Steelmaking Processes

Abstract

Bessemer proposed a production method of steelmaking within a very short time (15–20 min) by blowing air through molten pig iron without using fuel. Siemens brothers, Germany, developed temperature of the furnace around 1600 ℃ for the first time, on the principle of regeneration of heat from exhaust gases.

Appendices

Probable Questions

• 1. Hot metal contains 3.0% C, 1.25% Si, 0.8% Mn, 0.04% P and 0.03% S. Do you need a basic slag? Why?

• 2. ‘Si, Mn and C have good affinity for oxygen than Fe at steelmaking temperature; still Fe reacts first with oxygen’. Why?

• 3. What do you mean by ‘blow-hot condition’ in Bessemer converter? What will be the effect on steelmaking?

• 4. What happens in Bessemer steelmaking, if silicon content is less than 1.0% in hot metal?

• 5. What are the main sources of heat in acid and basic autogenous steelmaking processes?

• 6. ‘After drop of the big flame in basic Bessemer process, still air blowing is continued’ why? What is that known as?

• 7. What do you understand by ‘blown metal’?

• 8. ‘Total FeO content in acid slag is more than basic slag, but the activity of FeO in acid slag is lower than the basic slag’. Why?

• 9. What were the limitations of Bessemer steelmaking process?

• 10. What are the problems of using 100% solid pig iron or 100% hot metal as the charge in open-hearth furnace? What is the best combination of charge and why?

• 11. What is the basic concept of open-hearth process?

• 12. What is the other name of open-hearth process? Why it is called open-hearth furnace?

• 13. ‘After partly melt down of the solid charge in open-hearth furnace, the burning rate of fuel should be lower down’. Why?

• 14. Why hot metal is not charged in open-hearth furnace before the partial melting of scrap?

• 15. ‘Front flushing of slag, in open-hearth furnace, is advantages for steelmaking’. Explain. Why this is happened?

• 16. What are the difference between lime boil and ore boil? Which one is more vigorous than other?

• 17. How ‘blocking of heat’ is done in open-hearth furnace?

• 18. What are the basic concept of SIP and twin-hearth processes?

Examples

Example 14.1

Steel is being made in a Bessemer converter, using air to oxidize the impurities. The converter charge 20 tonnes of hot metal of composition: 3.5% C, 1.5% Si, 1.2% Mn. 3/4th of the carbon form CO and 1/4th form CO₂.

Calculate: (i) the volume of air used and (ii) the composition and total volume of product gases.

Solution

Assume: there is no loss of iron in slag.

• $\begin{aligned} & {\text{Reactions}}:\quad {\text{Si}} + {\text{O}}_{2} = {\text{SiO}}_{2} \quad {\text{Mn}} + 1/2\;{\text{O}}_{2} = {\text{MnO}} \\ & \quad \quad \quad \quad \quad\,\,\,\,28 \quad \,32\quad 60\quad \,\,\, \quad55\quad \quad \;\,16\quad \,\, 71 \\ \end{aligned}$

$$ \begin{array}{*{20}l} {{\text{C}} + {\text{O}}_{2} = {\text{CO}}_{2} \quad \quad \quad \& \quad \quad \quad {\text{C}} + 1/2\,{\text{O}}_{2} = {\text{CO}}} \hfill \\ {12\;\;32\quad \quad 44 \quad \quad \quad \quad \quad \quad \quad 12\quad 16\quad \quad \,\,\,28} \hfill \\ \end{array} $$

