Diffusion Lect#1

8/4/2019 Diffusion Lect#1

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MM 501Ashraf Ali NED University

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Diffusion in Solid State

MM-501: Phase Transformations in Solids

Prof. Dr. Ashraf Ali




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Diffusion Mechanisms

How 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 breakbonds and overcome lattice distortion.

There are many possible mechanisms but let’s consider the simple cases:1. Vacancy diffusion.2. Interstitial diffusion.




3

Vacancy Mechanism

Atoms can move from one site to another ifthere is sufficient energy present for theatoms 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 avacancy and the energy to move thevacancy.




4

Vacancy diffusion

- An atom adjacent to a vacant lattice site moves into it.

Essentially looks likean interstitial atom:lattice distortion

First, bonds with the neighboringatoms need to be broken

From Callister 6e resource CD.

To jump from lattice site to lattice site,

atoms need energy to break bondswith neighbors, and to cause thenecessary lattice distortions during

jump. This energy comes from thethermal energy of atomic

vibrations (Eav ~ kT)

Materials flow (the atom) is oppositethe vacancy flow direction.




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Interstitial atoms like hydrogen, helium,carbon, nitrogen, etc) must squeeze throughopenings between interstitial sites to diffuse

around in a crystal.

The activation energy for diffusion is theenergy required for these atoms to squeeze

through the small openings between thehost lattice atoms.

Interstitial Mechanism




6

Interstitial Diffusion

Migration 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 normallyweaker and there are many more interstitial sitesthan vacancy sites to jump to.


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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

timeareasurface

diffusingmass)(ormolesFlux

22J

dt

dM

A

l

At

M J

M =mass

diffused

time

J slope



9

Diffusion and Temperature

• Diffusion coefficient increases with increasing T .

D D o exp

÷ -

Q d

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

D o

Q d

R T

With conc. gradient fixed, higher D means higher flux of mass transport.



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• Diffusivity increases with T.

• Experimental Data:

D has exp. dependence on TRecall: Vacancy does also!

Dinterstitial >> Dsubstitutional

C in -Fe

C in -Fe Al in Al

Cu in Cu

Zn in Cu

Fe in -FeFe in -Fe

Diffusion and Temperature

ln D ln D0 -Qd R

1

T

log D log D0-Qd

2.3 R

1

T

Note:

pre-exponential [m2 /s]

activation energy

gas constant [8.31J/mol-K]

D D o Exp -

Q d

R T

diffusivity

[J/mol],[eV/mol]



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Steady-State Diffusion

dx

dC D J -

Fick’s first law of diffusion C1

C2

x

C 1

C 2

x 1 x 2

D diffusion coefficient(be careful of its unit)

Rate of diffusion independent of time

Flux proportional to concentration gradient =

dx

dC

12

12 linearifx x

C C

x

C

dx

dC

--



12

Diffusivity -- depends on:

1. Diffusion mechanism. Substitutional vs interstitial.

2. Temperature.3. Type of crystal structure of the host lattice.4. Type of crystal imperfections.

(a) Diffusion takes place faster along grainboundaries than elsewhere in a crystal.

(b) Diffusion is faster along dislocation linesthan through bulk crystal.

(c) Excess vacancies will enhance diffusion. 5. Concentration of diffusing species.



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Physical Aspect of D

1. D is the indicator of how fast atom moves.

2. In liquid state, D reaches similar level regardless of structure.3. In solid state, D shows high sensitivity to temperature and

structure.

4. Absolute temperature and Tm are what we should care about.



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Example: At 300ºC the diffusion coefficient andactivation energy for Cu in Si are

D (300ºC) = 7.8 x 10-11

m2

/sQ d = 41.5 kJ/mol

What is the diffusion coefficient at 350ºC?

-

-

1

01

2

021lnln and 1lnlnT R

Q D D T R

Q D D d d

---

121

212

11lnlnln

T T R

Q

D

D D D d

transformdata

D

Temp = T

ln D

1/ T



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Example (cont.)

