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ISO 10077-2:2012 Test case D.4 - Wood frame section and insulation panel

QuickField simulation of the test case "D.4. Wood frame section and insulation panel" presented in ISO 10077-2:2012*

Problem Type:
Plane-parallel problem of heat transfer.

Geometry:

EN ISO 10211:2007 test case 4
total 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 soft wood λ3 = 0.13 W/K·m.

Problem:
Calculate two-dimensional thermal conductance L2D in accordance with ISO 10077-2:2012*.
(Reference thermal conductance L2D = 0.346 W/(m*K) )

Solution:
Air contact surface thermal resistance Rs 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:

EN ISO 10077:2012 D4 heat flux

Two-dimensional thermal conductance L2D = Heat flux / Temperature difference = 6.97 / 20 = 0.349 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.

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* ISO 10077-2:2012 Thermal performance of windows, doors and shutters -- Calculation of thermal transmittance -- Part 2: Numerical method for frames.