Efficiency of Xogen Process
Longtime Xogen watcher vcrepair suggested that I look at the temperature chart provided as part of the report we reviewed yesterday. If we assume that this chart was built using the same test run that provided the 66.8 l of H2 for 18 Amp hours of current, then we can calculate two quantities that a lot of people comlained were not included in the report, the efficiency of the Xogen process and the Voltage associated with the 18 Amp hours.
Energy drawn from battery =
Change in heat energy in the water +
Change in chemical potential in the H2
Really, there would be other losses but I assume these are small. We can calculate both the change in the heat energy in the water and the change in the chemical potential from the data provided.
Change in heat energyThe chart states that the increase in heat in the water was 231 W for 37 minutes, and 254 W for 16 minutes. Simply adding these we get 12611 W minutes, which is 210 W hours or 757 kJ.
Change in chemical potentialThe report claims 66.8 l of H2. I think that's a little higher than it was, but I'll use it here anyway. At a reasonable lab temperature of 70 degrees, 66.8 litres is 2.77 moles. Multiplying by 285.83 kJ/mole, that makes for a 792 kJ increase in chemical potential.
Total Energy1550 kJ = 792 kJ (H2) + 757 kJ (heat)
Voltage24 V = 1550 kJ * 1 Wh/3600 J * 1 / 18 Ah
Efficiency51% = 792 kJ / 1550 kJ
Both these figures crosscheck when compared to the efficency one would compute if one assumes that the system tested was 8 cells connected in series to a 24 V power supply. In that case, the efficency would be 49% = 1.47 V / 3 V. Since we are working with only 2 significant figures in the charge reported by the ARC (18 Amps) this is reassuring agreement.