Ungraded HW Solutions for April 17 PDF

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Homework for April 17...

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9.61 Consider 3.5 kg of austenite containing 0.95 wt% C and cooled to below 727°C (1341°F). (a) What is the proeutectoid phase? (b) How many kilograms each of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid phase form? (d) Schematically sketch and label the resulting microstructure. Solution (a) The proeutectoid phase will be Fe 3C since 0.95 wt% C is greater than the eutectoid composition (0.76 wt% C). (b) For this portion of the problem, we are asked to determine how much total ferrite and cementite form. Application of the appropriate lever rule expression for the mass fraction of total

W =

CFe C - C0 3

CFe C - C

=

 yields

6.70 - 0.95 = 0.86 6.70 - 0.022

3

which, when multiplied by the total mass of the alloy, gives (0.86)(3.5 kg) = 3.01 kg of total ferrite. Similarly, for the mass fraction of total cementite,

WFe

C 3

=

C 0 - C 0.95 - 0.022 = = 0.14 C Fe C - C 6.70 - 0.022 3

And the mass of total cementite that forms is (0.14)(3.5 kg) = 0.49 kg. (c) Now we are asked to calculate how much pearlite and the proeutectoid phase (cementite) form. To determine the mass fraction of pearlite that forms (Wp), we apply Equation 9.22, in which

Wp =

C1¢ =0.95 wt% C :

6.70 - C1¢ 6.70 - 0.95 = 0.97 = 6.70 - 0.76 6.70 - 0.76

which corresponds to a mass of (0.97)(3.5 kg) = 3.4 kg. Likewise, from Equation 9.23, we calculate the mass fraction of proeutectoid cementite as follows:

WFe

3C¢

=

C1¢ - 0.76 0.95 - 0.76 = 0.03 = 5.94 5.94

which is equivalent to (0.03)(3.5 kg) = 0.11 kg of the total 3.5 kg mass. (d) Schematically, the microstructure would appear as follows:

9.62 Consider 6.0 kg of austenite containing 0.45 wt% C and cooled to less than 727°C (1341°F). (a) What is the proeutectoid phase? (b) How many kilograms each of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid phase form? (d) Schematically sketch and label the resulting microstructure. Solution (a) Ferrite is the proeutectoid phase since 0.45 wt% C is less than 0.76 wt% C. (b) For this portion of the problem, we are asked to determine how much total ferrite and cementite form. For the mass fraction of ferrite, application of the appropriate lever rule expression yields

W =

CFe C - C0 3

C Fe C - C

=

6.70 - 0.45 = 0.936 6.70 - 0.022

3

which corresponds to (0.936)(6.0 kg) = 5.62 kg of total ferrite. Similarly, for mass fraction of total cementite,

WFe C = 3

C 0 - C 0.45 - 0.022 = 0.064 = C Fe C - C 6.70 - 0.022 3

Or (0.064)(6.0 kg) = 0.38 kg of total cementite form. (c) Now consider the amounts of pearlite and proeutectoid ferrite. For pearlite we use Equation 9.20, as follows:

Wp =

C0¢ - 0.022 0.45 - 0.022 = 0.578 = 0.74 0.74

This corresponds to (0.578)(6.0 kg) = 3.47 kg of pearlite. Also, from Equation 9.21, we compute the mass fraction of proeutectoid ferrite as follows:

W ¢ =

0.76 - 0.45 = 0.420 0.74

Or, there are (0.420)(6.0 kg) = 2.52 kg of proeutectoid ferrite. (d) Schematically, the microstructure would appear as follows:

9.67 The microstructure of an iron–carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these two microconstituents are 0.174 and 0.826, respectively. Determine the concentration of carbon in this alloy. Solution To solve this problem we use Equation 9.20, which is the lever-rule expression for computation of the mass fraction of pearlite in a hypoeutectoid alloy. We are given the mass fraction of pearlite ( Wp), 0.826, which allows us to determine the concentration of carbon in the alloy,

Wp = And solving for

C0¢ leads to C0¢ =0.63 wt% C.

C0¢ . For this problem, Equation 9.20 is expressed as follows:

C0¢ - 0.022 = 0.826 0.74

9.78 For an iron–carbon alloy of composition 3 wt% C–97 wt% Fe, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: 1250°C (2280°F), 1145°C (2095°F), and 700°C (1290°F). Label the phases and indicate their compositions (approximate). Solution Below is shown the Fe-Fe 3C phase diagram with a vertical line constructed at 3 wt% C; also along this line noted temperatures of 1250°C, 1145°C, and 700°C.

Schematic microstructures for this iron-carbon alloy at the three temperatures are shown below; approximate phase compositions are also indicated.

