[ExI] extropy-chat Digest, Vol 96, Issue 36

Keith Henson hkeithhenson at gmail.com
Thu Sep 22 16:32:39 UTC 2011

On Thu, Sep 22, 2011 at 5:00 AM,  Eugen Leitl <eugen at leitl.org> wrote:


> Radiative losses go with T^4, so use a higher temperature.
> E.g. sodium-potassium cooled nuclear reactors in space need
> pretty small surface area to radiate nonnegligible power.

That's true, but not the whole story.  NASA got hung up by a hand
waving analysis of this kind back in 1962 and I don't think ever
considered it again.

Most of the time systems that need radiators are making power with the
temperature difference between a hot source and a cold sink.  The
Carnot efficiency is 1-Tcold/Thot, real engines are around 75% of
theory.  I.e., the colder the better for the heat sink.  Of course
colder radiators are larger and heavier.  So the question becomes,
what temperature is best (lowest mass per kW of output).

You can set this up as as an equation for the sum of radiator mass and
heat source mass then solve for the temperature where the derivative
is zero.  Or put the equations in a spread sheet and graph the

I did the latter using the assumption that solar heat collectors (one
side) and heat radiators (both sides) mass about the same.  Results


I can also send people the spreadsheet if anyone wants it.


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