Heat, Temperature, and Thermal FEM — From Q=mcΔT to Fourier's Law

Category: Physics Fundamentals | 2026-03-25 | サイトマップ
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Table of Contents
  1. Heat vs. Temperature: A Critical Distinction
  2. Specific Heat: Q = mcΔT
  3. Fourier's Law of Heat Conduction
  4. Thermal Resistance Networks
  5. Convection and Radiation
  6. Lumped Capacitance Method (Biot Number)
  7. FEM Thermal Analysis Fundamentals
  8. EV Battery Pack Thermal Management
  9. Cross-Topics

1. Heat vs. Temperature: A Critical Distinction

Temperature is a property of a material — it reflects the average kinetic energy of atoms or molecules. Heat is energy in transit — it flows from high temperature to low temperature regions. This distinction matters for CAE: FEM thermal analysis solves for the temperature field $T(x,y,z,t)$, not the heat content. Heat flows are derived quantities from the temperature gradients.

Three modes of heat transfer: conduction (through solid materials), convection (via moving fluid), and radiation (electromagnetic waves). FEM thermal analysis handles all three, but the mathematical treatment differs: conduction and convection are handled through boundary conditions, while radiation introduces geometric view factors and nonlinear temperature terms ($T^4$).

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For a CPU heat sink design, what matters more — the thermal conductivity of the fins, or the specific heat capacity of the metal?

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It depends on what you're trying to solve. For steady-state cooling — keeping the chip below 85°C under continuous full load — thermal conductivity is everything. High $k$ material (copper at 400 W/m·K vs. aluminum at 200) gets heat out faster. But for transient thermal management — handling a short power burst, like a GPU during a sudden render spike — specific heat matters because the fin material needs to absorb the pulse energy without a big temperature spike. EV batteries face both challenges: steady-state thermal conductivity keeps cells cool during sustained driving, but heat capacity absorbs the energy of fast-charging pulses. That's why battery thermal management systems use both high-$k$ cooling plates AND thermal interface materials with high volumetric heat capacity.

2. Specific Heat: Q = mcΔT

Specific heat $c_p$ (J/kg·K) is the energy required to raise 1 kg of material by 1 K:

$$Q = mc_p \Delta T$$

This is the fundamental relationship for energy storage in thermal systems. In FEM, it appears in the transient heat equation as the $\rho c_p \frac{\partial T}{\partial t}$ term.

MaterialThermal Conductivity k (W/m·K)Specific Heat cp (J/kg·K)Density ρ (kg/m³)
Copper4003858960
Aluminum 60611678962700
Steel (carbon)504907800
Silicon (chip)1507002330
Li-ion cell (approx.)1–3 (through-plane)~10002500
Thermal interface material3–10800–12002000–3500
Air (1 atm)0.02610051.21

3. Fourier's Law of Heat Conduction

Heat flows in the direction of decreasing temperature. Fourier's Law states the heat flux (W/m²) is proportional to the temperature gradient:

$$\mathbf{q} = -k\nabla T$$

The negative sign ensures heat flows from hot to cold. Combining with energy conservation gives the heat diffusion equation:

$$\rho c_p \frac{\partial T}{\partial t} = \nabla \cdot (k\nabla T) + \dot{q}'''$$

Where $\dot{q}'''$ is the volumetric heat generation rate (W/m³). For steady-state with no heat generation, this reduces to the Laplace equation: $\nabla^2 T = 0$.

This is structurally identical to other FEM field problems: replace $T$ with displacement $u$, and $k$ with elastic modulus, and you get the same mathematical structure. A structural FEM code can solve thermal problems with trivial input file changes.

Thermal Diffusivity

The rate at which temperature changes propagate through a material:

$$\alpha = \frac{k}{\rho c_p} \quad \text{(m²/s)}$$

High thermal diffusivity means temperature equilibrates quickly. Copper has $\alpha = 1.17 \times 10^{-4}$ m²/s; steel has $\alpha = 1.3 \times 10^{-5}$ m²/s — copper equilibrates about 9× faster. For transient thermal analysis, materials with very different diffusivities require different time step sizes to resolve accurately.

