Radiation - finding emissivity of a sphere

  • Thread starter JoeyBob
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  • #1
JoeyBob
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Homework Statement:
see attached
Relevant Equations:
dQ/dt = Ae(5.67*10^-8)*T^4
So using the above equation, e=dQ/dt / (A*5.67E-8*303.8^4)

The surface area of a sphere is 4(pi)r^2 and I get 136.8478 m^2. dQ/dt would be the net radiation (I think? Its in the correct units), 1074W.

Plugging everything in I get 0.01625, but the answer is 0.0524.

Now as I was writing this I figured out how to do it by trial and error but am a bit confused. So apparently the surrounding air is decreasing the net radiation? Is that because its of a lower temperature? So if the air was hotter would I add it instead of subtracting it? Below is the equation I used to solve.

dQ/dt = SA * constant * e * (T of sphere) ^4 - SA * constant * e * (T of air) ^4
 

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  • #2
haruspex
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The question setter seems to expect you to treat the surrounding air as a black body, so subtract the radiation the sphere absorbs from that. The fraction it absorbs will be governed by the same e value.
But it's nonsense. It takes a pretty thick layer of air to radiate much at all, and it certainly is not a black body (fortunately). Even outdoors, most of the received radiation would be from other solid bodies, such as the ground.
 
  • #3
JoeyBob
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The question setter seems to expect you to treat the surrounding air as a black body, so subtract the radiation the sphere absorbs from that. The fraction it absorbs will be governed by the same e value.
But it's nonsense. It takes a pretty thick layer of air to radiate much at all, and it certainly is not a black body (fortunately). Even outdoors, most of the received radiation would be from other solid bodies, such as the ground.
Am I understanding correctly that I subtract the absorbed radiation because its of a lower temperature? In other words, would I add it instead if the surrounding black body had a higher temperature than the sphere?
 
  • #4
haruspex
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Am I understanding correctly that I subtract the absorbed radiation because its of a lower temperature? In other words, would I add it instead if the surrounding black body had a higher temperature than the sphere?
No, you are told the net radiation from the sphere, so that's radiation out minus radiation in. And 'in' should only be what is absorbed, so depends on e.
 

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