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!