Sep-16

1

Phase Diagrams & Phase 

Tranformation

Microstructure - Phases

Ferrite

Plain C steel containing 0.44 wt. % C

Cementite

SEM micrograph

3000 x magnification

Basic Definitions• Alloy: A metallic substance that is composed of two or more elements.

• Component: A chemical constituent (element or compound) of an alloy, which may be used to specify its composition.

• Phase: A homogeneous portion of a system that has uniform physical and chemical characteristics.

• Equilibrium: The state of a system where the phase characteristics remain constant over indefinite time periods. 

– At equilibrium the free energy is a minimum.

• What equilibrium state do we get?

• If we specify...--a composition (e.g., wt% Cu - wt% Ni), and--a temperature (T )

then...How many phases do we get?

When we combine two elements 

y p gWhat is the composition of each phase?How much of each phase do we get?

Phase BPhase A

Nickel atomCopper atom

Phase Diagram

• Three externally controllable parameters that affect phase fraction and composition• Temperature • Pressure • CompositionComposition

• Phase diagram is constructed when various combinations of these parameters are plotted against one another

Unary (One component) phase diagram

Sep-16

2

Phase Equilibria: Solubility LimitIntroduction

– Solutions – solid solutions, single phase

– Mixtures – more than one phase

• Solubility Limit:

Max concentration forhi h l i l h

Sucrose/Water Phase Diagram

Solubility L

100

which only a single phase solution occurs.

Question: What is thesolubility limit at 20°C?

Answer: 65 wt% sugar.If Co < 65 wt% sugar: syrupIf Co > 65 wt% sugar: syrup + sugar.

65

Pur

e S

ugar

Tem

pe

ratu

re (

°C)

0 20 40 60 80 100Co =Composition (wt% sugar)

L

(liquid solution i.e., syrup)

yLimit L

(liquid) +

S

(solid sugar)20

40

60

80

Pur

e W

ater

Effect of T & Composition (Co)• Changing T can change # of phases:

D (100°C,90)2 phases

B (100°C,70)1 phase

path A to B.

• Changing Co can change # of phases: path B to D.

C)

100

80 L

(liquid)

A (20°C,70)2 phases

70 80 1006040200

Tem

pera

ture

(°C

Co =Composition (wt% sugar)

L

(liquid solution i.e., syrup)

20

40

60

0

(liquid) +

S

(solid sugar)

water-sugarsystem

Phase Equilibria

CrystalStructure

electroneg r (nm)

Ni FCC 1.9 0.1246

Simple solution system (e.g., Ni-Cu solution)

Cu FCC 1.8 0.1278

• Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume –Rothery rules) suggesting high mutual solubility.

• Ni and Cu are totally miscible in all proportions.

Phase Diagrams• Indicate phases as function of T, Co, and P. • For this course:

-binary systems: just 2 components.-independent variables: T and Co (P = 1 atm is almost always used).

• PhaseDiagram

• 2 phases:L (liquid)

1 00

1600

T(°C)

Diagramfor Cu-Nisystem

( q ) (FCC solid solution)

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

1500 L (liquid)

(FCC solid

solution)

1500

1600

T(°C)

L (liquid)

Cu-Ni

Phase Diagrams:# and types of phases

• Rule 1: If we know T and Co, then we know:--the # and types of phases present.

• Examples:

A(1100°C, 60): 1 phase:

C,3

5)

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

(FCC solid solution)

phasediagram

B(1250°C, 35): 2 phases: L +

B(1

250

° C

A(1100°C,60)

1300

T(°C)

L (liquid)

Cu-Ni system

Phase Diagrams:composition of phases

• Rule 2: If we know T and Co, then we know:--the composition of each phase.

