mass transport

EKC314:TRANSPORT PHENOMENACore Course for

B.Eng.(Hons.)-Chemical EngineeringSemester I (2014/2015)

Mohamad Hekarl Uzir, DIC.MSc.,PhD.(London)[email protected]

School of Chemical Engineering

Engineering Campus, Universiti Sains Malaysia

Seri Ampangan, 14300 Nibong Tebal

Seberang Perai Selatan, PenangEKC314-SCE – p. 1/57

Mass Transport

Diffusivity and the Mechanisms of Mass Transport:

The movement of one chemical species, A through abinary mixture of A and B due to a concentrationgradient of A is described by Fick’s Law of diffusion.

Fick’s Law is governed by the diffusivity constant, D inthe form given by;

mAy

A= ρDAB

xA0−0

Y

which can then be written as;

jAy = −ρDABdxA

dyEKC314-SCE – p. 2/57

Diffusivity and the Mechanisms ofMass Transport

Fick’s Law for Binary Diffusion (Molecular Mass Transport)

The term D is closely dependance on:temperature differencepressure difference

Consider a thin horizontal plate of area A andthickness Y . (to be discussed during lecture)

EKC314-SCE – p. 3/57



The given equation is the 1-dimensional form ofFick’s first law of diffusion

It is valid for any binary solid, liquid or gas solution,provided that jAy is defined as the mass flux relative tothe mixture velocity, vy.

During the operation, component A (gas) moves ratherslowly with a very small concentration. thus, vy isnegligibly different from 0, which is governed by;

vy = ωAvAy + ωBvBy

EKC314-SCE – p. 4/57



v is an average value with vA and vB, i.e. the massaverage velocity

The species velocity vA is not the instantaneousmolecular velocity of a molecule of A, but it is thearithmetic average velocities of all the molecules of Awithin a tiny volume element.

The mass flux is defined as;

jAy = ρωA(vAy − vy)

EKC314-SCE – p. 5/57



Similarly with the mass flux of component B, whichderivation leads into a conclusion that;

DAB = DBA

For the pair A-B, there is only ONE diffusivity with thefunction of pressure, temperature and composition.

EKC314-SCE – p. 6/57



The mass diffusivity can be correlated in the formsimilar to that of the momentum and thermal diffusivityas the Schmidt number;

Sc =ν

DAB=

µ

ρDAB

For gas mixture, Sc number: 0.2 to 3.0.

For liquid mixture, Sc number: ≥ 40, 000

EKC314-SCE – p. 7/57


Temperature and Pressure Dependence of Diffusivities

For binary gas mixture at low pressure, DAB is;1. inversely proportional to the pressure2. increased with the increase of temperature3. almost independent of the composition for the given

gas pair

An equation developed by combining the kinetic theoryand corresponding-states argument is given by;

pDAB

(pcApcB)1

3 (TcATcB)5

12

(

1MA

+ 1MB

)1

2

= a

(

T√TcATcB

)b

EKC314-SCE – p. 8/57


Temperature and Pressure Dependence of Diffusivities

Upon data analysis and correlation, the dimensionlessconstants obtained are:1. for non-polar gas pair: a = 2.745× 10−4 and

b = 1.823 (excluding He and H2)2. for pairs consisting of H2O and a non-polar gas:

a = 3.640× 10−4 and b = 2.334

If the gases A and B are non-polar and theirLennard-Jones parameters are known, thekinetic-theory method will give better accuracy.

EKC314-SCE – p. 9/57


Theory of Gas Diffusion in Gases at Low Density

For self-diffusion, DAA∗ the correlation is given by;

(cDAA∗)c = 2.96× 10−6

(

1

MA+

1

MA∗

)1

2 p2/3cA

T1/6cA

The above equation SHOULD NOT be used for He orH2 isotopes.

For binary-diffusion, the correlation expanded into theform of;

(cDAB)c = 2.96× 10−6

(

1

MA

+1

MB

)1

2 (pcApcB)1/3

(TcATcB)1/12EKC314-SCE – p. 10/57



Some results of the kinetic theory of gases were givenpreviously as;

the mean molecular speed relative to v;

u =

√

8κT

πm

the wall collision frequency per unit area in astationary gas;

Z =1

4nu

EKC314-SCE – p. 11/57



Some results of the kinetic theory of gases were givenpreviously as;

the mean free path;

λ =1√

2πd2n

EKC314-SCE – p. 12/57


Theory of Diffusion in Binary Liquids

The theory starts with the development of thehydrodynamic-theory from the Nernst-Einsteinequation given by;

DAB = κTuA

FA

where uA/FA is the mobility of a particle A[steady-state velocity of the particle attained under theaction of a unit force].

