The document discusses calculating the temperature at the boundary between a refractory brick layer and insulating material layer in a plane wall. Given information includes the thicknesses of each layer, thermal conductivities, and temperatures on both surfaces. The calculated temperature at the boundary is 867.75°C. Another example calculates the time to cool a glass sheet submerged in water from an initial to average final temperature. The calculated time is 130.21 hours.

1. UNIT OPERATIONS Group 2 4 ChE B Set 7 Bacal, Nathaniel Cueto, Ana De Vera, Bernadette Mercado, Rommel Nieva, Jared

2. # 7 A plane wall is composed of an 20 cm layer of refractory brick (k = 1.3W/mK) and a 5 cm layer of insulating material with k for the insulating material varying linearly as k = 0.034 + 0.00018 t where t is the temperature in °C. The inside surface temperature of the brick is 1100 °C and the outside surface temperature of the insulating material is 38 °C. Calculate the temperature at the boundary of the brick and insulation. 20 cm 5 cm 38 °C 1100 °C T’ = ? Refractory brick ; k = 1.3 W/m-k Insulating brick ; k = 0.034 + 0.00018t Basis: 1 m 2 of cross-sectional area Given: T 1 = 1100 °C ; X = 20 / 100 m T 2 T 1 T 2 = 38 °C ; X = 5 /100 m T’ = ?

3. Required: T’ = ? q = Σ Δ T = 1100 + 38 R T R1 + R2 Solution: R = Δ X KmAm R 1 = 20 / 100 = 0.1538 K / W (1.3) x (1m) 2 R 2 = 5 / 100 = ? (0.034 + 0.00018t) Eqn. 1 Since: q = q1 = q2 T’ - 38 = 1100 + 38 R2 R1 + R2 Eqn. 2 Assume : T’ = 600 °C ; T ave = (600 + 38) / 2 = 319 °C T ave = t Using Eqn. 1 : R2 = 5 / 100 = 0.5469 K / W (0.034 + [0.00018][319]) Using Eqn. 2 : T’ = 926.21 °C

4. % difference = (600 – 926.21) / 600 x 100 = 54.37% Assume : T’ = 926.21 °C ; T ave = (926.21 + 38) / 2 = 482.105 °C Using Eqn. 1 : R2 = 5 / 100 = 0.5469 K / W (0.034 + [0.00018][482.105]) Using Eqn. 2 : T’ = 867.75 °C % difference = (926.21 – 867.75) / 926.21 x 100 = 6.31% % difference is less than 10% so T’ ≈ 867.75 °C

7. Solution Using the Average Temperature Table ( Fig. 5.3-13 on page 377) At Y = 0.17, X = 0.63 T= 468750 s or 130.21 hr

8. Geankoplis 4.1-2 Determination of Thermal Conductivity. In determining the thermal conductivity of an insulating material, the temperatures were measured on both sides of a flat slab of 25 mm of the material and were 318.4 and 303.2 K. the heat flux was measured as 35.1 W/m2. Calculate the thermal conductivity in BTU/h-ft-ºF and in W/m-K q = 35.1 W/m 2 T 1 = 318.4 K T 2 = 303.2 K 25 mm

9. In W/m-K: 35.1 W/m 2 = - k (303.2–318.4) K 0.025 m 0.0577 W/m-K = k q = 35.1 W/m 2 In BTU/hr-ft- ºF: 0.0577 W 1 = 0.0333 BTU m-K 1.73 hr-ft- ºF T 1 = 318.4 K T 2 = 303.2 K 25 mm q = -k dT dx

10. Geankoplis 5.3-7 Cooling a Steel Rod. A long steel rod 0.305 m in diameter is initially at a temperature of 588 K. It is immersed in an oil bath maintained at 311 K. The surface convective coefficient is 125 W/m 2 -K. Calculate the temperature at the center of the rod after 1 hour. The average physical properties of the steel are k=38 W/m-K and α=0.0381 m 2 -h 0.305 m X X 1 = 0.1525

11. Given: D = 0.305 m x 1 = D/2 = 0.1525 m x = 0 T o = 588 K T 1 = 311 K h = 125 W/m 2 –K t = 1 hour k = 38 W/m-K α = 0.0381 m 2 -h m = k = 38 = 1.99 ~ 2 h x 1 (125)(0.1525) n = x = 0 = 0 x 1 0.1525 X = αt = (0.0381)(1) = 1.64 x 1 0.1525 Using Fig. 5.3-7 of Geankoplis (Gurney - Lurie Chart for cylinders) Y = 0.29 = T 1 – T = 311 – T T 1 - T o 311 – 588 T = 391.33 K , where T is the temperature at the center of the cylinder. 0.305 m x x 1 = 0.1525 m