Download - Diffusion Lecture#2
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MM 501Ashraf Ali NED University
Diffusion in Solid State
MM-501: Phase Transformations in Solids
Prof. Dr. Ashraf Ali
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Course OutlineDiffusion - how do atoms move through solids?
Diffusion mechanisms
Vacancy diffusion
Interstitial diffusion
Impurities
The mathematics of diffusion
Steady-state diffusion (Fick’s first law)
Nonsteady-State Diffusion (Fick’s second law)
Factors that influence diffusion
Diffusing species Host solid
Temperature Microstructure
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•Diffusion is easy in liquids and gases where atoms are relatively free to move around:
e.g. making chocolate milk. Really need to stir the milk to get that viscous chocolate syrup evenly mixed!
•Diffusion is difficult in solids due to bonding and requires, most of the time, external energy to mobilize the atoms.
•In solids, atoms are not fixed at its position but constantly moves (oscillates). The degree of movement depends on temperature (extrinsically) and structure (intrinsically) of the system
Inhomogeneous materials can become homogeneous by diffusion. For an active diffusion to occur, the temperature should be high enough to overcome energy barriers to atomic motion.
What is diffusion?•Material transport by atomic motion is called Diffusion.
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Diffusion •Diffusion refers to the process by which atoms/molecules intermingle as a result of their kinetic energy of random motion.
•Consider two containers of gas A and B separated by a partition. The molecules of both gases are in constant motion and make numerous collisions with the partition. If the partition is removed as in the lower illustration, the gases will mix because of the random velocities of their molecules. In time a uniform mixture of A and B molecules will be produced in the container.
•The tendency toward diffusion is very strong even at room temperature because of the high molecular velocities associated with the thermal energy of the particles.
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e.g., Materials processing
1. Chemical reactions/treatments require mass transfer (i.e. diffusion).
e.g. alloying, strengthening of materials via case hardening etc…
2. Many materials are heat treated to achieve necessary properties: prefer to develop methods that will allow relatively high diffusion rates to achieve an efficient/cost-effectively process, Homogenisation for example.
Why do we care about diffusion in solids?
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1. Hard Facing (Carburizing of Steels) Tough Tools and Parts.
Wear Facing of Gears, Wheels and Rails 2. Chemical Tempering of Glass and Ceramics
Toughened Ceramics (Corel Ware)Shard resistant safety glass
3. Thin Film Electronics (CMOS and Bipolar Transistors)
Doping of Semiconductors
Applications of Diffusion in Solids-1
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Diffusion Bonding -- (Adhesives and Cements - ceramic, metallic and polymer) a. Portland Cement as Bonding for Construction
b. Solvent Cements for PVC Polymeric Piping c. Thermocouple Junctions
Corrosion Protection Galvanizing, Electroplating, Anodizing, Inhibiting Gas (Chemical) Separation Processes - Diffusion membranes
Applications of Diffusion in Solids-2
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PROCESSING USING DIFFUSION (1)
• Case Hardening: --Diffuse carbon atoms into the host iron atoms at the surface. --Example of interstitial diffusion is a case hardened gear.
• Result: The "Case" is --hard to deform: C atoms "lock" planes from shearing. --hard to crack: C atoms put the surface in compression.
Fig. 5.0, Callister 6e.(Fig. 5.0 iscourtesy ofSurface Division, Midland-Ross.)
From Callister 6e resource CD.
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Diffusion Bonding• Diffusion bonding is a method of creating a
joint between similar or dissimilar metals, alloys, and nonmetals.
• Two materials are pressed together (typically in a vacuum) at a specific bonding pressure with a bonding temperature for a specific holding time.
PROCESSING USING DIFFUSION (2)
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• “Solid state” joining process
• Apply heat & pressure across interface to be joined
• Local plastic deformation +diffusion = joint!
• Useful for joining difficult to weld metals, dissimilar materials, metals and ceramics
B-18
Application of Diffusion Bonding
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• Doping of Silicon with P for n-type semiconductors:• Process:1. Deposit P rich layers on surface.
