Raw Materials

Abstract

Steel (i.e. refined impure iron) is the most used iron alloy; its production is more than 50 times total production of combined all other metals.

Appendices

Probable Questions

• 1. What are the types of iron ores? Name and discuss in brief.

• 2. Which factors decide the quality of an iron ore for its smelting in BF?

• 3. What do you mean by metallurgical coal?

• 4. Although Indian coking coal has high ash content, still it can be used for better environment friendly due to what?

• 5. What are the main functions of coke in a blast furnace?

• 6. What are the properties that depend value of coke (as a BF fuel)?

• 7. ‘Reactivity of coke has a significant influence on the reduction process’, explain.

• 8. What are the ideal burden qualities for hot metal production?

• 9. Discuss the basic principle of coke oven battery with stamp-charging.

• 10. Write short notes on (i) dry cooling of coke, (ii) blending with low ash coking coals and (iii) formed coke.

• 11. Why a BF cannot run without a critical amount of coke? Explain.

• 12. What do you mean by ‘insoluble in limestone’? What are the maximum limits?

• 13. Define the term ‘basicity’ and explain how is it different from V-ratio.

• 14. What are the problems with Indian raw materials for BF ironmaking?

• 15. What are the different processes of agglomeration? Discuss in brief.

• 16. What are the objects of sintering? What are the process variables for sintering?

• 17. ‘By charging super-flux sinter in the BF, limestone can be eliminated from the feed’. Why?

• 18. Write the basic principles of palletization process. Compare disc and drum pelletizers.

• 19. Compare between palletization and sintering processes.

• 20. What are the functions of a binder? What are the factors for selecting the binder?

• 21. What is the mechanism of ball formation in palletization process? State the factors that affect size of the pellets produced.

• 22. What do you mean by bonding mechanism for pellet formation?

• 23. What do you understand by cold-bonded pellets?

• 24. What do you mean by composite pellets? What are the merits offered, when they are used as burden material in ironmaking?

• 25. Explain briefly the different methods of testing of agglomerated products. Also give in brief the test procedures.

• 26. What are the tests you will recommend for the product of disc pelletizer, before charging to the BF?

• 27. What is reducibility? How it is evaluated?

Examples

Example 1.1

The initial weight of iron ore pellet is 1.325 g. The pellet is reduced by hydrogen at 750 °C for 30 min. After reduction, weight of pellet is decreased to 1.015 g. Find out the degree of reduction.

Given: Iron ore (Fe₂O₃) contains 65% total Fe.

Solution

Molecular weight of Fe₂O₃ = 2 × 56 + 3 × 16 = 160

Out of 160 part Fe₂O₃, iron content 112 parts

$$ 100\,{\text{part}}\quad \left[ {\frac{{\left( {112 \times 100} \right)}}{160}} \right] = 70\,{\text{parts i}} . {\text{e }}70\% \,{\text{Fe}} $$

Out of 160 part Fe₂O₃, oxygen content 48 parts

$$ 100\,{\text{part}}\quad \left[ {\frac{{\left( {48 \times 100} \right)}}{160}} \right] = 30\,{\text{part i}} . {\text{e}}\,30\% \,{\text{Oxygen}} $$

$$ \begin{aligned} {\text{Degree of reduction}}\left( \alpha \right) & = \left[ {\frac{{\left( {{\text{O}}_{2} {\text{ removed from the sample}}} \right) \times 100 }}{{\left( {{\text{Total removable O}}_{2} {\text{ present in the sample}}} \right)}}} \right] \\ & = \left[ {\frac{{\left( {\text{weight loss}} \right) \times 100 }}{{\left( {{\text{Total removable O}}_{2} {\text{ present in the sample}}} \right)}}} \right] \\ & = \left[ {\frac{{\left\{ {\left( {W_{1} - W_{2} } \right) \times 100} \right\} }}{{W_{\text{O}} }}} \right] \\ \end{aligned} $$

where

W ₁ :

initial weight of the sample,

W ₂ :

final weight of the sample,

W _O :

total removable O₂ present in the sample = W₁ × f_ore × ρ_ore × f_O

f _ore :

fraction of ore presents in sample,

ρ _ore :

fraction of purity of iron oxide in ore,

f _O :

fraction of oxygen present in iron oxide.

