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Plate with convection and radiation: inside temperature

Short answer

Heat flux from the surface: convection q_conv = 20×(300−20) = 5600 W/m², radiation q_rad = 0.9×5.67e-8×(573.15⁴−293.15⁴) = 5130 W/m², total q = 10730 W/m². Conduction temperature drop ΔT = qL/k = 10730×0.025/50 = 5.36 K. Inside plate temperature = 300 + 5.36 ≈ 305.4 °C.

A 2.5 cm thick plate (k = 50 W/mK) 50 cm by 75 cm is maintained at 300°C. Heat is lost from the plate surface by convection and radiation to the ambient air at 20°C. If the emissivity of the surface is 0.9 and the convection heat transfer coefficient is 20 W/m²K, determine the inside plate temperature. [σ = 5.67 × 10⁻⁸ W/m²K⁴]

Reading of the data. The 300 °C is taken as the temperature of the exposed outer surface, from which heat leaves by convection and radiation. The inside (other) face of the plate is hotter, because heat reaches the surface by conduction through the plate. Heat flow is steady and one-dimensional.

Given: L = 2.5 cm = 0.025 m, k = 50 W/m·K, A = 0.5 × 0.75 = 0.375 m², T_s = 300 °C = 573.15 K, T_∞ = 20 °C = 293.15 K, ε = 0.9, h = 20 W/m²·K, σ = 5.67×10⁻⁸ W/m²·K⁴.

Heat flux leaving the surface

Convection: q_conv = h (T_s − T_∞) = 20 × 280 = 5600 W/m²

Radiation (to surroundings at the air temperature): q_rad = εσ (T_s⁴ − T_∞⁴) = 0.9 × 5.67×10⁻⁸ × (573.15⁴ − 293.15⁴) = 0.9 × 5.67×10⁻⁸ × (1.0791×10¹¹ − 0.0739×10¹¹) = 5130 W/m²

Total: q = 5600 + 5130 = 10730 W/m²

Heat flow rate: Q = qA = 10730 × 0.375 = 4024 W

Temperature drop across the plate

Conduction: q = k (T_i − T_s)/L

T_i − T_s = qL/k = 10730 × 0.025/50 = 5.36 K

Inside temperature

T_i = 300 + 5.36

Answer: inside plate temperature T_i ≈ 305.4 °C.

✓ Answer checked by Claude Opus 5.5 (high effort)

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