Insulation Systems

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​Insulating Material Systems

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PIR/PUR

Due to its closed cell structure, this material has an excellent thermal conductivity of λ =0.022W/mK,In this way, we can use smaller insulation thicknesses to achieve the thermal transmittance that is established by current regulations.

XPS (extruded polystyrene)

The thermal conductivity (λ) of extruded polystyrene products basically depends on the foaming gas used. The conductivity obtained varies between0.029 and 0.036 W/m K 

EPS (expanded polystyrene)

Styrofoam material is an insulator derived from petroleum and natural gas, from which the plastic polymer styrene is obtained in the form of granules.

• conductivity coefficient of0.034 to 0.045 W/(m K)

Polyurethane foam

Polyurethane foam is known to be a very good performing insulating material.

• Thermal conductivity coefficient:0.023W/(mK)

wood fiber

• Thermal conductivity coefficient:0.04-0.06W/(mK)

rock wool

Rock wool is a thermal, incombustible and rot-proof insulating material.

• Thermal conductivity coefficient:0.034 to 0.041 W/(m K)

Glass wool

It is a glass wool in panels with higher density, waterproof and hygroscopic.

• Glass wool thermal conductivity coefficient:0.032- 0.044 W/(m K)

airgel

As a thermal insulator, airgel comes in flexible blankets.

• Conductivity coefficient:0.01W/(mK)

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low thermal conductivity

It indicates the capacity of the material to transmit heat, therefore, the lower it is, the more efficient it will be in preventing heat loss.

Insulating materials High thermal resistance

 It is the ability of a material to oppose the flow of heat, the higher it is, the more effective the insulation will be.

Resistance to the passage of water vapor

It has to do with the thickness and permeability of the material,   serves to prevent condensation.

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The calculation of the R value depends on the material itself, the  thickness of the material, in meters, is divided by the λ value (the thermal conduction coefficient). 

The higher the value, the better the insulation, a layer twice as thick will have twice the thermal resistance.   The formula is R = d / λ  where:

• R = thermal resistance in m 2 K/W

•  d = material thickness in m.

•  λ = thermal conduction coefficient in W/mK

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