# Document ```ME 210
Exam 2
Review Session
Assistant Professor
Re-cap of topics for Test #2
Chap 6 Fick’s Laws
 What are Fick’s Laws of Diffusion? Can you define the terms and units? (Know how to apply to
solve mathematical problems)
 How can the rate of diffusion be predicted for some simple cases?
 How does diffusion depend on structure and temperature?
Chap 6 Diffusion in Solids
 How does diffusion occur?
 How is activation energy calculated? (Know how to apply to solve mathematical problems)
 How is diffusion used in processing of materials?
Chap 10 Phase Diagrams
 What is a phase?
 What is thermodynamic equilibrium?
 What three components need to be used (established) to define the equilibrium?
 How to determine the composition and fraction of a phase in a two-phase regime (Lever Rule).
Chap 10 Microstructural development
 How can we use a phase diagram to predict microstructure?
 What is the consequence of solidification in an alloy (coring)?
Fick’s 1st Law for steady-state diffusion
dc
J  D
dx
►It tells you the flow rate (i.e., “flux”) of a
diffusing species due to a concentration
►J is the flux
►D is the diffusivity (e.g., m2/s, cm2/s, etc…)
►(dc/dx) is the concentration gradient – i.e.,
change in amount/distance (e.g., g/m, %/mm,
etc…)
Flux
► Flux is essentially the amount or rate of diffusion
moles (or mass) diffusing
mol
kg
J  Flux 

or
2
surface area time 
cm s m2s
M
l dM
J

At A dt
Diffusion Coefficient
• Refers to the amount of material flowing through an area over
a period of time.
 Qd 
D  Do exp  

 RT 
D = diffusion coefficient [m2/s]
Do = pre-exponential [m2/s]
Qd = activation energy [J/mol or eV/atom]
R = gas constant [8.314 J/mol-K]
T = absolute temperature [K]
degrees K  C  273
• This is an Arrhenius equation (very common to materials
science; used to describe the statically probability of an event)
A relevant problem:
A differential nitrogen pressure exists across a 2-mm-thick steel bulkhead. After
some time, steady-state diffusion of the nitrogen is established across the wall.
If the nitrogen concentration on the high-pressure surface of the wall is 2 kg/m3
and on the low-pressure surface is 0.2 kg/m3, what is the flow of nitrogen
through the wall (in kg/m2h)? The diffusion coefficient for nitrogen in this
steel is 1.0  10-10 m2/s at its operating temperature.
This is a flux problem… Draw the system… Then solve…
c
High
J x  D
2 kg/m3
x
c chigh  clow  2.0  0.2  kg/m
4




900
kg
/
m
x xhigh  xlow  0  2.0  103  m
Low
0.2 kg/m3
3
Make sure units are consistent
D  1.0  10
10
2
1.0  1010 m 2 3.6  103 s
7 m
m /s
 3.6  10
1
s
1
h
h
2
Substitute D into flux equation
2
c
kg 
kg