4. Thermal Resistance Networks

By analogy with Ohm's law ($V = IR$), heat flow can be analyzed using thermal resistance networks. The thermal resistance for conduction through a slab:

$$R_{\text{cond}} = \frac{L}{kA}, \qquad Q = \frac{\Delta T}{R_{\text{total}}}$$

Where $L$ is thickness, $k$ is conductivity, and $A$ is area. For convection: $R_{\text{conv}} = \frac{1}{hA}$ where $h$ is the convective heat transfer coefficient (W/m²·K).

CPU Thermal Resistance Budget

A typical desktop CPU cooling path from junction to ambient:

LayerMaterialThickness (mm)R_thermal (°C/W)
Silicon dieSi ($k$=150)0.50.003
SolderSnAgCu ($k$=60)0.10.002
Integrated heat spreaderCu ($k$=400)30.007
Thermal paste (TIM2)Phase-change ($k$=5)0.10.02
Heat sink base + finsAl ($k$=160)300.05
Convection (air cooling)Air flow0.10–0.30

Total R_th ≈ 0.18–0.38 °C/W. For a 200W CPU: ΔT = 200 × 0.38 = 76°C above ambient. At 25°C ambient: chip temperature ≈ 101°C — right at the thermal limit. This thermal resistance budget analysis guides heat sink and TIM selection long before FEM is needed.

5. Convection and Radiation

Newton's Law of Cooling (Convection)

$$Q_{\text{conv}} = hA(T_s - T_\infty)$$

Where $h$ is the convective heat transfer coefficient (W/m²·K) and $T_\infty$ is the fluid temperature. Typical values: natural convection air 5–25 W/m²·K, forced air convection 25–250, water cooling 500–10,000.

Stefan-Boltzmann Radiation Law

$$Q_{\text{rad}} = \varepsilon\sigma A(T_s^4 - T_{\text{surr}}^4)$$

Where $\varepsilon$ is emissivity (0 for perfect reflector, 1 for black body) and $\sigma = 5.67 \times 10^{-8}$ W/m²·K⁴. The $T^4$ dependence makes radiation dominant at high temperatures (>400°C) but negligible for room-temperature electronics. In FEM, the $T^4$ term makes the thermal equation nonlinear — requiring iterative solution.

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I'm modeling a steel furnace component at 800°C. Should I include radiation in the FEM analysis or can I skip it to save time?

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At 800°C you absolutely cannot skip radiation — it's the dominant heat transfer mechanism. Let's check: at 800°C (1073 K), radiation heat flux ≈ εσT⁴ = 0.8 × 5.67e-8 × 1073⁴ ≈ 60,000 W/m² = 60 kW/m². For comparison, natural convection at that temperature might only give 1–5 kW/m². Radiation is 10–60× stronger. Skipping radiation would give temperatures that are wildly wrong. In Abaqus, radiation boundary conditions are straightforward — you specify surface emissivity and ambient temperature. The only complication is view factors between multiple surfaces inside the furnace, which requires either analytical formulas or a separate view factor calculation step.

6. Lumped Capacitance Method (Biot Number)

For small objects with high thermal conductivity (metals cooled by gentle convection), the internal temperature may be nearly uniform — justifying the lumped capacitance assumption. The Biot number check:

$$\text{Bi} = \frac{hL_c}{k} \leq 0.1 \quad \Rightarrow \quad \text{lumped capacitance valid}$$

Where $L_c = V/A_s$ is the characteristic length (volume/surface area). When Bi ≤ 0.1, the temperature is essentially uniform throughout the object and the transient response is simply:

$$\frac{T(t) - T_\infty}{T_0 - T_\infty} = e^{-t/\tau}, \qquad \tau = \frac{\rho V c_p}{hA_s}$$

This is the thermal time constant — useful for estimating how long it takes to cool (or heat) a component without running FEM. For a 50mm aluminum cube with $h = 20$ W/m²K: $\tau = 7800/20 \times 0.05/3 = \ldots$ about 40 minutes. If your process cycle is 5 minutes, you're far from steady state and full transient FEM is needed.

