The walls of the house are usually built of several layers (for example, a load-bearing wall and insulation or load-bearing wall, insulation and facade walls) – each of these layers has a different resistance to heat escaping from the interior. Thermal resistance across the wall is the sum of the resistances of the individual layers. In order to check the thermal resistance of the wall, it is necessary to add to it the thermal resistance of its layers.arstw.

How to calculate the desired insulation thickness (Rd) needed to obtain a heat transfer coefficient of U = 0.2?

The thermal resistance of the wall with a heat transfer coefficient of 0.2 should be R = 5.0. From the expected wall resistance (Rn = 5.0) we deduct the resistance R of the existing wall (Rs) and the coefficient of heat inflow and outflow, which is 0.17. The result will indicate what resistance should a layer of insulation have. Using the data given below, you need to find out what insulation thickness corresponds to the result.

*Example 1 (Simplified): Load-bearing wall from hollow blocks (Porotherm 18.8 Dryfix), 18.8 cm thick. Thermal resistance of hollow blocks – 0,67. Clinker brick facade, 12 cm thick and 0.11 thermal resistance. Thermal resistance of air gap 2.5 – 5 cm = 0.18. How much insulation should be applied so that the wall resistance R is 5.0?*

* **Rd = 5,00 – 0,67 – 0,17 – 0,11 – 0,18 = 3,87*

*Thickness of insulation with a thermal conductivity coefficient of 0.040 should be 16 cm (then the thermal resistance of the insulation layer will be Rd = 4.0). The total wall thickness will be approximately 43.3 cm.*

*Thickness of thermal insulation with a thermal conductivity factor of 0.025 should be 10 cm (then insulation resistance is Rd = 4.0).*

*Example 2 (Simplified): Load-bearing wall from cellular concrete (Solbet Optimal Plus with a thickness of 24 cm). Thermal resistance of cellular concrete this thick, with density of 400 kg / m3 and a heat transfer coefficient of 0.105 is 2.29. How much thermal isolation do you need to get resistance R to be 5.0?*

* **Rd = 5,00 – 2,29 – 0,17 = 2,54*

*Thickness of thermal insulation with a thermal conductivity coefficient of 0.040 should be 11 cm (then the thermal resistance of the insulation layer will be Rd = 2.75).*

*Thickness of thermal insulation with a thermal conductivity of 0.025 should be 7 cm (then the thermal resistance of the insulation layer will be Rd = 2.8).*