my thoughts on the Stan Meyer "WFC"

From what I've read, Stan Meyer's (theoretical) electrolysis process involves several mechanisms ("theoretical" because I've never seen a working S. Meyer WFC replication despite the many attempts).

The first one is "voltage potential taking over". Stan Meyer talks a lot about this in his videos/papers, by
"voltage potential taking over" I think he means: voltage performing the work, instead of a current (free energy basically).

Voltage potential is the force (electrostatic) which moves electrons through a circuit performing work.

According to my understanding: The positive charge is created by "holes" (or positive ions) on the + side of the battery. These holes are waiting to be filled by electrons and they create a "sucking" force which pulls negative elementary charge or electrons through circuits, spark gaps, electrolytes, etc.

What if you could some how have this "sucking" action without current/ charge flow from the (-) side of a battery? For example, suck the "holes" H+ ions and/or valance electrons out of H2O molecules. Or somehow "inject" electrons into a potential or electric field, and make the injected electrons perform work?

Here's how it might apply to electrolysis:

Cathode (reduction): 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)

Anode (oxidation): 4OH⁻(aq) → O₂(g) + 2H₂O(l) + 4e⁻

These reactions occur at a ratio of 2:1 .. reduction to oxidation, respectively.

This is how it happens:

At the cathode two electrons "hit" two H2O molecules with a minimum energy (or velocity) of 1.23 eV (Volts), this causes the two H atoms to break free and combine to form H2, and 2OH(-) is left.. The 2OH(-) migrate (or propagate) across the cell toward the (+) side where the "sucking" action takes place.

After the reduction reaction occurs twice, you have 4OH(-) at the anode. The (+) charge in the battery continues to "suck" 4 negative charge/electrons (from within 4OH-) towards it to replace charge/electrons lost on the (-) side. (Note: These bound electrons have ~0 kinetic energy, unlike when they created H2). When the 4e- are sucked into the electrodes towards the battery O₂(g) + 2H₂O(l) are formed.



The second 'important' part of the Stan Meyer's "WFC" .. is what he called the "Electron Extraction Circuit" or EEC. From what I've read this is a key component to the functionality of the whole system. In a nutshell he talks about LED light injection and claims the wavelength required by his process is right around 300 nm (nanometers), he then talks about some sort of a "amp" consuming device attached to the circuit, and what he thinks is happening at the atomic level.. His descriptions of what happen at the atomic level probably got him laughed at a few times by scientists, and I doubt he actually knew what was occurring (his brother Steve Meyer admits this in an interview I heard on blogtalkradio) ..

So anyway, how could this photon injection actually contribute to his process? I thought about this, and this is my best guess:

There is an phenomenon in physics known as the "photoelectric effect", where electrons are emitted from matter (usually a metal) after the absorption of electromagnetic radiation (photons/light) of a specific (or minimum) energy, called the "work function". More info here: http://en.wikipedia.org/wiki/Work_function .. The work function of Chromium (20 percent of stainless steel) is 4.5 eV which translates to a photon with a wavelength of 275 nm (which is in the ultraviolet range).


I doubt either of these ideas could work to extract extra energy out of a cell, but it would be interesting to do some experiments with UV LEDs. And try to eject electrons out of your (-) electrode and see if they'll perform work (electrolysis). Steve Meyer also mentioned eventual electrode consumption/deteriation through the Stan Meyer process.

BBC leakage

Here is an image of the formation of Hydrogen and Oxygen on the outside of the positive and negative terminal plates.

This might be normal operation of a series neutral plate cell, but it might not. HHO didn't form on the outside of the plates with a weak NaOH electrolyte solution, or at least I didn't notice any. However, once a higher concentration of NaOH was added the outside HHO formation started again...

The cell is full of leaks and will have to be redone.

