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Chapter 5-
ISSUES TO ADDRESS...
• How does diffusion occur?
• Why is it an important part of processing?
• How can the rate of diffusion be predicted forsome simple cases?
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• How does diffusion depend on structureand temperature?
CHAPTER 5:DIFFUSION IN SOLIDS
Chapter 5-
Diffusion
Diffusion - Mass transport by atomic motion
Mechanisms• Gases & Liquids – random (Brownian) motion• Solids – vacancy diffusion or interstitial
diffusion
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Chapter 5-
Why Study Diffusion ?
• Diffusion plays a crucial role in…– Alloying metals => bronze, silver, gold
– Strengthening and heat treatment processes• Hardening the surfaces of steel
– High temperature mechanical behavior
– Phase transformations• Mass transport during FCC to BCC
– Environmental degradation• Corrosion, etc.
Chapter 5- 2
• Glass tube filled with water.• At time t = 0, add some drops of ink to one end
of the tube.• Measure the diffusion distance, x, over some time.• Compare the results with theory.
to
t1
t2
t3
xo x1 x2 x3
time (s)
x (mm)
DIFFUSION DEMO
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Chapter 5-
How do atoms move in Solids ?Why do atoms move in Solids ?
• Diffusion, simply, is atoms moving from one lattice site to another in a stepwise manner– Transport of material by moving atoms
• Two conditions are to be met:– An empty adjacent site
– Enough energy to break bonds and cause lattice distortions during displacement
• What is the energy source ?– HEAT !
• What else ?– Concentration gradient !
Chapter 5-
• Case Hardening:--Diffuse carbon atoms
into the host iron atomsat the surface.
--Example of interstitialdiffusion is a casehardened gear.
• Result: The "Case" is--hard to deform: C atoms
"lock" planes from shearing.--hard to crack: C atoms put
the surface in compression.
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Fig. 5.0, Callister 6e.(Fig. 5.0 iscourtesy ofSurface Division, Midland-Ross.)
PROCESSING USING DIFFUSION
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Chapter 5-
• Doping silicon with phosphorus for n-type semiconductors:• Process:
3. Result: Dopedsemiconductorregions.
silicon
Processing Using Diffusion
magnified image of a computer chip
0.5mm
light regions: Si atoms
light regions: Al atoms
2. Heat it.
1. Deposit P richlayers on surface.
silicon
Adapted from chapter-opening photograph, Chapter 18, Callister 7e.
Chapter 5-
• Interdiffusion: In an alloy, atoms tend to migratefrom regions of high conc. to regions of low conc.
Initially
Adapted from Figs. 5.1 and 5.2, Callister 7e.
Diffusion
After some time
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Chapter 5-
Chapter 5-
Diffusion Mechanisms (I)
Energy is needed to generate a vacancy, break bonds, cause distortions. Provided by HEAT , kT !Atom moves in the opposite direction of the vacancy !
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Chapter 5-
Diffusion Mechanisms (II)
Much faster than vacancy diffusion, why ? Smaller atoms like B, C, H, O. Weaker interaction with the larger atoms. More vacant sites, no need to create a vacancy !
Interstitial Diffusion
Chapter 5-
Diffusion• How do we quantify the amount or rate of diffusion?
• Measured empirically– Make thin film (membrane) of known surface area– Impose concentration gradient– Measure how fast atoms or molecules diffuse through the
membrane
sm
kgor
scm
mol
timearea surface
diffusing mass) (or molesFlux
22J
dt
dM
A
l
At
MJ M =
massdiffused
time
J slope
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Chapter 5-
• Concentration Profile, C(x): [kg/m3]
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• Fick's First Law:
Concentration
of Cu [kg/m 3]Concentration
of Ni [kg/m 3]
Position, x
Cu flux Ni flux
• The steeper the concentration profile, the greater the flux!Concentration gradient is the DRIVING FORCE !
Adapted from Fig. 5.2(c), Callister 6e.
J x D
dC
dx
Diffusion coefficient [m 2/s]
concentration
gradient [kg/m 4]
flux in x-dir.
[kg/m 2-s]
CONCENTRATION PROFILES & FLUX
Chapter 5-
Temperature Dependency !
Atom has to break bonds and “squeeze” thru => activation energy, Em ≈ 1 eV .
What is the probability to find a vacancy at a nearest site ?
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Chapter 5-
Temperature Effect !
