The document contains two engineering problems involving heat transfer calculations: 1) Calculating heat loss from a spherical furnace wall over 24 hours, finding the temperature at a radius of 1.1m is 543.64°C. 2) Modeling transient heat transfer in a slab, constructing a temperature profile table up to 4000 seconds using finite differences, with the temperature at 1500s being 150.8°C.

1. Assignment No. 2 Group #8 Rachelle Manalo Ronna Anne Oyco Jonas Jane Aguila Albert Jim Gonzales Joeward Manaig

4. PROBLEM NO. 1 - Solution a.) For a spherical section: A m = 4 π r 1 r 2 = 4 π x1x1.2 m = 15.0796m 2 q= T 1 - T 2 Δx kA m = [ (1100°C + 273) – (80°C + 273) ] K 0.2m 0.5 W x 15.0796m 2 m-K = 38453.09408 J x 3600s x 24hrs. s 1 hr = 3322347329 J ≈ 3.32 x 10 9 J

5. PROBLEM NO. 1 - Solution b.) At r= 1.1m, T=? q = 38453.09408 J/s A m 4 π (1)(1.1) m 2 = 2781.818182 J s-m 2 q = k T 1 - T 2 A m Δx 2781.818182 J = 0.5 W x [ (1100+273)-(T+273)] K s-m 2 m-K 0.1m T(°C) = 543.6363636 ≈ 543.64 °C

12. PROBLEM NO. 16 - Answer 152.1723 153.6640 160.4991 175.2437 197.6959 4000 151.4057 152.6833 158.8643 173.2194 196.0203 3500 150.7924 151.8146 157.1806 170.9653 194.0855 3000 150.3533 151.0852 155.4640 168.4258 191.7880 2500 150.0960 150.5248 153.7555 165.5284 188.9414 2000 150.0000 150.1600 152.1440 162.1856 185.1680 1500 150.0000 150.0000 150.8000 158.3200 179.6000 1000 150.0000 150.0000 150.0000 154.0000 170.0000 500 150 150 150 150 150 0 f 4 3 2 1 time (s)

15. PROBLEM NO. 16 – Solution

16. PROBLEM NO. 16 – Units

17. PROBLEM NO. 16 – Answer