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    Hall-Heroult Process

    Electrochemical Process to Reduce Alumina to Aluminum Alumina is dissolved in a molten fluoride solvent called cryolite

    2 Al2O3(cryolite) + 3 C(anode) 4 Al(liquid) + 3 CO2(g)12 e-

    T = 960oC

    Electrical work is needed:

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    = -E 4 V

    Electrolyte (cryolitic bath)

    Na3AlF6 + (AlF3)excess + CaF2 + Al2O3

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    Hall-Heroult Process

    Electrochemical Process to Reduce Alumina to Aluminum Alumina is dissolved in a molten fluoride solvent called cryolite

    Electrolyte (cryolitic bath)

    Na3AlF6 + (AlF3)excess + CaF2 + Al2O3

    CR = NaF / AlF3 (molar)

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    by CR and %Al2O3

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    Hall-Heroult Process

    Examples:

    1) Create a stream for a standard bath

    2) Phase Diagrams (CR = 2.2; 5% CaF2)

    3) Phase Diagrams (% Al2O3 vs xs-AlF3; 5%CaF2)

    4) Adding reactants to the electrolyte (Al2O3)

    5) Adding reactants to the electrolyte (Na2CO3)

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    ng reac an s o e e ec ro y e 37) Carbide formation in the cathode blocks

    8) Cathode / Collecting Bar / Refractory / Na(g)

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    Hall-Heroult Process

    Example #1: Create a stream for a standard bath

    Create a stream of the following electrolyte

    Na3AlF6+ 12.2 wt.% AlF3+ 5.0% CaF2+ 3.0% Al O

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    The stream will be constrained in

    enthalpy at 960oC.

    This stream will be used as reactant

    in several following examples.

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    Example #1: Create a stream for a standard bath

    FThall Database

    79.8% Na3AlF6

    + 12.2 wt.% AlF3+ 5.0% CaF2+ 3.0% Al2O3

    Hall-Heroult Process

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    Example #1: Create a stream for a standard bath

    Hall-Heroult Process

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    Example #1: Create a stream for a standard bath

    Hall-Heroult Process

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    Example #2: Phase Diagrams

    The cryolite (Na3AlF6) liquidus and the alumina (Al2O3) solubilityare very important for the cell operation.

    The formation of a side-ledge protects the cell

    lining from the corrosive electrolyte.

    Hall-Heroult Process

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    Alumina has to be transported in the bathto the reaction site under the anode.

    So a high solubility is better.

    A mapping of these phase diagram

    features will be useful.

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    Example #2: Phase Diagrams (CR = 2.2; 5% CaF2)

    FThall database

    CR = NaF / AlF3

    CR = 2.2

    (NaF)2.2(AlF3)1.0

    Hall-Heroult Process

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    File | Open HALLHEROULT_Isoplethal1

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    Example #2: Phase Diagrams (CR = 2.2; 5% CaF2)

    Hall-Heroult Process

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    Example #2: Phase Diagrams (CR = 2.2; 5% CaF2)

    X-axis

    Y-axis T = 800 1050oC

    Hall-Heroult Process

    ( ) ( )100

    111

    100

    3220.132.2

    32=

    ++ OAlCaFAlFNaF

    OAl

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    Constants P = 1 atm

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    ( ) ( )0.5

    111

    100

    3220.132.2

    2=

    ++ OAlCaFAlFNaF

    CaF

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    Example #2: Phase Diagrams (CR = 2.2; 5% CaF2)

    Hall-Heroult Process

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    Example #3: Phase Diagrams (% Al2O3 vs xs-AlF3; 5%CaF2)

    FThall database

    Na3

    AlF6

    AlF3(excess)

    CaF2Al2O3

    Hall-Heroult Process

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    T = 960oC

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    File | Open HALLHEROULT_Isothermal2

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    Example #3: Phase Diagrams (% Al2O3 vs xs-AlF3; 5%CaF2)

