Oxygen Steelmaking Processes

Abstract

Instate of blowing air from the bottom of the Bessemer converter, in LD process, blowing oxygen from top of the converter. Hot metal, contains high or medium phosphorus, is used for LD process, to remove phosphorus from the hot metal, lining of converter is always basic. In LD converter, reactions are basically slag-metal reactions. The mass transfers are occurring by force convection which is caused the turbulence of the bath.

Appendices

Probable Questions

• 1. What are the main features of LD process? What is LD stands for?

• 2. Discuss the advantages and disadvantages of eccentric throat of LD converter.

• 3. ‘Nozzles of lance in LD converter are made of copper and exposed to high temperature, and that are also not water cooled; but still that are not melted at all’ Why?

• 4. What are the advantages of multi-nozzle lance over a single nozzle lance in LD process?

• 5. What do you understand by ‘Joule-Thomson effect’ in LD converter?

• 6. Discuss the mechanism of jet-metal interaction in LD process. How lance height can affect the sequence of refining?

• 7. What do you mean by ‘depth of penetration of oxygen jet’? What are the factors on which depth of penetration depend? What is the co-relation between them? What are the conditions for de-carburization and de-phosphorization in terms of depth of penetration?

• 8. What is the mechanism of mass transfer in LD process?

• 9. Mechanism of refining in LD process depends on ‘forced convection’, not by ‘diffusion’. Explain.

• 10. What do you understand by ‘slopping’ and ‘splashing’?

• 11. ‘Steelmaking is an oxidizing process, de-sulfurization required reducing atmosphere, still 60% S can be removed in LD converter’. How?

• 12. ‘Foaming slag is an advantage in LD process, but it is disadvantage for open-hearth process’. Why?

• 13. Discuss the mechanism of refining in LD converter. How can you control the relative rates of de-carbonization and de-phosphorization in LD converter?

• 14. What are the factors that causes turbulence in LD converter? Discuss.

• 15. Explain the terms ‘foam’ and ‘emulsion’ with reference to LD process. Do they help refining in LD process? If so, how? Explain.

• 16. What do you mean by ‘slag-metal emulsion’? What is the mechanism of de-carburization in LD converter? Write the equation for overall de-carburization rate.

• 17. What do you understand by Mn-hump? How it occurs?

• 18. ‘The iron and phosphorous lines, on the Ellingham diagram, are so close to each other that the oxidation of phosphorous in steelmaking cannot be done without loss of iron’. How this problem can be overcome at the LD process?

• 19. How you can control of the carbon and phosphorus reactions in LD process?

• 20. How ‘blocking of heat’ is done in LD process?

• 21. Discuss the zone-lining and lining by slag splashing in LD converter.

• 22. ‘100% magnesite bricks are used at trunnion region in LD converter’. Why?

• 23. What do you mean by ‘Maxshutte tuyere’? How it functions?

• 24. ‘The outer tube of the ‘Maxshutte tuyere’ is made of stainless steel and the inner tube is made of copper’. Why?

• 25. ‘In OBM converter, tuyeres are placed in one side of the converter’. Why?

• 26. Discuss the basic principle of OBM process and state the main advantages of OBM process. Why iron loss in OBM process is less compared to LD process?

• 27. Explain the LD-AC process and its merits over that of conventional LD process.

• 28. Explain the basic principle of Kaldo and Rotor processes.

• 29. What advantage you get by rotating the Kaldo converter?

• 30. Why rotary steelmaking processes are not popular?

• 31. What is the function of secondary lance in Rotor converter?

• 32. Discuss the advantages of bottom blown process over that of top blowing process. How many types of modern oxygen steelmaking processes are developed?

• 33. ‘For purging of inert gas, porous plugs are preferred instead of tuyeres at the bottom of a converter/ladle’. Why?

• 34. What are the metallurgical improvements achieved by combined blowing processes?

• 35. What is ‘OLP’ stand for? Explain the basis for the development of OLP process of steelmaking. State the major operational steps that can enhance the refining rate.

• 36. What is KMS stands for? Discuss the basic principle of KMS process of steelmaking.

• 37. Discuss the LD-HC process.

Examples

Example 15.1

Chemistry of input and output materials for LD process is as follows:

Element, %	Hot metal	Scrap	Steel (to be produced)
C	3.50	0.20	0.15
Si	1.20	0.02	0.01
Mn	0.75	0.40	0.25
P	0.40	0.04	0.03
S	0.04	0.04	0.03

Calculate::

(i) Amount of hot metal to be charged per tonne of steel production.

(ii) Amount of slag produced and composition of slag.

Given data::

(i) Weight of steel scrap is 200 kg per tonne of steel production,

(ii) Weight of lime is 50 kg per ton of steel production (lime contain 94.5% CaO, 2.5% MgO, 1.5% SiO₂ and 1.5% Al₂O₃).

(iii)1.5% Fe loss with respect to steel production.

Solution

• Assumptions: (a) One tonne steel production,

• By balancing we get, 94.11%, 99.30% and 99.53% Fe contain in hot metal, steel scrap and produced steel, respectively.

