Fuel cells producing *liquid* water?

Robert Clark wrote:
I meant using cryogenic liquid hydrogen would make it easy to liquify
the water.
As noted by Cowan, 4 bar might be too high for a lightweight system. I
got this number from high performance fuel cells. They would work at 1
bar just not as efficiently.

Er, Bob, just how big is the H2 tank going to be?

In sci.physics, Tony Wesley

wrote
on 18 Aug 2006 16:22:55 -0700
.com:

Robert Clark wrote:
I meant using cryogenic liquid hydrogen would make it easy to liquify
the water.
As noted by Cowan, 4 bar might be too high for a lightweight system. I
got this number from high performance fuel cells. They would work at 1
bar just not as efficiently.

Er, Bob, just how big is the H2 tank going to be?


The fuel tank can be highly pressurized and the hydrogen fed through a
regular, presumably. I for one don't see that as a problem although the
regulator might get rather cold. :-) (Same issue as with air
conditioners.) Of course that might be advantageous, as one can then
route the exhaust past it.

--
#191,
Windows Vista. Because it's time to refresh your hardware. Trust us.

The Ghost In The Machine wrote:
In sci.physics, Tony Wesley

wrote
on 18 Aug 2006 16:22:55 -0700
.com:

Robert Clark wrote:

I meant using cryogenic liquid hydrogen would make it easy to liquify
the water.
As noted by Cowan, 4 bar might be too high for a lightweight system. I
got this number from high performance fuel cells. They would work at 1
bar just not as efficiently.

Er, Bob, just how big is the H2 tank going to be?



The fuel tank can be highly pressurized and the hydrogen fed through a
regular, presumably. I for one don't see that as a problem although the
regulator might get rather cold. :-) (Same issue as with air
conditioners.) Of course that might be advantageous, as one can then
route the exhaust past it.


By "highly pressurized", I assume you mean a contained energy density of
LESS THAN ONE PERCENT of gasoline.

http://www.tinaja.com/glib/energfun.pdf



--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email:

Please visit my GURU's LAIR web site at http://www.tinaja.com

In sci.physics, Don Lancaster

wrote
on Sat, 19 Aug 2006 10:56:45 -0700
:
The Ghost In The Machine wrote:
In sci.physics, Tony Wesley

wrote
on 18 Aug 2006 16:22:55 -0700
.com:

Robert Clark wrote:

I meant using cryogenic liquid hydrogen would make it easy to liquify
the water.
As noted by Cowan, 4 bar might be too high for a lightweight system. I
got this number from high performance fuel cells. They would work at 1
bar just not as efficiently.

Er, Bob, just how big is the H2 tank going to be?



The fuel tank can be highly pressurized and the hydrogen fed through a
regular, presumably. I for one don't see that as a problem although the
regulator might get rather cold. :-) (Same issue as with air
conditioners.) Of course that might be advantageous, as one can then
route the exhaust past it.


By "highly pressurized", I assume you mean a contained energy density of
LESS THAN ONE PERCENT of gasoline.

http://www.tinaja.com/glib/energfun.pdf


Well, my assumptions are as follows -- and yes, the energy
density is pretty bad. This is all pretty basic chemistry
stuff, but if I make any obvious errors, please let me
know. :-)

First, the typical form of hydrogen is diatomic hydrogen,
H2. (Monatomic hydrogen is theoretically possible but
would probably be highly unstable at room temperature.)
The bond energy of H2 is about 436 kJ/mol. The bond
energy of O2 is about 498 kJ/mol, and an H-O bond about
464 kJ/mol. Ergo, a reaction of 2 moles H2 and 1 mole O2
proceeds as follows:

H2 + 1/2 O2 = H2O + E

where E is the breakage of 2 H2 bonds, 1 O2 bond, and the formation
of 2 H-O bonds.

