Why is coupled analysis for Joule heating necessary? Isn't a simple current × resistance calculation sufficient?

As temperature increases, the resistivity of the metal also increases (approximately 0.4%/°C for copper), creating a positive feedback loop where the heating effect itself intensifies. To accurately determine the steady-state temperature, iterative coupling between electric field analysis and thermal analysis is required.

What is the fundamental equation for Joule heating?

The heat generated per unit volume is expressed as Q = J²/σ = σE², where J is current density, σ is electrical conductivity, and E is electric field strength. Due to the temperature dependence of resistivity, ρ(T) = ρ₀(1 + α(T − T₀)), the heat generation changes non-linearly as temperature rises.

When should weak coupling versus strong coupling be used in coupled analysis?

Weak (one-way) coupling is sufficient when temperature changes are small (approximately below 50°C) and changes in resistivity are negligible. For applications like EV busbars, where temperature rises exceeding 100°C are expected, strong coupling—which iteratively solves the electromagnetic and thermal fields—is essential.

What is the most critical point to consider in practical applications?

The accurate modeling of contact resistance. At bolted connections or press-fit interfaces, the contact resistance—which depends on contact pressure and surface roughness—becomes the dominant heat source. Without actual measurement data, it is impossible to correctly predict busbar temperature.

Electrothermal Coupled Analysis of Joule Heating

Category: 電磁-熱連成 | 更新 2026-04-12

Theory and Physics

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Joule heating is just about things getting hot when current flows, right? Why is coupled analysis specifically needed?

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Good question. "Things get hot when current flows" is correct, but the story doesn't end there. When temperature rises, the electrical resistivity of metals increases. For example, copper's resistance increases by about 0.4% per °C. This means that even with the same current, the heat generation increases further, causing the temperature to rise even more, and the resistance to increase even more... starting a positive feedback loop.

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Huh, does that mean the temperature rises infinitely?

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In reality, heat dissipation (convection, radiation, conduction) exists, so at some point heat generation and dissipation balance out, reaching a steady-state temperature. However, determining what that steady-state temperature will be in °C requires iteratively calculating the loop: "electromagnetic field → heat generation → temperature change → resistivity change → electromagnetic field." That's why electromagnetic-thermal coupled analysis is essential for accurate Joule heating analysis.

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Specifically, in what situations does this become a problem?

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A familiar example is the connection terminals for 800V inverters in EVs (electric vehicles). For busbars carrying large currents over 400A, it's necessary to check that the local temperature does not exceed 150°C. Exceeding this causes insulation material degradation and shortens lifespan. If you calculate ignoring the temperature dependence of resistivity, you get a temperature 20-30°C lower than reality, missing potential design flaws. Other typical examples include predicting fuse blow characteristics, substrate design for power semiconductors, and sag (deflection) calculations for power transmission lines.

Why is Coupled Analysis Necessary?

The reason Joule heating analysis cannot be simplified to a simple "P = I²R" calculation is summarized by the fact that electrical conductivity $\sigma$ is a function of temperature $T$. Even under constant current conditions, temperature rise changes resistivity, redistributing the current density. Analysis that ignores this bidirectional coupling deviates from actual measurements in the following ways:

Underestimation of Steady-State Temperature: Ignoring temperature dependence can lead to an underestimation of 20-30°C for copper busbars.
Missing Current Concentration: Localized resistance increases at cross-section changes or connection points can form hotspots.
Error in Transient Response: Temperature rise curves for transient large currents, like short-circuit currents, become nonlinear due to resistivity changes.

Governing Equations

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So, what equations are actually used to represent this?

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First, the electromagnetic field side. For steady current, the continuity equation comes from charge conservation. And the heat generation is determined by the dot product of current density and electric field:

Governing Equation for Electric Field (steady current approximation):

$$ \nabla \cdot \mathbf{J} = 0 \quad \text{(Current Continuity Equation)} $$

$$ \mathbf{J} = \sigma(T) \, \mathbf{E} = -\sigma(T) \, \nabla V \quad \text{(Ohm's Law)} $$

Joule Heating per Unit Volume:

$$ Q = \mathbf{J} \cdot \mathbf{E} = \frac{|\mathbf{J}|^2}{\sigma(T)} = \sigma(T) \, |\mathbf{E}|^2 $$

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Are $Q = J^2/\sigma$ and $Q = \sigma E^2$ the same thing? They look kind of opposite...