2. Heat it.
3. Result: Doped semiconductor regions.
silicon
silicon
magnified image of a computer chip
0.5mm
light regions: Si atoms
light regions: Al atoms
Fig. 18.0, Callister 6e.
From Callister 6e resource CD.
PROCESSING USING DIFFUSION (3)Thin Film Electronics (CMOS and Bipolar Transistors)
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DIFFUSION: THE PHENOMENON
Self-diffusion: Atoms within one material exchanging positions. (i.e. In an elemental solid, atoms also migrate).
Label some atoms After some time
A
B
C
DA
B
C
D
From Callister 6e resource CD.
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DIFFUSION: THE PHENOMENON
100%
Concentration Profiles0
Cu Ni
Interdiffusion: Atoms of one material diffusing into another and vice versa. For example, in an alloy, atoms tend to migrate from regions of large concentration (Impurity Diffusion).
Initially After some time
100%
Concentration Profiles0
Adapted from Figs. 5.1 and 5.2, Callister 6e.
From Callister 6e resource CD.
Consider a concentration gradient then atoms move from high conc. to low conc..
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Diffusion MechanismsHow do atoms move in a crystalline solid?
For diffusion to occur:1. Adjacent site needs to be empty (vacancy or
interstitial).2. Sufficient energy must be available to break
bonds and overcome lattice distortion.
There are many possible mechanisms but let’s consider the simple cases:
1. Vacancy diffusion.2. Interstitial diffusion.
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Vacancy Mechanism
Atoms can move from one site to another if there is sufficient energy present for the atoms to overcome a local activation energy
barrier and if there are vacancies present for the atoms to move into.
The activation energy for diffusion is the sum of the energy required to form a vacancy and the energy to move the vacancy.
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Vacancy diffusion- An atom adjacent to a vacant lattice site moves into it.
Essentially looks like an interstitial atom: lattice distortion
First, bonds with the neighboring atoms need to be broken
From Callister 6e resource CD.
To jump from lattice site to lattice site, atoms need energy to break bonds with neighbors, and to cause the necessary lattice distortions during jump. This energy comes from the thermal energy of atomicvibrations (Eav ~ kT)
Materials flow (the atom) is opposite the vacancy flow direction.
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Interstitial atoms like hydrogen, helium, carbon, nitrogen, etc) must squeeze through openings between interstitial sites to diffuse around in a crystal.
The activation energy for diffusion is the energy required for these atoms to squeeze through the small openings between the host lattice atoms.
Interstitial Mechanism
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Interstitial DiffusionMigration from one interstitial site to another (mostly for small atoms that can be interstitial impurities: (e.g. H, C, N, and O) to fit into interstices in host.
Carbon atom in Ferrite
Interstitial diffusion is generally faster than vacancy diffusion because bonding of interstitials to the surrounding atoms is normally weaker and there are many more interstitial sites than vacancy sites to jump to.
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Interstitial Diffusion-Animation
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Diffusivity -- depends on: 1. Diffusion mechanism. Substitutional vs interstitial.2. Temperature. 3. Type of crystal structure of the host lattice. Interstitial diffusion easier in BCC than in FCC.4. Type of crystal imperfections.
(a) Diffusion takes place faster along grain boundaries than elsewhere in a crystal.
(b) Diffusion is faster along dislocation lines than through bulk crystal.
(c) Excess vacancies will enhance diffusion.5. Concentration of diffusing species.
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MODELING DIFFUSION: FLUX (J)
Consider atoms, M, going through a plane of area, A, in time, t:
t = 0t = t’
2 atoms passed from left to right (+ direction)1 atom passed from right to left (- direction)
Net result:
'
1
tarea
atom
or
'At
M
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MODELING DIFFUSION: FLUXFlux (J): No. of Atoms diffusiong through Unit area and per Unit Time OR Materials diffusion through unit area per unit time.
J
1A
dMdt
kg
m2s
or
atoms
m2s
• Directional Quantity
• Flux can be measured for: --vacancies --host (A) atoms --impurity (B) atoms
J x
J y
J z x
y
z
x-direction
Unit area A through which atoms move.
In general: diffusion flux may or may not be the same over time
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MODELING DIFFUSIONWhat causes net flow of atoms?
