Solution Manual For Heat And Mass Transfer: Fundamentals And Applications, 5th Edition

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1-1Solutions ManualforHeat and Mass Transfer: Fundamentals & Applications5th EditionYunus A. Cengel & Afshin J. GhajarChapter 1INTRODUCTION AND BASIC CONCEPTS

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1-2Thermodynamics and Heat Transfer1-1CThermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state toanother. Heat transfer, on the other hand, deals with the rate of heat transfer as well as the temperature distribution within thesystem at a specified time.1-2C(a) The driving force for heat transfer is the temperature difference. (b) The driving force for electric current flow is theelectric potential difference (voltage). (a) The driving force for fluid flow is the pressure difference.1-3CThe caloric theory is based on the assumption that heat is a fluid-like substance called the "caloric" which is a massless,colorless, odorless substance. It was abandoned in the middle of the nineteenth century after it was shown that there is nosuch thing as the caloric.1-4CTheratingproblems deal with the determination of theheat transfer ratefor an existing system at a specifiedtemperature difference. Thesizingproblems deal with the determination of thesizeof a system in order to transfer heat at aspecified ratefor aspecified temperature difference.1-5CThe experimental approach (testing and taking measurements) has the advantage of dealing with the actual physicalsystem, and getting a physical value within the limits of experimental error. However, this approach is expensive, timeconsuming, and often impractical. The analytical approach (analysis or calculations) has the advantage that it is fast andinexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis.1-6CThe description of most scientific problems involves equations that relate thechangesin some key variables to eachother, and the smaller the increment chosen in the changing variables, the more accurate the description. Inthe limiting caseof infinitesimal changes in variables, we obtaindifferential equations, which provide precise mathematical formulations forthe physical principles and laws by representing the rates of changes asderivatives.As we shall see in later chapters, the differential equations of fluid mechanics are known, but very difficult to solveexcept for very simple geometries. Computers are extremely helpful in this area.1-7CModeling makes it possible to predict the course of an event before it actually occurs, or to study various aspects of anevent mathematically without actually running expensive and time-consuming experiments. When preparing a mathematicalmodel, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, andthe interdependence of these variables are studied. The relevant physical laws and principles are invoked, and the problem isformulated mathematically. Finally, the problem is solved using an appropriate approach, and the results are interpreted.

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1-31-8CThe right choice between a crude and complex model is usually thesimplestmodel which yieldsadequateresults.Preparing very accurate but complex models is not necessarily a better choice since such models are not much use toan analyst if they are very difficult and time consuming to solve. At the minimum, the model should reflect the essentialfeatures of the physical problem it represents.1-9CWarmer. Because energy is added to the room air in the form of electrical work.1-10CWarmer. If we take the room that contains the refrigerator as our system, we will see that electrical work is supplied tothis room to run the refrigerator, which is eventually dissipated to the room as waste heat.1-11CFor the constant pressure case. This is because the heat transfer to an ideal gas ismcpTat constant pressure andmcvTat constant volume, andcpis always greater thancv.1-12CThermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life.1-13CThe rate of heat transfer per unit surface area is called heat fluxq. It is related to the rate of heat transfer byAdAqQ.1-14CEnergy can be transferred by heat, work, and mass. An energy transfer is heat transfer when its driving force istemperature difference.

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1-41-15The filament of a 150 W incandescent lamp is 5 cm long and has a diameter of 0.5 mm. The heat flux on the surface ofthe filament, the heat flux on the surface of the glass bulb, and the annual electricity cost of the bulb are to be determined.AssumptionsHeat transfer from the surface of the filament and the bulb of the lamp is uniform.Analysis(a) The heat transfer surface area and the heat flux on the surface of the filament are2cm785.0)cm5)(cm05.0(DLAs26W/m101.9122W/cm191cm785.0W150ssAQq(b) The heat flux on the surface of glass bulb is222cm1.201cm)8(DAs2W/m750022W/cm75.0cm1.201W150ssAQq(c) The amount and cost of electrical energy consumed during a one-year period is$35.04/yrkWh)/.08kWh/yr)($0(438=CostAnnualkWh/yr438h/yr)8kW)(36515.0(nConsumptioyElectricittQ1-16EA logic chip in a computer dissipates 3 W of power. The amount heat dissipated in 8 h and the heat flux on the surfaceof the chip are to be determined.AssumptionsHeat transfer from the surface is uniform.Analysis(a) The amount of heat the chip dissipates during an 8-hour period iskWh0.024Wh24)h8)(W3(tQQ(b) The heat flux on the surface of the chip is2W/in37.52in08.0W3AQqLogic chipW3QQLamp150 W

