Termodinamika Metalurgi (MG-2112)

Teknik Metalurgi

MG-2211 Metallurgical ThermodynamicDr.-Ing. Zulfiadi Zulhan 2010

15. Phase Diagram

Dr.-Ing. Zulfiadi Zulhan

Teknik Metalurgi

Fakultas Teknik Pertambangan dan Perminyakan

Institut Teknologi Bandung

INDONESIA

Termodinamika Metalurgi (MG-2112)

Semester 3 – 2017/2018

Teknik Metalurgi

Dr.-Ing. Zulfiadi Zulhan 2017 MG-2112 Termodinamika Metalurgi

Course Content

1. Introduction and Definition of Terms

2. The First Law of Thermodynamics

3. The Second Law of Thermodynamics

4. Property Relationships

5. Equilibrium

6. Chemical Equilibrium and Ellingham Diagrams

7. Electrochemistry and Pourbaix Diagrams

8. Mid Exam

9. Ion Activity

10. Solutions

11. Gibbs-Duhem Equation and Application

12. Application of Electrochemical Methods to Calculate of Thermodynamics

Properties

13. Alternative Standard States

14. Activity Coefficient in Dilute Solutions Multi Components

15. Phase Diagram

16. Final Exam


Phases

A phase is a portion of matter that is uniform throughout, not only in

chemical but also in physical state.

Example 1: a mixture of ice and pure water → two phases mixture: solid

(ice) and liquid (water).

Example 2: ice – brine mixture exists in equilbrium with air (containing

water vapour) → system have three phases: ice, brine, and gas phase

containing water vapour and air (oxygen, nitrogen and other gases in

small quantities).


Phase Diagram Water


Components (Binary System)

One of the factors that determines the number of phases which may exist

in equilibrium is the number of components in a system.

The number of components (C) is defined as the number of chemical

species (N) less the number of indenpendent relationship among them

(R).

C = N - R

In binary solution, solution has two components.

Solution is made of two chemical entities/species (N=2). There is no

relations between two chemical species (R=0):

C = N - R = 2 – 0 = 2


Components (Binary System)

If now, two chemical entities A and B react to form a compound AB. In this

case, there are three chemical entities: A, B, AB (N = 3).

There is an equilibrium constant for the reaction A + B → AB, which relates

the thermodynamic activities of the three chemical entities at a given

temperature (R=1).

Number of component

C = N – R = 3 – 1 = 2

If there are multiple compound in A-B system, for each additional

compound, N increases → equilibrium contant increases → R increases.

C = N – R = 2


Specifying A System

Suppose that a system of interest has C components.

To determine the composition of a phase in the system, C-1 pieces of

information about that phase must be specified.

Example: for three-component system, we can specify composition of

phase A by establishing the mole fractions of components 1 and 2 in the

phase. Mole fraction of component 3 will be determined, because the sum

of the mole fractions must be one.

The number of variables to be specified (VAR) is the number of phases (P)

multiplied by (C-1) plus two overall system variables such as temperature

and total pressure of the system:

VAR = P (C-1) + 2


Equilibrium Conditions

For a system to be in equilibrium, the chemical potential (partial molar

Gibbs free energy) of a component must be the same throughout the

system.

If the system contains P phases, it yields P-1 independent equations:

P phase in i component of potential chemical is where

...

p

i

p

i

4

i

3

i

2

i

1

i

=====

p

i

1p

i

4

i

3

i

3

i

2

i

2

i

1

i ..., , , , ==== −

If the system contains C components, total number of relationship:

REL = C (P-1)


Gibbs Phase Rule

The degree of freedom available in system (F) is the difference between

the number of variables required to specify the system (VAR) and the

number of relationships required by the equilibrium condition (REL)

F = VAR – REL

F = P(C-1) + 2 – C(P-1)

P + F = C + 2

F = C – P + 2


One Component System

Pressure-Temperature Component

for a one component system (C=1) .

