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SOLUTIONFORCHAPTER7
1.FIND:LabelthephasefieldsinFigureHP7-1.
SOLUTION:Therearetworegionsofsinglephaseequilibriumseparatedbyaregionof
two-phaseequilibrium.Thelineseparatingthesinglephaseliquidfromthetwophase
liquidandsolidregioniscalledtheliquidus,andthelineseparatingthetwophaseliquid
andsolidfromthesinglephasesolidiscalledthesolidus.
SKETCH:
2.FIND:ThenamegiventothetypeofequilibriumdiagramshowninFig.HP7-1.
SOLUTION:Thediagramistermedisomorphous.Abovetheliquidusthereisonlyone
phase,namelytheliquid,andithasthesamestructureregardlessofcomposition.Below
thesolidusisthesolidphase,andaswellithasthesamestructureregardlessof
composition.
3.FIND:WhatisthecrystalstructureofcomponentB,why?
SOLUTION:Therearenophaseboundariesbelowthesolidusconsequentlythephase
andthereforethestructuremustbethesame.IfAisFCC,thenBandallcompositions
betweenAandBmustbethesamephaseandhencesamestructure.IfAisFCC,thenB
mustalsobeFCC.
4.FIND:SketchequilibriumcoolingcurvesforalloyX。andpurecomponentB.Explain
whytheyhavedifferentshapes.
SOLUTION:Theslopeofthetemperatureversustimebehaviorforthealloyinthe
singlephaseregioniscontrolledbythecoolingrateandtheheattransferredfromthe
liquid.Attheliquidustemperatureasmallamountofsolidisformed,releasingan
amountoflatentheatoffusionrelatedtothevolumeofsolidtransformed.Hence,atthe
liquidustemperaturetherewillbeachangeinslope.Sinceheatisbeinggeneratedthe
slopewillbelessthanthatabovetheliquidus.Asthetemperatureisreducedthroughthe
two-phaseregionasmallamountofsolidisformedandacorrespondingheatreleased;
oncethesolidustemperatureisreachednoadditionaltransformationtakesplaceand
thereisagainachangeinslope.Theslopeisgreaterasthesolidcoolsthanastheliquid
andsolidcooled.Forthepurecomponent,solidandliquidareinequilibriumat7B,and
hence,ahorizontallinewhenliquidtransformstosolidat7B.
SKETCH:
5.FIND:TheliquidustemperatureandthesolidustemperatureofalloyXo
SOLUTION:FromfigureHP7-1theliquidustemperatureisapproximately1110℃and
thesolidustemperatureisapproximately1070℃.
6.FIND:Determinecompositionsandphasefractionsofeachphaseinequilibriumat
1100℃foralloyXo.
SOLUTION:Atequilibrium,thetemperatureoftwophasesmustbethesame,andthe
compositionofthesolidisfoundwherethetielineintersectsthesolidusandtheliquidis
attheintersectionofthetielinewiththeliquidus.Fromthephasediagramthe
compositionofsolidis0.35Bandthatoftheliquidis0.55B.Usingtheleverrulethe
fractionofliquid,ft,andfractionofsolid,fsisdetermined.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
7.FIND:SketchyLand7sforalloyX。asthealloyiscooledunderequilibriumconditions
from1200℃toroomtemperature.
SOLUTION:Attheliquidustemperaturetheamountofliquidisalmost100%withonly
averysmallamountofsolid.Converselyatthesolidustemperature,theamountofsolid
isalmost100%withonlyaverysmallamountofliquid.Inatwo-phasefield:
fL+fs=l.
8.FIND:Changesinthecompositionsoftheliquidandsolidphasesduringquilibrium
coolingofalloyXothroughthetwo-phasefield.
SOLUTION:Attheliquidustemperature,thecompositionofthesolidisapproximately
0.30B.Asequilibriumsolidificationprogresses,thecompositionofthesolidincreasesto
0.5B,themaximumvalueitcanreach.Theliquidontheotherhandisinitiallythe
compositionofthealloyXo,andasequilibriumsolidificationprogressesthecomposition
increasesinBtothemaximumcompositionintheliquidofapproximately0.7B.
