Chapter 10: Problem 25
Cite two important differences between continuous cooling transformation diagrams for plain carbon and alloy steels.
Chapter 10: Problem 25
Cite two important differences between continuous cooling transformation diagrams for plain carbon and alloy steels.
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Get started for freeBriefly describe the simplest heat treatment procedure that would be used in converting a \(0.76 \mathrm{wt} \% \mathrm{C}\) steel from one microstructure to the other, as follows: (a) Spheroidite to tempered martensite (b) Tempered martensite to pearlite (c) Bainite to martensite (d) Martensite to pearlite (e) Pearlite to tempered martensite (f) Tempered martensite to pearlite (g) Bainite to tempered martensite (h) Tempered martensite to spheroidite
(a) For the solidification of iron, calculate the critical radius \(r^{*}\) and the activation free energy \(\Delta G^{*}\) if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are \(-1.85 \times 10^{9} \mathrm{~J} / \mathrm{m}^{3}\) and \(0.204\) \(\mathrm{J} / \mathrm{m}^{2}\), 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 \(0.292 \mathrm{~nm}\) for solid iron at its melting temperature.
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) \(100 \%\) fine pearlite (b) \(100 \%\) tempered martensite (c) \(50 \%\) coarse pearlite, \(25 \%\) bainite, and \(25 \%\) martensite
(a) Briefly describe the microstructural difference between spheroidite and tempered martensite. (b) Explain why tempered martensite is much harder and stronger.
Compute the rate of some reaction that obeys Avrami kinetics, assuming that the constants \(n\) and \(k\) have values of \(3.0\) and \(7 \times 10^{-3}\), respectively, for time expressed in seconds.
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