Heat transfer (convection & radiation)

AI Thread Summary
The discussion revolves around calculating the equilibrium temperature of two aluminum plates, one polished and the other painted black, subjected to specific heat transfer conditions. The polished plate's calculated temperature was significantly higher than expected, prompting a reevaluation of the emittance factor in radiation calculations. The correct approach involves considering both the incoming radiation and the emissivity of the surfaces when determining heat loss. After adjusting for these factors, the equilibrium temperatures were found to be 303.23K for the black plate and 285.06K for the polished plate. The importance of accurately accounting for emittance in thermal radiation is emphasized throughout the problem-solving process.
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Homework Statement


Two large thin aluminum plates, the first polished (emittance = 0.05 and the second painted black, are placed horizontally outdoors, where they are cooled by air at 283K. The heat transfer coefficient is 5 W/m^2-K on both the top and bottom. The top is irradiated with 750 W/m^2 and it radiates to the sky at 170K. The Earth below the plate is black at 283K. Find the equilibrium temperature of each plate.




Homework Equations


Q_conv = hA(Ts-Tenv)
Q_rad = e*sigma(Ts^4-Tsurr^)



The Attempt at a Solution


First I started by thinking about what equilibrium temperature is and I think its the change in heat transfer in the plate is zero.

Second, I drew a figure and I labeled the directions, we have the 750 W/m^2 going into the plate, radiation coming out, and convection going out from both the top and bottom faces.

I attempted it with the black plate and got an answer fairly close to the answer. The answer is 303.23K and I got about 310K. However, using the same method I got an answer much higher with the polished plate which leads me to believe my approach is wrong.


Thank you for any help.
 
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No radiation going towards the ground? :smile:
 
Mapes said:
No radiation going towards the ground? :smile:
Thank you! Using the temperature of the Earth as the isothermal heat sink for that direction of radiation, I got the correct answer!

But, my problem with the polish plate still exists where I am getting a higher temperature than the black plate. The correct answer should be 285.06K. I added the emittance to the radiation going towards the sky and also tried going towards the earth. My answer is 326K and 359K respectively.
 
Think of what happens when you shine 750 W m-2 at a mirror. It is all going to be absorbed, or do you need to adjust that value to take reflectivity into account?
 
I figured it out, I forgot to included emittance for the amount of heat it radiates out. Thanks again.
 
I missed that detail too until you reported that the first answer was wrong!
 
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