Question

Consider 3.5 kg of austenite containing 0.95 wt% c and 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?

Answer 1

A. The proeutectoid phase is Fe₃c because 0.95 wt/c  is greater than the eutectoid composition which is 0.76 wt/c

b.  We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is 
(0.14) × (3.5kg) = 0.49kg

c.  We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to 
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.