- 2074 Bhadra · 6 marks
Steel tube with asbestos: heat loss, interface temp
Short answer
Radial resistances per metre: R1=ln(2)/(2π×19)=0.00581, R2=ln(2.5)/(2π×0.2)=0.7292 °C/W. Q/L = (600−100)/0.7350 = 680 W/m. Interface temperature = 600 − 680×0.00581 = 596.1°C.
A thick-walled tube of stainless steel (k = 19 W/m°C) with 2 cm inside diameter and 1 cm thickness is covered with a 3 cm layer of asbestos insulation (k = 0.2 W/m°C). If the inside wall temperature of the pipe is maintained at 600°C and outside wall temperature of the insulation is maintained at 100°C, Calculate the heat loss per unit length. Also calculate the tube insulation interface temperature.
Given: stainless steel tube, k₁ = 19 W/m·°C, inner diameter 2 cm so r₁ = 0.01 m, thickness 1 cm so r₂ = 0.02 m. Asbestos, k₂ = 0.2 W/m·°C, thickness 3 cm so r₃ = 0.05 m. T₁ = 600 °C (inner wall), T₃ = 100 °C (outer surface of insulation). Per unit length (L = 1 m).
Resistances (per metre length)
R₁ = ln(r₂/r₁)/(2πk₁L) = ln(2)/(2π × 19) = 0.6931/119.38 = 0.00581 °C/W R₂ = ln(r₃/r₂)/(2πk₂L) = ln(2.5)/(2π × 0.2) = 0.9163/1.2566 = 0.7292 °C/W
R_total = 0.7350 °C/W
Heat loss per metre
Q/L = (T₁ − T₃)/R_total = 500/0.7350
Q/L = 680 W/m
Interface temperature
T₂ = T₁ − (Q/L) R₁ = 600 − 680.3 × 0.00581
T₂ = 596.1 °C
The steel is a good conductor, so the temperature drop across it is only about 4 °C; almost the full drop is across the asbestos. Check: T₂ − Q R₂ = 596.1 − 496.1 = 100 °C ✓.
✓ Answer checked by Claude Opus 5.5 (high effort)
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