# Isothermal transforamtion diagram - Austentite [closed]

A steel alloy is rapidly cooled to $$\pu{600^oC}$$, held for $$\pu{4 sec}$$, rapidly cooled to $$\pu{450^oC}$$, held for $$\pu{10 sec}$$ and quenched to room temperature. What phases are produced?

Why is the time reset back to $$0$$ when you cool to $$\pu{450^oC}$$? Is it because I am only considering the remaining austenite?

The given phase diagram provides some information on the time-temperature relationship for the formation of pearlite (P), bainite (B), and martensite (M) from austenite (A) form, which is known as known as gamma-phase iron ($$\gamma$$-$$\ce{Fe}$$). For example, It would take $$\pu{10 sec}$$ to 100% austenite to the 50% austenite-to-pearlite transformation to complete.

As following diagram indicate, rapid cooling of eutectoid composition of steel alloy (100% A) to $$\pu{600 ^\circ C}$$ stays as 100% A. If the temperature held at $$\pu{600 ^\circ C}$$ for $$\pu{4 sec}$$, austenite-to-pearlite transformation occur to ~25% (see the point noted at the diagram):

During the second rapid cooling from $$\pu{600 ^\circ C}$$ to $$\pu{450 ^\circ C}$$ (time $$\approx 0$$), there is no additional transformation (see the second point noted at the diagram with red line). At $$\pu{450 ^\circ C}$$, the time is start to count again at zero second, so that the time is moved to $$t=0$$ point (this is also because isotherm conversion will be started with unconverted austenite, which is now 75% of original specimen). While holding at $$\pu{450 ^\circ C}$$ for $$\pu{10 sec}$$, approximately 50% of this remaining unreacted 75% A (or ~37.5% of the original specimen) will transform to bainite (see the point noted in the diagram). Finally, upon cooling to room temperature, the remaining 37.5% A (of the original specimen) transforms to martensite. Thus, the final micro-structure of given steel alloy consists of about 25% pearlite, 37.5% bainite, and 37.5% martensite (no austenite).