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Thermal resistance expressions; three-layer composite wall

Short answer

Thermal resistance is opposition to heat flow, R_th = ΔT/Q. For plane wall: R = L/(kA); hollow cylinder: R = ln(r₂/r₁)/(2πkL); convection: R = 1/(hA). For three layers with surface temperatures T₁ and T₄, Q = (T₁−T₄)/(L₁/(k₁A)+L₂/(k₂A)+L₃/(k₃A)) by adding series resistances.

Define thermal resistance. Write down expressions of thermal resistance for plane wall, hollow cylinder and convection heat transfer. Derive an expression of heat transfer for a composite plane wall consisting of three layers using thermal resistance, inside and outside wall temperature.

Thermal resistance is the opposition a layer offers to heat flow. By analogy with Ohm's law (I = V/R), heat flow Q (like current) is driven by a temperature difference ΔT (like voltage) against a thermal resistance R_th:

Q = ΔT/R_th, so R_th = ΔT/Q (unit K/W or °C/W)

Expressions of thermal resistance

  • Plane wall (thickness L, area A, conductivity k): R = L/(kA)
  • Hollow cylinder (radii r₁ < r₂, length L): R = ln(r₂/r₁)/(2πkL)
  • Convection (film coefficient h, area A): R = 1/(hA)

Heat transfer through a three-layer composite plane wall

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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