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9 10 6 m2s table a 3 more accurate results are

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Unformatted text preview: rature at a specified location at a given time can be determined from the Heisler charts or one-term solutions. Here we will use the charts to demonstrate their use. Noting that the half-thickness of the plate is L 0.02 m, from Fig. 4–13 we have 1 Bi k hL t L2 100 W/m · °C 45.8 (120 W/m2 · °C)(0.02 m) (33.9 10 6 m2/s)(7 60 s) 35.6 (0.02 m)2 To Ti T T 0.46 Also, 1 Bi x L k hL L L 45.8 T To 1 T T 0.99 0.46 0.99 Therefore, T To T Ti T T T 0.455(Ti T To T Ti T T T) 500 0.455 and T 0.455(20 500) 282°C Therefore, the surface temperature of the plates will be 282°C when they leave the oven. Discussion We notice that the Biot number in this case is Bi 1/45.8 0.022, which is much less than 0.1. Therefore, we expect the lumped system analysis to be applicable. This is also evident from (T T )/(To T ) 0.99, which indicates that the temperatures at the center and the surface of the plate relative to the surrounding temperature are within 1 percent of each other. Egg Ti = 5°C h = 1200 W/ m 2·°C T = 95°C FIGURE 4–19 Schematic for Example 4–3. T = 500°C h = 120 W/m 2·°C 2 L = 4 cm Brass plate Ti = 20°C FIGURE 4–20 Schematic for Example 4–4. cen58933_ch04.qxd 9/10/2002 9:12 AM Page 226 226 HEAT TRANSFER Noting that the error involved in reading the Heisler charts is typically at least a few percent, the lumped system analysis in this case may yield just as accurate results with less effort. The heat transfer surface area of the plate is 2A, where A is the face area of the plate (the plate transfers heat through both of its surfaces), and the volume of the plate is V (2L)A, where L is the half-thickness of the plate. The exponent b used in the lumped system analysis is determined to be b hAs CpV h(2A) h Cp L Cp (2LA) 120 W/m2 · °C (8530 kg/m3)(380 J/kg · °C)(0.02 m) Then the temperature of the plate at t T (t ) T Ti T e bt → 7 min T (t ) 500 20 500 0.00185 s 1 420 s is determined from e (0.00185 s 1)(420 s) It yields T (t ) 279°C which is practically identical to the result obtained above using the Heisler charts. Therefore, we can use lumped system analysis with confidence when the Biot number is sufficiently small. EXAMPLE 4–5 T = 200°C h = 80 W/ m2 ·°C Stainless steel shaft Ti = 600°C D = 20 cm FIGURE 4–21 Schematic for Example 4–5. Cooling of a Long Stainless Steel Cylindrical Shaft A long 20-cm-diameter cylindrical shaft made of stainless steel 304 comes out of an oven at a uniform temperature of 600°C (Fig. 4–21). The shaft is then allowed to cool slowly in an environment chamber at 200°C with an average heat transfer coefficient of h 80 W/m2 · °C. Determine the temperature at the center of the shaft 45 min after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. SOLUTION A long cylindrical shaft at 600°C is allowed to cool slowly. The center temperature and the heat transfer per unit length are to be determined. Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the centerline. 2 The thermal properties of the shaft 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 stainless steel 304 at room temperature are k 14.9 W/m · °C, 7900 kg/m3, Cp 477 J/kg · °C, and 6 2 3.95 10 m /s (Table A-3). More accurate results can be obtained by using properties at average temperature. Analysis The temperature within the shaft may vary with the radial distance r as well as time, and the temperature at a specified location at a given time can cen58933_ch04.qxd 9/10/2002 9:12 AM Page 227 227 CHAPTER 4 be determined from the Heisler charts. Noting that the radius of the shaft is ro 0.1 m, from Fig. 4–14 we have 1 Bi 14.9 W/m · °C (80 W/m2 · °C)(0.1 m) k hro (3.95 t 10 ro2 6 m2/s)(45 (0.1 m)2 1.86 60 s) 1.07 To Ti T T 0.40 and To T 0.4(Ti T) 200 0.4(600 200) 360°C Therefore, the center temperature of the shaft will drop from 600°C to 360°C in 45 min. To determine the actual heat transfer, we first need to calculate the maximum heat that can be transferred from the cylinder, which is the sensible energy of the cylinder relative to its environment. Taking L 1 m, V ro2 L mCp(T Ti) 47,354 kJ m Qmax (7900 kg/m3) (0.1 m)2(1 m) 248.2 kg (248.2 kg)(0.477 kJ/kg · °C)(600 200)°C The dimensionless heat transfer ratio is determined from Fig. 4–14c for a long cylinder to be 1 1/Bi Bi h2 t k2 1 1.86 Bi2 0.537 (0.537)2(1.07) 0.309 Q Qmax 0.62 Therefore, Q 0.62Qmax 0.62 (47,354 kJ) 29,360 kJ which is the total heat transfer from the shaft during the first 45 min of the cooling. ALTERNATIVE SOLUTION We could also solve this problem using the one-term solution relation instead of the transient charts. First we find the Biot number Bi (80 W/m2 · °C)(0.1 m) 14.9 W/m · °C hro k 0.537 The coefficients 1 and A1 for a cylinder corresponding to this Bi are determined from...
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