70 part of Fe content in 100 part of iron ore (Fe₂O₃)

$$ 65\,{\text{part}}\quad \left[ {\frac{{\left( {100 \times 65} \right)}}{70}} \right] = 92.86\% $$

i.e. purity of iron oxide in ore = 92.86%

$$ \begin{aligned} {\text{So}}\,\rho_{\text{ore}} & = \left( {\frac{92.86}{100}} \right) = 0.9286 \\ f_{\text{O}} & = \left( {\frac{30}{100}} \right) = 0.3 \\ f_{\text{ore}} & = \left( {\frac{100}{100}} \right) = 1\left[ {{\text{since there is no binder in the pellet, so }}100\% \,{\text{iron ore}}} \right] \\ \end{aligned} $$

Since W_O = W₁ × f_ore × ρ_ore × f_O = 1.325 × 1 × 0.9286 × 0.3 = 0.3691 g.

Therefore, degree of reduction $\alpha = \left( {\frac{{\left( {1.325 - 1.015} \right) \times 100}}{0.3691}} \right) = {\mathbf{83}}.{\mathbf{98}}\%.$

Example 1.2

The initial weight of iron ore pellet is 1.5 g. The pellet is reduced by hydrogen at 750 °C for 30 min. After reduction, weight of the pellet comes down to 1.15 g. The pellet contains 3% binder, and iron ore contains 93% Fe₂O₃. Find out the degree of reduction?

Solution

Since degree of reduction $\left( \alpha \right) \, = \left[ {\frac{{\left\{ {\left( {W_{1} - W_{2} } \right) \times 100} \right\} }}{{W_{\text{O}} }}} \right]$

where

W ₁ :

initial weight of the sample,

W ₂ :

final weight of the sample,

W _O :

total removable O₂ present in the sample = W₁ × f_ore × ρ_ore × f_O

$$ \begin{aligned} f_{\text{ore}} & = {\text{fraction of ore presents in sample}} = \left[ {\frac{{\left( {100 {-} 3} \right)}}{100}} \right] = 0.97 \\ \rho_{\text{ore}} & = {\text{fraction of purity of iron oxide in ore}} = \left( {\frac{93}{100}} \right) = 0.93 \\ f_{\text{O}} & = {\text{fraction of oxygen present in iron oxide}} = 0.3 \\ \end{aligned} $$

Now W_O = W₁ × f_ore × ρ_ore × f_O = 1.5 × 0.97 × 0.93 × 0.3 = 0.4059 g.

Therefore, degree of reduction $\alpha = \left[ {\frac{{\left\{ {\left( {W_{1} - W_{2} } \right) \times 100} \right\} }}{{W_{\text{O}} }}} \right] = \left[ {\frac{{\left\{ {\left( {1.50\, - \, 1.15} \right) \times 100} \right\} }}{0.4059}} \right] = {\mathbf{86}}.{\mathbf{23}}\% .$

Example 1.3

Iron ore–coke composite pellet contains 12% coke and 3% binder. Initial weight of pellet is 1.55 g. Composite pellet undergoes reduction in nitrogen atmosphere at 1000 °C for 1 h. After reduction weight comes down to 1.27 g. Reduced pellet is then treated again with hydrogen gas at 750 °C for 30 min, after hydrogen treatment weight of reduced pellet further comes down to 1.13 g.

Calculate degree of reduction for composite pellet in nitrogen atmosphere.

Given: Ore contains 92% Fe₂O₃, and coke contains 75% carbon.

