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

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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

Engineering

1 answer:

pantera1 [306] · Answer 1 · 2026-02-22T19:23:25+03:00

Answer:

a) The phase before eutectoid is commonly referred to as cementite, with the chemical formula Fe₃C.

b) The total mass of ferrite obtained is 0.8311 kg.

The total cementite mass equals 0.1689 kg.

c) The total cementite mass accounts for 0.9343 kg.

Explanation:

Provided:

1 kg of austenite

a carbon content of 1.15 wt%

Cooled below 727°C

Questions:

a) Identify the proeutectoid phase.

b) Calculate the mass of total ferrite and cementite, Wf =?, Wc =?

c) Determine the mass of both pearlite and the proeutectoid phase, Wp =?

d) Create a schematic to illustrate the resulting microstructure.

a) The proeutectoid phase is referred to as cementite with the formula Fe₃C.

b) To find the total mass of formed ferrite:

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

With:

Ccementite = composition of cementite = 6.7 wt%

C₁ = composition of phase 1 = 0.022 wt%

C₂ = overall composition = 1.15 wt%

Inserting the values yields:

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

For the total mass of 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) The diagram illustrates the different compositions: (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)