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Sharp observation. Substituting the relation $\mathbf{J} = \sigma \mathbf{E}$ shows they are the same equation. In terms of usage, $J^2/\sigma$ is convenient when current density is known, and $\sigma E^2$ is convenient when electric potential is known. In practice, since electric potential is often solved as the unknown, it's frequently implemented in the form $\sigma |\nabla V|^2$.

Governing Equation for Thermal Field (energy conservation):

$$ \rho c_p \frac{\partial T}{\partial t} = \nabla \cdot (k \, \nabla T) + Q $$

Here, $\rho$ is density, $c_p$ is specific heat at constant pressure, $k$ is thermal conductivity, and $Q$ is the Joule heating amount (heat source term).

Physical Meaning of Each Term

$Q = \mathbf{J} \cdot \mathbf{E}$ (Joule Heating Term): The electric field accelerates charges, and their kinetic energy is converted to heat through collisions with lattice vibrations (phonons). Higher electron drift velocity and higher scattering frequency increase heat generation. 【Everyday Example】 Your smartphone charging cable feels warm to the touch—Joule heat is generated in the cable's internal resistance. Faster charging means higher current and more heat.
$\nabla \cdot \mathbf{J} = 0$ (Current Continuity Equation): Represents that charge does not accumulate in a steady state. At points where cross-sectional area changes, current density changes inversely proportionally, causing localized heat concentration. 【Practical Example】 A typical pattern where current density increases around bolt holes in busbars, forming hotspots.
$\rho c_p \partial T / \partial t$ (Thermal Storage Term): The delay in temperature change due to the material's heat capacity. Materials with large heat capacity like copper experience slower transient temperature rises. For short-circuit currents lasting a few seconds or less, this term has a significant influence.
$\nabla \cdot (k \nabla T)$ (Heat Conduction Term): Heat diffusion proportional to the temperature gradient. Copper has a very high thermal conductivity $k \approx 400$ W/(m·K), quickly homogenizing local temperature rises. With stainless steel ($k \approx 16$), hotspots remain pronounced.

Dimensional Analysis and Unit System

Variable	SI Unit	Typical Value (Copper, 20°C)
Electrical Conductivity $\sigma$	S/m	$5.96 \times 10^7$
Electrical Resistivity $\rho_e = 1/\sigma$	$\Omega$·m	$1.68 \times 10^{-8}$
Temperature Coefficient $\alpha$	1/K	$3.93 \times 10^{-3}$ (copper)
Current Density $J$	A/m²	$10^5$〜$10^7$ (busbar)
Heat Generation Density $Q$	W/m³	$10^4$〜$10^8$
Thermal Conductivity $k$	W/(m·K)	401 (copper), 16 (SUS304)

Temperature Dependence of Resistivity

The electrical resistivity of metals increases almost linearly with temperature (for metals, near room temperature):

$$ \rho(T) = \rho_0 \left[ 1 + \alpha (T - T_0) \right] $$

Here, $\rho_0$ is the resistivity at reference temperature $T_0$, and $\alpha$ is the temperature coefficient of resistance.

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How much does the value of $\alpha$ differ between materials?

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Summarizing representative values in a table makes it clear.

Material	$\rho_0$ [$\mu\Omega$·cm]	$\alpha$ [1/K]	Notes
Copper (Cu)	1.68	$3.93 \times 10^{-3}$	Most frequently used conductor
Aluminum (Al)	2.65	$4.29 \times 10^{-3}$	Used for busbars for weight reduction
Silver (Ag)	1.59	$3.80 \times 10^{-3}$	Contact material, high cost
Nichrome (NiCr)	108	$1.7 \times 10^{-4}$	Heater wire, α is extremely small
SUS304	72	$9.4 \times 10^{-4}$	Structural material, high resistance

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Nichrome's $\alpha$ is really small. Now I understand why it's used for heater wires—since the resistance doesn't change much with temperature, you get stable heat generation, right?

