---
title: "Steam pipe asbestos insulation thickness"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-pipe-asbestos-thickness-2-1kw"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-pipe-asbestos-thickness-2-1kw
site: BCT Engineering
---

# Steam pipe asbestos insulation thickness

Fundamental of Thermodynamics & Heat Transfer · Chapter 7: Introduction to Heat Transfer

- Asked in: 2072 Asoj, 6 marks
- Video: [Heat Loss Through Insulated Steam Pipe | Example 3-8 | Cylindrical Conduction Thermal Resistance Sol](https://www.youtube.com/watch?v=Z0XWz4upcN8) — Education Shop
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-pipe-asbestos-thickness-2-1kw
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7

## Question

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?

## 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).

## Answer

**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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which resistances are summed per metre length?
   A. Pipe wall and asbestos only
   B. Only the two films
   C. Only asbestos
   D. Inside film, pipe wall, asbestos, outside film
   Answer: D. All four are in series between steam and air. (Key point: Resistances in series)

2. What equation is solved for r3?
   A. ΣR = q
   B. q = ΔT x 2π r3
   C. ΔT/ΣR = q x 2π r3
   D. ΔT = q
   Answer: C. Q/L from resistances equals flux times outer area per metre, 2100 x 2π r3. (Key point: Condition)

3. The asbestos thickness is about
   A. 0.98 mm
   B. 98 mm
   C. 20 mm
   D. 9.8 mm
   Answer: D. r3 = 0.0898 m, so t = 0.0898 - 0.08 = 0.0098 m. (Key point: Result)

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