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Chapter 9: Problem 66

A steel alloy is known to contain \(93.8\) wt \(\%\) \(\mathrm{Fe}, 6.0 \mathrm{wt} \% \mathrm{Ni}\), and \(0.2 \mathrm{wt} \% \mathrm{C}\). (a) What is the approximate eutectoid temperature of this alloy? (b) What is the proeutectoid phase when this alloy is cooled to a temperature just below the eutectoid? (c) Compute the relative amounts of the proeutectoid phase and pearlite. Assume that there are no alterations in the positions of other phase boundaries with the addition of Ni.

Short Answer

Expert verified
The given steel alloy has a composition of approximately \(99.8 \mathrm{wt} \% \mathrm{Fe}\) and \(0.2 \mathrm{wt} \% \mathrm{C}\). The approximate eutectoid temperature for the alloy is \(727^{\circ} \mathrm{C}\). The proeutectoid phase for this alloy is ferrite (α-Fe), as its carbon content is less than the eutectoid composition. The relative amounts of proeutectoid ferrite and pearlite in the alloy are approximately \(73.68 \%\) and \(26.32 \%\) respectively.

Step by step solution

01

For this problem, we will assume that the presence of Ni does not affect the phase boundaries in the Fe-C phase diagram. Therefore, we only need to consider the \(\mathrm{Fe}\) and \(\mathrm{C}\) contents of the alloy. Given that the alloy contains \(93.8 \mathrm{wt} \% \mathrm{Fe}, 6.0 \mathrm{wt} \% \mathrm{Ni}\), and \(0.2 \mathrm{wt} \% \mathrm{C}\), we will consider the alloy to be \(99.8 \mathrm{wt} \% \mathrm{Fe}\) and \(0.2 \mathrm{wt} \% \mathrm{C}\) for the purpose of analyzing its behavior on the Fe-C phase diagram. **Step 2: Finding the eutectoid temperature**

On the Fe-C phase diagram, the eutectoid temperature is the temperature at which the eutectoid transformation from austenite (\(\gamma\)-Fe) to pearlite (ferrite and cementite phases) occur. This temperature can be found on the phase diagram at the intersection of ferrite-austenite boundary and ferrite-cementite boundary, which is approximately \(727 ^\circ \mathrm{C}\). **Step 3: Determining the proeutectoid phase**
02

The proeutectoid phase refers to the phase that forms just before the eutectoid transformation takes place when the alloy is cooled. If the carbon content in the alloy is less than the eutectoid composition (approximately \(0.76 \mathrm{wt} \% \mathrm{C}\)), the proeutectoid phase will be ferrite (α-Fe). In our case, the alloy contains \(0.2 \mathrm{wt} \% \mathrm{C}\), which is less than the eutectoid composition. Therefore, the proeutectoid phase is ferrite (α-Fe). **Step 4: Calculating the relative amounts of proeutectoid phase (ferrite) and pearlite**

Using the lever rule, we can calculate the relative amounts of ferrite and pearlite phases in the alloy. The lever rule is given by: \(\text{Phase fraction}_1 = \frac{C_2 - C_\text{avg}}{C_2 - C_1}\) and \\ \(\text{Phase fraction}_2 = \frac{C_\text{avg} - C_1}{C_2 - C_1}\) where \(\text{Phase fraction}_1\) and \(\text{Phase fraction}_2\) are the fractions of phase 1 and phase 2 respectively, \(C_\text{avg}\) is the average composition of the alloy, and \(C_1\) and \(C_2\) are the compositions of the phase boundaries for the two phases. In our case, the ferrite is the proeutectoid phase, and its boundary composition is \(0 \mathrm{wt} \% \mathrm{C}\) at the eutectoid temperature. The pearlite is composed of both ferrite and cementite phases, but we refer to the total composition of the pearlite phase: \(0.76 \mathrm{wt} \% \mathrm{C}\) at eutectoid temperature. The average composition is given by the \(\mathrm{C}\) content in the alloy: \(0.2 \mathrm{wt} \% \mathrm{C}\). Plugging in these values into the lever rule equations: \(\text{Ferrite fraction} = \frac{0.76 - 0.2}{0.76 - 0} = \frac{0.56}{0.76} = 0.7368\) \(\text{Pearlite fraction} = \frac{0.2 - 0}{0.76 - 0} = \frac{0.2}{0.76} = 0.2632\) So, the relative amounts of proeutectoid ferrite and pearlite in the alloy are approximately \(73.68 \%\) and \(26.32 \%\) respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eutectoid Temperature
When studying the fascinating world of metallurgy and materials science, one vital concept is the eutectoid temperature. This critical temperature is essential when dealing with alloys, like steel, as it represents the point at which a single solid phase transforms into two other distinct solid phases simultaneously during cooling.

