cen58933_ch04

# Solution a short cylinder is allowed to cool in

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Unformatted text preview: Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r-directions. 2 The thermal properties of the cylinder 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 brass at room temperature are k 110 W/m · °C and 33.9 10 6 m2/s (Table A-3). More accurate results can be obtained by using properties at average temperature. T = 25°C h = 60 W/ m2 ·°C x 0 L r Ti = 120°C ro Analysis (a) This short cylinder can physically be formed by the intersection of a long cylinder of radius ro 5 cm and a plane wall of thickness 2L 12 cm, as shown in Fig. 4–28. The dimensionless temperature at the center of the plane wall is determined from Figure 4–13a to be L t 2 L FIGURE 4–28 Schematic for Example 4–7. 1 Bi k hL 10 5 m2/s)(900 s) (0.06 m)2 8.48 110 W/m · °C (60 W/m2 · °C)(0.06 m) 30.6 (3.39 wall(0, t) T (0, t ) T Ti T 0.8 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 235 235 CHAPTER 4 Similarly, at the center of the cylinder, we have 10 5 m2/s)(900 s) (0.05 m)2 12.2 110 W/m · °C (60 W/m2 · °C)(0.05 m) 36.7 (3.39 t ro2 1 Bi k hro cyl(0, T (0, t ) T Ti T t) 0.5 Therefore, T (0, 0, t ) T Ti T wall(0, short cylinder t) cyl(0, t) 0.8 0.5 0.4 and T (0, 0, t ) T 0.4(Ti T) 25 0.4(120 25) 63°C This is the temperature at the center of the short cylinder, which is also the center of both the long cylinder and the plate. (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r 0), but at the outer surface of the plane wall (x L). Therefore, we first need to find the surface temperature of the wall. Noting that x L 0.06 m, x L 0.06 m 0.06 m 1 Bi k hL 1 110 W/m · °C (60 W/m2 · °C)(0.06 m) 30.6 T (L, t ) T To T 0.98 Then wall(L, t) T (L, t ) T Ti T T (L, t ) T To T To T Ti T 0.98 0.8 0.784 Therefore, T (L, 0, t ) T Ti T short cylinder wall(L, t) cyl(0, t) 0.784 0.5 0.392 25 0.392(120 25) 62.2°C and T(L, 0, t ) T 0.392(Ti T) which is the temperature at the center of the top surface of the cylinder. EXAMPLE 4–8 Heat Transfer from a Short Cylinder Determine the total heat transfer from the short brass cylinder ( kg/m3, Cp 0.380 kJ/kg · °C) discussed in Example 4–7. 8530 cen58933_ch04.qxd 9/10/2002 9:13 AM Page 236 236 HEAT TRANSFER SOLUTION We first determine the maximum heat that can be transferred from the cylinder, which is the sensible energy content of the cylinder relative to its environment: m Qmax ro2 L V mCp(Ti T ) (8530 kg/m3) (0.05 m)2(0.06 m) 4.02 kg (4.02 kg)(0.380 kJ/kg · °C)(120 25)°C 145.1 kJ Then we determine the dimensionless heat transfer ratios for both geometries. For the plane wall, it is determined from Fig. 4–13c to be 1 1/Bi Bi h2 t k2 1 30.6 2 0.0327 (0.0327)2(8.48) Bi Q Qmax 0.0091 0.23 plane wall Similarly, for the cylinder, we have Bi 1 1/Bi h2 t k2 1 36.7 2 0.0272 2 Bi (0.0272) (12.2) Q Qmax 0.0090 0.47 infinite cylinder Then the heat transfer ratio for the short cylinder is, from Eq. 4–28, Q Qmax short cyl Q Qmax 0.23 Q Qmax 1 0.47(1 1 2 Q Qmax 0.23) 0.592 1 Therefore, the total heat transfer from the cylinder during the first 15 min of cooling is Q EXAMPLE 4–9 0.592Qmax 0.592 (145.1 kJ) 85.9 kJ Cooling of a Long Cylinder by Water A semi-infinite aluminum cylinder of diameter D 20 cm is initially at a uniform temperature Ti 200°C. The cylinder is now placed in water at 15°C where heat transfer takes place by convection, with a heat transfer coefficient of h 120 W/m2 · °C. Determine the temperature at the center of the cylinder 15 cm from the end surface 5 min after the start of the cooling. SOLUTION A semi-infinite aluminum cylinder is cooled by water. The temperature at the center of the cylinder 15 cm from the end surface is to be determined. Assumptions 1 Heat conduction in the semi-infinite cylinder is twodimensional, and thus the temperature varies in both the axial x- and the radial r-directions. 2 The thermal properties of the cylinder and the heat transfer coefficient are constant. 3 The Fourier number is 0.2 so that the one-term approximate solutions are applicable. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 237 237 CHAPTER 4 Properties The properties of aluminum at room temperature are k 237 W/m · °C and 9.71 10 6 m2/s (Table A-3). More accurate results can be obtained by using properties at average temperature. Analysis This semi-infinite cylinder can physically be formed by the intersection of an infinite cylinder of radius ro 10 cm and a semi-infinite medium, as shown in Fig. 4–29. We will solve this problem using the one-term solution relation for the cylinder and the analytic solution for the semi-infinite medium. First we consider the infinitely long cylinder and evaluate the Biot number: 2 hro k Bi (120 W/m · °C)(0.1 m) 237 W/m · °C 0.05 The coefficients 1 and A1 for a cylinder corresponding to this Bi are determined from Table 4–1 to...
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## This note was uploaded on 01/28/2010 for the course HEAT ENG taught by Professor Ghaz during the Spring '10 term at University of Guelph.

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