Hi Radiant_Science,
Lets just agree to disagree on this one okay? For you want the Gas Processor to create the O3 and I want it to strip oxygen atoms of their electrons some four or more. Ozone is very bad as it eats up most types of seals. The GP is set up to not create O3 but lone O atoms these atoms are hit with electron collisions and photon bombardment in some of the absorption wavelengths of oxygen. Then this highly ionized air gas is mixed with the water mist that most of it will turn into vapor due to the phase diagram of water in that the water is about 85-90 degrees C under pressure and is injected into a low pressure zone.
As this mixture moves through the voltage zone the water molecules are charge each of the same polarity so they will repel each other and not form larger droplets, that overcomes the adhesive properties of water taking on the same charge. At around 20 kv the cohesive properties are overcome and the water droplet divides into two smaller droplets dividing the voltage between the two as it does, but as long as it is in the voltage zone it will be recharged and the process repeats until a critical volume is reached. Once at the critical volume and charged to the maximum degree again the molecule simply breaks down into it's component elements for this is an image charge with very little current not enough to ignite the mixture just to break it apart. Now why it is still in the voltage zone the voltage zone acts like the gas processor but only electrons are colliding with the freshly created oxygen and hydrogen atoms. When I ran my numbers on the speeds of the fuel injectors and the length of the voltage zone compared to a frequency of 50 kHz I got a total of 86-89 hits on the water molecules, already ionized air gases, and recirculated exhaust gases. So if it only take 40 hits to break down the water molecule into hydrogen and oxygen then the remaining hits will be on the hydrogen and oxygen atoms while still inside of the fuel injectors. Now raising the voltage will charge the water mist faster towards the dividing point.
As for the EEC on the injectors Meyer's only shows it as a amp consuming device inline with the positive side of the voltage zone. So that must be tested out to see just what it does.
The reason I wanted you to look at energy content calculations is if you did you would easily see that O3 + 6H does have more energy content than just plain 2H + O which is at 286 kJ/mol, but they both fall well short of gasolines 5088 kJ/mol (depending on grade). Only by stripping the oxygen atoms of 4 or more electrons will it surpass the energy content of gasoline. So if you don't allow the GP to strip these electrons off then you will not get the needed energy content to replace gasoline and if you allow the GP to produce O3 instead of striped oxygen atoms you will fall short on energy content. That is what I wanted you to see, the energy content in the reaction you are saying is taking place vs that of gasoline and then compare those numbers to what Meyer claims he was capable of doing. You will find that it wont match up to Meyer's claims.
You are right when you say it is unknown the chemical breakdown when the critical volume of water is reached: for is it H3O+(aq) + OH-(aq) = 4H (g) + 2O (g) or is it H2O(aq) = 2H (g) + O (g)? one of the two is happening, which one of the two as of now I can not answer. But yours seems more plausible due to the voltage stresses placed on the water molecules by the voltage zone.
Now what I think that quote you place on is saying is this; the unstable oxygen atoms have the necessary energy to reach in and take the hydrogen atoms from the water molecule if it comes into direct contact with the water molecules. I have run calculations showing that it does with a surplus of energy before on this thread.
Like I said before if you don't agree then lets just agree to disagree, okay? Thanks.
h2opower.
Lets just agree to disagree on this one okay? For you want the Gas Processor to create the O3 and I want it to strip oxygen atoms of their electrons some four or more. Ozone is very bad as it eats up most types of seals. The GP is set up to not create O3 but lone O atoms these atoms are hit with electron collisions and photon bombardment in some of the absorption wavelengths of oxygen. Then this highly ionized air gas is mixed with the water mist that most of it will turn into vapor due to the phase diagram of water in that the water is about 85-90 degrees C under pressure and is injected into a low pressure zone.
As this mixture moves through the voltage zone the water molecules are charge each of the same polarity so they will repel each other and not form larger droplets, that overcomes the adhesive properties of water taking on the same charge. At around 20 kv the cohesive properties are overcome and the water droplet divides into two smaller droplets dividing the voltage between the two as it does, but as long as it is in the voltage zone it will be recharged and the process repeats until a critical volume is reached. Once at the critical volume and charged to the maximum degree again the molecule simply breaks down into it's component elements for this is an image charge with very little current not enough to ignite the mixture just to break it apart. Now why it is still in the voltage zone the voltage zone acts like the gas processor but only electrons are colliding with the freshly created oxygen and hydrogen atoms. When I ran my numbers on the speeds of the fuel injectors and the length of the voltage zone compared to a frequency of 50 kHz I got a total of 86-89 hits on the water molecules, already ionized air gases, and recirculated exhaust gases. So if it only take 40 hits to break down the water molecule into hydrogen and oxygen then the remaining hits will be on the hydrogen and oxygen atoms while still inside of the fuel injectors. Now raising the voltage will charge the water mist faster towards the dividing point.
As for the EEC on the injectors Meyer's only shows it as a amp consuming device inline with the positive side of the voltage zone. So that must be tested out to see just what it does.
The reason I wanted you to look at energy content calculations is if you did you would easily see that O3 + 6H does have more energy content than just plain 2H + O which is at 286 kJ/mol, but they both fall well short of gasolines 5088 kJ/mol (depending on grade). Only by stripping the oxygen atoms of 4 or more electrons will it surpass the energy content of gasoline. So if you don't allow the GP to strip these electrons off then you will not get the needed energy content to replace gasoline and if you allow the GP to produce O3 instead of striped oxygen atoms you will fall short on energy content. That is what I wanted you to see, the energy content in the reaction you are saying is taking place vs that of gasoline and then compare those numbers to what Meyer claims he was capable of doing. You will find that it wont match up to Meyer's claims.
You are right when you say it is unknown the chemical breakdown when the critical volume of water is reached: for is it H3O+(aq) + OH-(aq) = 4H (g) + 2O (g) or is it H2O(aq) = 2H (g) + O (g)? one of the two is happening, which one of the two as of now I can not answer. But yours seems more plausible due to the voltage stresses placed on the water molecules by the voltage zone.
Now what I think that quote you place on is saying is this; the unstable oxygen atoms have the necessary energy to reach in and take the hydrogen atoms from the water molecule if it comes into direct contact with the water molecules. I have run calculations showing that it does with a surplus of energy before on this thread.
Like I said before if you don't agree then lets just agree to disagree, okay? Thanks.
h2opower.
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