Figures revised to "square" with George Wiseman's 3.8 x energy input (his figures
are based on 2 H2 + O2 , mine are based on H2 + 0.5 O) .
"During a Brown's Gas mon-atomic hydrogen (H) and mon-atomic oxygen (O) flame, we don't have to add any energy because the molecules are already in their simplest and highest energy atomic form. This means that "perfect" Brown's Gas can have 3.8 times the possible 'heat' energy that an "ordinary" H2 and O2 flame has (442.4 Kcal/115.7 Kcal).
Thus we can get 'plasma' type temperatures and effects as we weld, because the potential atomic energy is there, even if it doesn't show up as heat."
I wrote:
With 2.5 volts @ 0.4 amperes equal 1.0 watt = 1.0 joule/second (2.5e18 electrons/second ) flowing through the cell 2.5e18 > Hydrogen atoms are liberated at the cathode and1.25e18 Oxygen atoms are liberated at the cathode.When two Hydrogen atoms combine to form H2, 4.53 eV per bond (7.242e-19 joule)is liberated,and when two Oxygen atoms combine to form O2, 5.17 eV per bond (8.272e-19 joule)is liberated. H2 + 0.5 O2 ----> H2O + Energy = 228,619 joule/mole (18 grams).But, 2.5e18/2 x 7.242e-19 x sec = 1.0 joule = 1.0 watt for the H-H recombination,plus 2.5e18/4 x 8.272e-19 x sec = 0.517 joule = 0.517 watt forthe O-O recombination resulting in 1.517 watts of "Free energy"even before the H2 + 0.5 O2 ----> H2O recombination:(3.32e-5 grams O2 + 4.15e-6 grams H2 =3.735e-5 grams H2O x 12,700 joule/gram = watt-sec = 0.474 joules released.Net "Intrinsic OU " = ~ 1.517 joules Total heat output 1.517 + 0.474 = 1.991 Joulesor 4.2 times energy input.
