introduction to engineering materials engr2000 chapter 5...
TRANSCRIPT
Introduction to Engineering Materials
ENGR2000
Chapter 5: Diffusion
Dr. Coates
5.1 Introduction
• Many reactions & processes that are important in
the treatment of materials rely on the transfer of
mass either within a specific solid or from a liquid,
gas or another solid phase.
• Diffusion
– Material transport by atomic motion
Types of Diffusion
• Inter-diffusion or Impurity diffusion
– Atoms of one metal diffuse into another
– Net drift of atoms from high to low concentration
regions.
• Self-diffusion
– All atoms exchanging positions are of the same type:
occurs for pure metals.
You can observe compositional changes in order to
detect inter-diffusion, can you do the same to detect
self diffusion?
5.2 Diffusion Mechanisms
• Diffusion is the stepwise migration of atoms
from lattice site to lattice site.
• You are at a music concert, it’s very crowded
and you are exhausted. You are surrounded by
sweaty people and you can’t see the performance
very well. What physical conditions that would
exist ideally, so you would be motivated to move
from your current spot?
Necessary Conditions
empty adjacent sites
sufficient energy to break existing bonds
• The two conditions needed for diffusion to occur
are
Vacancy Diffusion
• Interchange of an atom from a normal lattice site
to an adjacent vacant lattice site.
Vacancy Diffusion
• How is the vacancy motion related to the diffusing
atom motion?
• Self-diffusion & inter-diffusion occur by this
mechanism.
Interstitial Diffusion
• Atoms migrate from an interstitial position to a
neighboring one that is empty.
• Atoms must be small enough to fit into the
interstitial positions.
• Would this apply to inter-diffusion or self diffusion?
Diffusion mechanisms in
metals & metal alloys
• Which would occur more rapidly, interstitial
diffusion or vacancy diffusion? Why?
Interstitial atoms are smaller & more mobile.
There are more empty interstitial positions than vacancies.
Rate of Mass transfer defined
• What are the units of J?
dt
dM
AJ
form aldifferenti In
At
MJ
(t) time unit per solidof A, area, sectionalcross
unit a to and through diffusing atoms) of number (or M,mass,
the as defined (J), flux Diffusion
1
Kg/m2s or #atoms/m2s
Fick’s First Law
dx
dCDJ
:lawfirst sFick'
t.coefficiendiffusion theis D where
:emperatureconstant tat
direction,x in the state)(steady atoms offlux diffusion Net
KmolJconstGasR
KetemperaturabsoluteT
atomeVormolJdiffusionforenergyactivationQ
RT
QDD
/31.8.
)(
//
exp
:dependence etemperaturt coefficienDiffusion
0
Steady State Diffusion
• Concentrations not changing with time
• What does this imply regarding
– Mass leaving/entering?
Does a constant diffusion flux
value (with time) imply steady
state?
yes
• What then are units of
D?
For diffusion problems it’s sometimes best to express
concentration, C, in terms of mass of diffusing
species per unit volume of solid
m2/s
dx
dCDJ
Fick’s First Law
• Fick’s first law
– restricted to problems for which the concentration
gradient does not change with time (steady state).
– the –ve sign indicates that mass flows from regions of
high concentration to regions of low concentration.
dx
dCDJ
Practical Examples of Diffusion
• The controlled diffusion of P-type or other
dopants into Si wafers.
• The controlled diffusion of oxygen and carbon
dioxide through a membrane in a heart-lung
machine permits surgeons to operate on the heart.
• Carburization of steels.
Example 5.1
• A thin plate of Fe is exposed to a carburizing
atmosphere on one side and a decarburizing
atmosphere on the other side at 700 0C. If a
condition of steady state is achieved, calculate the
diffusion flux of C through the plate if the
concentrations of C at positions of 5 mm & 10 mm
beneath the carburizing surface are 1.2 kg/m3 and
0.8 kg/m3 respectively. Assume a diffusion
coefficient of 3 x 10-11 m2/s at this temperature.
