Chapter 2

Chapter 2 Principles of Unsteady-State and Convective Mass Transfer

JUSTDepartment of Chemical Engineering Mass Transfer – ChE 461 Chapter 1-2

IntroductionIn previous chapter, we considered various mass transfer systems where the concentration or partial pressure at any point and the diffusion flux were constant with time, hence at steady state.

Before steady state can be reached, time must elapse after the mass transfer process is initiated for the unsteady-state conditions to disappear.

In heat transfer course, an unsteady-state equation was derived,

2

2

xT

tT

∂∂

=∂∂ α

In the same manner, an unsteady state mass transfer equation can be derived by applying a mass balance on component A.


x x + ΔxΔxΔz

Δy

Rate of inputxAxN

Consider the elemental solid shown in the drawing where A is diffusing the x-direction Rate of output

xxAxN ΔMass balance on component A in terms of moles without generation:

Rate of input = rate of output + rate of accumulation

xAxN inpute of rate =x

AAB x

cD∂∂

−=

xxAxN Δ+= output of ratexx

AAB x

cDΔ+∂

∂−=

Rate of accumulation is as follows for the volume ΔxΔyΔz m3:

tczyx A

∂∂

ΔΔΔ= )(onaccumulati of rate

Combining the three terms and dividing by ΔxΔyΔz:

⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∂∂

xcD

xtc A

ABA

xx

Ax

AAB

xx

AAB

xc

xxcD

xcD

Δ+

Δ+

∂∂

=Δ

∂∂−

∂∂

As Δx approaches zero:


2

2

xcD

tc A

ABA

∂∂

=∂∂For constant diffusivity DAB: cA = cA (x,t)

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

2

2

zc

yc

xcD

tc AAA

ABA

Similar derivation can be obtained for diffusion in all three direction:

Because of mathematical similarity between the equation for heat conduction, mathematical method used for solution of the unsteady-state heat-conduction can be used for unsteady-state mass transfer.


Diffusion in a Flat Plate with Negligible Surface Resistance

Consider the unsteady-state diffusion in the x direction for a plate of thickness 2x1.

x

c

0 x1 2x1

c0 @ t =0

c1 c2

c at t > 0(t = t)

For a very high mass-transfer coefficient outside the surface, resistance will be negligible and the concentration at the surface will be equal to that in the fluid, c1

Initial and boundary conditions:

c = c0 , t = 0, x = x,

⎥⎦

⎤⎢⎣

⎡⋅⋅⋅+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−−

=1

22

1

22

1

22

01

1

25sin

45exp

51

23sin

43exp

31

21sin

41exp

114

xxX

xxX

xxX

ccccY ππππππ

π

101

01 =−−

=ccccY

c = c1 , t = t, x = 0, 001

11 =−−

=ccccY

c = c1 , t = t, x = 2x1 , 001

11 =−−

=ccccY

01

1

ccccY

−−

=

Define dimensionless conc.:

2

2

xYD

tY

AB ∂∂

=∂∂

Solution is an infinite Fourier series:

where: 21xDtX =


Unsteady-State Diffusion in Various GeometriesSolution of unsteady-state mass transfer is time-consuming; convenient charts for various geometries are available.Previously, we discussed the case of negligible convective resistance at the surface. However, in many cases when a fluid is outside the solid, convective mass transfer is occurring at the surface.Convective mass-transfer coefficient (kc) is similar to convective heat transfer (h), and is defined as follows:

[compare with q = h (T1 – Ti )])( 1 LiLcA cckN −=

wherekc = mass transfer coefficient, in m/scL1 = bulk fluid concentration, in kg mol A/m3

cLi = fluid concentration just adjacent to the surface, in kg mol A/m3


Different interface conditions are shown in the drawing

Concentration drop across the fluid is cL1 – cLi

The concentration of the fluid adjacent to the surface, cLi, is related to the concentration of the fluid in the solid, ci, by the following equilibrium equation:

i

Li

ccK =

K: equilibrium distribution coefficient

(similar to Henry’s law coefficient for a gas and liquid)


1. Relation between mass and heat Transfer parameter.

Charts for unsteady heat transfer can be used to predict concentration in unsteady-state mass transfer by diffusion

