Introduction to transport phenomena Transport phenomena covers topics such asTransport phenomena covers topics such as

• Transport in the solid state by diffusion

•Transport of heat by conduction; transport of heat by

convection; transient heat flow; heat transport by thermal

radiation

• Engineering units and pressure in static fluidsg g p

• Momentum transport and laminar flow of Newtonian fluids

• Equations of continuity and conservation of momentum• Equations of continuity and conservation of momentum

• Fluid flow past submerged objects

T b l t fl h i l b l d it li ti•Turbulent flow; mechanical energy balance and its application

to fluid flow; mass transport in fluids.

WHY STUDY DIFFUSION?

DIFFUSION IN SOLIDSWHY STUDY DIFFUSION?

• Materials often heat treated to improve properties

• Atomic diffusion occurs during heat treatment• Depending on situation higher or lower diffusion• Depending on situation higher or lower diffusion rates desired

H t t ti t t d ti d h ti• Heat treating temperatures and times, and heating or cooling rates can be determined using the mathematics/physics of diffusionp y

Example: steel gears are “case-hardened” bydiffusing C or N to outer surface

DIFFUSION IN SOLIDS

At i h i f diff i

DIFFUSION IN SOLIDS

• Atomic mechanisms of diffusion

• Mathematics of diffusion Mathematics of diffusion

• Influence of temperature and diffusing species onDiffusion rateDiffusion rate

INTRODUCTION TO DIFFUSIONPhenomenon of material transport by atomic or particle

transport from region of high to low concentration

• What forces the particles to go from left to right?

• Does each particle “know” its local concentration?

• Every particle is equally likely to go left or right!• Every particle is equally likely to go left or right!

• At the interfaces in the above picture, there are more

particles going right than left this causes an average

“flux” of particles to the right!

DIFFUSION MECHANISMSDiffusion at the atomic level is a step-wise migration of atoms from lattice site to lattice site

Conditions for diffusion:

• there must be an adjacent empty site

atoms from lattice site to lattice site

• there must be an adjacent empty site• atom must have sufficient energy to break bonds with its neighbors and migrate to adjacent site (“activation” energy)

Higher the temperature, higher is the probability that an atom will have sufficient energy

h diff i t i ith t thence, diffusion rates increase with temperature

Types of atomic diffusion mechanisms:

• substitutional (through vacancies)• interstitial

A copper–nickel diff i l b fdiffusion couple before a high-temperature heat treatment.

Cu (red circles) andCu (red circles) andNi (blue circles)atom locations

ithi th diff iwithin the diffusioncouple.

Concentrations of copperand nickel as a function of position across the coupleposition across the couple.

(a) A copper–nickel diffusion coupleafter a high-temperature heatg ptreatment, showing the alloyeddiffusion zone

(b) Cu (red circles) and Ni (bluei l ) t l ti ithi thcircles) atom locations within the

couple.

(c) Concentrations of copper and(c) Concentrations of copper andnickel as a function of positionacross the couple.

• Interdiffusion: In an alloy or “diffusion couple”,DIFFUSION

y p ,atoms tend to migrate from regions of large to lowerconcentration. The copper atoms have migrated or diffusedinto the nickel, and that nickel has diffused into copper. Thisprocess, whereby atoms of one metal diffuse into another, istermed interdiffusion, or impurity diffusion., p y

Initially (diffusion couple)After some time

• Self-diffusion: In an elemental solid, atoms alsoDIFFUSION: PHENOMENA

,migrate. Diffusion occurs for pure metal but all atomsexchanging positions are of the same type this istermed as self diffusion. Self diffusion is not normallysubject to observation.

Label some atoms After some time

A

CC

DA

DA

B

D

BB

DIFFUSION MECHANISMSVacancy Diffusion:

• applies to substitutional impurities• applies to substitutional impurities• atoms exchange with vacancies• rate depends on:

b f i-- number of vacancies-- temperature-- activation energy to exchange.gy g

increasing elapsed time

ACTIVATION ENERGY FOR DIFFUSIONDIFFUSION

Initial state Final stateIntermediate stateInitial state Final stateIntermediate state

Energy Activation energy

INTERSTITIAL DIFFUSION

• Applies to interstitial pp

impurities.

