For
½O_{2(g)}=\left[O\right]_{(1 wt\% in Fe)}
\Delta G^\circ =-111,300-6.41T J (i)
and thus, for
Si_{(1 wt\% in Fe))}+2O_{(1 wt\% in Fe))}=SiO_{2(s)}
\Delta G^\circ =-610,800+242.32T J (ii)
From Equation (i) at 1600°C,
\frac{h_{O(1 wt\%)}}{p^{1/2}_{o_2}}=2.746\times 10^{3} (iii)
and, from Equation (ii) at 1600°C,
\frac{a_{SiO_2}}{h_{Si(1 wt\%)}h^{2}_{O(1 wt\%)}}=2.380\times 10^{4} (iv)
Thus, with p_{O_2}=5.57\times 10^{-12} atm and a_{SiO_2}=1 ,Equation (iii) gives
h_{O(1 wt\%)}=6.48\times 10^3 (v)
and Equation (iv) gives
h^{2}_{O(1 wt\%)}h_{Si(1 wt\%)}=4.0\times 10^{-5} (vi)
Division of Equation (vi) by h^{2}_{O(1 wt\%)} from Equation (v) gives
h_{Si(1 wt\%)} =1 (vii)
In the previous treatment the assumption that Si obeys Henry’ s law leads to the conclusion that
h_{Si(1 wt\%)} =wt\%Si=1
At 1600° C, from Table 13.1,
e^{Si}_{O}=-0.14 e^{O}_{O}=-0.2
e^{O}_{Si}=-0.25 e^{Si}_{Si}=-0.32
Thus, from Equation (v),
\log fo+\log \left[wt\% O\right]=\log (6.48\times 10^{-3})
or
-0.2\times \left[wt\% O\right]-0.14\times \left[wt\% Si\right]+\log \left[wt\% O\right]=-2.188 (viii)
and from Equation (vii),
\log f_{Si}+\log \left[wt\% Si\right]=\log (1)
or
0.32\left[wt\% Si\right]-0.25 \left[wt\% O\right]+\log \left[wt\% Si\right] =0 (ix)
Computer solution of Equations (viii) and (ix) gives
\left[wt\% Si\right]=0.631 and \left[wt\% O\right]=0.00798
In the example in Section 13.3, in which the effect of dissolved oxygen was ignored and it was assumed that f_{Si}=1 , the equilibrium weight percentage of Si in iron when a_{SiO_2}=1 and p_{O_2}=5.57\times 10^{-12} atm was 1.0. It is of interest to determine which of the two initial assumptions, that (1) e^{Si}_{Si}=0 and (2) e^{\circ }_{Si}=e^{Si}_{O}=0 , contributes more to the error in the initial calculation. Use e^{Si}_{Si}=0.32 and assume that e^{O}_{Si} and e^{Si}_{O} are zero. From Equation (ix):
0.32\left[wt\% Si\right]+\log [wt\% Si]=0
which gives [wt% Si] = 0.629, and from Equation (viii),
-0.20\times [wt\% O]-0.24\times 0.629+\log [wt\% O]=-2.188
which gives [wt% O] = 0.00797. The error introduced by ignoring the interaction between Si and O in solution in Fe is thus seen to be negligible in comparison with that introduced by assuming that Si obeys Henry’ s law over some initial range of composition.
table 13.1 Interaction Coefficients for Dilute Solutions of elements Dissolved in Liquid Iron at 1600° C |
element( i) |
element(j) |
Al |
C |
Co |
Cr |
H |
Mn |
N |
Ni |
O |
P |
S |
Si |
Al |
4.8 |
11 |
ــــ |
ــــ |
(34) |
ــــ |
(0.5) |
ــــ |
-160 |
ــــ |
4.9 |
6 |
C |
(4.8) |
22 |
1.2 |
-2.4 |
(72) |
ــــ |
(11.1) |
1.2 |
(-9.7) |
ــــ |
9 |
10 |
Co |
ــــ |
(6) |
ــــ |
ــــ |
(11) |
ــــ |
(4.7) |
ــــ |
(2.6) |
ــــ |
ــــ |
ــــ |
Cr |
ــــ |
(-10) |
ــــ |
ــــ |
(-11) |
ــــ |
(-16.6) |
ــــ |
(-13) |
ــــ |
(-3.55) |
ــــ |
H |
1.3 |
6 |
0.18 |
-0.22 |
0 |
-0.14 |
ــــ |
0 |
ــــ |
1.1 |
0.8 |
2.7 |
Mn |
ــــ |
ــــ |
ــــ |
ــــ |
(-7.7) |
ــــ |
(-7.8) |
ــــ |
(0) |
ــــ |
(-4.3) |
(0) |
N |
0.3 |
13 |
1.1 |
-4.5 |
ــــ |
-2 |
0 |
1 |
5 |
5.1 |
1.3 |
4.7 |
Ni |
ــــ |
(5.9) |
ــــ |
ــــ |
(0) |
ــــ |
(4.2) |
0 |
(2.1) |
ــــ |
(0) |
(1.0) |
O |
-94 |
-13 |
0.7 |
-4.1 |
ــــ |
0 |
(5.7) |
0.6 |
-20 |
7 |
-9.1 |
-14 |
P |
ــــ |
ــــ |
ــــ |
ــــ |
(34) |
ــــ |
(11.3) |
ــــ |
(13.5) |
ــــ |
(4.3) |
(9.5) |
S |
5.8 |
(24) |
ــــ |
-2.2 |
(26) |
-2.5 |
(3.0) |
0 |
(-18) |
4.5 |
-2.8 |
6.6 |
Si |
(6.3) |
24 |
ــــ |
ــــ |
(76) |
0 |
(9.3) |
0.5 |
(-25) |
8.6 |
(5.7) |
32 |
some interaction coefficients e_i ^j \times 10^2 for dilute solutions of elements dissolved in liquid iron at 1600° C. values in parentheses are calculated from e_i ^j = (MW_{i}/MW_{j}) e_{j}^{i}.(from J. f. elliott, m. Gleiser, and v. ramakrishna, Thermochemistry for Steelmaking , vol. 2, addison-wesley, reading, ma, 1963.) |