Baixe Solucionário Çengel Transferêcia de Calor- 4ª Edição - capítulo 03 e outras Notas de estudo em PDF para Engenharia Química, somente na Docsity! PROPRIETARY MATERIAL. © 2011 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. 3-1 Solutions Manual for Heat and Mass Transfer: Fundamentals & Applications Fourth Edition Yunus A. Cengel & Afshin J. Ghajar McGraw-Hill, 2011 Chapter 3 STEADY HEAT CONDUCTION PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw-Hill. PROPRIETARY MATERIAL. © 2011 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. 3-2 & re, mperature. Steady Heat Conduction in Plane Walls 3-1C The temperature distribution in a plane wall will be a straight line during steady and one dimensional heat transfer with constant wall thermal conductivity. 3-2C In steady heat conduction, the rate of heat transfer into the wall is equal to the rate of heat transfer out of it. Also, the temperature at any point in the wall remains constant. Therefore, the energy content of the wall does not change during steady heat conduction. However, the temperature along the wall and thus the energy content of the wall will change during transient conduction. 3-3C Convection heat transfer through the wall is expressed as )( ∞−= TThA ss . In steady heat transfer, heat transfer rate to the wall and from the wall are equal. Therefore at the outer surface which has convection heat transfer coefficient three times that of the inner surface will experience three times smaller temperature drop compared to the inner surface. Therefo at the outer surface, the temperature will be closer to the surrounding air te Q 3-4C The new design introduces the thermal resistance of the copper layer in addition to the thermal resistance of the aluminum which has the same value for both designs. Therefore, the new design will be a poorer conductor of heat. 3-5C (a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rod is the bottom or the top surface area of the rod, . (b) If the top and the bottom surfaces of the rod are insulated, the heat transfer area of the rod is the lateral surface area of the rod, 4/2DAs π= DLA π= . 3-6C The thermal resistance of a medium represents the resistance of that medium against heat transfer. 3-7C The combined heat transfer coefficient represents the combined effects of radiation and convection heat transfers on a surface, and is defined as hcombined = hconvection + hradiation. It offers the convenience of incorporating the effects of radiation in the convection heat transfer coefficient, and to ignore radiation in heat transfer calculations. 3-8C Yes. The convection resistance can be defined as the inverse of the convection heat transfer coefficient per unit surface area since it is defined as . )/(1 hARconv = 3-9C The convection and the radiation resistances at a surface are parallel since both the convection and radiation heat transfers occur simultaneously. PROPRIETARY MATERIAL. © 2011 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. 3-5 3-19 The two surfaces of a window are maintained at specified temperatures. The rate of heat loss through the window and the inner surface temperature are to be determined. Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity is constant. 4 Heat transfer by radiation is negligible. Properties The thermal conductivity of the glass is given to be k = 0.78 W/m⋅°C. Analysis The area of the window and the individual resistances are 2m 6.3m) 4.2(m) 5.1( =×=A L Glass T1 Q& C/W 04103.001111.000214.002778.0 C/W 01111.0 )m 6.3(C). W/m25( 222 2,o °= ° === conv Ah RR 11 C/W 002140 )m 6.3(C) W/m.78.0( m 006.0 C/W 02778.011 2,1, 2 1 glass °=++= ++= °= ° == °==== convglassconvtal RRRR Ak LR RR he steady rate of heat transfer through window glass is then )m 6.3(C). W/m10( 221 1,i ° conv Ah . to RglassRi Ro T T∞1 T∞2 W707= ° °−− = − = ∞∞ C/W 04103.0 C)]5(24[21 totalR TT Q& The inner surface temperature of the window glass can be determined from C4.4°=°−°=−=⎯→⎯−= ∞∞ C/W) 8 W)(0.0277707(C241,11 1, 11 conv conv RQTT R TT Q && PROPRIETARY MATERIAL. © 2011 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. 3-6 3-20 A double-pane window consists of two layers of glass separated by a stagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined. Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivities of the glass and air are constant. 4 Heat transfer by radiation is negligible. Air Properties The thermal conductivity of the glass and air are given to be kglass = 0.78 W/m⋅°C and kair = 0.026 W/m⋅°C. Analysis The area of the window and the individual resistances are 2m 6.3m) 4.2(m) 5.1( =×=A C/W 16924.0 01111.012821.0)00107.0(202778.02 C/W 01111.0 )m 6.3(C). W/m25( 11 C/W 12821.0 )m 6.3(C) W/m.026.0( 22 2 2 °= ° === air Ak RR m 012.0 C/W 00107.0 ) 6.3(C) W/m.78.0( m 003.0 C/W 02778.011 2,211, o 2o2 2 2,o 2 1 1 glass31 1,i °= +++=+++= ==== °= ° ==== °==== convconvtotal conv conv RRRRR Ah RR L Ak L RRR R The steady rate of heat transfer through window glass then becomes R1 R2 R3Ri Ro T∞1 T∞2R )m 6.3(C). W/m10( 221 °Ah m W154= ° °−− = − = ∞∞ C/W16924.0 C)]5(21[21 totalR TT Q& The inner surface temperature of the window glass can be determined from C16.7°°−°=−=⎯→⎯ − = ∞ ∞ =C/W)8 W)(0.0277154(C211,11 1, 11 conv conv RQTT R TT Q && PROPRIETARY MATERIAL. © 2011 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. 3-7 3-21 A double-pane window consists of two layers of glass separated by an evacuated space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined. Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. 2 Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. 3 Thermal conductivity of the glass is constant. 4 Heat transfer by radiation is negligible. Properties The thermal conductivity of the glass is given to be kglass = 0.78 W/m⋅°C. Analysis Heat cannot be conducted through an evacuated space since the thermal conductivity of vacuum is zero (no medium to conduct heat) and thus its thermal resistance is zero. Therefore, if radiation is disregarded, the heat transfer through the window will be zero. Then the answer of this problem is zero since the problem states to disregard radiation. Vacuum Discussion In reality, heat will be transferred between the glasses by radiation. We do not know the inner surface temperatures of windows. In order to determine radiation heat resistance we assume them to be 5°C and 15°C, respectively, and take the emissivity to be 1. Then individual resistances are 2m 6.3m) 4.2(m) 5.1( =×=A C/W 09505.0 01111.005402.0)00107.0(202778.02 C/W 01111.0 )m 6.3(C). W/m25( 11 C/W 05402.0 ]278288][278288)[m 6.3().K W/m1067.5(1 3222428 === °= ++× − RR K 1 1 C/W 00107.0 )m 6.3(C) W/m.78.0( m 003.0 C/W 02778.011 2,11, o 2o2 2 2,o 2 1 1 glass31 1,i °= +++=+++= = = °= ° ==== °==== convradconvtotal conv conv RRRRR Ah Ak L RRR R The steady rate of heat transfer through window glass then becomes R1 Rrad R3Ri Ro T∞1 T∞2R )m 6.3(C). W/m10( 221 °Ah ))(( 22 ++ = surrssurrs rad TTTTA R εσ W274= ° °−− = − = ∞∞ C/W09505.0 C)]5(21[21 totalR TT Q& The inner surface temperature of the window glass can be determined from C13.4°=°−°=−=⎯→⎯ − = ∞ ∞ C/W)8 W)(0.0277274(C211,11 1, 11 conv conv RQTT R TT Q && Similarly, the inner surface temperatures of the glasses are calculated to be 13.1 and -1.7°C (we had assumed them to be 15 and 5°C when determining the radiation resistance). We can improve the result obtained by reevaluating the radiation resistance and repeating the calculations. PROPRIETARY MATERIAL. © 2011 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. 3-10 3-25 A very thin transparent heating element is attached to the inner surface of an automobile window for defogging purposes, the inside surface temperature of the window is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the window is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 Thermal resistance of the thin heating element is negligible. Properties Thermal conductivity of the window is given to be k = 1.2 W/m · °C. Analysis The thermal resistances are Ah R i i 1 = Ah R o o 1 = and kA LR =win From energy balance and using the thermal resistance concept, the following equation is expressed: o o h i i −∞, RR TT Aq R TT + − =+ ∞ win ,11 & or )/(1)/()/(1 ,11, AhkAL TT Aq Ah TT o o h i i + − =+ − ∞∞ & o o h i i hkL TT q h TT /1//1 ,11, + − =+ − ∞∞ & C) W/m100/1(C) W/m2.1/m 005.0(C W/m15/1 2 +°⋅°⋅ )C 5( W/m1300C 22 121 °−−=+−° TT 2 °⋅ equation: (2 -T_1)/(1/15)+1300=(T_1-(-5))/(0.005/1.2+1/100) Discussion In actuality, the ambient temperature and the convective heat transfer coefficient outside the automobile vary with weather conditions and the automobile speed. To maintain the inner surface temperature of the window, it is necessary to vary the heat flux to the heating element according to the outside condition. Copy the following line and paste on a blank EES screen to solve the above 2 Solving by EES software, the inside surface temperature of the window is C 14.9 °=1T PROPRIETARY MATERIAL. © 2011 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. 3-11 3-26 A process of bonding a transparent film onto a solid plate is taking place inside a heated chamber. The temperatures inside the heated chamber and on the transparent film surface are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 Thermal contact resistance is negligible. Properties The thermal conductivities of the transparent film and the solid plate are given to be 0.05 W/m · °C and 1.2 W/m · °C, respectively. Analysis The thermal resistances are hA R 1conv = Ak L R f f f = and Aks s Using the thermal resistance conc LR s= ept, the llowing e uation is expressed: fo q s b f b R TT RR TT 2 conv − = + −∞ R inside the cham earranging and solving for the temperature ber yields ( ) b f f sss kLR ⎝/ b bf b T k L h TTTRRTTT +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎛ + − =++ − = 12 conv 2 ∞ C 127 °=°+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ °⋅ + °⋅°⋅ °− =∞ C 70C W/m05.0 m 001.0 C W/m70 1 C W/m2.1/m 013.0 C )5270( 2T he surfac temperature of the transparent film is T e s b f b R TT R TT 21 −=− b f f ss b bf s b T k L kL TTTR R TTT +⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛− =+ − = / 22 1 C 103 °=°+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ °⋅°⋅ °− = C 70 C W/m05.0 m 001.0 C W/m2.1/m 013.0 C )5270( 1T Discussion If a thicker transparent film were to be bonded on the solid plate, then the inside temperature of the heated chamber would have to be higher to maintain the temperature of the bond at 70 °C. PROPRIETARY MATERIAL. © 2011 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. 3-12 3-27 A power transistor dissipates 0.15 W of power steadily in a specified environment. The amount of heat dissipated in 24 h, the surface heat flux, and the surface temperature of the resistor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat is transferred uniformly from all surfaces of the transistor. Analysis (a) The amount of heat this transistor dissipates during a 24-hour period is Air, 30°C kWh 0.0036===∆= Wh6.3h) W)(2415.0(tQQ & (b) The heat flux on the surface of the transistor is 2 2 2 m 0001021.0m) m)(0.004 005.0( 4 m) 005.0( 2 4 2 =+= += π π ππ DLDAs Power Transistor 0.15 W 2 W/m1469=== 2m 0001021.0sA q& W15.0Q & (c) The surface temperature of the transistor can be determined from C111.6°= °⋅ +°=+=⎯→⎯−= ∞∞ )m 21C)(0.00010 W/m(18 W15.0C30)( 22 s sss hA QTTTThAQ & & 3-28 A circuit board houses 100 chips, each dissipating 0.06 W. The surface heat flux, the surface temperature of the chips, and the thermal resistance between the surface of the board and the cooling medium are to be determined. 