• Fe Balance: Fe input = Fe output

• Fe from HM + Fe from steel scrap = Fe goes to steel produce + Fe losses in slag

• Suppose weight of hot metal = W_HM

• 0.9411 W_HM + 0.993 × 200 = 0.9953 × 1000 + 0.015 × 1000

• Therefore, W_HM = 862.5 kg

• $\begin{aligned} & \quad \quad {\text{Fe}} + 1/2\,{\text{O}}_{2} = {\text{FeO}} \\ & \quad \quad 5 6\quad \quad \quad \quad \quad \;\;72 \\ \end{aligned}$

• $\begin{aligned} & \quad 56\,{\text{kg of Fe to form }}72\,{\text{kg of FeO}} \\ & \quad 15\,{\text{kg}}\quad \quad \quad \quad \quad \quad \left( {\frac{72}{56}} \right) \times 15 = 19.29\,{\text{kg of FeO}} \\ \end{aligned}$

• Si Balance: Si input = Si output

• Si from HM + Si from steel scrap + Si from lime

• = Si goes to steel produce + Si losses in slag (suppose weight of Si losses in slag = W_Si)

• $0.012 \times 862.5 + 0.0002 \times 200 + 0.015 \times 50 \times \left( {\frac{28}{60}} \right) = 0.0001 \times 1000 + W_{\text{Si}}$

• Therefore, W_Si = 10.64 kg

  • $\begin{aligned} & \quad \quad {\text{Si}} + {\text{O}}_{2} = {\text{SiO}}_{2} \\ & \quad \quad 28\quad \quad \quad \;\;60 \\ \end{aligned}$

  • $\begin{aligned} & \quad 28\,{\text{kg Si to form }}60\,{\text{kg of SiO}}_{2} \\ & \quad 10.64\,{\text{kg}}\quad \quad \quad \;\left( {\frac{60}{28}} \right) \times 10.64 = 22.8\,{\text{kg of SiO}}_{2} {\text{ in slag}} \\ \end{aligned}$

• Mn Balance: Mn input = Mn output

• Mn from HM + Mn from scrap = Mn in steel + Mn losses in slag

• $0.0075 \times 862.5 + 0.004 \times 200 = 0.0025 \times 1000 + W_{\text{Mn}}$

• Therefore, W_Mn = 4.77 kg

• $\begin{aligned} & \quad \quad {\text{Mn}} + 1/2\,{\text{O}}_{2} = {\text{MnO}} \\ & \quad \quad 5 5\quad \quad \quad \quad \quad \quad 71 \\ \end{aligned}$

• $\begin{aligned} & \quad 55\,{\text{kg Mn to form }}71\,{\text{kg of MnO}} \\ & \quad 4.77\,{\text{kg}}\quad \quad \quad \quad \left( {\frac{71}{55}} \right) \times 4.77 = 6.16\,{\text{kg of MnO in slag}}. \\ \end{aligned}$

• P Balance: P input = P output

• P from HM + P from scrap = P in steel + P goes to slag

• $\begin{aligned} & 0.004 \times 862.5 + 0.0004 \times 200 = 0.0003 \times 1000 + W_{\text{P}} \\ & \quad W_{\text{P}} = 3.23\,{\text{kg}} \\ \end{aligned}$

  • $\begin{aligned} & \quad \quad 2{\text{P}} + 5/2\,{\text{O}}_{2} = {\text{P}}_{2} {\text{O}}_{5} \\ & \quad \quad 2 \times 31\quad \quad \quad \quad \;142 \\ \end{aligned}$

  • $\begin{aligned} & \quad 62\,{\text{kg P to form }}142\,{\text{kg of P}}_{2} {\text{O}}_{5} \\ & \quad 3.23\,{\text{kg}}\quad \quad \quad \;\;\left( {\frac{142}{62}} \right) \times 3.23 = 7.4\,{\text{kg of P}}_{2} {\text{O}}_{5} \\ \end{aligned}$

• S Balance: S input = S output

• S from HM + S from scrap = S in steel + S goes to slag

• $0.0004 \times 862.4 + 0.0004 \times 200 = 0.0003 \times 1000 + W_{\text{S}}$

• Therefore, W_S = 0.125 kg

• $\begin{aligned} & \quad \quad {\text{CaO}} + {\text{S}} = {\text{CaS}} \\ & \quad \quad \;56\quad \;\;32\quad 72 \\ \end{aligned}$

• $\begin{aligned} & \quad 32\,{\text{kg S to form }}72\,{\text{kg of CaS}} \\ & \quad 0.125\,{\text{kg}}\quad \quad \quad \left( {\frac{72}{32}} \right) \times 0.125 = 0.28\,{\text{kg of CaS}} \\ \end{aligned}$

• $\begin{aligned} & \quad 72\,{\text{kg CaS formation CaO required }}56\,{\text{kg}} \\ & \quad 0.28\,{\text{kg}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( {\frac{56}{72}} \right) \times 0.28 = 0.22\,{\text{kg}} \\ \end{aligned}$

• CaO Balance: CaO input = CaO output

• CaO from lime = CaO in CaS + CaO in slag

• $\begin{aligned} & 0.945 \times 50 = 0.22 + W_{\text{CaO}} \\ & W_{\text{CaO}} = 47.03\,{\text{kg}} \\ \end{aligned}$

