nuby said:
All I can say is from my experience ... I've watched an electrolysis cell (plumbed into the vacuum system) with 10 stainless steel plates (304 grade, 16 ga.) performing 12V electrolysis on pure tap water (no electrolyte), and probably only producing 3 liters of (hydrogen/oxygen) gas per minute, apparently improve the fuel economy (city/highway average) by about 40 percent. This has been observed through about 4 tanks of regular unleaded gas.
I know the joules being produced by the hydrogen cell is around 3 liters per minute, and there is NO WAY that small amount of energy should increase power and mileage by 40 percent.. Which is why I think the faster combustion rate of hydrogen gas may be what helps performance..
My calculations show that indeed the contribution of the H2 combustion looks insufficient.
Gasoline supplies about 46 Mj per kg.
H2 supplies about 143Mj per kg at STP.
http://en.wikipedia.org/wiki/Fuel_efficiency
If 1 kg of gasoline supplies 1.3 liters and the car gets say 7 mpl (~27mpg), that suggests that it's getting about 9.3 mi/kg.
Now assuming for a moment that you did actually achieve a 40% increase in mileage, one must suppose that 40% more energy was utilized.
Working backward from the alleged result that would need to come to 13 mi/kg of gasoline.
If you are cruising at 60 mph, that would take you 13 minutes of travel time.
Assuming your claim of 3 liters of H2 generated at STP then that means you would have the contribution of 39 liters of H2 over the 13 minute period cruising period.
But 1 kg of H2 is 500 moles of H2 and at 22 liters per mole at STP that means that the device is only supplying 39/(500*22) parts of a kg or .0035 kg. In Mj this is then an additional .5 Mj.
This gives us a budget of 46 Mj for gas + .5 Mj from the H2 = 46.5Mj
But our requirement was for 40% more energy to do 40% more work.
We really needed to get 18.4 Mj from the H2 over that period or 64.4 Mj.
That 18 Mj shortfall looks like a clear violation of the conservation of energy.
Nota bene this calculation is based on the inherent Gibbs free energy of the fuel itself and is independent of the inefficiency of the motor. Neither did I deduct the additional inefficiency introduced in hydrolyzing the H2 by the alternator load supplying the battery.