Diagram Fasa (Phase Diagram)

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1 Diagram Fasa (Phase Diagram)
Lukhi Mulia S

2 Definisi dan Konsep Dasar
fasa Komponen Sistem

3 Definisi dan Konsep Dasar
Komponen : unsur atau senyawa yang terdapat didalam sistem. Fasa : bagian homogen dari suatu sistem yang memiliki sifat fisik dan kimia yang seragam.

4 Batas kelarutan (solubility limit)
Konsentrasi makismal suatu solute yang larut dalam solven Contoh : sistem larutan air-gula (C12H22O11-H2O) Batas kelarutan maksimal gula didalam air adalah 65%wt pada suhu 20°C Sehingga Jika komponen < 65wt% gula : sirup Jika komponen > 65wt% gula : sirup + kristal gula batas kelarutan meningkat dengan kenaikan temperatur

5 sugar–water (C12H22O11-H2O ) system.
Saturated syrup water Excess sugar

6 solubility limit Figure 9.1 The solubility of sugar (C12H22O11) in sugar–water syrup.

7 Effect of Temperature (T) & Composition (Co)
• Changing T can change # of phases: See path A to B. Changing Co can change # of phases: See path B to D. B (100°C,70) 1 phase D (100°C,90) 2 phases Source : A (20°C,70) 2 phases 70 80 100 60 40 20 Temperature (°C) Co =Composition (wt% sugar) L (liquid solution i.e, syrup) (liquid) + S (solid sugar) water- sugar system

8 ONE-COMPONENT (OR UNARY) PHASE DIAGRAMS/DIAGRAM FASA SISTEM SATU KOMPONEN
Komposisi konstan (pure substance) Ditentukan dalam tekanan P dan temperatur T tertentu.

9 ONE-COMPONENT (OR UNARY) PHASE DIAGRAMS/DIAGRAM FASA SISTEM SATU KOMPONEN
three different phases— solid, liquid, and vapor aO : equilibrium between solid and vapor phases bO : solid-liquid cO : liquid-vapor At 1 atm during heating the solid phase transform to the lquid phase (point 2) Liquid-vapor phase (point 3) And, finally, solid ice sublimes or vaporizes upon crossing the curve labeled aO.

10 Binary Phase Diagrams (Diagram Fasa Dua Komponen)
Variabel : temperatur, komposisi, dan tekanan Terdapat 3 daerah fasa yaitu : - daerah α - daerah L - α + L

11 Figure 9.3 (a) The copper–nickel phase diagram (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, Reprinted by permission of ASM International, Materials Park, OH.)

12 INTERPRETATION OF PHASE DIAGRAMS
three kinds of information are available in the binary system : the phases that are present the compositions of these phases the percentages or fractions of the phases

13 Determination of Phases Present
For example, an alloy of composition 60 wt% Ni–40 wt% Cu at 1100°C would be located at point A in Figure 9.3a;since this is within the α region, only the single phase will be present. On the other hand, a 35 wt% Ni–65 wt% Cu alloy at 1250°C (point B) will consist of both α and liquid phases at equilibrium.

14 Determination of Phases Present
wt% Ni 20 40 60 80 100 1000 1100 1200 1300 1400 1500 1600 T(ºC) L (liquid) a (FCC solid solution) L + liquidus solidus Cu-Ni phase diagram • Examples: A(1100ºC, 60 wt% Ni): 1 phase: a B (1250ºC,35) B (1250ºC, 35 wt% Ni): 2 phases: L + a A(1100ºC,60) Source :

15 Determination of Phase Compositions
• Examples: wt% Ni 20 1200 1300 T(ºC) L (liquid) a (solid) L + liquidus solidus 30 40 50 Cu-Ni system TA A Consider C0 = 35 wt% Ni tie line 35 C0 At TA = 1320ºC: Only Liquid (L) present CL = C0 ( = 35 wt% Ni) B TB 32 CL 4 C 3 At TD = 1190ºC: D TD Only Solid (a) present C = C0 ( = 35 wt% Ni) At TB = 1250ºC: Both  and L present Source : CL = C liquidus ( = 32 wt% Ni) C = C solidus ( = 43 wt% Ni)

16 Determination of Phase Amounts
If only one phase is present, the alloy is composed entirely of that phase; that is, the phase fraction is 1.0 or, alternatively, the percentage is 100%. If the composition and temperature position is located within a two-phase region  lever rule expression

17 lever rule expression 𝑊 𝐿 = 𝑆 𝑅+𝑆 𝑊 𝛼 = 𝑅 𝑅+𝑆
Lever rule expression for computation of liquid mass fraction Lever rule expression for computation of α-phase mass fraction 𝑊 𝐿 = 𝑆 𝑅+𝑆 𝑊 𝛼 = 𝑅 𝑅+𝑆 𝑊 𝐿 = 𝐶 𝛼 − 𝐶 0 𝐶 0 − 𝐶 𝐿 + 𝐶 𝛼 − 𝐶 0 𝑊 α = 𝐶 0 − 𝐶 𝐿 𝐶 0 − 𝐶 𝐿 + 𝐶 𝛼 − 𝐶 0 𝑊 𝐿 = 𝐶 𝛼 − 𝐶 0 𝐶 𝛼 − 𝐶 𝐿 𝑊 𝛼 = 𝐶 0 − 𝐶 𝐿 𝐶 𝛼 − 𝐶 𝐿

