工程材料科学与设计原书第2版课后习题答案4—8章.doc
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1、Solutions to Chapter 41. FIND: What material has a property that is hugely affected by a small impurity level?SOLUTION: Electrical conductivity spans a wide range. Incorporation of a few parts per million impurities can change electrical conductivity orders of magnitude. Small cracks in brittle mate
2、rials decrease their tensile strength by orders of magnitude. Small additions of impurity can change the color of gems. COMMENTS: These are but a few examples.2.COMPUTE: The temperature at which the vacancy concentration is one half that of 25oC.GIVEN: EQUATION:whereCv = vacancy concentrationQfv = a
3、ctivation energy for vacancy informationR = gas constant 8.314 J/mole-KT = absolute temperatureIn the present problemand T1 = 35 + 273 = 308KT2 = 25 + 273 = 298KalsoCv(35oC) = 2Cv(25oC)Thus,Solving for Qfv we get Qfv = 52893.5 J/mole.Using this value of Qfv, the Cv(25oC) can be calculatedThe problem
4、 requires us to calculate the temperature at which the vacancy concentration is Cv(25oC). Cv(25oC) = 2.675 x 10-10Thusfor solving T, we get:T = 288.63K or 15.63oC.3.COMPUTE:GIVEN:EQUATION:Dividing (1) by (2) we get:Solving for Q, we get:Q = 22033.56 J/mole= exp(-7.511)= 5.46 x 10-4The problem requir
5、es computing a temperature at which Cv = 3Cv(80oC).3Cv(80oC) = 3 x 5.46 x 10-4= 1.63 x 10-3solving for T, we get: T = 413.05K or 140.05oC4.5.FIND: Are Al and Zn completely soluble in solid solution?If Al-Zn system obeys all the Hume-Rothery rules. Then it is expected to show complete solubility.(i)T
6、he atomic radii of Al and Zn are 0.143nm and 0.133 nm respectively. The difference in their radii is 7.5% which is less than 15%.(ii)The electronegativities of Al and Zn are 1.61 an 1.65 respectively which are also very similar.(iii)The most common valence of Al is +3 and +2 for Zn.(iv)Al has an FCC
7、 structure where Zn has a HCP structure.It appears that Al-Zn system obeys 3 out of 4 Hume-Rothery rules. In this case they are not expected to be completely soluble.6.SHOW: The extent of solid solution formation in the following systems using Hume-Rothery Rules.(a) Al in NiSize: r(Ni) = 0.125nm; r(
8、Al) = 0.143nm difference = 14.4%Electronegativity: Al = 1.61; Ni = 1.91Most Common Valence: Al3+; Ni2+Crystal Structure: Al: FCC; Ni:FCCThe crystal structure of Al and Ni are the same and the most common valencies are also comparable. However, the size difference is close to 15% and the difference i
9、s electronegativities is rather significant.Based on this, it appears that Ni and Al would not form a solid solution over the entire compositional range.(b) Ti in NiSize: r(Ti) = 0.147 nm, r(Ni) = 0.125nm difference = 17.6%Electronegativity: Ti: 1.54; Ni: 1.91Valence:Ti4+; Ni2+Crystal Structure:Ti:H
10、CP; Ni FCCTi in Ni would not exhibit extensive solid solubility(c)Zn in FeSizer(Zn) = 0.133nm; r(Fe) - 0.124nm difference = 7.25%Electronegativity:Zn = 1.65; Fe = 1.83Most Common Valence: Zn2+; Fe2+Crystal Structure:An: HCP; Fe: BCCSince electronegativities and crystal structures are very different,
11、 Zn - Fe will not exhibit extensive solid solubility.(d)Si in AlSize r(Si) = 0.117 nm; r(Al) = 0.143nm; difference = 22.2%Electronegativity: (Si) = 1.90; Al = 1.61Valence: Si4+; A;3+Crystal Structure: Si: Diamond Cubic; Al: FCCSince the size difference is greater than 15%, and the crystal structures
