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.)

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

answer:

an example of solubility is salt or sugar dissolving in water.

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