5 diffusion.pdf

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ChapterChapter

55

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ChapterChapter

55

Topic ContentsTopic Contents

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ChapterChapter

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At the end of the lecture, students will be able:1. Name and describe the two atomic mechanisms of diffusion.

2. Distinguish between steady-state and nonsteady-state diffusion.

TOPIC OUTCOMES

3. (a) write Ficks firstand second laws in equation form,

and define all parameter

(b) Note the kind of diffusion for which each of these

equations is normally applied.

4. Write the solution to Ficks second law for diffusion into a semi-

infinite solid when the concentration of diffusing species at the

3

surface held constant. Define all parameters in this equation

5. Calculate the diffusion coefficient for some materials at a

specified temperature, given the appropriate diffusion constant

ChapterChapter

55

How does diffusion occur?

Why is it an important part of

processing?

How can the rate of diffusion be

predicted for some simple cases?

4

structure and temperature?


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ChapterChapter

55 Diffusion

Diffusion - Mass transport byatomic motion mi rate over aperiod of time

Mechanisms

Gases & Liquids random(Brownian) motion

5

Solids vacancy diffusion orinterstitial diffusion (similar

with defects/imperfection) Atomic Movement

Gas Liquid Solid

ChapterChapter

55

Interdiffusion: In an alloy, atoms tend to migratefrom regions of high conc. to regions of low conc.

Initially

Diffusion (solid)

After some time elevated T < Tm

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ChapterChapter

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Self-diffusion: In an elemental solid, atomsalso migrate.In solid, atomic movements are restricted due to a bonding equilibrium.

However, thermal vibrations occurring in solids do allow some atoms to move

Label some atomsAfter some time

CC

7

B

DA

B

ChapterChapter

55

Several different models for this atomic motionhave been proposed, of these possibilities,two dominate mechanisms for metallicdiffusion (similar concept with imperfection)

i. Vacancy Diffusion

ii. Interstitial Diffusion

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ChapterChapter

55

applies to substitutional

impurities

atoms exchange with

vacancies

Atom moves from one site to

another if:

(1) there is enough activation

9

(2) if there are vacancies orcrystal defects

increasing elapsed time

ChapterChapter

55

A lies to interstitial

impurities.

More rapid than

vacancy diffusion.

Size of diffusing atoms

must be relatively small

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ChapterChapter

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The diffusion rate in solid metal crystals is affected by 5 factors:

Type of diffusion mechanism (interstitial & substitutional) depending tosize of atom,ions)

Temperature of diffusion; normally when temp increase, diffusion ratealso increase

Concentration (quantity) of the diffusion species (concentration gradient);effect the solute atom

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Type of crystal structure (Lower atom packing order ex. BCC (0.68),FCC (0.74). Interatomic spaces between ion wider

Type of crystal imperfections present (Ex: grain boundaries less atomicpacking, excess vacancies will increase diffusion rate)

ChapterChapter

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Def: Energy required to produce the diffusive motion ofone mole of atoms.

Or atoms must overcome some energy barrier to move This ener barrier called activation ener

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ChapterChapter

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Diffusion is a time-dependentrocess

In a macroscopic sense thequantity of an element that istransported within another is afunction of time.

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ChapterChapter

55 Diffusion

How do we quantify the amountor rate of diffusion? J= Flux

Measured empirically

Make thin film (membrane) of known surface area

Impose concentration gradient (exp; diff pressure)

Measure how fast atoms or molecules diffuse

smkg

orscm

mol

timeareasurface

diffusingmass)(ormolesFlux

22J

14

through the membrane


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ChapterChapter

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Def: the mass (or,equivalently, thenumber of atoms) Mdiffusing through andperpendicular to areacross-sectional area of

x-direction

Unit area Athrough

15

so per un o me. whichatoms

move.

