Chapter-5-Nucleation-During-Phase-Transitions.ppt

1,Chapter5NucleationDuringPhaseTransitions,2,5.1General,UndercoolAphasetransformationoccursatatemperaturebelowtheequilibriumtemperature.Thisisespeciallytrueforthesolidphasetransformationprocess.NucleiVerysmallatomclustersformedinthestartingphase,theatomicarrangementwhichisequal,oratleastsimilartothestructureofthenewphase.Nucleation-growthprocess,Quasi-isotropic,3,CausesofhomogeneousnucleationFluctuationofenergy,atomicarrangement,andcompositionCausesofheterogeneousnucleationInterface,grainboundaries,impurityparticles,andsoon,4,5.2HomogeneousNucleationwithoutChangeofComposition,Thedrivingforceofthephasetransformationisstrictlygivenbythechangeinthestateofaggregation,orthechangeinthecrystallographicstructure.Theformationofphaseboundaryimprovestheenergyofthesystem,whichactsastheresistancetothephasetransformation.,1.Radiusofcriticalnuclei,5,ThetotalchangeinGibbsenergyofthesystemconnectedtothenucleationisgivenby,Ifthenucleusisspherical(thisisusuallycorrectinliquidsolidtransformations),thefollowingequationcanbeobtained,6,Theformationofanucleusabletogrow,byrandomfluctuationofthestructure,isfirstassociatedwithanincreaseoftheGibbsenergyofthesystem.,7,ThecriticalradiusrkisconnectedtothemaximumoftheG-rcurve.,Theradiusofthecriticalnucleuscanbeobtained:,临界形核功等于临界晶核表面能的三分之一，意味着液固两相的自由能差只能补偿临界晶核所需表面能的三分之二。另外三分之一从哪里来？,临界晶核表面能,8,AtTg,Gend=Gstart,andG=0,thusrk=Anundercooling,i.e.TTg,isnecessaryfortheformationofthenewphase.,9,平衡凝固时的自由能变化,因此,当过冷度T较小时，可以认为H和S与温度无关，可以得到：,10,由此可见，临界晶核半径和临界形核功随过冷度增大而减小。,11,液相中的固相晶核实际上是由一些短程有序的原子团簇演变而来的，这些团簇在瞬间具有与固相相同的原子排列。,根据统计热力学的原理，可以导出半径为r的球状团簇平均数量为：,式中，n0是系统中的原子总数，Gr为与团簇有关的附加自由能。当温度高于Tm时，上式适用于所有r值；低于Tm时则只适用于rrk的情况，因为大于rk的团簇是稳定的固体晶核，而不再是液相的一部分。由上式可见，一定尺寸的团簇数量随Gr升高成指数下降，而Gr随r增大迅速增大，因此，团簇的数量随尺寸的增大而迅速减小。,12,2.Rateofhomogeneousnucleation,若单位体积内有C0个原子，则达到临界尺寸的团簇数量为：,在这种团簇上再加上一个原子，它就可以成为晶核，设这种情况发生的概率为f0（可以看作常数），则均匀形核速率为：,（个m-3）,（个m-3s-1）,代人临界形核功：,13,2.Rateofhomogeneousnucleation,由于（T）2在幂指数中的作用，Nhom在一个很窄的温度范围内上升几个数量级。亦即均匀形核存在一个临界过冷度。对于大多数金属而言，该临界过冷度约为0.2Tm。,14,5.3HeterogeneousNucleation,Theformationofnucleidoesnotoccuratany,butatsomeveryspecificplacesinthevolumeofthestartingphase.Thesespecificplacesserveas“nucleationcatalysts”.Asaruleheterogeneousnucleationismoreprobablethanhomogeneousnucleation.,15,Assumethatacalottenucleusisformedonaplanebasis.,16,ThetotalchangeinGibbsenergyofthesystemconnectedtothenucleationisgivenby,Thevolumeofthecalotteis,Theareasofthephaseboundariesare,17,Theareaofthenewinterface/Fequalsthedecreaseoftheinterface/F.Thusthetotalchangeofboundaryenergyis,18,ThetotalchangeinGibbsenergyofthesystem:,Where,GhomisthetotalchangeinGibbsenergyofhomogeneousnucleation.,19,Set,Weobtain,and,均匀形核和非均匀形核的附加自由能,20,Since,If,then,Thus,thenucleationofnewphaseisveryeasy.,21,设体系中与模壁接触的原子数为n1，则临界晶核数量为：,非均匀形核速率：,式中，C1是单位体积内与模壁接触的原子数，f1为频率因子。