-----Original Message----- From: Michel Jullian > - but the 2 eV available > from loading alone without deuterium (contrast that to about .5 eV if the > hydrogen were burned in air) is a huge surprise -
MJ: Jones, where did you get that .5 eV figure? I did the maths and found about 1.5 eV instead, here is the Google calculator result; ((294.6 / 2) / 6.02e23) * kJ = 1.52719998 electron volts Michel, the half-eV figure is the common 'real world' estimate based on the maximum average temperature of the resultant steam - but even so, it appears you did not first deduct the dissociation energy of O2 and H2 and then later deduct the parasitic losses of NOx, peroxides etc. and the other losses that are expected in actual practice, for combustion in air? IOW there are lies, damn lies, and theoretical calculations ;) when trying to go from 'paper numbers' to actual practice. Kitamura's numbers were indicated to be actual practice (if they can be trusted) so it is fair to contrast those numbers with that which would happen if one were to actually burn H2 in air - and .5 eV is a fair estimate even if you discount the 80% of air which is nearly inert. Since water can be split into H2 and O2 with 1.23 volts - does it stand to reason that one could get 1.5 eV in return ? That was rhetorical; and of course this one of nature's built-in cases of "systemic overunity" - ... except for the damn lie that it simply does not work out that way in practice - but it does serve to contrast the large disparity of the "actual with the calculated". > Did I get it wrong? Well, let's say that you got it partly right and mostly wrong - if your intent was to suggest that hydrogen can be burned in air with resultant steam being formed at about 17,000 degrees K. Jones