ChapterChapter

55

Flux:

J M

At

dt

dM

A

l

At

MJ

M=

16

J

A dt g

m2s

or a oms

m2s

massdiffused

time

J slope


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ChapterChapter

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i. Steady State Diffusion. on - ea y a e us on

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ChapterChapter

55

Steady State: the concentration profile doesn'tchange with time (not vary with position!). (Ex; H2

diffuse through Pd non reacting gases)

Ficks first law of diffusionC1

C2

C1

C

Rate of diffusion independent of time

ux propor ona o concen ra on gra en =dx

Part A Part B

Thin Metal

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dxDJ

xx1 x2

D diffusion coefficient

12

12linearifxx

CC

x

C

dx

dC

-ve =diffusion from high to lower concentration-ve diffusion gradient


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ChapterChapter

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Steady State:Jx(right)Jx(left)

Apply Fick's First Law:(No change in system with time)

x e x r g

Concentration, C, in the box doesnt change w/time.

x

Jx D

dC

dx

D =

diffusivity

(Diffusion

coefficient)

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Result: the slope, dC/dx (concentration gradient) must be constant(i.e., slope doesn't vary with position)!

dC

dx

left

dC

dx

right

If Jx)left= Jx)right, then

ChapterChapter

55

Sometimes the term of driving force is used in thecontact of what compels (make) a reaction tooccur.

For diffusion reaction, several forces are possible

But when diffusion is according to Ficks first law,the concentration gradient is the driving force.

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dx

dCDJ


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ChapterChapter

55

A plate of iron is exposed to a carburizing

(carbon-rich) atm on one side and a

decarburizing (carbon-deficient) atm on the

other side at 700 C. if a condition of steady

state is achieved, calculated the diffusion flux

of carbon through the plate if the

concentration of carbon at positions of 5 and

10 mm (5 x 10-3 and 10-2 m) beneath the

21

. . ,

respectively. Assume a diffusion coefficient of

3 x 10-11 m2/s at this temperature.

ChapterChapter

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How to solve?

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ChapterChapter

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Steel plate at

700 C with

geometryC1

=1.2kg/m

3

C2=0

.8kg/m

3

Carbon Stead State =

Solution

s own:

richgas

10

Carbondeficient

gas

x1 x205

D=3x10-11m2/s

straight line!

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carbon transfers

from the rich tothe deficient side?

J DC2 C1x2 x1

2.4 109 kgm2s

m

ChapterChapter

55

Methylene chloride is a common ingredient of paintremovers. Besides being an irritant, it also may be

absorbed through skin. When using this paint

, .

If butyl rubber gloves (0.04 cm thick) are used, whatis the diffusive flux of methylene chloride through

the glove?

Data:

diffusion coefficient in butyl rubber:

24

= x - cm ssurface concentrations: C2 = 0.02 g/cm

3

C1 = 0.44 g/cm3


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ChapterChapter

55

CCdC glove

C

assuming linear conc. gradient

12

-xxdx

Dtb

6

2

C2

skinpaintremover

x1 x2

D= 110x 10-8 cm2/s

C2 =0.02 g/cm3

C1 =0.44 g/cm3

x2 x1 =0.04 cm

Data:

25

scm

g

10x16.1cm)04.0(

)g/cm44.0g/cm02.0(

/s)cm10x110( 25-

3328-

J

ChapterChapter

55

i. Diffusing Species

ii. Temperature Environments

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ChapterChapter

55

It depend on diffusion coefficient (D) ofspecific materials.

Example: Ionics species Li+, Na+, K+

Diffusion Rate: Li+>Na+>K+

why?

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Atom seize: Li+K+

ChapterChapter

55(cont).

Relative saiz for atoms and ions

Radius in nanometer unit

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ChapterChapter

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Diffusion coefficient increases with increasing T(major influence).

D Doexp

Qd

RT

=temp-independent in range for which equation is valid [m2/s]

= diffusion coefficient/diffusivity [m2/s]

= activation energy [J/mol or eV/atom]

D

Do

Qd

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= gas constant [8.314 J/mol-K]

= absolute temperature [K]

R

T

ChapterChapter

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Dhas exponential dependence on T

TC500

000

00

00

Dinterstitial >> Dsubstitutional

C in -FeC in -Fe

Al in AlFe in -FeFe in -Fe

D(m2/s)

10-14

10-81 1 3

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1000K/T0.5 1.0 1.510-20


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ChapterChapter

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The temperature dependence of thediffusion coefficients is:

D = Do exp (-Qd/RT)

In natural log:

ln D = ln Do (Qd/R)(1/T)

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In term of log base 10,

log D = log Do (Qd/2.3R)(1/T)

ChapterChapter

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At 300C the diffusion coefficient and activation energy for Cu in Siare

D(300C) = 7.8 x 10-11 m2/s

d .

