ms assigment

5
118. Chapter 4 / lmperfections in Solids lmportant Terms and Concepts alloy interstitial solid solution atomic vibration atom percent Boltzmann's constant Burgers vector composition dislocation line edge dislocation gram slze imperfection REFERENCES mlcloscopy microstructure mixed dislocation photomicrograph point defect scanning electron microscope (SEM) scanning probe microscope (SPM) screw dislocation self-interstitial solid solution solute solvent substitutional solid solution transmission electron mrcroscope (TEM) vacancy weight percent ASM Handbook, Vol. 9, Metallography and Mi- crostructures, ASM International, Materials Park, OH, 2004. Brandon, D., and W. D. Kaplan, Microstntchual Characterization of Materials, 2nd edition, Wiley, Hoboken, NJ, 2008. Clarke, A. R., and C. N. Eberhardt, Microscopy Techniqttes for Materials Sclence, CRC Press, Boca Raton, FL, 2002. Kelly,A., G.W Groves, and P Kidd, Crys tallography and Crystal Defeas, Wiley, Hoboken, NJ, 2000. QUESTIONS AND PROBIEMS Tilley, R. J. D., De/ects lir So/idg Wiley-Interscience, Hoboken, N! 2009. Van Bueren, H. G. .Imperfections in C/),.r/a/i North- Holland, Amsterdam (Wiley-Interscience, New York), 1960. Vander Voort, G. F., Metallography, Principles and Practice, ASM International, Materials Park, oH, 1984. Vacancies and Self-lnterttitials 4,1 Calculate the fraction of atom sites that are vacant tbr lead at its melting temperature of 327'C (600 K). Assume an energy for vacancy fbrmation of 0.52 ev/atom. 4.4 Atomic radius, crystal structure, electronega- tivity, and the most common valence are tabulated in the following table for several el- ements; for those that are nonmetals, only atomic radii are indicated. Atomic Radius Crystal Element (nm) Structure Electrc- negati|it! Valence +2 Cu c H o AI Co Cr Fe Ni Pd Pt Zn 0.12'78 0.071 0.046 0.060 0.7445 0.1431 0.12s3 0.1249 0.1241 0.1246 0.13'.76 0.1387 0.1332 FCC FCC HCP BCC BCC FCC FCC FCC HCP 1.9 1.5 1.8 1.6 1.8 1.8 2.2 2.2 1.6 +1 +3 +2 +3 +2 +2 +2 +2 +2 Which of these elements would you expect to form the following with copper: 4.2 Calculate the number of vacancies per cubic meter in iron at 855'C.The energy for vacancy formation is 1.08 ev/atom. Furthermore. the density and atomic weight for Fe are 7.65 g/cm- ano JJ.6) g/mot, 4.3 Calculate the activation energy for vacancy formation in aluminum, given that the equi- librium number of vacancies at 500'C (773 K) rs /.55 X l0 'm '. lhe atomic weighl and den- sity (at 500'C) for aluminum are, respectively, 26.98 g/mol and,2.62 glcm3 lmpuities in Sohds

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Page 1: MS assigment

118. Chapter 4 / lmperfections in Solids

lmportant Terms and Conceptsalloy interstitial solid solutionatomic vibrationatom percentBoltzmann's constantBurgers vectorcompositiondislocation lineedge dislocationgram slzeimperfection

REFERENCES

mlcloscopymicrostructuremixed dislocationphotomicrographpoint defectscanning electron microscope

(SEM)scanning probe microscope

(SPM)

screw dislocationself-interstitialsolid solutionsolutesolventsubstitutional solid solutiontransmission electron

mrcroscope (TEM)vacancyweight percent

ASM Handbook, Vol. 9, Metallography and Mi-crostructures, ASM International, MaterialsPark, OH, 2004.

Brandon, D., and W. D. Kaplan, MicrostntchualCharacterization of Materials, 2nd edition,Wiley, Hoboken, NJ, 2008.

Clarke, A. R., and C. N. Eberhardt, MicroscopyTechniqttes for Materials Sclence, CRC Press,Boca Raton, FL, 2002.