• $\begin{array}{*{20}l} {{\text{{\underline{Element}}}}\quad \quad {\text{{\underline{Amount\, in \, kg}}}}} \hfill & {{\text{{\underline{oxygen\, require\, in \, kg}}}}} \hfill \\ {{\text{Si}} \to ~20000 \times \left( {\frac{{1.5}}{{100}}} \right)~ = \;300\;{\text{kg}}} \hfill & {\left( {\frac{{32}}{{28}}} \right) \times 300 = 342.86\;{\text{kg}}} \hfill \\ {{\text{Mn}} \to 20000 \times \left( {\frac{{1.2}}{{100}}} \right) = 240\;{\text{kg}}} \hfill & {\left( {\frac{{16}}{{55}}} \right) \times 240 = 69.82\;{\text{kg}}} \hfill \\ {{\text{C}} \to 20000 \times \left( {\frac{{3.5}}{{100}}} \right) = 700\;{\text{k}}\;\;\quad ({\text{i}})\;3/4\;{\text{CO}}} \hfill & {\left( {\frac{{16}}{{12}}} \right) \times 3/4 \times 700 = 700.0\;{\text{kg}}} \hfill \\ {\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \,\,\,({\text{ii}})\;1/4\;{\text{CO}}_{2} } \hfill & {\left( {\frac{{32}}{{12}}} \right) \times 1/4 \times 700 = 466.67\;{\text{kg}}~} \hfill \\ \end{array}$

• Total oxygen required = 342.86 + 69.82 + 700.0 + 466.67 = 1579.35 kg

• 32 kg oxygen occupied volume 22.4 Nm³ at STP

• $\begin{aligned} {\text{Therefore}},1579.35\;{\text{kg}}\;{\text{oxygen}}\;{\text{occupied}}\;{\text{volume}} & = \left( {\frac{{22.4}}{{32}}} \right) \times 1579.35 \\ & = 1105.55\;{\text{Nm}}^{3} {\text{at}}\;{\text{STP}} \\ \end{aligned}$

• Since air contains 21% oxygen by volume

• $\begin{array}{*{20}l} {{\text{So}}\;{\text{total}}\;{\text{volume}}\;{\text{of}}\;{\text{air}} = 1105.55 \times ~\left( {\frac{{100}}{{21}}} \right) = ~5264.5\;{\text{Nm}}^{3} {\text{at}}\;{\text{STP}}} \hfill & {} \hfill \\ {{\text{Nitrogen}}\;{\text{contain}} = 5264.5 - 1105.55 = 4158.95\;{\text{Nm}}^{3} {\text{at}}\;{\text{STP}}} \hfill & {({\mathbf{76}}.{\mathbf{09}}\% )} \hfill \\ {{\text{CO}}\;{\text{form}} = \times 700 = 980.0\;{\text{Nm}}^{3} {\text{at}}\;{\text{STP}}} \hfill & {({\mathbf{17}}.{\mathbf{93}}\% )} \hfill \\ {{\text{CO}}_{2} \;{\text{form}} = \times 466.67 = 326.67\;{\text{Nm}}^{3} {\text{at}}\;{\text{STP}}} \hfill & {({\mathbf{5}}.{\mathbf{98}}\% )} \hfill \\ {{\text{Total}}\;{\text{volume}}\;{\text{of}}\;{\text{product}}\;{\text{gases}} = {\mathbf{5465}}.{\mathbf{62}}\;{\mathbf{Nm}}^{{\mathbf{3}}} {\text{at}}\;{\text{STP}}} \hfill & {({\mathbf{100}}\% )} \hfill \\ \end{array}$

Example 14.2

Chemistry of input and output materials for steel is being made in a basic Bessemer converter process which is as follows:

Element, %	Hot metal	Steel (to be produced)
C	3.40	0.15
Si	1.10	0.01
Mn	0.75	0.25
P	0.40	0.03
S	0.04	0.03

$\begin{aligned} {\text{Calculate: }} & \left( {\text{i}} \right)\,{\text{ Amount}}\,{\text{of}}\,{\text{hot}}\,{\text{metal}}\,{\text{to}}\,{\text{be}}\,{\text{charged}}\,{\text{per}}\,{\text{tonne}}\,{\text{of}}\,{\text{steel}}\,{\text{production}}{\text{.}} \\ & \left( {{\text{ii}}} \right)\,{\text{Amount}}\,{\text{of}}\,{\text{slag}}\,{\text{produced}}\,{\text{and}}\,{\text{composition}}\,{\text{of}}\,{\text{slag}}{\text{.}} \\\end{aligned}$

