"William Morse" wrote in message
...
Don Lancaster wrote in news:4l4oggF8btoU1
@individual.net:

(Don originally wrote)

You have to recognize that converting water vapor to liquid
consumes energy and has to be charged against the fuel cell
efficiency budget.

(snip in between)

What is the context about the statement "converting water vapor to
liquid consumes energy". That statement is in fact backwards. Water
vapor contains more energy than liquid water - about 9.8 kcal/mole, if
memory serves me.

The energy IS consumed and charged against the fuel cell.

The energy isn't consumed, unless you are stating that the fuel cell
originally produces liquid water and then subsequently turns it into
vapor - which AFAIK is not what happens.

It doesn't matter what the mechanism is, whether it first forms liquid and
evaporates it, or whether it initially forms vapor. Energy is a state
function, independent of path. What *does* matter is how you calculate the
denominator in the expression for the efficiency--i.e., whether you use the
reaction

H2 (g) + 1/2 O2 (g) -- H2O (g)

or the reaction

H2 (g) + 1/2 O2 (g) -- H2O (l)

to calculate the ideal amount of energy it can deliver. These two reactions
have very different deltaHs.


Conceivably, if the fuel cell system output liquid water, it would be
more efficient by the latent heat difference.

Which is one reason why stationary fuel cells are potentially a good
solution for some applications - e.g. hospitals that need heat, need a
backup source of power, and use enough energy to repay capital costs. I
say potentially because current stationary fuel cells are way too
expensive to repay the investment. And no, I don't think the fuel is
going to be hydrogen, at least not anytime soon.

Sort of the fuel cell equivalent of co-gen. Interesting idea. I agree with
other posters that comment that water vapor, at least at the T, P, and
concentration we're talking about, is relatively low-value heat, and
converting it to useful work would be difficult and inefficient. I think
Don's original point is that usually, to convert warm moist air to cool dry
air and liquid water, you need to use some sort of refrigeration--i.e., it
costs energy. High efficiency furnaces are able to recover some of the heat
of vaporization of water, but much of the water vapor still does go out the
stack, in 100 % RH air at ambient temperature. Converting it to electricity
would be extremely difficult and inefficient, since the ideal (Carnot)
efficiency is limited by the deltaT. However, in the fuel cell case, using
the water vapor to heat a building is a good idea, since you would then get
to take credit in the fuel cell efficiency not just for the heat you are
able to extract, but for replacing (essentially for free) all of the energy
content of the fuel you would otherwise burn to heat that building
(realizing that even combustion heating is never 100 % efficient).


Throwing away the heat --- throws it away.

Agreed, but it isn't charged against the fuel cell budget, it just isn't
credited to the budget. This might be considered a quibble, but there is
enough nonsense on this newsgroup without adding to the confusion by
stating that "converting water vapor to liquid consumes energy".

Agreed.

Eric Lucas