9.FIND:Fromthefollowingdataconstructaplausibleequilibriumphasediagram.
ComponentAmeltsat800CandBmeltsat1000C;AandBarecompletelysolublein
oneanotheratroomtemperature;andifsolidacontaining0.3Bisheatedunder
equilibriumconditions,thesolidtransformstoliquidhavingthesamecompositionat
500°C.
GIVEN:Meltingtemperaturesforthetwocomponents,congruentmeltingtemperature
andcomposition.
SOLUTION:Thesketchshownbelowsatisfiesalloftherequirementsstatedinthe
problem.ThemeltingtemperatureofpureAis8()0℃andthatforpureBis1()()()℃.The
congruentmeltingtemperatureisat5()()℃atacompositioncontaining().3B.Belowthe
congruentmeltingtemperaturethereisacontinuousasolidsolution.
1000
0
。
0
」nr
-B
a>d
E
a>
f-
1().FIND:Foralloyscontaining10,22,
25,27,and40wt%Ni,determinethenumberofphasespresentandthecompositionof
thephasesinequilibriumat1200℃.
SOLUTION:
AlloycompositionNumberofphasesinCompositionof
in%NiequilibriumPhases(%Ni)
101(Liquid)10%
221(Liquid)22%
252(Liquid+Solid)liquid:22%,solid:32%
272(Liquid+Solid)liquid:22%,solid:32%
401(Solid)40%
11.FIND:Beginningwithastatementofmassbalance,derivetheleverruleinatwo-phase
system.
SOLUTION:InatwophasefieldforanalloyofsomeoverallcompositionXo,the
soluteisdistributedinthetwophases:x0=faxa+fpxpwherethecompositionsare
expressedintermsofoneofthecomponents,L+f0=1.Then,f«=1-fp.Substituting:
Xo=(1-fp)Xa+fpXp
Xo=Xa-Xafp+Xpfp
InstallEquationEditoranddouble
clickheretoviewequation.
X。-Xa=(xp-Xa)fp
12.FIND:DiscusseachofthefactorsthatpermittheCu-Nisystemtobeisomorphousover
thetemperaturerange350-1000℃.
SOLUTION:TheempiricalrulesofHume-Rotheryidentifythecharacteristicsthattwo
elementsmusthaveincommonforextensivesolubility.Thisshouldrequirethat
a.thetwocomponentsmusthavethesamecrystalstructure
b.theatomicradiiofthetwoatomsmustbesimilar
c.thetwocomponentshavethecomparableelectronegativities,and
d.thetwocomponentshavethesimilarvalence.
13.FIND:Whatdoesthetemperature322°CrepresentinFigureHP7-2?
SOLUTION:322℃iscalledthecriticaltemperature.Atthecriticaltemperaturethere
isacorrespondingcriticalcomposition.Foranalloyofthiscomposition,coolingunder
equilibriumconditionsfromabovethecriticaltemperaturetobelowthistemperature
resultsintheformationoftwophasesfromonephaseofthecriticalcomposition.As
coolingcontinuesthetwophasesthatformhavedifferentcompositions.
14.FIND:InFigureHP7-2,area〕and0?differentcrystalstructures?
SOLUTION:a】anda?aretwophaseshavingthesamestructurebutdifferent
compositions.Thecompositionofthetwophasesaredeterminedasinanytwophase
systembyusingthetie-line.Wherethetie-lineattheequilibriumtemperatureintersects
thephaseboundariesdeterminesthecompositionofthetwophaseinequilibrium.
15.FIND:LocationoftheequilibriumphaseboundaryattemperatureT.
GIVEN:Alloy1containing30%Bandalloy2containing50%B,whenequilibratedat
temperatureTareinthesametwo-phase(L+S)region.Thefractionofliquidinalloy1
is0.8andthefractionofliquidinalloy2is0.4.
