On Feb 29, 10:36=A0am, Luke Campbell <lwc...@[EMAIL PROTECTED]
> wrote:
> On Feb 29, 6:08 am, CharlesRCap...@[EMAIL PROTECTED]
wrote:
>
> > The difference that I'd like to go with is that there is no reason I
> > can't take the heat that is radiated into the ~150K reservoir and then
> > compress it though a series of compressor/intercoolers until it is at
> > a high enough temperature to be rejected from a reasonable sized
> > radiator. The issue would be if the cost in waste heat and electrical
> > power would be too high to make that practical.
>
> There is a reason. =A0Suppose your reactor core produces 3 MJ of energy
> at 3000 K. =A0Because this is produced as heat, you also unavoidably end
> up with 1000 J/K of entropy. =A0I am going to simplify things for this
> discussion and assume all processes are reversible and heat/entropy
> exchanges take place at constant temperature. =A0Now, you can't just
> keep that 1000 J/K of entropy around - as your reactor produces more
> heat that entropy will pile up and eventually make things bad for
> you. =A0So you need to get rid of it. =A0Since entropy is associated
with
> heat, you get rid of entropy by dumping heat. =A0The amount of entropy S
> that is transferred along with an amount of heat Q at temperature T is
> S=3DQ/T or, rearranging, Q=3DST (more rigorously dQ =3D T dS, and you
need=
> to do an integration). =A0So if you expel the entropy at a radiator
> temperature of 150 K, you must also lose 1000 J/K * 150 K =3D 150 kJ of
> energy in the process. =A0If, however, you recompress the helium to
> radiate it at higher temperatures, say at 500 K, then you also lose
> more energy in the process, 1000 J/K * 500 K =3D 500 kJ. =A0Therefore,
the=
> hotter your radiator, the more energy you must radiate to get rid of
> the same amount of entropy. =A0This lowers your efficiency and makes you
> more visible. =A0Note that this accounting of entropy along with your
> energy is exactly what gives you the thermodynamics-limited
> efficiencies that Isaac mentioned.
Thank you for taking the time to write that.
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.
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.
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.
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?
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