Solution

As per Eq. (1.27):

$$ {\text{Degree of reduction}}\,\left( \alpha \right) = \left[ {\frac{{\left\{ {\left( {W_{\text{O}}^{i} - W^{\text{H}} } \right) \times 100} \right\} }}{{W_{\text{O}}^{i} }}} \right] $$

where $W_{\text{O}}^{i}$ = total removable O₂ present in the sample $= W_{1} \times \, f_{\text{ore}} \times \rho_{\text{ore}} \times f_{\text{O}}$

$$ W^{\text{H}} = W_{2} {-}W_{3} = 1.27{-}1.13 = 0.14\,{\text{g}} $$

Now $W_{\text{O}}^{i}$ = W₁ × f_ore × ρ_ore × f_O = 1.55 × 0.85 × 0.92 × 0.3 = 0.3636 g

$$ [{\text{Since}}\,f_{\text{ore}} = \left[ {\frac{{\left\{ {100 - \left( {12 + 3} \right)} \right\}}}{100}} \right] = \left[ {\frac{{\left\{ {100 - 15} \right\}}}{100}} \right] = \left( {\frac{85}{100}} \right) = 0.85] $$

$$ \begin{aligned} {\text{Degree of reduction (}}\alpha )& = \left[ {\frac{{\left\{ {\left( {W_{\text{O}}^{i} - W^{\text{H}} } \right) \times 100} \right\} }}{{W_{\text{O}}^{i} }}} \right] \\ & = \left[ {\frac{{\left\{ {\left( {0.3636 - 0.14} \right) \times 100} \right\}}}{0.3636}} \right] = {\mathbf{61}}.{\mathbf{5}}\% . \\ \end{aligned} $$

Example 1.4

Iron ore–coal composite pellet contains 20% coal and 3% binder. Initial weight of pellet is 1.6758 g. Composite pellet undergoes reduction in nitrogen atmosphere at 950 °C for 1 h. After reduction weight comes down to 1.370 g. During handling, pellet is broken; weight of broken pellet is 1.085 g before hydrogen treatment at 750 °C for 30 min, and after hydrogen treatment, weight of reduced pellet further comes down to 0.9568 g. Calculate degree of reduction for composite pellet in nitrogen atmosphere.

Given: Ore contains 85% Fe₂O₃, and coal contains 60% carbon.

Solution

As per Eq. (1.27):

$$ {\text{Degree of reduction}}\,\left( \alpha \right) = \left[ {\frac{{\left\{ {\left( {W_{\text{O}}^{i} - W^{\text{H}} } \right) \times 100} \right\} }}{{W_{\text{O}}^{i} }}} \right] $$

where $W_{\text{O}}^{i}$ = total removable O₂ in the sample = W₁ × f_ore × ρ_ore × f_O.

Since during handling pellet is broken, a correction factor (ε) will come.

$$ {\text{Correction factor}}\,\left(\upvarepsilon \right) = \left[ {\frac{{\left( {W_{3} - W_{4} } \right)}}{{W_{3} }}} \right] $$

where

W ₃ :

weight of broken pellet = 1.085 g

W ₄ :

weight of reduced pellet after hydrogen treatment = 0.9568 g.

Therefore, correction factor $\left( \varepsilon \right) = \left[ {\frac{{\left( {W_{3} - W_{4} } \right)}}{{W_{3} }}} \right] = \left[ {\frac{{\left( {1.085\, - \,0.9568} \right)}}{1.085}} \right] = 0.1182$

So W^H = W₂ × ε = 1.370 × 0.1182 = 0.1619 g.

Now $W_{\text{O}}^{i}$ = W₁ × f_ore × ρ_ore × f_O = 1.6758 × 0.77 × 0.85 × 0.3 = 0.329 g.

Therefore, degree of reduction $\left( \alpha \right) = \left[ {\frac{{\left\{ {\left( {W_{\text{O}}^{i} - W^{\text{H}} } \right) \times 100} \right\} }}{{W_{\text{O}}^{i} }}} \right] = \left[ {\frac{{\left\{ {\left( {0.329\, - \,0.1619} \right) \times 100} \right\} }}{0.329}} \right] = {\mathbf{50}}.{\mathbf{8}}\%.$

Example 1.5

Iron ore (blue dust)–coal composite pellet contains 15% coal and 5% binder. Initial weight of pellet is 1.85 g. Composite pellet undergoes reduction in nitrogen atmosphere at 1000 °C for 1 h. After reduction weight comes down to 1.29 g.

Calculate degree of reduction for composite pellet in nitrogen atmosphere.

