Diffusion in Solid Stateuotechnology.edu.iq/dep-materials/lecture/secondclass/Metals...Diffusion in Solid State (Diffusion in Solid State): - Diffusion Mechanism. It is known that

Diffusion in Solid State

-(Diffusion in Solid State):

Diffusion Mechanism

It is known that at high temperature, the Amplitude of vibrations of atoms leads to expansion, while keeping the (Frequency of Vibration) constant in general form

Diffusion by voids

RTHfNaNv /exp−=

The percentage of voids in metal lines of F. C. C to 2% at high temperatures

Wo, is the vibrations between the void and solvent A very far from the solute B W1, is the vibrations between the void and solvent A away, the closest distance from the void and solute B W2, are the frequencies between the void and solute B W3, are the frequencies between the void and solvent atoms A, where A is the closest distance

from the void. W4, are the frequencies between the void and solvent A away from the solute B. The relationship between these vibrations known (Correlation factor) (f).

34

21

34

1

)(722

)(72

WWW

WW

WWW

Wf

o

o

++

+=

Wo W4 =

321

31

27

27

WWW

WWf

++

+=

Interstitial atoms Diffusion by

RTQaD /2 exp−=ν ν Vibration frequency, , Q the activation energy necessary for diffusion, ΔHm

enthalpy

Fick,s first Law

dxdcDJ −=

Expressed in this law the following equation:

Where each of: J, is called flux. C, concentration of solute in the solvent D, diffusion coefficient To derive the law, we assume the existence of two surfaces adjacent the lattice A and B (each an alloy of copper and zinc), where the concentration of zinc in A, more than in the B,

The sum of the number of atoms of zinc, which can jump over the energy barrier per second, is: This is a net movement of atoms of zinc, down to include the focus from A to B. And therefore:

AnnJ BA

1).(6

−=ν ) ………2(

Where J is, flux. The A, sectional area, from the definition of C:

aAnC B

B .=

aAnC A

A .=

][ BA CCa

bJ −=

ν

وبما ان : dxdc

aCC BA −=

−

And CA, CB is composition of zinc in both A and B.

. :dxdc

baJ .

2ν−=

baD

2ν=

dxdcDJ −=

The negative sign indicates that the flux direction is the direction of decreasing n. In comparison with equation (1) we get: -

Fick,s second Law

Δv = A. Δx Where Δv is the size of the piece. The number of atoms inside the volume at a certain moment nΔv = n. A. Δx

xAdxdJJJ AA ∆−=′− .

xAdtdc

∆..JJ AA =′−

dxdJ

dtdc

−=

dxdcDJ −=

2

2

dxdcD

dtdc

=

Law second Fick,s Solution of

Dtxu

2=

)2

(.t

ududc

dtdu

dudc

dtdc

−==

dudc

Dtdxdu

dudc

dxdc .

21. ==

Dtdudc

dxdu

dudc

dxdc

41.)( 2

22

2

2

2

2

==

dududcd

dudc

dudcu

)(.2 2

2

==−

But :dudcy =

ydyudu =− 2

ln y = - u2 + cte 2uBey −=

B is a constant بما ان

dudcy =

اذن 2uBe

dudc −=

∫ += −u

u AdueBC0

2

is anther constant A

∫ −u

u due0

2

= erf (u).

Dtxu

2=

)2

(Dt

xBrefAC +=

function )u (erf

∫ −=u

u dueuerf0

22)(π

0 =u ∫ −u

u due0

2

0 = 0= erf (u) .

and

+∞=u ∫ −u

u due0

2

= 2π

1)2

(2)( ==∞π

πerf

1)2

(2)( −=−=−∞π

πerf

erf (-u) = -erf (u)

Q-1

Q-2

Q-3

Q-4

Q-5

Q-6

RTQoeDD /−= ) ..……15 (

"D": Diffusion coefficient Where Q represents the activation energy. The Do- diffusion constant , and be equal to ( 2aν ). Table some of the values of D, Do, Q, for some metals and alloys.