• Concentration Profile, C(x): [kg/m3]
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!
Adapted from Fig. 5.2(c), Callister 6e.
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Steady-State Diffusion: Fick’s First Law
Fick’s first law: the diffusion flux along direction x is proportional to the concentration gradient.
where D is the diffusion coefficient
The concentration gradient is often called the driving force in diffusion.
Fick’s first law applies to steady state flux in a uniform concentration gradient.
J x D
dCdx
Diffusion coefficient [m2/s]
concentration gradient [kg/m4]
flux in x-dir. [kg/m2-s]
The minus sign in the equation means that diffusion is down the concentration gradient. Concentration gradient: dC/dx (Kg m-4) is the
slope at a particular point on concentration profile.
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ExampleThe total membrane surface area in the lungs (alveoli) may be on the order of 100 square meters and have a thickness of less than a millionth of a meter, so it is a very effective gas-exchange interface.
Fick's law is commonly used to model transport processes in:• Foods, • Biopolymers,•Biomedical • Pharmaceuticals, • Porous soils, • Semiconductor doping process, etc.
Where can we use Fick’s Law?
CO2 in air has D~16 mm2/s, and, in water, D~ 0.0016 mm2/s
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Steady-State Diffusion: the concentration profile doesn't change with time.
Steady-State Diffusion
• 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
J x D
dCdx
dCdx
left
dCdx
right
• If Jx)left = Jx)right , then
From Callister 6e resource CD.
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• Steel plate at 7000C with geometry shown:
• Q: In steady-state, how much carbon transfers from the rich to the deficient side?
Adapted from Fig. 5.4, Callister 6e.C1
= 1.2kg/m3
C2 = 0.8kg/m3
Carbon rich gas
10mm
Carbon deficient
gas
x1 x205m
m
D=3x10-11m2/s
Steady State = straight line!
Example: Steady-state Diffusion
J DC2 C1
x2 x1
2.4 10 9 kg
m2s
Knowns: C1= 1.2 kg/m3 at 5mm (5 x 10–3 m) below surface.
C2 = 0.8 kg/m3 at 10mm (1 x 10–2 m) below surface.
D = 3 x10-11 m2/s at 700 C.
700 C
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• Concentration profile, C(x), changes w/ time.
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• To conserve matter: • Fick's First Law:
• Governing Eqn.:
Concentration, C, in the box
J (right)J (left)
dx
dCdt
=Dd2C
dx2
dx
dC
dtJ D
dC
dxor
J (left)J (right)
dJ
dx
dC
dt
dJ
dx D
d2C
dx2
(if D does not vary with x)
equate
Non-Steady-State DiffusionIn most real situations the concentration profile and the concentration gradient are changing with time. The changes of the concentration profile is given in this case by a differential equation, Fick’s second law.
Called Fick’s second law
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Fick's Second Law of Diffusion
In words, the rate of change of composition at position x with time, t, is equal to the rate of change of the product of the diffusivity, D, times the rate of change of the concentration gradient, dCx/dx, with respect to distance, x.
x dC d
D x d
d =
t dC d xx
Non-Steady-State Diffusion
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• Copper diffuses into a bar of aluminum.
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• General solution:
"error function"Values calibrated in Table 5.1, Callister 6e.
C(x,t) Co
Cs Co
1 erfx
2 Dt
pre-existing conc., Co 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.
