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Heat transfer (convection & radiation) 
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#1
Apr610, 10:12 PM

P: 4

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^2K 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(TsTenv) Q_rad = e*sigma(Ts^4Tsurr^) 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. 


#3
Apr710, 12:37 PM

P: 4

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. 


#4
Apr710, 12:49 PM

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Heat transfer (convection & radiation)
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?



#5
Apr710, 01:04 PM

P: 4

I figured it out, I forgot to included emittance for the amount of heat it radiates out. Thanks again.



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