diffusion lecture#2

1

MM 501Ashraf Ali NED University

Diffusion in Solid State

MM-501: Phase Transformations in Solids

Prof. Dr. Ashraf Ali

2


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

3


•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.

4


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.

5


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?

6


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

7


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

8


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.

9


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)

10


• “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

11


• 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.


PROCESSING USING DIFFUSION (3)Thin Film Electronics (CMOS and Bipolar Transistors)

12


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


13


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.


Consider a concentration gradient then atoms move from high conc. to low conc..

14


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.

15


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.

16


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


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.

17


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

18


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.

19


Interstitial Diffusion-Animation

20


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.

21


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

22


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

23


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.

24


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.

25


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

26


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


27


• 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

28


• Concentration profile, C(x), changes w/ time.

14

• 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

29


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

30


• Copper diffuses into a bar of aluminum.

15

• 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

31


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

32


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

33


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

34


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

35


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

36


Diffusion Data

37


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

38


• 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?

16

(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

39

MM 501Ashraf Ali NED University17

• 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

40


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.

41


42


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

43


44


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

45


• 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]

46


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?

47


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

48


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

49


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*

50


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


systemin molecules or atoms of number total = N E than greaterenergy withatoms of number = n where

e C = N

n

5-

total

*

kTE-

total

*

'

51


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


cal/mol or J/mol energy, activation = Q where,

e C = Reaction of Rate T R

Q-

52


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.

53


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

54


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

55


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

56


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

57


A Look at Diffusion Bonding

58


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.

59


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.

60



• 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.

61



• 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

62


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.

63


64


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

65


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

),(

66


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

67


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

68


• 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

69


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

70


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

71


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.

72


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

73


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

74


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”

75


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

76


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:

77


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)

78


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)]

79


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.

80


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.

81


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.

82


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.

83


Diffusion – Temperature Dependence (II)

Graph of log D vs. 1/T has slop of –Qd/2.3R, intercept of ln Do

84


Diffusion – Temperature Dependence (III)

Arrhenius plot of diffusivity data for some metallic systems

85


Diffusion of different species

Smaller atoms diffuse more readily than big ones, and diffusion is faster in open lattices or in open directions

86


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.

87


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.

88


89


90


91


92


93


94


95


96


97


98


99


100


101


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.

102


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

103


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.

104


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

105


Solution in the Class

106


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

107


108


109


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

110


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.

111


112


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.).

113


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

114


Next Class

Nucleation and Growth Kinetics

diffusion lecture#2

Documents