12、 are different, Si-Al would not exhibit extensive solid solubility.(e) Li in AlSize r(li): 0.152, r(Al): 0.143; difference - 6.29%Electronegativity: Li: 0.98; Al: 1.61Most Common Valence: Li1+; Al3+Crystal Structure: Li:BCC; Al: FCCSince electronegativity and crystal structures are very different, L
13、i-Al will not exhibit extensive solid solubility.(f)Cu in AuSize r(Cu) = 0.125nm; r(au) = 0.144nm; difference = 12.5%Electronegativity: Cu = 1.90; Au = 1.93Most Common Valence: Cu+; Au+Crystal Structure: Cu:FCC; Au:FCCCu-Au will exhibit extensive solid solubility.(g)Mn in FeSize r(Mn) = 0.112, r(Fe)
14、 = 0.124 difference = 10.71%Electronegativity: Mn 1.55; Fe 1.83Most Common Valence: Mn2+; Fe2+Crystal Structure: Mn:BCC; Fe BCCThe difference in electronegativity is high but Mn-Fe does obey the other 3 Hume-Rothery rules. Therefore, it will form solid solutions but not over the entire compositional
15、 range.(h)Cr in FeSize r(Cr) = 0.125nm, Fe = 0.144nm difference = 12.5%Electronegativity: Cr = 1.66; Fe = 1.83Most Common Valence: Cr3+; Fe2+Crystal Structure: Cr:BCC; Fe:BCCCr in Fe will exhibit extensive solid solubility but not over the entire compositional range since it obeys only 3 of 4 Hume-R
16、othery rules.(i)Ni in FeSize r(Ni) = 0.125nm, r(Fe) = 0.124nm difference = 0.8%Electronegativity: Ni: 1.91; Fe 1.83Most Common Valence: Ni3+; Fe3+Crystal Structure: Ni:FCC; Fe: BCCNi and Fe obeys 3 of the 4 Hume-Rothery rules therefore, extensive solid solution will be exhibited but not over the ent
17、ire compositional range.7.(a) When one attempts to add a small amount of Ni to Cu, Ni is the solute and Cu is the solvent.(b) Based on the relative sizes of Ni and Cu, radius of Ni = 0.128nm, radius of Cu = 0.125nm, these two are expected to form substitutional solid solutions.(c) Ni and Cu will be
18、completely soluble in each other because they obey all four Hume-Rothery rules.8. FIND: Predict how Cu dissolves in Al. DATA: Cu Alatomic radius (A)1.281.43electronegativity1.901.61valence1+,2+3+crystal structureFCCFCCSOLUTION: All of Hume-Rotherys rules must be followed for a substitutional solutio
19、n. In this case, the valences do not match. Cu will not go into substitutional positions in Al to a large extent. COMMENTS: This principle is often used to precipitation harden Al using Cu.9.What type of solid solution is expected to form when C is added to Fe?The radius of carbon atom is 0.077nm an
20、d that of an Fe atom is 0.124nm. The size difference between these two is 61% which is much grater than 15%. Thus, these two are not expected to form substitutional solid solution.If we compare the size ratio of C to Fe atoms with the size of tetrahedral and octahedral interstitial sites in BCC iron
21、, we find that C does not easily fit into either type of interstitial position. C, however, forms an interstitial solid solution with Fe but the solubility is limited.10. FIND: Calculate the activation for vacancy formation in Fe. GIVEN: The vacancy concentration at 727C = 1000K is 0.00022. SOLUTION
22、: We use equation 4.2-2 to solve this problem:Cv = exp (-Qfv/RT) Solving for Qfv: Qfv = -RT ln Cv = -(8.31 J/mole-K)(1000K) ln 0.00022 = 7.0 x 104 J/mole11.SHOW: A Schottky and Frenkel defect in MgF2 structuresA 2-D representation of the MgF2 structure containing a Schottky defect and a Frenkel defe
23、ct is shown below.12.Explain why the following statement is incorrect: In ionic solids the number of cation vacancies is equal to the number of anion vacancies.In ionic crystals, even in the presence of vacancies, the charge neutrality must be maintained. Therefore, single vacancies do not occur in
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