# HW07 Soln - Problem 10.3 J. F. Shackelford, Introduction to...

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Unformatted text preview: Problem 10.3 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 10.3 Although Section 10.1 concentrates on crystal nucleation and growth from a liquid, similar kinetics laws apply to solid-state transformations. For example, Equation 10.1 can be used to describe the rate of β precipitation upon cooling supersaturated α phase in a 10 wt% Sn -90 wt%Pb alloy. Given precipitation rates of 3.77×103 s–1 and 1.40×103 s–1 at 20°C and 0°C respectively, calculate the activation energy for this process. SOLUTION Following Example 10.1 on page 308-9 of the text, begin by calculating the ratio of the two precipitation rates to eliminate variables. The operating equation is (10.1) of the text on page 308, which describes growth rate as an Arrhenius expression of the form ˙ G = Ce−Q/RT where C is a constant, Q is the activation energy for diffusion, R is the universal gas constant, and T is the absolute temperature. Writing the ratio of the rates at both temperatures, [rate]20◦ Ce−Q/R[293 K] = [rate]0◦ Ce−Q/R[273 K] Substituting, Q 1 3.77 × 103 s−1 = e− 8.314 J/mol·K ( 293 K 3 s− 1 1.40 × 10 − 1 273 K ) simplifying, Q= −[8.314 J/mol · K] ln 3.77 ￿1 ￿ 1.40 1 − 273 K 293 K and solving, Q = 32.9 x 103 J/mol = 32.9 kJ/mol Problem 10.3 Solution Professor R. Gronsky page 1 of 1 Problem 10.13 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 10.13 (a) A eutectoid steel is cooled at a steady rate from 727 to 200°C in exactly 1 day. Superimpose this cooling curve on the TTT diagram of Figure 10.11. (b) From the result of your plot for part (a), determine at what temperature a phase transformation would first be observed. (c) What would be the first phase to be observed? (Recall the approximate nature os uning an isothermal diagram to represent a continuous cooling process.) SOLUTION The superposition of a cooling curve on a TTT diagram is complicated by the logarithmic scale of the abscissa, as seen in Figure 10.11. Begin by calculating the cooling rate given in the problem statement ∆T 727◦ C − 200◦ C = = 6.1 × 10−3 ◦ C/s ∆t 24 hr × 3600 s/hr Now note the resulting temperature after appropriate time intervals along the logarithmic axis of the cooling curve, tabulated below t (sec) 1 10 100 1,000 2,000 3,000 10,000 ΔT (°C) 0.006 0.06 0.6 6.1 12.2 18.3 61 T (°C) ≈ 727 726.94 726.4 720.9 715 709 666 Next plot the last data in the first and last columns on the TTT curve of Figure 10.11 and connect with a continuous curve… Problem 10.13 Solution Professor R. Gronsky page 1 of 2 Problem 10.13 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) (a) The resulting "cooling curve" appears as follows, with the parenthetical note in the problem statement acknowledging that such a "continuous" curve can only be approximated on a TTT plot, which is only appropriate for isothermal transformations. (b) The decomposition of austenite by the eutectoid reaction marks the start of the phase transformation. From the intersection of the cooling curve and the TTT diagram, the transformation begins at approximately 710°C. (c) From the TTT diagram, the "first phase" to appear is actually two phases in coarse pearlitic morphology, namely α + Fe3C. Problem 10.13 Solution Professor R. Gronsky page 2 of 2 Problem 10.15 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 10.15 Repeat Problem 10.13 for a steady rate of cooling in exactly 1 second. Reference: 10.13 (a) A eutectoid steel is cooled at a steady rate from 727 to 200°C in exactly 1 day. Superimpose this cooling curve on the TTT diagram of Figure 10.11. (b) From the result of your plot for part (a), determine at what temperature a phase transformation would first be observed. (c) What would be the first phase to be observed? (Recall the approximate nature os uning an isothermal diagram to represent a continuous cooling process.) SOLUTION The superposition of a cooling curve on a TTT diagram is complicated by the logarithmic scale of the abscissa, as seen in Figure 10.11. Begin by calculating the cooling rate given in the problem statement ∆T 727◦ C − 200◦ C = = 527◦ C/s ∆t 1s Now note the resulting temperature after appropriate time intervals