Worked U-value example: masonry and insulation
Lambda describes a material, R describes the resistance of a layer and U describes heat transmittance through an assembly. For homogeneous layers, record each thickness and conductivity, calculate the layer resistances and add the appropriate internal and external surface resistances.
Collect thickness, conductivity and boundary conditions
Record the layer order, product name and lambda source. Convert millimetres to metres before dividing by conductivity in W/(m·K). Use data applicable to the actual product and conditions. Surface resistance depends on heat-flow direction and boundary conditions; the example's values are stated assumptions, not defaults for every wall, roof or floor.
Calculate layers and the complete assembly
Ri = di / λi; Rtotal = Rsi + ΣRi + Rse; U = 1 / Rtotal
Layer and surface resistances are in m²·K/W; U is in W/(m²·K). Add the resistances, then take the reciprocal. US R-values and U-factors need conversion factors and cannot be mixed directly with SI values. The reciprocal of one insulation layer's R is not the U-value of an entire wall.
Example: masonry and insulation
Take a hypothetical 300 mm masonry layer at lambda 0.77 and 150 mm insulation at lambda 0.035. Their resistances are 0.3896 and 4.2857 m²·K/W. With explicitly assumed Rsi 0.13 and Rse 0.04, the total is 4.8453 and U is about 0.2064 W/(m²·K). These lambdas illustrate arithmetic; use declarations for your own products.
Enter all layers and compare consistent variants
Use the linked layered thermal calculator to enter all thicknesses, conductivity values and selected surface resistances. Keep boundary conditions consistent across variants and record the full input list with the result. Ground-contact floors, repeating and junction bridges, cavities and moisture effects need the appropriate additional method. A result near a project target deserves closer review.
Compare insulation thickness without changing the wall
Keep the masonry at 300 mm and lambda 0.77, insulation lambda 0.035, and surfaces 0.13 + 0.04. At 100 mm insulation, total R is about 3.4168 and U about 0.2927 W/(m²·K); at 200 mm, total R is about 6.2739 and U about 0.1594. For a measured 100 m² area, nominal insulation volumes are 10, 15 and 20 m³ at 100, 150 and 200 mm respectively. Packaging, openings and purchase allowance are separate inputs. These hypothetical cases show the effect of one changed input, not a required or available product thickness.
Worked examples
| Inputs and method | Result |
|---|---|
| 0.300 / 0.77 | R = 0.3896 m²·K/W |
| 0.150 / 0.035 | R = 4.2857 m²·K/W |
| 0.13 + 0.3896 + 4.2857 + 0.04 | Rtotal = 4.8453 m²·K/W |
| 1 / 4.8453 | U ≈ 0.2064 W/(m²·K) |
Result scope
This explains the homogeneous-layer calculation. It does not prove regulatory compliance, include all thermal bridges or replace a project-specific assessment. Verify product data and the applicable calculation method before specifying construction.