ME 210 Exam 2 Review Session Dr. Aaron L. Adams, Assistant Professor.
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Transcript of ME 210 Exam 2 Review Session Dr. Aaron L. Adams, Assistant Professor.
ME 210Exam 2
Review SessionDr. Aaron L. Adams,
Assistant Professor
Chap 6 Fick’s LawsWhat 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 SolidsHow 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 DiagramsWhat 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)?
Re-cap of topics for Test #2
Fick’s 1st Law for steady-state diffusion
►It tells you the flow rate (i.e., “flux”) of a diffusing species due to a concentration gradient.
►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…)
dcJ D
dx
Flux
► Flux is essentially the amount or rate of diffusion
sm
kgor
scm
mol
timearea surface
diffusing mass) (or molesFlux
22J
dt
dM
A
l
At
MJ
• Refers to the amount of material flowing through an area over a period of time.
• This is an Arrhenius equation (very common to materials science; used to describe the statically probability of an event)
Diffusion Coefficient
degrees K C 273
= pre-exponential [m2/s]
= diffusion coefficient [m2/s]
= activation energy [J/mol or eV/atom]
= gas constant [8.314 J/mol-K]
= absolute temperature [K]
D
Do
Qd
R
T
exp do TR
DQ
D
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/m2h)? The diffusion coefficient for nitrogen in this steel is 1.0 10-10 m2/s at its operating temperature.
xJ Dc
x
34
3
2.0 0.2 kg/m900 kg / m
0 2.0 10 mhigh low
high low
c
x x
c
x
c
10 2 3 210 2 71.0 10 m 3.6 10 m
1.0 10 m / 3.6 101 1 h h
ss
sD
Make sure units are consistent
Substitute D into flux equation2
7 44 2
kg m kg900 3.6 10 3.24 10
hm m hx
cJ D
x
Low0.2 kg/m3
High2 kg/m3
2 mm
This is a flux problem… Draw the system… Then solve…
A relevant problem:
exp do TR
DQ
D
Diffusion and Temperature
Adapted from Fig. 5.7, Callister & 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.)
D has exponential dependence on T
Dinterstitial >> Dsubstitutional
C in a-FeC in g-Fe
Al in AlFe in a-FeFe in g-Fe
1000 K/T
D (m2/s) C in a-Fe
C in g-Fe
Al in Al
Fe in a-Fe
Fe in g-Fe
0.5 1.0 1.510-20
10-14
10-8
T(C)15
00
1000
600
300
Interstitial atoms are smaller and more mobile.
Also there are more empty interstitial positions than vacancies.
This increases the probability of interstitial diffusion.
Interstitial diffusion is much faster. WHY?
Diffusion and Crystal Structure
Adapted from Fig. 5.7, Callister & 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.)
D has exponential dependence on T
DBCC >> DFCC
C in a-Fe C in g-Fe
1000 K/T
D (m2/s) C in a-Fe
C in g-Fe
Al in Al
Fe in a-Fe
Fe in g-Fe
0.5 1.0 1.510-20
10-14
10-8
T(C)15
00
1000
600
300
BCC structures are less dense (i.e., less close packed).
There is more room for interstitials to move.
Interstitials diffuse is much faster in BCC than FCC. WHY?
exp do TR
DQ
D
Diffusion DataConstant (aka ‘fixed’) Energy to cause diffusion
DiffusivityDepends on T(not fixed)
10
Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are:
D(300ºC) = 7.8 x 10-11 m2/sQd = 41.5 kJ/mol
What is the diffusion coefficient for Cu in to Si at 350ºC?
transform data
D
Temp = T
ln D
1/TTransform data for each temperature:
5
Plot and determine slope of line:
exp do TR
DQ
D
collect data
11
Plot and determine slope of line:
Y = ln D
X = 1/T
Y = MX + BM = slope = Y/X
2 1
2 1
ln lnslope of line
1 1dQD DY
X R
T T
2
1
Y
X
BASIC MATH
12
Example (cont.)
K 573
1
K 623
1
K-J/mol 314.8
J/mol 500,41exp /s)m 10 x 8.7( 211
2D
1212
11exp
TTR
QDD d
T1 = 273 + 300 = 573 K
T2 = 273 + 350 = 623 K
D2 = 15.7 x 10-11 m2/s
3
Solve for unknown value:
Remember to convert from °C to K
13
wt% Ni
20
1200
1300
T(oC)
L (liquid)
a(solid)L + a
liquidus
solidus
30 40 50
L + a
Cu-Ni system
Phase Diagrams:Determination of phase compositions
• If we know T and C0, then we can determine: -- the compositions of each phase. (Just read them)
• Examples:TA
A
35C0
32CL
At TA = 1320°C: Only Liquid (L) present CL = C0
( = 35 wt% Ni)
At TB = 1250°C: Both and L present CL = C liquidus ( = 32 wt% Ni) C = C solidus ( = 43 wt% Ni)
At TD = 1190°C: Only Solid (a) present C = C0
( = 35 wt% Ni)
Consider C0 = 35 wt% Ni
DTD
tie line
4C
3
Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).