Example: Non Steady-State Diffusion
t3>t2>t1
Fig. 6.5: Concentration profiles nonsteady-state diffsion taken at three different times
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T R
Q - D = D d
o exp
T R
Q - D = D d
olnln
WhereD is the Diffusivity or Diffusion Coefficient ( m2 / sec )Do is the prexponential factor ( m2 / sec )Qd is the activation energy for diffusion ( joules / mole )R is the gas constant ( joules / (mole deg) )T is the absolute temperature ( K )
Temperature Dependence of the Diffusion Coefficient
OR
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Carburizing or Surface Modifying System
Species A achieves a surface concentration of Cs
and at time zero the initial uniform concentration of species A in the solid is Co . Then the solution to
Fick's second law for the relationship between the concentration Cx at a distance x below the surface at
time t is given as
where Cs = surface concentration,
Co = initial uniform bulk concentration
Cx = concentration of element at distance x
from surface at time t.x = distance from surfaceD = diffusivity of diffusing species in host
latticet = time
Dt2
x erf - 1 =
C - C
C - C
os
ox
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Carburizing or Surface Modifying SystemSpecies A achieves a surface concentration of Cs and at time zero the initial uniform concentration of species A in the solid is Co . Then the solution to Fick's second law for the relationship between the concentration Cx at a distance x below the surface at time t is given as:
Dt2
x erf - 1 =
C - C
C - C
os
ox
where Cs = surface concentration, Co = initial uniform bulk concentrationCx = concentration of element at distance x from surface at time t.x = distance from surfaceD = diffusivity of diffusing species in host latticet = time
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Experimental Determination of Diffusion Coefficient
Tracer method• Radioisotopic tracer atoms are deposited at surface of solid by e.g. electro deposition• isothermal diffusion is performed for a given time t, often quartz ampoules are used (T<1600°C)• Sample is then divided in small slices either mechanically, chemically or by sputtering techniques• Mechanically: for diffusion length of > 10 µm; D>10-11 cm2/s• Sputtering of surface: for small diffusion length (at low temperatures) 2nm …10µm, for the range D = 10-21 …10-12 cm2/s
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Experimental Determination of Diffusion Coefficient
• Example:
Diffusion of Fe in Fe3Si• From those figures thediffusion constant can bedetermined with an accuracyof a few percent• Stable isotopes can be used
as well, when high resolution SIMS is used
• This technique is moredifficult
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Diffusion Data
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Non-Steady-State Diffusion: c(x,t)
Fick'sSecond"Law": c
t
xDc
x
Dc 2
x2
C(x,t) C0
Cs C0
1 erf x2 Dt
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• Copper diffuses into a bar of aluminum.• 10 hours processed at 600 C gives desired C(x).• How many hours needed to get the same C(x) at 500 C?
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(Dt)500ºC =(Dt)600ºCs
C(x,t) CoC Co
=1 erfx
2Dt
• Result: Dt should be held constant.
• Answer:Note: values of D are provided.
Key point 1: C(x,t500C) = C(x,t600C).
Key point 2: Both cases have the same Co and Cs.
t500(Dt)600
D500
110hr
4.8x10-14m2/s
5.3x10-13m2/s 10hrs
Processing Question
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• The experiment: we recorded combinations of t and x that kept C constant.
to
t1
t2
t3
xo x1 x2 x3
• Diffusion depth given by: xi Dti
C(xi,ti) CoCs Co
1 erfxi
2 Dti
= (constant here)
Diffusion Analysis
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Non steady-state diffusion
From Fick’s 1st Law: dx
dcDJ
Take the first derivative w.r.t. x:
dx
dcD
dx
d
dx
dJ
Conservation of mass:
i.e. flux to left and to right has to correspond to concentration change.
dx
dJ
dx
JJ
dt
dc lr
Sub into the first derivative:
dx
dcD
dx
d
dt
dcFick’s 2nd law
JrJl
dx c = conc. inside box
Partial differential equation. We’ll need boundary conditions to solve…
In most practical cases steady-state conditions are not established, i.e. concentration gradient is not uniform and varies with both distance and time. Let’s derive the equation that describes non steady-state diffusion along the direction x.
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If it is desired to achieve a specific concentration C1
i.e.
os
o
os
o
CC
CC
CC
CtxC 1),(constant
which leads to:
Dt
x
2constant
Known for given system
Specified with C1
1
1
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PROCESSING QUESTION• 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?
(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.
t500(Dt)600
D500
110hr
4.8x10-14m2/s
5.3x10-13m2/s 10hrs
Adapted from Callister 6e resource CD.
Dt2
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• Diffusivity increases with T.
• Experimental Data:
1000K/T
D (m2/s) C in -Fe
C in -Fe
Al in Al
Cu in Cu
Zn in Cu
Fe in -Fe
Fe in -Fe
0.5 1.0 1.5 2.010-20
10-14
10-8T(C)1
50
0
10
00
60
0
30
0
D has exp. dependence on TRecall: Vacancy does also!
