material science cht04 and cht08

Smith Foundations of Materials Science and Engineering 54

CHAPTER 4 SOLIDIFICATION, CRYSTALLINE IMPERFECTIONS, AND DIFFUSION IN SOLIDS

4.1 Describe and illustrate the solidification process of a pure metal in terms of the nucleation

and growth of crystals.

In general, the solidification of a pure metal involves: the formation of stable nuclei in the liquid melt; the growth of these nuclei into stable nuclei in the liquid melt; and the formation of a solidified structure containing grains formed from the crystals. These three stages are illustrated below.

4.2 Define the homogeneous nucleation process for the solidification of a pure metal.

In a homogeneous nucleation process, the liquid metal is sufficiently undercooled and thus able to provide the atoms to form nuclei.

4.3 In the solidification of a pure metal, what are the two energies involved in the

transformation? Write the equation for the total free-energy change involved in the transformation of liquid to produce a strain-free solid nucleus by homogeneous nucleation. Also illustrate graphically the energy changes associated with formation of a nucleus during solidification.

The two energies involved in the solidification of a pure metal are surface free energy and volume free energy. These energies contribute to the total free energy, ∆GT :

2 23 44 Volume free energy + Surface free energy

T V sG G G

r G rυπ π γ

∆ = ∆ + ∆

= ∆ +

=

Nuclei

Liquid

Crystals

Liquid

Grains

Grain Boundaries

r* Radius of particle, r

*rG∆

Retarding energy,

T V sG G G∆ = ∆ + ∆

Free

Ene

rgy

Chan

ge, ∆

G

Driving energy, VG∆

+

—


4.4 In the solidification of a metal, what is the difference between an embryo and a nucleus? What is the critical radius of a solidifying particle?

An embryo refers to a solidified cluster of atoms which is unstable and can thus redissolve. A nucleus is a comparatively larger stable cluster which can grow into a crystal. The dimension defining the boundary between an embryo and a nucleus is the critical radius.

4.5 During solidification, how does the degree of undercooling affect the critical nucleus

size? Assume homogeneous nucleation.

In general, the greater the degree of undercooling of a liquid melt, the smaller the critical radius of the nuclei formed.

4.6 Distinguish between homogeneous and heterogeneous nucleation for the solidification of

a pure metal.

In homogeneous nucleation of a liquid metal, the nuclei are formed from atoms of the metal solidifying. In heterogeneous nucleation, impurity atoms or container surfaces act as nucleating agents.

4.7 Calculate the size (radius) of the critically sized nucleus for pure platinum when

homogeneous nucleation takes place. The critical radius of nucleus is calculated as,

* 2 m

f

TrH T

γ−=∆ ∆

where 7 2 3240 10 J/cm , 2160 J/cm , 2045 Kf mH Tγ −= × ∆ = − = and the undercooling 0.2 0.2(2045 K) 409 K.mT T∆ = = = Substituting,

7 2* 7

32(240 10 J/cm )(2045 K) 1.11 10 cm

(-2160 J/cm )(409 K)r

−−− ×= = ×

4.8 Calculate the number of atoms in a critically sized nucleus for the homogeneous

nucleation of pure platinum.

The number of atoms in a critically sized nucleus is found as,

Volume of nucleus Volume of nucleusVolume per atom (Vol. of unit cell)(No. atoms per unit cell)

=

Using the critical radius result of Problem 4.7,


*3 7 3 21 3

3 9 3 29 3 23 3

23 3

4 4Vol. of critical-sized nucleus (1.11 10 cm) 5.73 10 cm3 3

Vol. of unit cell of Pt (0.32939 10 m) 3.574 10 m 3.574 10 cm

3.574 10 cmVol. per atom 8.94 atoms/FCC unit cell

r

a

π π − −

− − −

−

= = × = ×

= = × = × = ×

×= = 24 335 10 cm /atom−×

Substituting,

-21 3

-24 3Volume of nucleus 5.73 10 cmVolume per atom 8.935 10 cm /atom

×= =×

641 atoms

4.9 Calculate the size (radius) of the critical nucleus for pure iron when nucleation takes

place homogeneously.

For iron, 7 2 3204 10 J/cm , 2098 J/cm , 1808 Kf mH Tγ −= × ∆ = − = The amount of undercooling is then 0.2 0.2(1808 K) 361.6 K and mT T∆ = = = the critical radius becomes,

7 2*

32 2(204 10 J/cm )(1808 K)

( 2098 J/cm )(361.6 K)m

f

TrH T

γ −− − ×= = =∆ ∆ −

-89.72×10 cm

4.10 Calculate the number of atoms in a critically sized nucleus for the homogeneous

nucleation of pure iron.

The relevant volumes, based on the solution of Problem 4.9, are:

*3 8 3 21 3

3 9 3 29 3 23 3

23 3

4 4Vol. of critical-sized nucleus (9.72 10 cm) 3.85 10 cm3 3

Vol. of unit cell of Pt (0.28664 10 m) 2.355 10 m 2.355 10 cm

2.355 10 cmVol. per atom 1.12 atoms/BCC unit cell

r

a

π π − −

− − −

−

= = × = ×

= = × = × = ×

×= = 23 378 10 cm /atom−×

Thus, the number of atoms in a critically sized nucleus is:

-21 3

-23 3Volume of nucleus 3.85 10 cmVolume per atom 1.178 10 cm /atom

×= =×

327 atoms

4.11 Describe the grain structure of a metal ingot that was produced by slow-cooling the metal

in a stationary open mold. In general, equiaxed grains are formed adjacent to the cold mold wall where rapid


cooling occurs during solidification. Elongated columnar grains are formed in the metal ingot interior, in the direction of thermal gradients, due to slow cooling of the metal in the mold interior.

4.12 Distinguish between equiaxed and columnar grains in a solidified metal structure.

Equiaxed grain dimensions are approximately equal in all directions whereas columnar grains are elongated in a particular direction.

4.13 How can the grain size of a cast ingot be refined? How is grain refining accomplished

industrially for aluminum alloy ingots?

The grain size of a cast ingot can be refined by: solidifying the metal at a rapid rate; and adding grain refining agents (heterogeneous nucleating agents). Grain refining of aluminum ingots is accomplished through chill casting and by adding grain refining agents such as titanium and/or boron.

4.14 What special techniques must used to produce single crystals?

Single crystals can be produced by introducing a single crystal as a seed crystal. The seed continuously rotates as it is slowly lowered and then withdrawn from the melt.

4.15 How are large silicon single crystals for the semiconductor industry produced?

Large single crystals of silicon are produced using a pure silicon seed crystal with a pure silicon melt (Czochralski process).

4.16 What is a metal alloy? What is a solid solution?

A metal alloy is a mixture of two or more metals or of a metal (metals) and a non-metal (nonmetals). A solid solution is a type of alloy which is solid and consists of two or more elements atomically dispersed in a single phase structure.

4.17 Distinguish between a substitutional solid solution and an interstitial solid solution.

A substitutional solid solution is one in which the solute atoms of the elements replace those of the solvent atoms in the crystal lattice. An interstitial solid solution is one in which the solute atoms of the elements are positioned in the interstitial spaces between the solvent atoms of the crystal lattice.

4.18 What are the conditions that are favorable for extensive solid solubility of one element in

another?

Four conditions favor extensive solid solubility: 1. less than 15% difference between the atomic diameters of the elements forming

the solid solution;


2. identical valence of the elements; 3. similar electronegativities; 4. common crystal structure of the elements.

4.19 Using the data in the following table, predict the relative degree of solid solubility of the

following elements in aluminum: (a) copper (c) magnesium (e) silicon (b) manganese (d) zinc Use the scale very high, 70–100%; high, 30–70%; moderate, 10–30%;low, 1–10%; and very low, < 1%.

Element

Atom Radius

(nm)

Crystal

Structure

Electro-

negativity

Valence

Aluminum 0.143 FCC 1.5 +3 Copper 0.128 FCC 1.8 +2 Manganese 0.112 Cubic 1.6 +2, +3, +6, +7 Magnesium 0.160 HCP 1.3 +2 Zinc 0.133 HCP 1.7 +2 Silicon 0.117 Diamond Cubic 1.8 +4

(a) low (b) very low (c) moderate (d) high (e) low

4.20 Using the data in the following table, predict the relative degree of atomic solid solubility of the following elements in iron: (a) nickel (c) molybdenum (e) manganese (b) chromium (d) titanium Use the scale very high, 70–100%; high, 30–70%; moderate, 10–30%;low, 1–10%; and very low, < 1%.

