CHAPTER 4: IMPERFECTIONS IN SOLIDS

Diffusion in GasDIFFUSION: DEFINITIONSDiffusion is a process of mass transport that involves the atomic or molecular motion.In the simplest form, the diffusion can be defined as the random walk of an ensemble of particles from regions of high concentration to regions of lower concentration

Diffusion: Driving Force

In each diffusion process (heat flow, for example, is also a diffusion process ),the flux (of mater, heat, electricity, etc.) follows the general relation:

Flux = (Conductivity) x (Driving Force) In the case of atomic or molecular diffusion, the conductivity is referred to as thediffusivity or the diffusion constant, and is represented by the symbol D, which reflects the mobility of the diffusing species in the given environment. Accordingly one can assume larger values in gases, smaller ones in liquids, and extremely small ones in solids.

The driving force for many types of diffusionis the existence of a concentration gradient. The term gradient describes the variation ofa given property as a function of distance in the specified direction. Diffusion Flux (J) defines the mass transfer rate:

Directional Quantity Flux can be measured for: --vacancies --host (A) atoms --impurity (B) atoms

MODELING DIFFUSION: FLUXSTEADY-STATE DIFFUSION (Ficks First Law)

- Ficks first law

SSD takes place at constant rate. It means that throughout thesystem dC/dx=const and dC/dt=0DCDxDC/ Dx dC/dxThe diffusion flux is proportionalto the existing concentration gradientFIRST FICKs LAW

JxJyJzeyezexwhere D is the diffusion constant or diffusivity [m2/s] andC is a concentration [kg/m3]D=D0exp(-Qd/RT)here Q is the activation energy for the process: [J/mol];Do is temperature-independent pre-exponential constant: [m2/s] Concentration profile, C(x), changes with time: dC/dt0.NONSTEADY-STATE DIFFUSION

Example: diffusion from a finite volume through a membrane into a finite volume. The pressures in the reservoirs involved change with time as does, consequently, the pressure gradient across the membrane.

FICKs SECOND LAW

xConsider a volume element (between x andx+dx of unit cross section) in a diffusion system.The flux of a given material into a volume element (Jx) minus the flux out of the elementvolume (Jx+dx) equals the rate of material accumulation in the volume:

Using a Taylor series we can expand Jx+dx:Accordingly, as dx0 and using Ficks First Law:

And if D does not vary with x we have The formulation of Ficks Second Law:

Copper diffuses into a semi-infinite bar of aluminum. General solution:

EX: NON STEADY STATE DIFFUSIONBoundary conditions:For t=0, C=C0 at 0 x For t>0, C=Cs at x=0C=C0 at x =

C(x,t)-CoCs-Co=1-erfx2Dt Error FunctionThe terms erf and erfc stand for error function and complementary error function respectively - it is the Gaussian error function as tabulated (like trigonometric and exponential functions) in mathematical tables. (see Table 5.1, Callister 6e).

Its limiting values are:

THERMALLY ACTIVATED PROCESSES Arrhenius Equation Rate = Kexp(-Q/RT)

where K is a pre-exponential constant (independent of temperature),Q- the activation energy, R the universal gas constant and T the absolute temperatureExamples: rate of creep deformation, electrical conductivity in semiconductors, the diffusivity of elements in metal alloyEnergyAtom must overcome an activation energy q, to move from one stable position to anotherMechanical analog: the box must overcome an increase in potential energy DE, to move from one stable position to anotherTable 1

Table 2

First Ficks Law

Convection Stephans FluxDiffusion FluxBinary system

Maxwell-Stefan formula

19Josef Stefans evaporation diffusion tubeA vertical tube is partly filled with a liquid, which in turn evaporated and the vapor flowed out of the tube at the open end. The tube portion above the liquid level contained a (binary) mixture consisting of the surrounding gas (air) and the vapor, generated on the liquidgas interface. Due to evaporation, the liquid level fell, and the process was unsteady, even if all other parameters were kept constant. Under these conditions, Stefan derived equations for calculating the concentration distribution along tube length and the evaporation rate of the liquid.

The Stefan solution (1)Stefan started by applying the momentum and continuity equations to diffusion in the gaseous area above the liquid level. The gas mixture and its components were treated as ideal gases, and the total pressure and temperature are constant through out the whole system. The origin of the coordinate x is the plane of the upper tube end, and the position of the liquid level is denoted by L. The gas mixture contained two components, and the Stefan equations describing the diffusion process are:

0The indices1and 2 refer to the mixture components1(vapor) and2 (gas); t is the time,while A12 and A21 represent the resistance coefficients.

(14)(15)

The Stefan solution (2)Adding equations (15) and accounting (16):

0

Adding equations (14): (16)

(17)

(18)Adding equations (18):

and considering that the interface is impermeable to gas(component 2) and the liquid does not move, so c 2L=0, and u1L =0 , hence

(19)2GBoundary condition Eq. (14) for component 1 is written as:

And substituting c2u2 term from (20)

We can rewrite (14) as:

where, for reason of simplicity we have omitted some indices, and set:

Note that Eq. (20) is obtained from the boundary conditions at the interface, but due to Eq. (17) it is valid in the whole diffusion space.

