Charge calculations in pyrometallurgical processesmetalurji.mu.edu.tr/Icerik/metalurji.mu.edu.tr... · Complex charge calculation problems can be solved easily by simplifying assumptions

Charge calculations in pyrometallurgical processes

Charge calculations are carried out prior to operating a metallurgical process to determine the quantity of each type of raw material fed to the furnace in order to obtain the desired quantity of products

It is similar to stoichiometric problems but the engineer has to have a detailed knowledge on the internal working of the process in order to write the relevant reactions

Material balance by careful and detailed tracking of all elements in the input and output is the prerequisite of heat balance and complete definition of the system

Multiple reactions in metallurgical process makes it hard to keep track of all the chemical species in the reactants and products

Complex charge calculation problems can be solved easily by simplifying assumptions

e.g. It is safe to assume in iron blast furnace that all CaO, MgO and Al2O3 of the charge end up in the slagAlso molten pig iron can be considered to contain all Fe coming from the oreAll CO2 in the flue gases can be thought to originate from the reactions and air is simply O2

and N2

Charge calculation problems

Hints for effective material balance problem solving:1 – Read the question to understand the process, materials and unknowns2 – Draw a diagram3 – Define a base4 – Write down the independent equations and relations5 – Perform degree of freedom analysis6 – Do stoichiometric and materials balance calculations7 – Check your calculationsExample – Combustion of coal in furnace

Ultimate Analysis wt%

Material C H N S O

Coal 85 5 1 2 7

Base: 1000 kg coalC = 0.85*1000= 850 kgH = 0.05*1000= 5 kgN = 0.01*1000= 10 kgS = 0.02*1000= 20 kgO = 0.07*1000= 70 kg

Air = O2, N2

Flue gasesCO2

SO2

H2ON2

Reacting Gaseous Mixture

T, P

Chemical reactionsC + O2 = CO2

H2 + 1/2O2 = H2OS + O2 = SO2

Degree of freedom analysis

Material balance type – Initial input + generation = Final output + consumption

6 unknown labeled variables (VA,VG, XCO2, XH2O, XSO2, XN2)

- 4 independent atomic species balances that are involved in the reactions (C, H, S, O)

- 1 molecular balances on independent nonreactive species (N2)

- 1 other equation relating unknown variables (XCO2 + XH2O + XSO2 + XN2 =1)

= 0 degrees of freedom


VA Air 21% O2, 79% N2

VG Flue gasesXCO2 CO2

XSO2 SO2

XH2O H2OXN2 N2


T, P


H2 + 1/2O2 = H2OS + O2 = SO2

Calculate the volume of air necessary for complete combustion

Stoichiometry calculationnC = 850/12 = 70.83 kg-atom O2 consumption = 70.83 kg-molenH2 = 50/2 = 25 kg-mole O2 consumption = 12.5 kg-molenS = 20/32 = 0.625 kg-atom O2 consumption = 0.625 kg-molenO = 70/16 = 4.375 kg-atom nO2 = 2.1875 kg-mole O2 input from coal= 2.1875 kg-mole

O2 input from air= 81.7675 kg-moleO2 balance – Input + generation (0) = output (0) + consumption

Volume of air = 81.7675∗22.4

0.21= 8721 𝑚3 per 1000 kg coal


VA Air 21% O2, 79% N2

VG Flue gasesXCO2 CO2

XSO2 SO2

XH2O H2OXN2 N2


STP


H2 + 1/2O2 = H2OS + O2 = SO2

C + O2 = CO2

H2 + 1/2O2 = H2OS + O2 = SO2

Excess reactants

5000 m3 of regenerator gas of following composition is used to heat an open hearth furnace at 300 C per hour:

Air at 800 C is consumed 20% in excess of the theoretical requirement

Combustion reactionsCO + 1/2 O2 = CO2

H2 + ½ O2 = H2OCH4 + 2O2 = CO2 + 2H2O

Rational Analysis wt%

Material CO CO2 H2 CH4 H2O N2

Gaseous fuel 22 6 10 3 3 56

Gaseous fuel 5000 m3/hr56% N2, 22% CO, 10% H2, 6% CO2, 3% CH4, 3% H2O

V𝐴 Air 800 C79% N2, 21% O2

V𝐺 Flue gasesXCO2 CO2

XH2O H2OXN2 N2

XO2 O2

300 C, 1 atm

Degree of freedom analysis

Combustion reactions Material balance type – input + generation = output + consumptionCO + 1/2 O2 = CO2

H2 + ½ O2 = H2OCH4 + 2O2 = CO2 + 2H2O

6 unknown labeled variables ( V𝐴, V𝐺, XCO2, XH2O, XO2, XN2)

+ 3 independent chemical reactions

- 7 independent molecular species balances (CO, H2O, H2, CO2, CH4, O2, N2)

- 2 other equation relating unknown variables (XCO2 + XH2O + XO2 + XN2 =1, 1

6∗ 0.21 ∗ V𝐴 = V𝐺 ∗

𝑋𝑂2)



V𝐴 Air 800 C79% N2, 21% O2

V𝐺 Flue gasesXCO2 CO2

XH2O H2OXN2 N2

XO2 O2

300 C, 1 atm


Calculate the volume of air required to burn 1 m3 of regenerator gas per hour, then scale up

Basis: 1 m3/hr of regenerator gasGas composition @ 300 C Gas composition @ 0 C 0.22 m3 CO CO = 0.22*(273/573) = 0.105 m3

0.10 m3 H2 H2 = 0.10*(273/573) = 0.048 m3

0.03 m3 CH4 CH4 = 0.03*(273/573) = 0.014 m3

0.06 m3 CO2 CO2 = 0.06*(273/573) = 0.028 m3

0.03 m3 H2O H2O = 0.03*(273/573) = 0.014 m3

0.56 m3 N2 N2 = 0.56*(273/573) = 0.267 m3

O2 requirements from combustion reactionsCO + 1/2 O2 = CO2 O2 = 0.0525 m3

H2 + ½ O2 = H2O O2 = 0.024 m3

CH4 + 2O2 = CO2 + 2H2O O2 = 0.028 m3

Total O2 = 0.1045 m3


Air 800 C79% N2, 21% O2

Flue gasesCO2

H2ON2

O2

300 C, 1 atm

𝑃1𝑉1𝑇1

=𝑃2𝑉2𝑇2

𝑃1 = 𝑃2 = 1 𝑎𝑡𝑚

𝑉1 = 𝑉2𝑇1𝑇2

Volume of O2 at STP = 0.1045 m3

Volume of O2 at 800 C = 0.412 m3

Theoretical air volume = 0.421/0.21= 1.96 m3

Real air volume = 1.96*1.2 = 2.35 m3


Calculate the composition of flue gases

Basis: 1 m3 of regenerator gasGas composition @ STP Air composition @ STP

CO = 0.22*(273/573) = 0.105 m3 O2 = 0.1045*1.2 = 0.1255 m3

H2 = 0.10*(273/573) = 0.048 m3 N2 = 0.1255*(79/21) = 0.472 m3

CH4 = 0.03*(273/573) = 0.014 m3

CO2 = 0.06*(273/573) = 0.028 m3

H2O = 0.03*(273/573) = 0.014 m3

N2 = 0.56*(273/573) = 0.267 m3

Flue gas compositionCO2 = CO2(combustion1) + CO2(combustion3) + CO2(gas) = 0.105 + 0.014 + 0.028 = 0.147 m3 / 14.8%H2O = H2O(combustion2) + H2O(combustion3) + H2O(gas) = 0.048 + 0.028 + 0.014 = 0.090 m3 / 9.1%N2 = N2(gas) + N2(air) = 0.267 + 0.472 = 0.739 m3 / 74.0%O2 = O2(air) – O2(combustion1,2,3) = 0.1255 – 0.1045 = 0.021 m3 / 2.1%


