Ficks Laws

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Lecture #11

CHAPTER 6: DIFFUSION IN SOLIDS ISSUES TO ADDRESS... ADDRESS • What are Fick’s Laws of Diffusion? • How can the rate of diffusion be predicted for some simple cases?

Relevant Reading for this Lecture... • Pages 174-191. 1

Review of Previous Lecture • Diffusion is mass transport by atomic motion. • There are two main diffusion mechanisms in solids: – Substitutional diffusion by vacancies – Interstitial diffusion. Does not involve vacancies • Interstitial diffusion is more rapid than substitutional diffusion. WHY? • The rate of diffusion can be modeled using the Arrhenius equation

 Qd    Rate  D  Do exp  RT 

which allows us to determine: – Activation energy for the process – Diffusion coefficients for a given temperature 2

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HOW DO WE MODEL DIFFUSION PROCESSES? • Fick’s 1st Law for steady-state diffusion • Fick’s 2ndd Law for nonsteady-state diffusion

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FLUX • Every engineer must be familiar with different kinds of flux – Mass flux results in diffusion – Heat flux results in heat transfer – Momentum flux results in convection

• ALL these have similar units ( _______ per area per time) X Normalizing factors – takes out geometry and time!

• Mass flux is essentially the amount or rate of diffusion – Has units of mass per area per time

• For Diffusion, flux is defined as:

J  Flux 

moles (or mass) diffusing mol kg  or 2 2 cm s m s  surface area   time 

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FLUX • Essentially the amount or rate of diffusion

J  Flux 

moles (or mass) diffusing mol kg or 2  2 surface area time cm s m s

• Measured empirically – Make thin film (membrane) of known surface area – Impose concentration gradient – Measure how fast atoms or molecules diffuse through the membrane

J

M (mass diffusing)

M 1 dM  At A dt

J  slope time

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• Can be measured for vacancies, host (A) atoms, impurity (B) atoms

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Steady--State Diffusion Steady IF rate of diffusion is independent of time THEN Flux is proportional to concentration gradient = Initial (High conc.)

C1 C1

Fick’s first law of diffusion

C2 x1

if linear 6

dC dx

x

C

Final 2 (Low conc.)

J  D

dC dx

x2

dC C C2  C1   x2  x1 dx x NOTE: Initial concentration goes with initial position Final concentration goes with final position

D  diffusion coefficient previous lecture describes how to evaluate D 1

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EXAMPLE OF STEADY STATE DIFFUSION • Steel plate at 700°C with geometry Carbon shown: rich gas

3 g/m k 2 3 . =1 /m kg C1 8 . =0 C2 Steady State = straight line!

Carbon deficient gas D=3x10-11m2/s

Callister & Rethwisch 4e.

5m

m m 10 m

0 x1 x2

Adapted from Ex. Prob. 6.1 on p. 175 of

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

J  D

C2  C1   kg 2.4  10 9 x2  x1 m2s

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Example: Chemical Protective Clothing (CPC) • Methylene chloride (MC) is a common ingredient of paint removers. Besides being an irritant, it also may be g the skin. When using g this p paint absorbed through remover, 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? • Data: – diffusion coefficient of MC in butyl rubber: D = 110 x10-88 cm2/s – surface concentrations: C1 = 0.44 g/cm3 C2 = 0.02 g/cm3 8

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Example (con’t.) Solution – assuming linear conc. gradient (i.e., steady state) glove C1

tb 

paint remover

2 6D

J -D

skin C2 x1 x2 = 0.04

=0

J   (110 x 10-8 cm 2 /s)

dC C  C1  D 2 x2  x1 dx

Data:

D = 110 x 10-8 cm2/s C1 = 0.44 g/cm3 C2 = 0.02 g/cm3 x2 – x1 = 0.04 cm

(0.02 g/cm 3  0.44 g/cm 3 ) g  1.16 x 10-5 (0.04 cm) cm 2s

steady state means – nothing changes with time 9

In class example: A differential nitrogen pressure exists across a 2-mm-thick steel furnace wall. After some time, steady-state diffusion of the nitrogen is established across the wall. Given that the nitrogen concentration on the high-pressure surface of the wall is 2 kg/m3 and on the low-pressure surface is 0.2 kg/m3, calculate the flow of nitrogen through the wall (in kg/m2h) if the diffusion coefficient for nitrogen in this steel is 1.0  10-10 m2/s at the furnace operating temperature. Refer to previous lecture

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In class example: SOLUTION A differential nitrogen pressure exists across a 2-mm-thick steel furnace wall. After some time, steady-state diffusion of the nitrogen is established across the wall. Given that the nitrogen concentration on the high-pressure surface of the wall is 2 kg/m3 and on the low-pressure surface is 0.2 kg/m3, calculate the flow of nitrogen through the wall (in kg/m2h) if the diffusion coefficient for nitrogen in this steel is 1.0  10-10 m2/s at the furnace operating temperature.

