Phase Transformation Lecture 3
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Materials Engineering MM-501 Phase Transformation in Solids
MM-501 Phase Transformation in Solids
Lecture No: 03Diffusion : How do atoms move through solids?
What is diffusion?Material/Mass transport by atomic motion is called Diffusion. ORIt is a transport phenomenon caused by the motion of chemical species (molecules, atoms or ions), heat or similar properties of a medium (gas, liquid or solid) as a consequence of concentration (or, strictly, chemical potential) differences. In general, the species move from high concentration areas to low concentrations areas until uniform concentration is achieved in the medium.3
Diffusion is easy in liquids and gases where atoms are relatively free to move around:In solids, atoms are not fixed at its position but constantly moves (oscillates) . So, Diffusion is difficult in solids due to bonding and requires, most of the time, external energy to mobilize the atoms.
5Mass transport can generally involve:
fluid flow dominant in gases or liquids viscous flow flow of a viscous material, generally amorphous or semi crystalline (e.g. glasses and polymers) due to the forces acting on it at that moment;atomic diffusion principal mechanism in solids and in static liquids (as occurs in solidification).
Atomic diffusion occurs during important processes such as:solidification of materials .precipitation hardening, e.g. Al-Cu alloys annealing of metals to reduce excess vacancies & dislocations formed during workingmanufacture of doped silicon, e.g. as used in many electronic devices
For an active diffusion to occur, the temperature should be high enough to overcome energy barriers to atomic motionfor atom to jump into a vacancy site, it needs enough energy (thermal energy) to break the bonds and squeeze through its neighbors and take the new position. The energy necessary for motion is Em called the activation energy for vacancy motion.At activation energy Em has to be supplied to the atom so that it could break inter-atomic bonds and to move into the new position.
Figure: Schematic representation of the diffusion of an atom from its original position into a vacant lattice siteInhomogeneous materials can become homogeneous by diffusion.
12Diffusion MechanismsHow do atoms move between atomic sites?For diffusion to occur:Adjacent site needs to be empty (vacancy or interstitial).Sufficient energy must be available to break bonds and overcome lattice distortion.There are many diffusion mechanism to be observed but two possible mechanisms are considered:Vacancy diffusion.Interstitial diffusion.
131. Substitutional DiffusionDirect ExchangeRing Vacancy 2. Interstitial Diffusion
15 Vacancy Mechanism Atoms can move from one site to another if there is sufficient energy present for the atoms to overcome a local activation energy barrier and if there are vacancies present for the atoms to move into. The activation energy for diffusion is the sum of the energy required to form a vacancy and the energy to move the vacancy.
Vacancy diffusion- An atom adjacent to a vacant lattice site moves into it.
Essentially looks like an interstitial atom: lattice distortionFirst, bonds with the neighboring atoms need to be broken
From Callister 6e resource CD.To jump from lattice site to lattice site, atoms need energy to break bonds with neighbors, and to cause the necessary lattice distortions during jump. This energy comes from the thermal energy of atomic vibrations (Eav ~ o CT)Materials flow (the atom) is opposite the vacancy flow direction.
Interstitial atoms like hydrogen, helium, carbon, nitrogen, etc) must squeeze through openings between interstitial sites to diffuse around in a crystal.
The activation energy for diffusion is the energy required for these atoms to squeeze through the small openings between the host lattice atoms.
18Interstitial DiffusionMigration from one interstitial site to another (mostly for small atoms that can be interstitial impurities: (e.g. H, C, N, and O) to fit into interstices in host.
Carbon atom in FerriteInterstitial diffusion is generally faster than vacancy diffusion because bonding of interstitials to the surrounding atoms is normally weaker and there are many more interstitial sites than vacancy sites to jump to.
20How do we quantify the amount or rate of diffusion? Flux (J): No of atom s diffusing through unit area per unit time OR Materials diffusion through unit area per unit time.
Measured empiricallyMake thin film (membrane) of known surface areaImpose concentration gradientMeasure how fast atoms or molecules diffuse through the membrane
M =massdiffusedtimeJ slope
21Temperature Dependence of the Diffusion Coefficient : Diffusion coefficient increases with increasing T.