• MgO Balance: MgO from lime = MgO in slag

• $W_{\text{MgO}} = 0.025 \times 50 = 1.25\,{\text{kg}}$

• Al₂O₃ Balance: Al₂O₃ in lime = Al₂O₃ in slag

• $W_{{{\text{Al}}_{2} {\text{O}}_{3} }} = 0.015 \times 50 = 0.75$

• $\begin{aligned} {\text{Amount of slag}} & = W_{\text{FeO}} + W_{{{\text{SiO}}_{2} }} + W_{\text{MnO}} + W_{{{\text{P}}_{2} {\text{O}}_{5} }} + W_{\text{CaS}} + W_{\text{CaO}} + W_{\text{MgO}} + W_{{{\text{Al}}_{2} {\text{O}}_{3} }} \\ & = 19.29 + 22.8 + 6.16 + 7.4 + 0.28 + 47.03 + 1.25 + 0.75 = {\mathbf{104}}.{\mathbf{96}}\,{\mathbf{kg}} \\ \end{aligned}$

• Composition of slag: 18.38% FeO, 21.72% SiO₂, 5.87% MnO, 7.05% P₂O₅, 0.27% CaS, 44.81% CaO, 1.19% MgO and 0.71% Al₂O₃.

Example 15.2

Chemistry of input and output materials for LD steelmaking as follows:

Element/compound %	Hot metal	Scrap	DRI	Steel produced	Lime	Fe–Mn	Fe–Si
C	4.0	0.37	1.75	0.25		0.1	1.5
Si	1.0	0.27		0.2		2.0	70.0
Mn	0.5	0.61		0.25		72.5
P	0.4	0.05	0.03	0.03		0.1
S	0.04	0.05	0.01	0.03		0.03
CaO			1.1		95.0
MgO			0.8		2.5
SiO2			1.0		1.5
Al2O3			1.45		1.0
FeO			8.36
FeTol			92.0
Amount charged, kg	?	200	325		?	5	5

• Calculate (a) amount of H:M charged, (b) amount of lime charged, (c) amount of oxygen consumption and (d) amount of slag produced and composition of slag.

• Given: Fe loss in slag and fumes are 1.5 and 0.6% of liquid steel.

  • $\quad {\text{Basicity of slag}} = 3.0$

Solution

• Assumptions: One tonne steel production.

• By balancing we get, 94.06%, 98.65%, 99.24%, 28.5% and 25.27% Fe contain in hot metal, steel scrap, produced steel, Fe–Si and Fe–Mn, respectively.

• Fe Balance: Fe input = Fe output

• Fe from HM + Fe from steel scrap + Fe from DRI + Fe from Fe–Si + Fe from Fe–Mn = Fe goes to steel produce + Fe losses in slag + Fe losses in fumes

• Suppose weight of hot metal = W_HM

• $\begin{aligned} {\text{So}},\,0.9406 & \times W_{\text{HM}} + 0.9865 \times 200 + 0.92 \times 325 + 0.285 \times 5 + 0.2527 \times 5 \\ & = 0.9924 \times 1000 + 0.015 \times 1000 + 0.006 \times 1000 \\ \end{aligned}$

• Therefore, W_HM = 546.91 = 547 kg

• $\begin{aligned} & \quad \quad {\text{Fe}} + 1/2\,{\text{O}}_{2} = {\text{FeO}} \\ & \quad \quad 5 6\quad \quad 1 6\quad \quad \;\;72 \\ \end{aligned}$

• $\begin{aligned} & \quad 56\,{\text{kg of Fe to form }}72\,{\text{kg of FeO}} \\ & \quad 15\,{\text{kg}}\quad \quad \quad \quad \quad \quad \left( {\frac{72}{56}} \right) \times 15 = 19.29\,{\text{kg of FeO in slag}} \\ \end{aligned}$

• $\begin{aligned} & {\text{For}}\,56\,{\text{kg Fe oxygen required }}16\,{\text{kg}} \\ & \quad \;\;\;15\,{\text{kg}}\quad \quad \quad \quad \quad \quad \quad \quad \left( {\frac{16}{56}} \right) \times 15 = 4.29\,{\text{kg of oxygen}} \\ \end{aligned}$

• Si Balance: Si input = Si output

• Si from HM + Si from scrap + Si from Fe–Si + Si from Fe–Mn + Si from gange material in DRI + Si from lime = Si in steel + Si goes to slag

• $\begin{aligned} 0.01 & \times 547 + 0.0027 \times 200 + 0.7 \times 5 + 0.02 \times 5 + 0.01 \times 325 \times \left( {\frac{28}{60}} \right) + 0.015 \times W_{\text{L}} \times \left( {\frac{28}{60}} \right) \\ & = 0.002 \times 1000 + W_{\text{Si}} \\ \end{aligned}$

• Therefore, $W_{\text{Si}} = 9.13 + 7 \times 10^{ - 3} W_{\text{L}}$

• $\begin{aligned} & \quad \quad {\text{Si}} + {\text{O}}_{2} = {\text{SiO}}_{2} \\ & \quad \quad 28\quad 32\quad \;60 \\ \end{aligned}$

• 28 part Si form 60 part of SiO₂

• $$ \left( {9.13 + 7 \times 10^{ - 3} W_{\text{L}} } \right) \ldots \left[ {\left( {9.13 + 7 \times 10^{ - 3} W_{\text{L}} } \right) \times \left( {\frac{28}{60}} \right){\text{part of SiO}}_{2} } \right. $$