• More rapid than vacancy

diffusion (Why?).( y )

• Interstitial atoms smaller

d biland more mobile; more

number of interstitial sites

than vacancies

Diffusion flux (J) defined as the mass (or equivalentlyMODELING DIFFUSION: FLUX

Diffusion flux (J), defined as the mass (or, equivalently, the number of atoms) M diffusing through and perpendicular to a unit cross-sectional area of solid per unit of time. In

th ti l f J ∆M/(A ∆t)mathematical form, J = ∆M/(A ∆t)

J =

1A

dMdt

⇒kg

m2s

⎡

⎣ ⎢

⎤

⎦ ⎥ or

atoms

m2s

⎡

⎣ ⎢

⎤

⎦ ⎥

A dt m2s⎣ ⎦ m2s⎣ ⎦ • Directional Quantity

Jyy

Jx

y

Jz x

• Flux can be measured for:--vacancies

Jz xz

vacancies--host (A) atoms--impurity (B) atoms

Steady-State Diffusion: Fick's First Law of DiffusionDiffusion

For steady state conditions, the net flux of atoms is equal to the diffusivity times the concentration gradient.

d CJ = - DJ D d x

⎛ ⎞2

2

atom swhere J = flux or net flow of atom s * sm

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞2mD = diffusivity or diffusion coefficient s

d C t

⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞4

d C atom s = concentrationgradient d x m

⎛ ⎞⎜ ⎟⎝ ⎠

Rate of diffusion is independent of time; the diffusion flux does not change with

Steady-state diffusion across a thin plateRate of diffusion is independent of time; the diffusion flux does not change with time.

The concentration profile shows the concentration (C) vs the position within the solid (x); the slope at a particular point is the concentration gradient.

STEADY STATE DIFFUSION• Steady State: Steady rate of diffusion from one end to the other.

Implies that the concentration profile doesn't change with time.

Jx(left) = Jx(right)

Steady State:

Jx(right)Jx(left)

Concentration, C, in the box doesn’t change w/time.

x

dC• Apply Fick's First Law:

Jx = −DdC

dxdC⎛

⎜ ⎞ ⎟ =

dC⎛ ⎜

⎞ ⎟ • If J ) = J ) then

• Result: the slope, dC/dx, must be constant dx⎝

⎜ ⎠ ⎟ left

=dx⎝

⎜ ⎠ ⎟ right

• If Jx)left = Jx)right , then

Result: the slope, dC/dx, must be constant(i.e., slope doesn't vary with position)!

3

STEADY STATE DIFFUSION• Steel plate at

700C withgeometry C1 = 1.2kg/m

3

= 0.8kg/m3

g yshown: C2 = 0

Carbon rich gas Carbon

Steady State = straight line!

Carbon deficient

gas

D=3x10-11m2/s

• Q: How much carbon transfers

10mm

x1 x205m

mD 3x10 m /s

carbon transfers from the rich to the deficient side?

J = −DC2 − C1

x2 − x1

= 2.4 × 10−9 kg

m2s

Note: Steady state does not set in instantaneously.

Example: Chemical Protection Clothing

• Methylene chloride is a common ingredient of paint removers. Besides being an irritant, it also may be absorbed through skin. When using, protective gloves should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, what is the diffusive flux of methylene chloride through the glove?diffusive flux of methylene chloride through the glove?

• Data:– D in butyl rubber: D = 110 x10-8 cm2/s

surface concentrations: C = 0 02 g/cm3C = 0 44 g/cm3– surface concentrations:– Diffusion distance:

C2 = 0.02 g/cm3C1 = 0.44 g/cm3

x2 – x1 = 0.04 cmglove

CD

tb 6

2l=

C1

C

skinpaintremover = 1.16 x 10-5 g

cm2sJx = -D C2 - C1

x2 - x1C2x1 x2

2 1

Q. Consider a gas inside a container maintained at a

constant pressure P. Let the concentration of the gas

at the inner wall of the container be 10 kg/m3. If there is

vacuum out side the container, there will be diffusion

of atoms inside the container to outside, across theof atoms inside the container to outside, across the

wall of the container. If the thickness of the wall is 5

mm and diffusion coefficient of the atoms in themm and diffusion coefficient of the atoms in the

material of the container D=10-9 m2/s. What will be the

flux of the atoms?