• Examples:

TAA

At T A = 1320°C:

Only Liquid (L)

Co = 35 wt% Ni

BTB

tie line

wt% Ni20

1200

(solid)

30 40 5035Co

32CL

CL = Co ( = 35 wt% Ni)

At T B = 1250°C: Both and L CL = Cliquidus ( = 32 wt% Ni here)

C = Csolidus ( = 43 wt% Ni here)

At T D = 1190°C: Only Solid ( ) C = Co ( = 35 wt% Ni)

TB

DTD

4C3

Sep-16

3

Gibbs Phase Rule

F=C-P+2

Degrees of F d

Number of Number of Freedom components phases

How many number of intensive variables we have to specify to completely define/ specify the system

How many number of intensive variables thatcan be independently changed without changing the number of phases

Intensive variables of the system:• Temperature of each phase• Pressure of each phase• Composition of each phase = Relative concentration of

each component in that phase

Temperature of each phase is equal and is equal to the temperature of the system

Pressure of each phase is equal and is equal to the pressure of the system

To completely define the system means: specifying temperature, pressure and relative concentrations of each component in each phase of the system

Application of Gibbs Phase Rule: Unary phase diagram

F = ?

Gibbs Phase Rulewhen Pressure is constant

F=C-P+2F=C-P+1

If Pressure is fixed at a particular value

Degrees of Freedom

Number of components

Number of phases

a particular value

Phase diagrams for metals and ceramics are drawn at 1 atm. fixed pressure

F = ?

F=2-2+1=1Either T or wl or ws

Can vary only one of these without changing the

F=2-1+1=2Both T and wl (or, ws) can be varied without changing the number of phases

F=1-1+1=1Only T

changing the number of phases

F=1-2+1=0Invariant point

• Rule 3: If we know T and Co, then we know:--the amount of each phase (given in wt%).

• Examples:

At T A: Only Liquid (L)

WL = 100 wt%, W = 0

Co = 35 wt% Ni

Phase Diagrams:weight fractions of phases

1300

T(°C)

L (liquid)

Cu-Ni system

TAA

B

tie line

LAt T D: Only Solid ( )

WL = 0, W = 100 wt%

wt% Ni20

1200

(solid)

30 40 5035Co

32CL

BTB

DTD

4C3

R S

At T B: Both and L

wt% 733243

3543

= 27 wt%

WL S

R + S

W R

R + S

Sep-16

4

1500

1600

T(°C)

L (liquid)

Cu-Ni

Phase Diagrams:# and types of phases

• Rule 1: If we know T and Co, then we know:--the # and types of phases present.

• Examples:

A(1100°C, 60): 1 phase:

C,3

5)

wt% Ni20 40 60 80 10001000

1100

1200

1300

1400

(FCC solid solution)

phasediagram

B(1250°C, 35): 2 phases: L +

B(1

25

0° C

A(1100°C,60)

1300

L (liquid)T(°C)

A

L: 35wt%Ni

Cu-Nisystem

• Phase diagram:

Cu-Ni system.

• Consider

Co = 35 wt%Ni.

Ex:  Cooling in a Cu‐Ni Binary

4635

4332

L 32 t% Ni

B: 46 wt% NiL: 35 wt% Ni

C

D

wt% Ni20

1200

30 40 50110 0

(solid)

35Co

: 43 wt% Ni

L: 32 wt% Ni

L: 24 wt% Ni

: 36 wt% Ni

D

E

24 36

Non‐equilibrium/ Fast Cooling

• C changes as we solidify.• Cu-Ni case:

• Fast rate of cooling:

Cored structure• Slow rate of cooling:

Equilibrium structure

First to solidify has C = 46 wt% Ni.Last to solidify has C = 35 wt% Ni.

Cored vs Equilibrium Phases

Fi t t lidifUniform C:

First to solidify: 46 wt% Ni 35 wt% Ni

Last to solidify: < 35 wt% Ni

2 componentshas a special compositionwith a min. melting T.

Binary‐Eutectic Systems

• 3 single phase regions (L, )

• Limited solubility:

Ex.: Cu-Ag system

Cu-Agsystem

L (liquid)

L + L+

1200

T(°C)

800

1000

T 779°C

: Min. melting TE

• Eutectic transition

L(CE) (CE) + (CE)

: mostly Cu : mostly Ag

• TE : No liquid below TE

• CE

composition

Co , wt% Ag20 40 60 80 1000

200

400

600

800

CE

TE 8.0 71.9 91.2779 C

L + L+200

T(°C)

300 L (liquid)

183°C

• For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...

