- 2072 Asoj · 6 marks
Steam pipe asbestos insulation thickness
Short answer
With inner radius 0.06 m, outer 0.08 m, steam 150°C, air 20°C, and given coefficients, equate the required heat flux q = 2.1 kW/m² on the insulation’s outer surface to the total resistance per metre. Solving by trial gives outer radius r₃ ≈ 0.0898 m, so asbestos thickness ≈ 0.0098 m (9.8 mm).
A 150 mm steam pipe (k = 42 W/mK) has inside diameter of 120 mm and outside diameter of 160 mm. It is insulated at the outside with asbestos (k = 0.8 W/mK). The steam temperature is 150°C and the air temperature is 20°C. The heat transfer co-efficient for inner and outer surfaces are 100 W/m²K and 30 W/m²K. How thick should the asbestos to be provided in order to limit the heat loss to 2.1 kW/m²?
Reading of the data. The pipe is described as "150 mm", but the diameters given are 120 mm (inside) and 160 mm (outside), so the diameters are used. The limit "2.1 kW/m²" is a heat flux. The loss per metre length of the bare pipe in these conditions is only about 1.38 kW/m (calculated below), so the limit cannot be per metre of pipe length. It is taken as the heat flux q = 2.1 kW/m² on the outer surface of the insulation (the surface that loses heat to the air).
Given: r₁ = 0.06 m, r₂ = 0.08 m, k_pipe = 42, k_asb = 0.8 W/m·K; T_steam = 150 °C, T_air = 20 °C (ΔT = 130 K); h_i = 100, h_o = 30 W/m²·K. Unknown: outer radius r₃ of the asbestos.
Resistances per metre length
- Inside film: R_i = 1/(h_i 2π r₁) = 1/(100 × 2π × 0.06) = 0.02653 m·K/W
- Pipe wall: R_p = ln(r₂/r₁)/(2πk_p) = ln(1.3333)/(2π × 42) = 0.00109 m·K/W
- Asbestos: R_a = ln(r₃/r₂)/(2π × 0.8) = ln(r₃/0.08)/5.0265
- Outside film: R_o = 1/(h_o 2π r₃) = 1/(188.5 r₃)
(Bare pipe check: Q/L = 130/(0.02653 + 0.00109 + 0.06631) = 1384 W/m.)
Condition
Q/L = ΔT/(R_i + R_p + R_a + R_o), and Q/L = q × 2π r₃ = 2100 × 2π r₃
130/[0.02762 + ln(r₃/0.08)/5.0265 + 1/(188.5 r₃)] = 13 195 r₃
Solving by trial (r₃ in m):
| r₃ (m) | LHS (W/m) | RHS (W/m) |
|---|---|---|
| 0.085 | 1273 | 1122 |
| 0.090 | 1182 | 1188 |
| 0.0898 | 1185 | 1185 |
The two sides balance at r₃ = 0.0898 m. The heat loss is then Q/L = 1185 W/m.
Insulation thickness
t = r₃ − r₂ = 0.0898 − 0.08 = 0.0098 m
Answer: asbestos thickness ≈ 9.8 mm (about 1 cm).
✓ Answer checked by Claude Opus 5.5 (high effort)
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