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"G. R. L. Cowan" <gcowan@xxxxxxxx> wrote in message
news:42E2B887.BFE8F85D@xxxxxxxxxxx
>
> A real fuel cell, the Ballard Nexa,
> does 41 percent of delta 'G' at sea level at beginning of life.
> That's 39 percent of LHV.
>
> A vehicle prime mover made of ganged Nexa cells might
> include 30 of them, mass 390 kg, rated aggregate power
> 36 kW.
>
> Actual fuel cell vehicles use other versions with higher
> specific power (more kW/kg) and therefore lower
> conversion efficiency.
>
> We know it's somewhere far south of 39 percent.
> That's all we know; manufacturers never tell.
>
> (Actually, there is one other thing we know:
> the longest ranges attained by internal hydrogen combustion
> car prototypes in the 70s --
> http://www.hydrogen.org/h2cars/overview/main01.html
> -- no fuel cell car prototype has equalled or got anywhere
> near.)
>
>
> > This probably corresponds to a range of load
> > conditions encountered during driving, but I am
> > not sure. A tradeoff for higher efficiency vs
> > larger FC size would be possible. The optimum
> > may be to increase the size to achieve an efficiency
> > in the 60-75% range.
>
> No chance of that. As above said, 40-percent-efficient
> PEM fuel cells are way too big and heavy for cars,
> and the tradeoff can only go the other way.
>
> --- 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
So what's your take on this Siemens submarine FC?:
www.hdw.de/pdf/broschueren/Fuel_Cell_Plants_for_Non-nuclear_Submarines__en.pdf">http://www.hdw.de/pdf/broschueren/Fuel_Cell_Plants_for_Non-nuclear_Submarines__en.pdf
On the page marked '7' it shows 100% of the "fuel energy" divided up as
follows:
5% -- fuel cell auxilliary
65% -- electrical energy
15% -- desorption and evaporation of the hydrogen and oxygen
15% -- lost energy
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.
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