2. Chart for diffusion in various geometries

• Semi infinite solid (Fig 5.3-3)• Flat plate (Fig 5.3-5/6)• Long cylinder (Fig 5.3-7/8)• Sphere (Fig 5.3-9/10)• Average concentration (Fig 5.3- 13)


Example: Unsteady-State Diffusion in a Slab of Agar GelA solid slab of 5.15 wt% agar gel at 278 K is 10.16 mm thick and contains a uniform

concentration of urea of 0.1 kgmol/m3. Diffusion is only in the x-direction through two parallel flat surfaces 10.16 mm apart. The slab is suddenly immersed in pure turbulent water, so the surface resistance can be assumed to be negligible; that is, the convective coefficient kc is very large. The diffusivity of urea in the agar is 4.72 × 10-

10 m2/s.1) Calculate the concentration at the midpoint of the slab (5.08 mm from the surface) and

2.54 mm from the surface after 10 h.2) If the thickness of the slab is halved, what would be the midpoint concentration in 10

h?


CONVECTIVE MASS-TRANSFER COEFFICIENT

In previous chapter emphasis molecular diffusion in stagnant fluids or fluids in laminar flow. In many cases, the rate of diffusion is slow and more rapid transfer is desired.So, to achieved it the fluid velocity is increased until turbulent mass transfer occurs.

To have fluid in convective flow usually requires the fluid to be flowing past another immiscible fluid or a solid surface.Example: flowing in a pipe

Laminar flow: fluid flows in streamlines and its behavior can usually be described mathematically

Turbulent flow: no streamlines and large eddies or “chunks” of fluid exist moving rapidly in seemingly random fashion


Three regions of mass transfer can be visualized when a solute is dissolving from a solid to a fluid:

Laminar sub-layer: a thin viscous sublayer which is adjacent to the surface

• Characterized by molecular diffusion• No eddies present• Large concentration drop

Transition or buffer zone: a region adjacent to the laminar layer where gradual transition from molecular diffusion to mainly turbulent at the end occurs.

• Some eddies present• Mass transfer is the sum of turbulent + diffusion

Turbulent region: adjacent to the buffer zone where most of the transfer is by turbulent with very small diffusion

• concentration decrease very small • eddies motion

Typical plot for mass transfer of a dissolving solid from a surface to a turbulent fluid in a conduit

cA1Solid surface

Laminar sub-layer

Transition zone

Fluideddies

cA2

z

z

cA

cA1

cA2


Definition of Mass-Transfer coefficient• Previously, for molecular diffusion in stagnant fluid and fluid in laminar flow:

dzdcDJ A

ABA −=* ( )BAAA

ABA NNcc

dzdxcDN ++−=

• For turbulent flow, mass transfer is increased by eddy diffusivity, ε

(m/s)

( )dzdcDJ A

ABA ε+−=*

• ε

varied with distance, average ε

will be used; J*A1 is normally used which is

flux of A on surface area A1 (since the cress sectional area may vary) relative to the whole bulk surface.

)( 2112

*1 AA

MABA cc

zzDJ −

−+

−=ε

In terms of convective mass transfer coefficient:

12

'

zzDk MAB

c −+

=ε

)( 21'*

1 AAcA cckJ −=

where k’c is mass transfer coefficient: [Kgmol/s.m2.(kgmol/m3)]

or [m/s]


Mass-transfer coefficient for equimolar

counterdiffusion

∴ NA = -NB

( ) ( )BAAA

MABA NNcc

dzdxDcN +++−= ε

12

'

zzDk MAB

c −+

=ε

)( 21'

AAcA cckN −=

Since

where

Defining equation for the mass-transfer coefficient

• Other definitions for mass transfer coefficient depending on concentrations:

Gases: )( 21'

AAcA cckN −= )( 21'

AAG ppk −= )( 21'

AAy cyk −=Liquids: )( 21

'AAcA cckN −= )( 21

'AAL cck −= )( 21

'AAx xxk −=

• Mass transfer coefficients are related to each other, eg.

)( 21'

AAcA cckN −= )( 21'

AAy cyk −= ⎟⎠⎞

⎜⎝⎛ −=

cc

cck AA

y21'

ckk yc /'' =

( )21

'

AAy cc

ck

−=

Hence, (Also see Table 7.2-1 with corresponding units)


Mass-transfer coefficient for A diffusing through stagnant, nondiffusing

B

∴ NB = 0

12

'

zzDk MAB

c −+

=ε)( 21

'

AABM

cA cc

xkN −= where

• Rewriting using other units:Gases: )( 21 AAcA cckN −= )( 21 AAG ppk −= )( 21 AAy cyk −=Liquids: )( 21 AAcA cckN −= )( 21 AAL cck −= )( 21 AAx xxk −=

• Again, mass transfer coefficients are related to each other, eg.