NON STEADY STATE DIFFUSIONconcentration of diffusing species is a function of both time and position

• Concentration profile,C(x), changes

ith tiJ(right)J(left)

dx

with time.

• To conserve matter: • Fick's First Law:

Concentration, C, in the box

(right)(left)

• To conserve matter: • Fick s First Law:−

dx= −

dC

dt

J(left)J(right) J = −DdC

dx

dJ

dx= −

dC

dt

dJ

dx= −D

d2C

dx2

(if D does not vary with x)

• Governing Eqn.:dC d2C

dx dx )

equate

dCdt

= Dd C

dx2Fick’s second law

C t ti fil dx

Non-Steady-State Diffusion: another look• Concentration profile,

C(x), changes w/ time.J(right)J(left)

dx

Concentration, C, in the box• Rate of accumulation C(x)

∂C

∂tdx = Jx − Jx+dx →

∂C

∂tdx = Jx − (Jx +

∂Jx

∂xdx) = −

∂Jx

∂xdx

• Using Fick’s Law: ∂C

∂t= −

∂Jx

∂x= −

∂∂x

−D∂C

∂x⎛⎝⎜

⎞⎠⎟

Fick’s 2nd Law

Fick's Second"Law" ∂c∂t

= ∂

∂xD∂c

∂x

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟≈ D∂c 2

∂x2

⎛⎜⎜⎜⎜

⎞⎟⎟⎟• If D is constant:

∂t ∂x ∂x⎝ ⎠ ∂x⎝⎜

⎠

2

2

xcD

tc

∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

⎟⎠

⎞⎜⎝

⎛−=Dt

xerfBAtxc2

),(xt ∂⎠⎝ ∂ ⎠⎝ Dt2

Solution to 2o de with 2 constantsdetermined from Boundary Conditions and Initial Conditiondetermined from Boundary Conditions and Initial Condition

( )∫γ2( ) ( )∫ −=

πγ

0

2exp2 duuErf

Erf (∞) = 1Erf (-∞) = -1Erf (0) = 0f ( )Erf (-x) = -Erf (x)

EX: NON STEADY STATE DIFFUSION• Copper diffuses into a bar of aluminum.

pre existing conc C of copper atoms

Surface conc., Cs of Cu atoms bar

pre-existing conc., Co of copper atomss

1. Before diffusion, any of the diffusing solute atoms in the solid are uniformlydistributed with concentration of C

Csdistributed with concentration of C02. The value of x at the surface is zero and increases with distance into the solid.3 The time is taken to be zero the instant3. The time is taken to be zero the instant before the diffusion process begins.These boundary conditions are simply stated as

B.C. at t = 0, C = Co for 0 ≤ x ≤ ∞

at t > 0, C = CS for x = 0 (fixed surface conc.)

C C f⎟⎠

⎞⎜⎝

⎛−=Dt

xerfBAtxc2

),(C = Co for x = ∞⎠⎝ Dt2

EX: NON STEADY STATE DIFFUSION• Copper diffuses into a bar of aluminum.

pre existing conc C of copper atoms

Surface conc., Cs of Cu atoms bar

• General solution:C(x,t) − Co = 1 erf

x⎛ ⎝⎜

⎞ ⎠⎟

pre-existing conc., Co of copper atomss

CsC(x,t)

"error function”

( ) o

Cs − Co

= 1− erf2 Dt⎝ ⎜ ⎠ ⎟

Coposition, x

tot1

t2t3

error function

This equation is very useful for determining thet ti f th l t t i th lid tconcentration of the solute atom in the solid at

different depth, when the rod is exposed to thesolute atom for a known interval of timesolute atom for a known interval of time.

