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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
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Euromet 2006Influence 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 Tfus= 660.323°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(xi, T, P) • The limits of a two phase field aregiven by the common tangent of the G =f(xi)PT curves. • For a eutectic: • The three curves G=f(xi)p,T, have the same tangent.
Al - Sc • No data for the solidus…
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 + Al2O3 T =~ 660°C • Van’t Hoff law • dln(aAl)/dT = H/RT2 • aAl~ XAl • H = fusH(Al) = 10711.04J/mol • T = 0.452 K XAl= .9993316 • dT/dX = 676
Variation of the liquidus temperature versus of theyield of impurities • Tl = Tfus + i • Tl = temperature of the liquidus • Tfus 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*10-2ppm • = 466 Tl = 660.32°C
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.
Applications of Gibbs energies descriptions combined to kinetic Information