( ) )(/ 21'

AABMcA ccxkN −= )( 21 AAx xxk −= ⎟⎠⎞

⎜⎝⎛ −=

cc

cck AA

x21

ckxk xBMc //' =

( )21 AAx cc

ck

−=

Hence, (Also see Table 7.2-1 with corresponding units)

)()( 21

12AA

BM

avABA xx

xzzcDN −

−=Previously, the expression was derived

With eddy diffusivity, )()(

)(21

12AA

BM

avMABA xx

xzzcDN −

−+

=ε )(

)()(

2112

AABM

MAB ccxzz

D−

−+

=ε

or

kc ≡

Mass transfer coefficient for A diffusing through stagnant B

)( 21 AAc cck −=

)( 21

'

AABM

xA xx

xkN −=or )( 21 AAx xxk −=


Example: Vaporizing A and Convective Mass TransferA large volume of pure gas B at 2 atm pressure is flowing over a surface from which

pure A is vaporizing. The liquid A completely wets the surface, which is a blotting paper. Hence, the partial pressure of A at the surface is the vapor pressure of A at 298 K, which is 0.2 atm. The k’

y has been estimated to be 6.78 × 10-5 kgmol/s.m2.mol frac. Calculate NA , the vaporization rate, and also the values of ky anf kG .


Mass Transfer Coefficient for General Case of A and B diffusing and Convective flow using Film Theory

In this case mass transfer is assumed to occur through a thin film next to the wall of thickness δf and by molecular diffusion. The experimental value of k’

c for dilute solutions is used to determine δf:

( )BAAA

ABA NNxdzdxcDN ++−=

f

ABc

Dkδ

='

( ) ( )BAAA

MABA NNcc

dzdxDcN +++−= εSince

with molecular diffusion only:

Convective term

∫∫ +−−=

=

=

2

1 )(1

0

A

A

f x

x BAAA

Az

zAB NNxNdxdz

cD

δRearranging and integration:

( )( ) ⎥

⎦

⎤⎢⎣

⎡−+−+

+=

1

2'

//ln

ABAA

ABAAc

BA

AA xNNN

xNNNckNN

NN

• when NB = 0 ( )( )21' / AABMcA ccxkN −=⇒

xA1Solid surface

Fluid

xA2

z

δf


Mass Transfer Coefficient under High Flux ConditionsThe previous case assumes that film thickness is unaffected by high flux and bulk or convective flow (diffusion-induced convection)

Other definitions of the mass transfer coefficient which includes effect of diffusion-induced convection will be derived assuming stagnant nondiffusing B

( )BAAz

AABA NNx

dzdxcDN ++⎟

⎠⎞

⎜⎝⎛−=

=0 0Defining a mass-transfer coefficient in terms of the diffusion flux,

)( 210

0AAc

z

AAB xxck

dzdxcD −=⎟

⎠⎞

⎜⎝⎛−

=

( )1

210

1 A

AAcA x

xxckN−−

=

)( 21 AAcA xxckN −=In general, kc may be defined without regard to convective flow:

(1)

(2)

cAc kxk )1( 10 −=Combining (1) and (2):

kc0 and kc → for high flux kc

’ → for low flux


Relationship between kc0 and kc for high flux and kc

’ for low flux

)( 21

'

AABM

cA cc

xkN −=

The following expressions have previously derived using the film theory foe transfer of A by molecular diffusion and convective flow, with B being stagnant and nondiffusing

)( 21 AAc cck −=

BMc

c

xkk 1

' =Hence, 'x

x

kk

=

cAc kxk )1( 10 −=But,

BM

A

c

c

xx

kk 1

'

0 1−=

Note: correction factors obtained by film theory give results reasonably close to those using more complex theory, eg. penetration theory or boundary layer theory

'

0

x

x

kk

=

Correction factor for high flux


Example: High Flux Correction FactorsToluene A is evaporating from a wetted porous slab by having inert pure air at 1

atm flowing parallel to the flat surface. At a certain point the mass-transfer coefficient kc

’ for very low fluxes has been estimated as 0.2 lb mol/h.ft2. The gas composition at the interface at this point is xA1 = 0.65. Calculate the flux NA and the ratio kc /k’

c or kx /k’x and k0

c /k’c or k0

x /k’x to correct for high flux.


