Thermodynamics

Abstract

Success and efficiency of a process is depending on the thermodynamic calculation of the reactions. How much oxygen requires to oxidize impurities present in hot metal should be known before the actual process.

Appendices

Probable Questions

• 1. Discuss oxidizing power of slag.

• 2. What do you mean by sulphide capacity of slag?

• 3. Discuss thermodynamics of refining and carbon–oxygen equilibrium reaction.

• 4. Discuss thermodynamics of silicon reaction.

• 5. What do you mean by ‘phosphorous participation’? State the conditions that are required for de-phosphorization of liquid steel bath. Draw a plot showing dependence of phosphorous participation on the FeO content and V-ratio of steelmaking slag.

• 6. Discuss thermodynamics of phosphorous reaction.

• 7. Discuss ionic theory of manganese transfer.

• 8. Discuss ionic theory of sulphur reaction.

• 9. Discuss thermodynamics for oxygen in molten steel.

• 10. Discuss thermodynamics of chromium reactions for stainless steel production.

• 11. ‘De-carburization of the bath is not possible without chromium oxidation’. Why?

  What are the thermodynamic alternatives for that?

• 12. Discuss thermodynamic of vacuum degassing in steelmaking.

• 13. Discuss thermodynamics of vacuum degassing for hydrogen.

• 14. Discuss thermodynamics of degassing for oxygen.

• 15. Discuss de-sulphurization during vacuum.

• 16. Find out the correlation for equation of sulphur partition and sulphur capacity.

Examples

Example 20.1

In a steelmaking process to stop C–O₂ reaction at 0.1%C; (i) what is the equilibrium [O] presence in steel at 1600 °C? (ii) If the actual [O] presence is 0.025%, then what is the excess [O] presence in steel? (iii) What is the [C]% retain in steel at that [O] content (if all [O] consume by [C]) ?

${\text{Given:}}\log {k}_{\text{CO}} = \left( {\frac{1056}{T}} \right) + 2.131\;\left( {{\text{for}}\,{\text{C}} - {\text{O}}_{2} \,{\text{reaction}}} \right)$

Solution

• (i) $[{\text{C}}] + [{\text{O}}] = \{ {\text{CO}}\} ,$

  $$ \log {k}_{\text{CO}} = \left( {\frac{1056}{T}} \right) + 2.131 $$

T = 1600 + 273 = 1873 K

$\log {k}_{\text{CO}} = \left( {\frac{1056}{1873}} \right) + 2.131 = 2.6748$

Therefore, ${k}_{\text{CO}} = 495.22 = \left( {\frac{{p_{CO} }}{{[h_{{{\text{C}}]}} [h_{{{\text{O}}]}} }}} \right)$

Since [h_i] = f_i [wt% i].

Therefore, ${f}_{\text{c}} [{\text{wt}}\,\% {\text{C}}] \times {f}_{\text{o}} [{\text{wt}}\,\% {\text{O}}] = \left( {\frac{{p_{CO} }}{{k_{CO} }}} \right)$

Now putting, f_c = 1 and f_o = 1 [for dilute solution], $p_{\text{CO}} = 1\,{\text{atm}}$

$[{\text{wt}}\,\% {\text{C}}] \times [{\text{wt}}\,\% {\text{O}}] = \left( {1/{k}_{\text{CO}} } \right) = \left( {1/495.22} \right) = 2.019 \times 10^{ - 3}$

Since [wt% C] = 0.1.

Therefore, [wt% O]_eq = {(2.019 × 10⁻³)/0.1} = 0.02%.

• (ii) Excess [O] = [O]_ac − [O]_eq = 0.025 – 0.02 = 0.005%.

• (iii) Since [wt% C] x [wt% O] = 2.019 × 10⁻³ and [wt% O]_ac = 0.025%.

Therefore, [wt% C] = {(2.019 × 10⁻³)/0.025} = 0.081%.

Example 20.2

Find out activity of [O] content in liquid steel during reaction of Si at 1600 °C. [Si] content in liquid is 0.8%.

