![Details
• A more detailed approach considers the ratios of the
compositions of the phases in addition to the differences in
chemical potential [Martin et al., p10-11].
• Consider fig. 5.3b again. The triangle R-P-Q is similar
(mathematically) to the triangle R-T-U.
• The decrease in free energy per mole of precipitate is the
distance P-Q, which is smaller than the difference in
chemical potential for B in the supersaturated material at X0
and in the beta precipitate at Xbeta.
∆Gn = (µB
0 - µB
e){Xbeta-Xe}/{1-Xe}
= RT ln(B
0XB
0/B
eXB
e) {Xbeta-Xe}/{1-Xe} .
∆Gn = RT ln(X0/Xe) {Xbeta-Xe}/{1-Xe}.
• However, for precipitate that is pure B (or close to that) the
fraction in curly brackets, {Xbeta-Xe}/{1-Xe}, is close to one,
hence the result given on the previous slide.](https://image.slidesharecdn.com/phasetransformationsinsolids-220912150543-2a78ed4c/85/Phase-Transformations-in-solids-ppt-15-320.jpg)
Phase Transformations in solids.ppt
- 1. Phase Transformation - Basic Concepts
- 2. Figure 9-2 An interface is created when a solid forms from the liquid.
- 3. Figure 9-3 The total free energy of the solid-liquid system changes with the size of the solid. SL
- 4. Phase Transformation - Basic Concepts • Homogeneous nucleation – a sphere of radius r of a solid forms in a liquid GL: free energy per volume of liquid GS: free energy per volume of solid γ: solid-liquid Interface free energy per area of the A=4r2 V= 4/3 r3
- 5. Phase Transformation - Basic Concepts
- 6. Phase Transformation - Basic Concepts In addition, the phase transformation process often involves the diffusion of atoms which has its own activation energy, Qd.
- 7. -The total nucleation rate (nuclei per volume per second) Phase Transformation - Basic Concepts N
- 8. Phase Transformation - Basic Concepts
- 9. Driving Forces - Examples Driving Force Equation Approx. Value Force (MPa) Stored energy of deformation 1/2rmb2 r~1015 10 Grain boundaries 2sb/R sb~0.5 J/m2 10-2 Surface energy 2Dss/d D~10-3m 10-4 Chemical driving force R(T1- T0)c0lnc0 5%Ag in Cu @ 300°C 10+2 Magnetic field m0H2Dc/2 H~107A/m 10-4 Elastic energy t2(1/E1- 1/E2)/2 t~10 MPa 10-4 Temperature gradient DS 2lgradT/Wa l~5x10-10 10-5
- 10. Quantitative Approach • we can quantify the driving force for solidification (and allotropic phase change). • Assume that the free energies of the solid and liquid are approximately linear with temperature. Temperature Molar Free Energy GS GL TMelt ∆G ∆T
- 11. Driving force: solidification • Writing the free energies of the solid and liquid as: GS = HS - TSS GL = HL - TSL ∆G = ∆H - T∆S • At equilibrium, i.e. Tmelt, then the ∆G=0, so we can estimate the melting entropy as: ∆S = ∆H/Tmelt = L/Tmelt where L is the latent heat (enthalpy) of melting. • Ignore the difference in specific heat between solid and liquid, and we estimate the free energy difference as: ∆G L-T(L/Tmelt) = L∆T/Tmelt
- 12. Driving force: Precipitation • The other case of interest is precipitation. • This is more complicated because we must consider the thermodynamics of solids with variable composition. • Consider the chemical potential of component B in phase alpha compared to B in beta. This difference, labeled as ∆Gn on the right of the lower diagram is the driving force (expressed as energy per mole, in this case). • To convert to energy/volume, divide by the molar volume for beta: ∆GV = ∆Gn/Vm. R S T U N P Q Te
- 13. Driving force for nucleation • It is important to realize the difference between the driving force for the reaction as a whole, which is given by the change in free energy between the supersaturated solid solution and the two-phase mixture, shown as ∆G0 in 5.3b (the distance N-S). • Why a different free energy for nucleation? Because the first nuclei of beta to appear do not significantly change the composition of the parent material. Thus the free energy change for nucleation is the rate of change of free energy for the new, product phase (beta).