    X-axis

    Y-axis

    Hall-Heroult Process

    5001111

    100

    322363

    3=

    +++ OAlCaFAlFAlFNa

    AlF

    1001111

    100

    322363

    32=

    +++ OAlCaFAlFAlFNa

    OAl

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    Constants T = 960oC

    P = 1 atm

    Hall-Heroult Process - 14

    0.51111

    100

    322363

    2=

    +++ OAlCaFAlFAlFNaCaF

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    Example #3: Phase Diagrams (% Al2O3 vs xs-AlF3; 5%CaF2)

    Hall-Heroult Process

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    Example #4: Adding reactants to the electrolyte

    The thermal/heat balance on the electrolytic cell is very important

    The following additions are frequent:

    Al2O3 for the reduction reaction

    Hall-Heroult Process

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    Na2CO3 to increase CR3 Na2CO3 + 2 AlF3 6 NaF + Al2O3 + 3 CO2

    AlF3 to decrease CR

    CR = NaF / AlF3 (molar)

    All these reactants are much colder than the cell (10oC 200oC)

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    Hall-Heroult Process

    Example #4: Adding reactants to the electrolyte (Al2O3)

    File | Open HALLHEROULT_Add_Alumina

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    Hall-Heroult Process

    Example #4: Adding reactants to the electrolyte (Al2O3)

    1.0 ton of Bath

    (stream @ 960oC)

    Alumina @ 125o

    C-Al2O3-Al2O3Al2O3(H2O)

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    Na2CO3(s) (@ 20oC)

    Al(liq.) (@ 960oC)

    AlF3(s)

    (@ 50oC)

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    Hall-Heroult Process

    Example #4: Adding reactants to the electrolyte (Al2O3)

    1.0 ton of Bath

    (stream @ 960oC)

    + 5 kg Alumina-Al2O3-Al2O3Al2O3(H2O)

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    Adiabatic Addition

    Q = H = 0

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    Hall-Heroult Process

    Example #4: Adding reactants to the electrolyte (Al2O3)

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    Hall-Heroult Process

    Example #5: Adding reactants to the electrolyte (Na2CO3)

    File | Open HALLHEROULT_Add_Soda

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    Example #5: Adding reactants to the electrolyte (Na2CO3)

    1.0 ton of Bath

    (stream @ 960oC)

    Na2CO3(s) (@ 20

    o

    C)

    Al(liq.) (@ 960oC)

    Hall-Heroult Process

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    2 3

    added manually to thecell.

    A bag can partially

    touch the metal pad

    below the bath and

    react with it.

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    Example #5: Adding reactants to the electrolyte (Na2CO3)

    1.0 ton of Bath

    (stream @ 960oC)

    Na2CO3(s) (@ 20

    o

    C)

    Al(liq.) (@ 960oC)

    Hall-Heroult Process

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    2 3

    added manually to thecell = 10 kg

    A bag can partially

    touch the metal pad

    below the bath and

    react with it

    = 0 or 2 kg

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    Example #5: Adding reactants to the electrolyte (Na2CO3)

    Bags of Na2CO3 = 10 kg

    metal pad

    = 0 kg

    Result

    Hall-Heroult Process

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    T < 960oC

    4.2 kg gas

    (9.5 m3 gas)

    986.7 kg bath

    19.1 kg Na3AlF6

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    Example #5: Adding reactants to the electrolyte (Na2CO3)

    Bags of Na2CO3 = 10 kg

    metal pad

    = 2 kg

    Result

    Hall-Heroult Process

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    T > 960oC

    2.2 kg gas

    (7.8 m3 gas)

    1009.6 kg bath

    0.2 kg C

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    Hall-Heroult Process

    Example #6: Adding reactants to the electrolyte (AlF3)

    File | Open HALLHEROULT_Add_AlF3

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    H ll H l P

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    Hall-Heroult Process

    Example #6: Adding reactants to the electrolyte (AlF3)

    1.0 ton of Bath

    (stream @ 960oC)

    AlF3(s)

    (@ 50oC)

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    H ll H lt P

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    Hall-Heroult Process

    Example #6: Adding reactants to the electrolyte (AlF3)

    1.0 ton of Bath

    (stream @ 960oC)

    + 5 kg AlF3(s)

    (@ 50oC)

    Adiabatic Addition

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    Hall-Heroult Process - 28

    H ll H lt P

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    Hall-Heroult Process

    Example #6: Adding reactants to the electrolyte (AlF3)

    1.0 ton of Bath

    (stream @ 960oC)

    + 5 kg AlF3(s)

    (@ 50oC)

    T = 956.00oC

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    .