  (1)

• CaO Balance: CaO input = CaO output

• CaO from lime + CaO from gange material in DRI = CaO goes to slag

• $$ \begin{aligned} & 0.95 \times W_{\text{L}} + 0.011 \times 325 = W_{\text{CaO}} \\ & {\text{or}}, \; 0.95 \times W_{\text{L}} + 3.575 = W_{\text{CaO}} \\ \end{aligned} $$

  (2)
• $$ \begin{aligned} {\text{Since}}\,{\text{Basicity}} = 3 & = \left( {\frac{{\Sigma \,{\text{CaO in slag}}}}{{\Sigma \,{\text{SiO}}_{2} \,{\text{in slag}}}}} \right) \\ & = \left[ {\left( {0.95 \times W_{\text{L}} + 3.575} \right)/\left\{ {\left( {9.13 + 7 \times 10^{ - 3} W_{\text{L}} } \right) \times \left( {\frac{60}{28}} \right)} \right\}} \right] \\ \end{aligned} $$

• (from Eqs. 1 and 2)

• Therefore, W_L = 60.9 kg

• From Eq. 1: W_Si = 9.556 kg

• SiO₂ in slag = $W_{\text{Si}} \times \left( {\frac{60}{28}} \right) = 20.48\,{\text{kg}}$,

• $\begin{aligned} {\text{Out of this SiO}}_{2} {\text{ coming from oxidation}} & = \left[ {20.48{-}\left( {0.01 \times 325} \right){-}\left( {0.015 \times 60.9} \right)} \right] \\ & = 16.32\,{\text{kg}} \\ \end{aligned}$

• Therefore, oxygen required for SiO₂ formation by oxidation $= \left( {\frac{{\left( {32 \times 16.32} \right)}}{60}} \right) = 8.7\,{\text{kg}}$

• From Eq. 2: W_CaO = 61.43 kg.

• Mn Balance: Mn input = Mn output

• Mn from HM + Mn from scrap + Mn from Fe–Mn = Mn in steel + Mn goes to slag

• $0.005 \times 547 + 0.0061 \times 200 + 0.725 \times 5 = 0.0025 \times 1000 + W_{\text{Mn}}$

• Therefore, W_Mn = 5.08 kg

• $\begin{aligned} & [\;\;{\text{Mn}} + 1/2\,{\text{O}}_{2} = {\text{MnO}} \\ & \;\;\;\; 5 5\quad \quad 1 6\quad \quad \quad 7 1\quad ] \\ \end{aligned}$

• ${\text{MnO in slag}} = 5.08 \times \left( {\frac{71}{55}} \right) = {\mathbf{6}}{\mathbf{.56}}\,{\mathbf{kg}}$

• ${\text{Oxygen require}} = 5.08 \times \left( {\frac{16}{55}} \right) = {\mathbf{1}}{\mathbf{.48}}\,{\mathbf{kg}}$

• C Balance: C input = C output

• $\begin{aligned} & {\text{C from HM}} + {\text{C from scrap}} + {\text{C from Fe}}{-}{\text{Si}} + {\text{C from Fe}}{-}{\text{Mn}} + {\text{C from DRI}} \\ & \quad = {\text{C in steel}} + {\text{C goes as gases}} \\ \end{aligned}$

• $\begin{aligned} & 0.04 \times 547 + 0.0037 \times 200 + 0.015 \times 5 + 0.001 \times 5 + 0.0175 \times 325 \\ & \quad = 0.0025 \times 1000 + W_{\text{C}} \\ \end{aligned}$

• W_C = 25.89 kg

• Assume that C form gases 90% CO and 10% CO₂

• $\begin{aligned} & \quad {\text{C}} + 1/2\,{\text{O}}_{2} = {\text{CO}} \\ & \quad 1 2\quad \;16\quad \quad \;\;28 \\ \end{aligned}$

• Oxygen required for CO formation = $\left( {\frac{16}{12}} \right) \times 0.9 \times 25.89 = {\mathbf{30}}{\mathbf{.06}}\,{\mathbf{kg}}$

• $\begin{aligned} & \quad {\text{C}} + {\text{O}}_{2} = {\text{CO}}_{2} \\ & \quad 12\quad 32\quad \;44 \\ \end{aligned}$

• Oxygen required for CO₂ formation = $\left( {\frac{32}{12}} \right) \times 0.1 \times 25.89 = {\mathbf{6}}.{\mathbf{90}}\,{\mathbf{kg}}$

• P Balance: P input = P output

• $\begin{aligned} {\text{P from HM}} & + {\text{P from scrap}} + {\text{P from Fe}}{-}{\text{Mn}} + {\text{P from DRI}} \\ & = {\text{P in steel}} + {\text{P goes to slag}} \\ \end{aligned}$

• $\begin{aligned} 0.004 \times 547 & + 0.0005 \times 200 + 0.001 \times 5 + 0.0003 \times 325 \\ & = 0.0003 \times 1000 + W_{\text{P}} \\ \end{aligned}$

• Therefore, W_P = 2.09 kg

• $\begin{aligned} & \quad 2{\text{P}} + 5/2\,{\text{O}}_{2} = {\text{P}}_{2} {\text{O}}_{5} \\ & \quad 62\quad \quad 80\quad \quad \;142 \\ \end{aligned}$

