Wall U-value and insulation thickness
Estimate the additional homogeneous insulation needed for a chosen U-value. The method includes the existing layer resistance and surface resistances, but excludes repeating bridges, junction losses and moisture effects. Concrete, brick, aerated concrete, timber and steel walls require different checks.
Reviewed:
Calculator
Layer method and required insulation
Rtotal = Rsi + Σ(d / λ) + Rse; U = 1 / Rtotal; dins = max(0, λins × (1 / Utarget − Rexisting))
Rexisting must include all retained homogeneous layers and the chosen Rsi and Rse. Do not add them twice. Convert every thickness to metres before dividing by conductivity. The calculator returns zero if the entered existing resistance already meets the chosen target in this simplified model. Round upward to an available product thickness, then recalculate the full assembly.
Example target U = 0.25
With illustrative Rexisting = 0.50 m²·K/W and λins = 0.040 W/(m·K), U = 0.25 requires 140 mm of additional insulation. The required total resistance is 4.00. These inputs demonstrate the equation; the target is not asserted to be the legal limit for your location.
Example target U = 0.30
For the same Rexisting = 0.50 and λins = 0.040, U = 0.30 gives 113.3 mm. Changing only the target changes the required thickness. Compare available product sizes and the final recalculated U, rather than treating 113.3 mm as a specification.
Concrete, brick and aerated concrete walls
For an illustrative brick wall, 240 mm / 0.70 gives R = 0.343. Add illustrative surface resistance 0.17 and 140 mm / 0.040 insulation: Rtotal = 4.013 and U ≈ 0.249 W/(m²·K). Replace all example values with actual design values. Aerated concrete density and moisture can alter λ substantially; ordinary concrete must not inherit an aerated-concrete value. There is no best insulation for every masonry wall: check fire, moisture, fixing, compression and the complete facade system.
For another illustrative homogeneous example, 200 mm concrete at λ = 1.60 plus surface R = 0.17 gives existing R = 0.295; reaching U = 0.25 with λins = 0.040 needs 148.2 mm. A 240 mm aerated-concrete layer at illustrative λ = 0.12 gives existing R = 2.17 and needs 73.2 mm. Product-specific design values can change both results.
Timber and steel frames: thermal bridges
A cavity insulation value cannot represent the complete framed wall. Timber studs create parallel heat-flow paths; steel members can bridge the insulation strongly. Stud fraction, continuous insulation, fasteners and junction details need a suitable assembly method. Do not apply the homogeneous calculator as an effective U-value calculation for either frame. The linked NHBC source explains why repeating bridges belong in the element U-value.
Example target U = 0.20
For illustrative existing R = 0.50 m²·K/W including surfaces and λins = 0.040 W/(m·K), target U = 0.20 requires total R = 5.00 and an additional 180 mm. Recalculate after selecting an available product thickness. This target is a scenario, not an assertion of a legal requirement.
Wall U-value with EPS or mineral wool
U = 1 / (Rexisting + dins / λins)
Use the declared conductivity of the exact EPS or mineral-wool product. For illustrative existing R = 0.50 including surfaces and 150 mm insulation, λ = 0.038 gives U ≈ 0.225, while λ = 0.040 gives U ≈ 0.235 W/(m²·K). These values describe homogeneous examples. Fixings, bridges, cavities and installation details require the appropriate assembly assessment.
Masonry layers: editable U and R examples
R = Rsi + dwall / λwall + dins / λins + Rse; U = 1 / R
Select a wall scenario below or change every input. The examples use illustrative λ values: brick 0.70, concrete 1.60, aerated concrete 0.12, EPS 0.038 and mineral wool 0.040 W/(m·K). They are new arithmetic examples, not recovered product declarations. Replace these with design data. No cavity, plaster, repeating bridge or fastening correction is included; this model cannot calculate an effective timber- or steel-frame U-value.
Calculator — Masonry layers: editable U and R examples
Example target U = 0.35
With existing R = 0.50 including surfaces and insulation λ = 0.040, the illustrative requirement is 94.3 mm. This is an arithmetic target, not a legal limit.
Example target U = 0.40
With existing R = 0.50 including surfaces and insulation λ = 0.040, the illustrative requirement is 80 mm. Change the target in the calculator; check the final assembly separately.
Illustrative wall scenarios
Worked examples — illustrative inputs
| U [W/(m²·K)] | Rtotal [m²·K/W] | d: λ=0.040, Rexisting=0.50 [mm] |
|---|---|---|
| 0.25 | 4.000 | 140.0 |
| 0.30 | 3.333 | 113.3 |