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Not a perfect blackbody

  • Thread starter Silviu
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Homework Statement


Two parallel plates plates are maintained at temperatures ##T_L## and ##T_R## respectively and have emissivities ##\epsilon_L## and ##\epsilon_R## respectively. Given the Stephan-Boltzmann constant ##\sigma##, express the net energy transfer rate per area from the left plate (L) to the right plate (R).

Homework Equations




The Attempt at a Solution


The way they solve it is by calculating the transfer rate to the right and left like this: $$I_R=\epsilon_L\sigma T_L^4+(1-\epsilon_L)I_L$$ and $$I_L=\epsilon_R\sigma T_R^4+(1-\epsilon_R)I_R$$ and the net transfer is $$I_{net}=I_R-I_L$$ I am confused about the ##(1-\epsilon)I## term. Where does that come from? How can ##I_L## which is the energy leaving the right plate, contribute to the flux of energy TO the right plate? It is already there, within the right plate.
 

Answers and Replies

  • #2
Charles Link
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There are multiple reflections that occur if just the radiated energy is considered. The ## (1-\epsilon)I ## is a reflectivity term from the opposite surface. (By Kirchhoff's law, ## \rho+\epsilon=1 ##. Reflectivity ## \rho=1-\epsilon ##). ## \\ ## This makes it possible to avoid computing multiple reflections. Two equations and two unknowns allows also for solving for ## I_R ## and ## I_L ## separately. ## I_R ## and ## I_L ## are the incident intensities onto the right and left surface, respectively, and they each include energy that may result from multiple reflections. ## \\ ## I think if you compute it, you will find ## I_R \epsilon_R-\epsilon_R \sigma T_R^4=-(I_L \epsilon_L-\epsilon_L \sigma T_L^4)=I_R-I_L ##. The net flow from left to right is what is absorbed minus what is radiated by the right surface, and it is the minus of what is absorbed minus what is radiated by the left surface. And, yes, this result follows with just a little algebra on the two equations that you presented in part 3 above.
 
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