It was observed that the absolute threshold of hearing could be
obtained at **0.16Mw (rms),** and by calculation that there
was an amplitude of displacement of the eardrum of the order of
**10 to- eleventh power** and a corresponding amplitude of
the cochlear basilar membrane of** 10 to power - 13 meter**.
Corollary to this finding. I was able to achieve the absolute
reversible threshold of electrolysis at a power level of **0.16
mW (rms).**

By carrying out new calculations I was able to show that the water
was being vibrated with a displacement of the order of **1 A
= 10 to the power of -10 meters**. This displacement is of the
order of the diameter of the hydrogen atom. Thus it is possible
that the acoustic phonons generated by audio side bands of the
carrier are able to vibrate particle structures within the unit
water tetrahedron.

* * * * * * * *

We now turn to the measurement problem with respect to efficiency of electrolysis.

There are four means that can be used to measure the reactant product of water electrolysis .

For simple volume measurements one can use a precision nitro-meter such as the Pregl type.

For both volume and quantitative analysis one can use the gas chromatography with thermal conductivity detector.

For a continuous flow analysis of both volume and gas species the mass spectrometer is very useful.

Fore pure thermodynamic measurements the calorimeter is useful.

In our measurements , all four methods were examined, and it was found that the mass spectrometer gave the

most flexibility and the greatest precision . In the next section we will describe our measurement using the mass spectrometer.

**4 Methodology for the evaluation of the efficiency of water
decomposition by means of alternating current electrolysis.**

All systems used today for the electrolysis of water into hydrogen
as fuel, and oxygen as oxidant apply direct current to a strong
electrolyte solution. These systems range in efficiency from **50%
to 71%.** The calculation of energy efficiency in electrolysis
is defined as follows:

*" The energy efficiency is the ration of the energy released
from the electrolysis products formed ( when they are subsequently
used) to the energy required to effect electrolysis." *

*(1)*

The energy released by the exergonic process under standard conditions.

**H2(g)+ (1/2)O2(g) ------> H20(1) - 3 02.375 Btu which (1)**

**is 68.315 Kcal/mol. or, 286,021 Joules/mol, and is numerically
equal to the enthalphy charge (triangle H ) **

for the indicated process.

On the other hand the minimum energy ( or useful work input) required
at constant temperature and pressure for electrolysis equals the
Gibbs free energy change **( triangle G) 2**

Penner shows (Op.Cit.) that there is a basic relation derivable from the first and second laws of thermodynamics for isothermal changes which shows that

**triangle G = triangle H - T triangle S**

where triangle S represents the entropy change for the chemical reaction and T is the absolute temperature.

The Gibbs free energy change ( triangle G) is also related to the voltage (e) required to implement electrolysis by Faraday's equation,

**e=( triangle G/23.06 n ) volts **

Where triangle G is in Kcal/mol, and n is the number of electrons( or equivalents ) per mole of water electrolysed and has the numerical value 2 in the equation, ( endergonic process)

Therefore, according to equation (2) at atmospheric pressure,
and 300 degrees K , **triangle H=68.315 kcal/mol or h2O (1),
and triangle G = 56.620 kcal / mol of H2O (1) = 236,954 J/mol
H20 (1)** for the electrolysis of liquid water.

In view of these thermodynamic parameters for the electrolysis of water into gases, hydrogen and oxygen, we can establish by Eq.(2) numeric values where,

**triangle G (J/mol)**

**n= ------------------------- =<1**

**triangle G e (J/mol)**

where triangle** G e** is the electrical energy input to **h2O
(1)** in Joules,

and triangle G is the Gibbs free energy of H2O (1) . The conversion between the two quantities is One Watt second (Ws) = one Joule.

Or, in terms of gas volume, as hydrogen ,produced and measured ,

**measured H2 (c.c)**

**n= ---------------------- =<1**

**Ideal h2 (c.c)**

in accordance with these general principles we present the methodology
followed in evaluating the electrolytic of alternating current
on **H2O (1)** in producing the gases, hydrogen and oxygen.
No attempt has been made to utilize these gases according to the
process of** Eq(1).**

It is to be noted that the process

yields only **57.796 kcal /mol. Eq (7)** shows that per mole
of gases water formed at **300 degrees K**, the heat released
is reduced from the** 68.315 kcal/mol at Eq. (1)** by the molar
heat of evaporation of water at **300 degrees K (10.5 kcal)**
and the overall heat release is **57.796 kcal/mol if H2O (G**)
is formed at **300 degrees K**

In the following sections we describe the new method of electrolysis by means of alternating current, and the exact method and means used to measure the endergonic process of Eq.(4) and the governing Eq.(2) and Eq. (15)

*1. Op.Cit. Ref. (1) page 3. page 299ff.*

In order to properly couple **Component II** to a mass spectrometer
one requires a special housing around **Component II** that
it will capture the gases produced and permit these to to be drawn
under low vacuum into the mass spectometer. Therefore a stainless
steel and glass chamber was built to contain **Component II,**and
provision made to coiuple it directly through a CO2 watertrap
to ther mass spectrometer and **Component IV** were purged
with helum evacuated for a two hour period before any gas sample
were drawn.In this way contanination was minimized, The definitive
measurement were done at Gollob Analytical Services,Inc in Berkeley
Heights ,new Jersey

