diffusion chapter 8

Diffusion

Chapter 8

Selection of steel for gears

Wear resistance

0.8

1410

910

1150

725

0.02

67.6067.6

0 003

CCf CF

Fraction of cementitite by LEVER rule:

More carbon

More cementite

More wear resistance

Cementite is hard as well as brittle

Hardness or strength is desirable.

But brittleness is not.

Silica Glass is also hard and brittle

Selection of steel for gears

High Carbon Steel: Good wear resistance

but

Btittle

Low Carbon steel: Good ductility

but

poor wear resistance

Wear resistance is required only at the surface

High C steel on the surface

Mild steel inside

Q: How do you achieve this?

Ans: By case carburization

Case carburization

Pack a mild steel gear in carbon and heat at a high temperature in the austenite phase field for some time.

Carbon will enter into the mild steel to give a high-carbon wear resistant surface layer called case.

How do carbon enter into solid steel?

At what temperature and how long should we do the carburization?

The process why which Carbon enters into solid steel during Case carburization is an example of DIFFUSION

Diffusion is relative movement of atoms inside a solid

We can find appropriate time and tempearture for case carburization by solution of Fick’s second law

How do we create an n-p junction in silicon chip?

Ans: by DIFFUSION

Deposit n type element

Deposit p type element

Heat

Si substrate Si substrateSi substrate

Diffusion: flow of matter

Heat: flow of thermal energy

Electric current: flow of electric charge

Different kinds of flows in material

Heat: flow of thermal energy

Fourier’s law of heat conduction (1811)

x

Tq

q: heat flux (J m-2 s-1)

x

T

Gradient?

Temperature gradient

Thermal conductivity

Joseph Fourier (1768-1830)

Electric current: flow of charge

Ohm’s law of electrical conduction

(1827)

x

VEj

j : charge flux (C m-2 s-1), current density

x

V

Gradient?

Electric potential gradient, electric field E

electrical conductivity

Georg Simon Ohm

(1787-1854)

Diffusion: flow of mass

Fick’s first law of diffusion1855

x

cDj

j : mass flux (kg m-2 s-1, moles m-2 s-1)

x

c

Gradient?

concentration gradient, kg m-4

D: Diffusivity, m2 s-1

1829-1901

Temperature dependence of Diffusivity

RT

QDD exp0

D0 = preexponential factor

Q = activation energy

Empirical constants

Self-Diffusion in Amorphous Se (Problem 8.3)

0.00305 0.00315 0.00325

-34

-32

-30

-28

ln D

1/T

T (ºC) D (m2s-1)

35 7.7 x 10-16

40 2.4 x 10-15

46 3.2 x 10-14

56 3.2 x 10-13

D0 = 2 x 1027 m2 s-

1

Q = 250 kJ mol-1

RT

QDD exp0

TR

QDD

1lnln 0

x x+x

c c+c

j j + j

Mass in at x: min = A t j

Mass out at x+ x: mout = A t (j + j)

Mass accumulation between x and x+ x

m = min-mout

= A t ( j – j - j ) = -A t j

Change in concentration in a volume V = A x and time interval t :

m = -A t j

xA

mc

x x+x

c c+c

j j + j

Average rate of change of concentration between x and x + x in time interval t:

x

j

t

c

xA

jtA

x

jt

x

j

t

c

Instantaneous change in concentration at a time t, at a point x:

x

j

t

c

t

x

t

x

0

0

0

0 limlim

x

j

t

c

Fick’s 2nd Law

x

j

t

c

x

cD

xt

c

2

2

x

cD

t

c

If D is independent of x

Fick’s 2nd law

Using Fick’s First Law

Solution to Fick’s 2nd law:

2

2

x

cD

t

c

Solution depends on the boundary condition.

tD

xBAtxc

2erf),(

A and B : constants depending on the boundary conditions

erf (z) : Gaussian error function

The Gaussian Error Function

z

dzerf0

2 )(exp2

)(

0.2

0.4

0.6

0.8

1

exp (-2)

-3 -2 -1 1 2 3

0 z

Hatched area (2/) = erf (z)

erf (0) = 0, erf (-z) = - erf (z), erf (+ ) = +1, erf (- ) = -1

-3 -2 -1 1 2 3

-1

-0.5

0.5

1

z

dzerf0

2 )(exp2

)(

z

erf (z)0.85 0.770668

0.9 0.796908

0.95 0.820891

1. 0.842701

1.1 0.880205

1.2 0.910314

1.3 0.934008

1.4 0.952285

1.5 0.966105

1.6 0.976348

1.7 0.98379

1.8 0.989091

1.9 0.99279

2. 0.995322

2.2 0.998137

2.4 0.999311

2.6 0.999764

2.8 0.999925

0. 0.

