Quote:
Originally Posted by ChazInMT
Rustylugnut.....You need desperately to explain something to me. We have to assume that by adding 1 part in 100 to the fuel stream, (I'm just guessing that you are disassociating a quart of water for every 25 gallons of gas) that it is going to be exceedingly difficult increase the efficiency of the ICE. You can talk flame fronts and hydrogen seeding till yer blue in the face, but Gasoline IS HYDROGEN!! It isn't some sorta nerfgas, it is hydrogen loosely attached to carbon, that's it. The air is 21% oxygen, so when you add tiny amounts of hydrogen and oxygen that you have disassociated on board your car, you are really only adding a modicum of fuel to the stream.
Now it stands to reason that if it takes more energy to split the hydrogen and oxygen than you can get by recombining them....just alone! You can't get more energy when you recombine no matter what. And since an engine is 30% efficient, and an alternator is like 70% efficient, you only are getting 1 watt of energy in electricity for every 5 needed in fuel to power you hydrogen diffusing apparatus.
How the hell are you supposed to see any efficiency gain whatsoever when it takes 5 times the amount of energy to create the minuscule amount of extra fuel that requires 80% more energy to take apart than can be derived by burning it in your engine?
Where is your disconnect from reality here? Seriously, if you cannot explain how you are supposed to overcome these very basic principles of physics than you need to just go away and quit posting nonsensical drivel.
Numbers Rusty, show us numbers about where this magical energy comes from to improve a vehicles combustion enough to see any improvement at all when you are starting from a 5 to 1 energy deficit at the outset. How do you go from 5 to 1 to improvement? Tell us where the 500% efficiency gain is JUST TO BREAK-EVEN? Then explain where the 10,000% gains in efficiency are produced by adding tiny amounts of hydrogen and oxygen to the engines fuel stream which might be seen as an actual small improvement in MPG by someone who is very carefully testing their mileage?
Fact is you can't, because physics won't allow you to get more energy out of a system than you put into it, unless you're talking about nuclear power, which you clearly are not.
And by numbers you need to provide volumetric data on hydrogen produced and added to the fuel stream along with the volume of fuel in the form of gasoline on a like basis. Then explain what fantastic phenomena occurs at these volumes, which you are able to produce, that create so much extra energy. And tell me how much electric current/power is required to produce this hydrogen, and run that back through the alternator and engine to see how much fuel is required to generate the electricity that creates the hydrogen from water.
Here's a handy link to the Wiki page on the electrolysis of water to help you get a start.
Here's another handy link to determine the megajoules of energy in a gallon of gasoline (you'll need to know that in order to figure out it'll take at least 1.5 gallons of gas to perform electrolysis on a quart of water.) |
What you are speaking of is the simple viewpoint of the HIGH SCHOOL student. If I remember right, you came from a nuclear technology background so it should not come as a surprise to you that there are specialties in engineering .
I refer you to a book called "Combustion", by Glassman and Yetter. It is used in many undergraduate/graduate courses in combustion science. In the first part of the book the discussion centers on the complexity of combustion. Octane (C8H18) does not simply thermally de-polymerize and go directly to the combustion ashes of CO2 and H2O. There are many pathways and side reactions that the de-polymerization can take. This takes time. Thus you have a flame front and flame speed. One reaction pathway may release heat which is robbed by another pathway. A plethora of smaller chain molecules may exist at some point all competing for the energy or releasing energy as it may. This takes time. Yes, you can simply take the heats of formation of C8H18 and CO2 and H20 and get the total heat of combustion, but that is only part of the story. Time is the other. During de-polymerization, highly reactive radicals such as H+, OH- , OOH, HOOH and so forth are created. Their existence was theorized decades ago and only in the last decade has instrumentation and detection techniques allowed us to verify them. These radicals are important in that they "rip apart" the long chain hydrocarbons into shorter species. By adding a small amount of H2 and O3 before the ignition point free radicals can be formed via the heat addition of compression, radiation and turbulence since the energy of dissociation for H2 and O3 is quite low. This rich cloud of active radicals is now ready to pounce on the long chain hydrocarbons and release even more hydrogen and form even more radicals in a domino effect. The seeding of the fuel mix with these radicals "railroads" some of the side reactions resulting in a faster flame front. No more energy is created in this reaction, but the combustion TIME is shorter. If you have an engine that needs 30 degrees of ignition lead to produce a required amount of torque, it means you are wasting 30 degrees of pressure rise for that power stroke. If you can accelerate the flame rate and now only need 20 degrees of ignition lead, you are wasting much less of your pressure rise and can use it in the productive down stroke. You can now use less fuel.
All the above takes place in an engine that is in detonation. If your fuel mix is heated via intake turbulence, absorptive radiation from the engine mass and then by compression heating, you may just have produced the conditions for thermal de-polymerization to start BEFORE the ignition spark as free radicals are formed from the oxygen in the air, traces of exhaust gas and trapped active radicals from crevices in quench zones. If your conditions are sub-detonation, you can push the combustion in that (detonation) direction by seeding of hydrogen or ozone.
So, how much HHO is needed? That is not a simple question because all engines and power regimes are not the same. But, the best engines to respond to HHO addition are the same ones that want to detonate. Iron blocks with iron heads and high compression along with high swirl intakes need the least amount of HHO to see effect. I have built and tested to some degree, a 2.4 liter iron block and iron head engine that had a Siamese intake/exhaust configuration that heated the incoming air and thus the engine could run at very lean (28:1 AFR) and could produce useful power levels (20 hp) at cruise (1500 RPM) by using a 22:1 compression ratio. This engine was to power our AutoXprize entrant. By adding an exceedingly small amount of HHO, a reduction in ignition lead time as described above was seen. Thus a measurable fuel savings of a few percent could be gained depending on variables. So how small an amount? 1:750 - 1:2000 ( HHO:air). Since the engine cruised un-throttled, that means you would need only 1 liter per minute of HHO to only several hundred cubic centimeters per minute.
So how much electrical energy is that? I'll leave that to you. That is simple high school stuff.
Start with a Coulomb. Divide that into the moles to produce 1 liter of HHO gas then convert that back into amperes and apply your alternator efficiency. I've skipped some steps but I'm sure you can fill them in.