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1 atm): n = 16.90 ´ 10 -6 m 2 /s, k = 27.0 ´ 10 -3 W/m×K, a = 23.98 ´ 10 -6 m 2 /s, Pr = 0.706, b = 1/Tf. ANALYSIS: (a) For the vertical lateral (lat) and top horizontal (top) surfaces, the heat loss by radiation and convection is ( ) ( ) ( ) ( )( )2lat top lat r s top r sq q q h h DL T T h h D / 4 T Tp p¥ ¥= + = + - + + - where, from Eq. 1.9, the linearized radiation coefficient is ( ) ( )2 2r s sh T T T Tes ¥ ¥= + + ( ) ( )8 2 4 2 2 2 2rh 0.8 5.67 10 W / m K 320 300 K 320 300 K 5.41W/m K.-= ´ ´ × + + = × The free convection coefficient for the lateral and top surfaces is: Lateral-vertical plate: Using Eq. 9.26 with ( ) ( )( ) ( )33 2s 8 L 6 2 6 2 g T T H 9.8m/s 1/310K 47 27 K 0.5m Ra 1.950 10 16.90 10 m / s 23.98 10 m / s b na ¥ - - - - = = = ´ ´ ´ ´ ( ) L 2 1/6 L 8/279/16 0.387Ra Nu 0.825 1 0.492/Pr ì ü ï ïï ï= +í ý ï ïé ù+ï ïê úë ûî þ ( ) ( ) L 2 1/68 8/279/16 0.387 1.950 10 Nu 0.825 74.5 1 0.492/0.706 ì ü ´ï ïï ï= + =í ý ï ïé ù+ï ïê úë ûî þ L 2 lath Nu k / H 74.5 0.027W/m K/0.5m 4.02W/m K.= × = ´ × = × Continued \u2026.. PROBLEM 9.33 (Cont.) Top-horizontal plate: Using Eq. 9.30 with 2 c s D / 4 L A / P D / 4 0.075m D p p = = = = ( ) ( )( ) ( )323s c 5 L 6 2 6 2 g T T L 9.8m/s 1/310K 47 27 K 0.075m Ra 6.598 10 16.90 10 m / s 23.98 10 m / s b na ¥ - - - - = = = ´ ´ ´ ´ ( )L 1/41/4 5LNu 0.54Ra 0.54 6.598 10 15.39= = ´ = L 2 top ch Nu k / L 15.39 0.027W/m K/0.075m 5.54W/m K.= × = ´ × = × Hence, the heat loss by convection and radiation is ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 q 4.02 5.41 W / m K 0.30m 0.50m 47 20 K 5.54 5.41 W / m K 0.30 m / 4 47 20 K p p = + × ´ ´ - + + × ´ - ( )q 88.9 15.5 W 104W.= + = < (b) The effect of adding the vertical fins is to increase the area of the lateral surface to ( ) ( )wfA DH 30 t H 30 2 w Hpé ù= - × + ´ ×ë û ( ) ( )2 2wfA 0.30m 0.50m 30 0.005 0.500 m 30 2 0.075 0.500 mpé ù= ´ - ´ + ´ ´ê úë û [ ] 2 2 2wfA 0.471 0.075 m 2.25m 2.646m .= - + = where t and w are the thickness and width of the fins, respectively. Hence, the heat loss is now ( ) ( )lat top lat r wf s topq q q h h A T T q¥= + = + - + ( ) 2 2q 4.02 5.41 W / m 2.646m 20K 15.5W 515W.= + ´ ´ + = < Adding the fins to the lateral surface increases the heat loss by a factor of five. COMMENTS: Since the fins are not likely to have 100% efficiency, our estimate is optimistic. Further, since the fins see one another, as well as the lateral surface, the radiative heat loss is over predicted. PROBLEM 9.34 KNOWN: Surface temperature of a long duct and ambient air temperature. FIND: Heat gain to the duct per unit length of the duct. SCHEMATIC: ASSUMPTIONS: (1) Surface radiation effects are negligible, (2) Ambient air is quiescent. PROPERTIES: Table A-4, Air (Tf = (T¥ + Ts)/2 » 300K, 1 atm): n = 15.89 ´ 10 -6 m 2 /s, k = 0.0263 W/m×K, a = 22.5 ´ 10 -6 m 2 /s, Pr = 0.707, b = 1/Tf. ANALYSIS: The heat gain to the duct can be expressed as ( )( )s t b s t b sq 2q q q 2h H h W h W T T .¥¢ ¢ ¢ ¢= + + = × + × + × - (1) Consider now correlations to estimate s t bh , h , and h . From Eq. 9.25, for the sides with L º H, ( ) ( ) ( ) ( )33 2s 7 L 6 2 6 2 g T T L 9.8m/s 1/300K 35 10 K 0.2m Ra 1.827 10 . 15.89 10 m / s 22.5 10 m / s b na ¥ - - - - ´ = = = ´ ´ ´ ´ (2) Eq. 9.27 is appropriate to estimate sh , ( ) ( ) ( ) L 1/471/4 L 4/9 4/99/16 9/16 0.670 1.827 100.670Ra Nu 0.68 0.68 34.29 1 0.492/Pr 1 0.492/0.707 ´ = + = + = é ù é ù+ +ê ú ê úë û ë û L 2 sh Nu k / L 34.29 0.0263W/m K/0.2m 4.51W/m K.