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Chapter 5 -
Four Elements of Materials Science
OVERVIEW OF THE COURSE
Processing Structure Properties
Chemical Synthesis Melting Casting Annealing Sintering Diffusion ....
Atomic/Molecular St. Bond structure Crystal Structure Defect Structures Microstructure Energy Band Structure
Performance
Mechanical Electrical Optical Thermal Magnetic
Cost Reliability Efficiency Service Life ...
1 2
3 4
5
7
6
/
Chapter 5 -
Rate of Processes
Chemical Kinetics Transport Phenomenon
Matter Energy
Mass Transport Heat Transfer
DIFFUSSION
Chapter 5 - 3
ISSUES TO ADDRESS...
• How does diffusion occur?
• Why is it an important part of processing?
• How can the rate of diffusion be predicted for some simple cases?
• How does diffusion depend on structure and temperature?
Chapter 5: Diffusion
Chapter 5 -
Diffusion
Diffusion - Mass transport by atomic motion Mechanisms • Gases & Liquids – random (Brownian) motion • Solids – vacancy diffusion or interstitial diffusion
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.
DIFFUSION DEMO
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 - 8
• Interdiffusion: In an alloy, atoms tend to migrate from regions of high conc. to regions of low conc.
Initially
Adapted from Figs. 5.1 and 5.2, Callister & Rethwisch 8e.
Diffusion
After some time
Chapter 5 - 9
• Self-diffusion: In an elemental solid, atoms also migrate.
Label some atoms
Diffusion
A
B
C
D
After some time
A B
C
D
Chapter 5 -
More examples in 3-D !
Chapter 5 -
Diffusion Mechanisms
Energy is needed to generate a vacancy, break bonds, cause distortions. Provided by HEAT , kT ! Atom moves in the opposite direction of the vacancy !
Chapter 5 - 5
Substitutional Diffusion: • applies to substitutional impurities • atoms exchange with vacancies • rate depends on: --number of vacancies --activation energy to exchange.
Diffusion Mechanisms
Chapter 5 - 13
Diffusion Mechanisms
• Interstitial diffusion – smaller atoms can diffuse between atoms.
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 !
Adapted from Fig. 5.3(b), Callister & Rethwisch 8e.
Chapter 5 - 14
Adapted from chapter-opening photograph, Chapter 5, Callister & Rethwisch 8e. (Courtesy of Surface Division, Midland-Ross.)
• Case Hardening: -- Diffuse carbon atoms into the host iron atoms at the surface. -- Example of interstitial diffusion is a case hardened gear.
• Result: The presence of C atoms makes iron (steel) harder.
Processing Using Diffusion
Chapter 5 - 15
• Doping silicon with phosphorus for n-type semiconductors: • Process:
3. Result: Doped semiconductor regions.
silicon
Processing Using Diffusion
magnified image of a computer chip
0.5 mm
light regions: Si atoms
light regions: Al atoms
2. Heat it.
1. Deposit P rich layers on surface.
silicon
Adapted from Figure 18.27, Callister & Rethwisch 8e.
Chapter 5 - 16
Diffusion
• How do we quantify the amount or rate of diffusion?
( )( ) smkgor
scmmol
timearea surfacediffusing mass) (or molesFlux 22=≡≡J
Diffusion is a time-dependent process: the quantity of matter that is transported within another is a function of time.
Often it is necessary to know how fast diffusion occurs, or the rate of mass transfer .
Chapter 5 - 17
Diffusion • How do we quantify the amount or rate of diffusion?
( )( ) smkgor
scmmol
timearea surfacediffusing mass) (or molesFlux 22=≡≡J
J ∝ slope
dtdM
Al
AtMJ ==
M = mass
diffused time
• Measured empirically – Make thin film (membrane) of known surface area – Impose concentration gradient – Measure how fast atoms or molecules diffuse through the
membrane
Chapter 5 -
• Diffusion Flux:
10
• Directional Quantity (anisotropy ?)
