Steady state conditions of a heat-generating annulus

Question

Background: University heat transfer course, TA did a problem on the board involving a uniform heat-generating annular solid, cooled on the inside and outside by a coolant flow. The boundary conditions are given so that the inner surface and outer surface temperatures were the same ($T_i=T_o$). This ended up with a very clean solution to what the problem was looking for. Then I tried making $T_i\neq T_o$ and the calculations turned into a mess. Then I wondered if the steady state condition meant the temperatures necessarily had to equalize if there was uniform heat generation and cooling. I asked the TA and they couldn't give me a straight answer, and I don't know how to go about putting down a proof in one way or another. Searching online yielded some heat transfer information but nothing similar enough to the question I have.

Preamble: Consider a annular solid with $R_i < R_o$, uniform volumetric heat generation $\dot q$, uniform conductivity $k$, and cooled with the same coolant flow at both the inner and outer surfaces. The system is at steady state.

Proposition: $\boldsymbol{T_i = T_o}$ for a cross-sectional profile taken at any point along the length of the solid.

Thought process: The temperature profile will adjust itself in such a way that there is a uniform heat flux across each boundary in contact with the coolant flow. Taking advantage of the no-slip boundary condition we know that the only coolant particles that are receiving heat transfer is the differentially small volume spread across each surface, so we could say the contacting surface of the fluid is the same as that of the annular solid.

We are ignoring the effect the thickness of the annulus will have on coolant flow, as well as ignoring end effects due to heat transfer from the plane ends. Alternatively, consider the annular solid's (can I call it a pipe?) length to be infinite, I think they both lead to the same effect in the end.

So are my thoughts about this correct? If they are, how should I go about presenting it as a proof? If not, what is the actual (ideal) relation between $\boldsymbol{T_i}$ and $\boldsymbol{T_o}$?

Thank you for any help and insight you can provide regarding this matter.

Answer 1

No, your proposition is incorrect. There physical situation is perfectly well-defined, and the system's temperature profile just needs to adapt to it. The fact that the calculations got messy has nothing to do with whether the system is well-specified or not.

Now, the two-dimensional annular profile does make things more complicated, but to show that this is perfectly feasible, let me do the same thing but in one dimension, where the steady-state heat equation takes the form $$ -k\frac{\mathrm d^2T}{\mathrm dx^2}=\dot q, $$ with boundary conditions $T(x_i)=T_i$ and $T(x_o)=T_o$. The general solution of the differential equation reads $$ T(x) = -\frac{\dot q}{2k}x^2 +Ax+B, $$ and imposing the boundary conditions reduces it to $$ T(x) = -\frac{\dot q}{2k}(x-x_i)(x-x_o) + T_i\frac{x-x_o}{x_i-x_o} - T_o\frac{x-x_i}{x_i-x_o}, $$ proving that there is a single well-defined temperature profile for this configuration. (The details for the radial equation are slightly different, but structurally it is exactly the same.)

Finally, just to bolt things down completely, it's instructive to look at the heat flow at the edges: in general we have $$ k\frac{\mathrm dT}{\mathrm dx} = -\frac{\dot q}{2k}\left[(x-x_i)+(x-x_o)\right] + \frac{T_i- T_o}{x_i-x_o}, $$ which at the boundaries gives \begin{align} k\frac{\mathrm dT}{\mathrm dx}(x_i) & = -\frac{\dot q}{2k}(x_i-x_o) + \frac{T_i- T_o}{x_i-x_o} \quad \mathrm {and}\\ k\frac{\mathrm dT}{\mathrm dx}(x_o) & = -\frac{\dot q}{2k}(x_o-x_i) + \frac{T_i- T_o}{x_i-x_o}, \end{align} whose signs might differ depending on how big $(T_i- T_o)/(x_i-x_o)$, the externally-imposed temperature gradient, is when compared to the heating $\dot q$ over the given radii (and, in 2D, with some added complications coming from the geometry, but nothing major).

So yeah, there you have it. Linear differential equations tend to just produce superpositions of independent solutions, and that's all that's at play here.