材料科学基础(武汉理工大学,张联盟版)课后习题及答案 第七章

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材料科学基础课后习题答案

材料科学基础课后习题答案

《材料科学基础》课后习题答案第一章材料结构的基本知识4. 简述一次键和二次键区别答:根据结合力的强弱可把结合键分成一次键和二次键两大类。

其中一次键的结合力较强,包括离子键、共价键和金属键。

一次键的三种结合方式都是依靠外壳层电子转移或共享以形成稳定的电子壳层,从而使原子间相互结合起来。

二次键的结合力较弱,包括范德瓦耳斯键和氢键。

二次键是一种在原子和分子之间,由诱导或永久电偶相互作用而产生的一种副键。

6. 为什么金属键结合的固体材料的密度比离子键或共价键固体为高?答:材料的密度与结合键类型有关。

一般金属键结合的固体材料的高密度有两个原因:(1)金属元素有较高的相对原子质量;(2)金属键的结合方式没有方向性,因此金属原子总是趋于密集排列。

相反,对于离子键或共价键结合的材料,原子排列不可能很致密。

共价键结合时,相邻原子的个数要受到共价键数目的限制;离子键结合时,则要满足正、负离子间电荷平衡的要求,它们的相邻原子数都不如金属多,因此离子键或共价键结合的材料密度较低。

