
楼主 
发表于 2010624 13:48:05

显示全部楼层
(5) ThermodynanicMeasurement ~
In order to properly couple ComponentII to a mass spectrometer one requires a special housing around ComponentII that it will capture the gases produced and permit these to to be drawnunder low vacuum into the mass spectometer. Therefore a stainless steeland glass chamber was built to contain Component II, and provision madeto couple it directly through a CO2 watertrap to the mass spectrometerwith the appropriate stainless steel tubing. This chamber is designatedas Component IV. Both the mass spectrometer and Component IV were purgedwith helium and evacuated for a two hour period before any gas sampleswere drawn. In this way contamination was minimized. The definitive measurementwere done at Gollob Analytical Services in Berkeley Heights, New Jersey
We now describe the use of ComponentI and how its energy output to Component II is measured. The energyoutput of Component I is an amplitude modulated alternating current lookinginto a highly nonlinear load, i.e., the water solution. Component I isso designed that at peak load it is in resonance across the system Components I, II, and III  and the vector diagrams show that the capacitivereactance, and the inductance reactance are almost exactly 180° outof phase, so that the net power output is reactive (the dissipative poweris very small). This design ensures minimum power lossses across the entireoutput system. In the experiments to be described, the entire emphasisis placed on achieving the maximum gas yield (credit) in exchange for theminimum applied electrical energy.
The most precise way to measure theapplied energy from Component I to Component II and Component III is tomeasure the power, P, in watts, W. Ideally this should be done with a precisionwattmeter. But since we were interested in following the voltage and currentseparately, it was decided not to use the watt meter. Separate meters wereused to continuously monitor the current and the volts.
This is done by precision measurementof the volts across Component III as root mean square (rms) volts; andthe current flowing in the system as rms amperes. Precisely calibratedinstruments were used to take these two measurements. A typical set ofexperiments using water in the form of 0.9% saline solution 0.1540 molarto obtain high efficiency hydrolysis gave the following results:
rms Current = I = 25mA to 38 mA (0.025A to 0.038 A.)
rms Volts = E = 4 Volts to 2.6 Volts
The resultant ration between currentand voltage is dependent on many factors such as the gap distance betweenthe center and ring electrodes, dielectric properties of the water, conductivityproperties of the water, equilibrium states, isothermal conditions, materialsused, and even the pressure of clathrates. The above current and voltagevalues reflect the net effect of various combinations of such parameters.When one takes the product of rms current, and rms volts, one has a measureof the power, P in watts.
P = I x E = 25 mA x 4.0 volts =100mW (0.1 W)
and P = I x E =38 mA 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) asmeasured on a precision frequency counter. The wave form was monitoredfor harmonic content on an oscilloscope, and the nuclear magnetic relaxationcycle was monitored on an XY plotting oscilloscope in order to maintainthe proper hysteresis loop figure. All experiments were run so that thepower in watts, applied through Components I, II, and III ranged between98.8 mW to 100 mW.
Since by the International Systemof Units 1971 (ST), one Wattsecond (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, (1/2)O2,and this credit side was measured in two laboratories, on two kinds ofcalibrated instruments, namely gas chromatography machine, and mass spectrometermachine.
The volume of gases H2and (1/2)O2 was measured as produced under standard conditionsof temperature and pressure in unit time, i.e., in cubic centimeters perminute (cc/min), as well as the possibility contaminating gases,such asair oxygen, nitrogen and argon, carbon monoxide, carbon dioxide, watervapor, etc.
The electrical and gas measurementswere reduced to the common denominator of Joules of energy so that theefficiency accounting could all be handled in one currency. We now presentthe averaged results from many experiments. The standard error betweendifferent samples, machines, and locations is at +/ 10%, and we only usethe mean for all the following calculations.
II. Thermodynamic Efficiency forthe Endergonic Decomposition of Liquid Water (Salininized) to Gases UnderStandard Atmosphere ( 754 to 750 m.m. Hg) and Standard Isothermal Conditions@ 25° C = 77° F = 298.16° K, According to the Following Reaction:
H20 (1) _> H2(g)+ (1/2)O2(1)+ Delta G = 56.620 Kcal/mole (10)
As already described, Delta G isthe Gibbs function. We convert Kcal to our common currency of Joulesby the formula, One Calorie = 4.1868 Joules
Delta G = 56.620 Kcal x 4.1868 J= 236,954/J/mol of H2O where1 mole = 18 gr. (11)
Delta Ge = the electricalenergy required to yield an equivalent amount of energy from H2Oin the form of gases H2 and (1/2)O2.
To simplify our calculation we wishto find out how much energy is required to produce the 1.0 cc of H2Oas the gases H2 and (1/2)O2. There are (under standardconditions) 22,400 cc = V of gas in one mole of H2O. Therefore
Delta G / V = 236,954 J/ 22,400cc = 10.5783J/cc. (12)
We now calculate how much electricalenergy is required to liberate 1.0 cc of the H2O gases (whereH2 = 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 = 10.5783Joules, then
PV = 1 Js x 10.5783 J = 10.5783 Js,or, = 10.5783 Ws (13)
Since our experiments were run at100 mW ( 0.1 W) applied to the water sample in Component II, III, for 30minutes, we wish to calculate the ideal (100% efficient) gas productionat this total applied power level. This is,
0.1 Ws x 60 sec x 30 min = 180,00Joules (for 30 min.). The total gas production at ideal 100% efficiencyis 180 J/10.5783 J/cc = 17.01 cc H2O (g)
We further wish to calculate howmuch hydrogen is present in the 17.01 cc H2O (g).
17.01 cc H2O (g) x 0.666H2 (g) = 11.329 cc H2(g) (14)
17.01 cc H2O (g) x 0.333(1/2)O2 (g) = 5.681 cc (1/2) O2 (g)
Against this ideal standard of efficiencyof expected gas production, we must measure the actual amount of gas producedunder: (1) Standard conditions as defined above, and (2) 0.1 Ws powerapplied over 30 minutes. In our experiments, the mean amount of H2and (1/2)O2 produced, as measured on precision calibrated GC,and MS machines in two different laboratories, where SE is +/ 10%, is,
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, n, between the ideal yield,and measured yield,
Measured H2 (g) / IdealH2 (g) = 10.80 cc / 11.33 cc = 91.30%
(6) Alternative Methodology forCalculating Efficiency Based on the Faraday Law of Electrochemistry ~
This method is based on the numberof electrons that must be removed, or added to decompose, or form one moleof, a substance of valence one. In water H2O, one mole has thefollowing weight:
H = 1.008 gr /mol
H = 1.008 gr /mol
O = 15.999 gr/mol
Thus, 1 mol H2O = 18.015gr/mol
For a unvalent substance one grammole contains 6.022 x 1023 electrons = N = Avogadro's Number.If the substance is divalent, trivalent, etc., N is multiplied by the numberof 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 temperatureis 273.15° K = 0° C = T . Standard Pressure is one atmosphere= 760 mm Hg = P. 
