Question
Estimate the conductance for a surface of 5 m2 due to convection: Case Description a The air speed parallel to the surface is 4 m/s b At the windward side, wind speed equal to 6 m/s c At the leeward side, wind speed equal to 6 m/s d At the inside surface of a wall, the temperature difference between wall and air is 3 degree C e At the floor surface, the temperature is 17 degree C, and the air temperature is 21 degree C 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
Convective heat transfer coefficient a) hc(W/m2 K) at when air flowing parallel to surface with 4m/s :22(W/m2 K) conductance=hc*A=22*5=110W/K b)hc(W/m2 K) at when air flowing parallel to surface is 6m/sec:27(W/m2 K) conductance=hc*A=27*5=135W/K c)hc(W/m2 K) at when air flowing parallel to surface at leward side 6m/sec:14(W/m2 K) conductance=hc*A=14*5=70W/K d)at temperature difference 3C ha=2.64(W/m2 K) conductance=hc*A=2.64*5=13.2W/K
