2019秋九年级英语全册-Unit-3-Could-you-please-tell-

%A=Cb=xa%B=Ac=xb

%C=Ba=xccabxxa+xb+xc=AB=BC=CA.%A=Cb=xacabxxa+xb+xc.1bac’b’%A%C.bac’b’%A%C.2双线法abc双线法.双线法abc双线法.34Thefollowingfeaturesareofinterestinconnectionwithcompositionalaspectsofternarysystems.(1)IfastraightlinesuchasCDisdrawnfromonecornerofthetriangle(i.e.fromcomponentC)tointersecttheoppositeside(i.e.AB),thentheratiooftheothertwocomponents(i.e.AandB)isconstantinalloysrepresentedbypointslyingontheline.ConsideringalloycompositionsalongCD,

asthecompositionpointbecomesprogressivelyfurtherawayfromCthepercentageofCpresentinthealloysdecreases,buttherelativeamountsofBandAremainthesame.SomerulesPGH等比例规则：CD线上A、B组元的含量之比是一定值。.4Thefollowingfeaturesareof4Speciallineson线：A组元的含量相等。Am线：B,C两组元含量之比为一常数。.Speciallineson线：A组元的含量相等。.56

(3)Iftwoalloys(twophases)(e.g.dande)aremixedtogether,thecompositionoftheresultingmixtureliesonastraightlinejoiningthecompositionsofthetwoalloys(co-linearprinciple).Thetotalcompositionliesatthepointdividingthelineintosegmentsinverselyproportionaltotheratioofthequantitiesoftheoriginalalloys.共线规则（杠杆规则）.6(3)Iftwoalloys(twophase67M（a1b1c1)N(a2b2c2)P（xyz）.7M（a1b1c1)N(a2b2c2)78

Thisistheso-called'leverprinciple'or'leverrule'whichfindsanimportantapplicationin'tie-lines'whenanalloydecomposesintoatwo-phasemixture.Then,usingweightpercentages:杠杆规则.8Thisistheso-called'leve89

(4)Ifthreealloys(phases)

a,b,andcaremixedtogether,thecompositionofthemixture,o,lieswithinthetriangleformedbyjoiningthecompositionsofthealloysbystraightlines.Usingthe'centreofgravity'principle,thetrianglemaybeconsideredasbeingsupportedonapointfulcrumato

andhavingmassesatitscornersproportionaltotheamountsofthealloys.Then,inequilibrium,thepositionofoissuchthatthepercentagesofthethreealloysinthemixturearegivenby%a=a’o/aa’×100%%b=b’o/bb’×100%%c=c’o/cc’×100%Theapplicationofthisprinciplein'tie-triangles',representingthree-phasesinequilibrium.ABCαβLaa'cc'bb'o重心规则.9(4)Ifthreealloys(phases)910Procedure:

L:xα:yγ:z=1-x-yx×50%+y×85%+(1-x-y)×10%=40%x×40%+y×10%+(1-x-y)×20%=30%x=55.3%%;y=10.5%;z=34.2%Answer:~55%liquid,10%α,35%γ.(b)Procedure:L:0α:50%γ:50%A:aB:bC:c=1-a-ba=0×50%+0.5×85%+0.5×10%=47.5%b=0×40%+0.5×10%+0.5×20%=15%Answer:~48%A,15%B,37%C.InasystemABC,aternaryalloyofcomposition30wt%Band30wt%Cconsistsataparticulartemperatureofthreephasesofequilibriumcompositionsasfollows:Liquidphase50%A,40%B,10%Cαsolidsolution85%A,10%B,5%Cγsolidsolution10%