Example: Non-Steady-State DiffusionFCC iron carbon alloy initially containing 0 20 wt% C is carburized at an elevatedFCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface C content at 1.0 wt%.

If after 49.5 h the concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface, what temperature was treatment done?

)(f1f120.035.0),( xCtxC o ⎞⎜⎛−−

f( ) 0 8125)(erf12

erf120.00.1

),( zDtCC os

o −=⎠

⎞⎜⎝

⎛−=

−=

− ∴ erf(z) = 0.8125

z erf(z)find z where erf(z) = 0.8125.

0.90 0.7970z 0.81250.95 0.82097970.08209.0

7970.08125.090.095.0

90.0−−

=−

−z z = 0.93So,

Now solve for D Dtxz

2=

tzxD 2

2

4=

x2⎛⎜

⎞ (4 x 10−3m)2 1 h

∴D =

x4z2t

⎛

⎝⎜

⎞

⎠=

(4 x 10 m)(4)(0.93)2(49.5 h)

1 h3600 s

= 2.6 x 10−11 m2 /s

Solution (cont.):

• To solve for the temperature at which Dhas the above value, we use a rearranged form of Equation )lnln( DDR

QTo

d−

=

D=D0 exp(-Qd/RT)

• for diffusion of C in FCC Fe

Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol D = 2.6 x 10−11 m2 /s

T =148,000 J/mol

(8.314 J/mol-K)(ln 2.3x10−5 m2 /s − ln 2.6x10−11 m2 /s) ( )( )

T = 1300 K = 1027°C

Diffusion and Temperature

• Diffusivity increases with T exponentially

⎛ ⎞ QD = Do exp

⎛

⎝ ⎜

⎞

⎠ ⎟−

Qd

RT

= pre-exponential [m2/s]= diffusion coefficient [m2/s]D

Do

= activation energy [J/mol or eV/atom] = gas constant [8.314 J/mol-K]= absolute temperature [K]

Qd

RT absolute temperature [K]T

Activation energy - energy required to produce the movement of 1 mole of atoms by diffusion.

ln D = ln D0

−Q

dR

1

T

⎛⎝⎜

⎞⎠⎟ log D = log D

0− Q

d2.3R

1

T

⎛⎝⎜

⎞⎠⎟Note:

atoms by diffusion.

DIFFUSION AND TEMPERATURE

• Diffusivity increases with T.

pre-exponential [m2/s] (see Table 5.2, Callister 6e)activation energy

D= Doexp −Qd

⎛

⎝⎜

⎞

⎠⎟ diffusivity

[J/mol],[eV/mol] (see Table 5.2, Callister 6e)

gas constant [8.31J/mol-K]

D DoexpRT⎝

⎜ ⎠ ⎟ diffusivity ( , )

• Remember vacancy concentration: NV = N exp(-QV/kT)• QV is vacancy formation energy (larger this energy• QV is vacancy formation energy (larger this energy, smaller the number of vacancies)• Qd is the activation energy (larger this energy, smaller the diffusivity and lower the probability of atomic diffusion)

DIFFUSION AND TEMPERATUREpre-exponential [m2/s] (see Table 5.2, Callister 6e)

activation energy

D= D expQd

⎛ ⎜

⎞ ⎟ diffusivity

[J/mol],[eV/mol] (see Table 5 2 Callister 6e)

• Experimental Data:gas constant [8.31J/mol-K]

D= Doexp − dRT⎝

⎜ ⎠ ⎟ diffusivity (see Table 5.2, Callister 6e)

NOTE: log(D) = log(D0) Qd/(RT) Experimental Data:

C in

10-8T(C)1

50

0

10

00

60

0

30

0D has exp. dependence on T

NOTE: log(D) = log(D0) – Qd/(RT)

D (m2/s) C in α-Fe

in γ-FeZn10-14

p pRecall: Vacancy does also!Dinterstitial >> Dsubstitutional

C in α-Fe Cu in Cu

Al in Al

Cu in C

u

n in CuFe in α-F

Fe in γ-Fe

10 20

10 14C in γ-Fe Al in Al

Zn in Cu

Fe in α-FeFe in γ-Fe

1000K/TuFe

0.5 1.0 1.5 2.010-20

Adapted from Fig. 5.7, Callister 6e. (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.)