$\begin{aligned} {\text{Given data:}} & \,\left( {\text{i}}\, \right){\text{Weight}}\,{\text{of}}\,{\text{lime}}\,{\text{is}}\,{\text{50}}\,{\text{kg}}\,{\text{per}}\,{\text{tonne}}\,{\text{of}}\,{\text{steel}}\,{\text{production}}\,{\text{(lime}}\,{\text{contain}} \\ & \quad \;{\text{94}}{\text{.5\% }}\,{\text{CaO}},{\text{2}}{\text{.5\% MgO}},{\text{1}}{\text{.5\% SiO}}_{{\text{2}}} \,{\text{and}}\,{\text{1}}{\text{.5\% }}\,{\text{Al}}_{{\text{2}}} {\text{O}}_{{\text{3}}} {\text{)}} \\ & \left( {{\text{iii}}} \right)\,{\text{1}}{\text{.0\% }}\,{\text{Fe}}\,{\text{loss}}\,{\text{with}}\,{\text{respect}}\,{\text{to}}\,{\text{steel}}\,{\text{production}}{\text{.}} \\ \end{aligned}$

Solution

• Assumptions: (a) one tonne steel production,

• By balancing the chemistry of hot metal and produced steel, we get 94.31% and 99.53% Fe respectively.

• Fe balance: Fe input = Fe output

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

• $\begin{aligned} & {\text{Suppose}}\,{\text{weight}}\,{\text{of}}\,{\text{hot}}\,{\text{metal = W}}_{{{\text{HM}}}} \\ & 0.9431{\mkern 1mu} W_{{{\text{HM}}}} = 0.9953 \times 1000 + 0.01 \times 1000 \\ \end{aligned}$

• $\begin{gathered} {\text{Therefore, }}W_{{HM}} = {\mathbf{1065}}.{\mathbf{95}}{\mkern 1mu} {\mathbf{kg}} \hfill \\ \quad \quad \quad \,\,\quad {\mkern 1mu} {\text{Fe}} + 1/2\;{\text{O}}_{2} = {\text{FeO}} \hfill \\ \quad \quad \quad \,\,\quad {\mkern 1mu} 56\quad \quad \quad \quad \quad {\mkern 1mu} {\mkern 1mu} 72 \hfill \\ \quad \quad \quad \,\,\quad {\mkern 1mu} 56\;{\text{kg}}\;{\text{of}}\;{\text{Fe}}\;{\text{to}}\;{\text{form}}\;72\;{\text{kg}}\;{\text{of}}\;{\text{FeO}} \hfill \\ \quad \quad \quad \,\,\quad 10\;{\text{kg}}\quad \quad \quad \quad \quad \quad \left( {\frac{{72}}{{56}}} \right) \times 10 = 12.86\;{\text{kg}}\;{\text{of}}\;{\text{FeO}} \hfill \\ \end{gathered}$

• $\begin{aligned} & {\mathbf{Si}}\,{\mathbf{balance}}:{\text{Si}}\,{\text{input = Si}}\,{\text{output}} \\ & {\text{Si}}\;{\text{from}}\;{\text{HM}} + {\text{Si}}\;{\text{from}}\;{\text{lime}} = {\text{Si}}\;{\text{goes}}\;{\text{to}}\;{\text{steel}}\;{\text{produce}} \\ & \quad + {\text{Si}}\;{\text{losses}}\;{\text{in}}\;{\text{slag\, (Suppose}}\;{\text{weight}}\;{\text{of}}\;{\text{Si}}\;{\text{losses}}\;{\text{in}}\,{\text{slag}} = W_{\text{Si}} ) \\ & 0.011 \times 1065.95 + 0.015 \times 50 \times \left( {\frac{28}{60}} \right) = 0.0001 \times 1000 + {\text{W}}_{\text{Si}} \\ \end{aligned}$

• Therefore, W_Si = 11.98 kg

• $\begin{array}{*{20}l}& { \quad \quad {\text{Si}} + {\text{O}}_{2} = {\text{SiO}}_{2} } \hfill \\ & { \quad \quad 28\quad \quad \quad {\mkern 1mu} 60} \hfill \\ \end{array}$

• 28 kg Si to form 60 kg of SiO₂

• $11.98\;{\text{kg}}\quad \quad \left( {\frac{60}{28}} \right) \times 11.98 = 25.67\;{\text{kg}}\;{\text{of}}\;{\text{SiO}}_{2} \;{\text{in}}\;{\text{slag}}$

