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9.52 Compute the mass fractions of proeutectoid ferrite and pearlite that form in an iron–carbon alloy
containing 0.25 wt% C.
Solution
The mass fractions of proeutectoid ferrite and pearlite that form in a 0.25 wt% C iron-carbon alloy
are considered in this problem. From Equation 9.20
Wp =C0' ! 0.022
0.74= 0.25 ! 0.022
0.74= 0.31
And, from Equation 9.21 (for proeutectoid ferrite)
W!' =0.76 " C0'
0.74= 0.76 " 0.25
0.74= 0.69
9.54 The mass fractions of total ferrite and total cementite in an iron-carbon alloy are 0.88 and 0.12,
respectively. Is this a hypoeutectoid or hypereutectoid alloy? Why?
Solution
In this problem we are given values of Wα and WFe3C for an iron-carbon alloy (0.88 and 0.12,
respectively), and then are asked to specify whether the alloy is hypoeutectoid or hypereutectoid.
Employment of the lever rule for total α leads to
W! = 0.88 =CFe3C " C0CFe3C " C!
= 6.70 " C06.70 " 0.022
Now, solving for C0, the alloy composition, leads to C0 = 0.82 wt% C. Therefore, the alloy is
hypereutectoid since C0 is greater than 0.76 wt% C.
9.55 The microstructure of an iron-carbon alloy consists of proeutectoid ferrite and pearlite; the mass
fractions of these microconstituents are 0.20 and 0.80, respectively. Determine the concentration of
carbon in this alloy.
Solution
We are asked in this problem to determine the concentration of carbon in an alloy for which
W!' =
0.20 and Wp = 0.80. If we let
C0' equal the carbon concentration in the alloy, employment of the
appropriate lever rule expression, Equation 9.20, leads to
Wp =C0' ! 0.022
0.74= 0.80
Solving for
C0' yields
C0' = 0.61 wt% C.
9.56 Consider 2.0 kg of a 99.6 wt% Fe–0.4 wt% C alloy that is cooled to a temperature just below the
eutectoid.
(a) How many kilograms of proeutectoid ferrite form?
(b) How many kilograms of eutectoid ferrite form?
(c) How many kilograms of cementite form?
Solution
In this problem we are asked to consider 2.0 kg of a 99.6 wt% Fe-0.4 wt% C alloy that is cooled to
a temperature below the eutectoid.
(a) Equation 9.21 must be used in computing the amount of proeutectoid ferrite that forms. Thus,
W!' =0.76 " C0'
0.74= 0.76 " 0.40
0.74= 0.49
Or, (0.49)(2.0 kg) = 0.98 kg of proeutectoid ferrite forms.
(b) In order to determine the amount of eutectoid ferrite, it first becomes necessary to compute the amount of total ferrite using the lever rule applied entirely across the α + Fe3C phase field, as
W! =CFe3C " C0
’
CFe3C " C!= 6.70 " 0.406.70 " 0.022
= 0.94
which corresponds to (0.94)(2.0 kg) = 1.88 kg. Now, the amount of eutectoid ferrite is just the difference
between total and proeutectoid ferrites, or
1.88 kg – 0.98 kg = 0.90 kg
(c) With regard to the amount of cementite that forms, again application of the lever rule across the entirety of the α + Fe3C phase field, leads to
WFe3C =C0’ ! C"
CFe3C ! C"= 0.40 ! 0.0226.70 ! 0.022
= 0.057
which amounts to (0.057)(2.0 kg) = 0.114 kg cementite in the alloy.
9.57 Compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron–
carbon alloy.
Solution
This problem asks that we compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron-carbon alloy. This requires that we utilize Equation 9.23 with
C1' = 2.14 wt% C,
the maximum solubility of carbon in austenite. Thus,
WFe3C' =C1
' ! 0.765.94
= 2.14 ! 0.765.94
= 0.232
9.58 Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and
proeutectoid cementite are 0.846 and 0.049, respectively? Why or why not?
