Heat conduction through a slab with internal heat generation

In summary: Qin - a*Qvol = QmidQout = Qmid + b*QvolQin = (Tm - T1)*(2A*k)/LQmid = (Tm - T1)*(2A*k)/L - a*QvolQout = (T2 - Tm)*(2A*k)/L + (b - a/2)*QvolWhy 2A instead of A?2A because each side has length L/2 - so Ak/(L/2) = 2Ak/L.
  • #1
rafehi
49
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 seperately.

Any help?
 
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  • #2
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.
 
  • #3
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
 
  • #4
Why 2A instead of A?
 
  • #5
2A because each side has length L/2 - so Ak/(L/2) = 2Ak/L.

Is the reasoning right?
 
  • #6
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.)
 

1. How does heat conduction occur through a slab with internal heat generation?

Heat conduction through a slab with internal heat generation occurs when there is a temperature difference between the two sides of the slab. Heat will flow from the higher temperature side to the lower temperature side, generating heat within the slab.

2. What factors affect the rate of heat conduction through a slab with internal heat generation?

The rate of heat conduction through a slab with internal heat generation is affected by several factors such as the thickness and thermal conductivity of the slab, the temperature difference across the slab, and the amount of heat generated within the slab.

3. How does the thermal conductivity of the slab impact heat conduction through a slab with internal heat generation?

The thermal conductivity of the slab is a measure of how well it conducts heat. A higher thermal conductivity means heat can travel through the slab more easily, resulting in a faster rate of heat conduction.

4. What is the equation for calculating heat conduction through a slab with internal heat generation?

The equation for heat conduction through a slab with internal heat generation is q = kA(T2-T1)/L, where q is the heat flow rate, k is the thermal conductivity of the slab, A is the cross-sectional area, T2-T1 is the temperature difference across the slab, and L is the thickness of the slab.

5. What are some practical applications of heat conduction through a slab with internal heat generation?

Heat conduction through a slab with internal heat generation is relevant in many industries, such as in the design of building insulation, heat exchangers, and electronic devices. It is also important in understanding the Earth's internal heat transfer processes, such as in the Earth's crust and mantle.

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