Air intake Oxygen + Ionized 

 

I have Put Oxygen Nitrogen and Hydrogen into Separate = Btter Quality Oxygen on intake and Less Nitrogen Means

More Explosive Power , More Torque and less Emissions.

 

Sections as each has diferent rates of Energy (light Obsorption)  We can selecte diode that match the right wave length. 

So we only heat up (energiese ) the Gas we want to react and the ones we don want to react we try to avod energising or heating them up.

 

Ideally we want N and H to join together. or be energised enough so we can  strip of electron to make them very unstable in the air intake ionzer (gas processor)  As Stane Describes in his Patents. This mean when they Egnite they are explosive.

 

We USe DC and must have Basic Ionisation KNowledge of how to control ions in the airs space and direct them. 

 

So first we Shoot the right Light Wave length in NM  this energises the Target Gas  Signle Gas. Each is different.

 

Than We use the Carona Discharge ion emission, which we can focus and direct with  duty pule and +/i charges to the gas we are targeting 

this takes it futher to a higher energ state, we try to extract electrons and keep it unstable until blending block, where is mixes with other gases. 

 

Primarily we are target the  mixing and bond of H with N to make NH3 or have both  in very unstable states very along with Oxygen.

 

 

Oxygen 

 

 

If you ionize the oxygen atom positively with a high voltage DC field what's going to happen?

 

it will become positively charged and will be attracted to the negative side and be repelled by the positive side and them vice versa. I believe this was he talking about particle oscillation as energy generation. I believe that the higher the voltage the highest 

will be the impact between these molecules freeing up many electrons. 

==========================================

 

Ionisation Energies and electron affinity

 

The electron affinity of oxygen is 141  kJ mol-1. The ionisation energies of oxygen are given below.

 

Ionisation energy number Enthalpy /kJ mol-1

1st 1313.9

2nd 3388.3

3rd 5300.5

4th 7469.2

5th 10989.5

6th 13326.5

7th 71330

8th 84078.0

======================================

 

Ionisation Energies and electron affinity

 

The electron affinity of nitrogen is 7  kJ mol-1. The ionisation energies of nitrogen are given below.

 

Ionisation energy number Enthalpy /kJ mol-1

1st 1402.3

2nd 2856

3rd 4578.1

4th 7475.0

5th 9444.9

6th 53266.6

7th 64360

======================================

 

Ionisation Energies and electron affinity

The electron affinity of hydrogen is 72.8  kJ mol-1. The ionisation energies of hydrogen are given below.

Ionisation energy number Enthalpy /kJ mol-1

1st 1312.0

 

 

========================

 

Yeah, the whole goal is to remove 4 electrons from the oxygen atoms. As shown in Meyers tech brief drawings, and explained in his videos. And per Stan meyers switzerland video the LED's purpose is to aid the ionization.

Stan also states in his patent that the resonances (energy level/wavelengths)

of the oxygen atom can be determined by studying atomic physics.

 

Things to consider, if you research Oxygen ionization energies you'll find each successive ionization takes more energy.

 

1st ionization=13.61eV

2nd ionization=35.11eV

3rd ionization=54.93eV

4th ionization=77.41eV

 

I think the reason for this is as each electron is removed, the atom has a greater positive charge, thus it has a greater electric (opposite attraction) force to hold the remaining electrons in place.

 

This also means that for each electron removed, the energy levels required to excite the atom increase. In my research I don't believe there are any LED's which could excite even a singly ionized atom as it would have a greater energy than 35.11eV.

 

From all the research i have done the Oxygen absorption wavelengths are:

344nm, 360nm, 380nm, 477nm, 577nm, 630nm, 760nm.

