[email protected] ENGR-45_Lec-07_Diffusion_Fick-2.ppt 1 Bruce Mayer, PE Engineering-45:...
-
Upload
ira-davidson -
Category
Documents
-
view
218 -
download
2
Transcript of [email protected] ENGR-45_Lec-07_Diffusion_Fick-2.ppt 1 Bruce Mayer, PE Engineering-45:...
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt1
Bruce Mayer, PE Engineering-45: Materials of Engineering
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 45
Solid Solid StateState
Diffusion-Diffusion-22
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt2
Bruce Mayer, PE Engineering-45: Materials of Engineering
Learning Goals - DiffusionLearning Goals - Diffusion
How Diffusion Proceeds How Diffusion Can be Used in Material
Processing How to Predict The Rate Of Diffusion
Be Predicted For Some Simple Cases• Fick’s first and SECOND Laws
How Diffusion Depends On Structure And Temperature
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt3
Bruce Mayer, PE Engineering-45: Materials of Engineering
Recall Fick’s FIRST Law.Recall Fick’s FIRST Law. Fick’s 1st Law
Position, x
Cu flux Ni flux
x
C
dx
dCDJ
• Where– J Flux in kg/m2•s
or at/m2•s– dC/dx = Concentration
GRADIENT in units of kg/m4 or at/m4
– D Proportionality Constant (Diffusion Coefficient) in m2/s
In the SteadyState Case J = const• So dC/dx = const
– For all x & t
Thus for ANY two points j & k
jk
jk
xx
CC
x
C
dx
dC
Con
cen
., C
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt4
Bruce Mayer, PE Engineering-45: Materials of Engineering
NONSteady-State DiffusionNONSteady-State Diffusion In The Steady Case
tfC
tfJ
In The NONSteady, or Transient, Case the Physical Conditions Require
txfdxdCCJ ,,,
Liquid Source Vapor Bubble Saturation Transient (Calculated)
0.0
0.2
0.4
0.6
0.8
1.0
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
Radial Position Inside Bubble, r (mm)
Vap
or
Satu
rati
on
Fra
cti
on
, v
Pv(r,t) (t=0.01)
Pv(r,t) (t=0.04 s)
Pv(r,t) (t=0.10 s)
Pv(r,t) (t=0.15 s)
Pv(r,t) (t=0.2 s)
Pv(r,t) (t=0.3 s)
Pv(r,t) (t=0.35 s)
file = BubPv(t)2.xls
• Bubble Diameter = 4 mm
• Dv = 5 mm2/s
Increasing Time
222 /
01
sin121, ov rtDno
n
n
v errn
rrntr
• In The Above Concen-vs-Position Plot Note how, at x 1.5 mm, Both C and dC/dx CHANGEwith Time
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt5
Bruce Mayer, PE Engineering-45: Materials of Engineering
NONSteady State Diffusion MathNONSteady State Diffusion Math Consider the
Situation at Right Box Dimensions
• Width = x
• Height = 1 m
• Depth = 1 m– Into the slide
Box Volume, V = x•1•1 = x
Now if x is small
xCCC
C rightleft
2
avg
• Can Approximate C(x) as
Concentration,C, in the Box
J(right)J(left)
x
The Amount of Matl in the box, M
33
mm
kgVC M
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt6
Bruce Mayer, PE Engineering-45: Materials of Engineering
NONSteady State Diffusion contNONSteady State Diffusion cont or
Material ENTERING the Box in time t
rightleft JJ
For NONsteady Conditions
Concentration,C, in the Box
J(right)J(left)
dx
So Matl ACCUMULATES in the Box
xCxC M 11
tJJ
tAreaJM
leftleft
leftin
11
Material LEAVING the Box in time t
tJJ
tAreaJM
rightright
rightout
11
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt7
Bruce Mayer, PE Engineering-45: Materials of Engineering
NONSteady State Diffusion cont.2NONSteady State Diffusion cont.2 So the NET Matl
Accumulation
Adding (or Subtracting) Matl From the Box CHANGES C(x)
xxx
rightleft
x
CD
x
CD
t
Cx
ortJJCx11
With V = 1•1•x
Concentration,C, in the Box
J(right)J(left)
x
tJJM rightleft
CVMV
MC
V
M
V
MCCC oldnew
Partials Req’d asC = C(x,t)
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt8
Bruce Mayer, PE Engineering-45: Materials of Engineering