• $\begin{aligned} & 62\,{\text{kg P to form P}}_{2} {\text{O}}_{5} \,{\text{oxygen require}} = 80\,{\text{kg}} \\ & 2.09\,{\text{kg}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\;\; = \left( {\frac{80}{62}} \right) \times 2.09 = {\mathbf{2}}.{\mathbf{70}}\,{\mathbf{kg}} \\ \end{aligned}$

• P₂O₅ in slag = $\left( {\frac{142}{62}} \right) \times 2.09 = {\mathbf{4}}{\mathbf{.79}}\,{\mathbf{kg}}$

• S Balance: S input = S output

• $\begin{aligned} {\text{S from HM}} & + {\text{S from scrap}} + {\text{S from Fe}}{-}{\text{Mn}} + {\text{S from DRI}} \\ & = {\text{S in steel}} + {\text{S goes to slag}} \\ \end{aligned}$

• $\begin{aligned} 0.0004 \times 547 & + 0.0005 \times 200 + 0.0003 \times 5 + 0.0001 \times 325 \\ & = 0.0003 \times 1000 + W_{\text{S}} \\ \end{aligned}$

• Therefore, W_S = 0.053 kg

• $\begin{aligned} & \quad {\text{CaO}} + {\text{S}} = {\text{CaS}} \\ & \quad \quad \quad \;\; 3 2\quad \;72 \\ \end{aligned}$

• CaS in slag = $\left( {\frac{72}{32}} \right) \times 0.053 = 0.1188 = {\mathbf{0}}{\mathbf{.12}}\,{\mathbf{kg}}$

• MgO Balance: MgO input = MgO output

• MgO from DRI + MgO from lime = MgO goes to slag

• $\begin{aligned} & 0.008 \times 325 + 0.025 \times 60.9 = W_{\text{MgO}} \\ & {\mathbf{W}}_{{{\mathbf{MgO}}}} = {\mathbf{4}}.{\mathbf{12}}\,{\mathbf{kg}} \\ \end{aligned}$

• Al₂O₃ Balance: Al₂O₃ input = Al₂O₃ output

• Al₂O₃ from DRI + Al₂O₃ from lime = Al₂O₃ goes to slag

• $\begin{aligned} & 0.0145 \times 325 + 0.01 \times 60.9 = W_{{{\text{Al}}_{2} {\text{O}}_{3} }} \\ & W_{{{\text{Al}}_{2} {\text{O}}_{3} }} = {\mathbf{5}}.{\mathbf{32}}\,{\mathbf{kg}} \\ \end{aligned}$

• $\begin{aligned} {\mathbf{Weight}}\,{\mathbf{of}}\,{\mathbf{slag}} & = W_{\text{FeO}} + W_{{{\text{SiO}}_{2} }} + W_{\text{CaO}} + W_{\text{MnO}} + W_{{{\text{P}}_{2} {\text{O}}_{5} }} + W_{\text{CaS}} + W_{\text{MgO}} + W_{{{\text{Al}}_{2} {\text{O}}_{3} }} \\ & = 19.29 + 20.48 + 61.43 + 6.56 + 4.75 + 0.12 + 4.12 + 5.32 \\ & = {\mathbf{122}}.{\mathbf{07}}\,{\mathbf{kg}} \\ \end{aligned}$

• Composition of slag: 15.8% FeO, 16.78% SiO₂, 50.32% CaO, 5.37% MnO, 3.89% P₂O₅, 0.1% CaS, 3.37% MgO and 4.36% Al₂O₃.

• $\begin{aligned} {\mathbf{Oxygen}}\,{\mathbf{consumption}} & = {\text{O}}_{\text{FeO}} + {\text{O}}_{{{\text{SiO}}_{2} }} + {\text{O}}_{\text{MnO}} + {\text{O}}_{{{\text{CO}} }} + {\text{O}}_{{{\text{CO}}_{2} }} + {\text{O}}_{{{\text{P}}_{2} {\text{O}}_{5} }} \\ & = 4.29 + 8.70 + 1.48 + 30.06 + 6.90 + 2.70 = {\mathbf{54}}.{\mathbf{13}}\,{\mathbf{kg}} \\ \end{aligned}$

Example 15.3

LD converter is charged with 10 tonne of hot metal containing 3.5% C, 1.25% Si, 0.75% Mn and rest Fe. The impurities are oxidized and removed from HM by lancing pure oxygen into the converter. 40% C oxidize to CO₂ and 60% C to form CO. Find out (i) total amount of oxygen gas require (in N m³), (ii) the weight of slag produce.

Mention clearly, assumptions made. [Given: 1% Fe (w.r.t. HM) is lost.]

Solution

Assumptions::

(a) Steel contains 100% Fe,

(b) Oxygen used is pure form.

