---
title: "Composite wall conduction using finite elements"
subject: "Finite Element Method"
chapter: 3
key: "semester_8/finite_element_method:composite-wall-heat-conduction"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/composite-wall-heat-conduction
site: BCT Engineering
---

# Composite wall conduction using finite elements

Finite Element Method · Chapter 3: Direct Stiffness Method: Discrete Finite Elements

- Asked in: year not given, 6 marks
- Video: [1D FEM Composite wall](https://www.youtube.com/watch?v=bjFe0ptrR-0) — Dr Raed Sabri Hameed
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/composite-wall-heat-conduction
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-3

## Question

A furnace wall has an inner brick layer ($k = 1.2$ W/m K, thickness 250 mm) and an outer insulation layer ($k = 0.04$ W/m K, thickness 100 mm). The inner surface is at 800 °C and the outer surface is at 30 °C. Treating each layer as one linear one-dimensional heat conduction element (analogous to a bar element), find the interface temperature and the heat flow per square metre. Show the analogy with the bar element.

## Short answer

In the bar analogy, conductances are c₁=1.2/0.25=4.8 W/K (brick) and c₂=0.04/0.1=0.4 W/K (insulation). Assembly with T₁=800 °C, T₃=30 °C yields interface temperature 740.8 °C. The heat flux per square metre is q=4.8(800−740.8)=284.3 W/m².

## Answer

#### Analogy

| Bar problem | Heat conduction |
|---|---|
| Displacement $u$ | Temperature $T$ |
| Axial force $F$ | Heat flow $Q$ |
| Stiffness $AE/L$ | Conductance $kA/L$ |

Element equation: $\dfrac{kA}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} T_i \\ T_j \end{Bmatrix} = \begin{Bmatrix} Q_i \\ Q_j \end{Bmatrix}$.

#### Element conductances ($A = 1\ \text{m}^2$)

$$
c_1 = \frac{1.2}{0.25} = 4.8\ \text{W/K},\qquad c_2 = \frac{0.04}{0.10} = 0.4\ \text{W/K}
$$

```text
 800 C o-----brick-----o-----insulation-----o 30 C
       1       (1)     2         (2)         3
```

#### Assembly and solution

$$
\begin{bmatrix} 4.8 & -4.8 & 0 \\ -4.8 & 5.2 & -0.4 \\ 0 & -0.4 & 0.4 \end{bmatrix}\begin{Bmatrix} T_1 \\ T_2 \\ T_3 \end{Bmatrix} = \begin{Bmatrix} Q_1 \\ 0 \\ Q_3 \end{Bmatrix}
$$

With $T_1 = 800$ and $T_3 = 30$, the middle row (no heat source at node 2) gives
$$
-4.8(800) + 5.2\,T_2 - 0.4(30) = 0 \Rightarrow T_2 = \frac{3840 + 12}{5.2} = 740.77\ ^\circ\text{C}
$$

Heat flow through element 1: $q = 4.8(800 - 740.77) = 284.3$ W/m$^2$.

Check with element 2: $0.4(740.77 - 30) = 284.3$ W/m$^2$. Check with thermal resistances: $q = \dfrac{800 - 30}{0.25/1.2 + 0.10/0.04} = \dfrac{770}{2.7083} = 284.3$ W/m$^2$.

**Answer:** interface temperature $= 740.8\ ^\circ$C; heat flow $= 284.3$ W per m$^2$ of wall.

## Check yourself

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

1. In the heat conduction analogy, what corresponds to the bar stiffness AE/L?
   A. Conductance kA/L
   B. Temperature T
   C. Heat flow Q
   D. Thickness L
   Answer: A. Displacement maps to temperature, force to heat flow and stiffness to conductance kA/L. (Key point: Analogy)

2. What are the conductances of the brick and insulation layers (A = 1 m^2)?
   A. 0.2 and 2.5 W/K
   B. 4.8 and 0.4 W/K
   C. 1.2 and 0.04 W/K
   D. 4.8 and 2.5 W/K
   Answer: B. c1 = 1.2/0.25 = 4.8 and c2 = 0.04/0.10 = 0.4 W/K. (Key point: Conductance)

3. What equation gives T2 and what is its value?
   A. T2 = (800 + 30)/2 = 415 C
   B. T2 = 800 - 30 = 770 C
   C. Middle row (no source at node 2): -4.8(800) + 5.2 T2 - 0.4(30) = 0, T2 = 740.8 C
   D. 5.2 T2 = 800, T2 = 154 C
   Answer: C. With no heat source at node 2, Q2 = 0, giving T2 = 3852/5.2 = 740.77 C. (Key point: Interface temperature)

4. Which check confirms the heat flow of 284.3 W/m^2?
   A. q = 800 x 30
   B. q = 4.8 x 0.4
   C. q = 800/5.2
   D. q = (800 - 30) divided by the sum of thermal resistances 0.25/1.2 + 0.10/0.04
   Answer: D. 770 / 2.7083 = 284.3 W/m^2, same as 4.8(800 - 740.77) and 0.4(740.77 - 30). (Key point: Heat flow check)

## Related in Finite Element Method

- [Variational approach for one-dimensional heat conduction](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Avariational-approach-heat-conduction) — Finite Element Method, chapter 5
- [Linear shape functions for one-dimensional element](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Alinear-bar-shape-functions) — Finite Element Method, chapter 5

## Related in other subjects

- [Furnace wall: insulation thickness for heat loss](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Afurnace-fireclay-insulation-thickness) — Fundamental of Thermodynamics & Heat Transfer, chapter 7
- [Composite furnace wall heat loss](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Acomposite-furnace-wall-interface-temperatures) — Heat Transfer, chapter 2
- [Pipe insulation: extra layer to cut heat 40%](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Apipe-insulation-additional-layer-40-percent) — Fundamental of Thermodynamics & Heat Transfer, chapter 7
- [Furnace brickwork: red brick thickness without filling](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Afurnace-brickwork-diatomite-filling) — Fundamental of Thermodynamics & Heat Transfer, chapter 7