${\text{Given}}\log k_{{{\text{SiO}}_{2 } }} = \left( {\frac{ 2 9 , 7 0 0}{T}} \right) - 11.24.$

Solution

$$ [{\text{Si}}] + 2[{\text{O}}] = \left( {{\text{SiO}}_{2} } \right) $$

(1)

$$ k_{{{\text{SiO}}_{2} }} = \left( {\frac{{(a_{{SiO_{2 } )}} }}{{[a_{{{\text{Si}} ] }} [a_{{{\text{O}}]}}^{2 } }}} \right) $$

(2)

T = 1600 + 273 = 1873 K.

Therefore, log $k_{{{\text{SiO}}_{2 } }} = [( 2 9 , 7 0 0/1873){-}11.24] = 4.617$

$k_{{{\text{SiO}}_{2 } }} = 41391.77$

Considering formation of SiO₂ is pure form, so $\left( {a_{{{\text{SiO}}_{2 } }} } \right) = 1$ and

[a_Si] = f_Si [wt% Si] = [wt% Si] since for dilute solution, f_Si = 1.

From Eq. 2:

$\left[ {{a}_{\text{O}} } \right]^{2} = \left( {\frac{{(a_{{{\text{SiO}}_{2 } )}} }}{{k_{{{\text{SiO}}_{2 } }} \left[ {Wt\% \,{\text{Si}}} \right]}}} \right) = \left( {\frac{1}{{\left( {41391.77 \times 0.8} \right)}}} \right) = 3.0199 \times 10^{ - 5}$

Therefore, [a_O] = 5.495 × 10⁻³.

Example 20.3

Calculate the concentration of oxygen in molten iron at 1600⁰ C in equilibrium with pure FeO. Given log k_Fe = −(6372/T) + 2.73.

Solution

$$ {\text{FeO}}({\text{l}}) = {\text{Fe}}({\text{l}}) + {\text{O}}({\text{wt}}\% ) $$

(1)

$$ {\text{So}} k_{\text{Fe}} = \left\{ {\left[ {{h}_{\text{o}} } \right] \cdot \left[ {{a}_{\text{Fe}} } \right]/\left( {{a}_{\text{FeO}} } \right)} \right\} $$

(2)

Since a_FeO = 1 (for pure FeO) and a_Fe = 1 (assume pure Fe).

Again [h_o] = [f_o], [W_o] = [W_o] (if we consider for dilute solution, f_o = 1).

Now from Eq. (2), we get:

$$ {k}_{\text{Fe}} = \left[ {{W}_{\text{o}} } \right] $$

(3)

T = 1600 + 273 = 1873 K.

log k_Fe = −(6372/T) + 2.73 = [−(6372/1873) + 2.73] = −0.672.

Therefore, k_Fe = 0.213.

Hence, the concentration of oxygen in molten iron at 1600 ℃, [W_o] = 0.213%.

Example 20.4

Calculate the residual oxygen content of liquid iron containing 0.1 wt% Si in equilibrium with solid silica at 1600 ℃.

$\begin{aligned} & {\text{Given}}:\quad {\text{Si}}({\text{l}}) + {\text{O}}_{2} ({\text{g}}) = {\text{SiO}}_{2} ({\text{s}}),\Delta {G}^{\text{o}}_{1} = - 947.68 + 0.199\,{T}\,{\text{kJ}}/{\text{mol}} \\ & \quad \quad \quad \quad {\text{O}}_{2} ({\text{g}}) = 2[{\text{O}}](1\,{\text{wt}}\% \,{\text{std state}}),\Delta {G}^{\text{o}}_{2} = - 233.47{-}0.006\,{T}\,{\text{kJ}}/{\text{mol}} \\ & \quad \quad \quad \quad {\text{Si}}({\text{l}}) = [{\text{Si}}](1\,{\text{wt}}\% \,{\text{std state}}),\Delta {G}^{\text{o}}_{3} = - 119.24{-}0.053\,{T}\,{\text{kJ}}/{\text{mol}} \\ & \quad {e}_{\text{Si}}^{\text{Si}} = 0.32,{e}_{\text{Si}}^{\text{O}} = - 0.24,{e}_{\text{O}}^{\text{O}} = - 0.20,{\text{and}}\,{e}_{\text{O}}^{\text{Si}} = -0.14. \\ \end{aligned}$

