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How much PIR replaces EPS? Comparing insulation at equal thermal resistance

PIR boards and insulation

At a glance

PIR insulation is not invariably half as thick as EPS. The equivalent thickness depends on the thermal conductivity of both chosen products. This guide compares insulation layers at the same thermal resistance, using worked examples for 200 mm EPS and different PIR facings. It also explains why a calculated thickness must be checked against available board sizes, intended use, fixings and the complete wall build-up. Equal resistance does not mean equal moisture, fire or load performance. Use the calculation to compare viable options; it cannot justify substituting an arbitrary board or predict the energy bills of an entire building.

Two thin aluminium-faced PIR samples on a table

Compare the same quantities first

Express insulation thickness in metres. For a homogeneous layer, thermal resistance is calculated approximately as R = d / lambda. To retain the same resistance, PIR thickness equals EPS thickness × PIR lambda / EPS lambda. This compares the insulation layers alone; it is not yet the U-value of the complete wall.

Take both conductivity values from the selected products’ documentation. Do not compare one family’s best figure with an arbitrary, less favourable value for another. The comparison should reflect materials that can actually be used in the proposed construction.

A worked example for 20 cm EPS

Assume EPS with lambda 0.038 W/(m·K). At 200 mm, its resistance is approximately 5.26 m²·K/W. PIR with lambda 0.022 would need about 116 mm to match it; at lambda 0.019, the calculated thickness is 100 mm. These are mathematical results before selecting an available board thickness.

These conductivities are assumptions for this comparison. They do not mean that any PIR board can be bonded beneath any render. The calculation concerns heat conduction through a homogeneous layer. Use the actual product data when ordering; choosing its facing, fixing method and finishing system remains a separate task.

Better-performing EPS narrows the thickness difference

With EPS at lambda 0.031, the same 200 mm layer has a calculated resistance of approximately 6.45 m²·K/W. PIR at lambda 0.022 would need around 142 mm; at 0.019, approximately 123 mm. The EPS thickness has not changed, but the equivalent PIR thickness has changed substantially.

EPS at 0.031 W/(m·K) is a second calculation scenario here. That value alone does not establish suitability for a facade, roof or floor. Check intended use and mechanical performance independently. Good insulation does not make one product a substitute for every EPS type; permissible deformation also matters under load.

Use the correct conductivity for boards beneath render

For an illustrative glass-fleece-faced PIR board, assume lambda 0.025 W/(m·K). Matching 200 mm EPS at 0.038 requires approximately 132 mm PIR by the simple formula. That is a calculation, not a guarantee that this thickness is available. Select a suitable version and then check the entire insulation system.

Using 0.022 instead of the actual 0.025 would underestimate the required thickness. PIR identifies a core material, not one universal specification. Facings, thickness and the way performance is declared matter. Where a table gives R for the chosen size, cross-check the calculation against that value.

Equal R does not mean identical behaviour

Two layers can provide similar thermal resistance but differ in vapour transmission and reaction to fire. They may require different adhesives, fixings and protection. Floor applications add load requirements; roofs also require compatibility with waterproofing and the specified fixing arrangement.

CS(10) describes compressive stress at a specified deformation during testing. It is not blanket permission to apply an arbitrary load for many years. Substituting a material solely by converting centimetres can therefore be wrong. Return to the requirements of the actual building element.

Check what changes for the complete wall

If the masonry and other layers stay the same, similar insulation resistance produces a similar U-value in a simple section. It does not guarantee the same result for the entire building. Different fixings or window-reveal details can change heat loss at junctions.

Do not add surface resistances again when using a U-value already given for the complete assembly. Put every layer, thickness, parameter and assumption in one table, with the two options side by side. This makes mistakes easier to find and keeps the comparison fair. Do not compare insulation alone with a whole wall.

Compare the cost of achieving the same result

The price per square metre of boards with different thicknesses does not reveal which solution costs less. Calculate the cost of reaching the same thermal target, including fixings, finishing, transport and changes to flashings. Where space is limited, the space retained can also be relevant.

Thinner insulation can be justified at reveals, terraces or floors with limited build-up height. Where space is plentiful, the cost comparison may favour another option. The chosen solution must be buildable in accordance with its documentation. A useful conversion supports the discussion but cannot replace design work.

Frequently asked questions

Does 10 cm PIR always replace 20 cm EPS?

No. In this example, that ratio requires PIR at lambda 0.019 and EPS at 0.038. Other values produce a different ratio.

Does the calculated thickness need rounding?

Yes. Select an available thickness, then recalculate the performance of the complete assembly.

7 branches. Materials closer to your project.

  • SEROCK — Dzierżenin 34 E [mapa]
  • BŁONIE — ul. Sochaczewska 138A [mapa]
  • RADOM — ul. Grobickiego 8 [mapa]
  • ŁÓDŹ — Natolin 8 [mapa]
  • OLSZTYN — Barczewo, ul. Prosta 9 [mapa]
  • WROCŁAW — ul. Jerzmanowska 10C [mapa]
  • GORZÓW WLKP. — ul. Zielona 33/34 [mapa]
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