Point A is in gas region, (P =1),

degree of freedom:

F = C – P + 2

F = 1 - 1 + 2 = 2

In gas region, both temperature and

pressure may be arbitrarily fixed.



In two-phase regions, such as point B

on liquid vapor line:

F = C – P + 2

F = 1 - 2 + 2 = 1

On liquid vapor line, or on any other

of the two-phase lines, we may

arbitrarily fix either pressure or

temperature.

There is only one degree of freedom.

If temperature is fixed then pressure

is also fixed.



At triple point, where three phases:

solid, liquid and gas exist in

equilibrium, the degree of freedom:

F = C – P + 2

F =1 - 3 + 2 = 0

There is no degree of freedom, there

is only one combination of

temperature and pressure at which

all three phases may coexist in a

single-component system.


Example 1:

Consider the decomposition of calcium carbonate into calcium oxide and

carbon dioxide:

CaCO3 (s) → CaO (s) + CO2 (g)

C = N – R

Chemical species:

Relationship :

Number of “Component” = N – R =

Degree of freedom, F = C – P + 2 =

CaCO3(s), CaO(s), and CO2(g) = three chemical

species

a

a p K

3

2

CaCO

CaOCO= = one relationship

2

2 – 3 + 2 = 1

If T is fixed, then

pressure of carbon

dioxide is also fixed

(decomposition pressure

of calcium carbonate)


Example 2:

Suppose that an inert material, such as nitrogen, is added to the system

considered in example 1. Number of species will be:

Degree of freedom, F = C – P + 2 = 3 – 3 + 2 = 2

Temperature and total pressure can each be varied independently.

Pressure of carbon dioxide is fixed at a specific temperature, but the total

pressure of the system can be varied by varying the nitrogen pressure.

Four (CaO, CaCO3,

CO2 and N2)


Example 3:

Consider the equilibrium between solid carbon, carbon monoxide and

carbon dioxide:

CO2 (g) + C(s)→ 2CO (g)

There are two phases present: solid carbon and the gas.

Chemical species:

Relationship :

Number of “Component” = N – R =

Degree of freedom, F = C – P + 2 =

CO2(g), C (s), and CO(g) = three chemical species

( )

a p

p K

CCO

2

CO

2

= = one relationship

2

2 – 2 + 2 = 2 If T and P are fixed,

then the system is

specified.


Phase Rule for Condensed System

For condensed system, such as solids, it is common to assume that the

pressure of the system is one atmosphere.

The pressure variable might be ignored for small pressure changes, the

Phase Rule for condensed system:

F + P = C + 1

F = C – P + 1

F = C – P + 2 (temperature and pressure) (For gas)

F = C – P + 1 (temperature) (for condensed system, solid, liquid)


LIQUID

SOLID

Freezing Point Depression

At temperature Tm, the two phases, solid

and liquid, are in equilibrium.

Temperature interest (for freezing point

depression) is in the region below Tm,

pure solid A is taken as standard state.

At temperature lower than Tm:

T

Tm (melting temperature)

Phase diagram of single-

component system

ls AA =

==

pures,

purel,

s

l

a

a ln RT

f

f ln RTG

1 a if a ln RTG pures,purel, ==

meltingmeltingmeltingmelting S T - L S T - HG ==


LIQUID

SOLID


At temperature lower than Tm:T



component system

melting

meltingmeltingmelting

S T - L

S T - H G

=

=

L is latent heat of fusion. For

simplicity, assume that there is no

difference in heat capacity between

liquid and solid.

0 G e,temperatur melting At melting =


LIQUID

SOLID


T



component system

0 G e,temperatur melting At melting =

meltingm S T L =

( )

m

m

m

meltingT

TT L

T

T-1 L G

−=

=

( )

m

m

pures,

purel,

meltingT

TT L

a

a ln RT G

−=

=

meltingmeltingmelting S T - HG =


Addition of Material B to A

At temperature T’ (below Tm), activity of

A in an ideal A-B solution is a function of

composition.

