Re: Want help on thermodynamics question: internal energy of a gas
- From: Buddha Buck <blaisepascal@xxxxxxxxx>
- Date: Sun, 27 Apr 2008 14:04:41 +0000 (UTC)
On Apr 26, 10:26 am, Uncle Al <Uncle...@xxxxxxxxxxxxx> wrote:
Buddha Buck wrote:
I'm trying to convince a friend of mine that his "revolutionary energy
saving" scheme is physically ridiculous. He doesn't take at face
value my claim that the 2nd Law of Thermodynamics prevents it from
working as well as he wants. My explanation falters because I haven't
done the maths to show the ridiculousness of his plan. I'll be seeing
him again in a week, and I hope to have a written-out explanation of
where the energy goes in his device so I can walk him through the
maths without being all hand-wavy.
In doing the maths, I've come to realize that there is a part of
thermodynamics I don't understand well enough to model. I've checked
my old college physics texts, I've used Google to the best of my
ability, and before I came here I peeked at sci.physics (and decided I
probably wasn't going to get usable help there).
Here's what I'm trying to model: At one point in his device he has a
cylinder filled with nitrogen gas (well, air, which is close enough to
N2 for the level of precision I'm computing at). He compresses it
with a piston, and at peak compression injects a small quantity of
liquid nitrogen (at 77K) into the cylinder, using the high temperature
of the compressed air to vaporize the liquid nitrogen, thereby raising
the pressure in the cylinder, developing power as the piston is pushed
down. His claim is that the large expansion ratio of LN2 will allow a
large amount of power to be developed. My claim is that the extreme
cold of the LN2 will chill the gas in the cylinder such that there
isn't that large of a pressure increase -- and I wouldn't be surprised
if there was a net energy loss in the overall cycle.
My method of analysis falters at the injection of the LN2.
Here's what I've got:
Using the ideal gas law PV=nRT I can solve for n=PV/RT to get the
number of moles of N2 in the cylinder at the beginning of the cycle.
I can also, using the relationship PV=(\gamma-1)U, solve for U=PV/
(\gamma-1) to get the internal energy of the gas. For the compression
stage, I can use the relationship PV^\gamma=P'V'^\gamma to compute the
new pressure at the smaller volume, then either use PV/T = P'V'/T' or
P'V'=nRT' again to get the temperature. From there, I can compute the
new U' = P'V'/(\gamma-1), and \delta U = U'-U to get the energy needed
to compress the gas.
The next step in the cycle would be to inject a small amount of LN2.
My approach to this is to assume that \delta n moles of LN2 has an
internal energy of u, so after injection the cylinder would contain n'
= n+ \delta n moles of N2, have an internal energy of U'' = U'+u, a
volume of V', a pressure of P''=U''/((V'(\gamma-1)), and a temperature
T''=n'R/P''V'.
The only unknown is the internal energy u of \delta n moles of LN2.
Using U=nRT/(1-\gamma), and \gamma=1.4, I get the internal energy of
gaseous N2 at 77K to be 1.6kJ/mole. Wikipedia says the heat of
vaporization is 5.6kJ/mol, which would mean the internal energy of LN2
at 77K is -4.0kJ/mole, which is unrealistic. Obviously I'm doing
something wrong. The obvious answer is that \gamma is temperature
dependent and the internal energy of gaseous N2 at 77K is not as low
as 1.6kJ/mole. But I haven't been able to find a figure for it
anywhere on line.
So am I on the right track? If so, how do I compute the internal
energy from the LN2 injected into the cylinder?
Any help would be appreciated.
AS we say in the lab, KISS.
/_\P/_\V = energy, 101.325 jouls/liter-atm
Hmmm, I get 250 Joules/liter-atm, based on PV = (1.4-1)U. If I don't
put the 1.4-1, I get 101.325. Am I wrong in including the \gamma-1
factor?
1) Seal the cylinder with a piston at ambient pressure. No
problem.
A slight complication is that he isn't planning on starting with gas
at ambient pressure and temperature. He's running this thing with
waste gas from another process, so it'll be higher than ambient
pressure and temperature. But since he has a supply of the gas at
his initial working temperature and pressure, sealing the piston at
that pressure still isn't a problem.
2) Compress the gas. Losses due to friction and non-adiabatic
cylinder walls - for later.
I'm ignoring such losses right now.
3) Inject N2(liq). That's work - force through a distance for the
volume injected into hot high pressure.
I don't think he thought of that issue, I know I didn't. Using the
figures he's working with, it comes to about 20J. It's something to
keep in mind, but I'm not sure of it's a significant loss.
4) Vaporize; down stroke.
This is the part I'm concerned with computing. I know the rest of
this is working on general principles, but I feel I really need to
show him numbers and formulas all worked out. How much energy is used
in vaporizing the N2(liq) and bringing it to equilibrium temperature
with the gas already in the cylinder? What pressure and temperature
is in the cylinder after vaporization? This is what I don't know how
to compute. Once I know that, I can compute the pressure and
temperature of the gas in the cylinder prior to venting.
5) Vent excess gas; cylinder is now much cooler than at start for
vaporizing the liquid, work extracted, and venting. Loss of PV work
for venting. Repeat.
6) Everything quickly cools down and little excess happens.
As I said, in the actual device he's planning, he is using heated gas
from another process as input.
6) What of the work expended liquefying the nitrogen? All the
insulation on your cryogenic storage and plumbing?
I figure, based on a Carnot cycle, that to liquify 1 mole of N2 at
77K I need to pump 5.56kJ of heat out of the N2, and if I reject that
heat at 300K I'll need to use about 3 times that energy (or about
16.68kJ) to do it, assuming the theoretical highest efficiency.
That's not counting the work needed to cool the ambient N2 down to 77K
to begin with.
What of the work needed for that? He wants to use the rejected heat
(22.24kJ+/mol N2(liq)) for heating (domestic hot water, winter
heating, etc) -- and therefore free, since it displaces the energy
already used for that purpose. I can see his point, but I believe
that the amount of N2(liq) he'll get out of the process will be small
compared to the amount he'll need for the engine using it.
Never mind the insulation for the cryo storage and plumbing... I'm
wondering where he's going to get reliable injectors that'll take 77K
on one side and appromately 1000K on the other.
.
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