calcination roasting predominacearea

Extraction of metals

•Mining•Beneficiation (mineral processing)•Smelting (extraction of metals)•Refining

•Pyrometallurgy•Hydrometallurgy•Electrometallurgy

• In many cases pyro is cheaper than electro and hydro

• Employs inexpensive reducing agents• Higher temperatures lower the activation

energy barrier which opposes the progress of chemical reaction

• Rate of reactions almost double for every 10°C increase in temperature

Why pyro metallurgy

• Take advantage of the ability to shift the reaction equilibrium with temperature

• ZnO + C = Zn + CO(equilibrium is at left at 25°C. Shifts right at 1200°C)• M+2I2 MI4 Forward at 200°C, reverse at 1400°C

• Physical separation facilitated when products are in liquid or gaseous state(eg. Metal + slag ) or volatile chlorides

• Uses lesser space (higher/volume productivity)

Why pyro metallurgy

• Calcination ..Remove volatiles, decompose. • Roasting …Chemical conversion of ore (O2,Cl2)

• Smelting … Reduction of ore (C, H2, electric)• Calcination:

– CaCO3 = CaO +CO2 (DGT = 42300-37.7T) ..1000°C– Product may be leached (water/Acid/Alkali)

• Roasting(Oxidising, Volatilizing, chloridizing)– ZnS+1.5O2=ZnO+SO2 , 2NaCl+MS+2O2=Na2SO4+MCl2

– 4NaCl+2MO+S2+3O2 = 2Na2SO4+2MCl2

Pyrometallurgical steps

• Coal coke Oil and Gas• nH/nC ratio varies from less than 1 to 4• Net calorific value by Dulong’s formula• Coking (coking coals, low temp, high temp)

Expels CO2, H2O, C2H6, H2S, H2, NH3..)

• Gasification of Coal (Gas producers)

Fuel and Ore Preparation

DG° of some Carbon compounds

Gas producers

Gasification of coke and coalPartial combustion with air, O2, steam or CO2.

C + O2 = CO2 DG1° = -394100 – 0.8T J

(Dominates at low temp. and high O2)

2C + O2 = 2CO DG2° = -223400 – 175.3T J

Boudouard reactionCO2 + C = 2CO DG3° = +170700 – 174.5T J

With steamH2O + C = H2 + CO DG4° = +134700 – 142.5T J(All DG° values refer to graphite as standard state)

Gas producers

C + O2 = CO2

CO2 +C= 2CO

AIR + STEAM

Gas out

Height above grate

Coal or coke

H2O + C = H2 + CO N2

Air onlyAir+ Steam

Gas producers

Role of steam• Steam converts physical enthalpy to chemical

enthalpy• Increasing steam lowers temperature below the

required value• Alternate operation of gas producer

– Air blown– Steam blown (CO+H2, water gas)

• Alternate methods (preheat air, O2+H20, Electric)

Stoichiometry in gas producers

Stoichiometric calculations based on reactions between C, H2 and O2 provide

•Relative quantities of reactants used up•Quantity of products obtained•Composition of products

Calculation procedure – Choose the following reaction (shifted to the right at

1000°C)2C + O2 = 2CO

• Choose the basis (1 mole, kg or tonne)• Calculate moles of reactants in chosen basis• Calculate moles of other reactants and products (N2?)• Convert number of moles in suitable units (wt/vol)• Calculate the composition of product in suitable units


Case: Reaction between carbon and O2 + steam.

If the mixture has the mole ratio of nO2/nH2O = m,

2mC + mO2 = 2mCO …. (if nH2O = 1, then nO2 = m)

C + H2O = CO + H2

Which meansReactants ProductsC = 2m+1 moles CO = 2m+1 molesO2 = m moles H2 = 1 mole

H2O = 1 mole N2 (if O2 is from air)

(Also consider N2, if O2 is from air)


Case: Reactions in a mixture of H2O & CO in the temperature range 500-800°C.(Assume equilibrium)

t°C , Ptot (atm)

in presence of catalyst

n0CO

n0H2O

nH2OnCOnCO2nH2

nCH4

Stoichiometry calculations

Chemical reaction K @ 500°C K @ 800°C

CO + H2O = CO

2 + H

2K

1=5.8 K

1=1.2

CO + 3H2 = CH

4 + H

2O K

2=100 K

2=0.01

Assume:n0CO and n0H2O moles of CO & H2O enter the reactor

Effluent contains ni moles of CO, CO2, H2O, H2 and CH4

Mass balance from stoichiometerynCO = n0CO – x - ynCO2 = x

nH2O = n°H2O –x + y

nH2 = – x + y

nCH4 = y

ntot = n0CO + n°H2O - 2y


Where-x and y = moles of CO2 and CH4 respectively formed during reaction


ntot = total number of moles

ptot = Total pressure

Then pi = ni .(ptot/ntot)