NONSteady State Diffusion cont.3NONSteady State Diffusion cont.3 In Summary for
CONSTANT D
Now, And this is CRITICAL, by TAYLOR’S SERIES
xxx x
C
x
CD
t
Cx Concentration,
C, in the Box
J(right)J(left)
x
x
x
xCxCxC
xxxx
/
so
xxx
x
x
C
x
Cx
x
xCD
t
Cx
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt9
Bruce Mayer, PE Engineering-45: Materials of Engineering
NONSteady State Diffusion cont.4NONSteady State Diffusion cont.4 After Canceling
Now for very short t
Concentration,C, in the Box
J(right)J(left)
x
Finally Fick’s SECOND LAW for Constant Diffusion Coefficient Conditions
xx
x
x
CD
x
xCD
t
C2
2
2
2
2
2
0lim
x
CD
x
CD
t
C
t
C
xt
2
2
x
CD
t
C
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt10
Bruce Mayer, PE Engineering-45: Materials of Engineering
Comments of Fick’s 2nd LawComments of Fick’s 2nd Law The Formal
Statement
This Leads to the GENERAL, and much more Complicated, Version of the 2nd Law
Concentration,C, in the Box
J(right)J(left)
x
2
2
x
CD
t
C
This Assumes That D is Constant, i.e.;
txCDD , In many Cases
Changes in C also Change D
x
CD
xt
C
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt11
Bruce Mayer, PE Engineering-45: Materials of Engineering
Example – NonSS DiffusionExample – NonSS Diffusion Example: Cu Diffusing into a Long Al Bar
The Copper Concentration vs x & t
pre-existing conc., Co of copper atoms
Surface conc., Cs of Cu atoms bar
Co
Cs
position, x
C(x,t)
tot1
t2t3
The General Soln is Gauss’s Error Function, “erf”
Dt
xerf
CC
CtxC
oS
o
21
,
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt12
Bruce Mayer, PE Engineering-45: Materials of Engineering
Comments on the erfComments on the erf Gauss's Defining
Eqn
dyezerfz y
0
22
z is just a NUMBER
• Thus the erf is a (hard to evaluate) DEFINITE Integral
Treat the erf as any other special Fcn
Some Special Fcns with Which you are Familiar: sin, cos, ln, tanh• These Fcns used to
be listed in printed Tables, but are now built into Calculators and MATLAB
See Text Tab 5.1 for Table of erf(z)
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt13
Bruce Mayer, PE Engineering-45: Materials of Engineering
Comments on the erf cont.Comments on the erf cont. 1-erf(z) appears So
Often in Physics That it is Given its Own Name, The COMPLEMENTARY Error Function:
dyezerfcz y
0
221
Recall The erfc
Diffusion solution
Notice the Denom in this Eqn
Dt
xerfc
CC
CtxC
oS
o
4
,
dLDt 4 This Qty has SI
Units of meters, and is called the “Diffusion Length”• The Natural Scaling
Factor in the efrc
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt14
Bruce Mayer, PE Engineering-45: Materials of Engineering
SemiInfite Slab Diffusion vs Time
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
x (arbitrary units)
Rel
ativ
e C
on
cen
tati
on
ErfC t1; Ld = 0.2236
ErfC t2; Ld = 0.3873
ErfC t3; Ld = 0.6708
ErfC t4; Ld = 1.1619
ErfC t5; Ld = 2.0125
file = erfc_0401.xls
TIME
PARAMETERS• D = 0.05 (arb Units)
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt15
Bruce Mayer, PE Engineering-45: Materials of Engineering
Example Example D = f(T) D = f(T) Given Cu Diffusing
into an Al Bar At given point in the
bar, x0, The Copper Concentration reaches the Desired value after 10hrs at 600 °C• The Processing
Recipe
Get a New Firing Furnace that is Only rated to 1000 °F = 538 °C• To Be Safe, Set the
New Fnce to 500 °C
Need to Find the NEW Processing TIME for 500 °C to yield the desired C(x0) ChrsxC 600@100
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt16
Bruce Mayer, PE Engineering-45: Materials of Engineering
Example Example D = f(T) cont D = f(T) cont Recall the erf
Diffusion Eqn
For this Eqn to be True, need Equal Denoms in the erf
Dt
xerf
CC
CtxC
oS
o
41
,
