Problem E45 Set #7 Solutions 10.32 Consider 1.5 kg of a 99.7 wt% Fe0.3 wt% C alloy that is cooled to a temperature just below the eutectoid. (a) How many kilograms of proeutectoid ferrite form? (b) How many kilograms of eutectoid ferrite form? (c) How many kilograms of cementite form? Solution (a) Equation 10.21 must be used in computing the amount of proeutectoid ferrite that forms. Thus, ' 0.76 - C 0 0.76 - 0.30 = = 0.622 0.74 0.74
=
Or, (0.622)(1.5 kg) = 0.933 kg of proeutectoid ferrite forms. (b) In order to determine the amount of eutectoid ferrite, it first becomes necessary to compute the amount of total ferrite using the lever rule applied entirely across the + Fe3C phase field, as 0
=
3
3
=
6.70 - 0.30 = 0.958 6.70 - 0.022
which corresponds to (0.958)(1.5 kg) = 1.437 kg. Now, the amount of eutectoid ferrite is just the difference between total and proeutectoid ferrites, or 1.437 kg 0.933 kg = 0.504 kg (c) With regard to the amount of cementite that forms, again application of the lever rule across the entirety of the + Fe3C phase field, leads to 0 0.30 - 0.022 = = 0.042 6.70 - 0.022
3
=
3
which amounts to (0.042)(1.5 kg) = 0.063 kg cementite in the alloy.
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10.35 The mass fraction of eutectoid ferrite in an ironcarbon alloy is 0.71. On the basis of this information, is it possible to determine the composition of the alloy? If so, what is its composition? If this is not possible, explain why. Solution Yes, it is possible to determine the alloy composition; and, in fact, there are two possible answers. For the first, the eutectoid ferrite exists in addition to proeutectoid ferrite (for a hypoeutectoid alloy). For this case the mass fraction of eutectoid ferrite (W '') is just the difference between total ferrite and proeutectoid ferrite mass
fractions; that is W '' = W W ' Now, it is possible to write expressions for W (of the form of Equation 10.12) and W ' (Equation 10.21) in terms of C0, the alloy composition. Thus, =
0
3 3
0.76 - C 0 0.74
=
6.70 - C 0 0.76 - C 0 = 0.71 6.70 - 0.022 0.74
And, solving for C0 yields C0 = 0.61 wt% C. For the second possibility, we have a hypereutectoid alloy wherein all of the ferrite is eutectoid ferrite. Thus, it is necessary to set up a lever rule expression wherein the mass fraction of total ferrite is 0.71. Therefore,
=
0
3 3
=
6.70 - C 0 = 0.71 6.70 - 0.022
And, solving for C0 yields C0 = 1.96 wt% C.
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10.36 Often, the properties of multiphase alloys may be approximated by the relationship E (alloy) = EV + EV (10.24)
where E represents a specific property (modulus of elasticity, hardness, etc.), and V is the volume fraction. The subscripts and denote the existing phases or microconstituents. Employ the relationship above to determine the approximate Brinell hardness of a 99.75 wt% Fe0.25 wt% C alloy. Assume Brinell hardnesses of 80 and 280 for ferrite and pearlite, respectively, and that volume fractions may be approximated by mass fractions. Solution This problem asks that we determine the approximate Brinell hardness of a 99.75 wt% Fe-0.25 wt% C alloy, using a relationship similar to Equation 10.24. First, we compute the mass fractions of pearlite and proeutectoid ferrite using Equations 10.20 and 10.21, as
=
'0 - 0.022 0.25 - 0.022 = = 0.308 0.74 0.74
0.76 - C '0 0.76 - 0.25 = = 0.689 0.74 0.74
=