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1-51-17An aluminum ball is to be heated from 80C to 200C. The amount of heat that needs to be transferred to the aluminumball is to be determined.AssumptionsThe properties of the aluminum ball are constant.PropertiesThe average density and specific heat of aluminum are given to be= 2700 kg/m3andcp= 0.90 kJ/kgC.AnalysisThe amount of energy added to the ball is simply the change in its internal energy, and isdetermined from)(12transferTTmcUEpwherekg77.4m)15.0)(kg/m2700(66333DmVSubstituting,kJ515=C80)C)(200kJ/kgkg)(0.9077.4(transferETherefore, 515 kJ of energy (heat or work such as electrical energy) needs to be transferred to the aluminum ball to heat it to200C.1-18One metric ton of liquid ammonia in a rigid tank is exposed to the sun. The initial temperature is 4°C and the exposureto sun increased the temperature by 2°C. Heat energy added to the liquid ammonia is to be determined.AssumptionsThe specific heat of the liquid ammonia is constant.PropertiesThe average specific heat of liquid ammonia at (4 +6)°C/ 2 =5°C iscp= 4645 J/kgK (Table A-11).AnalysisThe amount of energy added to the ball is simply the change in its internal energy, and isdetermined from)(12TTmcQpwherekg1000tonmetric1mSubstituting,kJ9290=C)C)(2J/kgkg)(46451000(QDiscussionTherefore, 9290kJ of heat energy is required to transfer to 1 metric ton of liquid ammonia to heat it by 2°C.Also,the specific heat units J/kgºC andJ/kgK are equivalent, and can be interchanged.MetalballE

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1-61-19A 2 mm thick by 3 cm wide AISI 1010 carbon steel strip is cooled in a chamber from 527 to 127°C.The heat rateremoved from the steel strip is 100 kW and the speed it is being conveyed in the chamber is to be determined.Assumptions1Steady operating conditions exist.2The stainless steel sheet has constant properties.3Changes in potentialand kinetic energy are negligible.PropertiesFor AISI 1010 steel, the specific heat of AISI 1010 steel at (527 +127)°C/ 2 =327°C= 600 K is 685J/kg∙K(Table A-3), and the density is given as 7832 kg/m3.AnalysisThe mass of the steel strip being conveyed enters and exits the chamber at a rate ofVwtmThe rate of heat loss from the steel strip in the chamber is given as)()(outinoutinlossTTVwtcTTcmQppThus, the velocity of the steel strip being conveyed ism/s0.777K)127527(K)J/kg685)(m002.0)(m030.0)(kg/m7832(W10100)(33outinlossTTwtcQVpDiscussionA control volume is applied on the steel strip being conveyed in and out of the chamber.1-20EA water heater is initially filled with water at 50F. The amount of energy that needs to be transferred to the water toraise its temperature to 120F is to be determined.Assumptions1Water is an incompressible substance with constant specific.2No water flows in or out of the tank duringheating.PropertiesThe density and specific heat of water at 85ºF from Table A-9E are:=62.17 lbm/ft3and cp= 0.999 Btu/lbmR.AnalysisThe mass of water in the tank islbm7.498gal7.48ft1gal))(60lbm/ft17.62(33VmThen, the amount of heat that must be transferred to the water in the tank as it isheated from 50 to 120F is determined to beBtu34,874F)50F)(120Btu/lbmlbm)(0.9997.498()(12TTmcQpDiscussionReferring to Table A-9E the density and specific heat of water at 50ºFare:= 62.41 lbm/ft3and cp= 1.000Btu/lbmR and at 120ºF are:= 61.71 lbm/ft3and cp= 0.999 Btu/lbmR. We evaluated the water properties at an averagetemperature of 85ºF. However, we could have assumed constant properties and evaluated properties at the initial temperatureof 50ºF or final temperature of 120ºF without loss of accuracy.120F50FWater

