WHY STUDY Diffusion?
3. The time is taken to be zero the instant before the diffusion process begins
These boundary conditions are simply stated as
Application of these boundary conditions to Equation 5.4b yields the solution
(5.5) Cx"C0
Cs"C0 !1 " erfa x 21Dtb C ! C0 at x ! q
For t 7 0, C ! Cs1the constant surface concentration2 at x ! 0 For t ! 0, C ! C0 at 0 % x % q
C0. 0C
0t !D 0
2C 0x2 0C
0t ! 0 0x aD0C
0xb
5.4 Nonsteady-State Diffusion • 115
Fick’s second law
Fick’s second law—
diffusion equation for nonsteady-state diffusion (in one direction)
2A bar of solid is considered to be semi-infinite if none of the diffusing atoms reaches the bar end during the time over which diffusion takes place. A bar of length l is considered to be semi-infinite when l 7 101Dt.
Figure 5.5 Concentration profiles for nonsteady-state diffusion taken at three different times,t1,t2,and t3.
Distance
Concentration of diffusing species
t3 >t2 >t1
t2 t1
t3
Solution to Fick’s second law for the condition of constant surface concentration (for a semi-infinite solid)
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where represents the concentration at depth x after time t. The expression erf( ) is the Gaussian error function,3values of which are given in mathe-matical tables for various values; a partial listing is given in Table 5.1. The concentration parameters that appear in Equation 5.5 are noted in Figure 5.6, a concentration profile taken at a specific time. Equation 5.5 thus demonstrates the relationship between concentration, position, and time—namely, that being a function of the dimensionless parameter may be determined at any time and position if the parameters and D are known.
Suppose that it is desired to achieve some specific concentration of solute, in an alloy; the left-hand side of Equation 5.5 now becomes
This being the case, the right-hand side of this same expression is also a constant, and subsequently
(5.6a) x
21Dt!constant C1"C0
Cs"C0 !constant
C1, Cs,
C0,
x
&
1Dt,Cx,
x
&
21Dtx
&
21DtCx 116 • Chapter 5 / Diffusion
Table 5.1 Tabulation of Error Function Values
z erf(z) z erf(z) z erf(z)
0 0 0.55 0.5633 1.3 0.9340
0.025 0.0282 0.60 0.6039 1.4 0.9523
0.05 0.0564 0.65 0.6420 1.5 0.9661
0.10 0.1125 0.70 0.6778 1.6 0.9763
0.15 0.1680 0.75 0.7112 1.7 0.9838
0.20 0.2227 0.80 0.7421 1.8 0.9891
0.25 0.2763 0.85 0.7707 1.9 0.9928
0.30 0.3286 0.90 0.7970 2.0 0.9953
0.35 0.3794 0.95 0.8209 2.2 0.9981
0.40 0.4284 1.0 0.8427 2.4 0.9993
0.45 0.4755 1.1 0.8802 2.6 0.9998
0.50 0.5205 1.2 0.9103 2.8 0.9999
3This Gaussian error function is defined by
where x&21Dthas been replaced by the variable z.
erf1z2 ! 2
1p
!
0ze"y2 dyDistance from interface, x
Concentration, C
Cx – C0 C0
Cx Cs
Cs – C0
Figure 5.6 Concentration profile for nonsteady-state diffusion; concentration parameters relate to Equation 5.5.
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REVISED PAGES
or
(5.6b) Some diffusion computations are thus facilitated on the basis of this relation-ship, as demonstrated in Example Problem 5.3.
EXAMPLE PROBLEM 5.2
Nonsteady-State Diffusion Time Computation I
For some applications, it is necessary to harden the surface of a steel (or iron-carbon alloy) above that of its interior. One way this may be accomplished is by increasing the surface concentration of carbon in a process termed car-burizing;the steel piece is exposed, at an elevated temperature, to an atmo-sphere rich in a hydrocarbon gas, such as methane
Consider one such alloy that initially has a uniform carbon concentration of 0.25 wt% and is to be treated at 950 C (1750 F). If the concentration of carbon at the surface is suddenly brought to and maintained at 1.20 wt%, how long will it take to achieve a carbon content of 0.80 wt% at a position 0.5 mm below the surface? The diffusion coefficient for carbon in iron at this tem-perature is m2/s; assume that the steel piece is semi-infinite.
