Cylindrical thermal conduction resistance

Heat crosses cylindrical surfaces whose area 2πrL increases with radius. The temperature gradient therefore changes with radius for constant heat flow. Integrating Fourier’s law from r1 to r2 gives the integral dr/r and hence ln(r2/r1). A plane wall has constant area and a linear thickness term. The resistance model treats steady radial heat flow through a layer. Length, radius ratio and constant conductivity are part of its assumptions. Application: Calculate heat transfer through pipes, insulation and cylindrical walls. The numerical example For r1=0.02 m, r2=0.04 m, L=1 m and k=0.5 W/(m·K), Rth=ln(2)/π≈0.2206 K/W. With ΔT=40 K, Q≈181.3 W. Since kL has units W/K, resistance has units K/W anchors the idea in a concrete measurement case.

At steady state, the same heat rate passes through each concentric layer. Their thermal resistances are therefore in series and add: Rtotal=ΣRi. Contact resistance between layers and inner or outer convection are additional elements. Use each layer’s own radii and conductivity. The resistance model treats steady radial heat flow through a layer. Length, radius ratio and constant conductivity are part of its assumptions. The corresponding units are Rth in K/W; radii and L in m; k in W/(m·K). For r1=0.02 m, r2=0.04 m, L=1 m and k=0.5 W/(m·K), Rth=ln(2)/π≈0.2206 K/W. With ΔT=40 K, Q≈181.3 W. Since kL has units W/K, resistance has units K/W. This checks that the reference quantities are compatible.

A logarithm is defined for a dimensionless number. The ratio r2/r1 is dimensionless, whereas ln(r2) alone would depend on the length unit. Using both radii in metres or both in millimetres gives the same ratio. A physical layer also requires r2>r1. The resistance model treats steady radial heat flow through a layer. Length, radius ratio and constant conductivity are part of its assumptions. Scope: Integrating Fourier’s law over cylindrical area A(r)=2πrL gives Rth=ΔT/Q=ln(r2/r1)/(2πkL). It assumes steady one-dimensional radial flow, constant conductivity, no heat generation and negligible end effects. Concentric-layer resistances add; contact and convection resistances are separate. The logarithm takes a dimensionless radius ratio. A practical application is Calculate heat transfer through pipes, insulation and cylindrical walls.

Radial heat crosses area 2πrL, which grows with radius. Integrating Fourier’s law through the wall therefore produces ∫dr/r=ln(r₂/r₁). The ratio is dimensionless and the outer radius must exceed the inner one. A plane-wall expression Δr/(kA) is only an approximation when wall thickness is small relative to radius. The resistance model treats steady radial heat flow through a layer. Length, radius ratio and constant conductivity are part of its assumptions. Check the calculation with this case: For r1=0.02 m, r2=0.04 m, L=1 m and k=0.5 W/(m·K), Rth=ln(2)/π≈0.2206 K/W. With ΔT=40 K, Q≈181.3 W. Since kL has units W/K, resistance has units K/W. Keep the model limits and units in view.

For steady radial transfer through concentric layers, the same heat rate passes sequentially through every layer. Compute ln(rout/rin)/(2πkL) for each and add the resistances. Interface contact and inner or outer convection add separate resistances. Check that each layer shares the same length and that successive radii connect without gaps. The resistance model treats steady radial heat flow through a layer. Length, radius ratio and constant conductivity are part of its assumptions. The relation describes Steady radial conduction through a cylindrical layer has resistance proportional to the logarithm of its radius ratio. This helps distinguish its target quantity from neighbouring quantities.