中法charpter2a-化学恰当与燃烧温度

热能工程 授L‘EcoleCentraledeMechanicApplication–GasTurbine航空燃气涡轮发动Combustion&燃烧与燃Lecture2-CombustionAssoc.Prof.LIUYuying(SchoolofEnergyandPower热能工程 授MechanicApplication–GasTurbine第1章航空燃气涡第1章航空燃气涡轮发动机的基本工作原 10学的技术难点2•航空燃气涡轮发动机的基本工作4•压气机部件的工2•涡轮部件的工作2第2章的结构强度与振 10学引1叶片强22叶片振2转子振动与临界转2疲1第3章燃烧与燃烧 10学燃烧与1燃烧基4主燃41热能工程 授Part3:Combustionand燃烧与燃烧室概述（1学时燃烧基础知识（4学时火焰速度及可燃边主燃烧室（4学时加力燃烧室（1学时热能工程 授3.2CombustionFlameDropletevaporationandFlamespeedand3.4.5Flame热能工程 授TheaimofstoichiometryistodetermineexactlyhowmuchairmustbeusedtocompleyoxidizethefueltotheproductsCO2,H2O,N2,SO2.(完全氧化需要多Thecombustionintensitybetweenafuelandanoxidizerdependsontheirrelativeconcentrations. Whentheirconcentrationratioischemicallycorrectinthatallthereactantscanbetotallyconsumedinthereaction,thenthecombustionintensityisclosetothehighest,wecallthismodeofburningStoichiometriccombustion.（化学配平正好合适，强SomeSomeDefinitionsabout(Stoichiometricair(G0orL0)理论空气需(stoichiometric)fuel–airratio,F/A(f）空气比，油气(stoichiometric)air-fuelratioA/F(1/f)空气Equivalenceratioφ(φ） Toindicatethedeviationofamixture’sconcentrationExcessairratioα过量空气热能工程 授Stoichiometricair理论空C nm(O3.76N)nCOmHO3.76nm

4

4 ––IfCH4,thenStoichiometricair=––IfthenStoichiometricair=–IfH2,thenStoichiometricair=热能工程 授Fuel–airratioF/A油气比与化学恰Fuel–airtheratioofthemassoffueltothemassofairinthefF/Amfuel/Thestoichiometricfue-airratioF/Amfuel

(F/A)stoic

热能工程 授Air-FuelRatio(A/F)-空燃theratioofthemassofairtothemassoffuelintheA/F=mAir/m mair=massofairinthefeedmfuel=massoffuelinthefeedFuel-Airratio:A/F==化学恰当空燃比=理论空气需要CnHma(O23.76N2)nCO2(m/2)H2O((A/Fm1热能工程 授Richmorefuelthannecessary(A/F)mixture<(A/F)stoichLeanmoreairthan(A/F)mixture>&&&&f0aG0adEquivalenceratio(当量比 理论上： 实际上： 系数为mf/(ma/G0)，即热能工程 授Equivalenceratio(当量比F/(F/mf/mamf1/热能工程 授Equivalenceratioshowsthedeviationofanactualmixturefromstoichiometricconditions.(F/

(A/F

(F/ (A/F >1,fuelrich<1,fuellean=1,stoichiometric热能工程 授eExcessairratioα空气消耗系数,也称余气系数、空气过剩系数，用α表定义：实际空气供给量L(G)与理论需要量L0(G0L&f& &&fG0

热能工程 授ExcessairratioExcessairratiodefined (A/F(A/FExcessairisexpressedasapercentageoverthestoichiometricrequirementandisdefinedactualA/FratiostoichiometricA/FratiostoichiometricA/FExcessairwillalwaysreducetheefficiencyofcombustion热能工程 授绝热火焰温度-DefinitionIntheabsenceofanyheatloss（adiabatic）tothesurroundings,whencombustioniscomplete,thetemperatureoftheproductswillreacha um,whichiscalledtheadiabaticflametemperatureofthereaction umtemperatureencounteredinacombustionchamberislowerthanthetheoreticaladiabaticflame热能工程 授ysis:Flame火焰燃烧过程中热平衡燃燃烧温热量的 的化学热， 发热量Q 带入的物理热Q

热量的支燃烧产物含有的物理热Q产＝VnC产t燃烧产物传给周围物体的热量Q燃烧条件造成的不完全燃烧热损失Q热能工程 授Heating 发热量/热热量(单位：kJ/Kg或kJ/Nm3)HHV:HigherheatingValue（热值）,calculatedwiththeassumptionthatallofthewaterintheproductshascondensedtoliquid.（grossheatingvalue，QGWLHV:LowerHeatingValue（低位热值）,supposingnoneofthewateriscondensed.(Netheatingvalue,QDW)发热量典型值-

