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2 )/ln( oconv,pipetotal 2oconv, 2 12 pipe °=+=+= °=°== °=°== RRR Ah R kL rr R o \u3c0 \u3c0\u3c0 Water pipe Tair = -5°C Soil W658.1 C/W 3.0153 C)]5(0[ total 21 =° °\u2212\u2212=\u2212= \u221e\u221e R TT Q& kJ 57.83)s 3600J/s)(14 658.1( =×=\u394= tQQ & The amount of heat required to freeze the water in the pipe completely is kg 157.0)m 5.0()m 01.0()kg/m 1000( 232 ==== \u3c0\u3c1\u3c0\u3c1 Lrm V kJ 4.52kJ/kg) 7.333kg)( 157.0( === fgmhQ The water in the pipe will freeze completely that night since the amount heat loss is greater than the amount it takes to freeze the water completely . )4.5257.83( > PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-109 3-167E The surface temperature of a baked potato drops from 300°F to 200°F in 5 minutes in an environment at 70°F. The average heat transfer coefficient and the cooling time of the potato if it is wrapped completely in a towel are to be determined. Assumptions 1Thermal properties of potato are constant, and can be taken to be the properties of water. 2 The thermal contact resistance at the interface is negligible. 3 The heat transfer coefficients for wrapped and unwrapped potatoes are the same. Properties The thermal conductivity of a thick towel is given to be k = 0.035 Btu/h\u22c5ft\u22c5°F. We take the properties of potato to be those of water at room temperature, \u3c1 = 62.2 lbm/ft3 and cp = 0.998 Btu/lbm\u22c5°F. Analysis This is a transient heat conduction problem, and the rate of heat transfer will decrease as the potato cools down and the temperature difference between the potato and the surroundings decreases. However, we can solve this problem approximately by assuming a constant average temperature of (300+200)/2 = 250°F for the potato during the process. The mass of the potato is lbm 5089.0 )ft 12/5.1( 3 4)lbm/ft 2.62( 3 4 33 3 = = == \u3c0 \u3c0\u3c1\u3c1 rm V Rconv T\u221e RtowelTs Potato The amount of heat lost as the potato is cooled from 300 to 200°F is Btu 8.50F200)-F)(300Btu/lbm. lbm)(0.998 5089.0( =°°=\u394= TmcQ p The rate of heat transfer and the average heat transfer coefficient between the potato and its surroundings are Btu/h 6.609 h) 60/5( Btu 8.50 ==\u394= t QQ& F.Btu/h.ft 17.2 2 °=°\u2212=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e F)70250()ft (3/12 Btu/h 6.609 )( )( 2\u3c0TTA QhTThAQ so so && When the potato is wrapped in a towel, the thermal resistance and heat transfer rate are determined to be F/Btuh 6012.12539.03473.1 F/Btuh. 2539.0 ft )12/24.3(F).Btu/h.ft 2.17( 11 F/Btuh 3473.1 ft)12/5.1(ft]12/)12.05.1(F)[Btu/h.ft. 035.0(4 ft)12/5.1(ft]12/)12.05.1[( 4 convtoweltotal 222conv 21 12 towel °=+=+= °=°== °=+° \u2212+=\u2212= RRR hA R rkr rr R \u3c0 \u3c0\u3c0 Btu/h 4.112 F/Btuh 1.6012 F)70250( total =° °\u2212=\u2212= \u221e R TT Q s& min 27.1====\u394 h 452.0 Btu/h 4.112 Btu 8.50 Q Qt & This result is conservative since the heat transfer coefficient will be lower in this case because of the smaller exposed surface temperature. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-110 3-168E The surface temperature of a baked potato drops from 300°F to 200°F in 5 minutes in an environment at 70°F. The average heat transfer coefficient and the cooling time of the potato if it is loosely wrapped completely in a towel are to be determined. Assumptions 1Thermal properties of potato are constant, and can be taken to be the properties of water. 