Phase-Transformations-in-Metals-and-Alloys-solutions--.pdf

Diffmionless transformations

440

6

.

6

are two different but equivalent ways of producing the lattice invariant shear. Show exactly what is meant by this. What is the experimental proof of both types of shear? Draw a diagram to illustrate Bain's homogeneous deformation model for the fee -> bec diffusionless transformation. Assuming a7 = 3 56 A and flu = 2.86 A, and that cia for martensite is 1.15 calculate the maximum movement experienced by atoms during the transformation.

i

Assume that cia = 1.1.

i

Solutions to Exercises

Compiled by John C

.

6 7 .

.

Ion

What are the essential differences in martensite nucleation models based

(a) on changes at the core of a dislocation; (b) on dislocation strain field interaction? Discuss the advantages and disadvantages of both models in terms of the known characteristics of martensitic transformations.

8

6

.

6

9

.

Give possible reasons why the habit plane of martensite changes as a function of alloying content in steels and Fe-Ni alloys. What factors influence the retention of austenite in these alloys? What is the role of austenitic grain size in martensitic transformations? Is austenitic grain size important to the strength of martensite? What other factors are important to strength and roughness in technological

Chapter 1 1

1

.

Cp = 22 64 + 6.28 x HrTJ moP'K .

fT-C

Entropy increase AS =

- dT

,

hardened steels? 6

10 Suggest possible alloying and heat treatment procedures needed to design the following steels: (a) a quenched and tempered steel; (b) a dual phase steel; (c) a maraging steel; id) a TRIP steel. 11 How would you characterize the unique properties of alloys which can be utilized as 'memory metals'. How would you design a TiNi alloy for use as. e.g., a self-locking rivet? Give instructions on how it is to be

.

6

AC

f358 22.64 +:6.28 x lO T dT

_ 300-13SS -

MOO *

.

used.

/

H,L22 641n r+ 6 28 x HrT] .

.

40.83 J mol-'K'1 12 .

iquid Fe 1600 5-Fe y-Fe

9

i !

800 500a-Fe

o

l

£-Fe

-

300-

0

50

100

Pressure kbar ,

150

Solutions to exercises

Solutions to exercises

442

443

i

1

.

Liquid Fe

4

-Fe

a

y-Fe

Phases stable at low temperatures must have low enthalpies because the (-TS) term in the expression for G becomes negligible Phases .

stable at high temperatures, on the other hand have higher entropies to compensate for higher enthalpies. ,

y-Fe

15 .

J

At 800oC

At 1600oC

t 24

-

-

22

LL

0.0240

d = initial concentration.

4Ct)

r-7t2D/l

< 0.05 . --exp \ -p

J

K

D

(c) Assume that the time taken to remove 95% of all the hydrogen is so long that only the first term of the Fourier series is significant The hydrogen concentration at this stage will then be given by

Co

C{x, 0 = --sin y exp 1 i

.

CM

e

.

/2

.

I

as shown in the figure on page 458.

At the required time (tx) the shaded area in the figure will be 5% . of the area under the concentration line at / = 0 i.e. ,

V

C(xi, /,) dr = 0.05 C,,/ 0 31

0 2

4 X

4

4C0

/-7r2Df,\ ('

tvc

- -expl-sin y dx = 0.05 CJ

extrapolation of the tangents to the free energy curves at 1 and 2 to the corresponding sides of the free energy diagram, as shown above. All atoms diffuse so as to reduce their chemical potential. Therefore, A atoms will have a tendency to diffuse from a to P (\iaA >

2

which gives

459

Solutions to exercises

Solutions to exercises

458

= 0.282 t=0

\x%) and B atoms will have a tendency to diffuse from P to a

Co-

The resultant composition changes are indicated in the diagram.

T

Diffusion stops, and equilibrium is reached, when |iA = nil and \i% = ji . That this process results in a reduction in the total free energy of

c

the diffusion couple can be seen from the diagram below. The initial free energy Gj can be reduced to G? by a change in the compositions of the a and P phases to and A , the equilibrium compositions (provided < XBu]k < X%).

rr7777777777TT A 0 x

Note that this time is an order of magnitude larger than the time

derived in part (b). Consequently it is clearly justified to ignore all terms of the Fourier series but the first.

/-13400\

From Table 2.1. D = 0.1 exp i

e

.

.

P

RT

a CD c

Gi

Zl1\

2-1

D (20oC) = 4.08 x lO-4 mnrs ,

7

CD Z

.

f

Thus for / = 10 mm: f, = 19.2 h,

f

G2

for / = 100 mm: f, = 1920 h (80 days). 26 .

Mofar

free energy

A

XBulk

X°

B

xz P

1

2

2

7

.

Ha

Substituting into Darken's Equations (2.47) and (2.51) § Zn

/

5x

(equilibrium) (equilibrium)

Da

= XC

ll

DaZn

a

+ XZnD Cxi

we obtain

026 x lO"6 = (D n - Dacu) x 0.089 mm s

0

Ha

.

A

>

X

B

4

.

5 x 10~7

0

.

™ + 0.22 " 78 Dfin

.

From which

At the initial compositions 1 and 2 of a and P respectively the chemical potentials of A and B atoms in each phase can be found by

DaZn

5 1 x 10

Dr Cu

2

-

7

.

