Sponge Iron

Abstract

Processes that produce iron by reduction of iron ore in solid state are generally classified as direct reduction (DR) processes and the products also referred to as direct reduced iron (DRI) or sponge iron. Different processes of coal-based and gas-based are discussed in detail. Forms of sponge iron, characteristics of sponge iron and re-oxidation of sponge iron are described. Uses of sponge iron are also discussed.

Appendices

Probable Questions

• 1. What do you mean by sponge iron? How DR processes are classified? What do you mean by “solid-state reduction”?

• 2. Discuss the principle and process for production of sponge iron by coal-based process.

• 3. (i) Discuss the principle of rotary kiln process for sponge iron production. (ii) What are the problems of rotary kiln process and how to overcome them?

• 4. Discuss the salient features of rotary kiln process.

• 5. Why coal-based sponge iron cannot form HBI?

• 6. Discuss the principle of Midrex process for production of HBI.

• 7. ‘Sponge iron produced by gas-based process contains more carbon than by coal-based process’. Why?

• 8. Discuss the advantages of gas-based process over the coal-based process.

• 9. Discuss the HyL III process.

• 10. What do you understand by HBI? How that are produced? Discuss the advantages of HBI.

• 11. Discuss the characteristics of sponge iron.

• 12. ‘Sponge iron has an inherent tendency to re-oxidize back to its native stage’. Why?

• 13. What are prevention methods for re-oxidation of sponge iron?

• 14. Discuss the applications of sponge iron in iron and steel industries.

• 15. ‘Steelmakers are preferred gas-based sponge iron than coal-based sponge iron’. Why?

• 16. Discuss environmental benefits of sponge iron which acts as feed material in steelmaking furnaces.

• 17. Discuss the rotary hearth furnace processes.

• 18. Discuss the ITmk3 process.

• 19. Discuss iron carbide production.

Examples

Example 7.1

An iron ore has the following composition: 58.1% FeCO₃, 11.5% SiO₂, 11.7% CaCO₃, 10.5% Al₂O₃ and 8.2% H₂O. It is calcined in a rotary kiln furnace fired with coke containing 84% C, 13% ash and 3% H₂O. Enough air is supplied to burn the coke completely. Coke requires 80 kg per tonne of ore. Calculate (i) volume of air required at STP per tonne of ore and (ii) volume of gases generated per tonne of ore calcinated.

Solution

Since coke contains 84% C,

80 kg coke contains → 0.80 × 84 = 67.2 kg of carbon

$$ \begin{array}{*{20}c} {\text{C}} & + & {{\text{O}}_{2} } & = & {{\text{CO}}_{2} } \\ {12} & {} & {22.4\,{\text{m}}^{3} } & {} & {22.4\,{\text{m}}^{3} } \\ \end{array} $$

12 kg of C requires 22.4 m³ of O₂

$$ 67.2\quad \quad \left( {\frac{22.4 \times 67.2}{12}} \right) = 125.44\,{\text{m}}^{3} \,{\text{of O}}_{2} \,{\text{or}}\,{\text{CO}}_{2} \,{\text{form}} $$

21 m³ of O₂ content in 100 m³ of air

$$ 125.44\,{\text{m}}^{3} \,{\text{of O}}_{2} \quad \quad \left( {\frac{100 \times 125.44}{21}} \right) = {\mathbf{597}}.{\mathbf{33}}\,{\mathbf{m}}^{{\mathbf{3}}} \,{\mathbf{of}}\,{\mathbf{air}} $$

• Volume of air required at STP per tonne of ore = 597.33 m³

• Amount of N₂ = amount of air − amount of O₂ = 597.33 − 125.44 = 471.89 m³

• Amount of CO₂ form by burning of coke = 125.44 m³

• Iron ore contains 58.1% FeCO₃ and 11.7% CaCO₃.

