charge calculations in pyrometallurgical...
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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
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
VA Air 21% O2, 79% N2
VG Flue gasesXCO2 CO2
XSO2 SO2
XH2O H2OXN2 N2
Reacting Gaseous Mixture
T, P
Chemical reactionsC + O2 = CO2
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
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
VA Air 21% O2, 79% N2
VG Flue gasesXCO2 CO2
XSO2 SO2
XH2O H2OXN2 N2
Reacting Gaseous Mixture
STP
Chemical reactionsC + O2 = CO2
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)
= 0 degrees of freedom
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
Air at 800 C is consumed 20% in excess of the theoretical requirement
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
Gaseous fuel 5000 m3/hr56% N2, 22% CO, 10% H2, 6% CO2, 3% CH4, 3% H2O
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
Air at 800 C is consumed 20% in excess of the theoretical requirement
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%
Gaseous fuel 5000 m3/hr56% N2, 22% CO, 10% H2, 6% CO2, 3% CH4, 3% H2O
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)
+ 2 independent chemical reactions
- 3 independent molecular species balances (NH3, H2, N2)
- 1 other equation relating unknown variables (PV=m/MW*RT)
= 0 degrees of freedom
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
Rational Gas Analysis wt%
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
Rational Gas Analysis wt%
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)
+ 4 independent chemical reactions
- 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
Fuel 250 m3/ton MgO80% CH4
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
Fuel 250 m3/ton MgO80% CH4
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
Fuel 250 m3/ton MgO80% CH4
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
Rational Analysis wt%
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)
= 0 degrees of freedom
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
Rational Analysis wt%
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)
+ 7 independent chemical reactions
- 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)
= 0 degrees of freedom
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
Blast furnace gas 4200m3
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
Blast furnace gas 4200m3
58% N2, 26% CO, 12% CO2, 4% H2O
Slag
Pig iron 1000 kg93.8% Fe,4% C, 1.2% Si, 1% Mn
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
Blast furnace gas 4200m3
58% N2, 26% CO, 12% CO2, 4% H2O
SlagFe2O3, SiO2, MnO, Al2O3, CaO
Pig iron 1000 kg93.8% Fe,4% C, 1.2% Si, 1% Mn
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
Rational Analysis wt%
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,)
+ 3 independent chemical reactions
- 8 independent molecular species balances (Cu2S, FeS2, SiO2, Fe2O3, CaCO3, CaO, FeO, FeS)
- 0 other equation relating unknown variables
= 0 degrees of freedom
mS Slag35% SiO2
20% CaO45% FeO
mM Matte37.5% Cu2S62.5% FeS
Reverberatory furnace
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
Reverberatory furnace
Flux: CaCO3
Iron ore (80% Fe2O3, 20% SiO2)
1000 kg Copper ore17.5% Cu2S15% 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
Reverberatory furnace
Flux: CaCO3
Z kg Iron ore (80% Fe2O3, 20% SiO2)
1000 kg Copper ore17.5% Cu2S15% 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
Reverberatory furnace
Flux: CaCO3
Z kg Iron ore (80% Fe2O3, 20% SiO2)
1000 kg Copper ore17.5% Cu2S15% 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)
+ 6 independent chemical reactions
- 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)
= 0 degrees of freedom
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
25 tons Pig iron91% Fe3.5% C2% Si1% Mn2.5% P
Air500 m3/min
Flue gasCO, CO2, N2
SlagFe2O3, P2O5, SiO2
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
25 tons Pig iron91% Fe3.5% C2% Si1% Mn2.5% P
Air500 m3/min
Flue gasCO, CO2, N2
SlagFe2O3, P2O5, SiO2
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
25 tons Pig iron91% Fe3.5% C2% Si1% Mn2.5% P
Air500 m3/min
Flue gasCO, CO2, N2
SlagFe2O3, P2O5, SiO2
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
25 tons Pig iron91% Fe3.5% C2% Si1% Mn2.5% P
Air500 m3/min
Flue gasCO, CO2, N2
SlagFe2O3, P2O5, SiO2
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
25 tons Pig iron91% Fe3.5% C2% Si1% Mn2.5% P
Air500 m3/min
Flue gasCO, CO2, N2
SlagFe2O3, P2O5, SiO2
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
25 tons Pig iron91% Fe3.5% C2% Si1% Mn2.5% P
Air500 m3/min
Flue gasCO, CO2, N2
SlagFe2O3, P2O5, SiO2
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
Rational Analysis wt%
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
Air is blown at a rate of 100 m3/minute
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
3000 kg Flux75% SiO2
2% Cu2S5% FeS2
40 tons Matte34% Cu
Air100 m3/min
Slag 59% FeO, 8 CaO32% SiO2, 1% Al2O3
Blister copper
V Off-gasXSO2
XN2
Air is blown at a rate of 100 m3/minute
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
3000 kg Flux75% SiO2
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
Air is blown at a rate of 100 m3/minute
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
3000 kg Flux75% SiO2
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