.

2 x UT7

2

mm

s

2 s

mm

"1

1

a

mnr2 s

"

1

460

Solutions to exercises

Solutions to exercises

The expected variation of Dzn, Dcu and Da are shown schematically

461

1 1

below

D

'

D2n 0

A

a

b

d

c

B

XB

%Zn Cu

a

t

Solubility limit for a

Since Zn has a lower melting point than Cu it diffuses faster of the

LJ

.

two, and since increasing the zinc content reduces the liquidus temperature, all diffusivities can be expected to increase with increasing

0

Zn concentration. 28 .

JC

-

GB

P

U

Hb(2) r

Ml) P+7

a+(3

He

G

i M2)

Ml

A

b

c

B

462

Solutions to exercises

Solutions to exercises

463

(i) a

t

a

U C

s

B

f = 3C

GO

b

P

p

L 2

~

r

a

t b

Bulk

A

o9

B

.

05 .

Bulk composition B

The total distance the interface moves 5, can be calculated in terms of ,

(ii)

the total couple thickness, L, by writing down an equation describing the conservation of B, i.e.

0

3

.

2

.

a

= 5.6 x lO-2

'

.

9 |y + sj + ()[~ - s] - 0.5L

.

a

Chapter 3 29

3

.

.

1

Considering only nearest neighbours, if a surface atom has B 'broken* bonds, it will have an excess energy of B. e/2, where £ is the bond energy.

a

For f.c.c. crystals, each atom has twelve nearest neighbours in the f = 0

limation and NLl Avogadro's number (no. of atoms per mole). The surface energy per surface atom is therefore given by

P

A

bulk, so that 8 = Ls/6/Va, where Ls is the molar latent heat of sub-

XB

7SV

B

B Ls = -./Va

per surface atom

.

If each surface atom is associated with a surface area A, the surface

!

a

o

1

energy is B >0

iV

12A Na

per unit area

Q

A can be calculated in terms of the lattice parameter a\

09 .

Solutions to exercises

464

Solutions to exercises

Each surface atom is connected to two nearest neighbours in the

*

and £ sv

{220} plane. Therefore it must be connected to ten others out of the plane. Since the atoms are symmetrically disposed about the {220}

8

(cos 9 - sin 9)

which gives | -r]

27

{200)

{111}

9 < 0

~ "

\ d9 /o.o

{hki}

,

465

2?

{220S

*

At 9 = 0 there is a cusp in the £sv - 9 curve with slopes

Ge© a

V2

a

*

A

a

V3

\'2

2

60©

a

.

3

For a two-dimensional rectangular crystal with sides of lengths /j and A and surface energies ji and 72 respectively the total surface energy is given by ,

aV2

_

\2

3

a

2

2

G = 2(/17, + /:Y2) B

3

4

0 33

Ysv

.

0

Vsv

-

58[

0 67 .

5

The equilibrium shape is given when the differential of G equals

*

zero. i.e.

dG = 2(/ldy, + y /, + /2d72 + y.d/,) = 0

rf

0 42

LA/aJ

.

Assuming that 71 and 72 are independent of length gives 0 59

a2/V;

.

[ 2W ] a

7id/, + y2dl2 = 0

a

But since the area of the crystal A = IJ2 dA = /2d/2 + hdli = 0

plane, there must be five bonds above the plane of the paper and five below (giving a total of 12) + It can be shown that in general for f c c

Giving

.

.

.

.

constant

metals

h

4

ii

For the simple cases above however. A can be calculated directly ,

from a sketch as shown ,

3

4

.

.

(a) By measuring, the misorientation 9 = 11. (b) By constructing a Burger- :ircuit around a dislocation, the

Burgers vector is found to be 1.53 mm in the photograph (i e .

32

Ew

.

i

.

e

.

£

sv

= (cos 0 + sin |0|)

~

( de /o

2fl-

(-sin

For a low-angle grain boundary, the spacing of the dislocations is given by

2a

(cos 9 + sin 9) ~

de

.

one bubble diameter).

e>0

b D == -

sin 9

9 + cos e)~ 2a

*

1.53

D

8 0 mm .

°

sin 11 =o

2a2

which is very close to the mean dislocation spacing in the boundary.

Solutions to exercises

466 3

5

.

Solutions- to exercises

Like all other natural processes, grain boundary migration always results in a reduction in total free energy.

467

(b) For nucleus growth, reduction in free energy due to annihilation of dislocations must be greater than or equal to the retarding force due to grain boundary curvature.

Equating this with the driving force across a curved boundary

Grain growth i

During the process of grain growth all grains have approximately the same, low dislocation density, which remains unchanged during the

e

.

.

1.96 x l(f

r

grain growth process.

'

.

.

27 r

1

Grain "boundaries move towards their centres of curvature in this

case, because atoms tend to migrate across the boundaries in the opposite direction (from the high pressure side to the low pressure side), in order to reduce their free energy, or chemical potential. The process also results in a reduction of the total number of grains by the growth of large grains at the expense of smaller ones. The net result is a reduction in the total grain boundary area and total grain boundary energy.

27

.