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1-71-21A house is heated from 10C to 22C by an electric heater, and some air escapes through the cracks as the heated air inthe house expands at constant pressure. The amount of heat transfer to the air and its cost are to be determined.Assumptions1Air as an ideal gas with a constant specific heats at room temperature.2The volume occupied by the furnitureand other belongings is negligible.3The pressure in the house remains constant at all times.4Heat loss from the house to theoutdoors is negligible during heating.5The air leaks out at 22C.PropertiesThe specific heat of air at room temperature iscp= 1.007 kJ/kgC.AnalysisThe volume and mass of the air in the house are32m600m))(3m200(ght)space)(heifloor(Vkg9.747K)273.15+K)(10/kgmkPa287.0()mkPa)(6003.101(33RTPmVNoting that the pressure in the house remains constant during heating, the amount of heatthat must be transferred to the air in the house as it is heated from 10 to 22C isdetermined to bekJ9038C)10C)(22kJ/kgkg)(1.0079.747()(12TTmcQpNoting that 1 kWh = 3600 kJ, the cost of this electrical energy at a unit cost of $0.075/kWh is$0.195/kWh)kWh)($0.073600/9038(energy)ofcostused)(Unit(Energy=CostEnegyTherefore, it will cost the homeowner about 19 cents to raise the temperature in his house from 10 to 22C.22C10CAIR

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1-81-22An electrically heated house maintained at 22°C experiences infiltration losses at a rate of 0.7 ACH. The amount ofenergy loss from the house due to infiltration per day and its cost are to be determined.Assumptions1Air as an ideal gas with a constant specific heats at room temperature.2The volume occupied by the furnitureand other belongings is negligible.3The house is maintained at a constant temperature and pressure at all times.4Theinfiltrating air exfiltrates at the indoors temperature of 22°C.PropertiesThe specific heat of air at room temperature iscp= 1.007 kJ/kgC.AnalysisThe volume of the air in the house is32m450m))(3m150(ght)space)(heifloor(VNoting that the infiltration rate is 0.7 ACH (air changes per hour) and thus the air inthe house is completely replaced by the outdoor air 0.724 = 16.8 times per day, themass flow rate of air through the house due to infiltration iskg/day8485K)273.15+K)(5/kgmkPa287.0(day)/m045kPa)(16.86.89()ACH(33houseairairooooRTPRTPmVVNoting that outdoor air enters at 5C and leaves at 22C, the energy loss of this house per day iskWh/day40.4=kJ/day260,145C)5C)(22kJ/kg.007kg/day)(1.8485()(outdoorsindoorsairinfiltTTcmQpAt a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is$3.31/day0.082/kWh)kWh/day)($4.40(energy)ofcostused)(Unit(Energy=CostEnegy1-23Water is heated in an insulated tube by an electric resistance heater. The mass flow rate of water through the heater is tobe determined.Assumptions1Water is an incompressible substance with a constant specific heat.2The kinetic and potential energy changesare negligible,kepe0.3Heat loss from the insulated tube is negligible.PropertiesThe specific heat of water at room temperature iscp= 4.18 kJ/kg·C.AnalysisWe take the tube as the system. This is acontrol volumesince mass crosses the system boundary during the process.Weobservethat this is a steady-flow process since there is no change with time at any point and thus0and0CVCVEm, there is only one inlet and one exit and thusmmm21, and the tube is insulated. The energybalance for this steady-flow system can be expressed in the rate form as)(0)peke(since012ine,21ine,energiesetc.potential,kinetic,internal,inchangeofRate(steady)0systemmassandwork,heat,bynsferenergy tranetofRateTTcmWhmhmWEEEEEpoutinoutinThus,kg/s0.0266C)1506)(CkJ/kg4.18(kJ/s5)(12e,inTTcWmp5C0.7 ACH22CAIR15C60C5 kWWATER

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1-91-24Liquid ethanol is being transported in a pipe where heat is added to the liquid. The volume flow rate that isnecessary to keep the ethanol temperature below its flashpoint is to be determined.Assumptions1Steady operating conditions exist.2The specific heat and density of ethanol are constant.PropertiesThe specific heat and density of ethanol are given as 2.44 kJ/kg∙K and 789 kg/m3, respectively.AnalysisThe rate of heat added to the ethanol being transported in the pipe is)(inoutTTcmQpor)(inoutTTcQpVFor the ethanol in the pipe to be below its flashpoint, it is necessary to keepToutbelow 16.6°C. Thus, the volume flow rateshould beK)106.16)(KkJ/kg44.2)(kg/m789(kJ/s20)(3inoutTTcQpV/sm0.001573VDiscussionTo maintain the ethanol in the pipe well below its flashpoint, it is more desirable to have a much higher flow ratethan 0.00157 m3/s.