• That is, one tonne ore contains 581 kg FeCO₃ and 117 kg CaCO₃.

  $$ \begin{array}{*{20}c} {{\text{FeCO}}_{3} } & = & {\text{FeO}} & + & {{\text{CO}}_{2} } \\ {116} & {} & {} & {} & {22.4\,{\text{m}}^{3} } \\ \end{array} $$

116 kg of FeCO₃ forms 22.4 m³ of CO₂

$$ 581\,{\text{kg}}\quad \quad \left( {\frac{22.4 \times 581}{116}} \right) = 112.19\,{\text{m}}^{3} \,{\text{of}}\,{\text{CO}}_{2} $$

$$ \begin{array}{*{20}c} {{\text{CaCO}}_{3} } & = & {\text{CaO}} & + & {{\text{CO}}_{2} } \\ {100} & {} & {} & {} & {22.4{\text{m}}^{3} } \\ \end{array} $$

100 kg of CaCO₃ forms 22.4 m³ of CO₂

$$ 117\,{\text{kg}}\quad \quad \left( {\frac{22.4 \times 117}{100}} \right) = 26.21\,{\text{m}}^{3} {\text{ of CO}}_{2} $$

$$ \begin{aligned} {\text{Total CO}}_{2} \,{\text{form}} & = {\text{CO}}_{2} \,{\text{form coke}} + {\text{CO}}_{2} \,{\text{form FeCO}}_{3} + {\text{CO}}_{2} \,{\text{form CaCO}}_{3} \\ & = 125.44 + 112.19 + 26.21 = {\mathbf{263}}.{\mathbf{84}}\,{\mathbf{m}}^{{\mathbf{3}}} \\ \end{aligned} $$

18 kg of H₂O = 22.4 m³ of H₂O

$$ (0.03 \times 80){\text{kg of H}}_{2} {\text{O from coke}} = \left( {\frac{22.4 \times 2.4}{18}} \right){\text{m}}^{3} \,{\text{of H}}_{2} {\text{O}} = 2.99\,{\text{m}}^{3} \,{\text{of H}}_{2} {\text{O form coke}} $$

18 kg of H₂O = 22.4 m³ of H₂O

$$ 82\,{\text{kg}}\quad \quad \left( {\frac{22.4 \times 82}{18}} \right){\text{m}}^{3} \,{\text{of H}}_{2} {\text{O}} = 102.04\,{\text{m}}^{3} \,{\text{of H}}_{2} {\text{O form ore}}. $$

Total moisture = 2.99 + 102.04 = 105.03 m³

Therefore,

$$ \begin{aligned} {\text{total amount of gases }}({\text{N}}_{2} + {\text{CO}}_{2} + {\text{H}}_{2} {\text{O}}) & = 471.89 + 263.84 + 105.03 \\ & = {\mathbf{840}}.{\mathbf{76}}\,{\mathbf{m}}^{{\mathbf{3}}} \\ \end{aligned} $$

Example 7.2

Total iron content in ore and DRI are 65.1 and 90%, respectively. If the production rate of a rotary kiln furnace is 4.16 t/hr, what is the rate of iron ore charged to the rotary kiln furnace?

Solution

Fe Balance:

Fe from ore = Fe in DRI

$$ ({\text{T}}_{{{\text{Fe}},{\text{ore}}}} /100) \times {\text{W}}_{\text{ore}} = ({\text{T}}_{{{\text{Fe}},{\text{DRI}}}} /100) \times {\text{W}}_{\text{DRI}} $$

(1)

Since ${\text{W}}_{\text{DRI}} = 4.16\,{\text{t}}/{\text{hr}},{\text{T}}_{{{\text{Fe}},{\text{DRI}}}} = 90\% \,{\text{and}}\,{\text{T}}_{{{\text{Fe}},{\text{ore}}}} = 65.1\%$

So from Eq. (1): (65.1/100) × W_ore = (90/100) × 4.16 = 3.744

Therefore, ${\text{W}}_{\text{ore}} = (3.744/0.651) = 5.75\,{\text{t}}/{\text{hr}}$.

The rate of iron ore charged to the rotary kiln furnace is 5.75 t/hr.

Example 7.3

Find out the change of free energy $(\Delta G_{r,298}^{ \circ } )$ and minimum temperature for the following reactions:

• (a) $3{\text{Fe}}_{2} {\text{O}}_{3} + {\text{CO}} = 2{\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}}_{2}$

• (b) ${\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}} = 3{\text{FeO}} + {\text{CO}}_{2}$

Given:

$$ \begin{aligned} & 3{\text{Fe}}_{2} {\text{O}}_{3} = 2{\text{Fe}}_{3} {\text{O}}_{4} + 1/2{\text{O}}_{2} ,\Delta G_{r}^{ \circ } = 249.45{-}0.14\,{\text{T}}\,{\text{kJ}} \\ & {\text{Fe}}_{3} {\text{O}}_{4} = 3{\text{FeO}} + 1/2{\text{O}}_{2} ,\Delta G_{r}^{ \circ } = 312.21 - 0.125\,{\text{T}}\,{\text{kJ}} \\ & {\text{C}} + 1/2{\text{O}}_{2} = {\text{CO}},\Delta G_{r}^{ \circ } = - 111.71 - 0.088\,{\text{T}}\,{\text{kJ}} \\ & {\text{C}} + {\text{O}}_{2} = {\text{CO}}_{2} ,\Delta G_{r}^{ \circ } = - 394.13 - 8.4 \times 10^{ - 4} {\text{T}}\,{\text{kJ}} \\ \end{aligned} $$