9. 什么是单相组织?什么是两相组织?以它们为例说明显微组织的含义以及显微组织对性能的影响。

答:单相组织,顾名思义是具有单一相的组织。

即所有晶粒的化学组成相同,晶体结构也相同。

两相组织是指具有两相的组织。

单相组织特征的主要有晶粒尺寸及形状。

晶粒尺寸对材料性能有重要的影响,细化晶粒可以明显地提高材料的强度,改善材料的塑性和韧性。

单相组织中,根据各方向生长条件的不同,会生成等轴晶和柱状晶。

等轴晶的材料各方向上性能接近,而柱状晶则在各个方向上表现出性能的差异。

对于两相组织,如果两个相的晶粒尺度相当,两者均匀地交替分布,此时合金的力学性能取决于两个相或者两种相或两种组织组成物的相对量及各自的性能。

如果两个相的晶粒尺度相差甚远,其中尺寸较细的相以球状、点状、片状或针状等形态弥散地分布于另一相晶粒的基体内。

如果弥散相的硬度明显高于基体相,则将显著提高材料的强度,同时降低材料的塑韧性。

材料科学基础课后习题答案7

材料科学基础课后习题答案7

SOLUTION FOR CHAPTER 71. FIND: Label the phase fields in Figure HP7-1.SOLUTION: There are two regions of single phase equilibrium separated by a region of two-phase equilibrium. The line separating the single phase liquid from the two phaseliquid and solid region is called the liquidus, and the line separating the two phase liquid and solid from the single phase solid is called the solidus.SKETCH:2. FIND: The name given to the type of equilibrium diagram shown in Fig. HP7-1.SOLUTION: The diagram is termed isomorphous. Above the liquidus there is only one phase, namely the liquid, and it has the same structure regardless of composition. Below the solidus is the solid phase, and as well it has the same structure regardless ofcomposition.3. FIND: What is the crystal structure of component B, why?SOLUTION: There are no phase boundaries below the solidus consequently the phase and therefore the structure must be the same. If A is FCC, then B and all compositionsbetween A and B must be the same phase and hence same structure. If A is FCC, then B must also be FCC.4. FIND: Sketch equilibrium cooling curves for alloy X o and pure component B. Explainwhy they have different shapes.SOLUTION: The slope of the temperature versus time behavior for the alloy in thesingle phase region is controlled by the cooling rate and the heat transferred from theliquid. At the liquidus temperature a small amount of solid is formed, releasing anamount of latent heat of fusion related to the volume of solid transformed. Hence, at the liquidus temperature there will be a change in slope. Since heat is being generated theslope will be less than that above the liquidus. As the temperature is reduced through the two-phase region a small amount of solid is formed and a corresponding heat released;once the solidus temperature is reached no additional transformation takes place and there is again a change in slope. The slope is greater as the solid cools than as the liquid andsolid cooled. For the pure component, solid and liquid are in equilibrium at T B, and hence,a horizontal line when liquid transforms to solid at T B.SKETCH:5. FIND: The liquidus temperature and the solidus temperature of alloy X oSOLUTION: From figure HP7-1 the liquidus temperature is approximately 1110o C and the solidus temperature is approximately 1070o C.6. FIND: Determine compositions and phase fractions of each phase in equilibrium at1100o C for alloy X o.SOLUTION: At equilibrium, the temperature of two phases must be the same, and thecomposition of the solid is found where the tie line intersects the solidus and the liquid is at the intersection of the tie line with the liquidus. From the phase diagram thecomposition of solid is 0.35B and that of the liquid is 0.55B. Using the lever rule thefraction of liquid, f L, and fraction of solid, f s is determined.7. FIND: Sketch f L and f S for alloy X o as the alloy is cooled under equilibrium conditionsfrom 1200o C to room temperature.SOLUTION: At the liquidus temperature the amount of liquid is almost 100% with onlya very small amount of solid. Conversely at the solidus temperature, the amount of solidis almost 100% with only a very small amount of liquid. In a two-phase field:f L + f s = 1.8. FIND: Changes in the compositions of the liquid and solid phases during quilibriumcooling of alloy X o through the two-phase field.SOLUTION: At the liquidus temperature, the composition of the solid is approximately0.30B. As equilibrium solidification progresses, the composition of the solid increases to0.5 B, the maximum value it can reach. The liquid on the other hand is initially thecomposition of the alloy X o, and as equilibrium solidification progresses the composition increases in B to the maximum composition in the liquid of approximately 0.7 B.9. FIND:From the following data construct a plausible equilibrium phase diagram.Component A melts at 800︒C and B melts at 1000︒C; A and B are completely soluble in one another at room temperature; and if solid α containing 0.3 B is heated underequilibrium conditions, the solid transforms to liquid having the same composition at500︒C.GIVEN: Melting temperatures for the two components, congruent melting temperature and composition.SOLUTION: The sketch shown below satisfies all of the requirements stated in theproblem. The melting temperature of pure A is 800︒C and that for pure B is 1000︒C. The congruent melting temperature is at 500︒C at a composition containing 0.3 B. Below the congruent melting temperature there is a continuous α solid solution.10. FIND: For alloys containing 10, 22, 25, 27, and 40 wt% Ni, determine the number ofphases present and the composition of the phases in equilibrium at 1200o C.11. FIND: Beginning with a statement of mass balance, derive the lever rule in a two-phasesystem.SOLUTION: In a two phase field for an alloy of some over all composition X o, thesolute is distributed in the two phases: x o = fα xα + fβ xβ where the compositions areexpressed in terms of one of the components, fα + fβ = 1. Then, fα= 1 - fβ. Substituting:x o = (1 - fβ) xα + fβ xβx o = xα - xα fβ + xβfβx o - xα = (xβ - xα) fβ12. FIND: Discuss each of the factors that permit the Cu-Ni system to be isomorphous overthe temperature range 350-1000o C.SOLUTION: The empirical rules of Hume-Rothery identify the characteristics that two elements must have in common for extensive solubility. This should require thata. the two components must have the same crystal structureb. the atomic radii of the two atoms must be similarc. the two components have the comparable electro negativities, andd. the two components have the similar valence.13. FIND: What does the temperature 322O C represent in Figure HP7-2?SOLUTION: 322o C is called the critical temperature. At the critical temperature there is a corresponding critical composition. For an alloy of this composition, cooling under equilibrium conditions from above the critical temperature to below this temperatureresults in the formation of two phases from one phase of the critical composition. Ascooling continues the two phases that form have different compositions.14. FIND: In Figure HP7-2, are α1 and α2 different crystal structures?SOLUTION: α1 and α2 are two phases having the same structure but differentcompositions. The composition of the two phases are determined as in any two phasesystem by using the tie-line. Where the tie-line at the equilibrium temperature intersects the phase boundaries determines the composition of the two phase in equilibrium.15. FIND: Location of the equilibrium phase boundary at temperature T.GIVEN: Alloy 1 containing 30%B and alloy 2 containing 50% B, when equilibrated at temperature T are in the same two-phase (L + S) region. The fraction of liquid in alloy 1 is 0.8 and the fraction of liquid in alloy 2 is 0.4.SKETCH:SOLUTION: Since the fraction of liquid in alloy 1 is greater than that of solid, theliquidus is to the left of 0.3, and since the fraction of solid in alloy 2 is greater than that of liquid, the solidus must lie to the right of alloy 2. Using the lever rule:for alloy 1for alloy 20.8X L - 0.8X s = X s - 0.30.4X L - 0.4X s = X s - 0.5Solve for X s and X L,0.8X s - 0.8X L = X s - 0.30.8X s - 0.8X L = 2X s - 1.0____________________0 = -X s + 0.7, orX s = 0.7BTo find X L, substitute X s into one of the equations:0.4 - 0.56 = -0.8X L-0.16 = -0.8X L, orX L = 0.2BAs a check, use the other equation to calculate the fraction of liquid from thecompositions.16. FIND: Label all regions of the phase diagram and the boundaries of monovariantequilibrium for the diagram shown in Figure HP7-3.SKETCH/SOLUTION:17. FIND: Sketch an equilibrium cooling curve from above the eutectic to roomtemperature for an alloy of eutectic composition.SKETCH/SOLUTION:18. FIND: Explain why the equilibrium cooling curves for alloys on either side of theeutectic composition will be different than the equilibrium cooling curve for a eutecticalloy.SOLUTION: At the eutectic temperature, liquid of composition X E is in equilibriumwith two solids, Xα and Xβ . For a hypoeutectic alloy (composition to the left of theeutectic), the first phase to form at the liquidus temperature of the alloy is α. When the eutectic temperature is reached the liquid of the eutectic composition is in equilibriumwith the two solids, one of composition Xα and the other Xβ. Consequently, depending upon composition, the closer the overall composition of the alloy is to X E, the lessproeutectic α and the more eutectic. Similarly, for alloys to the right, proeutectic β will form. In terms of the phase rule at constant pressure, for the eutectic reactionF = C - P + 1 = 2 - 3 + 1 = 0. The eutectic is invariant, and solidifies under equilibriumconditions at one temperature, T E. For alloys on either side of the eutectic composition but between Xα and Xβ, the proeutectic phase forms and cooling occurs as it does in any two-phase s-l region. Once the eutectic isotherm is reached, the remaining liquid ofeutectic composition solidifies isothermally.19. FIND: The maximum solid solubility of B in A and of A in B in Figure HP7-3.SOLUTION: The maximum solubility of B in A occurs at 0.15 B and the maximumsolubility of A in B is 0.1 A.20. FIND: For an alloy of eutectic composition in Figure HP7-3, determine the compositionof the solid phases in equilibrium with the liquid.SOLUTION: The composition of the two solid phase α and β in equilibrium with liquid at T_ occurs at 0.15 B and 0.9 B.21. FIND: Plot f L, fα, and fβ as a function of temperature for the equilibrium cooling of analloy of eutectic composition.SOLUTION: At just above the eutectic temperature, the microstructure consists of allliquid of composition X E. (Determined from the phase diagram to be approximately0.63B.) Just below the eutectic temperature, the microstructure is a mixture of twophases, α and β. From the phase diagram, Xα ~ 0.15B and Xβ ~ 0.9B. To determine the fractions of α and β at temperatures below T E, use the lever-rule. Tabulated below arecompositions of the phases in equilibrium at several temperatures estimated from thephase diagram. These compositions are then used to calculate the amount of each phase present.Fraction of phases:At 600: f L = 1.0, fα = fβ = 0At 500: f L = 0Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.At 400: f L = 0Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.At 300: f L = 0Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.Instal l Equa tion E ditor a nd do uble-click h ere to view equat ion.22. FIND: For alloy X o (in Figure HP7-3), calculate the fraction of α that forms as primary αand the fraction of α that forms by eutectic decomposition when the alloy is cooled from 850o C to room temperature under equilibrium conditions.SOLUTIONS: The fraction of proeutectic α that forms under equilibrium conditionspresent in alloy X o cooled from 850o C to just above the eutectic is:Then the fraction of liquid which is of eutectic composition is:f L = 1 - fα = 1. - 0.69 = 0.31.The fraction of α that forms from liquid of eutectic composition, fα,E is :fα,E = (fraction of α formed from eutectic) (0.31)To check these results, the sum of proeutectic α plus that which forms from the eutectic =0.69 + 0.11 = 0.8. That amount should be the same as if we calculated directly fα that would be present just below the eutectic for alloy of composition X o.23. FIND: The total fraction of α at room temperature for alloy X o in Figure HP7-3.SOLUTION: Solubility of B in α at room temperature is approximately 0.02 A and the solubility of A in β is approximately 0.04 A (in terms of B 0.96 B).24. FIND: Equilibrium phase diagram given the information below.GIVEN: Component A melts at 900o C; component B melts at 1000o C; and there is aninvariant reaction at 600o C. The solubility of B in α is known to increase from almost nil at room temperature to a maximum of 10%. When an alloy containing 30% B is cooled under equilibrium conditions just above 600o C, a two-phase mixture is present, 50% αand 50% liquid. When the alloy is cooled just below 600o C, the alloy contains two solid phases, α and β. The fraction of α is 0.75. After cooling under equilibrium conditions to room temperature, the amount of α in the α + β mixture decreases to 68%.SOLUTION: At just above the invariant temperature there is a region of two phaseequilibrium, a mixture of 50% α and 50% L. Consequently if the overall composition is 50%β, and α is located at 10%β, then the liquid must be at ˜50% B. Just below theinvariant reaction, two solid phases are present, αat ˜10%B and β at some composition, Xβ. If fα = 0.75, then:At room temperature, if fα = 0.68, and the solubility of B in α is nil, thenThe invariant reaction is L _ α + β, a eutectic, hence the diagram shown below fits allconditions.25. FIND: A binary phase diagram consistent with the information given below.GIVEN: The binary system A-B with T B > T A is known to contain two invariantreactions of the type: L _ α + β at T1, and L _ β + γ at T2 where T1 < T A and T2 > T A ,and β (0.5 B) is a congruently melting phase at a temperature higher than T B.SKETCH/SOLUTION: Given T B > T A, and T1 < T2 and T2 > T A, then at the congruent temperature, solid transforms to liquid at the same composition, hence for the threephases, α, β and γ with some solubility, is shown below along with a similar phasediagram in which there is limited solubility of B in A and A in B and the congruent phase appears as a line compound.26. FIND: Which phase diagram in Figure HP7-4 is MgO-NiO and which is NiO-CaO?Label the regions on the diagram and identify the