96 x 106

Thus the smallest diameter = 1.0 fim

3

7

.

From the phase diagrams, the limit of solid solubility of Fe in Al is 04 wt% Fe, whereas that of Mg in Al is 17.4 wt% Mg. If one element is able to dissolve another only to a small degree, the extent of grain boundary enrichment will be large. (See for example Fig. 3 28, p. 138.) Thus grain boundary enrichment of Fe in dilute Al-Fe alloys would be expected to be greater than that of Mg in Al-Mg alloys. 0

.

.

Recrystailization

In this case, grain boundary energy is insignificant in comparison with the difference in dislocation energy density between recrystallized grain and surrounding deformed matrix. The small increase in total grain boundary energy that accompanies growth of a recrystailization nucleus is more than compensated for by the reduction in total dislocation energy. The boundaries of recrystailization nuclei can therefore migrate away from their centres of curvature.

3

.

6

3

8

.

See Fig. 3.35, p. 145.

If d(t < dp, then in general the dislocation spacing (D) will span n atom planes in the P phase and (n + 1) planes in the a phase, i.e.

D = nd$ = (n + 1) d, From the definition of 8 we have

(-ve)

(a) The pulling force acting on the boundary is equivalent to the free energy difference per unit volume of material.

dv = (l + S)da

2

If the dislocations have an energy of j- J m

and the

Substitution into the first equation gives

dislocation density is 1016 m~2, then the free energy per unit

n(l + 5) da = (n + 1) d(x

volume, G, is given by

1

10-x(0.28x , »x

G,10

4

Y

i e .

1%MJ|n.3

,

Thus the pulling force per unit area of boundary is 1.96 MN m~2.

.

n = o

and D = 5

"

7

Solutions to exercises

468 39

Solutions to exercises 1 26 - 1.43

2*2

.

469

.

Hence zone misfit

x 100%

1 43 .

2*1 -

71

11.89%.

When the misfit is less than 5%

strain energy effects are less important than interfacial energy effects and spherical zones minimize the total free energy. However when the misfit is greater than 5% the small increase in interfacial energy caused by choosing a disc shape is more than compensated by the reduction in coherency strain ,

72

,

72

,

energy.

The edges of the plate exert a force on the periphery of the broad

Thus the zones in Al-Fe alloys would be expected to be discshaped.

face equal to 72 . 4 . 2t2

This force acts over an area equal to (2X2)2

i

3

.

AP

Y2 ' 4 '

12

272

2 =

(2v2)2

x2

Assuming that the matrix is elastically isotropic, that both Al and Mg atoms have equal elastic moduli, and taking a value of 1/3 for Poisson's ratio, the total elastic strain energy ACS is given by:

AGS - 4 jiS2!/

|i = shear modulus of matrix; 5 = unconstrained misfit; V = volume of an Al atom.

\ 2x1 1 60 - 1.43 .

5

0 119 .

1 43 .

2x 2

V

= 4/3 * tt (1 43 x 10~!0)3 = 1.225 x lO"29 m3 \xAl = 25 GPa = 25 x lO9 Nm"2 .

AGS = 4 x 25 x 109 x (0.119)2 x 1.225 x 10~29 J atom = 1

2 -72 '2xx + 2-7, - 2x2

1

'2 2

_

, Yi h -

1 eV

3

.

3

11

Atomic radius of Al = 1.43 A Atomic radius of Fe = 1.26 A

1

1 .

.

6 x 10~19 J, thus

735 x lO"20

1

6 x 10~19

eV atom

-

1

.

1 eV atom"1

.

It is also implicitly assumed that individual Mg atoms are separated by large distances, so that each atom can be considered in isolation,

X2

See Section 3.4.1 (p. 143).

-

1000

0

.

10

.

kJ mol

1

AGS

(see also Exercise 3 3)

Xi

735 x HT20 x 6.023 x 1023

10.5 kJ m-r' Y2

From the Wulff theorem (p. 115) for an equilibrium plate shape: - =

.

AG s

The area of each edge is 2xx . 2x2 AF =

735 x HT20 J atom"1

.

In 1 mol there are 6.023 x lO23 atoms

The periphery of each edge is acted on by a force of magnitude

2y2-2xl +

i

-

i

e

.

.

dilute solutions.

The use of Equation 3.39 is also based on the assumption that the matrix surrounding a single atom is a continuum. 3

13

.

See Section 3.4.4. (p. 160).

3

Solutions to exercises

Solutions to exercises

470

crystal itions A are shifted to B the atoms above the glide plane in pos

14 When a Shockley partial dislocation passes through an

.

f

c c

.

.

3 16 .

.

positions, B into C positions, etc. i

e

.

.

A

B

A

v

A

B

C

A

C B

C

C

X

A

A

A

C

B

B

A

1

1

Plane B

B

B

B

A

A

A

A

A

C

C

C

C

C

B

B

B

B

B

1

8

B

B

B

B

B

C

C

C

C

C

C

A

A

A

A

A

A

B

B

B

B

- A-

-

Fee

/////////

B , '

C

C

B

B

B

A

A

C

C

c

C -

B

B

B

A

A

A

A

C

C

C

C

C

B

B

B

B

B

B

A

A

A

A

A

A

Twin

C

C

C

C

C

B

Fee

B

B

B

B

A

471

If a single atom in crystal I attempts to jump into a crystal II position a ring of dislocation and an unstable A upon A situation results. A Shockley partial dislocation in every {111} slip plane creates a glissile interface between two twinned crystals:

The above series of diagrams shows the twinning process. 3 15 .

