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

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Presentation transcript:

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

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

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 = °C

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.

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.

The whole process

Calculating phase diagrams

Al - Si

Al - B

Al - Pt

Al - Cr

Al - Ni

Al - Fe

Al - Cu

Al - Sc

No data for the solidus…

Al – O: not a good optimization

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.

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) = J/mol  T = K X Al = dT/dX = 676

Results

Comparison of results

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 = °C

Temperature of liquidus and solidus for a yield of impurity

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.

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

Al - Pt

Al - Ni

Al - Pt

Applications of Gibbs energies descriptions combined to kinetic Information

Creating Gibbs energies for each phase

Al - Si