Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Engineering 45 Solid StateDiffusion-2 Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Learning Goals - Diffusion • How Diffusion Proceeds • How Diffusion Can be Used in Material Processing • How to Predict The Rate Of Diffusion Be Predicted For Some Simple Cases • Fick’s first and SECOND Laws • How Diffusion Depends On Structure And Temperature

3. Fick’s 1st Law x C Recall Fick’s FIRST Law. Cu flux Ni flux Concen., C Position, x • In the SteadyState Case J = const • So dC/dx = const • For all x & t • Thus for ANY two points j & k • Where • J  Flux in kg/m2•s or at/m2•s • dC/dx = Concentration GRADIENT in units of kg/m4 or at/m4 • D  Proportionality Constant (Diffusion Coefficient) in m2/s

4. In The Steady Case NONSteady-State Diffusion • In The NONSteady, or Transient, Case the Physical Conditions Require • In The Above Concen-vs-Position Plot Note how, at x  1.5 mm, Both C and dC/dx CHANGEwith Time

5. Consider the Situation at Right x J J (left) (right) Concentration,C, in the Box NONSteady State Diffusion Math • Box Dimensions • Width = x • Height = 1 m • Depth = 1 m • Into the slide • Box Volume, V = x•1•1 = x • Now if x is small • Can Approximate C(x) as • The Amount of Matl in the box, M

6. or dx J J (left) (right) Concentration,C, in the Box NONSteady State Diffusion cont • Material ENTERING the Box in time t • For NONsteady Conditions • Material LEAVING the Box in time t • So Matl ACCUMULATES in the Box

7. So the NET Matl Accumulation x J J (left) (right) Concentration,C, in the Box NONSteady State Diffusion cont.2 • Adding (or Subtracting) Matl From the Box CHANGES C(x) • With V = 1•1•x • Partials Req’d asC = C(x,t)

8. In Summary for CONSTANT D x J J (left) (right) Concentration,C, in the Box NONSteady State Diffusion cont.3 • Now, And this is CRITICAL, by TAYLOR’S SERIES • so

9. After Canceling x J J (left) (right) Concentration,C, in the Box NONSteady State Diffusion cont.4 • Now for very short t • Finally Fick’s SECOND LAW for Constant Diffusion Coefficient Conditions

10. The Formal Statement x J J (left) (right) Concentration,C, in the Box Comments of Fick’s 2nd Law • This Leads to the GENERAL, and much more Complicated, Version of the 2nd Law • This Assumes That D is Constant, i.e.; • In many Cases Changes in C also Change D

11. Surface conc., C( x , t ) bar C of Cu atoms s C pre-existing conc., Co of copper atoms s t 3 t 2 t 1 t o C o position, x Example – NonSS Diffusion • Example: Cu Diffusing into a Long Al Bar • The Copper Concentration vs x & t • The General Soln is Gauss’s Error Function, “erf”

12. Gauss's Defining Eqn Comments on the erf • Some Special Fcns with Which you are Familiar: sin, cos, ln, tanh • These Fcns used to be listed in printed Tables, but are now built into Calculators and MATLAB • See Text Tab 5.1 for Table of erf(z) • z is just a NUMBER • Thus the erf is a (hard to evaluate) DEFINITE Integral • Treat the erf as any other special Fcn

13. 1-erf(z) appears So Often in Physics That it is Given its Own Name, The COMPLEMENTARY Error Function: Comments on the erf cont. • Notice the Denom in this Eqn • This Qty has SI Units of meters, and is called the “Diffusion Length” • The Natural Scaling Factor in the efrc • Recall The erfc Diffusion solution

15. Given Cu Diffusing into an Al Bar At given point in the bar, x0, The Copper Concentration reaches the Desired value after 10hrs at 600 °C The Processing Recipe Example  D = f(T) • Get a New Firing Furnace that is Only rated to 1000 °F = 538 °C • To Be Safe, Set the New Fnce to 500 °C • Need to Find the NEW Processing TIME for 500 °C to yield the desired C(x0)

16. Recall the erf Diffusion Eqn Example  D = f(T) cont • For this Eqn to be True, need Equal Denoms in the erf • Since CS and Co have NOT changed, Need • Since by the erf

17. Now Need to Find D(T) As With Xtal Pt-Defects, D Follows an Arrhenius Rln Example  D = f(T) cont.2 • Qd Arrhenius Activation Energy in J/mol or eV/at • R  Gas Constant = 8.31 J/mol-K = 8.62x10-5 eV/at-K • T  Temperature in K • Find D0 and Qd from Tab 5.2 in Text • For Cu in Al • D0= 6.5x10-5 m2/s • Qd = 136 kJ/mol • Where • D0 Temperature INdependent Exponential PreFactor in m2/s

18. Thus D(T) for Cu in Al Example  D = f(T) cont.3 • Thus for the new 500 °C Recipe • In this Case • D600 = 4.692x10-13 m2/s • D500 = 4.152x10-14 m2/s • Now Recall the Problem Solution • This is 10x LONGER than Before; Should have bought a 600C fnce

19. Recall The D(T) Rln Find D Arrhenius Parameters • Applied to the D(T) Relation • Take the Natural Log of this Eqn • This takes the form of the slope-intercept Line Eqn:

20. And, Since TWO Points Define a Line If We Know D(T1) and D(T2) We can calc D0 Qd Quick Example D(T) For Cu in Au at Upper Right y x Find D(T) Parameters cont • Slope, m = y/x • x = (1.1-0.8)x1000/K • = 0.0003 K-1 • y = ln(3.55x10-16) − ln(4x10-13) = −7.023

21. By The Linear Form Find D(T) Parameters cont.2 • in the (x,y) format • x1 = 0.0008 • y1 = ln(4x10-13) = −28.55 • So b • Now, the intercept, b • Finally D0 • Pick (D,1/T) pt as • (4x10-13,0.8)

22. Faster Diffusion for Open crystal structures Lower melting Temp materials Materials with secondary bonding Smaller diffusing atoms Cations Lower density materials Slower Diffusion for Close-packed structures Higher melting Temp materials Materials with covalent bonding Larger diffusing atoms Anions Higher density materials Diff vs. Structure & Properties

23. Diffusion Summarized • Phenomenon: Mass Transport In Solids • Mechanisms • Vacancy InterChange by KickOut • Interstitial “squeezing” • Governing Equations • Fick's First Law • Fick's Second Law • Diffusion coefficient, D • Affect of Temperature • Qd & D0 • How to Determine them from D(T) Data

24. WhiteBoard Work • Problem 5.28 • Ni Transient Diffusion into Cu