invariant reaction.SOLUTION: To determine whether the isomorphous system is NiO-MgO or NiO-CaO, one needs to know the ionic radii of the three cations. r(Ni2+) = 0.078 nm, r(Mg2+) =0.078 nm, and r(Ca2+) = 0.106 nm. Therefore, based upon size of the ions occupying thecation sublattice we would expect the NiO-MgO system to be the isomorphous system.27. FIND: What is the maximum solid solubility of Ag in Pt?GIVEN: Phase diagram in Figure HP7-5.SOLUTION: Maximum solid solubility of Ag in Pt is 10.5 wt% Ag.28. FIND: For an alloy of peritectic composition, what is the composition of the last liquidto solidify at 1186o C?SOLUTION: The composition of the last liquid to solidify is 66.3% Ag (the peritiectic composition is 42.4% Ag).29. FIND: What is the range of alloy compositions that will peritectically transform duringequilibrium cooling?SOLUTION: Compositions between 10.5 and 66.3 wt % Ag.30. FIND: Plot the fraction of liquid, f L, the fraction of α, fα, and the fraction of β, fβ, as afunction of temperature during equilibrium cooling from 1800 to 400o C. For an alloy of peritectic composition.SOLUTION: First, you must determine the compositions of the phase(s) that is (are) in equilibrium. If, at a particular temperature the alloy is in a single phase then thecomposition of the phase is the composition of the alloy and the microstructure contains 100% of that phase. When the alloy is in a two phase field, the compositions aredetermined in using the tie-lines, and the phase fractions are determined using the lever.If, however, the microstructure is equilibrated at an invariant temperature and thecomposition lies along the invariant line, then only the composition of the phases can be determined, not their relative amounts. For an alloy of 42.4% Ag cooled underequilibrium conditions the approximate compositions of the phases in equilibrium atseveral temperatures are summarized in the following table.Fractions of phases at 1400:At 1300:At 1200At 1100At 1000At 900At 800At 700 At 600At 500At 400Fraction of phases as a function of temperature are plotted below.31. FIND: Sketch a possible diagram given the information below.GIVEN: The two component system A-B with T A > T B contains two invariant reactions of the type L + α _ β at T1, and L _ β + γ at T2 (<T1).SKETCH/SOLUTION: Reaction at T1 is a peritectic, reaction at T2 a eutectic, T A > T B and T1 > T2.In the above figure some B is soluble in A and likewise some A is soluble in B. Thephase that melts incongruently at T1 has a change in solubility with temperature.Alternatively, we can construct a diagram where the solubility of the end members are nil and the incongruently melting phase, X, is a line compound. That diagram wouldresemble the diagram shown below.32. FIND: Label all phase fields and identify the invariant reactions in the Ag-Al Phasediagram shown in figure HP7-6.SKETCH/SOLUTION:33. FIND: Label the phase fields and identify the invariant reactions in the V2O5 - NiOphase diagram.SKETCH/SOLUTION:34. FIND: Apply the 1-2-1...rule to the V2O5 - NiO diagram at 600o C.SOLUTION: At 600o C the single-phase regions are V2O5, N-V, 2N-V, 3N-V and NiO.These narrow single-phase fields define the regions of two-phase equilibrium. Also, itshould be recalled that at a fixed temperature the composition of each phase in a two-phase field is constant. All that varies is the relative amount of each phase (i.e. the lever-rule).35. FIND: Label all phase fields and identify the invariant reactions in the V2O5 - Cr2O3system.SKETCH/SOLUTION:36. FIND: For compositions X and Y, plot the fraction of phases present as a function oftemperature.SOLUTION: From the phase diagram in Figure HP7-8 the relevant we find:Composition of alloy X ~ 0.47BComposition of alloy Y ~ 0.55BComposition of the liquid at the peritectic isotherm ~ 0.44BComposition of the eutectic ~ 0.21BLiquidus temperature of X ~ 1100o CLiquidus temperature of Y ~ 1400o CPeritectic temperature ~ 900o CEutectic temperature ~ 700o CFor alloy X, in the temperature range from the liquidus (1100o C) to just above theperitectic isotherm, f L , decreases from 1.0 to:while the fraction of the oxide increases from 0 to (1 - 0.95) = 0.05. In the temperature range just below the peritectic isotherm to just above the eutectic isotherm, the phase fractions change from:andtoandjust above the eutectic temperature.In the temperature range below the eutectic isotherm, the Instal l Equa tion E ditor a nd do uble -click h ere to view equat ion.areconstant and from the lever rule are:Similarly for alloy Y:In the temperature range from the liquidus temperature (1400o C) to just above theperitectic isotherm the fraction of liquid decreases from f L = 1 towhile the fraction of Cr 2O 3 increasing from 0 toBelow the eutectic isotherm the amount of Cr 2O 3⋅V 2O 5 and Cr 2O 3 are constant.37.FIND: Identify the invariant reactions occurring in the Cu-Pb system.SOLUTION: There are two invariant reactions in the Cu - Pb system, a monotectic at 955o C and a eutectic at 326o C. For a complete illustration examine how the reduction in solubility of the components in each other effect the monotectic systems shown in Figure 7.6-1.38.FIND: Phase diagram HP7-9 indicates no mutual solid solubility of Cu in Pb or Pb in Cu. Explain why this is not strictly true.SOLUTION: Although the phase diagram appears to indicate no solubility of Cu in Pb or Pb in Cu, as pointed out on page 274 of the text, the free energy of a pure component can always be reduced by small additions of a second component since the presence of the second component increases the randomness of the system. It then is a competition between the enthalpy and the entropy terms as to how much is added, since:∆G = ∆H - T ∆S39.FIND: If an alloy containing 63 wt% Pb is cooled under equilibrium conditions from 1100o C to room temperature, plot the fraction of phases present as a function of temperature.SKETCH/SOLUTION: When an alloy containing 63% Pb, is cooled from 1100o C to room temperature under equilibrium conditions the following phases form. Below 991o C and just above the monotectic isotherm, liquid phase separation occurs. The composition of the two liquids follow the liquidus boundary, and the amount is calculated by the lever-rule. Below the monotectic isotherm to just above the eutectic isotherm two phases are present, Cu with essentially no soluble Pb and liquid rich in Pb, the composition of which follows the liquidus. Below the eutectic isotherm there is a two-phase mixture of essentially pure Cu and pure Pb. At 991o C the microstructure is essentially all liquid containing 63% Pb. At 975o C the microstructure has separated into two liquids, one (L I ) containing approximately 80% Pb and the second (L II ) containing approximately 43% Pb.Just above the monotectic isotherm.Below the monotectic isotherm, the fraction of liquid decreases, while the fraction ofsolid Cu increases. Just below 955o CAt 900o CAt 800o CThere is so little solubility of Cu in the liquid that once the temperature drops below 600o C, the system is essentially solid Cu with liquid essentially composed of Pb. Below 326o C the microstructure is a mixture of solid copper and lead. The phase diagramsuggests that there is no measurable change in solubility of the components. Then,40. FIND: Most alloying elements used in commercial titanium alloys can be classified aseither alpha stabilizers or beta stabilizers. Figure HP7.10 contains the Ti-Al and Ti-Vequilibrium phase diagrams. Which alloying element would most likely be consideredthe alpha stabilizer? Explain.GIVEN: The Ti-Al and Ti-V phase diagrams.SOLUTION: Solid titanium exists in two crystal forms. β, the high temperatureallotrope which is bcc, and α, the low temperature form which is hcp. At 882 ︒C thesetwo phases are in equilibrium at 1 atm pressure for pure titanium. When alloyingelements are added the relative stability of the phases may change. For the Ti-Al system, adding aluminum to pure titanium increases the temperature at which α begins to form.Thus, aluminum is said toIn the Ti-V system, thetemperature at which the αphases formsdecrease with increasing V,said to be a βthus V is41. FIND: In the Ti-Al system, identify a phase and its composition that melts congruently.Estimate the congruent melting temperature from the phase diagram.GIVEN: The Ti-Al phase diagram and the definition of a congruently melting phase. A phase melts congruently, when the composition of the solid and the composition of theliquid for an alloy are the same.SOLUTION: Examination of the Ti-Al phase diagram shows that the composition near5 weight % ( ~ 10 atomic %) melts congruently at approximately 1730 ︒C.42.FIND: Label all the two-phase fields on the Ti-Al phase diagram.GIVEN: The Ti-Al phase diagram with all the single phase fields labeled.SOLUTION: To complete this problem, we need only recognize that in a binary system, phases appear across the diagram alternating 1-2-1-2-...-1. Included below is the Al-Ti phase diagram with all phase fields labeled.α + Ti Al 33Ti Al + TiAl TiAl + TiAl 2α + Ti Al 3β + TiAlβ + Lα + TiAlTiAl + LTiAl + δTiAl + δ2α700900110013001500T e m p e r a t u r e , °C23TiAl + βTiAl 23TiAl + αTiAl 3αTiAl + Al 3βTiAl + Al Allotropic phaseL + δα + β3δ + βTiAl 3βTiAl + L 170043. FIND: Label all the invariant reactions occurring in the Ti-Al system.GIVEN: The Ti-Al phase diagram.SOLUTION:See the sketch of the Ti-Al diagram below.44. FIND: From the phase diagram, estimate the temperature at which Ti3Al congruentlytransforms to α Ti.GIVEN: The Ti-Al phase diagram.SOLUTION: The phase Ti3Al transforms the α Ti of the same composition atapproximately 1200︒C and a composition of approximately 21 wt % Al (33 atomic % Al).45. FIND: When the line compound β TiAl3 is heated, what is the composition of the firstliquid that forms in equilibrium with β TiAl3? Does this compound melt congruently?Explain.GIVEN: The Ti-Al phase diagram.SOLUTION: TiAl3 appears as a line compound on the phase diagram. Atapproximately 1375︒C TiAl3 undergoes a peritectic reaction on heating underequilibrium conditions. The composition of the liquid in equilibrium with TiAl3 is notthe composition of TiAl3 and hence the phase does not melt congruently, but meltsincongruently. The composition of the liquid is approximately 63 wt % Al. Theperitectic reaction written on heating is:TiAl3→ liquid (63 wt % Al) + δ Ti.46. FIND: The presence of Ti3Al in Ti-Al alloys has a detrimental effect on the ductility. Tocontrol this problem, the amount of aluminum needs to be less than 6 wt. %.Consequently, all of the products that you are producing contain a maximum of 6 wt. %of aluminum. You have been told that additions of tin, zirconium, and oxygen (oftenpresent as an impurity) are all known to be alpha stabilizers. If there is a possibility thatsmall additions of oxygen may enter the system as your alloy is melted, should youreduce the amount of aluminum in your alloy or not worry about it if ductility is animportant property for your product? Explain your reasoning.GIVEN: The Ti-Al phase diagram.SOLUTION: Since both aluminum and oxygen are α stabilizers, a judicious choicewould be to reduce the amount of aluminum in your alloy. This would be a particularlywise choice since the critical amount is just at 6 % Al, the addition of any alloyingelement that might affect the amount of Ti3Al should be avoided. Therefore reducing the amount of Al should be reduced.47. FIND: Label all the two-phase fields in the Ti-V system.GIVEN: The Ti-V phase diagram with all the single phase fields labeled.SOLUTION: To complete this problem, we need only recognize that in a binary system, phase fields appear across the diagram alternating 1-2-1-2-...-1. Included below is asketch of the Al-V phase diagram with all phase fields labeled.48. FIND: Identify the invariant reaction that is occurring at 675︒C in the Ti-V system.GIVEN: The Ti-V phase diagram.SOLUTION: The invariant reaction that is occurring at 675︒C is a monotectoid reaction and can be written symbolically as:β1→α + β2.from 900︒C down to 500︒C, plot the fraction of phases present as a function of temperature.GIVEN: The Ti-V phase diagram.SOLUTION: The compositions and the fraction of each phase from the phase diagram are shown in the table below. From the table the sketch given below illustrates the relative fraction of phases present as a function of temperature.50.FIND: Consider an alloy containing 52 wt. % V. Describe the phases present and their compositions as the alloy is cooled under equilibrium conditions from 900︒C to 500︒C. GIVEN: The Al-V phase diagram.SOLUTION: An alloy containing 52 wt. % V at 900︒C is in the β Ti,V single phase field. When the alloy is cooled it passes through the critical temperature of a monotectoid at 850︒C. Cooling below this temperature results in the formation of a two phase mixture of V-poor and V-rich solids that have the β Ti,V structure. Decreasing the temperature results in two solid phases whose compositions continuously change. The composition of the two solids follow the solid-solid miscibility boundary as shown in the sketch below. Cooling to 675︒C results in the invariant monotectoid reaction at that temperature. Symbolically that reaction can be written as β1 → α + β2. Cooling below themonotectoid temperature results in the disappearance of the β1 phase. Since the relative solubilities of V in Ti and Ti in V decrease with temperature, cooling to 500︒C changes the compositions of the α and β2 phases following the solvus boundaries as shown in the sketch.51.FIND: If a titanium alloy containing 5 wt. % V is cooled under equilibrium conditions from 900︒C down to 500︒C, plot the fraction of phases present as a function of temperature. Which phase is richer in vanadium, α or β? GIVEN: The Ti-V phase diagram.SOLUTION: The β phase is richer in vanadium than the α phase.L + β ( Ti, V )β ( Ti, V ) + α ( Ti )β1β2+Comp. of β decreases in V along phase boundary12Comp. of β increases in V along phase boundaryMonotectoid isotherm。