• Mn balance: Mn input = Mn output

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

• 0.0075 × 1065.95 = 0.0025 × 1000 + W_Mn

• Therefore, W_Mn = 7.74 kg

$\begin{aligned} & \quad \quad {\text{Mn}} + 1/2\;{\text{O}}_{2} = {\text{MnO }} \\ & \quad \quad 55\quad \quad \quad \quad \quad \,\,71 \\ \end{aligned}$

55 kg Mn to form 71 kg of MnO

$7.74\;{\text{kg}}\quad \quad \left( {\frac{71}{55}} \right) \times 7.74 = 10.0\;{\text{kg}}\;{\text{of}}\;{\text{MnO}}\;{\text{in}}\;{\text{slag}}.$

• P balance: P input = P output

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

• 0.004 × 1065.95 = 0.0003 × 1000 + W_P

• W_P = 3.96 kg

  $\begin{array}{*{20}l} {2{\text{P}} + 5/2\;{\text{O}}_{2} = {\text{P}}_{2} {\text{O}}_{5} } \hfill \\ { 2 \times 31\quad \quad \quad \quad 142} \hfill \\ \end{array}$

• 62 kg P to form 142 kg of P₂O₅

• 3.96 kg $\left( {\frac{142}{62}} \right)$ × 3.96 = 9.07 kg of P₂O₅

• S balance: S input = S output

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

• 0.0004 × 1065.95 = 0.0003 × 1000 + W_S

• Therefore, W_S = 0.126 kg

  $\begin{array}{*{20}l} {{\text{CaO}} + {\text{S}} = {\text{CaS}}} \hfill \\ {56\quad \,\,\, 32\quad 72} \hfill \\ \end{array}$

• 32 kg S to form 72 kg of CaS

• 0.126 kg $\left( {\frac{72}{32}} \right)$ × 0.126 = 0.28 kg of CaS

• 72 kg CaS formation CaO required 56 kg

• 0.28 kg $\left( {\frac{56}{72}} \right)$ × 0.28 = 0.22 kg

• CaO balance: CaO input = CaO output

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

• $\begin{array}{*{20}l} {0.945\;{\text{x}}\;50 = 0.22 + {\text{W}}_{\text{CaO}} } \hfill \\ {{\text{W}}_{\text{CaO}} = 47.03{\text{kg}}} \hfill \\ \end{array}$

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

$$ \begin{aligned} & W_{{{\text{Al}}_{2} {\text{O}}_{3} }} = 0.015 \times 50 = 0.75 \\ & {\text{Amount}}\;{\text{of}}\;{\text{slag}} = {\text{W}}_{\text{FeO}} + {\text{W}}_{{{\text{SiO}}_{2} }} + {\text{W}}_{\text{MnO}} + {\text{W}}_{{{\text{P}}_{2} {\text{O}}_{5} }} + {\text{W}}_{\text{CaS}} + {\text{W}}_{\text{CaO}} + {\text{W}}_{\text{MgO}} + {\text{W}}_{{{\text{Al}}_{2} {\text{O}}_{3} }} \\ & \quad \quad \quad \quad \quad \quad = \, 12.86 + 25.67 + 10.0 + 9.07 + 0.28 + 47.03 + 1.25 + 0.75 = {\mathbf{106}}.{\mathbf{91}}\;{\mathbf{kg}} \\ \end{aligned} $$

Composition of slag: 12.03% FeO, 24.01% SiO₂, 9.35% MnO, 8.48% P₂O₃, 0.26% CaS, 43.99% CaO, 1.17% MgO and 0.7% Al₂O₃.

Example 14.3

400 kg scrap and 400 kg hot metal are charged into a basic open-hearth furnace. The composition of charge materials and product is as follows:

Materials	Composition (%)
	C	Mn	Si	P
Hot metal	3.6	1.9	0.9	0.15
Scrap	0.5	0.3	0.1	0.05
Steel produce	1.0	0.5	0.2	0.04
Iron ore	73% Fe2O3	12% MnO	15% SiO2	–

Slag contains 45% CaO, 20% FeO and 35% SiO₂ and MnO.

Find out: (1) weight of iron ore to be added, (2) weight of limestone to be added and (3) weight of slag produce.