A A

~

-

t

h

B "

-

'

C

i

A i

t

T

B

,D

I

E

F

C i

Let the interface CD move with a velocity v perpendicular to the

A

I

C

J. i

B

JL I

J

B

r

Consider unit area of interface perpendicular to ooth BC and CD. Mass flow perpendicular to BC = u x h. Mass flow perpendicular to CD = v x /. From the conservation of mass: u x h = v x /. u x h -

11

C

i

interface.

v

A .

I

C

B !

!

A -

X

A

.

Note, however, that as a result of the shape change produced by the transformation large coherency stresses will be associated with the interface (see Fig. 3.62a). Similar coherency stresses will arise as a result of the f.c.c./h.c.p. interface in Fig. 3.61. Strictly speaking. Fig. 3.60 is an incorrect representation of the stacking sequence that results from the passage of the partial dislocations. In layer 10, for example, there will not be a sudden change across the extra half-plane of A to B or B to C, but '

'

472

Solutions to exercises

Solutions to exercises

rather a gradual change associated with long range strain fields in

3 20 .

both crystals. 3

.

17

473

r2.

Solid/vapour interfaces and solid/liquid interfaces in non-metals are faceted and therefore migrate by ledge mechanisms .

Solid/liquid interfaces in metals are diffuse and migration occurs by random atom jumps. 3

.

18

See-Section 3.3.4 p. 130. ,

3 19 .

X

X,p

Suppose the alloy had reached equilibrium at a temperature T[ and consists of long plate-like precipitates. The bulk alloy composition is Xih the equilibrium concentrations at 7, and 7 are Xx and respectively, where 7% is the temperature to which the alloy is heated.

A

B

X

(ii) Interface control

(i) Diffusion control

From equations 1.41 and 1 43 we can write .

'

hb

= GB + RT In Yi*t

i

Hfc = Gb + RT\nycXc r

.

Aii =

a

- (is = RTlnycXc

/ = 0

a

X2Xo\

For ideal solutions: y; = y = 1 For dilute solutions (X « 1): y, = yc = constant (Henry's Law)

L

c

Distance

}

f = f1

such that in both cases X2

Aji'b = RT In

X

Xc

This can also be written

A k = RTln If the supersaturation is small

X

o

,

i.e.

(X,- - Xe) « Xe, then = RT X0

f = t3 > '2

Solutions to exercises

Solutions to exercises

474

475

Substituting Q - 1.6 x 10~29 m3 and y - 0.177 Jm : gives

(iii) Mixed control: similar to diffusion control except the interface

~

concentration in the a-matrix will be less than A , the equilibrium concentration at T2-

'

AGr - (5.435 x l(r20)nl 3 For 1 mm3, n{i - 6.25 x 1019 atoms = 10 atoms, nr - 9 x 1013 clusters mm 3: and when nQ = 60 atoms nx - 3 clusters mm-3; when n = 100 atoms nr = 4 x 10"8 clusters mm 3: or. alternatively. 1 cluster in 2.5 x 107 mm3 (251)

Therefore when n

Chapter 4

"

c

,

4

.

1

*

-47ir3 3

"

AGV + 47cr2YsL

c

,

.

Differentiating this equation with respect to r. dAGr

4

3

.

As the undercooling (AT) is increased there is an increasing contribution from AG in the equation ,

V

47i/-::*AGv + 87rr/SL

-

dr

4

AGr = --7cr3AG + 47rr2YS[ v

At the critical radius, / *, this expression is equal to zero "

0

= -4Trr

*2

'AG

v

whereas the interfacial energy is independent of AT Consequently

+ 8nr*ySL

.

for a given r. AG decreases with increasing AT, and the 'maximum' r

r

*

=

2ysl

cluster size increases somewhat.

AGy

4

.

4

From Equation 4

In order to calculate the critical value of AG, AG* at this radius,

13

.

oCoeXp{

the value of r* is substituted into the original equation where

ILlkT

7 = 7m- bJ

From which the following values are obtained: 167rysL 3(AGV)2

=

4

2

.

AT K

From Equation 4.10. at the equilibrium melting temperature Tm

iVh

om

m"3 s""1

180

07

200

8 x 106 1 x 1012

7 x lO"7

.

220

iVhom cm-3 s-1

8

1 x 10A

At the equilibrium melting temperature AGV = 0. so that Equation

Note the large change in N over the small temperature range (see

4 4 becomes

Fig. 4.6).

.

AGr(r= Tm) = 4Ttr2ySL For a cluster containing nQ atoms with an atomic volume Q we have

4

5

.