材科基

材科基
1、固体中扩散的特点 2、菲克定律(宏观现象) 菲克第一定律:稳态扩散 菲克第二定律:不稳态扩散 3、扩散系数是一个什么样的参数(D∝fr2) 4、扩散推动力(化学位梯度) 5、扩散系数的一般热力学关系式及其讨论 6、质点的扩散方式(五种、其中空位最常见,所需能量最小) 7、本征扩散、非本征扩散,及其相应的扩散系数D和扩散活化能 产生本征扩散与非本征扩散的原因与条件,分析、计算 非化学计量化合物中的扩散 8、萤石结构的ZrO2中Zr4+和O2-哪个扩散的活化能大?
NJ University of Technology
NJ University of Technology
NJ University of Technology
NJ University of Technology
NJ University of Technology
第六章、扩散与固相反应
பைடு நூலகம்
NJ University of Technology
第三章、非晶体固体
1、熔体的概念:不同聚合程度的各种聚合物的混合物 硅酸盐熔体的粘度与组成的关系 2、非晶态物质的特点 3、玻璃的通性(4个) 4、 Tg 、Tf ,相对应的粘度和特点 5、网络形成体、网络变化体、网络中间体
NJ University of Technology
第一章、晶体结构基础
1、晶体的基本概念 晶体的本质:质点在三维空间成周期性重复排列 晶体的基本性质:结晶均一性、各向异性、自限性、对称性、 最小内能性 2、对称的概念 物体中的相同部分作有规律的重复 对称要素:对称面、对称中心、对称轴(对称轴的类型和特点) 4次倒转轴不能被其他的对称要素及其组合取代 (L1、L2、L3、L4、L6、C、P、 L4i ) 对称操作:借助对称要素,使晶体的相同部分完全重复的操作 对称要素的组合必须满足晶体的整体对称要求,不是无限的

武汉理工 材料科学基础 课后答案 _第七章

武汉理工    材料科学基础  课后答案  _第七章

第七章答案7-1略7-2浓度差会引起扩散,扩散是否总是从高浓度处向低浓度处进行?为什么?解:扩散是由于梯度差所引起的,而浓度差只是梯度差的一种。

当另外一种梯度差,比如应力差的影响大于浓度差,扩散则会从低浓度向高浓度进行。

7-3欲使Ca2+在CaO中的扩散直至CaO的熔点(2600℃)时都是非本质扩散,要求三价离子有什么样的浓度?试对你在计算中所做的各种特性值的估计作充分说明。

已知CaO肖特基缺陷形成能为6eV。

解:掺杂M3+引起V’’Ca的缺陷反应如下:当CaO在熔点时,肖特基缺陷的浓度为:所以欲使Ca2+在CaO中的扩散直至CaO的熔点(2600℃)时都是非本质扩散,M3+的浓度为,即7-4试根据图 7-32查取:(1)CaO在1145℃和1650℃的扩散系数值;(2)Al2O3在1393℃和1716℃的扩散系数值;并计算CaO和Al2O3中Ca2+和Al3+的扩散活化能和D0值。

解:由图可知CaO在1145℃和1650℃的扩散系数值分别为,Al2O3在1393℃和1716℃的扩散系数值分别为根据可得到CaO在1145℃和1650℃的扩散系数的比值为:,将值代入后可得,Al2O3的计算类推。

7-5已知氢和镍在面心立方铁中的扩散数据为cm2/s和cm2/s,试计算1000℃的扩散系数,并对其差别进行解释。

解:将T=1000℃代入上述方程中可得,同理可知。

原因:与镍原子相比氢原子小得多,更容易在面心立方的铁中通过空隙扩散。

7-6在制造硅半导体器体中,常使硼扩散到硅单晶中,若在1600K温度下,保持硼在硅单晶表面的浓度恒定(恒定源半无限扩散),要求距表面10-3cm深度处硼的浓度是表面浓度的一半,问需要多长时间(已知D1600℃=8×10-12cm2/s;当时,)?解:此模型可以看作是半无限棒的一维扩散问题,可用高斯误差函数求解。

其中=0,,所以有0.5=,即=0.5,把=10-3cm,D1600℃=8×10-12cm2/s代入得t=s。

《材料科学基础》课后答案(1-7章)

《材料科学基础》课后答案(1-7章)

第一章8.计算下列晶体的离于键与共价键的相对比例(1)NaF (2)CaO (3)ZnS解:1、查表得:X Na =0.93,X F =3.98根据鲍林公式可得NaF 中离子键比例为:21(0.93 3.98)4[1]100%90.2%e ---⨯=共价键比例为:1-90.2%=9.8% 2、同理,CaO 中离子键比例为:21(1.00 3.44)4[1]100%77.4%e---⨯=共价键比例为:1-77.4%=22.6%3、ZnS 中离子键比例为:21/4(2.581.65)[1]100%19.44%ZnS e --=-⨯=中离子键含量共价键比例为:1-19.44%=80.56%10说明结构转变的热力学条件与动力学条件的意义.说明稳态结构与亚稳态结构之间的关系。

答:结构转变的热力学条件决定转变是否可行,是结构转变的推动力,是转变的必要条件;动力学条件决定转变速度的大小,反映转变过程中阻力的大小。

稳态结构与亚稳态结构之间的关系:两种状态都是物质存在的状态,材料得到的结构是稳态或亚稳态,取决于转交过程的推动力和阻力(即热力学条件和动力学条件),阻力小时得到稳态结构,阻力很大时则得到亚稳态结构。

稳态结构能量最低,热力学上最稳定,亚稳态结构能量高,热力学上不稳定,但向稳定结构转变速度慢,能保持相对稳定甚至长期存在。

但在一定条件下,亚稳态结构向稳态结构转变。

第二章1.回答下列问题:(1)在立方晶系的晶胞内画出具有下列密勒指数的晶面和晶向:(001)与[210],(111)与[112],(110)与 [111],(132)与[123],(322)与[236](2)在立方晶系的一个晶胞中画出(111)和 (112)晶面,并写出两晶面交线的晶向指数。

(3)在立方晶系的一个晶胞中画出同时位于(101). (011)和(112)晶面上的[111]晶向。

解:1、2.有一正交点阵的 a=b, c=a/2。

某晶面在三个晶轴上的截距分别为 6个、2个和4个原子间距,求该晶面的密勒指数。

材料科学基础(张联盟版)课后习题与答案

材料科学基础(张联盟版)课后习题与答案
由几何关系知: =0.054nm
比Li+的离子半径rLi+=0.078nm小,所以此时O2-不能互相接触。
晶胞参数 =0.373nm
Li2O的密度 g/cm3
2-18MgO和CaO同属NaCl型结构,而它们与水作用时则CaO要比MgO活泼,试解释之。
解:因为rMg2+与rCa2+不同,rCa2+>rMg2+,使CaO结构较MgO疏松,H2O易于进入,所以活泼。
2-6等径球最紧密堆积的空?
答:等径球最紧密堆积有六方和面心立方紧密堆积两种,一个球的周围有8个四面体空隙、6个八面体空隙。
2-7
答:n个等径球作最紧密堆积时可形成n个八面体空隙、2n个四面体空隙。
不等径球体进行紧密堆积时,可以看成由大球按等径球体紧密堆积后,小球按其大小分别填充到其空隙中,稍大的小球填充八面体空隙,稍小的小球填充四面体空隙,形成不等径球体紧密堆积。
所以钙钛矿结构遵守鲍林规则。
2-26硅酸盐晶体结构有何特点?怎样表征其学式?
答:硅酸盐晶体结构非常复杂,但不同的结构之间具有下面的共同特点:
(1)结构中的Si4+离子位于O2-离子形成的四面体中心,构成硅酸盐晶体的基本结构单元[SiO4]四面体。Si-O-Si是一条夹角不等的折线,一般在145°左右。
第二章答案
2-1略。
2-2(1)一晶面在
答:(1)h:k:l= =3:2:1,∴该晶面的晶面指数为(321);
(2)h:k:l=3:2:1,∴该晶面的晶面指数为(321)。
2-3在立方晶系晶胞中画出下列晶面指数和晶向指数:(001)与[
答:
2-4定性描述晶体结构的参量有哪些?定量描述晶体结构的参量又有哪些?

