chapter 5: diffusion - university of chapter 5: diffusion outline introduction diffusion mechanisms

Download Chapter 5: Diffusion - University of Chapter 5: Diffusion Outline Introduction Diffusion mechanisms

Post on 30-Apr-2020

0 views

Category:

Documents

0 download

Embed Size (px)

TRANSCRIPT

  • 1

    Chapter 5: Diffusion

    Outline Introduction Diffusion mechanisms Steady-state diffusion Nonsteady-state diffusion Factors that influence diffusion

    Introduction

    • Interdiffusion: In an alloy, atoms tend to migrate from regions of high concentration.

    Initially After some time

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

    100%

    Concentration Profiles 0

    Cu Ni 100%

    Concentration Profiles 0

    Diffusion: the phenomenon of material transport by atomic motion

  • 2

    Diffusion phenomenon (2)

    • Self-diffusion: In an elemental solid, atoms also migrate.

    Label some atoms After some time

    A

    B

    C

    D A

    B

    C

    D

    Diffusion mechanisms

    Vacancy diffusion

    Interstitial diffusion • small atoms

    Interstitial is more rapid than vacancy diffusion

  • 3

    Processing using diffusion

    Case Hardening: --Diffuse carbon atoms

    into the host iron atoms at the surface.

    --Example of interstitial diffusion is a case hardened gear.

    The presence of C atoms makes iron (steel) harder.

    Processing using diffusion

    Doping silicon with phosphorus for n-type semiconductors: Process:

    3. Result: Doped semiconductor regions.

    silicon

    magnified image of a computer chip

    0.5mm

    Dark regions: Si atoms

    light regions: Al atoms

    2. Heat it.

    1. Deposit P rich layers on surface.

    silicon

  • 4

    Steady-state diffusion

    Diffusion flux

    Steady-state diffusion: the diffusion does not change with time

    Concentration profile

    At MJ =( )( ) sm

    kgor scm

    mol timearea surface

    diffusing mass) (or molesFlux 22=≡≡J

    Flux vs concentration profile

  • 5

    Steady-state diffusion (continue)

    Concentration gradient=dC/dx

    Linear concentration gradient

    Driving force B

    B xx CC

    x C

    A

    A − −

    Δ Δ ==

    Examples (steady state)

    C1 = 1

    .2k g/m

    3

    C2 = 0

    .8k g/m

    3

    Carbon rich gas

    10m m

    Carbon deficient

    gas

    x1 x20 5m m

    D=3x10-11m2/s

    Steady State = straight line!

    • Steel plate at 7000C with geometry shown:

    • Q: How much carbon transfers from the rich to the deficient side?

    J = −D C2 − C1 x2 − x1

    = 2.4 ×10−9 kg

    m2s

  • 6

    Nonsteady-state diffusion

    Fick’s second law

    Non-steady state diffusion

    The concentration of diffucing 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 C

    ∂ ∂

    = ∂ ∂

    Fick’s Second Law

  • 7

    Non-steady state diffusion

    B.C. at t = 0, C = Co for 0 ≤ x ≤ ∞

    at t > 0, C = CS for x = 0 (const. surf. conc.)

    C = Co for x = ∞

    Copper diffuses into a bar of aluminum.

    pre-existing conc., Co of copper atoms

    Surface conc., C of Cu atoms bars

    Cs

    Solution

    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 y z 2

    0

    2 −∫π=

  • 8

    Nonsteady-state diffusion

    Examples (non-steady state)

    Sample Problem: 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 Eqn. 5.5

    ⎟ ⎠

    ⎞ ⎜ ⎝

    ⎛−= − −

    Dt x

    CC CtxC

    os

    o

    2 erf1),(

  • 9

    Solution (cont.)

    • t = 49.5 h x = 4 x 10-3 m • Cx = 0.35 wt% Cs = 1.0 wt% • Co = 0.20 wt%

    ⎟ ⎠

    ⎞ ⎜ ⎝

    ⎛−= − −

    Dt x

    CC C)t,x(C

    os

    o

    2 erf1

    )(erf1 2

    erf1 20.00.1 20.035.0),( z

    Dt x

    CC CtxC

    os

    o −=⎟ ⎠

    ⎞ ⎜ ⎝

    ⎛ −=

    − −

    = − −

    ∴ erf(z) = 0.8125

    Solution (cont.)

    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 90.0

    − −

    = −

    −z

    z = 0.93

    Now solve for D

    Dt xz

    2 =

    tz xD 2

    2

    4 =

    /sm 10 x 6.2 s 3600

    h 1 h) 5.49()93.0()4(

    m)10 x 4( 4

    211 2

    23

    2

    2 −

    − ==⎟

    ⎟ ⎠

    ⎞ ⎜ ⎜ ⎝

    ⎛ =∴

    tz xD

  • 10

    Factors that influence diffusion

    Diffusing species

    Diffusion and temperature

    Diffusion coefficient 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

  • 11

    D has exponential dependence on T

    Dinterstitial >> Dsubstitutional C in α-Fe C in γ-Fe

    Al in Al Fe in α-Fe Fe in γ-Fe

    1000K/T

    D (m2/s) C in α-Fe

    C in γ-Fe

    Al in Al

    Fe in α-Fe

    Fe in γ-Fe

    0.5 1.0 1.5 10-20

    10-14

    10-8 T(°C)15

    00

    10 00

    60 0

    30 0

    Diffusion and temperature

    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

    2 12

    11lnlnln TTR

    Q D DDD d

    transform data

    D

    Temp = T

    ln D

    1/T

  • 12

    Example (cont.)

    ⎥ ⎦

    ⎤ ⎢ ⎣

    ⎡ ⎟ ⎠ ⎞

    ⎜ ⎝ ⎛ −

    − = −

    K 573 1

    K 623 1

    K-J/mol 314.8 J/mol 500,41exp /s)m 10 x 8.7( 2112D

    ⎥ ⎦

    ⎤ ⎢ ⎣

    ⎡ ⎟⎟ ⎠

    ⎞ ⎜⎜ ⎝

    ⎛ −−=

    12 12

    11exp TTR

    QDD d

    T1 = 273 + 300 = 573K T2 = 273 + 350 = 623K

    D2 = 15.7 x 10-11 m2/s

    Diffusion FASTER for...

    • open crystal structures

    • materials w/secondary bonding

    • smaller diffusing atoms

    • lower density materials

    Diffusion SLOWER for...

    • close-packed structures

    • materials w/covalent bonding

    • larger diffusing atoms

    • higher density materials

    Summary

Recommended

View more >