Calculating Cooling Rate for a Black Body in a Vacuum with a Radiation Screen

AI Thread Summary
To calculate the cooling rate of a black body in a vacuum with a radiation screen, the primary heat transfer mechanism is black body radiation, which follows the Stefan-Boltzmann law, proportional to T^4. The equations for heat transfer should account for the radiative heat loss from both the black body and the radiation screen. The cooling rate can be expressed as dQ/dt = e*σ*4πr²(T^4_sphere - T^4_vacuum), where e is emissivity and σ is the Stefan-Boltzmann constant. The impact of the thin shell can be considered negligible in some cases, but its influence on the overall cooling rate should not be dismissed entirely. Understanding these principles is crucial for accurately determining the system's cooling dynamics.
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Homework Statement


A spherical black body with a radius r and a temperature T is placed in a spherical thin shell (radiation screen) with a radius R. The shell is also absolutely black on both sides. The space between the shell and the sphere and the space outer of the shell contain vacuum, i.e., no gas.
Find the cooling rate


Homework Equations



R = dQ / dT = h*A*dT


The Attempt at a Solution


The black body radiates heat with
h - some heat transfer constant
A - area of the radiating body

\frac{dQ}{dT}_{bb} = h_{vacuum} * 4 \Pi *r^{2} (T_{sphere} - T_{vacuum})

the same works for the radiation screen

\frac{dQ}{dT}_{rs} = h_{gas} * 4 \Pi *R^{2} (T_{vacuum} - T_{environment})

The point is that I'm not sure how to find the overall cooling rate of the system. Thanks.
 
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In a vacuum, there will be no conduction or convection heat transfer (your equations are for convection). The only heat transfer mode will be black body radiation which is proportional to T^4.
 
Upps, I mixed up radiation, conductivity...

So

dQ/dt = e*\sigma*4\Pi*r^{2}(T^{4}_{sphere}-T^{4}_{vacuum})

but what about the radiation screen? It is a thin shell. Can I neglect it?
 
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