83.0k views
3 votes
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.)

User MatG
by
5.0k points

1 Answer

7 votes

Answer:

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

Step-by-step 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

User Eranda
by
5.5k points