How do you calculate the required PIR thickness for a specific wall?

Also check the available space. The window reveal, eaves and plot boundary may limit the solution. A geometric constraint does not remove the thermal requirements. If the variant does not fit within the intended space, change the material or detail and reassess the whole assembly.

Sum the properly determined resistances of the layers and the surface resistances. For a homogeneous layer the relationship R = d/λ is used. The data must correspond to the actual materials and calculation conditions. Do not enter the lambda of the void alone as the parameter of the entire unknown wall.

In an old building, mark uncertain data. If a void occurs, do not treat it as a solid material of arbitrary thickness. A proper way of accounting for such a layer is needed. When the survey is poor, it is worth comparing variants and checking how the uncertainty affects the selection.

Assume a target U = 0.20 W/(m²·K). In a simplified assembly the required total resistance is then 1/0.20 = 5.00 m²K/W. Assume that the existing wall together with the surface resistances gives 0.80 m²K/W. The missing resistance of the new insulation is 4.20 m²K/W.

If a lambda of 0.022 W/(m·K) is assumed for the example, the thickness will be 4.20 × 0.022 = 0.0924 m, that is 92.4 mm. This is an illustrative calculation, without full corrections and thermal bridges. It does not constitute a ready-made selection for any wall nor a confirmation of legal requirements.

The calculation result must be translated into the actual variant. In the example given, a 100 mm board with the assumed lambda can be checked. Its resistance is approximately 4.55 m²K/W. After adding the assumed 0.80, we get approximately 5.35, which gives a simplified U of approximately 0.187 W/(m²·K).

However, do not stop at this number. The selected facing must suit the application. If the correct variant for plaster has a different lambda, the calculation must be repeated. Do not round the thickness down solely to make it easier to fit the board at the window frame.

It is also worth confirming the choice on the drawing. If the assumed 100 mm does not fit within the reveal, you must not keep the calculation result and enter 60 mm there. A local change requires its own assessment of temperature and connection to the frame.

A typical mistake is confusing millimetres with metres. Another is entering the U-value of a finished panel in the insulation lambda field. It also happens that an old layer is omitted or surface resistances are added twice. It is worth checking the data before comparing results to three significant figures.

Another problem is using a parameter from a different product variant. The type of facing and thickness can affect the declared properties. Therefore, record the full description of the assumed material next to the result. The name of the PIR group alone will not allow the calculation to be reproduced later.

Calculating the flat field does not describe all connections. Thermal bridges at ceilings and reveals require proper assessment. Fasteners and gaps may require corrections. Thicker insulation does not replace careful detailing of these locations.

For internal insulation, moisture and the temperature of the old wall must also be assessed. Even a very low U-value does not confirm that the assembly dries out safely. If the substrate is damp, the cause is determined first. The thickness calculator does not detect a leak or damage to the insulation at ground level.

Add the layer cross-section, facing type and assumed thickness to the schedule. Describe the solutions at windows, ceiling and floor. Indicate which assumptions require confirmation before work. The contractor will then know what is a fixed part of the design.

After changing the material, return to the calculation. A substitute of the same thickness may have a different resistance and different installation requirements. Acceptance should cover the conformity of the entire assembly, not just the number of centimetres. Only then does the calculation result relate to the completed wall.