Solution
Since this is a nonsteady-state diffusion problem in which the surface com-position is held constant, Equation 5.5 is used. Values for all the parameters in this expression except time t are specified in the problem as follows:
Thus,
We must now determine from Table 5.1 the value of z for which the error function is 0.4210. An interpolation is necessary, as
z erf(z)
0.35 0.3794
z 0.4210
0.40 0.4284
z "0.35
0.40 " 0.35! 0.4210 " 0.3794 0.4284 " 0.3794 0.4210 ! erfa62.5 s1&2
1t b Cx"C0
Cs"C0
! 0.80 " 0.25
1.20 " 0.25!1 " erf c 15 # 10"4 m2 2211.6 # 10"11 m2/s21t2d D ! 1.6 # 10"11 m2/s
x ! 0.50 mm ! 5 # 10"4 m Cx!0.80 wt% C
Cs!1.20 wt% C C0!0.25 wt% C 1.6 # 10"11
$
$
1CH42.
x2
Dt!constant
5.4 Nonsteady-State Diffusion • 117
carburizing
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118 • Chapter 5 / Diffusion
or
Therefore,
and solving for t,
EXAMPLE PROBLEM 5.3
Nonsteady-State Diffusion Time Computation II
The diffusion coefficients for copper in aluminum at 500 and 600 C are 4.8 and 5.3 m2/s, respectively. Determine the approximate time at 500 C that will produce the same diffusion result (in terms of concentra-tion of Cu at some specific point in Al) as a 10-h heat treatment at 600°C.
Solution
This is a diffusion problem in which Equation 5.6b may be employed. The composition in both diffusion situations will be equal at the same position (i.e., xis also a constant), thus
(5.7) at both temperatures. That is,
or
5.5 FACTORS THAT INFLUENCE DIFFUSION
Diffusing Species
The magnitude of the diffusion coefficient D is indicative of the rate at which atoms diffuse. Coefficients, both self- and interdiffusion, for several metallic systems are listed in Table 5.2. The diffusing species as well as the host material influence the diffu-sion coefficient. For example, there is a significant difference in magnitude between self-diffusion and carbon interdiffusion in iron at 500 C, the D value being greater for the carbon interdiffusion ( vs. ).This comparison also provides a contrast between rates of diffusion via vacancy and interstitial modes as discussed above. Self-diffusion occurs by a vacancy mechanism, whereas carbon diffusion in iron is interstitial.
Temperature
Temperature has a most profound influence on the coefficients and diffusion rates.
For example, for the self-diffusion of Fe in -Fe, the diffusion coefficient increases approximately six orders of magnitude (from 3.0 to 1.8 in rising temperature from 500 to 900 C (Table 5.2). The temperature dependence of$
m2/s2
#10"15
#10"21 a
m2/s 2.4 # 10"12 3.0 # 10"21 a $ t500! D600t600
D500 ! 15.3 # 10"13 m2/s2110 h2
4.8 # 10"14 m2/s !110.4 h D500 t500!D600 t600
Dt !constant
$
#10"13
#10"14 $
t !a62.5 s1&2 0.392 b
2
!25,400 s ! 7.1 h 62.5 s1&2
1t !0.392 z !0.392
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the diffusion coefficients is
(5.8) where
a temperature-independent preexponential
the activation energyfor diffusion (J/mol or eV/atom) R ! the gas constant, 8.31 J/mol-K or 8.62 eV/atom-K T ! absolute temperature (K)
The activation energy may be thought of as that energy required to produce the diffusive motion of one mole of atoms. A large activation energy results in a relatively small diffusion coefficient. Table 5.2 also contains a listing of and values for several diffusion systems.
Taking natural logarithms of Equation 5.8 yields
(5.9a) or in terms of logarithms to the base 10
(5.9b) Since , and R are all constants, Equation 5.9b takes on the form of an equa-tion of a straight line:
where y and x are analogous, respectively, to the variables log D and Thus, if log D is plotted versus the reciprocal of the absolute temperature, a straight line1
"
T.y ! b # mx Qd
D0,
log D ! log D0$ Qd 2.3Ra1
Tb ln D ! ln D0$Qd
R a1 Tb
Qd
D0
%10$5 Qd!
1m2/s2 D0!
D ! D0 exp a$Qd RTb
5.5 Factors That Influence Diffusion • 119 Table 5.2 A Tabulation of Diffusion Data
Diffusing Host Activation Energy Qd Calculated Values
Species Metal D0(m2/s) kJ/mol eV/atom T(!C) D(m2/s)
Fe !-Fe 2.8 % 10$4 251 2.60 500 3.0 % 10$21
(BCC) 900 1.8 % 10$15
Fe "-Fe 5.0 % 10$5 284 2.94 900 1.1 % 10$17
(FCC) 1100 7.8 % 10$16
C !-Fe 6.2 % 10$7 80 0.83 500 2.4 % 10$12
900 1.7 % 10$10
C "-Fe 2.3 % 10$5 148 1.53 900 5.9 % 10$12
1100 5.3 % 10$11
Cu Cu 7.8 % 10$5 211 2.19 500 4.2 % 10$19
Zn Cu 2.4 % 10$5 189 1.96 500 4.0 % 10$18
Al Al 2.3 % 10$4 144 1.49 500 4.2 % 10$14
Cu Al 6.5 % 10$5 136 1.41 500 4.1 % 10$14
Mg Al 1.2 % 10$4 131 1.35 500 1.9 % 10$13
Cu Ni 2.7 % 10$5 256 2.65 500 1.3 % 10$22
Source: E. A. Brandes and G. B. Brook (Editors), Smithells Metals Reference Book, 7th edition, Butterworth-Heinemann, Oxford, 1992.
