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Heat transfer (convection & radiation)

by fenix8o0
Tags: convection, heat, radiation, transfer
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fenix8o0
#1
Apr6-10, 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^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.
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Mapes
#2
Apr7-10, 06:13 AM
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No radiation going towards the ground?
fenix8o0
#3
Apr7-10, 12:37 PM
P: 4
Quote Quote by Mapes View Post
No radiation going towards the ground?
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.

Mapes
#4
Apr7-10, 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?
fenix8o0
#5
Apr7-10, 01:04 PM
P: 4
I figured it out, I forgot to included emittance for the amount of heat it radiates out. Thanks again.
Mapes
#6
Apr7-10, 01:21 PM
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I missed that detail too until you reported that the first answer was wrong!


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