em321 lesson 08a solutions ch5 - diffusion
TRANSCRIPT
Chapter 5.4-5.5: Diffusion (part 2)
EM321: Lesson 8a
Before class, pass out photocopy of Callister Tables 5.1 and 5.2
Review from Last Time• Types of diffusion
– Interstitial
– Vacancy
• Examples of diffusion
• Rate of diffusion
increasing elapsed time
Diffusion: Calculating Flux
• How do we quantify the amount or rate of diffusion?
sm
kgor
scm
mol
timearea surface
diffusing mass) (or molesFlux
22J
J slope
dt
dM
A
l
At
MJ
M =
mass
diffusedtime
Steady-State Diffusion: Calculating Flux
dx
dCDJ
Fick’s first law of diffusionC1
C2
x
C1
C2
x1 x2J = flux
D = diffusion coefficient (Table 5.2)
dC = change in concentration
dx = change in linear distance
For Steady-State Diffusion, the rate of diffusion is independent of time
Flux is proportional to concentration gradient =dx
dC
12
12 linear ifxx
CC
x
C
dx
dC
Steady-State Diffusion and Time
• Diffusion Coefficient, D, increases with increasing
Temperature, T
• The higher the Diffusion Coefficient, the more _______ atoms
will diffuse across a given concentration gradient, such that:
D1 t1 = D2 t2
Example: The diffusion coefficient for Copper in Aluminum is
D = 4.15 * 10-14 m2/s at 500oC and D = 4.69 * 10-13 m2/s at 600oC.
What time will be required at 600oC to produce the same diffusion result
(in terms of concentration at a specific point) as for 10 hours at 500oC?
t600 =D500t500
D600
= (4.15 10 14 m2 /s)(10h)
(4.69 10 13m2 /s)= 0.88 h
quickly
Diffusion and Temperature
Diffusion Coefficient, D, increases with increasing T.
D Do exp
Qd
RT
= pre-exponential [m2/s]
= diffusion coefficient [m2/s]
= activation energy [J/mol or eV/atom]
= gas constant [8.314 J/mol-K]
= absolute temperature [K]
D
Do
Qd
R
T
Diffusion and Temperature
D has an exponential dependence on T
Dinterstitial >> Dsubstitutional
C in -FeC in -Fe
Al in AlFe in -FeFe in -Fe
1000K/T
D (m2/s)
0.5 1.0 1.510-20
10-14
10-8
T( C)
15
00
10
00
60
0
30
0
From the graph of D vs. T, what can you
conclude about Dinterstitial vs. Dsubstitutional?
Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are
D(300ºC) = 7.8 x 10-11 m2/s
Qd = 41.5 kJ/mol
What is the diffusion coefficient at 350ºC?
1
01
2
02
1lnln and
1lnln
TR
QDD
TR
QDD dd
121
212
11lnlnln
TTR
Q
D
DDD d
transform dataD
Temp = T
ln D
1/T
Example: Diffusion and Temperature
Note: You were not given the
Pre-Exponential, Do
K 573
1
K 623
1
K-J/mol 314.8
J/mol 500,41exp /s)m 10 x 8.7( 211
2D
12
12
11exp
TTR
QDD d
T1 = 273 + 300 = 573K
T2 = 273 + 350 = 623K
D2 = 15.7 x 10-11 m2/s
Example (cont.): Diffusion and Temperature
Another Example: Diffusion Coefficient
Example: What is the Diffusion Coefficient, D, for the
interdiffusion of carbon in α-iron (BCC) at 900oC?
(Note: Use data in Table 5.2 to find the Pre-Exponential, Do.)
From Table 5.2, Do = ________m2/s for carbon in α-Fe
Qv = ____ kJ/mol = ___________ J/mol
) 1173)(K-J/mol 31.8(
J/mo 000,80exp/sm 10 6.2 = )( 27-
K
lD
Dα = 5.86 10-12 m2/s
80 80,000
6.2 * 10-7
Steady-State vs. Non-Steady State
Up until now, we have been looking at Steady-State diffusion
What does “Steady-State” mean?
