Final 2002
MATSE 259
Solutions to classwork #8
1. For a 99.65 wt% Fe – 0.35 wt% C alloy at a temperature just below the eutectoid, determine the following:
(a) The fractions of total ferrite and cementite phase.
(b) The fractions of the proeutectoid ferrite and pearlite.
(c) The fraction of eutectoid ferrite.
(a) This part of the problem is solved by application of the lever rule employing a tie line that extends all the way across the α + Fe3C phase field. Noting that the alloy composition is 99.65 wt% Fe – 0.35 wt% C, i.e. carbon concentration in the alloy, CO, is 0.35 wt%, the weight fraction of ferrite (α) is given as:
CFe C - Co
3
6.70 − 0.35
Wα = C
=
= 0.95
6.70 − 0.022
Fe3C - Cα and WFe3C = 1 − Wα = 0.05
(b) The fractions of proeutectoid ferrite and pearlite are determined by using the lever rule, and tie line that extends only to the eutectoid composition. Or
1
0.35 − 0.022
= 0.44
0.76 − 0.022
Wα' = 1 − Wp = 0.56
Wp =
and
(c) All ferrite is either proeutectoid or eutectoid. Therefore, the sum of these two ferrite fractions will equal the fraction of total ferrite, that is,
W α' + W αe = W α where Wαe denotes the fraction of the total alloy which is eutectoid ferrite. Values for Wα and Wα’ were determined in parts (a) and (b) as 0.95 and 0.56, respectively. Therefore,
Wαe = Wα − Wα' = 0.95 − 0.56 = 0.39
2. Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727°C.
(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?
(d) Schematically sketch and label the resulting microstructure.
This problem asks us to consider various aspects of 1.0 kg of austenite containing 1.15 wt% C that is cooled to below the eutectoid.
(a) The proeutectoid phase will be Fe3C since 1.15 wt% C is greater than the eutectoid (0.76 wt% C).
(b) For this portion of the problem, we are asked to determine how
Solutions to classwork #8
1. For a 99.65 wt% Fe – 0.35 wt% C alloy at a temperature just below the eutectoid, determine the following:
(a) The fractions of total ferrite and cementite phase.
(b) The fractions of the proeutectoid ferrite and pearlite.
(c) The fraction of eutectoid ferrite.
(a) This part of the problem is solved by application of the lever rule employing a tie line that extends all the way across the α + Fe3C phase field. Noting that the alloy composition is 99.65 wt% Fe – 0.35 wt% C, i.e. carbon concentration in the alloy, CO, is 0.35 wt%, the weight fraction of ferrite (α) is given as:
CFe C - Co
3
6.70 − 0.35
Wα = C
=
= 0.95
6.70 − 0.022
Fe3C - Cα and WFe3C = 1 − Wα = 0.05
(b) The fractions of proeutectoid ferrite and pearlite are determined by using the lever rule, and tie line that extends only to the eutectoid composition. Or
1
0.35 − 0.022
= 0.44
0.76 − 0.022
Wα' = 1 − Wp = 0.56
Wp =
and
(c) All ferrite is either proeutectoid or eutectoid. Therefore, the sum of these two ferrite fractions will equal the fraction of total ferrite, that is,
W α' + W αe = W α where Wαe denotes the fraction of the total alloy which is eutectoid ferrite. Values for Wα and Wα’ were determined in parts (a) and (b) as 0.95 and 0.56, respectively. Therefore,
Wαe = Wα − Wα' = 0.95 − 0.56 = 0.39
2. Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727°C.
(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?
(d) Schematically sketch and label the resulting microstructure.
This problem asks us to consider various aspects of 1.0 kg of austenite containing 1.15 wt% C that is cooled to below the eutectoid.
(a) The proeutectoid phase will be Fe3C since 1.15 wt% C is greater than the eutectoid (0.76 wt% C).
(b) For this portion of the problem, we are asked to determine how