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k T T h2 r T T 2 r T T ln r r p p e p s¥ - = - + - from which the total heat rate q¢ can then be determined. Using the IHT Correlations and Properties Tool Pads, the following results are obtained for the effect of the insulation thickness, with e = 0.85. 0 0.01 0.02 0.03 0.04 0.05 Insulation thickness, t(m) 20 50 80 110 140 170 S ur fa ce te m pe ra tu re , T s, 2( C ) 0 0.01 0.02 0.03 0.04 0.05 Insulation thickness, t(m) 0 100 200 300 400 500 600 700 800 H ea t l os s, q '(W /m ) The insulation significantly reduces Ts,2 and q¢ , and little additional benefits are derived by increasing t above 25 mm. For t = 25 mm, the effect of the emissivity is as follows. Continued... PROBLEM 9.55 (Cont.) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Emissivity, eps 34 36 38 40 42 44 S ur fa ce te m pe ra tu re , T s, 2( C ) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Emissivity, eps 48 49 50 51 52 53 H ea t l os s, q '(W /m ) Although the surface temperature decreases with increasing emissivity, the heat loss increases due to an increase in net radiation to the surroundings. PROBLEM 9.56 KNOWN: Dimensions and temperature of beer can in refrigerator compartment. FIND: Orientation which maximizes cooling rate. SCHEMATIC: ASSUMPTIONS: (1) End effects are negligible, (2) Compartment air is quiescent, (3) Constant properties. PROPERTIES: Table A-4, Air (Tf = 288.5K, 1 atm): n = 14.87 ´ 10 -6 m 2 /s, k =0.0254 W/m×K, a = 21.0 ´ 10 -6 m 2 /s, Pr = 0.71, b = 1/Tf = 3.47 ´ 10 -3 K -1 . ANALYSIS: The ratio of cooling rates may be expressed as ( ) ( ) sv v v h h s h T Tq h hDL . q h DL T T h p p ¥ ¥ - = = - For the vertical surface, find ( ) ( ) ( )( ) 2 3 1 s 3 3 9 3 L 6 2 6 2 g T T 9.8m/s 3.47 10 K 23 C Ra L L 2.5 10 L 14.87 10 m /s 21 10 m / s b na - - ¥ - - - ´ ´ ° = = = ´ ´ ´ ( )39 6LRa 2.5 10 0.15 8.44 10 ,= ´ = ´ and using the correlation of Eq. 9.26, ( ) ( ) L 2 1/66 8/279/16 0.387 8.44 10 Nu 0.825 29.7. 1 0.492/0.71 ´ = + = + ì ü ï ïï ï í ý ï ïé ù ê úï ïë ûî þ Hence L 2 L v k 0.0254W/m K h h Nu 29.7 5.03W/m K. L 0.15m ×= = = = × For the horizontal surface, find ( ) ( )3s 3 9 5D g T T Ra D 2.5 10 0.06 5.4 10 b na ¥-= = ´ = ´ and using the correlation of Eq. 9.34, ( ) ( ) D 2 1/65 8/279/16 0.387 5.4 10 Nu 0.60 12.24 1 0.559/0.71 ´ = + = + ì ü ï ïï ï í ý ï ïé ù ê úï ïë ûî þ D 2 D h k 0.0254W/m K h h Nu 12.24 5.18W/m K. D 0.06m ×= = = = × Hence v h q 5.03 0.97. q 5.18 = = < COMMENTS: In view of the uncertainties associated with Eqs. 9.26 and 9.34 and the neglect of end effects, the above result is inconclusive. The cooling rates are approximately the same. PROBLEM 9.57 KNOWN: Length and diameter of tube submerged in paraffin of prescribed dimensions. Properties of paraffin. Inlet temperature, flow rate and properties of water in the tube. FIND: (a) Water outlet temperature, (b) Heat rate, (c) Time for complete melting. SCHEMATIC: ASSUMPTIONS: (1) Negligible k.e. and p.e. changes for water, (2) Constant properties for water and paraffin, (3) Negligible tube wall conduction resistance, (4) Free convection at outer surface associated with horizontal cylinder in an infinite quiescent medium, (5) Negligible heat loss to surroundings, (6) Fully developed flow in tube. PROPERTIES: Water (given): cp = 4185 J/kg×K, k = 0.653 W/m×K, m = 467 ´ 10 -6 kg/s×m, Pr = 2.99; Paraffin (given): Tmp = 27.4°C, hsf = 244 kJ/kg, k = 0.15 W/m×K, b = 8 ´ 10 -4 K -1 , r = 770 kg/m 3 , n = 5 ´ 10 -6 m 2 /s, a = 8.85 ´ 10 -8 m 2 /s. ANALYSIS: (a) The overall heat transfer coefficient is i o 1 1 1 . U h h = + To estimate