Find the weight percent of austenite present in the steel:

A 0.4% C hypoeutectoid plain carbon steel is slowly cooled from 1540^oC to

(i) slightly above 723^oC and (ii) slightly below 723^oC.

Calculate the weight percent

1.      austenite present in the steel,

2.      ferrite present in the steel in case (i),

3.      proeutectoid ferrite prevent in the steel, and

4.      eutectoid ferrite and eutectoid cementite % present in the steel in case (ii).

Solution

Refer to Figure 8. Point 1 above the liquidus represents the state of liquid steel. The cooling occurs along the line xx and an equilibrium cooling is assumed. Freezing begins at point 2 which is intersection of liquidus and line xx. Temperature at 2 is 1510^oC. The steel solidifies completely at point 3 where temperature is 1471^oC. Now the entire alloy is composed of austenite (γ-phase) as indicated by first of Figure 19. No change occurs until point 4 on line A₃ is reached. At this point the precipitation of ferrite begins out of solid austenite. Further cooling increases the amount of ferrite and austenite decreases. The amount of austenite varies along with IK. The composition of the ferrite varies along with the line IL.

Calculation of % content will be made by lever rule.

The amount of austenite slightly above 723^oC is calculated from the line LK itself. i.e. taking LK as tie line.

 (a)       Weight % of austenite =   L₅/ L_K= ( 0.4 - 0.0250)/(0.8 - 0.025)

                                                       =  0.375/0.775  = 0.484 or 48.4%         ---------- (i)

                     Weight % of ferrite    = 5K / L_K       =     0.8 - 0.4/0.8 - 0.025

                                                               =     0.4/0.775 = 0.516 or 51.6%        -------- (ii)

 (b)       Weight % of proeutectoid ferrite slightly below 723^oC is same as that slightly above, that means 48.4%. ----------- (iii)

For calculating eutectoid ferrite, the weight of carbide will have to be subtracted form total mass of ferrite and cementite. Just below isothermal line LKM ferrite and pearlite are present and lever arm will extend up to ordinate representing 6.67% C.

Weight % of total (ferrite + cementite) just below 723^oC

=     6.67 - 0.4/6.67 - 0.025

 =   6.37/6.645

  = 0.96 or 96%

Weight % of Fe₃C just below 723^oC

=   0.4 - 0.025/6.67 - 0.025

=  0.375/6.645

= 0.0564 or 5.64%      

Weight % of eutectoid cementite = total ferrite - proeutectoid ferrite

                                                              = 96 - 51.6 = 44.4%

Weight % of eutectoid cementite (by difference)

= 100 - 48.4 - 5.64 - 44.4 = 1.56%                                         --------- (iv)