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cen58933_ch04

# First we consider the infinitely long cylinder and

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Unformatted text preview: be 1 0.3126 and A1 1.0124. The Fourier number in this case is 10 5 m2/s)(5 (0.1 m)2 (9.71 t ro2 60 s) 2.91 0.2 and thus the one-term approximation is applicable. Substituting these values into Eq. 4–14 gives cyl(0, 0 t) A1e 2 1 (0.3126)2(2.91) 1.0124e 0.762 The solution for the semi-infinite solid can be determined from 1 semi-inf(x, t) erfc x 2 t hx k exp h2 t k2 erfc h x 2 t t k First we determine the various quantities in parentheses: x 2 t h t k hx k h2 t k2 0.15 m 0.44 2 (9.71 10 5 m2/s)(5 60 s) (120 W/m2 · °C) (9.71 10 5 m2/s)(300 s) 237 W/m · °C (120 W/m2 · °C)(0.15 m) 237 W/m · °C h t k 0.086 0.0759 2 (0.086)2 0.0074 Substituting and evaluating the complementary error functions from Table 4–3, semi-inf(x, t) 1 erfc (0.44) exp (0.0759 0.0074) erfc (0.44 1 0.5338 exp (0.0833) 0.457 0.963 0.086) Now we apply the product solution to get T (x, 0, t ) T Ti T semi-infinite cylinder semi-inf(x, t) cyl(0, t) 0.963 0.762 0.734 Ti = 200°C T = 15°C h = 120 W/ m2 ·°C D = 20 cm x x = 15 cm 0r FIGURE 4–29 Schematic for Example 4–9. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 238 238 HEAT TRANSFER and T (x, 0, t ) T 0.734(Ti T) 15 0.734(200 15) 151°C which is the temperature at the center of the cylinder 15 cm from the exposed bottom surface. EXAMPLE 4–10 5°F Steak 1 in. FIGURE 4–30 Schematic for Example 4–10. 35°F Refrigerating Steaks while Avoiding Frostbite In a meat processing plant, 1-in.-thick steaks initially at 75°F are to be cooled in the racks of a large refrigerator that is maintained at 5°F (Fig. 4–30). The steaks are placed close to each other, so that heat transfer from the 1-in.-thick edges is negligible. The entire steak is to be cooled below 45°F, but its temperature is not to drop below 35°F at any point during refrigeration to avoid “frostbite.” The convection heat transfer coefficient and thus the rate of heat transfer from the steak can be controlled by varying the speed of a circulating fan inside. Determine the heat transfer coefficient h that will enable us to meet both temperature constraints while keeping the refrigeration time to a minimum. The steak can be treated as a homogeneous layer having the properties 74.9 lbm/ft3, Cp 0.98 Btu/lbm · °F, k 0.26 Btu/h · ft · °F, and 0.0035 ft2/h. SOLUTION Steaks are to be cooled in a refrigerator maintained at 5°F. The heat transfer coefficient that will allow cooling the steaks below 45°F while avoiding frostbite is to be determined. Assumptions 1 Heat conduction through the steaks is one-dimensional since the steaks form a large layer relative to their thickness and there is thermal symmetry about the center plane. 2 The thermal properties of the steaks and the heat transfer coefficient are constant. 3 The Fourier number is 0.2 so that the one-term approximate solutions are applicable. Properties The properties of the steaks are as given in the problem statement. Analysis The lowest temperature in the steak will occur at the surfaces and the highest temperature at the center at a given time, since the inner part will be the last place to be cooled. In the limiting case, the surface temperature at x L 0.5 in. from the center will be 35°F, while the midplane temperature is 45°F in an environment at 5°F. Then, from Fig. 4–13b, we obtain 0.5 in. 1 0.5 in. T (L, t ) T 35 To T 45 x L 5 5 0.75 1 Bi k hL 1.5 which gives h 1k 1.5 L 0.26 Btu/h · ft · °F 1.5(0.5/12 ft) 4.16 Btu/h · ft2 · °F Discussion The convection heat transfer coefficient should be kept below this value to satisfy the constraints on the temperature of the steak during refrigeration. We can also meet the constraints by using a lower heat transfer coefficient, but doing so would extend the refrigeration time unnecessarily. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 239 239 CHAPTER 4 The restrictions that are inherent in the use of Heisler charts and the oneterm solutions (or any other analytical solutions) can be lifted by using the numerical methods discussed in Chapter 5. TOPIC OF SPECIAL INTEREST Refrigeration and Freezing of Foods Control of Microorganisms in Foods Microorganisms such as bacteria, yeasts, molds, and viruses are widely encountered in air, water, soil, living organisms, and unprocessed food items, and cause off-flavors and odors, slime production, changes in the texture and appearances, and the eventual spoilage of foods. Holding perishable foods at warm temperatures is the primary cause of spoilage, and the prevention of food spoilage and the premature degradation of quality due to microorganisms is the largest application area of refrigeration. The first step in controlling microorganisms is to understand what they are and the factors that affect their transmission, growth, and destruction. Of the various kinds of microorganisms, bacteria are the prime cause for the spoilage of foods, especially moist foods. Dry and acidic foods create an undesirable environment for the growth of bacteria, but not for the growth of yeasts and molds. Molds are also encountered on moist surfaces, cheese, and spoiled foods. Specific viruses are encounte...
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