Solution

$$ \begin{aligned} & \quad [{\text{Si}}](1\,{\text{wt}}\% \,{\text{std state}}) = {\text{Si}}({\text{l}}) - \Delta {G}^{\text{o}}_{3} \\ & \quad 2[{\text{O}}](1\,{\text{wt}}\% \,{\text{std state}}) = {\text{O}}_{2} ({\text{g}}) - \Delta {G}^{\text{o}}_{2} \\ & \quad {\text{Si}}({\text{l}}) + {\text{O}}_{2} ({\text{g}}) = {\text{SiO}}_{2} ({\text{s}}),\quad \Delta {G}^{\text{o}}_{1} \\ & \quad [{\text{Si}}](1\,{\text{wt}}\% \,{\text{std state}}) + 2[{\text{O}}](1\,{\text{wt}}\% \,{\text{std state}}) = {\text{SiO}}_{2} ({\text{s}})\Delta {G}^{\text{o}}_{\text{r}} \\ \end{aligned} $$

$\begin{aligned} {\text{Therefore}}\,\Delta {G}^{\text{o}}_{\text{r}} & = \Delta {G}^{\text{o}}_{1} - \Delta {G}^{\text{o}}_{2} - \Delta {G}^{\text{o}}_{3} \\ & = ( - 947.68 + 0.199\,{T}){-}\{ ( - 233.47{-}0.006\,{T}) + ( - 119.24{-}0.053\,{T})\} ] \\ & = - 594.97 + 0.258\,{T}\,{\text{kJ}}/{\text{mol}}\quad \left[ {{T} = 1600 + 273 = 1873\,{\text{K}}} \right] \\ & = ( - 594.97 + 0.258 \times 1873) \times 1000\,{\text{J}}/{\text{mol}} \\ & = - 111736\,{\text{J/mol}} \\ \end{aligned}$

Since ΔG^o_r = −RT ln k.

Or −111,736 = −8.314 × 1873 × ln k.

Or ln k = 7.175.

$$ \begin{aligned} \log {f}_{\text{o}} & = {e}_{\text{o}}^{\text{O}} [{\text{wt}}\% \,{\text{O}}] + {e}_{\text{o}}^{\text{Si}} [{\text{wt}}\% \,{\text{Si}}] = ( - 0.2) \times ({\text{wt}}\% \,{\text{O}}) + ( - 0.14) \times (0.1) \\ & = - 0.2\,{\text{wt}}\% \,{\text{O}} - 0.014 \\ \end{aligned} $$

$$ \begin{aligned} \log {f}_{\text{Si}} & = {e}_{\text{Si}}^{\text{Si}} [{\text{wt}}\% \,{\text{Si}}] + {e}_{\text{Si}}^{\text{O}} [{\text{wt}}\% \,{\text{O}}] = 0.32 \times 0.1 + ( - 0.24) \times ({\text{wt}}\% \,{\text{O}}) \\ & = 0.032 - 0.24{\text{wt}}\% \,{\text{O}} \\ \end{aligned} $$

Equilibrium constant, k = [a_SiO2/(h_Si · h²_O)] = [1/(h_Si · h²_O)] (since a_SiO2 = 1).

Again h_i = f_i · wt% i.

Therefore, k = [1/{(f_Si · wt% Si) · (f_O · wt% O)²}].