Element

Atom Radius

(nm)

Crystal

Structure

Electro-

negativity

Valence

Iron 0.124 BCC 1.7 +2, +3 Nickel 0.125 FCC 1.8 +2 Chromium 0.125 BCC 1.6 +2, +3, +6 Molybdenum 0.136 BCC 1.3 +3, +4, +6 Titanium 0.147 HCP 1.3 +2, +3, +4 Manganese 0.112 Cubic 1.6 +2, +3, +6, +7

(a) high (b) very high (c) moderate (d) low (e) moderate


4.21 Calculate the radius of the largest interstitial void in the BCC α iron lattice. The atomic radius of the iron atom in this lattice is 0.124 nm, and the largest interstitial voids occur at the (¼, ½, 0); (½, ¾, 0); (¾, ½, 0); (½, ¼, 0), etc., type positions.

For BCC crystal structure,

4 4(0.124 nm) 0.286 nm3 3Ra = = =

Letting x = Fe atom radius + Interstitial void radius,

2 2 2 21 1 516 4 16

5 0.559 (0.559)(0.286 nm) 0.160 nm16

x a a a

x a a

= + =

= = = =

The interstitial void radius is thus,

0.160 nm 0.124 nmvoid FeR x R= − = − = 0.036 nm 4.22 Describe and illustrate the following types of point imperfections that can be present in

metal lattices: (a) vacancy, (b) divacancy, and (c) interstitialcy.

a) A vacancy, a point defect, is an atomic site which is missing an atom. b) A divacancy is a defect in a crystal lattice where two atoms are missing from

adjoining atomic sites. c) An interstitialcy is a point defect where an atom occupies an interstitial site between

surrounding atoms in normal sites.

4.23 Describe and illustrate the following imperfections that can exist in crystal lattices:

(a) Frenkel imperfection and (b) Schottky imperfection.

a) A Frenkel imperfection is a vacancy-interstitialcy pair which sometimes occurs in ionic crystals.

2a

4a

a RFe

x

Rvoid

Vacancy Interstitialcy


b) A Schottky imperfection is a cation-anion divacancy which sometimes occurs in ionic crystals.

4.24 Describe and illustrate the edge– and screw–type dislocations. What type of strain fields

surround both types of dislocations?

An edge dislocation is a line imperfection caused by an extra half plane of atoms between two normal planes of atoms. Whereas a screw dislocation is a line imperfection created by applying upward and downward shear stress to regions of a perfect crystal separated by a common plane.

The strain fields associated with the edge and screw dislocations are shown below:

Frenkel Imperfection Schottky Imperfection

Dislocation Line Screw Dislocation


4.25 Describe the structure of a grain boundary. Why are grain boundaries favorable sites for

the nucleation and growth of precipitates?

Grain boundaries are surface imperfections that separate grains of different orientations. The grain boundary, a narrow region between two grains, is approximately two to five atomic diameters in width and contains mismatched atoms from adjacent grains. Grain boundaries are favorable sites for the nucleation and growth of precipitates because the irregular atom arrangement at grain boundaries provides lower atomic packing and high energy. Atoms are thus able to diffuse more rapidly to form precipitates.

4.26 Why are grain boundaries easily observed in the optical microscope?

Grain boundaries can be easily observed under an optical microscope because they etch more rapidly than grains. Chemical etching thus produces tiny grooves along grain boundaries which appear as dark lines under an optical microscope because they reflect light less intensely.

4.27 How is the grain size of polycrystalline materials measured by the ASTM method?

In the ASTM method of measuring grain size of polycrystalline materials, the grain size number, n, is defined by the equation 12nN −= , where N is the number of grains per square inch, measured on a polished and etched surface at a magnification of 100x.

4.28 If there are 600 grains per square inch on a photomicrograph of a metal at 100x, what is

its ASTM grain-size number?

1600 2ln 600 ( 1)(ln 2)6.397 ( 1)(0.693)

9.23 1

nNnn

n

−= == −= −

= + = 10.23


4.29 If there are 400 grains per square inch on a photomicrograph of a ceramic material at 200x, what is its ASTM grain-size number?

11600 2

ln1600 ( 1)(ln 2)7.378 ( 1)(0.693)

10.64 1

nNn

nn

−= == −

= −= + = 11.64

4.30 Determine, by counting, the ASTM grain-size number of the low–carbon sheet steel

shown in Fig. P4.30. This micrograph is at 100x.

Estimating 40 grains/in2 from the micrograph,

4.31 Determine the ASTM grain-size number of the type 430 stainless steel micrograph shown in Fig. P4.31.This micrograph is at 200x.

Estimating 100 grains/in2 from the micrograph,

4.32 What is a thermally activated process? What is the activation energy for such a process?

A thermally active process is one which requires a definite amount of thermal energy to overcome an activation energy barrier and enter the reactive state.

4.33 Write an equation for the number of vacancies present in a metal at equilibrium at a

particular temperature and define each of the terms. Give the units for each term and use electron volts for the activation energy.

140 2ln 40 ( 1)(ln 2)3.689 ( 1)(0.693)

5.3 1

nNnn

n

−= == −= −

= + = 6.3

2

2

2

1

200No. of grains at 100 (100)100400 grains/in

400 2ln 400 ( 1)(ln 2)5.99 ( 1)(0.693)

8.64 1

nNn

nn

−

× =

=

= == −

= −= + = 9.64


v /v

E kTn NCe−=

v

v

where number of vacancies per cubic meter of metal total number of atom sites per cubic meter of metal activation energy to form a vacancy (eV) absolute temperature (K)

Boltzmann's

nN

ET

k

====

= 6constant 8.62 10 eV/K constantC

−= ×=

4.34 (a) Calculate the equilibrium concentration of vacancies per cubic meter in pure copper

at 850ºC. Assume that the energy of formation of a vacancy in pure copper is 1.00 eV. (b) What is the vacancy fraction at 800ºC?

a) In general, the equilibrium number of vacancies is v /

vE kTn NCe−= . For copper,

23 6 3

28 3Cu (6.02 10 atoms/at. mass)(8.96 10 g/m ) 8.49 10 atoms/mat. mass Cu (63.54 g/at. mass)

oNN ρ × ×= = = ×

Substituting and assuming Ev = 1.00 eV at 1123 K,

28 3-51.00 eV(8.49 10 atoms/m ) exp

(8.62 10 eV/K)(1123 K)nυ

= × − ×

= 24 32.77×10 vacancies/m

b) The vacancy fraction at 1073 K is,

10.81

-51.00 eVexp

(8.62 10 eV/K)(1073 K)n eNυ − −= = = ×

-52.02×10 vacancies/atom

4.35 (a) Calculate the equilibrium concentration of vacancies per cubic meter in pure silver at 750ºC. Assume that the energy of formation of a vacancy in pure silver is 1.10 eV. (b) What is the vacancy fraction at 700ºC?

a) The equilibrium number of vacancies is calculated as v /

v .E kTn NCe−= Thus for silver,

23 6 3

28 3Cu (6.02 10 atoms/at. mass)(10.5 10 g/m ) 5.86 10 atoms/mat. mass Cu (107.870 g/at. mass)

oNN ρ × ×= = = ×

Substituting and assuming Ev = 1.10 eV for vacancies formed at 1023 K,


28 3-51.10 eV(5.86 10 atoms/m ) exp

(8.62 10 eV/K)(1023 K)nυ

= × − ×

= 23 32.24×10 vacancies/m

b) The vacancy fraction at 973 K is,

13.12-51.10 eVexp

(8.62 10 eV/K)(973 K)n eNυ − −= = = ×

-62.01×10 vacancies/atom

4.36 Write the Arrhenius rate equation in the (a) exponential and (b) common logarithmic

forms.

/

10 10

a) Rate of reaction

b) log rate = log constant2.303

Q RTCeQ

RT

−=

−

4.37 Draw a typical Arrhenius plot of log10 of the reaction rate versus reciprocal absolute

temperature, and indicate the slope of the plot.