Solution

• (a)
  $$ \begin{array}{*{20}l} \begin{aligned} \quad 3{\text{Fe}}_{2} {\text{O}}_{3} & = 2{\text{Fe}}_{3} {\text{O}}_{4} + 1/2{\text{O}}_{2} ,\Delta G_{{{\text{Fe}}_{2} {\text{O}}_{3} }}^{ \circ } \\ \quad {\text{CO}} & = {\text{C}} + 1/2{\text{O}}_{2} , - \Delta G_{\text{CO}}^{ \circ } \\ \quad {\text{C}} + {\text{O}}_{2} & = {\text{CO}}_{2} ,\quad \Delta G_{{{\text{CO}}_{2} }}^{ \circ } \\ \end{aligned} \hfill \\ {\begin{array}{*{20}l} {\quad - - - - - - - - - - - - - - - - - - - } \hfill \\ {3{\text{Fe}}_{2} {\text{O}}_{3} + {\text{CO}} = 2{\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}}_{2} ,\Delta G_{f}^{ \circ } = (\Delta G_{{{\text{Fe}}_{2} {\text{O}}_{3} }}^{ \circ } - \Delta G_{\text{CO}}^{ \circ } + \Delta G_{{{\text{CO}}_{2} }}^{ \circ } )} \hfill \\ \end{array} } \hfill \\ \end{array} $$

Therefore,

$$ \begin{aligned} \Delta {\text{G}}_{f}^{ \circ } & = \left[ {(249.45{-}0.14\,{\text{T}}){-}( - 111.71{-}0.088\,{\text{T}}) + ( - 394.13 - 8.4 \times 10^{ - 4} {\text{T}})} \right] \\ & = - {\mathbf{32}}.{\mathbf{97}} - {\mathbf{0}}.{\mathbf{052}}\,{\mathbf{T}}\,{\mathbf{kJ}} \\ \end{aligned} $$

At equilibrium, $\Delta {\text{G}}_{f}^{ \circ } = 0$

Hence, −32.97 − 0.052 T = 0.

Therefore, T = −634 K = −361 ℃.

• (b) $\begin{array}{*{20}l} \begin{aligned} \quad {\text{Fe}}_{3} {\text{O}}_{4} & = 3{\text{FeO}} + 1/2{\text{O}}_{2} ,\quad \Delta G_{{{\text{Fe}}_{3} {\text{O}}_{4} }}^{ \circ } \\ \quad {\text{CO}} & = {\text{C}} + 1/2{\text{O}}_{2} ,\quad - \Delta G_{\text{CO}}^{ \circ } \\ \quad {\text{C}} + {\text{O}}_{2} & = {\text{CO}}_{2} ,\quad \Delta G_{{{\text{CO}}_{2} }}^{ \circ } \\ \end{aligned} \hfill \\ {\begin{array}{*{20}l} {\quad - - - - - - - - - - - - - - - - - - - } \hfill \\ {{\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}} = 3{\text{FeO}} + {\text{CO}}_{2} ,\Delta G_{f}^{ \circ } = (\Delta G_{{{\text{Fe}}_{3} {\text{O}}_{4} }}^{ \circ } - \Delta G_{\text{CO}}^{ \circ } + \Delta G^{ \circ }_{{{\text{CO}}_{2} }} )} \hfill \\ \end{array} } \hfill \\ \end{array}$

Therefore,

$$ \begin{aligned} \Delta G_{f}^{0} & = \left[ {(312.21{-}0.125\,{\text{T}}){-}( - 111.71{-}0.088\,{\text{T}}) + ( - 394.13 - 8.4 \times 10^{ - 4} {\text{T}})} \right] \\ & = {\mathbf{29}}.{\mathbf{79}}{-}{\mathbf{0}}.{\mathbf{037}}\,{\mathbf{T}}\,{\mathbf{kJ}} \\ \end{aligned} $$

At equilibrium, $\Delta G_{f}^{ \circ } = 0$

Hence, 29.79 − 0.037 T = 0.