1 Steady operating conditions exist. 2 Assumptions transferred uniformly from the entire front surface. Heat transfer from the back surface of the board is negligible. 2 Heat is nalysis (a The heat flux on the surface of the circuit board is A ) Chips Ts Q& T∞ 2m 0216.0m) m)(0.18 12.0( ==sA 2 W278=== 2m 0216.0sA q& /m× W)06.0100(Q & e sur ce temperature of the chips is (b) Th fa −= ∞ )( ss TThAQ& C67.8°= °⋅ × °=+= ∞ )m 0216.0)(C W/m10( W)06.0100(+C40 22 s s hA QTT & (c) The thermal resistance is C/W4.63°= °⋅ == )m 0216.0(C) W/m10( 11 22 s conv hA R PROPRIETARY MATERIAL. © 2011 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. 3-15 3-32 The roof of a house with a gas furnace consists of a concrete that is losing heat to the outdoors by radiation and convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 14 hour period are to be determined. Assumptions 1 Steady operating conditions exist. 2 The emissivity and thermal conductivity of the roof are constant. Q& Tin=20°C Tsky = 100 K Tair =10°C Properties The thermal conductivity of the concrete is given to be k = 2 W/m⋅°C. The emissivity of both surfaces of the roof is given to be 0.9. L=15 cm Analysis When the surrounding surface temperature is different than the ambient temperature, the thermal resistances network approach becomes cumbersome in problems that involve radiation. Therefore, we will use a different but intuitive approach. In steady operation, heat transfer from the room to the roof (by convection and radiation) must be equal to the heat transfer from the roof to the surroundings (by convection and radiation), that must be equal to the heat transfer through the roof by conduction. That is, rad+conv gs,surroundin toroofcond roof,rad+conv roof, toroom QQQQ &&&& === Taking the inner and outer surface temperatures of the roof to be Ts,in and Ts,out , respectively, the quantities above can be expressed as [ ]4,44282 , 224 , 4 , rad+conv roof, toroom K) 273(K) 27320()K W/m1067.5)(m 300)(9.0( C))(20m C)(300 W/m5()()( +−+⋅×+ °−°⋅=−+−= − ins insinsroominsroomi T TTTATTAhQ σε& m 15.0 )m 300)(C W/m2( ,,2,,cond roof, outsinsoutsins TT L TT kAQ − °⋅= − =& [ ]44,4282m 300)(9.0( + , 2244 ,, rad+conv surr, toroof K) 100(K) 273()K W/m1067.5)( C)10)(m C)(300 W/m12()()( −+⋅× °−°⋅=−+−= − outs outssurroutssurroutso T TTTATTAhQ σε& lving the quations above simultaneously gives he total amount of natural gas consumption during a 14-hour period is So e C1.2and , , out,, °−=°== sins TTQ C7.3 W37,440& T therms36.22 kJ 1080.080.080.0 ⎜ ⎜ ⎝ ===Q totalgas 5,500 therm1)s 360014)(kJ/s 440.37( =⎟⎟ ⎠ ⎞⎛×∆tQQ & inally, the money lost through the roof during that period is F $26.8== )therm/20.1$ therms)(36.22(lostMoney PROPRIETARY MATERIAL. © 2011 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. 3-16 3-33 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves will be determined. Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The furnace operates continuously. 4 The given heat transfer coefficient accounts for the radiation effects. Properties The thermal conductivity of the glass wool insulation is given to be k = 0.038 W/m⋅°C. Analysis The rate of heat transfer without insulation is Insulation L Ts Rinsulation Ro 2m 3m) m)(1.5 2( ==A W2340C)32110)(m 3(C) W/m10()( 22 =°−°⋅=−= ∞TThAQ s& In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be T∞ C/W 333.0C)32110( °=°−=∆=⎯→⎯∆= TRTQ total& W234 W234 W234010.0 =× QR Q total & & and in order to have this thermal resistance, the thickness of insulation must be = cm 3.4== °°⋅ m 034.0 )m C)(3 W/m.038.0()m C)(3 W/m10( 222 L °=+= +=+= C/W 333.01 1 conv L kA L hA RRR insulationtotal oting tha eat is saved at a rate of 0.9×2340 = 2106 W and the furnace operates continuously and thus 365×24 = 8760 h furnace efficiency is 78%, the amount of natural gas saved per year is N t h per year, and that the therms1.807 kJ 105,500 therm1 h 1 s 3600 0.78 h) kJ/s)(8760 106.2(SavedEnergy =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= ∆ = Efficiency tQsaved& he money aved is gy == he insulation will pay for its cost of $250 in T s Ener(savedMoney = year)(per 8.887$)1.10/therm therms)($1.807(energy) oft Saved)(Cos T yr 0.282=== $887.8/yr $250 savedMoney spent Money periodPayback which is equal to 3.4 months. PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-17 ∞TThAQ s& 3-34 An exposed hot surface of an industrial natural gas furnace is to be insulated to reduce the heat loss through that section of the wall by 90 percent. The thickness of the insulation that needs to be used is to be determined. Also, the length of time it will take for the insulation to pay for itself from the energy it saves will be determined. Assumptions 1 Heat transfer through the wall is steady and one-dimensional. 2 Thermal conductivities are constant. 3 The furnace operates continuously. 4 The given heat transfer coefficients accounts for the radiation effects. Properties The thermal conductivity of the expanded perlite insulation is given to be k = 0.052 W/m⋅°C. Analysis The rate of heat transfer without insulation is Insulation L Ts Rinsulation Ro 2m 3m) m)(1.5 2( ==A = W2340C)32110)(m 3(C) W/m10()( 22 =°−°⋅=− In order to reduce heat loss by 90%, the new heat transfer rate and thermal resistance must be T∞ C/W 333.0C)32110( °=°−=∆=⎯→⎯∆= TRTQ total& W234 W234 W234010.0 =× QR Q total & & and in order to have this thermal resistance, the thickness of insulation must be = cm 4.7== °⋅°⋅ m 047.0 )m C)(3 W/m052.0()m C)(3 W/m10( 222 L °=+= +=+= C/W 333.01 1 conv L kA L hA RRR insulationtotal oting tha eat is saved at a rate of 0.9×2340 = 2106 W and the furnace operates continuously and thus 365×24 = 8760 h furnace efficiency is 78%, the amount of natural gas saved per year is N t h per year, and that the therms1.807 kJ 105,500 therm1 h 1 s 3600 0.78 h) kJ/s)(8760 106.2(SavedEnergy =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= ∆ = Efficiency tQsaved& oney gy The m saved is Ener(savedMoney year)(per 8.887$)1.10/therm therms)($1.807(energy) oft Saved)(Cos= == he insulation will pay for its cost of $250 in T yr 0.282=== $887.8/yr $250 savedMoney spent Money periodPayback which is equal to 3.4 months. PROPRIETARY MATERIAL. © 2011 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. 3-20 4th wall with double pane windows: Rwall Rglass Rair Rglass Ri Ro C/W 023197.0000694.0020717.0001786.0 C/W 020717.0 27303.0 1511511 eqv °=⎯→⎯+=+= R R 033382.0 C/W 27303.0267094.0002968.022 C/W 002968.0 m)12.1)(C W/m78.0( m 005.0 C/W 033382.0 m)8.12.1(5)420( C/Wm 31.2 eqvtotal eqv windowwall airglass 22 glass glass 2 2 wall wall °=++=++= °=+×=+= °= ×°⋅ == °= ×−× °⋅ = − == oi RRRR RR RRR kA L R A valueR kA L R 8. C/W 267094.0 m)8.12.1)(C W/m026.0( m 015.0 2o2 air air °= ×⋅ == kA L R window Then W690= ° °− = − = ∞∞ C/W023197.0 C)824( total 21 R TT Q& The rate of heat transfer which will be saved if the single pane windows are converted to double pane windows is The amount of energy and money saved during a 7-month long heating season by switching from single pane to double pane windows become Money savings = (Energy saved)(Unit cost of energy) = (22,851 kWh)($0.08/kWh) = $1828 W45346905224 pane double pane singlesave =−=−= QQQ &&& kWh 22,851=h) 2430kW)(7 534.4( ××=∆= tQQ savesave & PROPRIETARY MATERIAL. © 2011 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. 3-21 ,− outsroomo TTAhQ& room and the refrigerated space can be expressed as 3-37 The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers of sheet metal. The minimum thickness of insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces is to be determined. Assumptions 1 Heat transfer through the refrigerator walls is steady since the temperatures of the food compartment and the kitchen air remain constant at the specified values. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation effects. Properties The thermal conductivities are given to be k = 15.1 W/m⋅°C for sheet metal and 0.035 W/m⋅°C for fiberglass insulation. Analysis The minimum thickness of insulation can be determined by assuming the outer surface temperature of the refrigerator to be 20°C. In steady operation, the rate of heat transfer through the refrigerator wall is constant, and thus heat transfer between the room and the refrigerated space is equal to the heat transfer between the room and the outer surface of the refrigerator. Considering a unit surface area, 1 mm L 1 mm insulation )(= W36=C)2024)(m 1(C) W/m9( 22 °−°⋅= Using the thermal resistance network, heat transfer between the Ri R1 Rins R3 Ro iinsulationmetalo refrigroom total refrigroom hk L k L h TT AQ R TT Q 121 / +⎟ ⎠ ⎞ ⎜ ⎝ ⎛+⎟ ⎠ ⎞ ⎜ ⎝ ⎛+ − = − = & & Troom Trefrig S g,ubstitutin C W/m4 1 C W/m035.0C W/m15.1 m 001.02 C W/m9 1 + C)224( W/m36 2 °−= 2222 °⋅ + °⋅ + °⋅ × °⋅ L Solv ing for L, the minimum thickness of insulation is determined to be L = 0.00875 m = 0.875 cm PROPRIETARY MATERIAL. © 2011 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. 3-22 3-38 Prob. 3-37 is reconsidered. The effects of the thermal conductivities of the insulation material and the sheet metal on the thickness of the insulation is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" k_ins=0.035 [W/m-C] L_metal=0.001 [m] k_metal=15.1 [W/m-C] T_refrig=2 [C] T_kitchen=24 [C] h_i=4 [W/m^2-C] h_o=9 [W/m^2-C] T_s_out=20 [C] "ANALYSIS" A=1 [m^2] “a unit surface area is considered" Q_dot=h_o*A*(T_kitchen-T_s_out) Q_dot=(T_kitchen-T_refrig)/R_total R_total=R_conv_i+2*R_metal+R_ins+R_conv_o R_conv_i=1/(h_i*A) R_metal=L_metal/(k_metal*A) R_ins=(L_ins*Convert(cm, m))/(k_ins*A) "L_ins is in cm" R_conv_o=1/(h_o*A) 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 kins [W/m-C] L i ns [ cm ] kins [W/m.C] Lins [cm] 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.4997 0.6247 0.7496 0.8745 0.9995 1.124 1.249 1.374 1.499 1.624 1.749 1.874 1.999 PROPRIETARY MATERIAL. © 2011 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. 3-25 3-40E A thin copper plate is sandwiched between two layers of epoxy boards. The effective thermal conductivity of the board along its 9 in long side and the fraction of the heat conducted through copper along that side are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one- dimensional since heat transfer from the side surfaces are disregarded 3 Thermal conductivities are constant. Copper Properties The thermal conductivities are given to be k = 223 Btu/h⋅ft⋅°F for copper and 0.15 Btu/h⋅ft⋅°F for epoxy layers. Analysis We take the length in the direction of heat transfer to be L and the width of the board to be w. Then heat conduction along this two-layer plate can be expressed as (we treat the two layers of epoxy as a single layer that is twice as thick) [ ] L Twktkt L TkA L TkA QQQ ∆ +=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∆+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∆= += epoxycopper epoxycopper epoxycopper )()( &&& Epoxy Ts ½ tepoxytcopper½ tepoxy Epoxy Heat conduction along an “equivalent” plate of thick ness t = tcopper + tepoxy and thermal conductivity keff can be expressed as Q LL ⎠⎝ epoxycoppereff board TwttkTkAQ ∆+=⎟⎞⎜⎛ ∆ )(& Setting the two relations above equal to each other and solving for the effective conductivity gives = epoxycopper epoxycopperepoxycopper tteffeff + Note that heat conduction is proportional to kt. Substituting, the epoxycopper )()()()()( ktkt kktktttk + =⎯→⎯+=+ fraction of heat conducted along the copper layer and the effective th rmal conductivity of the plate are determined to be and e FBtu/h. 93292.000375.09292.0)()()( FBtu/h. 00375.0ft) F)(0.15/12Btu/h.ft. 15.0(2)( FBtu/h. 9292.0ft) F)(0.05/12Btu/h.ft. 223()( epoxycoppertotal epoxy copper °=+=+= °=°= °=°= ktktkt kt kt F.Btu/h.ft 32.0 2 °= + ° = + + = ft )]12/15.0(2)12/05.0[( FBtu/h. 93292.0 t )()( epoxycopper epoxycopper t ktkt keff 99.6%==== 996.0 93292.0 9292.0 )( )( total copper copper kt kt f PROPRIETARY MATERIAL. © 2011 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. 