Dependence of the diffusion coefficient on temperature
activation energy
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2nd REVISE PAGES
should result, having slope and intercept of and log respectively. This is, in fact, the manner in which the values of and are determined experi-mentally. From such a plot for several alloy systems (Figure 5.7), it may be noted that linear relationships exist for all cases shown.
Concept Check 5.1
Rank the magnitudes of the diffusion coefficients from greatest to least for the fol-lowing systems:
N in Fe at 700°C Cr in Fe at 700°C N in Fe at 900°C Cr in Fe at 900°C
Now justify this ranking. (Note: Both Fe and Cr have the BCC crystal structure, and the atomic radii for Fe, Cr, and N are 0.124, 0.125, and 0.065 nm, respectively. You may also want to refer to Section 4.3.)
[The answer may be found at www.wiley.com/college/callister(Student Companion Site).]
Concept Check 5.2
Consider the self-diffusion of two hypothetical metals A and B. On a schematic graph of ln D versus , plot (and label) lines for both metals given that
and also that
[The answer may be found at www.wiley.com/college/callister(Student Companion Site).]
Qd1A2 7 Qd1B2.
D01A2 7 D01B2 1
&
TD0
Qd
D0,
"Qd
&
2.3R120 • Chapter 5 / Diffusion
1500 1200 1000 800 600 500 400 300
Diffusion coefficient (m2/s) 10–8
10–10
10–12
10–14
10–16
10–18
10–20
Temperature (°C)
Reciprocal temperature (1000/K)
0.5 1.0 1.5 2.0
Al in Al Zn in Cu
Cu in Cu C in –Fe!
Fe in –Fe!
C in –Fe"
Fe in –Fe"
Figure 5.7 Plot of the logarithm of the diffusion coefficient versus the reciprocal of absolute temperature for several metals.
[Data taken from E. A.
Brandes and G. B.
Brook (Editors), Smithells Metals Reference Book,7th edition, Butterworth-Heinemann, Oxford, 1992.]
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EXAMPLE PROBLEM 5.4
Diffusion Coefficient Determination
Using the data in Table 5.2, compute the diffusion coefficient for magnesium in aluminum at 550 C.
Solution
This diffusion coefficient may be determined by applying Equation 5.8; the values of and from Table 5.2 are 1.2 m2/s and 131 kJ/mol, respectively. Thus,
EXAMPLE PROBLEM 5.5
Diffusion Coefficient Activation Energy and Preexponential Calculations
In Figure 5.8 is shown a plot of the logarithm (to the base 10) of the diffusion coefficient versus reciprocal of absolute temperature, for the diffusion of cop-per in gold. Determine values for the activation energy and the preexponential.
Solution
From Equation 5.9b the slope of the line segment in Figure 5.8 is equal to , and the intercept at gives the value of log Thus, the activation energy may be determined as
where and are the diffusion coefficient values at and re-spectively. Let us arbitrarily take and
We may now read the corresponding log and log values from the line segment in Figure 5.8.
[Before this is done, however, a parenthetic note of caution is offered. The vertical axis in Figure 5.8 is scaled logarithmically (to the base 10); however, the actual diffusion coefficient values are noted on this axis. For example, for m2/s, the logarithm of D is not Furthermore, this log-arithmic scaling affects the readings between decade values; for example, at a location midway between and the value is not but, rather,
].
Thus, from Figure 5.8, at log while
for 1
#
T2!1.1 $ 10"3 1K2"1,1log#
T1D!2! "15.45,0.8 $ 10"3 1K2and the activation energy, as"1, D1! "12.40, 10"14.5!3.2 $ 10"15 10"14 10"15, 5 $ 10"1510"14.
"14.0 D !10"14
D2
D1
10"3 1K2D"11. D2 1
#
T1!0.8 $ 10"3 1K2"11#
T1 1#
T21!#
T1.1 $2, ! "2.3R£
log D1"log D2
1 T1
" 1 T2
§
Qd! "2.3R 1slope2 ! "2.3R £¢1log D2
¢a1 Tb
§ D0. 1
#
T !0"Qd
#
2.3R! 5.8 $ 10"13 m2/s
D !11.2 $ 10"4 m2/s2expc " 1131,000 J/mol2
18.31 J/mol-K21550 % 273 K2d
$ 10"4 Qd
D0
&
5.5 Factors That Influence Diffusion • 121
D0and Qdfrom Experimental Data
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