EKC314-SCE – p. 13/57



By applying the creeping flow equation of motion, withA is in spherical shape and slip condition applies, thefinal equation expands into;

uA

FA

=

(

3µB +RAβAB

2µB +RAβAB

)

1

6πµBRA

at the fluid-solid interface.

EKC314-SCE – p. 14/57



The limiting cases of βAB are of particular interest tothe system:1. βAB = ∞ (no-slip condition): At the fluid-solid

interface, the previous equation reduces intoStokes’s Law in the form given by;

DABµB

κT=

1

6πRA

EKC314-SCE – p. 15/57



The limiting cases of βAB are of particular interest tothe system:1. OR usually called the Stokes-Einstein equation. It

can be applied to the diffusion of a very largespherical molecules in solvents of low molecularweight and to suspended particles. It has also beenused to estimate the shapes of protein molecules.

2. βAB = 0 (complete slip condition): Similarly, theequation at the fluid-solid interface reduces into;

DABµB

κT=

1

4πRA

EKC314-SCE – p. 16/57



The limiting cases of βAB are of particular interest tothe system:2. If the molecules A and B are identical,

(self-diffusion) and they can be assumed to formcubic lattice with adjacent molecules, thus;

DAAµA

κT=

1

2π

(

NA

VA

)1

3

EKC314-SCE – p. 17/57



The formulae derived above only apply to dilutesolution of A and B.

EKC314-SCE – p. 18/57

Concentration Distributions in Solidsand in Laminar Flow

Shell Mass Balances; Boundary Conditions

1. The law of conservation of mass of species A in abinary system can be written over the volume of theshell as; (to be discussed)

2. A chemical species, ’A’ may leave or enter a system bydiffusion [molecular motion and convection], these areincluded in the term, NA.

3. Species ’A’ may also be produced or consumed byhomogeneous chemical reactions.

4. When the overall balance is complete, the shell is thenmade into an infinitesimally small thickness.

5. A system of differential equation is generated either inthe form of mass or molar flux.

EKC314-SCE – p. 19/57



6. The integration of the differential equation leads to anumber of constants which requires boundaryconditions for them to be determined.

(a) the concentration at surface can be specified:xA = xA0.

(b) mass flux at the surface can be specified:NAz = NA0. If the ratio of NA0/NAz is known, theconcentration gradient is already known.

(c) for diffusion occurs in a solid, at the solid surface,substance A is lost to the surrounding stream dueto;

NA0 = kc(cA0 − CAb)

EKC314-SCE – p. 20/57



6. The integration of the differential equation leads to anumber of constants which requires boundaryconditions for them to be determined.

(d) the rate of chemical reaction at the surface can bespecified.

EKC314-SCE – p. 21/57


Diffusion Through a Stagnant Gas Film

Consider a diffusion system where a liquid A isevaporating into gas A. Assuming that the liquidlevel is maintained at z = z1.

At the liquid-gas interface, the gas phase concentrationof A (in mole fraction) is xA1. This is the vapour

pressure of A divided by the total pressure, pvapA

p.

(Assuming that gas A and liquid B obey the idealgas mixture and that the solubility of gas B inliquid A is negligible)

A stream of gas mixture A-B with concentration xA2

flows past the top of the tube, maintaining the molefraction of A at xA2 for z = z2. EKC314-SCE – p. 22/57



System must be kept at constant pressure andtemperature and both gases A and B are assumed tobe ideal.

Neglecting the effect of dependency of gas velocity (ofthe z-component) with respect to the radius of thecontainer/cylinder.

At the steady state a balance equation of the formbelow is obtained;

NAz = −cDAB∂xA

∂z+ xA(NAz +NBz)

EKC314-SCE – p. 23/57



When NBz = 0, solving for NAz resulted into;

NAz = − cDAB

1− xA

dxA

dz

At a steady-state condition, for every increment of ∆z,the amount of A entering at plane z equals to theamount of A leaving at plane z +∆z, which leads to;

SNAz|z − SNAz|z+∆z = 0

where S is the cross-sectional area of thecolumn/cylinder.

