1. Euromet 2006 Influence of impurities on the melting temperature of Aluminum Pr. B. Legendre & Dr S. Fries EA 401

2. Impurities B, C, Cr, Cu, F, Fe, Mg, Mn, N, Ni, O, Pt S, Sc, Si, Ti, V, Zn. Al – X Optimized Al – Y Non optimized (or bad optimized), but the phase diagram is known Al – F no data

3. Aim Measurement of the slopes of the liquidus and of the solidus :  = (dT/dx)  l : for the liquidus  s : for the solidus The slope may be expressed in atomic fraction or in weight fraction. Determination of K : K =  l /  s For Al T fus = 660.323°C

4. Phase diagram For this subject we are interested only by a very small region in temperature and composition of binaries Al-X, but it is absolutely necessary to have a perfect knowledge of the full diagram. And to know the Gibbs function versus of T and x for each phase at a fixed pressure.

5. Phase diagram : Calphad Method A phase diagram is the graphical expression of phase equilibria, for each phase in a binary system G = f(x i, T, P) The limits of a two phase field are given by the common tangent of the G =f(x i ) PT curves. For a eutectic: The three curves G=f(x i ) p,T, have the same tangent.

6. The whole process

7. Calculating phase diagrams

8. Al - Si

11. Al - B

14. Al - Pt

18. Al - Cr

20. Al - Ni

22. Al - Fe

25. Al - Cu

28. Al - Sc

29. No data for the solidus…

30. Al – O: not a good optimization

32. Al - O Why it is not possible to have a good optimization of the Al-O binary? The oxygen forms with Al an oxide which is a thin film fixed on the surface of Al and then it is a kind of protection (aluminum pans are used on the gas in a kitchen…) after the fixation of a first layer the oxygen cannot go deeper in Al. There is no homogeneity and no equilibrium.

33. Al - O L  Al +  Al 2 O 3 T =~ 660°C Van’t Hoff law dln(a Al )/dT =  H/RT 2 a Al ~ X Al  H =  fus H(Al) = 10711.04J/mol  T = 0.452 K X Al =.9993316 dT/dX = 676

34. Results

35. Comparison of results

36. Variation of the liquidus temperature versus of the yield of impurities T l = T fus +   i T l = temperature of the liquidus T fus temperature of fusion of pure Al  = dT/dx x : weight fraction  i : yield of impurities in ppm*10 -6 Example : for Al-Cr  i = 9.9*10 -2 ppm  = 466 T l = 660.32°C

37. Temperature of liquidus and solidus for a yield of impurity

38. Conclusions It has been possible to determine the influence of impurities in Al for nearly all the selected elements. For O and S just an estimation has been possible. The most important modifications occurs with O and N It is possible to perform a calculation with two impurities.

39. Acknowledgement This work has been financially supported by LNE and BNM We are very grateful to these organisms for their contribution.

44. Al - Pt

47. Al - Ni

49. Al - Pt

50. Applications of Gibbs energies descriptions combined to kinetic Information

51. Creating Gibbs energies for each phase

52. Al - Si