AG* = i-F*-AG

For homogeneous nucleation it has been shown (see 4.1) that ,

47173 3

ncQ.

v

2TSL

* _

AGV

Therefore the expression for AGr becomes

Thus for a spherical nucleus

'

3QAic\2/3-r

V

4n7-*3

327tysL 3(AGV)3

Solutions to exercises

476 1

AG* - --

Solutions to exercises

16Try3SL

-AGv

All

Heterogeneous nucleus

2

3AG;

77777

L

This is identical to that derived in 4.1, and so the equation holds for homogeneous nucleation. For heterogeneous nucleation, it can be shown that

/

s

Size of

r

> = 2ysL

homogeneous

AGV

nucleus at same AT

The volume of a spherical cap on a flat mould surface is given by V

(same r')

\

(2 + cose)(l - cos9)2

\

,

nr

3

Thus v*

=

/2ysl\3 (2 + cos6)(l - cos e): \AGv/ 3

where 0 is the

'

wetting angle.

Substituting into the given equation

4 yIl AG* = - V*-AGV = -k-(2 + cose)(l - cose) i

2

The wetting angle between nucleus and mould wall (8) is fixed by the balance of surface tension forces (Equation 4.14). The activation energy barrier (AGhet) depends on the shape of the nucleus as determined by the angles a and 0. From Equation 4.23, for a given undercooling (AT), AGV and r* are constant, such that the following equalities apply .

7

j AGy

S

Writing the normal free energy equation for heterogeneous nucleation in terms of the wetting angle 0 and the cap radius r

AGhet = j- AGv + 4nr2YSLj

(2 + cos0)(l - cos0)2

i

e

.

.

het

hom

AGh*et

S

_ ~

AGh*hum

4

But from Equations 4.19 and 4.17 we have

AGh, he! AGh* horn

Volume of the heterogeneous nucleus Volume of a sphere with the same nucleus/ liquid interfacial radius

It can be seen that the shape factor {S) will decrease as a decreases and on cooling below Tm the critical value of AG will be reached at progressively lower values of AT, i.e. nucleation becomes easier. When a =S 90 - 0, S = 0 and there is no nucleation energy barrier. (It can be seen that a = 90 - 0 gives a planar solid/liquid interface, i e r = x even for a negligibly small nucleus volume.) Once nucleation has occurred, the nucleus can grow until it reaches the edge of the conical crevice. However, further growth into the liquid requires the solid/liquid interface radius to pass through a minimum of R (the maximum radius of the cone). This requires an undercooling given by ,

*

167ty|L (2 + cos0)(l - cos0)2

AG =W

4

which is identical to that obtained using AG* = tV *AG

v

46

See Section 4.1.3.

4

Consider a cone-shaped crevice with semiangle a as shown below:

.

t

'

>

-

.

7

.

.

<

2ysl R

h

LbT m

r

i

.

I i

e

.

Ar

2YsiTm RL

478 4

.

8

Solutions to exercises

Solutions to exercises

For conical crevices with a < 90 - 0 the solid/liquid interface can maintain a negative radius of curvature which stabilizes the solid above the equilibrium melting temperature (rm):

1 m

479

2

V

V 6

L

(1 - 6)

S

S

1 m

L

7777,

77777 Thus

G(II) = Cs(l - 8) + CL8 4- ySL + yLV (8 > 0) G(II) - Gs + 5(GL - Gs) + YSL + 7Lv (8 > 0)

S

LAT

At an undercooling AT below 7m, GL - Gs = -*

m

LAT

i

e

.

.

G(U) - Gs + --5 + 7sl + 7lv / rm

As the temperature is raised above rm the solid will melt back into the crevice to maintain equilibrium with a radius given by

LAT

or

G(II) = G(I) - Ay + - --5 Tm .

r

where Ay =

LAT

where { - AT) is now the superheat above r 4

4

.

9 10

.

- 7SL - 7Lv

This is shown in the figure below: m

.

If the situation described above is realized in practice it would explain the observed phenomena.

G(li

(a) The values of the three interfacial energies are as follows:

Gradient

LA 7

el

Solid-liquid = 0.132 J m"2; Liquid-vapour = 1.128 J m"2; Solid-vapour = 1.400 J mT2.

Ay

Thus the sum of the solid-liquid and liquid-vapour interfacial free energies is less than the solid-vapour free energy and there is no increase of free energy in the early stages of melting Therefore, it would be expected that a thin layer of liquid should form on the surface below the melting point because the dif,

.

,

ference in free energies could be used to convert solid into liquid (b) Imagine the system I below. The free energy of this system is

0

do

6 max

b

.

given by:

G(I) = Gs + y

sv

System II contains a liquid layer of thickness 5 and solid reduced to a height (1 - 8). (The difference in molar volume between liquid and solid has been ignored.)

Note that as 5 -> 0. G(II) G(I), which means that in practice ySL + Ylv ~> Ysv a result of an interaction between the SiL and LIV interfaces as they approach to within atomic dimensions of each other.

The optimum liquid layer thickness (5o) will be that giving a minimum free energy as shown. We cannot calculate this value

Solutions to exercises

Solutions to exercises

480

without a knowledge of the above interaction. However, it is reasonable to assume that the minimum will occur at a separation of a few atom diameters, provided 5max in the above diagram is at least a few atom diameters. 5max is defined by G(I) = G(II)

1

i

.

e

.