Efficiency and temperature Equations

Gibbs Energy of Formation (with variable temperature) = ΔG = (ΔH * (298.15/T1)) - T1ΔS

T1 = Temperature of cell Electrolyte (user input)
T2 = Temperature of HHO Gas coming out of cell (user input)
ΔH = -285.85 ( kJ / mole h2o liquid) at 25C (298.15 K) (constant)
ΔS = -163.34 (J / K) = (Entropy of Water) - (Entropy of HHO Gas) (constant)

For example, at a temperature electrolyte/cell temperature of 38C (311.15 K) the Gibbs Energy of formation might change to: ΔG = (-285850 J * (298.15 K / 311.15 K)) - (311.15 K * -163.34 J/K)
ΔG = -223.05 kJ / mol

So to disassociate 1 mole of H2O with a cell temperature of 38C, 223.05 kJ of electricity is needed.

This will produce 1 mole of H2 gas, and .5 mole of O2 gas:

22.4 L * (1.5 moles) = 33.6 L (@ 298 Kelvin)

If the gas temperature is 308.5 Kelvin (T2) the volume will be 33.6 * (T2 / 298.15) = 34.78 Liters

Convert to Watts/LPM:
223.05 kJ / 3600 s = 61.95 Watts

Convert 34.78 L/hr to LPM: 34.78 / 60 = .58 LPM

Convert to W/LPM: (1/.58) * 61.95 W = 106.81 W/LPM

Convert to MMW: 1000/106.8 W = 9.39 MMW

Calculate Voltage required from Faraday (below):

107.205 Amp/Hr will create 73.338 L (at 25C), and 75.92 L at * 38 C (73.338 * 308.5/298)

(61.95 W * 2) = 107.205 A * V
Volts = 1.15 V

Faraday Efficiency

Faraday electrolysis efficiency uses quantities of electrons (coulombs) and gas volume to determine the efficiency of an electrolytic cell.

According to Faraday law, 4 moles of electrons moving through a cell will create 2 moles of H2 gas, and 1 mole of O2 gas, at 100 percent Faraday Efficiency.

How to calculate 100 percent Faraday Efficiency:

First convert 4 moles of electrons into amps. (4 moles) * (avagadro's number) =
4 * 6.0221417e23 = 2.408856716e24 (electrons)

Then divide the total electrons by 1 Coulomb (quantity of electrons)

2.4088567e24 (electrons) / 6.24150947e18 (1 Coulomb) = 385941 Coulombs

Now calculate the Amp Hours:

Since (1 Amp) = (1 Coulomb * 1 Second)

(385941 C) / (3600 Seconds) = 107.205 Amp Hours (Ah)

Now figure out how much gas is in 2 moles of H2, and 1 mole of O2.
According to the ideal gas law, one mole of gas has a volume of 24.446 Liters at 25 C, 1 Atm. So:

(3 moles) * 24.446 L/mol = 73.338 Liters of HHO gas

So this is as exciting as Faraday Efficiency gets, it deals with Amps and Moles only (not energy or voltage).

So 107.205 Amps over one hour will generate 73.338 Liters of H2 O2 gas at 100 percent efficiency. That's it, 100 percent Faraday Efficiency.



It gets interesting when you throw the Gibbs Energy of Formation of water into the equation.
At 25 C the Gibbs Free Energy (see post below) 237.18 Joules are required to convert 1 mole of H2O into 1 mole of H2 gas and a 1/2 mole of O2 gas.


So from the above calculations we know: 107.205 Amps continuous for 1 Hour will create 3 moles of H2/O2 gas, which has a volume of 73.338 Liters.

If we multiply the Gibbs Free Energy (energy used to create 1.5 moles of gas) by 2, we should get the actual energy required to make the above quantity, 3 moles of gas.

237.18 kJ * 2 = 474.36 kJ

Convert 474.36 kJ to Watts:

474360 Joules / 3600 seconds = 131.7666 Watts

Now put everything together (amps + voltage)

Since: Watts = Amps * Volts

131.7666 Watts = 107.204 Amp * Volts

Volts = 1.229 V Which is considered the minimum voltage for electrolysis to occur at 25C, 101.325 kPa. Which is also 100 percent Faraday, and "Gibbs" Efficiency.


