For the FeC equilibrium phase diagram below Calculate at of

For the Fe-C equilibrium phase diagram below, Calculate at.% of carbon in a AISI 1030. What will be the overall composition of a steel (wt.% C) if the mass fraction of ferrite is the same as the mass fraction of pearlite (0.5)? For the composition of the steel in b). what is the mass fraction of cementite in pearlite? Is it possible to have a different steel with the same mass fraction of cementite in pearlite as in c)? Determine the composition of the steel if it exits.

Solution

a) For AISI 1030 => wt% of C will be = 0.30% and wt% of Fe = 99.70%

So at% of C will be = (%C/C_molweight)*100 / [(%C/C_molweight) + (%Fe/Fe_molweight)]

at% = (0.3/12)*100/[(0.3/12)+(99.7/55.845)] = 0.025*100/(0.025+1.785)

at% = 0.0138*100 = 1.38% Ans

b) Mass Fraction of Pearlite is same as Mass Fraction of ferrite W_f = W_p = 0.5

Using Lever Rule, W_f = (C₀-C_p) / (C_f – C_p)    { C_p = 0.76 and C_f = 0.022 from phase diagram }

c) Mass Fraction of cementite in Pearlite for 0.391 wt%C

By using Lever Rule = (C₀-C_f) / (C_c – C_f)       { C_c = 6.7 and C_f = 0.022 from phase diagram }

d) Mass fraction is 0.055 and value of C₀ for this will be

Using lever rule, 0.055 = (C₀-C_f) / (C_c – C_f)      

On solving , C₀ = 0.391 ans

So this is only steel having mass fraction of 0.055.