Consider 6.0 kg of austenite containing 0.45 wt % C and cooled to less than 727^∘ C(1341^∘ F).

step 1 Guys, today we have problem 62 in chapter 9 about phase diagrams. He wants to know in iron carbo

Consider 6.0 kg of austenite containing 0.45 wt % C and...

Consider $6.0 \mathrm{~kg}$ of austenite containing $0.45 \mathrm{wt} \%$ $\mathrm{C}$ and cooled to less than $727^{\circ} \mathrm{C}\left(1341^{\circ} \mathrm{F}\right)$. (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.

Consider $6.0 \mathrm{~kg}$ of austenite containing $0.45 \mathrm{wt} \%$ $\mathrm{C}$ and cooled to less than $727^{\circ} \mathrm{C}\left(1341^{\circ} \mathrm{F}\right)$.
(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.

Key Concepts

Austenite

Austenite is the high-temperature, face-centered cubic phase of iron that can dissolve a significant amount of carbon. It is the starting microstructure in many heat treatment processes and transforms into other phases such as pearlite and proeutectoid ferrite upon cooling below critical temperatures.

Fe-C Phase Diagram

The iron–carbon (Fe-C) phase diagram is an essential tool in metallurgy that maps the phases present in steel as a function of temperature and carbon composition. It shows critical points such as the eutectoid composition and temperature, guiding the prediction of phase transformations during the cooling process.

Lever Rule

The lever rule is a graphical method used with phase diagrams to calculate the relative amounts of different phases present at equilibrium. It is essential for determining the phase fractions such as proeutectoid ferrite, cementite, and pearlite in steels based on the overall composition and the phase boundaries.

Proeutectoid Phase

A proeutectoid phase is the phase that forms prior to the eutectoid reaction during cooling. In hypoeutectoid steels (with carbon content less than the eutectoid composition of 0.8 wt% C), the proeutectoid phase is ferrite, which forms before the remaining austenite transforms into pearlite.

Pearlite

Pearlite is a two-phased, lamellar microstructure consisting of alternating layers of ferrite and cementite. It forms during the eutectoid transformation of austenite when the temperature is lowered below the eutectoid point, and its formation and fraction are determined by the initial carbon content of the steel.

Transcript

00:01 Guys, today we have problem 62 in chapter 9 about phase diagrams.

00:06 He wants to know in iron carbon phase diagram what is a breed? toid phase we have a alloy of 0 .5 % of carbon that is cool to less than 727 degree.

00:23 And he also want to know how many kilograms of fluoride and cementite and how many kilograms of the light and the brete optoid phase format and they want the micro structure of this all so first we can get we can draw the line of our alloy it is at 0 .45 so actually it will be here here we have a ferrite breed octoid form so the first question number a we have ferrite alpha we have ferrite alpha to know how many kilograms in to know how many kilograms of right and cementite this line represent the mass or the percent of a percent of kilograms of right so we can get it to be 6 .7 minus 8 percent of the alloy 0 .45 this will be divided by the total distance 6 .7 minus 0 .02 can calculate this on calculator.

01:57 So we have 6 .7 minus 0 .45 over 6 .7 minus 0 .02.

02:06 So actually it will be 0 .93.

02:10 So we have a total mass of 6 kilograms.

02:16 So mass of right will be equal over.

02:22 Point nine three times six kilogram actually to be equal to 5 .58 kilograms as a mass of cementite would be equal 5 .58 minus six so it will be equal 5 .58 minus six so it will be equal 0 .42 kilogram.

03:14 All right this is the first three equations after that he want to know how many kilograms in each of the light and the blue tectoid phase form here we will work on this this period between 0 .02 and 0 .8 which is a tectoid point.

03:47 Alright, so to know the mass of pearl light, we actually should work on this period.

04:06 So mass of light will be equal 0 .0 .8 minus 0 .45 minus 0 .02 over 0 .8, minus 0 .02 so i have 0 .45 minus 0 .02 over 0 .8 minus 0 .02...

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