11750 m3/hr Air 800 C79% N2, 21% O2

4985 m3/hrFlue gasesCO2 14.8%H2O 9.1%N2 74.0%O2 2.1%

300 C, 1 atm

Nitriding Gas Treatment

Iron is nitrided by passing a mixture of gaseous ammonia and hydrogen through a furnace

ReactionsNH3 = ½ N2 + 3/2 H2

½ N2 = N

Calculate the amount of nitrogen, in gram/hr, that the iron picks up from the gas flowing in at a rate of 50 ml/min at 500 C

DOF analysis Material balance type – Input + generation - output - consumption = accumulation

2 unknown labeled variables ( V𝐺, 𝑚𝑁2)


- 3 independent molecular species balances (NH3, H2, N2)

- 1 other equation relating unknown variables (PV=m/MW*RT)


Rational Gas Analysis wt%

Material NH3 H2

Incoming gas 10 90

Outlet gas 7 93

3000 ml/hrGas input at 500 C90% H2, 10% NH3

𝑉𝐺 Gas output93% H2, 7% NH3

500 C, 1 atm

Fe N

Nitriding Gas Treatment

Iron is nitrided by passing a mixture of gaseous ammonia and hydrogen through a furnace

ReactionsNH3 = ½ N2 + 3/2 H2

½ N2 = N

Calculate the amount of nitrogen, in gram/ hr, that the iron picks up from the gas flowing in at a rate of 50 ml/min at 500 C

Basis: 3000 ml/hours of gas inputInput OutputNH3 = 300 ml NH3 = (300-x) ml where x is volume of consumed NH3

H2 = 2700 ml H2 = (2700+3/2x) ml where 3/2x is the volume of generated H2

Total = 3000 ml Total = 3000+1/2x%NH3 = 7/100 = (300-x)/(3000+x/2) volume of consumed NH3, x = 86.96 ml/hr

NH3 = ½ N2 + 3/2 H2 N2 generated per hour = 1/2x = ½*86.96 = 43.48 ml½ N2 = N N consumed in steel = 14 g/11200 ml N2 = 0.5475 g N / 43.48 ml N2


Material NH3 H2

Incoming gas 10 90

Outlet gas 7 93

Gas input90% H2, 10% NH3

Gas output93% H2, 7% NH3

500 C, 1 atm

Fe N

CalcinationCalcination is a thermal treatment process applied to ores and other solid materials in order to induce removal of volatile components like CO2 and H2O by thermal decomposition

Inputs – Solid ore, fuel gas, airOutputs – Solid calcine, off-gas

Calcination temperature is below the melting point of the components of the raw materialSolid ores are treated in the solid state and the product is also solid except the volatile components

Components of fuel gas are typically CO, hydrogen, oxygen and hydrocarbons which are the combustible components and CO2, N2 which are the diluents that do not take part in the combustion

Calcination exampleLimestone is not the preferred flux in various steel making processes since its decomposition is associated with a large amount of absorption of energyCharging of lime after calcination of limestone is more energy efficient

Rotary kiln is very often used to produce lime by calcination of limestone Rotary kilns are very long kilns that rotate 2 to 3 degree from the horizontal axis The feed enters and from other side, the calcine material discharges and they are frequently heated by an externals source of energy

Other commercial uses of rotary kiln is cement and the removal of water from alumina

Calcination furnace analysisMagnesium carbonate is decomposed to make MgO and CO2 by heating in a rotary kiln, using as fuel a natural gasCO2 formed by decomposition of magnesium carbonate mixes with the products of combustion to form the flue gas productFuel consumption is 250 m3/ton MgO at STP

ReactionsMgCO3 = MgO + CO2

CH4 + 2O2 = 2H2O + CO2

C2H6 + 7/2O2 = 3H2O + 2CO2

C3H8 + 5O2 = 4H2O + 3CO2


Material CH4 C2H6 C3H8 CO2 N2 O2 H2O

Natural gas 80 15 5

Flue gas 19.27 64.33 3.98 12.42

Ore MgCO3

AirFlue gases64.33 % N2

3.98 % O2

19.27 % CO2

12.42% O2

1000 kg/hr CalcineMgO

Fuel 250 m3/ton MgO80% CH4

15% C2H6

5% C3H8

Rotary Kiln

DOF analysis

ReactionsMgCO3 = MgO + CO2

CH4 + 2O2 = 2H2O + CO2 Material balance type – Input + generation = Output + consumptionC2H6 + 7/2O2 = 3H2O + 2CO2

C3H8 + 5O2 = 4H2O + 3CO2

3 unknown labeled variables ( V𝐴, V𝐺, 𝑚𝑂)


- 9 independent molecular species balances (MgCO3, MgO, H2O, C3H8, C2H6, CO2, CH4, O2, N2)

- 1 other equation relating unknown variables (E ∗ 0.21 ∗ V𝐴 = V𝐺 ∗ 0.0398)

= -3 degrees of freedom!

𝑚𝑂 OreMgCO3

𝑉𝐴 Air at STP 𝑉𝐺 Flue gases

64.33 % N2

3.98 % O2

19.27 % CO2

12.42% H2O

1000 kg/hrCalcineMgO


15% C2H6

5% C3H8

Rotary Kiln

Calculate the air consumption in m3 per ton of MgO produced per hour

Reactions CO2 balanceMgCO3 = MgO + CO2 kg-mole MgO = 1000/MWMgO = 25 kg-mole = CO2 from MgCO3

CH4 + 2O2 = 2H2O + CO2 kg-mole CH4 = 250*(80/100) = 200 m3/ton MgO = 8.93 kg-mole CH4

= 8.93 kg-mole CO2

C2H6 + 7/2O2 = 3H2O + 2CO2 kg-mole C2H6 = 250*(15/100) = 37.5 m3/ton MgO = 1.67 kg-mole C2H6

= 2*1.67 kg-mole C2H6 = 3.34 kg-mole CO2

C3H8 + 5O2 = 4H2O + 3CO2 kg-mole C3H8 = 250*(5/100) = 12.5 m3/ton MgO = 0.56 kg-mole C3H8

= 3*0.56 kg-mole C3H8 = 1.68 kg-mole CO2

Total kg-mole CO2 = 25 + 8.93 + 3.34 + 1.68 = 38.95 kg-moleTotal flue gas = 38.95*(100/19.27) = 202.155 kg-mole Total N2 = 202.155*(64.33/100) = 130.04 kg-mole N2

Since N2 in air = N2 in flue gas, Air consumption = 130.04*(100/79) = 164.6 kg-mole air/ ton MgO= 164.6*22.4 = 3687 m3 (STP) / ton MgO

Basis 1000 kg/hr of MgOOreMgCO3

Air 3687 m3 𝑉𝐺 Flue gases

64.33 % N2

3.98 % O2

19.27 % CO2

12.42% H2O Calcine1000 kg MgO


15% C2H6

5% C3H8

Rotary Kiln

Calculate the percent excess air

Excess O2 = 202.155*(3.98/100) = 8.055 kg-moleExcess air = 8.055*(100/21) = 38.36 kg-mole % Excess air = (38.36/(164.6-38.36))*100 = 30.38

