HW 8

Problem 10.2

a. To Find:

(a) The expression for total free energy change for a cubic nucleus, critical cube edge length,

a*, and G*

(b) Compare G* for a cube with that for a sphere

b. Given:

Nucleus is cubic; edge length of the cubic nucleus = ‘a’ units

c. Assumptions:

1. Nucleation is homogeneous/no impurities are present in the material.

2. We are asked to compare the G* for a cubic nucleus with the G* for spherical nucleus of

the same material. (This allows us to use the same values of Gv and for both cases.)

d. Solution:

(a)

Step 1:

Volume of a cubic nucleus of edge length ‘a’ = a3

Total surface area a cubic nucleus of edge length ‘a’ = 6a2 (since there are six faces each of

which has an area of a2).

Step 2:

Equation 10.1 for the case of the spherical nucleus of radius ‘r’ can be re-written as:

G = [(Volume of spherical nucleus (of radius r)) * Gv] + [(Surface area of spherical nucleus

(of radius r))*]

Analogously, for the case of the cubic nucleus of edge length ‘a’:

G = [(Volume of cubic nucleus (of edge length a)) * Gv] + [(Surface area of cubic nucleus (of

edge length a))*]

G = a3Gv 6a2 …(1)

Step 3:

Differentiating this expression with respect to a:

da

ad

da

Gad

da

Gd v )()( 23 6=

3a2Gv 12a

Step 4:

To calculate a*, we need to set this expression equal to zero.

0=da

Gd =>

3a2Gv 12a 0

Solving for a (= a*) :

a * = 4

Gv …(2)

Step 5:

Substituting the value of a* from equation (2) for ‘a’ in equation (1) gives the expression for

G* :

G * = (a*)3Gv 6(a*)2

4

Gv

3

Gv 6 4

Gv

2

*G

32 3

(Gv)2 …(3)

(b) From equation 10.3, for a spherical nucleus of radius ‘r’ :

*G2

3

)3(

16

vG

…(4)

Comparing (3) and (4), G* is greater for a cube than for a sphere.

This is because the surface-to-volume ratio for a cube is greater than that for a sphere.

(a) ; ; G*

(b) G*cubic nucleus >G*spherical nucleus since the surface-to-volume ratio for a cube is

greater than that for a sphere.

Problem 10.4

a. To Find:

(a) r*and G* for Fe.

(b) Number of atoms in a nucleus of critical size.

b. Given:

Heat of fusion, Hf = -1.85 X 109 J/m

3

= 0.204 J/m2

T= 295 0C = 295 K (please see ‘Note on T’ on Pages 3/ 4 to see why this is so)

Lattice parameter of pure Fe at melting temperature,’a’ = 0.292 nm

c. Assumptions:

(i) Homogeneous nucleation

(ii) Spherical nucleus

(iii) No other element is present

d. Solution:

(a)

Calculation of r*:

From equation 10.6 :

r * 2Tm

H f

1

Tm T

…(1)

This expression may be re-written as

TH

Tr

f

m 12*

…(1a)

NOTE ON T:

We note that the undercooling/supercooling/T = Tm – T

Tm is the equilibrium melting temperature.

T is the actual temperature at which the solidificaltion takes place.

From what we know, all temperatures in equations (1) and (2) are in Kelvin.

How about the T values in equations (1a) and (2a) ?

Table 10.1 lists the T values in degrees C. Do we need to add 273 to this value to convert it

to Kelvin???? The answer to that is NO!!!!

This is because T is the difference between Tm and T.

If both Tm and T are in Kelvin, we get a certain value of T.

If both Tm and T are in 0C, we get the same value of T that we got above!

Here’s an illustration of this:

Tm = 1538 0C for Fe. Let us suppose, that the solidification temperature, T = 1243

0C

If we just use the temperature values in 0C, T = 1538 – 1243 = 295

0C

Let’s now try calculating T when the values of Tm and T are in Kelvin.