热能工程 授发热量典型值-

热能工程 授发热量典型值-

热能工程 授热能工程 授ysis:Flame火焰温度分燃烧温度的宏观求解表Q＋Q＋QVC产T＋QQQ产 ＝Q低＋Q空＋Q燃－Q传－Q不－Q

T＝QT＝Q低＋Q空＋Q燃－QVn产

＝0，Q

热能工程 授Factors:AdiabaticFlame影响热热热热值高，理论燃烧温度也完全燃烧时，燃烧温度随余气系数增大而减小。这是因为空气消耗系数影响燃烧与空和空气的预热，增加了能量输入，有利于提高燃烧温空气中的氧富氧空气组织燃烧，可获得更高的燃烧温度，这是因为提高空气中的氧含量，氮气含量减少，即燃烧产物生成量减少，在燃烧过程中加热氮气所需能量热能工程 授热能工程 授热能工程 授热能工程 授反应物：C3H8热能工程 授AdiabaticFlameTemperatureK)（绝热火焰温度）ofFuelFuelEquivalenceRatioGaseousFuelFuelEquivalenceRatioGaseousEthane（乙烷Propane（丙烷Octane（辛烷LiquidOctane（辛烷No.2fuelDrySolid(25%Moisture)SolidFuels热能工程 授Calculation:AdiabaticFlame计算ChemicalEnergy（知识准备Standardenthalpyofformation（标准生成焓Absoluteenthalpy（绝对焓/总焓Enthalpyofcombustion(orJANAF定压绝热火焰温度与定容绝热火焰SecondlawChemical热能工程 授0Standardenthalpyofformation,hf0istheheatofreactionpermoleofproductformedisothermallyfromelementsintheirmoststableformsatthestandardreferencestatei.eP=1atmandT=25℃.Enthalpiesofformationarezerofortheelementsintheirnaturallyoccurringstate(orthemoststableform)atthereferencestate.fexample,hf热能工程 授Total(Absolute)enthalpy,h=h0+∆ Where∆hsisthesensibleenthalpy（显焓）correspondingtotemperaturechangefromTreftoT.热能工程 授Enthalpyof(combustion)reaction,∆hr∆hr=hf,prod-=Mihf,i-WhereMiisthemolenumberofeachproduct,andMjisthatofeachreactantHeatingvalue=-热能工程 授AboutJANAFJANAFistheacronymfor“JointArmy,NavyandJANAFisthemostcommonlyusedsourceofthermo-chemicalinformation,especiallyformilitaryandtheircontractors.AlmostallstudentsinUSAareaskedtobefamiliarwithJANAFTheinformationinJANAForiginateseitherfrommeasurementorfromcalculationsattributedtostatisticalmechanicsAsummaryisprovidedbyGlassman,orBorman,their热能工程 授Example1-生成Determinethetotalenthalpyofnitricoxide(NO)at298Kand1atm.Solution:sinceT=298kandP=1atm,sothetotalenthalpythestandardheatofThechemicalreactionforNOformationaccordingtotheJANAFtable∆h0 90.29/4.1868kcal/mol=21.57TherespectiveenergylevelsofN2,O2andNocanbesketchedbelow热能工程 授IttakesenergytocreateNOfromN2andMoregenerally,∆h0f>0meansenergyrequiredtoformthesubstance;while∆h0f<0meansenergyisreleasedtoformit热能工程 授Example2-Determinethetotalenthalpyofnitricoxide(NO)at1000Kand1atm.Totalenthalpy=standardheatofformation+sensible=90.29+22.23=112.52/4.1868=26.88Theenergylevel（能级）ofNO1000Kcanbesketched热能工程 授Example3-Determinethepowerextractedfromthefollowinghypotheticalreactor(1atm,completecombustion)；supposethevolumeflowrateofmethaneis2liter/sec(measuredat500K,1atm) First,anenergylevelysisis热能工程 授热能工程 授ThemajorproductsofthestoichiometriccombustionofmethanearegivenbyCH4+2(O2+3.76N2)→2H2O+CO2+7.52EnthalpiesfromJANAFaretabulatedasfT----0---00热能工程 授Sotheheatofreactioncorrespondingtothecombustionofmoleofmethaneinthereactor∆hr=Mihf,i-=2×(-193.73)+(-331.81)+7.52×-(-66.67)-2×6.09-7.52=-420.4552MJ/kgmolThatmeans,thereactantshaveahigherenergythantheproducts.（反应物具有更高的能量）Asaresult,theyreleaseenergyinthecombustionAreactionwith∆hr0iscalledan“exothermic（放热）whichisresponsiblefortheenergyreleasedbyAreactionwith∆hr0iscalledan“endothermic（吸热reaction,whichneedsasupplyofenergytoTheenergylevelsofthereactioncanbesketched热能工程 授Thepowerextractedfromthereactorcanbecalculated热能工程 授Power=-n(mole/sec)×∆hr=-PCH4×V/(RT)×∆hr(equationof=-NCH4/(NCH4+NO2+NN2)×Ptotal×=1/(1+2+7.52)×1.013×105×2×10-(8.314×500)×420.4552=1948热能工程 授**Calculationsfrom1stLawThe1stLaw:ConservationofEnergycanbechangedfromoneformtoanother,butitcannotbecreatedorTheenergyoftheuniverseisU=Q+U=changeinsystem’sinternalQ=W=热能工程 授Constantpressure/volumeadiabaticflame定压绝热火焰温度与定容绝热Constantpressureadiabaticflametemperature:thefuel-airmixtureburnsadiabaticallyatconstantpressure,suchasthesituationingasturbinecombustorsandfurnaces.Theabsoluteenthalpyofthereactantsattheinitialequalsthatoftheproductsatthefinalstate.（定压燃烧Constantvolumeadiabaticflametemperature:thefuel-airmixtureburnsadiabaticallyatconstantvolume,suchasthesituationininternalcombustionengines.Theabsoluteinternalenergyofthereactantsattheinitialstateequalsthatoftheproductsatthefinalstate.（定容燃热能工程 授Constantpressurevs.ConstantU=Q+Enthalpy:H=U+Atconstantpressure:ΔH=热能工程 授Constant-PressureHreac(Tinit,init)Hprod(Tad,Pf热能工程 授Constant-volumeUreac(Tinit,Pinit)Uprod(Tad,PfwhereUistheabsolute(orstandardized)internalenergyofthemixture.HreacH

