Training Conduction & Thermal Resistance Fourier's Law and Planar Conduction
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Fourier's Law and Planar Conduction

24 min Conduction & Thermal Resistance

Fourier’s Law and Planar Conduction

Heat conduction is the transfer of thermal energy through a material due to a temperature difference. Fourier’s law is the governing equation for all conduction problems.

Fourier’s Law of Heat Conduction

$$\dot{Q} = -kA\frac{dT}{dx}$$

For steady-state, 1-D, planar conduction through a slab of thickness $L$:

$$\dot{Q} = \frac{kA(T_1 - T_2)}{L}$$

$\dot{Q}$ = heat transfer rate (W), $k$ = thermal conductivity (W/(m·K)), $A$ = cross-sectional area (m²), $L$ = thickness (m).

Thermal Conductivity Reference

Copper: $k \approx 385$ W/(m·K) • Aluminum: $k \approx 205$ • Steel: $k \approx 50$ • Concrete: $k \approx 1.4$ • Fiberglass insulation: $k \approx 0.04$ • Air: $k \approx 0.026$

Example 1 — Window Pane

A glass window ($k = 0.8$ W/(m·K)) is 5 mm thick and 1.5 m² in area. Inside temperature is 22°C, outside is −5°C. Find the rate of heat loss.

$$\dot{Q} = \frac{kA(T_1 - T_2)}{L} = \frac{0.8 \times 1.5 \times (22 - (-5))}{0.005} = \frac{0.8 \times 1.5 \times 27}{0.005} = 6{,}480 \text{ W}$$

That’s 6.48 kW — this is why windows are a major source of heat loss!

Example 2 — Insulation Thickness

A furnace wall must limit heat loss to 500 W. Wall area = 2 m², $\Delta T = 800$°C, insulation $k = 0.05$ W/(m·K). Find the required thickness.

$$L = \frac{kA\Delta T}{\dot{Q}} = \frac{0.05 \times 2 \times 800}{500} = 0.16 \text{ m} = 16 \text{ cm}$$

Example 3 — Comparing Materials

A 10 cm wall conducts 1000 W/m² with $\Delta T = 50$°C. Find the thermal conductivity.

Heat flux $q = \dot{Q}/A = 1000$ W/m².

$$k = \frac{qL}{\Delta T} = \frac{1000 \times 0.10}{50} = 2.0 \text{ W/(m\cdot K)}$$

This is consistent with brick or stone.

Practice Problems

1. A copper plate ($k = 385$) is 2 cm thick, 0.5 m². $\Delta T = 100$°C. Find $\dot{Q}$.
2. An aluminum wall ($k = 205$) loses 4100 W through 1 m² with $\Delta T = 40$°C. Find thickness.
3. Heat flux through a 20 cm concrete wall ($k = 1.4$): $T_1 = 25$°C, $T_2 = 5$°C. Find $q$ in W/m².
4. A styrofoam cooler wall ($k = 0.033$) is 3 cm thick. Inside = 2°C, outside = 35°C. $A = 0.6$ m². $\dot{Q}$?
5. If you double the thickness of insulation, what happens to $\dot{Q}$?
6. Which material loses heat faster for the same geometry and $\Delta T$: steel ($k = 50$) or fiberglass ($k = 0.04$)?
Show Answer Key

1. $\dot{Q} = 385 \times 0.5 \times 100 / 0.02 = 962{,}500$ W = 962.5 kW

2. $L = 205 \times 1 \times 40 / 4100 = 2.0$ m

3. $q = 1.4 \times 20 / 0.20 = 140$ W/m²

4. $\dot{Q} = 0.033 \times 0.6 \times 33 / 0.03 = 21.8$ W

5. $\dot{Q}$ is halved (inversely proportional to $L$).

6. Steel — its $k$ is 1250× larger than fiberglass.

Thermal Resistance Networks