热力学第一定律Energy and the First Law of Thermodynamics.ppt

Chapter2EnergyandtheFirstLawofThermodynamics,深刻认识热力学第一定律的实质能量守恒了解热和功是系统与外界交换能量的两种方式，定义、特性及计算方法热力学第一定律能量方程的基本表达式运用热力学第一定律进行工程实践分析,本章要求,Workisdonebyasystemonitssurroundingsifthesoleeffectoneverythingexternaltothesystemcouldhavebeentheraisingofaweight.功是系统间相互作用而传递的能量当系统完成功时，其对外界的作用可以用在外界举起重物的单一效果来代替。注意：“举起重物”在热力学定义中是“过程产生的效果相当于重物举起，而不一定是真实举起。,Work,SignconventionandNotation,W0:workdonebythesystemW0,PerpetualMotionMachineoftheFirstKind(PMM1),Proof:ForacycleconsistsoftwoprocessesAandB,onecanwritecycdWadi=1-A-2dWadi+2-B-1dWadisincecycdWadi01-A-2dWadi-2-B-1dWadi1-B-2dWadiwhichviolatestheFirstLaw.ThereforePMM1isimpossible.,PMM1,Considertwocycleslinkingtwostates1and2ofasystemthatinvolvesbothworkandheat.SinceforacyclecycQ-cycW=0Therefore,1-A-2Q+2-B-1Q=1-A-2W+2-B-1W1-A-2Q+2-C-1Q=1-A-2W+2-C-1W,FirstLawforaProcessinvolvingbothWorkandHeat,and,2-B-1Q-2-C-1Q=2-B-1W-2-C-1Wor,2-B-1(QW)=2-C-1(QW)=E1E2Or,dE=QW;de=q-w,TheenergydefinedfromtheFirstLawconsistsofalldifferentformsofenergyasubstancemayhave.Theyinclude:E=U(internalenergy)+KE(kineticenergy)+PE(potentialenergy)+ChE(chemicalenergy)+NuE(nuclearenergy)+.KE=1/2(mV2)PE=mgh,Energy,Therefore,Q-W=dE=dU+dKE+dPE+dChE+dNuE+Forasystemcontainsonlyinternalenergy,(举例参见P44.)Q-W=dU;热力学第一定律的表达式Or,q-w=duperunitmass,Thefirstlawofthermodynamics,Energy,Energycanbetransformedfromoneformtoanotherandtransferredbetweensystems.Forclosedsystems,energycanbetransferredbyworkandheattransferInalltransformationsandtransfers,thetotalamountofenergyisconserved,Energy,Energyistheabilitytodowork.IthasunitsofJoules.Itisa“UnitofExchange”.Example1car=$20k1house=$100k5cars=1house,=,EnergyEquivalents,Whatisthecasefornuclearpower?1kgcoal42,000,000joules1kguranium82,000,000,000,000joules1kguranium2,000,000kgcoal!,KineticEnergy,KineticEnergyistheenergyofmotion.KineticEnergy=massspeed2,TheworkoftheforceFsasthebodymovesfromS1toS2alongthepathisexpressedbyfollowing,KineticEnergy,根据牛顿第二定律的推导，式2.6说明系统动能的变化等于合力F对系统做的功。,PotentialEnergy,Accordingly，thetotalworkequalsthechangeinkineticenergy.,式2.9说明，施加在物体上的所有力除了重力外，对物体所做的功，等于物体动能变化与势能变化的和。,PotentialEnergy,从一个特殊的例子来说明能量守恒定律：当物体仅仅受到重力时，那么2.9式的右边等于零，则有：,通过2.11式也表达了能量可以从一种形式转换成另外一种形式。,1meter,nail,Forasysteminvolvesmovingboundary,theworkinvolvedisgivenby:W=-FextdXworkdoneonthesystemFext=-pressurexAreaW=pAdX=pdV“+”dVworkdonebythesystem;“-”dVworkdoneonthesystem,ExpansionandCompressionWork,Forasystemsurroundedbyanatmosphericair,theusefulworkisW=(p-p0)dV=pdV-p0dVW1-2=pdV-p0dVOr,Example2.1,Problem:Agasinapiston-cylinderassemblyundergoesanexpansionprocessforwhichtherelationshipbetweenpressureandvolumeisgivenby,Whatwehaveknown:Initialpressureis3bars,Initialvolumeis0.1m3Finalvolumeis0.2m3Determine:Workfortheprocess,inKJ.a)n=1.5,b)n=1.0,c)n=0.,SchematicandGivenData,ThegasisaclosedsystemThemovingboundaryistheonlyworkmodeTheexpansionisapolytropicprocess,Assumptions,Analysis,在上述分析中，不同过程所做的功都可用p-V图上过程曲线下面的面积进行说明，面积大小与做功数值相符合。