I'm working on the numbers - hope this helps...
There are 55.346 moles in 1 liter of pure water... 18 g of H2O = 18 ml of water = 1 mole of water.
"The electrolysis of one mole of water produces a mole of hydrogen gas and a half-mole of oxygen gas in their normal diatomic forms." see below
A mole of water thus produces 22.4 liters of H2 at STP, and 11.2 liters of O2.
THUS: electrolyzing 18g = 18 ml = 1 mole of water will produce a 22.4 liters of H2 at STP, and 11.2 liters of O2 -- and this requires a TOTAL of 285.83 kJoules of energy = 285.83 kW-seconds. Round this to 286 kJ of energy, and then this agrees with the summary Mike posted.
THUS: electrolyzing 1.0 g = 1.0 ml of water will produce a 1.244 liters of H2 at room temp, and 0.622 liters of O2, or a total of 1.87 liters of gas -- and this requires a TOTAL of 15.9 kJoules of energy = 15.9 kW-seconds.
Continuing, 15.9 kW-seconds = 15,900 W-seconds divided by 3600 seconds/hour = 4.42 W-hrs to electrolyze 1.0 ml of water, producing 1244 ml of H2 gas and 622 ml of O2 gas.
In an experiment, one can measure the amount of water which was electrolyzed - weighing is probably the easiest way to do this. Then measure the amount of gas produced - using the graduated cylinder method I showed above in a photo is straightforward (a bit crude but sufficient to find roughly a +15% effect).
Also, measure the ENERGY input -- to do this is also straightforward - watts times seconds = Joules of energy input.
Then compare Energy In vs grams of water electrolyzed AND total gas volume produced in the process and see whether the gas produced agrees with the "standard model" above, or disagrees (and by how much).
PhysicsProf I think there is something wrong with your numbers. To produce 1 liter of HHO gas,and going by Faraday's MMW limit of 9.82,then it takes only 101.832 watt minutes to do so-or 6109.92 watt seconds. As you are dealing with 1.87 liters,we then multiply my value by 1.87,which gives us 11,425.5 watt seconds-where you have 15,900 watt seconds. So you have a value of 4.42 watt hours to electrolyze 1ml of water,and my value is only 3.173 watt hours. Going by your number's,Faradays MMW limit would only be 7.057--and this i(and many others already have) can break using just a DC current. To answer Grum's question (as far as i know),and using Faraday's MMW limit of 9.82,it takes 101.832 watts for 1 minute to produce 1 liter of HHO gas,at 1 atmosphere . Edit Faradays MMW limit is actually 9.28--not 9.82 as i stipulated. This limit now allows us to use 108 watt minutes to produce 1 liter of HHO,instead of the 101.8 watt minute's i quoted above. This is better for us Brad
« Last Edit: 2016-11-29, 08:36:47 by TinMan »
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