(see Table 6.2 Callister 6e)
Dinterstitial >> DsubstitutionalC 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.
Diffusion and Temperature
lnD lnD0 QdR
1T
logD logD0 Qd2.3R
1T
Note:
pre-exponential [m2/s]activation energy
gas constant [8.31J/mol-K]
D DoExp QdRT
diffusivity[J/mol],[eV/mol]
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Diffusion: Design ExampleDuring a steel carburization process at 1000oC, there is a drop in carbon concentration from 0.5 at% to 0.4 at% between 1 mm and 2 mm from the surface (g-Fe at 1000oC).
– Estimate the flux of carbon atoms at the surface.
Do = 2.3 x 10-5 m2/s for C diffusion in -Fe.
Qd = 148 kJ/mol
r-Fe = 7.63 g/cm3
AFe = 55.85 g/mol
– If we start with Co = 0.2 wt% and Cs = 1.0 wt% how long does it take to reach 0.6 wt% at 0.75 mm from the surface for different processing temperatures?
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T (oC) t (s) t (h)
300 8.5 x 1011 2.4x108
900 106,400 29.6
950 57,200 15.9
1000 32,300 9.0
1050 19,000 5.3
Need to consider factors such as cost of maintaining furnace at different T for corresponding times.
27782 yrs!
Diffusion: Design Example Cont’d
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Thin Film Electronics (CMOS and Bipolar Transistors)
Doping of Semiconductors Diffusion Bonding -- (Adhesives and Cements -
ceramic, metallic and polymer) Portland Cement as Bonding for Construction
Solvents Cements for PVC Polymeric Piping Solders and Welds for Thermocouple
Junctions Corrosion Protection Galvanizing, Electroplating, Anodizing, Inhibiting Gas (Chemical) Separation Processes - Diffusion membranes
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K)*J/(atom 101.38x = constant sBoltzmann = k
K e,temperatur absolute = T
e y Probabilit
23-
kT
E - E-*
'
Diffusion is a RATE PROCESS IN SOLIDS Probability of Finding an atom with energy E*
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Fraction of atoms or molecules having energies greater than E* which is itself much greater than the average energy E.
constant = C
eV/K 108.62x = constant sBoltzmann = k
K e,temperatur absolute = T
systemin molecules or atoms of number total = N E than greaterenergy withatoms of number = n where
e C = N
n
5-
total
*
kTE-
total
*
'
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Arrhenius' equation for the rate of many chemical reactions
constant rate = C
K) cal/(mol 1.986 or K)*J/(mol 8.314 = constant gas molar = R
K e,temperatur absolute = T
cal/mol or J/mol energy, activation = Q where,
e C = Reaction of Rate T R
Q-
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T R 2.303
Q - C = rate
T R
Q - C = rate
1010 loglog
lnln
Rewritten as linear functions of the reciprocal of the absolute temperature.
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Composition Specification
Definition: Amount of impurity (B) and host (A) in the system.
• Weight %
Two descriptions:
• Atom %
CB = mass of Btotal mass
x 100 C'B = # atoms of Btotal # atoms
x 100
• Conversion between wt % and at% in an A-B alloy:
CB = C'BAB
C'AAA + C'BABx 100 C'B =
CB/AB
CA/AA + CB/AB
• Basis for conversion:
mass of B = moles of B x AB
atomic weight of B
mass of A = moles of A x AA
atomic weight of A
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Example problem: conversion
• Determine the composition in at% of an alloy with 80wt% Al and 20wt% Cu
AAl = 26.98g/mol
ACu = 63.55g/mol
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Average alloy density
BA
BAavg VV
mm
volumetotal
masstotal
_
_ With
A
AA
mV
and
B
BB
mV
BBAA
BAavg mm
mm
//
Usually we know concentrations rather than total mass of each component
or100
BA
AA mm
mC 100
A
ABA C
mmm
BBAAavg CC
//
100
Also noting that:
B
A
B
A
m
m
C
C
Then we have: In terms of at% C’n
BBBAAA
BBAAavg ACAC
ACAC
// ''
''
BA
BA
A
ABBAA
A
A
avg
mm
CCmm
Cm
100
//
100
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Example problem: alloy density
• For the alloy with the composition given below, calculate number of Mo atoms/cm3.
