Practical Guige to Free Energy Devices

Against this ideal standard of efficiency of expected gas production, we must measure the actual amount of gas
produced under: (1) Standard conditions as defined above, and (2) 0.1 Ws power applied over 30 minutes. In
our experiments, the mean amount of H 2 and (1/2)O 2 produced, as measured on precision calibrated GC, and MS
machines in two different laboratories, where SE is +/- 10%, is,
Measured Mean = 10.80 cc H 2 (g)
Measured Mean = 5.40 cc (1/2) cc (1/2)O 2 (g)
Total Mean = 16.20 cc H 2 O (g)
The ratio, n, between the ideal yield, and measured yield,
Measured H 2 (g) / Ideal H 2 (g) = 10.80 cc / 11.33 cc = 91.30%
6. Alternative Method for Calculating Efficiency Based on the Faraday Law of Electrochemistry
This method is based on the number of electrons that must be removed, or added to decompose, or form one
mole of, a substance of valence one. In water (H 2 O), one mole has the following weight:
H = 1.008 gr /mol
H = 1.008 gr /mol
O = 15.999 gr/mol
Thus, 1 mol H 2 O = 18.015 gr/mol
For a univalent substance, one gram/mole contains 6.022 x 10-23 electrons = N = Avogadro's Number. If the
substance is divalent, trivalent, etc., N is multiplied by the number of the valence. Water is generally considered
to be of valence two.
At standard temperature and pressure (“STP”) one mole of a substance contains 22.414 cc, where Standard
temperature is 273.15 O K = 0 O C = T . Standard Pressure (one atmosphere) = 760 mm Hg = P.
One Faraday (“F”) is 96,485 Coulombs per mole (univalent).
One Coulomb (“C”) is defined as:
1 N / 1 F = 6.122 x 10 23 Electrons / 96,485 C = one C
The flow of one C/second = one Ampere.
One C x one volt = one Joule second (Js).
One Ampere per second @ one volt = one Watt = one Joule.
In alternating current, when amps (I) and Volts (E) are expressed in root mean squares (rms), their product is
Power in watts.
P = IE watts (Watts = Amps x Volts).
With these basic definitions we can now calculate efficiency of electrolysis of water by the method of Faraday’s
electrochemistry.
The two-electron model of water requires 2 moles of electrons for electrolysis (2 x 6.022 x 10 23 ), or two Faraday
quantities (2 x 96,485 = 192,970 Coulombs).
The amount of gas produced will be:
H 2 = 22,414 cc /mol at STP
(1/2)O 2 = 11,207 cc / mol at STP
Gases = 33.621 cc / mol H 2 O (g)
The number of coulombs required to produce one cc of gases by electrolysis of water:
193,970 C / 33621 C = 5.739567 C per cc gases.
A - 698