--the phases present: Pb-Snsystem

EX:  Pb‐Sn Eutectic System (1)

+ --compositions of phases:

CO = 40 wt% Sn

--the relative amount

C = 11 wt% SnC = 99 wt% Sn

L+

+

20018.3

C, wt% Sn20 60 80 1000

100

183 C61.9 97.8

of each phase:

150

40Co

11C

99C

SRW=C - CO

C - C

=99 - 4099 - 11

= 5988

= 67 wt%

SR+S

=

W =CO - C

C - C=R

R+S

=2988

= 33 wt%=40 - 1199 - 11

Sep-16

5

L+200

T(°C)

300 L (liquid)

L +

• For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find...

--the phases present: Pb-Snsystem

EX:  Pb‐Sn Eutectic System (2)

+ L--compositions of phases:

CO = 40 wt% Sn

--the relative amount 220SR

C = 17 wt% SnCL = 46 wt% Sn

L+

+

200

C, wt% Sn20 60 80 1000

100

183°Cof each phase:

W=CL - CO

CL - C=

46 - 40

46 - 17

=6

29= 21 wt%

WL =CO - C

CL - C=

23

29= 79 wt%

40Co

46CL

17C

SR

• Co < 2 wt% Sn• Result:

--at extreme ends--polycrystal of grains

i.e., only one solid phase.

Microstructures in Eutectic Systems: I

L+

T(°C)

300 L

400

L

L: Co wt% Sn

0

L+ 200

Co, wt% Sn10

2

20Co

100

30

+

(room T solubility limit)

TE

(Pb-SnSystem)

: Co wt% Sn

• 2 wt% Sn < Co < 18.3 wt% Sn• Result:

Initially liquid + then alone

finally two phases

Microstructures in Eutectic Systems: II

L +

T(°C)

300

L

400

L

L: Co wt% Sn

: C wt% Sn polycrystal fine -phase inclusions

Pb-Snsystem

200

Co , wt% Sn10

18.3

200Co

100

30

+

(sol. limit at TE)

TE

2(sol. limit at Troom)

: Co wt% Sn

• Co = CE• Result: Eutectic microstructure (lamellar structure)

--alternating layers (lamellae) of and crystals.

Microstructures in Eutectic Systems: III

Micrograph of Pb-Sn eutectic microstructure

T(°C)

300 LL: Co wt% Sn

160m

Pb-Snsystem

L

200

C, wt% Sn

20 60 80 1000

100

L +

183°C

40

TE

18.3

: 18.3 wt%Sn

97.8

: 97.8 wt% Sn

CE61.9

Lamellar Eutectic Structure

29

Eutectic Phase Diagram

Apply Gibbs Phase Rule

Sep-16

6

• If the Liquid composition is Euctectic composition• Result: Eutectic microstructure (lamellar structure)

--alternating layers (lamellae) of and crystals.

Microstructures in Eutectic Systems: III

Micrograph of Pb-Sn eutectic microstructure

T(°C)

300 LL: Co wt% Sn

160m

Pb-Snsystem

L

200

C, wt% Sn

20 60 80 1000

100

L +

183°C

40

TE

18.3

: 18.3 wt%Sn

97.8

: 97.8 wt% Sn

CE61.9

• 18.3 wt% Sn < Co < 61.9 wt% Sn

• Result: crystals and a eutectic microstructure

Microstructures in Eutectic Systems: IV

C = 18.3 wt% Sn

CL = 61.9 wt% SnS

• Just above TE :

Pb Sn

T(°C)

300 L

L: Co wt% Sn LL

WL = (1- W) = 50 wt%

SR + S

W= = 50 wt%

• Just below TE :C = 18.3 wt% Sn

C = 97.8 wt% SnS

R + SW= = 73 wt%

W = 27 wt%

Pb-Snsystem

18.3 61.9

SR

97.8

SR

primary eutectic

eutectic

L+200

Co, wt% Sn

20 60 80 1000

100

L+

40

+

TE

L+L+

+

200

Co, wt% Sn20 60 80 1000

300

100

L

TE

40

(Pb-Sn System)

Hypoeutectic & Hypereutectic

T(°C)

Co, wt% Sn20 60 80 1000 40

160 m

eutectic micro-constituent

hypereutectic: (illustration only)