Now, we compute the Brinell hardness of the alloy using a modified form of Equation 10.24 as + =
= (80)(0.689) + (280)(0.308) = 141 11.1 (a) Rewrite the expression for the total free energy change for nucleation (Equation 11.1) 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 11.2) and solve for both the critical cube edge length, a*, and also G*. (b) Is G* greater for a cube or a sphere? Why? Solution (a) This problem first asks that we rewrite the expression for the total free energy change for nucleation (analogous to Equation 11.1) for the case of a cubic nucleus of edge length a. The volume of such a cubic radius is a3, whereas the total surface area is 6a2 (since there are six faces each of which has an area of a2). Thus, the expression for G is as follows:
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= 3 + 62 Differentiation of this expression with respect to a is as
(a 3DGv ) d (6a 2 g) = + da da
= 32 + 12 If we set this expression equal to zero as 32 + 12 = 0 and then solve for a (= a*), we have
=
4
Substitution of this expression for a in the above expression for G yields an equation for G* as * = (a*) 3 DGv + 6 (a* ) 2 g 4 3 4 2 = + 6 32 3 ( ) 2
=
(b)
G
16 3 g3 = for a cubei.e., (32) is greater that for a spherei.e., v 3 (DG ) 2 (DG ) 2 v v
g3 (16.8) . The reason for this is that surface-to-volume ratio of a cube is greater than for a sphere. (DG ) 2 v
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11.2 (a) For the solidification of nickel, calculate the critical radius r* and the activation free energy G* if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are 2.53 10 9 J/ m3 and 0.255 J/m2, respectively. Use the supercooling value found in Table 11.1. (b) Now calculate the number of atoms found in a nucleus of critical size. Assume a lattice parameter of 0.360 nm for solid nickel at its melting temperature. Solution (a) This portion of the problem asks that we compute r* and G* for the homogeneous nucleation of the solidification of Ni. First of all, Equation 11.6 is used to compute the critical radius. The melting temperature for nickel, found inside the front cover is 1455 C; also values of H (2.53 x 109 J/m ) and (0.255 J/m2) are given
f 3
in the problem statement, and the supercooling value found in Table 11.1 is 319 C (or 319 K). Thus, from Equation 11.6 we have 2 gT 1 m * = T - T DH f m (2) (0.255 J / m2 ) (1455 + 273 K) 1 = - 2.53 109 J / m3 319 K = 1.09 10- 9 m = 1.09 nm
For computation of the activation free energy, Equation 11.7 is employed. Thus p g3T 2 16 1 m * = 3 DH 2 (T - T)2 fm
3 (16)(p ) (0.255 J / m2 ) (1455 + 273 K)2 1 = 2 2 (319 K) (3) (- 2.53 109 J / m3 )
= 1.27 10- 18 J
(b) In order to compute the number of atoms in a nucleus of critical size (assuming a spherical nucleus of radius r*), it is first necessary to determine the number of unit cells, which we then multiply by the number of atoms per unit cell. The number of unit cells found in this critical nucleus is just the ratio of critical nucleus and unit cell volumes. Inasmuch as nickel has the FCC crystal structure, its unit cell volume is just a3 where a is the
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unit cell length (i.e., the lattice parameter); this value is 0.360 nm, as cited in the problem statement. Therefore, the number of unit cells found in a radius of critical size is just 4 pr * 3 3 # unit cells / particle = a3
4 (p )(1.09 nm) 3 =3 = 116 unit cells (0.360 nm) 3