Theoretical air = 164.6-38.36 = 126.24 kg-mole

Basis 1000 kg of MgOOreMgCO3

Air 3687 m3Flue gases 202.155 kg-mole64.33 % N2

3.98 % O2

19.27 % CO2

12.42% H2OCalcine1000 kg MgO


15% C2H6

5% C3H8

Rotary Kiln

Roasting

Roasting is a preliminary step of metal extraction from sulphide oresThe process is partial or complete conversion of metal sulphide to oxide, sulphate or chlorides Oxide can be easily reduced; sulphate and chloride can be easily dissolved

Sulphide ores cannot be used to produce metal by pyrometallurgy

It is very difficult to reduce sulphide directly into the metalCarbon and hydrogen are not effective reducing agent to produce metal from sulphide as seen in the Ellingham

Another issue with direct reduction of metal sulphides is that there exist a mutual solubility between metal and sulphides which makes it difficult to extract the metal by pyrometallurgy

So the only route is to convert sulphide to oxide

Inputs – Sulphide ore, air, fuel if necessaryOutputs – Calcine, off-gas

Roasting is carried out below the melting point of the components of the ore By virtue of this, the roast product is in solid state in addition to the solid ore concentrate

Temperatures involved during roasting is of the order of 900 to 1100 degrees Celsius

Byproducts of roasting are rich in SO2 because sulphide ore has 20-30 % sulphur depending on the depositSo a large amount of a SO2, SO3 and nitrogen will be produced as the off-gasThese sulphurious gases are used to produce H2SO4

Oxidation of sulphides is exothermic and can supply all the energy needed for roasting to be self-sustaining

Heats of formation of some sulphides:Cu2S = -18950 kilocalories per kg moleZnS = -44000 kilocalories per kg moleFeS2 = -35500 kilocalories per kg mole Heat generated by oxidation reactionCuO = -37100 kilocalories per kg mole -136900 kilocalories per kg moleSO2 = -70940 kilocalories per kg mole SO3 = -93900 kilocalories per kg mole CO2 = -94450 kilocalories per kg mole CO = -26840 kilocalories per kg mole

If fuel is also used, there is also carbon dioxide and carbon monoxide in the off-gas

Cu2S + O2 = 2CuO + SO2

Types of roasting

Oxidizing roasting Sulphide ore is oxidized by passing air and providing an oxidizing atmosphereThe amount of oxidation must be controlled so that the formation of metal sulphate is avoided if it is not desirede.g. PbS + O2 = PbSO4 and PbOHigh temperature is required to break up the metal sulphate

In dead roasting all sulphur is eliminated However, if the extraction of metal is to be done through hydrometallurgical means, sulphateformation is preferred because sulphates dissolve easily in the solvent

Sulphatising roastingAs the name suggests the objective is to convert all sulphide into sulphate in an oxidizing atmosphere

Chloridizing roastingThe objective of chloridizing roasting is to convert a metal sulphide or oxide into chloridese.g. 2NaCl + MS + 2O2 = Na2SO4 +MCl2 direct chlorination4NaCl + 2MO + S2 + 3O2 = 2Na2SO4 + 2MCl2 indirect chlorination

Roasting furnace analysisPyrometallurgical extraction of ores rich in CuS, FeS2, ZnS is uneconomical due the difficulties involved in concentrating the oreRoasting is needed to remove all of the sulfur and subsequently to leach the ore in dilute sulfuric acid for the recovery of copper and zinc by hydrometallurgical methods

Copper, iron and zinc of the ore oxidize to CuO, Fe2O3 and ZnO

ReactionsCuS + 3/2O2 = CuO + SO2

2FeS2 + 11/2O2 = Fe2O3 + 4SO2

ZnS + 3/2O2 = ZnO + SO2

SO2 + 1/2O2 = SO3


Material Cu Fe Zn SiO2 S CaO,Al2O3, etc

SO2 SO3 O2, N2

Ore 6 25 4 20 33.6 11.4

Roast gases 2.5 0.4 97.1

Basis 1000 kg of copper oreOre4% ZnS6% CuS25% FeS2

20% SiO2

11.4% CaO, Al2O3, etc33.6% S

Air

Flue gases2.5% SO2

0.4% SO3

O2, N2

CalcineZnOCuOFe2O3

SiO2

CaO, Al2O3

Leaching

ReactionsCuS + 3/2O2 = CuO + SO2

2FeS2 + 11/2O2 = Fe2O3 + 4SO2

ZnS + 3/2O2 = ZnO + SO2

SO2 + 1/2O2 = SO3 Material balance type – Input + generation = output + consumption

DOF analysis10 unknown labeled variables (VA,VG, mc, XZnO, XCuO, XFe2O3, XSiO2, XCaO, Al2O3,etc, XO2, XN2)

- 5 independent atomic species balances that are involved in the reactions (Zn, Cu, Fe, S, O)

- 3 molecular balances on independent nonreactive species (N2, CaO, SiO2)

- 2 other equations relating unknown variables (XO2+ XN2=0.971, XZnO + XCuO + XFe2O3 + XSiO2 + XCaO

= 1)


1000 kg Ore4% Zn6% Cu25% Fe20% SiO2

11.4% CaO, Al2O3, etc33.6% S

VA Air at STP

VG Flue gases2.5% SO2

0.4% SO3

XO2 O2

XN2 N2

mc CalcineXZnO ZnOXCuO CuOXFe2O3 Fe2O3

XSiO2 SiO2

XCaO, Al2O3,etc CaO, Al2O3, etc

Leaching

Calculate the weight and approximate analysis of the calcine resulting from roasting 1 ton oreBase: 1000 kg oreInput 1000 kg OutputCu 60 kg CuO 60*(80/64) = 75 kgFe 250 kg Fe2O3 250*(160/112) = 357 kgZn 40 kg ZnO 40*(81/65) = 50 kgSiO2 200 kg SiO2 200 kgCaO, Al2O3, etc 114 kg CaO, Al2O3, etc 114 kgS 336 kgTotal 1000kg Total solids 796 kg

Analysis of calcineCuO = (75/796)*100 = 9.4% Fe2O3 = (357/796)*100 = 44.9%ZnO = (50/796)*100 = 6.3% SiO2 = (200/796)*100 = 25.1%Others (CaO, Al2O3, etc) = (114/796)*100 = 14.3%

Basis 1000 kg of copper oreOre4% Zn as ZnS6% Cu as CuS25% Fe as FeS2

20% SiO2

11.4% CaO, Al2O3, etc33.6% S

Air

Flue gases2.5% SO2

0.4% SO3

O2, N2

CalcineZnO SiO2

CuO CaO, Al2O3

Fe2O3

Leaching

Calculate the volume of flue gases per ton of oreReactions

CuS + 3/2O2 = CuO + SO2 VSO2 = 22.4 ∗ 1 ∗60∗

96

64

96= 21 𝑚3

2FeS2 + 11/2O2 = Fe2O3 + 4SO2 VSO2 = 22.4 ∗ 4 ∗250∗

120

112

120= 200 𝑚3 235 𝑚3

ZnS + 3/2O2 = ZnO + SO2 VSO2 = 22.4 ∗ 1 ∗40∗

97

65

97= 14 𝑚3

SO2 + 1/2O2 = SO3 VSO2 = 235 − x 𝑚3, VSO3 = x 𝑚3

%SO2 = 2.5/100 = (235-x)/Vfluegas

%SO3 = 0.4/100 = (x/Vfluegas)

