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The figure shows the tungsten filament with a constant diameter except for a piece of it which has half of the diameter as the rest of the wire. Assume the temperature is constant within each part and changes suddenly between the parts. If the temperature of the thick part is 2000K, the temperature of the thin part of the filament is as: enter image description here

I considered applying Fourier law hereby assuming the temperature of the middle part as $T$ and the thick part taking the midpoint of the thin part than from the midpoint of the thin part to end thick rod. After that assuming steady-state and approximating derivative as finite difference (*)

$$ \frac{dQ_{in} }{dt} = \frac{dQ_{out} }{dt}$$

By fouriers law:

$$ \frac{dQ}{dt} = - k A \frac{dT}{dx}$$

Hence,

$$ -k A \frac{\Delta T_{left}}{\Delta x} = -kA \frac{\Delta T_{right}}{\Delta x}$$

Now if I since I took the midpoint of the rod, the $\Delta x$ cancel, the areas cancel and conductivity cancels. This leaves me with:

$$ \Delta T_{left} =\Delta T_{right}$$

Now, the temperature difference from middle part to right (I.e: $ \Delta T_{right}$ )is given as $ 2000-T$ and for left part it is given as $ (\Delta T_{left})$ which is $ T-2000$:

$$ T-2000 = 2000 -T$$

Hence,

$$ T=2000$$

But... this is wrong?

Self-Research attempts:

In solutions from some websites, they equate the term which looks like a joules heating effect in current electricity to the Stefan Boltzmann law. I have written the equation they've used below:

$$ ( \frac{dQ}{dt})^2 R = \sigma A T^4$$

Where R is the thermal resistance

As far as I understand, the Stefan Boltzmann law is regarding radiation and the Fourier heat law is heat transfer through thermal conduction. I assume that they derived the equation of $ ( \frac{dQ}{dt} )^2 R$ using Fourier's law, this makes it even more confusing of how these two forms of heat transfer are equal.

*: I feel a bit uneasy doing this because it is written in the question that the temperature spike is sudden so the function must be discontinuous at that point

tryst with freedom
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1 Answers1

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If the filament is operating as part of an electrical circuit, then the electrical current I has to be the same in all three sections. If the inherent resistivity of the wire is $\rho$, then the resistance of the entire thin section is $\frac{\rho L}{A}$, where L is the length of the thin section and A is its cross sectional area. So the rate of heat generation in the thin part is $$Q=I^2\frac{\rho L}{A}$$This has to be equal to the rate of heat removal:$$I^2\frac{\rho L}{A}=\sigma \pi DLT^4$$So, $$T^4=\frac{I^2\rho }{\sigma \pi D A}$$The product DA is 8 times larger in the thick section than in the thin section. So $T^4$ is 8 times larger in the thin section than the thick section.

Chet Miller
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  • You had told that it was incorrect to say that the electric current heating is equivalent heating but then you wrote the same.. – tryst with freedom Oct 09 '20 at 14:39
  • In this case, instead of the internal energy of the filament increasing, the electrical energy enters and heat exits. – Chet Miller Oct 09 '20 at 14:41
  • So, what was different previously which made this idea fail? – tryst with freedom Oct 09 '20 at 14:44
  • I don't recall the example you are citing. – Chet Miller Oct 09 '20 at 14:46
  • You had written this : "They have assumed that the rate of heat generation from electrical current stays the same in the thin center section as in the two end sections. This can't be correct" in the comments of the original post – tryst with freedom Oct 09 '20 at 14:47
  • Or were you referring to my work when you said " this can't be correct" – tryst with freedom Oct 09 '20 at 14:47
  • The rate of heat generation per unit length of filament is inversely proportional to the cross sectional area, so it is 4x greater in the thin section than in the thick section; it is not the same in both sections. – Chet Miller Oct 09 '20 at 14:51
  • Oh that makes sense. One more question, you told there is no axial temperature flow, only radial. However wouldn't it be possible that there is heat flow from both thick sides into the middle thin part because it may be colder ( we don't know it's temperature). How did you deduce the direction should only be radial without knowing it's the actual temperature? – tryst with freedom Oct 09 '20 at 14:56
  • I think I am missing some point about steady state but I can't figure it out – tryst with freedom Oct 09 '20 at 14:57
  • In steady state, nothing is changing with time at any given spatial position. – Chet Miller Oct 09 '20 at 15:25