 

If you convert these to eV you'll get

 

760nm=1.63eV

630nm=1.96eV                                                  Basically we want to zero in on the nm number and match diode for O,H and N    

577nm=2.14eV                                                  once have reliable number for oxygen hydrogen and nitrogen we can target each

477nm=2.59eV                                                  in the air intake ionizer and energise the with the right wave length and make them 344nm=3.60eV    

380nm=3.26eV                                                  energize and weaken to loose a electron and heat up to react with n and h more 

360nm=3.44eV                                                  readily or voilently with ignition

344nm=3.60eV

 

Now these wavelengths will excite the oxygen atom, notice how each shorter wavelength has more energy (excites the atom to a higher energy). Alsonotice that the first ionization energy of oxygen is 13.61eV.

 

What these wavelengths do is aid the ionization process of the first electron. LED's are not available in a short enough wavelength to excite a singly ionized oxygen atom. They are not even available in a short enough wavelength to cause the first ionization. (13.61eV=91nm). 

 

Realistically LED's under 350nm tend to start getting expensive. I was given a quote on 300nm LED's for 100$ a piece. I honestly don't think stan used anything that expensive here. From all the research i have done thus far it seems that 760nm, 630nm, 577nm, 477nm, and possibly 380nm were used. more study will help. PS-the absorption spectrum of liquid oxygen I posted above is the wavelengths we need to study the most.

 

One thing I wonder, is why, and if for a reason Stan had 7LED's on his GP, as well as showing 7 energy levels in his Tech brief drawing? Perhaps he had a few in between, maybee 1268nm, and 1064 were also used, along with a possible wavelength in the low 400nm range.

 

Ionization energy is measured in units of -- umm -- energy (or total energy for a mass quantity) such as eV (or kJ/mol.) V/m is not a unit of energy, so there is no answer to your question; it's like asking for the distance from New Your to Chicago in pounds.

 

 

To start on the road to what you want, try this:

 

1) Start with the ionization energy in kJ/mole; I assume you're working in gaseous nitrogen, so you'll need the first ionization energy of the N2 molecule; 1402kJ/mol is the energy for atomic nitrogen. If you know where to find the energy for N2, please let me know. (It's what I was looking for when I stumbled across your question.)

 

2) Divide that by Avogadro's Number and multiply by 1000 to get J (per molecule.)

 

3) Divide by the magnitude of the electron charge to get eV.

 

Since one electron needs to be removed from the molecule, the energy in eV is numerically the same as the potential in V through which the electron must be moved. 

 

HYdrogen Notes for this Section

 

As an atom, it does not have a charge. H2, diatomic element 

If it loses the electron, it gets a positive charge, H+, as in acids  

If it gains an electron, it gets a negative charge, called hydride, H-

 

Since Hydrogen has one electron, when it becomes charged (when it becomes an ion), it will either 

loose an electron or gain one to become stable when it bonds, so it can have a postitve or negative charge.

 

========================================================================

 

Stan Cleary Says Stip off electron from Cell so that make the Hydrogen Unstable positive charge

 

========================================================================

 

Nitrogen Notes for this Section

 

If you mean ions which contain only Nitrogen ( without other elements ) ; They will be : 

[ N ]3- 

Nitride ion ; Having 3 Negative Charges 

[ N3 ]- 

Azide ion ; Having 1 Negative Charge 

 

======================================================

 

But here's the rub: Over what distance must the electron be moved?

 

I don't know how to determine this.

 

It probably has something to do with the size of the nitrogen atom, the size of the N2 molecule, and/or the spacing between molecules in the air.

 

If you know the applicable distance, then you can divide the potential computed above by that length to determine the field strength that will ionize the nitrogen.

 

Here's an easier way. According to Wikipedia, the dielectric strength of dry air at STP is about 3.3MV/m between spherical electrodes; it may be lower for other shapes.

 

That's the field strength needed to strike an arc.

 

Striking an arc is done by ionizing the air so that it becomes conductive.

 

So, the electric field needed to ionize N2 (most of air) is somewhere on the order of 3.3MV/m.

 

At 30kV/m I can be pretty sure you're not ionizing anything. If you're not striking an arc, you're not ionizing the gas.

 

Read more: http://www.physicsforums.com 

 

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