H2 bond = -436
1/2 O2 bond = -498/2
2 HO bonds = 2*464

Net: 243 kJ/mol

Darned good if one counts by *weight*; if one burns 1 kg
of gasoline one gets 45 MJ or so, but if one burns 1 kg
of diatomic hydrogen one gets 500 * 243 kJ = 121.5 MJ.
Therefore 1 kg of diatomic hydrogen can replace 1 gallon
(about 2.65 kg, since gasoline is about 0.70 kg/l) of gasoline.

However, hydrogen is a gas, making storage of large
quantities rather difficult. As you probably already
know, PV=nRT is a variant of the ideal gas law. T =
293 K on a spring day (68F); P = 101325 (1 atm), and V is
what we're trying to calculate. R = 8.314472 J/(mol K),
the gas constant.

If I want 1 kg of hydrogen, that's 500 moles; therefore

V = nRT/P = 500 * (8.314472) * (293) / (101325) = 12.02 cu m.

at 1 atm pressure. That's 12,020 liters or 3175 gallons.

Horrid.

If one assumes 4 atm for a fuel cell one still gets 793 3/4 gallons -- an
energy density by volume of 0.34%, when compared to gasoline,
if my computations are correct.

I don't consider 4 atm all that high a pressure though; my bicycle tires
take a higher pressure than that (about 90 psi, or 6 1/8 atm).

The best I can do pressurewise (there might be specialists
out there who can do higher) is SCBA/SCUBA gear. There
are probably a fair number of issues regarding seals and
regulators (though it's obviously doable since 3300psi is
not the pressure fed to one's lungs while diving!) but
if one assumes the pressure, instead of 101325 Pascal,
is 3300 psi = 22.75 MPa as specified in various tanks at
www.scuba.com, then one gets

V = nRT/P = 500 * (8.314472) * (293) / (22750000) = 53.54 liters,
or 14 gallons.

In other words, if I fill up my, say, 14 gallon "gas
tank" with standard liquid gasoline/petrol, I might get
420 miles, assuming 30 mpg. (Diesel is even slightly
better, though there are a number of issues since engine
compression for diesels is higher, among other things.)

If I fill up my "gas tank" with highly pressurized
hydrogen gas, I might get

420 / 14 * 121.5 / 45 = 81 miles.

(This is assuming, of course, that everything else is the
same: car size, car weight, car shape, driving habits,
engine power, and engine efficiency. There is a minute
possibility of extracting some of the energy from the
actual pressurization but that's not all that much.)

A prior poster mentioned the possibility of liquid
hydrogen; this is indeed possible but very problematic.
For starters, the boiling point of H2 is 20.28 K or
-252.87 C. Storage of liquid hydrogen would therefore
have to be in Dewar flasks or equivalent, and even then
some of the ambient heat would eventually leak in, which
would cause at least the following effects.

[1] Condensation on the tank, and at some point ice on
the tank. Since ice expands things could get nasty,
especially around the valve area.

[2] Loss of the hydrogen and displacement of the air,
if one's parked in an enclosed space. The best one can
hope for if a car's been sitting sufficiently long is
suffocation, but a H2/O2 mixture is highly explosive.

The good news: liquid H2 has a molar volume of 11.42 cm^3,
or 0.663 kg/gallon. The bad news: 14 gallons would have
1.128 GJ, or about 600 pounds of TNT, especially when it
leaks into the garage (that 14 gallons will fully displace
about 112 m^3 of space when evaporated, and will form a
dangerously explosive mixture for many times that amount).

Will insurance cover such explosions? I'd prefer walking
in that case... :-)

One final note: despite what one sees in the movies, according
to

http://www.intuitor.com/moviephysics/
(http://www.intuitor.com/moviephysics...tml#cigarettes)

it is very difficult to ignite gasoline with a
glowing cigarette, glowing (but not flame-lit) taper,
or metal spark. (If one does wish to test this, let me
suggest they know what they are doing first, and take all
appropriate safety precautions! Also, stay well downwind
of me... :-) ).

