phase diagrams

Post on 01-Feb-2016

73 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

DESCRIPTION

PHASE DIAGRAMS. STUDY OF PHASE RELATIONSHIPS IMPORTANT IN KNOWING PROPERTIES OF MATERIALS MAP OF TEMPERATURE, PRESSURE AND COMPOSITION BETWEEN PHASES IN EQUILIBRIUM IN A SYSTEM GIBBS PHASE RULE P + F = C + 2 Eg: states of matter- gas, liquid and solid – single phase - PowerPoint PPT Presentation

TRANSCRIPT

Element Groups (Families)

Alkali Earth Alkaline Earth Transition Metals

Rare Earth Other Metals Metalloids

Non-Metals Halogens Noble Gases

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1H He

-252.87 -268.6

2Li Be B C N O F Ne

1347 2970 2550 4827 -195.8 -183 -188.14 -246.1

3 Na Mg Al Si P S Cl Ar552.9 1107 2467 2355 280 444.6 -34.6 -186

4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr774 1484 2832 3287 3380 2672 1962 2750 2870 2732 2567 907 2403 2830 613 684.9 58.78 -153.4

5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe688 1384 3337 4377 4927 4612 4877 3900 3727 2927 2212 765 2000 2270 1750 989.8 184 -108.1

6Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn

678.4 1140 5400 5425 5660 5627 5027 4527 3827 2807 356.58 1457 1740 1560 962 337 -61.8

7Fr Ra ** Rf Db Sg Bh Hs Mt Uu

nUuu Uu

b

677 1737 ? ? ? ? ? ? ? ? ?

* La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

3469 3257 3127 3127 ? 1900 1597 3233 3041 2562 2720 2510 1727 1466 3315

** Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

3200 4790 ? 3818 3902 3235 2607 ? ? ? ? ? ? ? ?

PHASE DIAGRAMS• STUDY OF PHASE RELATIONSHIPS IMPORTANT IN KNOWING

PROPERTIES OF MATERIALS• MAP OF TEMPERATURE, PRESSURE AND COMPOSITION BETWEEN

PHASES IN EQUILIBRIUM IN A SYSTEM

GIBBS PHASE RULE

P + F = C + 2

Eg: states of matter- gas, liquid and solid – single phase Liquid mixture- oil and water- 2 phases In solid , several phases depending on crystal structure

• STUDY IMPORTANT IN ALLOYS• ALLOY- SUBSTANCE COMPOSED OF 2 OR MORE CHEMICAL

ELEMENTS

• MAIN CONSTITUENT- BASE METAL AND OTHERS ALLOYING ELEMENTS

Phase Rule

Temperature o C

0 100

Pressure

ICE

FUSION

WATER

WATER VAPOUR

SUBLIMATIONV

AP

OR

ISA

TIO

N

FUSION LINE ALMOST VERTICAL- VARIATION IN PRESSURE –NO EFFECT ON M.P. OF ICE

T

B

A

76cm

50

30 cm

• At ‘A’ , water vapour - 1 phase• At ‘B’ , water and water vapour co exist -2 phases• At ‘T’ , ice, water and water vapour exist – 3 phases• At ‘A’

• 1 + F = chemical compound H2O + 2

• F = 2 …. BIVARIANT• At ‘B’ • 2+ F = 1 +2, F = 1… UNIVARIANT• At ‘T’ • All three phases P = 3, 3 + F = 1 + 2; F = 0

INVARIANT

Equilibrium DiagramCase 1: Binary Alloy with COMPLETE SOLUBILITY IN BOTH

LIQUID AND SOLID PHASES in all compositions

Eg: Ag-Au Cu-Ni Ge-Si Al2O3-Cr2O3 Sb-Bi

Silver-Palladium Co-Ni Cu-Pt Fe-Pt Ni-Pt Ta-Ti

HUME ROTHERY’S RULE-

FICK’S LAWS OF DIFFUSION

FIRST LAW SECOND LAW

Elements A and B in a Binary Alloy

Cooling Curves & Phase Diagram

T1

Liquid

0 20 40 60 80 100

Composition, C (wt% B)

Tem

pe

ratu

re, T

(ºC

)