材料科学基础-课后习题及答案(武汉理工大学-张联盟版)

材料科学基础-课后习题及答案(武汉理工大学-张联盟版)

第二章答案2-1略。

2-2(1)一晶面在x、y、z轴上的截距分别为2a、3b、6c,求该晶面的晶面指数;(2)一晶面在x、y、z轴上的截距分别为a/3、b/2、c,求出该晶面的晶面指数。

答:(1)h:k:l==3:2:1,∴该晶面的晶面指数为(321);(2)h:k:l=3:2:1,∴该晶面的晶面指数为(321)。

2-3在立方晶系晶胞中画出下列晶面指数和晶向指数:(001)与[],(111)与[],()与[111],()与[236],(257)与[],(123)与[],(102),(),(),[110],[],[]答:2-4定性描述晶体结构的参量有哪些?定量描述晶体结构的参量又有哪些?答:定性:对称轴、对称中心、晶系、点阵。

定量:晶胞参数。

2-5依据结合力的本质不同,晶体中的键合作用分为哪几类?其特点是什么?答:晶体中的键合作用可分为离子键、共价键、金属键、范德华键和氢键。

离子键的特点是没有方向性和饱和性,结合力很大。

共价键的特点是具有方向性和饱和性,结合力也很大。

金属键是没有方向性和饱和性的的共价键,结合力是离子间的静电库仑力。

范德华键是通过分子力而产生的键合,分子力很弱。

氢键是两个电负性较大的原子相结合形成的键,具有饱和性。

2-6等径球最紧密堆积的空隙有哪两种?一个球的周围有多少个四面体空隙、多少个八面体空隙?答:等径球最紧密堆积有六方和面心立方紧密堆积两种,一个球的周围有8个四面体空隙、6个八面体空隙。

2-7n个等径球作最紧密堆积时可形成多少个四面体空隙、多少个八面体空隙?不等径球是如何进行堆积的?答:n个等径球作最紧密堆积时可形成n个八面体空隙、2n个四面体空隙。

不等径球体进行紧密堆积时,可以看成由大球按等径球体紧密堆积后,小球按其大小分别填充到其空隙中,稍大的小球填充八面体空隙,稍小的小球填充四面体空隙,形成不等径球体紧密堆积。

2-8写出面心立方格子的单位平行六面体上所有结点的坐标。

答:面心立方格子的单位平行六面体上所有结点为:(000)、(001)(100)(101)(110)(010)(011)(111)(0)(0)(0)(1)(1)(1)。

材料科学基础(张联盟版)课后部分习题

材料科学基础(张联盟版)课后部分习题

材料科学基础(张联盟版)课后部分习题材料基部分题目1.晶体结构1、解释下列概念晶系、晶胞、晶胞参数、空间点阵、米勒指数(晶面指数)、离子晶体的晶格能、原子半径与离子半径、配位数、离子极化、同质多晶与类质同晶、正尖晶石与反正尖晶石、反萤石结构、铁电效应、压电效应.2、(1)一晶面在x、y、z轴上的截距分别为2a、3b、6c,求出该晶面的米勒指数;(2)一晶面在x、y、z轴上的截距分别为a/3、b/2、c,求出该晶面的米勒指数。

3、在立方晶系的晶胞中画出下列米勒指数的晶面和晶向:(001)与[210],(111)与[112],(110)与[111],(322)与[236],(257)与[111],(123)与[121],(102),(112),(213),[110],[111],[120],[321]4、写出面心立方格子的单位平行六面体上所有结点的坐标。

5、已知Mg2+半径为0.072nm,O2-半径为0.140nm,计算MgO晶体结构的堆积系数与密度。

6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

7、从理论计算公式计算NaC1与MgO的晶格能。

MgO的熔点为2800℃,NaC1为80l℃, 请说明这种差别的原因。

8、根据最密堆积原理,空间利用率越高,结构越稳定,金钢石结构的空间利用率很低(只有34.01%),为什么它也很稳定?9、证明等径圆球面心立方最密堆积的空隙率为25.9%;10、金属镁原子作六方密堆积,测得它的密度为1.74克/厘米3,求它的晶胞体积。

11、根据半径比关系,说明下列离子与O2—配位时的配位数各是多?r o2-=0.132nm r Si4+=0.039nm r K+=0.133nm r Al3+=0.057nm r Mg2+=0.078n m12、为什么石英不同系列变体之间的转化温度比同系列变体之间的转化温度高得多?13、有效离子半径可通过晶体结构测定算出。

《材料科学基础》经典习题及答案

《材料科学基础》经典习题及答案

材料科学与基础习题集和答案第七章回复再结晶,还有相图的内容。

第一章1.作图表示立方晶体的()()()421,210,123晶面及[][][]346,112,021晶向。

2.在六方晶体中,绘出以下常见晶向[][][][][]0121,0211,0110,0112,0001等。

3.写出立方晶体中晶面族{100},{110},{111},{112}等所包括的等价晶面。

4.镁的原子堆积密度和所有hcp 金属一样,为0.74。

试求镁单位晶胞的体积。

已知Mg 的密度3Mg/m 74.1=mg ρ,相对原子质量为24.31,原子半径r=0.161nm 。

5.当CN=6时+Na 离子半径为0.097nm ,试问:1) 当CN=4时,其半径为多少?2) 当CN=8时,其半径为多少?6. 试问:在铜(fcc,a=0.361nm )的<100>方向及铁(bcc,a=0.286nm)的<100>方向,原子的线密度为多少?7.镍为面心立方结构,其原子半径为nm 1246.0=Ni r 。

试确定在镍的(100),(110)及(111)平面上12mm 中各有多少个原子。

8. 石英()2SiO 的密度为2.653Mg/m 。

试问: 1) 13m 中有多少个硅原子(与氧原子)?2) 当硅与氧的半径分别为0.038nm 与0.114nm 时,其堆积密度为多少(假设原子是球形的)?9.在800℃时1010个原子中有一个原子具有足够能量可在固体内移动,而在900℃时910个原子中则只有一个原子,试求其激活能(J/原子)。

10.若将一块铁加热至850℃,然后快速冷却到20℃。

试计算处理前后空位数应增加多少倍(设铁中形成一摩尔空位所需要的能量为104600J )。

11.设图1-18所示的立方晶体的滑移面ABCD 平行于晶体的上、下底面。

若该滑移面上有一正方形位错环,如果位错环的各段分别与滑移面各边平行,其柏氏矢量b ∥AB 。

材料科学基础(武汉理工大学,张联盟版)课后习题及答案 第七章

材料科学基础(武汉理工大学,张联盟版)课后习题及答案 第七章

材料科学基础(武汉理工大学,张联盟版)课后习题及答案第七章第七章答案7-1略7-2浓度差会引起扩散,扩散是否总是从高浓度处向低浓度处进行?为什么?解:扩散是由于梯度差所引起的,而浓度差只是梯度差的一种。

当另外一种梯度差,比如应力差的影响大于浓度差,扩散则会从低浓度向高浓度进行。

7-3欲使Ca在CaO中的扩散直至CaO的熔点(2600℃)时都是非本质扩散,要求三价离子有什么样的浓度?试对你在计算中所做的各种特性值的估计作充分说明。

已知CaO肖特基缺陷形成能为6eV。

解:掺杂M引起V’’Ca的缺陷反应如下:当CaO在熔点时,肖特基缺陷的浓度为:3+2+所以欲使Ca在CaO中的扩散直至CaO的熔点(2600℃)时都是非本质扩散,M的浓度为,即2+3+7-4试根据图7-32查取:(1)CaO在1145℃和1650℃的扩散系数值;(2)Al2O3在1393℃2+3+和1716℃的扩散系数值;并计算CaO和Al2O3中Ca和Al的扩散活化能和D0值。

解:由图可知CaO在1145℃和1650℃的扩散系数值分别为,Al2O3在1393℃和1716℃的扩散系数值分别为根据可得到CaO在1145℃和1650℃的扩散系数的比值为:,将值代入后可得,Al2O3的计算类推。