Calculate W/LPM and MMW:

73.338 L/Hr / 60 Minutes = 1.223 LPM
Requires 131.76 Watts.

1 LPM / 1.223 LPM = .81766
.81766 * 131.76 Watts = 107.73

107.73 Watts will generate 1 LPM at 100 percent efficiency, or 9.282 MMW, at 25C 101.325 kPa.

How to calculate Efficiency with Gibbs Energy of Formation

The actual energy efficiency formula considering the Gibbs energy of formationshould look like this:

To convert one mole of H2O into 1 mol H2 gas, and .5 mol O2 gas, according to the Gibbs Free energy, requires 237.18 kJ at 25 C, at 1 atm. Total volume of H2 O2 gas is 36.669 Liters at 25 C (ideal gas law: 1.5 moles * 24.446 Liters)

Convert 237.18 kJ to Watts ... 237180 Joules / 3600 seconds in hour = 65.883 Watts

33.6 Liters of HHO gas over 60 minutes = 33.6/60 = .56 LPM (Liters per minute)

So 100 percent efficiency at 25 C and 101.325 kPa .. at 65.883 Watts you'll be making .61115 LPM

Or at 1 LPM you'll be consuming 107.801 Watts, and have a MMW of 9.27 (at 25C water and gas temp, and 101.325 kPa pressure). Temperature of gas and water, and pressure, will change these numbers.

Efficiency - Gibbs Energy of Formation

The Gibbs Free Energy of a system is the thermodynamic potential that measures the "useful" work obtainable from a thermodynamic system. It can be thought of as the maximum amount of non-expansion work that can be extracted from a closed system. This maximum can be obtained from a completely reversible process. The reverse is known as Gibbs Energy of Formation.

The thermodynamic equation is:

G = U + pV - TS

or

G = H - TS or ΔG = ΔH - TΔS

G = Gibbs Energy
U= Internal Energy = (The total kinetic and potential energy associated with the vibrational and electric energy of atoms within molecules. Or the energy in all chemical bonds and kinetic energy of free conduction band electrons.) Measured in Joules.
P = Pressure. Measured in Pascals
V = Volume. Measured in cubic meters.
T = Temperature. Measured in Kelvins.
S = Entropy = Measured in (Joules/Kelvins) = The total energy of a system that is unable to perform work. Function of a quantity of heat which shows the possibility of conversion of heat to work.
H = Enthalpy = A thermodynamic potential which can be used to calculate the "useful" work obtainable from a closed system, under constant temperature and pressure.
Δ = change in x value


To calculate the internal energy U, you first need to calculate the work (energy) required to decompress one mole of H2O liquid into H2 + (.5)O2 gas.

Note: 1 mole of water creates 1 mole of hydrogen gas and .5 mole of oxygen gas, total (1.5 mol)
Also, one mole of an ideal gas creates a volume of 22.4 L, so (1.5 mol)*(22.4 L) = 33.6 L (at 0C, 1 atmosphere), one mole of water turned into HHO gas has a volume of 33.6 L.

W = work to decompress one mole H2O
T= 298.15 K
P = standard atmospheric pressure = 101.325 kPa * 1000 = 101325 Pa
V (in cubic meters) = (1.5 moles) * (22.4 L) * 10e-3 = .336 m³

W = PΔV = (101325 Pa) * (.336 m³/mol) * (298.15 K / 273 K) = 3715 Joules = 3.72 kJ

Using the enthalpy equation H = U+PV you can get:

ΔU = ΔH - PΔV

ΔH = 285.83 kJ (this value can be looked up here: http://www.lsbu.ac.uk/water/data.html)

So,
ΔU = 285.83 kJ - 3.72 kJ = 282.1 kJ

And TΔS needs to be calculated:

ΔQ = TΔS = Heat Generated by the system to maintain a constant temperature.
Note: Electrolysis is an endothermic reaction which would otherwise decrease the temperature of a system.