Solution
This problem asks if it is possible to have an iron-carbon alloy for which Wα = 0.846 and
WFe3 ! C
= 0.049. In order to make this determination, it is necessary to set up lever rule expressions for these two
mass fractions in terms of the alloy composition, then to solve for the alloy composition of each; if both
alloy composition values are equal, then such an alloy is possible. The expression for the mass fraction of
total ferrite is
W! =CFe3C " C0
CFe3C " C!= 6.70 " C0
6.70 " 0.022= 0.846
Solving for this C0 yields C0 = 1.05 wt% C. Now for
WFe3 ! C we utilize Equation 9.23 as
WFe3C' =C1
' ! 0.765.94
= 0.049
This expression leads to
C1' = 1.05 wt% C. And, since C0 =
C1' , this alloy is possible.
9.60 Compute the mass fraction of eutectoid ferrite in an iron-carbon alloy that contains 0.43 wt% C.
Solution
In order to solve this problem it is necessary to compute mass fractions of total and proeutectoid
ferrites, and then to subtract the latter from the former. To calculate the mass fraction of total ferrite, it is necessary to use the lever rule and a tie line that extends across the entire α + Fe3C phase field as
W! =CFe3C " C0
CFe3C " C!= 6.70 " 0.43
6.70 " 0.022= 0.939
Now, for the mass fraction of proeutectoid ferrite we use Equation 9.21 as
W!' =0.76 " C0
' "0.74
= 0.76 " 0.430.74
= 0.446
And, finally, the mass fraction of eutectoid ferrite Wα'' is just
Wα'' = Wα – Wα' = 0.939 –0.446 = 0.493
9.62 The mass fraction of eutectoid ferrite in an iron-carbon alloy is 0.82. On the basis of this
information, is it possible to determine the composition of the alloy? If so, what is its composition? If this
is not possible, explain why.
Solution
This problem asks whether or not it is possible to determine the composition of an iron-carbon
alloy for which the mass fraction of eutectoid ferrite is 0.82; and if so, to calculate the composition. Yes, it
is possible to determine the alloy composition; and, in fact, there are two possible answers. For the first,
the eutectoid ferrite exists in addition to proeutectoid ferrite. For this case the mass fraction of eutectoid ferrite (Wα'') is just the difference between total ferrite and proeutectoid ferrite mass fractions; that is
Wα'' = Wα – Wα' Now, it is possible to write expressions for Wα (of the form of Equation 9.12) and Wα' (Equation 9.21) in terms of C0, the alloy composition. Thus,
W!" =CFe3C " C0
CFe3C " C!"
0.76 " C00.74
= 6.70 ! C06.70 ! 0.022
! 0.76 ! C00.74
= 0.82
And, solving for C0 yields C0 = 0.70 wt% C.
For the second possibility, we have a hypereutectoid alloy wherein all of the ferrite is eutectoid ferrite. Thus, it is necessary to set up a lever rule expression wherein the mass fraction of total ferrite is 0.82. Therefore,
W! =CFe3C " C0CFe3C " C!
= 6.70 " C06.70 " 0.022
= 0.82
And, solving for C0 yields C0 = 1.22 wt% C.
9.66 A steel alloy is known to contain 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C.
(a) What is the approximate eutectoid temperature of this alloy?
(b) What is the proeutectoid phase when this alloy is cooled to a temperature just below the
eutectoid?
(c) Compute the relative amounts of the proeutectoid phase and pearlite.
Assume that there are no alterations in the positions of other phase boundaries with the addition of Ni.
Solution
(a) From Figure 9.34, the eutectoid temperature for 6.0 wt% Ni is approximately 650°C (1200°F).
(b) From Figure 9.35, the eutectoid composition is approximately 0.62 wt% C. Since the carbon
concentration in the alloy (0.2 wt%) is less than the eutectoid (0.62 wt% C), the proeutectoid phase is
ferrite. (c) Assume that the α–(α + Fe3C) phase boundary is at a negligible carbon concentration.
Modifying Equation 9.21 leads to
W!' =0.62 " C0'
0.62 " 0= 0.62 " 0.20
0.62= 0.68
Likewise, using a modified Equation 9.20
Wp =C0' ! 00.62 ! 0
= 0.200.62
= 0.32