It is hardly necessary to weigh the value of the World Energy bank account for any sophisticated person ,these days
It is grim. The oil reserves will dwindle away in a score of years or so, and the coal reserves will be gone in some twelve score years (1)
This is not to say that the outlook is hopeless. There is an abundance of alternative energy sources, but the economics of development and exploitation present an enormous short term strain on the world political and banking resources.
That it will be non toxic both in its short term, and in its long term, effects. That it will be renewable in that it can be used over and over again; That it will be safe to handle, and present minimal storage and transportation problems and costs.
And finally that it will be universally available anywhere on earth.
what is this magical fuel, and why is it not being used?
It can be used in its salt water form. It can be used in its brackish form. It can be used in its snow and ice form. When such water is decomposed by electrolytic fission into hydrogen and oxygen gases- it becomes a high energy fuel with three times the energy output which is available from an equivalent weight of high grade gasoline.
[1 The interested reader should refer to the special issue of National Geographic,ENERGY,February 1981.]
Please note because I am unable to print symbols when the word triangle is mentioned in a formula it is for the symbol triangle and when the words to the power of is mentioned it is a exponent power value. (web Ed)
The answer is simple .Because it costs too much with existing technology to convert water into gases hydrogen and oxygen. The basic cycle of using water for fuel is described in the following two equations, familiar to every high school student of Chemistry:
electrolysis
HO -------- +249.68 Btu h2 + (1/2)O2 per mole of water (1)
triangle G (1 mole = 18 gms. )
This means that 302.375 Btu of energy (heat or electricity) will be released when the gases, hydrogen and oxygen, combine. The end product (the exhaust ) from this reaction is water.
Note that more energy (under ideal conditions ) is released. from combining the gases than is used to free them from water.
It is know that under ideal conditions it is possible to get some 20% more energy out of reaction (2) above, then it takes to produce the gases of reaction (1) above. Therefore ,if reaction (1) could be carried out at 100% efficiency, the release of energy from reaction (2) in an optimally efficient engine( such as a low temperature fuel cell),there would be a net energy profit which would make the use of water as a fuel an economically feasible source of energy .
The cost of producing hydrogen is directly related to the cost of producing electricity. Hydrogen as produced today is generally a by product of off- peak-hour electrical production in either nuclear or hydro electrical plants. The electricity thus produced is the cheapest way of making hydrogen. We can compare the cost of production of electricity and the cost of producing hydrogen. The following is based on federal Power commission ,and American Gas Association Figures of 1970 and on a 1973 price evaluation (just before OPEC oil price escalation.)
Relative prices in dollars per 10000000 Btu . See Appendix 1 for definition of British Thermal units (a)
@9.1 mils/kWh
COST COMPONENT ELECTRICITY ELECTROLYTICALLY
PRODUCED HYDROGEN
Production 2.67 (b) 2.95 to 3.23 (b)
Transmission 0.61 0.52 ©
Distribution 1.61 0.34
Total Cost $4.89 $3.81 to $4.09
Table 1.
if we compare only the unit cost of production of electricity vs Hydrogen from the above table:
10000000 Btu H2 $3.23---------------------- = = 1.209 ,or 20.9% higher cost ,H2
10000000 Btu El. $2.67
It must also be noted that the price of natural gas is much cheaper than either electricity or hydrogen, but because of the price fluctuations due to recent de-regulation of gas. It is not possible to present a realistic figure. In the opinion of Penner (Op. Cit.),
if the hydrogen production cost component of its total cost could be reduced three fold, it would become a viable alternate energy source.
In order to achieve such a three-fold reduction in production costs, several major breakthroughs would have to occur.
By a combination of the breakthroughs outlined above ,1),2), and 3) utilized in a highly efficient engine to do work, it is possible to achieve a 15% to 20% surplus of energy return over energy input, theoretically.
It is of interest to record that a new invention is now being developed to realise the above outlined goal of cheap, clean renewable and high grade energy.
which produces hydrogen as fuel, and oxygen as oxidant, from ordinary or from sea water, eliminating the cost and hazard of liquefaction ,storage, transmission, and distribution. The saving of this aspect of the invention alone reduces the total cost of hydrogen by about 25%. This Thermodynamic Device is based on a new discovery- The efficient electrolytic fission of water into hydrogen gas and oxygen gas by the use of low frequency alternating currents as opposed to the conventual use of direct current, or ultra-high frequency current today. Such gas production from water by electrolytic fission approaches 100% efficiency under laboratory conditions and measurements.
This Thermodynamic Device has already been tested at ambient pressures and temperatures from sea level to an altitude of 10,000 feet above sea level without any loss of its peak efficiency. The device produces two types of gas bubbles; one type of bubble contains hydrogen gas; the other type contains oxygen gas. The two gases are thereafter easily separable by passive membrane filters to yield pure hydrogen gas, and pure oxygen gas.