= × = ´ × = × (3) For the top and bottom portions of the duct, L º As/P » W/2, (see Eq. 9.29), find the Rayleigh number from Eq. (2) with L = 0.1 m, RaL = 2.284 ´ 10 6 . From the correlations, Eqs. 9.31 and 9.32 for the top and bottom surfaces, respectively, find ( ) ( ) 1/31/3 6 2 t L k 0.0263W/m K h 0.15Ra 0.15 2.284 10 5.17W/m K. W / 2 0.1m × = ´ = ´ ´ = × (4) ( ) ( ) 1/46 2 b k 0.0263W/m K h 0.27 2.284 10 2.76W/m K. W / 2 0.1m × = ´ ´ ´ = × (5) The heat rate, Eq. (1), can now be evaluated using the heat transfer coefficients estimated from Eqs. (3), (4), and (5). ( )( )2 2 2q 2 4.51W/m K 0.2m 5.17W/m K 0.2m 2.76W/m K 0.2m 35 10 K¢ = ´ × ´ + × ´ + × ´ - q 84.8W/m.¢ = < COMMENTS: Radiation surface effects will be significant in this situation. With knowledge of the duct emissivity and surroundings temperature, the radiation heat exchange could be estimated. PROBLEM 9.35 KNOWN: Inner surface temperature and dimensions of rectangular duct. Thermal conductivity, thickness and emissivity of insulation. FIND: (a) Outer surface temperatures and heat losses from the walls, (b) Effect of insulation thickness on outer surface temperatures and heat losses. SCHEMATIC: ASSUMPTIONS: (1) Ambient air is quiescent, (2) One-dimensional conduction, (3) Steady-state. PROPERTIES: Table A.4, air (obtained from Properties Tool Pad of IHT). ANALYSIS: (a) The analysis follows that of Example 9.3, except the surface energy balance must now include the effect of radiation. Hence, cond conv radq q q\u2032\u2032 \u2032\u2032 \u2032\u2032= + , in which case ( )( ) ( ) ( )i s,1 s,2 s,2 r s,2 surk t T T h T T h T T\u221e\u2212 = \u2212 + \u2212 where ( )( )2 2r s,2 sur surs,2h T T T T\u3b5\u3c3= + + . Applying this expression to each of the top, bottom and side walls, with the appropriate correlation obtained from the Correlations Tool Pad of IHT, the following results are determined for t = 25 mm. Sides: Ts,2 = 19.3°C, h = 2.82 W/m2\u22c5K, hrad = 5.54 W/m2\u22c5K Top: Ts,2 = 19.3°C, h = 2.94 W/m2\u22c5K, hrad = 5.54 W/m2\u22c5K < Bottom: Ts,2 = 20.1°C, h = 1.34 W/m2\u22c5K, hrad = 5.56 W/m2\u22c5K With condq q\u2032\u2032 \u2032\u2032= , the surface heat losses may also be evaluated, and we obtain Sides: q\u2032 = 2H q\u2032\u2032 = 21.6 W/m; Top: q\u2032 = w q\u2032\u2032 = 27.0 W/m; Bottom: q\u2032 = w q\u2032\u2032 = 26.2 W/m < (b) For the top surface, the following results are obtained from the parametric calculations Continued... PROBLEM 9.35 (Cont.) 0 0.01 0.02 0.03 0.04 0.05 Insulation thickness, t(m) 15 25 35 45 Su rfa ce te m pe ra tu re , T s2 (C ) 0 0.01 0.02 0.03 0.04 0.05 Insulation thickness, t(m) 0 100 200 300 H ea t l os s, q '(W /m ) COMMENTS: Contrasting the heat rates of part (a) with those predicted in Comment 1 of Example 9.3, it is evident that radiation is significant and increases the total heat loss from 57.6 W/m to 74.8 W/m. As shown in part (b), reductions in Ts,o and \u2032q may be effected by increasing the insulation thickness above 0.025 W/m\u22c5K, although attendant benefits diminish with increasing t. PROBLEM 9.36 KNOWN: Electric heater at bottom of tank of 400mm diameter maintains surface at 70°C with engine oil at 5°C. FIND: Power required to maintain 70°C surface temperature. SCHEMATIC: ASSUMPTIONS: (1) Oil is quiescent, (2) Quasi-steady state conditions exist. PROPERTIES: Table A-5, Engine Oil (Tf = (T¥ + Ts)/2 = 310K): n = 288 ´ 10 -6 m 2 /s, k = 0.145 W/m×K, a = 0.847 ´ 10-7 m2/s, b = 0.70 ´ 10-3 K-1. ANALYSIS: The heat rate from the bottom heater surface to the oil is ( )s sq hA T T¥= - where h is estimated from the appropriate correlation depending upon the Rayleigh number RaL, from Eq. 9.25, using the characteristic length, L, from Eq. 9.29, 2A D / 4 D 0.4msL 0.1m. P D 4 4 p p = = = = = The Rayleigh number is ( ) 3g T T LsRaL b na - ¥= ( )2 3 1 3 39.8m/s 0.70 10 K 70 5 K 0.1 m 7Ra 1.828 10 .L 6 2 7 2288 10 m / s 0.847 10 m / s - -´ ´ - ´ = = ´- -´ ´ ´ The appropriate correlation is Eq. 9.31 giving ( )1 / 3hL 1/3 7Nu 0.15Ra 0.15 1.828 10 39.5L Lk= = = ´ = k 0.145W/m K 2h Nu 39.5 57.3W/m K.LL 0.1m ×= = ´ = ×