Equilibrium phase diagram

Pada umumnya logam tidak berdiri sendiri atau keadaan murni, tetapi lebih banyak dalam keadaan dipadu atau logam paduan dengan kandungan unsur-unsur tertentu sehingga struktur yang terdapat dalam keadaan setimbang pada temperatur dan tekanan tertentu akan berlainan.

Kombinasi dua unsur atau lebih yang membentuk paduan logam akan menghasilkan sifat yang berbeda dari logam asalnya.

Tujuan pemaduan = untuk memperbaiki sifat logam

Sifat yang diperbaiki adalah kekuatan, keuletan, kekerasan, ketahanan korosi, ketahanan aus, ketahanan lelah, dll.

1

2

Pada umumnya diagram fasa dibangun pada keadaan kesetimbangan (kondisinya adalah pendinginan yang sangat lambat). Diagram ini dipakai untuk mengetahui dan memprediksi banyak aspek terhadap sifat material.

Jenis pemaduan:1. Unsur logam + unsur logam Contoh: Cu + Zn; Cu + Al; Cu + Sn.2. Unsur logam + unsur non logam Contoh: Fe + C.

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Jika ditinjau dari posisi atom-atom yang larut, diperoleh dua jenis larutan padat:

1. Larutan padat substitusi

Adanya atom-atom terlarut yang menempati kedudukan atom-atom pelarut.

2. Larutan padat interstisi

Adanya atom-atom terlarut yang menempati rongga-rongga diantara kedudukan atom/sela antara.

Cu

Ni

FeC

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Pembentukan diagram fasa

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Garis liquidus = menunjukkan temperatur terendah dimana logam dalam keadaan cair atau temperatur dimana awal terjadinya pembekuan dari kondisi cair akibat proses pendinginan.

Garis solidus = menunjukkan temperatur tertinggi suatu logam dalam keadaan padat atau temperatur terendah dimana masih terdapat fasa cair.

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maximum solubility limit

The solubility of sugar (C12H22O11) in a sugar-water syrup.

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• Example:Phase Diagram of Water-Sugar System

Question: What is the solubility limit at 20°C?

• Solubility limit increases with T: e.g., if T = 100°C, solubility limit = 80wt% sugar

Pure

Sugar

Tem

pera

ture

(°C

)0 20 40 60 80 100

Co=Composition (wt% sugar)

L (liquid solution

i.e., syrup)

Solubility Limit L

(liquid)

+ S

(solid sugar)

65

20

40

60

80

100

Pure

W

ate

r

The Solubility Limit

Answer: 65wt% sugar If Co < 65wt% sugar:

If Co > 65wt% sugar:

syrupsyrup + sugar

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(a)

FIG. 3-50 (1) Heat pure metal to point FIG. 3-50 (1) Heat pure metal to point TTaa; (2) cooling of liquid metal ; (2) cooling of liquid metal a – ba – b; (3) at ; (3) at

point point bb, pure metal starts to , pure metal starts to precipitateprecipitate out of solution; (4) point out of solution; (4) point cc, pure metal , pure metal completely solid; curve from completely solid; curve from bb to to c c straight horizontal line showing constant straight horizontal line showing constant temperature temperature TTb-c b-c because thermal energy absorbed in change from liquid to solid; (5) because thermal energy absorbed in change from liquid to solid; (5)

more cooling of solid pure metal from more cooling of solid pure metal from c c to to d d and temperature begins to fall again. and temperature begins to fall again.

Cooling Curve for Pure Metal

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FIG. 3-50 (b) Cooling curve for pure iron. FIG. 3-50 (b) Cooling curve for pure iron.

(b)

Cooling Curve for Pure Iron

10FIG. 3-54 Allotropic forms of iron (three phases: bcc, fcc, bcc)FIG. 3-54 Allotropic forms of iron (three phases: bcc, fcc, bcc)

Allotropic Forms of Iron

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Cooling Curve for a Metal Alloy

(c)

FIG. 3-50 (c) Cooling curve for a metal alloy: (1) The alloy A-B heated to point FIG. 3-50 (c) Cooling curve for a metal alloy: (1) The alloy A-B heated to point aa (liquid phase, with both metals soluble in each other); (2) cooling of alloy(liquid phase, with both metals soluble in each other); (2) cooling of alloy in liquid in liquid phase; (3) point phase; (3) point bb, solidification begins; (4) point , solidification begins; (4) point cc, solidification complete; sloped , solidification complete; sloped b – c b – c due to changing from liquid to solid over the temperature range due to changing from liquid to solid over the temperature range TTbb toto T Tc c

because components because components A A and and BB have different melting/cooling temperatures; (5) have different melting/cooling temperatures; (5) further cooling from further cooling from c c to to d d of solid-state metal alloy.of solid-state metal alloy.