16.4wt% Mo (atomic weight = 95.94 g/mol)
83.6wt% W (atomic weight = 183.84 g/mol)
Densities of each element:
Mo = 10.22 g/cm3
W = 19.30 g/cm3
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A Look at Diffusion Bonding
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Introduction
• Diffusion bonding is a method of creating a joint between similar or dissimilar metals, alloys, and nonmetals.
• Two materials are pressed together (typically in a vacuum) at a specific bonding pressure with a bonding temperature for a specific holding time.
• Bonding temperature– Typically 50%-70% of the melting temperature of the most
fusible metal in the composition– Raising the temperature aids in the interdiffusion of atoms
across the face of the joint.
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How does diffusion bonding work?
• Bonding pressure– Forces close contact between the edges of
the two materials being joined.– Deforms the surface asperities to fill all of the
voids within the weld zone .– Disperses oxide films on the materials,
leaving clean surfaces, which aids the diffusion and coalescence of the joint.
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How does diffusion bonding work?
• Holding Time– Always minimized
• Minimizing the time reduces the physical force on the machinery.
• Reduces cost of diffusion bonding process.• Too long of a holding time might leave voids in the
weld zone or possibly change the chemical composition of the metal or lead to the formation of brittle intermetallic phases when dissimilar metals or alloys are being joined.
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How does diffusion bonding work?
• Sequence for diffusion bonding a ceramic to a metal– a) Hard ceramic and soft metal
edges come into contact.– b) Metal surface begins to
yield under high local stresses.– c) Deformation continues
mainly in the metal, leading to void shrinkage.
– d) The bond is formed
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Advantages of diffusion bonding
• Properties of parent materials are generally unchanged.
• Diffusion bonding can bond similar or dissimilar metals and nonmetals.
• The joints formed by diffusion bonding are generally of very high quality.
• The process naturally lends itself to automation.
• Does not produce harmful gases, ultraviolet radiation, metal spatter or fine dusts.
• Does not require expensive solders, special grades of wires or electrodes, fluxes or shielding gases.
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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 = 573 K
T2 = 273 + 350 = 623 K
D2 = 15.7 x 10-11 m2/s
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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
Dt
x
CC
CtxC
os
o
2erf1
),(
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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
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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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• 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
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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 breakthrough time (tb), i.e., how long could the gloves be used before methylene chloride reaches the hand?
• Data (from Table 22.5)– diffusion coefficient in butyl rubber:
D = 110 x10-8 cm2/s
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Example (cont).
