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deltaT<(1/4D)-(Xad/x)^2
deltaT=Pulse Width Duration in Nano-seconds (3-300ns)
D=the diffusion coefficient (cm^2 s^-1) 2.3x10^-5cm^2 s^-1
Xad= the density of Hydrogen ions on the cathode (cm^-2) 10^15cm-2 (for Platinum specifically, each electrode material is different, slightly)
X= the concentration of Hydrogen ions in the solution (cm^-3) 6x10^20cm^-3 for 1M of KOH solution.
Formula for pulsed DC electrolysis where by the application of current at Ultra-Short pulse width of 3-300nano-seconds at 25-40+Khz. The increasing of electrical frequency increases the applied current, pulse width must remain static in order to prevent the diffusion layer and electron double layers from forming on the the surface of the electrode. Thus instead Virtual breakdown occurs at the surface of the electrode and within the E-field generated surrounding the anode and cathodes respectively. The high frequency "impulse current" (high voltage of peak current @ 140V 0.3-1 amp per "cell"). If cell electrodes are 3cm apart, the device will produce H3O and OH(-+) radicals which will then decompose into H2 ions and 1/2O2. This is purely a field effect, and therefor the decomposition of the water is due to Virtual forces exerted upon the dipole nature of the H2O atoms.
What do you think?
deltaT=Pulse Width Duration in Nano-seconds (3-300ns)
D=the diffusion coefficient (cm^2 s^-1) 2.3x10^-5cm^2 s^-1
Xad= the density of Hydrogen ions on the cathode (cm^-2) 10^15cm-2 (for Platinum specifically, each electrode material is different, slightly)
X= the concentration of Hydrogen ions in the solution (cm^-3) 6x10^20cm^-3 for 1M of KOH solution.
Formula for pulsed DC electrolysis where by the application of current at Ultra-Short pulse width of 3-300nano-seconds at 25-40+Khz. The increasing of electrical frequency increases the applied current, pulse width must remain static in order to prevent the diffusion layer and electron double layers from forming on the the surface of the electrode. Thus instead Virtual breakdown occurs at the surface of the electrode and within the E-field generated surrounding the anode and cathodes respectively. The high frequency "impulse current" (high voltage of peak current @ 140V 0.3-1 amp per "cell"). If cell electrodes are 3cm apart, the device will produce H3O and OH(-+) radicals which will then decompose into H2 ions and 1/2O2. This is purely a field effect, and therefor the decomposition of the water is due to Virtual forces exerted upon the dipole nature of the H2O atoms.
What do you think?