7 m
J x   D    900 4   3.6  10
 3.24  104 2
x
h
m 
m h

2 mm
Diffusion and Temperature
300
600
1000
10-8
1500
D has exponential dependence on T
 Q 
D  Do exp   d 
 RT 
T(C)
Dinterstitial &gt;&gt; Dsubstitutional
D (m2/s)
C in a-Fe
C in g-Fe
10-14
Al in Al
Fe in a-Fe
Fe in g-Fe
Adapted from Fig. 5.7, Callister &amp; Rethwisch 8e. (Date for Fig. 5.7
taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals
Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)
10-20
0.5
1.0
1.5
1000 K/T
Interstitial diffusion is much faster. WHY?
Interstitial atoms are smaller and more mobile.
Also there are more empty interstitial positions than vacancies.
This increases the probability of interstitial diffusion.
Diffusion and Crystal Structure
300
600
1000
10-8
1500
D has exponential dependence on T
T(C)
 Q 
D  Do exp   d 
 RT 
DBCC &gt;&gt;
D (m2/s)
C in a-Fe
DFCC
C in g-Fe
10-14
Adapted from Fig. 5.7, Callister &amp; Rethwisch 8e. (Date for Fig. 5.7
taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals
Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)
10-20
0.5
1.0
1.5
1000 K/T
Interstitials diffuse is much faster in BCC than FCC. WHY?
BCC structures are less dense (i.e., less close packed).
There is more room for interstitials to move.
Diffusion Data
Constant (aka ‘fixed’)
Energy to cause diffusion
Diffusivity
Depends on T
(not fixed)
Example: At 300&ordm;C the diffusion coefficient and
activation energy for Cu in Si are:
 Q 
D  Do exp   d 
 RT 
D(300&ordm;C) = 7.8 x 10-11 m2/s
Qd = 41.5 kJ/mol
What is the diffusion coefficient for Cu in to Si at 350&ordm;C?
collect
data
transform
data
D
ln D
Temp = T
1/T
Transform data for each temperature:
Qd
lnD2 = lnD0 R
&aelig;1&ouml;
&ccedil;&ccedil; &divide;&divide;
&egrave; T2 &oslash;
and
Plot and determine slope of line:
D2
Qd
\ lnD2 - lnD1 = ln
=D1
R
10
Qd
lnD1 = lnD0 R
&aelig; 1 1&ouml;
&ccedil;&ccedil; - &divide;&divide;
&egrave; T2 T1 &oslash;
&aelig;1&ouml;
&ccedil;&ccedil; &divide;&divide;
&egrave; T1 &oslash;
5
BASIC MATH
Y = MX + B
M = slope = Y/X
2
Y = ln D
Y
X
1
X = 1/T
Plot and determine slope of line:
Qd
ln D2  ln D1
Y
 slope of line 

X
R
1 1
  
 T2 T1 
11
Example (cont.)
Solve for unknown value:
 Qd
D2  D1 exp 
 R
 1 1 
  
 T2 T1 
T1 = 273 + 300 = 573 K
Remember to convert from &deg;C to K
T2 = 273 + 350 = 623 K
D2  (7.8 x 10
11
  41,500 J/mol  1
1 
m /s) exp 