A typical Arrhenius plot of the logarithmic reaction rate is shown below for the SI absolute Kelvin temperature scale. The relationship between the log10 of the reaction rate and the inverse absolute temperature is linear with a slope of –Q/(2.303R). .

4.38 Describe the substitutional interstitial diffusion mechanisms in solid metals.

-11 , KT

Intercept = log10 (const.)

log 1

0 of r

eact

ion

rate

10

Slope 2.303(log rate)

(1/ )

QR

T

= −

∆=

∆

T, K


During substitutional diffusion of atoms in a solid alloy crystal lattice, solute atoms move into positions of solvent atoms in the matrix through a vacancy mechanism. In interstitial diffusion, small solute atoms move between the interstices of the solvent lattice.

4.39 Write the equation for Fick’s first law of diffusion, and define each of the terms in SI

units.

Fick’s first law of diffusion is given by:

2

2 3atoms m atoms 1 or in SI unit form,

s mm mdCJ Ddx s

= − = × ⋅

where J = flux or net flow of atoms; D = proportionality constant called the diffusivity (atomic conductivity) or diffusion coefficient;

dCdx

= concentration gradient.

4.40 What factors affect the diffusion rate in solid metal crystals?

The diffusion rate in solid metal crystals is affected by five factors:

1. Type of diffusion mechanism; 2. Temperature of diffusion; 3. Concentration of the diffusion species (concentration gradient); 4. Type of crystal structure; 5. Type of crystal imperfections present.

4.41 Write the equation for Fick’s second law of diffusion in solids and define each of the

terms.

Fick’s second law of diffusion in solids, written for the x-direction, is:

x xdC dCd Ddt dx dx

=

where rate of change of the concentration of the diffusing species in the x-direction;

concentration gradient of the diffusing species in the x-direction;

diffusion coefficient of the

x

x

dCdt

dCdx

D

=

=

= diffusing species.

4.42 Write the equation for the solution to Fick’s second law for the diffusion of a gas into the surface of a solid metal crystal lattice.


Fick’s second law of diffusion, for the diffusion of a gas into the surface of a solid metal crystal lattice is:

erf2

s x

s o

C C xC C Dt

− = −

where surface concentration of element in gas diffusing into the surface;

initial uniform concentration of element in solid; concentration of element at distance from surface at time ;

s

o

x

CCC x t

x

==== distance from surface;

diffusivity of diffusing solute element; time.

Dt

==

4.43 Describe the gas-carburizing process for steel parts. Why is the carburization of steel

parts carried out?

In the gas carburizing process for steel parts, the parts are placed in a furnace in contact with a gas rich in CO at about 927ºC. The carbon from the gas diffuses into the surface of the steel part and increases the carbon content of the outer surface region of the part. The higher carbon concentration at the surface makes the steel harder in this region. A steel part can thus be produced with a hard outer layer and a tough low carbon steel inner core. This duplex structure is important, for example, for many types of gears.

4.44 Consider the gas carburizing of a gear of 1018 steel (0.18 wt %) at 927ºC (1700ºF). Calculate the time necessary to increase the carbon content to 0.35 wt % at 0.40 mm below the surface of the gear. Assume the carbon content at the surface to be 1.15 wt % and that the nominal carbon content of the steel gear before carburizing is 0.18 wt %. D (C in λ iron) at 927ºC = 1.28 × 10 –11 m2/s.

The time required for this diffusion process is calculated using Fick’s second law,

erf2

s x

s o

C C xC C Dt

− = −

-4 11 2927

where: 1.15% 0.18% 0.35%

0.40 mm = 4 10 m 1.28 10 m / ss o x

C

C C C

x D −

= = =

= × = ×�

4

-11 2

1.15 0.35 4 10 mSubstituting erf1.15 0.18 2 (1.28 10 m / s)

55.900.8247 erf erf

t

zt

− − × =

− ×

= =


Interpolating from Table 4.5,

erf z

z

0.8209 0.95 0.8247 x 0.8427 1.0

4.45 The surface of a steel gear made of 1022 steel (0.22 wt % C) is to be gas-carburized at

927ºC (1700ºF). Calculate the time necessary to increase the carbon content to 0.30 wt % at 0.030 in. below the surface of the gear. Assume the carbon content of the surface to be 1.20 wt %. D (C in λ iron) at 927ºC = 1.28 × 10 –11 m2/s.

-4 11 2927

Given: 1.20% 0.22% 0.30%ft0.03 in.(0.3048 m/ft) = 7.62 10 m 1.28 10 m / s

12 in.

s o x

C

C C C

x D −

= = =

= × = ×

�

4

-11 2

1.20 0.30 7.62 10 merf1.20 0.22 2 (1.28 10 m / s)

106.490.9184 erf erf

s x

s o

C CC C t

zt

− − − × = =

− − ×

= =


erf z

z

0.9103 1.2 0.9184 x 0.9340 1.3

4.46 A gear made of 1020 steel (0.20 wt % C) is to be gas-carburized at 927ºC (1700ºF).

Calculate the carbon content at 0.90 mm below the surface of the gear after a 4.0-hour carburizing time. Assume the carbon content at the surface of the gear is 1.00 wt %. D (C in λ iron) at 927ºC = 1.28 × 10 –11 m2/s.

-4 11 2927

Given: 1.00% 0.20% ? 4 h = 14,400 s

0.09 mm = 9.0 10 m 1.28 10 m / ss o x

C

C C C t

x D −

= = = =

= × = ×�

0.8247 0.8209 0.95 0.9590.8427 0.8209 1.0 0.95

Thus,55.90 0.959

3397.7 s =

x x

zt

t

− −= =− −

= =

= 56.6 min.

2 2

0.9184 0.9103 1.2 1.2340.9340 0.9103 1.3 1.2

Thus,

106.49 106.49 7, 446.6 s = 1.234

x x

tz

− −= =− −

= = = 124 min.


4

-11 2

1.00 9.0 10 merf1.00 0.20 2 (1.28 10 m / s)(14,400 s)

1.25(1 ) erf (1.0482)

s x x

s o

x

C C CC C

C

− − − × = =

− − ×

− =


erf z

z

0.8427 1.00

x 1.0482 0.8802 1.10

4.47 A gear made of 1020 steel (0.20 wt % C) is to be gas-carburized at 927ºC (1700ºF). Calculate the carbon content at 0.04 in. below the surface of the gear after a 7.0-hour carburizing time. Assume the carbon content at the surface of the gear is 1.15 wt %. D (C in λ iron) at 927ºC = 1.28 × 10 –11 m2/s.

-3 11 2927

Given: 1.15% 0.20% ? 7 h = 25,200 s

0.040 in. = 1.02 10 m 1.28 10 m / ss o x

C

C C C t

x D −

= = = =

= × = ×�

3

-11 2

1.15 1.02 10 merf1.15 0.20 2 (1.28 10 m / s)(25,200 s)

1.15 erf (0.89798)0.95

s x x

s o

x

C C CC C

C

− − − × = =

− − ×

− =


erf z

z

0.7707 0.85

x 0.89798 0.7970 0.90

1.0482 1.00 0.8427 0.86081.10 1.00 0.8802 0.8427

Thus, 0.8608 erf (1.0482)

Substituting, 1.25(1 ) 0.8608x x

x x

C C

− −= =− −

=

− = = 0.311 wt %

0.89798 0.85 0.7707 0.79590.90 0.85 0.7970 0.7707

Thus, 0.7959 erf (0.89798)

Substituting, 1.15 0.7959

0.95x

x

x x

C C

− −= =− −

=

− = = 0.394 wt %


4.48 The surface of a steel gear made of 1018 steel (0.18 wt % C) is to be gas-carburized at

927ºC (1700ºF). Calculate the time necessary to increase the carbon content to 0.35 wt % at 1.00 mm below the surface. Assume the carbon content of the surface of the gear is 1.20 wt %. D (C in λ iron) at 927ºC = 1.28 × 10 –11 m2/s.

