1 dimensional heat transfer question

Click For Summary

Homework Help Overview

The discussion revolves around a one-dimensional heat transfer problem involving steady-state conditions, heat conduction, and heat generation. Participants are analyzing the equations governing the system and exploring the implications of boundary conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between heat conduction and heat generation, questioning the appropriateness of equating the two. There is also scrutiny of the assumption that heat generation is uniform and the implications of boundary conditions on the solution.

Discussion Status

Some participants have provided guidance on the need to consider boundary conditions and the continuity of derivatives. There is an acknowledgment of the original equations presented by the poster, with suggestions to express constants in terms of boundary values. Multiple interpretations of the equations and their implications are being explored.

Contextual Notes

There are indications that the problem may involve non-uniform heat generation, and participants are questioning the assumptions made regarding the constants in the equations. The original poster has boundary conditions provided, which are being referenced in the discussion.

theBEAST
Messages
361
Reaction score
0

Homework Statement


zNE6DL6.png


Homework Equations


d/dx(kdT/dx) + S = 0 (for steady state)
q_cond = -kAdT/dx
q_gen = S*Vol

The Attempt at a Solution


ksn9AyQ.jpg


So I tried to equate q_cond to q_gen but it seemed to give me the wrong answer. Does anyone know why? I believe to get the right answer I have to use the first equation I wrote in the relevant equations section.

Does my method not work because S is not uniform? But then it tells me that at x = 0, S = So so it shouldn't matter since we have a boundary condition?
 
Physics news on Phys.org
The quick reply would be that there is no reason to equate q_cond to q_gen: there are two mechanisms that cause temperature change, namely conduction and heat generation. The sum of the contributions to dT/dt has to cause a steady state, hence the = 0.
 
Last edited:
Could you also check q_gen = S * Vol ? This looks too much as if S is constant all over the place.
 
Hmm, I am not too sure about the equations you use for S, but I think you need to do something about the constants, C2 and C3 as well.

I would suggest looking at the boundary conditions. You have already been given an expression for the temperature, which you solved correctly. When T=T2 you get 1 (as you also did).

But the deriviative of the equation also has to be continuous over the boundary. (Note: can you explain why?)

You then have 2 equations with 2 unknowns. Remember that at this boundary x=L! (which you also wrote down).

So you could then express the constants C2 and C3 in terms of L. I think that is what you should start by doing.
EDIT: I solved it now (phew), but the equation you need would be:

$$ \frac{d^2T}{dx^2}+\frac{S}{k}=0 $$

Are you familiar with this expression, or?
 
Last edited:
This is exactly the same equation you wrote down (correctly) in your original post. So you shouldn't be too surprised that it gives you the right answer.
 
Chestermiller said:
This is exactly the same equation you wrote down (correctly) in your original post. So you shouldn't be too surprised that it gives you the right answer.

But not the one used in the attached paper :-) but yeah, then you must be familiar with it!
 

Similar threads

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