所计算出的功值取决于经历的具体过程和最终状态。关于多变过程的假设是非常重要的，如果给定的压力体积关系是从实验数据拟合获得的，那么仅当测量所得的压力等于施加在活塞截面上的压力时，pdV的数值才会提供一个比较可靠的做功估计值。,Comments,Quasistaticworkhasthefollowingfeatures:1.TheforceFdependsonlyonthestateofthesystemandisdependentofthedirectionofthedisplacementx.2.Theprocessisreversible.Thatistheinitialstateofthesystemcanberestoredbyreversingtheprocess.3.ThevalueofforceFremainsfiniteasdxapproacheszero.,Quasi-staticandNon-quasistaticWork,Fornon-quasistaticwork:1.Fdependsontherateofchangeofstate.2.Theworkinteractionisunidirectional.3.Fapproacheszeroasdxgoestozero.,系统在准平衡过程中完成的功量称为准静功准静功可以仅通过系统内部的参数来描述，而无须考虑外界的情况对于非平衡过程，系统完成的功量需要利用对系统进行实际测量来确定；而准平衡过程的概念对工程上获得可以接受的近似结果，带来很大的方便。,准静功,Thenumberofindependentpropertiesrequiredtodefineastateofasystemisequaltooneplusthenumberofpossiblequasistaticworkmodes.对于组成一定封闭系的给定平衡状态而言，可用N1个独立的状态参数来限定它。这里N是系统可能出现的准静功形式的数量，1则是考虑了系统与外界的热交换。,TheStatePrinciple,AsimplecompressiblesystemisdefinedasoneforwhichtheonlyrelevantquasistaticworkinteractionisboundarypdVwork.Forsuchasystemthenumberofindependentpropertiesrequiredtodefineastateistwo.对于简单可压缩系而言，热力系与外界交换的准静功只有气体的体积变化功（膨胀或压缩）一种形式，根据状态公理，决定该系统的平衡状态的独立状态参数只有2个,SimpleCompressibleSystem,dU=QW;or,du=qwperunitmassgeneraldifferentialformw=-pdv;onlytrueforasimplesystemu=qw;generalintegratedform=qpvtrueforasimplesystemSometextbookswillusedu=q+w;workdonebysystemin“-”u=q+w,FirstLawforaSimpleCompressibleSystem,Consideragasconfinedinacylinder-pistonarrangement,thepistonisloadedinsuchawaythatthepressureofthegasisconstant.FromtheFirstLaw,dE=Q-WForasimplesystem,dE=dUandW=pdVdU=QpdV;orQ=dU+pdV,EnthalpyandSpecificHeats,Butthepressureismaintainedconstant,dp=0.Therefore,Q=d(U+pV)DefineU+pVHEnthalpydQ=dH,Forasimplesystem,u=u(T,v)andh=h(T,p)du=(u/T)vdT+(u/v)Tdvdh=(h/T)pdT+(h/p)TdpDefine,Constantvolumespecificheatcvbycv=(u/T)v;du=cvdT+(u/v)TdvConstantpressurespecificheatcpbycp=(h/T)p;dh=cpdT+(h/p)Tdp,HeatCapacityforConstantVolumeProcesses(Cv),Heatisaddedtoasubstanceofmassminafixedvolumeenclosure,whichcausesachangeininternalenergy,U.Thus,Q=U2-U1=DU=mCvDTThevsubscriptimpliesconstantvolume,Heat,Qadded,m,m,DT,insulation,HeatCapacityforConstantPressureProcesses(Cp),Heatisaddedtoasubstanceofmassmheldatafixedpressure,whichcausesachangeininternalenergy,U,ANDsomePVwork.,CpDefined,Thus,Q=DU+PDV=DH=mCpDTThepsubscriptimpliesconstantpressureH,enthalpy.isdefinedasU+PV,soDH=D(U+PV)=DU+VDP+PDV=DU+PDVExperimentally,itiseasiertoaddheatatconstantpressurethanconstantvolume,thusyouwilltypicallyseetablesreportingCpforvariousmaterials.,JoulesExperiment,Jouleshowedthatmechanicalenergycouldbeconvertedintoheatenergy.,F,M,Dx,H2O,DT,W=FDx,Introduction,1843-1848年，英国酿酒商JamesPrescottJoule(1818-1889)以确凿无疑的定量实验结果为基础，论述了能量守恒和转化定律。焦耳的热功当量实验是热力学第一定律的实验基础。,Joule(1818-1889),TheFirstlawofaclosedsysteminacycle,容器、搅拌器和水组成一个热力系，这是一个闭口系统。让热力系从初始态经历一个循环过程而回到原态。例如：使容器绝热，让重物落下使搅拌器回转。此时有功加入到热力系中，依靠摩擦功转变成热，使水温升高。然后水对环境放热，温度下降而回到原态。