7. FEM Thermal Analysis Fundamentals

The FEM discretization of the heat equation produces a system analogous to the structural problem:

$$[\mathbf{C}]\dot{\mathbf{T}} + [\mathbf{K}_T]\mathbf{T} = \mathbf{Q}$$

Where $[\mathbf{C}]$ is the heat capacity matrix (analogous to structural mass matrix [M]), $[\mathbf{K}_T]$ is the thermal conductivity matrix (analogous to stiffness [K]), $\mathbf{T}$ is the nodal temperature vector, and $\mathbf{Q}$ is the nodal heat flux vector.

For steady-state: $[\mathbf{K}_T]\mathbf{T} = \mathbf{Q}$ (same form as static structural). For transient: the time integration uses the θ-method (generalization of backward/forward Euler). Abaqus thermal analysis uses the same solver architecture as structural — in fact, structural and thermal problems can be coupled sequentially (thermal loads driving thermal stresses) or fully (Abaqus Standard fully coupled thermal-structural elements).

Mesh Requirements for Thermal FEM

For steady-state conduction, element size requirements are less stringent than structural stress analysis — there are no stress concentrations. However, for transient problems, the thermal time scale must be resolved: element size $\Delta x$ and time step $\Delta t$ must satisfy $\Delta t > \frac{\rho c_p \Delta x^2}{2k}$ for explicit time integration, or use implicit schemes (θ=1 fully implicit) for unconditional stability.

8. EV Battery Pack Thermal Management

EV battery thermal management is one of the most demanding thermal CAE applications today. Lithium-ion cells generate heat during charge/discharge from three sources:

  • Ohmic (Joule) heating: $\dot{q} = I^2 R_{\text{internal}}$ — heat from electrical resistance
  • Entropic heat: Reversible heat from electrochemical reaction entropy change
  • Irreversible heat: From polarization overpotentials

The total heat generation in a 100 kWh EV battery pack during 2C fast charging (200 kW input):

$$\dot{Q}_{\text{total}} \approx \eta_{\text{loss}} \times P_{\text{charge}} = 0.05 \times 200{,}000 = 10{,}000 \text{ W} = 10 \text{ kW}$$

Maintaining cells below 45°C while rejecting 10 kW requires sophisticated cooling — typically aluminum cooling plates with water/glycol flowing between cell rows, thermally connected via thermal interface materials.

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For an EV battery pack thermal model, should I model every individual cell, or can I homogenize the pack into a bulk material?

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For pack-level thermal design, homogenization is standard practice. You model the pack as a block with equivalent orthotropic thermal properties: high conductivity along the cell axis (say 30 W/m·K for the through-plane cell direction) but much lower across cells (~1.5 W/m·K through-plane of the cells). The cooling plates are modeled explicitly because they carry most of the heat. This approach keeps the model at 100k–500k elements instead of billions. Save cell-level 3D FEM for local hot spot analysis — for example, after pack-level FEM identifies the hottest module, build a detailed 3D model of just that module with individual cells and TIM layer to verify temperature uniformity within the module.

9. Cross-Topics

TopicConnectionLink
Thermal ExpansionTemperature gradients cause thermal stress — thermo-structural couplingThermal Expansion
Electric CurrentJoule heating: electrical power dissipation → thermal sourceElectric Current, Voltage & Resistance
Fluid PressureConvective heat transfer requires fluid flow analysis (CFD-thermal coupling)Fluid Pressure and Buoyancy
Heat Transfer TheoryFull derivation of heat conduction FEM, convection, and radiationHeat Transfer
Heat Diffusion ToolInteractive 1D heat diffusion simulator with animationHeat Diffusion Simulator