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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 1 Chapter 6: Diffusion in Solids MSE 280: Engineering Materials Professor Lane W. Martin ([email protected])

MSE 280 Chap 6

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Page 1: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 1

Chapter 6: Diffusion in Solids

MSE 280: Engineering Materials

Professor Lane W. Martin

([email protected])

Page 2: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 2

Overview

ISSUES TO ADDRESS...

• How does diffusion occur?

• Why is it an important part of processing?

• How can the rate of diffusion be predicted for

some simple cases?

• How does diffusion depend on structure and

temperature?

Page 3: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 3

Mechanisms

• Gases & Liquids – random (Brownian) motion

• Solids – vacancy diffusion or interstitial diffusion

Importance of diffusion?

• All materials processing

• Energy storage (e.g. fuel cells, batteries…)

• Filtration

• Chemical reactions

• Semiconductor devices

• And more…

Diffusion

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 4

• 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 surface in compression

From Callister 6e resource CD.

Processing Using Diffusion (1)

Callister, Fig. 6.0

Page 5: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 5

• 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

Fig. 18.0, Callister 6e.

Processing Using Diffusion (2)

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 6

: Atoms of one material diffusing into another and

vice versa (e.g. In an alloy, atoms tend to migrate from regions of

large concentration)

Initially After some time

100%

Concentration Profiles0

Adapted from

Figs. 5.1 and

5.2, Callister 6e.

Diffusion: The Phenomenon

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 7

: Atoms within one material exchanging

positions. (i.e. In an elemental solid, atoms also migrate)

Label some atoms at t = 0 Observe at t > 0

A

B

C

D

Diffusion: The Phenomenon

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 8

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 different possible mechanisms but let’s

consider the simple/more common cases:

1. Vacancy diffusion

2. Interstitial diffusion

Diffusion Mechanisms

Page 9: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 9

Lattice

distortion

First, bonds with the neighboring

atoms need to be broken

1) Vacancy Diffusion

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 10

Typically more rapid than vacancy diffusion

SMALLER species diffuse FASTER

Callister Fig. 6.3

2) Interstitial Diffusion

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 11

Consider atoms (or mass, M) going through a plane

t = 0

t = 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

Modeling Diffusion: Flux

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 12

• :

• Directional Quantity

• Flux can be measured for: --vacancies

--host (A) atoms

--impurity (B) atoms

From Callister 6e resource CD.

In general: diffusion flux may or may not be

the same over time

Modeling Diffusion: Flux

Page 13: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 13

What causes net flow of atoms?

• Concentration Profile, C(x): [kg/m3]

• Fick's First Law:

Concentration

of Cu [kg/m3]

Concentration

of Ni [kg/m3]

Position, x

Cu flux Ni flux

• The steeper the concentration profile, the greater the flux!

Adapted from

Fig. 5.2(c),

Callister 6e.

From Callister 6e resource CD.

Derivation of Fick’s First Law (1)

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 14

Derivation of Fick’s First Law (1)

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 15

Derivation of Fick’s First Law (2)

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 16

: the concentration profile doesn't

change with time

• Apply Fick's First Law:

• Result: the slope, dC/dx, must be constant

(i.e., slope doesn't vary with position)!

Jx D

dC

dx

dC

dx

left

dC

dx

right

• If Jx,left = Jx,right , then

Steady State Diffusion

Page 17: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 17

dx

dCDJ

Fick’s first law of diffusion C1

C2

x

C1

C2

x1 x2

D diffusion coefficient

Rate of diffusion independent of time

Flux proportional to concentration gradient = dx

dC

12

12 linear ifxx

CC

x

C

dx

dC

Steady State Diffusion

Page 18: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 18

• Steel plate at 700°C

with geometry shown:

• Q: How much carbon

transfers from the rich to

the deficient side?

= 2.4x10-10 g/cm2s

Steady State Diffusion: Example

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 19

• Methylene chloride is a common ingredient of paint

removers. Besides being an irritant, it also may be

absorbed through the skin. When using this paint

remover, protective gloves should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, what is

the diffusive flux of methylene chloride through the

glove?

• Data:

– diffusion coefficient in butyl rubber:

D = 110 x10-8 cm2/s

– surface concentrations:

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

Example: Personal Protective Equipment

Page 20: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 20

scm

g 10 x 16.1

cm) 04.0(

)g/cm 44.0g/cm 02.0(/s)cm 10 x 110(

25-

3328-

J

12

12- xx

CCD

dx

dCDJ

Dtb

6

2

glove

C1

C2

skin paint

remover

x1 x2

• Solution – assuming linear conc. Gradient (i.e. steady-state)

D = 110 x 10-8 cm2/s

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

x2 – x1 = 0.04 cm

Data:

Compare this result to J = 2.4x10-10 g/cm2s calculated in the previous

example of C diffusion in Fe (at higher T and smaller diffusing species).

FASTER diffusion through materials with weak (secondary) bonds

Example: Personal Protective Equipment

Page 21: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 21

Example – Fick’s First Law

We can purify hydrogen gas by diffusing it through a palladium

sheet. Compute the number of kilograms of hydrogen that

pass / hour through a 6 mm thick sheet of palladium with an

area of 0.25 m2 at 600°C.

Assume:

• D = 1.7x10-8 m2 / s

• CH = 2.0 kg / m3

• CL = 0.4 kg / m3

• That steady state conditions have been achieved.