利用不同重物并多次测量后焦耳首先发现，加入的功量总是与放出的热量成比例：即：,1calorie=4.184Joules,Wheredidtheenergygo?,BytheFirstLawofThermodynamics,theenergyweputintothewater(eitherworkorheat)cannotbedestroyed.Theheatorworkaddedincreasedtheinternalenergyofthewater.Thisistheenergystoredintheatomsandmoleculesthatmakeupthewater;theymovefaster.,Example2.2,Aclosedsysteminitiallyatrestonthesurfaceoftheearthundergoesaprocessforwhichthereisanetenergytransfertothesystembyworkofmagnitude200Btu.Duringtheprocessthereisanetheattransferofenergyfromthesystemtoitssurroundingsof30Btu.Attheendoftheprocess,thesystemhasavelocityof200ft/satanelevationof200ft.Themassofthesystemis50lb,andthelocalaccelerationofgravityisg=32.0ft/s2.Determinethechangeofinternalenergyofthesystemfortheprocess,inBtu.,Solution,Known:ms=50lb,g=32.0ft/s2Duringtheprocess,Workdoneonthesystem,W=-200BtuHeattransferfromthesystem,Q=-30Btutheinitialstate,systemisatrest,therefore,V1=0thefinalstate,V2=200ft/s,h=200ftTofind:thechangeofinternalenergyofthesystemfortheprocess,inBtu.,Schematicdiagram,Assumptions,Aclosedsystemisunderconsideration.Attheendoftheprocess,thesystemismovingwithauniformvelocity.Thelocalaccelerationofgravityisconstantatg=32.0ft/s2.,Analysis,TheEquationofanenergybalance,Substitutingthevalueoftheparameterinabove,wecanevaluate:,Comments,ThepositivesignforUindicatesthattheinternalenergyofthesystemincreaseduringtheprocess.NotetheunitconversionswhenusingEnglishsystem:theenergy“balancesheet”,Comments,3.theenergy“balancesheet”Input:200Btu(work)Changeofsystemenergy+39.9Btu(KE)+12.8Btu(PE)Output:30Btu(heattransfer)+117.3Btu(internalenergy)+170.0BtuThenetinputexceedsthenetoutputby170Btu,andthesystemenergyincreasesbythisamount.,Example2.3,Consider5kgofsteam(watervapor)containedwithinapiston-cylinderassembly.Thesteamundergoesanexpansionfromstate1,wherethespecificinternalenergy(theinternalenergyperunitmass)isu1=2709.9kJ/kg,tostate2,whereu2=2659.6kJ/kg.Duringtheprocess,thereisaheattransferofenergytothesteamwithmagnitudeof80kJ.Also,apaddlewheeltransferenergytothesteambyworkintheamountof18.5kJ.Thereisnosignificantchangeinthekineticorpotentialenergyofthesteam.Determinetheamountofenergytransferbyworkfromthesteamtothepistonduringtheprocess,inkJ.,Solution,Known:ms=5kg,u1=2709.9kJ/kgu2=2659.6kJ/kgQ=+80kJWpw=-18.5kJTofind:Wpiston,Schematicdiagram,Assumptions,Thesteamisaclosedsystem.Thereisnochangeinthekineticorpotentialenergyofthesteam.,Analysis,Anenergybalancefortheclosedsystemillustratedby:KE+PE+U=Q-WThen,theenergyequationbecomes,Thenetworkfromthesystemconsistsoftwomodes:fromapaddlewheelandbythepiston.,Analysis,So,Incorporatingtheaboveexpressions,theresultsis,Substitutingvaluesintoabove,Wpiston=+350kJ,Comments,NoticethepositivesignforWpiston,whichimpliestheenergytransferisfromthesteamtothepistonasthesystemexpandsduringthisprocess.Theenergy“balancesheet”isasfollowing,InputOutput18.5(work,paddlewheel)350(work,piston)80.0(heattransfer)Total:98.5350,Example2.4,Example2.5,Therateofheattransferbetweenacertaineletricmotorand