SKETCH:
SOLUTION:Sincethefractionofliquidinalloy1isgreaterthanthatofsolid,the
liquidusistotheleftof0.3,andsincethefractionofsolidinalloy2isgreaterthanthatof
liquid,thesolidusmustlietotherightofalloy2.Usingtheleverrule:
InstallEquationEditoranddouble
clickheretoviewequation.
foralloy1
InstallEquationEditoranddouble
clickheretoviewequation.
foralloy2
0.8XL-0.8Xs=Xs-0.3
0.4XL-0.4Xs=Xs-0.5
SolveforXsandXL,
0.8Xs-0.8XL=Xs-0.3
0.8XS-0.8XL=2XS-1.0
0=-Xs+0.7,or
Xs=0.7B
InstallEquationEditoranddouble
clickheretoviewequation.
TofindXL,substituteXsintooneoftheequations:
0.4-0.56=-0.8XL
-0.16=-0.8XL,or
XL=0.2B
Asacheck,usetheotherequationtocalculatethefractionofliquidfromthe
InstallEquationEditoranddouble
clickheretoviewequation.
compositions.
16.FIND:Labelallregionsofthephasediagramandtheboundariesofmonovariant
equilibriumforthediagramshowninFigureHP7-3.
SKETCH/SOLUTION:
17.FIND:Sketchanequilibriumcoolingcurvefromabovetheeutectictoroom
temperatureforanalloyofeutecticcomposition.
SKETCH/SOLUTION:
18.FIND:Explainwhytheequilibriumcoolingcurvesfbralloysoneithersideofthe
eutecticcompositionwillbedifferentthantheequilibriumcoolingcurveforaeutectic
alloy.
SOLUTION:Attheeutectictemperature,liquidofcompositionXEisinequilibrium
withtwosolids,XaandXp.Forahypoeutecticalloy(compositiontotheleftofthe
eutectic),thefirstphasetoformattheliquidustemperatureofthealloyisa.Whenthe
eutectictemperatureisreachedtheliquidoftheeutecticcompositionisinequilibrium
withthetwosolids,oneofcompositionXaandtheotherXp.Consequently,depending
uponcomposition,theclosertheoverallcompositionofthealloyistoXE,theless
proeutecticaandthemoreeutectic.Similarly,foralloystotheright,proeutectic。will
form.Intermsofthephaseruleatconstantpressure,fortheeutecticreaction
F=C-P+l=2-3+l=0.Theeutecticisinvariant,andsolidifiesunderequilibrium
conditionsatonetemperature,TE.Foralloysoneithersideoftheeutecticcomposition
butbetweenXaandXp,theproeutecticphaseformsandcoolingoccursasitdoesinany
two-phases-1region.Oncetheeutecticisothermisreached,theremainingliquidof
eutecticcompositionsolidifiesisothermally.
19.FIND:ThemaximumsolidsolubilityofBinAandofAinBinFigureHP7-3.
SOLUTION:ThemaximumsolubilityofBinAoccursat0.15Bandthemaximum
solubilityofAinBis0.1A.
20.FIND:ForanalloyofeutecticcompositioninFigureHP7-3,determinethecomposition
ofthesolidphasesinequilibriumwiththeliquid.
SOLUTION:ThecompositionofthetwosolidphaseaandPinequilibriumwithliquid
atToccursat0.15Band0.9B.
21.FIND:Plot尢,启and/pasafunctionoftemperaturefortheequilibriumcoolingofan
alloyofeutecticcomposition.
