Thermal conduction resistance of a plane wall

A larger cross-sectional area provides more parallel paths for heat flow. At fixed thickness and temperature difference, resistance therefore falls as 1/A. Doubling area halves ideal conduction resistance while edge losses and geometry effects remain negligible. The lumped resistance assumes steady one-dimensional conduction and constant conductivity; resistances of layers in series add. Application: Compare layered insulation and walls under steady conduction. The numerical example For L=0.10 m, k=0.50 W/(m·K) and A=2 m², Rth=0.10 K/W; with a 5 K difference the ideal heat rate is 50 W anchors the idea in a concrete measurement case.

For layers in series, the steady heat rate is the same through each layer, so conduction resistances add: Rtotal=ΣLi/(kiAi). Interfaces may add contact resistance. Different areas or lateral heat flow can invalidate a simple one-dimensional series model. The lumped resistance assumes steady one-dimensional conduction and constant conductivity; resistances of layers in series add. The corresponding units are Rth in K/W; L in m; k in W/(m·K); A in m². For L=0.10 m, k=0.50 W/(m·K) and A=2 m², Rth=0.10 K/W; with a 5 K difference the ideal heat rate is 50 W. This checks that the reference quantities are compatible.

In steady operation, heat rate equals (Tinside−Toutside)/Rth. Greater resistance reduces heat flow at fixed temperature difference. K/W units make temperature difference divided by resistance equal watts. Transient heating also requires heat capacity and an unsteady model. The lumped resistance assumes steady one-dimensional conduction and constant conductivity; resistances of layers in series add. Scope: This resistance model summarizes Fourier conduction for constant conductivity, uniform area and steady one-dimensional flow: heat rate equals temperature difference divided by resistance. Contact resistance, edge losses, temperature-dependent properties and multidimensional flow are excluded. Resistances of plane layers in series add. A practical application is Compare layered insulation and walls under steady conduction.

Write the quantity and unit from the variable list next to every value: Rth in K/W; L in m; k in W/(m·K); A in m². Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example For L=0.10 m, k=0.50 W/(m·K) and A=2 m², Rth=0.10 K/W; with a 5 K difference the ideal heat rate is 50 W. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. The lumped resistance assumes steady one-dimensional conduction and constant conductivity; resistances of layers in series add.

The background states this starting condition: This resistance model summarizes Fourier conduction for constant conductivity, uniform area and steady one-dimensional flow: heat rate equals. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. The lumped resistance assumes steady one-dimensional conduction and constant conductivity; resistances of layers in series add.