(20)

(17)

(21)

The Stefan solution (3)Differentiating Eq.(21) with respect to x and combining with Eq. (15) for component 1 gives the relation :

With the boundary conditions:

This is known as the non-stationary Stefan diffusion equation

(21a)

(15)

(21)(22)Additional Boundary Conditions(22)

(21)

(23)

(20)

(24)(25)The Stefan solution (4)

0So we have:(21)

(25)

Stefan gave the solution of Eq. (21) in the form:

where the constants B and a are to be determined from the boundary conditions at x=L, and (26) becomes: and with (25)

And thus

(26)

(27)The Stefan solution (4)So we have:Stefan then proceeded to discuss the case when the molar liquid density was much larger than the gas density, namely, cL>>c, and b=(cL-c)/c becomes very large. He simplified the integral (27),

From this equation the following expression for the position of the interface was deduced :

Stefan compared this expression with an expression he derived in an earlier paper (Stefan, 1873) under the assumption of a constant evaporation rate (fixed interface):

Eq. (27) rests on the assumption c10=0, that is, p10=0. Assuming in this equation cL>>c1L, cL>>c1I, it becomes identical with Eq.(28). (27)

(27)

(28)Slattery & Mhetar SolutionLet us consider a vertical tube, partially filled with a pure liquid A. For time t < 0, this liquid is isolated from the remainder of the tube, which is filled with a gas mixture of A and B, by a closed diaphragm. The entire apparatus is maintained at a constant temperature and pressure (neglecting the very small hydrostatic effect). At time t = 0, the diaphragm is carefully opened, and the evaporation of A commences. Let us assume that A and B form an ideal-gas mixture. This allows us to say that the molar density c in gas phase is a constant throughout the gas phase. We also assume that B is insoluble in A.We wish to determine the concentration distribution of A in the gas phase as well as the position of the liquid-gas phase interface as functions of time.For simplicity, let us replace the finite gas phase with a semi-infinite gas that occupies all space corresponding to z2 > 0:

Z2Z30G(gas)L(liquid)at t =0 for all z2 > 0:

X (A)=X (Ao) (1)

X (A) - mole fraction of species A in the gas phase; X (Ao) - initial mole fraction of species A in the gas phase

The liquid gas phase interfaceis a moving plane

Z2 = h(t) (2)

and at z2 = h for all t > 0:

X(A)= X(A)e q. (3)Slattery & Mhetar Solution (2)Equations (1) and (2) suggest that we seek a solution to this problem of the form V1= V3 =0 V2= V2 (t, z2) (4)X(A)= X (A)(t, Z2) .V - molar averaged velocity of gas phase; V2 - Z2 component of the molar averaged velocityof gas phase. Since c can be taken to be a constant and since there is no homogeneous chemical reaction, the overall differential mass balance requires:

Z2Z30G(gas)L(liquid)U2V2Slattery & Mhetar Solution (3)Z2Z30G(gas)L(liquid)U2V2Slattery & Mhetar Solution (3)Slattery & Mhetar Solution (3)Slattery & Mhetar Solution (4)

Solution (5)

Solution (6)

(21)Notes:

(21)(22)Example (1)For Evaporation of MethanolT = 25,4 C p = 1.006 105 PaX (methanol)eq.= 0.172X(methanol)o = 0D (methanol,air) == 1.558 x 10 -5 m2/s(corrected from 1.325 10 -5 m2/s at 0C and 1 atm (Washburn, 1929, p. 62) using a popular empirical correlation (Reid et al., (1987) The Properties of Gases and Liquids, 4th Ed. McGraw-Hill, New York, U.S.A.1987, p. 587);c(L)= 24.6 kgmol/ 3 (Dean, J. A. (1979) Lange's Handbook of Chemistry, 12th ed. McGraw-Hill, New York, U.S.A. pp. 7 271. 10 89), c=0.0411 kg mol/m 3 (Estimated for air from Dean, 1979, p. 10- 92)).

The lower curve gives the position of the phase interface h (mm) as a function of t (s) for evaporation of methanol into air at T = 25.4C and p = 1.006 x 105 Pa. The upper curve is the same case derived by arbitrarily neglecting convection.Example (2)For Evaporation of Methyl Formate:T = 25,4 C p = 1.011 105 PaX (mformate)eq.= 0.784X(methanol)o = 0D (mformate,air) == 1.020 x 10 -5 m2/s(corrected from 0.872 10 -5 m2/s at C and 1 atm (Washburn, 1929, p. 62) using a popular empirical correlation (Reid et al., (1987) The Properties of Gases and Liquids, 4th Ed. McGraw-Hill, New York, U.S.A.1987, p. 587);c(L)= 16.1 kgmol/ 3 (Dean, J. A. (1979) Lange's Handbook of Chemistry, 12th ed. McGraw-Hill, New York, U.S.A. pp. 7 271. 10 89), c=0.0411 kg mol/m 3 (Estimated for air from Dean, 1979, p. 10- 92)).

The lower curve gives the position of the phase interface h (mm) as a function of t (s) for evaporation of methyl formate into air at T = 25.4C and p = 1.011 x 105 Pa. The upper curve is the same case derived by arbitrarily neglecting convection.