- 2072 Magh · 6 marks
Pipe insulation: extra layer to cut heat 40%
Short answer
Existing resistance per metre (insulation + convection) is 1.9594 m·K/W. For a 40% reduction, new resistance must be 3.2657 m·K/W. Adding a second layer (k = 0.08 W/m·K) and solving gives outer radius 0.1774 m, so extra insulation thickness = 0.0874 m (about 87 mm).
A 100 mm diameter pipe carrying steam is covered by a layer of insulation (K = 0.05 W/mK) having a thickness of 40 mm. The heat transfer coefficient between the outer surface of insulation and the ambient air is 20 W/m²K. Determine the required thickness of another insulating layer (K = 0.08 W/mK) that must be added to reduce the heat transfer rate by 40% assuming heat transfer coefficient remains the same.
Assumptions: the temperature difference between the steam (pipe surface) and the air is the same before and after; the resistance of the pipe wall and of the steam-side film is neglected. Per metre length.
Given: pipe outer radius r₁ = 50 mm = 0.05 m; first layer (k₁ = 0.05 W/m·K) thickness 40 mm so r₂ = 0.09 m; outer h = 20 W/m²·K; second layer k₂ = 0.08 W/m·K, outer radius r₃ unknown.
Existing resistance (per metre)
R_ins1 = ln(r₂/r₁)/(2πk₁) = ln(1.8)/(2π × 0.05) = 0.5878/0.31416 = 1.8710 m·K/W R_conv = 1/(h 2π r₂) = 1/(20 × 2π × 0.09) = 0.0884 m·K/W R_before = 1.9594 m·K/W
Required resistance
The heat loss falls by 40 %, so Q_new = 0.6 Q_old. With the same ΔT:
R_new = R_before/0.6 = 1.9594/0.6 = 3.2657 m·K/W
Added layer
R_new = R_ins1 + ln(r₃/r₂)/(2πk₂) + 1/(h 2π r₃)
3.2657 = 1.8710 + ln(r₃/0.09)/(0.5027) + 1/(125.66 r₃)
Solve by trial:
| r₃ (m) | Right side (m·K/W) |
|---|---|
| 0.15 | 2.940 |
| 0.17 | 3.183 |
| 0.1774 | 3.266 |
r₃ = 0.1774 m
Thickness of added layer = r₃ − r₂ = 0.1774 − 0.09 = 0.0874 m
Answer: the extra insulation must be about 87 mm (8.7 cm) thick.
✓ Answer checked by Claude Opus 5.5 (high effort)
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