175 m

hypoeutectic: Co = 50 wt% Sn 61.9eutectic

eutectic: Co =61.9 wt% Sn

You can draw Eutectic phase diagram stating from the eutectic reaction

Liquid

α

β

α+β

β

Other Phase Diagrams

Eutectoid

P it ti

+

Peritectic

Peritectoid α+

+ L

Peritectoid Phase diagram

Sep-16

7

Peritectic and Peritectoid Phase Diagrams

PeritectoidPeritectic

Eutectoid & Peritectic

Cu‐Zn Phase diagram

Peritectic transition + L

Eutectoid transition +

Eutectic

Eutectoid

Single Phase

Two solid Phases

Peritectic

PeritectoidSingle Phasesolid

Two Phases

Intermediate/ Intermetallic Compounds

• Not isomorphous with either of the components of the alloy

Mg2Pb

Note: intermetallic compound forms a line - not an area -because stoichiometry (i.e. composition) is exact.

of the alloy system

• Congruently melting intermediate phases Intermetallic phase

Cu-Zn : Intermediate Phases Fe-Cementite Phase Diagram

Sep-16

8

Few Principles about Phase Diagrams1. One phase regions may touch each other only at single points 

(point of congruent transformation), never along a boundary 

2. Adjacent one phase regions are seperated from each other by 2 phase regions involving the same 2 phases

3. Three 2 phase regions must originate upon three phase isotherm

4. Two three phase isotherms may be connected by a 2 phase region provided that there are 2 phases which are common to both of the three phase equilibria

5. All boundaries of 2 phase fields must project into 2 phase fields when they join a three phase isotherm

Fe-Cementite Phase Diagram

Iron‐Carbon (Fe‐C) Phase Diagram• 2 important

points

-Eutectoid (B):

+Fe3C

-Eutectic (A):

L +Fe3C

T(°C)

entit

e)

1600

1400

1200

1000

L

(austenite)

+L

+Fe3C

L+Fe3C

1148°CA

SR

Result: Pearlite = alternating layers of and Fe3C phases

120 m

(From Fig. 7.27, Callister Adapted Version.)

Fe 3

C (

cem

e

800

600

4000 1 2 3 4 5 6 6.7

Fe3C

+Fe3C

(Fe) Co, wt% C

727°C = Teutectoid

4.30

R S

0.76

Ceu

tect

oid

B

Fe3C (cementite-hard) (ferrite-soft)

Growth of Pearlite colony

Hypoeutectoid Steel

emen

tite)

1600

1400

1200

1000

L

(austenite)

+L

+ Fe3C

L+Fe3C

1148°C

T(°C)

(Fe-C System)

Fe 3

C (

ce800

600

4000 1 2 3 4 5 6 6.7

+ Fe3C

(Fe) Co, wt% C

727°C

C0

0.76

proeutectoid ferritepearlite

100 m Hypoeutectoidsteel

R S

w =S/(R+S)wFe3

C

=(1-w)

wpearlite = wpearlite

r s

w =s/(r+s)w =(1- w)

Hypereutectoid Steel

emen

tite)

1600

1400

1200

1000

L

(austenite)

+L

+Fe3C

L+Fe3C

1148°C

T(°C)

(Fe-C System)

srFe3C

Fe 3

C (

ce800

600

4000 1 2 3 4 5 6 6.7

+Fe3C

(Fe) Co, wt%C

0.76

Co

proeutectoid Fe3C

60 mHypereutectoid steel

pearlite

R S

w =S/(R+S)wFe3C =(1-w)

wpearlite = wpearlite

sr

wFe3C =r/(r+s)w =(1-w Fe3C)

3

Sep-16

9

Class Work/ Class TestDo it now on your own notebook

For an alloy, Fe‐0.40 wt% C at a temperature just below the eutectoid, determine the following

a) composition of Fe3C and ferrite ()b) h f bid ( i ) ib) the amount of carbide (cementite) in grams 

that forms per 100 g of steel

c) the amount of pearlite and proeutectoidferrite ()

Fe-Cementite Phase Diagram

Phase EquilibriaSolution:

100xCFe3

CCo

b) the amount of carbide (cementite) in grams that forms per 100 g of steel

a) composition of Fe3C and ferrite ()

CO = 0.40 wt% CC = 0.022 wt% CCFe C = 6.70 wt% C

3

1600

1400 L

T(°C)

g 3.94

g 5.7 CFe

g7.5100 022.07.6

022.04.0

100xCFe

3

CFe3 3

x

CC

Fe 3

C (

cem

entit

e)1200

1000

800

600

4000 1 2 3 4 5 6 6.7

(austenite)