7-5已知氢和镍在面心立方铁中的扩散数据为2cm/s和2cm/s,试计算1000℃的扩散系数,并对其差别进行解释。

解:将T=1000℃代入上述方程中可得。

,同理可知原因:与镍原子相比氢原子小得多,更容易在面心立方的铁中通过空隙扩散。

7-6在制造硅半导体器体中,常使硼扩散到硅单晶中,若在1600K温度下,保持硼在硅单晶-3表面的浓度恒定(恒定源半无限扩散),要求距表面10cm深度处硼的浓度是表面浓度的一半,问需要多长时间(已知D1600℃=8×10-12cm/s;当2时,)?解:此模型可以看作是半无限棒的一维扩散问题,可用高斯误差函数求解。

其中-122=0,,所以有0.5=s。

武汉理工大学材料科学基础课后习题答案 第二版 张联盟

武汉理工大学材料科学基础课后习题答案 第二版 张联盟
热缺陷——本征缺陷;固溶体——非本征缺陷;
分类
形成原因
形成条件
缺陷反应
化学式
溶解度、缺陷浓度
热缺陷
肖特基
弗伦克尔
热起伏
T>0k
O VM’’+ Vx··
MM Mi··+ VM’’
MX
MX
只受温度控制
固溶体
无限,有限,置换,间隙
搀杂
溶解
大小,电负性,电价,结构
无:受温度控制
有:搀杂量<固溶度 受温度控制
22、解:(1)有两种配位多面体,[SiO4],[MgO6],同层的[MgO6]八面体共棱,如59[MgO6]和49[MgO6]共棱75O2-和27O2-,不同层的[MgO6]八面体共顶,如1[MgO6]和51[MgO6]共顶是22O2-,同层的[MgO6]与[SiO4]共顶,如T[MgO6]和7[SiO4]共顶22O2-,不同层的[MgO6]与[SiO4]共棱,T[MgO6]和43[SiO4]共28O2-和28O2-;
14、解:
固溶体
机械混合物
化合物
形成原因
以原子尺寸“溶解”生成
粉末混合
原子间相互反映生成
相数
单相均匀
多相
单相均匀
化学计量
不遵守定比定律
/
遵守定比定律
化学组成
不确定
有几种混合物就有多少化学组成
确定
15、解:固溶体、晶格缺陷、非化学计量化合物都是点缺陷,是晶体结构缺陷,都是单相均匀的固体,结构同主晶相。
将化学式与组成式一一对应,求出待定参数x,由于O2-的量不同,将O2-的量化为1
Al0.16/1.16Mg0.92/1.16O Al2x/1+xMg1-2x/1+xO x=0.074,化学式Al0.148Mg0.852O1.074

《材料科学基础》经典习题及答案

《材料科学基础》经典习题及答案

《材料科学基础》经典习题及答案材料科学与基础习题集和答案第七章回复再结晶,还有相图的内容。

第一章102??,211?,?346?晶向。

1.作图表示立方晶体的?123?,?012?,?421?晶面及? ?,?2110??,1010??,1120??,1210?等。

2.在六方晶体中,绘出以下常见晶向?0001 3.写出立方晶体中晶面族{100},{110},{111},{112}等所包括的等价晶面。

4.镁的原子堆积密度和所有hcp金属一样,为。

试求镁单位晶胞的体积。

已知Mg的密度,相对原子质量为,原子半径r=。

?5.当CN=6时Na离子半径为,试问:1) 当CN=4时,其半径为多少?2) 当CN=8时,其半径为多少? 6. 试问:在铜的方向及铁(bcc,a=)的方向,原子的线密度为多少?7.镍为面心立方结构,其原子半径为rNi?。

试确定在镍的2,及平面上1mm中各有多少个原子。

3??SiOMg/m28. 石英的密度为。

试问:?mg?/m31) 1m中有多少个硅原子?2) 当硅与氧的半径分别为与时,其堆积密度为多少?10109.在800℃时个原子中有一个原子具有足够能量可在固体内移9动,而在900℃时10个原子中则只有一个原子,试求其激活能。

10.若将一块铁加热至850℃,然后快速冷却到20℃。

试计算处理前后空位数应增加多少倍。

3 11.设图1-18所示的立方晶体的滑移面ABCD平行于晶体的上、下底面。

若该滑移面上有一正方形位错环,如果位错环的各段分别与滑移面各边平行,其柏氏矢量b∥AB。

1) 有人认为“此位错环运动移出晶体后,滑移面上产生的滑移台阶应为4个b,试问这种看法是否正确?为什么?2)指出位错环上各段位错线的类型,并画出位错运动出晶体后,滑移方向及滑移量。

12.设图1-19所示立方晶体中的滑移面ABCD平行于晶体的上、下底面。

晶体中有一条位错线fed,de段在滑移面上并平行AB,ef段与滑移面垂直。

材料科学基础(张联盟版)课后习题与答案

材料科学基础(张联盟版)课后习题与答案
所以钙钛矿结构遵守鲍林规则。
2-26硅酸盐晶体结构有何特点?怎样表征其学式?
答:硅酸盐晶体结构非常复杂,但不同的结构之间具有下面的共同特点:
(1)结构中的Si4+离子位于O2-离子形成的四面体中心,构成硅酸盐晶体的基本结构单元[SiO4]四面体。Si-O-Si是一条夹角不等的折线,一般在145°左右。
所以 ,即O2-电价饱和。
2-21(1)画出O
解(1)略
(2)四面体空隙数与O2-数之比为2:1,八面体空隙数与O2-数之比为1:1
2-22根据电价规则,在下面情况下,空隙内各需填入何种价数的阳离子,并对每一种结构举出—个例子。(1)所有四面体空隙位置均填满;(2)所有八面体空隙位置均填满;(3)填满—半四面体空隙位置;(4)填满—半八面体空隙位置。
答:设晶胞的体积为V,相对原子质量为M,则晶胞体积 nm3
2-13根据半径比关系,说明下列离子与O
答:对于Si4+、K+、Al3+、Mg2+来说,其 依次是0.295、0.99、0.43、0.59;依据正离子配位数与正负离子半径比的关系知配位数为:Si4+4;K+8;Al3+6;Mg2+6。
2-14为什么石英不同系列变体之间的转化温度比同系列变体之间的转化温度高得多?
层状
3
平面层
[Si4O10]4-
4:10
架状
4
骨架
[SiO4]4-
[(AlxSi4-x)O8]x-
1:4
1:4
2-28下列硅酸盐矿物各属何种结构类型:Mg
答:分别为岛状;架状;单链;层状(复网);组群(双四面体)。
2-29根据Mg
解:(1)有两种配位多面体,[SiO4],[MgO6],同层的[MgO6]八面体共棱,如59[MgO6]和49[MgO6]共棱75O2-和27O2-,不同层的[MgO6]八面体共顶,如1[MgO6]和51[MgO6]共顶是22O2-,同层的[MgO6]与[SiO4]共顶,如T[MgO6]和7[SiO4]共顶22O2-,不同层的[MgO6]与[SiO4]共棱,T[MgO6]和43[SiO4]共28O2-和28O2-;

武汉理工大学材料科学基础课后习题和答案

武汉理工大学材料科学基础课后习题和答案

第一章绪论1、仔细观察一下白炽灯泡,会发现有多少种不同的材料每种材料需要何种热学、电学性质2、为什么金属具有良好的导电性和导热性3、为什么陶瓷、聚合物通常是绝缘体4、铝原子的质量是多少若铝的密度为2.7g/cm3,计算1mm3中有多少原子5、为了防止碰撞造成纽折,汽车的挡板可有装甲制造,但实际应用中为何不如此设计说出至少三种理由。

6、描述不同材料常用的加工方法。

7、叙述金属材料的类型及其分类依据。

8、试将下列材料按金属、陶瓷、聚合物或复合材料进行分类:黄铜钢筋混凝土橡胶氯化钠铅-锡焊料沥青环氧树脂镁合金碳化硅混凝土石墨玻璃钢9、 Al2O3陶瓷既牢固又坚硬且耐磨,为什么不用Al2O3制造铁锤第二章晶体结构1、解释下列概念晶系、晶胞、晶胞参数、空间点阵、米勒指数(晶面指数)、离子晶体的晶格能、原子半径与离子半径、配位数、离子极化、同质多晶与类质同晶、正尖晶石与反正尖晶石、反萤石结构、铁电效应、压电效应.2、(1)一晶面在x、y、z轴上的截距分别为2a、3b、6c,求出该晶面的米勒指数;(2)一晶面在x、y、z轴上的截距分别为a/3、b/2、c,求出该晶面的米勒指数。

3、在立方晶系的晶胞中画出下列米勒指数的晶面和晶向:(001)与[210],(111)与[112],(110)与[111],(322)与[236],(257)与[111],(123)与[121],(102),(112),(213), [110],[111],[120],[321]4、写出面心立方格子的单位平行六面体上所有结点的坐标。

5、已知Mg2+半径为,O2-半径为,计算MgO晶体结构的堆积系数与密度。

6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

7、从理论计算公式计算NaC1与MgO的晶格能。

MgO的熔点为2800℃,NaC1为80l℃, 请说明这种差别的原因。

8、根据最密堆积原理,空间利用率越高,结构越稳定,金钢石结构的空间利用率很低(只有%),为什么它也很稳定9、证明等径圆球面心立方最密堆积的空隙率为25.9%;10、金属镁原子作六方密堆积,测得它的密度为1.74克/厘米3,求它的晶胞体积。