--

-

K573

1

K623

1

K-J/mol314.8

J/mol500,41exp /s)m10x8.7( 211

2D

--

12

12 11exp T T R Q D D d

T1 = 273 + 300 = 573 K

T2 = 273 + 350 = 623 K

D 2 = 15.7 x 10-11 m2 /s



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• Steel plate at 7000C with geometry shown:

• Q: In steady-state, how much carbon transfersfrom the rich to the deficient side?

Adapted from Fig.5.4, Callister 6e .

Example: Steady-state Diffusion

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 with time.

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• To conserve matter: • Fick's First Law:

• Governing Eqn.:

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, dC x /dx, with respect to distance, x.

xd

C d D

xd

d =

t d

C d x x

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 .

Co

Cs

position, x

C(x,t)

tot1

t2 t3Adapted from Fig.5.5, Callister 6e .

Example: Non Steady-State Diffusion

t3>t2>t1

Fig. 6.5: Concentration profiles nonsteady-state diffusion taken at three different times

C0=Before diffusion

For t=0, C=C0 at 0x


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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 elevatedtemperature and in an atmosphere that gives asurface 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, determinethe temperature at which the treatment was carriedout.

• Solution tip: use Eqn.

-

--

Dt

x

C C

C t x C

o s

o

2erf1

),(



Solution (cont.):

– t = 49.5 h x = 4 x 10-3 m

– C x = 0.35 wt% C s = 1.0 wt%

– C o = 0.20 wt%

-

--

Dt

x

C C

C )t ,x ( C

o s

o

2erf1

)(erf12

erf120.00.1

20.035.0),(z

Dt

x

C C

C t x C

o s

o -

-

--

--

erf(z ) = 0.8125



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 0.8125

0.95 0.8209

7970.08209.0

7970.08125.0

90.095.0

90.0

--

-

-z

z 0.93

Now solve for D

Dt

x z

2

t z

x D

2

2

4

/sm10x6.2s3600

h1

h)5.49()93.0()4(

m)10x4(

4

2112

23

2

2-

-

t z

x D



MM 501Ashraf Ali NED University23

• To solve for the temperatureat which D has above value,

we use a rearranged form ofEquation (5.9a);

)lnln( D D R

Q T

o

d

-

from Table 5.2, for diffusion of C in FCC Fe

D o = 2.3 x 10-5 m2 /s Q d = 148,000 J/mol

/s)m10x6.2ln /sm10x3.2K)(ln-J/mol314.8(

J/mol000,14821125 --

-

T

Solution (cont.):

T = 1300 K = 1027°C




Example: Chemical Protective Clothing (CPC)

• Methylene chloride is a common ingredient of paintremovers.

• Besides being an irritant, it also may be absorbedthrough skin.

• When using this paint remover, protective gloves

should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, whatis the breakthrough time (t b ), i.e., how long could thegloves be used before methylene chloride reachesthe hand?

• Data (from Table 22.5):

– diffusion coefficient in butyl rubber: D = 110 x10-8 cm2/s




Example (cont).

Time required for breakthrough ca. 4 min

glove

C 1

C 2

skinpaintremover

x 1 x 2

• Solution – assuming linear conc. gradient

D t b

6

2

Equation 22.24

cm0.0412 - x x

D = 110 x 10-8 cm2 /s

min4s240 /s)cm10x110)(6(

cm)04.0(28-

2b t




T RQ - D = D d

o exp

T R

Q

- D = Dd

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




Carburizing or Surface Modifying System

Species A achieves a surface concentration of Cs and at time zero theinitial uniform concentration of species A in the solid is Co . Then the

solution to Fick's second law for the relationship between theconcentration Cx at a distance x below the surface at time t is given as:

Dt 2

x erf -1 =

C -C

C -C

os

o x

whereCs = surface concentration,Co = initial uniform bulk concentration

Cx = concentration of element atdistance x from surface at time t. x = distance from surface D = diffusivity of diffusing species inhost lattice

t = time




Experimental Determination ofDiffusion 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 areused (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




Experimental Determination ofDiffusion Coefficient

• Example:

Diffusion of Fe in Fe3Si

• From those figures the

diffusion constant can be

determined with an accuracy

of a few percent

• Stable isotopes can be usedas well, when high

resolution SIMS is used• This technique is more

difficult




Diffusion Data

Processing Question




• 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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• Result: Dt should be held constant.