= 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]DDoQdRTWith conc. gradient fixed, higher D means higher flux of mass transport.
22 Diffusivity increases with T. Experimental Data:
D has exp. dependence on TRecall: Vacancy does also!
Diffusion and Temperature
Note:pre-exponential [m2/s]activation energygas constant [8.31J/mol-K]
Ficks first law of diffusion
x1x2D diffusion coefficient(be careful of its unit)Rate of diffusion independent of timeFlux proportional to concentration gradient =
24Diffusivity -- depends on:1. Diffusion mechanism. Substitutional vs interstitial.2. Temperature. 3. Type of crystal structure of the host lattice. 4. Type of crystal imperfections. (a) Diffusion takes place faster along grain boundaries than elsewhere in a crystal. (b) Diffusion is faster along dislocation lines than through bulk crystal. (c) Excess vacancies will enhance diffusion.5. Concentration of diffusing species.
Microstructural Effect on Diffusion:If a material contains grains, the grains will act as diffusion pathways, along which diffusion is faster than in the bulk material.25
26Physical Aspect of DD is the indicator of how fast atom moves.In liquid state, D reaches similar level regardless of structure.In solid state, D shows high sensitivity to temperature and structure. Absolute temperature and Tm are what we should care about.
Example: At 300C the diffusion coefficient and activation energy for Cu in Si are
D(300C) = 7.8 x 10-11 m2/sQd = 41.5 kJ/mol
What is the diffusion coefficient at 350C?
DTemp = T
T1 = 273 + 300 = 573 K
T2 = 273 + 350 = 623 K
D2 = 15.7 x 10-11 m2/s
29 Steel plate at 7000C with geometry shown: Q: In steady-state, how much carbon transfers from the rich to the deficient side?Adapted from Fig. 5.4, Callister 6e.
Example: Steady-state Diffusion
Knowns: C1= 1.2 kg/m3 at 5mm (5 x 103 m) below surface.
C2 = 0.8 kg/m3 at 10mm (1 x 102 m) below surface.
D = 3 x10-11 m2/s at 700 C.
30 Concentration profile,C(x), changes with time.14 To conserve matter: Fick's First Law: Governing Eqn.:
Non-Steady-State DiffusionIn most real situations the concentration profile and the concentration gradient are changing with time. The changes of the concentration profile is given in this case by a differential equation, Ficks second law.Called Ficks second law
Fick's Second Law of Diffusion
In words, the rate of change of composition at position x with time, t, is equal to the rate of change of the product of the diffusivity, D, times the rate of change of the concentration gradient, dCx/dx, with respect to distance, x.
32 Copper diffuses into a bar of aluminum.15 General solution:"error function"Values calibrated in Table 5.1, Callister 6e.
Adapted from Fig. 5.5, Callister 6e.Example: Non Steady-State Diffusiont3>t2>t1Fig. 6.5: Concentration profiles nonsteady-state diffusion taken at three different timesC0=Before diffusionFor t=0, C=C0 at 0x
33Non-steady State DiffusionSample 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 tip: use Eqn.
t = 49.5 h x = 4 x 10-3 mCx = 0.35 wt%Cs = 1.0 wt%Co = 0.20 wt%
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 followszerf(z)0.900.7970z0.81250.950.8209
z = 0.93Now solve for D
To solve for the temperature at which D has above value, we use a rearranged form of Equation (5.9a);
from Table 5.2, for diffusion of C in FCC Fe Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol
Solution (cont.):T = 1300 K = 1027C
WhereD is the Diffusivity or Diffusion Coefficient ( m2 / sec )Do is the prexponential factor ( m2 / sec )Qd is the activation energy for diffusion ( joules / mole )R is the gas constant ( joules / (mole deg) )T is the absolute temperature ( K )Temperature Dependence of the Diffusion CoefficientOR
End of Lecture38
39Experimental Determination of Diffusion CoefficientTracer method Radioisotopic tracer atoms are deposited at surface of solid by e.g. electro deposition isothermal diffusion is performed for a given time t, often quartz ampoules are used (T 10 m; D>10-11 cm2/s Sputtering of surface: for small diffusion length (at low temperatures) 2nm 10m, for the range D = 10-21 10-12 cm2/s
40Experimental Determination of Diffusion Coefficient Example: Diffusion of Fe in Fe3Si From those figures thediffusion constant can bedetermined with an accuracyof a few percent Stable isotopes can be used as well, when high resolution SIMS is used This technique is moredifficult
42 Copper diffuses into a bar of aluminum. 10 hours processed at 600 C gives desired C(x). How many hours needed to get the same C(x) at 500 C?16
Result: Dt should be held constant. Answer:Note: values of D are provided.Key point 1: C(x,t500C) = C(x,t600C).Key point 2: Both cases have the same Co and Cs.