武汉理工大学材料科学基础各章节例题及答案

武汉理工大学材料科学基础各章节例题及答案

第二章晶体结构【例2-1】计算MgO和GaAs晶体中离子键成分的多少。

【解】查元素电负性数据得,则,,,MgO离子键%=GaAs离子键%=由此可见,MgO晶体的化学键以离子键为主,而GaAs则是典型的共价键晶体。

【提示】除了以离子键、共价键结合为主的混合键晶体外,还有以共价键、分子间键结合为主的混合键晶体。

且两种类型的键独立地存在。

如,大多数气体分子以共价键结合,在低温下形成的晶体则依靠分子间键结合在一起。

石墨的层状单元内共价结合,层间则类似于分子间键。

正是由于结合键的性质不同,才形成了材料结构和性质等方面的差异。

从而也满足了工程方面的不同需要。

【例2-2】 NaCl和MgO晶体同属于NaCl型结构,但MgO的熔点为2800℃, NaC1仅为80l℃,请通过晶格能计算说明这种差别的原因。

【解】根据:晶格能(1)NaCl晶体:N0=6.023×1023 个/mol,A=1.7476,z1=z2=1,e=1.6×10-19 库仑,,r0===0.110+0.172=0.282nm=2.82×10-10 m,m/F,计算,得:EL=752.48 kJ/mol (2)MgO晶体:N0=6.023×1023 个/mol,A=1.7476,z1=z2=2,e=1.6×10-19库仑,r0==0.080+0.132=0.212 nm=2.12×10-10 m,m/F,计算,得:EL=3922.06 kJ/mol则:MgO晶体的晶格能远大于NaC1晶体的晶格能,即相应MgO的熔点也远高于NaC1的熔点。

【例2-3】根据最紧密堆积原理,空间利用率越高,结构越稳定,但是金刚石的空间利用率很低,只有34.01%,为什么它也很稳定?【解】最紧密堆积的原理只适用于离子晶体,而金刚石为原子晶体,由于C-C共价键很强,且晶体是在高温和极大的静压力下结晶形成,因而熔点高,硬度达,很稳定。

武汉理工大学材料科学基础(第2版)课后习题和答案

武汉理工大学材料科学基础(第2版)课后习题和答案

武汉理工大学材料科学基础(第2版)课后习题和答案第一章绪论1、仔细观察一下白炽灯泡,会发现有多少种不同的材料?每种材料需要何种热学、电学性质?2、为什么金属具有良好的导电性和导热性?3、为什么陶瓷、聚合物通常是绝缘体?4、铝原子的质量是多少?若铝的密度为2.7g/cm3,计算1mm3中有多少原子?5、为了防止碰撞造成纽折,汽车的挡板可有装甲制造,但实际应用中为何不如此设计?说出至少三种理由。

6、描述不同材料常用的加工方法。

7、叙述金属材料的类型及其分类依据。

8、试将下列材料按金属、陶瓷、聚合物或复合材料进行分类:黄铜钢筋混凝土橡胶氯化钠铅-锡焊料沥青环氧树脂镁合金碳化硅混凝土石墨玻璃钢9、Al2O3陶瓷既牢固又坚硬且耐磨,为什么不用Al2O3制造铁锤?第二章晶体结构1、解释下列概念晶系、晶胞、晶胞参数、空间点阵、米勒指数(晶面指数)、离子晶体的晶格能、原子半径与离子半径、配位数、离子极化、同质多晶与类质同晶、正尖晶石与反正尖晶石、反萤石结构、铁电效应、压电效应.2、(1)一晶面在某、y、z轴上的截距分别为2a、3b、6c,求出该晶面的米勒指数;(2)一晶面在某、y、z轴上的截距分别为a/3、b/2、c,求出该晶面的米勒指数。

3、在立方晶系的晶胞中画出下列米勒指数的晶面和晶向:(001)与[210],(111)与[112],(110)与[111],(322)与[236],(257)与[111],(123)与[121],(102),(112),(213),[110],[111],[120],[321]4、写出面心立方格子的单位平行六面体上所有结点的坐标。

5、已知Mg2+半径为0.072nm,O2-半径为0.140nm,计算MgO晶体结构的堆积系数与密度。

6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

7、从理论计算公式计算NaC1与MgO的晶格能。

MgO的熔点为2800℃,NaC1为80l℃,请说明这种差别的原因。

武汉理工大学材料科学基础(第2版)课后习题和答案

武汉理工大学材料科学基础(第2版)课后习题和答案

武汉理工大学材料科学基础(第2版)课后习题和答案第一章绪论1、仔细观察一下白炽灯泡,会发现有多少种不同的材料?每种材料需要何种热学、电学性质?2、为什么金属具有良好的导电性和导热性?3、为什么陶瓷、聚合物通常是绝缘体?4、铝原子的质量是多少?若铝的密度为2.7g/cm3,计算1mm3中有多少原子?5、为了防止碰撞造成纽折,汽车的挡板可有装甲制造,但实际应用中为何不如此设计?说出至少三种理由。

6、描述不同材料常用的加工方法。

7、叙述金属材料的类型及其分类依据。

8、试将下列材料按金属、陶瓷、聚合物或复合材料进行分类:黄铜钢筋混凝土橡胶氯化钠铅-锡焊料沥青环氧树脂镁合金碳化硅混凝土石墨玻璃钢9、Al2O3陶瓷既牢固又坚硬且耐磨,为什么不用Al2O3制造铁锤?第二章晶体结构1、解释下列概念晶系、晶胞、晶胞参数、空间点阵、米勒指数(晶面指数)、离子晶体的晶格能、原子半径与离子半径、配位数、离子极化、同质多晶与类质同晶、正尖晶石与反正尖晶石、反萤石结构、铁电效应、压电效应.2、(1)一晶面在x、y、z轴上的截距分别为2a、3b、6c,求出该晶面的米勒指数;(2)一晶面在x、y、z 轴上的截距分别为a/3、b/2、c,求出该晶面的米勒指数。

3、在立方晶系的晶胞中画出下列米勒指数的晶面和晶向:(001)与[210],(111)与[112],(110)与[111],(322)与[236],(257)与[111],(123)与[121],(102),(112),(213),[110],[111],[120],[321]4、写出面心立方格子的单位平行六面体上所有结点的坐标。

5、已知Mg2+半径为0.072nm,O2-半径为0.140nm,计算MgO晶体结构的堆积系数与密度。

6、计算体心立方、面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

7、从理论计算公式计算NaC1与MgO的晶格能。

MgO的熔点为2800℃,NaC1为80l℃, 请说明这种差别的原因。

材料科学基础 第二版(张联盟 著)课后习题答案武汉理工大学出版

材料科学基础 第二版(张联盟 著)课后习题答案武汉理工大学出版

材料科学基础第二版(张联盟著)课后习题答案武汉理工大学出版材料科学基础第二版(张联盟著)课后习题大家找到答案了吗?下面是为大家推荐一些材料科学基础第二版(张联盟著)课后习题答案和下载地址,希望大家有用哦。

2-1名词解释晶系晶胞晶胞参数空间点阵晶面指数晶格能原子半径与离子半径配位数离子极化同质多晶与类质同晶正尖晶石与反正尖晶石反萤石结构铁电效应压电效应热释电效应电光效应2-2(1)一晶面在x、y、z轴上的截距分别为2a、3b、6c,求该晶面的晶面指数;(2)一晶面在x、y、z轴上的截距分别为a/3、b/2、c,求出该晶面的晶面指数。

2-3在立方晶系晶胞中画出下列晶面指数和晶向指数:(001)与[],(111)与[],()与[111],()与[236],(257)与[],(123)与[],(102),(),(),[110],[,[],[]2-4定性描述晶体结构的参量有哪些?定量描述晶体结构的参量又有哪些?2-5依据结合力的本质不同,晶体中的键合作用分为哪几类?其特点是什么?2-6等径球最紧密堆积的空隙有哪两种?一个球的周围有多少个四面体空隙、多少个八面体空隙?2-7n个等径球作最紧密堆积时可形成多少个四面体空隙、多少个八面体空隙?不等径球是如何进行堆积的?2-8写出面心立方格子的单位平行六面体上所有结点的坐标。

2-9计算面心立方、密排六方晶胞中的原子数、配位数、堆积系数。

2-10根据最密堆积原理,空间利用率越高,结构越稳定,金刚石结构的空间利用率很低(只有34.01%),为什么它也很稳定?2-11证明等径圆球六方最密堆积的空隙率为25.9%。

2-12金属镁原子作六方密堆积,测得它的密度为1.74g/cm3,求它的晶胞体积。

2-13根据半径比关系,说明下列离子与O2—配位时的配位数各是多少?已知rO2-=0.132nm,rSi4+=0.039nm,rK+=0.131nm,rAl3+=0.057nm,rMg2+=0.078nm。