The diffusion depends on temperature because;a- # of vacancies in the vicinityb- thermally activated successful jumps
Chapter 5-
Diffusion and Temperature
• Diffusion coefficient 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
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Chapter 5-
Diffusion and Temperature
Adapted from Fig. 5.7, Callister 7e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)
D has 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
10-8T(C)
1500
1000
600
300
Chapter 5-
• Diffusivity increases with T.
• Experimental Data:
1000K/T
D (m2/s) C in -Fe
C in -Fe
Al in A
l
Cu in Cu
Zn in CuFe in -Fe
Fe in -Fe
0.5 1.0 1.5 2.010 -20
10 -14
10 -8T(C)15
00
1000
600
300
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pre-exponential [m 2/s] (see Table 5.2, Callister 6e )activation energy
gas constant [8.31J/mol-K]
D Do exp QdRT
diffusivity
[J/mol],[eV/mol] (see Table 5.2, Callister 6e )
Dinterstitial >> Dsubstitutional
C in -FeC in -Fe Al in Al
Cu in Cu
Zn in Cu
Fe in -FeFe in -Fe
Adapted from Fig. 5.7, Callister 6e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)
DIFFUSION AND TEMPERATURE
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Chapter 5-
Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are
D(300ºC) = 7.8 x 10-11 m2/sQd = 41.5 kJ/mol
What is the diffusion coefficient at 350ºC?
101
202
1lnln and
1lnln
TR
QDD
TR
QDD dd
121
212
11lnlnln
TTR
Q
D
DDD d
transform data
D
Temp = T
ln D
1/T
D Do exp
Qd
RT
Chapter 5-
Example (cont.)
K 573
1
K 623
1
K-J/mol 314.8
J/mol 500,41exp /s)m 10 x 8.7( 211
2D
1212
11exp
TTR
QDD d
T1 = 273 + 300 = 573K
T2 = 273 + 350 = 623K
D2 = 15.7 x 10-11 m2/s
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Chapter 5-
Concentration Gradient
Chapter 5-
• Steady State: the concentration profile doesn't change with time.
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• Apply Fick's First Law:
• Result: the slope, dC/dx, must be constant(i.e., slope doesn't vary with position)!
J x(left) = J x(right)
Steady State:
Concentration, C, in the box doesn’t change w/time.
J x(right)J x(left)
x
Jx D
dC
dx
dC
dx
left
dC
dx
right
• If Jx)left = Jx)right , then
STEADY STATE DIFFUSION
Why is the minus sign ?
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Chapter 5-
• Steel plate at 700º C
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• Q: How much carbon transfers from the rich to the deficient side?
J DC2 C1
x2 x1
2 .4 10 9 kg
m2s
Adapted from Fig. 5.4, Callister 6e.
C 1 = 1.2kg/m3
C 2 = 0.8kg/m3
Carbon rich gas
10mm
Carbon deficient
gas
x1 x205mm
D=3x10 -11 m2/s
Steady State = straight line!
EX: STEADY STATE DIFFUSION
Chapter 5 -
HW: Chemical Protective Clothing (CPC)
• 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 diffusive flux of methylene chloride through the glove?
• Data:
– diffusion coefficient in butyl rubber: D = 110x10-8 cm2/s
– surface concentrations:
C2 = 0.02 g/cm3
C1 = 0.44 g/cm3
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Chapter 5 -
scm
g 10 x 16.1
cm) 04.0(
)g/cm 44.0g/cm 02.0(/s)cm 10 x 110(
25-
3328-
J
Example (cont).
12
12- xx
CCD
dx
dCDJ
D
tb 6
2
gloveC1
C2
skinpaintremover
x1 x2
• Solution – assuming linear conc. gradient
D = 110x10-8 cm2/s
C2 = 0.02 g/cm3
C1 = 0.44 g/cm3
x2 – x1 = 0.04 cm
Data:
Chapter 5-
Fick’s Second Law ; Non-steady state Diffusion
• In most practical cases, J (flux) and dC/dx (concentration gradient) change with time (t).– Net accumulation or depletion of species diffusing
• How do we express a time dependent concentration?
Concentration at a point xChanging with time
? Flux, J, changes at any point x !
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Chapter 5-
How do we solve this partial differential equation ?
• Use proper boundary conditions:– t=0, C = C0, at 0 ≤ x ≤ ∞
– t>0, C = Cs, at x = 0
C = C0, at x = ∞
Chapter 5 -
Non-steady State Diffusion
Adapted from Fig. 5.5, Callister 7e.