EKC314-SCE – p. 24/57



Division by S∆z and taking limit as ∆z → 0 leads to;

−dNAz

dz= 0

Combining with the previous equation gives;

d

dz

(

cDAB

1− xA

dxA

dz

)

= 0

EKC314-SCE – p. 25/57



For an ideal gas mixture;

p = cRT

and for gases, DAB is nearly independent of thecomposition, thus;

d

dz

(

1

1− xA

dxA

dz

)

= 0

Upon integration gives;

1

1− xA

dxA

dz= C1

EKC314-SCE – p. 26/57



With further integration leads to;

− ln (1− xA) = C1z + C2

By replacing C1 with − lnK1 and C2 with − lnK2,reduces the equation into;

1− xA = Kz1K2

EKC314-SCE – p. 27/57



Using boundary conditions:i. B.C. 1: at z = z1, xA = xA1

ii. B.C. 2: at z = z2, xA = xA2

the equation becomes;

(

1− xA

1− xA1

)

=

(

1− xA2

1− xA1

)

z−z1z2−z1

The profile for gas B can be determined usingxB = 1− xA

The slope of the profile, dxA/dz is not constantalthough NAz is.

EKC314-SCE – p. 28/57



The equation determined above can be used to obtainthe average values and mass fluxes at surfaces.

The average concentration of B in the region betweenz1 and z2 can be found using;

xB,avg

xB1

=

∫ z2z1( xB

xB1

)dz∫ z2z1

dz=

∫ 1

0(xB2

xB1

)ζdζ∫ 1

0dζ

=(xB2

xB1

)ζ

ln (xB2

xB1

)

∣

∣

∣

∣

∣

1

0

where ζ = (z − z1)/(z2 − z1) or can be rewritten as;

xB,avg =xB2 − xB1

ln xB2

xB1

EKC314-SCE – p. 29/57



With the previous equation, the mass transfer at theliquid-gas interface (rate of evaporation) can beobtained using;

NAz

∣

∣

∣

∣

∣

z=z1

=cDAB

1− xA1

dxA

dz

∣

∣

∣

∣

∣

z=z1

=cDAB

xB1

dxB

dz

∣

∣

∣

∣

∣

z=z1

=cDAB

z2 − z1ln

(

xB2

xB1

)

EKC314-SCE – p. 30/57



By combining the mass transfer equation with that ofthe logarithmic mean given by, xB,avg leads to;

NAz

∣

∣

∣

z=z1=

cDAB

(z2 − z1)(xB)ln(xA1 − xA2)

EKC314-SCE – p. 31/57


Diffusion with a Heterogeneous Chemical Reaction

Involved solid catalyst with either gas or liquid phasereactant.

Reaction only occurs at the catalyst surface whenreactant(s) diffuses towards the surface (externaldiffusion) OR diffuses into the porous catalyst (forinternal diffusion)

For a reaction involving a component A (reactant)producing component B (product), component A needsto diffuse into the surface of the catalyst at which thereaction will occur and the product B formed from thereaction will diffuse back out. The reaction is assumedto occur instantaneously.

EKC314-SCE – p. 32/57



Also assuming that the gas film is isothermal at thispoint.

For a reaction given by;

2A → B

at steady-state condition;

NBz = −1

2NAz

EKC314-SCE – p. 33/57



The substituted equations give;

NAz = − cDAB

1− 12xA

dxA

dz

Consider a thin slab of thickness ∆z in the gas film.The balanced equation gives;

dNAz

dz

EKC314-SCE – p. 34/57



Substitute with the above equation leads to;

d

dz

(

1

1− 12xA

dxA

dz

)

= 0

Upon integration w.r.t z resulted into;

−2 ln (1− 1

2xA) = C1z + C2 = −(2 lnK1)z − (2 lnK2)

EKC314-SCE – p. 35/57



Using boundary conditions of:1. B.C. 1: at z = 0 and xA = xA0

2. B.C. 2: at z = δ and xA = 0

Leads into;