G(I) - Ay +

G(I)

7m

-5

/

-Ke"

undercoolings we have max

nuclei m 2 s

oc exp LAT

Alternatively, AT

where k is approximately constant. Each time a cap is nucleated, it should grow rapidly across the interface to advance a distance h. It seems reasonable to suppose therefore that the growth rate will be proportional to N, i.e.

L5 max

If Smax = 10 nm (say), then AT = 16 K. It seems therefore that surface melting is theoretically possible a few degrees below 7m. 11

1

Ayrm

5 max

.

\

But. AG 3c AT,, the undercooling at the interface, so for small

,

4

481

v x exp

AT;

(a) Repeated surface nucleation (see Section 4.2.2, p. 198). (b) Very roughly. Equation 4.28 can be seen to be reasonable as

I

-

follows:

i;

Firstly, it is reasonable to suppose that the distance between successive turns of the spiral (L) will be linearly related to the

r

\

*

ii

r

h

L

t

S

minimum radius at the centre (r ) Thus we have .

L

r* * AT~]

Secondly, for small undercoolings, the lateral velocity of the steps (w) should be proportional to the driving force, which in turn is proportional to AT,

Suppose the edge of the cap nucleus is associated with an energy e (J m 1) Formation of such a cap will cause a free energy

u

ATj

-

.

Thus the velocity normal to the interface v is given by

change given by

uh

AG = -Kr2hAGv + lure The critical cap radius r* is given by

v = dr

= 0

e

i

.

e

.

where

4

r* =

.

12

r .

i

ATf

is the step height.

Equilibrium solidification (see Figs, 4.19 and 4.20)

MGV

From Fig. 4.19 the lever rule gives the mole fraction solid (/s) at T2

and

as

AG*

ne

2

XL - Xn

hAG V s

The rate at which caps nucleate on the surface should be proportional to exp I-I

XL - Xs

(Xs/k) - X0 (Xs/k) - Xs

kX0 s

1 - (1 - *)/s

482

Solutions to exercises

Solutions to exercises

This expression relates the composition of the solid forming at the interface at T2 to the fraction already solidified. For the case shown in Fig. 4.1°, it will be roughly as shown below:

483

No diffusion in solid perfect mixing in liquid (see Fig. 4.21). ,

Again, we have 2 ~~ T$

Xjy _

7\ - 7"}

X; - k-X

.

is now given by Equation 4.33 such that

Xo -

t t2-t3

.

i-k(i-fsy

k-»

_

Tx-Ts

(l-k)

where Tx > T2 > TE. For the phase diagram in Fig 4.21a. the following variation is therefore obtained (k - 0.47. the exact form of the curve depends on k, of course) .

0

4

The temperature of the interface (7%) as a function of the fraction solidified can be obtained using the following relationship which is apparent from Fig. 4.19 T2 - T3 r, - t3

To

X{) - Xs xi} - kXo

T3-

Substituting for Xs gives

0

This will be a curved line roughly as shown below for the case described for Fig. 4.19 (* - 0.47).

1

No diffusion in solid, no stirring in liquid (see Fig. 4.22)

t Q)

T2

Initial transient

1. (In this

No diffusion in solid, complete mixing in liquid

.

485

,

case, A- = 3.) The variation of composition along the bar can be calculated using Equation 4.33 i a)

e

.

.

Xs = kX(>(l -/s) .2

X

c

a

d2A7

2cX

- = -H

dAT,, r

;

A 7,)

Differentiating this equation with respect to X. V

491

Thus

a

d Ar,,

\4X*

1_\

4> 2/

:

_

A7n / J

6

a 14;. :

\

l(v. : ' _

is positive

dA

Hence undercooling is a minimum when /. = 2/.''

Substituting this value into the equation of the second differential d2v

2c

6ca*

4

.

20

The total change in molar free energy when liquid transforms into lamella a + (3 with a spacing a is given by Equation 4 37. i.e. .

_

dx3

=

>7 " "x12c

AG(a) = -AG( ) +

6ca*

A

4

16a * -c

8/,

Thus when /. = 2a*. the growth rate is a maximum.

The equilibrium eutectic temperature TE is defined bv X = * and G(x) =0. We can define a metastable equilibrium eutectic temperature at (7"E - ArE) such that at this temperature there is no change in free energy when L a + (3 with a spacing a. i.e. at TE - ATE ,

(ii) When the growth rate is fixed, the original equation may be

AG{X) = 0.

Also from Equation 4.38 at an undercooling of ArE

rewritten as follows

AHATe

AC( )

a = A To . r where

T,

Finally, then, combining these equations gives v

2ynfiVmTE

a

Li

Thus:

a

1

AT,)

\

AHX AH

Substituting: Yap - 0.4 J m-". rE = 1000-K..-p = 8 x 10x J m

k2

'

-

Differentiating with respect to /. -

a

Ar5

dATi)

1

2X*

dl

X2

X

dX

10

~

a \a2

6

-

AT,

X

3

i

r2

dAT.

gives

a3 /

e

.

.

for a - 0.2 \xm. ATE = 5 K X = 1.0 [im. ATB = 1 K

v . jn

Solutions to exercises

492

Solutions to exercises

Note that if these eutectics grow at the optimum spacing of 2X* the total undercooling at the interface during growth (Ar0) will be given by Equation 4.39 such that for A. = 0.2 urn.

