5.3
(a) Compare interstitial and vacancy atomic mechanisms for diffusion.
(b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.
Solution
(a) With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy.
Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism.
On the other
hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion
mechanism.
(b) Interstitial diffusion is normally more rapid than vacancy diffusion because:
atoms, being smaller, are more mobile;
(1) interstitial
and (2) the probability of an empty adjacent interstitial site is
greater than for a vacancy adjacent to a host (or substitutional impurity) atom.
5.10 Show that
Cx =
⎛ x2 ⎞
B
exp ⎜−
⎟
Dt
⎝ 4 Dt ⎠
is also a solution to Equation 5.4b. The parameter B is a constant, being independent of both x and t.
Solution
It can be shown that
Cx =
⎛ x2 ⎞
B
exp ⎜−
⎟
Dt
⎝ 4 Dt ⎠
is a solution to
∂C
∂ 2C
= D
∂t
∂x 2
simply by taking appropriate derivatives of the Cx expression. When this is carried out,
∂ 2C
B
∂C
= D
=
2
1/
∂t
∂x
2 D 2 t 3/ 2
⎛ x2
⎞
⎛ x2 ⎞
⎟⎟
⎜⎜
− 1 ⎟⎟ exp ⎜⎜−
⎝ 2 Dt
⎠
⎝ 4 Dt ⎠
5.11 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 ⎞
Cx − C0
0.45 − 0.20
=
= 0.2273 = 1 − erf ⎜
⎟
⎝ 2 Dt ⎠
1.30 − 0.20
Cs − C0
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 =
Now, from Table 5.2, at 1000°C (1273 K)
x
2 Dt
⎡
⎤
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
5.D3 The wear resistance of a steel shaft is to be improved by hardening its surface. This is to be
accomplished by increasing the nitrogen content within an outer surface layer as a result of nitrogen
diffusion into the steel. The nitrogen is to be supplied from an external nitrogen-rich gas at an
elevated and constant temperature. The initial nitrogen content of the steel is 0.002 wt%, whereas the
surface concentration is to be maintained at 0.50 wt%. For this treatment to be effective, a nitrogen
content of 0.10 wt% must be established at a position 0.40 mm below the surface. Specify appropriate
heat treatments in terms of temperature and time for temperatures between 475°C and 625°C. The
preexponential and activation energy for the diffusion of nitrogen in iron are 3 × 10-7 m2/s and 76,150
J/mol, respectively, over this temperature range.
Solution
This is a nonsteady-state diffusion situation; thus, it is necessary to employ Equation 5.5, utilizing
the following values for the concentration parameters:
C0 = 0.002 wt% N
Cs = 0.50 wt% N
Cx = 0.10 wt% N
Therefore
0.10 − 0.002
Cx − C0
=
0.50 − 0.002
Cs − C0
⎛ x ⎞
= 0.1968 = 1 − erf ⎜
⎟
⎝ 2 Dt ⎠
And thus
⎛ x ⎞
1 − 0.1968 = 0.8032 = erf ⎜
⎟
⎝ 2 Dt ⎠
Using linear interpolation and the data presented in Table 5.1
z
erf (z)
0.9000
0.7970
y
0.8032
0.9500
0.8209
y − 0.9000
0.8032 − 0.7970
=
0.8209 − 0.7970
0.9500 − 0.9000
From which
y =
x
= 0.9130
2 Dt
The problem stipulates that x = 0.40 mm = 4.0 × 10-4 m. Therefore
4.0 × 10−4 m
= 0.9130
2 Dt
Which leads to
Dt = 4.80 × 10-8 m2
Furthermore, the diffusion coefficient depends on temperature according to Equation 5.8; and, as
stipulated in the problem statement, D0 = 3 × 10-7 m2/s and Qd = 76,150 J/mol.
Hence
⎛ Q ⎞
Dt = D0 exp ⎜− d ⎟ (t) = 4.80 × 10 -8 m2
⎝ RT ⎠
(3.0
⎡
76,150 J/mol ⎤
−8
2
× 10 -7 m2 /s) exp ⎢−
⎥ (t) = 4.80 × 10 m
(8.31
J/mol - K)(T) ⎦
⎣
And solving for the time t
t (in s) =
0.160
⎛ 9163.7 ⎞
exp ⎜−
⎟
⎝
T ⎠
Thus, the required diffusion time may be computed for some specified temperature (in K). Below are
tabulated t values for three different temperatures that lie within the range stipulated in the problem.
