PHASE DIAGRAM · 2013. 11. 27. · L: 35wt%Ni α: 46wt%Ni C L: 35wt%Ni 46 43 32 24 35 L: 32wt%Ni...

29
BY PHASE DIAGRAM BY ENGR. MS TAYYABA BANO

Transcript of PHASE DIAGRAM · 2013. 11. 27. · L: 35wt%Ni α: 46wt%Ni C L: 35wt%Ni 46 43 32 24 35 L: 32wt%Ni...

  • BY

    PHASE DIAGRAM

    BY

    ENGR. MS TAYYABA BANO

  • Cu-Ni System

    • 2 phases: L (liquid) α (FCC solid solution)

    • 3 phase fields: 1400

    1500

    1600T(°C)

    L (liquid)

    • 3 phase fields:

    L

    L + α α

    wt% Ni20 40 60 80 10001000

    1100

    1200

    1300

    1400

    α (FCC solid solution)

    L +

    αliquidus

    solidus

  • • Rule 1: If we know T and Co, then we know:

    the # and types of phases present# and types of phases present# and types of phases present# and types of phases present.

    • Examples:

    1500

    1600T(°C)

    L (liquid)

    liquidus

    (1250,35) Cu-Ni

    phase

    A(1100, 60): 1 phase: α

    PHASE DIAGRAMS: # and types of phases

    wt% Ni20 40 60 80 10001000

    1100

    1200

    1300

    1400

    α (FCC solid solution)

    L +

    α

    liquidus

    solidus

    A(1100,60)B(1250,35)

    phase

    diagramB(1250, 35): 2 phases: L + α

  • Rule 2: If we know T and Co, then we

    know: the composition of each phase.

    • Examples:

    1300

    T(°C)

    L (liquid) liquid

    us

    solidu

    s

    TAA

    TBB

    tie line

    L + α

    Cu-Ni

    system

    At TA:

    Only Liquid (L) C = C ( = 35wt% Ni)

    Co = 35wt%Ni

    PHASE DIAGRAMS: composition of phases

    wt% Ni

    20

    1200

    α (solid)L +

    α

    solidu

    s

    30 40 50

    DTD

    TBB

    433532CoCL Cα

    CL = Co ( = 35wt% Ni)

    At TB:

    Both α and L CL = Cliquidus ( = 32wt% Ni here)

    Cα = Csolidus ( = 43wt% Ni here)

    At TD:

    Only Solid (α) Cα = Co ( = 35wt% Ni)

  • Rule 3: If we know T and Co, then we know:

    the amount of each phase (given in wt%).Cu-Ni

    system• Examples:

    At TA: Only Liquid (L)

    WL = 100wt%, Wα = 0At TD: Only Solid (α)

    Co = 35wt%Ni

    1300

    T(°C)

    L (liquid) liquid

    us

    solidu

    s

    TAA

    TBB

    tie line

    L + α

    PHASE DIAGRAMS: weight fractions of phases

    At TB: Both α and L

    At TD: Only Solid (α) WL = 0, Wα = 100wt%

    WL =S

    R +S

    Wα =R

    R +S

    %733243

    35430 wtCC

    CC

    L

    =−−=

    −−=

    α

    α

    = 27wt%

    wt% Ni

    20

    1200α

    (solid)L +

    α

    solidu

    s

    30 40 50

    DTD

    TBB

    433532CoCL Cα

    R S

    %27wtCC

    CC

    L

    o =−−=

    α

    α

  • • Sum of weight fractions:

    • Conservation of mass (Ni):

    • Combine above equations:

    WL + Wα = 1

    Co = WLCL + WαCα

    = RR +S

    Wα =Co −CLCα −CL

    = SR+ S

    WL= Cα −CoCα −CL

    THE LEVER RULE: A PROOF

    R S Cα −CLR SCα CL

    • A geometric interpretation:

    Co

    R S

    WαWL

    CL Cαmoment equilibrium:

    1−Wαsolving gives Lever Rule

    WLR = WαS

  • • Phase diagram:• Phase diagram:• Phase diagram:• Phase diagram:

    CuCuCuCu----Ni system.Ni system.Ni system.Ni system.

    • System is:• System is:• System is:• System is:

    --------binarybinarybinarybinaryi.e.i.e.i.e.i.e., 2 components:, 2 components:, 2 components:, 2 components:Cu and Ni.Cu and Ni.Cu and Ni.Cu and Ni.

    --------isomorphousisomorphousisomorphousisomorphous

    1300

    L (liquid)

    L +

    α

    T(°C)

    A

    D

    B

    L: 35wt%Ni

    α: 46wt%NiC

    L: 35wt%Ni

    46

    4332

    24

    35

    L: 32wt%Ni

    Cu-Ni

    system

    COOLING IN A Cu-Ni BINARY

    --------isomorphousisomorphousisomorphousisomorphousi.e., completei.e., completei.e., completei.e., complete

    solubility of onesolubility of onesolubility of onesolubility of one

    component incomponent incomponent incomponent in

    another; another; another; another; αααα phasephasephasephasefield extends fromfield extends fromfield extends fromfield extends from

    0 to 100wt% Ni.0 to 100wt% Ni.0 to 100wt% Ni.0 to 100wt% Ni.

    wt% Ni20

    1200

    30 40 501100

    α (solid)

    L +

    αD

    35Co

    E

    24 36

    α: 43wt%Ni

    L: 32wt%Ni

    L: 24wt%Ni

    α: 36wt%Ni

    • Consider

    Co = 35wt%Ni.

