10.1 Name the two stages involved in the formation of particles of a new phase. Briefly describe each.

The two stages involved in the formation of particles of a new phase are nucleation and growth. The nucleation process involves the formation of normally very small particles of the new phase(s) which are stable and capable of continued growth. The growth stage is simply the increase in size of the new phase particles. 10.2 (a) Rewrite the expression for the total free energy change for nucleation (equation 10.1) for the case of a cubic nucleus of edge length a (instead of a sphere of radius r). Now differentiate this expression with respect to a (per equation 10.2) and solve for both the critical cube edge length, a*, and also ∆𝐺𝐺*. (b) Is ∆𝐺𝐺* greater for a cube or a sphere? Why?

(a) The formula of total free energy change for nucleation assumed that each nucleus is spherical and has a radius r:

∆𝐺𝐺 =43

𝜋𝜋𝑟𝑟3∆𝐺𝐺𝐺𝐺 + 4𝜋𝜋𝑟𝑟2𝛾𝛾 By replacing it for the case in cubic nucleus of edge length a:

Δ𝐺𝐺 = 𝑎𝑎3Δ𝐺𝐺𝐺𝐺 + 6𝑎𝑎2𝛾𝛾

Differentiating it with respect to a:

𝑑𝑑∆𝐺𝐺𝑑𝑑𝑎𝑎

=𝑑𝑑(𝑎𝑎3Δ𝐺𝐺𝐺𝐺)

𝑑𝑑𝑎𝑎+

𝑑𝑑(6𝑎𝑎2𝛾𝛾)𝑑𝑑𝑎𝑎

𝑑𝑑∆𝐺𝐺𝑑𝑑𝑎𝑎 = 3𝑎𝑎2∆𝐺𝐺𝐺𝐺 + 12𝑎𝑎𝛾𝛾

Equating it to zero:

3𝑎𝑎2∆𝐺𝐺𝐺𝐺 + 12𝑎𝑎𝛾𝛾 = 0

𝑎𝑎* = −4𝛾𝛾

∆𝐺𝐺𝐺𝐺

Take note that a* = a. By substituting the expression which equivalent to a* in the equation of ∆𝐺𝐺 to yield the ∆𝐺𝐺*:

∆𝐺𝐺* = �−4𝛾𝛾

∆𝐺𝐺𝐺𝐺�

3

∆𝐺𝐺𝐺𝐺 + 6𝛾𝛾 �−4𝛾𝛾

∆𝐺𝐺𝐺𝐺�

2

Δ𝐺𝐺*=32𝛾𝛾3

(Δ𝐺𝐺𝐺𝐺)2

(b) ∆𝐺𝐺* of the cube is greater than the ∆𝐺𝐺* of the sphere because the surface-to-volume ratio of a cube is greater than for a sphere.

Δ𝐺𝐺*=32 � 𝛾𝛾3

(Δ𝐺𝐺𝐺𝐺)2� (for the cube) Δ𝐺𝐺*= 16𝜋𝜋3

� 𝛾𝛾3

(Δ𝐺𝐺𝐺𝐺)2� (for the sphere) 10.3 If copper (which has a melting point of 1085℃) homogeneously nucleates at 849℃, calculate the critical radius given values of −1.77 × 109 J/m3 and 0.200 J/m2, respectively, for the latent heat of fusion and the surface free energy. Given:

𝑇𝑇m=1085℃ + 273 Κ 𝑇𝑇=850℃ + 273 Κ 𝛾𝛾=0.200 J/m2

Δ𝐻𝐻f = − 1.77 × 109 J/m3 RTF: r* Eq’n Used:

𝑟𝑟* = �− 2𝛾𝛾𝛾𝛾m

Δ𝐻𝐻f� � 1

𝛾𝛾m−𝛾𝛾�

Solution: 𝑟𝑟* = �− 2�0.200 J/m2�(1085+ 273 Κ )

−1.77×109J/m3 � � 1

(1085 + 273 Κ )−(850 + 273 Κ )�

Answer:

𝑟𝑟* = 1.31 × 10−9m = 1.31 nm 10.4 (a) For the solidification of iron, calculate the critical radius r* and the activation free energy ∆𝐺𝐺* if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are −1.85 × 109 J/m3 and 0.204 J/m2, respectively. Use the supercooling value found in Table 10.1. (b) Now calculate the number of atoms found in a nucleus of critical size. Assume a lattice parameter of 0.295 nm for solid iron at its melting temperature. Given:

