5 Diffusion.pdf
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ChapterChapter
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ChapterChapter
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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
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How does diffusion occur?
Why is it an important part of
processing?
How can the rate of diffusion be
predicted for some simple cases?
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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
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Solids vacancy diffusion orinterstitial diffusion (similar
with defects/imperfection) Atomic Movement
Gas Liquid Solid
ChapterChapter
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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
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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
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applies to substitutional
impurities
atoms exchange with
vacancies
Atom moves from one site to
another if:
(1) there is enough activation
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(2) if there are vacancies orcrystal defects
increasing elapsed time
ChapterChapter
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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
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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
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so per un o me. whichatoms
move.
ChapterChapter
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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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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
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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
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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
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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
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. . ,
respectively. Assume a diffusion coefficient of
3 x 10-11 m2/s at this temperature.
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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
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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:
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= x - cm ssurface concentrations: C2 = 0.02 g/cm
3
C1 = 0.44 g/cm3
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ChapterChapter
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CCdC glove
C
assuming linear conc. gradient
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-xxdx
Dtb
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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
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i. Diffusing Species
ii. Temperature Environments
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ChapterChapter
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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
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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
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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
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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
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Find the activation energy (Qd) for material by experimentalmethod
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ChapterChapter
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ChapterChapter
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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
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General formulation: Ficks 2nd Law
C(x,t) CoCs Co
1 erf x2 Dt
,
"error function"
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Concentration, C
Cs
Cx
C0 - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
- -- - - - - - - -
- -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - -- -
Cs C0
Cx C0
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Distance from interface, x
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ChapterChapter
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ChapterChapter
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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
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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
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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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