Kelly,A., G.W Groves, and P Kidd, Crys tallographyand Crystal Defeas, Wiley, Hoboken, NJ, 2000.

QUESTIONS AND PROBIEMS

Tilley, R. J. D., De/ects lir So/idg Wiley-Interscience,Hoboken, N! 2009.

Van Bueren, H. G. .Imperfections in C/),.r/a/i North-Holland, Amsterdam (Wiley-Interscience, NewYork), 1960.

Vander Voort, G. F., Metallography, Principles andPractice, ASM International, Materials Park,oH, 1984.

Vacancies and Self-lnterttitials

4,1 Calculate the fraction of atom sites that arevacant tbr lead at its melting temperature of327'C (600 K). Assume an energy for vacancyfbrmation of 0.52 ev/atom.

4.4 Atomic radius, crystal structure, electronega-tivity, and the most common valence are

tabulated in the following table for several el-ements; for those that are nonmetals, onlyatomic radii are indicated.

AtomicRadius Crystal

Element (nm) StructureElectrc-

negati|it! Valence

+2CucHo

AICoCrFeNiPdPtZn

0.12'780.071

0.046

0.0600.74450.14310.12s30.12490.12410.12460.13'.76

0.13870.1332

FCCFCCHCPBCCBCCFCCFCCFCCHCP

1.9

1.5

1.8

1.6

1.81.82.22.21.6

+1+3+2+3+2+2+2+2+2

Which of these elements would you expect toform the following with copper:

4.2 Calculate the number of vacancies per cubicmeter in iron at 855'C.The energy for vacancyformation is 1.08 ev/atom. Furthermore. thedensity and atomic weight for Fe are 7.65g/cm- ano JJ.6) g/mot,

4.3 Calculate the activation energy for vacancyformation in aluminum, given that the equi-librium number of vacancies at 500'C (773 K)rs /.55 X l0 'm '. lhe atomic weighl and den-sity (at 500'C) for aluminum are, respectively,26.98 g/mol and,2.62 glcm3

lmpuities in Sohds

Page 2: MS assigment

(a) A substitutional solid solution havingcomplete solubility(b) A substitutional solid solution of incom-plete solubility

(c) An interstirial solid solution

4.5 For both FCC and BCC crystal structures,

there are two different types of interstitialsites. ln each case, one site is larger than theother and is normally occupied by impurltyatoms. For FCC, this larger one is located at

the center of each edge of the unit cell; it istermed an octaheilral interstitial site. On theother hand, with BCC the larger srte type ls

found at 0 ] J positions-that is,lying on {100}faces and situated midway between two unttcell edges on this face and one-quarter of the

distance between the other two unit celledges; it is termed a tetrahed.ral interstitial site.

For both FCC and BCC crystal structures,

compute the radius r of an impurity atom thatwill just fit into one of these sites in terms oIthe atomic radius R of the host atom.

S pecif icatio n of Co mposition

4,6 Derive the following equations:

(a) Equation 4.7a

(b) Equation 4.9a

(c) Equation 4.10a

(d) Equation 4.11b

4,7 What is the composition, in atom percent, ofan alloy that consists of 30 wto/" Zn a[d 70

wt% Cu?

4.8 What is the composition, in weight percent, ofan alloy that consists of 6 at% Pb and 94 at%Sn?

4.9 Calculate the composition, in weight percent,

of an alloy that contains 218.0 kg titanium'15 kg aluminum, and 10 kg vanadium

4.10 What is the composition, in atom percent, ofan alloy that contains 100 g tin and 68 g lead?

4.11 What is the composition, in atom percent, ofan alloy that contains 45.2 kg copper, 46.3 kg

zinc, and 0.95 kg lead?

4.12 What is the composition, in atom percent, of an

alloy that consists of 97 wt% Fe and 3 wt% Si?

