On Wed, 3 Jun 2015 21:57:44 +0000 (UTC), Skywise
wrote:

Larry Dighera wrote in
:

Hello Brian,

You seem to have firm grasp of the physics involved. Are you able to
suggest how efficient an electric power system powered by LH2 would have
to be to offset the power density difference from gasoline/kerosene?

A quick disclaimer.... I don't have any degrees in this stuff or
work with it. I just happen to be very interested in the sciences
in general and have taught myself a few things over the years.
Having said that, I try very hard to check my facts and figures
before I say anything. I have an allergy to foot-in-mouth disease.

On to your question....

Per numbers in http://www.tinaja.com/glib/energfun.pdf

Liquid H2 has an energy density of 2600 Watt-hours/liter.
Gasoline is 9000 Watt-hours/liter.

[I used energy density per volume rather than per mass
because that's the limiting factor in any vehicle, the
volume of the 'gas tank']

If we make an assumption for discussion's sake that an LH2
powered system were 100% efficient, then the gasoline system
would only need to be 2600/9000 = 29% efficient to reach
parity with LH2. But note that nothing is ever 100% efficient.
There are _always_ conversion losses. It's a matter of how
much.

Per:
http://en.wikipedia.org/wiki/Interna...rgy_efficiency

Engine efficiency is limited by thermodynamic laws. "Most steel
engines have a thermodynamic limit of 37%." Further, "most engines
retain an average efficiency of about 18%-20%."

Right away we see it's at least potentially possible for gasoline
to still beat out 100% efficient LH2. But let's go on the low
side and assume a gasoline engine is 18% efficient. Then we need
to figure out the efficiency required of an LH2 system to beat
gasoline:

9000 * 18% = 1620
1620 / 2600 = 62%

Therefore an LH2 system would have to be 62% efficient overall to
beat a typical gasoline engine.

Per the same Wikipedia article, "Electric motors are better still,
at around 85%-90% efficiency or more, but they rely on an external
power source (often another heat engine at a power plant subject to
similar thermodynamic efficiency limits)."

OK. So an electric motor _by itself_ is more than efficient, but
as stated it has to get it's electricity from somewhere else. We
are assuming an LH2 powered source.

Let's go with the high side of 90% on the electric motor. So we
have to now figure out what efficiency is required in converting
LH2 to electrity so a 90% efficient electric motor produces
1620 Wh/l of LH2...

1620/.90/2600 = 69%

Now that leaves us with finding out how efficiently LH2 can be
converted to electricity.

Per: http://energy.gov/eere/fuelcells/fuel-cells

"Fuel cells can operate at higher efficiencies than combustion
engines, and can convert the chemical energy in the fuel to
electrical energy with efficiencies of up to 60%."

So we may be coming up a bit short.

However, all my pondering here is surely a gross oversimplification.
And it's possible I goofed on my math or went astray with my logic.
And I imagine different sources will give different numbers. But I
hope it gives you some idea. There are surely other factors that
need to be taken into account. Some may make things work out better,
others may make things worse.

Brian


Hello Brian,

Thank you for your fair and conservative analysis. Very much appreciated.

So generally on a theoretical basis, it is within the realm of possibility that
using LH2 to generate electric power with a fuel-cell to power an electric
motor might be completive in terms of performance with today's General Aviation
internal combustion powered aircraft, because the efficiency of the
electrically powered system is potentially so much greater than the IC
technology, that it compensates for the reduced power density of the LH2 fuel
compared to petroleum. The laws of physics don't prohibit it.

Of course, for this to be realized, significant engineering remains to be
accomplished, but the path to electrically powered aircraft isn't a dead end
due to the laws of physics.

Larry