Modeling Large Air Cavities

[jackspearson]

I recall from training that you cannot simply stack or elongate air layers to achieve a large cavity within a system and I am wondering if anyone has a solution. I am working on a roof system that includes a 16" air cavity between the bottom of the roof sheathing (plywood) and the batt insulation laid in the TGI's. The largest air layer in the database is just shy of 6" and we were told not to adjust the width of air layers because the hygrothermal behavior change is not linear as in most other materials. Any thoughts?

[Thomas]

You can use the method described in standard EN 673 ("Glass in building - Determination of thermal transmittance (U value) - Calculation method") to compute the U-value for an air layer with the required thickness and then follow the rest of the procedure detailed in WUFI's online help (Reference | Material Data | Special Materials | Air Layers) to derive the effective material parameters for the air.

Kind regards,
Thomas

[Maria Arce]

Hi,

I am modelling a facade construction whose build-up comprises an SFS cavity (100mm deep) and a 129mm non-ventilated cavity void behind. This adds up a total non-ventilated cavity of 229mm. Wufi help guidelines indicate how to modify the thickness of the air layers (by means of changing the thermal conductivity and effective water vapor diffusion resistance factor). To do so, you have to interpolate values (Rnon-met and Rmet) from the table provided. However, the maximum values provided in the Table are for 150mm thick air layer. Please can you indicate me how can I model then a 229mm deep cavity.

Thanks

[Thomas]

Hi Maria Arce,

the formulas in standard EN 673 ("Glass in building - Determination of thermal transmittance (U value) - Calculation method") allow to compute the effective thermal conductivity and the effective mu-value for arbitrary air layer thicknesses. Future WUFI versions are expected to have these formulas directly implemented so that WUFI will be able to automatically determine the appropriate material parameters for all air layer thicknesses.

These formulas even allow to differentiate between various temperature levels. So you may choose between parameters for different mean temperatures Tm of the air layer and different temperature differences dT across the air layer. You may choose whichever situation is the most typical for your simualated case:

Air layer thickness: 0.229 m,
inclination of air layer: vertical,
long-wave emissivities of the bounding surfaces: 0.9 and 0.9:

Results:

Tm = 5 °C, dT = 5 °C: effective lambda = 1.275 W/mK, effective mu-value = 0.068
Tm = 5 °C, dT = 10 °C: effective lambda = 1.384 W/mK, effective mu-value = 0.052
Tm = 5 °C, dT = 20 °C: effective lambda = 1.525 W/mK, effective mu-value = 0.040

Tm = 10 °C, dT = 5 °C: effective lambda = 1.320 W/mK, effective mu-value = 0.070
Tm = 10 °C, dT = 10 °C: effective lambda = 1.427 W/mK, effective mu-value = 0.054
Tm = 10 °C, dT = 20 °C: effective lambda = 1.566 W/mK, effective mu-value = 0.041

Kind regards,
Thomas