Heat Conduction in Cylindrical Coordinates

Category: 熱解析 | Integrated 2026-04-06
CAE visualization for cylindrical conduction theory - technical simulation diagram
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Theory and Physics

Heat Conduction Equation in Cylindrical Coordinates

🧑‍🎓

Professor, doesn't thermal analysis for pipes and shafts work well with Cartesian coordinates?


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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


$$\frac{1}{r}\frac{d}{dr}\left(kr\frac{dT}{dr}\right) + \dot{q}_v = 0$$

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Compared to Fourier's law in Cartesian coordinates, there's an extra $1/r$ term.


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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.


Analytical Solution (No Internal Heat Generation)

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


$$T(r) = T_1 + \frac{T_2 - T_1}{\ln(r_2/r_1)}\ln\frac{r}{r_1}$$

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


$$q = \frac{2\pi k L (T_1 - T_2)}{\ln(r_2/r_1)}$$

🧑‍🎓

The $\ln$ appears because the area changes, right?


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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.


Case with Internal Heat Generation

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


$$T(r) = T_s + \frac{\dot{q}_v}{4k}(R^2 - r^2)$$

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


🧑‍🎓

This is the case for Joule heating in electrical wires, right?


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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.

Coffee Break Trivia

Origin of the Cylindrical Coordinate Laplacian

The heat conduction equation in cylindrical coordinates is expressed as ∂²T/∂r² + (1/r)∂T/∂r + (1/r²)∂²T/∂θ² + ∂²T/∂z² + q̇/λ = 0. The (1/r)∂T/∂r term in this Laplacian originates from the Jacobian of the coordinate transformation. Lame systematized the cylindrical coordinate Laplacian in elasticity theory in 1833, and it was applied to the Fourier heat equation about 20 years later, which is the historical background.

Physical Meaning of Each Term
  • Heat Storage Term $\rho c_p \partial T/\partial t$: Rate of thermal energy storage per unit volume. 【Everyday Example】 An iron frying pan is slow to heat up and cool down, while an aluminum pot heats up and cools down quickly—this is due to the difference in the product of density $\rho$ and specific heat $c_p$ (Heat Capacity). Objects with large heat capacity have slower temperature changes. Water has a very high specific heat (4,186 J/(kg·K)), so temperatures near the ocean are more stable than inland. In transient analysis, this term determines the rate of temperature change over time.
  • Heat Conduction Term $\nabla \cdot (k \nabla T)$: Heat conduction based on Fourier's law. Heat flux proportional to the temperature gradient. 【Everyday Example】 When you put a metal spoon in a hot pot, the handle gets hot—because metals have high thermal conductivity $k$, heat transfers quickly from the high-temperature side to the low-temperature side. A wooden spoon doesn't get hot because its $k$ is small. Insulation materials (like glass wool) have extremely small $k$, making heat transfer difficult even with a temperature gradient. This is a mathematical formulation of the natural tendency "heat flows where there is a temperature difference."
  • Convection Term $\rho c_p \mathbf{u} \cdot \nabla T$: Heat transport accompanying fluid motion. 【Everyday Example】 Feeling cool when a fan blows on you is because the wind (fluid flow) carries away the warm air near your skin and supplies fresh, cool air—this is forced convection. The ceiling area of a room becoming warm with heating is due to natural convection where heated air rises due to buoyancy. The fan in a PC's CPU cooler also dissipates heat via forced convection. Convection is an order of magnitude more efficient heat transport method than conduction.
  • Heat Source Term $Q$: Internal heat generation (Joule heat, chemical reaction heat, radiation absorption, etc.). Unit: W/m³. 【Everyday Example】 A microwave oven heats food via microwave absorption inside the food (volumetric heating). The heater wire in an electric blanket warms up via Joule heating ($Q = I^2 R / V$). Heat generation during lithium-ion battery charging/discharging and friction heat from brake pads are also considered as heat sources in analysis. Unlike boundary conditions that supply heat to the "surface" from outside, the heat source term represents energy generation "inside" the material.
Assumptions and Applicability Limits
  • Fourier's Law: Linear relationship where heat flux is proportional to temperature gradient (non-Fourier heat conduction is needed for extremely low temperatures or ultra-short pulse heating)
  • Isotropic Thermal Conductivity: Thermal conductivity does not depend on direction (anisotropy must be considered for composite materials, single crystals, etc.)
  • Temperature-Independent Material Properties (Linear Analysis): Assumption that material properties do not depend on temperature (temperature dependence is needed for large temperature differences)
  • Treatment of Thermal Radiation: View factor method for surface-to-surface radiation; DO method or P1 approximation for participating media
  • Non-Applicable Cases: Phase change (melting/solidification) requires consideration of latent heat. Thermal-stress coupling is essential for extreme temperature gradients
Dimensional Analysis and Unit Systems
VariableSI UnitNotes / Conversion Memo
Temperature $T$K (Kelvin) or CelsiusBe careful not to confuse absolute temperature and Celsius. Always use absolute temperature for radiation calculations.
Thermal Conductivity $k$W/(m·K)Steel: ~50, Aluminum: ~237, Air: ~0.026
Heat Transfer Coefficient $h$W/(m²·K)Natural convection: 5–25, Forced convection: 25–250, Boiling: 2,500–25,000
Specific Heat $c_p$J/(kg·K)Distinguish between specific heat at constant pressure and constant volume (important for gases)
Heat Flux $q$W/m²Neumann condition as a boundary condition

