Diagramas De Fases Y De Ellingham

11
Dispersion Strengthening by Phase
Transformations and Heat Treatment
11–2

Determine the constants c and n in Equation 11–2 that describe the rate of crystallization of polypropylene at 140 C. (See Figure 11–31)
Solution:

f

1

exp 1 ct n 2

T

140°C

413 K

We can rearrange the equation and eliminate the exponential by taking natural logarithms of both sides of therearranged equation. We can then note that
ln(1 f ) versus t is a power equation; if these terms are plotted on a log-log
plot, we should obtain a linear relationship, as the graph of the data below
indicates. Note that in setting up the equation for plotting, we switch the minus
sign from the right hand to the left hand side, since we don’t have negative
numbers on the log-log paper.
1 f exp 1 ctn 2
ln 1 1 f 2
ct n
ln 3 ln 1 1 f 2 4
ln 1 ct n 2
ln 3 ln 1 1 f 2 4
ln 1 c 2 n ln 1 t 2
A log-log plot of “ ln(1
f )” versus “t” is
shown. From the graph, we find that the slope
n 2.89 and the constant c can be found from
one of the points from the curve:
if f 0.5, t 55. Then
1 0.5 exp 3 c 1 55 2 2.89 4
c 6.47 10 6

f

t(min)

0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9

2837
44
50
55
60
67
73
86

ln(1

f)

0.1
0.22
0.36
0.51
0.69
0.92
1.20
1.61
2.302

123

The Science and Engineering of Materials

Instructor’s Solution Manual

2.0

1.0
− In (1 − f )

124

0.5
n = 2.89

0.2

0.1
5

11–3

10
t (min)

20

Determine the constants c and n in Equation 11-2 that describe the rate of
recrystallization of copper at 135C. (See Figure 11 – 2)
Solution:

f

1

exp 1 ct n 2

T

135°C

408 K

We can rearrange the equation and eliminate the exponential by taking natural logarithms of both sides of the rearranged equation. We can then note that
ln(1 f ) versus t is a power equation and should give a linear relationship in
a log-log plot. Note that in setting up the equation for plotting, we switch theminus sign from the right hand to the left hand side, since we don’t have
negative numbers on the log-log paper.
1 f exp 1 ct n 2
ln 1 1 f 2
ct n
ln 3 ln 1 1 f 2 4
ln 1 ct n 2
ln 3 ln 1 1 f 2 4
ln 1 c 2 ln 1 t 2

f

t (min)

0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9

5.0
6.6
7.7
8.5
9.0
10.0
10.5
11.5
13.7

ln(1
0.10
0.22
0.36
0.51
0.69
0.92
1.20
1.61
2.30f)

CHAPTER 11

Dispersion Strengthening by Phase Transformations and Heat Treatment

125

A log-log plot of “ ln(1 f )” versus “t” is shown. From the graph, we find
that the slope n 3.1 and the constant c can be found from one of the points
from the curve:
If f 0.6, then t 10. Then
1 0.6 exp 3 c 1 10 2 3.1 4
c 7.28 10 4.

4.0

2.0

− In (1 − f )

1.0

0.5
n = 2.89

0.20.1
30

11–4

50
t (min)

100

Determine the activation energy for crystallization of polypropylene, using the
curves in Figure 11 – 36.
Solution:

We can determine how the rate (equal to 1 t) changes with temperature:
rate

1t

c exp 1 Q RT 2

1 t 1s 12

1 1 9 min 21 60 s/min 2 1.85 10 3
1 1 55 min 21 60 s/min 2 3.03 10 4
1 1 316 min 21 60 s/min 2 5.27 10 5

1 1 1301 1 140
1 1 150

1 T 1K 12

273 2
273 2
273 2

2.48
2.42
2.36

10
10
10

3
3
3

The Science and Engineering of Materials

Instructor’s Solution Manual

From the semilog graph of rate versus reciprocal temperature, we find that the
slope is:

QR
Q

ln 1 10 3 2 ln 1 5 10 5 2
0.00246 0.00236
29,957
59,525 cal/mol

10−3

10−4
0.00246 − 0.00236

In (10−3) − In(5 × 10−5)

QR

Rate (s−1)

126

10−5
0.0023

11–16

0.0024
1/T (K−1)

0.0025

(a) Recommend an artificial age-hardening heat treatment for a Cu – 1.2% Be alloy
(see Figure 11 – 34). Include appropriate temperatures. (b) Compare the amount of
the g2 precipitate that forms by artificial aging at 400 C with the amount of the
precipitate that forms by natural aging.
Solution:...