Re: Want help on thermodynamics question: internal energy of a gas
- From: noshellswill <noshellswill@xxxxxxxxx>
- Date: Sun, 27 Apr 2008 14:04:42 +0000 (UTC)
On Sat, 26 Apr 2008 02:43:35 +0000, Gerard Westendorp wrote:
Buddha Buck wrote:
[..]
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.
This will work! But where do you get the liquid N2 from?
You need to do work to make liquid N2, part of which you get back when
you expand it.
Normally, we consider all thermal energy at room temperature to be
"waste", ie you cannot extract work from it. Pity, because there is a
huge amount of it.
But if you can find a free heat sink at 77K, suddenly you have solved
the world energy problem. Because you can then run heat engines
utilizing the heat at room temperature versus the heat sink at 77K.
The problem is, we have no easy access to a 77K heat sink, so we have to
make it using a cooling machine.
Perhaps it is confusing that the internal energy of liquid N2 is *less*
than that of ambient N2, yet you get work out of it, and you *need* work
to make it. When you make liquid N2, the energy of the cooling machine
goes to the surroundings, and the energy it extracts from the N2 goes
there too.
By the way, you don't need any gas in the cylinder. Just so doing an
*isothermal* expansion of the liquid : While expanding, get free energy
from the ambient by absorbing heat from it.
Gerard
G:
Yep, not lots different from a generic vapor-compression refrigeration
cycle. R134.a comes from the "throttle" as a L.T. liquid and vaporizes in
contact with a L.T. (but warmer) reservoir. If that evaporation
step happened in a piston it would do work.
nss
******
.
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- From: Buddha Buck
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- From: Gerard Westendorp
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