible work β minimum work required to do so). At sea level (101.325 kPa or P1) and constant pressure this equation simplifies to πππππ π π’πππ π = βπ1 (π2 β π1 ) Where π1 πππ π2 is the volume of gas to be placed in each balloon before compression and after compression respectively. For the work extracted by the generator, the buoyancy of the balloons forces a chain upwards which turns a cog connected to an electric generator. Assuming there are an equal number of balloons on each side of the chain (the chain side on its way up and the chain side on its way down) we can neglect the weight of the metal balloon casings. Therefore the net force experienced by the buoyant balloon on its way up is used to turn the generator. πΉπππ‘ = πΉπ΅π’ππ¦ππππ¦ β πΉπ€πππβπ‘ ππ πππππππ π ππ πππ β πΉπ΅π’ππ¦ππππ¦ So per balloon the maximum work extracted is (assuming a chain height of π₯ meters) π₯2 πππ₯π‘ππππ‘ππ = β« πΉπππ‘ ππ₯ π₯1 π₯ = πΉπ΅π’ππ¦ππππ¦ β« ππ₯ = ππ π1 π(π₯) = 9800π2 π₯ [π½ππ’πππ ] 0 In order for the compressed gas to displace water in each balloon, the pressure of the gas must be greater than the pressure of the water at the bottom of the chain. That is to say that π2 , the pressure of gas entering the bottom balloon must be at least: π2 = π1 + ππ ππ₯ = π1 + 9800π₯ Substituting for π1 = π2 π2 π1 and πππππ π π’πππ π = βπ1 (π2 β π1 ) πππππ π π’πππ π = βπ1 (π2 β π2 (π1 + 9800π₯) π1 + 9800π₯ ) = βπ1 π2 (1 β ) π1 π1 Bearing in mind πππ₯π‘ππππ‘ππ = 9800π2 π₯ Define system efficiency as π= β΄π= πππ₯π‘ππππ‘ππ πππππ π π’πππ π 9800π2 π₯ 9800π₯ = =1 π1 + 9800π₯ π1 π2 ( β 1) (π1 + 9800π₯ β π1 ) π1 Therefore it is not possible to run a buoyancy generator over-unity. Under absolutely ideal conditions, the best youβll get is as much energy out as you are pumping in using the compressor. While Iβm completely open to correction if someone can show me otherwise, Iβm fairly confident in the maths here. As of this moment, I do not believe it is possible for any buoyancy generator to operate in an over-unity fashion. Hope youβre having a wonderful week Ryan
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