Electronics and inhabitants of the International Space Station generate a signif

Question

Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station must get rid of. The only way that the station can exhaust thermal energy is by radiation, which it does using thin, 2.2m -by-3.4m panels that have a working temperature of about 6 degrees C.

How much power is radiated from each panel? Assume that the panels are in the shade so that the absorbed radiation will be negligible. Assume that the emissivity of the panels is 1.0.

Explanation / Answer

The emissivity ? simply characterizes something as a good absorber+radiator or a good reflector+insulator. It is a unitless constant which is ? = 1 for an optimal black-body radiator and which is ? = 0 for a perfect reflector. (Something which is implicit in the above statements: good radiators are also good absorbers. This is why good radiators are 'black bodies' -- they absorb whatever comes in perfectly. Shine a light on the Sun and it will absorb ~100% of that light.)

In your case, since these panels were designed explicitly to radiate out heat, I would assume an emissivity of 1 unless otherwise specified.

Q/dt=e*sigma*A*T^4
With:

A=7.48m^2
T=279k^4
sigma=7.48x10^-8 w/m^2*279K^4 = 453.229

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