along the logarithmic axis of the cooling curve, tabulated below t (sec) 0.1 0.5 1 ΔT (°C) 52.7 264 527 T (°C) 674 463 200 Next plot the last data in the first and last columns on the TTT curve of Figure 10.11 and connect with a continuous curve… Problem 10.15 Solution Professor R. Gronsky page 1 of 2 Problem 10.15 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) (a) The resulting "cooling curve" appears as follows, with the parenthetical note in the problem statement acknowledging that such a "continuous" curve can only be approximated on a TTT plot, which is only appropriate for isothermal transformations. (b) The decomposition of austenite by the martensitic reaction marks the start of the phase transformation. From the intersection of the cooling curve and the MS temperature on the TTT diagram, the transformation begins at approximately 225°C. (c) From the TTT diagram, the first phase to appear is martensite. Problem 10.15 Solution Professor R. Gronsky page 2 of 2 Problem 10.31 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 10.31 In heat-treating a complex-shaped part made from a 5140 steel, a final quench in stirred oil leads to a hardness of Rockwell C30 at 3 mm beneath the surface. This hardness is unacceptable, as design specifications require a hardness of Rockwell C45 at that point. Select an alloy substitution to provide this hardness, assuming the heat treatment must remain the same. SOLUTION The only data available in the text for a design analysis of this problem is that found in the hardenability curves for various steels with the same carbon concentration presented in Figure 10.24 (p. 326 of the text). By inspection of Figure 10.24, it is seen that the Jominy curve for a 5140 steel reports a hardness value of C30 at a distance of ≈ ⅝ inch from the water-quenched end. This is clearly deeper than the 3 mm (⅛ inch) specified in the problem so this hardness (or greater) will certainly have been achieved at 3 mm depth beneath the surface. Problem 10.31 Solution Professor R. Gronsky page 1 of 2 Problem 10.31 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) Now using the same figure to assess if other alloys might be capable of achieving the higher required hardness of C45 at the same distance from the quenched end (implying the same heat treatment), the answer is found. There are two candidates with hardness values of at least Rockwell C45. Alloys 9849 (C 50) and 4340 (C 52) will both meet the design specification at the same heat treatment. Problem 10.31 Solution Professor R. Gronsky page 2 of 2 Problem 10.33 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) 10.33 Specify an aging temperature for a 95 Al-5 Cu alloy that will produce a maximum of 5 wt% θ precipitate. SOLUTION The Al-Cu phase diagram is required for this problem, and it is found on page 278 of the text, Figure 9.27, reproduced here. Begin by finding the 95 Al – 5Cu alloy of the problem statement and note that an alloy of this composition will lie within the κ + θ two phase field below 500°C. Problem 10.33 Solution Professor R. Gronsky page 1 of 2 Problem 10.33 J. F. Shackelford, Introduction to Materials Science for Engineers, 7th Edition, Prentice Hall, New Jersey (2009) Next, apply the lever rule to assess the amount of θ phase at different temperatures. wt% θ = 5 − Cκ × 100% Cθ − Cκ Assume that the composition of the θ phase is ≈ 53 wt% Cu at all temperatures below 500°C, based upon its value of 52.5 wt% at the eutectic reaction isotherm. The calculation then becomes wt% θ = 5 − Cκ × 100% 53 − Cκ Now note that at RT (25°C), wt% θ = 5−0 × 100% = 9.45% 53 − 0 5−1 × 100% = 7.69% 53 − 1 At 300°C, wt% θ = At 400°C, wt% θ = 5 − 2. 5 × 100% = 4.95% 53 − 2.5 5−5 × 100% = 0% 53 − 5 while at 500°C, wt% θ = Consequently, within the error of these assumptions, it can be concluded a two phase microstructure containing no more than 5 wt% θ phase can be generated by aging at approximately 400°C. Problem 10.33 Solution Professor R. Gronsky page 2 of 2 ...
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## This note was uploaded on 01/21/2011 for the course E 45 taught by Professor Gronsky during the Fall '08 term at University of California, Berkeley.

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