![]() ![]() The energy loss corresponds to the work it would take to bring oxygenĪll the way up to your fuel cell (as you mentioned, oxygen is denser than air). The system is losing energy over time, and decreasing its total entropy. You are generating oxygen at altitude zero while consuming oxygen at 173 km altitude. ![]() Your machine wouldn’t be perpentual for the following reason: If you’ve made it this far and have the solution, email me and I’ll post the best answers in a new post.ĮDIT (): Leonard Bezinge replied with the following message: Why isn’t there a height where the gravitational potential energy of the waterĬould, in theory, exceed the losses from the electrochemical system? Why does this proposal violate thermodynamics? Long story short, chemistry $ \gg $ gravity.Īll that said, however, none of this answers the key question: Height requirement of water to break even = (energy required by turbine to break even) / (effective force provided by water) = $ 24 $ kJ / ($ 0.177 $ N ) = $ \textbf $ kmĮven with extremely generous assumptions, a truly ridiculous height is required to “break even”.Īnd well into outer space as defined by the Kármán line.Effective force provided by water = (turbine efficiency) * (mass of water) * (acceleration due to gravity) = $ 0.95 $ * $ 1 $ mol * $ 18 $ g/mol * $ 9.81 $ m/s$^2$ = $ 0.177 $ kg m / s$^2$ = $ 0.177 $ N.Energy required by turbine to break even = (Electrolysis energy required) - (fuel cell energy generated) = $ 249 $ kJ - $ 225 $ kJ = $ 24 $ kJ.Let’s very generously assume the electrolyzer, fuel cell, and turbineĪll have 95% energy efficiency (here let’s define turbine efficiencyĪs the ratio of electrical energy generated from gravitational potential energy). Water electrolysis requires 237 kJ/mol of electrical energy at a minimum,Īnd fuel cells produce 237 kJ/mol of electrical energy at a maximum. To get a sense for the absurdity of this idea in practice: Perhaps its biggest practical flaw is that the height of the “mountain” is truly massive. To be clear, this proposal has all sorts of practical problems-low electrolysis efficiencies, terminal velocity of water, etc. In effect, we are exploiting the phase behavior of the hydrogen and the water. This energy could compensate for the inefficiencies of the electrolyzer/fuel cell combo-and generate some extra on the side. So, if you were to place a turbine at the outlet of the water pipe, However, this setup has one key advantage: the water generated by the fuel cell at the top of the mountain can do work when it falls back to the electrolyzer on the ground. Of course, splitting and regenerating water is a useless exercise since each of these steps is inefficient you can never generate more energy from the fuel cell than what you put into the electrolyzer. The hydrogen and oxygen then combine in the fuel cell to create water,Īnd the electrical energy is fed back to the electrolyzer to split more water. You collect oxygen from the atmosphere at the top of the mountain. While the oxygen is denser than air, the hydrogen is not and thus rises to the fuel cell.įortunately oxygen is abundant in the atmosphere with the right oxygen membrane, You provide the electrolyzer with a spark, creating hydrogen and oxygen. On the ground, you have a tank of water connected to the inlet of the electrolyzer. Imagine you placed an electrolyzer on the ground, placed a fuel cell on top of a mountain,Īnd connected the inputs and outputs via tubes. The question is: why does this proposal violate thermodynamics? However, I have had an idea for a perpetual motion machine for a long time Perpetual motion machines are a thermodynamic impossibility. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |