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15 May, 16:51

Consider 8.0 kg of austenite containing 0.45 wt% C and cooled to less than 727°C (1341°F). (a) What is the proeutectoid phase? (b) How many kilograms each of total ferrite and cementite form? (c) How many kilograms each of pearlite and the proeutectoid phase form? (15 pts.)

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  1. 15 May, 17:21
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    a is formed above the eutecoid temperature as well (pro-eutecoid ferrite)

    Thus ferrite is proeutectoid phase since 0.45 wt% C is less than 0.76 wt% C

    b 7.487 kg of total ferrite

    0.51273 kg of total cementite

    c 4.62703 kg of total pearlite

    3.35135 kg of total proeutectoid ferrite

    Explanation:

    a) What is the proeutectoid phase

    is formed above the eutecoid temperature as well (pro-eutecoid ferrite)

    Thus ferrite is proeutectoid phase since 0.45 wt% C is less than 0.76 wt% C

    (b) How many kilograms each of total ferrite and cementite form

    Wα = CFe3C - Co = 6.70 - 0.45 = 0.93590

    CFe3C - Cα 6.70 - 0.022

    Which corresponds to 0.93590 x 8.0 kg = 7.487 kg of total ferrite

    Similarly for total cementite

    WCFe3C = Co - Cα = 0.45 - 0.022 = 0.006409

    CFe3C - Cα 6.70 - 0.022

    which corresponds to 0.006409 x 8. 0 kg = 0.51273 kg of total cementite

    c) How many kilograms each of pearlite and the proeutectoid phase form?

    considering the amount of pearlite and the proeutectoid phase ferrite formed

    Wp = Co¹ - 0.022 = 0.45 - 0.022 = 0.5784

    0.74 0.74

    which corresponds to 0.5784 x 8. 0 kg = 4.62703 kg of total pearlite

    Wα = 0.76 - 0.45 = 0.41892

    0.74

    which corresponds to 0.418.92 x 8. 0 kg = 3.35135 kg of total proeutetcoid ferrite
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