A,20%B,70%C.10Procedure:InasystemABC,10课堂练习1.确定合金I、II、III、IV的成分ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘II点：A%=60%B%=30%C%=10%.课堂练习1.确定合金I、II、III、IV的成分ABC11课堂练习ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘II1.确定合金I、II、III、IV的成分II点：A%=20%B%=50%C%=30%.课堂练习ABC9080706050403020101020312课堂练习ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘III1.确定合金I、II、III、IV的成分III点：A%=20%B%=20%C%=60%.课堂练习ABC9080706050403020101020313课堂练习ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘IV1.确定合金I、II、III、IV的成分IV点：A%=40%B%=0%C%=60%.课堂练习ABC9080706050403020101020314ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘2.标出

75%A+10%B+15%C

的合金课堂练习.ABC908070605040302010102030405153.标出50%A+20%B+30%C的合金ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘课堂练习.3.标出50%A+20%B+30%C的合金ABC90816课堂练习4.绘出A＝40%的合金ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘5.绘出C＝30%的合金.课堂练习4.绘出A＝40%的合金ABC908070617ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘6.绘出C

/B＝1/3的合金课堂练习7.绘出A

/C＝1/4的合金.ABC9080706050403020101020304051819共轭曲线及直线规则共轭曲线共轭连线固相中高熔点的组元含量通常比液相高。.19共轭曲线及直线规则共轭曲线共轭连线固相中高熔点的组元含量1920.20.20ABCsos1ss2s1s2l1l2lsos1s2l1l2lll2l1平衡结晶过程分析.ABCsos1ss2s1s2l1l2lsos1s2l1l2l2122

5.3TernaryeutecticphasediagramsTheclassificationoffourphasesequilibriuminaternarysystem:Eutecticreaction:L→α+β+γ（共晶）Peritecticreaction:L+α+β→γ(a)（包晶）or:L+α→β+γ(b)Monotecticreaction:L1→L2+α+β（偏晶）Eutectoidreaction:δ→α+β+γ（共析）Peritectoidreaction:δ+α+β→γ（包析）.225.3Ternaryeutecticphase2223单相区：4个两相区：3个三相区：4个四相区：1个.23单相区：4个.2324投影图ABC.24投影图ABC.24相图分析液相面固相面液相线固溶面固相面固溶面.相图分析液相面固相面液相线固溶面固相面固溶面.25三相平衡f=C-P+1=4-3=1，T确定，则f=0根据相区接触法则，三相区在L+α、L+β之下，α+β之上的空间中。ee'a'b'dc'.三相平衡f=C-P+1=4-3=1，T确定，则f=0ee26三相平衡时，每两个相都平衡，因此在等温截面上，三条共轭连线形成共轭三角形。共轭三角形的三个顶点表示三个平衡相的成分点。可用重心法则确定三个相的质量分数。ABC.三相平衡时，每两个相都平衡，因此在等温截面上，三条共轭连线形27LL+αL+βL+α+βαβα+βotete温度：

L0→α+β为三元系的共晶反应注意：是在一个温度范围内进行的。.LL+αL+βL+α+βαβα+βotete温度：.28LαβLαγL+α+βL+α+γL→α+βL+α→γ共晶型包晶型.LαβLαγL+α+βL+α+γL→α+βL+α→γ共晶型包29具有四相平衡反应的三元相图A-B，B-C，C-A两两组成共晶相图。组元在固态有限溶解的共晶相图.具有四相平衡反应的三元相图A-B，B-C，C-A两两组成共晶30

液相面固相面(组成)

面：二相共晶面三相共晶面溶解度曲面：6个两相区：6个区：单相区：4个三相区：4个四相区：1个

相图分析.相图分析.31..32..33四相平衡区f=C-P+1=3-4-1=0因此，四相平衡在三元相图中为一具有特定形状的水平面。αγβL.四相平衡区f=C-P+1=3-4-1=0αγβL.34αγβLL+α+γL+α+βL+β+γ发生四相共晶反应：

L→α+β+γ四相共晶反应.αγβLL+α+γL+α+βL+β+γ发生四相共晶反应：四相35投影图

液相区投影ACBαγβEe1e2e3ACBe1e2e3固相区投影AC