I have a confusion about how to calculate the thermal resistance of a solid rod ( not hollow).
The general formula is
$$R=\ln(r2/r1)/(2\pi Lk)$$
In my case, $r1=0$
Any hint would be highly appreciated.
[Physics] Thermal resistance of a rod
thermal conductivitythermodynamics
Related Solutions
As you increase the temperature, the electrons, which are a cold Fermi gas, get more excited. Only a thin skin of electrons around the Fermi energy, of width kT, can do anything at all, like conduct heat. As the temperature increases, the electrons can conduct more heat.
The heat conductivity of electrons is greater than the heat conductivity of phonons, and together these account for the entire thermal conductivity. What is going on is that as the temperature goes up, the higher thermal conductivity electrons are carrying a larger fraction of the heat, and this makes the thermal conductivity go down.
In the absence of electronic thermal conductivity, for an insulating material, the thermal conductivity would go down with temperature, and this is also true for the thermal conductivity of just the phonons in the metal. But the electron contribution leads to this otherwise paradoxical effect.
It always helps to remember that a metal is never classical, the electrons are always quantum. A metal is a like a gigantic chemical bond involving all the atoms in a metal nonlocally, this is the conduction band, and the shared electrons have classically paradoxical properties. This is why the Drude model is wrong and the Fermi model is right.
When a rod is heated, all the work is produced by the source of the heat and it goes into increasing the internal, thermal and elastic, energy of the rod and into heating of the surrounding air and other objects.
If the rod is not mechanically constrained, the only work it performs, or rather passes along, due to its expansion is against the atmospheric pressure, but this work represents a very small fraction of the total work performed by the source, so it could be neglected.
This is because the average pressure required to expand a steel rod by 0.1% should be (assuming Young's modulus for steel of $29000000psi$) on the order of $\frac {29000000psi\times 0.001} 2=14500psi$, which is about 1000 times greater than the atmospheric pressure, $14.7psi$. So, only about $0.01\%$ of the work performed (energy spent) by the heat source to expand the rod will be passed to the air.
If the rod was mechanically constrained, a greater fraction of the work or spent energy could be passed along and, given a specific load, we could calculate it.
For instance, if the rod was supporting a $100kg$ mass, its expansion by $1mm$ would lead to the increase of the potential energy of the mass by $mgh=1J$. Of course, a higher temperature and more energy would be needed to achieve the same $1mm$ expansion under load.
Best Answer
As the inner radius tends to towards zero the inner area through which the heat has to flow also tends to zero and this is the reason for the thermal resistance tending to infinity.
So you have the correct formula which correctly predicts that, as the inner radius tend to zero, the thermal resistance tends to infinity.