- 14. Driving force, contd. • Based on the expression for the chemical potential of a component (P&E 1.68): µB = GB + RT lnBXB • We can identify the chemical potential for B in the supersaturated phase as µB 0, and the chemical potential for B in precipitated beta in equilibrium with the alpha phase as µB e. Then the molar free energy difference, ∆Gn, (distance P-Q) is given by ∆Gn = µB 0 - µB e = RT ln(B 0XB 0/B eXB e). • We can simplify this for ideal solutions (=1) or for dilute solutions (=constant) to: ∆Gn = RT ln(X0/Xe).
- 15. Details • A more detailed approach considers the ratios of the compositions of the phases in addition to the differences in chemical potential [Martin et al., p10-11]. • Consider fig. 5.3b again. The triangle R-P-Q is similar (mathematically) to the triangle R-T-U. • The decrease in free energy per mole of precipitate is the distance P-Q, which is smaller than the difference in chemical potential for B in the supersaturated material at X0 and in the beta precipitate at Xbeta. ∆Gn = (µB 0 - µB e){Xbeta-Xe}/{1-Xe} = RT ln(B 0XB 0/B eXB e) {Xbeta-Xe}/{1-Xe} . ∆Gn = RT ln(X0/Xe) {Xbeta-Xe}/{1-Xe}. • However, for precipitate that is pure B (or close to that) the fraction in curly brackets, {Xbeta-Xe}/{1-Xe}, is close to one, hence the result given on the previous slide.
- 16. Examples • The driving force is very sensitive to the mole fractions of the precipitate, Xbeta,, supersaturated matrix, X0, and equilibrium matrix, Xe. • For a ratio Xe/X0 = 10-3, Xbeta=1, and a temperature of 300K (i.e. very low solubility at a low enough temperature), then the driving force is estimated to be - 17,000 J/mole. For the precipitation of pure Cu from Al, for example, this is a feasible estimate. • For many precipitation reactions, the volume fraction of the precipitate, Vbeta= {X0 -Xe}/ {Xbeta-Xe}, is of order 0.1, which means that the driving force is much less, ~ -1,700 J/mole. If the precipitate is not pure B, but incorporates a significant amount of A, as for example in Al6Mn, then there is a further reduction in the driving force.
- 17. Solid Solubility • Before we can proceed, however, we need to look at the variation in solubility with temperature. This variation can be quite complex, so we will simplify the situation (considerably) by considering a regular solid solution. See section 1.5.7 in P&E. In short, the solubility can often be approximated by an Arrhenius expression, where Q is the heat absorbed (enthalpy) when I mole of beta dissolves in A as a dilute solution. mB GB W 1 XB 2 RT ln XB From fig.1.36b, GB mB DGB XB XB e DGB RT ln XB e W 1 XB e 2 XB e 1 XB e exp DGB W RT exp DSB R exp DHB W RT A exp Q RT
- 18. Driving force as a function of temperature • Despite the complications of the previous analysis, the net result is startlingly simple. • The driving force is, to first order, proportional to the undercooling! DGn RTln X0 Xe RT ln Aexp DHB RT e Aexp DHB RT RTln exp DHB RTe RT ln exp DHB RT RT DHB RTe DHB RT DHBT T Te TTe DGn DHB Te DT
- 19. Nucleation Rate • Armed with this information about driving force (note that the proportionality to undercooling is the same as for solidification), we can now look at how the nucleation rate varies with temperature. • The energy barrier for nucleation varies as the square of the driving force: the nucleation rate is exponentially dependent on the barrier height.
- 20. Elastic energy effect • Elastic effects have the effect of offsetting the undercooling by ∆GS. • Diffusion limits the rate at which atoms can form a new phase and hence limits the nucleation rate at low temperatures.
- 21. Allotropic Phase Transformations • These are, for example, eutectoid transformations (or peritectoid). • The calculation of driving force is similar: – Notation (e.g. Fe-C system): X0:= alloy composition (e.g. C in Fe) X:= equilibrium austenite composition X:= equilibrium ferrite composition – Driving force: ∆Gn = {X - X}/X RT ln{(1-X/(1-X0)}. • For a typical case of T=1000K, X = 0.04, X = 0.001, X0:= 0.02 (equivalent to 0.4 weight % C), ∆Gn = -170 J/mole. This is much smaller than the previous estimates for precipitation. Homogeneous nucleation is essentially impossible!
- 22. Summary • For solidification, the driving force is proportional to the undercooling. • For the case of precipitation where the solid solubility is described by an ideal or regular solution, the driving force is also proportional to the undercooling. • The driving force is the information that we need in order to predict the nucleation rate.
- 23. Thanks! Any questions? You can find me at ● @hamzaahmed ● hamzaahmed0696@mail.me