Since CS and Co have NOT changed, Need
oS
oC
oS
oC
CC
CtxC
CC
ChrsxC
50006000 ,10,
Since by the erf
tD
xerf
hrsD
xerf
500
0
600
0
41
1041
500
600
500600
10
10
D
hrsDt
or
tDhrsD
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt17
Bruce Mayer, PE Engineering-45: Materials of Engineering
Example Example D = f(T) cont.2 D = f(T) cont.2 Now Need to
Find D(T) As With Xtal Pt-
Defects, D Follows an Arrhenius Rln
– Qd Arrhenius Activation Energy in J/mol or eV/at
– R Gas Constant = 8.31 J/mol-K = 8.62x10-5 eV/at-K
– T Temperature in K
Find D0 and Qd from Tab 5.2 in Text• For Cu in Al
– D0 = 6.5x10-5 m2/s
– Qd = 136 kJ/mol
• Where– D0 Temperature
INdependent Exponential PreFactor in m2/s
RT
QDD dexp0
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt18
Bruce Mayer, PE Engineering-45: Materials of Engineering
Example Example D = f(T) cont.3 D = f(T) cont.3 Thus D(T) for
Cu in Al Thus for the new
500 °C Recipe
In this Case• D600 = 4.692x10-13 m2/s
• D500 = 4.152x10-14 m2/s
Now Recall the Problem Solution
TKmolJ
molJD
/31.8
/136000exp105.6 5
500
600 10
D
hrsDt
hrshrs
t 0.113152.4
1092.46
This is 10x LONGER than Before; Should have bought a 600C fnce
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt19
Bruce Mayer, PE Engineering-45: Materials of Engineering
Find D Arrhenius ParametersFind D Arrhenius Parameters Recall The D(T) Rln
Applied to the D(T) Relation
Take the Natural Log of this Eqn
TR
QDD d 1
lnln 0
RT
QDD dexp0
This takes the form of the slope-intercept Line Eqn:
mxby
0ln
1
lny
Db
Tx
R-Qm
D
d
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt20
Bruce Mayer, PE Engineering-45: Materials of Engineering
Find D(T) Parameters contFind D(T) Parameters cont And, Since TWO
Points Define a Line If We Know D(T1) and D(T2) We can calc• D0
• Qd
Quick Example• D(T) For Cu in Au at
Upper Right
Slope, m = y/x
x
y
x = (1.1-0.8)x1000/K
• = 0.0003 K-1
y = ln(3.55x10-16) − ln(4x10-13) = −7.023
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt21
Bruce Mayer, PE Engineering-45: Materials of Engineering
Find D(T) Parameters cont.2Find D(T) Parameters cont.2 By The Linear Form
in the (x,y) format• x1 = 0.0008
• y1 = ln(4x10-13) = −28.55
So b
J/mol194500
0003.0023.731.8
d
d
d
d
Q
Q
xyRmRQ
orRQm
Now, the intercept, b
11 mxyb
orbmxy
Pick (D,1/T) pt as• (4x10-13,0.8)
819.9
0008.00003.0
023.755.28
b
b
Finally D0
smD
eeD b
250
819.90
1044.5
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt22
Bruce Mayer, PE Engineering-45: Materials of Engineering
Diff vs. Structure & PropertiesDiff vs. Structure & Properties Faster Diffusion for
• Open crystal structures
• Lower melting Temp materials
• Materials with secondary bonding
• Smaller diffusing atoms
• Cations• Lower density
materials
Slower Diffusion for• Close-packed
structures• Higher melting Temp
materials• Materials with
covalent bonding• Larger diffusing
atoms• Anions• Higher density
materials
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt23
Bruce Mayer, PE Engineering-45: Materials of Engineering
Diffusion SummarizedDiffusion Summarized Phenomenon: Mass Transport In Solids Mechanisms
• Vacancy InterChange by KickOut
• Interstitial “squeezing”
Governing Equations• Fick's First Law
• Fick's Second Law
Diffusion coefficient, D• Affect of Temperature
• Qd & D0
– How to Determine them from D(T) Data
[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt24
Bruce Mayer, PE Engineering-45: Materials of Engineering
WhiteBoard WorkWhiteBoard Work Problem 5.28
• Ni Transient Diffusion into CuProb 5.28: Ni Diffusion into Cu
0%
1%
2%
3%
4%
5%
0.0 0.5 1.0 1.5 2.0 2.5 3.0
x into Cu (mm)
CN
i (W
t% N
i in
Cu
)
file = erfc_0401.xls
500hrs @ 1000 ° 500hrs @ T2
Same Concentration