I'd be interested in whether someone has tested an H2/O2
mixture using similar methods -- with a very long pole,
of course. I do know that at least one experiment uses a
"bomb" of presumably rather thick glass with one or two
inlets and electrodes (though a better method of burning
hydrogen would be to use a fuel cell or a platinum (?)
catalyst). The "bomb" in this case is small and
protected, so it doesn't fly apart with the force of
the explosion within, and one gets droplets of water,
presumably, on the inside, after cooling. But scaling
upward in, say, one's garage, would get a little messier.

All in all, I'd be more in favor of biodiesel than a pure
hydrogen economy; it looks easier to handle.

--
#191,
Windows Vista. Because it's time to refresh your hardware. Trust us.

The Ghost In The Machine wrote:


If one assumes 4 atm for a fuel cell one still gets 793 3/4 gallons -- an
energy density by volume of 0.34%, when compared to gasoline,
if my computations are correct.


The supposedly higher energy density by weight of hydrogen is totally
useless for terrestral apps.

You have to consider the CONTAINED energy density by weight, which is
ALWAYS ridiculously less than gasoline.

As to energy density by volume, gasoline offers 9000 watthours per
liter, while STP hydrogen offers 2.7 watthours per liter electrically
recoverable or 3.3 watthours per liter total heat recovery.

At 4 BAR pressure, hydrogen offers 10.8 watthours per liter, or about
1/833rd that of gasoline. About 0.12 percent.

At 100 BAR pressure, hydrogen offers 270 watthours per liter, or about
1/33rd that of gasoline.

There is, of course, more hydrogen in a gallon of gasoline than there is
in a gallon of liquid hydrogen.

See http://www.tinaja.com/glib/energfun.pdf for a detailed analysis.

--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email:

Please visit my GURU's LAIR web site at http://www.tinaja.com

Don Lancaster wrote:

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


Wow, you got that completely backwards.

Water vapor gives up energy when it condenses to liquid. You could
harvest additional energy from the steam as it cools back to room
temperature.

Dave

Batteries are almost certainly more cost effective.

http://www.tinaja.com/glib/energfun.pdf


--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email:

Please visit my GURU's LAIR web site at http://www.tinaja.com

dave e wrote:
Don Lancaster wrote:


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



Wow, you got that completely backwards.


Not really.

Some of the extracted fuel cell energy is in the form of latent heat
that created the gaseous state.

If you cool to liquid water by removing heat and throwing it away in a
radiator or whatever, that energy is gone and not available for the
useful intended application.

Thus, the fuel cell efficiency goes DOWN unless you do something useful
with the latent heat extraction. The low delta-T and Carnot guarantees
useful extraction will be inefficient and enormously difficult.

But the whole project makes no sense whatsoever.


--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email:

Please visit my GURU's LAIR web site at http://www.tinaja.com

Don Lancaster wrote:
The Ghost In The Machine wrote:


If one assumes 4 atm for a fuel cell one still gets 793 3/4 gallons -- an
energy density by volume of 0.34%, when compared to gasoline,
if my computations are correct.


The supposedly higher energy density by weight of hydrogen is totally
useless for terrestral apps.

You have to consider the CONTAINED energy density by weight, which is
ALWAYS ridiculously less than gasoline.

As to energy density by volume, gasoline offers 9000 watthours per
liter, while STP hydrogen offers 2.7 watthours per liter electrically
recoverable or 3.3 watthours per liter total heat recovery.

At 4 BAR pressure, hydrogen offers 10.8 watthours per liter, or about
1/833rd that of gasoline. About 0.12 percent.

At 100 BAR pressure, hydrogen offers 270 watthours per liter, or about
1/33rd that of gasoline.

There is, of course, more hydrogen in a gallon of gasoline than there is
in a gallon of liquid hydrogen.

See http://www.tinaja.com/glib/energfun.pdf for a detailed analysis.