-3 11 2927

Given: 1.20% 0.18% 0.35%

1.0 mm = 1.0 10 m 1.28 10 m / ss o x

C

C C C

x D −

= = =

= × = ×�

3

-11 2

1.20 0.35 1.00 10 merf1.20 0.18 2 (1.28 10 m / s)

139.750.8333 erf erf

s x

s o

C CC C t

zt

− − − × = =

− − ×

= =


erf z

z

0.8209 0.95 0.8333 x 0.8427 1.0

4.49 A gear made of 1020 steel (0.20 wt % C) is to be gas-carburized at 927ºC (1700ºF). Calculate the carbon content at 0.95 mm below the surface of the gear after an 8.0–hour carburizing time. Assume the carbon content at the surface of the gear is 1.25 wt %. D (C in λ iron) at 927ºC = 1.28 × 10 –11 m2/s.

-4 11 2927

Given: 1.25% 0.20% ? 8 h = 28,800 s

0.95 mm = 9.5 10 m 1.28 10 m / ss o x

C

C C C t

x D −

= = = =

= × = ×�

4

-11 2

1.25 9.50 10 merf1.25 0.20 2 (1.28 10 m / s)(28,800 s)

1.25 erf (0.7823)1.05

s x x

s o

x

C C CC C

C

− − − × = =

− − ×

− =

2 2

0.8333 0.8209 0.95 0.9780.8427 0.8209 1.0 0.95

Thus,

139.75 139.750.978

20, 400 s =

x x

tz

t

− −= =− −

= =

= 340 min. = 5.67 h



erf z

z

0.7112 0.75

x 0.7823 0.7421 0.80

4.50 A gear made of 1018 steel (0.18 wt % C) is to be gas-carburized at 927ºC (1700ºF). If the carburizing time is 7.5 h, at what depth in millimeters will the carbon content be 0.40 wt %? Assume the carbon content at the surface of the gear is 1.20 wt %. D (C in λ iron) at 927ºC = 1.28 × 10 –11 m2/s.

11 2927

Given: 1.20% 0.18% 0.40%

7.5 h = 27,000 s 1.28 10 m / ss o x

C

C C C

t D −

= = =

= = ×�

-11 2

1.20 0.40 erf1.20 0.18 2 (1.28 10 m / s)(27,000 s)

0.7843 erf (850.52 ) erf

s x

s o

C C xC C

x z

− − = =

− − ×

= =


erf z

z

0.7707 0.85 0.7843 x1 0.7970 0.90

4.51 If boron is diffused into a thick slice of silicon with no previous boron in it at a

temperature of 1100ºC for 5 h, what is the depth below the surface at which the concentration is 1017 atoms/cm3 if the surface concentration is 1018 atoms/cm3? D = 4 × 10 –13 cm2/s for boron diffusing in silicon at 1100ºC.

18 3 17 3

4 13 21100

Given: 10 atoms/cm 10 atoms/cm 0.0

5.0 h = 1.8 10 s 4.0 10 cm / ss x o

C

C C C

t D −

= = =

= × = ×�

11

0.850.7843 0.7707 0.87590.7970 0.7707 0.90 0.85

Substituting,0.8759 850.520.00103 m = 1.03 mm

x x

z xx

−− = =− −

= ==

0.7823 0.75 0.7112 0.73120.80 0.75 0.7421 0.7112

Thus, 0.7312 erf (0.7823)

Substituting, 1.25 0.7312

1.05x

x

x x

C C

− −= =− −

=

− = = 0.48 wt %


18 17

18 -13 2 4

4

10 10 erf10 0 2 (4.0 10 cm / s)(1.80 10 s)

0.90 erf erf 1.697 10

s x

s o

C C xC C

x z−

− − = =

− − × ×

= = ×


erf z

z

0.8802 1.1 0.9000 x 0.9103 1.2

4.52 If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100ºC for 6 h, what is the depth below the surface at which the concentration is 1016 atoms/cm3 if the surface concentration is 1018 atoms/cm3? D = 2 × 10 –12 cm2/s for aluminum diffusing in silicon at 1100ºC.

18 3 16 3

4 12 21100


6.0 h = 2.16 10 s 2.0 10 cm / ss x o

C

C C C

t D −

= = =

= × = ×�

18 16

18 -12 2 4

4

10 10 erf10 0 2 (2.0 10 cm / s)(2.16 10 s)

0.99 erf erf 4.157 10

s x

s o

C C xC C

x z−

− − = =

− − × ×

= = ×


erf z

z

0.9891 1.8 0.9900 x 0.9928 1.9

11

4

4

1.10.9000 0.8802 1.1660.9103 0.8802 1.2 1.1

Substituting,

1.1661.697 10

1.98 10 cm

x x

xz

x

−

−

−− = =− −

= =×

= ×

11

4

4

1.80.9900 0.9891 1.8240.9928 0.9891 1.9 1.8

Substituting,

1.8244.157 10

7.58 10 cm

x x

xz

x

−

−

−− = =− −

= =×

= ×


4.53 Phosphorus is diffused into a thick slice of silicon with no previous phosphorus in it at a temperature of 1100ºC. If the surface concentration of the phosphorus is 1 × 10 18 atoms/cm3 and its concentration at 1 µm is 1 × 10 15 atoms/cm3, how long must the diffusion time be? D = 3.0 × 10 –13 cm2/s for P diffusing in Si at 1100ºC.

18 3 15 3

-4 13 21100


1.0 m = 1.0 10 cm 3.0 10 cm / ss x o

C

C C C

x Dµ −

= = =

= × = ×�

18 15 4

18 -13 2

10 10 10 cmerf10 0 2 (3.0 10 cm / s)

91.2870.999 erf erf

s x

s o

C CC C t

zt

−� �− −� �= =

− − � �×� �

� �= = � �


erf z

z

0.9981 2.2 0.9990 x 0.9993 2.4

4.54 If the diffusivity in Prob. 4.53 had been 1.5 × 10 –13 cm2/s, at what depth in micrometers

would the phosphorus concentration be 1 × 10 15 atoms/cm3?

Since 0.999, erf is still 2.35. Thus, for the same diffusion period,s x

s o

C C zC C

− =−

5-13 2

2.35, 7.07 10 cm = 2 (1.5 10 cm / s)(1509 s)

xz x −� �� = = = ×� �×� �

0.707 µm

4.55 Arsenic is diffused into a thick slice of silicon with no previous arsenic in it at 1100ºC. If

the surface concentration of the arsenic is 5 × 10 18 atoms/cm3, and its concentration at 1.2 µm below the silicon surface is 1.5 × 10 16 atoms/cm3, how long must the diffusion time be? (D = 3.0 × 10 –14 cm2/s for As diffusing in Si at 1100ºC.)

18 3 16 3

-6 -4 14 21100

Given: 5.0 10 atoms/cm 1.5 10 atoms/cm 0.0

1.20 10 m = 1.20 10 cm 3.0 10 cm / ss x o

C

C C C

x D −

= × = × =

= × × = ×�

2 2

0.9990 0.9981 2.2 2.350.9993 0.9981 2.4 2.2

Thus,

91.287 91.2872.35

1508 s =

x x

tz

t

− −= =− −

� � � �= =� � � ��

= 25.1 min.


18 16 4

18 -14 2

5.0 10 1.5 10 1.20 10 cmerf5.0 10 0 2 (3.0 10 cm / s)

346.40.9970 erf erf

s x

s o

C CC C t

zt

−� �− × − × ×� �= =

− × − � �×� �

� �= = � �


2 2

0.9970 0.9953 2.0 2.12 and,0.9981 0.9953 2.2 2.0

346.4 346.4 26,700 s = 2.12

x x

tz

− −= =− −

� � � �= = =� � � �� 7.42 h

4.56 Calculate the diffusivity D in square meters per second for the diffusion of nickel in FCC

iron at 1100ºC. Use values of D 0 = 7.7 × 10 –5 m2/s; Q = 280 kJ/mol; R = 8.314 J/ (mol· K). The diffusivity of the nickel into FCC iron at 1373 K is:

[ ]/ 5 2

5 2 -24.53

-280,000 J/mol(7.7 10 m / s) exp8.314 J/(mol K) (1373 K)

(7.7 10 m /s)(e )

Q RToD D e− −

−

� �� = = × � �� ⋅� � ��

= ×

= -15 21.71×10 m /s

4.57 Calculate the diffusivity in m2/s of carbon in HCP titanium at 700ºC. Use

D 0 = 5.10 × 10 – 4 m2/s; Q = 182 kJ/mol; R = 8.314 J/ (mol· K).

