材料科學基礎(chǔ)課后習題答案7

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

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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:

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foralloy1

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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

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clickheretoviewequation.

TofindXL,substituteXsintooneoftheequations:

0.4-0.56=-0.8XL

-0.16=-0.8XL,or

XL=0.2B

Asacheck,usetheotherequationtocalculatethefractionofliquidfromthe

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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,啟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

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At400:fL=o

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At300:fL=o

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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:

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Thenthefractionofliquidwhichisofeutecticcompositionis:

fL=l.fa=1.-0.69=0.31.

Thefractionofathatformsfromliquidofeutecticcomposition,九正is:

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fa,E=(fractionofaformedfromeutectic)(0.31)

Tochecktheseresults,thesumofproeutecticaplusthatwhichformsfromtheeutectic

=0.69+0.11=0.8.Thatamountshouldbethesameasifwecalculateddirectlyfathat

wouldbepresentjustbelowtheeutecticforalloyofcompositionXo.

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23.FIND:ThetotalfractionofaatroomtemperatureforalloyX。inFigureHP7-3.

SOLUTION:SolubilityofBinaatroomtemperatureisapproximately0.02Aandthe

solubilityofAinPisapproximately0.04A(intermsofB0.96B).

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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:

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InstallEquationEditoranddouble-

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InstallEquationEditoranddouble-

clickheretoviewequation.

InstallEquationEditoranddouble-

clickheretoviewequation.

InstallEquationEditoranddouble-

clickheretoviewequation.

InstallEquationEditoranddouble-

clickheretoviewequation.

InstallEquationEditoranddouble-

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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

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Fractionsofphasesat1400:

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At1300:

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At1200

InstallEquationEditoranddouble-

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InstallEquationEditoranddouble-

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At1100

InstallEquationEditoranddouble-

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At1000

InstallEquationEditoranddouble-

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InstallEquationEditoranddouble-

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InstallEquationEditoranddouble-

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At900

InstallEquationEditoranddouble-

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At800

InstallEquationEditoranddouble-

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InstallEquationEditoranddouble-

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At700

InstallEquationEditoranddouble-

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At600

InstallEquationEditoranddouble-

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InstallEquationEditoranddouble-

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At500

At400

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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