• HM 10 tonne = 10,000 kg

• 3.5% C in ${\text{HM}} = \left( {\frac{3.5}{100}} \right) \times 10 , 0 0 0= 350\,{\text{kg}}$

• 1.25% Si in ${\text{HM}} = \left( {\frac{1.25}{100}} \right) \times 10 ,000 = 125\,{\text{kg}}$

• 0.75% Mn in ${\text{HM}} = \left( {\frac{0.75}{100}} \right) \times 10 ,000 = 75\,{\text{kg}}$

• Fe loss = $\left( {\frac{1}{100}} \right) \times 10 ,000 = 100\,{\text{kg}}$

$\begin{aligned} & \quad {\text{Fe}} + 1/2\,{\text{O}}_{2} = {\text{FeO}} \\ & & \quad 5 6\quad \quad 11.2\quad \;\;72 \\ & 56\,{\text{kg of Fe to form }}72\,{\text{kg of FeO}} \\ & 100\,{\text{kg}}\quad \quad \quad \quad \quad \;\;\left( {\frac{72}{56}} \right) \times 100 = {\mathbf{128}}.{\mathbf{57}}\,{\mathbf{kg}}\,{\text{of FeO goes in slag}} \\ & {\text{For }}56\,{\text{kg of Fe to form FeO oxygen require }}11.2\,{\text{N}}\,{\text{m}}^{3} \\ & \quad \quad 100\,{\text{kg}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( {\frac{11.2}{56}} \right) \times 100 = {\mathbf{20}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \\ \end{aligned}$

$\begin{aligned} & \quad {\text{Si}} + {\text{O}}_{2} = {\text{SiO}}_{2} \\ & \quad 28\quad 22.4\quad 60 \\ & 28\,{\text{kg of Si to form }}60\,{\text{kg of SiO}}_{2} \\ & 125\,{\text{kg}}\quad \quad \quad \quad \quad \left( {\frac{60}{28}} \right) \times 125 = {\mathbf{267}}.{\mathbf{86}}\,{\mathbf{kg}}\,{\text{of SiO}}_{2} \,{\text{goes to slag}} \\ & 28\,{\text{kg of Si to form SiO}}_{2} \,{\text{oxygen require }}22.4\,{\text{N}}\,{\text{m}}^{3} \\ & 125\,{\text{kg}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( {\frac{22.4}{28}} \right) \times 125 = {\mathbf{100}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \\ \end{aligned}$

$\begin{aligned} & \quad {\text{Mn}} + 1/2\,{\text{O}}_{2} = {\text{MnO}} \\ & \quad 5 5\quad \quad 1 1. 2\quad \quad 7 1\\ & 55\,{\text{kg of Mn to form }}71\,{\text{kg of MnO}} \\ & 75\,{\text{kg}}\quad \quad \quad \quad \quad \quad \left( {\frac{71}{55}} \right) \times 75 = {\mathbf{96}}.{\mathbf{82}}\,{\mathbf{kg}}\,{\text{of MnO goes to slag}} \\ & {\text{For}}\,55\,{\text{kg of Mn to form MnO oxygen require }}11.2\,{\text{N}}\,{\text{m}}^{3} \\ & \quad \;\;75\,{\text{kg}}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( {\frac{11.2}{55}} \right) \times 75 = {\mathbf{15}}.{\mathbf{27}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \\ \end{aligned}$

• $\begin{aligned} & {\text{Out of }}350\,{\text{kg C}},40\% \,{\text{C to form CO}}_{2} = \left( {\frac{40}{100}} \right) \times 350 = 140\,{\text{kg}} \\ & \quad \quad \quad \quad \quad \quad \quad 60\% \,{\text{C to form CO}} = \left( {\frac{60}{100}} \right) \times 350 = 210\,{\text{kg}} \\ & \quad \quad \quad \quad \quad \quad {\text{C}} + {\text{O}}_{2} = {\text{CO}}_{2} \\ & \quad \quad \quad \quad \quad \quad 12\quad 22.4\;\;44 \\ \end{aligned}$

• Oxygen required for CO₂ formation = $\left( {\frac{22.4}{12}} \right) \times 140 = {\mathbf{261}}.{\mathbf{33}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}}$

• $\begin{aligned} & \quad \quad {\text{C}} + 1/2\,{\text{O}}_{2} = {\text{CO}} \\ & \quad \quad 1 2\quad 1 1. 2\quad \quad 2 8\\ \end{aligned}$

• Oxygen required for CO formation = $\left( {\frac{11.2}{12}} \right) \times 210 = {\mathbf{196}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}}$

• $\begin{aligned} {\text{Weight of slag produce}} & = {\text{weight of FeO}} + {\text{weight of SiO}}_{2} + {\text{weight of MnO}} \\ & = 128.57 + 267.86 + 96.82 = {\mathbf{493}}.{\mathbf{25}}\,{\mathbf{kg}} \\ \end{aligned}$

• $\begin{aligned} {\text{Total oxygen require}} & = {\text{for FeO}} + {\text{for SiO}}_{2} + {\text{for MnO}} + {\text{for CO}}_{2} + {\text{for CO}} \\ & = 20 + 100 + 15.27 + 261.33 + 196 = {\mathbf{592}}.{\mathbf{60}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \\ \end{aligned}$

Example 15.4

LD converter is charged with 20 tonnes of hot metal containing 4.0% C, 1.2% Mn, 1.4% Si and rest Fe. The impurities are oxidized and removed from hot metal by lancing oxygen. Three-fourths of carbon goes as CO and one-fourth goes as CO₂. Calculate: (i) oxygen required (in cubic metres), (ii) the composition and total volume of product gases. Mention clearly, the assumptions made in the above case.