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1-101-25A 2 mm thick by 3 cm wide AISI 1010 carbon steel strip is cooled in a chamber from 597 to 47°C to avoidinstantaneous thermal burn upon contact with skin tissue. The amount of heat rate to be removed from the steel strip is to bedetermined.Assumptions1Steady operating conditions exist.2The stainless steel sheet has constant specific heat and density.3Changesin potential and kinetic energy are negligible.PropertiesFor AISI 1010 carbon steel, the specific heat of AISI 1010 steel at (597 +47)°C/ 2 =322°C= 595 K is682J/kg∙K(by interpolation from Table A-3), and the density is given as 7832 kg/m3.AnalysisThe mass of the steel strip being conveyed enters and exits the chamber at a rate ofVwtmThe rate of heat being removed from the steel strip in the chamber is given askW176K)47597(K)J/kg682)(m002.0)(m030.0)(m/s1)(kg/m7832()()(3outinoutinremovedTTVwtcTTcmQppDiscussionBy slowing down the conveyance speed of the steel strip would reduce the amount of heat rate needed to beremoved from the steel strip in the cooling chamber. Since slowing the conveyance speed allows more time for the steel stripto cool.

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1-111-26Liquid water is to be heated in an electric teapot. The heatingtime is to be determined.Assumptions1Heat loss from the teapot is negligible.2Constant propertiescan be used for both the teapot and the water.PropertiesThe average specificheats are given to be 0.7 kJ/kg·K for theteapot and 4.18 kJ/kg·K forwater.AnalysisWe take the teapot and the water in it as the system, which isa closed system (fixed mass). The energy balance inthis casecan be expressedasteapotwatersysteminenergiesetc.potential,kinetic,internal,inChangesystemmassandwork,heat,bynsferenergy traNetUUUEEEEoutinThen the amount of energy needed to raise the temperature ofwater and theteapot from 15°C to 95°C iskJ3.429C)15C)(95kJ/kgkg)(0.7(0.5C)15C)(95kJ/kgkg)(4.18(1.2)()(teapotwaterinTmcTmcEThe 1200-W electric heating unit will supply energy at a rate of 1.2 kW or1.2 kJ per second. Therefore, the time needed forthis heater to supply429.3 kJ of heat is determined frommin6.0s358kJ/s2.1kJ3.429nsferenergy traofRatensferredenergy traTotaltransferinEEtDiscussionIn reality, it will take more than 6 minutes to accomplish this heating process since some heat loss is inevitableduring heating. Also, the specific heat units kJ/kg · °C and kJ/kg · K are equivalent, and can be interchanged.1-27It is observed that the air temperature in a room heated by electric baseboard heaters remains constant even though theheater operates continuously when the heat losses from the room amount to9000 kJ/h. The power rating of the heater is to bedetermined.Assumptions1Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of-141C and 3.77 MPa.2The kinetic and potential energy changes are negligible,kepe0.3The temperature of theroom remains constant during this process.AnalysisWe take the room as the system. The energy balance in this case reduces tooutineoutineoutinQWUQWEEE,,energiesetc.potential,kinetic,internal,inChangesystemmassandwork,heat,bynsferenergy traNet0sinceU=mcvT= 0 for isothermal processes of ideal gases. Thus,kW2.5kJ/h3600kW1kJ/h0009,outineQWAIRWe

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1-121-28The resistance heating element of an electrically heated house is placed in a duct. The air is moved by a fan, and heat islost through the walls of the duct. The power rating of the electric resistance heater is to be determined.Assumptions1Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141C and 3.77 MPa.2The kinetic and potential energy changes are negligible,kepe0.3Constant specific heats atroom temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications.PropertiesThe specific heat of air at room temperature iscp= 1.007 kJ/kg·C (Table A-15).AnalysisWe take the heating duct as the system. This is acontrol volumesince mass crosses the system boundary during theprocess. We observethat this is a steady-flow process since there is no change with time at any point and thus0and0CVCVEm. Also, there is only one inlet and one exit and thusmmm21. The energy balance for thissteady-flow system can be expressed in the rate form as)(0)peke(since012infan,outine,2out1infan,ine,energiesetc.potential,kinetic,internal,inchangeofRate(steady)0systemmassandwork,heat,bynsferenergy tranetofRateTTcmWQWhmQhmWWEEEEEpoutinoutinSubstituting, the power rating of the heating element is determined to bekW2.97C)C)(5kJ/kg7kg/s)(1.00(0.6+kW)3.0()kW25.0(ine,WWe300 W250 W