+L

+ Fe3C

+ Fe3C

L+Fe3C

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CFe C3C

Phase Equilibriac. the amount of pearlite and proeutectoid ferrite () 

Amount of pearlite = amount of  just above TECo = 0.40 wt% CC = 0.022 wt% CCpearlite = C = 0.76 wt% C 1600

1400 L

T(°C)

Co CC C

x 100 51.2 g

pearlite = 51.2 gproeutectoid = 48.8 g

Fe 3

C (

cem

entit

e)1200

1000

800

600

4000 1 2 3 4 5 6 6.7

(austenite)

+L

+ Fe3C

+ Fe3C

L+Fe3C

Co, wt% C

1148°C

T(°C)

727°C

CO

R S

CC

Hypoeutectoid Steel

emen

tite)

1600

1400

1200

1000

L

(austenite)

+L

+ Fe3C

L+Fe3C

1148°C

T(°C)

(Fe-C System)

Recap of µstructure evolution in steel

Fe 3

C (

ce800

600

4000 1 2 3 4 5 6 6.7

+ Fe3C

(Fe) Co, wt% C

727°C

C0

0.76

proeutectoid ferritepearlite

100 m Hypoeutectoidsteel

R S

w =S/(R+S)wFe3

C

=(1-w)

wpearlite = wpearlite

r s

w =s/(r+s)w =(1- w)

Hypereutectoid Steel

emen

tite)

1600

1400

1200

1000

L

(austenite)

+L

+Fe3C

L+Fe3C

1148°C

T(°C)

(Fe-C System)

srFe3C

Recap of µstructure evolution in steel

Fe 3

C (

ce800

600

4000 1 2 3 4 5 6 6.7

+Fe3C

(Fe) Co, wt%C

0.76

Co

proeutectoid Fe3C

60 mHypereutectoid steel

pearlite

R S

w =S/(R+S)wFe3C =(1-w)

wpearlite = wpearlite

sr

wFe3C =r/(r+s)w =(1-w Fe3C)

3

Sep-16

10

Phase Transformation

Phase transformations (change of the microstructure) can be divided into three categories:

• Diffusion-dependent with no change in phase composition or number of phases present (e.g. melting, solidification of pure metal, allotropic transformations, recrystallization, etc.)

• Diffusion-dependent with changes in phase compositions and/or number of phases (e.g. eutectic or eutectoid transformations)transformations)

• Diffusionless phase transformation - by cooperative small displacements of all atoms in structure, (e.g. martensitictransformation )

Diffusion-dependent phase transformations can be rather slow and the final structure often depend on the rate of cooling/heating.

A phase transformation occurs spontaneously only when G decreases in the p f p y ycourse of the transformation

The formation of a solid nucleus leads to a Gibbs free energy change of

SLvSLSLLv

SvS rGrAGGVG 23 4

3

4)(

Free energy change per unit volume. Below Tm, it is -ve

Sep-16

11

Homogeneous Nucleation & Energy Effects

GT = Total Free Energy

Surface Free Energy- Energy needed to make an interface)

SLS rG 24

= surface tension

r* = critical nucleus: nuclei < r* shrink; nuclei>r* grow (to reduce energy)

T gy= GS + GV

Volume (Bulk) Free Energy –releases of energy due to formation of solid

vV GrG 3

3

4

at r=r*, 0r

G

Homogeneous nucleation

SLvSLSLLv

SvS rGrAGGVG 23 4

3

4)(

at r=r*, 0r

G

v

SL

Gr

2*

23

*

3

16

v

SL

GG

TLTH

m

v

m

vv T

TL

T

THG

ngUndercooliTTT m

0, vvv LHGHv=Lv is latent heat of solidification

TH

T

Gr

v

mSL

v

SL

122*

Therefore,

22

23

2

3* 1

3

16

3

16

TH

T

GG

v

mSL

v

SL

Both r* and G* decrease with increasing undercooling

TH

T

Gr

v

mSL

v

SL

122*

22

23

2

3* 1

3

16

3

16

TH

T

GG

v

mSL

v

SL

Both r* and G* decrease with increasing undercooling

SLstrainvstrainSSLSLvS rGGrGVAGVG 23 43

4

strainv

SL

GGr

2*

3* 16

Homogeneous nucleation in solid

2

*

3

16

strainv

SL

GGG

Both r* and G* increase because of strain energy involved in solid state transformation