Factors that influence diffusion• The diffusing species, host material and temperature

influence the diffusion coefficient.• For example, there is a significant difference in magnitude p , g g

between self-diffusion and carbon interdiffusion in α iron at 500 °C.

Example: Comparing Diffuse in FeIs C in fcc Fe diff sing faster than C in bcc Fe?Is C in fcc Fe diffusing faster than C in bcc Fe?

(Table 6.2)• fcc-Fe: D0=2.3x10–5(m2/2)

Q=1 53 eV/atom T(°C)500

000

600

300

Q 1.53 eV/atomT = 900 C D=5.9x10–12(m2/2)

• bcc-Fe: D0=6.2x10–7(m2/2) Q 0 83 V/ t

D (m2/s)

10-8T( C)1 1 6 3

Q=0.83 eV/atomT = 900 C D=1.7x10–10(m2/2) 10-14

1000K/T0.5 1.0 1.510-20

• FCC Fe has both higher activation energy Q and D0 (holes larger in FCC).• BCC and FCC phase exist over limited range of T (at varying %C).• Hence, at same T, BCC diffuses faster due to lower Q.• Cannot always have the phase that you want at the %C and T you want!• Cannot always have the phase that you want at the %C and T you want!

which is why this is all important.

Example 2: At 300ºC the diffusion coefficient and activation energy for Cu in Si are:

D(300ºC) = 7.8 x 10-11 m2/sQd = 41,500 J/mol

Wh t i th diff i ffi i t t 350ºC?What is the diffusion coefficient at 350ºC?

Qtransform dataD ln D D = Doexp⎛ ⎝ ⎜ ⎞

⎠ ⎟−Qd

RT

⎞⎜⎛⎞

⎜⎛ 11 QQ

Temp = T 1/T

⎟⎠

⎞⎜⎜⎝

⎛−=⎟

⎠

⎞⎜⎜⎝

⎛−=

3500350

3000300

1lnln and 1lnlnTR

QDDTR

QDD dd

⎞⎜⎛350 11lll QDDD d

33⎠⎞

⎜⎝⎛ −−==−∴

300350300

350300350

11lnlnln TTR

QDDDD d

Example 2 (cont )Example 2 (cont.)⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎜⎝

⎛−−= 12

11exp TTR

QDD d

⎦⎣ ⎠⎝ 12 TTR

T1 = 273 + 300 = 573K

T2 = 273 + 350 = 623K

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

−= −

K 5731

K 6231

K-J/mol 314.8J/mol 500,41exp /s)m 10 x 8.7( 211

2D

D2 = 15.7 x 10-11 m2/s

34

Applications of Diffusion

Doping by Diffusion• Integrated circuits (ICs), found in

numerous electronic devices have been fabricated usinghave been fabricated using doping techniques.

• The base material for these ICs is silicon that has been “doped” with other materials. More precisely controlled• More precisely, controlled concentrations of impurities have been diffused into specific regions of the device to change the properties (improve electrical conductivity).conductivity).

PROCESSING USING DIFFUSION (2)• Doping Silicon with P for n-type semiconductors:

0 5mm• Process1. Deposit P rich

layers on surface.

