MSE 2101 HW VIII SOLUTIONS

THE IRON-CARBON PHASE DIAGRAM AND PHASE TRANSFORMATIONS

10.6 For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 1.7. If, after 100 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 99% completion?

SolutionThis problem calls for us to compute the length of time required for a reaction to go to 99% completion. It first becomes necessary to solve for the parameter k in Equation 10.17. In order to do this it is best manipulate the equation such that k is the dependent variable. We first rearrange Equation 10.17 as

and then take natural logarithms of both sides:

Now solving for k gives

And, from the problem statement, for y = 0.50 when t = 100 s and given that n = 1.7, the value of k is equal to

We now want to manipulate Equation 10.17 such that t is the dependent variable. The above equation may be written in the form:

And solving this expression for t leads to

Now, using this equation and the value of k determined above, the time to 99% transformation completion is equal t

10.9 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 TransformedTime (s)

0.212.6

0.828.2

SolutionThe 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:

Now taking natural logarithms

Or

which may also be expressed as

Now taking natural logarithms again, leads to

which is the form of the equation that we will now use. Using values cited in the problem statement, the two equations are thus

Solving these two expressions simultaneously for n and k yields n = 2.453 and k = 4.46 10-4.Now it becomes necessary to solve for the value of t at which y = 0.95. One of the above equationsviz

may be rewritten as

And solving for t leads to

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

10.10 The fraction recrystallizedtime data for the recrystallization at 600C 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 RecrystallizedTime (min)

0.2013.1

0.7029.1

SolutionThe 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:

Now taking natural logarithms

Or

which may also be expressed as

Now taking natural logarithms again, leads to

which is the form of the equation that we will now use. The two equations are thus

Solving these two expressions simultaneously for n and k yields n = 2.112 and k = 9.75 10-4.Now it becomes necessary to solve for y when t = 22.8 min. Application of Equation 10.17 leads to

10.11 (a) From the curves shown in Figure 10.11 and using Equation 10.18, 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 K1), and determine the activation energy for this recrystallization process. (See Section 5.5.)(c) By extrapolation, estimate the length of time required for 50% recrystallization at room temperature, 20C (293 K).

SolutionThis problem asks us to consider the percent recrystallized versus logarithm of time curves for copper shown in Figure 10.11.(a) The rates at the different temperatures are determined using Equation 10.18, which rates are tabulated below:

Temperature (C)Rate (min)-1

1350.1051194.4 10-21132.9 10-21021.25 10-2884.2 10-3433.8 10-5

(b) These data are plotted below.

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

Let us take 1/T1 = 0.0025 K-1 and 1/T2 = 0.0031 K-1; the corresponding ln rate values are ln rate1 = -2.6 and ln rate2 = -9.4. Thus, using these values, the slope is equal to

And, finally the activation energy is

= 94,150 J/mol

(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

which leads to

But since

10.12 Determine values for the constants n and k (Equation 10.17) for the recrystallization of copper (Figure 10.11) at 102C.

SolutionIn this problem we are asked to determine, from Figure 10.11, the values of the constants n and k (Equation 10.17) for the recrystallization of copper at 102C. One way to solve this problem is to take two values of percent recrystallization (which is just 100y, Equation 10.17) and their corresponding time values, then set up two simultaneous equations, from which n and k may be determined. 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:

Now taking natural logarithms

Or

which may also be expressed as

Now taking natural logarithms again, leads to

which is the form of the equation that we will now use. From the 102C curve of Figure 10.11, let us arbitrarily choose two percent recrystallized values, 20% and 80% (i.e., y1 = 0.20 and y2 = 0.80). Their corresponding time values are t1 = 50 min and t2 = 100 min (realizing that the time axis is scaled logarithmically). Thus, our two simultaneous equations become

from which we obtain the values n = 2.85 and k = 3.21 10-6.

10.18 Using the isothermal transformation diagram for an ironcarbon 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 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.(a) Cool rapidly to 700C (1290F), hold for 104 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 700C for 104 s, approximately 50% of the specimen has transformed to coarse pearlite. Upon cooling to room temperature, the remaining 50% transforms to martensite. Hence, the final microstructure consists of about 50% coarse pearlite and 50% martensite.

(b) Reheat the specimen in part (a) to 700C (1290F) for 20 h.

SolutionHeating to 700C for 20 h the specimen in part (a) will transform the coarse pearlite and martensite to spheroidite.

(c) Rapidly cool to 600C (1110F), hold for 4 s, rapidly cool to 450C (840F), hold for 10 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling to and holding at 600C for 4 s, approximately 50% of the specimen has transformed to pearlite (medium). During the rapid cooling to 450C no transformations occur. At 450C we start timing again at zero time; while holding at 450C for 10 s, approximately 50 percent of the remaining unreacted 50% (or 25% of the original specimen) will transform to bainite. And upon cooling to room temperature, the remaining 25% of the original specimen transforms to martensite. Hence, the final microstructure consists of about 50% pearlite (medium), 25% bainite, and 25% martensite.