Mass-Transfer Coefficients for Various Geometries Dimensionless Numbers

μυρL

=ReReynolds number L ≡

Dp for sphereL ≡

D for pipe

L ≡

L length for plateυ: mass average velocity if in pipeυ ≡ υ’: superficial velocity in the empty cross section of a packed bed (υ = υ’/ε)

ABDρμ

=ScSchmidt number)( transfer massfor y Diffusivit

)/(y diffusivitfor component shear ABD

ρμ=

AB

c

DLk '

Sc=Sherwood numberAB

BMc

DLyk

=AB

x

DL

ck '

= = . . .

υ

'

St ck=Stanton number

M

y

Gk '

=M

G

GPk '

= = . . . GM = υρ/Mav = υc

( ) ( ) ( )1/33/2'

3/2'

Sc Re/ScSc ===M

GSc

cD G

PkNkJυ

JD factor


Analogies among Mass, Heat and Momentum transferWhy analogies?

• Similarity of molecular diffusion equation for momentum, Fourier’s for heat and Fick for mass

• Data for pressure drop and heat transfer are available more than mass transfer

• In case of turbulent flow, the differential equations will contain time average velocities and in addition the eddy diffusivities of momentum (ευ

), mass (εD ), and heat transfer (εH ). The resulting equations can not be solved for lack of information about the eddy diffusivities, but one might expect results of the form:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

D

Scfεεψ υ,,

2Sc ReSh

1 ⎟⎠⎞

⎜⎝⎛=

νεψ DScf ,,

22

⎟⎟⎠

⎞⎜⎜⎝

⎛=

H

fεεψ υPr,,

2Sc ReNu

1 ⎟⎠⎞

⎜⎝⎛=

νεψ Hf Pr,,

22

• Equations or correlations for heat transfer can be used for mass transfer by replacing dimensionless number of the former by the later. See next slide


Corresponding dimensionless groups of mass and heat transfer

Mass Transfer Heat Transfer

12

1

AA

AA

cccc−−

μυρL

=Re

12

1

TTTT−−

Reynolds number: μυρL

=ReReynolds number:

ABAB DDν

ρμ

==ScSchmidt number:ανμ

==k

CpPrPrandtl number:

Sherwood number:k

hL=NuNusselt number:

AB

c

DLk '

Sc=AB

x

DL

ck '

=

Grashof number:23

Gr ⎟⎠

⎞⎜⎝

⎛Δ=

μρ

ρρgL

D Grashof number:2

3Gr ⎟⎠

⎞⎜⎝

⎛Δ=μρβ TgLH

Peclet:ABD

Lυ== ScRePeD Peclet:

αυυρ L

kLCp

H === Pr RePe

Stanton number:M

G

DD G

Pk '

PeSh

Sc ReShSt === Stanton number:

υρpHH C

h===

PeNu

Pr ReNuSt

2/3Sc St DDJ =JD factor: 2/3Pr St HHJ =JH factor:


Derivation of Mass-Transfer Coefficients in Laminar Flow

• When a fluid flowing in laminar flows and mass transfer by molecular diffusion is occurring, the equations are very similar to those for heat transfer by conduction in laminar flow.

• In theory it is not necessary to have experimental mass-transfer coefficients for laminar flow, since the equations for momentum transfer and diffusion can be solved

• Consider mass transfer of solute A into a laminar falling film as shown in the drawing

• Solute A in the gas is absorbed at the interface and then diffuses a distance into the liquid so that it does not penetrate the whole distance x = δ• Concentration profile at point z distance is shown in the drawing


0)1()1()1()1( =Δ+Δ−Δ+Δ Δ+Δ+ xNzNxNzN zzAzxxAxzAzxAx

Consider the mass balance on the elemental system:For steady-state: Rate of input = rate of output

convection zero +∂∂

−=xcDN A

ABAx

For dilute solution:

zAAz cN υ+= diffusion zero

2

2

xcD

zc A

ABA

z ∂∂

=∂∂υ

Substituting, dividing by ΔxΔz, letting ΔxΔz and approaches zero leads to:

