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. Suppose your reactor core produces 3 MJ of energy
at 3000 K. Because this is produced as heat, you also unavoidably end
up with 1000 J/K of entropy. I am going to simplify things for this
discussion and assume all processes are reversible and heat/entropy
exchanges take place at constant temperature. Now, 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. So you need to get rid of it. Since entropy is associated with
heat, you get rid of entropy by dumping heat. The amount of entropy S
that is transferred along with an amount of heat Q at temperature T is
S=Q/T or, rearranging, Q=ST (more rigorously dQ = T dS, and you need
to do an integration). So if you expel the entropy at a radiator
temperature of 150 K, you must also lose 1000 J/K * 150 K = 150 kJ of
energy in the process. If, 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 = 500 kJ. Therefore, the
hotter your radiator, the more energy you must radiate to get rid of
the same amount of entropy. This lowers your efficiency and makes you
more visible. Note that this accounting of entropy along with your
energy is exactly what gives you the thermodynamics-limited
efficiencies that Isaac mentioned.
Luke
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