- Concentration of diffusing species is a function of position only.
Not a function of time (does not change over time). C = C(x)
What do we do if our concentration gradient is “Non-Steady
State” (C is a function of both both position and time)?(e.g. when a material is first introduced to a concentration gradient and the
impurity atoms begin to diffuse through it)
- Use Fick’s Second Law (remainder of this lecture). Second
order equation with C = C(x,t)
Non-Steady State Diffusion
• The concentration of diffusing species is a function of both time and position C = C(x,t)
• In this case Fick’s Second Law is used
2
2
x
CD
t
CFick’s Second Law
Non-Steady State Diffusion
Boundary Conditions: at t = 0, C = _____ for 0 x
at t > 0, C = _____ for x = 0 (constant surface concentration)
C = _____ for x =
Pre-existing concentration, Co, of copper atoms
Surface concentration,
C , of Cu atoms Aluminum bars
Cs
Example: Copper diffusing into a bar of aluminum that already
has a low copper concentration, Co, within it
Co
Co
Cs
Solution to Fick’s Second Law:
C(x,t) = Conc. at point x at time t
erf (z) = error function
erf(z) values are given in Table 5.1
CS
Co
C(x,t)
Dt
x
CC
Ct,xC
os
o
2 erf1
dye yz 2
0
2
Using Table 5.1:
What value of z gives an error function of erf(z) = 0.4755?
What value of z gives an error function of erf(z) = 0.9592?
What value of z gives an error function of erf(z) = 0.3400?
z = 0.45
z = 1.45
z = 0.3112
Example: Non-steady State Diffusion
• Example: An FCC iron-carbon alloy initially containing 0.20 wt% C is
carburized at an elevated temperature and in an atmosphere that gives a
surface carbon concentration constant at 1.0 wt%. If after 49.5 h the
concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface,
determine the temperature at which the treatment was carried out.
• Solution: use
Dt
x
CC
CtxC
os
o
2erf1
),(
– t = 49.5 h x = 4 x 10-3 m
– Cx = 0.35 wt% Cs = 1.0 wt%
– Co = 0.20 wt%
)(erf12
erf120.00.1
20.035.0),(z
Dt
x
CC
CtxC
os
o
erf(z) = 0.8125
We must now determine from Table 5.1 the value of z for which the
error function is 0.8125. An interpolation is necessary as follows
z erf(z)
0.90 0.7970
z 0.8125
0.95 0.8209
7970.08209.0
7970.08125.0
90.095.0
90.0z
z 0.93
Now solve for D
Dt
xz
2 tz
xD
2
2
4
/sm 10 x 6.2s 3600
h 1
h) 5.49()93.0()4(
m)10 x 4(
4
2112
23
2
2
tz
xD
Example (cont.): Non-steady State Diffusion
• To solve for the temperature at
which D has the above value, we
use a rearranged form of the
Diffusion Coefficient equation )lnln( DDR
QT
o
d
from Table 5.2, for diffusion of C in FCC Fe
Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol
/s)m 10x6.2 ln/sm 10x3.2 K)(ln-J/mol 314.8(
J/mol 000,14821125
T
T = 1300 K = 1027ºC
Example (cont.): Non-steady State Diffusion
Example: Chemical Protective Clothing
• Methylene chloride is a common ingredient of paint removers. Besides
being an irritant, it also may be absorbed through skin. When using this
paint remover, protective gloves should be worn.
• If butyl rubber gloves (0.04 cm thick) are used, what is the breakthrough
time (tb), i.e., how long can the gloves be used before
methylene chloride reaches the hand?
• Data
– diffusion coefficient in butyl rubber:
D = 110x10-8 cm2/s
glove
C1
C2
skinpaint
remover
x1 x2
Discussion Questions:
How would you go about solving this problem? (Do not solve it.)
Do you need to know the surface concentration of methylene chloride in
order to solve this problem? Why or why not?
Conclusion
• Steady-State Diffusion over Time
• Diffusion and Temperature
– Calculating Diffusion Coefficient, D
• Non-Steady State Diffusion