ISO 10077-2:2012 Test case D.9 - PVC shutter profile - QuickField simulation example

Simulation problem

Geometry

Total width b = 57 mm

Given
Thermal conductivity of PVC, λ1 = 0.17 W/K-m.
Thermal conductivity of the gas filling, λ2 = 0.034 W/K-m.

Task
Calculate the two-dimensional thermal conductance L^2D of the frame in accordance with ISO 10077-2:2012*.
(Reference thermal conductance L^2D = 0.207 W/(m*K) )

Solution
Boundary conditions on the upper and bottom frame surfaces are thermal convection. The convection coefficient value is reciprocal to the surface resistance value R_s provided in the test case description:
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 the PVC shutter profile is shown below:

Two-dimensional thermal conductance L^2D = Heat flux per meter depth / Temperature difference = 4.139 / 20 = 0.207 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.9 - PVC shutter profile. Watch on YouTube
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• View simulation report in PDF

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