Engineering Board Exam Reviewer · Engineering Materials · hard
For a hypereutectoid steel with 1.2% C cooled slowly to just below 727 °C, what is the mass fraction of proeutectoid cementite? (eutectoid 0.76% C, cementite 6.70% C)
- 0.1791
- 0.9259
- 0.1764
- 0.0741
Answer
D. 0.0741
Solution
W Fe3C' = (1.2 - 0.76)/(6.70 - 0.76) = 0.44/5.94 = 0.0741.
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