(d) Cool rapidly to 400C (750F), hold for 2 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling to and holding at 400C for 2 s, no of the transformation begin lines have been crossed, and therefore, the specimen is 100% austenite. Upon cooling rapidly to room temperature, all of the specimen transforms to martensite, such that the final microstructure is 100% martensite.

(e) Cool rapidly to 400C (750F), hold for 20 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 400C for 20 s, approximately 40% of the specimen has transformed to bainite. Upon cooling to room temperature, the remaining 60% transforms to martensite. Hence, the final microstructure consists of about 40% bainite and 60% martensite.

(f) Cool rapidly to 400C (750F), hold for 200 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 400C for 200 s, the entire specimen has transformed to bainite. Therefore, during the cooling to room temperature no additional transformations will occur. Hence, the final microstructure consists of 100% bainite.

(g) Rapidly cool to 575C (1065F), hold for 20 s, rapidly cool to 350C (660F), hold for 100 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 575C for 20 s, the entire specimen has transformed to fine pearlite. Therefore, during the second heat treatment at 350C no additional transformations will occur. Hence, the final microstructure consists of 100% fine pearlite.

(h) Rapidly cool to 250C (480F), hold for 100 s, then quench to room temperature in water. Reheat to 315C (600F) for 1 h and slowly cool to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 250C for 100 s, no transformations will have occurredat this point, the entire specimen is still austenite. Upon rapidly cooling to room temperature in water, the specimen will completely transform to martensite. The second heat treatment (at 315C for 1 h)not shown on the above plotwill transform the material to tempered martensite. Hence, the final microstructure is 100% tempered martensite.

10.18 Using the isothermal transformation diagram for an ironcarbon 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 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.(a) Cool rapidly to 700C (1290F), hold for 104 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 700C for 104 s, approximately 50% of the specimen has transformed to coarse pearlite. Upon cooling to room temperature, the remaining 50% transforms to martensite. Hence, the final microstructure consists of about 50% coarse pearlite and 50% martensite.

(b) Reheat the specimen in part (a) to 700C (1290F) for 20 h.

SolutionHeating to 700C for 20 h the specimen in part (a) will transform the coarse pearlite and martensite to spheroidite.

(c) Rapidly cool to 600C (1110F), hold for 4 s, rapidly cool to 450C (840F), hold for 10 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling to and holding at 600C for 4 s, approximately 50% of the specimen has transformed to pearlite (medium). During the rapid cooling to 450C no transformations occur. At 450C we start timing again at zero time; while holding at 450C for 10 s, approximately 50 percent of the remaining unreacted 50% (or 25% of the original specimen) will transform to bainite. And upon cooling to room temperature, the remaining 25% of the original specimen transforms to martensite. Hence, the final microstructure consists of about 50% pearlite (medium), 25% bainite, and 25% martensite.

(d) Cool rapidly to 400C (750F), hold for 2 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling to and holding at 400C for 2 s, no of the transformation begin lines have been crossed, and therefore, the specimen is 100% austenite. Upon cooling rapidly to room temperature, all of the specimen transforms to martensite, such that the final microstructure is 100% martensite.

(e) Cool rapidly to 400C (750F), hold for 20 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 400C for 20 s, approximately 40% of the specimen has transformed to bainite. Upon cooling to room temperature, the remaining 60% transforms to martensite. Hence, the final microstructure consists of about 40% bainite and 60% martensite.

(f) Cool rapidly to 400C (750F), hold for 200 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 400C for 200 s, the entire specimen has transformed to bainite. Therefore, during the cooling to room temperature no additional transformations will occur. Hence, the final microstructure consists of 100% bainite.

(g) Rapidly cool to 575C (1065F), hold for 20 s, rapidly cool to 350C (660F), hold for 100 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 575C for 20 s, the entire specimen has transformed to fine pearlite. Therefore, during the second heat treatment at 350C no additional transformations will occur. Hence, the final microstructure consists of 100% fine pearlite.

(h) Rapidly cool to 250C (480F), hold for 100 s, then quench to room temperature in water. Reheat to 315C (600F) for 1 h and slowly cool to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 250C for 100 s, no transformations will have occurredat this point, the entire specimen is still austenite. Upon rapidly cooling to room temperature in water, the specimen will completely transform to martensite. The second heat treatment (at 315C for 1 h)not shown on the above plotwill transform the material to tempered martensite. Hence, the final microstructure is 100% tempered martensite.

10.18 Using the isothermal transformation diagram for an ironcarbon 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 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.(a) Cool rapidly to 700C (1290F), hold for 104 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 700C for 104 s, approximately 50% of the specimen has transformed to coarse pearlite. Upon cooling to room temperature, the remaining 50% transforms to martensite. Hence, the final microstructure consists of about 50% coarse pearlite and 50% martensite.

