The document describes a spherical furnace with an inner radius of 1m and outer radius of 1.2m. The wall has a thermal conductivity of 0.5 W/mK. The inner temperature is 1100°C and outer is 80°C. It asks to calculate the total heat loss over 24 hours and the heat flux and temperature at a radius of 1.1m. It then describes a 1m thick slab initially at 150°C that is exposed to 250°C fluid on one side with an insulated rear side. It asks to construct a temperature profile table using finite differences over 4000 seconds.

1. DE GUZMAN, John Wilbert HIZON, Donn Angelo M. PEDRIGAL, Ian Sygfryd REYES, Mervick Ann B. TINDUGAN, Farrah Kaye Z, 4 ChEB Group 8

3. Given: r i =1m r o =1.2m k=0.5 W/mK

4. a. For spherical section: Am = 4 π r i r o = 4 π (1) (1.2) = 15.079 m 2 Δ x = r o – r i = 1.2 – 1 = 0.2 m

5. q = 38451.534 W

6. b. = 2781.7873 W/m 2 T’ = 543.70 o C

8. Given: Δ x= 0.25 m Δ x 1 2 3 4 f insulation q T a = 250˚C α = 0.000025 m 2 /s κ= 20 W/ m•K h= 40 W/ m 2 •K

10. 1 2 3 4 f 0 s 500 s 1000 s 1500 s 2000 s 2500 s 3000 s 3500 s 4000 s

12. ANSWERS: 1 2 3 4 f 0 s 150 150 150 150 150 500 s 170 150 150 150 150 1000 s 178 154 150 150 150 1500 s 182.8 158 150.8 150 150 2000 s 186.32 161.52 152.08 150.16 150 2500 s 189.136 164.592 153.584 150.512 150.064 3000 s 191.4912 167.2992 155.1712 151.0368 150.2432 3500 s 193.5162 169.712 156.7699 151.705 150.5606 4000 s 195.2913 171.8844 158.3453 152.4891 151.0184

13. GEANKOPLIS: Problem 4.1-1 Insulation in a Cold Room. Calculate the heat loss per m 2 of surface area for a temporary insulating wall of a food cold storage room where the outside temperature is 299.9 K and the inside temperature is 276.5 K. The wall is composed of 25.4 mm of corkboard having a k of 0.0433 W/m • K Given: Δ x Δ x= 0.0254 m T 1 = 299.9 K T 2 = 276.5 K Basis: A=1 m 2 Req’d: q q q= 39.89 W/m 2

14. Problem 5.3-6 A flat brick wall 1.0 ft thick is the lining on one side of a furnace. If the wall is at uniform temperature of 100°F and one side is suddenly exposed to a gas at 1100°F, calculate the time for the furnace wall at a point 0.5 ft from the surface to reach 500°F. The rear side of the wall is insulated. The convection coefficient is 2.6 btu/h-ft 2 -°F. The physical properties of the brick are k=0.65 btu/h-ft-°F and α=0.02 ft 2 /h.

15. brick wall To=100F q required: time for the wall at a point 0.5ft from the surface to reach 500F 0.5ft 1.0 ft furnace T1= 1100F

17. Solution: m= k/(hx1) =0.65/(2.6 x 05) = 0.5 Y= (Ti-T)/(T1-To) = (1100-500)/(1100-100) = 0.6 Plotting these on Fig.5.3-6(Heisler), x≈ 0.4 x= αt/(x 1 ) 2 t = 0.4(x 1 ) 2 / α = 0.4(0.5) 2 /0.2 = 0.05h = 3mins