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On Using Thermo-Calc. Sourav Das, Researcher, Product Research Group, Research and Development Division, Tata Steel. What is thermodynamics??? Science of flow of heat. It is universal. We can find it in both organic (mitochondria, - PowerPoint PPT Presentation
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15/01/2011 1
On Using Thermo-Calc
Sourav Das,Researcher,
Product Research Group,Research and Development Division,
Tata Steel
2
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, UEnergy can be transformed (changed from one form to another), but can not created or destroyed.
2nd law : Defines entropy, SThe entropy of an isolated system which is not in equilibrium will tend to increase with time.
3rd law : Gives a numerical value to the entropyAs 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 namesU 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
ΔU = q – w
ordU = dq – dw, where, dw = PdV if P =
constant
q
AA
A
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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 = VT
U
Cp = , at constant pressurePT
H
H = U + PV dH = d(U + PV) = dU + d(PV) = dU + PdV + VdP = (dq – PdV) + (PdV + VdP) = dq + VdP
Specific heat at constant volume
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dTCHT
T
P2
1
assuming CP is constant
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
P P/2
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
2
1
T
T
P dTT
CSFor a reversible process, or
T
dqdS rev
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Entropy is a capacity property. Different entropies can be added together S1 + S2 + S3 = S4
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Entropy, S (again)
How does Entropy fit with the probability picture?
Through Boltzman’s law: S = k ln(w)
P
Possible arrangement, W = 1, Possible arrangement, W = very very large
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Probability of getting all the n mlecules at one side = 4 / [n(n-1)]
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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
A B
(1-x) x
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free energy of mechanical mixture
A B
Gib
bs
free e
nerg
y p
er
mole
Concentration x of B
G*
1-xx
0AG
0BG
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A B+
atomic solution
Solution
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free energy of solution
MG
free energy of mechanical mixture
A B
Gib
bs
free e
nerg
y p
er
mole
Concentration x of B
G*
x
0AG
0BG
G{x}
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Chemical potential
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.),,(.' constPTdndG AA
.),(.' constPTdndndG BBAA
xxdG BA )1(.
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Chemical potential
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A BxGib
bs f
ree e
nerg
y p
er
mole
Composition
G{x}
0AG
0BG
)1( xA
)(xB
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BA xxxG )1(}{
Conditions:
1.Under standard conditions (T = 298 K and P = 101.3 kPa)2.Without intermolecular interactions3.Natural isotope composition of elements
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AA BB ,
Equilibrium between two solutions
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Equilibrium between two solutions(contd……..)
wt% C
T = T1 (constant)
α
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Equilibrium between two solutions(contd……..)
wt% C
T = T2 (constant), T2 >T1
α
Xα Xα
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Equilibrium between two solutions(contd……..)
wt% C
α
T = T3 (constant), T3 > T2
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Phase diagram between two phases
T = T3
T = T2
T = T1
C
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γ
γ + α
α
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α
2x
2x
M1
M
M2
α
)(2
x )(2
x )(2
x )(2
x
α α +
+
Phase diagram among three phases
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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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dTCHT
T
P2
1
assuming CP is constant
2
1
T
T
P dTT
CS
For a reversible process, or T
dqdS rev
Heat capacity is a function of: 1. vibration of atoms 2. magnetic property of the atoms3. Electronic heat capacity due to electronic configuration4. Curling up of molecules etc etc……….
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CP = b1 + b2T + b3T2 + b4/T2
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α
Xα Xα
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12/07/2010 30
References:
1. Introduction to the Thermodynamics of Materials, 3rd edition, D. R. Gaskell
2. Online available course materials from Georgia State University, USA3. http://en.wikipedia.org4. Classroom video lectures on Thermodynamics from MIT, USA 5. Class room video lectures from University of Cambridge, UK6. Online course material from University of Texas at Austin, USA7. Thermodynamics in Materials Science, International Edition1993, R. T. DeHoff8. An Introduction to Metallurgy, 2nd edition, A. H. Cottrell9. Chemical Thermodynamics of Materials, C. H. P. Lupis
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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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