On Using Thermo-Calc
Sourav Das,
Researcher,
Product Research Group,
Research and Development Division,
Tata Steel
15/01/2011
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What is thermodynamics???
Science of flow of heat.
1.
It is universal. We can find it in both organic (mitochondria,
ATP etc) and inorganic (black holes, mechanical systems,
chemical reactions) objects.
2. It is based on macroscopic properties of matter.
3. Entirely empirical.
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Zeroth law : Defines temperature, T.
If A and C are both in thermal equilibrium with a third body B, then they are also in thermal equilibrium
with each other.
1st law
: Defines energy, U
Energy can be transformed (changed from one form to another), but can not created or destroyed.
2nd law
: Defines entropy, S
The entropy of an isolated system which is not in equilibrium will tend to increase with time.
3rd law
: Gives a numerical value to the entropy
As a system approaches to absolute zero, all the processes cease and the entropy of the system
approaches a minimum value
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Symbols
Full names
U
Internal energy
q
Quantity of heat
w
Work done by the system
V
Volume of the system
P
Pressure
T
Temperature in absolute scale
CP
Specific heat capacity at constant pressure
CV
Specific heat capacity at constant volume
H
Enthalpy
G
Gibbs free energy
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Internal Energy, U
A
A
A
q
ΔU = q – w
or
dU = dq – dw, where, dw = PdV if P =
constant
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Enthalpy, H
V
H = U + PV
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Specific heat capacities
Heat absorbed per unit change in temperature (dq/dT).
Since,
dq = dU + dw = dU + PdV (at constant pressure)
So, specific heat at constant volume, CV =
 U 


 T V
Specific heat at constant volume
H = U + PV
dH = d(U + PV) = dU + d(PV) = dU + PdV + VdP
= (dq – PdV) + (PdV + VdP)
= dq + VdP
Cp =
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 H,  at constant pressure


 T  P
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T2
H   C P dT
assuming CP is constant
T1
When the reaction A + B = C will be possible?
ΔH = Hfinal – Hinitial
= -ve
But, why is it only possible? Why will it not necessarily happen?
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Entropy, S
For a reversible process,
dq rev
dS 
T
T2
or
CP
S  
dT
T
T1
P/2
P
Why the reaction will happen in the direction of the arrow?
Why not in the opposite direction?
So, even if ΔH = 0, a reaction may spontaneously
happen if the ΔS > 0
Entropy is a capacity property. Different entropies can be added together
S1 + S2 + S3 = S4
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Entropy, S (again)
P/2
P
Possible arrangement, W = 1,
Possible arrangement, W = very very large
Probability of getting all the n mlecules at one side = 4 / [n(n-1)]
How does Entropy fit with the probability picture?
Through Boltzman’s law: S = k ln(w)
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Gibbs Free Energy, G
G = H – TS
or ΔG = ΔH - TΔS
Note: This is probably the most important parameter
in all thermodynamical calculations
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Equilibrium
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Allotropic Transition in Pure Iron
D. R. Gaskell,
Thermodynamics of Materials
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Mechanical Mixture
Gibbs free energy per mole
free energy of
mechanical mixture
GB0
G*
G A0
x
1-x
A
B
Concentration x of B
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Solution
Gibbs free energy per mole
free energy of
mechanical
mixture
G*
G A0
GM
G{x}
free energy of solution
A
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GB0
x
Concentration x of B
B
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Chemical potential
dG   A. dnA (T , P, const.)
'
dG
'
  A. dnA  B dnB
(T , P const.)
dG   A . (1  x)   B x
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Gibbs free energy per mole
Chemical potential
GB0
G A0
 A (1  x)
G{x}
B (x)
A
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x
Composition
B
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G{x}  (1 x)  A  x  B
Conditions:
1.Under standard conditions (T = 298 K and P = 101.3 kPa)
2.Without intermolecular interactions
3.Natural isotope composition of elements
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Equilibrium between two solutions

A
 
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
A
,

B
 

B
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Equilibrium between two solutions
(contd……..)
Gibbs free energy per mole
T = T1 (constant)

α
wt% C
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Gibbs free energy per mole
Equilibrium between two solutions
(contd……..)
T = T2 (constant), T2 >T1
α

Xα
Xα
wt% C
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Equilibrium between two solutions
(contd……..)
Gibbs free energy per mole
T = T3 (constant), T3 > T2
α

wt% C
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Phase diagram between two phases
Temperature
T = T3
γ
γ+α
T = T2
α
T = T1
C
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Phase diagram among three phases

α
α
M2
M1


α
M
x 2
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x 2
α
+


x2 (  ) x 2 ( ) x 2 ( )


+


x 2 (  )
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We have considered:
1. Ideal solution (random distribution of solute atoms) and
no change in binding energy when we mix the atoms
together
2. Binary solution
1. In regular solution, there will be excess free energy of
mixing and there may be liking or disliking among the atoms
(back up slide for enthalpy of mixing)
2. There can be 8-10 elements in a commercial alloy
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T2
H   C P dT
T1
For a reversible process,
assuming CP is constant
dqrev
dS 
T
T2
or
CP
S   dT
T
T1
Heat capacity is a function of:
1. vibration of atoms
2. magnetic property of the atoms
3. Electronic heat capacity due to electronic configuration
4. Curling up of molecules etc etc……….
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CP = b1 + b2T + b3T2 + b4/T2
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Gibbs free energy per mole
α

Xα
Xα
wt% C
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References:
1. Introduction to the Thermodynamics of Materials, 3rd edition,
D. R. Gaskell
2. Online available course materials from Georgia State University, USA
3. http://en.wikipedia.org
4. Classroom video lectures on Thermodynamics from MIT, USA
5. Class room video lectures from University of Cambridge, UK
6. Online course material from University of Texas at Austin, USA
7. Thermodynamics in Materials Science, International Edition1993,
R. T. DeHoff
8. An Introduction to Metallurgy, 2nd edition, A. H. Cottrell
9. Chemical Thermodynamics of Materials, C. H. P. Lupis
12/07/2010
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Thank you
for
kind attention
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Free energy of mixing
If we consider an ideal solution, the entropy of mixing will be:
If we consider a regular solution, there will be always a
change in bond energy and there will be excess free energy
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