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Exactly! Conversely, copper and aluminum have large $\alpha$, so their temperature dependence cannot be ignored in busbar or cable design. A 100°C temperature rise increases copper's resistance by about 40%. Analysis ignoring this is completely useless.

Feedback Structure of Coupling

Electromagnetic-thermal coupling in Joule heating has a clear positive feedback loop:

Heat generation $Q = J^2/\sigma(T)$ due to current density $\mathbf{J}$
Temperature $T$ rises
Resistivity $\rho(T)$ increases ($\sigma(T)$ decreases)
Current distribution changes for the same potential difference, altering local heat generation density
Return to 1

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Does the behavior differ between constant current sources and constant voltage sources?

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Sharp observation. For a constant voltage source, temperature rise → resistance increase → current decrease → heat generation decrease, acting in a self-limiting manner. Conversely, for a constant current source, temperature rise → resistance increase → heat generation $Q = I^2 R$ increases → temperature rises further, acting in a runaway direction. In power electronics, both are mixed depending on switching element drive conditions, so setting boundary conditions incorrectly can completely change the results.

Understanding Feedback with Familiar Examples

The filament (tungsten) of an incandescent light bulb has a resistivity of about 5.3 $\mu\Omega$·cm at room temperature, but jumps to about 15 times that at its operating temperature of 2500°C. This is why the inrush current immediately after power-on can be over 10 times the steady-state current. To reproduce this transient behavior in CAE analysis, the temperature dependence of resistivity and electromagnetic-thermal coupling are essential.

Coffee Break Trivia

180 Years Since Joule's Law—P = I²R Governs Modern ICs

In 1841, James Prescott Joule discovered through experiments passing current through iron wire that "heat generation is proportional to the square of the current and the resistance." This was an era when even the nature of electricity was a mystery. 180 years later, in cutting-edge 3nm semiconductor chips, billions of transistors generate over 100 W/cm² of Joule heat. This heat density is about 1/6th that of the sun's surface. The law discovered by Joule in his laboratory has become a modern bottleneck influencing the design of EVs, 5G base stations, and data centers.

Numerical Methods and Implementation

Weak Coupling and Strong Coupling

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I've heard of "weak coupling" and "strong coupling." How are they used differently?

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Coupling methods for Joule heating analysis can be broadly classified into three categories:

Method	Overview	Applicable Conditions	Accuracy
Uncoupled (One-Way)	Solve electromagnetic field → pass heat generation to thermal analysis (once only)	$\Delta T < 50$°C, $\alpha \Delta T \ll 1$	Low
Weak Coupling (Sequential Iteration)	Iterate electromagnetic→thermal→electromagnetic→thermal... until convergence	$\Delta T = 50$〜$200$°C	Medium~High
Strong Coupling (Monolithic)	Solve electromagnetic and thermal fields simultaneously as one system of equations	$\Delta T > 200$°C, strong nonlinearity	Highest

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Which one is most commonly used in practice?

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Overwhelmingly, weak coupling (sequential iteration method). There are three reasons. First, existing electromagnetic and thermal solvers can be reused as-is. Second, it's easier to debug—each physical field can be verified individually. Third, memory can be saved even for large-scale models. The monolithic method doubles the degrees of freedom, making memory usage severe.

Weak coupling iteration algorithm (Gauss-Seidel type):

Set initial temperature $T^{(0)}$ (usually ambient temperature)
Solve electric field equation using $\sigma(T^{(k)})$ to obtain $V^{(k+1)}$
Calculate $Q^{(k+1)} = \sigma(T^{(k)}) |\nabla V^{(k+1)}|^2$
Solve heat conduction equation with $Q^{(k+1)}$ as heat source term to obtain $T^{(k+1)}$
Convergence check: If $\|T^{(k+1)} - T^{(k)}\|_\infty < \varepsilon$, finish; otherwise, return to step 2

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What should the convergence criterion $\varepsilon$ be set to?

