1. 15/01/20111 On Using Thermo-Calc Sourav Das, Researcher, Product Research Group, Research and Development Division, Tata Steel

2. 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. 15/01/2011

3. 3 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. 1 st law : Defines energy, U Energy can be transformed (changed from one form to another), but can not created or destroyed. 2 nd law : Defines entropy, S The entropy of an isolated system which is not in equilibrium will tend to increase with time. 3 rd 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 15/01/2011

4. 4 SymbolsFull names UInternal energy qQuantity of heat wWork done by the system VVolume of the system PPressure TTemperature in absolute scale CPCP Specific heat capacity at constant pressure CVCV Specific heat capacity at constant volume HEnthalpy GGibbs free energy 15/01/2011

5. 5 Internal Energy, U ΔU = q – w or dU = dq – dw, where, dw = PdV if P = constant q A A A 15/01/2011

6. 6 Enthalpy, H V H = U + PV 15/01/2011

7. 7 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, C V = C p =, at constant pressure H = U + PV dH = d(U + PV) = dU + d(PV) = dU + PdV + VdP = (dq – PdV) + (PdV + VdP) = dq + VdP Specific heat at constant volume 15/01/2011

8. 8 assuming C P is constant When the reaction A + B = C will be possible? ΔH = H final – H initial = -ve But, why is it only possible? Why will it not necessarily happen? 15/01/2011

9. 9 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 For a reversible process, or 15/01/2011 Entropy is a capacity property. Different entropies can be added together S 1 + S 2 + S 3 = S 4

10. 10 Entropy, S (again) How does Entropy fit with the probability picture? Through Boltzman’s law: S = k ln(w) P Possible arrangement, W = 1, P/2 Possible arrangement, W = very very large 15/01/2011 Probability of getting all the n mlecules at one side = 4 / [n(n-1)]

11. 11 Gibbs Free Energy, G G = H – TS or ΔG = ΔH - TΔS Note: This is probably the most important parameter in all thermodynamical calculations 15/01/2011

12. 12 Equilibrium 15/01/2011

13. 13 Allotropic Transition in Pure Iron D. R. Gaskell, Thermodynamics of Materials 15/01/2011

14. 14 Mechanical Mixture 15/01/2011 free energy of mechanical mixture A B Gibbs free energy per mole Concentration x of B G* 1-x x

15. 15 Solution 15/01/2011 free energy of solution free energy of mechanical mixture A B Gibbs free energy per mole Concentration x of B G* x G{x}

16. 16 Chemical potential 15/01/2011

17. 17 Chemical potential 15/01/2011 A B x Gibbs free energy per mole Composition G{x}

18. 18 Conditions: 1.Under standard conditions (T = 298 K and P = 101.3 kPa) 2.Without intermolecular interactions 3.Natural isotope composition of elements 15/01/2011

19. 19, Equilibrium between two solutions 15/01/2011

20. 20 Equilibrium between two solutions (contd……..) Gibbs free energy per mole wt% C T = T 1 (constant)  α 15/01/2011

21. 21 Equilibrium between two solutions (contd……..) wt% C Gibbs free energy per mole T = T 2 (constant), T 2 >T 1 α  XαXα XαXα 15/01/2011

22. 22 Equilibrium between two solutions (contd……..) Gibbs free energy per mole wt% C  α T = T 3 (constant), T 3 > T 2 15/01/2011

23. 23 Phase diagram between two phases T = T 3 T = T 2 T = T 1 Temperature C 15/01/2011 γ γ + α α

24. 24   α M1M1 M M2M2   α α α+α+  ++  Phase diagram among three phases 15/01/2011

25. 2515/01/2011 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

26. 2615/01/2011 assuming C P is constant For a reversible process, or 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……….

27. 2715/01/2011 C P = b 1 + b 2 T + b 3 T 2 + b 4 /T 2

28. 2815/01/2011 wt% C Gibbs free energy per mole α  XαXα XαXα

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30. 12/07/201030 References: 1. Introduction to the Thermodynamics of Materials, 3 rd edition, D. R. Gaskell 2. Online available course materials from Georgia State University, USA 3. http://en.wikipedia.orghttp://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, 2 nd edition, A. H. Cottrell 9. Chemical Thermodynamics of Materials, C. H. P. Lupis

31. 3115/01/2011 Thank you for kind attention

32. 3215/01/2011 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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