Thermal conduction through layered materials

This page gives the UY1 working model/result for Thermal conduction through a compound slab. You reuse it when you turn thermal intuition into calculable models (energy balances, process reasoning, and kinetic/statistical links), with careful sign/units checks.

Module path: Thermodynamics (UY1)
Full routing: UY1 assessment map
Math toolkit: Mathematics for undergraduate physics

At a glance

Prerequisites: Fourier law for a single slab, steady-state heat flow.
Outcomes: solve multilayer slab and cylindrical insulation conduction.
Key result: thermal resistances in series add directly.
Common trap: assuming temperature drops equally across layers with different k and L.

Setup

Steady 1D conduction through two layers in series:

Same area A.
Layer 1: (k₁,L₁), layer 2: (k₂,L₂).
Boundary temperatures Tₕ > Tc.

At steady state, the same Q dot passes through each layer.

Physical interpretation:

A multilayer wall behaves like resistors in series: the same heat current must pass through every layer, and each layer “uses up” a temperature drop proportional to its resistance.
The layer with the largest Rᵢ = Lᵢ/(kᵢA) is the bottleneck and gets most of the temperature drop.

Bridge note:

O-Level intuition says “heat goes through both layers.”
UY1 upgrade treats each layer as a resistance and enforces equal heat current.

Core derivation / explanation

For each layer,

Q dot = (kᵢAΔ Tᵢ)/Lᵢ = (Δ Tᵢ)/Rᵢ, Rᵢ = Lᵢ/kᵢA.

Total temperature drop:

Δ T = Δ T₁ + Δ T₂ = Q dot (R₁ + R₂).

Hence

Q dot = (Δ T)/(R₁ + R₂) = (Tₕ-Tc)/(L₁/k₁A + L₂/k₂A).

Interface temperature Tᵢ follows from either layer drop.

For cylindrical insulation (inner radius a, outer radius b, length L):

Q dot = (2π kL(Tₕ-Tc))/(ln(b/a)).

Equivalent radial thermal resistance:

R_(th,cyl) = (ln(b/a))/(2π kL).

Units check:

[Rₜₕ] = K/W, Q dot = (Δ T)/Rₜₕ ⇒ [Q dot] = W.

Physical sanity check:

Adding insulation (larger Rₜₕ) decreases Q dot.
In cylindrical form, increasing b increases ln(b/a) and reduces heat leak.

Temperature drops are not “shared equally” unless the resistances are equal: Δ Tᵢ = Q dot Rᵢ. - The simple series result assumes the same area for all layers. If area changes (e.g. radial conduction), use the correct geometry-specific resistance. - For cylinders, the resistance is logarithmic in radius ratio ln(b/a), not proportional to thickness b-a. - Real walls can have contact resistances at interfaces; those add in series too.

Worked example

Heat flow through a two-layer wall

Problem

A wall has area A = 1.0 m². Layer 1 has thickness 0.10 m and conductivity 0.80 W m⁻¹K⁻¹; layer 2 has thickness 0.05 m and conductivity 0.04 W m⁻¹K⁻¹. The outer faces are held at 30°C and 0°C. Assuming steady one-dimensional conduction, constant conductivities and negligible contact resistance, find the heat-flow rate and identify which layer has the larger temperature drop.

Show full solution

Calculate each layer's resistance
Method
Replace each slab with its thermal resistance.
Reason
The layers carry the same heat-flow rate, so their resistances add in series.
Working
Answer
R₁ = 0.10/0.80(1.0) = 0.125 K W⁻¹, R₂ = 0.05/0.04(1.0) = 1.25 K W⁻¹.
Apply the total temperature drop
Method
Divide the face-to-face temperature difference by the total resistance.
Reason
A Celsius temperature difference has the same numerical value in kelvin, and the full 30 K drop spans both layers.
Working
Answer
Q dot = 30/(0.125 + 1.25) = 21.8 W.
Interpret the bottleneck
Working
Answer
The heat-flow rate is 21.8 W. Since Δ Tᵢ = Q dot Rᵢ, layer 2 has ten times the temperature drop of layer 1: about 27.3 K compared with 2.73 K. The thin insulating layer dominates the resistance.

Practice set

Check your understanding 1

Two equal-thickness slabs with constant conductivities form a wall. With the same face temperatures, area and ideal contact, does reversing their order reduce steady one-dimensional heat loss?

Show hint

Write the total resistance before and after reversing the slabs.

Show answer

No.

Rₜₒₜₐₗ = L/(k₁A) + L/(k₂A) is unchanged by swapping the terms, so

Q dot = Δ T/Rₜₒₜₐₗ is unchanged. The interface temperature can change.

Check your understanding 2

A wall has total resistance 1.375 K W⁻¹ and a fixed 30 K face-to-face temperature difference. A third layer adds 0.50 K W⁻¹ in series. Find the new heat-flow rate.

Show hint

Add the new resistance to the existing total before dividing the temperature difference.

Show answer

Q dot = 30/(1.375 + 0.50) = 16.0 W. This is smaller than the original

21.8 W because the same temperature difference now drives heat through a greater resistance.

Check your understanding 3

Why must the heat-flow rate entering an interface equal the rate leaving it at steady state?

Show hint

Consider what unequal rates would do to the energy stored near the interface.

Show answer

Unequal rates would accumulate or remove internal energy near the interface, changing its temperature. Steady state requires no such accumulation, so the rates must match.

Summary + next steps

Compound conduction is best solved with thermal-resistance networks.
Same heat rate through all series layers is the central steady-state condition.
Cylindrical geometry introduces logarithmic resistance.

Next steps:

Compare with boundary-layer transport in UY1: convection.
Then use combined convection-conduction resistances in UY1: convective heat transfer.
Revisit the single-layer building block in UY1: conduction and compare radiative exchange in UY1: radiation.

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