3-26 3-41 Warm air blowing over the inner surface of an automobile windshield is used for defrosting ice accumulated on the outer surface. The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the windshield is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. 5 The automobile is operating at 1 atm. Properties Thermal conductivity of the windshield is given to be k = 1.4 W/m · °C. Analysis The thermal resistances are Ah R i i 1 = Ah R o o 1 = and kAwin From energy balance and using the thermal res LR = istance oncept, th following equation is expressed: c e i i o o RR TT R TT + − = − ∞∞ win ,11, win 1,o∞ or ,1 RR TT TT R o i − − − = ∞ i k LTT i −⎟ ⎞ ⎜ ⎛− = ∞ 11 ,1 hTTh oo ⎟⎠⎜⎝−∞ 1, ection i For the ice to begin melting, the outer surface temperature of the windshield ( 1T ) should be at least 0 °C. The conv heat transfer coefficient for the warm air is C W/m112 2 °⋅= ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ °⋅ −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ °⋅°−− °− = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = − −1 ∞ ∞ 1 2 1, ,1 C W/m4.1 m 005.0 C W/m200 1 C )010( C )250( 1 k L hTT TT h oo i i Discussion In practical situations, the ambient temperature and the convective heat transfer coefficient outside the automobile vary with weather conditions and the automobile speed. Therefore the convection heat transfer coefficient of the warm air necessary to melt the ice should be varied as well. This is done by adjusting the warm air flow rate and temperature. PROPRIETARY MATERIAL. © 2011 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. 3-27 3-42 The thermal contact conductance for an aluminum plate attached on a copper plate, that is heated electrically, is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 Heat transfer by radiation is negligible. Properties The thermal conductivity of the aluminum plate is given to be 235 W/m · °C. Analysis The thermal resistances are kA LR =cond and hA From energy balance and using the thermal res R 1= istance concept, the following equation is expressed: conv convcond 1 elec / RRAR TT Aq c ++ − = ∞& or )/(1)/(/ 1 elec hAkALAR TT Aq ++ − = ∞& c Rearranging the equation and solving for the contact resistance yields C/Wm102586 CW/m67 1 C W/m235 025.0C )20100( 2 − °− = m W/m5300 1 25 2 elec 1 °⋅×= °⋅ − °⋅ −− − = − ∞ . hk L q TT Rc & herma contact conductance is Discussion By comparing the value of the thermal contact conductance with the values listed in Table 3-2, the surface conditions of the plates appear to be milled. The t l C W/m16000 2 °⋅== cc Rh /1 PROPRIETARY MATERIAL. © 2011 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. 3-30 3-51 Two cylindrical aluminum bars with ground surfaces are pressed against each other in an insulation sleeve. For specified top and bottom surface temperatures, the rate of heat transfer along the cylinders and the temperature drop at the interface are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional in the axial direction since the lateral surfaces of both cylinders are well-insulated. 3 Thermal conductivities are constant. Bar Bar Interface Properties The thermal conductivity of aluminum bars is given to be k = 176 W/m⋅°C. The contact conductance at the interface of aluminum-aluminum plates for the case of ground surfaces and of 20 atm ≈ 2 MPa pressure is hc = 11,400 W/m2⋅°C (Table 3-2). Ri Rglass RoAnalysis (a) The thermal resistance network in this case consists of two conduction resistance and the contact resistance, and they are determined to be T1 T2 C/W 0447.0 /4]m) (0.05C)[ W/m400,11( 11 22 c contact °= °⋅ == πcAh R C/W 4341.0 /4]m) (0.05C)[ W/m(176 m 15.0 plate == LR 2 °= °⋅ πkA hen the rate of heat transfer is determined to be T W142.4= ° ×++ )4341.020447.0(2 barcontacttotal RRR °− = ∆ = ∆ = C)20150(TTQ& C/W e rate of heat transfer through the bars is 142.4 W. ) The temperature drop at the interface is determined to be Therefore, th (b C6.4°=°==∆ C/W) W)(0.04474.142(contactinterface RQT & PROPRIETARY MATERIAL. © 2011 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. 3-31 Generalized Thermal Resistance Networks 3-52C Two approaches used in development of the thermal resistance network in the x-direction for multi-dimensional problems are (1) to assume any plane wall normal to the x-axis to be isothermal and (2) to assume any plane parallel to the x- axis to be adiabatic. 3-53C The thermal resistance network approach will give adequate results for multi-dimensional heat transfer problems if heat transfer occurs predominantly in one direction. 3-54C Parallel resistances indicate simultaneous heat transfer (such as convection and radiation on a surface). Series resistances indicate sequential heat transfer (such as two homogeneous layers of a wall). 3-55 A typical section of a building wall is considered. The average heat flux through the wall is to be determined. Assumptions 1 Steady operating conditions exist. Properties The thermal conductivities are given to be k23b = 50 W/m⋅K, k23a = 0.03 W/m⋅K, k12 = 0.5 W/m⋅K, k34 = 1.0 W/m⋅K. Analysis We consider 1 m2 of wall area. The thermal resistances are C/Wm 1.0 C) W/m0.1( m 1.0 C/Wm 1032.1 0.005)C)(0.6 W/m50( m 005.0m) 08.0( )( C/Wm 645.2 0.005)C)(0.6 W/m03.0( m 6.0m) 08.0( )( C/Wm 02.0 C) W/m5.0( m 01.0 2 34 34 34 25 23b b 2323 2 23a a 2323 2 12 12 12 °⋅= °⋅ == °⋅×= +°⋅ = + = °⋅= +°⋅ = + = °⋅= °⋅ == − k t R LLk L tR LLk L tR k t R ba b ba a The total thermal resistance and the rate of heat transfer are C/Wm 120.01.0 1032.1645.2 1032.1645.202.0 2 5 5 34 2323 2323 12total °⋅=+⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ×+ × += +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + += − − R RR RR RR ba ba 2 W/m125= ⋅ °− = − = C/Wm 0.120 C)2035( 2 total 14 R TT q& PROPRIETARY MATERIAL. © 2011 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. 3-32 3-56 A wall consists of horizontal bricks separated by plaster layers. There are also plaster layers on each side of the wall, and a rigid foam on the inner side of the wall. The rate of heat transfer through the wall is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer by radiation is disregarded. Properties The thermal conductivities are given to be k = 0.72 W/m⋅°C for bricks, k = 0.22 W/m⋅°C for plaster layers, and k = 0.026 W/m⋅°C for the rigid foam. Analysis We consider 1 m deep and 0.28 m high portion of wall which is representative of the entire wall. The thermal resistance network and individual resistances are R7R6 R5 R4 R3 R2 R1Ri T∞1 T∞2 C/W 737.4179.0804.0)325.0(2747.2357.02 C/W 804.0 45.45 1 833.0 1 45.45 11111 C/W 179.0 )m 128.0(C) W/m20( 11 C/W 833.0 )m 125.0(C) W/m72.0( m 15.0 C/W45.45 )m 1015.0(C) W/m22.0( m 15.0 C/W 325.0 )m 128.0(C) W/m22.0( m 02.0 ==== LRRR C/W 747.2 )m 128.0(C) W/m026.0( C/W 357.0 )m 1 11 21 543 2 2 2,o 24 253 262 21 221, °=++++=++++= °=⎯→⎯++=++= °= ×°⋅ === °= ×°⋅ === °= ×°⋅ ==== °= ×°⋅ °= ×°⋅ === °= × === omiditotal mid mid conv brick ocenter plaster side plaster foam convi RRRRRR R RRRR Ah RR kA LRR Ah LRRR kA kA RR RR The steady rate of heat transfer through the wall per is m 02.0 28.0(C) W/m10(1 °⋅ L Ah 2m 28.0 W49.5 C/W737.4 C)]4(22[( 21 = ° °−− = − = ∞∞ totalR TT Q& Then steady rate of heat transfer through the entire wall becomes W470=×= 2 2 m 28.0 m)64( W)49.5(totalQ& PROPRIETARY MATERIAL. © 2011 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. 3-35 3-59 A wall is to be constructed of 10-cm thick wood studs or with pairs of 5-cm thick wood studs nailed to each other. The rate of heat transfer through the solid stud and through a stud pair nailed to each other, as well as the effective conductivity of the nailed stud pair are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer can be approximated as being one-dimensional since it is predominantly in the x direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance between the two layers is negligible. 4 Heat transfer by radiation is disregarded. Properties The thermal conductivities are given to be k = 0.11 W/m⋅°C for wood studs and k = 50 W/m⋅°C for manganese steel nails. Analysis (a) The heat transfer area of the stud is A = (0.1 m)(2.5 m) = 0.25 m2. The thermal resistance and heat transfer rate through the solid stud are W2.2= ° ° = ∆ = °= °⋅ )m 25.0(C) W/m11.0( 2kA == C/W 636.3 C8 C/W 636.3m 1.0 studR TQ LR & (b) The thermal resistances of stud pair and nails are in parallel Stud stud T1 T2 L Q& W4.7= ° ° = ∆ = °=⎯→⎯+=+= °= −°⋅ == °= °⋅ == = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ == C/W 70.1 C8 C/W 70.1 18.365.3nailsstudtotal RRR 11111 C/W 65.3 )m 000628.025.0(C) W/m11.0( m 1.0 C/W 18.3 )m 000628.0(C) W/m50( m 1.0 m 000628.0 4 m) 004.0(50 4 50 2 2 2 22 stud total stud nails nails R TQ R kA LR kA LR DA & ππ c) The effective conductivity of the nailed stud pair can be determined from Rstud T1 T2 ( C W/m.0.235 °= ° = ∆ =⎯→⎯ ∆ = )m C)(0.258( m) 1.0 W)(7.4( 2TA LQk L TAkQ effeff & & PROPRIETARY MATERIAL. © 2011 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. 3-36 3-60E A wall is to be constructed using solid bricks or identical size bricks with 9 square air holes. There is a 0.5 in thick sheetrock layer between two adjacent bricks on all four sides, and on both sides of the wall. The rates of heat transfer through the wall constructed of solid bricks and of bricks with air holes are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer. Properties The thermal conductivities are given to be k = 0.40 Btu/h⋅ft⋅°F for bricks, k = 0.015 Btu/h⋅ft⋅°F for air, and k = 0.10 Btu/h⋅ft⋅°F for sheetrock. Analysis (a) The representative surface area is . The thermal resistance network and the individual thermal resistances if the wall is constructed of solid bricks are 2ft 3906.0)12/5.7)(12/5.7( ==A RoR5 R4 R3 R2 R1Ri T∞1 T∞2 F/Btuh 57.308 ft)]12/5.0()12/7(F)[ftBtu/h 10.0( ft 12/9 F/Btuh 288 ft)]12/5.0()12/5.7(F)[ftBtu/h 10.0( ft 12/9 F/Btuh 0667.1 )ft 3906.0(F)ftBtu/h 10.0( ft 12/5.0 F/Btuh 7068.1 )ft 3906.0(F)ftBtu/h 5.1( 11 2o3 22 251 22 °⋅= ×⋅⋅ === °⋅= ×°⋅⋅ === °⋅= °⋅⋅ ==== °= °⋅⋅ == kA LRR kA LRR kA LRRR Ah R plaster plaster plaster i i Btu/h 6971.4 F/Btuh 5804.9 F)3580( F/Btuh 5804.94267.00667.13135.50667.17068.1 F/Btuh 3135.5 51.557.308288432 °⋅=⎯→⎯++=++= mid mid R RRRR 111111 F/Btuh )ft 3906.0(F)ftBtu/h 6( F/Btuh 51.5 ft)]12/7()12/7(F)[ftBtu/h 40.0( ft 12/9 21 51 24 = °⋅ °− = − = °⋅=++++=++++= °⋅ °⋅⋅ °⋅= ×°⋅⋅ === ∞∞ total omiditotal o brick R TT Q RRRRRR Ah kA LRR & en ste y of heat transfer through entire wall becomes 4267.011 22 ===oR 1 Th ad rate Btu/h 3610== 2m 3906.0 ft) 10(ft) 30(Btu/h) 6971.4(totalQ& (b a ) The thermal resistance network and the individual thermal resistances if the wall is constructed of bricks with air holes re ft 1406.0)12/5.1()12/5.1(9 =×=airholesA T∞1 Ri R1 R2 R3 R4 R5 R6 Ro T∞2 22 ft 1997.01406.0ft) 12/7( =−=bricksA 2 PROPRIETARY MATERIAL. © 2011 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. 