Transient radiative cooling of a high-conductivity slab
Nonlinear heat transferPASS · <0.01%Nonlinear transient cooling of a thermally lumped block (Bi ≪ 1) from T₀ = 900 °C over 600 s by Stefan–Boltzmann radiation (ε = 0.85) into a 20 °C ambient. Temperature at t = 600 s and cumulative radiated energy are verified against the exact radiative cooling ODE ρ·c_p·L·dT/dt = −εσ(T⁴ − T_∞⁴).Nonlinear transient cooling of a thermally lumped block (Bi ≪ 1) from T₀ = 900 °C over 600 s by Stefan–Boltzmann radiation (ε = 0.85) into a 20 °C ambient. Temperature at t = 600 s and cumulative radiated energy are verified against the exact radiative cooling… show full text
- Tolerance:
- 5%
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Open & Solve in BrowserProblem statement
- Geometry
- Block L = 50 mm, cross-section 10 × 10 mm; tetrahedral mesh, element size 5 mm.
- Material
- High conductivity k = 10⁵ W/(m·K) (uniform temperature across length, Bi < 10⁻⁴), ρ = 7800 kg/m³, c_p = 460 J/(kg·K).
- Boundary conditions
- Initial temperature T₀ = 900 °C; surface number 2 (x = 50 mm) radiating with ε = 0.85 to T_∞ = 20 °C; remaining faces insulated. Crank–Nicolson θ = 0.5 with Newton–Raphson at each step, Δt = 5 s, t_end = 600 s.
- Measured quantity
- Temperature at t = 600 s compared against the exact ODE solution (695.576 °C) and global energy balance residual |ΔU + E_rad| / E_scale.
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units mm
material E=2e11 nu=0.3 density=7800 kappa=1e5 specific_heat=460
primitive box dx=50 dy=10 dz=10
mesh tet size=5
thermal surface number 2 radiation=0.85 ambient=20
solve thermal transient dt=5 tend=600 theta=0.5 t0=900 frames=3
probe temperature x=0 time=600 id="T600"Results comparison
| Reference value | T(600 с) = 695.576 °Cbalance residual ≤ 1e-9 |
| Our result | T(600 с) = 695.581 °Cresidual 5.19e-10 |
| Deviation | <0.01% |
| Tolerance | 5% |
| Status | PASS · <0.01% |
| Mesh details | tet4, size 5 mm (195 nodes); 120 steps of Δt = 5 с, θ = 0.5, 240 Newton iterations (max 2/step) |
Updated: 2026-10-27 · version not recorded