egas-calculations.pdf

Electrolytic Gas

Peter E. W. Lowrie

Preamble

This paper sets out to describe why an experimental apparatus , namely:

● Mitzubishi Cyclone

engine(s) runs on electrolytic 1 gas. That is not so strange in itself as it is well known that

Hydrogen is a fuel*. The benefit of using electrolytic gas is that the Hydrogen has with it,

its own oxidiser – Oxygen. Already perfectly proportioned, no gas mixing is required amd

so complete comb ustion is accomplished without the need for additional air. Here's the

rub; the car engine uses a separate belt driven three phase marine alternator in a “Y”

winding with output rated at 150 Amps at 24 Volts which is fed into three electrolysis

cells, each cell gets a single phase. Beginning at 12 Volts the cells are heated partly by hot

exhaust gas and partly by way of voltage on the plates within acting as heating elements.

When the cells get up to temperature (about 75° C) the alternator tickle supply is reduced

to a range between 1.24 to 2.00 volts which then serves to increase electrolysis efficiency

in accordance with Faraday's Law(s) of electrolysis and thus efficiencies in the order of

97.5% are achieved the cells consu ming circa 600 Amps each the process becomes

endother mic and provides gas more than sufficient to fuel the engine.

Thus the engine generates its own fuel (and oxidiser) with ample power to spare. Most

argue that this is an impossible situation; at best the engine becomes a dynamic brake

and at worst it just won't work. The explanation being that you can't get more energy out

than what you put in and in citing various texts, at first glance appears quite correct. The

fact of the matter – as this paper will prove – is that the texts are either wrong or fail to

supply all of the information.

Electroche mical -V- Physical reactions

There are two bonds on the molecule – one for each Hydrogen atom (of course).

Therefore for one molecule of H 2 O 870kj will break apart the water molecule and the

equivalent energy, it is said, will initiate the gas recombination. Gas and oxidiser atoms

never being in isolation, there being billions of them in n volume they undergo a chain

reaction until all available atoms are recombined releasing enorm ous energy in quick

time. The plasma speed is 3.9kM/sec. The ash is water. Few texts referenced to date

accounts for the energy release during recombinatio n, more information about this

apparent dichoto my will be discussed.

The heat flame (plasma) total energy is calculated by multiplying the respective atoms'

ionisation energies for the process. The ionisation energies 2 of H = 1312.06kj and O =

1313.95kj ie 1 , (3388.33kj ie 2 , 5300.51kj ie 3 ) upon combustion the (outer in Oxygen) electron

orbits interact, merge and settle down to stable orbits:

Also called: Water Gas, Rhodes Gas, Browns Gas.

www.ktf- split.hr/ periodni /e n

Page: 1 of 16

H ie 1 + H ie 1 O ie 1 = 3,447,962.47kj

(1)

Where ie is Ionisation Energy and n the energy layer (three in Oxygen) and only one in

Hydrogen.

In the discussion of thermo dy na mics relating to the subject matter the question of heat

is a very important one, there being two types of heat. On the one hand is heat due to

molecular motion and on the other is the heat of photons goming and going during

reactions

In order of energy strength...The formation of Hydrogen is a nuclear reaction there being

two well known modes of it's coming into existance; primordial Hydrogen and neutron

decay. Neither of which are relevent to this discussion. Oxidisation (burning) of

Hydrogen is a physical reaction of the middle energy order. And electrolytic

decomp osition is a[n] [electro]chemical reaction. The notion of water being a fuel rests

in Electrolysis, of the low energy order – a chemical reaction and Combustion – a

physical reaction of the middle energy order being vastly disproportionate. This can be

shown to be the case; let  = 870kj be the total energy consu me d in splitting H 2 O and 

be the oxidisation expenditure in equation (1) of [combustion] energy.

 /  = 3963.18

(2)

This does not describe some “hidden” energy, nor do special- pleading of over- unity claims

have any relevance . The calculation reflects the starting position of the two gases. Even if

the bond strength is subtracted from both sides:

( - ) /  = 3962.18

(3)

...it is apparent that the combustion product is highly energetic. So which is it? The

chemical equation given:

H 2 O (l) +435kj --> H + H + O (g) +435kj --> H 2 O (l)

(5)

is clearly mistaken because, a) only half the required energy is given and b) there is a

continuing addition of energy on both sides and appears not to factor- in any energy

release. In any case equation (5) plugs in the correct values to the formula provided in

the text (footnote #2) and equation (6) enlarges to include the energy release.

H 2 O (l) + 435kj --> HO + H + 435kj --> H + H + O (g) [+870kj --> H 2 O (l) ]

(4)

On the one hand energy is expended to break the bonds and on the other hand energy

must be expended to remake them and clearly this imbalanced situation leaves a lot to

be desired. What actually happens is that 870kj/mole -1 is expended to begin the

recombination process and 3962kj/mole -1 is liberated as a net gain. The reason for this is

that the ionisation process in electrolysis is [electro]chemical, the initiating energy is

[electro]che mical however the resultant ionisation (comb ustion) process is physical.

Page: 2 of 16

H+H (g) +O (g) +870kj --> (H 2 O (ion) -3962kj) -->H 2 O (l)

(5)

...and then multiply this reaction by the gas volume denoted by Avogadro's nu mber.