Vfluegas = 8103.5 𝑚3, VSO3 = x = 32.4 𝑚3, VSO2 = 202.6 𝑚3

Basis 1000 kg of copper oreOre40 kg Zn 60 kg Cu 250 kg Fe200 kg SiO2

114 kg CaO, Al2O3, etc336 kg S

Air

Flue gases2.5% SO2

0.4% SO3

O2, N2

CalcineZnO SiO2

CuO CaO, Al2O3

Fe2O3

Leaching

Calculate the volume roasting air, percent excess and the composition of the flue gasesRoasting reactions Volume of O2 consumed

CuS + 3/2O2 = CuO + SO2 VO2 = 22.4 ∗3

2∗

60∗96

64

96= 31.5 𝑚3

2FeS2 + 11/2O2 = Fe2O3 + 4SO2 VO2 = 22.4 ∗11

2∗

250∗120

112

120= 275 𝑚3

ZnS + 3/2O2 = ZnO + SO2 VO2 = 22.4 ∗3

2∗

40∗97

65

97= 21 𝑚3

SO2 + 1/2O2 = SO3 VO2 = 32.4 ∗1

2= 16.2 𝑚3

Theoretical air input Vair-th = (343.7/0.21) = 1636.67 𝑚3

Percentage of excess air has to be calculated from O2+N2 balance in order to obtain actual air V𝑉𝑂2+𝑁2 = 𝑉𝑎𝑖𝑟−𝑡ℎ + 𝑉𝑒𝑥𝑐𝑒𝑠𝑠 − 𝑉𝑂2𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑

7868.5 = 1636.67 + 1636.67 ∗ 𝑦 − 343.7𝑦 = 4.02, 𝑉 = 1636.67 + 1636.7 ∗ 4.02 = 8216.07 𝑚3

𝑉𝑂2𝑒𝑥𝑐𝑒𝑠𝑠 = 8216.07 ∗ 0.21 − 343.7 = 1381.67 𝑚3, 𝑉𝑁2 = 8216.07 ∗ 0.79 = 6490.69 𝑚3

Basis 1000 kg of copper oreOre40 kg Zn 60 kg Cu 250 kg Fe200 kg SiO2

114 kg CaO, Al2O3, etc336 kg S

Air

Flue gases 8103.5 𝑚3

SO2 202.6 𝑚3

SO3 32.4 𝑚3

O2, N2 7868.5 𝑚3

CalcineZnO SiO2

CuO CaO, Al2O3

Fe2O3

Leaching

343.7 𝑚3

SmeltingIt is a unit process similar to roasting, to heat a mixture of ore concentrate above the melting pointThe objective is to separate the gangue mineral from liquid metal or matte The state of the gangue mineral in case of smelting is liquid which is the main difference between roasting and smelting

Inputs – Ore, flux, fuel, airOutput – Metal or Matte, slag

When metal is separated as sulphide from smelting of ore, it is called Matte smeltinge.g. Cu2S and FeSWhen metal is separated as liquid, it is called reduction smelting e.g. Ironmaking

Density of liquid metal or matte is around 5-5.5 g/cm3

Density of slag is around 2.8-3 g/cm3

The additives and fluxes serve to convert the waste or gangue materials in the charge into a low melting point slag which also dissolves the coke ash and removes sulphur

IronmakingAbout 1 billion tonnes of iron is produced in the world annually by blast furnacesBlast furnace economics are such that larger units have lower unit production costs, hence modern blast furnaces are bigger and produce more than 10000 tonnes per day

The blast furnace is a counter-current reactor in which the descending column of burden materials reacts with the ascending hot gasesThe process is continuous with raw materials being regularly charged to the top of the furnace

and molten iron and slag being tapped from the bottom of the furnace at regular intervals

free moisture is driven off from the burden materials and hydrates and carbonates are disassociated indirect reduction of the iron oxides by

carbon monoxide and hydrogen occursthe burden starts to soften and melt, at 700-1,000 Cdirect reduction of the iron and other oxidesand carbonization by the coke occurs at 1,000-1,600 CMolten iron and slag start to drip combustion air that is preheated through to the bottom of the furnace to 900-1,300 C and often enriched with

oxygen is blown into the furnace in the combustion zone at 1,850-2,200 C, coke reacts with the oxygen and steam in the blast to form carbon monoxide and hydrogen as well as heat iron and slag melt completely

The blast furnace itself is a steel shaft lined with fire resistant, refractory materialsThe hottest part of furnace - where the walls reach a temperature >300 °C - are water cooled

Coke is a principle source of thermal energy and as well as chemical energy in ironmakingCarbon of the coke reduces iron oxide to ironThe combustion of carbon of coke also provides a thermal energy

Hot blast air is introduced through the tuyere so a counter current against the descending burden is created by the gases travelling upward

In any counter current heat and mass exchange reactor, which consists of gas and solid, the permeability of the bed and the distribution of the burden are very important issues

For the smooth operation of the blast furnace, the upward rising gases should travel unhindered They should also transfer their heat and mass to the descending burdenThe burden distribution should be homogeneous so that it constitutes a uniform distribution of iron and facilitate smooth movement of burden gases

Carbon of coke reacts with O2 at the tuyere level because of availability of oxygenC + O2 + 3.76N2 = CO2 + 3.76N2CO2 + 3.76N2 + C = 2CO + 3.76N2

Upward gas rising consists of CO, CO2 and nitrogen

A temperature approximately around 1900-2100 C is created as a result of reaction of carbon of coke with oxygen at the tuyere level The exit temperature of the gas is approximately somewhere between 200 to 250 C during the discharge from the top of the furnace

The following reactions do not require very high percentage of carbon monoxideSo they can occur towards the upper region

3Fe2O3 + CO = 2Fe3O4 + CO2Fe3O4 + CO = 3FeO + CO2

whereas, the reaction FeO + CO = Fe + CO2 requires high concentration of COSo it occurs near the middle of the furnace where the concentration of CO is highAt around 900 C, the equilibrium concentration of CO in the CO-CO2 mixture is around 65 to 70 percent for FeO to be able to reduce to ironSome iron oxide is also reduced directly by carbon This reduction is endothermic in nature, whereas all other reactions are exothermic reduction.

Iron blast furnace analysis

Consider an iron blast furnace charged with iron ore, limestone and coke of following analyses:

The ultimate analysis of the pig iron gives 93.8% Fe, 4% C, 1.2% Si, 1% MnFor every ton of pig iron produced, 1750 kg of iron ore and 500 kg limestone are used and 4200 m3 of flue gas is producedThe rational analysis of flue gases gives 58% N2, 26% CO, 12% CO2, 4% H2O

ReactionsFe2O3 + 3CO = 2Fe + 3CO2 SiO2 + 2C = Si + 2COCaCO3 = CaO + CO2 MnO + C = Mn + COC + 1/2O2 = CO CO2 + C = 2CO


Material Fe2O3 SiO2 MnO Al2O3 H2O C CaCO3

Ore 80 12 1 3 4

Limestone 4 1 95

Coke 10 90

Basis 1000 kg of pig iron

Ore 1750 kgLimestone 500 kgCoke

Air

Blast furnace gas 4200m3

Slag

Pig iron 1000 kg

ReactionsFe2O3 + 3CO = 2Fe + 3CO2 SiO2 + 2C = Si + 2COCaCO3 = CaO + CO2 MnO + C = Mn + COC + 1/2O2 = CO CO2 + C = 2COC + O2 = CO2

DOF analysis

8 unknown labeled variables ( V𝐴, 𝑚𝐶 , 𝑚𝑆, XFe2O3, XSiO2, XMnO, XAl2O3, XCaO)


- 14 independent molecular species balances (Fe2O3, SiO2, MnO, Al2O3, H2O, CaCO3, C, CO, CO2,

