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Conduction and Thermal Resistance

Fourier's law rearranged is Ohm's law, so the whole of circuit analysis carries over — including the case where adding insulation increases the heat loss.

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Fourier's law rearranged gives Q = ΔT/R_th, structurally identical to Ohm's law — so composite walls, surface films and geometry all become a resistance network, including the case where adding pipe insulation increases the heat loss.

Fourier's law as a resistance

Q = −kA·(dT/dx) rearranges to Q = ΔT/R_th with R_th = L/kA. Temperature plays the role of voltage and heat flow that of current, so series and parallel combination, dividers and network analysis all carry over unchanged.

The minus sign in Fourier's law is the second law in one character: heat flows down the gradient, never up it. Conductivity spans four orders of magnitude, from 0.02 W/mK for still air to 400 for copper.

The resistances in a real wall

ElementResistanceComment
Plane wallL / kAProportional to thickness
Cylinderln(r₂/r₁) / 2πkLOnly logarithmic in radius
Sphere(1/r₁ − 1/r₂) / 4πkApproaches a finite limit as r₂ → ∞
Surface film1 / hANewton's law of cooling, as a resistance
Contact between layersR_c / AOften the largest and most overlooked

In a double-glazed unit the glass contributes almost nothing — the trapped air and its surface films are the insulation. Still air has h ≈ 5 W/m²K against a stiff wind's 50, which is why wind chill is a real effect and not a perception.

Series, parallel and thermal bridges

Layers in a wall are resistances in series, so the largest dominates. A high-conductivity path through the insulation — a steel stud, a balcony slab, a window frame — is a parallel resistance that short-circuits it.

That is a thermal bridge, and the circuit analogy predicts its effect exactly: a small parallel resistance across a large series one carries most of the flow. It is why modern building regulations specify bridging separately from the wall build-up.

The critical radius

For a cylinder, adding insulation raises conduction resistance only as ln(r₂/r₁) while increasing the outer area linearly, which lowers the convection resistance. Two effects pull opposite ways.

Below the critical radius r_crit = k/h, the area effect wins and insulation makes things worse. For a thin wire it can be several millimetres — exploited deliberately, since cable sheathing helps conductors run cooler. For any normal pipe, r_crit is well inside the pipe and insulation always helps.

Fins

Since convection is usually the dominant resistance, adding area on that side is the effective move. But heat must conduct along a fin to reach its tip, so a long fin's far end sits near ambient and contributes little.

Fin efficiency measures that fraction, which is why heatsinks use many short fins rather than a few long ones — until the spacing chokes the airflow between them.

The Biot number

Bi = hL/k compares internal conduction resistance with external convection resistance. Below 0.1 the body is nearly isothermal and cools as a single exponential — the lumped capacitance model, mathematically identical to an RC circuit. Above it, internal gradients matter.

The numbers you will be asked for

Fourier's law

Q = −kA · dT/dx

Plane wall resistance

R = L / kA

Cylindrical resistance

R = ln(r₂/r₁) / (2πkL)

Convection resistance

R = 1 / hA

Critical radius

r_crit = k / h

cylinder; 2k/h for a sphere

Biot number

Bi = hL_c / k

L_c = V/A

Watch it work

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Check yourself

question 1 / 4

One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

What does rearranging Fourier's law into Q = ΔT/R_th buy you?
In a double-glazed window, what is doing the insulating?
Adding insulation to a thin wire increases its heat loss. How?
A steel stud runs through the insulation in a timber-frame wall. What does the analogy predict?

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