4.13 Convert the atom percent composition inProblem 4.11 to weight Percent.

Questions and Problemr f 19

4.14 Calculate the number ol atoms per cubicmeter in aluminum.

4.15 The concentralion of carbon in an iron-carbon alloy is 0.15 wt%. What is the con-centration in kilograms of carbon per cubicmeter of all

Determine the approximate density of a high-leaded brass that has a composition of 64.5

a solid solution consisting of two elements

(designated as 1 and 2), sometimes it is desir-able to determine the number of atoms percubic centimeter of one element in a solidsolution, Nl, given the concentration of thatelement specified in weight percent, C1. Thiscomputation is possible using the followingexpresslon:

where

NoCt(4.18)

Na : Avogadro's numberp1 and p, = densities of the two elements

A1 = the atomic weight of element 1

Derive Equation 4.18 using Equation 4.2 and

expressions contained in Section 4.4.

4.20 Gold forms a substitutional solid solutionwith silver. Compute the number of gold

atoms per cubic centimeter for a silver-goldalloy that contains 10 wt% Au and 90 wt%Ag. The densities of pure gold and silver are

1,9.32 and 10.49 g/cm3, respectively.

4.21 Germanium forms a substitutional solid solu-tion with silicon. Compute the number ofgermanium atoms per cubic centimeter for a

4.t6

94t+L@o c,)

4.17 Calculate the unit cell edge length for an 85

wt% Fe-15 wt% V alloy. All of the vanadiumis in solid solution, and at room temperaturethe crystal structure for this alloy is BCC.

Some hypothetical alloy is composed of 12.5

wt% of metal A and 87.5 wt% of metal B. Ifthe densities of metals A and B are 4.25 and

6.35 g/cm3, respectively, whereas their respec-

tive atomic weights are 61.5 and 125.7 g/mol,determine whether the crystal structure forthis alloy is simple cubic, face-centered cubic,or body-centered cubic. Assume a unit celledge length of 0.395 nm.

Page 3: MS assigment

f 2O ' Chapter 4 I lnpertedionr in Solids

germanium_silicon alloy that contains 15 wt%Ge and 85 wt% Si. The densities of puregerrnanium and silicon are 5.32 and 2.33 g/cmr,respectively.

4.22 Sometimes it is desirable to determine theweight percent of one element, Cl, that willproduce a specified concentration in terms ofthe number of atoms per cubic centimeter, Nl,for an alloy composed of two types of atomi.This computation is possible using the fofIowing expression:

lnterfacial Defech

4.27 For-an FCC single crystal, would you expect thesurface

_energy for a (100) plane to be greateror less than rhat for a { | l | ) plane? Whyl-( Norc,you may \^,ant lo consult lhe solution to prob_lem 3.54 at the end of Chapter 3.)

4.28 For^a BCC single crystal, would you expect thesurface energy for a (100) plane to be greatero-r, less than that for a (110) plane? Why?(Nore; You may want to consult the solutionto Problem 3.55 at the end of Chapter 3.)

4.29 (a) For a given material, would you expect thesudace energy to be greater than, the iame as,or less than the grain boundary energy? Why?(b) The grain boundary energy of a smalfangle grain boundary is less tliin for a high_angle one. Why is this so?

4.30 (a) Briefly describe a tw.in and a twinboundary.

(b) Cite the difference between mechanicaland annealing twins.

4.31 For each of the following stacking sequenceslound in FCC metals, cite the type of planardefect that exists:(a)...ABCABCBACBA...(b)...ABCABCBCABC..,Noq copy the stacking sequences and indi_cate.the position(s) of planar defect(s) with avertical dashed line.

G rain S ize Dete rminatio n

4.32 (a) Using the intercept method, determine theaverage grain size, in millimeters, of the spec_rmen whose microstructure is shown in Fig_ure 4.I4(b); use at least seven straightlinesegments.

(b) Estimate the ASTM grain size number forthis material.

4.33 (a) Employing the intercept technique, deter_mrne the average grain size for the steel speci_men whose microstructure is shown il Figure9.25(c);use at least seven straighrline segmlnts.(b) Estimate the ASTM grain size number forthis material.