4317 The experiment: we recorded combinations of t and x that kept C constant.
Diffusion depth given by:
= (constant here)Diffusion Analysis
43Typically, we let the experiment described on slide 2 run during the class; at this point in the class, the results are available for the class to look at.
44Non steady-state diffusionFrom Ficks 1st Law:
Take the first derivative w.r.t. x:
Conservation of mass:i.e. flux to left and to right has to correspond to concentration change.
Sub into the first derivative:
Ficks 2nd law
c = conc. inside boxPartial differential equation. Well need boundary conditions to solveIn most practical cases steady-state conditions are not established, i.e. concentration gradient is not uniform and varies with both distance and time. Lets derive the equation that describes non steady-state diffusion along the direction x.
45EX: NON STEADY-STATE DIFFUSION Copper diffuses into a bar of aluminum (semi infinite solid). General solution:"error function"Values calibrated in Table 5.1, Callister 6e.
Adapted from Fig. 5.5, Callister 6e.From Callister 6e resource CD.CoCs
At to, C = Co inside the Al bar
toAt t > 0, C(x=0) = Cs and C(x=) = Co
If it is desired to achieve a specific concentration C1
constantwhich leads to:
Known for given system
Specified with C1
48PROCESSING QUESTION Copper diffuses into a bar of aluminum. 10 hours at 600C gives desired C(x). How many hours would it take to get the same C(x) if we processed at 500C?
Result: Dt should be held constant. Answer:Note: valuesof D areprovided here.Key point 1: C(x,t500C) = C(x,t600C).Key point 2: Both cases have the same Co and Cs.
Adapted from Callister 6e resource CD.
49Diffusion: Design ExampleDuring a steel carburization process at 1000oC, there is a drop in carbon concentration from 0.5 at% to 0.4 at% between 1 mm and 2 mm from the surface (g-Fe at 1000oC).
Estimate the flux of carbon atoms at the surface.Do = 2.3 x 10-5 m2/s for C diffusion in g-Fe.Qd = 148 kJ/molrg-Fe = 7.63 g/cm3AFe = 55.85 g/mol
If we start with Co = 0.2 wt% and Cs = 1.0 wt% how long does it take to reach 0.6 wt% at 0.75 mm from the surface for different processing temperatures?
50T (oC)t (s)t (h)3008.5 x 10112.4x108900106,40029.695057,20015.9100032,3009.0105019,0005.3
Need to consider factors such as cost of maintaining furnace at different T for corresponding times.
27782 yrs!Diffusion: Design Example Contd
51A Look at Diffusion Bonding
52IntroductionDiffusion bonding is a method of creating a joint between similar or dissimilar metals, alloys, and nonmetals.
Two materials are pressed together (typically in a vacuum) at a specific bonding pressure with a bonding temperature for a specific holding time.
Bonding temperatureTypically 50%-70% of the melting temperature of the most fusible metal in the compositionRaising the temperature aids in the interdiffusion of atoms across the face of the joint.
53How does diffusion bonding work?Bonding pressureForces close contact between the edges of the two materials being joined.Deforms the surface asperities to fill all of the voids within the weld zone .Disperses oxide films on the materials, leaving clean surfaces, which aids the diffusion and coalescence of the joint.
54How does diffusion bonding work?Holding TimeAlways minimizedMinimizing the time reduces the physical force on the machinery.Reduces cost of diffusion bonding process.Too long of a holding time might leave voids in the weld zone or possibly change the chemical composition of the metal or lead to the formation of brittle intermetallic phases when dissimilar metals or alloys are being joined.