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1-131-29A room is heated by an electrical resistance heater placed in a short duct in the room in 15 min while the room is losingheat to the outside, and a 300-W fan circulates the air steadily through the heater duct. The power rating of the electric heaterand the temperature rise of air in the duct are to be determined.Assumptions1Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141C and 3.77 MPa.2The kinetic and potential energy changes are negligible,kepe0.3Constant specific heats atroom temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications.3Heat loss from the duct is negligible.4The house is air-tight and thus no air is leaking in or out of the room.PropertiesThe gas constant of air isR= 0.287 kPa.m3/kg.K (Table A-1). Also,cp= 1.007 kJ/kg·K for air at roomtemperature (Table A-15) andcv=cp–R= 0.720 kJ/kg·K.Analysis(a) We first take the air in the room as the system. This is a constant volumeclosed systemsince no mass crosses thesystem boundary. The energy balance for the room can be expressed as)()()(1212outinfan,ine,outinfan,ine,energiesetc.potential,kinetic,internal,inChangesystemmassandwork,heat,bynsferenergy traNetTTmcuumtQWWUQWWEEEvoutinThe total mass of air in the room iskg284.6)K288)(K/kgmkPa0.287()m240)(kPa98(m240m865331133RTPmVVThen the power rating of the electric heater is determined to bekW4.93s)60C)/(181525)(CkJ/kg0.720)(kg284.6()kJ/s0.3()kJ/s200/60(/)(12infan,outine,tTTWQWmcv(b) The temperature rise that the air experiences each time it passes through the heater is determined by applying the energybalance to the duct,TcmhmWWhmQhmWWEEpoutininfan,ine,20out1infan,ine,0)peke(sinceThus,C6.2)KkJ/kg1.007)(kg/s50/60(kJ/s)0.34.93(infan,ine,pcmWWT568 m3We300 W200 kJ/min

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1-141-30The ducts of an air heating system pass through an unheated area, resulting in a temperature drop of the air in the duct.The rate of heat loss from the air to the cold environment is to be determined.Assumptions1Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141C and 3.77 MPa.2The kinetic and potential energy changes are negligible,kepe0.3Constant specific heats atroom temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications.PropertiesThe specific heat of air at room temperature iscp= 1.007 kJ/kg·C (Table A-15).AnalysisWe take the heating duct as the system. This is acontrol volumesince mass crosses the system boundary during theprocess. Weobservethat this is a steady-flow process since there is no change with time at any point and thus0and0CVCVEm. Also, there is only one inlet and one exit and thusmmm21. The energy balance for thissteady-flow system can be expressed in the rate form as)(0)peke(since021out21energiesetc.potential,kinetic,internal,inchangeofRate(steady)0systemmassandwork,heat,bynsferenergy tranetofRateTTcmQhmQhmEEEEEpoutoutinoutinSubstituting,kJ/min272C3CkJ/kg1.007kg/min90outTcmQp90 kg/minAIRQ·

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Solution Manual For Heat And Mass Transfer: Fundamentals And Applications, 5th Edition - Page 16 preview image

1-151-31Air is moved through the resistance heaters in a 900-W hair dryer by a fan. The volume flow rate of air at the inlet andthe velocity of the air at the exit are to be determined.Assumptions1Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141C and 3.77 MPa.2The kinetic and potential energy changes are negligible,kepe0.3Constant specific heats atroom temperature can be used for air.4The power consumed by the fan and the heat losses through the walls of the hair dryerare negligible.PropertiesThe gas constant of air isR= 0.287 kPa.m3/kg.K (Table A-1). Also,cp= 1.007 kJ/kg·K for air at roomtemperature (Table A-15).Analysis(a) We take the hair dryer as the system. This is acontrol volumesince mass crosses the system boundary during theprocess. We observe that this is a steady-flow process since there is no change with time at any point and thus0and0CVCVEm, and there is only one inlet and one exit and thusmmm21. The energy balance for thissteady-flow system can be expressed in the rate form as)(0)peke(since012ine,2010infan,ine,energiesetc.potential,kinetic,internal,inchangeofRate(steady)0systemmassandwork,heat,bynsferenergy tranetofRateTTcmWhmQhmWWEEEEEpoutoutinoutinThus,kg/s0.03575C)2550)(CkJ/kg1.007(kJ/s0.912e,inTTcWmpThen,/sm0.03063)/kgm0.8553)(kg/s0.03575(/kgm0.8553kPa100)K298)(K/kgmkPa0.287(31133111vVvmPRT(b) The exit velocity of air is determined from the conservation of mass equation,m/s5.5224322222233222m1060)/kgm0.9270)(kg/s0.03575(1/kgm0.9270kPa100)K323)(K/kgmkPa0.287(AmVVAmPRTvvvT2=50CA2= 60 cm2P1= 100 kPaT1= 25CWe=900 W·

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