O2, N2, Mn, Si, Fe)

- 1 other equation relating unknown variables (XFe2O3+ XSiO2+ XMnO+ XAl2O3+ XCaO =1)


Basis 1000 kg/hr of pig ironOre 1750 kg/hr80% Fe2O3, 12% SiO2, 1% MnO, 3% Al2O3, 4% H2O

Limestone 500 kg/hr95% CaCO3, 1% H2O, 4% SiO2

Coke 𝑚𝐶10% SiO2, 90% C

Air


58% N2, 26% CO, 12% CO2, 4% H2O

Slag 𝑚𝑆Fe2O3, SiO2, MnO, Al2O3, CaO

Pig iron 1000 kg93.8% Fe,4% C, 1.2% Si, 1% Mn

ReactionsFe2O3 + 3CO = 2Fe + 3CO2 SiO2 + 2C = Si + 2COCaCO3 = CaO + CO2 MnO + C = Mn + COC + 1/2O2 = CO CO2 + C = 2CO

Calculate the quantity of coke used per ton of pig ironCarbon balance:Ccoke + Climestone = C pig iron + Cflue gas let x be the weight of coke

0.9𝑥

12+0.95 ∗ 500 ∗ 12/100

12=0.04 ∗ 1000

12+4200 ∗ 0.26

22.4+4200 ∗ 0.12

22.4

0.075x + 4.75 = 3.333 + 71.25, x= 69.833/0.075=931 kg coke per ton of pig iron

Calculate the air consumption per ton of pig ironN2 balance:N2(air) = N2(flue gas) = 4200*0.58 = 2436 m3, Air consumption = 2436*(100/79) = 3083.5m3

Basis 1000 kg of pig ironOre 1750 kg80% Fe2O3, 12% SiO2, 1% MnO, 3% Al2O3, 4% H2O

Limestone 500 kg95% CaCO3, 1% H2O, 4% SiO2

Coke10% SiO2, 90% C

Air


58% N2, 26% CO, 12% CO2, 4% H2O

Slag


Calculate the composition of the slagFe2O3 balance:Fe2O3(ore) = Fe2O3(pig iron) + Fe2O3(slag)

1750*0.8 = 0.938*1000*(160/112) + Fe2O3(slag), Fe2O3(slag)=1400-1340= 60 kgSiO2 balance:SiO2(ore) + SiO2(limestone) + SiO2(coke) = SiO2(pig iron) + SiO2(slag)

1750*0.12 + 500*0.04 + 931*0.1 = 0.012*1000*(60/28) + SiO2(slag), SiO2(slag)=210+20+93.1-25.7MnO balance: =297.4 kgMnO(ore) = MnO(pig iron) + MnO(slag)

1750*0.01 + 0.01*1000*(71/55) + MnO(slag)

MnO(slag) = 17.5 – 12.9 = 4.6 kgAl2O3 balance:Al2O3(ore) = Al2O3(slag) = 1750*0.03 = 52.5 kgCaO balance:CaO(limestone) = CaO(slag) = 500*(56/100)*0.95 = 266 kg

Basis 1000 kg of pig ironOre 1750 kg80% Fe2O3, 12% SiO2, 1% MnO, 3% Al2O3, 4% H2O

Limestone 500 kg95% CaCO3, 1% H2O, 4% SiO2

Coke10% SiO2, 90% C

Air


58% N2, 26% CO, 12% CO2, 4% H2O

SlagFe2O3, SiO2, MnO, Al2O3, CaO


Total slag composition:Fe2O3 60 kg 8.82%SiO2 297.4 kg 43.70%MnO 4.6 kg 0.67%Al2O3 52.5 kg 7.71%CaO 266 kg 39.10%Total 680.5 kg

SmeltingIt is a unit process similar to roasting, to heat a mixture of ore concentrate above the melting pointThe objective is to separate the gangue mineral from liquid metal or matte The state of the gangue mineral in case of smelting is liquid which is the main difference between roasting and smelting

Inputs – Ore, flux, fuel, airOutput – Metal or Matte, slag, off-gas

When metal is separated as sulphide from smelting of ore, it is called Matte smeltinge.g. Cu2S and FeSWhen metal is separated as liquid, it is called reduction smelting e.g. Ironmaking

Density of liquid metal or matte is around 5-5.5 g/cm3

Density of slag is around 2.8-3 g/cm3

The additives and fluxes serve to convert the waste or gangue materials in the charge into a low melting point slag which also dissolves the coke ash and removes sulphur

Matte SmeltingAdvantages of matte smelting • Low melting point of matte so that less amount of thermal energy is required by converting

the metal of the ore in the form of sulphide and then extracting the metal e.g. melting point of Cu2S and FeS is around 1000 degrees Celsius

• Cu2S which is contained in the matte, does not require any reducing agentIt is converted to oxide by blowing oxygen

• Matte smelting is beneficial for extraction of metal from sulphide ore, particularly when sulphide ore is associated with iron sulphide which forms eutectic point with Cu and Ni

The grade of the matte is defined as the copper grade of matteA matte of 40 percent means, it has 40% copper, so matte is always given in terms of copper, because it is used to produce copper not iron

Slag in matte smelting is mixture of oxidese.g. in smelting of copper ore concentrate the slag may contain SiO2, Al2O3, calcium oxide, FeO, Fe2O3, Fe3O4

The desirable properties of slag are low viscosity, solubility, low melting point

Typical reactions in Cu matte smelting:6CuO + 4FeS = 3 Cu2S + 4FeO + SO2

or if the O2 pressure is high 6CuO + 4FeS = 2CuSO4 + 2FeS2CuSO4 + 2FeS = Cu2S + 2FeO + 3SO2

Cu2O + FeS = Cu2S + FeO

Oxygen has greater affinity for iron than copper:10Fe2O3 + FeS = 7Fe3O4 + SO23Fe3O4 + FeS = 10FeO + SO2

In the ideal condition matte contains only Cu2S and FeS, plus little amount of Fe3O4 if oxygen is dissolved

Higher oxides of iron are difficult to remove by the slagRoasting has to be controlled in order to minimize the formation of Fe2O3 or Fe3O4, which may enter the matte during smelting

Off-gas consists of SO2, nitrogen, oxygen if excess amount of air is used and sometimes SO3

depending on the reactionIf fuel is used, CO and CO2 may also be present depending upon the state of combustion

Flash Smelting

Conventionally smelting is carried out in reverberatory furnaces, fired with coal or oilNowadays reverberatory furnaces are being replaced by flash smelting furnaces that have been developed in recent years

The advantages of flash smelting is that it combines both converting and smelting, whereas in the reverberatory furnace the ore has to be smelted first and then it is transferred to reverberatory furnace for converting purposes The reason for this combination is the economical processing of large amount of sulphurdioxide that is created especially in roastingCollecting the concentrated off-gas from flash smelting and converting to H2SO4 is much more feasible

Other advantages of flash smelting:• Very fine particles of ore concentrates are injected, so the reaction is extremely rapid and

very high temperatures are created • Heat generated is sufficient to carry out the smelting