SOLUTION:Atjustabovetheeutectictemperature,themicrostructureconsistsofall
liquidofcompositionXE.(Determinedfromthephasediagramtobeapproximately
0.63B.)Justbelowtheeutectictemperature,themicrostructureisamixtureoftwo
phases,aandp.Fromthephasediagram,Xa〜().15BandXp〜0.9B.Todeterminethe
fractionsofaandpattemperaturesbelowTE,usethelever-rule.Tabulatedbeloware
compositionsofthephasesinequilibriumatseveraltemperaturesestimatedfromthe
phasediagram.Thesecompositionsarethenusedtocalculatetheamountofeachphase
present.
TemperatureCompositionofCompositionofCompositionof
(℃)Lap
(fractionofB)(fractionofB)(fractionofB)
6000.63
5000.150.9
4000.080.92
3000.050.95
Fractionofphases:
At600:ft=1.0,fa=fp=0
At500:fL=0
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At400:fL=o
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At300:fL=o
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
Temperature
fLfafp
(℃)
600℃1.
500℃0.360.64
400℃0.340.66
3()()℃0.360.64
22.FIND:ForalloyXo(inFigureHP7-3),calculatethefractionofathatformsasprimary
aandthefractionofathatformsbyeutecticdecompositionwhenthealloyiscooled
from850℃toroomtemperatureunderequilibriumconditions.
SOLUTIONS:Thefractionofproeutecticathatformsunderequilibriumconditions
presentinalloyXocooledfrom850°Ctojustabovetheeutecticis:
InstallEquationEditoranddouble
clickheretoviewequation.
Thenthefractionofliquidwhichisofeutecticcompositionis:
fL=l.fa=1.-0.69=0.31.
Thefractionofathatformsfromliquidofeutecticcomposition,九正is:
InstallEquationEditoranddouble
clickheretoviewequation.
fa,E=(fractionofaformedfromeutectic)(0.31)
Tochecktheseresults,thesumofproeutecticaplusthatwhichformsfromtheeutectic
=0.69+0.11=0.8.Thatamountshouldbethesameasifwecalculateddirectlyfathat
wouldbepresentjustbelowtheeutecticforalloyofcompositionXo.
InstallEquationEditoranddouble
clickheretoviewequation.
23.FIND:ThetotalfractionofaatroomtemperatureforalloyX。inFigureHP7-3.
SOLUTION:SolubilityofBinaatroomtemperatureisapproximately0.02Aandthe
solubilityofAinPisapproximately0.04A(intermsofB0.96B).
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
24.FIND:Equilibriumphasediagramgiventheinformationbelow.
GIVEN:ComponentAmeltsat900℃;componentBmeltsat1000℃;andthereisan
invariantreactionat600℃.ThesolubilityofBinaisknowntoincreasefromalmostnil
atroomtemperaturetoamaximumof10%.Whenanalloycontaining30%Biscooled
underequilibriumconditionsjustabove600°C,atwo-phasemixtureispresent,50%a
and50%liquid.Whenthealloyiscooledjustbelow600℃,thealloycontainstwosolid
phases,aandp.Thefractionofais0.75.Aftercoolingunderequilibriumconditionsto
roomtemperature,theamountofainthea+尸mixturedecreasesto68%.
SOLUTION:Atjustabovetheinvarianttemperaturethereisaregionoftwophase
equilibrium,amixtureof50%aand50%L.Consequentlyiftheoverallcompositionis
50%p,andaislocatedat10%P,thentheliquidmustbeat〜50%B.Justbelowthe
invariantreaction,twosolidphasesarepresent,aat〜10%Band0atsomecomposition,
Xp.Iffa=0.75,then:
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
Atroomtemperature,iffa=0.68,andthesolubilityofBinaisnil,then
TheinvariantreactionisL_a+P,aeutectic,hencethediagramshownbelowfitsall
conditions.
25.FIND:Abinaryphasediagramconsistentwiththeinformationgivenbelow.
GIVEN:ThebinarysystemA-BwithTB>7Aisknowntocontaintwoinvariant
reactionsofthetype:L_a+0at7卜andL_。+yat“whereT\<T\andT?>T人,
and0(0.5B)isacongruentlymeltingphaseatatemperaturehigherthanTB.
