Thermal resistance of composite wall, electric analogy

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)

Electrical analogy

Resistance of a plane wall (conduction)

Fourier's law for a plane wall of thickness L, area A and conductivity k:

Q = −kA dT/dx = kA (T₁ − T₂)/L = (T₁ − T₂)/(L/kA)

So R_cond = L/(kA).

Resistance of a surface in convection

Newton's law: Q = hA (T_s − T_∞) = (T_s − T_∞)/(1/hA), so R_conv = 1/(hA).

Composite wall (resistances in series)

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.

If hot and cold fluids at T_∞1 and T_∞2 touch the surfaces with coefficients h₁ and h₂, add the film resistances 1/(h₁A) and 1/(h₂A) in series:

Q = (T_∞1 − T_∞2)/[1/(h₁A) + L₁/(k₁A) + L₂/(k₂A) + L₃/(k₃A) + 1/(h₂A)]