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    Hall Heroult Process

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    Example #7: Carbide formation in the cathode blocks

    The lifetime of the cell is very dependent on the cathode block(graphite) lifetime.

    The cathode blocks can be degraded by mechanical erosion, but alsoby chemical reactions with the penetrating sodium vapors andliquid electrolyte.

    Hall-Heroult Process

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    Sodium is generated at the metal

    bath interface with

    3 NaF + Al AlF3 + 3 Na

    Residual N2 and CO can be

    present in the carbon pores init.

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    Na

    Hall Heroult Process

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    Example #7: Carbide formation in the cathode blocks

    Hall-Heroult Process

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    Hall Heroult Process

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    Example #7: Carbide formation in the cathode blocks

    Hall-Heroult Process

    Bath; CR=2.2

    C with smallamounts of N & O

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    File | Open HALLHEROULT_Cathode1

    Penetrating Na

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    Example #7: Carbide formation in the cathode blocks

    FACT53 Gas species

    NaCN(liq.)

    C

    Fthall

    Hall-Heroult Process

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    Metal

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    Hall-Heroult Process

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    Example #7: Carbide formation in the cathode blocks

    X-axis

    Y-axis

    Hall-Heroult Process

    10..0100

    32=

    OAl

    ( ) ( )01.0..0

    111

    1

    3220.132.2

    =

    ++ OAlCaFAlFNaF

    Na

    00.01 kg Na / kg of bath

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    Constants T = 960oC & P = 1 atm

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    ( ) ( )5

    111

    100

    3220.132.2

    2=

    ++ OAlCaFAlFNaF

    CaF

    1113220.132.2

    ++ OACaFA FNaF

    ( ) ( )5

    111

    1

    3220.132.2

    =

    ++ OAlCaFAlFNaF

    C

    CaF2 is 5% of the mass of the bath

    The carbon to bath mass ratio is fixed at 5.0(arbitrary)

    0 10% Al2O3 in the bath

    Hall-Heroult Process

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    Example #7: Carbide formation in the cathode blocks

    Hall Heroult Process

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    Hall-Heroult Process

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    Hall Heroult Process

    Example #8: Cathode / Collecting Bar / Refractory / Na(g)

    The current collecting bar (steel) entrapped inside the cathode block(C-graphite);

    a SiO2/Al2O3 layer is beneath it;

    Na(vapors) percolated through the

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    Cathode block from the metal pad;

    Unwanted reaction products can

    increase the electrical resistivity

    $$

    T 900-960oC

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    Hall-Heroult Process

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    Hall Heroult Process

    Example #8: Cathode / Collecting Bar / Refractory / Na(g)

    FACT53

    Gas species FToxid

    Refractory Mat.

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    Steel

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    File | Open HALLHEROULT_Cathode_CollectingBar

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    Hall-Heroult Process

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    Example #8: Cathode / Collecting Bar / Refractory / Na(g)

    PNa is variedfrom 10-10 atm to10-1 atm

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    Hall-Heroult Process

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    Example #8: Cathode / Collecting Bar / Refractory / Na(g)

    Extract & Plot

    Results

    Na(g)

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    Hall-Heroult Process

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    Example #8: Cathode / Collecting Bar / Refractory / Na(g)

    X-axis : log10(PNa)

    Na(g) is species #13

    Y-axis : log10(grams)

    Select all s ecies with

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    Mole (max) > 0

    Hall-Heroult Process

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    Example #8: Cathode / Collecting Bar / Refractory / Na(g)

    At low log10(PNa)

    C+ Austenite(fcc)

    + Refractory (SiO2 + mullite)

    At intermediate log10(PNa)

    C

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    + Austenite(fcc)

    + Albite (NaAlSi3O8) + SiO2

    Higher:+ Slag+ Nepheline (NaAlSiO4)

    At high log10(PNa)

    C+ Ferrite(bcc)+ Slag (Na2SiO3 + NaAlO2-NaAlSiO4(s.s.))