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Composite wall conduction using finite elements

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.

A furnace wall has an inner brick layer (k=1.2k = 1.2 W/m K, thickness 250 mm) and an outer insulation layer (k=0.04k = 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.

Analogy

Bar problemHeat conduction
Displacement uuTemperature TT
Axial force FFHeat flow QQ
Stiffness AE/LAE/LConductance kA/LkA/L

Element equation: kAL[1−1−11]{TiTj}={QiQj}\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 m2A = 1\ \text{m}^2)

c1=1.20.25=4.8 W/K,c2=0.040.10=0.4 W/Kc_1 = \frac{1.2}{0.25} = 4.8\ \text{W/K},\qquad c_2 = \frac{0.04}{0.10} = 0.4\ \text{W/K} 
 800 C o-----brick-----o-----insulation-----o 30 C
       1       (1)     2         (2)         3

Assembly and solution

[4.8−4.80−4.85.2−0.40−0.40.4]{T1T2T3}={Q10Q3}\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 T1=800T_1 = 800 and T3=30T_3 = 30, the middle row (no heat source at node 2) gives

−4.8(800)+5.2 T2−0.4(30)=0⇒T2=3840+125.2=740.77 ∘C-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)=284.3q = 4.8(800 - 740.77) = 284.3 W/m2^2.

Check with element 2: 0.4(740.77−30)=284.30.4(740.77 - 30) = 284.3 W/m2^2. Check with thermal resistances: q=800−300.25/1.2+0.10/0.04=7702.7083=284.3q = \dfrac{800 - 30}{0.25/1.2 + 0.10/0.04} = \dfrac{770}{2.7083} = 284.3 W/m2^2.

Answer: interface temperature =740.8 ∘= 740.8\ ^\circC; heat flow =284.3= 284.3 W per m2^2 of wall.

✓ Answer checked by Claude Opus 5.5 (high effort)

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