ih , find D 6 4m 4 0.1kg/s Re 10,906 D 0.025m 467 10 kg/s mp m p - ´= = = ´ ´ ´ × & and noting the flow is turbulent, use the Dittus-Boelter correlation ( ) ( )4/5 0.34/5 0.3D DNu 0.023Re Pr 0.023 10,906 2.99 54.3= = = 2D i Nu k 54.3 0.653W/m K h 1418W/m K. D 0.025m ´ ×= = = × To estimate oh , find ( ) ( ) ( ) ( )32 4 13s D 6 2 8 2 9.8m/s 8 10 K 55 27.4 K 0.025mg T T D Ra 5 10 m / s 8.85 10 m / s b na - - ¥ - - ´ -- = = ´ ´ ´ 6 DRa 7.64 10= ´ and using the correlation of Eq. 9.34, ( ) D 2 1/6 D 8/279/16 0.387Ra Nu 0.60 35.0 1 0.559/Pr = + = + ì ü ï ïï ï í ý ï ïé ù ê úï ïë ûî þ D 2 o k 0.15W/m K h Nu 35.0 210W/m K. D 0.025m ×= = = × Alternatively, using the correlation of Eq. 9.33, Continued \u2026.. PROBLEM 9.57 (Cont.) n D DDNu CRa with C 0.48, n 0.25 Nu 25.2= = = = 2 o 0.15W/m K h 25.2 151W/m K. 0.025m ×= = × The significant difference in ho values for the two correlations may be due to difficulties associated with high Pr applications of one or both correlations. Continuing with the result from Eq. 9.34, 3 2 i o 1 1 1 1 1 5.467 10 m K / W U h h 1418 210 -= + = + = ´ × 2U 183W/m K.= × Using Eq. 8.46, find m,o 2m,i p T T DL 0.025m 3m W exp U exp 183 T T mc 0.1kg/s 4185J/kg K m K p p¥ ¥ æ ö- æ ö´ ´= - = -ç ÷ ç ÷ç ÷- ´ × ×è øè ø& ( ) ( )m,o m,iT T T T 0.902 27.4 27.4 60 0.902 C¥ ¥ é ù= - - = - - °ë û m,oT 56.8 C.= ° < (b) From an energy balance, the heat rate is ( ) ( )p m,i m,oq mc T T 0.1kg/s 4185J/kg K 60 56.8 K 1335W= - = ´ × - =& < or using the rate equation, ( ) ( ) ( )2m 60 27.4 K 56.8 27.4 K q U A T 183W/m K 0.025m 3m 60 27.4 n 56.8 27.4 p - - - = D = × - - l l q 1335W.= (c) Applying an energy balance to a control volume about the paraffin, in stE E= D 2 sf sfq t V h L WH D / 4 hr r pé ù× = = -ê úë û ( ) ( ) 3 2 2 5770kg/m 3mt 0.25m 0.025m 2.44 10 J/kg 1335W 4 p´ é ù= - ´ê úë û 4t 2.618 10 s 7.27h.= ´ = < COMMENTS: (1) The value of oh is overestimated by assuming an infinite quiescent medium. The fact that the paraffin is enclosed will increase the resistance due to free convection and hence decrease q and increase t. (2) Using 2oh 151W/m K= × results in 2 m,oU 136W/m K,T 57.6 C,= × = ° q = 1009 W and t = 9.62 h. PROBLEM 9.58 KNOWN: A long uninsulated steam line with a diameter of 89 mm and surface emissivity of 0.8 transports steam at 200°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. FIND: (a) The heat loss per unit length for a calm day when the ambient air temperature is 20°C; (b) The heat loss on a breezy day when the wind speed is 8 m/s; and (c) For the conditions of part (a), calculate the heat loss with 20-mm thickness of insulation (k = 0.08 W/m\u22c5K). Would the heat loss change significantly with an appreciable wind speed? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Calm day corresponds to quiescent ambient conditions, (3) Breeze is in crossflow over the steam line, (4) Atmospheric air and large surroundings are at the same temperature; and (5) Emissivity of the insulation surface is 0.8. PROPERTIES: Table A-4, Air (Tf = (Ts + T\u221e)/2 = 383 K, 1 atm): \u3bd = 2.454 × 10-5 m2/s, k = 0.03251 W/m\u22c5K, \u3b1 = 3.544 × 10-5 m2/s, Pr = 0.693. ANALYSIS: (a) The heat loss per unit length from the pipe by convection and radiation exchange with the surroundings is b cv radq q q\u2032 \u2032 \u2032= + ( ) ( )4 4b D b s,b b b bs,bq h P T T P T T P D\u3b5 \u3c3 \u3c0\u221e \u221e\u2032 = \u2212 + \u2212 = (1,2) where Db is the diameter of the bare pipe. Using the Churchill-Chu correlation, Eq. 9.34, for free convection from a horizontal cylinder, estimate Dh ( ) D 2 1/ 6 b D 8/ 279 /16 0.387 Rah DNu 0.60 k 1