材料科学基础第七章习题及答案

材料科学基础第七章习题及答案

1页 1.已知某二元合金的共晶反应为:(1) 试求含50%B 的合金完全结晶后,初晶α与共晶(α+β)的重量%,α相与β相的重量%;共晶体中α相与β相的重量%。

(2) 若测出显微组织中β初晶与(α+β)共晶各占一半时,试求该合金成分。

2. 已知在A-B 二元合金中,A (熔点600℃)与B (熔点500℃)在液态无限互溶,固态时A 在B 中的最大固溶度(质量分数)为w A =0.30,室温时为w A =0.10;但B 在固态和室温时均不溶于A 。

在300℃时,含w B =0.40的液态合金发生共晶反应。

试绘出A-B 合金相图;并分析w A =0.20,w A =0.45,w A =0.80的合金在室温下组织组成物和相组成物的相对量。

3. 试根据含碳量3.5%亚共晶白口铁的平衡组织,计算其中各组织组成物的相对含量。

答案1. 解:(1)根据杠杆定律可得(2) 设该合金中B 的重量%为wB ,则 2. 解:(1)A -B 合金相图如下图所示(2)合金为0.2A -0.8B 时,室温下相组成物为A 和β相,其相对量为室温下组织组成物为β+A Ⅱ,其相对量与相组成物相同,即(3)合金为0.45A -0.55B 时,室温下相组成物为A 和β相,其相对量为 室温下组织组成物为β初+(A+β)共晶+A Ⅱ,在共晶反应刚完成时,冷却至室温时,将由β初’和(A+β)共晶的β中析出A Ⅱ。

由于共晶β中析出的A Ⅱ与共晶A 连接在一起,故略去不计。

由β初’中析出的A Ⅱ的相对量为%所以,室温下β初的相对量为 '%%%50%11.11%38.89%A ββ=-=-=Ⅱ初初该合金室温下组织组成物的相对量为(4)合金为0.8A -0.2B 时,室温下相组成物为A 和β相,其相对量为 室温下组织组成物为A+(A+β)共晶,其相对量为3. 解:含碳量3.5%的亚共晶白口铁的平衡组织为P+Fe3C Ⅱ+Ld’。

共晶反应刚完成时,室温下组织组成物的相对量为 4. 解:(1) 冷却曲线如图所示。

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第七章答案
7-1略
7-2浓度差会引起扩散,扩散是否总是从高浓度处向低浓度处进行?为什么?
解:扩散是由于梯度差所引起的,而浓度差只是梯度差的一种。

当另外一种梯度差,比如应力差的影响大于浓度差,扩散则会从低浓度向高浓度进行。

7-3欲使Ca2+在CaO中的扩散直至CaO的熔点(2600℃)时都是非本质扩散,要求三价离子有什么样的浓度?试对你在计算中所做的各种特性值的估计作充分说明。

已知CaO肖特基缺陷形成能为6eV。

解:掺杂M3+引起V’’Ca的缺陷反应如下:
当CaO在熔点时,肖特基缺陷的浓度为:
所以欲使Ca2+在CaO中的扩散直至CaO的熔点(2600℃)时都是非本质扩散,M3+的浓度为
,即
7-4试根据图7-32查取:(1)CaO在1145℃和1650℃的扩散系数值;(2)Al2O3在1393℃和1716℃的扩散系数值;并计算CaO和Al2O3中Ca2+和Al3+的扩散活化能和D0值。

解:由图可知CaO在1145℃和1650℃的扩散系数值分别为,Al2O3在1393℃和1716℃的扩散系数值分别为
根据可得到CaO在1145℃和1650℃的扩散系数的比值为:
,将值代入后可得,Al2O3的计算类推。

7-5已知氢和镍在面心立方铁中的扩散数据为cm2/s和
cm2/s,试计算1000℃的扩散系数,并对其差别进行解释。

解:将T=1000℃代入上述方程中可得,同理可知。

原因:与镍原子相比氢原子小得多,更容易在面心立方的铁中通过空隙扩散。

7-6在制造硅半导体器体中,常使硼扩散到硅单晶中,若在1600K温度下,保持硼在硅单晶表面的浓度恒定(恒定源半无限扩散),要求距表面10-3cm深度处硼的浓度是表面浓度的
一半,问需要多长时间(已知D1600℃=8×10-12cm2/s;当时,)?解:此模型可以看作是半无限棒的一维扩散问题,可用高斯误差函数求解。

其中=0,,所以有0.5=,即=0.5,把=10-3cm,D1600℃=8×10-12cm2/s代入得t=s。

7-7 Zn2+在ZnS中扩散时,563℃时的扩散系数为3×10-4cm2/s;450℃时的扩散系数为1.0×10-4cm2/s,求:(1)扩散活化能和D0;(2)750℃时的扩散系数;(3)根据你对结构的了解,请从运动的观点和缺陷的产生来推断活化能的含义;(4)根据ZnS和ZnO相互类似,预测D随硫的分压而变化的关系。

解:(1)参考7-4得=48856J/mol,D0=3×10-15cm2/s;
(2)把T=1023K代入中可得=cm2/s;
7-8实验测得不同温度下碳在钛中的扩散系数分别为2×10-9cm2/s(736℃)、5×10-9cm2/s
(782℃)、1.3×10-8cm2/s(838℃)。

(1)请判断该实验结果是否符合;(2)请计算扩散活化能,并求出在500℃时碳的扩散系数。

解:(1)设=2×10-9cm2/s,=5×10-9cm2/s,=1.3×10-8cm2/s,=1009K,=1055K,=1111K。

将,和,代入并按照7-4所用方法得
=2342787J/mol,同理代入,和,得=2342132J/mol。

,可以认为该实验符合;(2)由上步可知=2342787J/mol;(3)将T=773K代入
得=cm2/s。

7-9在某种材料中,某种粒子的晶界扩散系数与体积扩散系数分别为D gb=2.00×10-10exp(-19100/RT)cm2/s和D v=1.00×10-4exp(-38200/RT)cm2/s,试求晶界扩散系数和体积扩散系数分别在什么温度范围内占优势?
解:当晶界扩散系数占优势时有D gb>D v,即>
,所以有T<1455.6K;当T>1455.6K时体积扩散系数占优势。

7-10假定碳在α-Fe(体心立方)和;γ-Fe(面心立方)中的扩散系数分别为:
Dα=0.0079exp[-83600/RT]cm2/s;Dγ=0.21exp[-141284/RT]cm2/s,计算800℃时各自的扩散系数,并解释其差别。

解:将T=1073K代入题中两式分别得Dα1073=cm2/s Dγ1073=cm2/s。

原因:扩散介质结构对扩散有很大影响。

α-Fe为体心立方,而γ-Fe为面心立方,体心立方较面心立方疏松。

结构疏松,扩散阻力小而扩散系数大。

7-11碳、氮、氢在体心立方铁中的扩散活化能分别为84kJ/mol、75kJ/mol和13kJ/mol,试对此差异进行分析和解释。

解:碳、氮、氢的原子半径依次减小,原子半径越小就越更容易在体心立方的铁中通过空隙扩散,扩散活化能相应也就越低。

7-12MgO、CaO、FeO均具NaCl结构,在各晶体中它们的阳离子扩散活化能分别为:Na+在NaCl中为41kcal/mol,Mg2+在MgO中为83kcal/mol,Ca2+在CaO中为77kcal/mol,Fe3+在FeO中为23kcal/mol,试解释这种差异的原因。

7-13试分析离子晶体中,阴离子扩散系数-般都小于阳离子扩散系数的原因。

解:离子晶体一般为阴离子作密堆积,阳离子填充在四面体或八面体空隙中。

所以阳离子较易扩散。

如果阴离子进行扩散,则要改变晶体堆积方式,拆散离子晶体的结构骨架,阻力就会较大。

故离子晶体中,阴离子扩散系数-般都小于阳离子扩散系数。

7-14试从结构和能量的观点解释为什么D表面>D晶面>D晶内。

解:固体表面质点在表面力作用下,导致表面质点的极化、变形、重排并引起原来的晶格畸变,表面结构不同于内部,并使表面处于较高的能量状态。

晶体的内部质点排列有周期性,每个质点力场是对称的,质点在表面迁移所需活化能较晶体内部小,则相应的扩散系数大。

同理,晶界上质点排列方式不同于内部,排列混乱,存在着空位、位错等缺陷,使之处于应力畸变状态,具有较高能量,质点在晶界迁移所需的活化能较晶内小,扩散系数大。

但晶界上质点与晶体内部相比,由于晶界上质点受两个晶粒作用达到平衡态,处于某种过渡的排列方式,其能量较晶体表面质点低,质点迁移阻力较大因而D晶界<D表面。

7-15试讨论从室温到熔融温度范围内,氯化锌添加剂对NaCl单晶中所有离子(Zn、Na、Cl)的扩散能力的影响。

7-16试推测在贫铁的Fe3O4中氧分压和铁离子扩散的关系;试推测在铁过剩的Fe3O4中氧分压和氧扩散的关系。

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