X* = 0.1 \im.

and X = 1.0 (im,

= 0.5 \xm.

A To = 10 K

493

Chapter 5 5 1 .

AG,, = Rrf ln (a) By direct substitution into the above equation

AT{) = 2 K

AGU = 420.3 J mor1

4 21 .

h

(b) Applying the lever rule to the system at equilibrium (X - X )

U>-d -J

Mole fraction of precipitate = j-

- = 0 08 .

(Xp - XQ)

Assuming the molar volume is independent of composition P

P

P

a

will also be the volume fraction

,

this

.

(c)

50 nm

Rod-like eutectic

Lamellar eutectic

For a lamellar eutectic the total interfacial area per unit volume of eutectic is given by: 2/X. irrespective of volume fraction of p. For the rod eutectic, considering rods of unit length, and diameter d the area of a/(3 interface per unit volume of eutectic is given by

Assuming a regular cubic array with a particle spacing of 50 nm the number of particles per cubic metre of alloy =

.

1

.

nd

X

(50 x w y

2nd

3/2

X2V3

For the rod eutectic to have the minimum interfacial energy, then 2nd

X

2

< T' X '

.

e.

d

x3

A

71

21

Let all the particles be of equal volume and spherical in shape with a radius r. Then the total volume of particles in 1 m3 of alloy =

7 < ,

8 *10

_

8 x 1021 x 7tr3

d depends on the volume fraction of p. (/)

Equating this with the volume fraction of precipitate

f~ 4 /

8 x 1021 x nr = 0.08 m3

2

v3

From which / < /c = - 2n

4 22

See Sections 4.4 and 4.5.

4 23

See Section 4.5.

.

.

0 28. .

r = 13.4 nm.

Thus in 1 m3 of alloy the total interfacial area = 8 x 1021 x 47U-2 = 1.8 x 107 m2

I:

Solutions to exercises

Solutions to exercises

494

(d) If Tap = 200 mJ m

2

495

From Equation ] .40

"

GB GB Mb GA HX = GA

total interfacial energy = 200 x 1.8 x 107 mJ m 3 alloy = 3 6 x 106 J m"3 alloy "

.

= 36 J mol

"

1

+ + + +

RT}nXq + D(l - Z,,)2 KTlnXe + na - Xe)2 RT\n(l - X0) + QJCo2 RT\n (1 - Xe) + nx2

36 9%

(e) The fraction remaining as interfacial energy

Combining the above equations gives

420.3

ir

-

(f) When the precipitate spacing is 1 im; 1

No of particles per m3

(1 x 10~6)3

53 .

(a) AGn

RTln - per mole of precipitate

1 x 1018m"3

Thus for a precipitate with X{)

Using the same method as in (c), the particle radius is found to Thus in 1 m3 of alloy the total interfacial area =

(b) Assuming that the nucleus is spherical with a radius

7

1 x 10i8 x 4n x (2.67 x 10 )2 = 8

orientation, the total free energy change associated with nucieation may be defined as

96 x 105 m2

Total interfacial energy =1.8 x 10 J irT3 alloy .

and ignor-

ing strain energy effects and the variation of y with interface

.

= 1

1 andZe - 0.02 at 600 K:

.

AGn = 8.0 kJ mol

be 267 nm.

~

0

4

AG = --7W3-ACV + 4nr2y

8 J mol"1

Fraction remaining as interfacial energy = 0.4%

where AGv is the free energy released per unit volume. Differentiation of this equation yields the critical radius r*

52 .

m

r

G

0 50 nm .

AG n

a

(c) The mean precipitate radius for a particle spacing of 50 nm was calculated as 13.4 nm = 27 r*. For a 1 jam dispersion the precipitate radius, 267 nm = 534 r*.

Go

03

E3

54 .

Mi

Gr

CQ

a

O

P A

X

,

B

US

AG0 = G0 - Gf

G0 = XQv.l + (1 - Z0)nO Gf = Zon| + (1 - x0) \i%

Xe

Xo

496

Solutions to exercises

Solutions to exercises

From Equation 1.68

n£ = GB + = Gb

i

5

f)

.

497

(a)

rinyoZo

V

+ RT\njcXc

where y0 and ye are the activity coefficients for alloy compositions X{) and Xe respectively

I

AGn

0

e

Co

dti YO O yeXe

For ideal solutions y0 = ye = 1

For dilute solutions yo = Ye = constant (Henry's Law) In both cases i

AG n

RT\n

t

t

(b) 5

5

.

(a) Consider equilibrium of forces at the edge of the precipitate:

a -

X

full

L

v

8 x

Using the simplified approach, above, the carbon concentration gradient in the austenite, - may be expressed as due

For unit area of interface Yaa

L

= 2ya(3COSe

For unit area of interface to advance a distance dr, a volume of

9 = cos"17""

= 53 1

material 1. dx must be converted from y containing CY to a containing Ca moles of carbon per unit volume, i.e. (Cv - Ca)dbc

°

.

moles of carbon must be rejected by diffusion through the y. . . The flux of carbon through unit area in time dt is given by V

(b) The shape factor 5(0) is defined as 5(8) = ~(2 + cos8)(l - cos0)2 = 0 208 .

