Heat transfer (convection & radiation)

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Discussion Overview

The discussion revolves around a homework problem involving heat transfer through convection and radiation between two aluminum plates with different surface finishes. Participants explore the calculation of equilibrium temperatures for the plates under specified conditions, including external irradiation and environmental temperatures.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant describes their approach to finding the equilibrium temperature, noting the importance of heat transfer balance.
  • Another participant questions the initial assumption regarding radiation going towards the ground, suggesting that the Earth's temperature should be considered as a heat sink.
  • A participant reports achieving the correct answer for the black plate after incorporating the Earth's temperature but struggles with the polished plate, indicating discrepancies in their calculations.
  • One participant suggests considering the reflectivity of the polished surface when calculating absorbed radiation from the 750 W/m² input.
  • Another participant acknowledges a mistake in their calculations related to the emittance of the polished plate, which they later correct.
  • A participant admits to missing the same detail regarding emittance, indicating a shared challenge in the problem-solving process.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating the equilibrium temperature, particularly regarding the polished plate. There is no consensus on the correct methodology or final answers, as participants continue to refine their calculations.

Contextual Notes

Participants highlight limitations in their initial assumptions, particularly regarding the treatment of radiation and emittance for the polished plate. The discussion reflects ongoing adjustments to their calculations based on these considerations.

Who May Find This Useful

Students and educators interested in heat transfer concepts, particularly in the context of convection and radiation, may find this discussion relevant.

fenix8o0
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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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