Nonlinear heat conduction: temperature-dependent conductivity κ(T) (Kirchhoff transform)
Nonlinear heat transferPASS · <0.01%Steady heat conduction through a bar with linearly temperature-dependent conductivity κ(T) = κ₀(1 + βT) = 20·(1 + 0.01·T) W/(m·K) between 0 °C and 100 °C. While constant κ would yield 50 °C at mid-span, the exact Kirchhoff integral transform gives T(L/2) = 58.1139 °C.Steady heat conduction through a bar with linearly temperature-dependent conductivity κ(T) = κ₀(1 + βT) = 20·(1 + 0.01·T) W/(m·K) between 0 °C and 100 °C. While constant κ would yield 50 °C at mid-span, the exact Kirchhoff integral transform gives T(L/2) = 58.… show full text
- Tolerance:
- 5%
Problem statement
- Geometry
- Prismatic bar 100 × 10 × 10 mm; tetrahedral mesh, element size 5 mm.
- Material
- Tabular thermal conductivity κ(T) = 20·(1 + 0.01·T) W/(m·K) specified via kappa_table="-50:10,400:100".
- Boundary conditions
- Surface number 1 (x = 0) fixed at 0 °C; surface number 2 (x = 100 mm) fixed at 100 °C; lateral faces insulated.
- Measured quantity
- Mid-span temperature T(x = 50 mm) compared against the exact Kirchhoff root (−1 + √(1 + β(100 + 5000β)))/β = 58.1139 °C.
WebCAE Terminal
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units mm
material E=2e11 nu=0.3 kappa=20 kappa_table="-50:10,400:100"
primitive box dx=100 dy=10 dz=10
mesh tet size=5
thermal surface number 1 temperature=0
thermal surface number 2 temperature=100
solve thermal
probe temperature x=50 id="mid"Results comparison
| Reference value | T(50 mm) = 58.1139 °C |
| Our result | T(50 mm) = 58.1139 °C |
| Deviation | <0.01% |
| Tolerance | 5% |
| Status | PASS · <0.01% |
| Mesh details | tet4, size 5 mm (366 nodes); 4 Newton iterations, residual 2.68e+1 → 2.85e-11 W |
Updated: 2026-10-27 · version not recorded