 8.314 J/mol - K  623 K 573 K 
2
D2 = 15.7 x 10-11 m2/s
3
12
Phase Diagrams:
Determination of phase compositions
• If we know T and C0, then we can determine:
-- the compositions of each phase. (Just read them)
Consider C0 = 35 wt% Ni
At TA = 1320&deg;C:
Only Liquid (L) present
CL = C0 ( = 35 wt% Ni)
At TD = 1190&deg;C:
Only Solid (a) present
Ca = C0 ( = 35 wt% Ni)
At TB = 1250&deg;C:
Both a and L present
CL = C liquidus ( = 32 wt% Ni)
Ca = C solidus
( = 43 wt% Ni)
13
C0 = C
Cu-Ni system
T(oC)
• Examples:
alloy
( = 35 wt% Ni)
A
TA
1300
tie line
L (liquid)
B
TB
1200
TD
20
D
3032 35
CL C0
wt% Ni
a
(solid)
40 4 3
50
Ca
Adapted from Fig. 10.3(a), Callister &amp; Rethwisch 4e. (Fig.
10.3(a) is adapted from Phase Diagrams of Binary Nickel
Alloys, P. Nash (Ed.), ASM International, Materials Park, OH
(1991).
Phase Diagrams:
Determination of phase weight fractions
(We use the Lever Rule)
• If we know T and C0, then can determine:
-- the amount (i.e., weight fraction) of each phase. (Calculate them)
• Examples:
Consider C0 = 35 wt% Ni
S
R +S
Wa 
R
R +S
14
43 - 35
=
= 0.73
43 - 32
= 0.27
A
TA
At TA : Only Liquid (L) present
WL = 1.00, Wa = 0
At TD : Only Solid ( a) present
WL = 0, W a = 1.00
At TB : Both a and L present
WL 
Cu-Ni system
T(oC)
1300
tie line
L (liquid)
B
TB
S
R
1200
TD
20
a
(solid)
D
3032 35
CL C0
wt% Ni
40 4 3
Adapted from Fig. 10.3(a), Callister &amp; Rethwisch 4e. (Fig.
10.3(a) is adapted from Phase Diagrams of Binary Nickel
Alloys, P. Nash (Ed.), ASM International, Materials Park,
OH (1991).
Ca
50
The Lever Rule
(gives us the ‘amounts’ of phases that are present)
► Tie line – connects the phases in equilibrium with
each other – also sometimes called an isotherm
T(oC)
What fraction of each phase?
Think of the tie line as a lever
(i.e., teeter-totter)
tie line
1300
L (liquid)
B
TB
a
(solid)
1200
R
20
30CL
Ma
ML
S
C0 40 Ca
wt% Ni
R
50
Callister &amp; Rethwisch 4e.
Ca - C0
ML
S
WL =
=
=
ML + Ma R + S Ca - CL
S
Ma  S  M L  R
C0 - CL
R
Wa =
=
R + S Ca - CL
WL  Wa  1 (i.e., 100%)
15
Binary Phase Solidification
Solubility limits can change as function of temperature. This affects microstructure.
Alloys above solubility
limit but below max
solubility (far from
the eutectic). The
solidification path is:
L: C0 wt% Sn
T(oC)
400
L
300
L+a
LL+aa
aab
100
a: C0 wt% Sn
a
200
TE
L
a
a
b
a+ b
0
10
2
(sol. limit at T room )
20
C0
Pb-Sn
system
30
C, wt% Sn
18.3
(Max sol. limit at TE)
16
Adapted from Fig. 10.12, Callister &amp; Rethwisch 4e.
Eutectic Solidification
T(oC)
Pure
Elements
Eutectic
Pure Elements
L: C0 wt% Sn
300
L
Pb-Sn
system
a
200
L+a
Lb b
183&deg;C
TE
100
Callister &amp; Rethwisch 4e.
ab
0
Pb
20
18.3
40
b: 97.8 wt% Sn
a: 18.3 wt%Sn
60
CE
61.9
80
100
97.8
Sn
C, wt% Sn
L  a + b ; a saturated with Sn and b saturated with Pb must
form at the same time. Requires diffusion.
Mix of A &amp; B17has lower melting point than pure A or pure B.
Microstructural Developments
• For alloys for which 18.3 wt% Sn &lt; C0 &lt; 61.9 wt% Sn
• Result: a phase particles and a eutectic microconstituent
T(oC)
L: C0 wt% Sn
L
300
Pb-Sn
system
a
200
R
TE
α
L
L+ a
L+β b
S
S
R
0
Pb
20
18.3
primary a
eutectic a
eutectic b
40
60
61.9
C, wt% Sn
10.16, Callister &amp; Rethwisch 4e.
• Just above TE :
Ca = 18.3 wt% Sn
CL = 61.9 wt% Sn
S
= 0.50
=
Wa
R+S
WL = (1- Wa) = 0.50
• Just below TE :
a+b
100
a
L
80
100
97.8
Ca = 18.3 wt% Sn
Cb = 97.8 wt% Sn
Wa = S = 0.73
R+S
Wb = 0.27
Hypoeutectic &amp; Hypereutectic
T(oC)
300
Callister &amp; Rethwisch 4e.
Binary Phase Diagrams,
2nd ed., Vol. 3, T.B.
Massalski (Editor-in-Chief),
ASM International,
Materials Park, OH, 1990.)
a
200
L+ a
a+b
100
20
40
hypoeutectic: C0 = 50 wt% Sn
a
a
L+b b
TE
0
(Figs. 10.14 and
10.17 from Metals
Handbook, 9th ed.,
Vol. 9,
Metallography and
Microstructures,
American Society for
Metals, Materials
Park, OH, 1985.)
(Pb-Sn System)
L
a
60
80
eutectic
61.9
hypereutectic: (illustration only)
b
b
a
Fig. 10.17, Callister &amp;
Rethwisch 4e.
19
C, wt% Sn
eutectic: C0 = 61.9 wt% Sn
a a
175 mm
100
b
b b
b
160 mm
eutectic micro-constituent
Callister &amp; Rethwisch 4e.
Callister &amp; Rethwisch 4e.
(Illustration only)
Understand how Microstructures Evolve
Hypo-eutectoid alloys form
primary a on prior grain
boundaries. These alloys
have low to medium strength,
and good ductility.
Hyper-eutectoid alloys form primary Fe3C on
prior grain boundaries. These alloys are brittle,
but strong. They also have excellent wear
resistance.
Summary
► Study your notes, assessments, and in book example
problems.
► Exam #2 will have 7 questions.
► There will be NO true/false or multiple choice questions.
► All calculations AND short answer.
► Know how to do the types of problems that we’ve
covered and you can do quite well.
► Good luck!
```