ΔS = Entropy of a chemical reaction = (Entropy of 1 mole H2O liquid) - [ (Entropy of H2 gas) + (Entropy of .5 * O2 gas)]

Values can be looked up:

Quantity	H2O	H2	0.5 O2	Change
Enthalpy	-285.83 kJ	0	0	ΔH = 285.83 kJ
Entropy	69.91 J/K	130.68 J/K	0.5 x 205.14 J/K	TΔS = 48.7 kJ

ΔS = 69.91 J/K mol - [ 130.68 J/K mol + (.5 * 205.4 J/K mol) ] = -163.34 J/K mol

Now figure out what TΔS is.

T = 298K

298 K * -163.34 J/K = 48675 J = 48.7 kJ

So now we can use the first equation to figure out the Gibbs Free Energy:
ΔG = ΔH - TΔS

ΔG = 285.83 kJ - 48.7 kJ = 237.1 kJ

Efficiency - Heat generation and loss

While an electrolysis cell is producing H2 O2 gas, most of the time it is producing heat as well. You might be thinking the amount of heat lost in a cell isn't a useful measurement. But it can be useful for the following reasons.

Thermal measurements can be used to determine your cells overall efficiency. For example:

(H2 Gas Energy in Joules) + (Heat Generated in Joules)
/ (Electrical Energy in Joules)

Should give you an accurate measurement of your cells efficiency.

(Ambient air temperature) / (Cell temperature) may work as well to determine efficiency.

If you are approaching 100 percent efficiency, the heat output can be used to determine if any excess heat/energy is being generated (more energy out than into cell), this is a big consideration in cold fusion experiments.

Heat loss can occur the following ways:

1. Through the outside cell material (HDPE, gaskets, etc), or electrolyte exposed to ambient air.
2. From generated gas
3. Through a water circulation system (Bubblers, Pumps/Tanks, etc)

To calculate heat loss through cell construction material (or electrolyte exposed to air), several variables are required (once the cell warmed up).

- Air temperature
- Surface temperature of external cell material
- Surface area of each external cell material exposed to ambient air (plates, acrylic, electrolyte, etc)
- The Time which you are measuring your cell's efficiency and gas output.
- Thermal conductivity of materials




Q = thermal energy transfered in in Joules
t = Time (seconds)
h = thermal conductivity of material (W/(m·K)
A = surface area inside cell of conducting material (meters^2)
T(0) = outside temperature of material (C)
T(env) = temperature of environment (air)

NOTE: The Thermal conductivity (h) of a material are measured in: W/(m·K)
The English version: Btu·ft/(h·ft²·°F) can be converted with the following:
1 Btu·ft/(h·ft²·°F) = 1.730735 W/(m·K).

A list of Thermal conductivity measurements can be found here: http://en.wikipedia.org/wiki/List_of_thermal_conductivities
(these measurements were taken around room temperature. Temperature isn't a huge factor here imo)
A few examples:

Stainless Steel: 16.3 (W·m⁻¹·K⁻¹)
Acrylic: 0.2 (W·m⁻¹·K⁻¹)
Air: 0.025 (W·m⁻¹·K⁻¹)
High-Density Polymer: .33 (W·m⁻¹·K⁻¹)
Low-density Polymer: .16 (W·m⁻¹·K⁻¹)
UHMW-PE: 5.053 (W·m⁻¹·K⁻¹)

For example:

You have a 12" x 6" x 7" UHMW-PE box cell. 2.5833 sq ft
The outside temperature of the box is 110F
The outside air temperature is 75F.
And you want to determine the heat energy loss of the cell (running at constant temp) over 60 seconds, do the following:

Feet to meters conversion => 1 foot = 0.3048 meter
2.5833 (sq ft) * .3048 = .78738 m^2

Fahrenheit to Celsius conversion => subtract 32, and divide by 1.8
Outside box temperature = (110F - 32) / 1.8 = 43.33 C
Outside air temperature = (75F - 32_ / 1.8 = 23.88 C

Heat Loss Equation:

(Q/t) = h * A * (T.cell - T.air)

--> (Q/60) = 5.053 * .7873 * (43.33 - 23.88)
--> Q/60 = 77.37
--> Q = 4642 Joules

So the cell material is losing 4642 Joules every 60 seconds in heat energy. Or 77.43 Watts