Layer resistance
R = L / (kA)
L is thickness, k is conductivity and A is heat-transfer area.
Existing engineering calculator
Calculate steady-state conduction through a plate or pipe wall with an insulation layer.

Select the material layers, then enter thickness, heat-transfer area and temperature difference.
The typical thermal conductivity is filled from the selected material and can be manually adjusted.
The insulation conductivity can also be manually adjusted when verified project data is available.
Engineering knowledge
Conduction moves heat from a hotter side to a colder side through a solid. A higher thermal conductivity increases heat flow, while a thicker layer or lower conductivity increases its thermal resistance.
Use the result as a conduction-only check; convective films and radiation can be important in a complete heat-loss calculation.
Read the heat-transfer guide →R = L / (kA)
L is thickness, k is conductivity and A is heat-transfer area.
Rₜ = R₁ + R₂
Individual layer resistances add in series.
Q = ΔT / Rₜ
Q is reported as a positive rate for a positive temperature difference.
It provides a preliminary plane-layer approximation. Detailed insulated-pipe design should use cylindrical radial conduction and suitable inside and outside film coefficients.
Increasing insulation thickness raises its thermal resistance and therefore reduces conduction for the same temperature difference and area.