We now describe the use of Component I and how it energy output
to **Component II** is measured. The energy output of **Component
I** is an amplitude modulated alternating current looking into
a highly non-linear load, i.e ,The water solutiuon **Component
I** is so designed that at peak load it is in resonance across
the sysytem - **Components I,II,** and** III** , and the
vector diagrams show that the capacitive reactance,and the inductance
reactance are almost exactly **180 deg out of phase**,so that
the net power output is reactive ( the dissipative power is very
small).This design insures **minimum power lossses across the
entire output system.**

In the experiments to be described,the entire emphasis is placed
on achieving the maximum gas yield (credit) in exchange for the
minimum applied electrical energy, The most precise way to measure
the applied energy from **Component I** to **Component II**
and **Component III** is to measure The power, P, in watts,W.
Ideally this should be done with a precision wattmeter. But since
we were interested in following the voltage and current separately,it
was decided not to use the watt meter. Separate meters were used
to continuously monitor the current and the volts,

This is done by precision measurement of the volts across **Component
III** as root mean square (**rms**) volts;and the current
flowing in the system as rms amperes.Precisely calibrated instruments
were used to take these two measurements. A typical set of experiments
[ using water in the form of **0.9** saline solution **0.1540
molar**] to obtain high efficiency hydrolysis gave the following
results.

**rms Current = I = 25mA to 38 mA (0.025 A TO 0.038 A.)**

**rms Volts =E = 4 Volts to 2.6 Volts**

The resultant ration between current and voltage is dependent
on many factors such as the **gap distance** between the center
and ring electrodes,dielectric properties of the water,conductivity
properties of the water,equilibrium states,isothermal conditions,materials
used,and even the presuure of clathrates. The above current and
voltage values reflect the net effect of various combinations
of such parameters.When one takes the product of rms current,and
rms volts one has a measure of the power, P in watts

**P= I x E = 25 mA x 4.0 volts =100 mW (0.1 W)**

**and P = I x E =38uA x 2.6 volts = 98.8 mW (0.0988 W)**

At these power levels (with load), the **resonant frequency of
the system is 600 Hz (plus or minus 5 Hz)** as measured on a
precision frequency counter. The wave form was monitored for harmonic
content on an oscilloscope, and the nuclear magnetic relaxation
cycle was monitored on an X-Y plotting oscilloscope in order to
maintain the proper hysteresis loop figure. *All experiments
were run so that the power iin watts, aplied through Components
I,II, and III ranged between 98.8 mW to 100mW.*

Since by the International System of Units -1971 (ST),

**One Watt-second (Ws) is exactly equal to one Joule (J),**our
measurements of efficiency used these two yardsticks (1 Ws = 1J)
from the debit side of the measurement.

The energy output of the system is,of course, the two gases,Hydrogen(H2) and Oxygen (1/2O), and this credit side was measured in two laboratories,on two kinds of calibrated Instruments, namely

**Gas Chromatography Machine, and, Mass Spectrometer Machine.**

The volume of gases **H2 and (1/2)O2** was measured as produced
under standard conditions of temperature and pressure in unit
time,i.e, in cubic centimeters per minute (cc/nin), as well as
the possibility contaminating gases,such as air oxygen, nitrogen
and argo, carbon momoxide, carbon dioxide, water vapor etc.

The electrical and gas measurements were reduced to the common
denominator of Joules of energy so that the efficiency accounting
could all be handled in one currency. We now present the averaged
results from many experiments. The standard error between different
samples,machines,and locations is at **minus or plus 10 percent**
and we only use the mean for all the following calculations.

II. Thermodynamic Efficiency for the Endergonic Decomposition
of liquid water (salininized) to gases under standard atmosphere
(**754 to 750 m.m. Hg**)' and standard Isothermal conditions
**@ 25 deg C = 77 deg F = 298.16 deg K,** according to the
following reaction.

**H20 (1) _> h2(g) + (1/2)O2(1) + triangle G 56.620 Kcal /mole**
(10)

As already described triangle G is the Gibbs function.

we convert Kcal to our common currency of Joules by the formula

**One Calorie =4.1868 Joules**

**Triangle G = 56.620 Kcal x 4.1868 J = 236,954/J/mol of H2)
where**

1 mole =18 gms.

Triangle G2 the electrical energy required to yield an equivalent amount of energy from H2O in the form of gases H2 and (1/2)O2.

To simplify our calcualtion we wish to find out how much energy is required to produce the 1.0 c.c. of H2O as the gases H2 and (1/2)02.