Numerical Methods and Implementation

Axisymmetric FEM Model

🧑‍🎓

When solving cylindrical heat conduction with FEM, is a 3D model necessary?


🎓

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


Model TypeDegrees of FreedomComputation TimeAccuracy
3D Full~1 millionMinute orderBaseline
3D 1/4 Symmetry~250kSecond orderEquivalent
2D Axisymmetric~1000InstantEquivalent
🧑‍🎓

Running a 3D model when 2D axisymmetric is sufficient is wasteful, isn't it?


🎓

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


Discretization with Finite Difference Method

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Central differencing in cylindrical coordinates requires caution.


$$\frac{1}{r_i}\frac{k(r_{i+1/2}(T_{i+1}-T_i)/\Delta r - r_{i-1/2}(T_i-T_{i-1})/\Delta r)}{\Delta r} + \dot{q}_v = 0$$

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


$$2k\frac{T_1 - T_0}{\Delta r^2} + \dot{q}_v = 0$$

.


🧑‍🎓

The singularity at the center needs special treatment, huh.


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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.


Solution for Multilayer Cylinders

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


$$q = \frac{T_{\text{in}} - T_{\text{out}}}{\sum_{i=1}^{n} \frac{\ln(r_{i+1}/r_i)}{2\pi k_i L}}$$

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


🧑‍🎓

It's the same addition as series resistances in an electrical circuit, right?


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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.

Coffee Break Trivia

Solution Steps for Logarithmic Temperature Distribution

The steady-state temperature distribution of an infinitely long hollow cylinder (inner radius ri, outer radius ro) is T(r)=Ti + (Ti−To)·ln(r/ro)/ln(ri/ro), which is logarithmic. This solution is completed in three steps: separation of variables → ODE integration → determining C with two boundary conditions. Similar problems appear in Japanese university entrance exams, and domestic heat engineering textbooks always treat this as a chapter-opening example problem.

Linear Elements vs. Quadratic Elements

In heat conduction analysis, linear elements often provide sufficient accuracy. For areas with steep temperature gradients (thermal shock, etc.), quadratic elements are recommended.

Heat Flux Evaluation

Calculated from the temperature gradient within an element. Smoothing may be necessary, similar to nodal stresses.

Convection-Diffusion Problem

When the Peclet number is high (convection-dominated), upwinding stabilization (SUPG, etc.) is needed. Not required for pure heat conduction problems.

Time Step for Transient Analysis

Set a time step sufficiently smaller than the characteristic diffusion time $\tau = L^2 / \alpha$ ($\alpha$: Thermal Diffusivity). Automatic time step control is effective for rapid temperature changes.

Nonlinear Convergence

Nonlinearity due to temperature-dependent material properties is often mild, and Picard iteration (direct substitution method) is often sufficient. Newton's method is recommended for strong nonlinearity like radiation.

Steady-State Analysis Convergence Criterion

Convergence is determined when the temperature change at all nodes falls below a threshold (e.g., $|\Delta T| / T_{max} < 10^{-5}$).

Analogy for Explicit and Implicit Methods

Explicit method is like "a weather forecast predicting the next step using only current information"—fast computation but unstable with large time steps (misses storms). Implicit method is like "a prediction considering future states"—stable even with large time steps but requires solving equations at each step. For problems without rapid temperature changes, using the implicit method with larger time steps is more efficient.

Practical Guide

Pipe Heat Loss Calculation

🧑‍🎓

Please tell me typical practical situations where cylindrical heat conduction is used.


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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).


Layer$r$ [mm]$k$ [W/(m K)]Thermal Resistance [m K/W]
Pipe Wall48.6→57.1516.30.00160
Insulation57.15→107.150.052.016
Internal Convection0.00033
External Convection0.0149

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


🧑‍🎓

The thermal resistance of the pipe wall is almost negligible, huh.


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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.


Log Mean Radius

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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:


$$A_{lm} = \frac{A_2 - A_1}{\ln(A_2/A_1)} = \frac{2\pi L(r_2 - r_1)}{\ln(r_2/r_1)}$$

🧑‍🎓

So you can reuse the flat plate formula by using the logarithmic mean, right?


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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.


Mesh Generation Notes

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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

🧑‍🎓

Biased mesh is because the heat flux is larger on the inner surface, right?


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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.


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