ISO 10077-2:2012 Test case D.10 - Wood frame section with glazing - QuickField simulation example

Simulation problem

Geometry

Given
Thermal conductivity the of gas filling, λ1 = 0.034 W/K-m.
Thermal conductivity of EPDM, λ2 = 0.25 W/K-m.
Thermal conductivity of glass λ3 = 1 W/K-m.
Thermal conductivity of polysulfide λ4 = 0.4 W/K-m.
Thermal conductivity of silica gel λ5 = 0.13 W/K-m.
Thermal conductivity of soft wood λ6 = 0.13 W/K-m.

Task
Calculate two-dimensional thermal conductance L_Ψ^2D in accordance with ISO 10077-2:2012*.
(Reference thermal conductance L_Ψ^2D = 0.481 W/(m*K) )

Solution
Air contact surface thermal resistance R_s is caused by convention mechanism of heat transfer (contrary to conductivity mechanism in case of two solid bodies in contact).
The convection coefficient value is reciprocal to the surface resistance value:
α₂₀ = 1/0.13 W/(K·m²),
α₀ = 1/0.04 W/(K·m²).

Thermal conductivities of air gaps are adjusted to include the convection and radiation heat transfers in real situation, and correlated with the actual temperatures on the gap contours according to the recommendation of ISO 10077-2:2012. Equivalent air gap conductivity calculations are automated in the accompanying Excel spreadsheet.

Results
Heat flux and temperature distribution in frame section profile:

Two-dimensional thermal conductance L_Ψ^2D = Heat flux per meter depth / Temperature difference = 9.47 / 20 = 0.474 W/(m*K).
Difference between the calculated and reference values is less than 3%. This simulation accuracy complies with the requirements of ISO 10077-2:2012.

Video

• ISO 10077-2:2012 Test case D.10 - Wood frame section with glazing. Watch on YouTube
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• View simulation report in PDF

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