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function of IHT, the following results are obtained for the temperature history of the plate. Continued \u2026.. PROBLEM 9.21 (Cont.) The time for the plate to cool to 100°C is t 2365s\u2248 < COMMENTS: (1) Although the plate temperature is comparatively large and radiation emission is significant relative to convection, much of the radiation leaving one plate is intercepted by the adjoining plate if the spacing between plates is small relative to their width. The net effect of radiation on the plate temperature would then be small. (2) Because of the increase in \u3b2 and reductions in \u3bd and \u3b1 with increasing t, the Rayleigh number decreases only slightly as the plate cools from 300°C to 100°C (from 4.72 × 109 to 4.48 × 109), despite the significant reduction in (T - T\u221e). The reduction in h from 7.2 to 5.6 W/m2\u22c5K is principally due to a reduction in the thermal conductivity. 0 5 0 0 1 0 0 0 1 5 0 0 2 0 0 0 2 5 0 0 Tim e , t(s ) 1 0 0 1 4 0 1 8 0 2 2 0 2 6 0 3 0 0 P la te te m pe ra tu re , T( C ) PROBLEM 9.22 KNOWN: Thin-walled container with hot process fluid at 50°C placed in a quiescent, cold water bath at 10°C. FIND: (a) Overall heat transfer coefficient, U, between the hot and cold fluids, and (b) Compute and plot U as a function of the hot process fluid temperature for the range 20 \u2264 T h\u221e, \u2264 50°C. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Heat transfer at the surfaces approximated by free convection from a vertical plate, (3) Fluids are extensive and quiescent, (4) Hot process fluid thermophysical properties approximated as those of water, and (5) Negligible container wall thermal resistance. PROPERTIES: Table A.6, Water (assume Tf,h = 310 K): \u3c1h = 1/1.007 × 10-3 = 993 kg/m3, cp,h = 4178 J/kg\u22c5K, \u3bdh = µh/\u3c1h = 695 × 10-6 N\u22c5s/m2/993 kg/m3 = 6.999 × 10-7 m2/s, kh = 0.628 W/m\u22c5K, Prh = 4.62, \u3b1h = kh/\u3c1hcp,h = 1.514 × 10-7 m2/s, \u3b2h = 361.9 × 10-6 K-1 ; Table A.6, Water (assume Tf,c = 295 K): \u3c1c = 1/1.002 × 10-3 = 998 kg/m3, cp,c = 4181 J/kg\u22c5K, \u3bdc = µc/\u3c1c = 959 × 10-6 N\u22c5s/m2/998 kg/m3 = 9.609 × 10-7 m2/s, kc = 0.606 W/m\u22c5K, Prc = 6.62, \u3b1c = kc/\u3c1ccp,c = 1.452× 10-7 m2/s, \u3b2c = 227.5 × 10-6 K-1. ANALYSIS: (a) The overall heat transfer coefficient between the hot process fluid, ,hT\u221e , and the cold water bath fluid, ,cT\u221e , is ( ) 1h cU 1 h 1 h \u2212= + (1) where the average free convection coeffieicnts can be estimated from the vertical plate correlation Eq. 9.26, with the Rayleigh number, Eq. 9.25, ( ) 2 1/ 6 LL 8 / 279 /16 0.387Ra Nu 0.825 1 0.492 Pr = + + \uf8f1 \uf8fc\uf8f4 \uf8f4\uf8f4 \uf8f4\uf8f2 \uf8fd\uf8f4 \uf8f4\uf8ee \uf8f9\uf8ef \uf8fa\uf8f4 \uf8f4\uf8f0 \uf8fb\uf8f3 \uf8fe 3 L g TL Ra \u3b2 \u3bd\u3b1 \u2206 = (2,3) To affect a solution, assume ( )s ,h ,iT T T 2 30 C 303K\u221e \u221e= \u2212 = =$ , so that the hot and cold fluid film temperatures are Tf,h = 313 K \u2248 310 K and Tf,c = 293 K \u2248 295 K. From an energy balance across the container walls, ( ) ( )h ,h s c s ,ch T T h T T\u221e \u221e\u2212 = \u2212 (4) the surface temperature Ts can be determined. Evaluating the correlation parameters, find: Hot process fluid: ( ) ( )32 6 1 9 L,h 7 2 7 2 9.8m s 361.9 10 K 50 30 K 0.200m Ra 5.357 10 6.999 10 m s 1.514 10 m s \u2212 \u2212 \u2212 \u2212 × × \u2212 = = × × × × Continued... PROBLEM 9.22 (Cont.) ( ) ( ) 21/ 69 L,h 8 / 279 /16 0.387 5.357 10 Nu 0.825 251.5 1 0.492 4.62 × = + = + \uf8f1 \uf8fc\uf8f4 \uf8f4\uf8f4 \uf8f4\uf8f2 \uf8fd\uf8f4 \uf8f4\uf8ee \uf8f9\uf8ef \uf8fa\uf8f4 \uf8f4\uf8f0 \uf8fb\uf8f3 \uf8fe 2 2hL,hh h h Nu 251.5 0.628 W m K 0.200m 790 W m K L = = × \u22c5 = \u22c5 Cold water bath: ( ) ( )32 6 1 9 L,c 7 2 7 2 9.8m s 227.5 10 K 30 10 K 0.200m Ra 2.557 10 9.609 10 m s 1.452 10 m s \u2212 \u2212 \u2212 \u2212 × × \u2212 = = × × × × \ufffd ( ) ( ) 21 69 L,c 8 / 279 16 0.387 2.557 10 Nu 0.825 203.9 1 0.492 6.62 × = + = + \uf8f1 \uf8fc\uf8f4 \uf8f4\uf8f4 \uf8f4\uf8f2 \uf8fd\uf8f4 \uf8f4\uf8ee \uf8f9\uf8ef \uf8fa\uf8f4 \uf8f4\uf8f0 \uf8fb\uf8f3 \uf8fe 2 ch 203.9 0.606 W m K 0.200 m 618 W m K= × \u22c5 = \u22c5 From Eq. (1) find ( ) 1 2 2U 1 790 1 618 W m K 347 W m K\u2212= + \u22c5 = \u22c5 < Using Eq.