• Answer: Note: values of D areprovided.

Key point 1: C(x,t500C) = C(x,t600C).Key point 2: Both cases have the same Co and Cs.

Processing Question




• The experiment: we recorded combinations of

t and x that kept C constant.

• Diffusion depth given by:

C(xi, t i )-Co

Cs -Co 1- erf

xi

2 Dt i

= (constant here)

Diffusion Analysis




Non steady-state diffusion

From Fick’s 1st

Law: dx

dc D J -

Take the first derivative w.r.t. x:

-dx

dc D

dx

d

dx

dJ

Conservation of mass:

i.e. flux to left and to right has to correspond to concentration change.

dx

dJ

dx

J J

dt

dc lr

-

--

Sub into the first derivative:

dx

dc D

dx

d

dt

dcFick’s 2nd law

J r J l

dx c = conc.

inside bo

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 notuniform and varies with both distance and time. Let’s

derive the equation that describes non steady-statediffusion along the direction x.




EX: NON STEADY-STATE DIFFUSION

• Copper diffuses into a bar of aluminum (semi infinite solid).

• General solution:

"error function"

Values calibrated in Table 5.1, Callister 6e .

Adapted fromFig. 5.5,Callister 6e .

From Callister 6e resource CD.

Co

Cs

position, x

C(x ,t)At to, C = Co inside the Al bar

t o

At t > 0, C(x=0) = Cs and C(x=∞) = Co

t 1 t 2

t 3




If it is desired to achieve a specific concentration C 1

i.e.

--

--

os

o

os

o

C C C C

C C C t xC 1),( constant

which leads to:

Dt

x

2constant

Known for given system

Specified with C 1

1

1







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?

• 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.

Adapted from Callister 6e resource CD.

Dt 2




Diffusion: Design Example

During a steel carburization process at 1000oC, there is a drop incarbon concentration from 0.5 at% to 0.4 at% between 1 mm and2 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 ittake to reach 0.6 wt% at 0.75 mm from the surface for differentprocessing temperatures?




T (oC) t (s) t (h)

300 8.5 x 1011 2.4x108

900 106,400 29.6

950 57,200 15.91000 32,300 9.0

1050 19,000 5.3

Need to consider factors such as cost of maintaining furnace at different T forcorresponding times.

27782 yrs!

Diffusion: Design Example Cont’d




A Look at Diffusion Bonding




Introduction

• Diffusion bonding is a method of creating a joint betweensimilar 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 aspecific holding time.

• Bonding temperature

– Typically 50%-70% of the melting temperature of the mostfusible metal in the composition

– Raising the temperature aids in the interdiffusion of atomsacross the face of the joint.

H d diff i b di




How does diffusion bondingwork?

• Bonding pressure

– Forces close contact between the edges ofthe two materials being joined.

– Deforms the surface asperities to fill all of thevoids within the weld zone .

– Disperses oxide films on the materials,

leaving clean surfaces, which aids thediffusion and coalescence of the joint.

H d diff i b di




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• Holding Time – Always minimized

• Minimizing the time reduces the physical force onthe machinery.

• Reduces cost of diffusion bonding process.

• Too long of a holding time might leave voids in theweld zone or possibly change the chemicalcomposition of the metal or lead to the formation of

brittle intermetallic phases when dissimilar metalsor alloys are being joined.

H d diff i b di




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• Sequence for diffusionbonding a ceramic to ametal

– a) Hard ceramic and soft metaledges come into contact.

