Re: Slow Stealth
- From: Luke Campbell <lwcamp@xxxxxxxxx>
- Date: Fri, 29 Feb 2008 14:42:30 -0800 (PST)
On Feb 29, 1:36 pm, CharlesRCap...@xxxxxxxxx wrote:
On Feb 29, 10:36 am, Luke Campbell <lwc...@xxxxxxxxx> wrote:
I'm afraid that I didn't understand all of it however. Basically what
you are saying is that thinking of the waste as heat is problematic
and I should rather be thinking of the waste as units of entropy
(Joules per degree Kelvin) 1000 of them in this example.
It can simplify things.
I also understand that you are saying that regardless of any
efficiencies of compressors or other machinery, that it is more
expensive (in free energy, is that the correct term?) to get rid of
that entropy at a higher temperature. You have to radiate more heat to
get rid of the same amount of entropy.
Yes. Exactly.
Free energy has some technical meanings in thermodynamics, referring
to energy-like quantities of systems in thermal or particle
equilibrium with other systems, but I understand what you are talking
about.
Okay, assuming I am correct in understanding what you said, then it's
not really a difference in type of problem, but more a difference in
the scope of the problem. It's going to be significantly more
expensive to get rid of the entropy by increasing the temperature.
Expensive in terms of energy expenditure, yes. It can end up being
less expensive in terms of mass if you end up with smaller radiators.
One question I have is how you derived the 1000 K/J value for entropy
in the example system? What did you have to do to determine that?
I divided the heat produced by the reactor core (3 MJ) by the assumed
temperature of the reactor core and its coolant (3000 K). Things get
more complex if the heat is not exchanged at constant temperature.
Luke
.
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