Tm = (1538 + 273 ) K; T = (1243+273) K

T = (1538+273) – (1243+273) = 295 K

Hence, we can conclude that T (0C) = T (K)

Tm = 1538 0C = (1538+273) K = 1811 K

T = 295 K

r*

(2)(0.204 J /m2) (1538 273 K)

1.85 109 J /m3

1

295 K

=

1.35 109m 1.35 nm

Calculation of G*

Similarly, equation 10.7:

G * 16 3Tm

2

3H f2

1

(Tm T)2 …(2)

may be re-written as 22

23

)(

1

3

16*

TH

TG

f

m

…(2a)

G*

(16)()(0.204 J /m2)

3(1538 273 K)2

(3)(1.85 109 J /m3)2

1

(295 K)2

= 1.56 X 10

-18 J

(b)

The critical nucleus contains several unit cells of Fe.

Number of unit cells in nucleus of critical size = Volume of nucleus / Volume of unit cell

Volume of spherical nucleus of radius r = 4/3 * π * r3

Volume of unit cell of lattice parameter/edge length ‘a’ = a3

Number of unit cells in the critical nucleus = 4/3 * π * (1.35 X 10-9

)3

/ (0.292 X 10-9

)3

Number of unit cells in the critical nucleus = 414

Upon solidification, liquid Fe (L) transforms to -Fe which is BCC.

Hence, there are 2 atoms per unit cell

Number of atoms in the critical nucleus = Number of atoms per unit cell * Number of

unit cells = 2 * 414 = 828 atoms.

(a) r* = 1.35 nm

G* = 1.56 X 10 -18

J

(b) 828 atoms

Problem 10.5a

a. To Find:

r*, n* for T = 200 K and T = 300 K

b. Given:

1. Problem concerning the solidification of Fe

2. n* = 106 when T = 295 K (for explanation, see ‘Notes on T ‘ on Pages 3/4)

3. Heat of fusion, Hf = -1.85 X 109 J/m

3 = 0.204 J/m

2 (From Prob. 10.4)

c. Assumptions:

(i) Homogeneous nucleation

(ii) Spherical nucleus

(iii) No other element is present

d. Solution:

Calculation of r* at given undercoolings/supercoolings:

The procedure for this is exactly the same as the procedure to calculate r* in Prob. 10.4.

From equation 10.6 :

r * 2Tm

H f

1

Tm T

…(1)

This expression may be re-written as

TH

Tr

f

m 12*

…(1a)

At T = 200 K, from equation (1a) :

r*=r*200

(2)(0.204 J /m2) (1538 273 K)

1.85 10 9 J /m3

1

200 K

= 2.00 10-9 m = 2.00 nm

At T = 300 K, from equation (1a) :

r* =

r300*

(2)(0.204 J /m2) (1538 273 K)

1.85 10 9 J /m3

1

300 K

= 1.33 10-9 m = 1.33 nm

Calculation of n* at given undercoolings/supercoolings:

Equation 10.8 is used to determine the number of stable nuclei:

n* = K1 exp(-G*/kT) …(2)

Step1: Calculation of temperature T, given the undercooling.

T = Tm – T => T = Tm - T

Tm (K) = 1538 + 273 = 1811 K

In Problem 4, T = 295 0C => T = 295 K {We know that T (

0C) = T (K) …(see ‘A Note on

T’ on Pages 3 /4 for explanation)}=> T = 1811 – 295 = 1516 K

In this problem, when T = 200 K, T = 1811 – 200 = 1611 K

When T = 300 K, T = 1811 – 300 = 1511 K

Step 2: Calculation of constant K1

Re-arranging equation 10.8 , we have

K1 n *

exp G *

kT

…(2a)

n* = 106 when T = 295 K, T = 1516 K. From Problem 10.4, G* = 1.56 X 10

-18 J at this

undercooling. Substituting these values in (2a) allows us to determine K1.

