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Technical Physics

Heat conduction: Fourier's law and thermal resistance

Fourier's law and material conductivities, the plane wall with proof, thermal resistance, the composite wall and transmittance with an example, the cylinder, critical radius, contact resistance.

Thermodynamics says how much heat is exchanged, heat transfer says how fast. There are three mechanisms: conduction, convection and radiation. In this lesson we study conduction, the passage of heat through a stationary material, introduce thermal resistance, which makes thermal problems similar to electric circuits, and apply it to walls, pipes and insulation.

Fourier’s law

The heat flux per unit area is proportional to the temperature gradient and flows from hot to cold:

    \[ \mathbf{q} = -k\,\nabla T \]

where k is the thermal conductivity, in W/(m·K). Its values vary widely:

Materialk in W/(m·K)
copper400
aluminium237
steel15-50
glass1
brick0.5-0.9
water0.6
rock wool, polystyrene0.03-0.04
still air0.026

Insulators work by trapping still air in small cells, because air conducts very little as long as it does not move.

The plane wall

Theorem. In a plane wall of thickness L, area A and constant conductivity k, in steady state, temperature varies linearly between the two faces and the heat crossing it is

    \[ \dot Q = \frac{k A}{L}\,(T_1 - T_2) = \frac{T_1 - T_2}{R},\qquad R = \frac{L}{kA} \]

Proof. In steady state and with no internal generation, the heat entering each thin layer equals the heat leaving it, so the flux q is the same at every depth x. By Fourier’s law dT/dx = -q/k is constant: temperature is linear, and integrating from 0 to L gives T_1 - T_2 = qL/k. Multiplying by the area gives the statement. ∎

The formula is the thermal Ohm’s law: the temperature difference is the “voltage”, the heat flow the “current”, R the thermal resistance in K/W. All the circuit rules follow: resistances of layers in series add up, those of parallel paths combine like conductances.

The composite wall

A house wall is made of different layers, and on each face there is convective exchange with the air, described by a resistance 1/(hA) (next lesson).

Two-layer composite wall with the temperature profile and the thermal resistance network
The temperature drops a little in the indoor air, a little in the brick, a lot in the insulating layer and again a little in the outdoor air. Below, the same wall as a series of four resistances: indoor convection, first layer, second layer, outdoor convection.

Per square metre of wall:

    \[ R_{tot} = \frac{1}{h_i} + \frac{L_1}{k_1} + \frac{L_2}{k_2} + \frac{1}{h_e},\qquad U = \frac{1}{R_{tot}} \]

U is the thermal transmittance, in W/(m²·K), the figure limited by building energy-efficiency regulations.

Example. A 25 cm brick wall (k = 0.7), coefficients h_i = 7.7 and h_e = 25 W/(m²·K): R = 0.13 + 0.357 + 0.04 \approx 0.53 m²·K/W, U \approx 1.9 W/(m²·K). Adding 8 cm of insulation (k = 0.035, R = 2.29): R \approx 2.81, U \approx 0.36 W/(m²·K). The heat lost falls more than fivefold: with a 20 °C difference, from 38 to 7 W per square metre. It also shows in the temperature profile: almost the whole drop happens in the insulation, the largest resistance in the series.

The cylinder and pipe insulation

For a cylindrical layer (a pipe or its lagging) between radii r_1 and r_2, of length L:

    \[ R = \frac{\ln(r_2/r_1)}{2\pi kL} \]

Proof. In steady state the same heat \dot Q crosses every cylindrical surface of radius r, of area 2\pi rL: \dot Q = -k\,2\pi rL\,dT/dr. Separating variables, dT = -\frac{\dot Q}{2\pi kL}\frac{dr}{r}, and integrating from r_1 to r_2: T_1 - T_2 = \frac{\dot Q}{2\pi kL}\ln\frac{r_2}{r_1}. ∎

The logarithmic dependence has a curious consequence: adding insulation increases the conduction resistance, but it also increases the outer surface exchanging by convection. For very thin pipes and cables, below a critical radius r_c = k/h (about 1 cm for a plastic insulator in still air), a thin coating makes them lose more heat rather than less. For electrical wires this is an advantage: the sheath helps cool them.

Contact resistance and internal generation

Two solid surfaces in contact never touch perfectly: air remains between the asperities, creating a contact resistance. In electronics it matters a great deal: between a component and its heat sink a thermal paste or thermal pad is placed, filling the gaps and cutting this resistance by up to ten times.

When heat is generated inside the material (a resistor carrying current, a winding, a chip) the temperature profile is no longer linear but parabolic, with its maximum where heat finds it hardest to get out. That is why the hottest point of a component, the junction, is inside it and cannot be measured directly: it is computed with thermal resistances, as we will see in the next lesson.

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