Taking both side logs:

$\begin{aligned} \log {k} & = - \left[ {\left\{ {\log {f}_{\text{Si}} + \log ({\text{wt}}\% \,{\text{Si}})} \right\} + 2\left\{ {\log {f}_{\text{O}} + \log ({\text{wt}}\% \,{\text{O}})} \right\}} \right] \\ & = - \left[ {\left\{ {(0.032 - 0.24\,{\text{wt}}\% \,{\text{O}}) + \log (0.1)} \right\} + 2\left\{ {( - 0.2\,{\text{wt}}\% \,{\text{O}} - 0.014) + \log ({\text{wt}}\% \,{\text{O}})} \right\}} \right] \\ & = - \left[ {\left\{ {(0.032 - 0.24\,{\text{wt}}\% \,{\text{O}}) + ( - 1)} \right\} + \left\{ {( - 0.4\,{\text{wt}}\% \,{\text{O}} - 0.028) + 2\log ({\text{wt}}\% \,{\text{O}})} \right\}} \right] \\ & = \left[ {0.996 + 0.64{\text{wt}}\% \,{\text{O}} - 2\log ({\text{wt}}\% \,{\text{O}})} \right] \\ \end{aligned}$

Since value of wt% O is very small, value of log (wt% O) can be taken as negligible.

[since ln x = 2.303 log x; ln k = 7.175 = 2.303 log k,

Therefore, log k = (7.175/2.303) = 3.1155].

So log k = 0.996 + 0.64 wt% O = 3.1155.

Therefore, wt% O = 3.31.

Example 20.5

Find out the oxygen concentration in liquid steel after degassing at 1600 ℃ and 1.0 torr. [wt% C] = 0.2.

Given:

• (i) log k_O = (6120/T) + 0.15.

• (ii) log k_CO = (1168/T) + 2.07.

Solution

${T} = 1600 + 273 = 1873\,{\text{K}},760\,{\text{torr}} = 1\,{\text{atm}},{\text{so}}\,1.0\,{\text{torr}} = (1/760)\,{\text{atm}}$

$$ 2[{\text{O}}] = {\text{O}}_{2} $$

(1)

Equilibrium constant,

$$ {k}^{{\prime }}_{\text{O}} = \left[ {{p}_{{{\text{O}}2}} /\left( {{h}_{\text{O}} } \right)^{2} } \right] $$

(2)

Since h_i = f_i × wt% i = wt% i, due to f_i = 1.

From Eq. (2):

$$ {\text{wt}}\% \,{\text{O}} = \surd \left( {{p}_{{{\text{O}}2}} /{k}^{{\prime }}_{\text{O}} } \right) = {k}_{\text{O}} \,\times \,\surd {p}_{{{\text{O}}2}} $$

(3)

Since k_O = √(1/k^′_O).

Again log k_O = (6120/T) + 0.15 = (6120/1873) + 0.15 = 3.417.

Therefore, k_O = 2615.082; now putting this value to Eq. (3):

wt% O = k_O × √p_O2 = 2615.08 × √(1/760) = 94.86%.

This high value of oxygen content in steel is impossible; hence, oxygen cannot be removed by this way. Therefore, oxygen in steel can be removed by reacting with carbon.

$$ [{\text{C}}] + [{\text{O}}] = \{ {\text{CO}}\} $$

(4)

Equilibrium constant,

$$ {k}_{\text{CO}} = \left[ {{p}_{\text{CO}} /\left( {{h}_{\text{C}} \times {h}_{\text{O}} } \right)} \right] = \left[ {{p}_{\text{CO}} /({\text{wt}}\% \,{\text{C}} \times {\text{wt}}\% \,{\text{O}})} \right] $$

(5)

Again log k_CO = (1168/T) + 2.07 = (1168/1873) + 2.07 = 2.693.

Therefore, k_CO = 493.853; since [wt% C] = 0.2 and p_CO = (1/760).

Putting these values in Eq. (5):

$\begin{aligned} {\text{Therefore}}\,{\text{wt}}\% \,{\text{O}} & = \left[ {{p}_{\text{CO}} /\left( {{\text{wt}}\% \,{\text{C}} \times {k}_{\text{CO}} } \right)} \right] = \left[ {(1/760)/(0.2 \times 493.853)} \right] \\ & = {\mathbf{1}}{\mathbf{.332 \times 10}}^{{{\mathbf{ -5}}}} {\mathbf{\% }} \\ \end{aligned}$

Example 20.6

Find out wt% O in liquid steel to remove C from HM (content 3% C, 1.2% Si, 0.8% Mn, 0.04% S, 0.4% P) at 1600 °C.

