What phases in what proportions approximately would be present in a small sectio

Question

What phases in what proportions approximately would be present in a small section of 1080 steel at the end of the following sequence: 1) soak at 750°C for 2 h; 2) quench to 600° C and hold for 10 seconds; 3) qunch to room temperature (25°C). 4. The "small section" implies rapid transitions between temperatures are possible (i.e., instant quenches) After 1) the steel is 100% -austenite. After 2), the microstructure is fully transformed to pearlite. After 3, the pearlite is still stable. e +earbide r End of 4001 M. Sart of transformation 200 Hence, the phases are -ferrite and Fe3C (cementite/carbide). We use the lever rule to determine proportions. From memory, Fe3C is at 6.7 w/o C, a-ferrite is at 0.022 w/o C Y+M 1.0 10 100 1000 Tine, sec (log s : %-ferrite = (6.7-0.8)/(6.7-0.022). 100%-88.3%, %FeaC= 10.7%

Explanation / Answer

1 It can be clearly seen from the phase diagram that 1080 Steel will remain as Pearlite at the room temperature. The temperature at which steel will convert into Martensite will be at around -20oC. So, there is no doubt that it will be a Pearlite at the room temperature. Other way of forming Martensite is through very rapid Quenching.

2. The values 6.7 and 0.022 represents the weight %(concentration) of Carbon that is present in the Fe3C and Alpha-ferrite. This value is fixed and can be seen from the Fe-C phase Diagram.

In the lever rule we use the values of the composition of the Carbon and this has to be remembered!

In case the value is not remembered, then it can be looked in any material science book which has Fe-C Phase Diagram.

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