- 2075 Baisakh · 6 marks
Furnace wall: insulation thickness for heat loss
Short answer
Use q = (T_in − T_out)/(L1/k1 + L2/k2). Total resistance per unit area = (900−25)/1200 = 0.7292 m²·K/W. Brick resistance 0.3/1.2 = 0.25 m²·K/W gives insulation resistance 0.4792 m²·K/W. Thickness = 0.4792 × 0.05 = 0.024 m (≈24 mm).
A furnace is made of fireclay brick of thickness 0.3 m and thermal conductivity of 1.2 W/m.K. The outside surface is to be insulated by an insulating material with the thermal conductivity of 0.05 W/m.K. Determine the thickness of the insulating layer in order to limit the heat loss per unit area of the furnace wall to 1200 W/m². When the inside surface of wall is at 900°C and the outside surface is at 25°C.
Given: fireclay brick L₁ = 0.3 m, k₁ = 1.2 W/m·K; insulation k₂ = 0.05 W/m·K (thickness L₂ unknown); q = 1200 W/m²; T_in = 900 °C, T_out = 25 °C. Surface films are not given, so the temperatures are taken as the surface temperatures of the wall.
Method
For a two-layer plane wall, heat flux per unit area:
q = (T_in − T_out)/(L₁/k₁ + L₂/k₂)
Total resistance required
R_total (per m²) = ΔT/q = (900 − 25)/1200 = 875/1200 = 0.7292 m²·K/W
Brick resistance
R₁ = L₁/k₁ = 0.3/1.2 = 0.25 m²·K/W
Insulation resistance and thickness
R₂ = R_total − R₁ = 0.7292 − 0.25 = 0.4792 m²·K/W
L₂ = R₂ k₂ = 0.4792 × 0.05 = 0.02396 m
Answer: insulation thickness ≈ 0.024 m (about 24 mm).
Check: R = 0.25 + 0.02396/0.05 = 0.7292; q = 875/0.7292 = 1200 W/m² ✓.
✓ Answer checked by Claude Opus 5.5 (high effort)
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