How much PIR insulation gives the same thermal resistance as EPS?

Layers with the same thermal resistance have a similar effect on heat flow. Load capacity, water absorption, fire performance and vapour permeability can still differ. Check each property required by the design before substituting one insulation material for another, even if the R-value stays the same.

For a homogeneous layer, R = d/λ. To keep R constant, multiply the original thickness by the new material’s conductivity and divide by the original conductivity. In symbols, d(new) = d(original) × λ(new) / λ(original). The two thicknesses can use the same unit in this comparison. When calculating R itself, enter thickness in metres.

For example, 20 cm of EPS with an assumed λ of 0.036 W/(m·K) gives an R-value of about 5.56 m²K/W. Approximately 12.2 cm of insulation at λ = 0.022 or 14.4 cm at λ = 0.026 gives the same resistance. These thicknesses apply to the conductivities in this example. Other PIR and EPS grades require their own calculation.

A calculated thickness such as 12.2 cm may be unavailable as a standard board. Select an available size and recalculate its resistance. Rounding down lowers R and falls short of the original target. Also check whether the product’s declared conductivity changes with thickness.

At λ = 0.031, a 20 cm EPS layer gives about 6.45 m²K/W. Matching it at λ = 0.022 now takes roughly 14.2 cm. Although the EPS thickness in the two examples is identical, its thermal resistance differs. A claim that PIR needs just half the thickness would be misleading here.

The figures compare individual layers. They exclude whole-wall corrections and the rounding used in a specific product’s declared thermal resistance.

PIR facings matter when selecting conductivity data. A particular foil-faced board may declare λ = 0.022 W/(m·K). Boards designed to receive render may have values of 0.025–0.027, depending on grade and thickness. Use the figure for the board actually specified.

If the calculated thickness is unavailable, use the data for an available board and check the result again. Simply rounding down to fit beneath a sill could reduce performance below the target. The junction detail or insulation system may need changing. Confirm the required performance before resolving the space constraint.

The U-value accounts for the resistance of every layer and the relevant surface resistances. Fixings and gaps in insulation may require corrections. Window reveals, lintels and floor junctions also need assessment. Revisit them if a substitution changes the construction at these points.

For internal insulation, also consider vapour resistance. An aluminium facing can greatly restrict inward drying. Two layers with identical thermal resistance can behave differently when the wall gets wet. A thickness comparison alone cannot resolve this question.

Thinner insulation may allow the existing flashings to stay or leave more usable room inside. Elsewhere, there may be little benefit if the eaves need rebuilding anyway. Compare the full cost of systems that provide a similar thermal result, allowing for the thickness each system actually needs.

Add insulation, fixing, finishing and alterations around openings. The total shows whether a slimmer system is economical. To compare environmental impacts, use comparable product declarations and a stated service life. Conductivity alone cannot establish an environmental advantage.