Heat Conduction in Cylindrical Coordinates

For cylindrical shapes, Cartesian coordinates lead to inefficient shape representation and complex meshing. Using cylindrical coordinates $(r, \phi, z)$ allows leveraging symmetry to reduce dimensions. The one-dimensional radial steady-state heat conduction equation is

This term reflects the change in area because the volume element is proportional to $r$. This is also the physical cause of the critical insulation radius.

The solution for the case $\dot{q}_v = 0$ is as follows.

The temperature distribution is logarithmic, different from the linear distribution in a flat plate. The heat flow rate is

Correct. Since the cross-sectional area $A = 2\pi r L$ is proportional to $r$, for a constant heat flow rate $q$, the heat flux $q'' = q/A$ is inversely proportional to $r$. The heat flux is larger on the inner surface.

The solution for the case with uniform internal heat generation $\dot{q}_v$ is

The maximum temperature at the center $r=0$ is $T_{\max} = T_s + \dot{q}_v R^2/(4k)$.

Exactly. For an AWG18 copper wire (diameter 1.02mm) carrying 10A, the heat generation is about 10 W/m. $\dot{q}_v \approx 1.2 \times 10^7$ W/m$^3$, and the center temperature rise can be estimated from this formula.

If symmetric in the circumferential direction, a 2D axisymmetric model is sufficient. Computational cost becomes less than 1/100.

Leveraging symmetry is fundamental for computational efficiency. In Ansys Mechanical, use PLANE55 (KEYOPT(3)=1); in Abaqus, use DCAX4 (axisymmetric 4-node element).

Central differencing in cylindrical coordinates requires caution.

At $r = 0$ (central axis), $1/r$ becomes a singularity, so apply L'Hôpital's rule to get

.

Similarly, handling the central axis is important in FEM as well. Axisymmetric elements automatically perform correct processing at nodes where $r = 0$, but caution is needed when using wedge elements on the axis in a 3D model.

For multilayer cylinders (pipe + insulation + cladding, etc.), connect the analytical solutions for each layer.

To include contact thermal resistance between layers, add $1/(2\pi r_i h_{c,i} L)$ to the denominator.

Exactly, it's the concept of a thermal resistance network. The strength of cylindrical coordinates is that even for 5 or more layers, hand calculations are possible.

The most common is heat loss calculation for plant piping. Here's an example for a steam pipe (outer diameter 114.3mm, SUS304, insulation 50mm).

Total thermal resistance $R_{\text{total}} = 2.033$ m K/W. The insulation accounts for 99% of the total.

Yes. The thermal resistance of metal pipes is less than 1/1000th of the insulation, so there's virtually no temperature drop. Therefore, whether the pipe material is copper or SUS has almost no effect on heat loss.

The thermal resistance per unit length for a cylinder $R = \ln(r_2/r_1)/(2\pi k)$ can be expressed as an equivalent flat plate's $R = t/(k \cdot A_{lm})$. $A_{lm}$ is the log mean area:

If $r_2/r_1 < 2$, the arithmetic mean $(r_1+r_2)/2$ gives an error below 4%. For simplified calculations, the arithmetic mean is often sufficient.

Here are points to note for axisymmetric mesh generation for cylinders.

• Use biased mesh in the radial direction (finer on the inner side)
• Minimum of 3 elements in the radial direction for thin-walled pipes
• For multilayer structures, share nodes between layers or use Tied Contact

Exactly. The heat flux is inversely proportional to the radius, so it's larger on the inner surface. Using a finer mesh there improves accuracy for heat flux evaluation.