Heat conduction with variable conductivity

[SC83]

I want to reduce the heat conduction through a steel rod by locally reducing the cross-section of the steel rod. However, it seems that the conduction through the rod is not dependent on the position of the reduced cross-section (i.e. near the hot or cold surface). I'm trying to explain this mathematically (see attachment Link), but I'm stuck. Can somebody help me?

Boundary conditions:
[ul]
[li]The temperature through the rod goes from 300 K to 20 K.[/li]
[li]The rod has a reduced cross-section A2 << A1. A2 has a length L2. The cross-section A1 = A3.[/li]
[li]The total length of the rod doesn't change, i.e. L1 + L2 + L3 = L. The length of L2 is constant, L1 and L2 can vary to position the reduced cross-section near 300 K or 20 K.[/li]
[li]The thermal conductivity changes with the temperature, i.e. k(T) = k0 + aT.[/li]
[/ul]

Question: why is Q not changing with L1 and L3 and how can I prove this mathematically (see attachment Link)?

 

[IRstuff]

Why do you think it would change?

From the simple perspective of Q = T * S, where S is the thermal conductivity in W/K, the equation doesn't care how you got the conductivity to go down.

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[georgeverghese]

Does the k value increase considerably at 20K in comparison to that at 300K?

 

[mskds545]

In your pdf, the next step is to substitute k1 = k0 + a/2*(T1+T2), k2 = k0 + a/2*(T2+T3), k3 = k0 + a/2*(T3+T4). Because thermal conductivity is linear with temperature, you can use the average temperature. Whether you can use this to show analytically that Q is independent of L1, or if it's worth the effort to do so, is up to you.

By the way, it is not only true that Q is independent of L1 if k is linear with temperature. It's true for any arbitrary function k(T), even a discontinuous one. You can demonstrate this by numerically integrating Fourier's law for any k(T).

-mskds545