I have the steady state equation with an internal source as

$$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2k\frac{\partial T}{\partial r}\right)+Q=0$$

which has the analytic solution


where I know $C_1$ and $C_2$ based on some boundary conditions.

However, if I would like to find at what depth a particular temperature ($T_i$) occurs, should I go about this using a numerical approximation such as the midpoint method or I could solve analytically, recovering three different solutions for the depth:

$$r_i=\left[-C_1,\pm\sqrt{-\frac{6k(T_i-C_2-C_1)}{Q}} \right]$$

Analytical solutions are always best, so how do I know which one to use for this? Should I make sure that the temperature I am estimating the depth for falls between the boundary conditions?


1 Answer 1


The steady-state heat equation (Fourier) for a spherical geometry (with full symmetry) with internal heat generation $Q$ ($\mathrm{Wm^{-3}}$) can be written as: $$\frac{1}{r^2}[kr^2 T'(r)]'+Q=0$$

where $T(r)$ is the temperature in function of $r$ and $k$ is the thermal diffusivity of the material. We get: $$\frac{1}{r^2}(2kr T'+kr^2T'')+Q=0$$ $$\frac{2T'}{r}+kT''+Q=0$$ $$krT''+2kT'+Qr=0$$

This solves indeed (Wolfram alpha confirmed) to:

$$T(r)=-\frac{Qr^2}{6k}+\frac{C_1}{r}+C_2$$ where $C_1$ and $C_2$ are integration constants, to be determined by the boundary conditions (BCs).

Boundary conditions:

Firstly, look at the term $C_1/r$. It makes $T(r) \to \infty$ for $r \to 0$. Such a singularity can of course not be accepted and we must assume that $C_1=0$, a very common assumption in similar cases.

So that:


Resolving $r$ for a give $T(r)$ now becomes easy:


As regards the second BC, we're quite limited in our choice.

A BC containing the term $\Big(\frac{\text{d}T}{\text{d}r}\Big)_{r=R}$ would be related heat flowing off the boundary, $r=R$.

Or, assuming the sphere loses heat through convection, then with Newton's Cooling Law:


where $V$ is the volume of the sphere, $A$ its surface area, $h$ the heat transfer coefficient and $T_e$ the temperature of the environment. The equation relies on 'heat in = heat out'.

$T(R)$ can easily be extracted from $(2)$.

So the simplest BC is $[R,T(R)]$. Inserted into $(1)$ this will yield $C_1$.

Note that maintaining $[R,T(R)]$ means constant cooling of the boundary at $r=R$ is needed, because of the constant heat flux $Q$.


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