On 07/20/2014 04:58 PM, Robert Bruninga wrote: > 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?
Well, here's mine again: Stefan-Boltzmann law: P = A*e*s*T^4, where P = radiated thermal power in watts A = radiating area in m^2 e = emissivity (dimensionless) s = Stefan-Boltzmann constant: 5.6703e-8 W/(m^2 K^4) T = absolute temperature, kelvins Solving for T: T = (P / (A*e*s)) ** (1/4) The solar constant at 1 AU is 1367.5 W/m^2. For a 10 cm cube with a = e = 1.0 (blackbody) at 1 AU, with the sun normal on one surface, the absorbed solar thermal power is 1367.5 * .01 = 13.675 W. The radiating surface is six times the absorbing area, or .06 m^2, because it radiates from all six sides (including the side facing the sun). Plugging into the previous formula: T = (13.675 / (.06 * 1.0 * 5.6703e-8)) ** (1/4) = 251.8 K = -21.4 C. Shape matters a very great deal. If you have a blackbody sphere instead of a cube, then the ratio of radiating to absorbing area is only 4:1 instead of 6:1 and the equilibrium temperature rises to 278.7K = +5.5C. A thin flat plate (normal to the sun) is only 2:1 so it's even warmer: 331.38K = +58.2 C (quite toasty). Here's why we may get different answers. If the sun shines on an edge of the cube instead of normal to one surface, the effective absorbing area increases to sqrt(2) times that of a single face. In that case, the equilibrium temperature increases to 274.5K = +1.4C. And if the sun shines directly on a corner, the absorbing area becomes (3/4) * sqrt(3) times the area of a single face and the equilibrium temperature is 268.8 K = -4.3C. But without active attitude control you can't count on *any* specific orientation with respect to the sun, and that's precisely the problem. Even a random average is not necessarily useful if the cube is small (with a low thermal mass) and turning very slowly. Add the ~90 min periodic changes due to eclipses and the much slower periodic changes with the seasons and you can see why the thermal design of a 1U unstabilized cubesat is such a challenging problem. It's much less of a problem on a larger spacecraft because you don't have to cover every surface with solar cells to maximize power generation, you have a much greater thermal mass to smooth out eclipse (but not seasonal) variations, and you have room for insulation to thermally isolate the external surfaces from the interior components. --Phil _______________________________________________ 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