$\begin{aligned} & {\text{Given}}\,\log {k}_{\text{CO}} = \left[ {(1056/{T}) + 2.131} \right] \\ & {e}_{\text{c}}^{\text{c}} = 0.14,{e}_{\text{c}}^{\text{Mn}} = - 0.012,{e}_{\text{c}}^{\text{P}} = 0.051,{e}_{\text{c}}^{\text{S}} = 0.046,{e}_{\text{c}}^{\text{Si}} = 0.08 \\ & {e}_{\text{o}}^{\text{c}} = - 0.13,{e}_{\text{o}}^{\text{Mn}} = - 0.021,{e}_{\text{o}}^{\text{P}} = 0.07,{e}_{\text{o}}^{\text{S}} = - 0.133,{e}_{\text{o}}^{\text{Si}} = - 0.131. \\ \end{aligned}$

Solution

T = 1600 + 273 = 1873 K.

So, log k_CO = [(1056/1873) + 2.131] = 2.6948.

Therefore, k_CO = 495.22.

$\begin{aligned} {\text{Again}},\log {f}_{\text{c}} & = {e}_{\text{c}}^{\text{S}} [{\text{wt}}\% \,{\text{S}}] + {e}_{\text{c}}^{\text{C}} [{\text{wt}}\% \,{\text{C}}] + {e}_{\text{c}}^{\text{Si}} [{\text{wt}}\% \,{\text{Si}}] + {e}_{\text{c}}^{\text{Mn}} [{\text{wt}}\% \,{\text{Mn}}] \\ & \quad + {e}_{\text{c}}^{\text{P}} [{\text{wt}}\% \,{\text{P}}] \\ & = 0.046 \times 0.04 + 0.14 \times 3.0 + 0.08 \times 1.2 + ( - 0.012) \times 0.8 + 0.051 \times 0.4 \\ & = 0.5286 \\ \end{aligned}$

Therefore, f_c = 3.378.

$\begin{aligned} {\text{Similarly}},\log {f}_{\text{o}} & = {e}_{\text{o}}^{\text{S}} [{\text{wt}}\% \,{\text{S}}] + {e}_{\text{o}}^{\text{C}} [{\text{wt}}\% \,{\text{C}}] + {e}_{\text{o}}^{\text{Si}} [{\text{wt}}\% \,{\text{Si}}] + {e}_{\text{o}}^{\text{Mn}} [{\text{wt}}\% \,{\text{Mn}}] \\ & \quad + {e}_{\text{o}}^{\text{P}} [{\text{wt}}\% \,{\text{P}}] \\ & = ( - 0.133) \times 0.04 + ( - 0.13) \times 3.0 + ( - 0.131) \times 1.2 + ( - 0.021) \times 0.8 \\ & \quad + 0.07 \times 0.4 \\ & = - 0.54132 \\ \end{aligned}$

Therefore, f_o = 0.2875.

We know that k_CO = {p_co/h_c × h_o}.

Therefore, h_c × h_o = {p_co/k_CO} = {1/k_CO} {since p_co = 1 and h_i = f_i × [wt% i]}.

So, f_c × [wt% C] × f_o × [wt% O] = {1/k_CO}.

$\begin{aligned} {\text{Therefore}},[{\text{wt}}\% \,{\text{O}}] & = \left\{ {1/{k}_{\text{CO}} } \right\} \times \left\{ {1/\left( {{f}_{\text{o}} \times {f}_{\text{c}} \times [{\text{wt}}\% \,{\text{C}}]} \right)} \right\} \\ & = \left\{ {1/\left( {495.22 \times 0.2875 \times 3.378 \times 3.0} \right)} \right\} \\ & = 1/1442.8359 = {\mathbf{6}}{\mathbf{.93 \times 10}}^{{{\mathbf{ -4}}}} \\ \end{aligned}$

Example 20.7

Calculate the chemical potential of nitrogen in liquid steel at 1600 °C. Steel contents 0.5% C, 0.5% Mn, 0.2% P and 0.01% N.

Given: [wt% N] f_N = k_N (p_N2)^1/2, where log k_N = {(−188.1/T) − 1.246}.

e^C_N = 0.25, e^N_N = 0, e^P_N = 0.051, e^Mn_N = −0.02.