– b) Metal surface begins toyield under high local stresses.

– c) Deformation continues

mainly in the metal, leading tovoid 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 metalsand nonmetals.

• The joints formed by diffusion bonding are generally ofvery 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 ofwires or electrodes, fluxes or shielding gases.




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Summary II

1. Diffusion is just one of many mechanisms formass 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 chemicalheterogeneities

– 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

G i B d Diff i




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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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Diffusion Thermally Activated




56

Diffusion – Thermally Activated

Process (I) In order for atom to jump into avacancy site, it needs to possesenough energy (thermal energy) tobreak the bonds and squeezethrough its neighbors. The energynecessary for motion, Em, is calledthe activation energy for vacancymotion.

At activation energy Em has to besupplied to the atom so that it couldbreak inter-atomic bonds and tomove into the new position.

Schematic representation of the

diffusion of an atom from its original

position into a vacant lattice site.

Diffusion Thermally Activated




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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 energyEm (~ 1 eV/atom) and a large fluctuation in energy (when theenergy 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 byequation that is attributed to Swedish chemist

Arrhenius :

where R0 is an attempt frequency proportional to the frequency of

atomic vibrations.

Diffusion Thermally Activated Process (III)




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Diffusion – Thermally Activated Process (III)

For the vacancy diffusion mechanism the probability for any atom in a solid tomove 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.

Diffusion – Temperature Dependence (I)




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Diffusion – Temperature Dependence (I)

Diffusion coefficient is the measure of

mobility 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 8.62x10-5 /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.

Diffusion – Temperature Dependence (II)




60

Diffusion Temperature Dependence (II)

Graph of log D vs. 1/T has

slop of – Qd /2.3R, interceptof ln Do

Diffusion Temperature Dependence (III)




61

Diffusion – Temperature Dependence (III)

Arrhenius plot of diffusivity data for some metallic systems

Diff i f diff t i




62

Diffusion of different species

Smaller atoms diffuse more readily than big ones, and diffusion is

faster in open lattices or in open directions

Diff i R l f h i (I)




63

Diffusion: Role of the microstructure (I)

Self-diffusion coefficients

for Ag depend on the

diffusion path.

In general, the diffusivity is

greater through less

restrictive structural regions

– grain boundaries,

dislocation cores, externalsurfaces.

Diffusion: Role of the




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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 by

lines.

We can see the difference between atomic

mobility in the bulk crystal and in the grain

boundary region.




65

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 carbonreaching the free surface reacts with CO2 gas to form CO, which is

then rapidly pumped away.

Determine the carbon concentration, C 2

, 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,

C eq, is 0.1wt% for the CO/CO2 ratio established at the surface of the

steel.

Exercise




66

Exercise Pe

l D

1.76The Péclet number is

Note: The value of the Péclet number suggests mixed kinetic behavior is expected.

C 2 C eq C 0 -C eq

1 l

D

0.11.5- 0.1

1

1.76

, [wt%]

C 2 0.61wt%.

The carbon concentration in

the steel at the free surface,C 2, is

The steady-state flux is J ss = 1.51 10-6 [wt% C cm/s]

J ss = 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




67

Exercise

2. Two steel billets — a slab and a solid cylinder — contain 5000ppmresidual 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.




68

Exercise 1 5

c m

2 h = 1 0 c m

2h=10 cm

1 5 c m

10 cm

2h = 10 cm

2 h

= 1 0 c m

Rectangular and cylindrical slabs of steel

10 cm




69

Given:t=24 hr=86400 s

Co= 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 alsoX (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

andFor Cylinder:C1= (Co-Cs) x (flong x fshort) + Cs




70




71




72

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

Factors that Influence Diffusion:




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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.




74




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Concepts to remember

• Diffusion mechanisms and phenomena. – Vacancy diffusion.

– Interstitial diffusion.

• Importance/usefulness of understandingdiffusion (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.).




76

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




77

Next Class

Nucleation and Growth Kinetics



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Diffusion Lect#1