Examples – In a copper ore, chalcopyrite (CuFeS2) is 34%, pyrite (FeS2) is 30% and SiO2 is 36%a) Determine the % Cu and % gangue in the oreb) What % Fe in the ore concentrate is to be removed to make 40% matte? Consider Cu2Sc) If only excess S is eliminated in the ore concentrate, what is the composition of the

resulting matte?AWCu= 64, AWFe= 56, AWS= 32

a- ore = Cu2S + gangue% Cu = 34 * (64/184) = 11.83 %Gangue = 100 – 11.83*(160/128) = 85.21 %, % Cu2S = 14.78 %

b –40

100=

11.83%

14.78%+%𝐹𝑒𝑆

0.4(14.78+%FeS) = 11.83, % FeS = 14.795 after removal of FeO, % Fe = 14.795*(56/88) = 9.415Initial % Fe = 34*(56/184)+30*(56/120) = 24.35 %% Fe to be removed = 24.35 – 9.415 = 14.935 %

c – CuFeS2 decomposes according to the reaction 2CuFeS2 = Cu2S + 2FeS + SFeS2 decomposes according to the reaction FeS2 = FeS + S% FeS = 24.35*(88/56) = 38.26 %

Matte grade = %𝐶𝑢

% 𝐶𝑢2𝑆+% 𝐹𝑒𝑆∗ 100 = 22.2 %

Examples – A copper matte may be represented as mCu2S.nFeS with no fixed values of m and nCalculate m and n for a matte grade of 38 %

0.38 =𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐶𝑢

𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐶𝑢2𝑆 + 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝐹𝑒𝑆

=160𝑚 ∗ 128/160

160𝑚 + 88𝑛

60.8𝑚 + 33.44𝑛 = 128𝑚

𝑚

𝑛=33.44

67.2≈ 0.5

Matte may be represented as Cu2S.2FeS or 2Cu2S.4FeS or 3Cu2S.6FeS

Example – Copper ore is smelted in a reverberatory furnace together with a copper concentrate. The fluxes are pure CaCO3 and iron ore. (Neglect off-gases for simplicity)

Reactions: CaCO3 = CaO + CO2

FeS2 + O2 = FeS + SO2 3Fe3O4 + FeS = 10FeO + SO2


Material Cu2S FeS2 SiO2 Fe2O3

Copper ore 17.5 67.5 15

Copper concentrate 35 25 40

Iron ore 20 80

Slag

Matte

Reverberatory furnace

Flux: CaCO3

Iron ore

Copper ore

Copper concentrate

Reaction: CaCO3 = CaO + CO2

DOF analysis5 unknown labeled variables (mC, mo, mF, mS mS,)


- 8 independent molecular species balances (Cu2S, FeS2, SiO2, Fe2O3, CaCO3, CaO, FeO, FeS)

- 0 other equation relating unknown variables


mS Slag35% SiO2

20% CaO45% FeO

mM Matte37.5% Cu2S62.5% FeS


mF Flux: CaCO3

mo Iron ore (80% Fe2O3, 20% SiO2)

1000 kg Copper ore17.5% Cu2S15% SiO2

67.5% FeS2

mC Copper concentrate35% Cu2S40% SiO2

25% FeS2

Reaction: CaCO3 = CaO + CO2

Calculate the quantities of concentrate, iron ore and flux in order to smelt 1000 kg of copper ore and obtain a matte grade of 30% Cu and a slag with the composition 35% SiO2 , 20% CaO, 45% FeO

Let X be the quantity of Cu concentrateLet Y be the quantity of matteLet Z by the quantity of iron oreLet U be the quantity of slag

Four equations are needed to solve for the four variables X, Y, Z, U

Slag35% SiO2

20% CaO45% FeO

Matte37.5% Cu2S62.5% FeS


Flux: CaCO3

Iron ore (80% Fe2O3, 20% SiO2)


67.5% FeS2

Copper concentrate35% Cu2S40% SiO2

25% FeS2

Cu2S balance: SiO2 balance:Cu2S(in ore) + Cu2S(in Cu conc.) = Cu2S(in matte) SiO2(Cu ore) + SiO2(Cu conc.) + SiO2(iron ore)=Cu2S(in ore) = 17.5% * 1000 = 175 kg SiO2(slag)Cu2S(in Cu-conc.) = 0.35 * X 0.15% * 1000 + 0.40X + 0.20Z = 0.35UCu2S(in matte) = 37.5% * Y = 0.375 Y Equation 2: 0.40X + 0.20Z – 0.35U = -150Equation 1: 175+ 0.35X = 0.375Y

Fe Balance:Fe(Cu ore) + Fe(Cu conc.) + Fe(iron ore) = Fe(matte) + Fe(slag)Fe(Cu ore) = 67.5% * 1000 * (56/120) = 315 kg Fe(Cu conc.) = 25% * X * (56/120) = 0.117X kgFe(iron ore) = 80% * Z * (112/160) = 0.56Z kg Fe(matte) = 62.5% * Y * (65/88) = 0.398Y kgFe(slag) = 45% * U * (56/72) = 0.35U kg Equation 3: 315+0.117X+0.56Z=0.398Y+0.35U

U kg Slag35% SiO2

20% CaO45% FeO

Y kg Matte37.5% Cu2S62.5% FeS


Flux: CaCO3

Z kg Iron ore (80% Fe2O3, 20% SiO2)


67.5% FeS2

X kg Copper concentrate35% Cu2S40% SiO2

25% FeS2

Sulphur balance:S(Cu ore) + S(Cu conc.) = S(matte), S(flue gas)≈0S(Cu ore) = (0.175 * (32/160) * 1000) + (0.675 * (64/120) * 1000) = 395 kgS(Cu conc.) = (0.35 * X * (32/160)) + (0.25 * X * (64/120)) = 0.203X kgS(matte) = (0.375 * (32/160) * Y) + (0.625 * (32/88) * Y) = 0.3023Y kgEquation 4: 395+0.203X = 0.3023YEquation 1: 175+ 0.35X = 0.375YEquation 2: 0.40X + 0.20Z – 0.35U = -150Equation 3: 315+0.117X+0.56Z=0.398Y+0.35U

CaO in slag: 20% * 7463.4 = 1492.7 kg CaO, 1492.7 * (100/56) = 2665.5 kg CaCO3

U kg Slag35% SiO2

20% CaO45% FeO

Y kg Matte37.5% Cu2S62.5% FeS


Flux: CaCO3

Z kg Iron ore (80% Fe2O3, 20% SiO2)


67.5% FeS2

X kg Copper concentrate35% Cu2S40% SiO2

25% FeS2

X=3210.2 kg Cu conc.Y=3462.7 kg Matte

Z=589.6 kg Iron oreU=7463.4 kg Slag

ConvertingLiquid metal or matte coming from the smelting furnace with impurities is converted to high purity metal in oxidizing environmentsEither steady air, blown air or blown oxygen are utilized to oxidize the gangue speciesGangue oxide minerals are removed with the initially forming slag

Inputs – Pig iron, cast iron for steel converting, Cu-Fe matte for copper converting, flux, airOutputs – Slag, steel or blister copper, off-gas

Furnaces usedHearthsPuddling furnacesCementation furnacesBessemer furnacesOpen Hearth furnacesBasic oxygen furnacesElectric arc furnaces

Converting Pig IronWrought or worked iron was the main malleable iron used in rails and structures until large scale, commercial production of steelIt contained low amount of carbon (0.04 to 0.08%) and was worked by hand into bars and various shapes due to its malleabilitySlag up to 2% is mixed in its microstructure in the form of fibrous inclusions like woodPig iron and cast iron were initially converted to wrought iron in hearths in ancient times then in puddling furnaces during 18th centuryIn these processes the charge was heated to melting temperature by burning charcoal and oxidized by airPuddling process involves manually stirring the molten pig iron, which decarburizes the iron. As the iron is stirred, globs of wrought iron are collected into balls by the stirring rod and those are periodically removed by the puddler

