---
title: "Furnace wall: insulation thickness for heat loss"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:furnace-fireclay-insulation-thickness"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/furnace-fireclay-insulation-thickness
site: BCT Engineering
---

# Furnace wall: insulation thickness for heat loss

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

- Asked in: 2075 Baisakh, 6 marks
- Video: [Problem on heat conduction in furnace wall](https://www.youtube.com/watch?v=v5ydSfrArNo) — Ashish Chaudhari
- 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/furnace-fireclay-insulation-thickness
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7

## Question

A furnace is made of fireclay brick of thickness 0.3 m and thermal conductivity of 1.2 W/m.K. The outside surface is to be insulated by an insulating material with the thermal conductivity of 0.05 W/m.K. Determine the thickness of the insulating layer in order to limit the heat loss per unit area of the furnace wall to 1200 W/m². When the inside surface of wall is at 900°C and the outside surface is at 25°C.

## Short answer

Use q = (T_in − T_out)/(L1/k1 + L2/k2). Total resistance per unit area = (900−25)/1200 = 0.7292 m²·K/W. Brick resistance 0.3/1.2 = 0.25 m²·K/W gives insulation resistance 0.4792 m²·K/W. Thickness = 0.4792 × 0.05 = 0.024 m (≈24 mm).

## Answer

**Given:** fireclay brick L₁ = 0.3 m, k₁ = 1.2 W/m·K; insulation k₂ = 0.05 W/m·K (thickness L₂ unknown); q = 1200 W/m²; T_in = 900 °C, T_out = 25 °C. Surface films are not given, so the temperatures are taken as the surface temperatures of the wall.

#### Method
For a two-layer plane wall, heat flux per unit area:

q = (T_in − T_out)/(L₁/k₁ + L₂/k₂)

#### Total resistance required
R_total (per m²) = ΔT/q = (900 − 25)/1200 = 875/1200 = 0.7292 m²·K/W

#### Brick resistance
R₁ = L₁/k₁ = 0.3/1.2 = 0.25 m²·K/W

#### Insulation resistance and thickness
R₂ = R_total − R₁ = 0.7292 − 0.25 = 0.4792 m²·K/W

L₂ = R₂ k₂ = 0.4792 × 0.05 = 0.02396 m

**Answer: insulation thickness ≈ 0.024 m (about 24 mm).**

Check: R = 0.25 + 0.02396/0.05 = 0.7292; q = 875/0.7292 = 1200 W/m² ✓.

## Check yourself

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

1. How is the required total resistance per m^2 found?
   A. R_total = ΔT/q = 875/1200
   B. R_total = ΔT x q
   C. R_total = k/L
   D. R_total = q/ΔT
   Answer: A. q = ΔT/R gives R_total = 0.7292 m^2 K/W. (Key point: Required total resistance)

2. The resistance of the brick layer is
   A. 0.3 x 1.2 = 0.36
   B. 0.3/0.05 = 6
   C. k1/L1 = 4
   D. L1/k1 = 0.3/1.2 = 0.25
   Answer: D. R1 = L/k = 0.25 m^2 K/W. (Key point: Brick resistance)

3. How is the insulation thickness obtained?
   A. L2 = R1 x k2
   B. L2 = (R_total - R1) x k2 = 0.4792 x 0.05 = 0.024 m
   C. L2 = R_total / k2
   D. L2 = R_total x k1
   Answer: B. R2 = L2/k2, so L2 = R2 k2 = 0.024 m (about 24 mm). (Key point: Insulation thickness)

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