Therefore, T = 805 K = 532 ℃.

Example 7.4

Calculate the minimum per cent of carbon monoxide required to reduce FeO at 727 ℃ and (i) one atmospheric pressure and (ii) 1.5 atm. The reduction is taking place as follows:

$$ {\text{FeO}} + {\text{CO}} = {\text{Fe}} + {\text{CO}}_{2} \quad \Delta G^{ \circ } = - 22802.8 + 28.45\,{\text{T}}\,{\text{J}}/{\text{mol}} $$

Solution

• T = 727 + 273 = 1000 K

• $\Delta G^{ \circ } = - 22802.8 + 28.45 \times 1000 = 5647.2\,{\text{J}}/{\text{mol}}$

• Since ∆G^° = −RT ln k

• Therefore, $5647.2 = - 8.314 \times 1000 \times \ln k$

• ln k = −0.6792

• Therefore, k = 0.507.

• Again, $k = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right)$

• Case I

• $p_{{{\text{CO}}_{2} }} + p_{\text{CO}} = 1\,{\text{atm}}$

• Let p_CO = x atm, $p_{{{\text{CO}}_{2} }}$ = 1 − x

Therefore, $\left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right) = \left( {\frac{1 - x}{x}} \right) = k = 0.507$

• Therefore, x = 0.6635 atm.

• Per cent of CO = 66.35%

• Case II

• $p_{{{\text{CO}}_{2} }} + p_{\text{CO}} = 1.5\,{\text{atm}}$

• Let p_CO = x atm, $p_{{{\text{CO}}_{2} }}$ = 1.5 − x

Therefore, $\left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right) = \left( {\frac{1.5 - x}{x}} \right) = K = 0.507$

• Therefore, x = 0.9953 atm.

• Per cent of CO = 99.53%

Example 7.5

Find the change in the percentage of hydrogen in a mixture of 75% H₂ and 25% CH₄ in equilibrium with carbon if the total pressure changes from 1 to 0.1 atm. What is the temperature of this equilibrium at 1 atm pressure?

Given: ${\text{CH}}_{4} = {\text{C}} + 2\,{\text{H}}_{2} ,\quad \Delta G^{\circ} = 90165.2{-}109.45\,{\text{T}}\,{\text{J}}$

Solution

Equilibrium constant, $k = \left( {\frac{{p_{{{\text{H}}_{2} }}^{2} }}{{p_{{{\text{CH}}_{4} }} }}} \right)$

Total pressure = 1 atm = p_H2 + p_CH4

Since 75% H₂ and 25% CH₄, i.e. p_H2 = 0.75, and p_CH4 = 0.25

Hence, $k = \left( {\frac{{p_{{{\text{H}}_{2} }}^{2} }}{{p_{{{\text{CH}}_{4} }} }}} \right) = \left( {\frac{{0.75^{2} }}{0.25}} \right) = 2.25$, and ln k = 0.8109.

Since $\Delta G^{\circ} = - {\text{RT}}\ln k = - 8.314 \times T \times \ln k$

Therefore, $90165.2{-}109.45\,{\text{T}} = - 8.314 \times {\text{T}} \times 0.8109 = - 6.74\,{\text{T}}$

Hence, T = 877.88 K = 604.88 ℃.

• Now, total pressure changes from 1 atm to 0.1 atm.

• So, $p_{{{\text{H}}_{2} }} + p_{{{\text{CH}}_{4} }} = 0.1\,{\text{atm}}$

• Let $p_{{{\text{H}}_{2} }} = x$, $p_{{{\text{CH}}_{4} }} = (0.1{-}x)$

• Since ΔG^o = −RT ln k

• Therefore, $90165.2{-}109.45\,{\text{T}} = - 8.314 \times {\text{T}} \times \ln k$

• Or $90165.2{-}109.45 \times 878 = - 8.314 \times 878 \times \ln k$ (since T = 878 K)

• Or ln k = 0.813 or k = 2.25

• Since $k = \left( {\frac{{p_{{{\text{H}}_{2} }}^{2} }}{{p_{{{\text{CH}}_{4} }} }}} \right) = \left( {\frac{{x^{2} }}{0.1 - x}} \right) = 2.25$

Or $x^{2} + 2.25x{-}0.225 = 0$

$$ {\text{Roots}} = \left( {\frac{{\left\{ { - b \pm \surd \left( { b^{2} {-} 4ac} \right)} \right\}}}{2a}} \right) $$

Therefore, $x = \left( {\frac{{\left\{ { - 2.25 \pm \surd \left( { 2.25^{2} {-} 4.1.( - 0.225)} \right)} \right\}}}{{2{\text{x}}1}}} \right)$

Hence, x = 0.096, so H₂ = 9.6%.

