山东大学:《工程热力学》课程PPT教学课件(双语)第五章 M,E analysis of control system

Chap4 SummaryGeneral:Ow,=Pdy,P-VdiagramThework associatedBoundary workwithexpansionandIsobaricprocess,Pvn=const PolytropicprocesscompressionisothermalprocessofanidealgasW,=P,V,ln(V2/V)AE.(KJ)E-EouEnergyBalancesyatenNet enengy tnansferChange in intemal,kioctic.foranysystemandanyprocessbyhealwortandmasspoteatial,tc,enengis1stlawforclosedsystem=WWWWQ=Qnetin=0-0aet.ououtEnergyBalanceforclosedsystemO-W=AEThe energyrequired to raise T of 1kg ofa substanceby 1K,kJ/(kg.k)ahSpecific heatCv atconstant1Cp atconstantPCp22aTIdealgases:Cp=Cv+R;k=Cp/CvForsolidsand liquids,Cp=Cv=cU,h fromtable.Ah=h-hc,(T)dTepavg(T-T)Changes ofu, h for ideal gasesByusingCvorCpUsingaveragespecificheats.Cv,avg=Cv(T2/2+T1/2)Cp,avg=Cp(T2/2+T1/2)Forimcompressiblec(T)dT=Cav(T2-T)Ah=Au+VAPChanges of u,h for solid/fluidsubstances
Chap4 Summary 1 The work associated with expansion and compression 1st law for closed system General: δWb=Pdv, P-V diagram U,h from table. Specific heat Changes of u, h for ideal gases Changes of u, h for solid/fluid Boundary work Energy Balance for any system and any process Isobaric process, PVn=const Polytropic process isothermal process of an ideal gas Wb=P1V1ln(V2/V1) The energy required to raise T of 1kg of a substance by 1K, kJ/(kg.k) By using Cv or Cp Using average specific heats. Cv,avg=Cv(T2/2+T1/2) Cp,avg=Cp(T2/2+T1/2) Energy Balance for closed system Cv at constant V Cp at constant P Ideal gases: Cp=Cv+R; k=Cp/Cv For solids and liquids, Cp=Cv=C For imcompressible substances

Chap5 SummaryConservationofmassprinciple:thenetmasstransfertoorfroma controlvolumeduringatimeintervalisequaltotheConservationofmassnetchangeinthetotal masswithinthecontrol volumedmcymoutAmcmindtoutinFlow energy:Totalenergyofaflowingfluidof1kg2(kJ/kg)Waow=PV=h+ke+pe=hgzEnergyofflowingfluid2V2Rate of energy transportE=mo=gzIm2V=大mMassbalanceIncompressibleOUDUinSingle streamPVA=PVA2nmSteadyflowProcess/systemV=V,-VA=VA2IncompressibleSinglestreamIFor△KE=0,△PE=0q-W=h2-hTurbines,compressorsNozzles,diffusersHeatexchangersSteadyflowdevicesPipeand ductflowThrottling valvesMixingchambers2
Chap5 Summary 2 Conservation of mass principle: the net mass transfer to or from a control volume during a time interval is equal to the net change in the total mass within the control volume. Energy of flowing fluid Nozzles, diffusers Steady flow Process/system Steady flow devices Conservation of mass Mass balance Flow energy: Total energy of a flowing fluid of 1kg Enthalpy is associated with the energy pushing the fluid into or out of CV Single stream Incompressible Incompressible Single stream Energy balance for general steady flow systems For For single stream △KE=0, △PE=0 Turbines, compressors Throttling valves Mixing chambers Heat exchangers Pipe and duct flow Rate of energy transport

5-1 Conservation of massOxygen reacts with2kg16kg18kg+hydrogenH202H,OWaterduringan18kg2kg16kg+H2H,Oelectrolysisprocess02Massisconservedduring a process
5-1 Conservation of mass • Oxygen reacts with hydrogen • Water during an electrolysis process • Mass is conserved during a process. 3

5-1 Conservation of massMassisconservedduringaprocessMass and energy can be converted to each otheraccording to the formula(质能方程) proposed by AlbertEinstein.E = mc2However, during the energy interactions in practice: thechange of mass is very small and can be neglectede.g: 1kg H20 is formed by H2 and O2, release energy 15879kJ (1.76*10-10kg)
5-1 Conservation of mass • Mass is conserved during a process. • Mass and energy can be converted to each other according to the formula(质能方程) proposed by Albert Einstein. • However, during the energy interactions in practice: the change of mass is very small and can be neglected. e.g: 1kg H2O is formed by H2 and O2 , release energy 15879kJ (1.76*10-10kg) 4

5-1 Conservation of mass: Closed system: no change on mass of thesystemControl volume (open system): mass cancross the boundary. We need to know themass in and out of the system
5-1 Conservation of mass • Closed system: no change on mass of the system • Control volume (open system): mass can cross the boundary. We need to know the mass in and out of the system. 5

5-1 Conservation of massMass flowrate m : the amount of mass flowingthrough a cross section per unit timeAcrosssmallareaOm=pV,dASm(kg/s)pV,dA.mAcross entire areaAControlsurfacedAUse Vavg, then(kg/s)mDVavgA
5-1 Conservation of mass • Mass flow rate : the amount of mass flowing through a cross section per unit time. 6 Across small area Across entire area Use Vavg, then

5-1 Conservation of mass· Volume flow ratee y : the volume of fluid flowingthrough a cross section per unit time(m/s)7Vn dA.= VavgA,= VAcVavgAverageVolumeflowratevelocityV=VavgAcCrosssectionRelations of m andVm = pV=
5-1 Conservation of mass • Volume flow rate : the volume of fluid flowing through a cross section per unit time. • Relations of and 7 Average velocity Volume flow rate

5-1 Conservation of mass Conservation of mass principle: the netmass transfer to or from a control volumeduring a time intervalis egual to the netchange in the total mass within the controlvolume.Total mass enteringTotal massleavingNetchangein massthe CV during tthe CV during AtwithintheCVduringAtmin-mout=dmcv/dtmin- mout= AmcyIn rate formMass balance
5-1 Conservation of mass • Conservation of mass principle: the net mass transfer to or from a control volume during a time interval is equal to the net change in the total mass within the control volume. 8 Mass balance In rate form

mout=dmcy/dtminChange in CV:EFordifferential volume:1dm0dm=pdvdAControlvolume(CV)Total mass within the CV :Controlsurface(CS)dymeRate of change of mass in CV:dmeydpdvdtdt'Cv
• Change in CV: • For differential volume: • Total mass within the CV : • Rate of change of mass in CV: 9

=dmcv/dtmoutmiMass flowin or out of CV:VEFordifferentialmassflow rate:ndmSm=pVndAVn=Vcose=V.n0dAControlSm=pV,dA=p(Vcos0)dA=p(V.n)dAvolume(CV)Controlsurface(CS)Netmassflowrate(overtheentirecontrosurface) :o(V.n)dASmpV,dAmnetCS'CsCsmoutmin10
• Mass flow in or out of CV: • For differential mass flow rate: • Net mass flow rate (over the entire control surface) : 10
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