D{dCtdx) dt, where D is the diffusion coefficient. Equating the two expressions gives

(Cy - Ca) dx

dt

Solutions to exercises

Solutions to exercises

498

* =

dt

d(\dx )

(D ~ ?v

l

\;

L

)*s

-

(C7 - Ca) where / is the volume fraction of austenite

Thus using the simple concentration profile obtained earlier

dr

,

:

X

'

.

L

.

3 D *max = (1 - f}.;>),

UCy - Co) 2

2(C0 - Ca)x

D

(Cy - Co)

(The same answer is obtained for any polyhedron.) Approximately, f-, is given by

Substituting for L in the rate equation

D(C, - Cp)2

dx

.

J (CY-Cn:)

The width of the diffusion zone L may be found by noting that conservation of solute requires the two shaded areas in the diagram to be equal (Co - Ca)x

499

_

0

XfL

Xy

- Xa

"

dr

2(Co - Ca)(CY - Ca)x

Y

Assuming that the molar volume is constant, the concentrations may be replaced by mole fractions (X = CV ). Integration of the rate equation gives the half-thickness of the boundary

In the present case /7 *max

slabs as X

= 0

.

43. such that for D - 300 jim;

= 36 5 fim .

This value will be approached more slowly than predicted by the (Xy-X

V

parabolic equation, as shown schematically in the diagram below.

Dt)

(xt) - xay*{xy - xa)™ 40-

(c) The mole fractions in the above equation can be replaced approximately by weight percentages. For ferrite precipitation from austenite in an Fe-0.15 wt% C alloy at 800 0C we have ,

"1

O? .

xn

0 15; .

giving x

1

.

Parabolic equation

\ Real variation (schematic) 20-

.

3 x HT12

% 304

i

:

0 02;

Dl

Maximum half-thickness

"

*

m

49 x 10""6

"

t

s

- I

1/2

10-

(d) The previous derivation of x(t) only applies for short times. At longer times the diffusion fields of adjacent slabs begin to overlap reducing the growth rate. The lever rule can be used to calculate the maximum half-thickness that is approached for long times. Assume the grains are spherical with diameter D. When the transformation is complete the half-thickness of the ferritic slabs Omax) is given by

0 0

100

200

300

400

500

600

700

Time (S)

The exact variation would require a more exact solution to the

Solutions to exercises

Solutions to exercises

500

diffusion problem. However, the approximate treatment leading to the parabolic equation should be applicable for short times. 57

501

59 .

v

.

E

a

0)

u

X

Nucleation or growth rate

Consider unit area of interface perpendicular to the diagram: Civilian transformations that are induced by an increase in temperature show increasing nucleation and growth rates with increasing superheat above the equilibrium temperature (Tc). This is because both driving force and atom mobility (diffusivity) increase with increasing AT.

Mass flow in the direction of u = u x h: Mass flow in the direction of v"= v x X. From the conservation of mass: u x h = v x X u x h v

5

.

8

5

.

10

(a)

G = XAGA + XBGB + QXAXB + RKXA\nXA + XBlnXB) GA = GB = 0 gives: G = QXaXq + RT(XA\nXA + XBlnXB)

/= 1 - exp(-tfO

dG = ClXAdXB + nXBdXA

At short times this equation becomes

+ RT[dXA + dXB + \nXAdXA + \nXBdXB}

n

/= Kt

but

(a) Pearlitic nodules grow with a constant velocity, v. The voiume fraction transformed after a short time t is given by

XA + XB - 1 -

.

d A + dXB = 0 dG

/4tcv3\ 3

47t(v03

/

.

_

d e

-

b) + RT(\nXB - \nXA)

d2G

e

.

= Q( A

. -

2Q + RT

K =

-r 3 3d

.

n = 3

{ + A) xB

dZg d2G

RT

dXi

XAX]B

-

2Q

(b) For short times, slabs growing in from the cube walls will give

(b) This system has a symmetrical miscibilitv gap with a maximum at . /6v\

i2-vt 6d-

f i e .

d3

_

Ui

=

XA = XB = 0.5 for which t

d2G ART - in

.

6v

K = a

.

n = 1.

It can be seen that as T increases

d2G

d

changes from negative to

Solutions to exercises

502

Solutions to exercises

positive values. The maximum of the solubilitv gap (T = Tc)

(d) The locus of the chemical spinodal is given by

d2G

corresponds to

503

0

dA-B .

2R

e

.

RT

2O = 0

-

dG

A A A'b

(c) Equating -to zero in the equations gives dX B

T

= 4

-

Q(XA - XB) + RT(\nXQ - \nXA)

0

Putting Q = 2RTC gives T _

B(1

b)

-

This is also shown in the figure.

2(1 - 2XB) 5 11 .

.

G o + AA ) = G(Xn) + - (AX) +

-T1 + . . .

G(X{) - AX) = G(X,) + (-AX) +

This equation can be used to plot the coordinates of the miscibility gap as shown below:

!-p + . . .

Total free energy of an alloy with parts of composition (Xt) + AX) and (Xu - AX) is given by -

G(XI) + AX)

,

G(Xn - OX)

Miscibility

lnr,Y