The problems of converting to a hydrogen economy include the higher
energy of getting it out of easily available sources such as
hydrocarbons than the energy available in the hydrogen and that of
storage and transport.
Of the two I consider the problem of storage and transport to be
solvable near term.
One method being investigated is using glass microspheres to store it
at high pressu

AU researchers looking at hydrogen in tiny glass beads as fuel source
for cars.
http://www.fuelcellsworks.com/Supppage1764.html

This uses microspheres of diameters of a few microns to store it at up
to 100 MPa. It has been found the microspheres will infuse the hydrogen
when heated to high temperatures at high pressure. Then store it at
normal pressure and temperature. The hydrogen can be released again at
high temperature and pressure conditions. The latest research shows
that using lasers can speed up the speed at which the hydrogen is
released.
The Department of Energy has set the ultimate goal for hydrogen energy
storage to be superior to that of gasoline as above 10 MJ energy stored
per kg of total weight and 10 MJ per L of total volume. At an energy
content of hydrogen at 142 MJ per kg, this means about .07 kg of H2 per
kg of total storage system weight and .07 kg of H2 per liter of total
storage system volume.
A material that might be able to reach these criteria is "tetrahedral
amorphous diamond" if used in the form of microspheres.
This report gives an average tensile stength of 7.3 GPa when tested on
micron-scale samples:

Young’s modulus, Poisson’s ratio and failure properties of
tetrahedral amorphous diamond-like carbon for MEMS devices.
J. Micromech. Microeng. 15 (2005) 728–735
doi:10.1088/0960-1317/15/4/009
http://ej.iop.org/links/q03/3NXzoBo,...jmm5_4_009.pdf

The thickness to radius ratio of a spherical pressurized tank is given
by:

h/r = Δp/(2σ)

where h is the wall thickness, r the radius of the sphere, Δp the
overpressure, and σ the tensile strength of the material.

This page gives properties of hydrogen at various pressures and
temperatures (there is deviation from the ideal gas law at very high
pressures):

Hydrogen Properties Package.
http://www.inspi.ufl.edu/data/h_prop_package.html

At a temperature of 300 K, a pressure of 6000 bar gives a density of
72 kg/m^3, or .072 kg/l.
Using a tensile strength of 7.2 GPa = 72,000 bar for the tetrahedral
amorphous diamond and 6000 bar pressure for the hydrogen, the thickness
to radius ratio for a spherical tank would be h/r = 1/24.
The volume for a sphere is V = (4/3)Pi*r^3. For a wall thickness small
compared to the radius, we can take the volume of the wall to be 4*h*
Pi*r^2, which equals (1/6)*Pi*r^3, when h/r = 1/24.
Since the volume of the tank and the wall both have r to the third
power, the radius will cancel when calculating the ratio of the
hydrogen mass to the mass of the tank wall material. So I'll take r =
1. Then the mass of the hydrogen in the tank would be 72*(4/3)*Pi =
301.6 kg.
I'll take the density of tetrahedral amorphous diamond to be that of
diamond, 3500 kg/m^2. Then the mass of the container would be:
3500*(1/6)*Pi = 1885 kg. Then the ratio of the mass of hydrogen to the
container wall mass would be 301.6/1885 = 0.16.
The tetrahedral amorphous diamond is amorphous as is glass. So it may
be that heat and laser irradiation could also allow hydrogen to be
infused and/or released.


Bob Clark

Don Lancaster wrote:

dave e wrote:
Don Lancaster wrote:

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

Wow, you got that completely backwards.

Not really.

Yes you did. I spotted it too and if you're trying to weasel your way out of it,
expect to be challenged. Energy is *liberated* when steam condenses.

Graham

On Sun, 20 Aug 2006 23:02:42 +0100, Eeyore wrote:


Don Lancaster wrote:

dave e wrote:
Don Lancaster wrote:

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

Wow, you got that completely backwards.

Not really.

Yes you did. I spotted it too and if you're trying to weasel your way out of it,
expect to be challenged. Energy is *liberated* when steam condenses.


But its lost in his proposed application unless he cooks potatoes
with it or something.



--
Posted via a free Usenet account from http://www.teranews.com