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Heat transfer (convection & radiation)

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    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.




    2. Relevant equations
    Q_conv = hA(Ts-Tenv)
    Q_rad = e*sigma(Ts^4-Tsurr^)



    3. 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.
     
  2. jcsd
  3. Apr 7, 2010 #2

    Mapes

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    No radiation going towards the ground? :smile:
     
  4. Apr 7, 2010 #3
    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.
     
  5. Apr 7, 2010 #4

    Mapes

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    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?
     
  6. Apr 7, 2010 #5
    I figured it out, I forgot to included emittance for the amount of heat it radiates out. Thanks again.
     
  7. Apr 7, 2010 #6

    Mapes

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    I missed that detail too until you reported that the first answer was wrong!
     
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