(1− 1

2xA) = (1− 1

2xA0)

1− zδ

which then gives the molar flux of reactant through thefilm;

NAz =2cDAB

δln

(

1

1− 12xA0

)

EKC314-SCE – p. 36/57


Diffusion with a Homogeneous Chemical Reaction

Consider a reaction involving component A (gasphase) and B (liquid phase) with reaction following,A(g) + B(l) → AB(l)

Assuming that the formation of AB does not affect thediffusion process (pseudo-binary assumption) andupon mass balance on species A over a thickness ∆zof the liquid phase;

NAz

∣

∣

zS −NAz

∣

∣

z+∆zS − k

′′′

1 cAS∆z = 0

where k′′′

1 is the first order rate constant for thedecomposition of A, with S as the cross-sectional areaof the liquid. EKC314-SCE – p. 37/57



The division by S∆z and taking limit ∆z → 0 gives;

dNAz

dz+ k

′′′

1 cA = 0

For a very small concentration of A, the byapproximation,

NAz = −DABdcAdz

EKC314-SCE – p. 38/57



And therefore leads to;

DABd2cAdz2

− k′′′

1 cA = 0

The above equation can be solved using the boundaryconditions:1. B.C. 1: at z = 0 and cA = cA0

2. B.C. 2: at z = L and NAz = 0 OR dcAdz

= 0

EKC314-SCE – p. 39/57



With the dimensionless variable as;

d2Γ

dζ2− φ2Γ = 0

the dimensionless variables include:1. dimensionless concentration: Γ = cA

cA0

2. dimensionless length: ζ = zL

3. dimensionless Thiele Modulus: φ =

√

k′′′

1L2

DAB

EKC314-SCE – p. 40/57



Using the necessary boundary conditions:1. B.C. 1: at ζ = 0, Γ = 1

2. B.C. 2: at ζ = 1, dΓdζ

= 0

which then gives the solution of the form of;

Γ = C1 coshφζ + C2 sinhφζ

Solving for the constants leads to;

Γ =coshφ coshφζ − sinhφ sinhφζ

coshφ=

cosh [φ(1− ζ)]

coshφ

EKC314-SCE – p. 41/57



Substitute back to the original notation resulted into;

cAcA0

=cosh

√

k′′′

1L2

DAB(1− z

L)

cosh

√

k′′′

1L2

DAB

The above equation can be used to determine theaverage concentration in the liquid phase;

cA,avg

cA0

=

∫ L

0( cAcA0

)dz∫ L

0dz

=tanhφ

φ

EKC314-SCE – p. 42/57



The molar flux at the plane z = 0 can also be foundusing;

NAz

∣

∣

∣

z=0= −DAB

dcAdz

∣

∣

∣

z=0=

(

cA0DAB

L

)

φ tanhφ

EKC314-SCE – p. 43/57


Diffusion into a Falling Liquid Film (Gas Absorption)

Consider a system of forced-convection mass transferin which viscous flow and diffusion occur under suchconditions that the velocity field can be considered asvirtually unaffected by the diffusion.

Specifically, consider consider the absorbtion of gas Aby a laminar falling film of liquid B.

Material A is only slightly soluble in B, so that theviscosity of the liquid is unaffected.

The diffusion also takes place very slowly in the liquidfilm that component A (gas) will not penetrate very farinto the film (the penetration distance will be small incomparison with the film thickness)

EKC314-SCE – p. 44/57



Let the system is the absorption of O2 in H2O.

Consider the momentum transfer of the falling film(refer to momentum transfer example), which resultedinto the velocity profile in z-direction given by;

vz(x) = vmax

[

1−(x

δ

)2]

ignoring the end effect

EKC314-SCE – p. 45/57



Consider mass balance on component A that changesw.r.t thickness ∆x as well as ∆z which resulted into;

NAz

∣

∣

∣

zW∆x−NAz

∣

∣

∣

z+∆zW∆x

+NAx

∣

∣

∣

xW∆z −NAx

∣

∣

∣

x+∆xW∆z = 0

Dividing the above equation with W∆x∆z and theusual limiting process as volume element becomes→ 0 gives;

∂NAz

∂z+

∂NAx

∂x= 0

EKC314-SCE – p. 46/57



But, NAz and NAx are given by;

NAz = −DAB∂cA∂z

+ xA(NAz +NBz) ≈ cAvz(x)

and

NAx = −DAB∂cA∂x

+ xA(NAx +NBx) ≈ DAB∂cA∂x

respectively.

