---
title: "Steel tube with asbestos: heat loss, interface temp"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steel-tube-asbestos-heat-loss"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steel-tube-asbestos-heat-loss
site: BCT Engineering
---

# Steel tube with asbestos: heat loss, interface temp

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

- Asked in: 2074 Bhadra, 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/steel-tube-asbestos-heat-loss
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7

## Question

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.

## 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.

## Answer

**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 ✓.

## Check yourself

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

1. With 2 cm inside diameter, 1 cm steel thickness and 3 cm asbestos, the radii r1, r2, r3 are
   A. 0.02, 0.04, 0.07 m
   B. 0.01, 0.01, 0.03 m
   C. 0.01, 0.02, 0.05 m
   D. 0.02, 0.03, 0.06 m
   Answer: C. r1 = 1 cm, r2 = 1 + 1 = 2 cm, r3 = 2 + 3 = 5 cm. (Key point: Radii)

2. Which layer has the larger resistance?
   A. They are equal
   B. Steel
   C. Neither, they are zero
   D. Asbestos (0.729 vs 0.0058 °C/W)
   Answer: D. Steel is a good conductor (k = 19), so most of the drop is across the asbestos. (Key point: Dominant resistance)

3. Heat loss per metre is
   A. ΔT/R_total = 500/0.735 = 680 W/m
   B. 500/0.729 = 686
   C. 500 x 0.735
   D. 600/0.735
   Answer: A. Q/L = (T1 - T3)/(R1 + R2) = 680 W/m. (Key point: Heat loss)

4. How is the interface temperature found?
   A. T2 = T1 - (Q/L) R2
   B. T2 = T1 - (Q/L) R1 = 596.1 °C
   C. T2 = (T1 + T3)/2
   D. T2 = T3 + (Q/L) R1
   Answer: B. Only the steel drop of about 4 °C is subtracted from 600 °C. (Key point: Interface temperature)

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