What is the diffusion coefficient (D) at 350C?

transformdata

D ln D

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1

01

2

02

1lnlnand

1lnln

TR

QDD

TR

QDD dd

121

212

11lnlnln

TTR

Q

D

DDD d

Temp = T 1/T


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ChapterChapter

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12

1211expTTR

QDD d

11

-

J/mol500,41exp/s)m10x8.7( 2112D

T1 = 273 + 300 = 573K

T2 = 273 + 350 = 623K

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.

D2 = 15.7 x 10-11 m2/s

ChapterChapter

55 Problem 4:

Find the activation energy (Qd) for material by experimentalmethod

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ChapterChapter

55 Solution 4:

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ChapterChapter

55

Most practical diffusion situations

us on ux an e concen ra ongradient at some particular point in asolid vary with time

Follow Ficks Second Law

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ChapterChapter

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Copper diffuses into a bar of aluminum.

pre-existing conc., Co of copper atoms

Surface conc.,Cs of Cu atoms

bar

Co

Cs

,

tot1

t2t3Cx

t3>t2>t1

X

3715

General formulation: Ficks 2nd Law

C(x,t) CoCs Co

1 erf x2 Dt

,

"error function"

ChapterChapter

55

Concentration, C

Cs

Cx

C0 - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

- -- - - - - - - -

- -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - -- -

Cs C0

Cx C0

38

Distance from interface, x


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ChapterChapter

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ChapterChapter

55

Consider one such alloy that initially has auniform carbon concentration of 0.25 wt% and

.of carbon at the surface is suddenly brought toand maintained at 1.20 wt.%, how long will ittake to achieve a carbon content of 0.80 wt.%at a position 0.5 mm below the surface? The

40

temperature is 1.6 x 10-11 m2/s; assume thatthe steel piece is semi-infinite.


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ChapterChapter

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Cs = 1.2 wt.% C Co = 0.25 wt. % C Cx = 0.80 wt.% C X = 0.50 mm = 5 x 10-4 m = -11 2.

Using formula: Ficks 2nd Law

Cx Co = 1 erf (x/2Dt)Cs - Co

erf (x/2Dt) = 0.4210or

erf (z) = 0.4210

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From table 5.1, get the z value (use ratio concept)

z erf (z)

0.35 0.3794 z - 0.35 = 0.4210 0.3794

z 0.4210 0.40 - 0 .35 0.4284 - 0 .3794

0.40 0.4284

z = 0.392

ChapterChapter

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z = 0.392

0.392 = (x/2Dt)

Rearranged equation include x and D value from question

t = 25400 s = 7.1 h

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ChapterChapter

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Diffusion FASTER for...

Diffusion SLOWER for...

-

materials w/secondarybonding

smaller diffusing atoms

materials w/covalentbonding

larger diffusing atoms

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ChapterChapter

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Applications

Case hardening of steel by gas carburizing

Impurity diffusion into silicon wafers forintegrated circuits

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ChapterChapter

55 Exercise 1

Consider the gas carburizing of gear steela . a cu a e me n m nu esnecessary to increase the carbon contentto 0.40% at 0.50 mm below the surface.Assume that the carbon content at thesurface is 0.90% and that the steel has anominal carbon content of 0.20%

Given D927C = 1.28 x 10-11 m2/s

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ChapterChapter

55

Exercise 2

From exercise 1, calculate the carboncon en a . mm enea e sur aceof the gear after 5 h carburizing time.Assume that the carbon content of thesurface of the gear is 0.90% and the steelhas a nominal carbon content of 0.20%

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ChapterChapter

55 Answers

Exercise 1: 143 minutes

Exercise 2: 0.52%

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