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Publicado: 27 de junio de 2012
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 exp1 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 the rearranged equation. We can thennote 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 exp1 ct n 2 ln11 f 2 ct n ln1ct n 2 ln3ln11 f 2 4 ln3 ln11 f 2 4 ln1c2 n ln1t2 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 c1552 2.89 4 c 6.47 10 6 f 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 t(min) 28 37 44 50 55 60 67 73 86 ln(1 0.1 0.22 0.36 0.51 0.69 0.92 1.20 1.61 2.302 f)
123 124
The Science and Engineering of Materials
Instructor’s Solution Manual
2.0
1.0 − In (1 − f )
0.5 n = 2.89
0.2
0.1 5 10 t (min) 20
11–3
Determine the constants c and n in Equation 11-2 that describe the rate of recrystallization of copper at 135 C. (See Figure 11–2) Solution: f 1 exp1 ct n 2 T 135°C 408 K
We can rearrange the equation and eliminate the exponentialby 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 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 exp1 ct n 2 ln11 f 2 ct n ln3ln11 f 2 4 ln1ct n 2 ln3 ln11 f 2 4 ln1c2 ln1t2 f 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 t (min) 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 2.30 f)
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 ccan be found from one of the points from the curve: If f 0.6, then t 10. Then 1 0.6 exp 3 c1102 3.1 4 c 7.28 10 4.
4.0
2.0
1.0 − In (1 − f )
0.5 n = 2.89
0.2
0.1 30 50 t (min) 100
11–4
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 exp1 Q RT2 1 t 1s 1 2 1 1130 1 1140 1 1150 1 T 1K 1 2 2732 2732 2732 2.48 2.42 2.36 10 10 10
3 3 3
1 19 min2160 s/min2 1.85 10 3 1 155 min2160 s/min2 3.03 10 4 1 1316 min2160 s/min2 5.27 10 5
126
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: Q R Q R Q ln110 3 2 ln1510 5 2 0.00246 0.00236 29,957 59,525 cal/mol
10−4 0.00246 − 0.00236
10−5 0.0023 0.0024 1/T (K−1) 0.0025
11–16
(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 naturalaging. Solution: (a) For the Cu–1.2% Be alloy, the peritectic temperature is 870 C; above this temperature, liquid may form. The solvus temperature is about 530 C. Therefore: 1) Solution treat between 530 C and 870 C (780 C is typical for beryllium copper alloys) 2) Quench 3) Age below 530 C (330 C is typical for these alloys) (b) We can perform lever law calculations at 400 C and at room...
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 exp1 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 the rearranged equation. We can thennote 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 exp1 ct n 2 ln11 f 2 ct n ln1ct n 2 ln3ln11 f 2 4 ln3 ln11 f 2 4 ln1c2 n ln1t2 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 c1552 2.89 4 c 6.47 10 6 f 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 t(min) 28 37 44 50 55 60 67 73 86 ln(1 0.1 0.22 0.36 0.51 0.69 0.92 1.20 1.61 2.302 f)
123 124
The Science and Engineering of Materials
Instructor’s Solution Manual
2.0
1.0 − In (1 − f )
0.5 n = 2.89
0.2
0.1 5 10 t (min) 20
11–3
Determine the constants c and n in Equation 11-2 that describe the rate of recrystallization of copper at 135 C. (See Figure 11–2) Solution: f 1 exp1 ct n 2 T 135°C 408 K
We can rearrange the equation and eliminate the exponentialby 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 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 exp1 ct n 2 ln11 f 2 ct n ln3ln11 f 2 4 ln1ct n 2 ln3 ln11 f 2 4 ln1c2 ln1t2 f 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 t (min) 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 2.30 f)
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 ccan be found from one of the points from the curve: If f 0.6, then t 10. Then 1 0.6 exp 3 c1102 3.1 4 c 7.28 10 4.
4.0
2.0
1.0 − In (1 − f )
0.5 n = 2.89
0.2
0.1 30 50 t (min) 100
11–4
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 exp1 Q RT2 1 t 1s 1 2 1 1130 1 1140 1 1150 1 T 1K 1 2 2732 2732 2732 2.48 2.42 2.36 10 10 10
3 3 3
1 19 min2160 s/min2 1.85 10 3 1 155 min2160 s/min2 3.03 10 4 1 1316 min2160 s/min2 5.27 10 5
126
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: Q R Q R Q ln110 3 2 ln1510 5 2 0.00246 0.00236 29,957 59,525 cal/mol
10−4 0.00246 − 0.00236
10−5 0.0023 0.0024 1/T (K−1) 0.0025
11–16
(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 naturalaging. Solution: (a) For the Cu–1.2% Be alloy, the peritectic temperature is 870 C; above this temperature, liquid may form. The solvus temperature is about 530 C. Therefore: 1) Solution treat between 530 C and 870 C (780 C is typical for beryllium copper alloys) 2) Quench 3) Age below 530 C (330 C is typical for these alloys) (b) We can perform lever law calculations at 400 C and at room...
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