Isothermal transformation diagram (picture included)

[qwerty0123]

using the isothermal transformation diagram provided of an iron-carbon alloy at eutectoid composition, determine the final microconstituants and approximate percentages of a small specimen that has been subjected to the following time-temperature heat treatments. in each case assue the specimen is initially 800C.


A)Cool rapidly to 400C, hold for 2seconds and then quench to room temperature.
b) Rapidly cool to 575C, hold for 20seconds, rapidly to 350C, hold for 100seconds and quench to room temperature.


diagram:
http://imageshack.us/photo/my-images/143/img0142lk.jpg/

have no idea what I am supposed to be doing :/

 

[KataKoniK]

Thank you for your post. I am a scientist and I would be glad to help you with your question. Based on the information provided, I understand that you are trying to determine the final microconstituents and approximate percentages of a small specimen that has been subjected to different heat treatments using the isothermal transformation diagram of an iron-carbon alloy at eutectoid composition.

To begin, let us first understand what an isothermal transformation diagram is. This diagram shows the microstructural changes that occur in an alloy as it is cooled at a constant temperature. In this case, the alloy is an iron-carbon alloy at eutectoid composition, which means it contains 0.77% carbon. The diagram shows the different phases that form at different temperatures and the time required for these transformations to occur.

Now, let us look at the first heat treatment condition: cooling rapidly to 400C, holding for 2 seconds and then quenching to room temperature. According to the diagram, at 800C, the alloy is in the austenite phase. As it is cooled to 400C, it will transform into pearlite, which is a mixture of ferrite and cementite. This transformation takes place in a very short time, as indicated by the 2 seconds holding time. Finally, when the alloy is quenched to room temperature, the pearlite phase will be retained. Therefore, the final microconstituents in this case will be pearlite and the approximate percentage will be 100%.

Now, let us consider the second heat treatment condition: rapidly cooling to 575C, holding for 20 seconds, rapidly cooling to 350C, holding for 100 seconds and quenching to room temperature. At 800C, the alloy is in the austenite phase. As it is cooled to 575C, it will partially transform into pearlite and partially into bainite. This transformation takes place in 20 seconds, as indicated by the holding time. When the alloy is further cooled to 350C, the remaining austenite will transform into bainite. This transformation takes place in 100 seconds. Finally, when the alloy is quenched to room temperature, the bainite phase will be retained. Therefore, the final microconstituents in this case will be bainite and the approximate percentage will be 100%.

I hope this helps you understand the process of determining