Steady-State Conduction with Internal Heat Generation
Theory and Physics
Physics of Internal Heat Generation
Professor, in what situations does internal heat generation occur?
It appears in various forms such as Joule heating from electrical resistance, decay heat from nuclear fuel, and chemical reaction heat. All are expressed as the volumetric heat generation rate $\dot{q}_v$ [W/m$^3$].
Governing Equation for a Flat Plate
The governing equation for a flat plate (thickness $2L$, both surfaces at the same temperature $T_s$) with uniform internal heat generation $\dot{q}_v$ is
The solution is a parabolic distribution.
The center temperature is $T_{\max} = T_s + \dot{q}_v L^2 / (2k)$.
It makes intuitive sense that it becomes a parabola. The center is the hottest.
The important point is that $T_{\max} \propto L^2$. Doubling the plate thickness quadruples the center temperature rise. Making it thinner is the most effective cooling measure.
Case of a Cylinder
For a cylinder of radius $R$ (outer surface temperature $T_s$)
Center temperature $T_{\max} = T_s + \dot{q}_v R^2 / (4k)$. Compared to $2k$ for the plate, it's $4k$ for the cylinder, meaning the cylinder cools more efficiently.
Case of a Sphere
For a sphere (radius $R$)
The denominator is even larger at $6k$. The larger the surface area/volume ratio, the higher the cooling efficiency.
| Shape | $T_{\max} - T_s$ | Surface Area/Volume |
|---|---|---|
| Flat Plate | $\dot{q}_v L^2/(2k)$ | $1/L$ |
| Cylinder | $\dot{q}_v R^2/(4k)$ | $2/R$ |
| Sphere | $\dot{q}_v R^2/(6k)$ | $3/R$ |
So the sphere has the best cooling efficiency.
That's because a sphere has the maximum surface area for the same volume. However, the story becomes a bit more complex when convection conditions are included.
Governing Equation for Internal Heat Generation
The steady-state heat conduction equation with internal heat generation is ∇²T + q̇/k = 0. The heat generation density spans over three orders of magnitude, from nuclear fuel pellets (q̇≈10⁸ W/m³) to Li-ion battery cells (q̇≈10⁵ W/m³). The importance of this equation surged in the 1940s with nuclear reactor design.
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 heats up and cools down slowly, 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)), which is why 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 metal has a high thermal conductivity $k$, heat transfers quickly from the hot side to the cold 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 term mathematically expresses the natural tendency of "heat flowing 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 under a fan 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 where heat is supplied externally to the "surface," 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 or 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: Surface-to-surface radiation uses the view factor method; for participating media, the DO method or P1 approximation is applied
- Non-applicable Cases: Phase change (melting/solidification) requires consideration of latent heat. Extreme temperature gradients necessitate thermal-stress coupling
Dimensional Analysis and Unit Systems
| Variable | SI Unit | Notes / Conversion Memo |
|---|---|---|
| Temperature $T$ | K (Kelvin) or Celsius | Be careful not to confuse absolute and Celsius temperatures. 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 constant pressure and constant volume specific heat (important for gases) |
| Heat Flux $q$ | W/m² | Neumann condition as a boundary condition |
Numerical Methods and Implementation
Implementation in FEM
How do you set up internal heat generation in FEM?
Set the volumetric heat generation rate for the elements. In FEM formulation, the following is added to the element heat load vector:
where $N_i$ is the shape function. For uniform generation, $\dot{q}_v \cdot V_e / n_{\text{node}}$ is equally distributed to each node.
How do you set it in Ansys?
Set it using the BFE (Body Force on Element) command.
```
BFE,ALL,HGEN,,1e6 ! Apply 1e6 W/m3 to all elements
```
In the Workbench GUI, apply the Internal Heat Generation condition to the Body. In Abaqus, set it with *DFLUX, BF$(\dot{q}_v)$.
Handling Non-Uniform Heat Generation
In real problems, the heat generation rate is often spatially non-uniform.
| Example | Heat Generation Distribution | Modeling |
|---|---|---|
| Electrical Resistor | Uniform ($I^2R/V$) | Constant |
| Nuclear Fuel Rod | Cosine distribution (axial) | Table/Function input |
| Electronic Circuit Board | Local (IC parts only) | Define Body per component |
| Induction Heating | Concentrated at skin depth | Coupled with electromagnetic analysis |
Induction heating requires coupling with electromagnetic analysis, I see.
Calculate the eddy current density using Ansys Maxwell or COMSOL AC/DC module, then input $\dot{q}_v = J^2/\sigma$ (Joule heating density) into the thermal analysis. In Ansys, data transfer is automated via Workbench's coupling function.
Mesh Considerations
In internal heat generation problems, if the mesh is too coarse, temperature peaks get averaged out.
So sufficient mesh is needed to accurately capture the center temperature.
Correct. Concentrate mesh in areas with the maximum temperature gradient (near boundaries). For a cylinder, the gradient is zero near the center so it can be coarse, but the area near the outer surface should be fine.
Analytical Solution for Heat Generation in a Flat Plate
For a flat plate of thickness 2L with uniform generation q̇, the center temperature is Tmax = Ts + q̇L²/(2k). For a 10mm thick silicon wafer (k=150 W/m·K) with q̇=10⁶ W/m³, the center temperature rise is only 0.08K—a calculation example showing its high uniformity.
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 the element. Smoothing may be required, similar to nodal stresses.
Convection-Diffusion Problem
When the Peclet number is high (convection-dominated), upwind 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 thermal 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 Determination
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 "predicting the next step using only current information, like a weather forecast"—calculation is fast but unstable with large time steps (misses storms). Implicit method is like "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
Application Example: Joule Heating in an Electrical Wire
I'd like to see a concrete calculation example.
Let's consider an AWG18 copper wire (diameter 1.02mm, $\rho_e = 1.7 \times 10^{-8}$ Ωm) with a 10A current.
| Parameter | Value |
|---|---|
| Cross-sectional area $A$ | $8.17 \times 10^{-7}$ m$^2$ |
| Resistance $R/L$ | 0.0208 Ω/m |
| Heat generation $I^2R/L$ | 2.08 W/m |
| $\dot{q}_v$ | $2.55 \times 10^6$ W/m$^3$ |
| $T_{\max} - T_s$ | $\dot{q}_v R^2/(4k) = 0.0016$℃ |
It only rises by 0.0016℃?
That's because copper's $k = 398$ W/(m K) is very large. The temperature difference is dominated not by the wire interior but by the insulation and external convection. In other words, the important factor in wire thermal design is the outer surface temperature of the insulation; the temperature distribution inside the copper can be considered almost uniform.
Application Example: Nuclear Fuel Rod
For a nuclear reactor fuel rod (UO$_2$ pellet, $k = 3$ W/(m K), $\dot{q}_v = 4 \times 10^8$ W/m$^3$, radius 5mm)
A 833℃ temperature difference? That's on a completely different scale.
That's because UO$_2$ has low $k$ and $\dot{q}_v$ is orders of magnitude larger. Ensuring the fuel center temperature does not exceed the melting point (about 2800℃) is the core of safety design.
Verification Points
Verify internal heat generation problems by checking the following.
- Shape of temperature distribution: Parabolic distribution for uniform generation
- Location of maximum temperature: Is it at the symmetry center?
- Energy balance: Total heat generation $\dot{q}_v \times V$ = Heat dissipation from surfaces
- Comparison with theoretical solution: Check calculations for simple shape parts
Checking the energy balance seems the most reliable.
In Ansys, check the Reaction Summary in the t
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