B.C. at t = 0, C = Co for 0 x
at t > 0, C = CS for x = 0 (const. surf. conc.)
C = Co for x =
• Copper diffuses into a bar of aluminum.
pre-existing conc., Co of copper atoms
Surface conc., C of Cu atoms bar
s
Cs
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Chapter 5 -
• Copper diffuses into a bar of aluminum.
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• General solution:
“Gaussian error function"
pre-existing conc., C o of copper atoms
Surface conc., Cs of Cu atoms bar
Co
Cs
position, x
C(x,t)
tot1
t2t3 Adapted from
Fig. 5.5, Callister 6e.
NON STEADY STATE DIFFUSION
C(x, t ) Co
Cs Co
1 erfx
2 Dt
C (x, t) = concentration at any time and position !
Chapter 5 -
• Concentration profile,C(x), changesw/ time.
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• To conserve matter: • Fick's First Law:
• Governing Eqn.:
Concentration, C, in the box
J (right)J (left)
dx
d C
dt= D
d 2C
dx 2
dx
d C
dtJ D
d C
dxor
J (left)J (right)
d J
dx
d C
dt
d J
dx D
d2 C
dx 2
(if D does not vary with x)
equate
NON STEADY STATE DIFFUSION
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Chapter 5 -
Solution:
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
Chapter 5 -
Non-steady State Diffusion• Sample Problem: 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 Eqn. 5.5
• Determine z value from Table 5.1
Dt
x
CC
CtxC
os
o
2erf1
),(
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Chapter 5 -
Solution (cont.):
– t = 49.5 h x = 4 x 10-3 m
– Cx = 0.35 wt% Cs = 1.0 wt%
– Co = 0.20 wt%
Dt
x
CC
C)t,x(C
os
o
2erf1
)(erf12
erf120.00.1
20.035.0),(z
Dt
x
CC
CtxC
os
o
erf(z) = 0.8125
Chapter 5 -
Solution (cont.):
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 0.7970z 0.81250.95 0.8209
7970.08209.0
7970.08125.0
90.095.0
90.0
z
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
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Chapter 5 -
• To solve for the temperature at which D has above value, we use a rearranged form of Equation (5.9a);
)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
Solution (cont.):
T = 1300 K = 1027°C
Chapter 5-
• Copper diffuses into a bar of aluminum.• 10 hours at 600C gives desired C(x).• How many hours would it take to get the same C(x)
if we processed at 500C?
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(Dt) 500ºC =(Dt) 600ºCs
C (x, t ) CoC Co
= 1 erfx
2Dt
• Result: Dt should be held constant.
• Answer:Note: valuesof D areprovided here.
Key point 1: C(x,t500C) = C(x,t600C).Key point 2: Both cases have the same Co and Cs.
t 500 (Dt )600
D500
110 hr
4.8x10 -14 m2/s
5.3 x10 -13 m2/s 10hrs
Example 5.3
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Chapter 5-
Size Impact on Diffusion
Smaller atoms diffuse faster
Chapter 5-
Fast Tracks for diffusion !eg. self-diffusion of Ag :-Areas where lattice is pre-strained can allow for faster diffusion of atoms-Less energy is needed to distort an already strained lattice !
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Chapter 5-
Important
• Temperature - diffusion rate increases with increasing temperature (WHY ?)
• Diffusion mechanism – interstitials diffuse faster (WHY ?)
• Diffusing and host species - Do, Qd is different forevery solute - solvent pair
• Microstructure - grain boundaries and dislocationcores provide faster pathways for diffusing species, hence diffusion is faster in polycrystalline vs. single crystal materials. (WHY ?)
Chapter 5- 20
Diffusion FASTER for...
• open crystal structures
• lower melting T materials
• materials w/secondarybonding
• smaller diffusing atoms
• cations
• lower density materials
Diffusion SLOWER for...
• close-packed structures
• higher melting T materials
• materials w/covalentbonding
• larger diffusing atoms
• anions
• higher density materials
SUMMARY:STRUCTURE & DIFFUSION
WHY ?
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Chapter 5-
HW 6:
Solve EXAMPLE PROBLEMS: 5.1, 5.2, 5.3, 5.4, 5.5
HW 7: Chemical Protective Clothing (CPC)
Chapter 5-
QUIZ 4
a.Compare interstitial and vacancy atomic mechanisms for diffusion.
b.Which one is more rapid?
c.Why?