3-37 F/Btuh 389.9 )ft F)(0.1997ftBtu/h 40.0( ft 12/9 F/Btuh 62.355 )ft F)(0.1406ftBtu/h 015.0( ft 12/9 25 24 °⋅= °⋅⋅ === °⋅= °⋅⋅ === kA LRR kA LRR brick airholes Btu/h 492.3 F/Btuh 885.12 F)3580( F/Btuh 885.124267.00667.1618.80667.17068.1 F/Btuh 618.8 389.9 1 62.355 1 57.308 1 288 111111 21 61 5432 = °⋅ °− = − = °⋅=++++=++++= °⋅=⎯→⎯+++=+++= ∞∞ total omiditotal mid mid R TT Q RRRRRR R RRRRR & Then steady rate of heat transfer through entire wall becomes Btu/h 2680== 2ft 3906.0 ft) 10(ft) 30(Btu/h) 492.3(totalQ& PROPRIETARY MATERIAL. © 2011 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. 3-40 3-63 A coat is made of 5 layers of 0.15 mm thick synthetic fabric separated by 1.5 mm thick air space. The rate of heat loss through the jacket is to be determined, and the result is to be compared to the heat loss through a jackets without the air space. Also, the equivalent thickness of a wool coat is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the jacket is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer. Properties The thermal conductivities are given to be k = 0.13 W/m⋅°C for synthetic fabric, k = 0.026 W/m⋅°C for air, and k = 0.035 W/m⋅°C for wool fabric. Analysis The thermal resistance network and the individual thermal resistances are R1 R2 R3 R4 R5 R6 R7 R8 R9 Ro Ts1 T∞2 C/W 2214.00320.00462.040009.0545 C/W 0320.0 )m 25.1(C) W/m25( 11 C/W 0462.0 )m 25.1(C) W/m026.0( m 0015.0 C/W 0009.0 )m 25.1(C) W/m13.0( m 00015.0 22 28642 297531 °=+×+×=++= °= °⋅ == °= °⋅ ====== °= °⋅ ======= oairfabrictotal o air fabric RRRR hA R kA LRRRRR kA LRRRRRR and W113= °C/W 2214.0totalR °− = − ∞ C)025(21s TTQ& If the jacket is made of a single layer of 0.75 mm thick synthetic fabric, the rate of heat transfer would be = W685= °+×+× C/W )0320.00009.05(5 ofabrictotal RRR °− = − = − = ∞∞ C)025(2121 ss TTTTQ& he thickn s of a wool fabric that has the same thermal resistance is determined from T es mm 8.29==⎯→⎯+ °⋅ =° +=+= m 00829.00320.0 )m 25.1(C) W/m035.0( C/W 2214.0 1 2 fabric wooltotal LL hAkA LRRR o PROPRIETARY MATERIAL. © 2011 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. 3-41 3-64 A coat is made of 5 layers of 0.15 mm thick cotton fabric separated by 1.5 mm thick air space. The rate of heat loss through the jacket is to be determined, and the result is to be compared to the heat loss through a jackets without the air space. Also, the equivalent thickness of a wool coat is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the jacket is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer. Properties The thermal conductivities are given to be k = 0.06 W/m⋅°C for cotton fabric, k = 0.026 W/m⋅°C for air, and k = 0.035 W/m⋅°C for wool fabric. Analysis The thermal resistance network and the individual thermal resistances are R1 R2 R3 R4 R5 R6 R7 R8 R9 Ro T1 T∞2 C/W 2268.00320.00462.04002.0545 C/W 0320.0 )m 25.1(C) W/m25( 11 C/W 0462.0 )m 25.1(C) W/m026.0( m 0015.0 C/W 002.0 )m 25.1(C) W/m06.0( m 00015.0 22 2o8642 297531cot °=+×+×=++= °= °⋅ == °= ⋅ ====== °= °⋅ ======= oairfabrictotal o air ton RRRR hA R kA LRRRRR kA LRRRRRR and W110= °C/W 2268.0totalR °− = − ∞ C)025(21s TTQ& If the jacket is made of a single layer of 0.75 mm thick cotton fabric, the rate of heat transfer will be = W595= °+×+× C/W )0320.0002.05(5 ofabrictotal RRR °− = − = − = ∞∞ C)025(2121 ss TTTTQ& he thickn s of a wool fabric for that case can be determined from T es mm 8.52==⎯→⎯+ °⋅ =° +=+= m 00852.00320.0 )m 25.1(C) W/m035.0( C/W 2268.0 1 2 fabric wooltotal LL hAkA LRRR o PROPRIETARY MATERIAL. © 2011 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. 3-42 3-65 In an experiment, the convection heat transfer coefficients of (a) air and (b) water flowing over the metal foil are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional. 3 Thermal properties are constant. 4 Thermal resistance of the thin metal foil is negligible. Properties Thermal conductivity of the slab is given to be k = 0.023 W/m · K and the emissivity of the metal foil is 0.02. Analysis The thermal resistances are kA LR =cond hA R 1conv = and Ah R rad rad 1 = From energy balance and using the thermal resistance concept, the following equation is expressed: cond 21 elec rad 1surr conv 1 R TT Aq R TT R T−∞T −=+ − + & or 1 elec rad 1surr cond 21 conv 11 TT Aq R TT R TT R −⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − − − = ∞ & 1 elec 1surr21 1q TTTT h ⎟ ⎞ ⎜ ⎛ − − − − = & rad/1/ TThkL −⎟⎠⎜⎝ ∞ ) For air flowing over the metal foil, the radiation heat transfer coefficient is ))(( 2 surr 2 surr 2 rad ⋅= ++= TTTTh ssεσ (a K )293423(K )293423)(K W/m1067.5)(02.0( 222428 ++⋅×= − K W/m215.0 The convection heat transfer coefficient for air flowing over the metal foil is K W/m37.3 2 ⋅= −⎥⎦ ⎤ ⎢⎣ ⎡ − ⋅ − − ⋅ − = K )15020( 1 W/m5000 W/m215.0/1 K )15020( K W/m023.0/m 025.0 K )20150( 2 2 K h (b) For water flowing over the metal foil, the radiation heat transfer coefficient is ))(( K W/m1201.0 K )293303(K )293303)(K W/m1067.5)(02.0( 2 ⋅= ++⋅×= The convection heat transfer coefficient for water flowing over the metal foil is 222428 surr 2 surr 2 ++= − TTTTh ssεσ rad K W/m499 2 ⋅= −⎥⎦ ⎤ ⎢⎣ ⎡ − ⋅ − − ⋅ − = K )3020( 1 W/m5000 W/m1201.0/1 K )3020( K W/m023.0/m 025.0 K )2030( 2 2 K h Discussion If heat transfer by conduction through the slab and radiation on the metal foil surface is neglected, the convection heat transfer coefficient for the case with air flow would deviate by 3.2% from the result in part (a), while the convection heat transfer coefficient for the case with water flow would deviate by 0.2% from the result in part (b). PROPRIETARY MATERIAL. © 2011 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. 3-45 ho [W/m2.C] Qtotal [W] 5 54834 10 70939 15 78670 20 83212 25 86201 30 88318 35 89895 40 91116 45 92089 50 92882 5 10 15 20 25 30 35 40 45 50 50000 55000 60000 65000 70000 75000 80000 85000 90000 95000 ho [W/m 2-C] Q to ta l [W ] PROPRIETARY MATERIAL. © 2011 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. 3-46 3-68E The thermal resistance of an epoxy glass laminate across its thickness is to be reduced by planting cylindrical copper fillings throughout. The thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the plate is one-dimensional. 3 Thermal conductivities are constant. Properties The thermal conductivities are given to be k = 0.10 Btu/h⋅ft⋅°F for epoxy glass laminate and k = 223 Btu/h⋅ft⋅°F for copper fillings. Analysis The thermal resistances of copper fillings and the epoxy board are in parallel. The number of copper fillings in the board and the area they comprise are 2 2 22 2 2 ft 7606.007272.08333.0 ft 07272.0 4 ft) 12/02.0(333,33 4 fillings)copper of(number 333,33 ft) 12/06.0(ft) 12/06.0( ft 8333.0 ft 8333.0ft) 12/12(ft) 12/10( =−=−= === == == coppertotalepoxy copper copper total AAA DnA n A ππ Rcopper Repoxy The thermal resistances are evaluated to be F/Btuh 0548.0 )ft 7606.0(F)ftBtu/h 10.0( ft 12/05.0 F/Btuh 000257.0 )ft 07272.0(F)ftBtu/h 223( 2°⋅⋅kAcopper ft 12/05.0 2 °⋅= °⋅⋅ == °⋅=== kA LR LR epoxy hen the thermal resistance of the entire epoxy board becomes T F/Btuh 0.000256 °⋅=⎯→⎯+=+= board epoxycopperboard R RRR 0548.0 1 000257.0 1111 PROPRIETARY MATERIAL. © 2011 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. 3-47 Heat Conduction in Cylinders and Spheres 3-69C When the diameter of cylinder is very small compared to its length, it can be treated as an infinitely long cylinder. Cylindrical rods can also be treated as being infinitely long when dealing with heat transfer at locations far from the top or bottom surfaces. However, it is not proper to use this model when finding temperatures near the bottom and the top of the cylinder. 3-70C No. In steady-operation the temperature of a solid cylinder or sphere does not change in radial direction (unless there is heat generation). 3-71C Heat transfer in this short cylinder is one-dimensional since there will be no heat transfer in the axial and tangential directions. PROPRIETARY MATERIAL. © 2011 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. 3-50 3-74 Prob. 3-73 is reconsidered. The effect of the thickness of the insulation on the rate of heat loss from the steam and the temperature drop across the insulation layer are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_infinity_1=280 [C] T_infinity_2=5 [C] k_steel=15 [W/m-C] D_i=0.05 [m] D_o=0.055 [m] r_1=D_i/2 r_2=D_o/2 t_ins=3 [cm] k_ins=0.038 [W/m-C] h_o=22 [W/m^2-C] h_i=80 [W/m^2-C] L=1 [m] "ANALYSIS" A_i=pi*D_i*L A_o=pi*(D_o+2*t_ins*Convert(cm, m))*L R_conv_i=1/(h_i*A_i) R_pipe=ln(r_2/r_1)/(2*pi*k_steel*L) R_ins=ln(r_3/r_2)/(2*pi*k_ins*L) r_3=r_2+t_ins*Convert(cm, m) "t_ins is in cm" R_conv_o=1/(h_o*A_o) R_total=R_conv_i+R_pipe+R_ins+R_conv_o Q_dot=(T_infinity_1-T_infinity_2)/R_total DELTAT_pipe=Q_dot*R_pipe DELTAT_ins=Q_dot*R_ins Tins [cm] Q [W] ∆Tins [C] 1 2 3 4 5 6 7 8 9 10 174.9 109 83.44 69.64 60.93 54.88 50.41 46.95 44.18 41.91 227.2 249.6 257.8 261.9 264.4 266 267.2 268.1 268.7 269.2 1 2 3 4 5 6 7 8 9 10 40 60 80 100 120 140 160 180 220 230 240 250 260 270 tins [cm] Q [ W ] ∆ T i ns [ C ] PROPRIETARY MATERIAL. © 2011 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. 3-51 3-75 A 50-m long section of a steam pipe passes through an open space at 15°C. The rate of heat loss from the steam pipe, the annual cost of this heat loss, and the thickness of fiberglass insulation needed to save 90 percent of the heat lost are sulation. 6 The combined heat transfer coefficient on the outer surface remains on is given to be k = 0.035 W/m⋅°C. lysis (a) The rate of heat loss from the steam /m 22 °−°⋅ ) The amount of heat loss per year is he amount of gas consumption from the natural gas furnace that has an efficiency of 75% is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one- dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal conductivity is constant. 4 The thermal contact resistance at the interface is negligible. 5 The pipe temperature remains constant at about 150°C with or without in constant even after the pipe is insulated. Properties The thermal conductivity of fiberglass insulati Ana pipe is 2m 71.15m) 50(m) 1.0( === ππDLAo W20()( =−= airsobare TTAhQ& W42,412=C)15150)(m 71.15(C) (b kJ/yr 10337.1s/yr) 360024kJ/s)(365 412.42( 9×=××=∆= tQQ & T therms/yr903,16 kJ 105,500 therm1kJ/yr 10337.1 9 ⎜⎜ ⎛× =gasQ 75.0 =⎟⎟ ⎠ ⎞ ⎝ lost is therm)/52.0($) therms/yr(16,903= (c) In order to save 90% of the heat loss and thus to reduce it to 0.1×42,412 = 4241 W, the thickness of insulation needed is The annual cost of this energy = energy) ofcost used)(UnitEnergy (costEnergy $8790/yr= determined from kL rr Ah TT RR TT Q oo airs insulationo airs insulated π2 )/ln(1 12+ − = + − =& Ro Tair Rinsulation Ts Substituting and solving for r2, we get m 0692.0 )m 50(C) W/m035.0(2 )05.0/ln( )]m 50(2(C)[ W/m20( 1 C)15150( W4241 2 2 2 2 =⎯→⎯ °⋅ + °⋅ °− = r r r ππ Then the thickness of insulation becomes cm 1.92=−=−= 592.612 rrtinsulation PROPRIETARY MATERIAL. © 2011 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. 3-52 3-76 An electric hot water tank is made of two concentric cylindrical metal sheets with foam insulation in between. The fraction of the hot water cost that is due to the heat loss from the tank and the payback period of the do-it-yourself insulation kit are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one- dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal resistances of the water tank and the outer thin sheet metal shell are negligible. 