4A Group 8 - Problem Set #2
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The document contains several engineering problems related to heat transfer and thermodynamics. Problem 16 asks the student to use a numerical finite difference method to construct a temperature profile table for a 1m thick slab initially at 150°C that is suddenly exposed to 250°C fluid on one face and insulated on the other, using 4 slices and time steps up to 4000s. Problem 4.1-1 asks the student to calculate the heat loss per square meter for a temporary insulating wall with given temperatures, thickness, and material properties. Problem 5.3-6 asks the student to determine the time for a furnace wall lining initially at 100°F to reach 500°F at a depth of 0.5ft, given properties
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The document describes a problem calculating the heat loss from a furnace with inner dimensions of 1m x 1.2m x 0.75m and 11cm thick refractory walls. It provides the inner and outer surface temperatures and conductivity and calculates the total heat loss as 881798399.8 J over 24 hours by conduction through the walls. A second problem calculates the heat loss from a duct with ceramic and insulation layers given temperatures, thicknesses, and conductivities. It uses an iterative process to calculate the interface temperature and heat loss.
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The document provides information about the dimensions and material properties of an electric furnace. The furnace has firebrick walls that are 2m thick with a thermal conductivity of 1.12 W/mK. It also has a quartz observation window that is 5x5x0.6 cm with a thermal conductivity of 0.07 W/mK. Given the inner surface temperature is 1100°C and outer is 121°C, the heat loss from the furnace per unit time is calculated to be 150.523W.
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The document provides information about the heat loss calculation of a furnace with inner dimensions of 1m x 1.2m x 0.75m and wall thickness of 11 cm made of refractory material. The inner and outer surface temperatures are 200°C and 38°C respectively. The total heat loss by conduction in 24 hours is calculated to be 801.65 MJ.
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2- C?>,cllblm,cvblkjbvclkbjlcjblkjlbkjcvlkbjonduction.pdf
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CH EN 3453 Heat Transfer 2014 Fall Utah Homework HW 03 Assignment
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- 1. CHUATOCO, Anthony DALANON, Farlene DONIEGO, Joanne FULGUERAS, Alyssa Marie LAGRADA, Rabbie Group 8 Unit Operations 4 ChE - A
- 2. Problem 4.1-1 Problem #16 (Click on the following to view the solutions) Problem Set 2 Geankoplis Problem 5.3-6 Problem #8 GROUP 8
- 4. Problem #8 Given: Required: 1 m 1.2 m 1.1 m 80 ° C 1100 ° C 0.5 W/m · K T a) total heat loss (q) for 24 hrs b) heat flux (q/A) & T at 1.1m radius
- 5. Solution: q = q i = q o = - Σ T R T R T = Δ x K m A m = T i - T R i = T - T o R o A m = 4 π · r i · r o Problem #8
- 6. Solution: R T = Δ x K m A m = 1m – 1.2m = Problem #8 - 0.0265 K/W q = - Σ T R T = - (1100°C – 80°C) - 0.0265 K/W = 38453.09 W or J/s q = 3.322 x 10 9 J a) total heat loss (q) for 24 hrs
- 7. Solution: Problem #8 b) heat flux (q/A) & T at 1.1m radius q = q i = T i - T R i 1m – 1.1m ( 0.5 W/m·K) [ 4 π (1m) (1.1m)] - ( 1100°C – T ) 38453.09 W = T = 543.64 °C q/A = 2781.82 W/m 2
- 8. Problem #16 (US) A large slab 1m thick is initially at a uniform temperature of 150 o C . Suddenly its front face is exposed to a fluid maintained at 250 ° C , but its rear face remained insulated. The fluid has a convective coefficient of 40 W/m 2 · K . Assume the solid has a thermal diffusivity of 0.000025 m 2 /s and a thermal conductivity of 20 W/m · K . Using a numerical finite difference method with M=5 and 4 slices , construct a table for the temperature profile of the slab up to a time of 4000 sec .
- 9. Problem #16 Given: Required: Slab: Thickness = 1m 0 T n = 150 °C Front Face: T a = 250 °C h = 40 W/m 2 K κ = 20 W/mK α = 0.000025 m 2 /s Rear Face: Insulated t = 4000s 4 slices M = 5 Δ x = 1m/4 = 0.25 m table for the temperature profile Using a numerical finite difference method
- 10. Problem #16 Solution: Δ t = 500s N = 0.5 q 1 2 3 4 f Δ x = 0.25m insulation T a = 250 °C
- 11. Problem #16 Solution: Front Face: for n=1 General: for n=2 to n=4 Rear Face: for n=f
- 12. Problem #16 Solution: t n
- 13. 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 276.5 K . The wall is composed of 25.4 mm of corkboard having a k of 0.0433 W/m∙K.
- 14. Problem 4.1-1 Given: Required: Wall: T o = 299.9 K T i = 276.5 K Side = 25.4 mm k = 0.0433 W/m∙K heat loss (q) per m 2
- 15. Solution: q = - Σ T R T R T = Δ x K m A m Problem 4.1-1 Basis: 1 m 2 q = 299.9 K -276.5 K R T = 0.0254m (0.0433 W/m∙K)(1m 2 ) = - 0.587 K/W - 0.587 K/W q = 39.89 W/ m 2
- 16. Problem 5.3-6 Unsteady-State Conduction in a Brick Wall: 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 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 and the physical properties of the brick are k =0.65 Btu/h·ft ° F and 0.02 ft 2 /h .
- 17. Problem 5.3-6 Given: Required: Flat brick wall: T = 500 °F T 1 = 1100 °F h = 2.6 Btu/h·ft 2 F κ = 0.65 Btu/h·ft ° F α = 0.02 ft 2 /h t = ? (at 0.5 ft) T 0 = 100 °F 0.5 ft
- 18. Solution: Problem 5.3-6 From the graph (Heisler): X = 6.5 t = 81.25 hrs