18 Determination of Phase Amounts
• Examples: wt% Ni 20 1200 1300 T(°C) L (liquid) a (solid) L + liquidus solidus 3 4 5 Cu-Ni system T A 35 C o 32 B D tie line R S C o = 35 wt% Ni At T A : Only Liquid (L) W L = 100 wt%, W a = 0 At T D : Only Solid ( a ) W L = 0, W = 100 wt% At T B : Both a and L = 27 wt% WL = S R + Wa Source :

19 Contoh Tentukan komposisi fasa α dan L paduan 35 wt% Ni – 65 wt% Cu pada suhu 1250°C.

20 Penyelesaian : Pada Fig. 9.3b diperoleh C0 = 35 wt% Ni, Cα = 42,5 wt% Ni dan CL = 31,5 wt% Ni 𝑊 𝐿 = 42,5−35 42,5−31,5 =0,68 𝑊 𝐿 = 𝐶 𝛼 − 𝐶 0 𝐶 𝛼 − 𝐶 𝐿 𝑊 𝛼 = 𝐶 0 − 𝐶 𝐿 𝐶 𝛼 − 𝐶 𝐿 𝑊 𝛼 = 35−31,5 42,5−31,5 =0,32

21 Ex: Cooling of a Cu-Ni Alloy
• Phase diagram: Cu-Ni system. T(ºC) L (liquid) L: 35wt%Ni Cu-Ni system • Consider microstuctural changes that accompany the cooling of a C0 = 35 wt% Ni alloy A (1300°C) B (1260°C) C 1250°C) D (1220°C) a: 46 wt% Ni L: 35 wt% Ni 130 a A + L B 46 35 C 43 32 a: 43 wt% Ni L: 32 wt% Ni D 24 36 L: 24 wt% Ni a: 36 wt% Ni a + 120 L E a (solid) 110 20 3 35 4 5 Source : C0 wt% Ni

22 BINARY EUTECTIC SYSTEMS
three single-phase regions : α,, and L A liquid phase tranformed into the two solid α and  phases is called eutectic reaction The eutectic reaction can be written as follows:

23 BINARY EUTECTIC SYSTEMS
CE : the eutectic composition TE : the eutectic temperature Figure 9.7 The copper–silver phase diagram. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), Reprinted by permission of ASM International, Materials Park, OH.]

24 EX 1: Pb-Sn Eutectic System
• For a 40 wt% Sn-60 wt% Pb alloy at 150ºC, determine: -- the phases present Pb-Sn system Answer: a + b L + a b 200 T(ºC) 18.3 C, wt% Sn 20 60 80 100 300 L (liquid) 183ºC 61.9 97.8 -- the phase compositions Answer: Ca = 11 wt% Sn Cb = 99 wt% Sn -- the relative amount of each phase Answer: 150 R S 11 C 40 C0 99 C W  = C - C0 C - C 59 88 = 0.67 S R+S W  = C0 - C C - C R R+S 29 88 = 0.33 Source :

25 Exercise Cite the phases that are present and the phase compositions for the following alloys: 15 wt% Sn–85 wt% Pb at 100°C

26 THE GIBBS PHASE RULE P + F = C + N P is the number of phases present
F is the number of degrees of freedom or the number of externally controlled variables (e.g., temperature, pressure, composition) C is the number of components in the system. N is the number of noncompositional variables (e.g., temperature and pressure).

27 Tekanan konstan ( 1 atm), N = 1 P + F = C + 1 jumlah komponen, C = 2 ( Cu and Ag) P + F = = 3 F = 3 – P jika ada satu fasa, maka P = 1 F = 3 – P F = 3 -1 = 2 jika ada 2 fasa maka F = 3 – P = 3 – 2 = 1 Jika ada 3 fasa, maka F = 3 – P = 3 – 3 = 0

28 Diagram fasa dasar dari sistem dua komponen
Larut sempurna dalam keadaan cair dan membentuk eutektik dengan kelarutan padat terbatas. Contoh Sistim paduan timah-timbal

29 Diagram fasa dasar dari sistem dua komponen
2. Larut sempurna dalam keadaan cair dan membentuk eutektik dengan komponen murni

30 3. Larut sempurna dalam keadaan cair dan dalam keadaan padat A dan B larut sempurna dalam keadaan cair dan dalam keadaan padat untukk setiap komposisi

31 4. Larut sempurna dalam keadaan cair dan mempunyai senyawa dengan titik maksimum
Cairan e1 = fasa padat A + fasa AmBn Cairan e2 = fasa AmBn + fasa padat B

32 5. larut sempurna dalam keadaan cair dan mempunyai senyawa dengan titik peritektik
Cairan c + kristal primer B (d)  senyawa AmBn

33 Larut sempurna dalam keadaan cair dan mempunyai peritektik dengan kelarutan terbatas
Kristal primer (d) + cairan c  larutan padat α(p)

34 7. Pemisahan dua fasa cairan
Cairan L2 (m)  cairan L1(c) + fasa padat (d)