0.5mm

siliconmagnified image of a computer chip

2. Heat it.3. Result: Doped

semiconductor light regions: Si atoms

regions.

silicon light regions: Al atoms

• Hot-dip galvanizing is a form of galvanization. It is the process of coating iron, steel, or aluminum with a thin zinc layer, by steel, or aluminum with a thin zinc layer, by passing the metal through a molten bath of zinc at a temperature of around 860 °F (460 °C). • When exposed to the atmosphere, the pure zinc (Zn) reacts with oxygen (O ) to form zinczinc (Zn) reacts with oxygen (O2) to form zinc oxide (ZnO), which further reacts with carbon dioxide (CO2) to form zinc carbonate (ZnCO3), a dull grey, fairly strong material. g y y g• In many environments, the steel below the coating will be protected from further corrosion. •Galvanized steel is widely used in applications where rust resistance is neededapplications where rust resistance is needed. A hot-dip galvanizing 'kettle' with fume hood

Galvanized steel and coils popular for applications inpopular for applications in industrial goods, automobile components, precision tubes, consumer durable and many more.

Galvanized i-beams.

PROCESSING USING DIFFUSION (1)• Case Hardening:

-- Example of interstitialdiffusion is a casehardened gear.

-- Diffuse carbon atomsinto the host iron atomsat the surface.

• Result: The "Case" is--hard to deform: C atoms

"lock" planes from shearinglock planes from shearing.--hard to crack: C atoms put

the surface in compression.

PROCESSING QUESTION• Copper diffuses into a bar of aluminum.• 10 hours at 600C gives desired C(x).• How many hours would it take to get the same C(x)y g ( )

if we processed at 500C, given D500 and D600?Key point 1: C(x,t500C) = C(x,t600C).Key point 2: Both cases have the same Co and Cs

(Dt) (Dt)

• Result: Dt should be held constant.Key point 2: Both cases have the same Co and Cs.

⎞⎛CtC )( (Dt)500ºC =(Dt)600ºC

13 2

⎟⎠⎞

⎜⎝⎛−=

−−

Dtxerf

CCCtxC

s 21),(

0

0

• Answer:Note: valuesof D aret500

= (Dt)600D

= 110hr

5.3x10-13m2/s 10hrs

provided here.500 D5004.8x10-14m2/s

A steel rod has a uniform concentration 0.25W% ofA steel rod has a uniform concentration 0.25W% of

carbon. One end of the rod is kept in contact with an

atmosphere of carbon concentration of 1.2W%. The

rod is heated at 950 0C. What will be the concentration

of carbon at a depth of 0.5mm after a time interval of 7 0hours? Diffusion coefficient of carbon in steel at 950 0C

is 1.6x10-11 m2/sec and also find how long will it take to g

get a concentration of 0.7W% at a depth of 0.5 mm?

The diffusivity of gallium in silicon is 8x10-13 cm2/sec atThe diffusivity of gallium in silicon is 8x10 cm /sec at

1100 0C and 1x10-10 cm2/see at 1300 0C (a) Determine

the value of diffusion coefficient D0 and activation

energy Q, assuming the Arrhenius rate law is obeyedenergy Q, assuming the Arrhenius rate law is obeyed

for this diffusion system (b) Calculate the diffusivity of

the system at 1200 0C?

Consider the gas carburizing of a gear of 1018Consider the gas carburizing of a gear of 1018

steels (0.18wt%) at 927 0C. Calculate the time

necessary to increase the carbon content to

0 30wt% at 0 60 mm below the surface of the gear0.30wt% at 0.60 mm below the surface of the gear.

Assume the carbon content at the surface to be

1.00 wt% and that the nominal carbon content of

the steel gear before carburizing is 0.18wt%the steel gear before carburizing is 0.18wt%

where D at 927 0C =1.28x10-11 m2/sec

SUMMARY:STRUCTURE & DIFFUSION

Diffusion FASTER for... Diffusion SLOWER for...

