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# Rocket Science/Thermal

• Subject: [amsat-bb] Rocket Science/Thermal
• From: Bob Bruninga <bruninga@xxxxxxxx>
• Date: Sat, 7 Apr 2001 00:12:05 -0400 (EDT)

```For you space/thermal/rocket scientists, I have a question:

Lets say I have a 1 foot cube satellite with 100W per sqft of solar
radiation shining on one side and the other sides all facing Black Space
and the absorbtivity and emissitivity of our solar panels on all sides are
such as to give us a thermal equilibrium of say 0 degrees C.

Now, lets say our solar panels are 10% efficient, so 10 watts of
electrical power is being delivered internally to the spacecraft.  But
this doesn't matter, since it also becomes heat which is no different
than if the solar panels produced no current and converted 100% of the
solar radiation to heat in the first place.

BUT, Now assume I shunt all of that 10 watts to an EXTERNAL resistor
thermally isolated from the satellite and radiating to black space.  It
seems to me that the satellite must then cool down due to the loss of the
10W no longer remaining in the satellite itself.  Will it be 10% cooler?
At 270 deg Kelvin, that could be as much as 27 degrees cooler?

Or am I overlooking something?

We are mounting the "10W" resistor on a standoff, painting it white for
maximum radiation and minimum absorption, and then placing a highly
reflective surface between it and the satellite so none of its radiated
heat gets back to the satellite.  We do understand that for every inch of
"reflective isolation" coating we have on the satellite body, we are also
then losing that much radiative cooling to black space, so it must be kept
small.

Just wanna make sure this experiment is worth doing...

de WB4APR, Bob

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