5 Heat loss from the top and bottom surfaces is negligible. Properties The thermal conductivities are given to be k = 0.03 W/m⋅°C for foam insulation and k = 0.035 W/m⋅°C for fiber glass insulation Analysis We consider only the side surfaces of the water heater for simplicity, and disregard the top and bottom surfaces (it will make difference of about 10 percent). The individual thermal resistances are 2m 885.1m) 5.1(m) 40.0( === ππ LDA ii C/W 0106.0 )m 885.1(C). W/m50( 11 22 °=° == ii i Ah R Tw Ro T∞2 RfoamRi 2m 168.2m) 5.1(m) 46.0( == =o ππ LDA o C/W 0384.0 )m 168.2(C). W/m12( 11 22 °=° == oo o Ah R C/W 5433.04943.00384.00106.0 C/W 4943.0 )m 5.1(C) W/m03.0(22 2 °⋅foam kL ππ )20/23ln()/ln( 12 °=++=++= °=== foamoi RRRR rrR total The rate of heat loss from the hot water tank is W74.60 C/W 0.5433 C)2760(2 = ° °− = − = ∞ total w R TTQ& The amount and cost of heat loss per year are 5h/yr) 24kW)(365 06074.0( =×=∆= tQQ & kWh/yr 1.32 ofost 15.2%=== == 152.0 280$ 57.42$ 57.42$kWh)/08.0($kWh) 1.(532=cost)it energy)(Un ofAmount (Energy C f If 3 cm thick fiber glass insulation is used to wrap the entire tank, the individual resistances becomes 2m 450.2m) 5.1(m) 52.0( === ππ LDA oo Tw Rfiberglass Ro T∞2 RfoamRi C/W0340.0 )m 450.2(C) W/m12( 2o2 ⋅oo o Ah 11 °===R C/W 9106.03717.04943.00340.00106.0 C/W 3717.0 )m 5.1(C) W/m035.0(22 22 °= °⋅ ==fiberglass Lk R ππ )23/26 C/W 4943.0 )m 5.1(C) W/m03.0(2 )20/23ln( 2 )/ln( 2 1 12 °=++=+++= °= °⋅ = fiberglassfoamoitotal RRRRR Lk rr ππ The rate of heat loss from the hot water heater in this case is =foamR ln()/ln( 23 rr + W24.36 C/W 0.9106 C)2760(2 = ° °− = − = ∞ total w R TT Q& The energy saving is saving = 60.74 - 36.24 = 24.5 W The time necessary for this additional insulation to pay for its cost of $30 is then determined to be months 21≈==⎯→⎯ == days 638hours 306,15 period Time 30$kWh)/08.0$period)( kW)(Time 0245.0(Cost PROPRIETARY MATERIAL. © 2011 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. 3-55 3-79 Steam flows in a steel pipe, which is insulated by gypsum plaster. The rate of heat transfer from the steam and the temperature on the outside surface of the insulation are be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one- dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. Properties (a) The thermal conductivities of steel and gypsum plaster are given to be 50 and 0.5 W/m⋅°C, respectively. Steam L Insulation Analysis The thermal resistances are Ri Ti Rins To Rsteel Ro C/W0004974.0 m) m)(20 (0.16C) W/m200( 11 C/W011032.0 m) C)(20 W/m5.0(22 ins ins °⋅ππ Lk )8/16ln()/ln( C/W0000458.0 m) C)(20 W/m50(2 )6/8ln( 2 )/ln( C/W0003316.011 2 23 steel 12 steel °= °⋅ == °=== °= °⋅ == °=== π ππ oo o Ah R DD R Lk DD R he total thermal resistance and the rate of heat transfer are R m) m)(20 (0.06C) W/m800( 2 °⋅ πii i Ah T C/W011907.00004974.0011032.00000458.00003316.0inssteeltotal °=+++=+++= oi RRRRR W15,957= ⋅ °− = − = C/Wm 0.011907 C)10200( 2 totalR TT Q oi& (b) The temperature at the outer surface of the insulation is determined from C17.9°=⎯→⎯ °⋅ °− =⎯→⎯ − = s s o os T T R TT Q C/Wm 0.0004974 C)10( W957,15 2 & PROPRIETARY MATERIAL. © 2011 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. 3-56 ft 916.0ft) 1(ft) 12/5.3( = === ππ LDA i The individual resistances are 3-80E A steam pipe covered with 2-in thick fiberglass insulation is subjected to convection on its surfaces. The rate of heat loss from the steam per unit length and the error involved in neglecting the thermal resistance of the steel pipe in calculations are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one- dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. Properties The thermal conductivities are given to be k = 8.7 Btu/h⋅ft⋅°F for steel and k = 0.020 Btu/h⋅ft⋅°F for fiberglass insulation. Analysis The inner and outer surface areas of the insulated pipe are 2 2 ft 094.2ft) 1(ft) Ri T∞1 Rinsulation Ro T∞2 Rpipei 12/8(== ππ LDA oo F/Btuh 65.5096.0516.5002.0036.021 ⋅=+++=+++= oitotal RRRRR F/Btuh 096.0 )ft 094.2(F).Btu/h.ft 5( 11 F/Btuh 516.5 )ft 1(F)Btu/h.ft. 020.0(2 )2/4ln( 2 )/ln( F/Btuh 002.0 )ft 1(F)Btu/h.ft. 7.8(2 )75.1/2ln( 2 )/ln( F/Btuh 036.0 )ft 916.0(F).Btu/h.ft 30( 11 2o2 2 23 2 1 12 1 22 ° °⋅=== °⋅= ° === °⋅= ° === °⋅= ° == oo o insulation pipe ii i Ah R Lk rr RR Lk rr RR Ah R ππ ππ hen the st ady rate of heat loss from the steam per ft. pipe length becomes T e Btu/h 69.91=°−=−= ∞∞ F)55450(21 TTQ& °⋅ F/Btuh 5.65totalR alue of total thermal resistance will be If the thermal resistance of the steel pipe is neglected, the new v F/Btuh 648.5096.0516.5036.02 °=++=++= oitotal RRRR or involved in calculations becomes Then the percentage err 0.035%=× ° °− = 100 F/Btuh 65.5 F/Btuh)648.565.5(%error which is insignificant. PROPRIETARY MATERIAL. © 2011 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. 3-57 3-81 Hot water is flowing through a 15-m section of a cast iron pipe. The pipe is exposed to cold air and surfaces in the basement. The rate of heat loss from the hot water and the average velocity of the water in the pipe as it passes through the basement are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one- dimensional since there is thermal symmetry about the center line and no variation in the axial direction. 3 Thermal properties are constant. Properties The thermal conductivity and emissivity of cast iron are given to be k = 52 W/m⋅°C and ε = 0.7. Analysis The individual resistances are 2 2 m 168.2m) 15(m) 046.0( m 885.1m) 15(m) 04.0( === === ππ ππ LDA LDA oo ii C/W 00003.0 )2/3.2ln()/ln( )m 885.1(C). W/m120( 12 22 °=== ° rr R Ah ii )m 15(C) W/m.52(22 C/W 00442.011 1 ° °=== ππ Lk R pipe i The outer surface temperature of the pipe will be somewhat below the water temperature. Assuming the outer surface temperature of the pipe to be 80°C (we will check this assumption later), the radiation heat tran efficient is determined to be 222428 =+×= − mperature, the radiation and convection heat transfer coefficients can be added and the result can be taken as the combined heat transfer coefficient. Then, Ri Rpipe Ro T∞1 T∞2 sfer co .K W/m167.5283)+353]()K 283()K 353)[(.K W/m1067.5)(7.0( ))(( 2 22 2 ++= surrsurrrad TTTTh εσ Since the surrounding medium and surfaces are at the same te C/W 02732.002287.000003.000442.0 C/W 02287.0 )m 168.2(C). W/m17.20( 11 22 °=° ==o Ah R C. W/m.2015167.5 22, °=++=++= °=+=+= opipeitotal ocombined convradcombined RRRR hhh es 17 The rate of heat loss from the hot water pipe then becom W2928= ° °− = − = ∞∞ C/W 0.02732 C)1090(21 totalR TTQ& For a temperature drop of 3°C, the mass flow rate of water and the average velocity of water must be m/s 0.186===⎯→⎯= = °° = ∆ =⎯→⎯∆= 4 m) 04.0()kg/m 1000( kg/s 2335.0 kg/s 2335.0 C) C)(3J/kg. (4180 J/s 2928 2 3 πρ ρ c c p p A mVVAm Tc QmTcmQ & & & &&& Discussion The outer surface temperature of the pipe is C77.0= C/W0.00003)+(0.00442 C)90( W928 2 1 °→ ° °− =→ + − = ∞ s s pipei s TT RR TTQ& which is close to the value assumed for the surface temperature in the evaluation of the radiation resistance. Therefore, there is no need to repeat the calculations. PROPRIETARY MATERIAL. © 2011 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. 3-60 3-84E Steam exiting the turbine of a steam power plant at 100°F is to be condensed in a large condenser by cooling water flowing through copper tubes. For specified heat transfer coefficients and 0.01-in thick scale build up on the inner surface, the length of the tube required to condense steam at a rate of 120 lbm/h is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one- dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 Heat transfer coefficients are constant and uniform over the surfaces. Properties The thermal conductivities are given to be k = 223 Btu/h⋅ft⋅°F for copper tube and be k = 0.5 Btu/h⋅ft⋅°F for the mineral deposit. The heat of vaporization of water at 100°F is given to be 1037 Btu/lbm. Analysis When a 0.01-in thick layer of deposit forms on the inner surface of the pipe, the inner diameter of the pipe will reduce from 0.4 in to 0.38 in. The individual thermal resistances are 2 2 ft 157.0ft) 1(ft) 12/6.0( ft 099.0ft) 1(ft) 12/38.0( === === ππ ππ LDA LDA oo ii Ri T∞1 Rpipr Ro T∞2 Rdeposit F/Btuh 3095.000425.001633.000029.02886.0 F/Btuh 00425.0 )ft 157.0(F).Btu/h.ft 1500( 11 F/Btuh. 01633.0 )ft 1(F)Btu/h.ft. 5.0(22 2 1 °= ° == depdeposit Lk R ππ )19.0/2.0ln()/ln( F/Btuh 00029.0 f 1(F)Btu/h.ft. 223(2 )2/3ln( 2 )/ln( F/Btuh 2886.0 )ft 9 11 22 12 2 °=+++=+++= °= ° == °= ° == °=== odepositpipeitotal oo o pipe RRRRR Ah R rr kL rr R R ππ The heat transfer rate per ft length of the tube is 09.0(F).Btu/h.ft 35( 2 °ii i Ah )t Btu/h 9.96 F/Btu 0.3095 F)70100(21 = ° °− = − = ∞∞ totalR TT Q& The total rate of heat transfer required to condense steam at a rate of 120 lbm/h and the length of the tube required can be determined to be ft 1284=== === 9.96 440,124length Tube Btu/h 440,124Btu/lbm) 7lbm/h)(103 120( Q Q hmQ total fgtotal & & && PROPRIETARY MATERIAL. © 2011 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. 3-61 3-85E Prob. 3-83E is reconsidered. The effects of the thermal conductivity of the pipe material and the outer diameter of the pipe on the length of the tube required are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" T_infinity_1=100 [F] T_infinity_2=70 [F] k_pipe=223 [Btu/h-ft-F] D_i=0.4 [in] D_o=0.6 [in] r_1=D_i/2 r_2=D_o/2 h_fg=1037 [Btu/lbm] h_o=1500 [Btu/h-ft^2-F] h_i=35 [Btu/h-ft^2-F] m_dot=120 [lbm/h] "ANALYSIS" L=1 [ft] “for 1 ft length of the tube" A_i=pi*(D_i*Convert(in, ft))*L A_o=pi*(D_o*Convert(in, ft))*L R_conv_i=1/(h_i*A_i) R_pipe=ln(r_2/r_1)/(2*pi*k_pipe*L) R_conv_o=1/(h_o*A_o) R_total=R_conv_i+R_pipe+R_conv_o Q_dot=(T_infinity_1-T_infinity_2)/R_total Q_dot_total=m_dot*h_fg L_tube=Q_dot_total/Q_dot kpipe [Btu/h.ft.F] Ltube [ft] 10 1176 30.53 1158 51.05 1155 71.58 1153 92.11 1152 112.6 1152 133.2 1151 153.7 1151 174.2 1151 194.7 1151 215.3 1151 235.8 1150 256.3 1150 276.8 1150 297.4 1150 317.9 1150 338.4 1150 358.9 1150 379.5 1150 400 1150 0 50 100 150 200 250 300 350 400 1145 1150 1155 1160 1165 1170 1175 1180 kpipe [Btu/h-ft-F] L t ub e [f t] PROPRIETARY MATERIAL. © 2011 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. 3-62 Do [in] Ltube [ft] 0.5 1154 0.525 1153 0.55 1152 0.575 1151 0.6 1151 0.625 1150 0.65 1149 0.675 1149 0.7 1148 0.725 1148 0.75 1148 0.775 1147 0.8 1147 0.825 1147 0.85 1146 0.875 1146 0.9 1146 0.925 1146 0.95 1145 0.975 1145 1 1145 0.5 0.6 0.7 0.8 0.9 1 1145.0 1147.5 1150.0 1152.5 1155.0 D o [in] L t ub e [f t] PROPRIETARY MATERIAL. © 2011 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. 3-65 3-88 An electric wire is tightly wrapped with a 1-mm thick plastic cover. The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one- dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 The thermal contact resistance at the interface is negligible. 5 Heat transfer coefficient accounts for the radiation effects, if any. Properties The thermal conductivity of plastic cover is given to be k = 0.15 W/m⋅°C. Analysis In steady operation, the rate of heat transfer from the wire is equal to the heat generated within the wire, W104)A 13)(V 8( ==== IWQ e V&& Rconv T∞2 Rplastic T1The total thermal resistance is C/W 2746.00490.02256.0 C/W 0490.0 )m 14(C) W/m.15.0(2 )1.1/1.2ln( 2 )/ln( C/W 2256.0 m)] m)(14 (0.0042C)[. W/m24( 11 plasticconvtotal 12 plastic 2conv °=+=+= °= ° == °= ° == RRR kL rrR Ah R oo ππ π Then the interface temperature becomes C58.6°=°+°=+=⎯→⎯−= ∞ total1 totalR ∞ )C/W 2746.0)( W104(C3021 RQTTTTQ && The critical radius of plastic insulation is mm 25.6m 00625.0 C. W/m24 2 == ° == h rcr Doubling the thickness of the plastic cover will increase the outer radius of the wire to 3 mm, which is less than the critical radius of insulation. Therefore, doubling C W/m.15.0 °k the thickness of plastic cover will increase the rate of heat loss and decrease the interface temperature. PROPRIETARY MATERIAL. © 2011 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. 