0.025 0.0282036

0.05 0.056372

0.1 0.112463

0.15 0.167996

0.2 0.222703

0.25 0.276326

0.3 0.328627

0.35 0.379382

0.4 0.428392

0.45 0.475482

0.5 0.5205

0.55 0.563323

0.6 0.603856

0.65 0.642029

0.7 0.677801

0.75 0.711156

0.8 0.742101TABLE 8.1

Mistakes in the textbook in the boxed values

Carburisation of steel

cs

Distance in steel from surface

c (wt% C)

xc0

Surface concentration

initial concentration

Concentration profile after carburization for time t at a temperature T

Boundary conditions:

1. c=c0 at x>0 , t=0

2. c=cs at x=0 , t>0

Carburising atmosphere Steel

Carburisation of steel (contd.)

tD

xBAtxc

2erf),(

Boundary conditions:

2. c(x,t) = c0 at x > 0, t = 0

1. c(x,t) = cs at x = 0, t > 0

B.C. 1 cs = A – B erf(0) = A

B.C. 2 c0 = A – B erf(+) = A-B

A = cs

B = cs – c0

tD

xccctxc ss

2erf)(),( 0

Case carburization of steel Problem 8.4

Initial concentration c0 = 0.2 wt% C

Surface concentration cs = 1.4 wt % C

Temperature = 900 ºC = 1173 K

Desired concentration c = 1.0 wt% C at x = 0.2 mmAt 900 ºC the equilibrium phase of steel is austenite ()

Diffusivity data for C in austenite:

D0 = 0.7 x 10-4 m2s-1

Q = 157 kJ mol-1

RT

QDD exp0 = 7.13688 x 10-12

m2s-1

Carburization of steels

0.1 0.2 0.3 0.4 0.5

0.2

0.4

0.6

0.8

1.0

1.2

1.4

100 s

1000 s

10000 s

Distance in steel from surface, mm

wt% C

tD

xccctxc ss

2erf)(),( 0

tD

x

2erf)2.04.1(4.10.1

3333.02.1

4.0

2.04.1

0.14.1

2erf

tD

x

z erf(z)

0.30 0.3286

0.35 0.3794

0.305 0.3333

)305.0(erf3333.02

erf

tD

x

305.02

tD

x

305.0 10 7.136882

102.012-

3

t

t = 15062 s = 4 h 11 min

This is reasonable. If not, change D by changing T

Ans

Atomic Mechanism of Diffusion

How does C enter into solid steel?

is INTERSTITIAL solid solution of C in FCC Fe

C occupies octahedral voids in FCC Fe

Maximum solubility of C in austenite () is 2.14 wt% at 1150 ºC.

C at% 9

85.5598

122

122

Cwt%2

Thus 9 out of 91 OH voids are occupied.

90 % of OH voids are empty

C atoms can jump from one interstitial site to another vacant interstitial site.

This is interstitial diffusion.

For OH voids the void size is 0.414 R but the window through which C atoms can jump outside is only 0.155 R.

Thus to jump out of an interstitial OH site the C atoms will have to displace neighbouring Fe atoms. This will increase the energy of the system

OH void

OH void

Potential energy

Hm

A carbon atom can jump to a neighbouring site if it has sufficient energy Hm. It can gain this energy only through random thermal vibration.

If thermal vibration frequency is then it makes attempts per second.

Only a fraction

RT

Hmexp of these attempts

will have an energy Hm and will be successful.

1 2

C1 C2

No. of successful jumps per second from plane 1 to plane 2,

n1->2 = A c1 exp(- Hm/RT) pNo. of successful jumps per second from plane 2 to plane 1,

n2->1 = A c2 exp(- Hm/RT) pNet jumps per second from

plane 1 to plane 2

n = n1->2 - n2->1 = A (c1-c2) exp(- Hm/RT) p

122 exp

cc

RT

HpA m

Flux:x

c

RT

Hp

A

nj m

exp2

x

c

RT

Hpj m

exp2 Fick’s 1st Law

RT

HpD mexp2

20 pD mHQ

An atom making a successful jump may remain in plane 1, go to the back plane or jump to forward plane 2. Thus only a fraction p of successful jumps are from plane 1 to plane 2. This factor has been omitted in the textbook.

Initially After some time

Adapted from Figs. 5.1 and 5.2, Callister 6e.

100%

0

Ni

x

Wt % Ni

100%

0

increasing elapsed time

How is diffusion taking place in a substitutional solid solution ?

Mechanism of substitutional diffusion

Vacancy mechanism of substitutional diffusion

However, only a very small fraction of the order of 10-4 to 10-30 are vacant.

A jump can only be successful if the neighbouring site is vacant.

Probability of finding a vacant site

= fraction of vacant site

RT

H

N

n fexp

x

c

RT

H

RT

Hpj fm

expexp2

fmsubs HHQ 2,0 pD subs

Sbstitutional diffusion is usually slower than interstitial diffusion due to difficulty of finding a vacant site.

Lattice diffusion

Grain boundary diffusion

Surface diffusion

Experimentally

Qsurface < Qgrain boundary < Qlattice

Other Diffusion Paths