Co, Cynthia Eniceo, Raniel Maramba, Paulo Reyes, Charmaine Ramos, Hesed GROUP # 3
4.3-2 A wall of furnace 0.244m thick is constructed of material having a thermal conductivity of 1.30W/mK. The wall will be insulated on the outside material having an average k of 0.346 W/mK, so the heat loss from the furnace will ve equal or less than 1830 W/m2. the inner surface temperature is 1588K and the outer 299K. Calculate the thickness of insulation required.
0.244m x X=0.179m T1=1588K T2=299K kA=1.3W/mK kB=0.346W/mK 4.3-2
5.3-3. Cooling a slab of Aluminum A large piece of Aluminum that can be considered a semi-infinite solid initially has a uniform temperature of 505.4K. The surface is suddenly exposed to an environment at 338.8K with a surface convection coefficient of 455 W/m2K. Calculate the time in hours for the temperature to reach 388.8 K at a depth of 25.4 mm. the average physical properties are ά = 0.340 m2.h and k=208 W/mK.
X1= 25.4mm=0.0254m To =505.4 K T1 = 338.8 K H= 455 W/m2K t=? If T=388.88K ά=0.340 m 2 /h K=208 W/mK X= άt/X1 2 Y= T1-T/ T1-To = [(338.8-388.8)/(338.8-505.4)] = 0.30 M= k/hX1 = 208 / (455 x0.0254) =18
Problem # 3 Compute the heat loss per square meter of surface for a furnace wall 23cm thick. The inner and outer surface temperature are 315 o C and 38 o C respectively. The variation of the thermal conductivity in W/mL, with temperature in o C is given by the following relation : k=0.006T-1.4x10 -6 T 2 k=bT + cT 2 Km=bTm + cT 2 m Km = bTm + c/3 (T 1 2 + T 1 T 2 + T 2 2 ) KM = 1.059 – 0.05256 Km=1.0064 W/mK Rt = 0.23 / (1 X 1.0064) = 0.2285 Q= (315-38)/0.2285 Q = 1212.25 W
Problem # 11 An insulated steam pipe having an outside diameter of 0.0245m is to be covered with 2 layers of insulation each having a thickness of 0.0245m. The average thermal conductivity of one material is approximately four times that of the other. Assuming that the inner and outer surface temperature of the composite insulation are fixed, how much will the heat be reduced when the better insulating material is next to the pipe then when it is the outer layer?
#11 D1=0.0245m D2=0.0245+0.0245=0.049m D3=0.49+0.0245=0.735m DL; L=1m A1=0.077m 2 A2=0.154m 2 A3=0.231m 2 Case1 Case2 k 4k k 4k Case1 Case2 ratio between case 1 and case 2 is 0.73. the better insulating material should be placed at the outer most layer. B A 2 B