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Insulating layer: minimum outside heat transfer coefficient

Short answer

Conduction flux through insulation: q = 0.8 × (300−100)/0.05 = 3200 W/m². Convection flux q = h × (100−25) = 75h. Equating gives h = 3200/75 ≈ 42.7 W/m²·K, the minimum outside heat transfer coefficient.

The inside surface of an insulating layer is at 300°C and the outside surface is dissipating heat by convection into air at 25°C. The insulating layer has a thickness of 5cm and thermal conductivity of 0.8W/mK. What is the minimum heat transfer coefficient at the outside surface if the outside surface temperature should not exceed 100°C?

Given: inside surface T₁ = 300 °C, outside surface T₂ ≤ 100 °C, air T_∞ = 25 °C, thickness L = 5 cm = 0.05 m, k = 0.8 W/m·K. Steady, one-dimensional flow. We need the minimum h so that the outside surface does not exceed 100 °C.

Heat flux through the layer (conduction)

q = k (T₁ − T₂)/L = 0.8 × (300 − 100)/0.05 = 160/0.05 = 3200 W/m²

Heat flux by convection from the outside surface

q = h (T₂ − T_∞) = h (100 − 25) = 75 h

Equate the fluxes

75 h = 3200

h = 3200/75 = 42.67 W/m²·K

Answer: the minimum outside heat transfer coefficient is h ≈ 42.7 W/m²·K.

A smaller h would force the surface temperature above 100 °C to dissipate the same heat.

✓ Answer checked by Claude Opus 5.5 (high effort)

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