From
Section 3, it can be concluded that precipitate types after tempering are mainly Cr
7C
3, Cr
23C
6. Experimental results in
Figure 6 show the morphology of carbides nucleated on oxides. By thermodynamics, the initial generation temperatures of carbides and oxide inclusions are calculated to explain this phenomenon. The solidification of molten steel can be divided into three stages, namely, liquid phase, solid–liquid dual phase, and solid phase. The liquidus temperature (T
l) and solidus temperature (T
s) of test steel are calculated to be 1365 K and 1682 K by the Empirical formula [
24], respectively. The results are similar to the calculation of the Thermo-Calc software.
where
w is the percentage of mass concentration for the element.
4.1. Formation of Oxides and Carbides in the Liquid Phase
Depending on the composition of molten steel, the possible forms of oxides in the steel include Cr
2O
3, Al
2O
3, MnO, SiO
2, V
2O
3, and MoO
2. The Gibbs free energy of formation of oxides in the liquid steel is obtained through the following chemical reaction, as shown in
Table 4.
In order to make the reaction (1~6) spontaneously proceed to the right, the condition to be reached is ∆G < 0.
where
,
,
are the activities of oxides M
xO, M, and O, respectively; [%M] and [%O] are the mass fractions of metal M and oxygen in molten steel, respectively;
f[M] and
f[O] are the activity coefficients of the metal elements M and oxygen, respectively, calculated by the Wagner model.
According to
Table 1, except for the high content of Cr, the content of other alloying elements is relatively low. So the mass fraction of the alloying elements is regarded as tending to zero, and the solvent is iron liquid whose concentration is close to 1. The Wagner model can be characterized by a single interaction coefficient, in which the activity coefficient expression of each element is
where
is the first interaction coefficient between the solute elements,
k is any solute component of the multiphase, and
j is the 2, 3, ..., n solute component. The first interaction coefficient of each element in the molten steel for O, Cr, Al, Mn, Si, V, Mo is shown in
Table 5 [
25].
According to Equation (4) and
Table 5, the activity coefficient of each element was obtained, which are
fC = 1.02,
fO = 0.06,
fCr = 0.71,
fAl = 1.45,
fMn = 0.82,
fSi = 2.14,
fV = 0.35, and
fMo = 0.76.
In the liquid phase (T > 1682 K), each activity coefficient and concentration are substituted into Equation (3), and the Gibbs free energy of the oxide is generated. The relationship diagram of ∆G ~ T is shown in the liquid zone of
Figure 10a. It can be seen that only Al
2O
3 is formed in the liquid phase, and other oxides cannot be formed in the liquid phase.
When the actual solubility product of the precipitate-forming element is larger than the equilibrium solubility product, carbide can be formed. The solubility product of Cr
7C
3 or Cr
23C
6 in liquid steel is deduced as follows:
According to the principle of precipitation thermodynamics, the equilibrium solubility product of M
xC
y in liquid steel can be expressed as
According to Equations (5)–(7), the solubility products of Cr
23C
6 and Cr
7C
3 in liquid steel are obtained as follows:
Put
fC = 1.02,
fCr = 0.71 into Equations (8) and (9), and the relationship between ln([%Cr]
x[%C]
y) and T is shown in the liquid zone of
Figure 8b. It can be seen that Cr
23C
6 and Cr
7C
3 cannot be formed in liquid steel.
4.2. Formation of Oxides and Carbides in Solid–Liquid Dual Phase
Due to the element segregation in the solidification front, the Scheil formula is introduced in the solidification process as follows:
where [%M] and [%N] are the mass fraction of the metal element M and the nonmetal element N in the solid–liquid phase during solidification;
[%M0] and [%N0] are the mass fraction of the metal element M and the nonmetal element N in the liquid steel before solidification;
K
M and K
N are the equilibrium solute partition coefficients for M and N, respectively. In the literature [
27], the equilibrium partition coefficients of solute elements oxygen, carbon, vanadium, chromium, molybdenum, and aluminum in the solidification process are 0.02, 0.17, 0.90, 1, 0.80, and 0.60, respectively;
T is the temperature of the system during solidification, K;
Tm is the melting point of pure iron, 1809 K;
Tl, Ts are the liquidus and solidus temperatures of steel, 1682 K and 1365 K, respectively.
