12, Describe the simplest heat treatment procedure that would be used in convert
ID: 592794 • Letter: 1
Question
12, Describe the simplest heat treatment procedure that would be used in converting a 0.76 wt%C steel from martensite to spheroidite. A. Rapidly cool to 600°C (1110°F), hold at this temperature for 7 s, rapidly cool to 45°C (840"F), hold at this B. C. D. E. temperature for 4 s, then quench to room temperature. Heat to a temperature in the vicinity of 700°c for on the order of 24 h, cool to room temperature Rapidly cool to 350°C, hold for 300 s, then quench to room temperature Rapidly cool to 675 'C, hold for 7 s, then quench to room temperature None of the above are the correct heat treatment procedureExplanation / Answer
If we simply consider the TTT diagram of 0.76 wt % C steel we find that for martensite to spheroidite heat to a temperature in the vicinity of 700 deg centrigade (i,e below eutectoid temperature for the order of 24 h) .The driving force for the formation of spheroidite is the net reduction in ferrite- cementite phase boundary area.
If we rapidly cool the specimen to 600 eg centrigade , 50% of the specimen transformed to Pearlite.during rapid cooling no transformation occurs but if we hold at this tepm for 4 s , 25% of the specimen is converted into bainite. Afterthat if we quench to room temperature 25% of the original specimen transforms into martensite.
Therefore final microstructure consists of 50% pearlite(medium), 25% bainite, 25% martensite.
After cooling & holding at 350 deg centrigade for 300 s, the entire specimen has transformed to bainite.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.