Question

Consider 6.9 kg of austenite containing 0.40 wt% C, cooled to below 727˚C (1341˚F).

(a) What is the proeutectoid phase?
(b) How many kilograms each of total ferrite and cementite form?
(c) How many kilograms each of pearlite and the proeutectoid phase form?

See Animated Figure 9.24

Answer 1

(a) Ferrite is the proeutectoid phase since 0.40 wt% C is less than 0.76 wt% C

(b) we are asked to determine how much total ferrite and cementite form. Application of lever rule yields:
W_α=C_Fe3C-C_o/C_Fe3C-C_α =6.7-0.40/6.7-0.022 =0.9433

which corresponds to (0.9433)(6.9 kg) = 6.508 kg of total ferrite.

Thus, total cementite is given as:

W_Fe3C=1-Wα =1-0.9433 = 0.0567

or (0.0567)(2.5 kg) = 0.391 kg of total cementite formed.

(c) Now consider the amounts of pearlite and proeutectoid ferrite. The fraction of pearlite is given as:

W_p=CO'-0.022/0.76-0.022 =0.40-0.022/0.76-0.022 =0.5121

Thus, amount of proeutectoid ferrite is given as:

W_α' = 1 – W_P = 1-0.5121 =0.4879

or, (0.4879)(6.9 kg) = 3.3665 kg of proeutectoid ferrite