--- title: "Insulating layer: minimum outside heat transfer coefficient" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 7 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:insulating-layer-min-outside-h-300c" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/insulating-layer-min-outside-h-300c site: BCT Engineering --- # Insulating layer: minimum outside heat transfer coefficient Fundamental of Thermodynamics & Heat Transfer · Chapter 7: Introduction to Heat Transfer - Asked in: 2068 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/insulating-layer-min-outside-h-300c - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7 ## Question 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? ## 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. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What principle is used to find h? A. Heat flux by conduction through the layer equals heat flux by convection from the outside surface B. Outside temperature equals air temperature C. Resistance of the layer equals 1/h D. Heat flux inside is zero Answer: A. In steady state the same flux leaves the layer by conduction and goes to the air by convection. (Key point: Equating fluxes) 2. What is the conduction flux through the layer? A. 0.8 x 275/0.05 = 4400 W/m2 B. 0.8 x 200/0.05 = 3200 W/m2 C. 0.8 x 200 x 0.05 = 8 W/m2 D. 0.8 x 75/0.05 = 1200 W/m2 Answer: B. q = k(T1 - T2)/L with T1 - T2 = 300 - 100 = 200 K and L = 0.05 m. (Key point: Conduction flux) 3. Why is the calculated h the minimum value? A. A larger h would raise the surface above 100 °C B. A smaller h gives a lower surface temperature C. A smaller h would force the surface above 100 °C to dissipate the same heat D. h has no effect on surface temperature Answer: C. Since q = h(T2 - T_inf), smaller h needs a larger T2 for the same q, exceeding the 100 °C limit. 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