V(initPf)热能工程 授Example:constantpressureflameEstimatetheconstantpressureadiabaticflametemperatureforthecombustionofastoichiometricCH4-airmixtureattheinitialstateof1atmand298K.Usethefollowingassumptions(定比热容忽略离解简化算法CompletecombustionandnoTheproductmixtureenthalpiesareestimatedusingconstantspecificheatsevaluatedat1200K(≈0.5(Ti+Tad),Tadisguessedtobeabout2100K).Solution:mixtureCH4+2(O2+3.76N2)→CO2+2H2O+CH4+2(O2+3.76N2)→CO2+2H2O+CH4+2(O2+3.76N2)→CO2+2H2O+Enthalpyofformationat298KSpecificheatat1200K00Hprod=Σnjhj0+ΣnjCpj(Tad-(2)[-241845+43.87(Tad-298)] Tad=2318

CommentsTad=2318

热能工程 授Comparingtheaboveresultswiththeexperimentpositionbasedcomputationshownasbefore(2250K)showsthatthesimplifiedapproachoverestimates（高估）Tadbyslightlylessthan100K.Removingassumption2andre-calculatingTadusingvariableCp(orfindthesensibleenthalpyfromJANAFHprod=Σnh0+Σ

CwillyieldTad=2328

j∫Thenweconcludethatthe~100Kdifferenceistheresultofneglectingdissociation.CH4化学恰当燃烧时的火焰温度计算算热能工程 授ExampleconstantvolumeadiabaticflameEstimatetheconstantvolumeadiabaticflametemperaturethecombustionofastoichiometricCH4-airmixtureusingthesameassumptionsasinthelastexampleattheinitialstateof1atmand298K.Thesamecompositionandpropertiesusedinthelastexampleapplyhere.Wenote,however,thattheCp,ivaluesshouldbeevaluatedatatemperaturesomewhatgreaterthan1200K,sincetheconstantvolumeTadwillbehigherthantheconstantpressureTad.Nevertheless,wewillusethesamevaluesasFirstlaw:Ureact(Tinit,Pinit)=Uprod(Tad,Pf),i.e.,Hreact–VPinit=Hprod–VPf, Hreact–Hprod–Ru(nreactTinit–nprodTad)=

