• Support PF! Buy your school textbooks, materials and every day products Here!

Heat transfer (convection & radiation)

  • Thread starter fenix8o0
  • Start date
  • #1
4
0

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.
 

Answers and Replies

  • #2
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
17
No radiation going towards the ground? :smile:
 
  • #3
4
0
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.
 
  • #4
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
17
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
4
0
I figured it out, I forgot to included emittance for the amount of heat it radiates out. Thanks again.
 
  • #6
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
17
I missed that detail too until you reported that the first answer was wrong!
 

Related Threads for: Heat transfer (convection & radiation)

Replies
1
Views
792
Replies
1
Views
4K
Replies
0
Views
4K
Replies
12
Views
10K
Replies
1
Views
858
Replies
1
Views
916
Replies
1
Views
3K
Replies
3
Views
2K
Top