Inasmuch as 4 atoms are associated with each FCC unit cell, the total number of atoms per critical nucleus is just
(116 unit cells / critical nucleus)(4 atoms / unit cell) = 464 atoms / critical nucleus
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11.3 (a) Assume for the solidification of nickel (Problem 11.2) that nucleation is homogeneous, and the number of stable nuclei is 106 nuclei per cubic meter. Calculate the critical radius and the number of stable nuclei that exist at the following degrees of supercooling: 200 K and 300 K. (b) What is significant about the magnitudes of these critical radii and the numbers of stable nuclei? Solution (a) For this part of the problem we are asked to calculate the critical radius for the solidification of nickel (per Problem 11.2), for 200 K and 300 K degrees of supercooling, and assuming that the there are 10 6 nuclei per meter cubed for homogeneous nucleation. In order to calculate the critical radii, we replace the Tm T term in Equation 11.6 by the degree of supercooling (denoted as T) cited in the problem statement. For 200 K supercooling, 2 gT m 1 * = 200 DH DT f (2)(0.255 J / m2 ) (1455 + 273 K) 1 = - 2.53 10 9 J / m3 200 K = 1.74 10-9 m = 1.74 nm For 300 K supercooling, (2)(0.255 J / m2 ) (1455 + 273 K) 1 * = 300 - 2.53 10 9 J / m3 300 K = 1.16 10-9 m = 1.16 nm In order to compute the number of stable nuclei that exist at 200 K and 300 K degrees of supercooling, it is necessary to use Equation 11.8. However, we must first determine the value of K1 in Equation 11.8, which in turn requires that we calculate G* at the homogeneous nucleation temperature using Equation 11.7; this was done in Problem 11.2, and yielded a value of G* = 1.27 10-18 J. Now for the computation of K1, using the value of n* for at the homogenous nucleation temperature (106 nuclei/m 3):
1 =
* DG * exp kT
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=
106 m3 / 1.27 10- 18 J exp - 23 J / atom - K) (1455 K - 319 K) (1.38 10 = 1.52 1041 nuclei/m 3
Now for 200 K supercooling, it is first necessary to recalculate the value G* of using Equation 11.7, where, again, the Tm T term is replaced by the number of degrees of supercooling, denoted as T, which in this case is 200 K. Thus 3 2 * = 16 p g Tm 1 200 3 DH 2 (DT ) 2 f (16)(p )(0.255 J / m2 )3 (1455 + 273 K)2 1 = (200 K)2 (3) (- 2.53 109 J / m3 )2 = 3.24 10-18 J And, from Equation 11.8, the value of n* is DG* 200 200 = 1 exp kT 3.24 10- 18 J = (1.52 10 41 nuclei / m3 ) exp - 23 J / atom - K) (1455 K - 200 K) (1.38 10 = 8.60 10-41 stable nuclei Now, for 300 K supercooling the value of G* is equal to (16)(p ) (0.255 J / m2 )3 (1455 + 273 K)2 1 * 300 = (3)(- 2.53 109 J / m3 )2 (300 K)2 = 1.44 10-18 J
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from which we compute the number of stable nuclei at 300 K of supercooling as DG* * 300 300 = K 1 exp kT 1.44 10- 18 J = (1.52 10 41 nuclei / m3 ) exp - 23 J / atom - K) (1455 K - 300 K) (1.38 10
= 88 stable nuclei (b) Relative to critical radius, r* for 300 K supercooling is slightly smaller that for 200 K (1.16 nm versus 1.74 nm). [From Problem 11.2, the value of r* at the homogeneous nucleation temperature (319 K) was 1.09 nm.] More significant, however, are the values of n* at these two degrees of supercooling, which are dramatically different8.60 10-41 stable nuclei at T = 200 K, versus 88 stable nuclei at T = 300 K!