Solution

T = 1600 + 273 = 1873 K.

$\log {k}_{\text{N}} = \left\{ {( - 188.1/1873){-}1.246} \right\} = - 1.3464$

Therefore, k_N = 0.045.

$\begin{aligned} {\text{Again}},\log {f}_{\text{N}} & = {e}_{\text{N}}^{\text{C}} {\text{wt}}\% {\text{C}} + {e}_{\text{N}}^{\text{N}} {\text{wt}}\% {\text{N}} + {e}_{\text{N}}^{\text{P}} {\text{Wt}}\% {\text{P}} + {e}_{\text{N}}^{\text{Mn}} {\text{Wt}}\% {\text{Mn}} \\ & = 0.25 \times 0.5 + 0 \times 0.01 + 0.051 \times 0.2 + ( - 0.02) \times 0.5 \\ & = 0.1252 \\ \end{aligned}$

Therefore, f_N = 1.334.

Since k_N (p_N2)^1/2 = [wt% N] f_N = 0.01 × 1.334 = 0.01334.

Therefore, (p_N2)^1/2 = 0.01334/0.045 = 0.296.

So, p_N2 = 0.0876.

$\begin{aligned} {\text{Chemical potential of nitrogen in liquid steel}} & = {\upmu }_{{{\text{N}}2}} = {\text{RT}}\ln {\text{p}}_{{{\text{N}}2}} \\ & = 8.314 \times 1873 \times \ln (0.0876) \\ & = - {\mathbf{37}}{\mathbf{.865}}\,{\mathbf{kJ}}/{\mathbf{mol of N}}_{2} \\ \end{aligned}$

Example 20.8

Calculate the oxygen potential of liquid steel in contact with a molten slag at 1600 °C. Assume equilibrium partitioning of oxygen between slag and metal.

Given:

• (1) [O]_wt% + Fe (l) = FeO (l), k₁ = 4.35 at 1600 °C.

• (2) O₂ (g) = 2 [O]_wt%, k₂ = 6.89 × 10⁶ at 1600 °C.

Solution

T = 1600 + 273 = 1873 K.

k₁ = [a_FeO/(a_O · a_Fe)] = (1/h_O) [since a_Fe = 1 and a_FeO = 1, pure at std state].

Therefore, h_O = (1/k₁) = (1/4.35) = 0.23.

Again k₂ = [(h_O)²/P_O2].

Or P_O2 = [(h_O)²/k₂] = [(0.23)²/(6.89 × 10⁶)] = 7.67 × 10⁻⁹.

$\begin{aligned} {\text{Therefore}},{\text{oxygen potential of liquid steel}} & = {\mu }_{{{\text{O}}2}} = {\text{RT}}\ln {p}_{{{\text{O}}2}} \\ & = 8.314 \times 1873 \times \ln \left( {7.67 \times 10^{ - 9} } \right) \\ & = - 290979.63\,{\text{J}}/{\text{mol}} \\ & = - {\mathbf{290}}{\mathbf{.98}}\,{\mathbf{kJ}}/{\mathbf{mol}} \\ \end{aligned}$

Example 20.9

Calculate and compare thermodynamic efficiency of de-sulphurization of molten steel by pure CaO and molten slag of CaO–SiO₂–Al₂O_3. Also find out residual sulphur content in metal if residual oxygen content is 0.001 wt%.

$\begin{aligned} & {\text{Given}}:\frac{{\left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}} = 3 \times 10^{ - 2} ,{\text{for X}}_{\text{CaO}} = 0.6, X_{{SiO_{2} }} = 0.1\,{\text{and}}\,X_{{Al_{2} O_{3} }} = 0.3\,{\text{at}}\,1650\,{^\circ {\text{C}}}. \\ & \quad \quad {\text{C}}_{\text{S}}^{{\prime }} = 7.5 \cdot {\text{C}}_{\text{S}} ,{\text{slag}},{\text{C}}_{\text{S}} = 0.2. \\ \end{aligned}$

Solution

Assume, wt% of S in slag as 2. i.e. (w_S) = 2.