2 The heat transfer coefficients for wrapped and unwrapped potatoes are the same. Properties The thermal conductivity of a thick towel is given to be k = 0.035 Btu/h\u22c5ft\u22c5°F. The thermal conductivity of air is given to be k = 0.015 Btu/h\u22c5ft\u22c5°F. We take the properties of potato to be those of water at room temperature, \u3c1 = 62.2 lbm/ft3 and cp = 0.998 Btu/lbm\u22c5°F. Analysis This is a transient heat conduction problem, and the rate of heat transfer will decrease as the potato cools down and the temperature difference between the potato and the surroundings decreases. However, we can solve this problem approximately by assuming a constant average temperature of (300+200)/2 = 250°F for the potato during the process. The mass of the potato is lbm 5089.0 )ft 12/5.1( 3 4)lbm/ft 2.62( 3 4 33 3 = = == \u3c0 \u3c0\u3c1\u3c1 rm V Rconv T\u221e Rair Ts Potato Rtowel The amount of heat lost as the potato is cooled from 300 to 200°F is Btu 8.50F200)F)(300Btu/lbm. lbm)(0.998 5089.0( =°\u2212°=\u394= TmcQ p The rate of heat transfer and the average heat transfer coefficient between the potato and its surroundings are Btu/h 6.609 h) 60/5( Btu 8.50 ==\u394= t QQ& F.Btu/h.ft 17.2 2 °=°\u2212=\u2212=\u23af\u2192\u23af\u2212= \u221e\u221e F)70250()ft (3/12 Btu/h 6.609 )( )( 2\u3c0TTA QhTThAQ so so && When the potato is wrapped in a towel, the thermal resistance and heat transfer rate are determined to be F/Btuh 1195.22477.03134.15584.0 F/Btuh. 2477.0 ft )12/28.3(F).Btu/h.ft 2.17( 11 F/Btuh 3134.1 ft)12/52.1(ft]12/)12.052.1(F)[Btu/h.ft. 035.0(4 ft)12/52.1(ft]12/)12.052.1[( 4 F/Btuh. 5584.0 ft)12/50.1(ft]12/)02.050.1(F)[Btu/h.ft. 015.0(4 ft)12/50.1(ft]12/)02.050.1[( 4 convtowelairtotal 222conv 32 23 towel 21 12 air °=++=++= °=°== °=+° \u2212+=\u2212= °=+° \u2212+=\u2212= RRRR hA R rkr rr R rkr rr R \u3c0 \u3c0\u3c0 \u3c0\u3c0 Btu/h 9.84 F/Btuh. 2.1195 F)70250( total =° °\u2212=\u2212= \u221e R TT Q s& min 35.9====\u394 h 598.0 Btu/h 9.84 Btu 8.50 Q Qt & This result is conservative since the heat transfer coefficient will be lower because of the smaller exposed surface temperature. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-111 3-169 An ice chest made of 3-cm thick styrofoam is initially filled with 45 kg of ice at 0°C. The length of time it will take for the ice in the chest to melt completely is to be determined. Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time. 2 Heat transfer is one-dimensional. 3 Thermal conductivity is constant. 4 The inner surface temperature of the ice chest can be taken to be 0°C at all times. 5 Heat transfer from the base of the ice chest is negligible. Properties The thermal conductivity of styrofoam is given to be k = 0.033 W/m\u22c5°C. The heat of fusion of water at 1 atm is . kJ/kg 7.333=ifh Rconv T\u221e Rchest Ts Ice chest Analysis Disregarding any heat loss through the bottom of the ice chest, the total thermal resistance and the heat transfer rate are determined to be C/W 6678.107508.05927.1 C/W 07508.0 )m 74.0(C). W/m18( 11 C/W 5927.1 )m 5708.0(C) W/m.033.0( m 03.0 m 74.0)5.0)(4.0()5.0)(3.0(2)4.0)(3.0(2 m 5708.0)06.05.0)(06.04.0()06.05.0)(03.03.0(2)06.04.0)(03.03.0(2 convchesttotal 22conv 2chest 2 2 °=+=+= °=°== °=°== =++= =\u2212\u2212+\u2212\u2212+\u2212\u2212= RRR hA R kA LR A A o i o i W79.16 C/W 1.6678 C)028( total =° °\u2212=\u2212= \u221e R TT Q s& The total amount of heat necessary to melt the ice completely is kJ 685,16kJ/kg) kg)(333.7 50( === ifmhQ Then the time period to transfer this much heat to the cooler to melt the ice completely becomes days 11.5==×===\u394 h 276s 10937.9 J/s 79.16 J 000,685,16 5 Q Qt & PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-112 3-170 A wall is constructed of two large steel plates separated by 1-cm thick steel bars placed 99 cm apart. The remaining space between the steel plates is filled with fiberglass insulation. The rate of heat transfer through the wall is to