Solution

Assumptions::

(i) Steel is almost pure Fe and impurities are completely oxidized,

(ii) No loss of Fe during the process.

• Si content in HM: $\left( {\frac{1.4}{100}} \right) \times 20 ,000 = 280\,{\text{kg}}$

• $\begin{aligned} & \quad {\text{Si}} + {\text{O}}_{2} = {\text{SiO}}_{2} \\ & \quad 28\quad 22.4 \\ & 28\,{\text{kg of Si react with }}22.4\,{\text{N}}\,{\text{m}}^{3} \,{\text{O}}_{2} \\ & 280\quad \quad \quad \quad \quad \quad \quad \left( {\frac{22.4}{28}} \right) \times 280 = {\mathbf{224}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \,{\mathbf{O}}_{{\mathbf{2}}} \\ \end{aligned}$

• Mn content in HM: $\left( {\frac{1.2}{100}} \right) \times 20 ,000 = 240\,{\text{kg}}$

• $\begin{aligned} & \quad 2{\text{Mn}} + {\text{O}}_{2} = 2{\text{MnO}} \\ & \quad 2\times 5 5\quad 2 2. 4\\ & 110\,{\text{kg of Mn react with }}22.4\,{\text{N}}\,{\text{m}}^{3} \,{\text{O}}_{2} \\ & 240\quad \quad \quad \quad \quad \quad \quad \quad \left( {\frac{22.4}{110}} \right) \times 240 = {\mathbf{48}}.{\mathbf{87}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \,{\mathbf{O}}_{{\mathbf{2}}} \\ \end{aligned}$

• Carbon content in HM: $\left( {\frac{4.0}{100}} \right) \times 20 ,000 = 800\,{\text{kg}}$

• C form CO: $800 \times \left( {\frac{3}{4}} \right) = 600\,{\text{kg}}$

• C form CO₂: $800 \times \left( {\frac{1}{4}} \right) = 200\,{\text{kg}}$

• $\quad \begin{array}{*{20}c} {2{\text{C}} + {\text{O}}_{2} = 2{\text{CO}}} \\ { 2\times 1 2\; 2 2. 4\; 2\times 2 2. 4} \\ \end{array}$

• $\begin{aligned} & 24\,{\text{kg of C react with }}22.4\,{\text{N}}\,{\text{m}}^{3} \,{\text{O}}_{2} \,{\text{to form CO}} \\ & 600\quad \quad \quad \quad \quad \quad \quad \left( {\frac{22.4}{24}} \right) \times 600 = {\mathbf{560}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \,{\mathbf{O}}_{{\mathbf{2}}} {\mathbf{for CO}} \\ & 600\quad \quad \quad \quad \quad \quad \quad \left( {\frac{2 \times 22.4}{24}} \right) \times 600 = {\mathbf{1120}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \,{\mathbf{CO}} \\ \end{aligned}$

• $\begin{aligned} & \quad {\text{C}} + {\text{O}}_{2} = {\text{CO}}_{2} \\ & \quad 12\;22.4\;\;\;22.4 \\ \end{aligned}$

• $\begin{aligned} & 12\,{\text{kg of C react with }}22.4\,{\text{N}}\,{\text{m}}^{3} \,{\text{O}}_{2} \,{\text{to form CO}}_{2} \\ & 200\quad \quad \quad \quad \quad \quad \quad \left( {\frac{22.4}{12}} \right) \times 200 = {\mathbf{373}}.{\mathbf{33}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \,{\mathbf{O}}_{{\mathbf{2}}} \,{\mathbf{for}}\,{\mathbf{CO}}_{{\mathbf{2}}} \\ & 200\quad \quad \quad \quad \quad \quad \quad \left( {\frac{22.4}{24}} \right) \times 200 = {\mathbf{373}}.{\mathbf{33}}\,{\mathbf{N}}\,{\mathbf{m}}^{{\mathbf{3}}} \,{\mathbf{CO}}_{{\mathbf{2}}} \\ \end{aligned}$

• Total oxygen required: 224 + 48.87 + 560 + 373.33 = 1206.2 N m³ O₂

• Total volume of gases: 373.33 + 1120 = 1493.33 N m³

• %CO =  $\left( {\frac{1120}{1493.33}} \right)$  × 100 = 75%

• %CO₂ = $\left( {\frac{373.33}{1493.33}} \right)$ × 100 = 24.9998 = 25%.

Example 15.5

Calculate the temperature of the liquid steel rise if 20 kg of ferrosilicon is added to 50 t ladle at 1600 °C. Ferrosilicon contains 80% Si. Assuming half of silicon added reacts with dissolved oxygen and rest half simply dissolved in molten steel.

• Given: Specific heat of molten steel 44 J/mol K, H₁₈₇₃ − H₂₉₈ for silicon is 33.5 kJ/mol, heat of solution of silicon in liquid steel is −119.3 kJ/mol and heat of reaction for [Si] + 2[O] = SiO₂ (s) is −594.6 kJ/mol. All data are at 1873 K.

