Illustration .1A 0.4% C hypo-eutectoid plain carbon steel is gradually cooled from 1540^oC to:

(i) Slightly above 723^oC and

(ii) Slightly below 723^oC.

Determine the weight percent

(a)        Austenite present in the steel,

(b)        Ferrite present in the steel in case (i),

(c)        pro-eutectoid ferrite avoid in the steel, and

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

Solution

Refer to Figure of Iron-Carbon Phase Diagram. Point 1 above the liquids illustrates the state of liquid steel. The cooling arises along the line xx and an equilibrium cooling is unspecified.

Freezing initiates at point 2 that is intersection of liquidus and line xx. Temperature at 2 is 1510^oC. The steel solidifies wholly at point 3, here temperature is 1471^oC. The entire alloy is here composed of austenite (g-phase) as denoted with first of Figure of Cooling Scheme. No change arises till point 4 on line A3 is reached. At such point the ferrite precipitation begins out of solid austenite. Further cooling raises the amount of austenite and ferrite reduces. The amount of austenite varies with IK. The composition of ferrite varies with the line IL.

Determine of % content will be made via lever rule.

The austenite amount slightly above 723^oC is determined from the line LK itself.

That is, taking LK as tie line.

(a)   Weight % of austenite  =  L5/ LK

= (0.4 - 0.025)/(0.8 - 0.025)

= 0.0375/0.0775

= 0.0484 or 48.4%.........................Eq(1)

Weight % of ferrite= 5K/ LK

=   (0.8 - 0.4)/ (0.8 - 0.025)

=0.4/0.775

= 0.516 or 51.6%................Eq(2)

(b)   Weight % of pro-eutectoid ferrite slightly underneath 723^oC is similar as that slightly above that is 48.4%. . . . Eq(3)

For determining eutectoid ferrite, the weight of carbide will have to be subtracted from whole mass of cementite and ferrite. Only below isothermal line LKM ferrite and pearlite are shows and lever arm will extend upto ordinate demonstrating 6.67% C.

Weight % of total (ferrite + cementite) merely underneath 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 (via difference)

= 100 - 48.4 - 5.64 - 44.4 = 1.56%