[amsat-bb] ANS-199 AMSAT News Service Special Bulletin - AMSAT Fox-1C Launch Opportunity Announced
bruninga at usna.edu
Sun Jul 20 23:58:26 UTC 2014
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?
On Sun, Jul 20, 2014 at 6:59 AM, Phil Karn <karn at ka9q.net> wrote:
> On 07/19/2014 09:23 PM, Robert Bruninga wrote:
> > I cannot believe that. The equilibrium of a nominally black (solar
> > 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
> > when you DO have attitude control and have things that are not equally
> > 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.
> 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
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