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11.5 The kinetics of the austenite-to-pearlite transformation obey the Avrami relationship. Using the fraction transformedtime data given here, determine the total time required for 95% of the austenite to transform to pearlite: Fraction Transformed 0.2 0.6 Time (s) 280 425
Solution For this problem, we are given, for the austenite-to-pearlite transformation, two values of y and two values of the corresponding times, and are asked to determine the time required for 95% of the austenite to transform to pearlite. The first thing necessary is to set up two expressions of the form of Equation 11.17, and then to solve simultaneously for the values of n and k. In order to expedite this process, we will rearrange and do some algebraic manipulation of Equation 11.17. First of all, we rearrange as follows: 1 = exp (- kt n )
Now taking natural logarithms ln (1 - y) = - kt n
Or ln (1 - y) = kt n
which may also be expressed as 1 n ln = kt 1 - y
Now taking natural logarithms again, leads to
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ln ln
1 1 -
= ln k + n ln t y
which is the form of the equation that we will now use. Using values cited in the problem statement, the two equations are thus 1 ln = ln k + n ln (280 s) 1 - 0.2 1 ln = ln k + n ln ( 425 s) 1 - 0.6 Solving these two expressions simultaneously for n and k yields n = 3.385 and k = 1.162 10-9. Now it becomes necessary to solve for the value of at t which y = 0.95. One of the above equationsviz ln (1 - y) = kt n
may be rewritten as
=
ln (1 - y) k
And solving for t leads to ln (1 - y) 1/n = k
Now incorporating into this expression values for n and k determined above, the time required for 95% austenite transformation is equal to ln (1 - 0.95) 1/3.385 = = 603 s 1.162 10- 9
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11.6 (a) From the curves shown in Figure 11.11 and using Equation 11.18, determine the rate of recrystallization for pure copper at the several temperatures. (b) Make a plot of ln(rate) versus the reciprocal of temperature (in K 1), and determine the activation energy for this recrystallization process. (See Section 6.5.) (c) By extrapolation, estimate the length of time required for 50% recrystallization at room temperature, 20C (293 K). Solution This problem asks us to consider the percent recrystallized versus logarithm of time curves for copper shown in Figure 11.11. (a) The rates at the different temperatures are determined using Equation 11.18, which rates are tabulated below: Temperature (C) 135 119 113 102 88 43 Rate (min) 0.105 4.4 10-2 2.9 10-2 1.25 10-2 4.2 10-3 3.8 10-5
-1
(b) These data are plotted below.
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The activation energy, Q, is related to the slope of the line drawn through the data points as
= ( )
where R is the gas constant. The slope of this line is equal to = ln rate1 - ln rate 2 D ln rate = 1 1 1 D T1 T2 T
Let us take 1/T1 = 0.0025 K-1 and 1/T2 = 0.0031 K-1; the corresponding ln rate values are ln rate 1 = -2.6 and ln rate2 = -9.4. Thus, using these values, the slope is equal to = - 2.6 - (- 9.4) = - 1.133 10 4 K 0.0025 K-1 - 0.0031 K-1
And, finally the activation energy is
= ()( ) = - (- 1.133 10 4 K-1 ) (8.31 J/mol - K)
= 94,150 J/mol
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(c) At room temperature (20C), 1/T = 1/(20 + 273 K) = 3.41 10-3 K-1. Extrapolation of the data in the plot to this 1/T value gives () 12.8
which leads to (12.8) = 2.76 106 () 1
But since =
1 0.5
0.5 =
1 1 = 2.76 10- 6 (min) - 1
= 3.62 105 min = 250 days
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Metastable Versus Equilibrium States
11.7 (a) Briefly describe the phenomena of superheating and supercooling. (b) Why do these phenomena occur? Solution (a) Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation. (b) These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. supercooling. The driving force is enhanced during superheating or
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11.8 Suppose that a steel of eutectoid composition is cooled to 675C (1250F) from 760C (1400F) in less than 0.5 s and held at this temperature. (a) How long will it take for the austenite-to-pearlite reaction to go to 50% completion? To 100% completion? (b) Estimate the hardness of the alloy that has completely transformed to pearlite. Solution (a) From Figure 11.23, a horizontal line at 675C intersects the 50% and reaction completion curves at about 80 and 300 seconds, respectively; these are the times asked for in the problem statement. (b) The pearlite formed will be coarse pearlite. From Figure 11.31(a), the hardness of an alloy of composition 0.76 wt% C that consists of coarse pearlite is about 205 HB (93 HRB).
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11.9 What is the driving force for the formation of spheroidite? Solution The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area.
