Heat Transfer fourier's law Problem

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The discussion revolves around solving a heat transfer problem using Fourier's law, Stefan-Boltzmann law, and convection principles. Participants are trying to determine the rear surface temperature of a wall given its front surface temperature, thermal conductivity, and surrounding conditions. Key equations for radiative and convective heat fluxes are established, emphasizing that these fluxes are in parallel with the conductive flux through the wall. Some users express confusion over calculations, particularly regarding temperature conversions and the relationship between the heat fluxes. Ultimately, the conversation highlights the importance of correctly applying heat transfer equations to find the rear surface temperature accurately.
  • #31
guppygould said:
Radiative heat flux as 17470 & convective heat flux as 2600 and I've tried both ways with the temperatures because I had it in mind from one of your earlier posts. Thanks for all your help and comments CHET I can see I'm probably getting to be a pain now. Is the correct equation for the conductive heat flux Q=-K(deltaT) though?

q = -k (ΔT/Δx)

Also, recheck your radiative flux calculation. It seems too high by a factor of about 10.
 
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  • #32
Ok I will do tomorrow! Thanks again
Leo
 
  • #33
And as if by magic I have an answer that looks reasonable, I can't thank you enough for your help CHET!
 
  • #34
Hi, I've been working through this same question and found an answer of 439 centigrade - which seems far too high for me, i assumed the back wall would be a lower temperature than the front wall, and so must be less than 152 (the front wall) and larger than 0 obviously
 
  • #35
my working is as follows:
Qconduction - Qconvection - Qradiation = 0

Qconduction = (0.65 x (425 - T2)) / 0.05 = 5525 - 13T2 (the 425 being 152 + 273 to convert into kelvin)

Qconvection = 20 x (152 - 22) = 2600

Qradiation = 0.8 x 5.67x10-8 x (4254 - 2954) = 1136.36

so,
(5525 - 13T2) - 2600 - 1136.36 = 0

1788.64 = 13T2
137.58 = T (kelvin)
-135.41 = T (celsius)
which is clearly wrong
 
  • #36
i also tried changing the T1 and T2 in the conduction equation:
Qconduction = (0.65 x (T1 - 425)) / 0.05 = 13T1 - 5525
and so the final answer will be 439.5 celcius - which seems to high to me
 
  • #37
Let me try to help you get back on the right track.

The three temperatures of interest are:
  • Inside wall temperature 152
  • Outside wall temperature T2
  • Air temperature 20
Which two of these temperatures determine the conductive heat flux?
Which two of these temperatures determine the convective heat flux?
Which two of these temperatures determine the radiative heat flux?

Based on the answers to these three questions, write an equation for each of the three heat fluxes in terms of the appropriate temperatures.

Are the convective and radiative heat fluxes in series or in parallel with one another?
Is the conductive heat flux in series or in parallel with the combination of radiative and convective heat fluxes?

Chet
 
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  • #38
Ah ok, thanks a lot :)
 

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