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Layered wall heat flow
Combine material and surface-film resistances and inspect interface temperatures.
A clear answer. Here’s how.
Adds 1/hinside, each layer thickness/conductivity and 1/houtside as area-normalised resistances. Heat flux=(Tinside−Toutside)/Rtotal; interface temperatures step down by flux×resistance in supplied layer order. Supports 1–100 uniform layers. Assumes one-dimensional steady conduction and constant properties; thermal bridges, moisture, radiation exchange details, cavities and condensation are not modelled.
See a worked example
Worked example
A fixed demonstration, separate from the values you enter above.
- CSV: Layer,Thickness_mm,Conductivity
Layer,Thickness_mm,Conductivity Plasterboard,12.5,0.25 Insulation,100,0.04 Brick,100,0.77
- Inside air temperature (°C)
- 20
- Outside air temperature (°C)
- 0
- Wall area (m²)
- 20
- Inside film coefficient (W/m²·K)
- 8
- Outside film coefficient (W/m²·K)
- 25
Example result
| Boundary / layer | Resistance (m²·K/W) | Temperature after boundary (°C) |
|---|---|---|
| Inside air | 20 | |
| Inside film | 0.125 | 19.1212252631 |
| Plasterboard | 0.05 | 18.7697153683 |
| Insulation | 2.5 | 1.19422062952 |
| Brick | 0.12987012987 | 0.28120791582 |
Total area-normalised resistance 2.8448701298701295 m²·K/W; U-value 0.35150989477528477 W/(m²·K); heat flux 7.030197895505696 W/m². Positive flow is inside to outside.