⇒ υz is needed and has been derived in fluid mechanics: ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

2

max, 1δ

υυ xzz

Also υz,max = (3/2)υz,av

• If solute A penetrates only a short distance into the fluid: ⇒ υz = υz,max = υmax


2

2

max xcD

zc A

ABA

∂∂

=∂∂

∴υBoundary conditions:z = 0 cA = 0x = 0 cA = cA0 x = ∞

cA = 0

⎟⎟⎠

⎞⎜⎜⎝

⎛=

max0 /4erf

υzDx

cc

ABA

ASolution by Laplace transform:

00)(

== ∂

∂−=

x

AABxAx x

cDzN

Local molar flux at the surface x = 0 at position z:

zDc AB

A πυmax

0=

Total moles of A transferred per second to the liquid over the entire length:

∫ ==⋅L

xAxA dzNLN0

0)()1()1( LDcL AB

A πυmax

04)1.(=∫ ⎟

⎠⎞

⎜⎝⎛=

LAB

A dzz

Dc0 2/1

2/1max

01)1(

πυ

where L/υmax = tL ≡

time of exposure of the liquid to the solute A in the gas

∴ 5.05.0 /1 and transfermass of Rate LAB tD∝ Basis for penetration theory in turbulent mass transfer


Mass Transfer for Laminar Flow inside Pipes

• For laminar flow of a liquid or gas inside a pipe: 2100Re <=μρυD

• Experimental data for mass transfer from the wall for gases are presented graphicallycA : exit concentrationcA0 : inlet concentrationcAi : concentration at the interface between the gas and the gas

W: flow in kg/s

L: length of mass transfer section

• For liquids with small values of DAB :3/2

0

0 5.5−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−

LDW

cccc

ABAAi

AA

ρ


Mass Transfer for Turbulent Flow inside Pipes

• For laminar flow of a liquid or gas inside a pipe: 2100Re >=μρυD

• For gases or liquids flowing inside a pipe:

31

ScRe023.0)(Sh 83.0'

===AB

BMc

AB

c

DDpk

DDk Re > 2100

0.6 <Sc < 3000

Mass Transfer for Flow inside Wetted-Wall Towers• When a gas is flowing inside the core of a wetted-wall tower, the same correlations that are used for mass transfer of a gas in laminar or turbulent flow in a pipe are applicable

• For the mass transfer in the liquid film flowing down the wetted-wall tower use:

zDczN AB

AxAx πυmax

00)( == LDcLLN AB

AA πυmax

04)1.()1( =⋅and


Example: Mass Transfer Inside a TubeA tube is coated on the inside with naphthalene and has an inside diameter of 20 mm and a length of 1.1 m. Air at 318 K and an average pressure of 101.3 kPa flows through this pipe at a velocity of 0.8 m/s. Assuming that the absolute pressure remains essentially constant, calculate the concentration of naphthalene in the exit air.


Mass Transfer in Flow Outside Solid SurfaceFor gases or evaporation of liquids in the gas phaseLaminar region; ReL= Lυρ/μ < 15, 000

In terms of Sherwood number:

5.0Re664.0 −= LDJ

3/15.0'

ShRe664.0Sh −== LAB

c

DLk

For gases and 15,000 < ReL < 300,000:2.0Re036.0 −= LDJ

For liquids and 600 < ReL < 50,000:5.0Re99.0 −= LDJ

L

NA

υ

A


Example: Mass Transfer from a Flat Plate

A large volume of pure water at 26.1oC is flowing parallel to a flat plate of solid benzoic acid, where L = 0.244 m in the direction of flow. The water velocity is 0.061 m/s. The solubility benzoic acid in water is 0.02948 kgmol/m3. The diffusivity of benzoic acid is 1.245 × 10-9 m2/s. Calculate the mass transfer coefficient kL and the flux NA .


