Chap 4 Solns - s3.amazonaws.com

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,


 Q 
Nv
0.55 eV / atom
= exp − v  = exp −

 kT 
N
 (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
AAl
(6.022 × 10 23 atoms / mol)(2.62 g / cm3)
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,
Q
ln N v = ln N − v
kT
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and, after some algebraic manipulation
N 
Qv = − kT ln  v 
 N 
 7.57 × 10 23 m−3 
= − (8.62 × 10 -5 eV/atom - K) (500°C + 273 K) ln 

 5.85 × 10 28 m−3 
= 0.75 eV/atom
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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
'
=
C Zn
=
C Zn ACu
× 100
C Zn ACu + CCu AZn
(30)(63.55 g / mol)
× 100
(30)(63.55 g / mol) + (70)(65.41 g / mol)
= 29.4 at%
'
CCu
=
=
CCu AZn
× 100
C Zn ACu + CCu AZn
(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 = 2 n−1
= 28−1 = 128 grains/in.2
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(b) Now it is necessary to compute the value of N for no magnification. In order to solve this problem it is
necessary to use Equation 4.17:
 M 2
n−1
NM
 =2
 100 
where NM = the number of grains per square inch at magnification M, and n is the ASTM grain size number.
Without any magnification, M in the above equation is 1, and therefore,
 1 2
N1 
 = 28−1 = 128
 100 
And, solving for N1, N1 = 1,280,000 grains/in.2.
5. The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute
the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of
0.20 m2 at 500°C. Assume a diffusion coefficient of 1.0 × 10-8 m2/s, that the concentrations at the high- and lowpressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state
conditions have been attained.
Solution
This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first becomes
necessary to employ both Equations 5.1a and 5.3. Combining these expressions and solving for the mass yields
M = JAt = − DAt
∆C
∆x
 0.6 − 2.4 kg / m3 
= − (1.0 × 10 -8 m2 /s)(0.20 m2 ) (3600 s/h) 

 5 × 10−3 m 
= 2.6 × 10-3 kg/h
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6. Determine the carburizing time necessary to achieve a carbon concentration of 0.45 wt% at a position 2 mm into
an iron–carbon alloy that initially contains 0.20 wt% C. The surface concentration is to be maintained at 1.30 wt%
C, and the treatment is to be conducted at 1000°C. Use the diffusion data for γ-Fe in Table 5.2.
Solution
In order to solve this problem it is first necessary to use Equation 5.5:
C x − C0
 x 
= 1 − erf 

 2 Dt 
Cs − C0
wherein, Cx = 0.45, C0 = 0.20, Cs = 1.30, and x = 2 mm = 2 × 10-3 m. Thus,
 x 
0.45 − 0.20
Cx − C0
=
= 0.2273 = 1 − erf 

 2 Dt 
Cs − C0
1.30 − 0.20
or
 x 
erf 
 = 1 − 0.2273 = 0.7727
 2 Dt 
By linear interpolation using data from Table 5.1
z
erf(z)
0.85
0.7707
z
0.7727
0.90
0.7970
z − 0.850
0.7727 − 0.7707
=
0.900 − 0.850 0.7970 − 0.7707
From which
z = 0.854 =
x
2 Dt
Now, from Table 5.2, at 1000°C (1273 K)
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

148, 000 J/mol
D = (2.3 × 10 -5 m2 /s) exp −

 (8.31 J/mol- K)(1273 K) 
= 1.93 × 10-11 m2/s
Thus,
0.854 =
2 × 10−3 m
(2) (1.93 × 10−11 m2 /s) (t)
Solving for t yields
t = 7.1 × 104 s = 19.7 h
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