The author of this paper is not in isolation insofar as these apparently ano malous results

are concerned, others have addressed the issue. 3 " The smallest amount of energy

needed to electrolyse one mole of water is 65.3 Wh at 25 degrees Celsius (77

degrees F). When the Hydrogen and Oxygen are recombined into water during

combustion 79.3 Wh of energy is released. 14 Wh more energy is released in

burning Hydrogen and Oxygen than is required to split water. This excess must be

absorbed from the surrounding media(environment) in the form of heat during

electrolysis." [...] "At 25 degrees celcius, for voltages of 1.23 to 1.47 V, the

electrolysis reaction ABSORBS HEAT. At over 1.47 V at 25 degrees celcius, the

reaction gives off heat. "

The electrolysis cell voltage, overvoltage may be 1.3V therefore 1.47V + 1.3V = 2.77V is the

voltage supplied.

One mole of water weighs 18 grams.

1000 Grams water = 1 Liter

1000 grams / 18grams = 55.55moles

Therefore 1 Liter H 2 O produces 55.55 Moles of Hydrogen and 27.775 Moles of Oxygen.

870Kj H 2 O produces as above which is then equal to 48328.50kj per Liter H 2 O.

Conversion to kWhr divide by 3600 ( or x by .0002778) = 3.658 kWhr per Liter H 2 O

1 Mole of Gas = 24.450 liters of gas at room temperature, “T” and atmosp heric pressure

“P”. 55.55 x 24.450 = 1,358.3 liters of Hydrogen (and 679.15 liters of Oxygen) from 1 Liter

of H 2 O.

Gibbs Free Energy

Do the rules comprising Gibbs Free Energy reconcile with the energy shown in equation

(1 - 7)? No. Provided there are energy co- efficients involved does GFE get turned on it's

head? The Wien effect does it to Ohms Law. Why not here as well? Some reactions are

spontaineo us because they are exother mic  H <0, and others are spontaineous because

they are entropic  S >0. The combustion of electrolytic gas is exother mic but the

question of entropy is a moot point. Two disordered gases ignited combine to make

water which is not an increasing entropy proposition, therefore in our calculation  S< 1.

Provided the GFE of a system at the end of the reaction is defined as the enthalpy of the

system less the product of the temperature multiplied by it's entropy.

H = Enthalpy

S = Entropy

G =

H -

( TS )

(x)

"FUEL FROM WATER" M. A.Peavey. Merit, Inc. LCCCN 88- 188956 ISBN 0-945516- 04- 5 Page 22.

Page: 3 of 16

The usual standard- state free energy of this system  G ˚ cannot be applied herein due to

the rapidly changing temperatures involved in the reaction and the Enthalpy is:

 G =

H g + H g + O g (1S) = -318.402 kcal/mol

=  H

H g + H g + O g (1S) = -151.81

= S x 6000˚

=  H - -910,860kcal/ m ol

(x)

Following combustion of the gases at 6000º C the resulting oxide cools until it reaches

ambient temperature, the molecules return to their lowest energy and the amou n t of

energy released persuant to GFE is close to the energy taken by the other route in earlier

equations, in other words – a lot more than the Hydrogen bond strength.

Alas it is not that simple though. One cannot say that the “heat” of electrolysis is

equivalent to this heat of comb ustion their being crossovers of photonic and molecular

heat during the exchanges.

Faraday's Laws 4

(8)

where:

V = volume of the gas [L],

R = ideal gas constant = 0.0820577 L*atm/( mol*K), = current [A],

T = temperature [°K],

t = time [s],

F = Faraday’s constant = 96485.31 As/mol,

p = ambient pressure [atm],

z = num ber of excess electrons (2 for H 2 , 4 for O 2 ).

Assume that STP (Standard Temperature and Pressure) conditions and the electrolyzer

runs at one Amp for one hour:

T = 0

C = 273.15° K

p = 1 atm

t = 3600 seconds

Courtesy Tero Ranta

Page: 4 of 16

I = 1 Amp

Total Oxygen- Hydrogen volume is Hydrogen volume + Oxygen volume:

0820577

273

3600

0820577

273

3600

418151

209075

627226

96485

This correspo n d s to about 0.627 liters per hour per Amp or 1.595Ah/l per cell.

If for example you have 7 cells in series and put 11A through the electrolyzer, according

to Faraday’s Law you would produce:

0.627l/Ah*11A*7 = ~48.3 Liters per hour

(10)

at STP conditions.

Note, however, that this applies only at a certain temperature (0°C) and pressure (1 atm).

The produced gas volume will scale with ratio of temperatures in Kelvins (higher

temperature = higher volume) and inversely with the ratio of pressures (lower pressure =

higher volume).

If at 0°C (273.15°K) the production rate is 0.627 l/Ah, then at 25°C:

273.15°K+25°K=298.15K

(11)

the production rate is:

298.15/273.15 = ~109%

(12)

larger or about 0.685 l / Ah. With 7 cells and 11A this would be 52.5 Liters per hour.

On the other hand is the output gas has a temperature of 40°C while it is being measured

and the ambient pressure of 0.75 atm (about 1.5km elevation above sea level), the

electrolyzer that produces 48.3 liters per hour at STP will produce:

313.15°K/273.15°K*1atm/0.75at m*48.3l/hr = 73.8l/hr

(13)

So even though the volume of the gas is larger at higher temperature and lower pressure,

the energy contained in the gas or the energy required to electrolyze it is the same. If you

produce the gas at 40C and 0.75ATM and bring it to 0C and 1ATM, the volume will

reduce by about 35%. Thus it is very important to include the pressure and temperature

in calculations.

What may not be immediately clear is that different experimenters report various results

that are prima facae out of step with cherished laws. Some experimenters report

efficiencies that appear in excess of Faraday's Law yet others no matter how hard they

try attain only low levels of efficiency. Are the over unity claims due to errors in metho d

Page: 5 of 16

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