I have a closed cylinder with a gas inside at 3,000 K.

I want to find the temperature profile (over time) through the cylinder wall, something like figure (b) shown below:

In my case, however, $T_\infty$ does not stay fixed at 3,000 K over time but changes (cooling) due to heat transfer through the cylinder wall.

Unfortunately I was not able to find any solved problem like that in textbooks because any existing solved problem considers the temperature inside the bulk of gas as a constant.

How can I model this problem?

  • 1
    $\begingroup$ It looks like you would need to solve the heat equation with a convection boundary condition to obtain these temperature profiles. $\endgroup$
    – Paul
    Commented Dec 8, 2015 at 3:30
  • $\begingroup$ It would be helpful if you can state your initial and boundary conditions more completely. There are analytic solutions for many problems like this, but I can't say which, if any, would apply without knowing the specifics. $\endgroup$
    – Dan
    Commented Dec 9, 2015 at 11:40
  • $\begingroup$ Maybe my answer to this question will help. It is about a composite material, but the equations are not any different in your case. Specifically check out Simulating a composite material, it even provides a working Python script :) $\endgroup$
    – nluigi
    Commented Dec 10, 2015 at 21:43

1 Answer 1


I don't know the whole solution, but I can show you a way you can probably go.

As @nluigi pointed out in his comment, the governing equations of your problem are given in Modeling Transient Heat Transfer between two 1-D materials:

$$ \frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2} $$

with temperature $T$, thermal diffusivity $\alpha=\frac{k}{\rho \, c_P}$, time $t$, space $x$, thermal conductivity $k$, density $\rho$ and specific heat capacity $c_P$.

The fundamental solution of this differential equation is given by

$$ H(x,t) = \frac{1}{(4 \pi \alpha t)^{\frac{n}{2}}} \, \exp \left( - \frac{|x|^2}{4 \alpha t} \right) $$

where $n$ is the dimension of your space coordinate $x$ and $|x|^2 = \sum_{i=1}^n x_i^2$ is the square of the Euklid norm (with $x_i$ being the coordinates of $x$ in a cartesian coordinate system). Please note that $x$ may be a scalar value for one-dimensional problems ($n=1$), a two-dimensional vector for two-dimensional problems ($n=2$) or a three-dimensional vector for three-dimensional problems ($n=3$).

An example for a one-dimensional problem is heat transfer through a wall. Your problem is (approximately) two-dimensional. A three dimensional problem would be a hot or cold sphere within a large room.

To calculate a solution for a specific problem, you need the initial conditions, given as

$$ T(x,t=0) = T_0(x) $$

and calculate the convolution between the fundamental solution and the initial conditions:

$$ T(x,t) = (H * T_0) (x,t) = \int\limits_{{\mathbb R}^n} H(x-y,t) \, T_0(y) \, \mathrm{d}y \\ T(x,t) = \frac{1}{(4 \pi \alpha t)^{\frac{n}{2}}} \int\limits_{{\mathbb R}^n} \exp \left( - \frac{|x-y|^2}{4 \alpha t} \right) T_0(y) \, \mathrm{d}y $$

When I understand your problem correctly, you can use

$$ T_0(x) = \left\{ \begin{array}{ll} T_I & \mbox{for } |x| < x_B \\ 0 & \mbox{otherwise} \end{array}\right. $$

with $T_I$ being your inner temperature at the beginning and $x_B$ being the coordinate of your wall. Note that since the differential equation is linear, you can add any baseline temperature to the solution as well as multiply the solution with any scale. Thus for the sake of mathematical solution, you could calculate with $T_I=1$ and scale the result. However, I'm an engineer and it doesn't hurt much to just drag $T_I$ along.

I recently had the one-dimensional problem, and this one can still be handled with some effort:

$$ T(x,t) = \frac{1}{\sqrt{4 \pi \alpha t}} \int\limits_{-\infty}^{+\infty} \exp \left( - \frac{|x-y|^2}{4 \alpha t} \right) T_0(y) \, \mathrm{d}y \\ = \frac{T_I}{\sqrt{4 \pi \alpha t}} \int\limits_{-x_B}^{+x_B} \exp \left( - \frac{|x-y|^2}{4 \alpha t} \right) \, \mathrm{d}y \\ = \frac{T_I}{2} \left( \mathrm{erf} \left( \frac{x+x_B}{\sqrt{4 \alpha t}} \right) - \mathrm{erf} \left( \frac{x-x_B}{\sqrt{4 \alpha t}} \right) \right) $$

with $\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int\limits_0^x \exp\left( -t^2 \right) \, \mathrm{d}t$ being the error function.

For $\alpha=1$, $T_I=1$, $x_B=1$ the following graph shows the solution for some $t$:


However, for higher dimensions of the initial conditions given above, the equation needs to be transformed into spherical coordinates first. For two dimensions you get

$$ T(x,t) = \frac{1}{4 \pi \alpha t} \int\limits_{{\mathbb R}^2} \exp \left( - \frac{|x-y|^2}{4 \alpha t} \right) T_0(y) \, \mathrm{d}y \\ \Rightarrow\quad T(x,t) = \frac{T_I}{4 \pi \alpha t} \int\limits_{0}^{r_B} \left( \int\limits_0^{2\pi} \exp \left( - \frac{|x-y|^2}{4 \alpha t} \right) \, \mathrm{d}\varphi(y) \right) r(y) \, \mathrm{d}r(y) $$

giving (at least me) considerable headaches when trying to evaluate. Probably somebody out there knows what

$$ \int\limits_0^{2\pi} \exp \left( - |x-y|^2 \right) \, \mathrm{d}\varphi(y) $$

evaluates to, but I currently don't. Probably there also is a much cleaner approach to evaluate the convolution integral in two dimensions.

  • $\begingroup$ You can probably find the 2D and 3D solution in this book: Crank J (1956) The mathematics of diffusion. Oxford University Press, London $\endgroup$
    – Robin
    Commented Dec 18, 2015 at 10:08

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