First off, let's take a look at Hess's Law:
"The heat change (ΔH) that accompanies a given chemical reaction is the same whether the reaction occurs in one step or in several steps."
(http://www.science.uwaterloo.ca/~cchieh ... /hess.html
What does this mean for w-4-g, oxyhdrogen and PICC? Just what it says. If you start with 2 C8H18 + 25 O2 and end up with 16 CO2 and 18 H2O, there will be the same total change in energy, no matter what path you take to get there. It's like the change in temperatures (or changes in displacements)... if it's 22°C today and was 24°C yesterday, then that is a change of 2 Celsius degrees -- it doesn't matter that it dropped to 16°C overnight... it could have dropped to -40°C overnight and it would *still* be a net change of 2 Celsius degrees from day-to-day. Back to combustion, it doesn't matter what reactions are undergone to get from the isooctane and oxygen to the carbon dioxide and water. You can take many paths to get there (most of which will require the *input* of extra energy, which is subsequently re-released), but the net energy released from that net reaction will always be the same. There is no magic additive to get the gasoline to release more energy than it already does. The standard combustion reaction always releases the same amount of energy. Always.
Yes. That reaction doesn't always take place -- it is the ideal.
But let's look at possible gains using published EPA numbers.
Not all the gasoline is combusted, that's true. But about 99% is, according to EPA estimates... the best gain possible in that regard is about 1%. You can't burn more than the 100% of what's going into the cylinder.
Now, one of the products that the catalytic converter deals with is HxCx unburned hydrocarbons (HC). Ignoring CARB limits and more modern standards, even the 1994 limits on HC emissions were 0.25 g/mile. A quick bit of simple math will tell us that a 25mpg car (2650 grams/gallon for gasoline * 1/25 gallons/mile = 106 grams of gasoline per mile) must only be emitting 0.25g/106g = 0.0024 = 0.24% HC at the maximum. Now, granted most cars will be well under this at the tailpipe, as it is only the upper limit set by Congress, but the catalytic converter is also helping to oxidize the initially higher levels of HC. I would think people would be hard pressed to prove that any modern engine is producing much more than 4% HC, seeing as how engines in the 50's were only producing on the order of 7% HC (2650 g/gallon * 1/15 gallons/mile = 177 g/mile... 13g/mile HC produced / 177 g/mile gasoline used = 7.3%), long before any reductions started being built in to the engine operations and before any exhaust treatments started being employed. So, we're still not finding room for such a major breakthrough in "efficiency". And even then, forming the smaller HxCx hydrocarbons is releasing *some* energy -- just not as much as if complete combustion had taken place for those molecules.
So, the only other place for "better combustion" is for more CO2 to be formed, and less CO. Let's look at that. Again the 1994 standards are 3.4 grams per mile CO, down from 87 grams per mile in the 50's. If the combustion was complete, one gallon of gasoline would produce 8875 grams of CO2. So 8875g CO2/gallon * 1/25 gallon/mile = 355g/mile CO2 produced by the modern 25mpg engine. 8875g/gal * 1/15 gal/mile = 592g/mile CO2 produced by the 1950's engine. The current emission of CO over CO2 is around 1% and the historical value produced was around 20%. Even if the engine still is producing 20% CO and 80% CO2 today (which it isn't), what would be the difference in the energy released? Forming CO2 releases 3.5 times the energy that forming CO does. So, forming 100% CO2 would be releasing 17% more energy (0.2 + 0.8*3.5 = 3, 3(CO/CO2 mix)/3.5(pure CO2) = 1.17).
So the absolute most you could hope for is an overall 20% increase in the combustion efficiency. The actual value you can manage is considerably lower than that, but even assuming the worst of your engine to start with, there just isn't room for more than a 20% gain in the combustion efficiency of gasoline. And oddly enough, the auto makers have tried to release every drop of energy that they can out of the gasoline -- they're not going to leave a 20% gain just sitting there unattended.
Without totally redesigning the car and lowering both the weight and the drag, there is no possible way to do *anything* to the gasoline to double the mileage it produces with an existing car. There is no possible way to even get a 50% gain. 5%, possibly.
The major inefficiency of the internal combustion engine is limited by the theoretical Carnot efficiency. There is very little that can be done to raise the engine's efficiency from 25-30%. Again, this is thermodynamics. So, you can't get the gain by making the gasoline "burn better", and you can't improve on the losses from the engine just by modifying the fuel a little bit. Where do people think these magic gains are coming from?
The only real way you can improve the mileage of an existing vehicle is through more efficient driving.
And again, it takes 105 watts input to the alternator shaft for every 100 watts of electricity it outputs. When you turn on the rear defroster wires or operate power windows, it *does* put more load on the engine. The alternator is never producing surplus electricity. So, your "little bit" of electricity is putting a larger load on the engine. It is. You try to draw more electricity off of it, and it gets harder to turn. This is a regular demonstration at science centers.
So, with the efficiency of the engine and the electrolytic cell, you need the engine to consume about 560 watts (from either gasoline or a gasoline/hydrogen mixture) to produce 100 watts "worth" of hydrogen. See? It takes more energy to make the gas, than what you get back burning it. Always.
So, producing the gas is a net loss of energy, and can't possibly have the effect on combustion that is claimed for it. It doesn't need to be tested. I also don't need to test that I can't fit one gallon of gas into a 1 cup measuring cup. The math says it is not possible. The math says there is no room for it.