Skip to main content

Conduction through composite cylinder, three layers

Short answer

For three concentric layers with radii r₁–r₄ and conductivities k₁–k₃, the steady radial heat flow is Q = 2πL (T₁ − T₄) / [ln(r₂/r₁)/k₁ + ln(r₃/r₂)/k₂ + ln(r₄/r₃)/k₃]. Each layer behaves as a thermal resistance R = ln(r_out/r_in)/(2πkL); these resistances add in series, yielding Q = ΔT / ΣR.

Derive an expression for conduction heat transfer through a composite cylinders consisting of three layers of different materials.

Steady, radial heat conduction through a long composite cylinder of three layers (for example a pipe with two insulation layers) is derived with Fourier's law.

Setup

A long cylinder has three concentric layers in perfect contact: radii r₁ < r₂ < r₃ < r₄ and conductivities k₁, k₂, k₃. The inner surface (r₁) is at T₁ and the outer surface (r₄) at T₄; interface temperatures are T₂ and T₃. Heat flows radially, steadily, through length L, with the same Q through each layer.

 Cross-section (layers outward from the axis)
   axis | layer 1 | layer 2 | layer 3 |
        r1        r2        r3        r4
   T1 --[R1]-- T2 --[R2]-- T3 --[R3]-- T4

Derivation

For a single cylindrical layer (Fourier's law with A = 2πrL, Q = −k·2πrL·dT/dr, integrated between the radii):

Q = 2πk L (T_in − T_out)/ln(r_out/r_in)

Apply to each layer:

T₁ − T₂ = Q ln(r₂/r₁)/(2πk₁L) T₂ − T₃ = Q ln(r₃/r₂)/(2πk₂L) T₃ − T₄ = Q ln(r₄/r₃)/(2πk₃L)

Adding, the interface temperatures cancel:

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

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

In resistance form, Q = (T₁ − T₄)/(R₁ + R₂ + R₃), with R_n = ln(r_{n+1}/r_n)/(2π k_n L) in series.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Fundamental of Thermodynamics & Heat Transfer chapter 7 questions →