Chapter 5.4-5.5: Diffusion (part 2)

EM321: Lesson 8a

Before class, pass out photocopy of Callister Tables 5.1 and 5.2

Review from Last Time• Types of diffusion

– Interstitial

– Vacancy

• Examples of diffusion

• Rate of diffusion

increasing elapsed time

Diffusion: Calculating Flux

• How do we quantify the amount or rate of diffusion?

sm

kgor

scm

mol

timearea surface

diffusing mass) (or molesFlux

22J

J slope

dt

dM

A

l

At

MJ

M =

mass

diffusedtime

Steady-State Diffusion: Calculating Flux

dx

dCDJ

Fick’s first law of diffusionC1

C2

x

C1

C2

x1 x2J = flux

D = diffusion coefficient (Table 5.2)

dC = change in concentration

dx = change in linear distance

For Steady-State Diffusion, the rate of diffusion is independent of time

Flux is proportional to concentration gradient =dx

dC

12

12 linear ifxx

CC

x

C

dx

dC

Steady-State Diffusion and Time

• Diffusion Coefficient, D, increases with increasing

Temperature, T

• The higher the Diffusion Coefficient, the more _______ atoms

will diffuse across a given concentration gradient, such that:

D1 t1 = D2 t2

Example: The diffusion coefficient for Copper in Aluminum is

D = 4.15 * 10-14 m2/s at 500oC and D = 4.69 * 10-13 m2/s at 600oC.

What time will be required at 600oC to produce the same diffusion result

(in terms of concentration at a specific point) as for 10 hours at 500oC?

t600 =D500t500

D600

= (4.15 10 14 m2 /s)(10h)

(4.69 10 13m2 /s)= 0.88 h

quickly

Diffusion and Temperature

Diffusion Coefficient, D, increases with increasing T.

D Do exp

Qd

RT

= 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

Diffusion and Temperature

D has an exponential dependence on T

Dinterstitial >> Dsubstitutional

C in -FeC in -Fe

Al in AlFe in -FeFe in -Fe

1000K/T

D (m2/s)

0.5 1.0 1.510-20

10-14

10-8

T( C)

15

00

10

00

60

0

30

0

From the graph of D vs. T, what can you

conclude about Dinterstitial vs. Dsubstitutional?

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

D(300ºC) = 7.8 x 10-11 m2/s

Qd = 41.5 kJ/mol

What is the diffusion coefficient at 350ºC?

1

01

2

02

1lnln and

1lnln

TR

QDD

TR

QDD dd

121

212

11lnlnln

TTR

Q

D

DDD d

transform dataD

Temp = T

ln D

1/T

Example: Diffusion and Temperature

Note: You were not given the

Pre-Exponential, Do

K 573

1

K 623

1

K-J/mol 314.8

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

2D

12

12

11exp

TTR

QDD d

T1 = 273 + 300 = 573K

T2 = 273 + 350 = 623K

D2 = 15.7 x 10-11 m2/s

Example (cont.): Diffusion and Temperature

Another Example: Diffusion Coefficient

Example: What is the Diffusion Coefficient, D, for the

interdiffusion of carbon in α-iron (BCC) at 900oC?

(Note: Use data in Table 5.2 to find the Pre-Exponential, Do.)

From Table 5.2, Do = ________m2/s for carbon in α-Fe

Qv = ____ kJ/mol = ___________ J/mol

) 1173)(K-J/mol 31.8(

J/mo 000,80exp/sm 10 6.2 = )( 27-

K

lD

Dα = 5.86 10-12 m2/s

80 80,000

6.2 * 10-7

Steady-State vs. Non-Steady State

Up until now, we have been looking at Steady-State diffusion

What does “Steady-State” mean?

- Concentration of diffusing species is a function of position only.

Not a function of time (does not change over time). C = C(x)

What do we do if our concentration gradient is “Non-Steady

State” (C is a function of both both position and time)?(e.g. when a material is first introduced to a concentration gradient and the

impurity atoms begin to diffuse through it)

- Use Fick’s Second Law (remainder of this lecture). Second

order equation with C = C(x,t)

Non-Steady State Diffusion

• The concentration of diffusing species is a function of both time and position C = C(x,t)

• In this case Fick’s Second Law is used

2

2

x

CD

t

CFick’s Second Law

Non-Steady State Diffusion

Boundary Conditions: at t = 0, C = _____ for 0 x

at t > 0, C = _____ for x = 0 (constant surface concentration)

C = _____ for x =

Pre-existing concentration, Co, of copper atoms

Surface concentration,

C , of Cu atoms Aluminum bars

Cs

Example: Copper diffusing into a bar of aluminum that already

has a low copper concentration, Co, within it

Co

Co

Cs

Solution to Fick’s Second Law:

C(x,t) = Conc. at point x at time t

erf (z) = error function

erf(z) values are given in Table 5.1

CS

Co

C(x,t)

Dt

x

CC

Ct,xC

os

o

2 erf1

dye yz 2

0

2

Using Table 5.1:

What value of z gives an error function of erf(z) = 0.4755?

What value of z gives an error function of erf(z) = 0.9592?

What value of z gives an error function of erf(z) = 0.3400?

z = 0.45

z = 1.45

z = 0.3112

Example: Non-steady State Diffusion

• Example: An FCC iron-carbon alloy initially containing 0.20 wt% C is

carburized at an elevated temperature and in an atmosphere that gives a

surface carbon concentration constant 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,

determine the temperature at which the treatment was carried out.

• Solution: use

Dt

x

CC

CtxC

os

o

2erf1

),(

– t = 49.5 h x = 4 x 10-3 m

– Cx = 0.35 wt% Cs = 1.0 wt%

– Co = 0.20 wt%

)(erf12

erf120.00.1

20.035.0),(z

Dt

x

CC

CtxC

os

o

erf(z) = 0.8125

We must now determine from Table 5.1 the value of z for which the

error function is 0.8125. An interpolation is necessary as follows

z erf(z)

0.90 0.7970

z 0.8125

0.95 0.8209

7970.08209.0

7970.08125.0

90.095.0

90.0z

z 0.93

Now solve for D

Dt

xz

2 tz

xD

2

2

4

/sm 10 x 6.2s 3600

h 1

h) 5.49()93.0()4(

m)10 x 4(

4

2112

23

2

2

tz

xD

Example (cont.): Non-steady State Diffusion

• To solve for the temperature at

which D has the above value, we

use a rearranged form of the

Diffusion Coefficient equation )lnln( DDR

QT

o

d

from Table 5.2, for diffusion of C in FCC Fe

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

/s)m 10x6.2 ln/sm 10x3.2 K)(ln-J/mol 314.8(

J/mol 000,14821125

T

T = 1300 K = 1027ºC

Example (cont.): Non-steady State Diffusion

Example: Chemical Protective Clothing

• Methylene chloride is a common ingredient of paint removers. Besides

being an irritant, it also may be absorbed through skin. When using this

paint remover, protective gloves should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, what is the breakthrough

time (tb), i.e., how long can the gloves be used before

methylene chloride reaches the hand?

• Data

– diffusion coefficient in butyl rubber:

D = 110x10-8 cm2/s

glove

C1

C2

skinpaint

remover

x1 x2

Discussion Questions:

How would you go about solving this problem? (Do not solve it.)

Do you need to know the surface concentration of methylene chloride in

order to solve this problem? Why or why not?

Conclusion

• Steady-State Diffusion over Time

• Diffusion and Temperature

– Calculating Diffusion Coefficient, D

• Non-Steady State Diffusion