Rate of Homogeneous nucleation

As T T G*

exp(-(G*)/kT)

2* 1

TG

TTT m

exp( (G )/kT)

d

Rate of Homogeneous nucleation

Sep-16

12

Heterogeneous nucleation

SMSM

SSLL

MLML

MLSMSMSMSLSLvS AAAGVG

cosSLMLSM

Heterogeneous nucleation

SMSM

SSLL

MLML

MLSMSMSMSLSLvS AAAGVG

cosSLMLSM cos12 2 rASL

SrVS3

3

4

SrGrG SLv

23 4

3

4

Heterogeneous nucleation

SMSM

SSLL

MLML

SrGrG SLvhet

23 4

3

4 SLvhet

3

Solidification: Nucleation Processes

• Homogeneous nucleation• Homogeneous nucleation– nuclei form in the bulk of liquid metal

– requires supercooling (typically 80-300°C max)

• Heterogeneous nucleation– Nuclei form on mold wall or inoculants– allows solidification with only 0.1-10ºC supercooling

Nucleation rate

0.5

0.75

1

hete

ro/

G* ho

mo

→

G*hetero (0o) = 0

no barrier to nucleation

G*hetero (90o) = G*

homo/2

G*hetero (180o) = G*

homo

no benefit

Variation of S() with

0

0.25

0 30 60 90 120 150 180 (degrees) →

G

* h

Complete wetting No wettingPartial wetting

SL

SMMLCos

Sep-16

13

kT

G

eII

*homo

0homohomo

kT

G

eII

*hetero

0heterohetero

= f(number of nucleation sites) = f(number of nucleation sites)

Nucleation rate: Homogeneous vs. Heterogeneous

~ 1042 ~ 1026

BUTthe exponential term dominates

Ihetero > Ihomo

Rate of Phase Transformation(K

) →

99% = finish

Increasing % transformation

Time – Temperature – Transformation (TTT) diagram → phase transformation

t (sec) →

T

1% = start

Sep-16

14

Transformations & Undercooling

• For transformation to occur, must cool to below 727°C

• Eutectoid transformation (Fe-Fe3C system): + Fe3C

0.76 wt% C0.022 wt% C

6.7 wt% C

1600

1400 L

L

T(°C)

Fe 3

C (

cem

entit

e)

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

(austenite)

+L

+Fe3C

+Fe3C

L+Fe3C

(Fe) C, wt% C

1148°C

ferrite

727°C

Eutectoid:Equil. Cooling: Ttransf. = 727ºC

T

Undercooling by Ttransf. < 727C

0.76

0.02

2

Isothermal Transformation Diagrams2 solid curves are plotted:  one represents the time required at each temperature for the start of the transformation;

the other is for transformation completion.

The dashed curve corresponds to 50% completion.

The austenite to pearlite transformation will occur only if the alloy is supercooled to below the eutectoid temperature (727˚C).

Time for process to complete depends on the temperature.

Next class onwards Prof. Sumantra Mandalwill teach at the same class timings

Solidification of pure metal:• Cooling curve• Concept of supercooling• Homogeneous and heterogeneous nucleationheterogeneous nucleation processes, 

• Microstructure of pure metals

Cooling Curve for pure metal Dendritic structureDevelopment of Thermal Dendrites

a) Spherical Nucleus

b) The interface becomes unstable

c) Primary arms develop in crystallographic directions <100> in cubic crystals

d) Secondary and tertiary arms develop

Sep-16

15

Solidification

heat

Grains can be ‐ equiaxed   (roughly same size in all directions)

‐ columnar   (elongated grains)

~ 8 cm

Columnar in area with less undercooling

Shell of equiaxed grains due to rapid cooling (greater T) near wall

flow

Typical Cast Structure

Fine grain

Columnar grain

Equiaxed grain

Cast Grain Structure

Schematic Diagram

Cast Microstructure

Typical Cast Structure

Fine grain

Columnar grain

Equiaxed grain