Heat Transfer fourier's law Problem

Click For Summary

Homework Help Overview

The discussion revolves around a heat transfer problem involving a wall with specified thickness, thermal conductivity, and surface emissivity. Participants explore the application of Fourier's law, Stefan-Boltzmann law, and convection principles to determine the rear surface temperature of the wall based on given front surface conditions and surrounding temperatures.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relevance of the wall area in calculations and whether it can be assumed or omitted. There are attempts to derive equations for radiative and convective heat fluxes and how they relate to conductive heat transfer through the wall. Questions arise regarding the correct application of temperature conversions and the impact of steady-state conditions on heat flux calculations.

Discussion Status

Several participants have provided insights into the relationships between different heat transfer mechanisms, suggesting that the power conducted through the wall must equal the power lost through radiation and convection. There is ongoing exploration of the equations needed, with some participants expressing uncertainty about their calculations and assumptions.

Contextual Notes

Participants note constraints such as the lack of numerical values for certain parameters and the challenges of using mobile devices for typing equations. There is also mention of potential errors in calculations leading to unexpected temperature results.

  • #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.
 
Physics news on Phys.org
  • #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 celsius - 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
 
  • Like
Likes   Reactions: timmayy
  • #38
Ah ok, thanks a lot :)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K