The diffusivity of carbon into HCP titanium is:

[ ]/ 4 2

4 2 -22.49

-182,000 J/mol(5.10 10 m /s) exp8.314 J/(mol K) (973 K)

(5.10 10 m /s)(e )

Q RToD D e− −

−

� �� = = × � �� ⋅� � ��

= ×

= -14 28.64×10 m /s

4.58 Calculate the diffusivity in m2/s for the diffusion of zinc in copper at 350ºC. Use D 0 =

3.4 × 10 –5 m2/s; Q = 191 kJ/mol.

The diffusivity of zinc into copper at 623 K is:


[ ]/ 4 2

5 2 -36.88

-191,000 J/mol(3.40 10 m /s) exp8.314 J/(mol K) (623 K)

(3.40 10 m /s)(e )

Q RToD D e− −

−

� �� = = × � �� ⋅� � ��

= ×

= -21 23.29×10 m /s

4.59 The diffusivity of manganese atoms in the FCC iron lattice is 1.5 × 10 –14 m2/s at 1300ºC

and 1.5 × 10 –15 m2/s at 400ºC. Calculate the activation energy in kJ/mol for this case in this temperature range. Data: R = 8.314 J/(mol·K).

The activation energy may be calculated using the Arrhenius type equation,

1300 C 2

1 2 1400 C

exp( / ) 1 1expexp( / )

D Q RT QD Q RT R T T

� �� − −= = −� �� − � �

�

�

1 2where 400 C = 673 K and 1300 C = 1573 K. Substituting,T T= =� �

14 2

15 2

4

4

1.5 10 m /s 1 1exp8.314 J/(mol K) 1573 K 673 K1.5 10 m /s

10 exp (1.0226 10 )

ln(10) (1.0226 10 )22,518 J/mol =

Q

Q

QQ

−

−

−

−

� �× − � �= −� �� ⋅× � �

� �= ×� �

= ×= 22.5 kJ/mol

4.60 The diffusivity of copper atoms in the aluminum lattice is 7.5 × 10 –13 m2/s at 600ºC and

2.5 × 10 –15 m2/s at 400ºC. Calculate the activation energy for this case in this temperature range. [R = 8.314 J/(mol·K).]

The activation energy associated with copper diffusing into aluminum for this temperature range is,

600 C

2 1400 C

1 2

1 1exp

where 400 C = 673 K and 600 C = 873 K. Substituting,

D QD R T T

T T

� �� −= −� ��

= =

�

�

� �


13 2

15 2

5

5

7.5 10 m /s 1 1exp8.314 J/(mol K) 873 K 673 K2.5 10 m /s

300 exp (4.094 10 )

ln(300) (4.094 10 )139,320 J/mol =

Q

Q

QQ

−

−

−

−

� �× − � �= −� �� ⋅× � �

� �= ×� �

= ×= 139.3 kJ/mol

4.61 The diffusivity of iron atoms in the BCC iron lattice is 4.5 × 10 –23 m2/s at 400ºC and

5.9 × 10 –16 m2/s at 800ºC. Calculate the activation energy in kJ/mol for this case in this temperature range. [R = 8.314 J/(mol·K).] The activation energy associated with iron diffusing into BCC iron for this temperature range is,

800 C

2 1400 C

1 2

1 1exp

where 400 C = 673 K and 800 C = 1073 K. Substituting,

D QD R T T

T T

−= −

= =

�

�

� �

16 2

23 2

5

5

5.9 10 m /s 1 1exp8.314 J/(mol K) 1073 K 673 K4.5 10 m /s

13,111,111.1 exp (6.662 10 )

ln(13,111,111.1) (6.662 10 )246,007 J/mol =

Q

Q

QQ

−

−

−

−

× − = − ⋅×

= ×

= ×= 246 kJ/mol

Smith Foundations of Materials Science and Engineering Solution Manual 168

CHAPTER 8 PHASE DIAGRAMS

8.1 Define (a) a phase in a material and (b) a phase diagram.

(a) A phase in a material is a microscopic region that differs in structure and/or composition from another region.

(b) A phase diagram is a graphical representation of the phases present within a materials system for a range of temperatures, pressures and compositions.

8.2 In the pure water pressure-temperature equilibrium phase diagram (Fig. 8.1) what phases

are in equilibrium for the following conditions: (a) Along the freezing line (b) Along the vaporization line (c) At the triple point

(a) Along the freezing line, liquid and solid phases are in equilibrium. (b) Along the vaporization line, liquid and vapor phases exist in equilibrium. (c) At the triple point, all three phase – vapor, liquid and solid – coexist.

8.3 How many triple points are there in the pure iron pressure-temperature equilibrium phase

diagram of Fig. 8.2? What phases are in equilibrium at each of the triple points?

Three triple points can be identified having the following phases in equilibrium:

1. vapor, liquid and δ Fe 2. vapor, δ Fe, and γ Fe 3. vapor, γ Fe, α Fe

8.4 Write the equation for Gibbs phase rule and define each of the terms.

The equation for the Gibbs phase rule is:

P + F = C +2 where P = the number of phases that coexist within a specific system F = the degrees of freedom for the system C = the number of components in the system

8.5 Refer to the pressure-temperature equilibrium phase diagram for pure water (Fig. 8.1) and answer the following: (a) How many degrees of freedom are there at the triple point? (b) How many degrees of freedom are there along the freezing line?


(a) At the triple point, there are zero degrees of freedom. (b) Along the freezing line of pure water, there is one degree of freedom.

8.6 What is a binary isomorphous alloy system?

The binary isomorphous alloy system is a two-component system in which the two elements are completely soluble in each other in the liquid and solid states and form a single type of crystal structure for all compositions.

8.7 What are the four Hum-Rothery rules for the solid solubility of one element in another?

The four Hum-Rothery rules for the solid solubility of one element in another are:

1. The crystal structure of each element of the solid solution must be the same. 2. The size of the atoms of each of the two elements must not differ by more

than fifteen percent. 3. The elements should not form compounds with each other; there should be no

appreciable difference in the electronegativities of the two elements. 4. The elements should have the same electron valence.

8.8 A number of elements along with their crystal structures and atomic radii are listed in the

following table. Which pairs might be expected to have complete solid solubility in each other?

Crystal

Structure Atomic radius (nm)

Crystal Structure

Atomic radius (nm)

Silver FCC 0.144 Lead FCC 0.175 Palladium FCC 0.137 Tungsten BCC 0.137 Copper FCC 0.128 Rhodium FCC 0.134 Gold FCC 0.144 Platinum FCC 0.138 Nickel FCC 0.125 Tantalum BCC 0.143 Aluminum FCC 0.143 Potassium BCC 0.231 Sodium BCC 0.185 Molybdenum BCC 0.136

Pairs of these elements which may be expected to have complete solid solubility in each other are:

Silver–Palladium Palladium–Platinum Tantalum–Molybdenum Silver–Gold Palladium–Rhodium Tantalum–Tungsten Silver-Rhodium Palladium–Gold Gold–Platinum Copper–Nickel Rhodium–Platinum

8.9 Derive the lever rule for the amount in weight percent of each phase in two-phase regions of a binary phase diagram. Use a phase diagram in which two elements are completely soluble in each other.


The lever-rule equations can be derived by first recognizing that the sum of the weight fractions of the liquid and solid phases which must equal 1.

1l sX X+ =

Considering the weight balance of B in the alloy as a whole and the sum of B in the two phases, we arrive at:

0 l l s sw X w X w= +

Combining these two equations gives

0 (1 )s l s sw X w X w= − +

Solving for the Xs gives the first lever-rule:

Wt fraction of solid phase o ls

s l

w wXw w

−= =−

Similarly, the second lever-rule is found to be:

0Wt fraction of liquid phase sl

s l

w wXw w

−= =−

8.10 Consider an alloy containing 70 wt % Ni and 30 wt % Cu (see Fig. 8.3).

(a) At 1350ºC make a phase analysis assuming equilibrium conditions. In the phase analysis include the following:

(i.) What phases are present? (ii.) What is the chemical composition of each phase? (iii.) What amount of each phase is present?