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A practical guideline is 0.1°C to 1°C based on absolute temperature. For cases like EV busbars where the allowable temperature margin is only about 10°C, use 0.1°C. For heater design with margins of tens of °C, 1°C is sufficient. The number of iterations typically converges in 5-15 cycles.

Finite Element Formulation

Weak Form for Electric Field:

$$ \int_\Omega \sigma(T) \, \nabla N_i \cdot \nabla V \, d\Omega = \int_{\Gamma_N} J_n \, N_i \, d\Gamma $$

Here, $N_i$ is the shape function, $J_n$ is the normal current density on the boundary.

Weak Form for Thermal Field:

$$ \int_\Omega \rho c_p N_i \frac{\partial T}{\partial t} \, d\Omega + \int_\Omega k \, \nabla N_i \cdot \nabla T \, d\Omega = \int_\Omega Q \, N_i \, d\Omega + \int_{\Gamma_c} h(T_\infty - T) N_i \, d\Gamma $$

The second term on the right side is the convective boundary condition ($h$: heat transfer coefficient, $T_\infty$: ambient temperature).

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Do you use the same mesh for the electric and thermal fields?

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For Joule heating analysis, since the electric and thermal fields exist in the same region, the same mesh is basically used. This is a major difference from fluid-structure interaction (FSI), eliminating the need to worry about interpolation errors in data transfer. In COMSOL or Ansys Workbench, coupled calculations are automatically performed on the same mesh.

Time Integration and Convergence Control

For transient analysis, time integration requires ingenuity because the response times of the electric field ($\tau_e \sim \varepsilon/\sigma \sim 10^{-18}$ s) and the thermal field ($\tau_T \sim \rho c_p L^2/k \sim 1$〜$100$ s) are extremely different.

Parameter	Recommended Value	Notes
Electric Field Analysis	Steady-State Solution	Since $\tau_e \ll \tau_T$, use steady-state solution at each timestep
Thermal Field Timestep	$\Delta t \leq \tau_T / 10$	Backward Euler method (unconditionally stable) recommended
Coupled Iteration Convergence Criterion	$\Delta T < 0.1$〜$1$ °C	Absolute temperature criterion
Coupled Iteration Relaxation Factor	0.5〜0.8	Prevents divergence; Aitken relaxation also effective
Maximum Coupled Iterations	20〜30	If exceeded, reduce relaxation factor

Improving Computational Efficiency Using Time Scale Separation

The response speed of the electromagnetic field (femtosecond order) is orders of magnitude faster than that of the thermal field (second~minute order). This means that at each timestep of the thermal field, the electromagnetic field can be assumed to reach steady-state instantaneously. This "quasi-steady approximation" allows omitting transient electromagnetic analysis, requiring only one steady-state electric field solution per thermal timestep. This dramatically reduces computational cost. For AC current, the average heat generation over one electromagnetic cycle is calculated and passed to the thermal field.

Practical Guide

Analysis Flow and Boundary Conditions

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When actually starting a coupled Joule heating analysis, what should I do first?

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The practical analysis flow is as follows:

Geometry Modeling: Extract conductor parts from CAD data. Do not omit bolt holes, chamfers, or fillets (they affect current concentration).
Material Definition: Input temperature-dependent tables for $\rho(T)$, $k(T)$, $c_p(T)$. Especially $\rho(T)$ is mandatory.
Electrical Boundary Conditions: Set current density or total current value on the current inlet surface, and GND ($V = 0$) on the outlet surface.
Thermal Boundary Conditions: Set natural convection ($h = 5$〜$15$ W/m²K) on external surfaces, and radiation ($\varepsilon = 0.1$〜$0.9$) as needed.
Coupling Settings: Set iteration count and convergence criteria for weak coupling.
Post-Processing: Check maximum temperature, current density distribution, heat flux distribution.

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How do you determine the convective heat transfer coefficient $h$? Textbooks say 5-25, but that range is too broad...