4.34 For an ASTM grain size of g, approximatelyhow many grains would there bJ per squarecentimeter

100(4.le)

, Nl.Pz p2

NtAt pl

4.23 Molybdenum forms a substitutional solid so_lution with tungsten. Compute the weight per_cent of molybdenum that must be added totun^gsten to yield an alloy that contains 1.0 X10" Mo atoms per cubic centimeter. The den-

where

No : Avogadro's numberp1 and, p2 = densities of the two elements

A, and Ar: the atomic weights of the twoelements

Derive Equation 4.19 using Equation 4.2 andexpressions contained in Section 4.4.

sities of pure Mo and W are 10.22 and 19.30

4.25 Silver and palladium both have the FCC crys_tal slruclure. and pd forms a substitutionalsolid solution for all concentrations at roomtemperature. Compute the unit cell edgelength for a 75 wt% Ag-25 wt% pd alloy. Theroom-temperature density ofpd is 12.02 g/cm3,and its atomic weight and atomic radius are106.4 g/mol and 0.138 nm, respectively.

D islocatio w - Linear D efeck

4.26 Cite the relative Burgers vector_dislocationline orientations for edge, screw, and mixeddislocations.

4.24 Niobium forms a substitutional solidl6iiE6iwith vanadium. Compute the weight percentof niobium that must be added to vanadiumto yield an alloy that contains 1.55 X 1022 Nbatoms per cubic centimeter. The densities ofpure Nb and V are 8.57 and 6.10 g/cm3,

Page 4: MS assigment

PIEftolese

Ithe

I

Itool

lheus-

tr'I

t

Quertions and Problems ' 145

QUESTIONS AND PROBLEMS

f.o

.iroduction

5.1 Brielly explain the difference between self-diffusion and interdiftusion.

5l Self diftusion involves the motion of atomsthat are all of the same type: therefore, it is

not subject to observation by compositionalchanges, as with interdiffusion. Suggest onervay in which sell-diffusion may be monitored.

Diffusion Mechanisms

5J (a) Compare interstitial and vacancy atomicmcchanisms for diffusion.(b) Cite two reasons why interstitial diflusionis normally more rapid than vacancy diffusion.

Steady - State D iff u t ion

5.{ Briefly explain the concept of steady state as

it applies to diffusion.

55 (a) Briefly explain the concept of a drivingtorce.

(b) What is the driving force for steady-statediffusion?

The purification of hydrogen gas by diffusiontluough a palladium sheet was discussed inSeclion 5.3. Compute the number of kilo-grams of hydrogen that pass per hour througha 5-mm-thick sheet of palladium having anarea of 0.25 m2 at 500'C. Assume a diflusioncoefficient of 1.0 x 10 8 m2ls, that the con-centrations at the high- and low-pressuresides of the plate are 2.4 and 0.6 kg ofhydrogen per cubic meter of palladium, and

conditions have been attained.

of steel 1.8 mm thick has nitrogenatmospheres on both sides at 1200"C and is per-mitted to achieve a steady-state diffusion con-dition. The diffusion coelficient for nitrogenin steel at this temperature is 6 x 10-rr m2/s,

and the diffusion flux is found to be 1.2 x10 7 kg/m'? . s. Also, it is known that the con-cenlraliun of nirrogen in lhe \teel at lhe hiEh-

pressure surface is 4 kg/mr. How lar into thesheet from this high-pressure side will theconcentration be 2.0 kg/mr? Assume a linearconcentration profile.

1 mm thick was exposed

and a decarburizing atmosphere on the otherside at 725"C. After reaching steady state, theiron was quickly cooled to room temperature.The carbon concentrations at the two sudacesof the sheet were determined to be 0.012 and0.0075 wt%. Compute the diffusion coetTicientif the diffusion llux is 1.5 x 10 8 kg/m2.s.Hint: Use Equation 4.9 to convert the con-centrations from weight percent to kilogramso[ carhon per cubic meter of iron.

ere othydrogen gas, the concentration of hydrogenin the iron, Cn (in weight percent), is a func-tion of hydrogen pressure. pH- (in MPa), andabsolute temperature (I) according to

cH:1.34 x 1o '?tf[exp ft+#*)(5.14)

Furthermore, the values of D6 and Q,1 for thisdiffusion system are 1.4 x 10-7 m2/s and13,400 J/mol, respectively. Consider a thiniron membrane I mm thick that is at 250'C.Compute the diffusion flux through this mem-brane if the hydrogen pressure on one side ofthe membrane is 0.15 MPa, and on the otherside 7.5 MPa.