Mass Transfer for Flow Past Single Sphere

Previously, the following equation was derived for stagnant medium

( )2112

AAp

ABA cc

DDN −=∴

⇒ Mass-transfer coefficient kc, which is k’c for a dilute solution is:

p

ABc D

Dk 2' =

Rearranging, 0.2Sh'

==AB

pc

DDk Only for very low Re= Dpυρ/μ

Gases; 0.6 < Sc < 2.7; 1 < Re = Dpυρ/μ < 48 000 3/153.0 ScRe552.00.2Sh +=

Liquids; 2 < Re = Dpυρ/μ < 2000 3/15.0 ScRe95.00.2Sh +=

Liquids; 2,000 < Re = Dpυρ/μ < 17,000 3/162.0 ScRe347.0Sh=


Example: Mass Transfer from a Sphere

Calculate the value of the mass-transfer coefficient and the flux for mass transfer from a sphere of naphthalene to air at 45oC and 1 atm flowing at a velocity of 0.305 m/s. The diameter of the sphere is 25.4 mm. The diffusivity of naphthalene in air at 45oC is 6.92×10-6 m2/s and the vapor pressure of solid naphthalene is 0.555 mm Hg.


Mass Transfer to Packed BedsTypical processing operation including packed beds

Drying operationsAdsorption or desorption of gases or liquids by solid particles, eg. charcoal Mass transfer of gases and liquids to catalyst particles

Void fraction of the bed, ε, is an important parameter in the bed:

0.2Shmin bed, theof volumeTotal

min pace, voidsof volume3

3

===ε

Gases in packed bed of spheres; 10 < Re < 10,000: 4069.0Re4548.0 −==εHD JJ

where,μυ'

Re pD=

υ’: superficial velocity (mass average velocity in the empty tube without packing


Liquid in packed bed; 0.0016 < Re < 55 ; 165 < Sc < 70 000:

3/2Re09.1 −=εDJ

Liquid in packed bed; 55 < Re < 1 500 ; 165 < Sc < 10 690:

31.0Re250.0 −=εDJ

Gases or Liquid in Fluidized bed of spheres; 10 < Re < 4 000:

4069.0Re4548.0 −=εDJ

Liquid in Fluidized bed; 1 < Re < 10 :

72.0Re1068.1 −=εDJ

Correction factor is introduced for non-spherical particles


Calculation Method for Packed BedsSelect proper correlation for JD and then calculate kc in m/s. Calculate the total external surface area A m2 of the solids

bp

b VD

aVA )1(6 ε−==

)]/()ln[()()(

21

21

AAiAAi

AAiAAicA cccc

ccccAkAN−−−−−

=

a: m2 surface area/m3 total volumeVb : total volume of the bed, in m3

Calculate the log mean driving force at the inlet and outlet of the bed:

cAi : concentration at the surface of the solid, kg mol/m3;cA1 : inlet bulk fluid concentration, kg mol/m3;cA2 : outlet bulk fluid concentration, kg mol/m3;


Example: Mass Transfer of a Liquid in a Packed BedPure water at 26.1oC flows at the rate of 5.514×10-7 m3/s through a packed bed of benzoic-

acid spheres having a diameter of 6.375 mm. The total surface area of the spheres in the bed is 0.01198 m2 and the void fraction is 0.436. The tower diameter is 0.0667 m. The solubility of benzoic acid in water is 2.948×10-2 kg mol/m3.

1. Predict the mass-transfer coefficient kc . Compare with the experimental value of 4.665×10-6 m/s.

2. Using the experimental value of kc , predict the outlet concentration of benzoic acid in the water


Mass Transfer to Suspensions of Small Particles

ExamplesLiquid-liquid hydrogenation -hydrogen diffuse from gas bubbles through an

organic liquid then to small suspended catalyst particles.Fermentation-oxygen diffuses from small gas bubbles through the aqueous

medium then to the small suspended particles.

1. Mass Transfer to small particles <0.6 mmTo predicts mass transfer coefficients from small gas bubbles such as oxygen or air to the liquid phasefrom liquid phase to the surface of small catalyst particles, microorganisms, solids or liquids.

31

23/2Sc31.02' ⎟⎟

⎠

⎞⎜⎜⎝

⎛ Δ+= −

c

c

P

ABL p

gDDk ρμ Δρ = ρc - ρp

ρc : density of continuous phaseMolecular diffusion Due to free fall or rise of the sphere due to gravitation


Example: Mass Transfer from Air Bubbles in Fermentation

Calculate the maximum rate of absorption of O2 in a fermenter from air bubbles at 1 atm pressure having diameters of 100 μm at 37oC into water having a zero concentration of dissolved O2 . The solubility of O2 from air in water at 37oC is 2.26×10-7 g mol O2 /cm3 liquid or 2.26×10-4 kg mol O2 /m3. The diffusivity of O2 in water at 37oC is 3.25×10-9 m2/s. Agitation is used to produce the air bubbles.