The separate gases are now ready to be combined in a chemical fusion with a small activation energy such as That from a catalyst or an electrical spark, and yield energy in the form of heat, or steam, or electricity- as needed .When the energy is released by the chemical fusion of hydrogen and oxygen, the exhaust product is clean water. The water exhaust can be released into nature .and then renewed in its energy content by natural processes of evaporation, solar irradiation in cloud form, an subsequent precipitation as rain on land or sea ,and then collected again as a fuel source. Or, the exhaust water can have its energy content pumped up by artificial processes such as through solar energy acting through photocells. Hence, the exhaust product is both clean and renewable. The fuel hydrogen, and the oxidant oxygen, can be used in any form of heat engine as an energy source- If economy is not an important factor. But the practical considerations of maximum efficiency dictate that a low temperature FUEL CELL with its direct chemical fusion conversion from gases to electricity offers the greatest economy and efficiency from small power plants (less than 5 kilowatts).
For large power plants ,steam and gas turbines are the ideal heat engines for economy and efficiency. With the proper engineering effort ,automobiles could be converted rather easily to use water as the main fuel source.
2 AN ELEMENTARY INTRODUCTION TO THE DESIGN AND OPERATION OF THE THERMODYNAMIC DEVICE TO ELECTROLYSE WATER WITH A.C.
The Thermodynamic Device (TD) is made up of three principal components: AN electrical function generator, Component I, that energizes a water cell, the TD ,Component II and Component III , a weak electrolyte.
COMPONENT I. The Electrical Function Generator See Fig 1.
This electronic device has a complex alternating current output consisting of an audio frequency (range 20 to 200 Hz) amplitude modulation of a carrier wave (range - 200 to 100,000 Hz)
The output is connected by two wires to Component II at the center electrode, and at the ring electrode. See fig1. The impedance of this output signal is continuously being matched to the load which is the water solution in Component II.
The TD is fabricated of metals and ceramic in the geometric form of a coaxial cylinder made up of a centered hollow tubular electrode which is surrounded by a larger tubular steel cylinder.
These two electrodes comprise the coaxial electrode system. energised by Component I. The space between the two electrodes is, properly speaking, Component III which contains the water solution to be electrolysed. The center hollow tubular electrode carries water into the cell, and is further separated from the outer cylindrical electrode by a porous ceramic vitreous material. The space between the two electrodes contains two lengths of tubular Pyrex glass, shown in Figures 2 and 3. the metal electrode surface in contact with the water solution are coated with a nickel alloy.
This consists of the water solution ,the two glass tubes, and the geometry of the containing wall of Component II. It is the true load for Component I, and its electrode of Component II.
The Component III water solution is more properly speaking ,ideally a 0.1540 Molal Sodium Chloride solution, and such is a weak electrolyte. In figure 4 we show the hypothetical tetrahedral structure of water molecule, probably in the form in which the complex electromagnetic waves of Component I to see it. The center of mass of this tetrahedral form is the oxygen atom. The geometric arrangement of the p electrons of oxygen probably determine the vectors I(L1) and I (L2) and I (H1) and I(H2) which in turn probably determine the tetrahedral architecture of the water molecule. The p electron configuration of oxygen is shown in figure 5 . reference to Figure 4 shows that the diagonal of the right side of the cube has at its corner terminations the positive charge hydrogen (h+) atoms; and that the left side of the cube diagonal has at its corners the lone pair electrons,(e). It is to be further noted that this diagonal pair has an orthonormal relationship.
we will now portray the complex electromagnetic wave as the tetrahedral water molecule sees it. The first effect felt by the water molecule is in the protons of the vectors, I (H1) and I (H2). These protons feel the 3 second cycling of the amplitude of the carrier frequency and its associated side bands as generated by Component I. This sets up a rotation moment of the proton magnetic moment which one can clearly see on the X Y plot of an oscilloscope, as an hysteresis loop figure. However, it is noted that this hysteresis loop does not appear in the liquid water sample until all the parameters of the three components have been adjusted to the configuration which is the novel basis of this device. The hysteresis loop gives us a vivid portrayal of the nuclear magnetic relaxation cycle of the proton in water.
The next effect felt by the water molecule is the Component I carrier resonant frequency, Fo. At the peak efficiency for electrolysis the value of Fo is 600 Hz plus or minus 5 Hz.
This resonance however is achieved through control of two other factors. The first is the molal concentration of salt in the water . this is controlled by measuring the conductivity of the water through the built in current meter of Component I. There is maintained an idea ratio of current to voltage I/E =0.01870 which is an index to the optimum salt concentration of 0.1540 Molal.
The second fact/or which helps to hold the resonant which helps to hold the resonant frequency at 600 Hz is the gap distance of Y, between the centre electrode, and the ring electrode of Component II.
This gap distance will vary depending on the size scale of Component II, but again the current flow, I, is used to set it to the optimal distance when the voltage reads between 2.30 (rms) volts, at resonance Fo, and at molal concentration ,0.1540. The molal concentration of the water is thus seen to represent the electric term of the water molecule and hence its conductivity.
The amplitude modulation of the carrier gives rise to side bands in the power spectrum of the carrier frequency distribution. It is these side bands which give rise to an acoustic vibration of the liquid water, and it is believed to the tetrahedral water-molecule. The importance of the phonon effect.- The acoustic vibration of water in electrolysis was discovered in a round -about way. research work with Component I had earlier established that it could be used for the electro stimulation of hearing in humans. When the output of Component I is comprised of flat circular metal plates applied to the head of normal hearing humans, it was found that they could hear pure tones and speech. Simultaneously acoustic vibration could also be heard by an outside observer with a stethoscope placed near on of the electrodes on the skin.
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