热能工程 授Hreact=(1)(-74831)+(2)(0)+(7.52)(0)=-74831kJHprod=(1)[-393546+56.21(Tad-298)]+(2)[-241845+43.87(Tad-298)](7.52)[0+33.71(Tad-=-877236+397.5(Tad-298) Ru(nreactTinit–nprodTad)=8.315×10.52×(298-Tad)wherenreact=nprod=10.52kmol.Finally,theTadcanbesolvedtoTad=2889热能工程 授Forthesameinitialconditions,constantvolumecombustionresultsinmuchhighertemperatures(571Khigherinthisexample)thanforconstantpressurecombustion.Thisisaconsequenceofthepressureforcesngnowork（不作功）whenthevolumeisfixed.Thefinalpressureiswellabovetheinitialpressure:Pf=Pinit(Tad/Tinit)=9.69atm热能工程 授**Calculationsfrom Law为什基本思路是什热能工程 授ChemicalThermodynamicsalonecan’tdeterminewhatspeciesbeintheWhentheproductshavereachedchemicalequilibrium,thecompositionoftheproductscanbedetermined.Onlywhenthecompositionofproductsisdetermined,thethermodynamicpropertiesofthemixture,suchasu,h,etc.,maybecalculatedTadrequiresknowledgeofthecompositionofthecombustionproducts.ChemicalChemicalInhightemperaturecombustionprocesses,theproductsofcombustionarenotasimplemixtureofidealproducts.Themajorspeciesdissociate,producingaminorTheidealcombustionproductsforburningahydrocarbonwithairareCO2,H2O,O2,andN2.Dissociationofthesespeciesyieldsthespecies:H2,OH,CO,H,O,N,NO,andpossiblyChemicalChemicalObjective&ObjectiveTodeterminethemolefractionsofalloftheproductspeciesatagiventemperatureandChemicalequilibriumisachievedforconstanttemperatureandpressuresystemswhentherateofchangeofconcentrationgoestozeroforallspecies（0）热能工程 授ChemicalChemicalequilibriumhasitsrootsinthe2ndlawofthermodynamics;dS=0或GH-equilibriumconstant化学平衡热能工程 授DeterminationofequilibriumConsideringatypicalreactioninaA+bB⇌cC+dDInequilibrium, (dG)=Gprod–Greact=0TheGibbs 函数）foramixtureofidealcanbeexpressedGmix=ΣniForamixtureofidealgas,theGibbsfunctionfortheithspeciesisgivenby（见《工程热力学》中，热力学一般关系式）:gi,T=g0i,T+RuTwhereg0i,TistheGibbsfunctionofthepurespeciesatstandard-statepressureandcanbefoundinJANAFtablesandPiisthepartialpressure,Ru=8.314kJ/(kgmol-k)istheuniversalgasconstant（普适气体常数）热能工程 授ThenthechangeofGibbsfunctionofthesystemafterreactiongi,T=g0,T+RTgi,T=g0,T+RTln(P/Piu =cgc,T+dgd,T–aga,T–b=c +d –a –b RuT d Pb)+TTln

Accordingtothedefinitionofequilibrium Kp=PcPd andsupposeP0=1atm,wheninequilibrium,theaboveequationcanbereducedto:c +d –a –b +RTlnK= lnKp=ag0a,T/RuT+b /RT–c /RT–d /R 热能工程 授Example:equilibriumEqualmolesofH2andO2reacttoproduceH2O,H2,andO2at2500Kand1atm.FindthepercentbyvolumeofH2,O2,andH2OduetotheequilibriumreactionH2O=H2+0.5O2usingthemethodofequilibriumconstants.LetzmolesofH2reactwithzmolesofO2toformSupposethemolefractionofeachspeciestobeχH2O,χH2,andχO2,then,thereactioncanbewrittenaszH2zO2HOH2OH

H2O2Ahydrogenatombalance2z=2χH2O+2热能工程 授Similarly,anoxygenatombalancegives:2z=χH2O+2χO2Combiningthetwoequationsresults2χH2O+2χH2=χH2O+2 Obviously,thesumofthemolefractionsoftheproductsshouldbe1:χH2O+χH2+χO2=Accordingtoprevious

H2OH2O=H2+lnKp=ag0a,T/RuT+bg0b,T/cg0c,T/RuT–dg0d,T/=g0H2O/RuT–g0H2/RuT–0.5g0O2/UsingthedatafromJANAF热能工程 授lnKp=(-106416-0–0)/8.314/2500=- Kp=accordingtothedefinitionof Kp=PcPd/Pa sinceP=1atm,

)0.5P0.5/

χH2(χO2)0.5/χH2O= solvingequations(1),(2),and(3)simultaneouslyχH2O=

=

=Foragivenglobalreaction,ifTisgiven,thenKpcould热能工程 授Example:SecondlawConsiderthefollowingcombustioninafixedvolume,vesselinwhichafixedmassofreactantsformCO+0.5O2→Ifthefinaltemperatureishighenough,CO2willSupposingtheproductstoconsistonlyofCO2,CO,andO2,thenwecanwrite:CO+0.5O2→(1-α)CO2+αCO+0.5αwhereαisthefractionoftheCO2WecancalculatetheadiabaticflametemperatureTfasfunctionofα,accordingtothefirstWhatconstraintsareimposedbythesecondlawonthereaction?(dS)U,V,m= （熵增为

热能工程 授Theentropyoftheproductmixturecanbecalculatedbysummingtheproductspeciesentropies,i.e.,Smix=Σnisi=(1-α)sCO2+αsCO+0.5Andtheindividualspeciesentropyareobtainedfromsi=s0(T )+∫c /TdT–Rln(P/P) =s0(T)–Rln(P/P Wheres0(T)canbefoundinJANAFtables,andP Obviously,Themixtureentrop