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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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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
- 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
: 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
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
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?
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)
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
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
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
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
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
temp of E1
const T line
temp of P
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