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 <k...@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 > _______________________________________________ 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