Question
An apartment is heated by a radiator with the surface 0.8 m2. The total area of the other surfaces of the. room (floor, walls, ceiling) is 50 m3. Calculate the net radiation from the radiator to the other surfaces for a radiator temperature of 40 degree C and other surface temperature equal to 21 degree C. How much more heat is leaving the radiator if the radiator temperature is increased to 80 degree C? The emissivity for the radiator is 0.97 and for the other surfaces it is 0.90. 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
net heat radiation=0.97*(273+40)4*0.8-(0.9*(273+21)4*50)=93 W
b)when radiator temp increased to 80 C
net heat radiation=0.97*(273+80)4*0.8-(0.9*(273+21)4*50)=351 W