With the aid of clear sketches, highlight two major difference between homogeneous and heterogeneous nucleation.
b) The iron–iron carbide eutectoid reaction is and is fundamental to the development of microstructure in steel alloys. On cooling, austenite, with an intermediate carbon content, transforms to a ferrite phase, having a much lower carbon content, and also cementite, with much higher carbon. Figure 4.1 is a temperature-timetransformation (TTT) diagram for the said process at 675℃.
i. Determine how long it takes for austenite to transform to pearlite at 675℃.
ii. How long would it takes austenite to transform to pearlite at 600℃ and 700℃.
iii. What can you deduce from solution in (i) and (ii) abov
For second part please provide the figure 4.1 as mentioned in the question. Without figure the problem can't be answered. So I humbly request you to upload the question with figure so that you can be helped with the answer.
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With the aid of clear sketches, highlight two major difference between homogeneous and heterogeneous nucleation. b)...
please answer question 2,3,4 A critical feature of steel is that a considerable amount of carbon can be dissolved in the austenite, phase, (up to 2.14 w/o at 1147"C), whereas carbon is essentially insoluble in ferrite, Cooling from point d to e, just above the eutectoid but still in the an increased fraction of the +y region, will produce phase and a microstructure similar to that shown: the particles precinitates out in the form af an intermetallic.compound called iron-carbide...