Consider that A and B are immiscible in

the solid state, but form ideal solution in

liquid state.

Liquid of composition XA,l (where activity aA,l =1) is in equilibrium with pure

solid A at Temperature T’.

( )

m

m

pures,

purel,

meltingT

TT L

a

a ln RT G

−=

=

Activity of pure liquid A is greater than

one at temperatures below Tm (Solid A

was considered as standard state).



Consider: dissolving of pure liquid

A in the liquid solution:

Dissolution of pure solid A in

liquid solution is the sum of:

melting of pure A and dissolution

of pure liquid A in liquid solution

solutionl,pure,l A A =

=

purel,

solutionl,

a

a ln RT G

solutionl,pure,s A A =

If liquid solution is in equilibrium

with pure solid, then G=0

( )solutionl,A,

m

m X ln -RTT

TT L =

−

If solution is ideal

If T is close to Tm

( )( )2m

msolutionl,A,

T R

TTL-X ln

−=

( )0

a

a ln RT

T

TT L G

purel,

solutionl,

m

m =

+

−=



Note: XA = 1 - XB

( )( )2m

mB

T R

TTL-)X-(1 ln

−=

If variable z is small: ln (1-z) = -z

( )( )2m

mB

T R

TTLX

−=

T = Tm – T (melting point

depression)


Melting Point Depression

( )( )2m

mB

T R

TTLX

−=

T = Tm – T (melting point

depression)

This expression can be plotted

on a phase in which temperature

is ordinate and composition in

absisca.

In two phase region labelled

“L+S” (liquid plus solid), pure

solid A is in equilibrium with a

liquid solution.


Simple Eutectic Phase Diagram


Example

Calculate the lowering of melting point of silver caused by addition of one

mole percent of lead. Although there is small solubility of lead in solid

silver, it was assummed that silver lead system follows the following

equation:

For silver: Tm = 1234 K, and L = 11 300 J/mol

( )( )2m

mB

T R

TTLX

−=

T = 11.2 K


http://www.ques10.com


Lever Rule

At T1, the phase diagram tells us that the

equilibrium liquid composition is XB,l. The

equilibrium solid composition is XB,S.

Relative quantities or fractions can be

calculated using mass balance

If Fl = fraction liquid and FS =

fraction solid, then Fl + FS = 1

Based on a mass balance for B:

sB,slB,lB X F X FX +=

sB,slB,lBSl X F X FX )FF( +=+

)X (X F )XX(F sB,BSBlB,l −=−

Liquid


Lever Rule

Lever rule can be expressed as a ratio of

fraction liquid to the fraction of solid:

For fraction of liquid:

XX

XX

F

F

BlB,

sB,B

s

l

−

−=

XX

XXF

sB,lB,

sB,B

l−

−=

Note: Fl + FS = 1


Lever Rule: Example

Calculate the fraction of liquid at 500K in a lead-tin binary alloy containing

50 mol% tin. At 500 K, the equilibrium composition of the liquid is 59

mol% tin, and the composition of the solid is 24 mol% tin.

74.0Fl =


Simple Eutectic Diagram

Liquid = solid A + solid BConsider: material A

and B are immiscible

in solid state,

completely miscible in

liquid state.

Addition of B to A

lowers the melting

point of A. addition of

A to B lowers the

melting point of B


Cooling Curve (pure material)

If a pure material (pure A), is

cooled from a temperature Tm

(its melting temperature) to

below Tm by removing thermal

energy at a constant rate,

temperature of material as a

function of time follows a pattern

illustrated in the figure 9.6.

During solidification, liquid A and

solid A are in equilibrium,

temperature of the system does

not change (thermal arrest).

Once all material A has solidified, temperature decrease continue.