(b) Make a similar phase analysis at 1500ºC. (c) Sketch the microstructure of the alloy at each of these temperatures by using circular

microscopic fields.

(a) (i) The phases present are the liquid and solid (L + α). (ii) The chemical composition of liquid is wl = 62 wt % Ni while that of the

solid is ws = 74 wt % Ni. (iii) The weight percent of solid and liquid are:

74 70Wt % of liquid phase 100%74 62

−= × =−

33.3%

70 62Wt % of solid phase 100%74 62

−= × =−

66.67%


(b) At 1500ºC, the alloy is 100% liquid. (c) The microstructure of the alloy at these temperatures would look similar to the

following sketches.

8.11 Describe how the liquidus and solidus of a binary isomorphous phase diagram can be

determined experimentally.

The liquidus and solidus of a binary isomorphous phase diagrams can be determined experimentally by measuring cooling rate for several specific alloy compositions and plotting the corresponding liquid-solid curves. The phase diagram can then be constructed by plotting the liquidus and solidus temperatures versus composition of the alloys.

8.12 Explain how a cored structure is produced in a 70% Cu-30% Ni alloy.

A cored structure is produced in a 70% Cu-30% Ni alloy when the alloy is cooled rapidly; without sufficient time for complete solid-state diffusion, concentration gradients remain in the alloy structure.

8.13 How can the cored structure in a 70% Cu-30% Ni alloy be eliminated by heat treatment?

The cored structure can be eliminated in ingots and castings by heat treating at elevated temperatures. This homogenization process accelerates the required solid-state diffusion and thus produces a homogeneous structure in the alloy.

8.14 Explain what is meant by the term liquation. How can a liquated structure be produced in

an alloy? How can it be avoided?

Liquation is the localized melting which occurs if an alloy is heated to a temperature greater than the lowest melting temperature of the alloy’s constituents. The result of such overheating is a liquated structure, in which grain boundaries may be melted. To avoid liquation, the heat treatment should be performed such that the melting temperature is approached slowly but never exceeded.

100% Liquid Solid α phase

Liquid phase L

1350ºC 1500ºC


Figure 8.31 The copper-silver phase diagram.

8.15 Consider the binary eutectic copper-silver phase diagram in Fig. 8.31. Make phase analyses of an 88 wt % Ag-12 wt % Cu alloy at the temperatures (a) 1000ºC, (b) 800ºC, (c) 780 C + T∆o , (d) 780 C T.− ∆o In the phase analysis, include:

(i) The phases present (ii) The chemical compositions of the phases (iii) The amounts of each phase (iv) Sketch the microstructure by using 2 cm diameter circular fields.

(a) At 1000ºC:

Phases present: liquid Compositions of phases: 100%

(b) At 800ºC, Phases present: liquid beta Compositions of phases: 78% Ag in liquid phase 93% Ag in β phase


Amounts of phases:

93 88Wt % liquid phase 100%93 78

88 78Wt % beta phase 100%93 78

−= × =−

−= × =−

33.3%

66.6%

(c) At 780 C + T∆o ,

Phases present: liquid beta Compositions of phases: 71.9% Ag in liquid phase 91.2% Ag in β phase Amounts of phases:

91.2 88Wt % liquid phase 100%

91.2 71.9

88 71.9Wt % beta phase 100%91.2 71.9

−= × =−

−= × =−

16.6%

83.4%

(d) At 780 C T,− ∆o

Phases present: alpha beta Compositions of phases: 7.9% Ag in α phase 91.2% Ag in β phase Amounts of phases:

91.2 88Wt % alpha phase 100%91.2 7.9

88 7.9Wt % beta phase 100%91.2 7.9

−= × =−

−= × =−

3.84%

96.16%

8.16 If 500g of a 40 wt % Ag-60 wt % Cu alloy is slowly cooled from 1000ºC to just below

780ºC (see Fig. 8.31): (a) How many grams of liquid and proeutectic alpha are present at 850ºC? (b) How many grams of liquid and proeutectic alpha are present at 780 C + T∆o ? (c) How many grams of alpha are present in the eutectic structure at 780 C T?− ∆o (d) How many grams of beta are present in the eutectic structure at 780 C T?− ∆o

(a) At 850ºC,

40 7.9Wt % liquid 100% 72.8%52 7.9

−= × =−


52 40Wt % proeutectic 100% 27.2%52 7.9

Weight of liquid phase 500 g 0.728Weight of proeutectic 500 g 0.272

α

α

−= × =−

= × == × =

364 g136 g

(b) In the eutectic structure at 780 C + T∆o ,

40 7.9Wt % liquid 100% 50.2%71.9 7.9

−= × =−

71.9 40Wt % proeutectic 100% 49.8%71.9 7.9

Weight of liquid phase 500 g 0.502Weight of proeutectic 500 g 0.498

α

α

−= × =−

= × == × =

251 g249 g

(c) In the eutectic structure at 780 C T,− ∆o the number of grams of α present is,

91.2 40Wt % total 100% 61.5%91.2 7.9

Weight of total 500 g 0.615

α

α

−= × =−

= × = 307.5 g

(d) In the eutectic structure at 780 C T,− ∆o the number of grams of β present is,

40 7.9Wt % total 100% 38.5%91.2 7.9

Weight of 500 g 0.385

β

β

−= × =−

= × = 192.5 g

8.17 A lead-tin (Pb-Sn) alloy consists of 60 wt % proeutectic β and 60 wt % eutectic α β+ at

183 C .T− ∆o Calculate the average composition of this alloy (see Fig. 8.11).

Since the alloy contains 60 wt % proeutectic β, the wt % Sn must lie between 61.9 wt % and 97.5 wt %:

61.9% proeutectic 0.6097.5 61.9

0.6(35.6) 61.9 83.3%

x

x

β −= =−

= + =

Thus, the alloy consists of 83.3 % Sn and 16.7 % Pb.

8.18 A Pb-Sn alloy (Fig. 8.11) contains 40 wt % β and 60 wt % α at 50ºC. What is the average composition of Pb and Sn in this alloy?


At 50ºC, the phase compositions are 100% Sn for β and approximately 2% Sn for α. Thus,

100.0% 0.60, 100 0.6(98.0) 41.2%

100.0 2.0x xα −= = = − =

−

The alloy consists of 41.2 % Sn and 58.8 % Pb.

8.19 An alloy of 30 wt % Pb and 70 wt % Sn is slowly cooled from 250ºC to 27ºC (see Fig. 8.11). (a) Is this alloy hypoeutectic or hypereutectic? (b) What is the composition of the first solid to form? (c) What are the amounts and composition of each phase that is present at 183 C ?T+ ∆º (d) What is the amount and composition of each phase that is present at 183 C ?T− ∆º (e) What are the amounts of each phase present at room temperature?

(a) This alloy is hypereutectoid; the composition lies to the right of the eutectic point. (b) The first solid to form is solid solution β containing approximately 98 % Sn.

(c) At 183 C ,T+ ∆º the compositions of the phases present are 61.9% Sn in liquid phase

and 19.2% Sn in beta phase. The amounts of the respective phases present are:

97.5 70Wt % liquid 100%97.5 61.9

70 61.9Wt % beta 100%97.5 61.9

−= × =−

−= × =−

77.2%

22.8%

(d) At 183 C ,T− ∆º the compositions of the phases present are 19.2 % Sn in α phase and 97.5 % Sn in β phase. The amounts of the respective phases present are:

97.5 70Wt % total 100%

97.5 19.2

70 19.2Wt % total beta 100%97.5 19.2

α −= × =−

−= × =−

35.1%

64.8%

(e) As the alloy is cooled below the eutectic temperature, the tin content in the alpha

phase and the lead content in the beta phase are further reduced. However, at room temperature (20ºC), equilibrium is not achieved because the diffusion rate is so slow. Referring to Fig. 8.11, if the solvus line is extrapolated to 20ºC, the approximate composition of alpha and beta are 2.0% and 100.0 %, respectively. Thus,


100 70Wt % total 100%100 2

70 2Wt % total beta 100%100 2

α −= × =−

−= × =−

30.6%

69.4%

Figure 8.32 The iridium-osmium phase diagram. 8.20 Consider the binary peritectic iridium-osmium phase diagram of Fig. 8.32. Make phase

analyses of a 70 wt % Ir-30 wt % Os at the temperatures (a) 2600ºC (b) 2665 C ,T+ ∆o and (c) 2665 C .T− ∆o In the phase analyses include:

(i.) The phases present (ii.) The chemical compositions of the phases (iii.) The amounts of each phase (iv.) Sketch the microstructure by using 2 cm diameter circular fields.