• Flux can be measured for: --vacancies --host (A) atoms --impurity (B) atoms
MODELING DIFFUSION: FLUX
Material
Diffusion is a time-dependent process !
RATE OF MATERIAL TRANSPORT
Chapter 5 -
• When concentration C is plotted vs position (or distance) plotted with in the solid: Concentration Profile, the slope will give us concentration gradient. It is sometimes convenient to express conc. In terms of mass of diffusing species per unit volume of a solid. C(x): [kg/m3]
11
• Fick's First Law:
Concentration of Cu [kg/m3]
Concentration of Ni [kg/m3]
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.
CONCENTRATION PROFILES & FLUX
Chapter 5 -
Concentration Gradient
Chapter 5 -
• Steady State: the concentration profile doesn't change with time.
12
• Apply Fick's First Law:
• Result: the slope, dC/dx, must be constant (i.e., slope doesn't vary with position)!
Jx = −D
dCdx
dCdx⎛
⎝ ⎜
⎞
⎠ ⎟ left
=dCdx
⎛
⎝ ⎜
⎞
⎠ ⎟ right
• If Jx)left = Jx)right , then
STEADY STATE DIFFUSION
Why is the minus sign ?
Chapter 5 -
• Steel plate at 700º C
13
• Q: How much carbon transfers from the rich to the deficient side?
Adapted from Fig. 5.4, Callister 6e.
EX: STEADY STATE DIFFUSION
Chapter 5 - 23
Example: 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 = 110 x10-8 cm2/s – surface concentrations:
C2 = 0.02 g/cm3
C1 = 0.44 g/cm3
Chapter 5 - 24
scmg 10 x 16.1
cm) 04.0()g/cm 44.0g/cm 02.0(/s)cm 10 x 110( 2
5-33
28- =−
−=J
Example (cont).
12
12- xxCCD
dxdCDJ
−
−−≅=
Dtb 6
2=
glove C1
C2
skin paint remover
x1 x2
• Solution – assuming linear conc. gradient
D = 110 x 10-8 cm2/s
C2 = 0.02 g/cm3
C1 = 0.44 g/cm3
x2 – x1 = 0.04 cm
Data:
Chapter 5 -
Effect of Temperature
D = Do exp ⎛ ⎝ ⎜
⎞ ⎠ ⎟
- Qd R T
= 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
interstitial
Chapter 5 - 26
Diffusion and Temperature
• Diffusion coefficient increases with increasing T.
D = Do exp ⎛ ⎝ ⎜
⎞ ⎠ ⎟
- Qd R T
= 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
Chapter 5 - 27
Diffusion and Temperature
Adapted from Fig. 5.7, Callister & Rethwisch 8e. (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
D interstitial >> D substitutional C in α-Fe C in γ-Fe
Al in Al Fe in α-Fe Fe in γ-Fe
1000 K/T
D (m2/s)
0.5 1.0 1.5 10-20
10-14
10-8 T(°C) 15
00
1000
600
300
Chapter 5 - 28
Diffusion and Temperature
Adapted from Fig. 5.7, Callister & Rethwisch 8e. (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
D interstitial >> D substitutional C in α-Fe C in γ-Fe
Al in Al Fe in α-Fe Fe in γ-Fe
1000 K/T
D (m2/s)
0.5 1.0 1.5 10-20
10-14
10-8 T(°C) 15
00
1000
600
300
Chapter 5 - 8
• APF for a body-centered cubic structure = 0.68
ATOMIC PACKING FACTOR: BCC
a
a 2
a 3
a R
Chapter 5 -
Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell
a
10
• APF for a body-centered cubic structure = 0.74
ATOMIC PACKING FACTOR: FCC
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 !
Chapter 5 - 32
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?
!!"
#$$%
&'=!!
"
#$$%
&'=
101
202
1lnln and 1lnlnTR
QDDTR
QDD dd
⎟⎟⎠
⎞⎜⎜⎝
⎛−−==−∴
121
212
11lnlnln TTR
QDDDD d
transform data
D
Temp = T
ln D
1/T
Chapter 5 - 33
Example (cont.)