Solution

• Basis of calculation: (1) one tonne of steel production and (2) limestone is pure.

• Fe balance:

• Fe from HM + Fe from scrap + Fe from ore = Fe in steel + Fe loss in slag

• $$ \begin{aligned} & \left( {\frac{93.45}{100}} \right) \times 400 + \left( {\frac{99.05}{100}} \right) \times 400 + \left( {\frac{112}{160}} \right) \times \left( {\frac{73}{100}} \right) \times W_{\text{slag}} \times W_{\text{ore}} = \left( {\frac{98.26}{100}} \right) \times 1000 \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \,\, + \left( {\frac{20}{100}} \right) \times \left( {\frac{56}{72}} \right) \times W_{\text{slag}} \\ \end{aligned} $$

• Or 373.8 + 396.2 + 0.511 W_ore = 982.6 + 0.156 W_slag

• Or

  $$ 0.511\;W_{\text{ore}} + 0.156\;W_{\text{slag}} = 212.6 $$

  (1)

• Si balance:

• Si from HM + Si from scrap + Si from ore = Si in steel + Si loss in slag

• $$ \left( {\frac{0.9}{100}} \right) \times 400 + \left( {\frac{0.1}{100}} \right) \times 400 + \left( {\frac{15}{100}} \right) \times \left( {\frac{28}{60}} \right) \times W_{\text{ore}} = \left( {\frac{0.2}{100}} \right) \times 1000 + W_{{{\text{Si}}\;{\text{in}}\;{\text{slag}}}} $$

• Or 3.6 + 0.4 + 0.07 W_ore = 2.0 + W_{Si in slag}

• Or W_{Si in slag} = 2.0 + 0.07 W_ore

• $$ {\text{Therefore}},W_{{{\text{SiO}}2\;{\text{in}}\;{\text{slag}}}} = \left( {2.0 + 0.07\;W_{\text{ore}} } \right) \times \left( {\frac{60}{28}} \right) = 4.29 + 0.15\;W_{\text{ore}} $$

  (2)

• Mn balance:

• Mn from HM + Mn from scrap + Mn from ore = Mn in steel + Mn loss in slag

• $$ \left( {\frac{1.9}{100}} \right) \times 400 + \left( {\frac{0.3}{100}} \right) \times 400 + \left( {\frac{12}{100}} \right) \times \left( {\frac{55}{71}} \right) \times W_{\text{ore}} = \left( {\frac{0.5}{100}} \right) \times 1000 + W_{{{\text{Mi}}\;{\text{in}}\;{\text{slag}}}} $$

• Or 7.6 + 1.2 + 0.093 W_ore = 5.0 + W_{Mn in slag}

• Or W_{Mn in slag} = 3.8 + 0.093 W_ore

• $$ {\text{Therefore}},W_{{{\text{MnO}}\;{\text{in}}\;{\text{slag}}}} = \left( {3.8 + 0.093\;W_{\text{ore}} } \right) \times \left( {\frac{71}{55}} \right) = 4.91 + 0.12\;W_{\text{ore}} $$

  (3)

• CaO balance:

• CaO from limestone = CaO in slag

• (56/100) × W_LS = W_{CaO in slag}

• Or

  $$ W_{{{\text{CaO}}\;{\text{in}}\;{\text{slag}}}} = 0.56\;W_{\text{LS}} $$

  (4)

• Weight of slag = W_{FeO in slag} + $W_{{{\text{SiO}}_{2\,{\text{in\, slag}} } }}$ + W_{MnO in slag} + W_{CaO in slag}

• W_slag = 0.2 W_slag + (4.29 + 0.15 W_ore) + (4.91 + 0.12 W_ore) + 0.56 W_LS

• Or

  $$0.8\;W_{\text{slag}} = 0.27\;W_{\text{ore}} + 0.56\;W_{\text{LS}} + 9.2 $$

  (5)

• Again, W_{CaO in slag} = 0.56 W_LS = 0.45 W_slag

• $$ {\text{Therefore}},W_{\text{LS}} = 0.8\;W_{\text{slag}} $$

  (6)