Temperature
(°C)
Time
s
h
500
22,500
6.3
550
11,000
3.1
600
5800
1.6
5.22 The diffusion coefficients for silver in copper are given at two temperatures:
T (°C)
D (m2/s)
650
5.5 × 10–16
900
1.3 × 10–13
(a) Determine the values of D0 and Qd.
(b) What is the magnitude of D at 875°C?
Solution
(a) Using Equation 5.9a, we set up two simultaneous equations with Qd and D0 as unknowns as
follows:
ln D1 = lnD0 −
Qd ⎛ 1 ⎞
⎜ ⎟
R ⎜⎝ T1 ⎟⎠
ln D2 = lnD0 −
Qd ⎛ 1 ⎞
⎜ ⎟
R ⎜⎝ T2 ⎟⎠
Solving for Qd in terms of temperatures T1 and T2 (923 K [650°C] and 1173 K [900°C]) and D1 and
D2 (5.5 × 10-16 and 1.3 × 10-13 m2/s), we get
Qd = − R
= −
ln D1 − ln D2
1
1
−
T1 T2
[
]
(8.31 J/mol - K) ln (5.5 × 10 -16) − ln (1.3 × 10 -13)
1
1
−
1173 K
923 K
= 196,700 J/mol
Now, solving for D0 from Equation 5.8 (and using the 650°C value of D)
⎛Q ⎞
D0 = D1 exp ⎜⎜ d ⎟⎟
⎝ RT1 ⎠
⎡
⎤
196, 700 J/mol
= (5.5 × 10 -16 m2 /s) exp ⎢
⎥
⎣ (8.31 J/mol - K)(923 K) ⎦
= 7.5 × 10-5 m2/s
(b) Using these values of D0 and Qd, D at 1148 K (875°C) is just
⎡
⎤
196, 700 J/mol
D = (7.5 × 10 -5 m2 /s) exp ⎢−
⎥
⎣ (8.31 J/mol - K)(1148 K) ⎦
= 8.3 × 10-14 m2/s
Note: this problem may also be solved using the “Diffusion” module in the VMSE software. Open
the “Diffusion” module, click on the “D0 and Qd from Experimental Data” submodule, and then do the
following:
1. In the left-hand window that appears, enter the two temperatures from the table in the book
(converted from degrees Celsius to Kelvins) (viz. “923” (650ºC) and “1173” (900ºC), in the first two
boxes under the column labeled “T (K)”. Next, enter the corresponding diffusion coefficient values
(viz. “5.5e-16” and “1.3e-13”).
3. Next, at the bottom of this window, click the “Plot data” button.
4. A log D versus 1/T plot then appears, with a line for the temperature dependence for this
diffusion system.
At the top of this window are give values for D0 and Qd; for this specific problem
these values are 7.55 × 10-5 m2/s and 196 kJ/mol, respectively
5. To solve the (b) part of the problem we utilize the diamond-shaped cursor that is located at
the top of the line on this plot. Click-and-drag this cursor down the line to the point at which the entry
under the “Temperature (T):” label reads “1148” (i.e., 875ºC). The value of the diffusion coefficient
at this temperature is given under the label “Diff Coeff (D):”. For our problem, this value is 8.9 ×
10-14 m2/s.
5.23 Below is shown a plot of the logarithm (to the base 10) of the diffusion coefficient versus
reciprocal of the absolute temperature, for the diffusion of iron in chromium. Determine values for
the activation energy and preexponential.
Solution
This problem asks us to determine the values of Qd and D0 for the diffusion of Fe in Cr from the
plot of log D versus 1/T. According to Equation 5.9b the slope of this plot is equal to −
Qd
2.3 R
(rather
Q
than − d since we are using log D rather than ln D) and the intercept at 1/T = 0 gives the value of
R
log D0. The slope is equal to
slope =
log D1 − log D2
Δ (log D)
=
⎛1 ⎞
1
1
−
Δ⎜ ⎟
T1
T2
⎝T ⎠
Taking 1/T1 and 1/T2 as 0.65 × 10-3 and 0.60 × 10-3 K-1, respectively, then the corresponding values of
D1 and D2 are 2.81 × 10-16 and 1.82 × 10-15, as noted in the figure below.
