Question
The exterior wall consists of light-weight concrete with an external insulation of mineral wool with external rendering. The interior surface is painted. Calculate the U-value for the structure. In order to improve the thermal performance an extra insulation (cellular plastic) is placed at the inside surface, which is in turn covered with a gypsum board. How thick must the extra insulation be in order to reach the U-value 0.24 W/m3K. The original wall structure looks like: With the extra insulation we get two additional layers. The thermal conductivity of the cellular plastic is 0.033 W/mK, and the thermal resistance of the gypsum board is 0.06 m2K/W. The surface heat transfer coefficients at the interior and exterior surfaces are 8 and 25 W/m2K respectively. Concrete: 1500 W, light-weight concrete: 100 W. cellular plastic: 50 W, wood: 1 000 W Before retrofitting: 2 080 W, after: 1702 W 40.1 W, -4.3 degree C Temperatures from the outside to the inside, boundary temperatures, surface temperatures and temperatures at material interfaces: -10 -8.7 -6.6 + 19.6 20 degree C 5.15 W/m a) 110 W/K b) 134.8 W/K c) 70 W/K d) 13.2 W/K e) 2.5 W/K H.7: 370 W, the same Heat flow with solar radiation: 512.6 W, without: 1 584 W, temperature with solar radiation: 14.95 degree C. without: 0.22 degree C 1-glass: -196 W, 2-glasses: -348 W, 3-glasses: -346 W, Temperature on roof: 40.7 degree C, south facade: 35.5 degree C, other facades: 25.1 degree C, the internal surface temperatures are the same since the thermal resistance of the metal structure can be neglected a) 443 W b) 106 W c) 60W From 93 to 351 W -7.4 degree C -22 degree C With low emissivity coating: 1.54 W/m2K, without: 2.98 W/m2K From 0.34 W/m2K to 0.24 W/m2K Day time: -0.009 W/m2K, regular U-value: 0.24 W/m2K Vertical: 0.164 m2K/W. Horizontal: 0.174 W/m2K With solar radiation: -14.64 W/m2K, heat flow -392 W, without; 2.96 W/m2K, heat flow 66.7 W L67 W, 5 glasses (4 air gaps)
Explanation / Answer
U value=1/(1/.015+0.036/0.050+0.11/.150)=0.343=0.34 now to make it 0.24 0.24=1/(1/.015+0.036/0.050+0.11/.150+0.033/t)+0.06 0.18=1/(1/.015+0.036/0.050+0.11/.150+0.033/t) t=0.0054 5.4 mm