SKETCH/SOLUTION:GivenTB>TA,andT|<T2andT9>TA,thenatthecongruent
temperature,solidtransformstoliquidatthesamecomposition,henceforthethree
phases,a,Pandywithsomesolubility,isshownbelowalongwithasimilarphase
diagraminwhichthereislimitedsolubilityofBinAandAinBandthecongruentphase
appearsasalinecompound.
26.FIND:WhichphasediagraminFigureHP7-4isMgO-NiOandwhichisNiO-CaO?
Labeltheregionsonthediagramandidentifytheinvariantreaction.
SOLUTION:TodeterminewhethertheisomorphoussystemisNiO-MgOorNiO-CaO,
oneneedstoknowtheionicradiiofthethreecations.r(Ni2+)=0.078nm,r(Mg2+)=
0.078nm,andr(Ca2+)=0.106nm.Therefore,baseduponsizeoftheionsoccupyingthe
cationsublatticewewouldexpecttheNiO-MgOsystemtobetheisomorphoussystem.
27.FIND:WhatisthemaximumsolidsolubilityofAginPt?
GIVEN:PhasediagraminFigureHP7-5.
SOLUTION:Maximumsolidsolubi出yofAginPtis10.5wt%Ag.
28.FIND:Foranalloyofperitecticcomposition,whatisthecompositionofthelastliquid
tosolidifyat1186℃?
SOLUTION:Thecompositionofthelastliquidtosolidifyis66.3%Ag(theperitiectic
compositionis42.4%Ag).
29.FIND:Whatistherangeofalloycompositionsthatwillperitecticallytransformduring
equilibriumcooling?
SOLUTION:Compositionsbetween10.5and66.3wt%Ag.
30.FIND:Plotthefractionofliquid,/L,thefractionofandthefractionof0、加,asa
functionoftemperatureduringequilibriumcoolingfrom1800to400℃.Foranalloyof
peritecticcomposition.
SOLUTION:First,youmustdeterminethecompositionsofthephase(s)thatis(are)in
equilibrium.If,ataparticulartemperaturethealloyisinasinglephasethenthe
compositionofthephaseisthecompositionofthealloyandthemicrostructurecontains
100%ofthatphase.Whenthealloyisinatwophasefield,thecompositionsare
determinedinusingthetie-lines,andthephasefractionsaredeterminedusingthelever.
If,however,themicrostructureisequilibratedataninvarianttemperatureandthe
compositionliesalongtheinvariantline,thenonlythecompositionofthephasescanbe
determined,nottheirrelativeamounts.Foranalloyof42.4%Agcooledunder
equilibriumconditionstheapproximatecompositionsofthephasesinequilibriumat
severaltemperaturesaresummarizedinthefollowingtable.
TemperatureCompositionofCompositionofCompositionof
(℃)L(%B)a(%B)P(%B)
160042.4
TemperatureCompositionofCompositionofCompositionof
(℃)L(%B)a(%B)P(%B)
1550(〜liquidustemp.)42.45.()
140057.07.0
130062.59.0
12006610.3
110()7.552
10005.562
9005.()72
8004.076
7003.082
6002.585
5002.088
4002.091
InstallEquationEditoranddouble
clickheretoviewequation.
Fractionsofphasesat1400:
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clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At1300:
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At1200
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At1100
InstallEquationEditoranddouble
clickheretoviewequation.
At1000
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At900
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At800
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At700
InstallEquationEditoranddouble
clickheretoviewequation.
At600
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
InstallEquationEditoranddouble
clickheretoviewequation.
At500
At400
InstallEquationEditoranddouble
clickheretoviewequation.
Temperature
fLfafp
(℃)
16001.
1550〜~o.+
14000.70.3
13000.620.38
12000.580.42
110()0.210.79
10000.350.65
9000.440.56
8000.470.53
7000.500.50
6000.520.48
5000.530.47
Temperature
fLfafp
(℃)
4000.540
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