10.1 Name the two stages involved in the formation of particles of a new phase. Briefly describe each. Answer 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. 10.4 (a) For the solidification of nickel, calculate the critical radius r* and the activation free energy ΔG* 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 10.1. (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 10.6 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  109 J/m3) and  (0.255 J/m2) are given f

in the problem statement, and the supercooling value found in Table 10.1 is 319 C (which is also 319 K because this value is really Tm - T, which is the same for Celsius and Kelvin). Thus, from Equation 10.6 we have

æ 2 T ö æ 1 ö m r * =ç ÷ç ÷ è H f ø è Tm - T ø é (2) (0.255 J/m2 ) (1455 + 273 K)ù æ 1 ö =ê úç ÷ - 2.53 109 J/m3 êë úû è 319 K ø

=1.09 10- 9 m =1.09 nm For computation of the activation free energy, Equation 10.7 is employed. Thus

æ 16p  3T 2 ö 1 m G * =ç ÷ 2 çè 3Hf ÷ø (Tm - T) 2 3 ù é ù (16)(p ) ( 0.255 J/m2 ) (1455 + 273 K)2 ú é 1 ê = ú ê 2 2 ú ê 9 J/m3 êë (319 K) úû (3) ( ) 2.53 10 û ë

=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 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 p r *3 3 # unit cells/particle = a3 æ 4ö (p )(1.09 èç 3 ÷ø

=

nm)3

(0.360 nm)3

=116 unit cells

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

10.10 The kinetics of the austenite-to-pearlite transformation obeys the Avrami relationship. Using the fraction transformed–time 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

The first thing necessary is to set up two expressions of the form of Equation 10.17, 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 10.17. First of all, we rearrange as follows:

( )

1 - y =exp - ktn Now taking natural logarithms

ln (1 - y) =- ktn Or

- ln (1 - y) =ktn

(10.17c)

which may also be expressed as

æ 1 ö =kt n ln ç è 1 - y÷ø Now taking natural logarithms again of both sides of this equation, leads to

é æ 1 öù ln ê ln ç ÷ ú =ln k + n lnt ë è 1 - yø û which is the form of the equation that we will now use. Using values cited in the problem statement, the two equations are thus

ì é 1 ùü ln í ln ê ú ý = ln k + n ln(280 s) îï ë 1 - 0.2 û þï ì é 1 ùü ln í ln ê ú ý = ln k + n ln(425 s) ïî ë 1 - 0.6 û ïþ

Solving these two expressions simultaneously for n and k yields n = 3.385 and k = 1.162  10-9 for time in seconds. Now it becomes necessary to solve for the value of t at which y = 0.95. Equation 10.17c given above may be rewritten as follows:

tn = -

ln (1 - y) k

And solving for t leads to 1/n

é ln (1 - y) ù t = êú k û ë

Now incorporating into this expression values for n and k determined above, the time required for 95% austenite transformation is equal to

é ln (1 - 0.95) ù t = êú êë 1.162 10- 9 úû

1/3.385

= 603 s

10.19 Using the isothermal transformation diagram for an iron–carbon alloy of eutectoid composition (Figure 10.22), 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 time–temperature treatments. In each case assume that the specimen begins at 760°C (1400°F) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Cool rapidly to 350°C (660°F), hold for 103 s, then quench to room temperature. (b) Rapidly cool to 625°C (1160°F), hold for 10 s, then quench to room temperature. (c) Rapidly cool to 600°C (1110°F), hold for 4 s, rapidly cool to 450°C (840°F), hold for 10 s, then quench to room temperature. (d) Reheat the specimen in part (c) to 700°C (1290°F) for 20 h. (e) Rapidly cool to 300°C (570°F), hold for 20 s, then quench to room temperature in water. Reheat to 425°C (800°F) for 103 s and slowly cool to room temperature. (f) Cool rapidly to 665°C (1230°F), hold for 103 s, then quench to room temperature. (g) Rapidly cool to 575°C (1065°F), hold for 20 s, rapidly cool to 350°C (660°F), hold for 100 s, then quench to room temperature. (h) Rapidly cool to 350°C (660°F), hold for 150 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. determinations. (a) 100% bainite

Figure 10.22 is used in these

(b) 50% medium pearlite and 50% martensite (c) 50% fine pearlite, 25% bainite, and 25% martensite (d) 100% spheroidite (e) 100% tempered martensite (f) 100% coarse pearlite (g) 100% fine pearlite (h) 50% bainite and 50% martensite 10.20 Make a copy of the isothermal transformation diagram for an iron–carbon alloy of eutectoid composition (Figure 10.22) and then sketch and label time–temperature paths on this diagram to produce the following microstructures: (a) 100% coarse pearlite (b) 50% martensite and 50% austenite (c) 50% coarse pearlite, 25% bainite, and 25% martensite Solution Below is shown the isothermal transformation diagram for a eutectoid iron-carbon alloy, with timetemperature paths that will yield (a) 100% coarse pearlite; (b) 50% martensite and 50% austenite; and (c) 50% coarse pearlite, 25% bainite, and 25% martensite....