⎥⎦
⎤⎢⎣
⎡⎟⎠
⎞⎜⎝
⎛−
−= −
K 5731
K 6231
K-J/mol 314.8J/mol 500,41exp /s)m 10 x 8.7( 211
2D
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
1212
11exp TTR
QDD d
T1 = 273 + 300 = 573 K T2 = 273 + 350 = 623 K
D2 = 15.7 x 10-11 m2/s
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 x Changing with time
? Flux, J, changes at any point x !
Chapter 5 - 35
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
xCD
tC
∂
∂=
∂
∂Fick’s Second Law
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 - 37
Non-steady State Diffusion
Adapted from Fig. 5.5, Callister & Rethwisch 8e.
B.C. at t = 0, C = Co for 0 ≤ x ≤ ∞
at t > 0, C = CS for x = 0 (constant surface 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
C s
Chapter 5 - 38
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)
( )⎟⎠
⎞⎜⎝
⎛−=
−
−
Dtx
CCCt,xC
os
o
2 erf1
dye yz 2
0
2 −∫π=
Adapted from Fig. 5.5, Callister & Rethwisch 8e.
Chapter 5 - 39
Non-steady State Diffusion ⎟⎠
⎞⎜⎝
⎛−=
−
−
Dtx
CCCtxC
os
o
2erf1),(
Chapter 5 - 40
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 ⎟⎠
⎞⎜⎝
⎛−=
−
−
Dtx
CCCtxC
os
o
2erf1),(
Chapter 5 - 41
Non-steady State Diffusion ⎟⎠
⎞⎜⎝
⎛−=
−
−
Dtx
CCCtxC
os
o
2erf1),(
Chapter 5 - 42
Chapter 5 - 43
Solution (cont.):
– t = 49.5 h x = 4 x 10-3 m – Cx = 0.35 wt% Cs = 1.0 wt% – Co = 0.20 wt%
⎟⎠
⎞⎜⎝
⎛−=
−
−
Dtx
CCC)t,x(C
os
o
2erf1
)(erf12
erf120.00.120.035.0),( z
Dtx
CCCtxC
os
o −=⎟⎠
⎞⎜⎝
⎛−=
−
−=
−
−
∴ erf(z) = 0.8125
Chapter 5 - 44
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 0.7970 z 0.8125 0.95 0.8209
7970.08209.07970.08125.0
90.095.090.0
−
−=
−
−z
z = 0.93
Now solve for D
Dtxz
2=
tzxD 2
2
4=
/sm 10 x 6.2s 3600
h 1h) 5.49()93.0()4(
m)10 x 4(4
2112
23
2
2−
−==
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=∴
tzxD
Chapter 5 - 45
• To solve for the temperature at which D has the above value, we use a rearranged form of Equation (5.9a);
)lnln( DDRQTo
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 000,148
21125 −− −=T∴
Solution (cont.):
T = 1300 K = 1027ºC
Chapter 5 - 17
• The experiment: we recorded combinations of t and x that kept C constant.
• Diffusion depth given by:
C(xi,ti)−CoCs −Co
= 1− erf xi2 Dti
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ = (constant here)
DIFFUSION DEMO: ANALYSIS
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?
16
• Result: Dt should be held constant.
• Answer: Note: values of D are provided here.
Key point 1: C(x,t500C) = C(x,t600C). Key point 2: Both cases have the same Co and Cs.
Example 5.3
Chapter 5 -
Size Impact on Diffusion
Smaller atoms diffuse faster
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 for every solute - solvent pair
• Microstructure - grain boundaries and dislocation cores provide faster pathways for diffusing species, hence diffusion is faster in polycrystalline vs. single crystal materials. (WHY ?)
Chapter 5 - 50
Diffusion FASTER for... • open crystal structures • materials w/secondary bonding • smaller diffusing atoms • lower density materials
Diffusion SLOWER for... • close-packed structures • materials w/covalent bonding • larger diffusing atoms • higher density materials
Summary
