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"Don W." included:
>
>
> ... If I understand correctly, the HHV of hydrogen is ALL the potential
> chemical energy contained in the hydrogen, including the fraction that is
> unrecoverable except as low grade heat because the "oxidized hydrogen" is
> formed as a gas and not as a liquid. The LHV of hydrogen is the five
> sixths of the energy that is not necessarily "exothermic" (MIGHT be
> recovered in some form other than heat.) Is this understanding correct?
>
> IF that is correct, then it seems proper (to me, anyway) to use HHV and say
> that the maximum possible efficiency of the fuel cell is 83% instead of
> using the LHV and saying that the maximum possible efficiency of the fuel
> cell is 100% even though there is heat in excess of that 100% dissipated in
> or slightly after the fuel cell.
>
> Others disagree. Here is a link with some good information that says
> Ballard is right to use HHV in calculating the efficiency of their fuel
> cells:
> http://planetforlife.com/h2/h2fuelcell.html
I tend to use the delta 'G', -228.3 kJ/mol
for hydrogen oxidation yielding water vapour,
-237.15 kJ/mol for hydrogen oxidation yielding liquid water.
This small difference reflects the fact that water
at 25 C isn't very far from its boiling point,
where the difference would be zero.
If there were a perfectly efficient electrolyser
it would put in exactly 228.3 kJ of DC electricity
to get a mole of hydrogen out of a mole of water vapour ...
or 237.15 kJ to get one out of liquid water.
A little low-grade heat would be absorbed here and re-released here;
looking only at DG, one doesn't see this and doesn't have to
deal with ideal efficiencies of 83 percent or 120 percent.
Instead, ideal machine efficiency is simply 100 percent.
A perfectly efficient fuel cell would get all of that 237 kJ
or all of the 228 kJ back.
The only real air-breathing ones for which I know of published
data that can be made to yield an efficiency get 93.5 kJ/mol.
But having the oxygen come pure from a tank makes a big difference,
so that the submarine claims might be enthusiastic
by only five percentage points, or ten. 15, tops. For sure, tops.
Fairly sure.
When air-breathing, you have to pump the air;
oxygen from a LOX tank pumps itself,
and if you don't count the cold vapour's expansion potential
as part of the onboard energy, you get the sort of
efficiency number a brochure writer likes.
I have no precise efficiency measurement for any specific
pure-oxygen fuel cell, so I suppose they too are about
10 points below the range that is commonly quoted,
low 60s to 70 percent, without, of course,
saying percent of what.
I did once look at how much a 60-percent-of-DG cell,
if one existed, would lose if it had to purify its own
oxygen. Hydrogen takes 252.0 g oxygen per kilowatt-hour
of -DG. By the above hypothesis a cell can make 600 Wh
with this. Oxygen from zeolite PSA oxygen purifiers
costs 0.645 Wh/g, so the 252.0 g costs 162.5 Wh,
dropping us to 440 Wh, 44 percent.
The same 16-percentage-point deduction would apply
to whatever real efficiency real pure-oxygen fuel cells have.
--- Graham Cowan, former hydrogen fan
www.eagle.ca/~gcowan/Paper_for_11th_CHC.html">http://www.eagle.ca/~gcowan/Paper_for_11th_CHC.html --
boron as energy carrier: real-car range, nuclear cachet
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