Time required for breakthrough ca. 4 min
gloveC1
C2
skinpaintremover
x1 x2
• Solution – assuming linear conc. gradient
Dtb 6
2
Equation 22.24
cm 0.04 12 xxD = 110 x 10-8 cm2/s
min 4 s 240/s)cm 10 x 110)(6(
cm) 04.0(28-
2bt
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Summary II
1. Diffusion is just one of many mechanisms for mass transport.
2. Electrical field can produce mass transport.
3. Magnetic field can produce mass transport.
4. Combination of fields can produce mass transport such as electrochemical transport.
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Application: Homogenization time
Solidification usually results in chemical heterogeneities– Represent it with a sinusoid of wavelength, λ– Composition should homogenize when, x > λ/2– The approximate time necessary is:
Homogenization time
- increases with λ2
- decreases exponentially with T
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Application:Service Life of a Microelectronic Device
•Microelectronic devices– have built-in heterogeneities– Can function only as long as these doped regions survive
• To estimate the limit on service life, ts– Let doped island have dimension, λ– Device is dead when, x ~ λ/2, hence
Service life- decreases with miniaturization (λ2)- decreases exponentially with T
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Influence of Microstructure on Diffusivity
Interstitial species– Usually no effect from microstructure– Stress may enhance diffusion
Substitutional species– Raising vacancy concentration increases D
• Quenching from high T• Solutes• Irradiation
– Defects provide “short-circuit” paths• Grain boundary diffusion• Dislocation “core diffusion”
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Adding Vacancies Increases D
• Quench from high T– Rapid cooling freezes in high cv– D decreases as cv evolves to equilibrium
• Add solutes that promote vacancies– High-valence solutes in ionic solids
• Mg++ increases vacancy content in Na+Cl-• Ionic conductivity increases with cMg
– Large solutes in metals– Interstitials in metals
• Processes that introduce vacancies directly– Irradiation– Plastic deformation
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Grain Boundary Diffusion
• Grain boundaries have high defect densities– Effectively, vacancies are already present– QD ~ Qm
• Grain boundaries have low cross-section– Effective width = δ– Areal fraction of cross-section:
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Jumps of Interstitial Atoms
• Atom motion– Must overcome barrier ΔGm to move from site to site (~ 1eV)– Attempts with vibrational frequency νe ~ 1014/sec
• Number of jumps per unit timeω = (# attempts/time)(probability of jump/attempt)
(of the order of 105/sec at room T)
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Jumps of Substitutional Atoms
• For a Substitutional atom to jump– There must be a neighboring vacancy to permit the jump– The atom must overcome its barrier and jump
ω = [P(vacant site)][P(jump given vacant site)]
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Diffusion – Thermally Activated Process (I)
In order for atom to jump into a vacancy site, it needs to posses enough energy (thermal energy) to break the bonds and squeeze through its neighbors. The energy necessary for motion, Em, is called the activation energy for vacancy motion.
At activation energy Em has to be supplied to the atom so that it could break inter-atomic bonds and to move into the new position.
Schematic representation of the diffusion of an atom from its original position into a vacant lattice site.
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Diffusion – Thermally Activated Process (II)
The average thermal energy of an atom (kBT = 0.026 eV for room temperature) is usually much smaller that the activation energy Em (~ 1 eV/atom) and a large fluctuation in energy (when the energy is “pooled together” in a small volume) is needed for a jump.
The probability of such fluctuation or frequency of jumps, Rj, depends exponentially from temperature and can be described by equation that is attributed to Swedish chemist
Arrhenius :
where R0 is an attempt frequency proportional to the frequency of atomic vibrations.
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Diffusion – Thermally Activated Process (III)
For the vacancy diffusion mechanism the probability for any atom in a solid to move is the product of the probability of finding a vacancy in an adjacent lattice site (see Chapter 4):
and the probability of thermal fluctuation needed to overcome the energy barrier for vacancy motion
The diffusion coefficient, therefore, can be estimated as
Temperature dependence of the diffusion coefficient, follows the Arrhenius dependence.
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Diffusion – Temperature Dependence (I)Diffusion coefficient is the measure ofmobility of diffusing species.
D0 – temperature-independent preexponential (m2/s)Qd – the activation energy for diffusion (J/mol or eV/atom)R – the gas constant (8.31 J/mol-K or 8.62x/atom-KT – absolute temperature (K)
The above equation can be rewritten as
The activation energy Qd and preexponential D0, therefore, can be estimated by plotting lnD versus 1/T or logD versus 1/T. Such plots are Arrhenius plots.
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Diffusion – Temperature Dependence (II)
Graph of log D vs. 1/T has slop of –Qd/2.3R, intercept of ln Do
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Diffusion – Temperature Dependence (III)
Arrhenius plot of diffusivity data for some metallic systems
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Diffusion of different species
Smaller atoms diffuse more readily than big ones, and diffusion is faster in open lattices or in open directions
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Diffusion: Role of the microstructure (I)
Self-diffusion coefficients for Ag depend on the diffusion path.
In general, the diffusivity is greater through lessrestrictive structural regions – grain boundaries, dislocation cores, external surfaces.
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Diffusion: Role of the microstructure (II)
The plots (opposite) are from the computer simulation by T. Kwok, P. S. Ho, and S. Yip.
Initial atomic positions are shown by the circles, trajectories of atoms are shown bylines.
We can see the difference between atomic mobility in the bulk crystal and in the grain boundary region.