عملیة االنتشار المذیب في المذاب

Dثانیة 2متر / oم 500

Dثانیة 2متر / oم 1000

Doثانیة 2متر /

Q x 10 3 جول / مول

Cu في Cu 10 -13 10 –7 2 x 10 -5 196 Fe ) فيα ( Fe 10 –14 10 –7,5 118 x 10 –4 281 C ) فيα ( Fe 10 –5 10 –3 0,008 x 10 -4 83 C ) فيγ ( Fe 10 –8,3 10 –4,5 0,7 x 10 -4 157

Mn ) فيγ ( Fe 10 –17 10 –10 4 x 10 -4 205 Ni ) فيγ ( Fe 10 –17 10 –9,6 2,6 x 10 -4 295

Zn في Cu 10 –10 10 –7 0,72 x 10 -4 170 Cu في Al 10 –7,5 0.25 منصھر x 10 -4 121

Which can be written as the following:

ko TR

QDD 1.lnln −=

If a line chart decreed between ln D and inverted absolute temperature as in Figure () will get a straight line equal to his inclination, and since R is the gas constant, we will have a Q value easily.

Since the intersection with the vertical coordinates, represents a ln Do, logarithmic tables, and we can get the value of Do.

معامل االنتشار الذاتي كالفن ToK درجة الحرارة D ثانیة 2سم /

kT/1 1كالفن / Log D

700 1.9 x 10 -11 1.42 x 10 -3 -1.72 800 5 x 10 –10 1.25 x 10 –3 -9.20 900 6.58 x 10 –9 1.11 x 10 –3 -8.12 1000 5 x 10 –8 1 x 10 –3 -7.2 1100 2,68 x 10 -7 0.91 x 10 –3 -6.57

Figure () shows the relationship between the diffusion coefficient and the absolute temperature in some metals and alloys.

7 Q-

8 Q-

9 Q-

-n: types of transactions diffusio

diffution-Self-1 It is intended that the diffusion of radioactive metal dissolved in the same non-radioactive metal, such as Ni * in Ni, and Ag * in Ag. in general ,the self-diffusion equation as follows:

RTQ

selfself eDD /*0

* *

−=

1- grain boundaries Diffusion – The diffusion in fine g,b is greater than large g,b . Equation can be written as follows:

RTbQgbogbg eDD /...

−=

In general, the value of activation energy for the grain boundaries diffusion is half of the value of the Self-diffusion .

1- grain boundaries Diffusion –

selfbg QQ21

. ≅

From this we can deduce that the diffusion coefficient in the g.b will be larger than the diffusion coefficient of Self-diffusion.

selfbg DD >.

3 - surface diffusion: - Which is diffusion along the surface of crystals and is usually faster than the g.b diffusion and the Self-diffusion

RTQ

ossDDs /−=

We note that:

selfbgs QQQ << .

selfbgs DDD >> ,

Phenomenon of Krkindal: -

pair diffusible of( copper – nickel) alloy, was heated at 900oc for a period of 40 hours,. If one of alloys was containing 40% = a% nickel in place is located to the right of the surface interval, and the other alloy was containing 50% = b% nickel in place is located to the left of the interface ,find

A – The diffusion equation of copper – nickel alloy B - the diffusion coefficient ( note that at a point about 0.2 cm from the boundary is 42.5% )? C - the time required to get the same concentration, but just 0.4 cm from the boundary?

A - The first condition, at the beginning of the experiment time will be zero, and the concentration, is % b when X is negative., So the diffusion equation would be:

02%

~D

xBerfAb −+=

)(% −∞+= BerfAb BAb −=%

The second condition, at the beginning of the experiment also be time zero, and the concentration is% a when X is positive, so the diffusion equation would be:

02%

~D

xBerfAa ++=

)(% ∞+= BerfAa BAa +=%

2%% baA +

=

2%% baB −

=

NNi =

2%% ba + +

2%% ba −

tD

xerf.2

~

NNi = 45% - 5% tD

xerf.2

~

B - Upon passage of 40 hours, the concentration became 42.5%) at distance of 0.2 cm from the boundary. So:

tD

xerf.2

%5%45%5.42~

−=

5.0

05.0025.0

.2==

tDxerf

erf ( 0.477 ) = 0.5

477.0360040.2

2.0=

xD

sec10563.2

27 cmxD −=

C - Since the concentration is fixed,

~

1

1

.2%5%45%5.42

tD

xerf−=

2

~

2

.2%5%45%5.42

tD

xerf−=

: اذا

2

~

2

1

~

1

.2.2 tD

xerftD

xerf =

2

2

1

1

tx

tx

=