Put the oxygen concentration and the metal element concentration calculated by the Scheil equation into Equation (2), and the relationship of oxides between ∆G and T can be obtained, as shown at the solid–liquid two-phase zone of
Figure 10a. In the same way, the relationship of carbides between ln([%Cr]
x[%C]
y) and T is shown in
Figure 10b. It can be seen from
Figure 10a that when the temperature is in the solid–liquid two-phase region, the Gibbs free energy of Al
2O
3 is much smaller than other oxides, so Al
2O
3 is easily formed. Due to the segregation of solidification, the concentration of the liquid phase in the solid–liquid two phases is greater than the solid phase.
The equilibrium solubility product curves of Cr
7C
3 and Cr
23C
6 in the liquid phase of the solidification front have intersections with the actual solubility product curves. With the decrease of temperature, the actual solubility product becomes greater than the equilibrium solubility product, and the precipitation thermodynamic condition is reached. The intersection points are the initial precipitation temperatures of precipitates in the solid–liquid two-phase zone, and the theoretical precipitation temperature of Cr
7C
3 and Cr
23C
6 are 1425 K and 1375 K, respectively, which can be judged by the intersection point [
28].
4.3. Formation of Carbides in Solid Phase
Since the Gibbs free energy of element dissolution in solid-phase ferrite has not been studied, empirical data in austenite is still used in the calculation process. During the solidification process, high-carbon 8 mass% Cr tool directly enters the austenite zone and has no solute in the high-temperature ferrite zone according to
Figure 2. The chemical reaction of Cr forming carbides in austenite can be obtained by Equations (13) and (14).
where δ represents high-temperature ferrite and γ represents austenite.
According to the equilibrium solid solubility formula of graphite C in austenite,
The free energy of dissolving graphite C into austenite is
According to Equations (13)–(16), the solubility products of Cr
23C
6 and Cr
7C
3 in austenite are obtained as follows:
Combined with the solubility product formula, the relationship between the solubility product and the temperature can be obtained for Cr
23C
6 and Cr
7C
3, as shown in
Figure 10b. It can be seen that Cr
23C
6 and Cr
7C
3 begin to precipitate. The temperature is lower and the difference between the actual solubility product and the equilibrium solubility product is larger. The carbide is easier to be formed. This is consistent with the statistical results in
Section 3.1, that is, the amount of carbides and carbides nucleated oxides in B is greater than in A. The thermodynamic calculation results of the precipitation of oxides and carbides are in agreement with the observations in
Section 3.2. Because Al
2O
3 precipitates in the liquid phase, Cr
7C
3 and Cr
23C
6 begin to precipitate in the solid–liquid two-phase region. Therefore, the phase nucleation of Al
2O
3 is precipitated first, and Cr
7C
3 and Cr
23C
6 are attached to the edge as the core.
4.4. The Improvement of Wear Resistance
Al
2O
3 inclusion could be taken as heterogeneous nuclei for carbide formation [
29,
30,
31]. Since the surface on the Al
2O
3 inclusion is a carbide, it is assumed that in hardness, the carbide nucleated on oxide is the same as the carbide. In the following discussion, both carbides and carbides nucleated on oxides are referred to as particles. S. Wei et al. reported carbides’ dispersed distribution can resist microcutting of hard abrasive particles [
23].
The distance between carbides decreases and it is more difficult for abrasives to cut into the matrix, so the wear resistance is better [
2]. To clearly compare the distance between adjacent particles, related parameters are listed in
Table 6. The calculation method is shown in Equations (19) and (20).
where S
i is the total area occupied by the particles,
i = 1 represents carbides, and
i = 2 represents carbides nucleated on oxides. N
i and d
j can be obtained by INCA Steel. Assume that the particles are homogeneously distributed in the test steel with equal diameter D
i, as shown by the solid circle in
Figure 11. The distance Z between the actual adjacent particles is calculated as follows:
where S (mm
2) is the area of the measured region. According to
Section 2.3, the measured area is 4 × 4 mm.
Figure 11 is a schematic diagram of carbides in steel and hard particles during the wear process. According to Equations (1) and (2), the distance Z between the actual adjacent particles was obtained for samples A and B, as shown in
Table 6 and
Figure 11.
The solid circle represents the actual particle (i.e., carbide of the carbide nucleated oxide) and the dotted circle is the auxiliary line to acquire Z in
Figure 11. As can be seen from
Figure 11, the distance Z between the actual adjacent particles is 201.36 μm for sample A, and Z for sample B is 121.30 μm. The triangle represents hard particle (i.e., silicon carbide particle). The length of hard particle embedded in the matrix for B is shorter than for A. So the resistance of sample B is prior to sample A. The conclusion is consistent with Zhang’s opinion [
2].