𝑇𝑇m=1538 + 273 Κ (front page of Material Science and Engineering, 8th ed by Callister & Rethwisch) ∆𝑇𝑇=568 Κ 𝛾𝛾=0.204 J/m2 Δ𝐻𝐻f = − 1.85 × 109 J/m3 𝑎𝑎 = 0.295 nm

RTF:

(a) r* and ∆𝐺𝐺*

(b) # unit cells/particle

Eq’ns Used:

(a) 𝑟𝑟* = �− 2𝛾𝛾𝛾𝛾m

Δ𝐻𝐻f� � 1

∆𝛾𝛾�

Δ𝐺𝐺*= 16𝜋𝜋

3�𝛾𝛾3𝛾𝛾m2

∆𝐻𝐻f 2� � 1

(∆𝛾𝛾)2�

(b) # unit cells/particles = 43𝜋𝜋𝑟𝑟* 3

𝑎𝑎3 Solutions:

(a) 𝑟𝑟* = �− 2�0.204 J/m2�(1538 + 273 Κ )

−1.85×109 J/m3 � � 1

568 Κ � Δ𝐺𝐺*= 16𝜋𝜋

3��0.204 J/m2�

3(1538 + 273 Κ )2

(−1.85×109 J/m3 )2 � � 1(568 Κ) 2

�

Answers: 𝑟𝑟* = 7.03 × 10−10m = 0.703 nm Δ𝐺𝐺* =4.23 × 10−19 J

(b) # unit cells/particles = 43𝜋𝜋(0.703 nm)3

(0.295 nm)3

= 57 unit cells

Inasmuch as four atoms are associated with each FCC unit cell, the total number of atoms per critical nucleus is just �57 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

𝑐𝑐𝑟𝑟𝑢𝑢𝑢𝑢𝑢𝑢𝑐𝑐𝑎𝑎𝑐𝑐 𝑢𝑢𝑢𝑢𝑐𝑐𝑐𝑐𝑐𝑐𝑢𝑢� �4 𝑎𝑎𝑢𝑢𝑎𝑎𝑎𝑎𝑐𝑐

𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐� = 228 𝑎𝑎𝑢𝑢𝑎𝑎𝑎𝑎𝑐𝑐

𝑐𝑐𝑟𝑟𝑢𝑢𝑢𝑢𝑢𝑢𝑐𝑐𝑎𝑎𝑐𝑐 𝑢𝑢𝑢𝑢𝑐𝑐𝑐𝑐𝑐𝑐𝑢𝑢

10. 5 Assume for the solidification of iron (Problem 10.4) that nucleation is homogeneous. Calculate the critical radius that exist at the following degrees of supercooling: 200K and 300K. Given:

𝑇𝑇m=1538 + 273 Κ (front page of Material Science and Engineering, 8th ed by Callister & Rethwisch) ∆𝑇𝑇=200 Κ ; ∆𝑇𝑇=300 Κ 𝛾𝛾=0.204 J/m2 (using from 10.4) Δ𝐻𝐻f = − 1.85 × 109 J/m3 (using from 10.4)

RTF:

(a) r* at 200 (b) r* at 300

Eq’ns Used:

𝑟𝑟* = �− 2𝛾𝛾𝛾𝛾m

Δ𝐻𝐻f� � 1

∆𝛾𝛾�

Solutions:

(a) 𝑟𝑟* = �− 2�0.204 J/m2�(1538 + 273 Κ )

−1.85×109 J/m3 � � 1

200 Κ � (b) 𝑟𝑟* = �− 2�0.204 J/m2�(1538 + 273 Κ )

−1.85×109 J/m3 � � 1

300 Κ �

Answers: 𝑟𝑟* = 2.00 × 10−9m = 2.00 nm 𝑟𝑟* = 1.33 × 10−9m = 1.33 nm 10.6 For some transformation having kinetics that obey the Avrami equation. The parameter n is known to have a value of 1.7. If, after 100s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 99% completion? Given: n = 1.7 t = 100 s y = 0.5 RTF: t (y = 0.99) Eq’n Used:

𝑒𝑒−(𝑘𝑘𝑢𝑢𝑛𝑛) = (1 − 𝑦𝑦)

Solutions:

𝑒𝑒−�𝑘𝑘(100)1.7� = (1 − 0.5) 𝑘𝑘 = 2.76 × 10−4 𝑒𝑒−�2.76×10−4(𝑢𝑢)1.7� = (1 − 0.99)

