Describe what transformation happen in red line, green line, blue line and yellow line!
Solution.
a. (Red) The specimen is cooled rapidly to 433 K and left for 20 minutes. The cooling rate is too rapid for pearlite to form at higher temperatures; therefore, the steel remains in the austenitic phase until the Ms temperature is passed, where martensite begins to form. Since 433 K is the temperature at which half of the austenite transforms to martensite, the directquench converts 50% of the structure to martensite. Holding at 433 K forms only a small quantity of additional martensite, so the structure can be assumed to be half martensite and half retained austenite.
b. (Green) The specimen is held at 523 K for 100 seconds, which is not long enough to form bainite. Therefore, the second quench from 523 K to room temperature develops a martensitic structure.
c. (Blue) An isothermal hold at 573 K for 500 seconds produces a half-bainite and half-austenite structure. Cooling quickly would result in a final structure of martensite and bainite.
d. (Orange) Austenite converts completely to fine pearlite after eight seconds at 873 K. This phase is stable and will not be changed on holding for 100,000 seconds at 873 K. The final structure, when cooled, is fine pearlite.
source : http://www.sv.vt.edu/classes/MSE2094_NoteBook/96ClassProj/examples/kimttt.html