Horizontal (lower) and vertical (upper) cross-sections of a single puddling furnace. A. Fireplace grate; B. Firebricks; C. Cross binders; D. Fireplace; E. Work door;

F. Hearth; G. Cast iron retaining plates; H. Bridge wall

Commercial production of low carbon, low impurity steel was limited to inefficient and expensive process of adding carbon to carbon-free wrought iron between 17th and 19th

centuries

The manufacturing process, called cementation process, consisted of heating bars of wrought iron in a furnace in between powdered charcoal layers at about 7000 C for about a weekCarbon slowly diffuses into iron and dissolves in the iron, raising the carbon percentageSteel obtained from this process is called “blister steel” due to the blister-like marks formed on the surface due to the evolved gases during the manufacturing process

Up to 3 tons of coke was burnt for each ton of steel producedThe fuel and labor costs resulted in a small scaleproduction of steel that was about 8 times more expensive

The Bessemer process reduced the time needed to make steel of this quality to about half an hour while onlyrequiring coke to melt the pig iron initially

The Bessemer process - Henry Bessemer patented the process in 1855 The process is carried out in a large ovoid steel container lined with clay or dolomiteThe capacity of a converter is from 8 to 30 tons of molten iron

The key principle is removal of impurities from the iron by oxidation with air being blown through the molten iron The oxidation process removes impurities such as silicon, manganese, and carbon as oxidesThese oxides either escape as gas or form a solid slagThe oxidation also raises the temperature of the iron mass and keeps it moltenThe refractory lining of the converter also plays a role in the conversion—the clay lining is used in the acid Bessemer, in which there is low phosphorus in the raw materialDolomite, limestone or magnesite are used when the phosphorus content is high in the basic Bessemer

Once the converter is charged with molten pig iron, a strong thrust of air is blasted across the molten mass through tuyeres provided at the bottom of the vessel

The conversion process called the "blow" is typically completed in around twenty minutes

During this period the progress of the oxidation of the impurities is judged by the appearance of the flame issuing from the mouth of the converter since there is not enough time to make material analyses

The blow may be interrupted at certain periods to avoid the oxidation of certain impurities

Required amount of flux is added at the beginning of each period to produce the slag of desired composition and amount

At the end of the process all traces of the silicon, manganese, carbon, phosphorus and sulphur are oxidized, leaving the converter with pure iron

In order to give the steel the desired properties, other impurities can be added to the molten steel when conversion is complete

Steel converter analysis

A basic pneumatic steel converter is charged with 25 tons of pig iron containing various impuritiesIn addition to the removal of all of the C, Si, Mn and P, iron equivalent to 5% of the weight of charged iron oxidizes at a constant rate throughout the bessemerizing operationEnough lime is added to obtain a slag containing 35% CaO2/3 of the carbon in steel oxidizes to CO and 1/3 goes to CO2

Air compressor delivers air at a rate of 500 m3/min for specific periods of time

Bessemer converter

FluxCaO

Pig iron

Air

Flue gasCO, CO2, N2

SlagFe2O3, P2O5, SiO2

MnO, CaO

Steel

Ultimate Analysis wt%

Material Fe C Si Mn P

Pig iron 91 3.5 2 1 2.5

DOF analysis

12 unknown labeled variables ( 𝑉𝐴, 𝑉𝐹, mF, mS, mSt, XFe2O3, XP2O5, XSiO2, XMnO, XCO, XCO2, XN2)


- 14 independent molecular species balances (Fe, C, Si, Mn, P, SiO2, Fe2O3, CaO,, MnO, P2O5, CO,

CO2, N2, O2)

- 4 other equation relating unknown variables (XFe2O3+ XP2O5+ XSiO2+ XMnO = 0.65, XCO+ XCO2+ XN2

5% Fe oxidizes, 1/3 C oxidizes to CO2)


Bessemer converter

mF FluxCaO

25 tons Pig iron91% Fe3.5% C2% Si1% Mn2.5% P

𝑉𝐴 Air500 m3/min

𝑉𝐹 Flue gasCO, CO2, N2

mS SlagFe2O3, P2O5, SiO2

MnO, 35% CaO

mSt Steel

Calculate the volume of air required for the operationBasis 25 tons of pig iron

Oxidation step 1: Si + O2 = SiO2 Oxidation step 3: C + 1/2O2 = CO, C + O2 = CO2

Weight of Si in Pig iron = 0.02 * 25000 = 500 kg Weight of C in iron = 0.035 * 25000 = 878 kgnSi= 500/28 = 17.857 kg-atom nC = 878/12 = 72.917 kg-atomnO2= 17.857 kg-mole nC(for CO) = (2/3) * 72.917 = 48.611 kg-atom

nO2 = (½) * 48.611 = 24.306 kg-moleOxidation step 2: 2Mn + O2 = 2MnO nC(for CO2) = (1/3) * 72.917 = 24.306 kg-atomWeight of Mn in Pig iron = 0.01 * 25000 = 250 kg nO2= nC(for CO2) = 24.306 kg-atomnMn= 250/55 = 4.545 kg-atomnO2= 4.545/2 = 2.273 kg-mole Oxidation step 4: 4P + 5O2 = 2P2O5

Weight of P in iron = 0.025 * 25000 = 625 kgnP=625/31 = 20.161 kg-atomnO2=(5/4) * nP=25.201 kg-mole

Bessemer converter

FluxCaO


Air500 m3/min

Flue gasCO, CO2, N2


MnO, 35% CaO

Steel

Calculate the volume of air required for the operationBasis 25 tons of pig iron

Total O2 used during Si, Mn, C, P oxidation = 93.94 kg-mole

Considering small amount of Fe oxidizing in all steps:nFe = 1137.5/56 = 20.312 kg-atom2Fe + 3/2 O2 = Fe2O3

nO2 = (¾)*nFe = (¾)*20.312 = 15.23 kg-mole

Total O2 used = 2.273 + 17.857 + 24.306 + 24.306 + 25.201 + 15.23 = 109.17 kg-mole O2

Volume of air required = (109.17/0.21) * 22.4 = 11644.8 m3/25 ton of Pig ironTotal blowing time = 11644.8/500 = 23.29 minutes

Bessemer converter

FluxCaO


Air500 m3/min

Flue gasCO, CO2, N2


MnO, 35% CaO

Steel

Calculate the durations of each blowing period

Total O2 used = 2.273 + 17.857 + 24.306 + 24.306 + 25.201 + 15.23 = 109.17 kg-mole O2

Volume of air required for period 1= (2.273/0.21) * 22.4 = 242.45 m3/500 kg SiVolume of air required for period 2= (17.857/0.21) * 22.4 = 1904.75 m3/250 kg MnVolume of air required for period 3= (48.162/0.21) * 22.4 = 5137.3 m3/878 kg CVolume of air required for period 4= (25.201/0.21) * 22.4 = 2688.1 m3/625 kg PVolume of air distributed in all periods= (15.23/0.21) * 22.4 = 1624.5 m3/1138 kg FeTotal blowing time = 11644.8/500 = 23.29 minutes

Bessemer converter

FluxCaO


Air500 m3/min

Flue gasCO, CO2, N2


MnO, 35% CaO

Steel

FeFe2O3

SiSiO2 MnMnO CCO, CO2 PP2O5

0 4.43 4.99 17.04 23.29Time (min)