N onsteady - State Diffusion

5.10 Show that

is also a solution to Equation 5.4b. Theparameter B is a constant, being independentol both r and r.

5.11 Determine the carburizing time necessary toachieve a carbon concentration of 0.45 wtTo ata position 2 mm into an iron{arbon alloy thatinitially contains 0.20 wtl" C. The surface con-centration is to be maintained at 1.30 wt% C,

and the treatment is to be conducted at 1000'C.Use the diffusion data for 7-Fe in Thble 5.2.

5.12 An FCC iron carbon alloy initially containing0.35 wt% C is exposed to an oxygen-rich andvirtually carbon-free atmosphere at 1400 K(1127'C). Under these circumstances the carbondiffuses from the alloy and reacts at the surface

p /.1 \c': ?Dt*P ( +o,-J

that

to a carburizing gas atmosphere on one side

Page 5: MS assigment

Chapter 22.3 (a) L.66 x 10-'?4 g/amu;

(b) 6.022 x 102r atoms/g-molz.r4

/ a Yu-'r,r: \"8 )A-BLo-

/a1t't-'' /e1'Lt "'

\ia ) \,8 /2.L5 (c) ro : 0.279 nm, Eo : -4.57 eY2.19 63.2% for TiOz; 1.0% for InSb

Chapter 33.2 Vc = 6.62 x 1'0 2e m3

3.8R:0.136nm3.ll (t'1 Vg: 1.40 x 10-28 m3'

(b) a:0.323 nm,c:0.515 nm3.14 Metal B: face-centered cubic3.16 (a) n : 8.0 atoms/unit cell;

(h) p:4.96g1cm33.!9 Vc:8.07 x 10 2 nm3

3.22 000,100, 110,010,001, 101, 111, 011,1 ;0,iLt,tLL,oli,Lo |, ano I t]

3,29 Direction 1: [012]3.31 Direction ,4: [01!;

Direction C: [112]3.32 Dreclion B:12321;

Direction D: [136]3.33 (b) [T10], [10], and [110]3,35 Direction A: [1011]3.41 Plane B: (1!2) or (712)!,42 Plane A: (312)3.43 Plane B: (221)3.aa (c) [010] or [010]3.a6 (a) (010) and (100)3.s0 (b) (10T0)

1

3.53 (b) LDr11(W) : 3.65 x 10'm-'1

3.54 (a) PDlu :2R2\,5

3.55 (b) PDlro(V) : L.522 x 10Le m 2

3.59 70 : 81.24"

3.60 drle:0.2862nm3.62 (a\ dzr: 0.1520 nm;

(b) R:0.2463nm3.64 d1q : 0.2072 nm, a = 0.2845 nm

Chapter 4.4.1 N,/N : 4.3 x 10-s4.3 .Q, = 0.75 eV/atom4.5 For FCC, r : 0.41R4.7 C2" : 29.4 at"/", CL": 70.6 at'/.4.8 Cp5 = 10.0 wt%, Cs, : 90.0 wt%

4.10 C!, : 71..9 at%", Cpr : 28.1 at'/.4.14 N61 : 6.05 x 1028 atoms/ml4.17 a:0.289nm4.20 NA" = 3.36 X 1021 atoms/cm34.24 CNy : 35.2 \\r"/"4,32 (a) d = 0.066 mm4.34 (b) NM : 198,300 grains/cm2

4.Dl CLi : 2.06 wt"/"

Chapter 55.6 M:3.2 x 10-3 kgh5.8 D:4.23 x 10-11 mzls

5.IIt:1.9.7h5.15r:40h5.18 Z: 1.152K (879"C)

5.21 (a) Qd : 252.4 kI lmol,Do:2.2x 10 5 m2ls;

(b) D : 5.4 x 10 15 m2ls

s.24 T:10s1 K (778'C)5.29 .r : 1.52 mm5,33 to : 47.4 min

5.D1 Not possible3.52 (a) LDlep =

2R\/t