The document contains two physics problems involving heat transfer calculations. The first problem involves calculating the required thickness of insulation needed to limit heat loss from a furnace wall to 1830 W/m^2, given the wall thickness, material properties, and inner and outer surface temperatures. The second problem involves calculating the time required for the temperature at a depth of 25.4 mm to reach 388.8 K as a slab of aluminum cools at its surface.

1. Co, Cynthia Eniceo, Raniel Maramba, Paulo Reyes, Charmaine Ramos, Hesed GROUP # 3

2. 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.

3. 0.244m x X=0.179m T1=1588K T2=299K kA=1.3W/mK kB=0.346W/mK 4.3-2

4. 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.

5. X1= 25.4/2 = 12.7mm 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 x 0.0127) = 36

6. X= άt/X 1 2 52 = 0.340 t / 0.0127 2 t=0.025 hrs.

7. 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 + cT2 Km=bTm + cT2m Km = 0.006[(315+38)/2] + -1.4x10-6 [(315+38)2/4] Km= 1.01539 W/mK Rt = 0.23 -------- (1m2) (1.01539) = 0.2265 Q= (315-38)/0.2265 Q = 1222.86 W

8. 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?

9. #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 Case2-Case1=1.48k(T1-T3) If the better insulating material is next to the pipe, heat transfer will be reduced by 1.48k(T1-T3) B A 2 B