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"Don Lancaster" <don@xxxxxxxxxx> wrote in message
news:3ksnd3Fvuc8oU1@xxxxxxxxxxxxxxxxx
> Don W. wrote:
<snip>
> >
> > Even if they wrongly used the LHV as 100%, that would only bring them
down
> > to 48% electrical efficiency (and bring up the "lost energy" figure to
> > 32%.)
> >
> > I'd like to know definitively which figure should be used for
calculating
> > electrical efficiency of a fuel cell. This fuel cell supposedly
operates
> > at about 80 degrees C -- well under the boiling point of water, but at
the
> > actual point where oxygen and hydrogen are combined is the product
steam or
> > liquid water? If steam, then about 17% of the potential energy in the
> > hydrogen cannot be recovered as electricity because it's 'lost' as heat
in
> > creating higher energy water molecules than liquid water. Even if the
> > water is condensed before leaving the fuel cell, the energy is still
> > disipated as heat and not as electricity. Is my understanding correct?
Is
> > there any logical justification for using LHV for electrical efficiency
> > calculation?
> >
> > I wonder about the "5% fuel cell auxilliary" energy. Is this heat
energy
> > that goes to some constructive use in the fuel cell?
> >
> > Don W.
> >
> >
>
> The lower value should be used.
>
> Electrolysis can be up to one sixth endothermic.
> Thus, a hydrogen fuel cell at most MUST be at least one sixth exothermic.
>
>
Do you mean that the higher value should be used (HHV)? If 100% of the
energy in a kilogram of hydrogen is assumed to be 142 MJ, then fuel cell
efficiency will appear to be lower than if 100% of the energy in a kilogram
of hydrogen is assumed to be 120 MJ.
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
Don W.
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