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Lecture

ENG 2002 Lecture Notes - Jmol, Linear Interpolation, Carburizing

6 Pages
73 Views
Winter 2014

Department
Engineering
Course Code
ENG 2002
Professor
Hany E.Z.Farag

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PROBLEM SOLUTIONS
1. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume
an energy for vacancy formation of 0.55 eV/atom.
Solution
In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation
4.1. As stated in the problem, Qv = 0.55 eV/atom. Thus,
N
v
N= exp Q
v
kT
= exp 0.55 eV/atom
(8.62 × 10
5
eV/atom- K)(600 K)
= 2.41 × 10-5
2. Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium number of
vacancies at 500
°
C (773 K) is 7.57
×
1023 m-3. The atomic weight and density (at 500
°
C) for aluminum are,
respectively, 26.98 g/mol and 2.62 g/cm3.
Solution
Upon examination of Equation 4.1, all parameters besides Qv are given except N, the total number of
atomic sites. However, N is related to the density, (ρAl), Avogadro's number (NA), and the atomic weight (AAl)
according to Equation 4.2 as
N = N
A
ρ
Al
A
Al
=
(
6.022 × 10
23
atoms /mol
)(
2.62 g /cm
3
)
26.98 g /mol
= 5.85 × 1022 atoms/cm3 = 5.85 × 1028 atoms/m3
Now, taking natural logarithms of both sides of Equation 4.1,
ln N
v
= ln N Q
v
kT
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
and, after some algebraic manipulation
Q
v
= kT ln N
v
N
=
(
8.62 × 10-5 eV/atom- K
)
(500°C + 273 K) ln 7.57 × 1023 m3
5.85 × 1028 m3
= 0.75 eV/atom
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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Description
PROBLEM SOLUTIONS 1. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy for vacancy formation of 0.55 eV/atom. Solution In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation 4.1. As stated in the problem, Q= 0.55 eV/atom. Thus, v N v  Q v  0.55 eV/atom  = exp   = exp  −5  N  kT   (8.62 × 10 eV/atom- K )(600 K)  = 2.41 × 10-5 2. Calculate the activation energy for vacancy formati on in aluminum, given that the equilibrium number of vacancies at 500 °C (773 K) is 7.57 × 10 23m . The atomic weight and density (at 500 °C) for aluminum are, respectively, 26.98 g/mol and 2.62 g/cm . Solution Upon examination of Equation 4.1, all parameters besides Q are given except N, the total number of v atomic sites. However, N is related to the density, ( ρ ), Avogadro's number ( N ), and the atomic weight ( A ) Al A Al according to Equation 4.2 as N A Al N = AAl (6.022 × 10 23 atoms/mol )(2.62 g/cm 3) = 26.98 g/mol = 5.85 × 1022 atoms/cm = 5.85 × 10 28 atoms/m 3 Now, taking natural logarithms of both sides of Equation 4.1, lnN = lnN − Q v v kT Excerpts from this work may be reproduced by instructors for distribution on a not -for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook hasAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. and, after some algebraic manipulation N  Q = − kT ln  v  v  N  -5  7.57 × 10 23 m −3 = − (8.62 × 10 eV/atom - K)(500°C + 273 K) ln 28 −3  5.85 × 10 m  = 0.75 eV/atom Excerpts from this work may be reproduced by instructors for distribution on a not -for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook haAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3. What is the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu? Solution In order to compute composition, in atom percent, of a 30 wt% Zn -70 wt% Cu alloy, we employ Equation 4.6 as ' CZnACu CZn = C A +C A × 100 Zn Cu Cu Zn (30)(63.55 g/mol) = (30)(63.55 g/mol) + (70)(65.41 g/mol) 100 = 29.4 at% C ' = CCu Zn × 100 Cu C Zn Cu +CCu Zn (70)(65.41 g/mol) = × 100 (30)(63.55 g/mol) + (70)(65.41 g/mol) = 70.6 at% 4. For an ASTM grain size of 8, approximately how many grains would there be per square inch at (a) a magnification of 100, and (b) without any magnification? Solution (a) This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 8 at a magnification of 100 ×. All we need do is solve for the parameter N in Equation 4.16, inasmuch as n = 8. Thus n−1 N = 2 8−1 2 = 2 = 128 grains/in. Excerpts from this work may be reproduced by instructors for distribution on a not -for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textboAny other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (b) Now it is necessary to compute the value of N for no magnification. In order to s
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