Then

and

CO + H2O = CO

2 + H

2

CO + 3H2 = CH

4 + H

2O

DryingH2O(l) = H2O(g) DH0 = 43.9kJ

CalcinationCaCO3 CaO + CO2 DH0 = 177.8kJ

MgCO3 MgO + CO2

FeCO3 FeO + CO2

Fe(OH)2 FeO + H2O

Calcination and drying

900 ° C

417 ° C

400 ° C

200 ° C

Calcination and drying

CaCO3 + cokeGas

1

2

3

Air CaO

Preheating zone

Reaction zone

Cooling zone

SolidGas

AirSolid

Heat balance calculations in tutorial

1000°C500°C

100°C

340°C

800°C 900°C

• Preheating zoneSolids preheated to 800°C in counter curret by hot

furnace gases. No calcination. No combustion.• Reduction ZoneCoke burning and CaCO3 decomposition takes place.

Avg 1000°C. Gasses leave at 900°C.• Cooling ZoneBurnt line is cooled in counter current with cold air.

Burnt lime is cooled to 100°C

Heat balance for shaft furnace

Assumptions (deviations can be easily corrected for)• Excess air used to assure complete consumption (say

25%)• Limestone contains 100% CaCO3 (no impurities)• Coke contains 100% carbon (no impurities)• Disregard heat losses from walls• At room temp. assume rO2 = rN2 = rAir

• Calculate amount of Carbon required to calcine per unit weight of CaCO3

Heat balance for shaft furnaceCalculations


Heat requirement per mole of CaCO3

Reaction : CaCO3(800°C) = CaO(1000°C) + CO2(900°C)

Input kcal Output kcalCaCO3 20.8 CaO 12.0Heat balance 43.8 CO2 10.1

CaCO3=CaO + CO2

42.5

Sum 64.6 Sum 64.6

Heat available per mole of CarbonReaction : C (800°C) +(1.25O2 + 4.7N2)(500°C) = (CO2+ 0.25O2+ 4.7N2)(900°C) ….1.25 *(79/21)= 4.7

** Combustion air preheated to 500°C

Input kcal Output kcalC 3.2 CO2 10.11.25O2 ** 4.4 0.25O2 1.74.7N2 15.5 4.7N2 30.0C+O2 = CO2 97.0 Available heat 78.3Sum 120.1 Sum 120.1


• Combustion air temperature may be calculated from heat balance of the cooling zone.

• Calculations show that Carbon required per mole of Calcium carbonate is 43.8/78.3 = 0.56 mole.

• Theoretical requirement = 0.067kg of C per kg CaCO3

• Add 0.1kg to cover heat losses through walls. • Find heat surplus which goes out as top gas temp.• Compute heat balance for cooling zone for the required product

temperature • Tutorial problem: Work out total heat balance for all the three

zones, for 1kg (10 moles) of CaCO3


Heat balance for calcination shaft furnace

Roasting of sulfides

Roasting is the oxidation of metal sulfides to metal oxide and sulfur dioxideZnS + 3O2 = 2ZnO + 2SO2

2FeS2 + 5.5O2 = Fe2O3 + 4SO2

Additionally. Formation of SO3,metal sulfates and complex oxides such as ZnFe2O4

Typical ores – sulfides of Cu, Zn & PbRoasting below MP of sulfides & oxides(usually below 900-1000°C, but above 500-600°C … for sufficient velocity of reaction)


Thermodynamics of roasting•3 components, hence max of 5 phases. (i.e. 4 condensed + 1 gas phase)•Gas phase normally contains SO2, O2 also SO3, S2 •Following equilibria exist in gaseous components

S2+ 2O2 = 2SO2 ………………. (1)

2SO2 + O2 = 2SO3 ………………. (2)•For a given temp., gas composition is defined by partial pressure of the gasses.•Also for a fixed gas composition, the composition of condensed phases is fixed.•Thus phase relations in a ternary system at constant temperature can be defined by a 2D diagram…

