Given: Thermal conductivity of slab λ1 = 1.15 W/K·m.
Thermal conductivity of insulation λ2 = 0.12 W/K·m.
Thermal conductivity of air λ3 = 0.029 W/K·m.
Thermal conductivity of metal λ4 = 230 W/K·m.
Problem: The concrete slab is insulated from the metal beam by the insulation layer. Determine the flux and the temperature distribution in the reference points (every corner).
Solution: Only half of the model is presented due to symmetry (as recommended in the ISO standard). Air contact surface thermal resistance Rs is caused by the convection.
The convection coefficient value is reciprocal to the surface resistance value:
αsurf.ext = 1/0.06 W/(K·m2),
αsurf.int = 1/0.11 W/(K·m2).
Results: Temperature distribution in media:
Temperature calculated in QuickField is presented in the table below (reference temperatures are in the brackets). To comply with the ISO 10211:2007 the difference between calculated and reference temperature should be less than 0.1°K. This simulation accuracy complies with the requirements of ISO 10211:2007.