Temperature Dependence of CaO Desulfurization Mechanism in Molten Ni-Base Superalloy

Abstract

Desulfurization phenomenon of Ni-base superalloy using a solid CaO was identified by adjusting the temperature of melt. Ni-base superalloy and NiS powder were heated together at 1400 °C, 1500 °C, and 1600 °C. Dense and porous CaO rods were inserted into the melts individually. After the rods were pulled out, CaO, CaS, and calcium aluminates were detected on the rods by X-ray diffraction analysis. Electron probe microanalysis showed that only Ca, O, Al, and S distributed at the melt/CaO interfaces. At 1500 °C and 1600 °C, Al and S were also detected at particle boundaries in the rods. S contents in the alloys decreased as the desulfurization time passed and the temperature was raised. There was no prominent correlation between the rod porosity and the S contents. The desulfurization reaction was suggested to be that CaO, Al, and S react to generate CaS and calcium aluminates. When the temperature is high enough, the calcium aluminates form a solid–liquid coexisting state. Effective diffusion coefficient, which shows the S diffusivity in the generated layer at the melt/CaO interface, depends on the temperature and can be expressed by the Arrhenius equation. It has been supposed that the desulfurization reaction mainly occurs on CaO, and a macroscale surface area controls the desulfurization rate.

Appendix

Derivation of Thermodynamic Equilibrium Calculation

Thermodynamics equilibrium in Figure 13 is explained as follows. Gibbs free energies at 1873 K (1600 °C) are calculated by references.[28,29,30]

$$ 3 {\text{CaO }}\left( {\text{solid}} \right) + 2 {\text{Al }}\left( {\text{liquid}} \right) + 3/2{\text{S}}_{ 2} \left( {\text{gas}} \right) = 3 {\text{CaS }}\left( {\text{solid}} \right) + {\text{Al}}_{ 2} {\text{O}}_{ 3} \left( {\text{solid}} \right) $$

(A-1)

$$ \Delta G_{1}^{0} \, = \, - \,821{,}506 \, \left( {{\text{J}}/{\text{mol}}} \right) $$

(A-2)

$$ {\text{Al }}\left( {\text{liquid}} \right) = {\text{Al }}\left( {\text{in Ni}} \right) $$

(A-3)

$$ \Delta G_{2}^{0} = - 190,935 \, \left( {{\text{J}}/{\text{mol}}} \right) $$

(A-4)

$$ \frac{1}{2}S_{ 2} \, = \,S \, \left( {\text{in Ni}} \right) $$

(A-5)

$$ \Delta G_{3}^{0} \, = \, - \,90,214 \, \left( {{\text{J}}/{\text{mol}}} \right) $$

(A-6)

Therefore, the Gibbs free energy of the desulfurization reaction at 1873 K is

$$ 3 {\text{CaO }}\left( {\text{solid}} \right)\, + \, 2 {\text{Al }}\left( {\text{in Ni}} \right)\, + \, 3 {\text{S }}\left( {{\text{in}}\,{\text{Ni}}} \right)\, = \, 3 {\text{CaS }}\left( {\text{solid}} \right)\, + \,{\text{Al}}_{ 2} {\text{O}}_{ 3} \left( {\text{solid}} \right) $$

(A-7)

$$ \Delta G_{4}^{0} = \Delta G_{1}^{0} - 2 \times \Delta G_{2}^{0} - 3 \times \Delta G_{3}^{0} = - 1 6 8{,}9 9 3 { }\left( {{\text{J}}/{\text{mol}}} \right) $$

(A-8)

The Equilibrium constant K₁ is

$$ \Delta G_{4}^{0} = - RT\ln K_{1} $$

(A-9)

$$ K_{1} = \, 51,821 $$

(A-10)

K₁ is written as

$$ K_{1} = \frac{{a_{\text{CaS}}^{3} \times a_{{{\text{Al}}_{ 2} {\text{O}}_{ 3} }} }}{{a_{\text{CaO}}^{3} \times a_{\text{Al}}^{2} \times a_{\text{S}}^{3} }} $$

(A-11)

Therefore, logarithm of Al activity: ${ \log }(a_{\text{S}} )$ is

$$ a_{\text{S}}^{3} = \frac{{a_{\text{CaS}}^{3} \times a_{{{\text{Al}}_{ 2} {\text{O}}_{ 3} }} }}{{a_{\text{CaO}}^{3} \times a_{\text{Al}}^{2} \times K_1}} $$

(A-12)

$$ { \log }\left( {a_{S}^{3} } \right) {\text{ = log}}\left( { \frac{{a_{\text{CaS}}^{3} \times a_{{{\text{Al}}_{2} {\text{O}}_{3} }} }}{{a_{\text{CaO}}^{3} \times a_{\text{Al}}^{2} \times K_1}} } \right) $$

(A-13)

$$ { \log }\left( {a_{\text{S}} } \right) = - \frac{2}{3}{ \log }\left( {a_{\text{Al}} } \right) + \frac{1}{3}{ \log }\left( { \frac{{a_{\text{CaS}}^{3} \times a_{{{\text{Al}}_{2} {\text{O}}_{3} }} }}{{a_{\text{CaO}}^{3} \times K_1}} } \right) $$

(A-14)

Using the value of $a_{\text{CaO}}$ = 0.99, $a_{{{\text{Al}}_{2} {\text{O}}_{3} }}$ = 0.0095,[31] and assuming that $a_{\text{CaS}}$ = 1, Eq. [A-14] is written as

$$ { \log }\left( {a_{\text{S}} } \right) = - \frac{2}{3}\log \left( {a_{\text{Al}} } \right) - 2.24 $$

