Heat conduction through a slab with internal heat generation

Click For Summary

Homework Help Overview

The discussion revolves around heat conduction through a slab with internal heat generation. The problem involves a slab with specified temperatures at both ends and a point source of heat generation located in the center. Participants are exploring how to model the heat transfer through the material given these conditions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are considering the implications of modeling the heat generation as a point source versus a distributed source. There is discussion about dividing the heat flux between the left and right sides of the slab and establishing relationships between the temperatures and heat transfers at various points.

Discussion Status

Some participants are providing insights into how to set up the equations based on the heat transfer principles, while others are questioning the definitions and calculations presented. There is an ongoing exploration of the relationships between the heat transfers and temperatures, with no explicit consensus reached yet.

Contextual Notes

Participants are navigating potential assumptions about the heat generation and its impact on the overall heat transfer, as well as the boundary conditions applied to the problem. There are also considerations regarding the definitions of variables and the physical interpretation of the equations being used.

rafehi
Messages
48
Reaction score
1

Homework Statement


There's a slab of a material with temperature T1 on the left and the T2 on the right. The thickness of the material is l with area A. In the centre, there is heat generation Qvol in the centre, which is a thin rod.

Find the heat transfer Q through the material.

Homework Equations





The Attempt at a Solution


My issue is with modeling the Qvol. If it was heat generation through the material, I'd just use the DE
D2T/dt2 = -(1/k)Qvol
with BCs T(0) = T1
T(l) = T2
which is easily solved,

But it's a 'point' source of heat, and unless I'm mistaken, I can't use this. The lecturer seemed to hint that it could be solved with looking at each side of the slab separately.

Any help?
 
Physics news on Phys.org
I think the lecturer gave a good hint. Assume that the heat flux from Qvol is divided up between the left and right sides. For each side, you have a known temperature on one side and an unknown fraction of Qvol entering from the other side. Also, the fractions must add up to one. And finally, the temperature at the middle is the same in each equation. That should give you an equal number of equations and unknowns to solve the problem.
 
So basically treat it as a series question?

Defining the temps as
TL, TM and TR, (left, mid, right),
Qin as the Q coming in from the left,
Qmid as the heat through the middle,
Qout as the heat from the right,
a*Qvol as the fraction going into the left,
a*Qvol as the fraction going into the right,
L a the length with each side = to L/2:

Qin - a*Qvol = Qmid
Qout = Qmid + b*Qvol

Qin = (Tm - T1)*(2A*k)/L
so that
Qmid = (Tm - T1)*(2A*k)/L - a*Qvol

and (Fourier's Equation for the right side):
Qmid = (T2 - Tm)*(2A*k)/L

Solving the above two equations to get rid of Tm:
Qmid = (T2 - T1)*(A*k)/L - (a/2)*Qvol

So,
Qout = (T2 - T1)*(A*k)/L + (b - a/2)*Qvol

Qout = (T2 - T1)*(A*k)/L + (1 - 3a/2)*Qvol
 
Why 2A instead of A?
 
2A because each side has length L/2 - so Ak/(L/2) = 2Ak/L.

Is the reasoning right?
 
Ah, got it. That part's OK, then. It's just that your last line looks a little off and I'm looking for problems in the calculations.

Why do you have Qin = (Tm - T1)*(2A*k)/L when Qin is defined as the heat transfer coming in? If Tm>T1, this predicts positive flux coming in from the left, which doesn't make sense.

Then you have Qmid = (T2 - Tm)*(2A*k)/L; shouldn't this be Qout? (And again there seems to be a minus sign missing.)
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K