?
/103700
/8.0
/2.1
10
5
700
:Given
2110
3
3
J
smCD
mkgC
mkgC
mmx
mmx
CT
B
A
B
A
smkg
xx
CCD
dx
dCDJ
BA
BA
29 /104.2
:lawfirst sFick' Using
5.4 Non Steady-State Diffusion
• Fick’s first law
– restricted to problems for which the concentration
gradient does not change with time.
• Diffusion through a solid bar…
Fick’s Second Law
x
txJ
xx
txJtxJx,tC
xxdAdxdAdV
dAtxJdAtxJdVx,tC
dAtxJdAtxJdVtt
txCtxC
inout
outoutinin
inout
outoutinin
outoutinin
ii
ii
,,,
t
: Using
,,t
,,,,
out rate flow -in rate flow on accumulati of rate
:dV volumeelementalan in solute theof balance' mass' Using
1
1
Fick’s Second Law
.(geometry) conditionsboundary and
conditions initial theon depends solution - ,
t
:law second sFick'
,,:)dependence time(withlaw first s Fick'Using
2
2
x
txCD
x,tC
x
txCDtxJ
Fick’s Second Law solution
(semi-infinite solid)
law. second sFick' osolution t -
21
,
:source dreplenishely continuous awith
and plate, thick a intodiffusion 0,
x0 ion,concentrat initialconstant 0,
surface, at theion concentratconstant 0,0
:constant held is surface at the conc.
hein which t solid infinite semi a intodiffusion For
0
0
0
0
Dt
xerf
CC
CtxC
CtxC
CtxC
CtxC
s
s
z
dyyzerf0
2exp2
What are dimensions of erf (z)?
Questions?
• What if you want to attain a fixed composition ?
–𝑥2
𝐷𝑡= constant, Prove this!
• What about a fixed composition at a particular
distance?
– 𝐷𝑡 = constant, Prove this!
Example 5.2
• For some applications, it is necessary to harden
the surface of a steel (Fe & C alloy) above that of
its interior. One way this may be accomplished is
by increasing the surface concentration of C in a
process termed carburizing; the steel is exposed at
an elevated temperature, to an atmosphere rich in a
hydrocarbon gas such as methane.
Cs
C0
• Consider one such alloy that initially has a
uniform C concentration of 0.25 wt. % and is to be
treated at 950 0C. If the concentration of the C at
the surface is suddenly brought to and maintained
at 1.20 wt. %, how long will it take to achieve a C
content of 0.80 wt. % at a position 0.5 mm below
the surface? The diffusion coefficient for C in Fe
at this temperature is 1.6 x 10-11 m2/s. Assume that
the steel plate is semi-infinite.
Example 5.2
Diffusion of C into a large slab of Fe…
Example 5.2
?
/106.1
950
5.0
%80.0,
%2.1
%25.0
:Given
211
0
0
t
smTD
CT
mmx
txC
C
C
s
hst
Dt
x
Dt
xerf
Dt
xerf
Dt
xerf
CC
CtxC
s
1.7400,25
392.02
:5.1 Table Using
4215.02
21
25.020.1
25.080.0
21
,
:law second sFick' oSolution t
0
0
Example 5.3
• The diffusion coefficients for Cu in Al at 500 0C
and 600 0C are 4.8 x 10-14 m2/s and 5.3 x 10-13
m2/s, respectively. Determine the approximate
time at 500 0C 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 0C.
How do you know whether this is a
steady state scenario or not?
Example 5.3
?
