ISO 10077-2:2012 Test case D.2 - Aluminium clad wood frame section and insulation panel - QuickField simulation example

Simulation problem

Geometry

Frame width b = 110 mm

Given
Thermal conductivity of EPDM, λ1 = 0.25 W/K-m.
Thermal conductivity of insulation panel, λ2 = 0.035 W/K-m.
Thermal conductivity of aluminium, λ3 = 160 W/K-m.
Thermal conductivity of soft wood, λ4 = 0.13 W/K-m.

Task
Calculate two-dimensional thermal conductance L^2D in accordance with ISO 10077-2:2012*.
(Reference thermal conductance L^2D = 0.263 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 distribution in frame section profile:

Two-dimensional thermal conductance L^2D = Heat flux per meter depth / Temperature difference = 5.33 / 20 = 0.266 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.2 - Aluminium clad wood frame section and insulation panel. Watch on YouTube
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• View simulation report in PDF

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