• Again ($W_{{{\text{SiO}}_{2} }}$ _{in slag} + W_{MnO in slag}) = 0.35 W_slag

• Or (4.29 + 0.15 W_ore) + (4.91 + 0.12 W_ore) = 0.35 W_slag (from Eqs. 2 to 3)

• Or

  $$ 0.27\;{\text{W}}_{\text{ore}} + 9.2 = 0.35\;{\text{W}}_{\text{slag}} \;{\text{or}}\;{\text{W}}_{\text{ore}} = 1.3\;{\text{W}}_{\text{slag}} {-}9.2 $$

  (7)

• From Eq. (1): 0.511 W_ore + 0.156 W_slag = 212.6

• Or 0.511 (1.3 W_slag – 9.2) + 0.156 W_slag = 212.6 (from Eq. 7)

• Or 0.66 W_slag – 4.7 + 0.156 W_slag = 212.6 or 0.82 W_slag = 212.6 + 4.7 = 217.3

• Therefore, $\varvec{W}_{{{\mathbf{slag}}}} = {\mathbf{265}}\;{\mathbf{kg}}$.

• From Eq. 7: $\varvec{W}_{{{\mathbf{ore}}}} = 1.3\;W_{\text{slag}} {-}9.2 = \left( {1.3 \times 265} \right){-}9.2 = {\mathbf{335}}.{\mathbf{3}}\;{\mathbf{kg}}$

• From Eq. 6: $\varvec{W}_{{{\mathbf{LS}}}} = 0.8\;W_{\text{slag}} = 0.8 \times 265 = {\mathbf{212}}\;{\mathbf{kg}}.$

Example 14.4

Composition of combustion gas burned at the open-hearth furnace is 39.8% H₂, 36.9% N₂, 7.3% CH₄, 8.2% CO, 5.5% CO₂ and 2.3% O₂. Composition of exit gas is 11.6% H₂, 77.5% N₂, 7.1% CO, 3.1% CO₂ and 0.7% O₂. How much heat is utilized by the furnace and how much air required for combustion of m³ gas?

Given: Calorific values of gases (J/m³): 10,932.79 for H₂, 12,811.41 for CO, 36,091.18 for CH₄.

Solution

• $$ \begin{aligned} & {\text{Heat}}\;{\text{generate}}\;{\text{from}}\;{\text{burning}}\;{\text{gases}}:{\text{H}}_{2} :\left( {\frac{39.8}{100}} \right)\,{\text{x}}\,10{\text{,}}932.79 = 4{\text{,}}351.25\;{\text{J}} \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {\text{CO}}:\left( {\frac{8.2}{100}} \right)\,{\text{x}}\,12{\text{,}}811.41 = 1{\text{,}}050.54\;{\text{J}} \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {\text{CH}}_{4} : \left( {\frac{7.3}{100}} \right)\,{\text{x}}\,36{\text{,}}091.18 = 2{\text{,}}634.66\;{\text{J}} \\ & {\text{Total}}\;{\text{heat}}\;{\text{generated}}\;{\text{by}}\;{\text{burning}}\;{\text{gases}} = 4{\text{,}}351.25 + 1{\text{,}}050.54 + 2{\text{,}}634.66 = 8{\text{,}}036.45\;{\text{J}} \\ \end{aligned} $$
• $$ \begin{aligned} & {\text{Heat}}\;{\text{carries}}\;{\text{away}}\;{\text{by}}\;{\text{exit}}\;{\text{gases}}:{\text{H}}_{2} :\left( {\frac{11.6}{100}} \right)\,{\text{x}}\,10{\text{,}}932.79 = 1{\text{,}}268.20\;{\text{J}} \\ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {\text{CO}}:\left( {\frac{7.1}{100}} \right)\,{\text{x}}\,12{\text{,}}811.41 = 909.61\;{\text{J}} \\ & {\text{Total}}\;{\text{heat}}\;{\text{carries}}\;{\text{away}}\;{\text{by}}\;{\text{exit}}\;{\text{gases}} = 1{\text{,}}268.20 + 909.61 = 2{\text{,}}177.81\;{\text{J}} \\ \end{aligned} $$

