Given: Thermal conductivity of AAC λ_{1} = 0.7 W/K·m.
Thermal conductivity of insulation λ_{2} = 0.04 W/K·m.
Thermal conductivity of brick λ_{3} = 1.0 W/K·m.
Thermal conductivity of concrete slab λ_{4} = 2.5 W/K·m.
Thermal conductivity of plaster λ_{5} = 1 W/K·m.

Problem: The floor slab penetrates the wall near the corner and forms the cold bridge. Determine the flux and the thermal coupling coefficients.

Solution: Air contact surface thermal resistance R_{s} is caused by the convection. The convection coefficient value is reciprocal to the surface resistance value:
α = 1 / R_{s} [W/(K·m^{2})].

Thermal coupling coefficients (reference values are in the brackets).

External

First floor

Second floor

External

(1.000)

(0.000)

(0.000)

First floor

0.375 (0.378)

0.401 (0.399)

0.223 (0.223)

Second floor

0.329 (0.331)

0.216 (0.214)

0.455 (0.455)

To comply with the ISO 10211:2007 the difference between calculated and reference heat flux should be less than 2% (0.1°K - for temperature). This simulation accuracy complies with the requirements of ISO 10211:2007.