The eutectoid temperature in the Fe-C (iron-carbon) phase diagram is specifically important for steel, as it determines when and how the steel's microstructure will change. For instance, in the context of our exercise, the eutectoid temperature for a steel alloy is a fixed point, normally at around 727°C, where the transformation of austenite (\f\(\textbf{\textgamma}-Fe\f\)) to a mixture known as pearlite occurs.

This temperature is a crucial pivot point in heat treatment processes, reflecting the alloy's behaviors such as hardness, strength, and ductility will be determined by how it is cooled through this temperature. As the carbon content plays a significant role in determining the steel's properties, the eutectoid temperature helps provide insights into the expected mechanical properties of the finished product.
Proeutectoid Phase
The proeutectoid phase is a part of the Fe-C phase diagram narrative that often perplexes students. To shed some light, 'proeutectoid' refers to the phase that forms first, as an alloy is cooled down just before the eutectoid transformation happens.

For an alloy like the one mentioned in our exercise with carbon content below the eutectoid composition, the proeutectoid phase will be ferrite (\f\(\textbf{\textalpha}-Fe\f\)). This softer and more ductile phase forms in the steel as it cools from high temperatures down to the eutectoid temperature, where it subsequently undergoes a transformation to pearlite.

Understanding the proeutectoid phase is key to predicting the characteristics of steel. For example, in hypo-eutectoid steels (less carbon), ferrite forms on the grain boundaries of austenite, effectively outlining where the subsequent pearlitic transformation will occur. The right amount and distribution of this phase is essential for certain steel applications, affecting its weldability and toughness.
Lever Rule
Another intriguing concept in the context of phase diagrams is the lever rule, a mathematical tool used to determine the proportions of different phases present in an alloy at equilibrium. It's like a seesaw – balancing the concentrations on either side of the overall composition.

To apply the lever rule, as we did in the step-by-step solution, we consider the weight percent of carbon at the boundaries of the phases in question and the overall carbon content (\f\(\textbf{\textC}_{\textbf{\textavg}}\f\)) of the alloy. It's all about leverage – calculations relying on the distances between these concentrations on the phase diagram.

In the exercise provided, we used the lever rule to calculate the amounts of proeutectoid ferrite and pearlite, identifying their fractions within a steel alloy. This tells us how much of each phase we have, significantly impacting the alloy's properties. For students and professionals alike, mastering the lever rule paves the way to better control over material properties, knowing precisely what you'll get out of the furnace.
Pearlite
Finally, let's talk about pearlite, a two-phased, lamellar (layered) structure composed of alternating layers of ferrite and cementite that forms in steel during the cooling process, specifically at the eutectoid temperature of 727°C.

Pearlite is fundamental to steel's strength and hardness. Observing it under the microscope reveals its characteristic appearance, much like mother of pearl, hence the name. The thinner and more numerous the layers, the harder and stronger the steel becomes. Such intricacies can be controlled through heat treatment processes.

In our exercise, pearlite forms when the steel alloy we are analyzing is cooled to and through the eutectoid temperature. The proportion of pearlite in the alloy affects its overall hardness and tensile strength, making it a critical phase for engineers to understand. By adjusting the cooling rate and other treatment parameters, metallurgists can manipulate the amount and characteristics of pearlite in steel, tailoring the material to its intended application.

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Most popular questions from this chapter

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

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).

For \(11.20 \mathrm{~kg}\) of a magnesium-lead alloy of composition \(30 \mathrm{wt} \% \mathrm{~Pb}-70 \mathrm{wt} \% \mathrm{Mg}\), is it possible, at equilibrium, to have \(\alpha\) and \(\mathrm{Mg}_{2} \mathrm{~Pb}\) phases having respective masses of \(7.39 \mathrm{~kg}\) and \(3.81 \mathrm{~kg}\) ? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, explain why.

A hypothetical A-B alloy of composition \(55 \mathrm{wt} \%\) B-45 \(\mathrm{wt} \% \mathrm{~A}\) at some temperature is found to consist of mass fractions of \(0.5\) for both \(\alpha\) and \(\beta\) phases. If the composition of the \(\beta\) phase is \(90 \mathrm{wt} \%\) B-10 \(\mathrm{wt} \% \mathrm{~A}\), what is the composition of the \(\alpha\) phase?

Two intermetallic compounds, \(\mathrm{AB}\) and \(\mathrm{AB}_{2}\), exist for elements \(\mathrm{A}\) and \(\mathrm{B}\). If the compositions for \(\mathrm{AB}\) and \(\mathrm{AB}_{2}\) are \(34.3 \mathrm{wt} \% \mathrm{~A}-65.7\) \(\mathrm{wt} \% \mathrm{~B}\) and \(20.7 \mathrm{wt} \% \mathrm{~A}-79.3 \mathrm{wt} \% \mathrm{~B}\), respectively, and element \(\mathrm{A}\) is potassium, identify element B.
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