Properties of common insulating materials

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All lambda are given in W / (mK)

Calculation of the thermal resistance of a wall or a wall

The thermal resistance is given by the following formula:

R = e / lambda

With e = wall thickness in m
and lambda = coefficient of heat loss of the materials used.

For walls made up of several layers of materials the thermal resistances add up, thus for a wall made up of 2 materials 1 and 2:

Total R = R1 + R1 = e1 / lambda1 + e2 / lambda2

We thus obtain R in m2.K / W which is not very significant, on the other hand its inverse: W / m2.K it is: it is the thermal transmission in Watts (in other words: the thermal losses) of the wall per degree of difference in T ° and m2 of the wall.

A thermal resistance at least equal to 2 is considered as standard insulation.

Example for an insulated wall with a resistance of 3:

- R = 3
- Outdoor temperature: 5 ° C
- Indoor temperature: 19 ° C

The losses by conduction in the wall will therefore be equal to (19-5) * 1 / 3 = 4,66 W per m2 of surface of this wall. If the wall is 40m2 the heating power to be supplied is therefore 4,66 * 50 = 233 Watts to maintain the interior temperature at 19 ° C, assuming that this is the only heat loss from this room.

This is a first simple approach to carry out the thermal balance of a house (other losses occur by convection, air renewal, thermal bridges, carpentry, etc.).

Learn more see the 2005 Thermal Regulation: RT2005

Lambda conduction coefficients for common materials

• Gemstones

• Concretes

• Plasters

• The terracotta

• Wood and materials of plant origin

• Industrial insulating materials for petrochemicals

• Natural soils and floor coverings

• The most common metals

• Other materials and gases

More:
- Forum insulation
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