2. Mass Transfer to Large Gas Bubbles > 2.5 mm

For large bubbles or liquid drops >2.5 mm, the mass coefficient can be predicted by

Large bubbles are produced when pure liquids are aerated in mixing vessels and sieve plate columns.

31

25.0Sc42.0' ⎟⎟

⎠

⎞⎜⎜⎝

⎛ Δ= −

c

cL

gkρρμ

3. Mass Transfer to Particles in Transition Region

In mass transfer in the transition region between small and large bubbles in the size range 0.6-2.5 mm, the mass coefficient can be approximated by assuming that it increases linearly with bubbles diameter.


4. Mass Transfer to Particles in Highly Turbulent Mixers

When agitation power is increased beyond that needed for suspension of solid or liquids particles and the turbulence force become larger than the gravitational forces, so mass transfer coefficient

( )4

1

23/2 /13.0Sc' ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

c

cL

gVPkρ

μ P/V: power input per unit volume


Molecular Diffusion Plus Convection and Chemical Reaction

J*A: molar flux of A in kg mol A/s.m2 relative to molar average velocity υM

NA: molar flux of A relative to stationary coordinate

AAA υρ=n (kg A/s.m2) AAA υcN = (kg mol A/s.m2) or

Assuming υ to be the mass average velocity of the stream relative to stationary coordinate and can be obtained from υA and υB as:

AA υυυ AA ww +=

where wA = ρA /ρ (weight fraction of A)

υA : velocity of A relative to stationary coordinate

BA υυυ BAM xx +=

Molar average velocity υM in m/s relative to stationary coordinate is:

BA υυcc

cc BA +=

BA υρρυ

ρρ BA +=


Molar diffusion flux relative to the molar average velocity υM is:

)(*MAAA cJ υυ −= (kg mol A/s.m2)

)(*MAAAj υυρ −= (kg A/s.m2)or

Molar diffusion flux relative to the mass average velocity υM is:

)( υυ −= AAA cJ (kg mol A/s.m2)

)( υυρ −= AAAj (kg A/s.m2)or

Fick’s law as given relative to υM:

dzdxcDJ A

ABA −=*

Fick’s law to terms of a mass flux relative to υ:

dzdwDj A

ABA ρ−=

Prove that jA + jB = 0

)( υυρ −= BBBj)( υυρ −= AAAj

=+⇒ BA jjυρυρυρυρ BBBAAA −+−

)( AABBAA ρρυυρυρ +−+=

BA υρρυ

ρρυ BA +=Use:

Identity is proved


Equation of Continuity for a Binary MixtureObjective: Derive general equation for a binary mixture of A and B for diffusion

and convection including unsteady state diffusion and chemical reaction

Consider the element ΔxΔyΔx fixed in spaceGeneral mass balance on A is:

⎟⎠

⎞⎜⎝

⎛=⎟

⎠

⎞⎜⎝

⎛

+⎟⎠

⎞⎜⎝

⎛−⎟

⎠

⎞⎜⎝

⎛

onaccumulatiof rate

A mass of generationof rate

outA massof rate

inA massof rate

dtdmzyxr

yxnzxnzynyxnzxnzyn

A

zzAzyyAyxxAxzAzyAyxAx

=ΔΔΔ+

ΔΔ−ΔΔ−ΔΔ−ΔΔ+ΔΔ+ΔΔ Δ+Δ+Δ+

nAz⏐z

Δx

Δy

ΔznAx⏐x nAx⏐x+Δx

x

z

y

nAz⏐z+Δz

Divide by ΔxΔyΔz and letting ΔxΔyΔz approach zero:


AAzAyAxA rz

ny

nx

nt

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂ρ

In vector notation: ( ) AAA rn

t=⋅∇+

∂∂ρ

AAzAyAxA Rz

Ny

Nx

Nt

c=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂

Dividing both sides by MA :

RA : kmol A/s.m3

Substituting NA and Fick’s law from:MA

AABAz c

dzdxcDN υ+−=

( ) ( ) AAABMAA RxcDct

c=∇⋅∇−⋅∇+

∂∂ υ Final General equation


Equation for constant c and DAB

:

