[Pony Exp.305]

Very Interesting>>>
http://water4gas.com/2books.htm
or
http://runyourcarwithwater.com/

(IMG:http://www.frrax.com/rrforum/style_emoticons/default/bs.gif) (IMG:http://www.frrax.com/rrforum/style_emoticons/default/bs.gif) (IMG:http://www.frrax.com/rrforum/style_emoticons/default/bs.gif) ???

[z28tt]

The systems disassociate water into Brown's gas?

Wikipedia to the rescue:
http://en.wikipedia.org/wiki/Brown%27s_gas#Brown.27s_gas

When ignited, the gas mixture converts to water vapor and releases energy, which sustains the reaction: 241.8 kJ (235 BTU) of energy (LHV) for every mole of H2 burned.

The idea is to split water (2*H2O = 2 H2 + 02, about 80-90% efficient according to http://en.wikipedia.org/wiki/Electrolysis_...ter#Efficiency) through electrolysis and then recombine it again?!! How can that possibly generate energy, when nothing is ever 100% efficient.

If a gallon of gas has 112,000 BTU, and your fuel rate at cruise is 15mpg @ 60 mph, that's 4 gallons per hour, or 448 kBTU's. To increase the mileage by 10%, you need to generate 45 kBTU's of h2 per hour, or 192 moles (or grams, if a mole equals a gram) per hour. Adding 96 moles of 02 at 32 grams/mole, gives you 3072 grams of 02.
192+3072=3264 kg (3.264 Liters) of water/hr needed to electolyze to make enough h2 needed to supplement 10% of the gasoline

If you need 4 electrons to split a water atom into 2 h2's and one o2, need 96 moles of water/hr, then you need 384 moles of electrons/hr, or 2.3E26 electrons. Since it's 1.6E-19 electrons per coulomb, you need 3.68E7 coulombs/hr.

An Amp-Hour is 3600 coulombs, so you'll need 10,214 AH to feed the electrolysis reaction.

A volt is 1 joule per coulomb. With a 12.6 volt battery, you'll need 3.68E7 coulombs/hr * 12.6 Joules/coulomb = 4.63E8 Joules/hr
Since a watt is 1 joule/sec, that works out to be 128,700 Watts.

That's a pretty damn big power draw on the alternator!

Much better to just accept the 35% loss in burning gasoline, than add another 2 steps with their inefficiencies

http://en.wikipedia.org/wiki/Water-fuelled_car

I supposed I could have saved a bit of time calc'ing formulas if I just converted the 45,000 BTU's needed for a 10% gain in mileage * 1055 into Joules and divided by 3600 into Watt/hr (which is 13,129 Watts/hr, so I'm off by a factor of ten with the crazy calcs above, which may either be a mistake, or somehow show the inefficiency if it's correct). (IMG:http://www.frrax.com/rrforum/style_emoticons/default/smile.gif) Regardless, 13,000 Watts is a pretty big load too...

Maybe if you used solar cells to charge a battery for the water electrolysis! (IMG:http://www.frrax.com/rrforum/style_emoticons/default/wink.gif)

This post has been edited by z28tt: Jun 11 2008, 04:31 PM