(A-15)

The Gibbs free energy of following reaction at 1873 K is[32]

$$ 3 {\text{CaO }}\left( {\text{solid}} \right) + {\text{Al}}_{ 2} {\text{O}}_{ 3} \left( {\text{solid}} \right) = 3 {\text{CaO}}\cdot \text{Al}_{ 2} {\text{O}}_{ 3} \left( {\text{solid}} \right) $$

(A-16)

$$ \Delta G_{5}^{0} = - \,76,613 \, \left( {{\text{J}}/{\text{mol}}} \right) $$

(A-17)

The following desulfurization reaction at 1873 K can be also calculated as

$$ 6 {\text{CaO }}\left( {\text{solid}} \right) + 2 {\text{Al }}\left( {\text{in Ni}} \right) + 3 {\text{S }}\left( {\text{in Ni}} \right) = 3 {\text{CaS }}\left( {\text{solid}} \right) + 3 {\text{CaO}}\cdot \text{Al}_{ 2} {\text{O}}_{ 3} \left( {\text{solid}} \right) $$

(A-18)

$$ \Delta G_{6}^{0} \, = \,\Delta G_{1}^{0} \, - \,2\, \times \,\Delta G_{2}^{0} \, - \,3\, \times \,\Delta G_{3}^{0} \, + \,\Delta G_{5}^{0} \, = \, - \,245{,}606 \, \left( {{\text{J}}/{\text{mol}}} \right) $$

(A-19)

The equilibrium constant K₂ is

$$ K_{ 2}=7{,}109{,}430 $$

(A-20)

Therefore, ${ \log }(a_{\text{S}} )$ is

$$ { \log }\left( {a_{\text{S}} } \right) = - \frac{2}{3}{ \log }\left( {a_{\text{Al}} } \right) + \frac{1}{3}{ \log }\left( { \frac{{a_{\text{CaS}}^{3} \times a_{{3{\text{CaO}} \cdot {\text{Al}}_{2} {\text{O}}_{3} }} }}{{a_{\text{CaO}}^{6} \times K_2}} } \right) $$

(A-21)

The values are $a_{\text{CaO}}$ = 1 and $a_{{{\text{Al}}_{2} {\text{O}}_{3} }}$ = 0.06,[32] therefore $a_{{3{\text{CaO}} \cdot {\text{Al}}_{2} {\text{O}}_{3} }}$ = 0.06 calculated by Eqs. [A-16] and [A-17]. The value of $a_{{3{\text{CaO}} \cdot {\text{Al}}_{2} {\text{O}}_{3} }}$ is assumed as 0.01 and 0.1 in this calculation. Assuming that $a_{\text{CaS}}$ = 1, Eq. [A-21] is

$$ { \log }\left( {a_{\text{S}} } \right)\,{ = }\, - \,\frac{2}{3}\log \left( {a_{\text{Al}} } \right)\, - \,2.95\quad \left( {{\text{When}} \, = \,0.0 1} \right) $$

(A-22)

$$ { \log }\left( {a_{\text{S}} } \right)\, = \, - \,\frac{2}{3}\log \left( {a_{\text{Al}} } \right)\, - \,2.62\quad \left( {{\text{When}}\, = \,0. 1} \right) $$

(A-23)

Derivation of log(a _Al) and log(a _S)

Logarithm of Al activity: a_Al can be written by activity coefficient: f_Al, and content of Al in the melt: [Al pct].

$$ a_{\text{Al}} = f_{\text{Al}} \left[ {{\text{Pct}}\,{\text{Al}}} \right] $$

(A-24)

In this paper, only Al and S in molten Ni are considered to derive a_Al. log(a_Al) is written by interaction parameters:$e_{i}^{j}$ as

$$ \log \left( {a_{\text{Al}} } \right) = e_{\text{Al}}^{\text{Al}} \left[ {{\text{Pct}}\,{\text{Al}}} \right] + e_{\text{Al}}^{\text{S}} \left[ {{\text{Pct}}\,{\text{S}}} \right] + \log \left[ {{\text{Pct}}\,{\text{Al}}} \right] $$

(A-25)

Similar to log(a_Al), log(a_S) is written as

$$ \log \left( {a_{S} } \right) = e_{\text{S}}^{\text{Al}} \left[ {{\text{Pct}}\,{\text{Al}}} \right] + e_{\text{S}}^{\text{S}} \left[ {{\text{Pct}}\,{\text{S}}} \right] + \log \left[ {{\text{Pct}}\,{\text{S}}} \right] $$

(A-26)

log(a_Al) and log(a_S) are calculated by these values: $e_{\text{Al}}^{\text{Al}}$ = 0.08,[29] $e_{\text{S}}^{\text{Al}}$ = 0.133,[30] $e_{\text{S}}^{\text{S}}$ = − 0.035,[31] and

$$ e_{\text{Al}}^{\text{S}} \, = \,e_{\text{S}}^{\text{Al}} \, \times \,\frac{{M_{\text{Al}} }}{{M_{\text{S}} }}\, + \,0.434\, \times \,10^{ - 2} \, \times \,\frac{{M_{\text{S}} - M_{\text{Al}} }}{{M_{\text{S}} }}\, = \,0. 1 1 3 $$

(A-27)