3-66 3-89 To avoid condensation on the outer surface, the necessary thickness of the insulation around a copper pipe that carries liquid oxygen is to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Thermal contact resistance is negligible. Properties The thermal conductivities of the copper pipe and the insulation are given to be 400 W/m · °C and 0.05 W/m · °C, respectively. Analysis From energy balance and using the thermal resistance concept, the following equation is expressed: combined , convccond,icond,combined ,, R TT RRRR TT soio −= +++ − ∞∞∞ AhhALk DD Lk DD Ah 1223 11 2 )/ln( 2 )/ln(1 +++ ππ TTTT soio ,,, −= − ∞∞∞ ci combinedcombined LDh TT L DπhLkLkLDh TT ci 3combined1 12 3combined 22 ππππ − +++ − DDDD soio , 23 ,, 11)/ln()/ln(1 = ∞∞∞ Rearranging yields ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ += − ∞∞ 3combined ,, ln(1 Dh TT io ++ −∞ 1 1223 , 1 2 )/ln( 2 )/ hDk DD k DD TT ciso ⎥ ⎥ ⎦ ⎤ °⋅ + °⋅ + ⎢ ⎣ ⎡ °⋅ °⋅+= °− °+ )m 020.0)(C W/m120( 1 )C W/m400(2 )m 020.0m/ 025.0ln( )C W/m05.0(2 )m 025.0/ln( )C W/m20(1 C )1020( C )20020( 2 3 3 2 DD Copy the following line and paste on a blank EES screen to solve the above equation: (20+200)/(20-10)=1+20*D_3*(ln(D_3/25e-3)/(2*0.05)+ln(25/20)/(2*400)+1/(120*20e-3)) Solving by EES software, the outer diameter of the insulation is The thickness of the insulation necessary to avoid condensation on the outer surface is m 0839.03 =D m 0.0295=−= − > 2 m 025.0m 0839.0 2 23 DDt Discussion If the insulation thickness is less than 29.5 mm, the outer surface temperature would decrease to the dew point at 10 °C where condensation would occur. PROPRIETARY MATERIAL. © 2011 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. 3-67 n ent. Critical Radius of Insulation 3-90C In a cylindrical pipe or a spherical shell, the additional insulation increases the conduction resistance of insulation, but decreases the convection resistance of the surface because of the increase in the outer surface area. Due to these opposite effects, a critical radius of insulation is defined as the outer radius that provides maximum rate of heat transfer. For a cylindrical layer, it is defined as hkr /= where k is the thermal conductivity of insulation and h is the external convectio heat transfer coeffici rc 3-91C For a cylindrical pipe, the critical radius of insulation is defined as . On windy days, the external convection heat transfer coefficient is greater compared to calm days. Therefore critical radius of insulation will be greater on calm days. hkrcr /= 3-92C Yes, the measurements can be right. If the radius of insulation is less than critical radius of insulation of the pipe, the rate of heat loss will increase. 3-93C No. 3-94C It will decrease. 3-95E An electrical wire is covered with 0.02-in thick plastic insulation. It is to be determined if the plastic insulation on the wire will increase or decrease heat transfer from the wire. Assumptions 1 Heat transfer from the wire is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal properties are constant. 4 The thermal contact resistance at the interface is negligible. Properties The thermal conductivity of plastic cover is given to be k = 0.075 Btu/h⋅ft⋅°F. Analysis The critical radius of plastic insulation is in) 0615.0(in 36.0ft 03.0 F.Btu/h.ft 5.2 22 °h cr Since the outer radius of the FBtu/h.ft. 075.0 =>== ° == rkr wire with insulation is smaller than critical radius of insulation, plastic insulation will increase eat transfer from the wire. Wire Insulation h PROPRIETARY MATERIAL. © 2011 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. 3-70 Heat Transfer from Finned Surfaces 3-99C Fins should be attached to the outside since the heat transfer coefficient inside the tube will be higher due to forced convection. Fins should be added to both sides of the tubes when the convection coefficients at the inner and outer surfaces are comparable in magnitude. 3-100C Increasing the rate of heat transfer from a surface by increasing the heat transfer surface area. 3-101C The fin efficiency is defined as the ratio of actual heat transfer rate from the fin to the ideal heat transfer rate from the fin if the entire fin were at base temperature, and its value is between 0 and 1. Fin effectiveness is defined as the ratio of heat transfer rate from a finned surface to the heat transfer rate from the same surface if there were no fins, and its value is expected to be greater than 1. 3-102C Heat transfer rate will decrease since a fin effectiveness smaller than 1 indicates that the fin acts as insulation. 3-103C Fins enhance heat transfer from a surface by increasing heat transfer surface area for convection heat transfer. However, adding too many fins on a surface can suffocate the fluid and retard convection, and thus it may cause the overall heat transfer coefficient and heat transfer to decrease. 3-104C Effectiveness of a single fin is the ratio of the heat transfer rate from the entire exposed surface of the fin to the heat transfer rate from the fin base area. The overall effectiveness of a finned surface is defined as the ratio of the total heat transfer from the finned surface to the heat transfer from the same surface if there were no fins. 3-105C Fins should be attached on the air side since the convection heat transfer coefficient is lower on the air side than it is on the water side. 3-106C Welding or tight fitting introduces thermal contact resistance at the interface, and thus retards heat transfer. Therefore, the fins formed by casting or extrusion will provide greater enhancement in heat transfer. 3-107C If the fin is too long, the temperature of the fin tip will approach the surrounding temperature and we can neglect heat transfer from the fin tip. Also, if the surface area of the fin tip is very small compared to the total surface area of the fin, heat transfer from the tip can again be neglected. 3-108C Increasing the length of a fin decreases its efficiency but increases its effectiveness. PROPRIETARY MATERIAL. © 2011 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. 3-71 3-109C Increasing the diameter of a fin increases its efficiency but decreases its effectiveness. 3-110C The thicker fin has higher efficiency; the thinner one has higher effectiveness. 3-111C The fin with the lower heat transfer coefficient has the higher efficiency and the higher effectiveness. 3-112 A relation is to be obtained for the fin efficiency for a fin of constant cross-sectional area , perimeter p, length L, and thermal conductivity k exposed to convection to a medium at ∞ with a heat transfer coefficient h. The relation is to be simplified for circular fin of diameter D and for a rectangular fin of thickness t. cA T Assumptions 1 The fins are sufficiently long so that the temperature of the fin at the tip is nearly T . 2 Heat transfer from the fin tips is negligible. ∞ Analysis Taking the temperature of the fin at the base to be T and using the heat transfer relation for a long fin, fin efficiency for long fins can be expressed as b ph kA LhpL hpkA TThA TThpkA cc bfin bc 1 )( )( re temperatubaseat fin were entire theif fin thefrom ratefer heat trans Ideal fin thefrom ratefer heat trans Actual fin == − − = = ∞ ∞ η h, T∞ D p= πD Ac = πD2/4 Tb This relation can be simplified for a circular fin of diameter D and rectangular fin of thickness t and width w to be h kt Lwh wtk Lhtw wtk Lph kA L h kD LhD Dk Lph kA L c c 2 1 2 )(1 )(2 )(11 2 1 )( )4/(11 rrectangulafin, 2 circularfin, =≅ + == === η π π η 3-113 The maximum power rating of a transistor whose case temperature is not to exceed 80 is to be determined. s isothermal at 80 operties bient thermal resistance is given to be Analysis The maximum power at which this transistor can be operated safely is °C Assumptions 1 Steady operating conditions exist. 2 The transistor case i °C. Pr The case-to-am 20 T∞ R Ts °C / W. W1.8= ° °− = − = ∆ = − ∞ − C/W 25 C )3580( ambientcase case ambientcase R TT R TQ& PROPRIETARY MATERIAL. © 2011 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. 3-72 3-114 A fin is attached to a surface. The percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption is to be determined. Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction only (normal to the plate). 3 The heat transfer coefficient is constant and uniform over the entire fin surface. 4 The thermal properties of the fins are constant. 5 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivity of the aluminum fin is given to be k = 237 W/m⋅°C. Analysis The expressions for the heat transfer from a fin under infinitely long fin and adiabatic fin tip assumptions are )tanh()( )( tipins. fin long mLTThpkAQ TThpkAQ bc bc ∞ ∞ −= −= & & L = 10 cm D = 4 mm The percent error in using long fin assumption can be expressed as 1 )tanh( 1 )tanh()( )tanh()()( Error % tipins. tipins.fin long −= − −−− = − = ∞ ∞∞ mLmLTThpkA mLTThpkATThpkA Q QQ bc bcbc & && where 1- 2 2 m 116.7 4/m) 004.0()C W/m.237( m) 004.0(C). W/m12( = ° ° ==m π π ckA hp ubstitutin S g, [ ] 63.5%==−=−= 635.01m) 10.0)(m 116.7(tanh 11 1 )tanh( Error % 1-mL his result shows that using infinitely long fin assumption may yield results grossly in error. fins varies in one direction only (normal to the ce. 4 The thermal properties of the fins he fins. Properties The thermal conductivity of the fin is given to be k = 200 W/m⋅°C. Analysis The fin temperature at a distance of 5 cm from the base is determined from T 3-115 A very long fin is attached to a flat surface. The fin temperature at a certain distance from the base and the rate of heat loss from the entire fin are to be determined. Assumptions 1 Steady operating conditions exist. 2 The temperature along the plate). 3 The heat transfer coefficient is constant and uniform over the entire fin surfa are constant. 5 The heat transfer coefficient accounts for the effect of radiation from t C29.8°=⎯→⎯= − −− −−∞ TTT mx c 14(20⎯→⎯ = ×° ×+×° == Te T kA hp m )05.0)(3. 1- 2 2 2040 m 3.14 )m001.005.0)(C W/m.200( 0.001)m20.05C)(2. W/m20( The rate of heat loss from this very long fin is 20°C 40°C = − ∞ e Tb W2.9= −××+×= −= ∞ )2040()001.005.0(200(0.001)20.05)(220( )(fin long TThpkAQ bc& PROPRIETARY MATERIAL. © 2011 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. 3-75 3-118 A commercially available heat sink is to be selected to keep the case temperature of a transistor below 90°C in an environment at 20°C. Assumptions 1 Steady operating conditions exist. 2 The transistor case is isothermal at 90°C. 3 The contact resistance between the transistor and the heat sink is negligible. Analysis The thermal resistance between the transistor attached to the sink and the ambient air is determined to be C/W 1.75 °=°−= − =⎯→⎯ ∆ = ∞− C)2090(transistor ambientcase TT RTQ& le 3-6 reveals that HS6071 in vertical position, HS5030 nd HS6115 in both horizontal and vertical position can be selected. available heat sink is to be selected to keep the case temperature of a transistor below 55°C in an hermal transistor and the heat sink is between the transistor attached to the sink and the ambient air is determined to be T∞ RTs − W40ambientcase QR & The thermal resistance of the heat sink must be below 1.75°C/W. Tab a 3-119 A commercially environment at 18°C. Assumptions 1 Steady operating conditions exist. 