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CHAPTER6
DIFFUSION
PROBLEM SOLUTIONS
Introduction
6.1 Self-diffusion is atomic migration in pure metals--i.e., when all atoms exchanging positions are of the
same type. Interdiffusion is diffusion of atoms of one metal into another metal.
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6.2 Self-diffusion may be monitored by using radioactive isotopes of the metal being studied. The motion
of these isotopic atoms may be monitored by measurement of radioactivity level.
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Diffusion Mechanisms
6.3 (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: (1) interstitial atoms,
being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a
vacancy adjacent to a host (or substitutional impurity) atom.
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Steady-State Diffusion
6.4 Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to
the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion
flux is independent of time.
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6.5 (a) The driving force is that which compels a reaction to occur.
(b) The driving force for steady-state diffusion is the concentration gradient.
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6.6 This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first
becomes necessary to employ both Equations 6.1a and6.3. Combining these expressions and solving for the mass
yields
M = JAt = DAt
C
x
= (1.7 10-8 m2/s)(0.25 m2)(3600 s/h)0.4 2.0 kg/m3
6 103 m
= 4.1 x 10-3 kg/h
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6.7 We are asked to determine the position at which the nitrogen concentration is 0.5 kg/m3. This problem
is solved by using Equation 6.3 in the form
J = DCA CBx
A
xB
If we take CA to be the point at which the concentration of nitrogen is 2 kg/m3, then it becomes necessary to solve
forxB, as
xB = xA + DCA CB
J
AssumexA is zero at the surface, in which case
xB = 0 + (1.2 10-10 m2/s)
2 kg/m3 0.5 kg/m3
1.0 107 kg/m2 - s
= 1.8 x 10-3 m = 1.8 mm
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6.8 This problem calls for computation of the diffusion coefficient for a steady-state diffusion situation.
Let us first convert the carbon concentrations from weight percent to kilograms carbon per meter cubed using
Equation 6.9a. For 0.015 wt% C
CC"
=
CCCC
C+
CFe
Fe
103
=0.015
0.015
2.25 g/cm3+
99.985
7.87 g /cm3
103
1.18 kg C/m3
Similarly, for 0.0068 wt% C
CC" =
0.0068
0.0068
2.25 g/cm3+
99.9932
7.87 g /cm3
103
= 0.535 kg C/m3
Now, using a rearranged form of Equation 6.3
D = JxA xBCA CB
= (7.36 10-9 kg/m2 - s) 2 103 m
1.18 kg/m3 0.535 kg/m3
= 2.3 x 10-11 m2/s
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6.9 This problems asks for us to compute the diffusion flux of nitrogen gas through a 1.5-mm thick plate
of iron at 300C when the pressures on the two sides are 0.10 and 5.0 MPa. Ultimately we will employ Equation
6.3 to solve this problem. However, it first becomes necessary to determine the concentration of hydrogen at each
face using Equation 6.13. At the low pressure (or B) side
CN(B) = (4.90 10-3) 0.10 MPa exp
37,600 J /mol
(8.31 J /mol - K)(300 + 273 K)
5.77 x 10-7 wt%
Whereas, for the high pressure (or A) side
CN(A) = (4.90 10-3) 5.0 MPa exp
37,600 J /mol
(8.31 J /mol - K)(300 + 273 K)
4.08 x 10-6 wt%
We now convert concentrations in weight percent to mass of nitrogen per unit volume of solid. At face B there are
5.77 x 10-7 g (or 5.77 x 10-10 kg) of hydrogen in 100 g of Fe, which is virtually pure iron. From the density of iron
(7.87 g/cm3), the volume iron in 100 g (VB) is just
VB =100 g
7.87 g /cm3= 12.7 cm3 = 1.27 10-5 m3
Therefore, the concentration of hydrogen at the B face in kilograms of N per cubic meter of alloy [ ] is justCN(B)''
CN(B)'' =
CN(B)
VB
=5.77 1010 kg
1.27 105 m3= 4.54 x 10-5 kg/m3
At the A face the volume of iron in 100 g (VA) will also be 1.27 x 10-5 m3, and
CN(A)
'' =CN(A)
VA
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=
4.08 109 kg
1.27 105 m3= 3.21 x 10-4 kg/m3
Thus, the concentration gradient is just the difference between these concentrations of nitrogen divided by the
thickness of the iron membrane; that is
C
x=
CN(B) CN(A)
xB xA
=4.54 x 105 kg /m3 3.21 x 104 kg /m3
1.5 103 m= 0.184 kg/m4
At this time it becomes necessary to calculate the value of the diffusion coefficient at 300C using Equation 6.8.