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Klasifikasi Diagram Kesetimbangan Fasa (Cair-Padat)

1. Larut sempurna dalam keadaan cair dan padat.

2. Larut sempurna dalam keadaan cair, tidak larut dalam keadaan padat (reaksi eutektik).

3. Larut sempurna dalam keadaan cair, larut sebagian dalam keadaan padat (reaksi eutektik).

4. Larut sempurna dalam keadaan cair, larut sebagian dalam keadaan padat (reaksi peritektik).

5. Larut sempurna dalam keadaan cair, tidak larut dalam keadaan padat dan membentuk senyawa.

6. Larut sebagian dalam keadaan cair (reaksi monotektik).

7. Tidak larut dalam keadaan cair maupun padat.

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

• 3 phase fields: L L +

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

• 2 phases:– L (liquid)– (FCC

solid solution)

• 2 lines (phase boundaries):– The liquidus

line (L/L+)– The solidus

line (/L+)

• 3 phase fields:– L– L + –

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• example:

A (1100°C, 60wt% Ni):

1 phase:

B (1250°C, 35wt% Ni):

2 phases: L +

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

L +

liquidus

solid

us

A(1100,60)

B(1

250,3

5)

Rules of Determining Number & Types of Phases

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The lever arm rule

x adalah komposisi paduan yang akan dihitung persentase fasa-fasanya pada temperatur T, maka tarik garis yang memotong batas kelarutannya (garis L-S).

Jika x = wo; L = wl dan S = ws

maka % fasa cair dan padat :

%100xww

wwL

ls

os

%100xww

wwS

ls

lo

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wt% Ni20

1200

1300

T(°C)

L (liquid)

(solid)

30 40 50

TAA

DTD

TBB

tie line

433532CoCL C

• contoh: C0 = 35 wt%Ni

At TA:Only Liquid (L)CL = C0 = 35 wt%Ni

At TD:Only Solid ()C = C0 = 35 wt%Ni

At TB:Both and LCL = CLiquidus = 32 wt%Ni

C = CSolidus = 43 wt%Ni

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500

1600T(°C)

L (liquid)

(FCC solid solution)

A(1100,60)

B(1

250,3

5)

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%7,72

%1003243

3543

L

xL

%3,27

%1003243

3235

S

xS

Contoh lain: pada wo= 53% Ni

wl (32%) wo(35%) ws(43%)

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% fasa cair dan padat:

%38

%1004558

5358

L

xL

%62

%1004558

4553

S

xS

wl (45%) wo(53%) ws(58%)

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For the alloys listed below:

60 wt% Ni-40 wt% Cu at 1100°C

35 wt% Ni-65 wt% Cu at 1250°C

(1) Phase(s) that are present

(2) The composition of each phase

Example: Determine the phase(s) that are present Example: Determine the phase(s) that are present and the composition of the phase(s)and the composition of the phase(s)

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60 wt% Ni-40 wt% Cu at 1100°C

(L)(1) Determine the phase(s) that are present

Point A:

phase

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60 wt% Ni-40 wt% Cu at 1100°C (Point A):

(2) Determine the composition of each phase

C = C0 = 60 wt% Ni

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35 wt% Ni-65 wt% Cu at 1250°C

(L)(1) Determine the phase(s) that are present

Point B

+ L phases

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35 wt% Ni-65 wt% Cu at 1250°C (Point B):

+ L

(2) Determine the composition of each phase

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• 35 wt% Ni-65 wt% Cu at 1250°C (Point B): in two phase ( + L) region

(2) Determine the composition of each phase

42.53531.5

Tie Line

Composition (wt% Ni)CL C0 C

Draw a tie line

Composition of L: intersection /+L — CL = 31.5 wt% Ni

Composition of a: intersection L/+L — C = 42.5wt% Ni

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• ConsiderCo = 35wt%Ni

• Upon cooling– L

35wt% 32wt% 24wt%

– 46wt% 43wt% 36wt%

– Equilibrium cooling

Sufficiently slow cooling rate gives enough time for composition readjustments

wt% Ni20

1200

1300

30 40 501100

L (liquid)

(solid)

L +

L +

T(°C)

A

D

B

35Co

L: 35wt%Ni

: 46wt%Ni

C

E

L: 35wt%Ni

464332

24

35

36: 43wt%Ni

L: 32wt%Ni

L: 24wt%Ni

: 36wt%Ni

Equilibrium Cooling in a Cu-Ni Binary System