How this calculation works
Layers in series carry the same steady heat flow while each contributes its own temperature drop. A thick conductive layer can add less resistance than a thin insulating layer, so thickness alone does not describe a wall’s thermal behavior.
Rtotal = Rinside + Σ(thickness/conductivity) + Routside; U = 1/Rtotal; heat = U × area × temperature differenceA worked example
The example’s 100 mm layer at conductivity 0.5 contributes 0.2 m² K/W. Its 50 mm insulation at 0.04 contributes 1.25. Adding surface resistances 0.13 and 0.04 gives 1.62 m² K/W. U is therefore about 0.617284 W/m² K. Ten square metres under a twenty-degree difference transfer about 123.457 W, and the interior surface is about 18.395 °C.
Good to know
Steady series conduction only; no thermal bridges, transient storage, moisture or regulatory U-value certification. Air cavities and reflective systems cannot generally be represented by arbitrary bulk conductivity alone. Use appropriate assembly data and engineering judgment for real construction.
Reference: MIT — thermal resistance circuits (opens in a new tab) ↗
What the model answers
Layers in series carry the same steady heat flow while each contributes its own temperature drop. A thick conductive layer can add less resistance than a thin insulating layer, so thickness alone does not describe a wall’s thermal behavior.
This page keeps entered surface resistances separate from material layers. The resulting U value applies to this ideal one-dimensional assembly, rather than a whole building with junctions, windows and heat bridges.
The Surface condensation and dew point calculator compares an entered surface temperature with the air’s vapor state. This wall’s interior-surface estimate is a thermal input to that separate check, not a condensation prediction by itself.
Measure and enter consistently
List layers from indoors toward outdoors. Thickness is millimetres and conductivity is W/(m K); the calculation converts thickness to metres before dividing. Surface resistances use m² K/W directly and are not additional material thicknesses.
Use an area and boundary temperatures that describe the same section. Conductivity and film resistances are fixed inputs. The page does not choose values from a material name or infer wind-dependent exterior conditions.
Interpret the result
Temperature drops follow each resistance’s share of the total. The plotted horizontal positions are the ordered boundaries, not scaled thickness distances. A large drop on a narrow plotted interval therefore does not imply a steep measured physical gradient.
The heat-transfer metric retains its sign when outdoors is warmer. Every boundary should remain between the two ambient temperatures in this series model. A surface-temperature result can inform a separate dew-point comparison but is not a moisture-transport analysis.
Work through the numbers
The example’s 100 mm layer at conductivity 0.5 contributes 0.2 m² K/W. Its 50 mm insulation at 0.04 contributes 1.25. Adding surface resistances 0.13 and 0.04 gives 1.62 m² K/W.
U is therefore about 0.617284 W/m² K. Ten square metres under a twenty-degree difference transfer about 123.457 W, and the interior surface is about 18.395 °C.
Where this model stops
Steady series conduction only; no thermal bridges, transient storage, moisture or regulatory U-value certification.
Air cavities and reflective systems cannot generally be represented by arbitrary bulk conductivity alone. Use appropriate assembly data and engineering judgment for real construction.
A few questions, answered
Why is the plot not drawn to layer thickness?
It shows the sequence of boundary temperatures. Its x axis is an index so thin films and thick materials can be inspected together.
Can I use the result for an entire house?
Only for the entered ideal wall area. Other envelope sections, bridges and air exchange need their own treatment.