Theoretical MINIMUM Amps/Liter/Minute/Cell?

[danconkincs]

I am just trying to check my numbers from my calculations for Theoretical MINIMUM Amps/Liter/Minute/Cell? ZeroFossilFuels said at the November Expo that it was about 7 amps and the best he had done was 7.5. According to my numbers and other people's actual generator results, he must have been talking about a multi-cell stack (he didn't elaborate) since the more cells you have in series driven by the same current, the more HHO you can generate per minute, assuming of course that you still have enough voltage to drive each cell for effective electrolysis.

Thanks ahead of time for any direction on this,
Dan

[danconkincs]

Shane,

Thanks for your response. In theory, though, every molecule of water that gets broken down in a cell takes transfer of two electrons, as long as the voltage is high enough to cause disassociation of the water molecules first (~.74 volts) and then elements (about 1.22 volts). If you take the number of atoms of H and O in a liter of HHO, you should be able to divide that by the number of charges in a coulomb. Then, since 1 coulomb per second of current is 1 amp, you should be able to determine:

the minimum Amps/liter/minute/cell. Any additional current would be from losses, such as electrochemical leakage between cells. Isn't that correct? I have done that two different ways, but the numbers are significantly different. I am trying to get some independent verification of at least one of them hopefully.

Dan

[Johnh]

Hi Dan
Although this figure will give you the Amps needed to disassociate the molecules it doesn't take into account the energy needed to expand the gasses and overcome the electrical resistance in the circuit
This paper explains it from a total energy persective
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html
John

[danconkincs]

John,

It has been a while since my last Thermodynamics class, so thanks for the reference to a concise reminder of the parameters and relationships involve.

In electrolysis, though (ignoring leakage currents), an "Energy Increase" would have to be from just an increase in driving potential (i.e. Voltage), wouldn't it? The electrons and charges have to go somewhere; i.e. they have to complete the electrochemical circuit!

Dan