Answer: 𝑡𝑡 = 305 𝑠𝑠

10.7 Compute for the rate of reaction that obeys Avrami kinetics, assuming that the constants n and k have values of 5.0 and 8 × 10−3 respectively, for time expressed in seconds. Given: n = 5.0 k = 8 × 10−3 y = 0.5 (Using from 10. 6) RTF: r Eq’n Used:

𝑒𝑒−(𝑘𝑘𝑢𝑢𝑛𝑛) = (1 − 𝑦𝑦) 𝑟𝑟 = 1

𝑢𝑢

Solutions:

𝑒𝑒−�8×10−3(𝑢𝑢)5.0� = (1 − 0.5) 𝑡𝑡 = 2.44 𝑠𝑠 𝑟𝑟 = 1

2.44

Answer: 𝑟𝑟 = 0.41 𝑠𝑠-1

10.8 It is known that the kinetics of recrystallization for some alloy obey the Avrami equation and that the value of n in the exponential is 2.5. If, at some temperature, the fraction recrystallized is 0.5 after 200 min, determine the rate of recrystallization at this temperature. Given: n = 2.5 t = 200 min y = 0.5 RTF: r Eq’n Used:

𝑟𝑟 = 1𝑢𝑢

Solutions:

𝑟𝑟 = 1

200

Answer: 𝑟𝑟 = 5 × 10−3 𝑚𝑚𝑚𝑚𝑚𝑚-1

10.9 The kinetics of the austenite-to-pearlite transitions obey the Avrami relationship. Using the fraction transformed-time table give here, determine the total time required for the 95% of the austenite to transform to pearlite:

Fraction Transformed Time (s) 0.2 12.8 0.8 28.5

Given: Data from table above RTF: t (y = 0.95) Eq’n Used:

𝑒𝑒−(𝑘𝑘𝑢𝑢𝑛𝑛) = (1 − 𝑦𝑦)

Solutions:

𝑒𝑒−[𝑘𝑘(12.8)𝑛𝑛] = (1 − 0.2) 𝑒𝑒−[𝑘𝑘(28.5)𝑛𝑛] = (1 − 0.8) −𝑘𝑘 = ln(1−0.2)

12.8𝑛𝑛 → 𝑒𝑒𝑞𝑞′𝑚𝑚. 1 −𝑘𝑘 = ln(1−0.8)28.5𝑛𝑛 → 𝑒𝑒𝑞𝑞′𝑚𝑚. 2

ln(1 − 0.2)

12.8𝑢𝑢 =ln(1 − 0.8)

28.5𝑢𝑢 𝑚𝑚 = 2.5

𝑒𝑒−�𝑘𝑘(12.8)2.5� = (1 − 0.2)

𝑘𝑘 = 4.13 × 10−4

𝑒𝑒−�4.13×10−4(𝑢𝑢)2.5� = (1 − 0.95)

Answer: 𝑡𝑡 = 36.66 𝑠𝑠

10.10 The fraction recrystallized at 600oC of a previously deformed steel are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of 22.8 min.

Fraction Transformed Time (min) 0.2 13.2 0.7 29.3

Given: Data from table above RTF: y (t = 22.8) Eq’n Used:

𝑒𝑒−(𝑘𝑘𝑢𝑢𝑛𝑛) = (1 − 𝑦𝑦)

Solutions:

𝑒𝑒−[𝑘𝑘(13.2)𝑛𝑛] = (1 − 0.2) 𝑒𝑒−[𝑘𝑘(29.3)𝑛𝑛] = (1 − 0.7) −𝑘𝑘 = ln(1−0.2)

13.2𝑛𝑛 → 𝑒𝑒𝑞𝑞′𝑚𝑚. 1 −𝑘𝑘 = ln(1−0.7)29.3𝑛𝑛 → 𝑒𝑒𝑞𝑞′𝑚𝑚. 2

ln(1 − 0.2)

13.2𝑢𝑢 =ln(1 − 0.7)

29.3𝑢𝑢 𝑚𝑚 = 2.11

𝑒𝑒−�𝑘𝑘(13.2)2.11`� = (1 − 0.2)

𝑘𝑘 = 9.55 × 10−4

𝑒𝑒−�9.55×10−4(22.8)2.11� = (1 − 𝑦𝑦)

Answer: 𝑦𝑦 = 0.49