What of the level of precision you needs?

Thank you for ur response. How to estimate the fraction of formation using reaction peak. I have tried by integrating the peak.Is it the right way. Please suggest.

The most direct way to estimate the fraction of transformation (FT) from DSC is to knowledge about specific heat of crystallization of precipitated phases (must be in a reference book if this phase to be known and stable) and normalized of result of integrating the peak on specific heat.

Hai Sergey...Al 6061 has Mg2Si precipitates..the evolution sequence is GP zones, Beta double prime, beta prime and beta. In this beta double prime is in coherence with the matrix,, but it is unstable. Its formation results exothermic peak. If you dont mind could you please look at the article that i have attached here. In this article figure 8 shows the transformation fraction.

Thanks for the article - it will be usefull for me. How I can understand the sample is full crystalline and multiphase. In this case only direct structural methods (PXRD or TEM) can revealed the real transformed fraction (because in article all kinetic curves are normalized on 100% transformed fraction where the samples are multiphases).

You can find the document "Starink M J. Analysis of aluminium based alloys by calorimetry: quantitative analysis of reactions and reaction kinetics [J]. International Materials Reviews, 2004, 49: 3-4." That must apply faor

Or, you can find other articles from the same autor

Perhaps this article can help

hartwin weber

you could check, Analytical characterization of Aluminum, Steel, ans superalloys by D. Scott Mac Kenzie and G. E. Totten, in this book there is a chapter about DSC in aluminum alloys, including topics about kinetics, and precipitation.

Dear Mr Rao I have a little knowledge about the material u r working with..But the most general way to find out the fraction transformed in any kinetic study through dsc is to find out the ratio of fractional area at a particular temperature to the total area under the peak.