There are (under standard conditions ) **22,400 c.c. - V of gas
in one mole of H2O.** Therefore

**triangle G 236,954 j**

**-------- = --------- = 10.5783 j/cc**

**V 22,400 cc**

We now calculate how much electrical energy is required to *liberate
1.0 cc of the H2O gases (where H2 = 0.666 parts, and (1/2)O2 =
0.333 parts by volume )* from liquid water. Since **P= 1 Ws=
1 Joule , and V= 1.0 cc of gas = 10P:5783 Joules**, then

**PV =1 Js x 10.5783 J = 10.5783 Js or**

**=10.5783 Ws**

Since our experiments were run at **100 mW( 0.1 W)** applied
to the water sample in Component II,III, for **30 minutes ,**we
wish to calculate the ideal (**100% **efficient) gas production
at this total applied poer level. This is,** 0.1Ws x 60 sec x
30 min =180,00 Joules ( for 30 min )**

The total gas production at Ideal **100%** efficiency is **180.00
J/10.5783 J/cc =17.01 cc H2O (g)**

We further wish to calculate how much hydrogen is present in the
**17.01 cc H2O (g).**

**17.01 cc H2O (g) x 0.666 H2 (g) = 11.329 cc H2(g)**

**17.01 cc H2O (g) x 0.333(1/2) C2 (g) = 5.681 cc (1/2)O2 (g)**

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. 2) 0.1 Ws power applied
over 30 minutes. In our experiments , the mean amount of H2 and
(1/2)O2 produced ,as measured on precision calibrated GC, and
MS machines in two different Laboratories, wher S.E is plus or
minus 10%*

**Measured Mean = 10.80 cc H2 (g)**

**Measured Mean - 5.40 cc (1/2) cc (1/2) O2 (g)**

**Total Mean 16.20 cc H2O (g)**

**The ratio N1 between the ideal yield, and measured yield,**

**Measured H2 (g) 10.80 cc**

**N1 = --------------------- ----------- 91.30%**

**Ideal H2 (g) 11.33 cc **

This method is based on the number of electrons that must be removed, or added decompose ,or form one mole of, a substance of valence one. In water H2O,one mole has the following weight:

**H= 1.008 gm /mol**

**H= 1.008 gm /mol**

**O = 15.999 gm/mol**

**Thus, 1 mol H2O = 18.015 gm/mol**

*For a unvalent substance one gram mole contains 6.022 x 10
to 23 third power 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 (STP) temperature and pressure one mole of a substance
contains 22.414 c.c., Where Standard temperature is 273.15 deg
k = ) 0 deg C. =T . Standard Pressure is One atmosphere =760 mm.Hg
=P*

*One Faraday (1F) is 96,485 Coulombs per mole (univalent).*

One Coulomb is defined as:

**1N 6.122 x 10 to the 23 power Electrons**

**------- = --------------------------------------------- One
Coulomb**

**1F 96.485 Coulombs**

*the flow of one coulomb per second = One Ampere*

**One Coulomb x One volt = one Joule second**

In alternating current when amps (I) and Volts (E) are expressed iin root mean squares (rms) their product is Power.

**P= IE watts**

With these basic definitions we can now calculate efficiency of electrolysis of water by the method of Faraday is electrochemistry.

The two-electron model of water requires 2 moles of electrons
for electrolysis **(2x6..022 x 10 to the 23** power),or two
Faraday quantities **(2x 96,485 = 192,970 c.).**

The amount of gas produced will be:

**H2 = 22,414 c.c. /mol at STP**

**1/2O2 = 11,207 c.c. / mol at STP**

**Gases = 33.621 c.c. /mol H2O (g)**

The number of coulombs required to produce one c.c of gases by electrolysis of water

**193,970 C.**

**-------------------- = 5.739567 Coulombs per c.c. gases**

**33,621 c.c **

Then ** 5,739 Coul /cc /sec =5.739 amp/sec/cc.**

How many cc of total gases will be produced by 1 A/sec?

**0.1742291709 cc.**

How many cc of total gases will be produced by 1 A/min ?

What does this represent as the gases H2 and O2 ?`

**1/2C2 =3.136438721 cc / Amp/min.**

**H2 =6.2728 cc/Amp /min**

We can now develop a Table for values of current used in some of our experiments,and disregarding the voltage as is done conventionally.

**I calculations for 100 mA per minute**

**Total Gases 1.04537 cc/mim**

**H2 = .6968 cc /min**

**1/2O2 = .3484 cc/min**

**30 min. H2 =20.9054 cc/ 30 minutes**

*II Calculations for 38 mA per minute*

**Total Gases = 0.3972 cc/ 30 minutes**

**H2 = 0.2645 cc/ min**

**1/2O2 = 0.1323 cc/min**

**30 min. H2 7.9369 cc/min**

*III Calculations for 25mA per minute*

**30 min. H2 = 5.2263 cc/ minute**

**Figure 6 and 7** show two of the many energy production systems
that may be configured to include renewable sources and the present
electrolysis technique. **Figure 6** shows a prposed photovoltaic
powered system using a fuel cell as the primary battery. Assuming
optimum operating conditions using .**25 watt seconds of energy
from the photovoltaic array would enable .15 watt seconds to be
load.**

**Figure 7** depicts several renewable sources operating in
conjuncction with the electrolysis device to provide motive power
for an automobile.

**Text Supplied and original diagrams From Fred Epps Thanks
**

**Some drawings have been reproduced by Geoff Egel**

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