vi . diffusion in solids - university of houstonlsun3/exemplarylectures/mece3345.pdfv. diffusion in...

V. Diffusion in Solids

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MECE 3345 Materials Science

VI . Diffusion in Solids

copyright © 2008 by Li Sun


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Examples• Ink in water • Biology

blood

air

Transpiration

• Smell


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Definition

Diffusion: The dispersal of material by random atom or molecularmovement from regions of high concentration to those of lower concentration.

The result of diffusion is to decrease the concentration non-uniformity.


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Ni-CuNi Cu

Cu in Ni Ni in Cu NiCu alloyNi Cu

x = 0

conc

entra

tion

t = 0

0

100

t = ∞t = t1


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Diffusion MechanismVacancy diffusion

Interstitial diffusion


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Activation Energy

∆G

STHG ∆−∆=∆Vibration frequency: ν

Ener

gy

Nearest neighbor sites: z

: FrequencyJumpAtom position

mmm STHG ∆−∆=∆

mm STH ∆>>∆

Interstitial diffusion Vacancy diffusion

⎟⎠

⎞⎜⎝

⎛ ∆−=Γ

RTG

vzX mV exp

=Γ

⎟⎠⎞

⎜⎝⎛ ∆−=

RTHDD mexp0


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Vacancy Diffusion

• Applies to substitutional impurities• Atoms must hop to open vacancy• Rate depends on probability of overcoming migration activation

energy, Qm

pQ = Bexp(-Hm/kT)And• Probability of a vacancy

pv = Nv/N = exp(-Hv/kT)• Here is Diffusion Coefficient is:

D = D0 exp(-Hd/kT)where Hd = Hm + Hv


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Interstitial Diffusion

• Applies to interstitial impurities (tend to be low concentration)• Most interstitial sites are not occupied• More rapid than vacancy diffusion• Activation energy involves overcoming the migration activation

energy, Hm.• Rate is defined by the Diffusion Coefficient

D = D0exp(-Hd/kT)

where D0 is a pre-exponential factor (m2/s) and Hd = Hm


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DataInterstitial diffusion in Fe Self-diffusion data for metals

solute ∆Hm(kJmol-1) D0(10-6m2s-1)

C 84 2.0

N 76 0.3

H 13 0.1

metal Tm(K)

Q (kJmol-1)

D0 (10-6m2s-1)

D(Tm) (10-12m2s-1)

Al 933 142 170

31

190

49

Cu 1356 200

1.9

0.59

0.65Ni 1726 279

γ-Fe 1805 284 0.29

x = 0

isotope


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Other Diffusion Mechanisms

Direct exchange

Zener ring


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Fick’s First Law

n2

a

x

dxdnann += 12 FrequencyJumpv :

1221 →→ −= JJJ

dxdna

nnaJ

τ

ττ

61

61

61

2

21

−=

⎟⎠⎞

⎜⎝⎛ −=

τ61: directionxalongFrequencyJump

site lattice at thestay atome if tuneaverage : 1v

=τn1


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Geometric Factor and Diffusion Coefficient

dxdnDJ −= 1221 →→ −= JJJ

τα

2aD =

Diffusion constant Geometric factor

Average stay time at interstitial site

Lattice parameter

2161 λτ

=D distance jump :λEinstein formula

Simple cubic FCC

a22

=λ

121

=α

BCC

a23

=λ

81

=α

a=λ

61

=α


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Steady State Diffusion

Steady state: the state where the concentration profile does not change with time, which means at any time the concentration at certain spot in space is a constant.

Position x

c

Fick’s first law:

conc

entra

tion

A

B

dxdnDJ −= n: number density

Position x

J


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Fick’s Second Law

n2n1

Position x

Position x

c

t1A

B

conc

entra

tion

xc

t2A

B

21→J 12→J

conc

entra

tion

A

x

∆x

J


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Diffusion into a Semi-infinite Solid

t ≤ 0, before diffusion startn(x) = n0, 0≤ x≤ ∞

A B

x = 0

n0

x = ∞0

ns

t > 0, diffusion start n (x=0) = ns , n (x = ∞) = n0

x = 0

n0

x = ∞

A B

0

ns

Solution to the Fick’s second law

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−

−

Dtxerf

nnnn

s

tx

21

0

0)(, erf: Gaussian error function

Position

conc

entra

tion

x

n

0

t=0

t increases

n0

ns

Fixed surface concentration.


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Error function ( ) ξπ

ξ dexerfx∫ −=0

22


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Examples

Steady state diffusion : Gas purification

Hydrogen gas purification can be achieved by pass mixed gas through palladium sheet.

Pump Pump Assume the there is a hydrogen purification system with cross-sectional area of 0.2m2, the palladium film thickness is 10mm hydrogen concentration on the two sides are 3kg/m3 and 0.5kg/m3. If the room temperature hydrogen diffusion coefficient in Pd is 1.0×10-15m2/s and the activation energy is 27.8kJ/mol. Calculate the pure hydrogen generation speed (kg/hour) when the system works at 500oC.


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xA0

C1

C2

diffusion direction

C1 C2

xB

What we know:C1= 3kg/m3, C2 = 0.5kg/m3

Palladium film thicknessd=10mm = 0.01mDiffusion area A=0.2m2

D @ 300K = 1.0×10-11m2/sQd = 27.8kJ/mol

Calculate how much hydrogen can diffuse through in 1hour at 500oC.

I.


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Diffusion into a Semi-infinite SolidII. Calculate the carburizing time to achieve 0.45 wt% C at a position of 2mm into a 0.2 weight percent steel. Sample was maintained at 1000oC with 1.3 wt% carbon atmosphere.

Cs C0

c

0.2 wt %1.3 wt %

Position

conc

entra

tion

x0

t=0C0

Cs

t2 > t1t = t1

0.0 0.5 1.0 1.5 2.0

0.0

0.2

0.4

0.6

0.8

1.0

erf(z

)


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Diffusion Coefficient and Thermally Activated Process

Thermally activated processes: The processes have exponentially increasing rates versus temperature.

Vacancy formation, electrical conductivity, creep and diffusion are thermally activated processes.

⎟⎠⎞

⎜⎝⎛−=

RTQDD dexp0

D0: temperature independent parameter (m2/s).Qd: activation energy for diffusion (J/mol).R: gas constant 8.31J/mol⋅KT: absolute temperature


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Arrhenius Plot

⎟⎠⎞

⎜⎝⎛−=

RTQCrate exp

Arrhenius Plot: log (rate) vs. T1

TRQCrate 1)log()log( −=

From the slope the thermal activation energy can be calculated

(1/K)

10-8

D (m

2 /s)

10-12

10-15

10-18

10-21

10-24

0 0.4 0.8 1.2 1.6 2.0 2.4

T1000


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Diffusion and Temperature

D has exponential dependence on T

Dinterstitial >> Dsubstitutional

C in α-FeC in γ-Fe

Al in AlFe in a-FeFe in g-Fe

1000K/T

D (m2/s) C in α-Fe

C in γ-Fe

Al in Al

Fe in α-Fe

Fe in γ-Fe

0.5 1.0 1.510-20

10-14

10-8T(°C)15

00

1000

600

300

vi . diffusion in solids - university of houstonlsun3/exemplarylectures/mece3345.pdfv. diffusion in...

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