Predominance area diagrams

logpO2

logpSO2


Lines that describe equilibrium between the three phases are defined by equations-MS2 (c) = ½ S2(g) …….(M is divalent)

½ S2(g) + O2(g) = SO2(g)

MS + 3/2 O2 = MO + SO2

SO2 + ½ O2 = SO3

MO + SO3 = MSO4

MSO4 = MO.ySO3 + (1-y)SO3

MO.ySO3 = MO + ySO3

• ½ S2(g) + O2(g) = SO2(g) ……………(1)

• MS + 3/2 O2 = MO + SO2 ……………(2)

At roasting temp., above reactions have large -ve free energy changes and SO2/O2 equilibrium

is far right. At 1100K, the equilibrium constant for equation (1) is 1.2 x

1013

\ At this temperature, SO2 is stable phase, even at very small pO2

Roasting reactions

• ½ S2(g) + O2(g) = SO2(g) ……………(1)

• MS + 3/2 O2 = MO + SO2 ……………(2)

At roasting temp., above reactions have large -ve free energy changes and SO2/O2 equilibrium

is far right. E.g. in case of FeS, for reaction (2)DG° (cal)= -113000 + 9.5T logT - 1.965(10-3T2) - 6.7TAt 1100K, the equilibrium constant, K is 1.2 x 1018

\ sulfide is converted to oxide at relatively small pO2

Roasting reactions


For the given Me-O-S system diagram-Me + SO2 = MeS + O2

2Me + O2 = 2MeO

2MeS + 3O2 = 2MeO

2MeO+2SO2+O2 = 2MeSO4

MeS+2O2 = MeSO4

Additional equations may be required if-MeS2, Me2O3, Me2(SO4)3, MeO.MeSO4 exist

Me + SO2 = MeS + O2 …………..(3)

2Me + O2 = 2MeO …………..(4)

2MeS + 3O2 = 2MeO + 2SO2 …………..(5)

2MeO + 2 SO2 + O2 = 2 MeSO4 …………..(6)

MeS + 2O2 = MeSO4 …………..(7)

The equilibria are given by the expressionsLog pO2 – log pSO2 = log K3

Log pO2 = –log K4

2log pSO2 – 3log pO2 = log K5

2log pSO2 + log pO2 = log K6

2 log pO2 = - log K7



pS2 = 1atm pSO3 = 1atm

logpSO2

logpO2

• For a given metal, the form of equilibrium expression is the same for all metals

• The slope of corresponding curves in the diagram are same

• Only value of k’s may differ• This means that the positions of lines may

change• Consequently size and position of areas may

change

Why Predominance area

• These areas are called predominance area for the particular phase

• In this area, phases exist even though SO2 and O2 may be changed independent of each other

• There are 2 deg., 1 deg. and invariant regions• Lines for reactions for formation of SO2 & SO3

S2+ 2O2 = 2SO2 ………………. (1)

2SO2 + O2 = 2SO3 ………………. (2)

are given by the expressions2 log pSO2 – 2 log pO2 = log K1 + log pS2

2 log pSO2 +2 log pO2 = - log K2 + 2log pSO3


Creating predominance area diagrams

Ni-S-O system at 1000K• Condensed phases have Ni, NiO, Ni3S2, NiSO4

• Gas phase SO2, O2 (some SO3 & S2)• Consider reactions such as• NiS(c) + 1.5 O2(g) = NiO(c) + SO2(g) ….DG = 104.8kcal

For which K’ = pSO2/pO23/2

and log pSO2 = 1.5 logpO2 + logK’ ……….(CD)

• Likewise – Ni3S2(c) + 3.5 O2(g) = 3NiO(c) + 2SO2

DG=308.4kcal, and K’’ = pSO22/pO2

7/2

\ logpSO2 = 7/4 logpO2 + ½ log k’’ ……….(BC)

List all reactions and their equilib. Constants• NiS + 1.5O2 = NiO + SO2

• Ni3S2+3.5O2 = 3NiO + 2SO2

• Ni + ½O2 = NiO (DG = -44.5kcal)..(AB)

• NiS + 2O2 = NiSO4 (DG = -125.8kcal)..(DG)

• 2NiO + 2SO2+O2= 2NiSO4 (DG = -32.6kcal) ..(DH)

