Heat Transfer: Conduction

Concept

Conduction is the transfer of heat through a stationary medium (solid, liquid, or gas) due to a temperature gradient. In solids, it occurs via atomic lattice vibrations and the movement of free electrons. The foundational principle governing conduction is Fourier's Law, which states that the rate of heat transfer is proportional to the temperature gradient and the cross-sectional area through which the heat flows. The proportionality constant is the thermal conductivity of the material. For steady-state conduction (where temperatures do not change with time), engineers often use an electrical analogy, treating heat flow as "current" driven by a temperature difference "voltage" across a "thermal resistance."

Formula & Method

Fourier's Law for one-dimensional steady-state heat conduction through a plane wall of thickness $L$ and area $A$ is:

$q = -k A \frac{dT}{dx} = k A \frac{T_1 - T_2}{L}$

Where $T_1$ is the higher temperature and $T_2$ is the lower temperature.

Using the electrical analogy, the thermal resistance for conduction ( $R_{cond}$ ) of a plane wall is defined as:

$R_{cond} = \frac{L}{k A}$

Thus, the heat transfer rate ( $q$ ) can be simply written as:

$q = \frac{\Delta T}{R_{cond}} = \frac{T_1 - T_2}{R_{cond}}$

For a composite wall made of multiple layers (materials A, B, C) in series, the total thermal resistance is the sum of the individual resistances:

$R_{total} = R_A + R_B + R_C = \frac{L_A}{k_A A} + \frac{L_B}{k_B A} + \frac{L_C}{k_C A}$

$q = \frac{T_{inner} - T_{outer}}{R_{total}}$

Variables & Units

$q$ = Rate of heat transfer, in Watts (W) or J/s.
$k$ = Thermal conductivity of the material, in W/(m·K) or W/(m· $^\circ$ C).
$A$ = Cross-sectional area perpendicular to heat flow, in square meters (m $^2$ ).
$T_1, T_2$ = Temperatures at the respective boundaries, in Kelvin (K) or Celsius ( $^\circ$ C).
$\Delta T$ = Temperature difference driving the heat flow, in K or $^\circ$ C.
$L$ = Thickness of the material layer, in meters (m).
$R_{cond}$ = Thermal resistance to conduction, in K/W or $^\circ$ C/W.

Worked Example

Problem: A furnace wall consists of two layers: an inner layer of refractory firebrick ( $L_A = 0.20 \text{ m}$ , $k_A = 1.4 \text{ W/(m·K)}$ ) and an outer layer of common brick ( $L_B = 0.10 \text{ m}$ , $k_B = 0.6 \text{ W/(m·K)}$ ). The inner surface temperature of the firebrick is $1000^\circ\text{C}$ and the outer surface temperature of the common brick is $50^\circ\text{C}$ . The wall has a cross-sectional area of $2.0 \text{ m}^2$ . Calculate the total thermal resistance and the steady-state heat loss through the wall.

Calculation:

Identify the parameters: $A = 2.0 \text{ m}^2$ , $T_1 = 1000^\circ\text{C}$ , $T_2 = 50^\circ\text{C}$ , $\Delta T = 950 \text{ K}$ .
Calculate the thermal resistance of the firebrick layer ( $R_A$ ): $R_A = \frac{L_A}{k_A A} = \frac{0.20}{1.4 \times 2.0} = \frac{0.20}{2.8} = 0.0714 \text{ K/W}$
Calculate the thermal resistance of the common brick layer ( $R_B$ ): $R_B = \frac{L_B}{k_B A} = \frac{0.10}{0.6 \times 2.0} = \frac{0.10}{1.2} = 0.0833 \text{ K/W}$
Calculate the total thermal resistance ( $R_{total}$ ): $R_{total} = R_A + R_B = 0.0714 + 0.0833 = 0.1547 \text{ K/W}$
Calculate the heat transfer rate ( $q$ ): $q = \frac{\Delta T}{R_{total}} = \frac{950}{0.1547} = 6140 \text{ W} = 6.14 \text{ kW}$

Engineering Meaning

The concept of thermal resistance makes analyzing complex, multi-layered systems (like building insulation or insulated pipes) straightforward. Materials with low thermal conductivity ( $k$ ) are insulators and will have a high thermal resistance, severely bottlenecking the flow of heat. Conversely, materials like copper or aluminum have very high thermal conductivities, presenting almost zero resistance to heat flow, making them ideal for heat sinks.

Engineering Check

When calculating heat transfer, ensure that the thermal conductivity ( $k$ ) used is appropriate for the expected temperature range, as $k$ often varies with temperature. Additionally, remember that true 1-D conduction assumes the cross-sectional area $A$ is constant and edge effects are negligible. For cylindrical systems (like pipes), the area $A$ increases with the radius, so a different logarithmic resistance formula must be used.

Explicit Exclusions

This foundational article explicitly excludes transient (time-varying) conduction, heat generation within the medium, multi-dimensional conduction (2D/3D), radial conduction (cylinders and spheres), and the mechanisms of convection and radiation.