The same type of cooling curve is observed if one cools a liquid of

eutectoic composition


Cooling Curve (pure material)


Cooling Curve (other than eutectic composition)


Complete Miscibility

Consider two component system A-B, where A and B are completely

miscible in both solid and liquid state, and form ideal solution in both.

( ) p

s

p

l

m

mmelting -

T

TT L G =

−=

( )

m

mp

l

p

sT

TT L -

−=


Phase Diagram


Fe-Ti Diagram


Phase Diagram


Complete Miscibility, Ideal Solution

A

o

AA

rel

A a ln RT G G G =−=

i

nnS,V,innS,P,innV,T,innP,T,iijijijij

n

U

n

H

n

F

n

G=

=

=

=

i = chemical potential of component i

BBAA X V X V V +=

Bn,P,TA

A

n

VV

=

l,A

p

lA,l,A a ln RT += solution) (ideal Xln RT l,A

p

lA,l,A +=

l,BBl,AAl X XG +=

) Xln RT ( X ) Xln RT (XG l,B

p

lB,Bl,A

p

lA,Al +++=


Graphical Representation: Ideal Solution

AABA

PT,B

A VXVX dX

VdX :1 −=

ABBB

PT,B

B VXVX dX

VdX :2 −=

BBAA X V X V V :3 +=

B

PT,B

A V dX

VdXV :31 =

++

A

PT,B

B V dX

VdXV :23 =

−−

10. Solutions



s,BBs,AAs X XG +=

( )

m

mp

l

p

sT

TT L -

−=

) Xln X Xln X(RTG l,BBl,AAl +=

) Xln RT ( X ) Xln RT (XG s,B

p

sB,Bs,A

p

sA,As +++=

−−

−−

+=

)TT(T

LX)TT(

T

LX

) Xln X Xln X(RTG

B,m

B,m

Bs,BA,m

A,m

As,A

s,Bs,Bs,As,As



T > Tm,A and T > Tm,B

solid

liquid

Tm,A = 900 K

Tm,B = 1300 K

L/Tm = 9 J/K (for A

and B)



T < Tm,A and T < Tm,B

solid

liquid

Tm,A = 900 K

Tm,B = 1300 K

L/Tm = 9 J/K (for A

and B)



T = Tm,A and T < Tm,B

solid

liquid

Tm,A = 900 K

Tm,B = 1300 K

L/Tm = 9 J/K (for A

and B)



solid liquid

Tm,B > T > Tm,A

liquid solidliquid

+ solid

xB1 xB2

Tm,A = 900 K

Tm,B = 1300 K

L/Tm = 9 J/K (for A

and B)



Phase diagram: complete range

of liquid and solid miscibiliy, ideal

solution

solid

liquid

Tm,A

Tm,B


Example (cont.): Complete Miscibility, Ideal

Solution


Complete Miscibility, Non Ideal Behavior


p

i

4

i

3

i

2

i

1

i ... =====

at certain T


Specific Gibbs

free energy of

pure A and B at

different levels

IDEAL SOLUTIONS:

10.

So

luti

on

s


Ph

ase

mis

cib

ility

ga

p in

so

lid s

tate

: tw

o

so

lid s

olu

tio

ns (

an

d

)


Zone Refining

Liquid Co is in equilibrium

with solid kCo which has a

lower solute content.

On soldification of an

impure metal with initial

concentration Co, the first

solid formed has

composition kCo.

Enrichment of liquid stops

after the composition of

solute in molten zone Co/k.


Zone Refining


Teknik MetalurgiThank you for your attention!

Dr.-Ing. Zulfiadi Zulhan

Department of Metallurgical Engineering

Institute of Technology Bandung

Jl. Ganesha No. 10

Bandung 40132

INDONESIA

Telefon : +62 (0) 22 250 2239

Fax : +62 (0) 22 250 4209

Mobile : +62 (0) 813 22 93 94 70

E-Mail: [email protected]

[email protected]

mailto:[email protected]



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