10
,
/103.5600
/108.4500
:Given
500
600
0
2130
2140
t
ht
samex
sametxC
sameC
sameC
smCD
smCD
s
ht
tDtD
constDt
constDt
x
constDt
xerf
Dt
xerf
CC
CtxC
:law secondsFick' to Solution
s
4.110
2
2
21
,
500
600600500500
0
0
5.5 Factors that Influence Diffusion
- Temperature
constant. gas universal theis
andKelvin in re temperatu theis
constant, a is
diffusion,for energy activation theis where
exp
:t coefficienDiffusion
0
0
R
T
D
Q
RT
QDD d
5.5 Factors that Influence Diffusion
- Diffusing Species
– Diffusing species/host
• Crystal structure (APF)
• Atomic radius
– Temperature
Compare the self-diffusion of Fe and the
inter-diffusion of C in Fe at 500 0C
Example 5.4
- Using the data in Table 5.2, compute the
diffusion coefficient for Mg in Al at 550 0C.
Example 5.4
sm
KKmolJ
molJsmCD
RT
QDD
/108.5
273550./31.8
/000,131exp/102.1550
exp
:t coefficienDiffusion
213
240
0
Design Example 5.1
• The wear resistance of a steel gear is to be improved byhardening its surface. This is to be accomplished byincreasing the carbon content within an outer surface layeras a result of C diffusion into steel; the C is to be suppliedfrom an external C-rich gaseous atmosphere at an elevatedand constant temperature. The initial C content of the steelis 0.20 wt%, whereas the surface concentration is to bemaintained at 1.00 wt %. For this treatment to be effective,a C content of 0.60 wt% must be established at a position0.75 mm below the surface. Specify an appropriate heattreatment in terms of temperature and time fortemperatures between 900 C and 1050 C. Use data inTable 5.2 for the diffusion of C in -iron.
?
?
1050900
75.0
%60.0,
%00.1
%20.0
Fe-in C -
:Given
00
0
t
T
CTC
mmx
txC
C
C
s
27
0
0
1024.6
4747.02
:5.1 Table Using
5.02
21
,
:law second sFick' oSolution t
mDt
Dt
x
Dt
xerf
Dt
xerf
CC
CtxC
s
range. given the in etemperatur
specified somefor calculated bemay time required the -
s
T
t
mRT
QtDDt
RT
QDD
:tcoefficien Diffusion
molJQ
smD
:5.2 Table Using
d
d
d
810,17exp
0271.0
104.6exp
exp
/000,148
/103.2
270
0
250
What R should we use
(units)?
Diffusion in Semi-Conductors
• Semiconductor Integrated Chips (IC)
– Base material is single crystal silicon
– Two heat treatments
• Predisposition Step: impurity atoms (gas phase) are diffused
under pressure where surface composition remains constant
• Drive in diffusion: used to transport impurity atoms farther
without increasing impurity content
– Higher temperature
– Oxide layer on surface (few impurity atoms diffuse out)
Diffusion in Semi-Conductors
law second sFick' osolution t -
4exp,
:source) dreplenishely continuous a(not consumed issupply theif
and plate, thick a intodiffusion ,
ion,concentrat initialconstant 0,
surface, at theion concentrat ,0
:plate thick a intodiffusion For
20
0
0
Dt
x
Dt
QtxC
CtxC
CtxC
CtxC s
pp
s
tDC
Q
2Q where
tors.semiconduc of doping -
.)(mass/area species diffusing theofsupply theis where
0
0
Surface Concentration
Diffusion coefficient for the predisposition step
predisposition treatment time
5.7 Other Diffusion Paths
• Atomic migration can also occur along
– dislocations
– Grain boundaries
– External surfaces
• Short-circuit diffusion paths
– Faster than bulk (normal) diffusion
– their contribution is insignificant because the cross-
sectional areas of these paths are small.
Example
• The diffusion of H in FCC Fe
– r(H) << r(Fe)
– interstitial mechanism
• The self-diffusion in FCC Fe
– vacancy mechanism
– higher activation energy
Example
• Explain each of these observations
(b) The activation energy for the diffusion of H in BCC
Fe is less than that for diffusion in FCC Fe.
Example
• The diffusion of H in BCC Fe
– interstitial mechanism
– BCC structure is more open (lower APF)
– low activation energy
• The diffusion of H in FCC Fe
– interstitial mechanism
– FCC structure is a close-packed structure (high APF)
– high activation energy