1. chapter 5: diffusion 2 introduction material transport by atomic motion diffusion couple: eg.,...
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Chapter 5: Diffusion 2
Introduction
•Material transport by atomic motion
•Diffusion couple:
eg., Cu-Ni in close contact; hold at elevated temperature
for extended period and cool to room temperature.
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Chapter 5: Diffusion 3
Before After
Introduction continue…
•Interdiffusion or impurity diffusion •Self diffusion: same type of atoms; no composition change
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Diffusion mechanismsVacancy diffusion
Interstitial diffusion
1.Vacancy Diffusion
Atom from normal lattice position changes position with
an adjacent vacancy (vacancy lattice site). So, the atoms
and vacancies travel in opposite directions. Both self-
diffusion and inter (impurity)-diffusion can occur thus
Chapter 5: Diffusion
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Diffusion mechanisms continue …
Source: William Callister 7th edition, chapter 5, page 112, figure 5.3(a)
Chapter 5: Diffusion
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Diffusion mechanisms continue …
2) Interstitial Diffusion
Atoms move from one interstitial site to another (vacant) interstitial site.
Source: William Callister 7th edition, chapter 5, page 112, figure 5.3(b)
Chapter 5: Diffusion
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Steady state DiffusionJ=M/At
If J is constant, steady-state diffusion exists.
Where,
J= rate of mass transfer with time, kg/m2-sec or atoms/m2-sec
A= Area across which diffusion is occurring
t= elapsed time, sec
Chapter 5: Diffusion
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Steady state Diffusion continue….
Steady-state diffusion Concentration profileSource: William Callister 7th edition, chapter 5, page 113, figure 5.4
Chapter 5: Diffusion
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Steady state Diffusion continue….
C: Concentration of diffusing species, kg/m3 or gm/cm3
x: Position
Hydrogen gas across palladium plate
Concentration Gradient: Slope at a particular point on the curve
=dC/dx
BA
BAXXCCΔC/ΔX
Chapter 5: Diffusion
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Steady state Diffusion continue….
Fick’s first law of diffusion
D = Diffusion coefficient, m2/sec
dc/dx = Concentration gradient is the main driving force for diffusion
(-) = Concentration gradient decrease
xdcdDJ
Chapter 5: Diffusion
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Problem
Carbon diffusing through a plate of iron
Steady state Diffusion continue….
5mm 10mm
Carbon deficient
Carbon rich
Concentration: 1.2 kg/m3 0.8 kg/m3
= -
Chapter 5: Diffusion
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Problem continue
Diffusion coefficient: 3 x 10-11 m2/sec
-
= 2.4 x 10-9 kg/m2-sec
= -
)10(10-)10(5
0.8)(1.2/sec)m10(3 33
211
BA
BAXXCCDJ
Steady state Diffusion continue….
Chapter 5: Diffusion
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Non-steady state Diffusion
•Diffusion flux and concentration gradient vary with time;
net accumulation or depletion of diffusing species results
………… Fick’ second law
……… Modified Fick’s second law
Chapter 5: Diffusion
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Semi-Infinite solid
Surface concentration at the other end is constant. eg, Bar
of length, l > 10Dt , i.e., none of the diffusing atoms reach
the bar end during the time-period of diffusion
Assumptions:
1. Co = Concentration before diffusion
2. x = Distance; at surface it is 0. It increases into the solid
3. t = Time; zero(0) at the instant diffusion starts
Chapter 5: Diffusion
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Semi-Infinite solid continue….
We have, for t=0, C=Co at 0≤x≤
For t>0, C=CS (Constant surface concentration) at x = 0
Also, C= CO at x=
This equation shows relationship between concentration, position and time
Error functionChapter 5: Diffusion
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Semi-Infinite solid continue….
•Concentration at distance x, CX is a function of
•If time (t) and position (x) are known and CO, CS and D
are given, CX can be determined
erf
Where, CX=Concentration at depth x after time t.
erf =erf2
Chapter 5: Diffusion
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Semi-Infinite solid continue….
=constant
Therefore =constant
=constant
If CX=C1 at a specific concentration of solute,
Chapter 5: Diffusion
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Semi-Infinite solid continue….
Surface concentration
Concentration C at distance x
Concentration before diffusion
Chapter 5: Diffusion
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Problem
Carburization of steel using methane (CH4) at 950°C
(1750°F)
Steel: 0.25 wt% Carbon. Using CH4, carbon at surface is
suddenly brought to and maintained at 1.2 wt% carbon.
How long will it take to achieve a carbon content of
0.80% carbon at a position 0.5 mm below the surface?
Chapter 5: Diffusion
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Problem continue…
D= 1.6 x 10-11 m2/sec
Chapter 5: Diffusion
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Problem continue…
Chapter 5: Diffusion
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Problem
The diffusion coefficients for copper in aluminum at 500
and 600 C are 4.8x10-14 and 5.3x10-13 m2/s, respectively.
Determine the approximate time at 500 C that will
produce the same diffusion result (in terms of
concentration of Cu at some specific point in Al) as a 10-h
heat treatment at 600°C.
To produce the same effect at 500°C, how long will it
take?
Chapter 5: Diffusion
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Problem continue…
Dt= constant
Chapter 5: Diffusion
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Factors in diffusion
•Temperature
•Time
D increases 5 orders of magnitude with temperature
Chapter 5: Diffusion
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Factors in diffusion continue…
Do = temperature independent pre-exponential (m2/sec)
Qd = the activation energy for diffusion (J/mol,cal/mol
and ev/atom)
R = gas constant, 8.31 J/mol-K or 8.62 eV/atom-K
T = absolute temperature (K)
)RT
dQexp(DD o
Chapter 5: Diffusion
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Factors in diffusion continue…
Chapter 5: Diffusion
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Summary
•Self Diffusion
•Inter-Diffusion
•Steady state J=M/At Fick’s First law
•Non-steady state Fick’s second law
•Temperature effect
•Activation energy
Chapter 5: Diffusion