Thus,
D = D0 exp Qd
RT
= (3.0 107 m2 /s) exp 76,150 J /mol(8.31 J /mol K)(300 + 273 K)
= 3.40 x 10-14 m2/s
And, finally, the diffusion flux is computed using Equation 6.3 by taking the negative product of this diffusion
coefficient and the concentration gradient, as
J = DC
x
= (3.40 10-14 m2/s)( 0.184 kg/m4) = 6.26 10-15 kg/m2 - s
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Nonsteady-State Diffusion
6.10 It can be shown that
Cx = BDtexp x
2
4Dt
is a solution to
C
t= D
2C
x2
simply by taking appropriate derivatives of the Cx expression. When this is carried out,
C
t= D
2C
x2=
B
2D1/ 2t3/ 2
x2
2Dt 1
exp
x2
4Dt
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6.11 We are asked to compute the carburizing (i.e., diffusion) time required for a specific nonsteady-state
diffusion situation. It is first necessary to use Equation 6.5:
Cx
C0Cs C0
= 1 erfx
2 Dt
wherein, Cx = 0.30, C0 = 0.10, Cs = 0.90, andx = 4 mm = 4 x 10-3 m. Thus,
Cx C0Cs C0
=0.30 0.10
0.90 0.10= 0.2500 = 1 erf
x
2 Dt
or
erf
x
2 Dt
= 1
0.2500 = 0.7500
By linear interpolation using data from Table 6.1
z erf(z)
0.80 0.7421
z 0.7500
0.85 0.7707
z 0.800
0.850 0.800=
0.7500 0.7421
0.7707 0.7421
From which
z = 0.814 =x
2 Dt
Now, from Table 6.2, at 1100C (1373 K)
D = (2.3 10-5 m2/s) exp 148,000 J /mol
(8.31 J /mol- K)(1373 K)
= 5.35 x 10-11 m2/s
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Thus,
0.814 =4 103 m
(2) (5.35 1011 m2 /s) (t)
Solving fortyields
t= 1.13 x 105 s = 31.3 h
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6.12 This problem asks that we determine the position at which the carbon concentration is 0.25 wt% after
a 10-h heat treatment at 1325 K when C0 = 0.55 wt% C. From Equation 6.5
Cx
C0Cs C0
=0.25 0.55
0 0.55= 0.5455 = 1 erf
x
2 Dt
(a)
Thus,
erfx
2 Dt
= 0.4545
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Alternatively,
erf2
x
xC A B
Dt
=
Using initial condition ( ), 0 0.55C x = ,
0.55 A B=
Using boundary condition ( )0, 0C t = ,
0 A=0.55B =
For the given concentration ,0.25x
C =
0.25 0.55 erf 2
x
Dt
= +
or
erf 0.45452
x
Dt
=
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Using data in Table 6.1 and linear interpolation
z erf (z)
0.40 0.4284
z 0.4545
0.45 0.4755
z 0.40
0.45 0.40=
0.4545 0.4284
0.4755 0.4284
And,
z = 0.4277
Which means thatx
2 Dt= 0.4277
And, finally
x = 2(0.4277) Dt = (0.8554) (4.3 1011 m2 /s)(3.6 104 s)
= 1.06 x 10-3m = 1.06 mm
Note: this problem may also be solved using the Diffusion module in the VMSEsoftware. Open the Diffusion
module, click on the Diffusion Design submodule, and then do the following:
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(b)
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1. Enter the given data in left-hand window that appears. In the window below the label D Value enter
the value of the diffusion coefficientviz. 4.3e-11.
2. In the window just below the label Initial, C0 enter the initial concentrationviz. 0.55.
3. In the window the lies below Surface, Cs enter the surface concentrationviz. 0.
4. Then in the Diffusion Time t window enter the time in seconds; in 10 h there are (60 s/min)(60
min/h)(10 h) = 36,000 sso enter the value 3.6e4.
5. Next, at the bottom of this window click on the button labeled Add curve.
6. On the right portion of the screen will appear a concentration profile for this particular diffusion
situation. A diamond-shaped cursor will appear at the upper left-hand corner of the resulting curve. Click and drag
this cursor down the curve to the point at which the number below Concentration: reads 0.25 wt%. Then read
the value under the Distance:. For this problem, this value (the solution to the problem) is 1.05 mm.
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6.13 This problem asks us to compute the nitrogen concentration (Cx) at the 2 mm position after a 25 h
diffusion time, when diffusion is nonsteady-state. From Equation 6.5
Cx
C0
Cs C0=
Cx
0
0.2 0 = 1 erfx
2 Dt
= 1 erf2 103 m
(2) (1.9 1011 m2 /s) (25 h)(3600 s /h)
= 1 erf (0.765)
Using data in Table 6.1 and linear interpolation
z erf (z)
0.750 0.7112
0.765 y
0.800 0.7421
0.765 0.750
0.800 0.750=
y 0.7112
0.7421 0.7112
from which
y = erf (0.765) = 0.7205
Thus,
Cx
0
0.2 0= 1.0 0.7205
This expression gives
Cx = 0.056 wt% N
Note: this problem may also be solved using the Diffusion module in the VMSEsoftware. Open the Diffusion
module, click on the Diffusion Design submodule, and then do the following:
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1. Enter the given data in left-hand window that appears. In the window below the label D Value enter
the value of the diffusion coefficientviz. 1.9e-11.
2. In the window just below the label Initial, C0 enter the initial concentrationviz. 0.
3. In the window the lies below Surface, Cs enter the surface concentrationviz. 0.2.
4. Then in the Diffusion Time t window enter the time in seconds; in 25 h there are (60 s/min)(60
min/h)(25 h) = 90,000 sso enter the value 9e4.
5. Next, at the bottom of this window click on the button labeled Add curve.
6. On the right portion of the screen will appear a concentration profile for this particular diffusion
situation. A diamond-shaped cursor will appear at the upper left-hand corner of the resulting curve. Click and drag
this cursor down the curve to the point at which the number below Distance: reads 2.00 mm. Then read the
value under the Concentration:. For this problem, this value (the solution to the problem) is 0.06 wt%.
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y 0.150
0.200 0.150=
0.2000 0.1680
0.2227 0.1680
from which
y = 0.1793 =206.4 s
t
And, solving fortgives
t= 1.33 x 106 s = 368 h = 15.3 days
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6.14 (a) The general solution is
( ), erf 2
xC x t A B
Dt
=
Using the first initial condition, i.e.
( ) 2,0 for 0C x C x= > in the above expression,
2C A= B
Using the second initial condition, i.e.
( ) 1,0 for 0C x C x=