STRUCTURE & DIFFUSION

• open crystal structures

• lower melting T materials

• close-packed structures

• higher melting T materials• lower melting T materials

• materials w/secondarybonding

• higher melting T materials

• materials w/covalentbondingbonding

• smaller diffusing atoms

bonding

• larger diffusing atoms

• lower density materials • higher density materials

• Surface of Ni plate at 10000C contains

Example: Diffusion of radioactive atoms Surface of Ni plate at 1000 C contains

50% Ni63 (radioactive) and 50% Ni (non-radioactive). • 4 microns below surface Ni63 /Ni = 48:52• Lattice constant of Ni at 1000 C is 0.360 nm.• Experiment shows that self-diffusion of Ni is 1.6 x 10-9 cm2/sec

What is the flux of Ni63 atoms through a plane 2 µm below surface?g µ

C2 =(4Ni /cell )(0.48Ni63 /Ni)

(0.36x10−9m)3 /cell

41 1 1027Ni63 / 3

C1 =(4Ni /cell )(0.5Ni63 /Ni)

(0.36x10−9m)3 /cell27 63 3

J = −DC2 − C1 =

= 41.15x1027Ni63 /m3 = 42.87x1027Ni63 /m3

= −(1.6x10−13m2 /sec)(41.15 − 42.87)x1027Ni63 /m3

6J Dx2 − x1

(4 − 0)x10−6m

= +0.69x1020Ni63 /m2 •s

DIFFUSION DEMO: ANALYSIS• The experiment: record combinations of

t and x that kept C constant. toto

t1

t2

t3

x o x 1 x 2 x 3

⎛ ⎞

C(xi, t i ) − CoCs − Co

= 1− erf xi2 Dt i

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = (constant here)

• Diffusion depth given by:

xi ∝ Dtii i

Diffusivity -- the proportionality constant between flux and concentration gradient depends on:

1 Diff sion mechanism S bstit tional s interstitial1. Diffusion mechanism. Substitutional vs interstitial.2. Temperature. 3. Type of crystal structure of the host lattice. Interstitial diffusion easier in BCC than in FCC.4. Type of crystal imperfections.

(a) Diffusion takes place faster along grain(a) Diffusion takes place faster along grain boundaries than elsewhere in a crystal.

(b) Diffusion is faster along dislocation lines than through bulk crystalthan through bulk crystal.

(c) Excess vacancies will enhance diffusion.5. Concentration of diffusing species.

S d d ti l diff ti l tiSecond order partial differential equations are nontrivial and difficult to solve.

Consider diffusion in from a surface where the concentration of diffusing species is always constant This solution applies to gas diffusion intoconstant. This solution applies to gas diffusion into a solid as in carburization of steels or doping of semiconductors.

Boundary Conditions

For t = 0, C = Co at 0< x For t > 0 C = Cs at x = 0 and

C = C at x =ooC = Co at x =oo

Q. The diffusion coefficient for Li in Ge at 500 0C is

of the order of 10-10 m2/sec. What is its

i t di t f t ti i 1 h ?approximate distance of penetration in 1 hour?

Q. The diffusion coefficient for Li in Ge at 500 0C

is of the order of 10-10 m2/sec. How much time it

should take to penetrate a distance of 2 mm?should take to penetrate a distance of 2 mm?

Q. What is the diffusion coefficient for Cu in Al

at 550 0C assuming Q=121kJ/mol and

D0=0 25x 10-10 m2/sec What is approximateD0=0.25x 10-10 m2/sec. What is approximate

distance of penetration of Cu atoms in 1 h?

Q Th i ti i iQ. The vacancy migration energy in copper is

0.8 eV. The self diffusion coefficient at 700 K

and 1000 K are 3.43x10-15 and 1.65x10-11 m2/sec

ti l D t i threspectively. Determine the vacancy

concentration at these two temperatures.

The wear resistance of a steel gear is to be improvedby hardening its surface. This is to be accomplished byincreasing the carbon content within an outer surfacelayer as a result of carbon diffusion into the steel; thecarbon is to be supplied from an external carbon richcarbon is to be supplied from an external carbon-richgaseous atmosphere at an elevated and constanttemperature. The initial carbon content of the steel istemperature. The initial carbon content of the steel is0.20 wt%, whereas the surface concentration is to bemaintained at 1.00 wt%. For this treatment to beeffective, a carbon content of 0.60 wt% must beestablished at a position 0.75 mm below the surface.S if i t h t t t t i t fSpecify an appropriate heat treatment in terms oftemperature and time for temperatures between 900 0Cand 1050 0Cand 1050 C.