FW: Hydrogen Reforming

["Roger Stockton" <stockton@racalcanada.com>]

Hi all,

More clarification on hydrogen fuel cells and reformers
courtesy of Jonathan Dodge.

Thanks to Jonathan and all who've helped clear this up for me!

Roger.

 -----Original Message-----
From:	jldodge@uswest.net [SMTP:jldodge@uswest.net]
Sent:	Thursday, August 26, 1999 9:01 PM
To:	stockton@racalcanada.com
Subject:	Hydrogen Reforming

Hy Roger.

I'm replying directly to you since I'm not on the EV List.
Feel free to pass this along.  I just prefer to follow the
archives than deal with a huge daily download into in my
in-box.

So for any chemical reaction, there is a certain amount of
energy liberated (or absorbed).  In the case of methanol to
carbon dioxide and water, 727 kiloJoules per mole of
methanol is released.  Now, for the ICE (with assumption of
perfect combustion) and the fuel cell with reformer, the net
reaction is the same.  So for each mole of methanol
consumed, 727 kiloJoules of energy is available.  Now, here
is my point about efficiency.  The ICE harnesses this energy
thermally; it is limited by the Carnot heat cycle, which
means the power out depends on the difference between two
temperatures.  (An ICE cannot produce as much power on a hot
day as on a cold day since the difference between ambient
and exhaust temperature is smaller.)  The fuel cell
harnesses the energy electrochemically.  It produces
electricity.  It is not bound by the heat cycle, so ambient
temperature doesn't matter for power out (except when
considering freezing or boiling of liquids involved).

Anyway, the ICE throws all but 18% of the energy from the
reaction out the tail pipe in the form of heat.  Only 18% of
the reaction energy goes to turning a shaft.

The reformer throws away 30% of the reaction energy just
keeping itself hot enough to sustain the endothermic
reforming reaction.  The fuel cell throws away 45% in the
form of heat and support electronics etc.  Some of the heat
by the way could be at least partially recovered and used.
The rest of the reaction energy is available in the form of
electricity.

You could say that the fuel cell system needs an additional
loss in the form of electromechanical energy conversion (a
motor and controller).  You would only lose about 10% there
for an overall efficiency that is still twice that of the
ICE.

The same argument holds for any fuel.  The energy avaible
from the net chemical reaction is the same for each.  They
just harness the energy in different forms.

So finally, an ICE will consume 2.14 times the amount of
fuel for the same power output as a reformer/fuel cell
system, regardless of which type of fuel you choose.  If you
run both on gasoline, the ICE will consume over twice the
amount of fuel as the reformer/fuel cell system.  If you run
both on methanol, the ICE will still consume over twice the
amount of fuel.

Thanks for a great discussion!

Jonathan Dodge