55How does diffusion bonding work?Sequence for diffusion bonding a ceramic to a metala) Hard ceramic and soft metal edges come into contact.b) Metal surface begins to yield under high local stresses.c) Deformation continues mainly in the metal, leading to void shrinkage.d) The bond is formed
56Advantages of diffusion bondingProperties of parent materials are generally unchanged.
Diffusion bonding can bond similar or dissimilar metals and nonmetals.
The joints formed by diffusion bonding are generally of very high quality.
The process naturally lends itself to automation.
Does not produce harmful gases, ultraviolet radiation, metal spatter or fine dusts.
Does not require expensive solders, special grades of wires or electrodes, fluxes or shielding gases.
58Summary IIDiffusion is just one of many mechanisms for mass transport.
Electrical field can produce mass transport.
Magnetic field can produce mass transport.
Combination of fields can produce mass transport such as electrochemical transport.
60Application: Homogenization time
Solidification usually results in chemical heterogeneities Represent it with a sinusoid of wavelength, Composition should homogenize when, x > /2 The approximate time necessary is:
Homogenization time- increases with 2- decreases exponentially with T
61Application:Service Life of a Microelectronic Device
Microelectronic devices have built-in heterogeneities Can function only as long as these doped regions survive To estimate the limit on service life, ts Let doped island have dimension, Device is dead when, x ~ /2, hence
Service life- decreases with miniaturization (2)- decreases exponentially with T
62Influence of Microstructure on Diffusivity Interstitial species Usually no effect from microstructure Stress may enhance diffusion
Substitutional species Raising vacancy concentration increases D Quenching from high T Solutes Irradiation Defects provide short-circuit paths Grain boundary diffusion Dislocation core diffusion
63Adding Vacancies Increases D
Quench from high T Rapid cooling freezes in high cv D decreases as cv evolves to equilibrium
Add solutes that promote vacancies High-valence solutes in ionic solids Mg++ increases vacancy content in Na+Cl- Ionic conductivity increases with cMg Large solutes in metals Interstitials in metals
Processes that introduce vacancies directly Irradiation Plastic deformation
64Grain Boundary Diffusion
Grain boundaries have high defect densities Effectively, vacancies are already present QD ~ Qm
Grain boundaries have low cross-section Effective width = Areal fraction of cross-section:
67Diffusion Thermally Activated Process (I)In order for atom to jump into a vacancy site, it needs to posses enough energy (thermal energy) to break the bonds and squeeze through its neighbors. The energy necessary for motion, Em, is called the activation energy for vacancy motion.
At activation energy Em has to be supplied to the atom so that it could break inter-atomic bonds and to move into the new position.
Schematic representation of the diffusion of an atom from its original position into a vacant lattice site.
68Diffusion Thermally Activated Process (II)The average thermal energy of an atom (kBT = 0.026 eV for room temperature) is usually much smaller that the activation energy Em (~ 1 eV/atom) and a large fluctuation in energy (when the energy is pooled together in a small volume) is needed for a jump.
The probability of such fluctuation or frequency of jumps, Rj, depends exponentially from temperature and can be described by equation that is attributed to Swedish chemistArrhenius :
where R0 is an attempt frequency proportional to the frequency of atomic vibrations.
69Diffusion Thermally Activated Process (III)For the vacancy diffusion mechanism the probability for any atom in a solid to move is the product of the probability of finding a vacancy in an adjacent lattice site (see Chapter 4):
and the probability of thermal fluctuation needed to overcome the energy barrier for vacancy motion
The diffusion coefficient, therefore, can be estimated as
Temperature dependence of the diffusion coefficient, follows the Arrhenius dependence.
70Diffusion Temperature Dependence (I)
Diffusion coefficient is the measure ofmobility of diffusing species.
D0 temperature-independent preexponential (m2/s)Qd the activation energy for diffusion (J/mol or eV/atom)R the gas constant (8.31 J/mol-K or 8.62x10-5 /atom-KT absolute temperature (K)
The above equation can be rewritten as
The activation energy Qd and preexponential D0, therefore, can be estimated by plotting lnD versus 1/T or logD versus 1/T. Such plots are Arrhenius plots.