,a）临界形核功随过冷度的变化；b）形核功相同时的形核速率,22,23,固态相变时也需要形核，并且由于固体结构致密、原子键合力较强，比容变化会导致晶格畸变，由此引起的畸变能不能忽视。因此，其相变阻力更大，需要的过冷度也较大。,固相均匀形核与液相均匀形核,相同点：1.都是均匀形核（单位时间单位体积的形核数相同）；2.都有形核功，都靠能量起伏提供；3.都存在界面能Ag。不同点：在固相中均匀形核存在应变能DGS,在液相中不存在。,24,固相非均匀形核与液相非均匀形核,相同点：都借助于形核位置形核。不同点：1）液相在外来的形核位置（模壁、杂质、形核剂）形核；固相在晶体缺陷（空位、位错、层错、晶界、相界）形核；2）液相形核时自由能变化为：DGVDGV+Ag固相形核时自由能变化为：DGV（DGV-DGS）+AgDGdGS为应变能，Gd为缺陷形核释放的能量,25,Grainboundariesandlatticedefectsarethefavorablepositionsforthenucleationofanewphaseinsolidphasetransition.Insolidphasetransformation,latticedeformationsresultfromthevolumedifferenceofthestartingandnewphase.ThisincreasestheGibbsenergyduringnucleation,andthustheshapeofthenuclei.,Shapesofnucleiinsolidphasetransformations,26,IftheincreaseofGibbsenergyispredominatelygivenbytheboundaryenergy,thensphericalnucleiarepreferable.IftheincreaseofGibbsenergyispredominatelygivenbythelatticedeformation,thenplatenucleiarepreferable.Mostrealnucleiareinbetween.,27,Precipitationsinanas-agedCu-Ni-SnalloyNotethattheseprecipitationsareverysmall.,28,Nucleationofmelting,熔化过程是不需要过热的。,其原因在于以下关系：,S,L,V,由右图可见：,所以必然有：,因此，液相的临界形核功为0，不需要过热。,29,5.4HomogeneousNucleationwithChangeinComposition,Ifthenewphasehasadifferentcompositionthanthestarting,then,atnucleation,thechangeinGibbsenergyduetothechangeincompositionhastobetakenintoaccount.,30,Assumethatthenewphasehasthesamecrystalstructureasthestartingphase.Theequilibriumcompositionsaregivenbythecontactpointsofthedoubletangent.,31,Set:N=numberofatomsinthetotalconsideredvolumen=numberofatomsinthenucleusx0=compositionofthestartingphase=compositionofthematrixafternucleusformation=compositionofthenucleus=averagevolumeofanatomG(x0),G(),G()=Gibbsenergiesperunitvolumeatthecorrespondingcompositions,32,ThechangeofGibbsenergyresultingfromtheformationofnucleuscomposedofnatomsis:,Ifx0,thefollowingequationcanbereached:,33,ThelineB1B2isthechangeofGibbsenergyatthenucleationof.,34,TheincreaseinboundaryGibbsenergyisinwhich,Fistheinterfacearea,sthenumberofatomsperunitsurface,Ktheinteractionparameter,andzthecoordinationnumber.ThetotalchangeinGibbsenergyis,35,Thethermodynamicmechanismforthelaminarstructureofpearlite.,36,5.5SpinodalDecomposition,Spinodaldecompositionisaspontaneousprocesswithoutnucleation.Inflectionpoint:Thesignofthecurvatureofacurvechangesatthispoint.Thesecondorderderivativeatthispointiszero.,37,Thermodynamice