Question
Calculate the heat flow through a window' with the area 2.5 m2, and single-, double- and triple-glazed window. The solar radiation intensity (perpendicular to the window surface) is 300 W/m2. The indoor temperature is 20 degree C the the outdoor one is -10 degree C. The transmittance for the different windows can be found in Figure 3.33. The thermal resistance for the three types of windows, including surface resistances, is 0.17, 0.35 and 0.53 m2K/W. 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
for single layer
q= ka(t2-t1)/d
k= 0.17
a=2.5
d=300
t2-t1= 30
q=0.17*2.2*30/0.300= 37.4j
for two layers
q= ka(t2-t1)/d
k= 0.35
a=2.5
d=300
t2-t1= 30
q=0.35*2.2*30/0.300= 77j
for triple layer
q= ka(t2-t1)/d
k= 0.53
a=2.5
d=300
t2-t1= 30
q=0.53*2.2*30/0.300= 116.6j