The values of log D1 and log D2 are –15.60 and –14.74, and therefore,
Qd = − 2.3 R (slope)
Qd = − 2.3 R
log D1 − log D2
1
1
−
T1
T2
⎡
⎤
− 15.60 − (−14.74)
= − (2.3)(8.31 J/mol- K) ⎢
⎥
⎣ (0.65 × 10−3 − 0.60 × 10−3 ) K−1 ⎦
= 329,000 J/mol
Rather than trying to make a graphical extrapolation to determine D0, a more accurate value is
obtained analytically using Equation 5.9b taking a specific value of both D and T (from 1/T) from the
plot given in the problem;
for example, D = 1.0 × 10-15 m2/s at T = 1626 K (1/T = 0.615 × 10-3 K-1).
Therefore
⎛Q ⎞
D0 = D exp ⎜ d ⎟
⎝ RT ⎠
⎡
⎤
329,000 J/mol
= (1.0 × 10 -15 m2 /s) exp ⎢
⎥
(8.31
J/mol
K)(1626
K)
⎣
⎦
= 3.75 × 10-5 m2/s
7.
Core 9.01
8.
Core 9.02
9. (A)
Conc. of Al
10E-18/cm3
0
t
(B) D = D0 exp(−
E
)
RT
ln D = ln D0 + (− E / RT )
Solve this equation at both 1100oC and 1000oC,
E = 5.12 × 105 J / mol
Here we use E and provided diffusivity at 1300oC to calculate D at 900oC
D900 = 1.91× 10 −16 cm 2 / sec
9.1 Consider the sugar–water phase diagram of Figure 9.1.
(a) How much sugar will dissolve in 1500 g water at 90°C (194°F)?
(b) If the saturated liquid solution in part (a) is cooled to 20°C (68°F), some of the sugar will
precipitate out as a solid. What will be the composition of the saturated liquid solution (in wt% sugar)
at 20°C?
(c) How much of the solid sugar will come out of solution upon cooling to 20°C?
Solution
(a) We are asked to determine how much sugar will dissolve in 1000 g of water at 90°C. From
the solubility limit curve in Figure 9.1, at 90°C the maximum concentration of sugar in the syrup is
about 77 wt%. It is now possible to calculate the mass of sugar using Equation 4.3 as
Csugar (wt%) =
77 wt% =
msugar
msugar + mwater
msugar
msugar + 1500 g
× 100
× 100
Solving for msugar yields msugar = 5022 g
(b) Again using this same plot, at 20°C the solubility limit (or the concentration of the saturated
solution) is about 64 wt% sugar.
(c) The mass of sugar in this saturated solution at 20°C (m' sugar ) may also be calculated using
Equation 4.3 as follows:
64 wt% =
m' sugar
m' sugar + 1500 g
× 100
which yields a value for m'sugar of 2667 g. Subtracting the latter from the former of these sugar
concentrations yields the amount of sugar that precipitated out of the solution upon cooling m"sugar ;
that is
m" sugar = msugar − mÕsugar = 5022 g − 2667 g = 2355 g
9.11 A copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu is slowly heated from a temperature
of 1300°C (2370°F).
(a) At what temperature does the first liquid phase form?
(b) What is the composition of this liquid phase?
(c) At what temperature does complete melting of the alloy occur?
(d) What is the composition of the last solid remaining prior to complete melting?
Solution
Shown below is the Cu-Ni phase diagram (Figure 9.3a) and a vertical line constructed at a
composition of 70 wt% Ni-30 wt% Cu.
(a) Upon heating from 1300°C, the first liquid phase forms at the temperature at which this
vertical line intersects the α-(α + L) phase boundary--i.e., about 1345°C.
(b) The composition of this liquid phase corresponds to the intersection with the (α + L)-L phase
boundary, of a tie line constructed across the α + L phase region at 1345°C--i.e., 59 wt% Ni;
(c) Complete melting of the alloy occurs at the intersection of this same vertical line at 70 wt%
Ni with the (α + L)-L phase boundary--i.e., about 1380°C;
(d) The composition of the last solid remaining prior to complete melting corresponds to the
intersection with α-(α + L) phase boundary, of the tie line constructed across the α + L phase region at
1380°C--i.e., about 79 wt% Ni.