At 05:15 PM 10/10/2014, Alan Fletcher wrote:
b) If it were perfectly transparent, then we can treat the outside of the inner cylinder as the source.The energy per square can be calculated, but the area is smaller (as r^2)But what's the emissivity of the inner cylinder? Or can we assume that it's radiating as a pure black body?
The inner cylinder will be in thermal equilibrium, so I think it would have an emmissivity of 1.0
In that case the power/area goes up a lot, and the area goes down a little. So maybe the current analysis IS the lower of the two.