BTB
C0 = C alloy ( = 35 wt% Ni)
14
• If we know T and C0, then can determine: -- the amount (i.e., weight fraction) of each phase. (Calculate them)• Examples:
At TA : Only Liquid (L) present WL = 1.00, Wa = 0
At TD : Only Solid ( ) present
WL = 0, Wa
= 1.00
Phase Diagrams:Determination of phase weight fractions(We use the Lever Rule)
wt% Ni
20
1200
1300
T(oC)
L (liquid)
a(solid)L + a
liquidus
solidus
30 40 50
L + a
Cu-Ni system
TAA
35C0
32CL
BTB
DTD
tie line
4Ca
3
R S
At TB : Both and L present
= 0.27
WL= S
R + S
Wa= R
R + S
Consider C0 = 35 wt% Ni
Adapted from Fig. 10.3(a), Callister & Rethwisch 4e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).
15
► Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm
The Lever Rule(gives us the ‘amounts’ of phases that are present)
What fraction of each phase? Think of the tie line as a lever (i.e., teeter-totter)
ML M
R S
LSM M R wt% Ni
20
1200
1300
T(oC)
L (liquid)
a(solid)L + a
liquidus
solidus
30 40 50
L + aB
T B
tie line
C0CL C
SR
Adapted from Fig. 10.3(b), Callister & Rethwisch 4e.
1 (i.e., 100%)LW W
Binary Phase Solidification
Alloys above solubility limit but below max solubility (far from the eutectic). The solidification path is:
L L + a a
a a + b
Adapted from Fig. 10.12, Callister & Rethwisch 4e.
Pb-Snsystem
L + a
200
T(oC)
C, wt% Sn10
18.3
200C0
300
100
L
a
30
a+ b
400
(Max sol. limit at TE)
TE
2(sol. limit at Troom)
La
L: C0 wt% Sn
ab
a: C0 wt% Sn
16
liquidussolidus
solvus
Solubility limits can change as function of temperature. This affects microstructure.
Adapted from Fig. 10.13, Callister & Rethwisch 4e.
Pb-Snsystem
L
a
200
T(oC)
C, wt% Sn
20 60 80 1000
300
100
L
a b
L+ a
183°C
40
TE
18.3
: 18.3 wt%Sn
97.8
: 97.8 wt% Sn
CE61.9
L: C0 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 & B has lower melting point than pure A or pure B.
Eutectic Solidification Pure ElementsEutecticPure
Elements
17
Pb Sn
18
• For alloys for which 18.3 wt% Sn < C0 < 61.9 wt% Sn• Result: a phase particles and a eutectic microconstituent
Microstructural Developments
18.3 61.9
SR
97.8
SR
primary aeutectic a
eutectic b
WL = (1-Wa) = 0.50
Ca = 18.3 wt% Sn
CL = 61.9 wt% SnS
R + SWa = = 0.50
• Just above TE :
• Just below TE :C
= 18.3 wt% Sn
C = 97.8 wt% Sn
SR + S
W = = 0.73
W = 0.27
Adapted from Fig. 10.16, Callister & Rethwisch 4e.
Pb-Snsystem
L+β200
T(oC)
C, wt% Sn
20 60 80 1000
300
100
L
a b
L+
40
+
TE
L: C0 wt% Sn LL
α
Pb
19
L+L+b
+
200
C, wt% Sn20 60 80 1000
300
100
L
TE
40
(Pb-Sn System)
Hypoeutectic & Hypereutectic
Adapted from Fig. 10.8, Callister & Rethwisch 4e. (Fig. 10.8 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), ASM International, Materials Park, OH, 1990.)
160 mm
eutectic micro-constituentAdapted from Fig. 10.14, Callister & Rethwisch 4e.
hypereutectic: (illustration only)
b
bb
bb
b
Adapted from Fig. 10.17, Callister & Rethwisch 4e. (Illustration only)
(Figs. 10.14 and 10.17 from Metals Handbook, 9th ed.,Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.)
175 mm
a
a
a
aa
a
hypoeutectic: C0 = 50 wt% Sn
Adapted from Fig. 10.17, Callister & Rethwisch 4e.
T(oC)
61.9eutectic
eutectic: C0 = 61.9 wt% Sn
Hypo-eutectoid alloys form primary 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.
Understand how Microstructures Evolve
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!