71Diffusion Temperature Dependence (II)
Graph of log D vs. 1/T has slop of Qd/2.3R, intercept of ln Do
72Diffusion Temperature Dependence (III)
Arrhenius plot of diffusivity data for some metallic systems
73Diffusion of different species
Smaller atoms diffuse more readily than big ones, and diffusion is faster in open lattices or in open directions
74Diffusion: Role of the microstructure (I)
Self-diffusion coefficients for Ag depend on the diffusion path.
In general, the diffusivity is greater through lessrestrictive structural regions grain boundaries, dislocation cores, external surfaces.
75Diffusion: Role of the microstructure (II)
The plots (opposite) are from the computer simulation by T. Kwok, P. S. Ho, and S. Yip.
Initial atomic positions are shown by the circles, trajectories of atoms are shown bylines.
We can see the difference between atomic mobility in the bulk crystal and in the grain boundary region.
76Exercise1. A thick slab of graphite is in contact with a 1mm thick sheet of steel. Carbon steadily diffuses through the steel at 925C. The carbon reaching the free surface reacts with CO2 gas to form CO, which is then rapidly pumped away.
Determine the carbon concentration, C2, adjacent to the free surface, and the find the carbon flux in the steel, given that the reaction velocity for C+CO22CO is =3.010-6cm/sec.
At 925C, the solubility of carbon in the steel in contact with graphite is 1.5wt% and the diffusivity of carbon through steel is D=1.710-7cm2/sec. The equilibrium solubility of carbon in steel, Ceq, is 0.1wt% for the CO/CO2 ratio established at the surface of the steel.
The Pclet number isNote: The value of the Pclet number suggests mixed kinetic behavior is expected.
The carbon concentration in the steel at the free surface, C2, is
The steady-state flux is Jss = 1.51 10-6 [wt% C cm/s]Jss = 1.18 10-7 [g/ cm2-s]Divide by the density of steel, =12.8 cm3/100g to obtain the steady-state flux of carbon
78Exercise2. Two steel billetsa slab and a solid cylindercontain 5000ppm residual H2 gas. These billets are vacuum annealed in a furnace at 725C for 24 hours to reduce the gas content. Vacuum annealing is capable of maintaining a surface concentration in the steel of 10ppm H2 at the annealing temperature.
Estimate the average residual concentration of H2 in each billet after vacuum annealing, given that the diffusivity of H in steel at 725C is DH=2.2510- 4 cm2/sec.
15 cm15 cm2h = 10 cm2h = 10 cm
Rectangular and cylindrical slabs of steel10 cm
80Given:t=24 hr=86400 sCo= Initial Concentration= 5000 ppmCs= Surface concentration= 10 ppmDH= 2.25x10-4 cm2/sC1= average residual concentration=?We know that:(C1-Co) / (Cs-Co) = Constant(z) and also X (Dt) or x = Constant x (Dt) or Constant(z) = (Dt) / x2Now we can write:(C1-Co) / (Cs-Co) = (Dt) / x2or C1= (Co-Cs) x (f) + Cs
Therefore, For slab:C1= (Co-Cs) x (flong x fshort x fshort) + Csand For Cylinder:C1= (Co-Cs) x (flong x fshort) + Cs
83STRUCTURE & DIFFUSIONDiffusion FASTER for...
open crystal structures
lower melting T materials
materials with secondary bonding
smaller diffusing atoms
lower density materialsDiffusion SLOWER for...
higher melting T materials
materials with covalent bonding
larger diffusing atoms
higher density materials
84Factors that Influence Diffusion: Summary" Temperature - diffusion rate increases very rapidly with increasing temperature" Diffusion mechanism - interstitial is usually fasterthan vacancy" Diffusing and host species - Do, Qd is different forevery solute, solvent pair" Microstructure - diffusion faster in polycrystalline vs. single crystal materials because of the accelerated diffusion along grain boundaries and dislocation cores.
85Concepts to rememberDiffusion mechanisms and phenomena.Vacancy diffusion.Interstitial diffusion.Importance/usefulness of understanding diffusion (especially in processing).Steady-state diffusion.Non steady-state diffusion.Temperature dependence.Structural dependence (e.g. size of the diffusing atoms, bonding type, crystal structure etc.).