2 The transistor case is isot at 55°C. 3 The contact resistance between the negligible. Analysis The thermal resistance C/W 1.5 °=°−= − =⎯→⎯ ∆ = ∞− − W25 C)1855(transistor ambientcase ambientcase Q TT R R TQ & & T∞ RTs The thermal resistance of the heat sink must be below 1.5°C/W. Table 3-6 reveals that HS5030 in both horizontal and ertical positions, HS6071 in vertical position, and HS6115 in both horizontal and vertical positions can be selected. v PROPRIETARY MATERIAL. © 2011 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. 3-76 3-120 A turbine blade is exposed to hot gas from the combustion chamber. The heat transfer rate to the turbine blade and the temperature at the tip are to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. 4 The cross-sectional area of the turbine blade is uniform. Properties The thermal conductivity of the turbine blade is given as 17 W/m · °C. Analysis The turbine blade can be treated as a uniform cross section fin with adiabatic tip. The heat transfer rate to the turbine blade can be expressed as mLTThpkAQ bc tanh)(blade −= ∞& where 366.4)m 053.0( )m 1013.5)(C W/m17( )m 11.0)(C W/m538( 5.0 24 25.0 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ×°⋅ °⋅ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = − L kA hpmL c C W/7184.0)m 1013.5)(C W/m17)(m 11.0)(C W/m538( 242 °=×°⋅°⋅= −chpkA to the turbine blade is & perature distribution is expressed as The heat transfer rate W376=°−°= )366.4tanhC( )450973)(C W/7184.0( bladeQ For adiabatic tip, the tem mL xLm TT TxT cosh )(cosh)( − = − − ∞ b ∞ he temperature at the tip of the turbine blade is T C 960 °=°+°−=+ − = ∞ ∞ C 973 366.4cosh C )973450( cosh T mL TT T bL Discussion The tolerance of the turbine blade to high temperature can be increased by applying Zirconia based thermal barrier coatings (TBCs) on the blade surface. PROPRIETARY MATERIAL. © 2011 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. 3-77 3-121 Pipes used for transporting superheated vapor are connected together by flanges. The temperature at the base of the flange and the rate of heat loss through the flange are to be determined. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal properties are constant. 3 Heat transfer by radiation is negligible. 4 The flanges profile is similar to circular fins of rectangular profile. Properties The thermal conductivity of the pipes is given as 16 W/m · °C. Analysis The heat transfer rate through the pipe wall is equal to the heat transfer rate through the flanges: or fQQ && =pipe )()/ln( 2 ∞−= − TThA DD TT tk bff io bi ηπ Rearranging the equation yields )/ln( 2 ff DD tkhA πη + )ln( 2 io i io ff b T D tkThA T πη + = ∞ From Table 3-3, for circular fins of rectangular profile we have / D m 055.0 2 m 02.0 2 m 09.02/22 =+=+= trr c 1 2 2 m 055.0[(2)(2 =−= ππ rrA cf 2222 m 0151.0])m 2/05.0() =− m 025.0 22c m 02.0m 06.0m 09.02/ =+−=+= tLL Hence, 2m 0005.0)m 02.0)(m 025.0( === tLA cp 1398.0 )m 0005.0)(C W/m16( C W/m10)m 025.0( 2/1 2 2 2/3 2/1 2/3 = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ °⋅ °⋅ =⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = p c kA hLξ 83.1 m 030.0 m 055.0/ 12 ==rr c Using Figure 3-44, the fin efficiency is 97.0≈fη . The temperature at the base of the flange is C 148 °= °⋅ +°⋅ ° °⋅ +°°⋅ = )50/60ln( )C W/m16)(m 020(2)m 0151.0)(C W/m10)(97.0( )C 150( )50/60ln( )C W/m16)(m 020(2)C 25)(m 0151.0)(C W/m10)(97.0( 22 22 π π . . Tb The rate of heat loss through the flange is Discussion The flanges act as extended surfaces, which enhanced heat transfer from the pipes. W18=°−°⋅=−= ∞ C )25148)(m 0151.0)(C W/m10)(97.0()( 22TThAQ bfff η& PROPRIETARY MATERIAL. © 2011 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. 3-80 3-124E The handle of a stainless steel spoon partially immersed in boiling water extends 7 in. in the air from the free surface of the water. The temperature difference across the exposed surface of the spoon handle is to be determined. Assumptions 1 The temperature of the submerged portion of the spoon is equal to the water temperature. 2 The temperature in the spoon varies in the axial direction only (along the spoon), T(x). 3 The heat transfer from the tip of the spoon is negligible. 4 The heat transfer coefficient is constant and uniform over the entire spoon surface. 5 The thermal properties of the spoon are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the spoon. Properties The thermal conductivity of the spoon is given to be k = 8.7 Btu/h⋅ft⋅°F. Analysis Noting that the cross-sectional area of the spoon is constant and measuring x from the free surface of water, the variation of temperature along the spoon can be expressed as cosh )(cosh)( mL xLm TT TxT b − = − − ∞ ∞ h, T∞ Tb L = 7 in 0.08 in 0.5 in where 2ft 000278.0)ft 12/08.0ft)( 12/5.0( ft 0967.0)ft 12/08.0ft 12/5.0(2 == =+= cA p 1- 2 2 ft 95.10 )ft 000278.0)(FBtu/h.ft. 7.8( °ckA Noting that x = L )ft 0967.0)(F.Btu/h.ft 3( = ° == hpm = 7/12=0.583 ft at the tip and substituting, the tip temperature of the spoon is etermined to be d F75.4= 296 1)75(200+F75= )583.095.10cosh( 0cosh)75(200+F75= cosh )(cosh )()( − −+= ∞∞ LLm TTTLT b ° −° × −° mL Therefore, the temperature difference across the exposed section of the spoon handle is F124.6°=°−=−=∆ F)4.75200(tipTTT b PROPRIETARY MATERIAL. © 2011 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. 3-81 3-125E The handle of a silver spoon partially immersed in boiling water extends 7 in. in the air from the free surface of the water. The temperature difference across the exposed surface of the spoon handle is to be determined. Assumptions 1 The temperature of the submerged portion of the spoon is equal to the water temperature. 2 The temperature in the spoon varies in the axial direction only (along the spoon), T(x). 3 The heat transfer from the tip of the spoon is negligible. 4 The heat transfer coefficient is constant and uniform over the entire spoon surface. 5 The thermal properties of the spoon are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the spoon.. Properties The thermal conductivity of the spoon is given to be k = 247 Btu/h⋅ft⋅°F. Analysis Noting that the cross-sectional area of the spoon is constant and measuring x from the free surface of water, the variation of temperature along the spoon can be expressed as cosh )(cosh)( mL xLm TT TxT b − = − − ∞ ∞ where A p h, T∞ Tb L = 7 in 0.08 in 0.5 in 2ft 000278.0)ft 12/08.0ft)( 12/5.0( ft 0967.0)ft 12/08.0ft 12/5.0(2 == =+= c 1- 2 ft 055.2)ft 0967.0)(F.Btu/h.ft 3( =°== kA hpm 2 )ft 000278.0)(FBtu/h.ft. 247( °c Noting that x = L = 0.7/12=0.583 ft at the tip and substituting, the tip temperature of the spoon is determined to be F144.1= 81.1 1)75(200+F75= −° )583.0055.2cosh( 0cosh)75(200+F75= cosh )(cosh )()( ° × −° − −+= ∞∞ mL LLm TTTLT b erefore, the temperature difference across the exposed section of the spoon handle is Th F55.9°=°−=−=∆ C)1.144200(tipTTT b PROPRIETARY MATERIAL. © 2011 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. 3-82 3-126E Prob. 3-124E is reconsidered. The effects of the thermal conductivity of the spoon material and the length of its extension in the air on the temperature difference across the exposed surface of the spoon handle are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" k_spoon=8.7 [Btu/h-ft-F] T_w=200 [F] T_infinity=75 [F] A_c=(0.08/12*0.5/12) [ft^2] L=7 [in] h=3 [Btu/h-ft^2-F] "ANALYSIS" p=2*(0.08/12+0.5/12) a=sqrt((h*p)/(k_spoon*A_c)) (T_tip-T_infinity)/(T_w-T_infinity)=cosh(a*(L-x)*Convert(in, ft))/cosh(a*L*Convert(in, ft)) x=L "for tip temperature" DELTAT=T_w-T_tip kspoon [Btu/h.ft.F] ∆T [F] 5 124.9 16.58 122.6 28.16 117.8 39.74 112.5 51.32 107.1 62.89 102 74.47 97.21 86.05 92.78 97.63 88.69 109.2 84.91 120.8 81.42 132.4 78.19 143.9 75.19 155.5 72.41 167.1 69.82 178.7 67.4 190.3 65.14 201.8 63.02 213.4 61.04 225 59.17 0 45 90 135 180 225 50 60 70 80 90 100 110 120 130 kspoon [Btu/h-ft-F] ∆ T [F ] PROPRIETARY MATERIAL. © 2011 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. 3-85 A 3-128 A circuit board houses 80 logic chips on one side, dissipating 0.04 W each through the back side of the board to the surrounding medium. The temperatures on the two sides of the circuit board are to be determined for the cases of no fins and 864 copper pin fins on the back surface. Assumptions 1 Steady operating conditions exist. 2 The temperature in the board and along the fins varies in one direction only (normal to the board). 3 All the heat generated in the chips is conducted across the circuit board, and is dissipated from the back side of the board. 4 Heat transfer from the fin tips is negligible. 5 The heat transfer coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant. 7 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivities are given to be k = 20 W/m⋅°C for the circuit board, k = 386 W/m⋅°C for the copper plate and fins, and k = 1.8 W/m⋅°C for the epoxy adhesive. Analysis (a) The total rate of heat transfer dissipated by the chips is 2 cm W2.3 W)04.0(80 =×=Q& The individual resistances are = 2m 0216.0m) 18.0(m) 12.0( = C/W 89031.0 )m 0216.0(C). W/m52( 11 C/W 00617.0 )m 0216.0(C) W/m.30( m 004.0 22conv 2board °= ° == °= ° == hA R kA LR Rboard T1 RAluminum Rconv T∞2 Repoxy T2 C/W 8965.089031.000617.0convboardtotal °=+=+= RRR The temperatures on the two sides of the circuit board are C42.9 C42.9 °≅°=°−°=−=⎯→⎯ − = °≅°=°+°=+=⎯→⎯ − = ∞ ∞ C85.42C/W) 00617.0)( W2.3(C87.42 C87.42C/W) 8965.0)( W2.3(C40 board12 board 21 total21 total 21 RQTT R TTQ RQTT R TTQ && && Therefore, the board is nearly isothermal. (b) Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be 1- 2 2 m 74.18)m 0025. 0()C W/m.237( )C. W/m52(44 4/ = ° ° ==== kD h Dk Dh kA hpm c π π 956.0 m02.0m 74. 1- × The fins can be assumed to be at base temperature provided that the fin area is modified by multiplying it by 0.956. Then the various thermal resistances are 18 )m 02.0m 74.18tanh(tanh -1 fin = × == mL mLη C/W 00514.0 )m 0216.0(C) W/m.8.1( m 0002.0 2epoxy °=° == kA LR C/W 00024.0 )m 0216.0(C) W/m.386( m 002.0 2Al °=° == kA LR 2 unfinnedfinnedfinswith total, 2 22 unfinned 2 finfinned m) 02.0(m) 0025.0(864956.0 =×== DLnA ππη m 1471.00174.01297.0 m 0174.0 4 )0025.0(8640216.0 4 8640216.0 m 1297.0 =+=+= =×−=−= AAA DA ππ C/W 1307.0 )m 1471.0(C). W/m52( 11 22 finswith total, conv °= ° == hA R C/W 1423.01307.000024.000514.000617.0convaluminumepoxyboardtotal °=+++=+++= RRRRR Then the temperatures on the two sides of the circuit board becomes C40.4 C40.5 °≅=°−°=−=⎯→⎯ − = °≅°=°+°=+=⎯→⎯ − = ∞ ∞ 44.40C/W) 00617.0)( W2.3(C46.40 C40.46C/W) 1423.0)( W2.3(C40 board12 board 21 total21 total 21 RQTT R TTQ RQTT R TTQ && && PROPRIETARY MATERIAL. © 2011 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. 3-86 3-129 A hot plate is to be cooled by attaching aluminum pin fins on one side. The rate of heat transfer from the 1 m by 1 m section of the plate and the effectiveness of the fins are to be determined. Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction only (normal to the plate). 3 Heat transfer from the fin tips is negligible. 4 The heat transfer coefficient is constant and uniform over the entire fin surface. 