• Therefore, heat utilized by the furnace = 8,036.45 – 2,177.81 = 5,858.64 J

• $$ \begin{gathered} {\text{H}}_{2} + 1/2\;{\text{O}}_{2} = {\text{H}}_{2} {\text{O}} \to {\text{O}}_{2} {\mkern 1mu} {\text{require}} = \left( {\frac{{39.8}}{{100}}} \right){\text{x}}1/2 = 0.199\;{\text{m}}^{3} {\text{ }} \hfill \\ {\text{CO}} + 1/2\;{\text{O}}_{2} = {\text{CO}}_{2} \to {\text{O}}_{2} {\mkern 1mu} {\text{require}} = \left( {\frac{{8.2}}{{100}}} \right){\text{x}}1/2 = 0.041\;{\text{m}}^{3} {\text{ }} \hfill \\ {\text{CH}}_{4} + 2{\text{O}}_{2} = {\text{CO}}_{2} + 2{\text{H}}_{2} {\text{O}} \to {\text{O}}_{2} {\mkern 1mu} {\text{require}} = \left( {\frac{{7.3}}{{100}}} \right){\text{x}}2 = 0.146{\text{m}}^{3} {\text{ }} \hfill \\ {\text{Total}}\;{\text{O}}_{2} {\mkern 1mu} {\text{require}} = 0.199 + 0.041 + 0.146 = 0.386\;{\text{m}}^{3} \hfill \\ {\text{O}}_{2} \;{\text{present}}\;{\text{in}}\;{\text{the}}\;{\text{combustion}}\;{\text{gas}} = \left( {\frac{{2.3}}{{100}}} \right) = 0.023{\text{m}}^{3} \hfill \\ \end{gathered} $$

• $\begin{aligned} & {\text{Total}}\;{\text{O}}_{2} \,{\text{require}} = 0.199 + 0.041 + 0.146 = 0.386\;{\text{m}}^{3} \\ & {\text{O}}_{2} \;{\text{present}}\;{\text{in}}\;{\text{the}}\;{\text{combustion}}\;{\text{gas}} = \left( {\frac{{2.3}}{{100}}} \right) = 0.023{\text{m}}^{3} \\ \end{aligned}$

• Hence, actual O₂ required for combustion of gas = 0.386 − 0.023 = 0.363 m³

• Air contains 21% O₂,

So, ${\text{air}}\;{\text{required}}\;{\text{for}}\;{\text{combustion}}\;{\text{of}}\,{\text{gas}} = {\mathbf{1}}.{\mathbf{73}}\;{\mathbf{m}}^{{\mathbf{3}}} /{\mathbf{m}}^{{\mathbf{3}}} {\mathbf{of}}\;{\mathbf{gas}}$.

Problems

Problem 14.1

Steel is being made in a Bessemer converter, using air to oxidize the impurities. The converter charges 10 tonnes of hot metal of composition: 4.0% C, 1.4% Si, 1.2% Mn. 3/4th of the carbon form CO and 1/4th form CO₂.

Calculate: (i) the volume of air used, (ii) the composition and total volume of product gases and (iii) weight of slag produce.

[Ans: (i) Volume of air: 2871.93 Nm³ at STP, (ii) Total volume of product gases: 3015.49 Nm³ at STP; 75.24% nitrogen,18.57% CO and 6.19% CO₂ and (iii) Wt of slag: 454.91 kg].

Problem 14.2

800 kg hot metal (4.0% C, 1.2% Si, 1.0% Mn, 0.4% P) is charged with scrap (0.25% C, 0.25% Si, 0.5% Mn, 0.04% P) into a basic open-hearth furnace. Limestone is added to maintain slag basicity 2.5. Find out (i) amount of scrap, and (ii) amount of limestone to be charged also and (iii) amount of slag produced with composition.

Given: (i) Product steel contains 0.75% C, 0.2% Si, 0.4% Mn, 0.04% P; (ii) Limestone contains 2% SiO₂, 3% Al₂O₃; and (iii) 1.5% Fe loss in slag.

[Ans: (i) 256.57 kg, (ii) 127.44 kg, (iii) 131.36 kg (14.68% FeO, 20.55% SiO₂, 5.19% MnO, 5.05% P₂O₃, 51.61% CaO, and 2.91% Al₂O₃)].