Bob Boyce 101 efficiency details

[newbie123]

you need to use the same voltage he did.

This isn't accurate...  As you just said,  Faraday's law has nothing to do with watts/voltage... just amps. (Btw, Faraday never accomplished 100 percent "Faraday" Efficiency, he just calculated it)

its not watts nor does he state watt or joule or anything like that, its AMPERE.

Right. Did you read any of my posts? 

Also faraday's law only applies to those using his voltage.

The "100 percent efficiency voltage" will actually change with temperature and pressure... But 100 percent current (Faraday) efficiency will remain the same regardless.

[TheNOP]

its not watts nor does he state watt or joule or anything like that, its AMPERE. 

if you can't see the relations between those units you should go back to school.

yes, Faraday's law require amperes unit for electrolisis efficiency calculations.
but that does not mean you can't calculate the amps from the watts or the amps from the joules.

in fact, you can convert any units to any other quantifier units, if you have the values you need to do the convertion.

[TheNOP]

The "100 percent efficiency voltage" will actually change with temperature and pressure... But 100 percent current (Faraday) efficiency will remain the same regardless.

in other words.

Faraday's law of electrolysis is using mol in its formula.
a mol is not a unit of volume, it is a unit of amount of substance.(atoms, molecules, etc...)

this mean, you have to compensate for temperature and pressure when you measure your gases output by volume.
but Faraday's law does not require to be compensate.

[CrazyEwok]

if you can't see the relations between those units you should go back to school.

yes, Faraday's law require amperes unit for electrolisis efficiency calculations.
but that does not mean you can't calculate the amps from the watts or the amps from the joules.

in fact, you can convert any units to any other quantifier units, if you have the values you need to do the convertion.

Really you can calculate the amps from simply having the watts can you??? i have 4watts of power being produced how many amp am i outputting???
Oh wait you can't tell me the amp because you are missing some information... the voltage produced!!!

and all conversions of power have relevant losses so no you can't do a simple converstion, because this would involve having to calculate your losses... Also in order to measure your amps there needs to be a voltage as your electrical charge won't move without a voltage difference between 2 points. And your saying i need to go to school... for the perdantic or those not quite understanding.
Watts = Volts x Amps
Faraday states that it takes X amps to breakdown the bonds of water.
he doesn't state what voltage you use. i bet i can get crap loads more gas out of 10amps at 2kv then i can out of 10amps at 12v. same ampere different voltage.

So no unless you go back and look at how many volts he used in his calculations you or anyone else quoting his work is wasting their and everyone elses time. If you find the voltage then that can in turn be converted into Watts, this is a calculation that can be used by everyone... so off you go convert your little calculation or stick to your "amps = Gas" notation....

[newbie123]

i bet i can get crap loads more gas out of 10amps at 2kv then i can out of 10amps at 12v. same ampere different voltage.

Wrong....    If you ran a cell at 2kv 10amps,  you'd  probably get crap loads more heat and steam (w/ a very low production efficiency), but the hydroxy gas would be the same as 12V 10A..

So no unless you go back and look at how many volts he used in his calculations you or anyone else quoting his work is wasting their and everyone elses time. If you find the voltage then that can in turn be converted into Watts, this is a calculation that can be used by everyone... so off you go convert your little calculation or stick to your "amps = Gas" notation....

Again, Faraday efficiency has NOTHING to do with volts or even energy, his "volts" are totally irrelevant!