• 3NiS + O2 = Ni3S2 + SO2 (DG = -72.3kcal) ..(FC)

• Ni3S2 + 2O2 = 3Ni + 2SO2 ..(EB)



C

D

B

G

H

AE

F NiO

NiSO4NiS

Ni3S2

Ni

1000 K

log pO2

log

pSO

2

Kellog diagram for Copper and Iron Sulfides

log pO2

log

pSO

2

Fe3O

4

FeO

Fe

CuCu 2S

FeS

Fe3O 4

Cu 2SCuS

pS2 = 1atmFe

3O4

Fe2O

3Cu Cu

2O

Cu2O

CuO

Fe2O

3

FeSO4

FeS 2

Fe2(SO4)3CuSO4

FeSO4

CuSO4

CuOCuO.CuSO

4

T=700°C

pSO3 = 1atm

Effect of temperature on roasting equilibria

To bring nonferrous metals into water soluble form• MeS + 2NaCl + 2O2 = Na2SO4 + MeCl2 (10%NaCl/5-600°C)

To convert nonferrous metals into volatile chlorides• MeO + CaCl2 = MeCl2 (g) +CaO (1250°C)

• MeS +CaCl2 + ½ O2 = MeCl2(g) + CaO + SO2

• MeO + Cl2 = MeCl2 (g) + ½ O2

• MeS + Cl2 + O2 = MeCl2(g) + SO2 (900-1000°C)

Iron oxide is not easily chloridised• 3MeO + 2FeCl3 = 3MeCl2 +Fe2O3

This reaction is strongly shifted to the right

Chloridizing roast

Oxidation of Cu2S in Air: Ref temp=25°C = 298°K.

Cu2S(1300°C) + (O2+3.76N2)(25°C)=(2Cu+SO2+3.76N2)(1250°C)

Heat balance during roasting

Input Kcal/mole Output Kcal/mole

Cu2S(l) 30.0 2Cu 22.8

O2 + 3.76N2 0.0 SO2 15.1

S(rh)*+O2=SO2 -DH°298 71.0 3.76N2 35.3

Cu2S=2Cu+

S(rh)*

DH°298

19.6

Heat surplus 8.2

Total Input 101.0 Total output 101.0

Oxidation of Cu2S in Air: Ref temp= 1300°C = 1573°K..

Cu2S(1300°C) + (O2+3.76N2)(25°C)=(2Cu+SO2+3.76N2)(1250°C)



2Cu 0.8 O2 10.3

SO2 0.7 3.76N2 36.8

3.76N2 1.5Cu2S=2Cu+

½S*(g)33.8

½S2*+O2=SO2 86.1 Heat surplus 8.2

Total input 89.1 Total output 89.1

Technology of roasting• Multiple hearth furnace• Flash roasting• Fluidized bed roasting• Sinter roasting


Multiple hearth roasting

Dwight Lloyd sintering machine

Oxidation of Cu2S in Air: Ref temp= 25°C. (Surplus heat)

(FeS2 + 4.13O2 +15.5N2)(25°C)=(0.5 Fe2O3+2SO2+1.38O2+15.5N2)(900°C)

Heat balance for pyrite(FeS2) roasting


FeS2 25°C 0 0.5Fe2O3

H(1173)15.5

4.13O2+15.5N2

02SO2 + 15.5N2 +1.38O2

123.125°C

Fe+0.75O2=0.5Fe2O3

99 FeS2=Fe+2S DH298 36.0-DH298

2S + O2=SO2-DH298 142 Heat surplus 56.9

Total input 241 Total output 241

Reduction smelting

• Mineral + gangue + reducing agent + flux = metal/matte + slag + gas

• Fluxes• Slags• Reducing agents

– C, H2, Reactive metals(Al, Ca)

slags

• Vitreous by-product of smelting• Collects unreduced gangue in a separate layer• Provides a medium to collect impurities

– Synthetic slags• Should have large difference in sp.gv • Should be fluid enough• Contains oxides, silicates, metals and sometime sulfides• Metal loss in slag• Basic(O2-, SiO4

4-)/Neutral(SiO44-)/ Acidic(chain silicate anions- Si2O6

7-, Si3O108-

& ring silicate anions Si4O128-

• Silicate degree = moles of acidic O2 from SiO2/ moles of basic oxygen from CaO, MgO, FeO

(<1) (=1) (>1)

calcination roasting predominacearea

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