Therefore, the change in the percentage of hydrogen in a mixture of 75% H₂ and 25% CH₄ in equilibrium with carbon, if the total pressure changes from 1 atm to 0.1 atm, is 9.6%.

Example 7.6

Calculate the equilibrium CO: CO₂ ratio and %CO at 900 ℃ according to reaction:

$$ {\text{FeO}} + {\text{CO}} = {\text{Fe}} + {\text{CO}}_{2} $$

Given:

$$ \begin{aligned} & {\text{C}} + {\text{O}}_{2} = {\text{CO}}_{2} \quad \Delta G^{ \circ } = - 394,100{-}0.84\,{\text{T}}\;{\text{J}} \\ & {\text{C}} + 1/2{\text{O}}_{2} = {\text{CO}}\quad \Delta G^{ \circ } = - 111,000{-}87.65\,{\text{T}}\;{\text{J}} \\ & {\text{Fe}} + 1/2{\text{O}}_{2} = {\text{FeO}}\quad \Delta G^{ \circ } = - 259,600 + 62.55\,{\text{T}}\;{\text{J}} \\ \end{aligned} $$

Solution

$$ \begin{array}{*{20}l} \begin{aligned} \quad {\text{FeO}} = {\text{Fe}} + 1/2{\text{O}}_{2} \quad \Delta G^{ \circ } = 259,600 - 62.55\,{\text{T}} \hfill \\ \quad {\text{CO}} = {\text{C}} + 1/2{\text{O}}_{2} \quad \Delta G^{ \circ } = 1111,000 + 87.65\,{\text{T}} \hfill \\ \quad {\text{C}} + {\text{O}}_{2} = {\text{CO}}_{2} \quad \Delta G^{ \circ } = - 394,100 - 0.84\,{\text{T}} \hfill \\ \end{aligned} \hfill \\ \hline {\begin{array}{*{20}c} { - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - } \\ {{\text{FeO}} + {\text{CO}} = {\text{Fe}} + {\text{CO}}_{2} \quad \Delta G^{ \circ } = - 23500 + 24.26\,{\text{T}} = - 23500 + 24.26 \times 1173 = 4956.98\,{\text{J}}} \\ \end{array} } \hfill \\ \end{array} $$

• Since ∆G^° = −RT ln k

• Therefore, 4956.98 = −8.314 × 1173 × ln k.

• ln k = −508

• Therefore, k = 0.6015.

Again, $k = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right)$

So, $\left( {\frac{{p_{{CO_{2} }} }}{{p_{CO} }}} \right) = \left( {\frac{1}{k}} \right) = \left( {\frac{1}{0.6015}} \right) = {\mathbf{1}}.{\mathbf{66}}$

Therefore, the equilibrium CO: CO₂ ratio is 1.66.

• Let p_CO = x, since $p_{\text{CO}} + p_{{{\text{CO}}_{2} }} = 1$.

• Therefore, $p_{{{\text{CO}}_{2} }} = 1 - x$.

• Hence, $k = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right) = \left( {\frac{1 - x}{x}} \right) = 0.6015$

• Therefore, $x = \left( {\frac{1}{1.6015}} \right)$ = 0.624, i.e. CO = 62.4%.

Example 7.7

Find out whether FeO can be reduced by H₂/H₂O mixture containing 60% H₂ and 40% H₂O at 727 ℃.

$$ \begin{aligned} & {\text{Fe}} + 1/2{\text{O}}_{2} = {\text{FeO}}\quad \Delta G^{ \circ } = - 259,600 + 62.55\,{\text{T}}\;{\text{J}} \\ & {\text{H}}_{2} + 1/2{\text{O}}_{2} = {\text{H}}_{2} {\text{O}}\quad \Delta G^{ \circ } = - 246,000 + 54.8\,{\text{T}}\;{\text{J}} \\ \end{aligned} $$

Solution

$$ \begin{array}{*{20}l} \begin{aligned} {\text{FeO}} & = {\text{Fe}} + 1/2{\text{O}}_{2} \quad \Delta G^{ \circ } = 259,600 - 62.55\,{\text{T}} \\ {\text{H}}_{2} + 1/2{\text{O}}_{2} & = {\text{H}}_{2} {\text{O}}\quad \Delta G^{ \circ } = - 246,000 + 54.8\,{\text{T}} \\ \end{aligned} \hfill \\ \hline {{\text{FeO}} + {\text{H}}_{2} = {\text{Fe}} + {\text{H}}_{2} {\text{O}}\quad \Delta G^{ \circ } = 13600{-}7.75\,{\text{T}} = 13600{-}7.75 \times 1000 = 850\,{\text{J}}} \hfill \\ \end{array} $$

• ∆G° = −RT ln k

• Therefore, 5850 = −8.314 × 1000 × ln k.