J x  D

c x

2 kg/m3

3 c chigh  clow  2.0  0.2  kg/m    900 kg / m 4 3 x xhigh  xlow  0  2.0  10  m

0.2 kg/m3

2 mm 0

Make sure units are consistent !!

D  1.0  1010 m 2 / s 

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1.0  10 10 m 2 3.6  103 s m2  3.6  107 1 s 1 h h

Substitute:

J x  D 11

P. 175

c kg  m2 kg     900 4   3.6  107  3.24  104 2 x m  h m h 

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Non--steady State Diffusion Non

• When the concentration of diffusing species is a function of both time and position C = C(x,t) • We must use Fick’s Second Law:

Don’t let this scare you. Diff Differential i l equations i are not needed for this course!

C1

Fick’s 2nd Law

C  2C D 2 t x

tb 

2 6D

C2 x1 x2

Most practical situations are non-steady state! i.e. diffusion flux varies with time (resulting in accumulation or depletion of diffusing species) 12

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Non--steady State Diffusion Non • Consider Cu diffusion into a semi-infinite* bar of aluminum. Surface conc., Cs of Cu atoms

bar

(assume constant CS)

pre-existing conc., Co of copper atoms

high conc.

Low conc.

Cs Adapted from Fig. 6.5, Callister & Rethwisch 8e.

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Boundary Conditions: at t = 0, C = Co for 0  x   at t > 0, C = CS for x = 0 (constant surface conc.) C = Co for x = 

Solution: C  x , t   Co  x   1  erf   Cs  Co  2 Dt  (Gaussian error function – describes the shape of the curve)

C(x,t) = Conc. at point x at time t erf (z) = error function



2 



z 0

e  y dy 2

erf(z) values are given in Table 6.1.

CS

C(x,t) Co

Adapted from Fig. 6.5, Callister & Rethwisch 8e.

Just look up erf(z) to find z. 14

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One more thing that the Error Function does  x  C(x, t)  Co  1  erf   Cs  Co 2 Dt 

Partial table of Error Function values. Full table 6.1 appears on p. 177 of text and on next slide.

z

erf(z)

Note since erf(0.5) ≈ 0.5 → midway between c2 and c1 is given by  x     0.5  x  Dt  2 Dt  15

To achieve a 2-fold increase in concentration requires a 4-fold increase in time!

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In class example:

Non-steady State Diffusion • Sample Problem: An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature p and in an atmosphere p that g 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. 6.5

C ( x, t )  C o  x   1  erf   Cs  Co  2 Dt 

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Solution (cont.): We know:

– t = 49.5 h – Cx = 0.35 0 35 wt% – Co = 0.20 wt%

C( x , t )  Co  x   1  erf   Cs  Co  2 Dt 

x = 4 x 10-3 m Cs = 1 1.0 0 wt% We don’t know D

C ( x, t )  Co 0.35  0.20  x    1  erf    1  erf ( z ) Cs  Co 1.0  0.20  2 Dt   erf(z) = 0.8125

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Solution (cont.): We must now determine from Table 6.1 the value of z for which the error function is 0.8125. An interpolation is necessary as follows (linear) [erf(z) from table]

z

erf(z) f( )

0.90 z 0.95

0.7970 0.8125 0.8209

Now solve for D

z  0.90 0.8125  0.7970  0.95  0.90 0.8209  0.7970

z  0.93

z

x 2 Dt

D

x2 4 z 2t

3 2  x2  1h   ( 4 x 10 m) D    2.6 x 10 11 m2 /s  4z 2t  ( 4)(0.93)2 ( 49.5 h) 3600 s  

Not done yet! What is the temperature?

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Last lecture!

Solution (cont.): • To solve for the temperature at which D has the above value, we can use Eqn (6.9a) and solve for T.

 Qd   D  Do exp  RT 

T 

Qd R (lnDo  lnD )

from Table 6.2, for diffusion of C in FCC Fe Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol



T

148,000J/mol (8.314J/mol- K)(ln2.3x105 m2/s  ln 2.6x1011 m2/s)

T = 1297 K = 1024ºC 3

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Summary • Mass flux (J) is the movement of mass (atoms, grams, moles) per unit area per unit time • Fick’s Fick s first law: Diffusion down a concentration gradient; independent of time

J  D

dC dx

• Fick’s second law: Diffusion function of position and time

C  2C D 2 t x - One solution based semi-infinite boundary conditions is

C  x , t   Co  x   1  erf   Cs  Co  2 Dt  21

Outcome is

 x     0.5  x  Dt  2 Dt  To achieve a 2-fold increase in concentration requires a 4fold increase in time!

What have we learned so far about diffusion….. Diffusion FASTER for...

Diffusion SLOWER for...

• open crystal structures

• close-packed structures

• smaller diffusing atoms

• larger diffusing atoms

• lower density materials

• higher density materials

• materials w/secondary bonding (e.g., liquids)

• materials w/covalent or ionic bonding

(W k Bonds) (Weaker B d)

(St (Stronger Bonds) B d)

Diffusion goes down a concentration gradient: Fick’s 1st law Diffusion 2x distance requires 4x time: Fick’s 2nd law 22

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