Hummh,
We get an equilibrium of a cube to be about 55F (13C) when exposed to the sun on one side and all the other sides radiating to cold space. (assuming they are thermally connected).
I wonder why the big difference between our calculations? Bob, WB4aPR
On Sun, Jul 20, 2014 at 6:59 AM, Phil Karn karn@ka9q.net wrote:
On 07/19/2014 09:23 PM, Robert Bruninga wrote:
I cannot believe that. The equilibrium of a nominally black (solar
panels
on all sides) spacecraft is something like about 0 to 30 C (32F to 90F) a very benign operational range. The only time you DO have thermal issues
is
when you DO have attitude control and have things that are not equally
over
time seeing the sun and dark sky.
See Dick's paper for the details; I'm just quoting his results. I know the basic physics of heat transfer in space but I would never call myself an expert. He is.
But I can do a back-of-the-envelope calculation that tells me he's right.
The solar cells they're using have an absorptivity and emissivity that is both 0.98, as I recall, so a cubesat covered with them is essentially a perfect blackbody.
A blackbody cube with one face normal to the sun at 1 AU will reach an equilibrium temperature of -21.35 C. The problem is that the ratio of radiating area to absorbing area for a cube is 6:1 (with the sun normal to one surface). A sphere would be warmer because its ratio of radiating to absorbing area is only 4:1. A thin flat plate normal to the sun (like a solar wing) would be even warmer -- 2:1.
And that -21.35 C figure is for continuous sunlight. Throw in eclipses and things get much worse. Yes, it would be a little better when the sun shines on a corner rather than normal to a face, and Earth albedo and IR radiation will warm things a little, but not enough to matter.
--Phil
PS: Temperature of 10 cm blackbody cube at 1 AU:
Area facing sun: .01 m^2 Solar constant: 1367.5 W/m^2 Absorbed power = 13.675 W
Total radiating area: .06 m^2 Emissivity = 1.0 (perfect blackbody) Stefan-Boltzmann constant = 5.6703e-8 W/(m^2K^4)
T = (13.675 W / (5.6703e-8 * 1.0 * .06)) ** (1/4) = 251.8K == -21.35 C _______________________________________________ Sent via AMSAT-BB@amsat.org. Opinions expressed are those of the author. Not an AMSAT-NA member? Join now to support the amateur satellite program! Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb