The document provides data from a test on a flat-walled furnace with inner and outer brick walls of unknown conductivity. It gives the temperatures measured at various points within the furnace walls. It then calculates the percentage of heat loss that would be saved by adding 5cm of magnesia insulation to the furnace outer wall. Without the insulation, the heat loss is calculated to be 1612.87W, and with it added the heat loss is reduced to 542W, saving 33.61% of the heat loss.

1. Unit Operations Assignment #2 Abaricia, Emmanuel Caddarao, Patrick Janda, Maria Rosario Sebastian, Leah Jessica

2. Problem Set #2 5. The following data was obtained in a test on a flat-walled furnace, the linings of which consist of a 11.5 cm. non-corrosive brick of unknown conductivity and the outer wall of 20 cm clay brick, also of unknown conductivity. The temperature of the inner wall (flame side) was found to be 735°C and that of the outer wall 185°C. This furnace was lagged with 5 cm of magnesia (k=0.07 W/mK) thermocouples inserted at various points and the ff. data taken: Temp of inner wall (flame side)735°C; Temp at the junction of brick layers700°C Temp at the junction of ordinary brick and magnesia475°C Temp of the outer surface of magnesia88°C Calculate the percentage of heat loss that is saved by the lagging. Solution: Without Lagging 735 °C 185 °C Basis: A=1m2 q = Σ ∆T/Rt= 735-185 q=q1=q2 R1 + R2 k1 = 0.69 (W/mK) R1 = (11.5/100) = 0.167 (0.69)(1) Find k2: k(W/mK) T (°C) 1.00 200 1.47 600 1 2 11.5 20

3. Iteration 1: Iteration 2: Let T = 215 °C Let T = 484.73 °C Tave= 200 °C Tave= 334.87 °C k2 = 1 (W/mK) k T R2 = (20/100) = 0.2 1.00 200 (1)(1) x 334.87 1.47 600 q = 735-185 __ = 735 – T 334.87-200 = x-1 0.167 + 0.2 0.167 600-200 1.47-1 T = 484.73 °C x = 1.16 R2 = 0.172 q = 735-185 __ = 735 – T 0.167 + 0.172 0.167 T = 464.06 °C Iteration 3: Let T = 464.06 °C Tave= 324.53 °C k T 1.00 200 x 324.53 1.47 600

4. 324.53-200 = x-1 600-200 1.47-1 x = 1.15 R2 = 0.174 q = 735-185__ = 735 – T 0.167 + 0.174 0.167 T = 465.65 °C q = 735 – 465.65 = 1612.87 W 0.167 With Lagging: q = q1 = q2 = q3 q = Σ∆T/Rt using q = q3 735- 88 = 475-88 R3 = ( 5/100) = 0.714 Rt R3 (0.07) (1) q = 475-88 = 542 W 0.714 % heat loss saved = (542/1612.87) x (100) = 33.61 % 1 2 3 700 0 C 475 0 C 735 0 C 88 0 C 11.5 20cm 5cm Magnesia k = 0.07 (W/mK)

5. .13 ( US) What length of time is required to heat a 2.5 cm diameter, 5 cm long copper cylinder ( k= 380 W/mk) from 16deg C to 93 deg C in a furnace whose temperature is 315 deg C? The value of h at the cylinder surface is 204 KJ/hm 2 K and its thermal diffusivity is 0.186 m 2 /s. Given: D= 2.5cm X 1 = .0125 m L=5cm ; k= 380 W/mk To= 16 deg C T 1 = 315 deg C T=93deg C h= 204 KJ/hm 2 K or 56.67 W/m 2 K; α = 0.186 m 2 /s Required t in sec. Solution: Solution: Y = T1- T = 315-93 = 0.74 m = k__ = (380) _________ = 536.43 T1 – To 315-16 Һ x 1 ( 56.67 ) (.0125) Using Heisler chart: X = the obtained slope does not fit any of the available given chart t = X x 1 2 = ( 2.8) (.0125) 2 α (0.186)

6. Geankoplis 4.2-3. Removal of Heat from a Bath. Repeat Problem 4.2-2 but for a cooling coil made of 308 stainless steel having an average thermal conductivity of 15.23 W/mK. Problem 4.2-2. A cooling coil of 1.0 ft of 304 stainless-steel tubing having an inside diameter of 0.25 in and an outside diameter of 0.40 in is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40 °F and is 80 °F on the outside. Calculate the heat removal in btu/s and watts. Given: do = 0.40 in or 0.03 ft di = 0.25 in or 0.02 ft L = 1.0 ft km = 15.23 W/mK or 8.8 btu/h ft °F q = ∆T R = ∆X R kmAm Am = ((π)(1)(0.03-0.02))/(ln (0.03/0.02)) ∆ X = [(.4/12)- (.25/12)] / 2 = 6.25x10-3 q = (80 – 40) __ (6.25x10-3)_______ (8.8) ((π)(1)(0.03-0.02)) ln (0.03/0.02) 1 btu = 1054.368 J q = 4705.67 btu x 1h = 1.3071 btu/s or 1378.16 W h 3600s

7. 5.3-5.Cooking a Slab of Meat . A slab of meat 25.44 mm thick originally at a uniform temperature of 10 °C is to be cooked from both sides until the center reaches 121 °C in an oven at 177 °C. The convection coefficient can be assumed constant at 25.6 W/m2K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/mK and thermal diffusivity 5.85 x 10-4m2/h. x1 = 25.4/2 = 12.7 mm Given: To = 10 °C T1 = 177 °C T = 121 °C x = 0 Һ = 25.6 W/m2K k = 0.69 W/mK α = 5.85 x 10 -4 m2/h Required: t in s Solution: Y = T1- T = 177-121 = 0.34 m = k__ = (0.69) _________ = 2.12 T1 – To 177-10 Һ x1 (25.6) (12.7/1000) Using Heisler chart: X = 2.8 t = X x 1 2 = ( 2.8) (12.7/1000) 2 α (5.85x10-4/3600) t = 2779.15 s