• ln k = −0.7036

• $k = 0.4948 = \left( {\frac{{p_{{{\text{H}}_{2} {\text{O}}}} }}{{p_{{{\text{H}}_{2} }} }}} \right),p_{{{\text{H}}_{2} {\text{O}}}} + p_{{{\text{H}}_{2} }} = 1\,{\text{atm}},{\text{Let }}p_{{{\text{H}}_{2} }} = x$

• Therefore, 0.4948 = $\left( {\frac{1 - x}{x}} \right)$

• x = 0.6689 atm

• Pct of hydrogen = 66.89%

Since 60% H₂ is present in mixture, equilibrium H₂ requires minimum 66.89%; hence, FeO cannot be reduced by H₂ and H₂O mixture containing 60% H₂ and 40% H₂O at 727 ℃.

Another thing should be noted that ∆G⁰ value is positive. Hence, from that value also it can be concluded that reduction is not feasible at all at 727 ℃.

Example 7.8

Iron sample is heated at 1000 ℃ in an atm of hydrogen which contains some moisture. Find out the maximum permissible water vapour in hydrogen to avoid oxidation of iron sample.

Given: ${\text{FeO}} + {\text{H}}_{2} = {\text{Fe}} + {\text{H}}_{2} {\text{O}}\quad \Delta G^{ \circ } = 12761.2{-}7.03{\text{T}}\;{\text{J}}/{\text{mol}}$

Solution

• T = 1000 ℃ = 1273 K

• $\Delta G^{ \circ } = 12761.2{-}7.03\,{\text{T}} = 12761.2{-}7.03 \times 1273 = 3812.01\,{\text{J}}/{\text{mol}}$

• Since ∆G^° = −RT ln k

• Therefore, $3812.01 = - 8.314 \times 1273 \times \ln k$

• ln k = −0.3602

• $k = 0.6975 = \left( {\frac{{p_{{{\text{H}}_{2 } {\text{O}}}} }}{{p_{{{\text{H}}_{2} }} }}} \right) = \left( {\frac{1 - x}{x}} \right)$ [let $p_{{{\text{H}}_{2} }} = x$]

Therefore, x = 0.5891, i.e. H₂ = 58.91%.

• So, ${\text{H}}_{2} {\text{O}} = 100{-}58.91 = 41.09\%$

• Hence, the maximum permissible water vapour in hydrogen to avoid oxidation of iron sample is 41%.

Problems

Problem 7.1

Total iron content in ore and DRI are 66% and 92%, respectively. If the production rate is 5 t/hr, what is the rate of iron ore charged to the rotary kiln furnace?

[Ans: 6.97 t/hr.]

Problem 7.2

Hydrogen is passed through a furnace at 800 ℃, containing FeO (solid), which is undergoing reduction to form Fe. Find out the percentage of utilization of H₂ gas, assuming that equilibrium prevails.

Given: $\Delta G_{r}^{\circ} = 8368\,{\text{J}}/{\text{mol at }}800^\circ {\text{C}}$.

[Ans: 71.87%]

Problem 7.3

Find out whether FeO can be reduced by H₂ and H₂O mixture containing 60% H₂ and 40% H₂O at 600 ℃.

$$ \begin{aligned} & {\text{Fe}} + 1/2{\text{O}}_{2} = {\text{FeO}}\quad \Delta G^{ \circ } = - 259,600 + 62.55\,{\text{T}}\;{\text{J}} \\ & {\text{H}}_{2} + 1/2{\text{O}}_{2} = {\text{H}}_{2} {\text{O}}\quad \Delta G^{ \circ } = - 246,000 + 54.8\,{\text{T}}\;{\text{J}} \\ \end{aligned} $$

[Ans: Since 60% H₂ is present in mixture, equilibrium H₂ requires minimum 71.94%; hence FeO cannot be reduced by H₂ and H₂O mixture containing 60% H₂ and 40% H₂O at 600 ℃.]