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11.10 Using the isothermal transformation diagram for an ironcarbon alloy of eutectoid composition (Figure 11.23), specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following timetemperature treatments. In each case assume that the specimen begins at 760C (1400F) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (c) Cool rapidly to 665C (1230F), hold for 103 s, then quench to room temperature. . Solution This problem asks us to determine the nature of the final microstructure of an iron-carbon alloy of eutectoid composition, that has been subjected to various isothermal heat treatments. Figure 11.23 is used in these determinations. (c) 100% coarse pearlite
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11.11
Using the isothermal transformation diagram for a 1.13 wt% C steel alloy (Figure 11.49),
determine the final microstructure (in terms of just the microconstituents present) of a small specimen that has been subjected to the following timetemperature treatments. In each case assume that the specimen begins at 920C (1690F) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to 775C (1430F), hold for 500 s, then quench to room temperature. (b) temperature. Rapidly cool to 700C (1290F), hold at this temperature for 105 s, then quench to room
Solution We are asked to determine which microconstituents are present in a 1.13 wt% C iron-carbon alloy that has been subjected to various isothermal heat treatments. These microconstituents are as follows: (a) Proeutectoid cementite and martensite (b) Spheroidite
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11.18 Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite. Solution The hardness and strength of iron-carbon alloys that have microstructures consisting of -ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.
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8.1 To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm2. Suppose that all the dislocations in 1000 mm3 (1 cm3) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to 109 mm2 by cold working. What would be the chain length of dislocations in 1000 mm3 of material? Solution The dislocation density is just the total dislocation length per unit volume of material (in this case per cubic millimeters). Thus, the total length in 1000 mm3 of material having a density of 105 mm-2 is just
(105
mm-2 ) (1000 mm3 ) = 108 mm = 105 m = 62 mi
Similarly, for a dislocation density of 109 mm-2, the total length is
(109 mm-2 )(1000
mm3 ) = 1012 mm = 109 m = 6.2 105 mi
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8.3 (a) Compare planar densities (Section 3.15 and Problem W3.46 [which appears on the books Web site]) for the (100), (110), and (111) planes for FCC. (b) Compare planar densities (Problem 3.44) for the (100), (110), and (111) planes for BCC. Solution (a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation 3.12 as 110 (FCC) = 1 4 R2 2 0.177 R2
=
Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem W3.46, which are as follows: 100 () = 1 = 0.25 R2 0.29 R2
4 2 1
111 (FCC) =
2 R2
= 3
(b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in Homework Problem 3.44, which are as follows: 100 () = 3 = 0.19 R2 0.27 R2
16 2 3
110 (BCC) =
8 R2
= 2
Below is a BCC unit cell, within which is shown a (111) plane.
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(a) The centers of the three corner atoms, denoted by A, B, and C lie on this plane. Furthermore, the (111) plane does not pass through the center of atom D, which is located at the unit cell center. The atomic packing of this plane is presented in the following figure; the corresponding atom positions from the Figure (a) are also noted.
(b) Inasmuch as this plane does not pass through the center of atom D, it is not included in the atom count. One sixth of each of the three atoms labeled A, B, and C is associated with this plane, which gives an equivalence of onehalf atom. In Figure (b) the triangle with A, B, and C at its corners is an equilateral triangle. And, from Figure ( b), the area of this triangle is . The triangle edge length, x, is equal to the length of a face diagonal, as indicated 2 in Figure (a). And its length is related to the unit cell edge length, a, as
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2 = 2 + 2 = 22
or
= 2
For BCC, = 4 (Equation 3.3), and, therefore, 3 4 3 2
=
Also, from Figure (b), with respect to the length y we may write 2 2 + = 2 2
which leads to =
3 . And, substitution for the above expression for x yields 2 = 3 4 2 3 4 2 = = 2 3 2 2
Thus, the area of this triangle is equal to 1 4 2 4 2 8 2 1 = = 2 2 3 2 3
=
And, finally, the planar density for this (111) plane is 0.5 atom 8 R2 3 3 16 R 2 0.11 R2
111 (BCC) =
=
=
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8.4 One slip system for the BCC crystal structure is {110} 111 . In a manner similar to Figure 8.6b, sketch a {110}-type plane for the BCC structure, representing atom positions with circles. Now, using arrows, indicate two different 111 slip directions within this plane. Solution Below is shown the atomic packing for a BCC {110}-type plane. The arrows indicate two different 111 type directions.
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