Meaning of line he i line shared by phase fields of b l c l b c l
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H. B. C. b. c. E. F. G. a. A. meaning of line HE : i) line shared by phase fields of B+L, C+L, B+C+L ii) comp path of in the phase field of B+C+L iii) along this line, B and C are in with the melt (L) AFEG : phase field of A

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meaning of line HE : i) line shared by phase fields of B+L, C+L, B+C+L

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Meaning of line he i line shared by phase fields of b l c l b c l

H

B

C

b

c

E

F

G

a

A

  • meaning of line HE :

    i) line shared by phase fields of B+L, C+L, B+C+L

    ii) comp path of in the phase field of B+C+L

    iii) along this line, B and C are in with the melt (L)

    AFEG : phase field of A

    BHEF : " B

    CGEH : " C

    for a melt whose comp falls within the primary phase field of (for example) B, the first (primary) crystal to appear on cooling is B

  • projection


Meaning of line he i line shared by phase fields of b l c l b c l

congruent melting

  • binary joins and Alkemade lines

    - binary join : a line connecting pts representing comps of crystals in the ternary (why binary join? ∵behaving like a system)

    - Alkemade line : a join having a boundary line

joins : A-B, B-BC, BC-C, B-C, A-C, A-BC

Alkemade lines : A-B (wE), B-BC (tE), BC-C (sD), A-C (rD), A-BC (ED)

* B-C join is not an Alkemade line (no common boundary)

→ implying that there is no phase field of

 Alkemade theorem

: the intersection of a boundary line (or extension) with its corresponding Alkemade line (or extension) represents on the boundary line while on the Alkemade line


Meaning of line he i line shared by phase fields of b l c l b c l

Fig. 6.19. System A-B-C with isotherms added.

Fig. 6.20. isothermal section at 700°C

Fig. 6.21. Isothermal section at 600°C.

Fig. 6.22. isothermal section at 400°C

Fig. 6.23. isothermal section at 300°C

isothermalsections


Meaning of line he i line shared by phase fields of b l c l b c l

cooling path: study in the ternary with a congruently melting binary AB compound


Meaning of line he i line shared by phase fields of b l c l b c l

P2

P1

1) X belonging to ∆ (comp triangle) A-C-AB, final solidification at E1

L → (at TX) L+C → (along e1E1) L+C+A → A+C+AB at E1

2) Y within ∆ A-C-AB, finally at E1

L → (at TY) L+AB → (along e2E1) L+AB+A → AB+A+C at E1

3) Z within ∆ B-C-AB, finally at E2

L → (at TZ) L+B → (along e3E2) L+B+AB → B+AB+C at E2

cf) at E1 and E2, three boundary lines are approaching within a 360° range and thus the ternary liq disappears L → A+C+AB (E1) or L → B+C+AB (E2) thereby becoming invariant pts

if the lines are within a 180° range → one of the phases will dissolve (or resorb) and becoming a invariant pt

in an isoplethal analysis of the ternary

i) the comp is in what primary field? or on a boundary line or join?

ii) the comp is in which comp triangle? or on a join which is one side of a comp triangle?

iii) what invariant pt is associated with a particular comp triangle?


Meaning of line he i line shared by phase fields of b l c l b c l

cooling path: isoplethal study in the ternary with a congruently melting AB compound but a join

P

E


Meaning of line he i line shared by phase fields of b l c l b c l

cooling path: isoplethal study in the ternary with a congruently melting AB compound but a non-binary join

P

E


Meaning of line he i line shared by phase fields of b l c l b c l

  • why non-binary?: the primary phase field of B exists on AB-C, however, it cannot be described by a combination of AB and C

  • according to Alkemade theorem → intersection of EP with C-AB being located on an extension of EP → T decreases from P to E

    X : L → (at TX) L+B → (along e4P) L+B+C → (at P) B+C+AB

    Y : L → (at TY) L+B → (along e4P) L+B+C → (along PE) L+C+AB → (at E) C+A+AB

    check points !!!!

    1) draw a vertical pseudo-binary diagram along C-AB

    2) what happens on cooling at the point, P

    (explain the answer: L → C+AB+L → A+C+AB)


Meaning of line he i line shared by phase fields of b l c l b c l

  • ternary cooling path for an melting binary compound AB

the comp pt of the incongruently melting binary compound falling outside of its primary field

so, the first solid (Xtal) on cooling at comp BC is not a BC phase but a phase of , in this case


Meaning of line he i line shared by phase fields of b l c l b c l

[I]

X : belong to Δ A-C-BC, thus final solidification at D (A-C-BC)

L → (at TX) L+C → (along HD) L+A+C → (at D) A+C+BC

Solid starting from C → moving to K via M → finally jumping from K to its original comp X (A+C+BC) when L reached D, invariant peritectic


Meaning of line he i line shared by phase fields of b l c l b c l

[II]

Y : belonging to Δ A-C-AB, thus final solidification at E (A-C-AB)

L → (at TY) L+B → (along GP) L+B+C → (along PE) L+C+AB → finally L+C+AB ⇒ A+C+AB at E

Solid starting from B → moving to R → jumping from R to J then moving to M via K → finally jumping to original comp Y (A+C+AB) when L reached E, invariant eutectic


Meaning of line he i line shared by phase fields of b l c l b c l

[III]

Z : ∈ Δ A-B-BC comp. triangle, thus final solidification at E (A-B-BC)

L → (at TZ) L+C → (along KR) L+C+BC → (at R) its corresponding solid is now BC only (all the C portion disappeared, so-called resorb) → (on RG) L+BC → (along GE) L+A+BC → finally L+A+BC ⇒ A+B+BC at E

Solid starting from C → moving to BC → staying at BC for a while (when L is on RG) → moving from BC to F (on A-BC) → finally jumping to original comp Z (A+B+BC) when L reached E, invariant eutectic


Meaning of line he i line shared by phase fields of b l c l b c l

A

[IV]

K1

K2

K3

E1

K : comp on an Alkemade line of A-BC

L →(at TK)L+C → (along RP) L+A+C → at P, the solid part composed of A+C and the rxn of A+C+L ⇒ A+BC (zero amount of C) occurring

K

K1

K3

K2

temp of E1

const T line

temp of P


Meaning of line he i line shared by phase fields of b l c l b c l

[V]

recurrent crystallization : very strong curvature of pP, exaggerated case of [III]

X : L → (at TX) L+B → (along KR) L+AB+B (at R, B completely resorbed) → (on an extension of AB-R-R’) L+AB → (at R’) contact with AB+B again (due to high curvature) ∴ (along R’P) L+AB+B → (along PE) L+AB+C → finally at E, L+AB+C ⇒ AB+C+A


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