EKC314-SCE – p. 47/57



Upon substitution gives;

vmax

[

1−(x

δ

)2]

∂cA∂z

= DAB∂2cA∂x2

with boundary conditions;1. B.C. 1: at z = 0, cA = 0

2. B.C. 2: at x = 0, cA = CA0

3. B.C. 3: at x = δ, ∂cA∂x

= 0

EKC314-SCE – p. 48/57



The first B.C.: the film consists of pure B at the top(z = 0)

The second B.C.: at the liquid-gas interface, theconcentration of A is determined by the solubility of Ain B (that is cA0)

The third B.C.: A cannot diffuse through the solid wall

Due to this reasons, the equation needs to be modifiedsuch that the B.C. is valid thus the new equationbecomes;

vmax

∂cA∂z

= DAB∂2cA∂x2

EKC314-SCE – p. 49/57



With the new boundary conditions;1. B.C. 1: at z = 0, cA = 0

2. B.C. 2: at x = 0, cA = CA0

3. B.C. 3: at x = ∞, cA = 0

By applying the method of combination of variablesleads to;

cAcA0

= 1− 2√π

∫

x√

4DABzvmax

0

exp (−ξ)2dξ

EKC314-SCE – p. 50/57



Or it can also be written in the form of;

cAcA0

= 1− erfx

√

4DABzvmax

= erfcx

√

4DABzvmax

This will give the local mass flux at the gas-liquidinterface using;

NAx

∣

∣

∣

x=0= −DAB

∂cA∂x

∣

∣

∣

x=0= cA0

√

DABvmax

πz

EKC314-SCE – p. 51/57



Then the total molar flow of A across the surface atx = 0 is given by;

WA =

∫ W

0

∫ L

0

NAx

∣

∣

∣

x=0dzdy

= WLcA0

√

4DABvmax

πL

EKC314-SCE – p. 52/57


Diffusion into a Falling Liquid Film (Solid Dissolution)

Consider a liquid B flowing in laminar motion down avertical wall.

The film begins far enough up the wall such that vzdepends only on y for z ≥ 0.

For 0 ≤ 0 ≤ L, the wall is made of a species A that isslightly soluble in B.

For a short distances downwards, species A will notdiffuse very far into the falling film.

A is present only in a very thin boundary layers nearthe solid surface.

EKC314-SCE – p. 53/57



The diffusing A molecules will experience a velocitydistribution similar to that of the falling film next to thewall at y = 0 where;

vz =ρgδ2 cos β

2µ

[

1−(x

δ

)2]

When cosβ = 1 and x = δ − y, thus;

vz =ρgδ2

2µ

[

1−(y

δ

)2]

=ρgδ2

2µ

[

2(y

δ

)

−(y

δ

)2]

EKC314-SCE – p. 54/57



At and adjacent to the wall;

(y

δ

)2

≪(y

δ

)

therefore, the velocity vz ca be approximated to;

vz =(ρ)

EKC314-SCE – p. 55/57

Equations of Change forMulticomponent Systems

Equation of Continuity for a Multicomponent Mixture:

To establish the equation of continuity for variousspecies in a multicomponent mixture (using massbalance)

To obtain diffusion equations (in various forms) byinserting the mass flux equation of continuity

To combine all equations of change for mixture forproblem solving.

EKC314-SCE – p. 56/57

Equations of Change forMulticomponent Systems

Equation of Continuity for a Multicomponent Mixture:

i. rate of increase of mass, α in the volume element,(

∂ρα∂t

)

∆x∆y∆z

ii. rate of addition of mass, α across face at x, nαx|x∆y∆z

iii. rate of removal of mass, α across face at x+∆x,nαx|x+∆x∆y∆z

iv. rate of production of mass, α by chemical reaction,rα∆x∆y∆z

EKC314-SCE – p. 57/57

mass transport

Documents

mechanisms of mass transport

mass transportdiffusivity

diffusivity constant

ficks law of diffusion

mass flux of component

law of diffusionit

dab dxadyekc314sce

dabfor gas mixture