(a) At 2600ºC, (i) Phases Present: Liquid and alpha phases (ii) Compositions of Phases: 16% Os in liquid phase; 38% Os in alpha phase (iii) Amounts of phases:

30 16Wt % alpha 100%38 16

38 30Wt % liquid 100%38 16

−= × =−

−= × =−

63.6%

36.4%

Solid α phase

Liquid phase L


(b) At 2665 C ,T+ ∆o (i) Phases Present: Liquid and beta phases (ii) Compositions of Phases: 23% Os in liquid phase; 61.5% Os in β phase (iii) Amounts of phases:

30 23Wt % beta 100%61.5 23

61.5 30Wt % liquid 100%61.5 23

−= × =−

−= × =−

18.2%

81.8%

(c) At 2665 C ,T− ∆o (i) Phases Present: Liquid and alpha phases (ii) Compositions of Phases: 23% Os in liquid phase; 43% Os in α phase (iii) Amounts of phases:

30 23Wt % alpha 100%43 23

43 30Wt % liquid 100%43 23

−= × =−

−= × =−

35.0%

65.0%

8.21 Consider the binary peritectic iridium-osmium phase diagram of Fig. 8.32. Make phase analyses of a 40 wt % Ir-60 wt % Os at the temperatures (a) 2600ºC (b) 2665 C ,T+ ∆o and (c) 2665 C ,T− ∆o (d) 2800ºC. Include in the phase analyses the four items listed in Prob. 8.20.

(a) At 2600ºC,

(i) Phases Present: Alpha and beta phases (ii) Compositions of Phases: 43% Os in alpha phase; 61.5% Os in beta phase (iii) Amounts of phases:

61.5 60Wt % alpha 100%61.5 43

60 43Wt % beta 100%61.5 43

−= × =−

−= × =−

8.1%

91.9%

(b) At 2665 C ,T+ ∆o (i) Phases Present: Liquid and beta phases (ii) Compositions of Phases: 23% Os in alpha phase; 61.5% Os in beta phase (iii) Amounts of phases:

61.5 60Wt % liquid phase 100%61.5 23

60 23Wt % beta phase 100%61.5 23

−= × =−

−= × =−

3.9%

96.1%

Solid β phase

Liquid phase L

Solid α phase

Liquid phase L

Solid α phase

Beta phase

Solid β phase

Liquid phase


(c) At 2665 C ,T− ∆o (i) Phases Present: Alpha and beta phases (ii) Compositions of Phases: 43% Os in liquid phase; 61.5% Os in beta phase (iii) Amounts of phases:

61.5 60Wt % alpha 100%61.5 43

60 43Wt % beta 100%61.5 43

−= × =−

−= × =−

8.1%

91.9%

(d) At 2800ºC, (i) Phases Present: Alpha and beta phases (ii) Compositions of Phases: 45% Os in liquid phase; 85% Os in beta phase (iii) Amounts of phases:

85 60Wt % liquid 100%85 45

60 45Wt % beta 100%85 45

−= × =−

−= × =−

62.5%

37.5%

8.22 Describe the mechanism that produces the phenomenon of surrounding in a peritectic

alloy which is rapidly solidified through the peritectic reaction.

Surrounding in a rapidly solidified peritectic alloy is a nonequilibrium phenomenon in which the alpha phase is encased by the beta phase during the peritectic reaction. As a result, the solid beta phase acts as a barrier to alpha diffusion and the peritectic reaction rate decreases continuously.

8.23 Can coring and surrounding occur in a peritectic-type alloy which is rapidly solidified?

Explain.

In a rapidly solidified peritectic alloy, coring can occur during the formation of the primary alpha phase and subsequently, the cored alpha phase can be surrounded by the beta phase during the peritectic reaction.

8.24 Consider an Fe-4.2 wt % Ni alloy (Fig. 8.16) that is slowly cooled from 1550 to 1450ºC.

What weight percent of the alloy solidifies by the peritectic reaction?

Referring to Fig. 8.16 on the following page, during the peritectic reaction, liquid and δ phases react to form solid γ:

4.2 4.0Wt % 100%4.3 4.0

γ −= × =−

66.7%

At the end of the reaction, there is an excess of δ phase having a wt % of 33.3 %.

Solid α phase

Beta phase

Solid α phase

Beta phase


Figure 8.16 The peritectic region of the iron-nickel phase diagram. The peritectic point is located at 4.3% Ni and 1517°C, which is point c.

8.25 Consider an Fe-5.0 wt % Ni alloy (Fig. 8.16) that is slowly cooled from 1550 to 1450ºC.

What weight percent of the alloy solidifies by the peritectic reaction?

During the peritectic reaction, liquid solidifies to form solid γ:

5.4 5.0Wt % 100%5.4 4.3

γ −= × =−

36.4%

8.26 Determine the weight percent and composition in weight percent of each phase present in

an Fe-4.2 % Ni alloy (Fig. 8.16) at 1517 C .T+ ∆o

Just above the eutectic temperature, the alloy composition is: 4.0 wt % Ni in the δ phase and 5.4 wt % Ni in the liquid phase. The weight percentages of these phases are:

4.2 4.0Wt % liquid 100%5.4 4.0

5.4 4.2Wt % 100%5.4 4.0

δ

−= × =−

−= × =−

14.3%

85.7%


8.27 Determine the composition in weight percent of the alloy in the Fe-Ni system (Fig. 8.16) that will produce a structure of 40 wt % δ and 60 wt % γ, just below the peritectic temperature.

For a 40 wt % δ composition to exist at 1517 C ,T− ∆o

4.3Wt % 0.40 4.18%4.3 4.0

x xδ −= = =−

Thus, the alloy contains 4.18% Ni and 95.82% Fe.

8.28 What is a monotectic invariant reaction? How is the monotectic reaction in the copper-lead system important industrially?

A monotectic invariant reaction is one in which a liquid phase reacts isothermally to form a solid phase and a new liquid phase. The monotectic reaction is important industrially to the copper-lead system because it can produce a nearly pure lead phase in copper-zinc brasses which improves the machining properties of the alloys; the lead sufficiently reduces the ductility of the alloys to cause machined chips to naturally break away from the workpiece.

8.29 In the copper-lead (Cu-Pb) system (Fig. 8.23) for an alloy of Cu-10 wt % Pb, determine the amounts and compositions of the phases present at (a) 1000ºC (b) 955 C ,T+ ∆o (c) 955 C ,T− ∆o and (d) 200ºC.

(a) At 1000ºC,

Compositions of Phases: 100% Cu, 0% Pb in α phase; 81% Cu, 19% Pb in L1 phase Amounts of Phases:

19 10Wt % 100%19 0

α −= × =−

47.4% 110 0Wt % 100%19 0

L −= × =−

52.6%

(b) At 955 C ,T+ ∆o


36 10Wt % 100%36 0

α −= × =−

72.2% 110 0Wt % 100%36 0

L −= × =−

27.8%

(c) At 955 C ,T− ∆o

Compositions of Phases: 100% Cu, 0% Pb in α phase; 13% Cu, 87% Pb in L2 phase


Amounts of Phases:

87 10Wt % 100%87 0

α −= × =−

88.5% 210 0Wt % 100%87 0

L −= × =−

11.5%

(d) At 200ºC,

Compositions of Phases: 99.995% Cu, 0.005% Pb in α phase; 0.007% Cu, 99.993% Pb in β phase Amounts of Phases:

99.99 10Wt % 100%99.99 0

α −= × =−

90% 10 0Wt % 100%99.99 0

β −= × =−

10%

Figure 8.23 The copper-lead phase diagram.


8.30 For an alloy Cu-70 wt % Pb (Fig. 8.23), determine the amounts and compositions in weight percent of the phases present at (a) 955 C ,T+ ∆o (b) 955 C ,T− ∆o and (c) 200ºC.