Given: Blue dust contains 95.7% Fe₂O₃, 1.6% SiO₂ and 1.8% Al₂O₃. Coal contains 49.7% fixed carbon, 32.5% VM and 11.6% ash.

Solution

As per Eq. (1.34):

$$ \alpha = \left( {\frac{4}{7}} \right) \times \left[ {\frac{{f_{\text{wl}} - (f_{\text{coal}} \, \times \,f_{\text{vm}} )}}{{\left( {f_{\text{ore}} \,x\,\rho_{\text{ore}} \,x\,f_{\text{O}} } \right)}}} \right] \times 100 $$

Now $W_{\text{O}}^{i}$ = W₁ × f_ore × ρ_ore × f_O = 1.85 × 0.8 × 0.957 × 0.3 = 0.425 g.

Since f_ore = [{100 – (15 + 5)}/100] = 0.8

$$ {\text{Fractional}}\,{\text{weight}}\,{\text{loss}} = {f}_{\text{wl}} = \left( {\frac{{\Delta W_{t} }}{{W_{1} }}} \right) = \left[ {\left( {W_{1} {-}W_{2} } \right)/W_{1} } \right] = \left[ {\left( {1.85{-}1.29} \right)/1.85} \right] = 0.303 $$

Therefore, $\alpha = \left( {\frac{4}{7}} \right) \times \left[ {\frac{{f_{\text{wl}} - (f_{\text{coal}} \, \times \,f_{\text{vm}} )}}{{\left( {f_{\text{ore}} \, \times \,\rho_{\text{ore}} \, \times \,f_{\text{O}} } \right)}}} \right] \times 100 = \left( {\frac{4}{7}} \right) \times \left[ {\frac{0.303 - (0.15\, \times \,0.325)}{{\left( {0.8\, \times \,0.957\, \times \,0.3} \right)}}} \right] \times 100$

$$ = \left( {\frac{4}{7}} \right) \times \left[ {\frac{0.303 - (0.049)}{{\left( {0.8\,\times\,0.957\,\times\,0.3} \right)}}} \right] \times 100\left( {\frac{4}{7}} \right) \times \left[ {\frac{0.254}{{\left( {0.23} \right)}}} \right] \times 100 = {\mathbf{63}}.{\mathbf{11}}\% . $$

Example 1.6

Iron ore (blue dust)–coal composite pellet contains 25% coal and 5% binder. Initial weight of pellet is 1.82 g. Composite pellet undergoes reduction in nitrogen atmosphere at 1000 °C for 1 h. After reduction weight comes down to 1.15 g.

Calculate degree of reduction for composite pellet in nitrogen atmosphere.

Given: Blue dust contains 95.7% Fe₂O₃, 1.6% SiO₂ and 1.8% Al₂O₃. Coal contains 42% fixed carbon, 24% VM and 28% ash.

Solution

As per Eq. (1.34):

$$ \alpha = \left( {\frac{4}{7}} \right) \times \left[ {\frac{{f_{\text{wl}} - (f_{\text{coal}} \, \times \,f_{\text{vm}} )}}{{\left( {f_{\text{ore}} \, \times \,\rho_{\text{ore}} \, \times \,f_{\text{O}} } \right)}}} \right] \times 100 $$

Now $W_{\text{O}}^{i}$ = W₁ × f_ore × ρ_ore × f_O = 1.82 × 0.7 × 0.957 × 0.3 = 0.366 g.

Since f_ore = [{100 – (25 + 5)}/100] = 0.7.

$$ {\text{Fractional}}\,{\text{weight}}\,{\text{loss}} = {\text{f}}_{\text{wl}} = \left( {\frac{{\Delta W_{t} }}{{W_{1} }}} \right) = \left[ {\left( {{W}_{1} {-}{W}_{2} } \right)/{W}_{1} } \right] = \left[ {\left( {1.82{-}1.15} \right)/1.82} \right] = 0.368 $$

