Homework%202%20solution - IE370 Homework 2. Due Feb 2nd...

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IE370 Homework 2. Due Feb 2 nd name: 1. Describe the relative ability of iron to dissolve carbon in solution when in the form of austenite ( the elevated temperature phase) and when in the form of ferrite at room temperature (2’) Ans: At an elevated temperature, pure iron assumes a face centered cubic structure known as austenite. The key features of austenite include high solubility of carbon, at most 2.11% of carbon can be dissolved in austenite. At room temperature, there is almost no carbon in ferrite, as the temperature increase to eutectic temperature, ferrite can hold about 0.02% carbon in a solid solution. 2. At 0.77% Carbon, austenite transforms to pearlite (mixture of two phases: ferrite and cementite:Fe 3 C) when it slowly cools from 727C 0 below 725C 0 , what is the element composition of these two phases in pearlite), and compositions of each of the two phases. (3’). At 727 ° C, the composition is γ with 100% At 725 ° C, the compositions are α plus 3 Fe C . Ratio of : (6.7-0.77)/(6.7-0.02)=88.8% 3 Fe C : (0.77-0.02)/(6.7-0.02)=11.2% 3. For hypoeutectoid steel (C% less than 0.77%), what structure will it assume upon slow
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This note was uploaded on 05/05/2011 for the course IE 370 taught by Professor Chunghorng,r during the Spring '08 term at Purdue University-West Lafayette.

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Homework%202%20solution - IE370 Homework 2. Due Feb 2nd...

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