( ) ( ) ( ) AAABAMMAA RcDcct

c=∇−⋅∇+⋅∇+

∂∂ 2υυ

Equimolar

counterdiffusion

for gases:

dzdcDN A

ABAz −= (or υM = 0)

Assuming DAB = constant and RA = 0

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

2

2

zc

yc

xcD

tc AAA

ABA (previously derived)


Diffusion and Chemical Reaction at the Boundary

Consider gas A diffusing from the bulk gas phase to the catalyst surface, where it reacts instantaneously and reversibly as follows:

A → 2BGas B then diffuses back, see the drawing δ

zz1 z2

NANB

catalystsurface

At steady state, NA = - 2NB

)( BAAA

ABAz NNxdzdxcDN ++−=Since,

)2( BAAA

ABAz NNxdzdxcDN −+−=

Rearranging and integration, ∫∫ +−=

=

=

2

1

2

1 10

A

A

x

x A

AAB

z

zA x

dxcDdzNδ

or,2

1

11ln

A

AABA x

xcDN++

=δ

Instantaneous ⇒ xA2 = 0


For Slow surface reaction

2

1

11ln

A

AABA x

xcDN++

=δ

If the reaction A → 2B is not instantaneous, but is rather slow and describes by the first order kinetic:

AAAz cxkckN '1

'1 ===δ

''12 // cAzAzA kNckNx δδ == ==

where k’1 is first order homogeneous reaction velocity constant in m/s.

In this case the flux is still given by:

But replacing xA2 by

ckNxcDNA

AABA '

1

1

/11ln++

=δ


Example: Diffusion and Chemical Reaction at a Boundary

Pure gas A diffuses from point 1 at a partial pressure of 101.32 kPa to point 2 a distance 2.0 mm away. At point 2 it undergoes a chemical reaction at the catalyst surface and A → 2B. Component B diffuses back at steady state. The total pressure is P = 101.32 kPa. The temperature is 300 K and DAB = 0.15×10-4 m2/s.

(a) For instantaneous rate of reaction, calculate xA2 and NA .(b) For a slow reaction where k’

1 = 5.63×10-3 m/s, calculate xA2 and NA .


Diffusion and Homogeneous Reaction in a Phase

AAzAyAxA Rz

Ny

Nx

Nt

c=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂

Consider component A undergoes an irreversible chemical reaction in the homogeneous phase B while diffusing, as follows: A → C.

0+−=dzdcDN A

ABAz

Assuming A is very dilute in phase B, which can be a gas or a liquid, thus at steady state the bulk term is dropped (no movement):

Since for this case,

AA

AB Rdz

dcDdzd

=⎟⎠⎞

⎜⎝⎛−

For first order reaction (RA = -k’cA ) and constant DAB : AAB

A cDk

dzcd '

2

2

=⇒

For the boundary conditions cA = cA1 @ z = 0 and cA = cA2 @ z = L:( ) ( )

( )LDkzLDkczDkc

cAB

ABAABAA /sinh

)(/sinh/sinh'

'1

'2 −+

= cA = cA (z) for reactionsin gases, liquids, or solids


Multi-component Diffusion of GasesMulticomponent component diffusion occurs when three or more components A, B, C, … are present.

Example: diffusion of A through a stagnant nondiffusing mixture of B, C, D, …

For NB = NC = … = 0

)()( 21

12AA

iM

AmA pp

pzzRTPDN −

−= (Stefan-Maxwell method)

where pim is the log mean of pi1 = P – pA1 and Pi2 = P – pA2

Also,⋅⋅⋅++

=ACCABB

Am DxDxD

//1

''

where x’B = mol B/mole inerts:

,1

'

A

BB x

xx−

= ⋅⋅⋅−

=A

CC x

xx1

'


Example: Diffusion of A Through Nondiffusing

B and CAt 298 K and 1 atm total pressure, methane (A) is diffusing at steady state

through nondiffusing argon (B) and helium (C), At z = 0, the partial pressures in atm are pA1 = 0.4, pB1 = 0.4, and pC1 = 0.2, and at z2 = 0.005 m, pA2 = 0.1, pB2 = 0.6, and pC2 = 0.3. The binary diffusivities from Table 6.2-1 are DAB = 2.02×10-5 m2 /s, DAC = 6.75×10-5 m2 /s, DBC = 7.29×10-5 m2 /s. Calculate NA .

Chapter 2

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