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Radial conduction through hollow cylinder; resistance

Short answer

From Fourier's law in radial coordinates, integrating yields steady heat conduction Q = 2πkL (T₁−T₂)/ln(r₂/r₁). The thermal resistance is R_th = ln(r₂/r₁)/(2πkL); the temperature profile is logarithmic.

Derive an expression for steady state radial heat conduction through a hollow cylinder. Also derive expression for its thermal resistance.

Consider a long hollow cylinder with inner radius r₁, outer radius r₂, length L and constant conductivity k. The inner surface is at T₁ and the outer surface at T₂ (T₁ > T₂). Heat flows radially and steadily, with no heat generation.

Heat conduction equation

Fourier's law at radius r, where the area is A = 2πrL:

Q = −k (2πrL) dT/dr

In steady state Q is the same at every radius. Separate the variables:

Q dr/r = −2πkL dT

Integrate from r₁ to r₂ and from T₁ to T₂:

Q ln(r₂/r₁) = 2πkL (T₁ − T₂)

Q = 2πkL (T₁ − T₂)/ln(r₂/r₁)

The temperature at any radius r follows from integrating from r₁ to r:

T(r) = T₁ − (T₁ − T₂) ln(r/r₁)/ln(r₂/r₁)

This is a logarithmic temperature profile, unlike the linear one in a plane wall.

Thermal resistance

From Q = (T₁ − T₂)/R_th:

R_th = ln(r₂/r₁)/(2πkL)

This can also be written in the plane-wall form R = (r₂ − r₁)/(k A_m). The form is exact when A_m is the logarithmic mean area A_lm = (A₂ − A₁)/ln(A₂/A₁); for a thin wall, the arithmetic mean area 2π r_m L, with r_m = (r₁ + r₂)/2, is a good approximation.

✓ Answer checked by Claude Opus 5.5 (high effort)

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