- 2073 Bhadra · 6 marks
Heat flow through composite plane wall, three layers
Short answer
For three layers in series, each resistance R_i = L_i/(k_i A). Total resistance is sum, so heat flow Q = (T1 − T4) / [L1/(k1 A) + L2/(k2 A) + L3/(k3 A)].
Derive a heat flow equation through a composite plane wall consisting of three layers of different materials.
The heat flow through a composite plane wall of three layers is derived below using Fourier's law for each layer.
Setup
Consider a plane wall of area A made of three layers in perfect contact: thicknesses L₁, L₂, L₃ and conductivities k₁, k₂, k₃. The outer surfaces are at T₁ (hot side) and T₄ (cold side); the interface temperatures are T₂ and T₃. Heat flow is steady and one-dimensional, so the same Q passes through every layer.
T1 | k1 | T2 | k2 | T3 | k3 | T4
-->| L1 | | L2 | | L3 |--> Q
T1 --[R1]-- T2 --[R2]-- T3 --[R3]-- T4
Derivation
Fourier's law for each layer:
Q = k₁A (T₁ − T₂)/L₁ , Q = k₂A (T₂ − T₃)/L₂ , Q = k₃A (T₃ − T₄)/L₃
Write each as a temperature drop:
T₁ − T₂ = Q L₁/(k₁A) T₂ − T₃ = Q L₂/(k₂A) T₃ − T₄ = Q L₃/(k₃A)
Add the three equations; the interface temperatures cancel:
T₁ − T₄ = Q [L₁/(k₁A) + L₂/(k₂A) + L₃/(k₃A)]
Q = (T₁ − T₄)/(L₁/(k₁A) + L₂/(k₂A) + L₃/(k₃A)) = (T₁ − T₄)/(R₁ + R₂ + R₃)
where R = L/(kA) for each layer. The resistances are in series, so they add, as in an electric circuit.
The interface temperatures follow from T₂ = T₁ − Q R₁ and T₃ = T₂ − Q R₂.
For equal layer areas the wall can also be written with an overall coefficient: U = 1/(L₁/k₁ + L₂/k₂ + L₃/k₃) and Q = U A (T₁ − T₄).
✓ Answer checked by Claude Opus 5.5 (high effort)
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