2019秋九年級英語全冊-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雙線法abc雙線法.雙線法abc雙線法.34Thefollowingfeaturesareofinterestinconnectionwithcompositionalaspectsofternarysystems.(1)IfastraightlinesuchasCDisdrawnfromonecornerofthetriangle(i.e.fromcomponentC)tointersecttheoppositeside(i.e.AB),thentheratiooftheothertwocomponents(i.e.AandB)isconstantinalloysrepresentedbypointslyingontheline.ConsideringalloycompositionsalongCD,

asthecompositionpointbecomesprogressivelyfurtherawayfromCthepercentageofCpresentinthealloysdecreases,buttherelativeamountsofBandAremainthesame.SomerulesPGH等比例規(guī)則：CD線上A、B組元的含量之比是一定值。.4Thefollowingfeaturesareof4Speciallineson線：A組元的含量相等。Am線：B,C兩組元含量之比為一常數(shù)。.Speciallineson線：A組元的含量相等。.56

(3)Iftwoalloys(twophases)(e.g.dande)aremixedtogether,thecompositionoftheresultingmixtureliesonastraightlinejoiningthecompositionsofthetwoalloys(co-linearprinciple).Thetotalcompositionliesatthepointdividingthelineintosegmentsinverselyproportionaltotheratioofthequantitiesoftheoriginalalloys.共線規(guī)則（杠桿規(guī)則）.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:杠桿規(guī)則.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重心規(guī)則.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課堂練習(xí)1.確定合金I、II、III、IV的成分ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘II點：A%=60%B%=30%C%=10%.課堂練習(xí)1.確定合金I、II、III、IV的成分ABC11課堂練習(xí)ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘II1.確定合金I、II、III、IV的成分II點：A%=20%B%=50%C%=30%.課堂練習(xí)ABC9080706050403020101020312課堂練習(xí)ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘III1.確定合金I、II、III、IV的成分III點：A%=20%B%=20%C%=60%.課堂練習(xí)ABC9080706050403020101020313課堂練習(xí)ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘IV1.確定合金I、II、III、IV的成分IV點：A%=40%B%=0%C%=60%.課堂練習(xí)ABC9080706050403020101020314ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘2.標(biāo)出

75%A+10%B+15%C

的合金課堂練習(xí).ABC908070605040302010102030405153.標(biāo)出50%A+20%B+30%C的合金ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘課堂練習(xí).3.標(biāo)出50%A+20%B+30%C的合金ABC90816課堂練習(xí)4.繪出A＝40%的合金ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘5.繪出C＝30%的合金.課堂練習(xí)4.繪出A＝40%的合金ABC908070617ABC908070605040302010102030405060708090102030405060708090←A%B%↗C%↘6.繪出C

/B＝1/3的合金課堂練習(xí)7.繪出A

/C＝1/4的合金.ABC9080706050403020101020304051819共軛曲線及直線規(guī)則共軛曲線共軛連線固相中高熔點的組元含量通常比液相高。.19共軛曲線及直線規(guī)則共軛曲線共軛連線固相中高熔點的組元含量1920.20.20ABCsos1ss2s1s2l1l2lsos1s2l1l2lll2l1平衡結(jié)晶過程分析.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單相區(qū)：4個兩相區(qū)：3個三相區(qū)：4個四相區(qū)：1個.23單相區(qū)：4個.2324投影圖ABC.24投影圖ABC.24相圖分析液相面固相面液相線固溶面固相面固溶面.相圖分析液相面固相面液相線固溶面固相面固溶面.25三相平衡f=C-P+1=4-3=1，T確定，則f=0根據(jù)相區(qū)接觸法則，三相區(qū)在L+α、L+β之下，α+β之上的空間中。ee'a'b'dc'.三相平衡f=C-P+1=4-3=1，T確定，則f=0ee26三相平衡時，每兩個相都平衡，因此在等溫截面上，三條共軛連線形成共軛三角形。共軛三角形的三個頂點表示三個平衡相的成分點?？捎弥匦姆▌t確定三個相的質(zhì)量分?jǐn)?shù)。ABC.三相平衡時，每兩個相都平衡，因此在等溫截面上，三條共軛連線形27LL+αL+βL+α+βαβα+βotete溫度：

L0→α+β為三元系的共晶反應(yīng)注意：是在一個溫度范圍內(nèi)進行的。.LL+αL+βL+α+βαβα+βotete溫度：.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具有四相平衡反應(yīng)的三元相圖A-B，B-C，C-A兩兩組成共晶相圖。組元在固態(tài)有限溶解的共晶相圖.具有四相平衡反應(yīng)的三元相圖A-B，B-C，C-A兩兩組成共晶30

液相面固相面(組成)

面：二相共晶面三相共晶面溶解度曲面：6個兩相區(qū)：6個區(qū)：單相區(qū)：4個三相區(qū)：4個四相區(qū)：1個

相圖分析.相圖分析.31..32..33四相平衡區(qū)f=C-P+1=3-4-1=0因此，四相平衡在三元相圖中為一具有特定形狀的水平面。αγβL.四相平衡區(qū)f=C-P+1=3-4-1=0αγβL.34αγβLL+α+γL+α+βL+β+γ發(fā)生四相共晶反應(yīng)：

L→α+β+γ四相共晶反應(yīng).αγβLL+α+γL+α+βL+β+γ發(fā)生四相共晶反應(yīng)：四相35投影圖

液相區(qū)投影ACBαγβEe1e2e3ACBe1e2e3固相區(qū)投影AC