Therefore,

$$ \begin{aligned} \alpha & = \left( {\frac{4}{7}} \right) \times \left[ {\frac{{f_{\text{wl}} - (f_{\text{coal}} \, \times \,f_{\text{vm}} )}}{{\left( {f_{\text{ore}} \, \times \,\rho_{\text{ore}} \, \times \,f_{\text{O}} } \right)}}} \right] \times 100 = \left( {\frac{4}{7}} \right) \times \left[ {\frac{0.368 - (0.25\, \times \,0.24)}{{\left( {0.7\, \times \,0.957\, \times \,0.3} \right)}}} \right] \times 100 \\ & = \left( {\frac{4}{7}} \right) \times \left[ {\frac{0.368 - (0.06)}{{\left( {0.7\, \times \,0.957\, \times \,0.3} \right)}}} \right] \times 100 = \left( {\frac{4}{7}} \right) \times \left[ {\frac{0.308}{{\left( {0.2} \right)}}} \right] \times 100 = {\mathbf{88}}.{\mathbf{0}}\%. \\ \end{aligned} $$

Problems

Problem 1.1

The initial weight of iron ore pellet is 1.52 g. The pellet contains 3% binder. The pellet is reduced by hydrogen at 700 °C for 30 min. After reduction, weight of pellet is decreased to 1.315 g. Find out the degree of reduction.

Given: Ore contains 94% Fe₂O₃.

[Ans: 49.3%]

Problem 1.2

Iron ore pellet contains 2% bentonite. The initial weight of iron ore pellet is 2.435 g. The pellet is reduced by hydrogen at 800 °C for 60 min. After reduction, weight of pellet is decreased to 1.84 g. Find out the degree of reduction.

Given: Iron ore (Fe₂O₃) contains 64% total Fe.

[Ans: 90.9%]

Problem 1.3

Iron ore–char composite pellet contains 15% char and 2% binder. Initial weight of pellet is 1.546 g. Composite pellet undergoes reduction in nitrogen atmosphere at 1000 °C for 1 h. After reduction weight comes down to 1.269 g. Reduced pellet is then treated again with hydrogen gas at 750 °C for 30 min, after hydrogen treatment weight of reduced pellet further comes down to 1.125 g.

Calculate degree of reduction for composite pellet in nitrogen atmosphere.

Given: Ore contains 90% Fe₂O₃, and char contains 80% carbon.

[Ans: 58.44%]

Problem 1.4

Iron ore–char composite pellet contains 20% coal and 5% binder. Initial weight of pellet is 1.75 g. Composite pellet undergoes reduction in nitrogen atmosphere at 1000 °C for 1 h. After reduction weight comes down to 1.42 g. During handling pellet is broken, weight of broken pellet is 1.42 g before hydrogen treatment at 750 °C for 30 min and after hydrogen treatment weight of reduced pellet further comes down to 0.965 g. Calculate degree of reduction for composite pellet in nitrogen atmosphere.

Given: Ore contains 66.5% total Fe, and coal contains 65% carbon.

[Ans: 38.94%]

Problem 1.5

Iron ore (blue dust)–coal composite pellet contains 9.8% coal and 8.2% binder. Initial weight of pellet is 1.8056 g. Composite pellet undergoes reduction in nitrogen atmosphere at 1000 °C for 1 h. After reduction weight comes down to 1.467 g.

Calculate degree of reduction for composite pellet in nitrogen atmosphere.

Given: Blue dust contains 95.7% Fe₂O₃, 1.6% SiO₂ and 1.8% Al₂O₃. Coal contains 49.7% fixed carbon, 32.5% VM and 11.6% ash.

[Ans: 37.86%]

Problem 1.6

Iron ore (blue dust)–coal composite pellet contains 9.6% coal and 10.4% binder. Initial weight of pellet is 2.048 g. Composite pellet undergoes reduction in nitrogen atmosphere at 1000 °C for 1 h. After reduction weight comes down to 1.687 g.

Calculate degree of reduction for composite pellet in nitrogen atmosphere.

Given: Blue dust contains 95.7% Fe₂O₃, 1.6% SiO₂ and 1.8% Al₂O₃. Coal contains 42% fixed carbon, 24% VM and 28% ash.

[Ans: 38.13%]