Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 18

What is the driving force for the formation of spheroidite?

Short Answer

Expert verified
Answer: The driving force for the formation of spheroidite is the reduction in the system's free energy through the minimization of interface energy between the ferrite and cementite phases. This is achieved by the diffusion of carbon atoms at high temperatures, allowing the formation of round cementite particles within the ferrite matrix during the slow cooling process of heat treatment.

Step by step solution

01

Understand Spheroidite Formation

Spheroidite forms during the heat treatment of steel, specifically when steel is heated to a temperature that is above the lower critical temperature and then cooled slowly. At these temperatures, the steel undergoes several phase transformations, and spheroidite is one microstructure that can form as a result of the slow cooling process.
02

Identify the Components of Spheroidite

Spheroidite has a distinct microstructure that consists of round cementite particles, which are a type of iron carbide (Fe3C) dispersed within a ferrite matrix. Ferrite is a solid solution of iron and carbon that has a body-centered cubic crystal structure, while cementite is an intermetallic compound with an orthorhombic crystal structure.
03

Understand the Thermodynamic Driving Force

The driving force for the formation of any microstructure is the reduction of the system's free energy. In the case of spheroidite, the arrangement of the cementite particles within the ferrite matrix reduces the overall energy of the system. This is because at the microstructural level, the spheroidite structure minimizes the interface energy (surface energy) between the ferrite and cementite phases.
04

Relate the Formation to Diffusion

At high temperatures during the heat treatment process, carbon atoms within the steel can diffuse relatively easily. The slow cooling process allows sufficient time for the carbon atoms to diffuse and form the round cementite particles within the ferrite matrix. This diffusion process of carbon atoms, along with the reduction in interface energy, drives the formation of spheroidite.
05

Summarize the Driving Force for Spheroidite Formation

The driving force for the formation of spheroidite is the reduction in the system's free energy through the minimization of interface energy between the ferrite and cementite phases. This is achieved by the diffusion of carbon atoms at high temperatures, allowing the formation of round cementite particles within the ferrite matrix during the slow cooling process of heat treatment.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at the following rates: (a) \(0.005^{\circ} \mathrm{C} / \mathrm{s}\) (b) \(0.05^{\circ} \mathrm{C} / \mathrm{s}\) (c) \(0.5^{\circ} \mathrm{C} / \mathrm{s}\) (d) \(5^{\circ} \mathrm{C} / \mathrm{s}\)

Using the isothermal transformation diagram for an iron-carbon alloy of eutectoid composition (Figure \(10.22\) ), specify the nature of the final microstructure (in terms of microconstituents present and approximate percentages of each) of a small specimen that has been subjected to the following time-temperature treatments. In each case assume that the specimen begins at \(760^{\circ} \mathrm{C}\) \(\left(1400^{\circ} \mathrm{F}\right)\) and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Cool rapidly to \(350^{\circ} \mathrm{C}\left(660^{\circ} \mathrm{F}\right)\), hold for \(10^{3} \mathrm{~s}\), then quench to room temperature. (b) Rapidly cool to \(625^{\circ} \mathrm{C}\left(1160^{\circ} \mathrm{F}\right)\), hold for \(10 \mathrm{~s}\), then quench to room temperature.(c) Rapidly cool to \(600^{\circ} \mathrm{C}\left(1110^{\circ} \mathrm{F}\right)\), hold for \(4 \mathrm{~s}\), rapidly cool to \(450^{\circ} \mathrm{C}\left(840^{\circ} \mathrm{F}\right)\), hold for \(10 \mathrm{~s}\), then quench to room temperature. (d) Reheat the specimen in part (c) to \(700^{\circ} \mathrm{C}\) \(\left(1290^{\circ} \mathrm{F}\right)\) for \(20 \mathrm{~h}\). (e) Rapidly cool to \(300^{\circ} \mathrm{C}\left(570^{\circ} \mathrm{F}\right)\), hold for \(20 \mathrm{~s}\), then quench to room temperature in water. Reheat to \(425^{\circ} \mathrm{C}\left(800^{\circ} \mathrm{F}\right)\) for \(10^{3}\) s and slowly cool to room temperature. (f) Cool rapidly to \(665^{\circ} \mathrm{C}\left(1230^{\circ} \mathrm{F}\right)\), hold for \(10^{3} \mathrm{~s}\), then quench to room temperature. (g) Rapidly cool to \(575^{\circ} \mathrm{C}\left(1065^{\circ} \mathrm{F}\right)\), hold for \(20 \mathrm{~s}\), rapidly cool to \(350^{\circ} \mathrm{C}\left(660^{\circ} \mathrm{F}\right)\), hold for \(100 \mathrm{~s}\), then quench to room temperature. (h) Rapidly cool to \(350^{\circ} \mathrm{C}\left(660^{\circ} \mathrm{F}\right)\), hold for \(150 \mathrm{~s}\), then quench to room temperature.

(a) For the solidification of nickel, calculate the critical radius \(r^{*}\) and the activation free energy \(\Delta G^{*}\) if nucleation is homogeneous. Values for the latent heat of fusion and surface free energy are \(-2.53 \times 10^{9} \mathrm{~J} / \mathrm{m}^{3}\) and \(0.255 \mathrm{~J} / \mathrm{m}^{2}\), respectively. Use the supercooling value found in Table 10.1. (b) Now, calculate the number of atoms found in a nucleus of critical size. Assume a lattice parameter of \(0.360 \mathrm{~nm}\) for solid nickel at its melting temperature.

Briefly describe the simplest heat treatment procedure that would be used in converting a 0.76 wt% C steel from one microstructure to the other, as follows: (a) Martensite to spheroidite (b) Spheroidite to martensite (c) Bainite to pearlite (d) Pearlite to bainite (e) Spheroidite to pearlite (f) Pearlite to spheroidite (g) Tempered martensite to martensite (h) Bainite to spheroidite

Name the two stages involved in the formation of particles of a new phase. Briefly describe each.
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Famous Physicists

Read Explanation

Work Energy and Power

Read Explanation

Solid State Physics

Read Explanation

Electricity

Read Explanation

Turning Points in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free