Solution

Si in ferrosilicon = 20 × 0.8 = 16 kg.

$$ {\text{Si}}\,\left( {\text{s}} \right)\,{\text{at}}\,298 \to {\text{Si}}\,\left( {\text{s}} \right)\,{\text{at}}\,1873,H_{1873} - H_{298} = 33.5\,{\text{kJ/mol}} = \left( {33.5/28} \right) \times 1000 = 1196.43\,{\text{kJ/kg}}\,{\text{Si}} $$

$$ {\text{Si}}\,\left( {\text{s}} \right)\,{\text{at}}\,1873 \to \left[ {\text{Si}} \right]\,{\text{at}}\,1873,\Delta H = - 119.3\,{\text{kJ/mol}} = \left( { - 119.3/28} \right) \times 1000 = - 4260.71\,{\text{kJ/kg}}\,{\text{Si}} $$

———————————————————————————————————

$$ {\text{Si}}\,\left( {\text{s}} \right)\,{\text{at}}\,298 \to \left[ {\text{Si}} \right]\,{\text{at}}\,1873,\Delta H_{\text{s}} = - 3064.28\,{\text{kJ/kg}}\,{\text{Si}} $$

• Heat generated for 16 kg Si dissolved in molten steel = 16 × −3064.28 = −49,028.48 kJ

$$ \left[ {\text{Si}} \right] + 2\left[ {\text{O}} \right] = {\text{SiO}}_{2} \,\left( {\text{s}} \right)\,\Delta H_{\text{r}} = - 594.6\,{\text{kJ/mol}} = - 21,235.71\,{\text{kJ/kg}}\,{\text{Si}} $$

• Now half of silicon added reacts with dissolved oxygen = 16/2 = 8 kg

• So, heat generated for 8 kg Si reacts with dissolved oxygen = −169,885.71 kJ

• Total heat generated for 16 kg Si dissolved in molten steel and 8 kg Si reacts with dissolved oxygen = −49,028.48 + (−169,885.71) = −218,914.19 kJ

• Specific heat of molten steel 44 J/mol K = 44/56 = 0.786 kJ/kg K

• Now heat required for 50 t molten steel to rise 1 K = 0.786 × 50,000 = 39,300 kJ

  • 39,300 kJ heat required for 50 t molten steel to rise temperature 1 K

    $$ \begin{aligned} {\text{Therefore}},\,218 ,914.19\,{\text{kJ heat rise temperature for }}50\,{\text{t molten steel}} & = \left( {218 ,914.19/39 ,300} \right) \\ & = 5.57\,{\text{K}} \\ \end{aligned} $$

• The temperature of the liquid steel rise by 5.6 K.

Problems

Problem 15.1

Chemistry of input and output materials for LD process is as follows:

Element, %	Hot metal	Scrap	Steel (to be produced)
C	3.25	0.22	0.15
Si	1.15	0.02	0.01
Mn	0.80	0.40	0.30
P	0.40	0.04	0.03
S	0.04	0.04	0.03

Calculate::

(i) Amount of hot metal to be charged per tonne of steel production.

(ii) Amount of slag produced and composition of slag.

Given data::

(i) Weight of steel scrap is 250 kg per tonne of steel production,

(ii) Weight of lime is 50 kg per tonne of steel production (lime contains 95% CaO, 2.0% MgO, 1.5% SiO₂ and 1.5% Al₂O₃),

(iii) 1.5% Fe loss with respect to steel production.

• [Ans: (i) 807.12 kg, (ii) 101.85 kg (18.94% FeO, 20.16% SiO₂, 5.68% MnO, 6.81% P₂O₅, 0.27% CaS, 46.43% CaO, 0.98% MgO and 0.74% Al₂O₃)]

Problem 15.2

LD converter is charged with 10 tonne of hot metal containing 4.0% C, 1.5% Si, 1.0% Mn and rest Fe. The impurities are oxidized and removed from HM by lancing pure oxygen into the converter. One-quarter of the carbon is oxidized to CO₂ and three quarters to form CO. Find out (i) total amount of oxygen gas require (in N m³), (ii) the volume of product gases.

• Mention clearly, assumptions made.

• [Ans: (i) 607 N m³, (ii) 747 N m³ (25% CO₂ + 75% CO)]

Problem 15.3

Chemistry of input and output materials for LD steelmaking is as follows:

Element/compound %	Hot metal	Scrap	DRI	Steel produced	Lime	Fe–Mn	Fe–Si
C	3.5	0.25	1.7	0.2		0.1	1.5
Si	1.1	0.27		0.2		2.0	70.0
Mn	0.5	0.61		0.25		72.5
P	0.4	0.05	0.03	0.03		0.1
S	0.04	0.05	0.01	0.03		0.03
CaO			1.1		95.0
MgO			0.8		2.5
SiO2			1.0		1.5
Al2O3			1.45		1.0
FeO			8.36
FeTol			92.0			25.27	28.5
Amount charged, kg	550	200	?		55	5	5

• Calculate (a) amount of DRI charged, (b) amount of oxygen consumption, and (c) amount of slag produced and composition of slag.

• Given: Fe loss in slag is 1.5 of liquid steel

• [Ans: (a) Wt of DRI = 313.37 kg, (b) Amount of oxygen = 51.82 kg, (c) Amount of slag = 116.9 kg (16.5% FeO, 18.4% SiO₂, 47.6% CaO, 5.6% MnO, 4.1% P₂O₅, 0.1% CaS, 3.3% MgO and 4.4% Al₂O₃.)]