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2 years ago

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Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 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? (d) Schematically sketch and label the resulting microstructure

1 answer:

sergejj [24]2 years ago

6 0

Answer:

a) The proeutectoid phase is known like cementite and its chemical formula is Fe₃C

b) The kilograms of total ferrite and is 0.8311 kg

The kilograms of total cementite is 0.1689 kg

c) The kilograms of total cementite is 0.9343 kg

Explanation:

Given:

1 kg of austenite

1.15 wt% C

Cooled to below 727°C

Questions:

a) What is the proeutectoid phase?

b) How many kilograms each of total ferrite and cementite form, Wf = ?, Wc = ?

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

d) Schematically sketch and label the resulting microstructure

a) The proeutectoid phase is known like cementite and its chemical formula is Fe₃C

b) To get the mass of the total ferrite form:

$W_{f} =\frac{C_{cementite}-C_{2} }{C_{cementite}-C_{1} }$

Here,

Ccementite = composition of cementite = 6.7 wt%

C₁ = composition of phase 1 = 0.022 wt%

C₂ = composition of alloy = 1.15 wt%

Substituting values:

$W_{f} =\frac{6.7-1.15}{6.7-0.022} =0.8311kg$

To get the mass of the total cementite:

$W_{c} =\frac{C_{2}-C_{1}}{C_{cementite}-C_{1} } =\frac{1.15-0.022}{6.7-0.022} =0.1689kg$

c) The mass of pearlite:

$W_{p} =\frac{6.7-1.15}{5.94} =0.9343kg$

d) In the diagram you can see the different compositions (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)

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