K1

)1516(/1038.1

1056.1exp

/10

)( 23

18

36

KatomJ

J

mnuclei= 2.42 X 10

38 nuclei/m

3 …(3)

Step 3: Calculation of G* at the given undercoolings/supercoolings

Equation 10.7:

G * 16 3Tm

2

3H f2

1

(Tm T)2 …(2)

may be re-written as 22

23

)(

1

3

16*

TH

TG

f

m

…(2a)

From equation (2a) , at T = 200 K:

G* = G*200

(16)()(0.204 J /m2)3 (1538 273 K)2

(3)(1.85 109 J /m3)2

1

(200 K)2

= 3.41 10-18 J

From equation (2a) , at T = 300 K:

G* =

G300*

(16)()(0.204 J /m2)3 (1538 273 K)2

(3)(1.85 109 J /m3)2

1

(300 K)2

= 1.51 10-18 J

Step 4: Determining n*

A) At 200K

T = 1611 K; G* = G*200 = 3.41 10-18 J

n*200 = K1 exp(-G*200 / kT) = 2.42 X 1038

exp (-3.41 10-18 / (1.38 X 10-23

* 1611))

n*200 = 5.89 10-29

stable nuclei

B) At 300K

T = 1511 K; G* = G*300 = 1.51 10-18 J

n*300 = K1 exp(-G*300 / kT) = 2.42 X 1038

exp (-1.51 10-18 / (1.38 X 10-23

* 1511))

n*300 = 8.59 106 stable nuclei

[Note: The ‘200’ and ‘300’ in the subscripts of r*, G* and n* denote the undercooling]

r*200 = 2.00 nm

r*300 = 1.33 nm

n*200 = 5.89 10-29

stable nuclei

n*300 = 8.59 106 stable nuclei

Problem 10.9

a. To Find:

Total time required for 95% transformation of austenite to pearlite.

b. Given:

1. The kinetics of this transformation can be described by the Avrami relationship.

2.

y =Fraction Transformed t=Time (s)

0.2 12.6

0.8 28.2

c. Assumptions:

(i) The kinetics of this transformation can be truly described by the Avrami relationship.

(ii) The data in the table is accurate

d. Solution:

The Avrami relationship, i.e., Equation 10.17 :

y = 1 – exp (-ktn)

This may be re-written as :

1 – y = exp (-ktn)

Taking the natural log on both sides:

ln(1-y) = -ktn

...(1)

Using the 2 sets of data given in the table, it is possible to evaluate the two constants k and n.

For y = 0.2, t = 12.6s, equation (1) becomes:

ln (1 – 0.2) = - k (12.6)n

ln(0.8) = - k (12.6)n

...(2)

For y = 0.8, t = 28.2s, equation (1) becomes:

ln (1-0.8) = - k (28.2)n

ln(0.2) = - k (28.2)n

…(3)

Dividing equation (2) by (3):

ln(0.8) / ln (0.2) = [-k (12.6)n

]/ [- k (28.2)n

]

0.1386 = (12.6/28.2)n

0.1386 = (0.4468)n

Taking natural log (log to the base 10 is also ok, the idea is to sole for n) on both sides:

ln (0.1386) = ln (0.4468)n

ln (0.1386) = n ln (0.4468)

n = ln (0.1386) / ln (0.4468) = 2.45 …(4)

Substituting the value of n in equation (2) (it’s ok to substitute it in equation (3) , the idea is to

solve for k. The units of k depend on the units of time and the value of n) :

ln (0.8 ) = - k (12.6)2.45

k = 4.49 X 10-4

…(5)

Substituting the values of k, n and y=0.95 in equation (1) (or in the original form of the Avrami

equation, equation 10.7) to solve for the time for 95% (= 95/100 = 0.95) transformation of

austenite to pearlite:

ln(1-y) = -ktn

ln (1-0.95) = - 4.49 X 10-4

(t)2.45

(t)2.45

= ln(0.05) / (- 4.49 X 10-4

)

(t)2.45

= 6672 (you make take natural log or log to the base 10 on both sides or …

t = (6672)1 / 2.45

= 36.38 s

36.38 s

Problem 10.12

a. To Find:

The constants k, n for the Avrami relationship (equation 10.17) governing the recrystallization of

copper at 102 0C.

b. Given:

(i) The Avrami relationship describes the recrystallization kinetics of copper

(ii) Temperature = 102 0C

(iii) The variation of y=% recrystallization with t=time (min) is plotted in Fig. 10.11.

c. Assumptions:

(i) The kinetics of this transformation can be truly described by the Avrami relationship.

