Cousider a double-glazed window with the area 3 m2 and an air gap width of 2 cm.
ID: 1825663 • Letter: C
Question
Cousider a double-glazed window with the area 3 m2 and an air gap width of 2 cm. The interior temperature is 20 degree C, and the exterior one is -5 degree C. The solar radiation intensity is 300 W/m2. The thermal resistance of the glass can be neglected. The emissivity of the glass is 0.92. Calculate the net heat loss through the window with and without solar radiation. What is the effective U-value (see also Exercise H 17) for the two cases? The surface heat transfer coefficient for convection and radiation at the exterior, alpha e. and interior, alphai, is 25 and 8 W/m2K respectively. The air between the glasses can be considered stagnant due to the small distance. The thermal conductivity of air is 0.026 W/mK. The solar transmittance is 0.76. Neglect absorption of solar radiation in the glasses. 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) 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
according to my knowledge this is the possible way to solve the problem.. plz rate me first Q = KA(T2 - T1)/d total area of the building 8*12*2.4= 326.4(m^3) q=k*48*(20-0)/8, for roof, where k is 6 q=k*32*(20-0)/12, for height of walls, where k is 5 q=k*24*(20-0)/2.4, for the width i.e walls, where k is 0...