Liquidus curve

Solidus curve

Composition, C (% wt of B)

L + α

Phase (Equilibrium) Diagram

WAWB

Pa b

WA / WB = b/a

LEVER RULE

• With Fulcrum at P, weights WA and WB at the end of a lever, for equilibrium, the lever rule states:

P

XYLiquid + Solid

Liquid

Solid

16 37 58

For P: SS/LS = (37-16)/(58-37)= 1/1

P1

47.5

For P1: SS/LS = 31.5/ 10.5= 3/1

P

XYLiquid + Solid

Liquid

Solid

16 37 58

= (37-16)/(58-37)

P1

47.5

31.5/ 10.5= 3

The structures shown are at NON EQUILIBRIUM CONDITIONS

1453

1083

If fl and fs are the liquid and solid fractions,

There are Three variables, one of these can be chosen as independent

Case 2: Binary Alloy with COMPLETE SOLUBILITY IN LIQUID STATE in all

compositions, but COMPLETELY INSOLUBLE IN THE SOLID

STATE

• A very doubtful situation in practice, since most solid metals appear to dissolve small quantities of other metals

• In Bismuth-Cadmium, mutal solid solubility is negligible.

• Bi- heavy, brittle- positioned near to non metals in periodic table- Rhombic type structure-covalent bond

• Cadmium- HCP-

Bismuth- Cadmium Equilibrium Diagram

When two metals show complete solubility in liquid state, and complete insolubility in the solid state,they

do so by crystallising out as alternate layers of the two pure metals.

This laminated structure termed as EUTECTIC

40Cd/60Bi

Te

When two metals show complete solubility in liquid state, and complete insolubility in the solid state, they do so by crystallising out as alternate layers of the two pure metals.

This laminated structure termed as EUTECTIC

40Cd/60Bi

Te (EUTECTIC Temperature)

INVARIANT REACTION

At E, solid Cadmium (40%) and solid Bismuth(60%) co-exist EUTECTIC

A: Molten homogeneous alloy – 1 phase with 2 components, Bi and Cd1+F = 2 +1 (only temperature is the variable, not pressure) , F=2

•B: 2 + F = 2 + 1, F= 1

•C: 3 + F = 2 + 1, F=0

Eutectic is considered as

an intimate mixture of two metals

Phase Rule applied, P+F = C+ 1

3 + F = 2 +1, F = 0

Cooling curve for Eutectic (similar to pure metal)

For compositions to left /right of Eutectic

Te

mp

era

ture

o C

Time

HUME ROTHERY’S RULE

Gold- Silver, Copper- Nickel, Germanium- Silicon, Antimony- Bismuth,

Aluminium Oxide- Chromium Oxide etc. are examples

FICK’S LAWS OF DIFFUSION

Mass Flow Process by which atoms (molecules) change their positions relative to their neighbours in a given phase under the influence of thermal energy and gradient

:

FICK’S LAWS OF DIFFUSION

FIRST LAW

dn/dt = no. of moles of B atoms crossing per unit time

D= Diffusion coefficient

A= Planar area

dc/dx= concentration gradient

If J = flux flow / unit area per unit time,

SECOND LAW

If D is independent of concentration,

INVARIANT REACTIONS

Case 3: Two metals completely soluble in all proportions in liquid state, but partially

soluble in solid state

• Melting Point of Lead:3270C• Melting Point of Tin: 2320C• Eutectic Temperature: 1830C• Eutectic Composition: 62% Sn, 38%Pb• Max. solid solubility tin in lead at 1830C: 19.5% tin• Max. Solid solubility of lead in tin at 1830C: 2.6% lead• Eutectic of two solid solutions α and β (instead of two

metals) form

Melting Point of Tin (Pb) : 2320C Melting Point of Lead (Sn) :3270C Eutectic Temperature: 1830CEutectic Composition: 38%Pb, 62% Sn Max. solid solubility tin in lead at 1830C: 19.5% tinMax. Solid solubility of lead in tin at 1830C: 2.6% lead

Liquid solubility of salt in water & partial solid solubility of one metal in another- ( (similarity schematically represented)

top related