(ii) Fig.10.11 is accurate

(iii) Temperature is exactly 102 0C (so that we can read use the curve corresponding to

this temperature in Fig.10.11)

(iv) The material at hand is pure copper, i.e., no impurities/ alloying elements are present

(so that we can use Fig.10.11)

d. Solution:

If we obtain 2 sets of data from the curve corresponding to 102 0C from Fig. 10.11, this problem

becomes identical to Prob. 10.9.

In reading the points off the graph, we must remember that :

1. The time axis is scaled logarithmically.

2. The numbers noted on this axis (the time-axis) are the actual values and not the log values.

Page 135 of the text-book outlines the procedure to estimate values on the log-scale.

If there is a plot mid-way between 101 and 10

2, on the log-scale, it's value is 10

1.5 = 31.6 and not

5 X 101

.

Set 1: y = 20% = 0.2 t = 101+0.7

= 50.12 mins

(Estimating the time: the distance between 101 and 10

2 , measured out with a ruler, ≈ 2.7 cm.

The time-co-ordinate corresponding to y=0.2 is ≈1.9 cm away from 101

. Thus, it is located at a

distance that is 1.9/2.7 = 0.70 or 70% of the distance between 101 and 10

2 . Hence, t = 10

1+0.7 )

Set 2: y = 80% = 0.8 t = 100 mins

NOTE: It’s ok to use any 2 sets of data obtained from this curve. Several values of y and

corresponding values of t, as obtained from the curve for 102 0C (Fig. 10.11) are given below.

y t(mins)

10% = 0.1 10^(1+(1.6/2.7)) = 10^(1+0.59) = 10^1.59=38.90

20% = 0.2 10^(1+(1.9/2.7)) = 10^(1+0.7)=10^1.7 = 50.12

30% = 0.3 10^(1+(2.1/2.7)) = 10^(1+0.78) = 10^1.78= 60.26

40% = 0.4 10^(1+(2.2/2.7)) = 10^(1+0.81) = 10^1.81= 64.56

50% = 0.5 10^(1+(2.3/2.7)) = 10^(1+0.85) = 10^1.85= 70.79

60% = 0.6 10^(1+(2.4/2.7)) = 10^(1+0.89) = 10^1.89= 77.62

70% = 0.7 10^(1+(2.5/2.7)) = 10^(1+0.92) = 10^1.92= 83.18

80% = 0.8 10^2 = 100

90% = 0.9 10^(2+(0.1/2.7)) = 10^(1+0.04) = 10^2.04= 109.65

Table 1. y,t obtained from the curve for 102 0C (Fig. 10.11)

The Avrami relationship, i.e., Equation 10.17 :

y = 1 – exp (-ktn)

This may be re-written as :

1 – y = exp (-ktn)

Taking the natural log on both sides:

ln(1-y) = -ktn

....(1)

For y = 0.2, t = 50.12 mins, equation (1) becomes:

ln (1 – 0.2) = - k (50.12)n

ln(0.8) = - k (50.12)n

...(2)

For y = 0.8, t = 100 mins, equation (1) becomes:

ln (1-0.8) = - k (100)n

ln(0.2) = - k (100)n

…(3)

Dividing equation (2) by (3):

ln(0.8) / ln (0.2) = [-k (50.12)n

]/ [- k (100)n

]

0.1386 = (50.12/100)n

0.1386 = (0.5012)n

Taking natural log (log to the base 10 is also ok, the idea is to sole for n) on both sides:

ln (0.1386) = ln (0.5012)n

ln (0.1386) = n ln (0.5012)

n = ln (0.1386) / ln (0.5012) = 2.86 …(4)

Substituting the value of n in equation (2) (it’s ok to substitute it in equation (3) , the idea is to

solve for k. The units of k depend on the units of time and the value of n) :

ln (0.8 ) = - k (50.12)2.86

k = 3.06 X 10-6

…(5)

Note – It’s not necessary for the time to be converted from minutes to seconds because the units

of ‘k’ reflect the units of time and also depend on the value of n.

It is not incorrect if you did so! If time is in seconds,, you’d get the same value of n, i.e., 2.86.

The value of k would be different: k = - ln (0.8) / (60*50.12)2.86

= 2.52 X 10-11

.

This is perfectly ok. If you’ve converted time to hours, months or years, that’s ok, too!!

n = 2.86

k = 3.06 X 10-6