Period 1 Period 2 Period 3 Period 4

Calculate the weight of CaO added to the converter

Si + O2 = SiO2 4P + 5O2 = 2P2O5

Weight of Si in Pig iron = 0.02 * 25000 = 500 kg Weight of P in iron = 0.025 * 25000 = 625 kgnSi= 500/28 = 17.857 kg-atom nP=625/31 = 20.161 kg-atomnSiO2= 17.857 kg-mole nP2O5= 20.161/2 = 10.081 kg-moleWeight of SiO2 in slag = 17.857*60 = 1071.4 kg Weight of P2O5 in slag = 17.081*142 =

1431.5 kg2Mn + O2 = 2MnOWeight of Mn in Pig iron = 0.01 * 25000 = 250 kg 2Fe + 3/2 O2 = Fe2O3

nMn= 250/55 = 4.545 kg-atom nFe = 1137.5/56 = 20.312 kg-atomnMnO= 4.545 kg-mole nFe2O3 = 20.312/2 = 10.156 kg-moleWeight of MnO in slag = 4.545*71 = 322.7 kg Weight of Fe2O3 in slag = 10.156*160 =

1625kgMnO + SiO2 + P2O5 + Fe2O3 = 4450.6 kg, 35% CaO= (4450.6/0.65)*0.35 = 2396.5 kg CaO in slag

Bessemer converter

FluxCaO


Air500 m3/min

Flue gasCO, CO2, N2


MnO, 35% CaO

Steel

Calculate the weight and composition of the slag

Component Weight (kg) Weight %SiO2 1071.4 15.64%MnO 322.7 4.71 %P2O5 1431.5 20.90%Fe2O3 1625 23.75%CaO 2396.5 35.00%Total 6848 100.00%

Bessemer converter

FluxCaO


Air500 m3/min

Flue gasCO, CO2, N2


MnO, 35% CaO

Steel

Calculate the weight of CaO added to the converter in each blowing period

Total CaO added as flux = 2396.5 kg CaO

CaO consumed each minute = 2396.5/23.29 = 102.9 kg CaO consumed in period 1 = 102.9 * 4.43 = 455.8 kgCaO consumed in period 2 = 102.9 * 0.56 = 58.02 kgCaO consumed in period 3 = 102.9 * 11.95 = 1229.4 kgCaO consumed in period 4 = 102.9 * 6.25 = 643.3 kg

Bessemer converter

FluxCaO


Air500 m3/min

Flue gasCO, CO2, N2


MnO, 35% CaO

Steel

FeFe2O3

SiSiO2 MnMnO CCO, CO2 PP2O5

0 4.43 4.99 17.04 23.29Time (min)

Period 1 Period 2 Period 3 Period 4

Copper convertingLiquid matte from the smelting process is oxidized in a bessemer or basic oxygen furnace by blowing air or oxygenThe difference between steel converting and copper converting is that the value mineral Cu2S is oxidized as well in the latter process

Cu2S + O2 = Cu l + SO2

FeS2 in the matte is oxidized initially due to its higher oxidation free energyExcess S in the matte may also oxidize preferentially prior to reduction of copper

Blowing, fluxing and slagging may be done periodically due to convenienceFlux is commonly added in batches due to the high amount of charge material and the limited space of furnaces The amount of flux batches and blowing rate affects the time taken to produce slag in periods

SO2 on the surface of the copper evaporate and form blisters on the solidifying copperBlister copper purity is around 99% and electrolysis treatment is needed to obtain pure copperConverted blister copper is considered as 100% pure for convenience in material balance

Copper converter analysis

40 tons of matte carrying 34% Cu is charged in a bessemer converterThe flux is added in batches of 3000 kg, the converter is blown after each addition to obtain slag of the given compositionBlister copper is produced after the removal of slags formed using partially added fluxes Air is blown at a rate of 100 m3/minute

Bessemer converter

Flux

Matte

Air

Slag

Blister copper


Material FeO CaO SiO2 Al2O3 Cu2S FeS2

Slag 59 8 32 1

Flux ? 75 ? 2 5

V Off-gasXSO2

XN2

Air is blown at a rate of 100 m3/minute

DOF analysis

9 unknown labeled variables (ms , mbc ,V , XFe, XS, XCaO, XAl2O3, XSO2, XN2)

- 4 independent atomic species balances that are involved in the reactions (Cu, Fe, S, O)

- 4 molecular balances on independent nonreactive species (N2, CaO, SiO2, Al2O3)

- 3 other equation relating unknown variables (XFe + XS =0.66, XCaO+ XAl2O3=0.18, XSO2+ XN2=1)

= -2 degrees of freedom!

Bessemer converter

3000 kg Flux75% SiO2

2% Cu2S5% FeS2

XCaO

XAl2O3

40 tons Matte34% CuXfe

XS

Air100 m3/min

ms Slag 59% FeO, 8% CaO, 32% SiO2, 1% Al2O3

mbc Blister copper

V Off-gasXSO2

XN2


Calculate the time of each partial blow

Cu2S in matte = 40000 * 0.34 * (160/128) = 17000 kg, FeS=23000 kg

Let X be the weight of slag Let Y be the weight of FeS oxidized in one blowSiO2 balance: Fe balance:75% * 3000 = 0.32 X (56/88) * Y + 5% * 3000 * (56/120) = 59% * 7031 * (56/72)X = 7031 kg Y = 4963 kg = 56.4 kg-moles

Oxygen required for 1 blow:FeS + 3/2O2 = FeO + SO2 FeS2 + 5/2O2 = FeO + 2SO2

O2 required = 56.4 * (3/2) = 84.6 kg-moles O2 required = 1.25 * (5/2) = 3.125 kg-molesTotal O2 required = 87.725 kg-moles

Time for 1 blow = 87.725

0.21∗

22.4

100= 93.57 minutes

Bessemer converter


2% Cu2S5% FeS2

40 tons Matte34% Cu

Air100 m3/min

Slag 59% FeO, 8 CaO32% SiO2, 1% Al2O3

Blister copper

V Off-gasXSO2

XN2


Calculate the number of partial blows and the weight of flux to be added for the last partial blow to completely remove FeO in the slag

Time for 1 blow = 87.725

0.21∗

22.4

100= 93.57 minutes

The weight of FeS oxidized in one blow = 4963 kg = 56.4 kg-molesNumber of partial blows = 23000/4963 = 4.63 ≈ 5

FeS oxidized in the 5th blow = 23000 – (4*4963) = 3148 kgLet Z be the amount of flux batch added in the last periodO2 balance:(3148/88) *(3/2) + (0.05Z /120) *(5/2) = 87.725 * 3148/4963 Z= 1903 kg

Bessemer converter


2% Cu2S5% FeS2

40 tons Matte17 tons Cu2S23 tons FeS

Air100 m3/min

7031 kg Slag 59% FeO, 8 CaO32% SiO2, 1% Al2O3

Blister copper

V Off-gasXSO2

XN2


Calculate the total time for blowing the charge to convert to blister copper

Blow Period (min)1 93.572 93.573 93.574 93.575 3148/4963 * 93.57 = 59.40 minutesTotal time to remove Fe in the matte and flux completely = 431.68 minutes

Cu2S + O2 = Cu l + SO2Total Cu2S = 17000/160 = 108 kg-moles, Total O2 required = 108 kg-moles Total air required = (108/0.21) * 22.4 = 11520 m3, Time required to convert Cu = 115.2 minutesTotal time of operation = 546.88 minutes

Bessemer converter


2% Cu2S5% FeS2

40 tons Matte17 tons Cu2S23 tons FeS

Air100 m3/min

7031 kg Slag 59% FeO, 8 CaO32% SiO2, 1% Al2O3

Blister copper

V Off-gasXSO2

XN2

Documents

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