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Chapter 9: Problem 57

Compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron-carbon alloy.

Short Answer

Expert verified
Answer: The maximum mass fraction of proeutectoid cementite in a hypereutectoid iron-carbon alloy is approximately 0.231 (23.1%).

Step by step solution

01

Understand the iron-carbon phase diagram and hypereutectoid alloys

Hypereutectoid iron-carbon alloys have a carbon content greater than the eutectoid composition (0.77 wt% C) and less than approximately 2.14 wt% C. In these alloys, proeutectoid cementite (Fe3C) forms before the eutectoid reaction occurs. After the eutectoid reaction, the remaining austenite is converted into pearlite, which is a mixture of ferrite (α-Fe) and cementite (Fe3C).
02

Computation of maximum mass fraction of proeutectoid cementite

We need to find the maximum mass fraction of proeutectoid cementite for an iron-carbon alloy. In order to do this, we can use the lever rule. The lever-rule formula for mass fraction is given by: \(mx_{cementite} = \frac{C_{total} - C_{eutectoid}}{C_{cementite} - C_{eutectoid}}\) where, \(mx_{cementite}\) - mass fraction of proeutectoid cementite \(C_{total}\) - total carbon content of the alloy \(C_{eutectoid}\) - eutectoid carbon content (0.77 wt% C) \(C_{cementite}\) - composition of the cementite phase (6.7 wt% C)
03

Calculate the mass fraction for maximum proeutectoid cementite

Let's assume the total carbon content of our hypereutectoid alloy is the maximum, i.e., 2.14 wt% C. Now, we can plug these values into the lever rule formula for mass fraction: \(mx_{cementite} = \frac{2.14 - 0.77}{6.7 - 0.77}\) \(mx_{cementite} = \frac{1.37}{5.93}\) \(mx_{cementite} \approx 0.231\) The maximum mass fraction of proeutectoid cementite in a hypereutectoid iron-carbon alloy is approximately 0.231 (23.1%).

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Most popular questions from this chapter

For alloys of two hypothetical metals \(\mathrm{A}\) and B, there exist an \(\alpha\), A-rich phase and a \(\beta\), Brich phase. From the mass fractions of both phases for two different alloys provided in the following table (which are at the same temperature), determine the composition of the phase boundary (or solubility limit) for both \(\alpha\) and \(\beta\) phases at this temperature. $$ \begin{array}{lcc} \hline \text { Alloy Composition } & \text { Fraction } & \text { Fraction } \\\ & \alpha \text { Phase } & \beta \text { Phase } \\ \hline 60 \mathrm{wt} \% \mathrm{~A}-40 \mathrm{wt} \% \text { B } & 0.57 & 0.43 \\ 30 \mathrm{wt} \% \text { A-70 wt \% B } & 0.14 & 0.86 \\ \hline \end{array} $$

For a copper-silver alloy of composition 25 \(\mathrm{wt} \% \mathrm{Ag}-75 \mathrm{wt} \% \mathrm{Cu}\) and at \(775^{\circ} \mathrm{C}\left(1425^{\circ} \mathrm{F}\right)\), do the following: (a) Determine the mass fractions of \(\alpha\) and \(\beta\) phases. (b) Determine the mass fractions of primary \(\alpha\) and eutectic microconstituents. (c) Determine the mass fraction of eutectic \(\alpha\).

What is the principal difference between congruent and incongruent phase transformations?

Consider \(2.5 \mathrm{~kg}\) of austenite containing \(0.65\) wt \(\%\) C, cooled to below \(727^{\circ} \mathrm{C}\left(1341^{\circ} \mathrm{F}\right)\). (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.

For an iron-carbon alloy of composition 5 wt \% C-95 wt \(\%\) Fe, make schematic sketches of the microstructure that would be observed for conditions of very slow cooling at the following temperatures: \(1175^{\circ} \mathrm{C}\left(2150^{\circ} \mathrm{F}\right), 1145^{\circ} \mathrm{C}\) \(\left(2095^{\circ} \mathrm{F}\right)\), and \(700^{\circ} \mathrm{C}\left(1290^{\circ} \mathrm{F}\right)\). Label the phases and indicate their compositions (approximate).
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