5 The thermal properties of the fins are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivity of the aluminum plate and fins is given to be k = 237 W/m⋅°C. Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be 3 cm 0.6 cm D=0.25 cm 1- 2 2 m 37.15 )m 0025.0)(C W/m.237( )C. W/m35(44 4/ = ° ° ==== kD h Dk Dh kA hpm c π π 935.0 m 03.0m 37.15 1- fin = × == mL η The number of fins, finned and unfinned surface areas, and heat transfer )m 03.0m 37.15tanh(tanh -1 ×mL rates from those areas are 777,27 m) 006.0(m) 006.0( m 1 2 ==n W2107 C)30100)(m 86.0)(C. W/m35()( 2unfinnedunfinned °=−= ∞TThAQ b& W300,15 C)30100)(m 68.6)(C. W/m35(935.0 )( m 86.0 4 )0025.0( 277771 4 277771 m 68.6 4 )0025.0()03.0)(0025.0(27777 4 27777 2 22 finfinmaxfin,finfinned 2 22 unfinned 2 22 fin = °− = °−°= −== = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ −=⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −= = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ += ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ += ∞TThAQQ DA DDLA b && ηη ππ π πππ The rate of heat transfer if there were no fin attached to the plate would be 22 fin nofin =°−°=−= ∞TThAQ b& hen the fin effectiveness becomes Then the total heat transfer from the finned plate becomes kW 17.4=×=+=+= W1074.12107300,15 4unfinnedfinnedfintotal, QQQ &&& no m 1)m 1)(m 1( 2fin no ==A W2450C)30100)(m 1)(C. W/m35()( T 7.10=== 2450 400,17 fin no fin fin Q Q & & ε PROPRIETARY MATERIAL. © 2011 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. 3-87 3-130 A hot plate is to be cooled by attaching copper pin fins on one side. The rate of heat transfer from the 1 m by 1 m section of the plate and the effectiveness of the fins are to be determined. Assumptions 1 Steady operating conditions exist. 2 The temperature along the fins varies in one direction only (normal to the plate). 3 Heat transfer from the fin tips is negligible. 4 The heat transfer coefficient is constant and uniform over the entire fin surface. 5 The thermal properties of the fins are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivity of the copper plate and fins is given to be k = 386 W/m⋅°C. Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be 3 cm 0.6 cm D=0.25 cm 1- 2 2 m 04.12 )m 0025.0)(C W/m.386( )C. W/m35(44 4/ = ° ° ==== kD h Dk Dh kA hpm c π π 959.0 m 03.0m 04.12 1- fin ×mL The number of f )m 03.0m 04.12tanh(tanh -1 = × == mLη ins, finned and unfinned surface areas, and heat transfer rates from those areas are 27777 m) 006.0(m) 006.0( m 1 2 ==n W2107C)30100)(m 86.0)(C W/m35()( W700,15 C)30100)(m 68.6)(C. W/m35(959.0 2 maxfin,finfinned = )( m 68.6 4 )0025.0()03.0)( 4 2o2 unfinnedunfinned 2 finfin 22 2 2 fin =°−=−= = °−° −== ⎥⎦⎢⎣⎠⎝ =⎥ ⎤ ⎢⎣ + ⎥⎦⎢⎣ ∞ ∞ TThAQ TThAQ b b & & ηη π Then the total heat transfer from the finned plate becomes The rate of heat transfer if there were no fin attached to the plate would be Then the fin effectiveness becomes 0025.0(2777727777 2 ⎢ ⎡ =⎥ ⎤ ⎢ ⎡ += DDLA πππ ⎥⎦ m 86.0 4 )0025.0( 277771 4 277771 2unfinned =⎥ ⎤ ⎢ ⎡ −=⎟ ⎟ ⎞ ⎜ ⎜ ⎛ −= DA ππ Q& kW 17.8=×=+=+= W1078.12107700,15 4unfinnedfinnedfintotal, QQQ &&& W2450C)30100)(m 1)(C. W/m35()( m 1)m 1)(m 1( 22 fin nofin no 2 fin no =°−°=−= == ∞TThAQ A b & 7.27=== 2450 17800 fin no fin fin Q Q & & ε PROPRIETARY MATERIAL. © 2011 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. 3-90 & by S=1/(kR). Heat Transfer in Common Configurations 3-133C Under steady conditions, the rate of heat transfer between two surfaces is expressed as )( 21 TTSk −= where S is the conduction shape factor. It is related to the thermal resistance Q 3-134C It provides an easy way of calculating the steady rate of heat transfer between two isothermal surfaces in common configurations. 3-135 Hot and cold water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between the pipes is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two- dimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant. Properties The thermal conductivity of concrete is given to be k = 0.75 W/m⋅°C. Analysis The shape factor for this configuration is given in Table 3-7 to be m 59.14 )m 06.0)(m 06.0(2 )m 06.0()m 06.0()m 4.0(4cosh )m 12(2 2 4cosh 2 222 1 21 2 2 2 1 2 1 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −− = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −− = − − π π DD DDz LS T2 = 15°C D = 6 cm T1 = 60°C L = 12 m z = 40 cm Then the steady rate of heat transfer between the pipes becomes W492=°−°=−= C)1560)(C W/m.75.0)(m 59.14()( 21 TTSkQ& PROPRIETARY MATERIAL. © 2011 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. 3-91 3-136 Prob. 3-135 is reconsidered. The rate of heat transfer between the pipes as a function of the distance between the centerlines of the pipes is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "GIVEN" L=12 [m] D_1=0.06 [m] D_2=D_1 z=0.40 [m] T_1=60 [C] T_2=15 [C] k=0.75 [W/m-C] "ANALYSIS" S=(2*pi*L)/(arccosh((4*z^2-D_1^2-D_2^2)/(2*D_1*D_2))) Q_dot=S*k*(T_1-T_2) z [m] Q [W] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1158 679 555 492.3 452.8 425.1 404.2 387.7 374.2 362.9 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 300 400 500 600 700 800 900 1000 1100 1200 z [m] Q [ W ] PROPRIETARY MATERIAL. © 2011 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. 3-92 3-137E A row of used uranium fuel rods are buried in the ground parallel to each other. The rate of heat transfer from the fuel rods to the atmosphere through the soil is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant. Properties The thermal conductivity of the soil is given to be k = 0.6 Btu/h⋅ft⋅°F. Analysis The shape factor for this configuration is given in Table 3-7 to be ft 5298.0 )ft 12/8( )ft 15(2 sinh )ft 12/1( )ft 12/8(2 ln )ft 3(2 4 2sinh2ln ⎟⎞⎜⎛ πzw 24 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×= ⎠⎝ ×= π π π π π wD LS hen the steady rate of heat transfer from the fuel rods becomes pe of a district heating system is buried in the soil. The rate of heat loss from the pipe is to be ns exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant. Analysis Since z >1.5D, the shape factor for this onfiguration is given in Table 3-7 to be T2 = 60°F 8 in T1 = 350°F 15 ft D = 1 in L = 3 ft total T Btu/h 92.2=°−°=−= F)60350)(FBtu/h.ft. 6.0)(ft 5298.0()( 21total TTkSQ& 3-138 The hot water pi determined. Assumptions 1 Steady operating conditio Properties The thermal conductivity of the soil is given to be k = 0.9 W/m⋅°C. c m 44.20 )]m 08.0/()m 8.0(4ln[ )m 12(2 )/4ln( 2 === ππ Dz LS Then the steady rate of heat transfer from the pipe becomes W1067=°−=−= C)260)(C W/m.9.0)(m 44.20()( o21 TTSkQ& 60°C L = 12 m D = 8 cm 2°C 80 cm PROPRIETARY MATERIAL. © 2011 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. 3-95 m 3142.0)m 2)(m 05.0( 2 2 =°− === DLA ππ idering the shape factor, the heat loss for vertical part of the tube can be determined from 3-141 Hot water is flowing through a pipe that extends 2 m in the ambient air and continues in the ground before it enters the next building. The surface of the ground is covered with snow at 0°C. The total rate of heat loss from the hot water and the temperature drop of the hot water in the pipe are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the ground is constant. 4 The pipe is at the same temperature as the hot water. Properties The thermal conductivity of the ground is given to be k = 1.5 W/m⋅°C. Analysis (a) We assume that the surface temperature of the tube is equal to the temperature of the water. Then the heat loss from the part of the tube that is on the ground is 0)(C. W/m22( )( 2 °= −= ∞TThAQ ss& W518C)580)(m 3142. s Cons m 44.3 )m 05.0( ⎦⎣⎠⎝ D )m 3(4ln )m 3(2 4ln 2 = ⎥ ⎤ ⎢ ⎡ = ⎟ ⎞ ⎜ ⎛ = ππ L LS W/m.5.1)(m 44.3()( 21 =°=−= TTSkQ& The shape factor, and the rate of heat loss on the horizontal part that is in the ground are 5°C 3 m 80°C -3°C 20 m C)]3(80)[C °−− W428 m 9.22 )m 05.0( ⎥⎦ ⎢ ⎣⎠⎝ D )m 3(4ln )m 20(2 4ln 2 = ⎤⎡ = ⎟ ⎞ ⎜ ⎛ ππ z LS 5.1)(m 9.22()( 21 =°−−=−= TTSkQ& talQ (b) Using the water properties at the room temperature, the temperature drop of the hot water as it flows through this 25-m section of the wall becomes = W2851C)]3(80)[C W/m.° and the total rate of heat loss from the hot water becomes & W3797=++= 2851428518 to C0.31°= ° ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ====∆ ∆= )CJ/kg. 4180( 4 )m 05.0()m/s 5.1)(kg/m 1000( J/s 3797 )()( 23 πρρ pcpp p cVA Q c Q cm QT TcmQ & & & & & && V PROPRIETARY MATERIAL. © 2011 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. 3-96 3-142 The walls and the roof of the house are made of 20-cm thick concrete, and the inner and outer surfaces of the house are maintained at specified temperatures. The rate of heat loss from the house through its walls and the roof is to be determined, and the error involved in ignoring the edge and corner effects is to be assessed. Assumptions 1 Steady operating conditions exist. 2 Heat transfer at the edges and corners is two-or three-dimensional. 3 Thermal conductivity of the concrete is constant. 4 The edge effects of adjoining surfaces on heat transfer are to be considered. Properties The thermal conductivity of the concrete is given to be k = 0.75 W/m⋅°C. Analysis The rate of heat transfer excluding the edges and corners is first determined to be 2total m 7.403)2.06)(4.012(4)4.012)(4.012( =−−+−−=A W167,18C)315( m 2.021L The heat transfer rate th )m 7.403)(C W/m.75.0()( 2 total =°− ° =−TT kA Q& rough the edges can be determined using the shape factor relations in Table 3-7, = m 04.26m) 0.54(124+m) 2.0(15.04 54.0415.04edges4corners4edges+corners =××= ×+×=×+×= wLS m 04.26()( 21edges+cornersedges+corners =°−°=−= TTkSQ& Q& e rate of heat transfer is determined from W234C)315)(C W/m.75.0 3°C 15°C L L )( and kW 18.4=×=+= W10840.1234167,18 4total Ignoring the edge effects of adjoining surfaces, th 2total m 432)6)(12(4)12)(12( =+=A kW 4.19101.94C)315( m 2.0 C W/m.75.0()( 21 total °=− )m 432)( 4 2 =×=°−TT L kA Q& The percentage error involved in ignoring the effects of the edges then becomes = 5.4%=×−= 100 4.18 4.184.19%error PROPRIETARY MATERIAL. © 2011 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. 3-97 3-143 The inner and outer surfaces of a long thick-walled concrete duct are maintained at specified temperatures. The rate of heat transfer through the walls of the duct is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant. Properties The thermal conductivity of concrete is given to be k = 0.75 W/m⋅°C. Analysis The shape factor for this configuration is given in Table 3-7 to be m 7.896 25.1ln785.0 )m 25(2 ln785.0 241.125.1 16 20 == ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⎯→⎯<== ππ b a LS b a Then the steady rate of heat transfer through the walls of the duct becomes kW 47.1=×=°−°=−= W10.714C)30100)(C W/m.75.0)(m 7.896()( 421 TTSkQ& 100°C 30°C 20 cm 16 cm 3-144 A spherical tank containing some radioactive material is buried in the ground. The tank and the ground surface are maintained at specified temperatures. The rate of heat transfer from the tank is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the ground is constant. Properties The thermal conductivity of the ground is given to be k = 1.4 W/m⋅°C. T2 =15°C T1 = 140°C z = 5.5 m D = 3 m Analysis The shape factor for this configuration is given in Table 3-7 to be m 83.21 m 5.5 m 325.01 )m 3(2 25.01 2 = − = − = ππ z D DS Then the steady rate of heat transfer from the tank becomes W3820=°−°=−= C)15140)(C W/m.4.1)(m 83.21()( 21 TTSkQ&