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9.19  Development of Microstructure in Iron–Carbon Alloys  •  343


                          Solution
                          (a)   This part of the problem is solved by applying the lever rule expressions using a tie
                             line that extends all the way across the a + Fe 3 C phase field. Thus, C  0  is 0.35 wt% C,
                             and

                                                            6.70 - 0.35
                                                     W a =            = 0.95
                                                           6.70 - 0.022
                             and

                                                            0.35 - 0.022
                                                    W Fe 3 C =         = 0.05
                                                            6.70 - 0.022
                          (b)   The fractions of proeutectoid ferrite and pearlite are determined by using the lever rule
                             and a tie line that extends only to the eutectoid composition (i.e., Equations 9.20 and 9.21).
                             We have
                                                           0.35 - 0.022
                                                     W p =            = 0.44
                                                           0.76 - 0.022
                             and

                                                            0.76 - 0.35
                                                     W a  =            = 0.56
                                                           0.76 - 0.022

                          (c)  All ferrite is either as proeutectoid or eutectoid (in the pearlite). Therefore, the sum of
                             these two ferrite fractions equals the fraction of total ferrite; that is,

                                                         W a  + W ae = W a
                             where W ae  denotes the fraction of the total alloy that is eutectoid ferrite. Values for W a  and
                             W a¿  were determined in parts (a) and (b) as 0.95 and 0.56, respectively. Therefore,

                                                W ae = W a - W a  = 0.95 - 0.56 = 0.39




                                 Nonequilibrium Cooling
                                 In this discussion of the microstructural development of iron–carbon alloys, it has been
                                                                                        3
                                 assumed that, upon cooling, conditions of metastable equilibrium  have been continu-
                                 ously maintained; that is, sufficient time has been allowed at each new temperature for
                                 any necessary adjustment in phase compositions and relative amounts as predicted
                                 from the Fe–Fe 3 C phase diagram. In most situations these cooling rates are impracti-
                                 cally slow and unnecessary; in fact, on many occasions nonequilibrium conditions are
                                 desirable. Two nonequilibrium effects of practical importance are (1) the occurrence of
                                 phase changes or transformations at temperatures other than those predicted by phase
                                 boundary lines on the phase diagram, and (2) the existence at room temperature of
                                 nonequilibrium phases that do not appear on the phase diagram. Both are discussed in
                                 Chapter 10.




              3 The term metastable equilibrium is used in this discussion because Fe 3 C is only a metastable compound.
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