At 1650 °C, for CaO,

$\frac{{\left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}} = 3 \times 10^{ - 2} \,({\text{given}})$

For slag, C_S = 0.2.

Now,

${C}_{\text{S}}^{{\prime }} = \frac{{\left( {{w}_{\text{S}} } \right) \cdot \left[ {{h}_{\text{O}} } \right]}}{{\left[ {{h}_{\text{S}} } \right]}}$ (Eq. 20.155)

${C}_{\text{S}}^{{\prime }} = \frac{{\left( {{w}_{\text{S}} } \right) \cdot \left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}}$ (Taking [h_O] = [w_O] and [h_S] = [w_S] as approximation).

$\frac{{{\text{C}}_{\text{S}}^{{\prime }} }}{{\left( {{\text{w}}_{\text{S}} } \right)}} = \frac{{\left[ {{\text{w}}_{\text{O}} } \right]}}{{\left[ {{\text{w}}_{\text{S}} } \right]}}$

$\frac{{7.5 \cdot {\text{C}}_{\text{S}} }}{2} = \frac{{\left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}}$

$\frac{(7.5) \cdot (0.2)}{2} = \frac{{\left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}}$

$\frac{{\left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}} = 0.75$

$\frac{{\left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}}$ is used as index of thermodynamic efficiency.

As given, w_O = 0.001, then

$\frac{{\left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}} = 0.75$

$\frac{0.001}{{\left[ {{w}_{\text{S}} } \right]}} = 0.75$

$\left[ {{w}_{\text{S}} } \right] = \frac{0.001}{0.75}$

${\mathbf{w}}_{{\mathbf{S}}} = {\mathbf{1}}{\mathbf{.33 \times 10}}^{{{\mathbf{ - 3}}}}$ (answer for slag)

If W_O = 0.001, then

$\frac{{\left[ {{w}_{\text{O}} } \right]}}{{\left[ {{w}_{\text{S}} } \right]}} = 3 \times 10^{ - 2}$

$\left[ {{w}_{\text{S}} } \right] = \frac{0.001}{{3 \times 10^{ - 2} }}$

$\left[ {{w}_{\text{S}} } \right] = 0.33$ for pure CaO.

The thermodynamic efficiency,

$$ \left( {\frac{slag}{CaO}} \right) = \frac{0.75}{{3 \times 10^{ - 2} }} = 25 $$

So, the slag is 25 times more effective as a de-sulphurizer as compared to pure lime.

Problems

Problem 20.1

Find out activity of [O] content in liquid steel during reaction of Mn at 1600 °C. Mn content in liquid steel is 0.8%.

${\text{Given}}:\log {k}_{\text{Mn}} = (12440/{T}){-}5.33.\quad [{\text{Ans}}:0.0609]$

Problem 20.2

Estimate the activity of S in liquid steel containing 0.04% S, 0.8% C, 0.1% Si, 0.5% Mn and 0.04% P at 1600 °C in 1 wt% standard state.

Given: e^C_S = 0.11, e^Si_S = 0.063, e^S_S = −0.028, e^Mn_S = -0.026 and e^P_S = 0.029.

[Ans: 0.048]

Problem 20.3

Carbon–oxygen reaction in steelmaking:

$[{\text{C}}] + [{\text{O}}] = \{ {\text{CO}}\}$

At 0.2% C, and p_CO = 10⁻⁵ atm; find out the [wt% O] at 1600 °C.

${\text{Given}}:\log {k} = \left( {\frac{1056}{T}} \right) + 2.131\quad \left[ {{\text{Ans}}: 10^{ - 7} {\text{wt}}\% \,{\text{O}}} \right]$

Problem 20.4

Solution of nitrogen in liquid iron may be assumed to obey Sievert’s law. Nitrogen content in liquid iron at 1873 K in equilibrium with 1 atm pressure of nitrogen is measured as 0.044 (mass%). What will be the equilibrium nitrogen content in liquid iron (mass%) if nitrogen is reduced to 0.25 atm? [Ans: 0.022%].