Conduction through a composite cylinder

A composite cylinder is a cylinder made of several concentric layers of different materials in perfect contact (for example a pipe with insulation layers). Heat flows steadily in the radial direction.

Single cylindrical layer

For a cylindrical layer of inner radius r₁, outer radius r₂, length L and conductivity k, with temperatures T₁ at r₁ and T₂ at r₂ (steady, radial, no heat generation), the area changes with radius, A = 2πrL. Fourier's law:

Q = −k (2πrL) dT/dr

Separate variables and integrate:

Q ∫{r₁}^{r₂} dr/r = −2πkL ∫{T₁}^{T₂} dT

Q ln(r₂/r₁) = 2πkL (T₁ − T₂)

Q = 2πkL (T₁ − T₂)/ln(r₂/r₁), so R_th = ln(r₂/r₁)/(2πkL)

Composite cylinder of n layers

Let the layers have radii r₁, r₂, ..., r_{n+1} and conductivities k₁, ..., k_n. The inner surface (r₁) is at T₁ and the outer surface (r_{n+1}) at T_{n+1}. The same Q passes through every layer. For layer j:

T_j − T_{j+1} = Q ln(r_{j+1}/r_j)/(2π k_j L)

Add all layers; the interface temperatures cancel:

T₁ − T_{n+1} = (Q/2πL) Σ ln(r_{j+1}/r_j)/k_j

Q = 2πL (T₁ − T_{n+1})/Σ [ln(r_{j+1}/r_j)/k_j]

For three layers: Q = 2πL (T₁ − T₄)/[ln(r₂/r₁)/k₁ + ln(r₃/r₂)/k₂ + ln(r₄/r₃)/k₃].

The layers act as thermal resistances R_j = ln(r_{j+1}/r_j)/(2π k_j L) in series, so Q = ΔT_total/ΣR_j. If convection at the inner and outer surfaces is present, add 1/(h_i 2π r₁ L) and 1/(h_o 2π r_{n+1} L) to the sum.