  • 2 componentshas a special composition

    with a min. melting T.

    • 3 single phase regions

    (L, α, β) • Limited solubility:

    α: mostly Cu

    Ex.: Cu-Ag system

    L (liquid)

    α L + α L+β β

    1200 T(°C)

    800

    1000

    TE 779°C

    Cu-Ag

    system

    BINARY-EUTECTIC SYSTEMS

    α: mostly Cu β: mostly Ni • TE: No liquid below TE

    • CE: Min. melting T

    composition

    L+β β

    α + β

    Co, wt% Ag 20 40 60 80 100 0

    200

    400

    600

    800

    CE

    TE 8.0 71.9 91.2 779°C

  • L (liquid)

    α L + α L+β β

    120 0 T(°C)

    800

    1000

    TE 8.0 71.9 91.2 779°C

    α + β

    Co, wt% Ag 20 40 60 80 100 0

    200

    400

    600

    CE

  • IRON-CARBON/ SYSTEM/ PHASE DIAGRAMPHASE DIAGRAM

  • • 2 important points

    -Eutectic (A):

    -Eutectoid (B):

    L⇒ γ +Fe3C

    γ ⇒ α +Fe3C

    C (cementite)

    1600

    1400

    1200

    1000

    800

    L

    γ (austenite)

    γ+L

    γ+Fe3Cα+γ

    L+Fe3C

    δ

    727°C = Teutectoid

    1148°C

    T(°C)

    A

    B

    SR

    γ γγγ

    α

    IRON-CARBON (Fe-C) PHASE DIAGRAM

    Result: Pearlite = alternating layers of α and Fe3C phases.

    120µm

    Fe3C (cementite)

    800

    600

    4000 1 2 3 4 5 6 6.7

    α+Fe3C

    (Fe) Co, wt% C0.77 4.30

    727°C = TeutectoidBR S

    Fe3C (cementite-hard)

    α (ferrite-soft)

    αCeutectoid

  • DEVELOPMENT OF MICROSTRUCTURESDEVELOPMENT OF MICROSTRUCTURES

    � Austenite precipitates Fe3C at Eutectoid Transformation Temperature (727°C).

    γγγγ αααα + Fe3C Cooling

    Heating

    � When cooled slowly, forms Pearlite, which is a micro-contituent made of ferrite (αααα) and Cementite (Fe3C),

    Heating

  • HYPOEUTECTOID STEEL

    (Fe-C System)

    C (cementite)

    1600

    1400

    1200

    1000

    L

    γ (austenite)

    γ+L

    γ+Fe3C

    L+Fe3C

    δ

    1148°C

    T(°C)

    γ γγγ

    γγγ γ

    α

    Adapted from

    Fig. 9.27,Callister6e. (Fig. 9.27 courtesy Republic Steel Corporation.)

    Co

    Fe3C (cementite)

    800

    600

    4000 1 2 3 4 5 6 6.7

    α+Fe3C

    Co, wt% C0.77

    727°C

    R Sαγγ γ

    γ r s

    wα =s/(r+s)wγ =(1-wα)

    wα =S/(R+S)wFe3C =(1-wα)

    wpearlite = wγ

    ααα

    αα

    α pearlite

    100µm Hypoeutectoid steel

  • HYPOHYPO--EUTECTOIDEUTECTOID

    “Proeutectoid” means it formed ABOVE or BEFORE the Eutectoid Temperature!

  • How to calculate the relative amounts of proeutectoid

    phase (α or Fe3C) and pearlite?Eutectoid ,0.76% CEutectoid ,0.76% C

    2.14

    HypoEutectoidHyperEutectoid

    2.14

  • � Fraction of pearlite:

    74.0

    022.0'

    022.076.0

    022.0' −=−−=

    += CoCo

    UT

    TWp 74.0022.076.0

    =−

    =+

    =UT

    Wp

  • � Fraction of Proeutectoid :

    74.0

    '76.0

    022.076.0

    '76.0 CoCo

    UT

    UW

    −=−−=

    +=α

  • � Fraction of pearlite: (for Hypereutectoid alloys)

    94.5

    '170.6

    76.070.6

    '170.6 CC

    XV

    XWp

    −=−−=

    +=

    94.576.070.6XVWp =−

    =+

    =

  • � Fraction proeutectoid cementite: (Hypereutectoid)

    76.0'176.0'1 −=−== CCVW94.576.070.63

    =−

    =+

    =XV

    W CFe

  • (Fe-C System)

    C (cementite)

    1600

    1400

    1200

    1000

    L

    γ (austenite)

    γ+L

    γ+Fe3C

    L+Fe3C

    δ

    1148°C

    T(°C)

    γ γγγ

    γγγ γ

    Fe3C

    HYPEREUTECTOID STEEL

    Co

    Fe3C (cementite)

    800

    600

    4000 1 2 3 4 5 6 6.7

    α+Fe3C

    Co, wt% C0.77

    R Sαs

    wFe3C =r/(r+s)wγ =(1-wFe3C)

    wα =S/(R+S)wFe3C =(1-wα)

    wpearlite = wγpearlite

    60µm Hypereutectoid steel

    r

    γγγ γ

    Fe3C

  • HYPERHYPER--EUTECTOIDEUTECTOID

    “Proeutectoid” means it formed ABOVE or BEFORE the Eutectoid Temperature!

  • THANK YOU