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 22

D Do exp

Qd

R T

= pre-exponential [m2/s] = diffusion coefficient [m2/s]

= activation energy [J/mol or eV/atom] = gas constant [8.314 J/mol-K]

= absolute temperature [K]

D Do

Qd

R T

Note: Same type of T-dependence as

vacancies and interstitials

Temperature Dependence of Diffusion

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 23

Adapted from Fig. 6.7, Callister & Rethwisch 3e. (Date for Fig. 6.7

taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals

Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)

D interstitial >> D substitutional

C in α-Fe C in γ-Fe

Al in Al Fe in α-Fe Fe in γ-Fe

1000 K/T

D (m2/s)

0.5 1.0 1.5 10-20

10-14

10-8

T(C) 1

50

0

10

00

600

300

Again, SMALLER

species diffuse FASTER

Temperature Dependence of Diffusion

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 24

Example: At 300ºC the diffusion coefficient and activation energy for

Cu in Si are

D(300ºC) = 7.8 x 10-11 m2/s

Qd = 41.5 kJ/mol

What is the diffusion coefficient at 350ºC?

1

01

2

02

1lnln and

1lnln

TR

QDD

TR

QDD dd

121

212

11lnlnln

TTR

Q

D

DDD d

transform data D

Temp = T

ln D

1/T

Temperature Dependence of Diffusion

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K 573

1

K 623

1

K-J/mol 314.8

J/mol 500,41exp /s)m 10 x 8.7( 211

2D

12

12

11exp

TTR

QDD d

T1 = 273 + 300 = 573 K

T2 = 273 + 350 = 623 K

D2 = 15.7 x 10-11 m2/s

Temperature Dependence of Diffusion

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1000 K/T

D (m2/s)

0.5 1.0 1.5 10-20

10-14

10-8

T(C) 1

50

0

10

00

600

30

0

C diffusion in γ-Fe

(FCC) is slower than

in α-Fe (BCC) even at

higher T!

More OPEN STRUCTURES allow FASTER diffusion.

Crystal Structure and Diffusion

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 27

Note that we’d have to remove carbon from the right

side and add to the left side to keep a constant flux.

Possible in the previous example of processing using

gas source that can be added and removed but in many

cases this may not be possible….

Non steady-state diffusion

And now for the complication…

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• The concentration of diffusing species is a function of both time and position C = C(x,t)

• In this case is used

2

2

x

CD

t

C

Non-Steady State Diffusion

Page 29: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 29

From Fick’s 1st Law:

dx

dcDJ

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

dx

dcD

dx

d

dx

dJ

Conservation of mass:

Jr Jl

i.e. fluxes to left and to right have to correspond to concentration change.

dx

dJ

dx

JJ

dt

dc lr

dx

Sub into the first derivative:

dx

dcD

dx

d

dt

dcFick’s 2nd law

c = conc.

inside box

Partial differential equation. We’ll need boundary conditions to solve…

Non-Steady State Diffusion

Page 30: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 30

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

C o

C s

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

Example Non-Steady State Diffusion

Page 31: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 31

C(x,t) = Conc. at point x at time t

erf (z) = error function

erf(z) values are given in Table 6.1 in text

CS

Co

C(x,t)

Dt

x

CC

Ct,xC

os

o

2 erf1

dye yz 2

0

2

Solution

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Suppose it is desired to achieve a specific concentration

(C1) in an alloy, therefore:

os

o

os

o

CC

CC

CC

CtxC 1),(constant

which leads to:

Dt

x

2constant

Known for given system

Specified with C1

Non-Steady State Diffusion Cont.

Page 35: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 35

• Copper diffuses into a bar of aluminum

• 10 hours at 600°C 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: values

of D are

provided here

Key point 1: C(x,t500C) = C(x,t600C).

Key point 2: Both cases have the same Co and Cs.

Dt2

Example Processing Question

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 36

• During 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 (γ-Fe at 1000oC).

• Estimate the flux of carbon atoms at the surface given:

Do = 2.3 x 10-5 m2/s for C diffusion in γ-Fe.

Qd = 148 kJ/mol

ργ-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?

Diffusion Design Example

Page 37: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 37

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

27,782 yrs!

0.1

10

1000

105

107

109

1011

200 400 600 800 1000 1200

tim

e (

hours

)

T (K)

Diffusion Design Example

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 38

Fick’s 2nd Law Example

Determine the carburizing time necessary to achieve a carbon

concentration of 0.30 wt% at a position 4 mm into an iron-

carbon alloy that initially contains 0.10 wt% C. The surface

concentration is to be maintained at 0.90 wt% C and the

treatment is to be conducted at 1100°C.

Assume: D = 5.3x10-11 m2 / s

Page 39: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 39

Need to consider coulomb interactions between ions…

+ - - - - - - + + + +

+ - - - - - + + + + +

+ - - - - - - + + + +

+ - - - - - + + + +

+ - - - - - - + + + +

+ - - - - - + + + + +

Diffusion in Ionic Solids

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 40

Which do you expect to diffuse faster, cations

or anions?

•Smaller cations will usually diffuse faster

•But analogous to charge neutrality required for

defect formation, each ion will need counter

charge to move with it (e.g. vacancy, impurity or

free electrons or holes).

Diffusion in Ionic Solids

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MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 41

• Open crystal structures

• Lower melting T materials

• Materials w/secondary

bonding

• Smaller diffusing atoms

• Lower density materials

• Close-packed structures

• Higher melting T materials

• Materials w/covalent

bonding

• Larger diffusing atoms

• Higher density materials

Structure and Diffusion

Page 42: MSE 280 Chap 6

MSE 280: Engineering Materials ©Lane W. Martin Spring 2011 42

• Diffusion mechanisms and phenomena – Vacancy diffusion

– Interstitial diffusion

• Importance/usefulness of understanding diffusion (especially in processing)

• Steady-state diffusion

• Temperature dependence – similarity to vacancies and interstitials

• Non steady-state diffusion

• Structural dependence (e.g. size of the diffusing atoms, bonding type, openness of medium)

Concepts to Remember