(4), find the resulting surface temperature ( ) ( )2 2s s790 W m K 50 T K 618 W m K T 30 K\u22c5 \u2212 = \u22c5 \u2212 sT 32.4 C= $ Which compares favorably with our assumed value of 30°C. (b) Using the IHT Correlations Tool, Free Convection, Vertical Plate and following the foregoing approach, the overall coefficient was computed as a function of the hot fluid temperature and is plotted below. Note that U increases almost linearly with ,hT\u221e . 20 30 40 50 Hot process fluid temperature, Tinfh (C) 100 200 300 400 O ve ra ll c oe ffi cie nt , U (W /m ^2 .K ) COMMENTS: For the conditions of part (a), using the IHT model of part (b) with thermophysical properties evaluated at the proper film temperatures, find U = 352 W/m\u22c5K with Ts = 32.4°C. Our approximate solution was a good one. (2) Because the set of equations for part (b) is quite stiff, when using the IHT model you should follow the suggestions in the IHT Example 9.2 including use of the intrinsic function Tfluid_avg (T1,T2). PROBLEM 9.23 KNOWN: Height, width, emissivity and temperature of heating panel. Room air and wall temperature. FIND: Net rate of heat transfer from panel to room. SCHEMATIC: ASSUMPTIONS: (1) Quiescent air, (2) Walls of room form a large enclosure, (3) Negligible heat loss from back of panel. PROPERTIES: Table A-4, Air (Tf = 350K, 1 atm): n = 20.9 ´ 10 -6 m 2 /s, k = 0.03 W/m×K, a = 29.9 ´ 10-6 m2/s, Pr = 0.700. ANALYSIS: The heat loss from the panel by convection and radiation exchange is ( ) ( )4 4s s surq hA T T A T T .es¥= - + - With ( ) ( ) ( )( ) ( )( ) 33 2 s 9 L 12 4 2 g T T L 9.8m/s 1/350K 100K 1m Ra 4.48 10 20.9 29.9 10 m / s b an ¥ - - = = = ´ ´ and using the Churchill-Chu correlation for free convection from a vertical plate, ( ) L 2 1/6 L 8/279/16 0.387RahL Nu 0.825 196 k 1 0.492/Pr ì ü ï ïï ï= = + =í ý ï ïé ù+ï ïê úë ûî þ 2h 196k/L 196 0.03W/m K/1m 5.87W/m K.= = ´ × = × Hence, ( ) ( ) ( ) ( ) 2 2 4 48 2 4 2 q 5.86W/m K 0.5m 100K 0.9 5.67 10 W / m K 0.5m 400 300 K- = × é ù+ ´ ´ × -ê úë û q 293W 447 W 740W.= + = < COMMENTS: As is typical of free convection in gases, heat transfer by surface radiation is comparable to, if not larger than, the convection rate. The relative contribution of free convection would increase with decreasing L and Ts. PROBLEM 9.24 KNOWN: Initial temperature and dimensions of an aluminum plate. Condition of the plate surroundings. Plate emissivity. FIND: (a) Initial cooling rate, (b) Validity of assuming negligible temperature gradients in the plate during the cooling process. SCHEMATIC: ASSUMPTIONS: (1) Plate temperature is uniform, (2) Chamber air is quiescent, (3) Chamber surface is much larger than that of plate, (4) Negligible heat transfer from edges. PROPERTIES: Table A-1 , Aluminum (573K): k = 232 W/m×K, cp = 1022 J/kg×K, r = 2702 kg/m 3; Table A-4 , Air (Tf = 436K, 1 atm): n = 30.72 ´ 10 -6 m2/s, a = 44.7 ´ 10-6 m2/s, k = 0.0363 W/m×K, Pr = 0.687, b = 0.00229 K-1. ANALYSIS: (a) Performing an energy balance on the plate, ( ) ( ) [4 4s sur st pq 2A h T T T T E Vc dT/dtes r¥é ù- = - - + - = =ê úë û& ( ) ( )4 4sur pdT/dt 2 h T T T T / wces r¥é ù= - - + -ê úë û Using the correlation of Eq. 9.27, with ( ) ( ) ( )33 2 1i 8 L 6 2 6 2 g T T L 9.8m/s 0.00229K 300 27 K 0.5m Ra 5.58 10 30.72 10 m / s 44.7 10 m / s b na - ¥ - - - ´ - = = = ´ ´ ´ ´ ( ) ( ) ( ) 1/481/4 L 4/9 4/99/16 9/16 0.670 5.58 100.670Rak 0.0363 h 0.68 0.68 L 0.5 1 0.492/Pr 1 0.492/0.687 ì ü ì ü ´ï ï ï ïï ï ï ï= + = +í ý í ý ï ï ï ïé ù é ù+ +ï ï ï ïê ú ê úë û ë ûî þ î þ 2h 5.8W/m K.= × < Hence the initial cooling rate is ( ) ( ) ( )( ) ( ) 4 42 8 2 4 3 2 5.8W/m K 300 27 C 0.25 5.67 10 W / m K 573K 300KdT dt 2702kg/m 0.016m 1022J/kg K -× - + ´ ´ × - = - × é ù ê úë û o dT