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Engineering 45. Solid State Diffusion-2. Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. Learning Goals - Diffusion. How Diffusion Proceeds How Diffusion Can be Used in Material Processing
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Engineering 45 Solid StateDiffusion-2 Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu
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
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
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
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
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
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)
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
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
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
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”
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
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
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)
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
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
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
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:
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
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)
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
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
WhiteBoard Work • Problem 5.28 • Ni Transient Diffusion into Cu