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Exercise1. A thick slab of graphite is in contact with a 1mm thick sheet of steel. Carbon steadily diffuses through the steel at 925C. The carbon reaching the free surface reacts with CO2 gas to form CO, which is then rapidly pumped away.
Determine the carbon concentration, C2, adjacent to the free surface, and the find the carbon flux in the steel, given that the reaction velocity for C+CO22CO is =3.010-6cm/sec.
At 925C, the solubility of carbon in the steel in contact with graphite is 1.5wt% and the diffusivity of carbon through steel is D=1.710-7cm2/sec. The equilibrium solubility of carbon in steel, Ceq, is 0.1wt% for the CO/CO2 ratio established at the surface of the steel.
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ExercisePe
lD
1.76The Péclet number is
Note: The value of the Péclet number suggests mixed kinetic behavior is expected.
C2 Ceq C0 Ceq
1 lD
0.11.5 0.111.76
, [wt%]
C2 0.61wt%.
The carbon concentration in the steel at the free surface, C2, is
The steady-state flux is Jss = 1.51 10-6 [wt% C cm/s]
Jss = 1.18 10-7 [g/ cm2-s]
Divide by the density of steel, =12.8 cm3/100g to obtain the steady-state flux of carbon
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Exercise2. Two steel billets—a slab and a solid cylinder—contain 5000ppm residual H2 gas. These billets are vacuum annealed in a furnace at 725C for 24 hours to reduce the gas content. Vacuum annealing is capable of maintaining a surface concentration in the steel of 10ppm H2 at the annealing temperature.
Estimate the average residual concentration of H2 in each billet after vacuum annealing, given that the diffusivity of H in steel at 725C is DH=2.2510- 4 cm2/sec.
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Exercise15 cm
2h=
10 c
m
2h=10 cm
15 cm
10 cm
15 cm
15 cm
2h = 10 cm
2h =
10
cm
Rectangular and cylindrical slabs of steel
10 cm
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Solution in the Class
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Given:t=24 hr=86400 sCo= Initial Concentration= 5000 ppmCs= Surface concentration= 10 ppmDH= 2.25x10-4 cm2/sC1= average residual concentration=?
We know that:(C1-Co) / (Cs-Co) = Constant(z) and also
X (Dt) or x = Constant x (Dt)
or Constant(z) = (Dt) / x2
Now we can write:(C1-Co) / (Cs-Co) = (Dt) / x2 or C1= (Co-Cs) x (f) + Cs
Therefore, For slab:C1= (Co-Cs) x (flong x fshort x fshort) + Cs
and For Cylinder:C1= (Co-Cs) x (flong x fshort) + Cs
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STRUCTURE & DIFFUSIONDiffusion FASTER for...
• open crystal structures
• lower melting T materials
• materials with secondary bonding
• smaller diffusing atoms
• lower density materials
Diffusion SLOWER for...
• close-packed structures
• higher melting T materials
• materials with covalent bonding
• larger diffusing atoms
• higher density materials
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Factors that Influence Diffusion: Summary
Temperature - diffusion rate increases very rapidly with increasing temperature
Diffusion mechanism - interstitial is usually faster
than vacancy
Diffusing and host species - Do, Qd is different for
every solute, solvent pair
Microstructure - diffusion faster in polycrystalline vs. single crystal materials because of the accelerated diffusion along grain boundaries and dislocation cores.
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Concepts to remember
• Diffusion mechanisms and phenomena.– Vacancy diffusion.– Interstitial diffusion.
• Importance/usefulness of understanding diffusion (especially in processing).
• Steady-state diffusion.• Non steady-state diffusion.• Temperature dependence.• Structural dependence (e.g. size of the diffusing
atoms, bonding type, crystal structure etc.).
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Summary
Activation energy
Concentration gradient
Diffusion
Diffusion coefficient
Diffusion flux
Vacancy diffusion
Make sure you understand language and concepts:
Driving force
Fick’s first and second laws
Interdiffusion
Interstitial diffusion
Self-diffusion
Steady-state diffusion
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Next Class
Nucleation and Growth Kinetics