(b) Reheat the specimen in part (a) to 700C (1290F) for 20 h.

SolutionHeating to 700C for 20 h the specimen in part (a) will transform the coarse pearlite and martensite to spheroidite.

(c) Rapidly cool to 600C (1110F), hold for 4 s, rapidly cool to 450C (840F), hold for 10 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling to and holding at 600C for 4 s, approximately 50% of the specimen has transformed to pearlite (medium). During the rapid cooling to 450C no transformations occur. At 450C we start timing again at zero time; while holding at 450C for 10 s, approximately 50 percent of the remaining unreacted 50% (or 25% of the original specimen) will transform to bainite. And upon cooling to room temperature, the remaining 25% of the original specimen transforms to martensite. Hence, the final microstructure consists of about 50% pearlite (medium), 25% bainite, and 25% martensite.

(d) Cool rapidly to 400C (750F), hold for 2 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling to and holding at 400C for 2 s, no of the transformation begin lines have been crossed, and therefore, the specimen is 100% austenite. Upon cooling rapidly to room temperature, all of the specimen transforms to martensite, such that the final microstructure is 100% martensite.

(e) Cool rapidly to 400C (750F), hold for 20 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 400C for 20 s, approximately 40% of the specimen has transformed to bainite. Upon cooling to room temperature, the remaining 60% transforms to martensite. Hence, the final microstructure consists of about 40% bainite and 60% martensite.

(f) Cool rapidly to 400C (750F), hold for 200 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 400C for 200 s, the entire specimen has transformed to bainite. Therefore, during the cooling to room temperature no additional transformations will occur. Hence, the final microstructure consists of 100% bainite.

(g) Rapidly cool to 575C (1065F), hold for 20 s, rapidly cool to 350C (660F), hold for 100 s, then quench to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 575C for 20 s, the entire specimen has transformed to fine pearlite. Therefore, during the second heat treatment at 350C no additional transformations will occur. Hence, the final microstructure consists of 100% fine pearlite.

(h) Rapidly cool to 250C (480F), hold for 100 s, then quench to room temperature in water. Reheat to 315C (600F) for 1 h and slowly cool to room temperature.

SolutionBelow is Figure 10.22 upon which is superimposed the above heat treatment.

After cooling and holding at 250C for 100 s, no transformations will have occurredat this point, the entire specimen is still austenite. Upon rapidly cooling to room temperature in water, the specimen will completely transform to martensite. The second heat treatment (at 315C for 1 h)not shown on the above plotwill transform the material to tempered martensite. Hence, the final microstructure is 100% tempered martensite.

10.23 Name the microstructural products of eutectoid ironcarbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at the following rates:(a) 200C/s,(b) 100C/s, and(c) 20C/s.

SolutionWe are called upon to name the microstructural products that form for specimens of an iron-carbon alloy of eutectoid composition that are continuously cooled to room temperature at a variety of rates. Figure 10.27 is used in these determinations.(a) At a rate of 200C/s, only martensite forms.(b) At a rate of 100C/s, both martensite and pearlite form.(c) At a rate of 20C/s, only fine pearlite forms.

10.25 Cite two important differences between continuous cooling transformation diagrams for plain carbon and alloy steels.

SolutionTwo important differences between continuous cooling transformation diagrams for plain carbon and alloy steels are: (1) for an alloy steel, a bainite nose will be present, which nose will be absent for plain carbon alloys; and (2) the pearlite-proeutectoid noses for plain carbon steel alloys are positioned at shorter times than for the alloy steels.

10.29 On the basis of diffusion considerations, explain why fine pearlite forms for the moderate cooling of austenite through the eutectoid temperature, whereas coarse pearlite is the product for relatively slow cooling rates.

SolutionFor moderately rapid cooling, the time allowed for carbon diffusion is not as great as for slower cooling rates. Therefore, the diffusion distance is shorter, and thinner layers of ferrite and cementite form (i.e., fine pearlite forms).

10.33 Briefly explain why the hardness of tempered martensite diminishes with tempering time (at constant temperature) and with increasing temperature (at constant tempering time).

SolutionThis question asks for an explanation as to why the hardness of tempered martensite diminishes with tempering time (at constant temperature) and with increasing temperature (at constant tempering time). The hardness of tempered martensite depends on the ferrite-cementite phase boundary area; since these phase boundaries are barriers to dislocation motion, the greater the area the harder the alloy. The microstructure of tempered martensite consists of small sphere-like particles of cementite embedded within a ferrite matrix. As the size of the cementite particles increases, the phase boundary area diminishes, and the alloy becomes softer. Therefore, with increasing tempering time, the cementite particles grow, the phase boundary area decreases, and the hardness diminishes. As the tempering temperature is increased, the rate of cementite particle growth also increases, and the alloy softens, again, because of the decrease in phase boundary area.

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