(a) At 955 C ,T+ ∆o

Compositions of Phases: 64% Cu, 36% Pb in L1 phase 13% Cu, 87% Pb in L2 phase; Amounts of Phases:

270 36Wt % 100%87 36

L −= × =−

66.7% 187 70Wt % 100%87 36

L −= × =−

33.3%

(b) At 955 C ,T− ∆o


87 70Wt % 100%87 0

α −= × =−

19.5% 270 0Wt % 100%87 0

L −= × =−

80.5%

(c) At 200ºC,

Compositions of Phases: 99.995% Cu, 0.005% Pb in α phase; 0.007% Cu, 99.993% Pb in β phase Amounts of Phases:

99.99 70Wt % 100%99.99 0

α −= × =−

30% 70 0Wt % 100%99.99 0

β −= × =−

70%

8.31 What is the average composition (weight percent) of a Cu-Pb alloy that contains 30 wt %

L1 and 70 wt % α at 955 C ?T+ ∆o

For a 30 wt % L1 composition to exist at 955 C ,T+ ∆o

10Wt % 0.30 10.8%

36 0xL x−= = =

−

Thus, the alloy contains 10.8% Pb and 89.2% Cu.


8.32 Write equations for the following invariant reactions: eutectic, eutectoid, peritectic, and peritectoid. How many degrees of freedom exist at invariant reaction points in binary phase diagrams?

Eutectic Reaction: cooling + L α β→ Eutectoid Reaction: coolingα β γ→ + Peritectic Reaction: coolingLα β+ → Peritectoid Reaction: coolingα β γ+ → There are zero degrees of freedom at the invariant reaction points in binary phase diagrams.

8.33 How are eutectic and eutectoid reactions similar? What is the significance of the –oid

suffix? The eutectic and eutectoid reactions are similar in that they both involve the decomposition of a single phase into two solid phases. The –oid suffix indicates that a solid, rather than liquid, phase is decomposing.

8.34 Distinguish between (a) a terminal phase and (b) an intermediate phase.

A terminal solid solution phase occurs at the end of a phase diagram, bordering on pure components. Whereas, an intermediate solid solution phase occurs within a composition range inside the phase diagram and is separated from other phases in a binary diagram by two-phase regions.

8.35 Distinguish between (a) an intermediate phase and (b) an intermediate compound.

Intermediate phases, which may occur in binary metal or ceramic phase diagrams, represent a range of solid solution compositions. Conversely, an intermediate compound has a fixed composition and definite stoichiometry at room temperature and is formed between two metals or a metal and a nonmetal.

8.36 What is the difference between a congruently melting compound and an incongruently

melting one?

A congruently melting compound maintains its composition right up to its melting point. Whereas an incongruently melting compound undergoes peritectic decomposition upon heating; the solid compound decomposes into a liquid and another solid solution.

8.37 Consider the Cu-Zn phase diagram of Fig. 8.25.

(a) What is the maximum solid solubility in weight percent of Zn in Cu in the terminal solid solution α?

(b) Identify the intermediate phases in the Cu-Zn phase diagram.


(c) Identify the three-phase invariant reactions in the Cu-Zn diagram. (i) Determine the composition and temperature coordinates of the invariant

reactions. (ii) Write the equations for the invariant reactions. (iii) Name the invariant reactions.

(a) The maximum solid solubility in weight percent of zinc in copper in the solid solution

α is 39%. (b) The intermediate phases are β, γ, δ, and ∫. (c) The three-phase invariant reactions are:

1. Peritectic reaction at 903ºC, 36.8% Zn (32.5% Zn) + (37.5% Zn) (36.8% Zn)Lα β→

2. Peritectic reaction at 835ºC, 59.8% Zn

(56.5% Zn) (59.8% Zn) (59.8% Zn)Lβ γ+ → 3. Peritectic reaction at 700ºC, 73% Zn

(69.8% Zn) (80.5% Zn) (73% Zn)Lγ δ+ →

4. Peritectic reaction at 598ºC, 78.6% Zn (76.5% Zn) (89% Zn) (78.6% Zn)Lδ + → ò

5. Peritectic reaction at 424ºC, 97.3% Zn

(87.5% Zn) (98.3% Zn) (97.3% Zn)L η+ →ò

6. Eutectoid reaction at 558ºC, 73% Zn (73% Zn) (69.8% Zn) (78.6% Zn)δ γ→ + ò

7. Eutectoid reaction at 250ºC, 47% Zn

(47% Zn) (37% Zn) (59% Zn)β α γ′ → +

8.38 Consider the aluminum-nickel (Al-Ni) phase diagram of Fig. 8.33 on the following page. For this phase diagram: (a) Determine the coordinates of the composition and temperature of the invariant

reactions. (b) Write the equations for the three-phase invariant reactions and name them. (c) Label the two-phase regions in the phase diagram.

(a) and (b):

1. Eutectic reaction at 639ºC, 0.1% Ni

3(0.1% Ni) Al(0.05% Ni) Al Ni (42% Ni)L → +


2. Peritectic reaction at 854ºC, 42% Ni

3 2 3(28% Ni) Al Ni (55% Ni) Al Ni (42% Ni)L + →


3 2(44% Ni) AlNi(63% Ni) Al Ni (59% Ni)L + →

4. Peritectoid reaction at 700ºC, 81% Ni

3 3 5AlNi (77% Ni) AlNi (86% Ni) Al Ni (81% Ni)+ →


3(87% Ni) AlNi(83% Ni) AlNi (86% Ni)L + →

6. Eutectic reaction at 1385ºC, 86% Ni

3(87% Ni) Ni (90% Ni) AlNi (86% Ni)L → +

(c) The two-phase regions are identified in Fig. 8.33 below.


Figure 8.34 Nickel-vanadium phase diagram.

8.39 Consider the nickel-vanadium (Ni-V) phase diagram of Fig. 8.34. For this phase diagram repeat questions (a), (b), and (c) of Prob. 8.38.

(c) The two-phase regions are identified in Fig. 8.34 above. (a) and (b):

1. Eutectoid reaction at 906ºC, 29.5% V

3 2Ni (29.5% V) Ni V (25% V) Ni V(30.5% V)→ +

2. Eutectoid reaction at 890ºC, 35.2% V

2Ni (35.2% V) Ni V (32% V) (51% V)σ ′→ +

3. Eutectic reaction at 1202ºC, 47.5% V (47% V) Ni (40% V) (51% V)L σ ′→ +

4. Peritectic reaction at 1280ºC, 64% V


(58.1% V) V (73% V) (64% V)L σ ′+ → + 5. Peritectoid reaction at 900ºC, 75.3% V

3(70.7% V) V (88.1% V) NiV (75% V)σ ′ + → +

Figure 8.35 Titanium-aluminum phase diagram.

8.40 Consider the titanium-aluminum (Ti-Al) phase diagram of Fig. 8.35. For this phase diagram, repeat questions (a), (b), and (c) of Prob. 8.38.

(c) The two-phase regions are identified in Fig. 8.35 above. (a) and (b):

1. Peritectoid reaction at 1285ºC, 32% Al Ti (30% Al) TiAl (35% Al) Ti (32% Al)β α+ →

2. Eutectoid reaction at 1125ºC, 27% Al

3Ti (27% Al) TiAl(34% Al) Ti Al(26% Al)α → +


3. Peritectic reaction at 1380ºC, 58% Al (61% Al) TiAl(56% Al) (58% Al)L δ+ →

4. Peritectic reaction at 1350ºC, 63% Al

3(69% Al) (60% Al) TiAl (63% Al)L δ+ →

5. Peritectoid reaction at 1240ºC, 55% Al

2TiAl(51% Al) (57% Al) TiAl (55% Al)δ+ →

6. Eutectoid reaction at 1150ºC, 58% Al

2 3(58% Al) TiAl (55% Al) TiAl (63% Al)δ → +

7. Peritectic reaction at 665ºC, 99% Al

3(100% Al) TiAl (68% Al) Al(99% Al)L + →

8.41 What is the composition of point y in Fig. 8.29? As indicated in the ternary diagram figure below, the composition of point y is 20% A, 30% B and 50% C.

30%

20%

50%

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