MechanicalHeat TransfermediumFE ExamCoursework

Heat Flow Through a Composite Wall

A wall of area $A = 1.0\ \text{m}^2$ consists of two layers in series:

Brick: $200\ \text{mm}$ thick, $k = 0.70\ \text{W/m·K}$
Insulation: $50\ \text{mm}$ thick, $k = 0.040\ \text{W/m·K}$

The temperature difference across the whole wall is $\Delta T = 25\ \text{K}$ . Neglect surface convection resistances.

Find the rate of heat transfer through the wall.

Given

L1	200 mm	Brick thickness
k1	0.7 W/m*K	Brick conductivity
L2	50 mm	Insulation thickness
k2	0.04 W/m*K	Insulation conductivity
ΔT	25 K	Overall temperature difference

Hint 1

Treat this as a resistance network: layers in series add.

Hint 2

Thermal resistance of a slab is R = L/(kA).

Hint 3

Convert thicknesses to metres, then Q = ΔT/R_total.

Worked solution — try the problem first

Layers in series add their thermal resistances, exactly like resistors in a circuit:

$R = \frac{L}{kA}, \qquad R_{total} = R_1 + R_2, \qquad Q = \frac{\Delta T}{R_{total}}$

Brick: $R_1 = \frac{0.200}{0.70 \times 1.0} = 0.2857\ \text{K/W}$

Insulation: $R_2 = \frac{0.050}{0.040 \times 1.0} = 1.250\ \text{K/W}$

Total: $R_{total} = 0.2857 + 1.250 = 1.5357\ \text{K/W}$

Heat flow: $Q = \frac{25}{1.5357} = 16.28\ \text{W}$

$\boxed{Q = 16.3\ \text{W}}$

Note the dominance of the insulation. At one quarter the thickness, it contributes 81% of the total resistance — because its conductivity is 17.5× lower. This is why a thin insulating layer transforms a wall's performance.

Concepts:ConductionThermal resistanceComposite wallsFourier's law

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