Prof. R. Shanthini 27 Feb & 05 Mar 2012 Content: Mass transfer: concept and theory PM3125: Lectures 1 to 5 uploaded at

Prof. R. Shanthini 27 Feb & 05 Mar 2012

Content:

Mass transfer: concept and theory

PM3125: Lectures 1 to 5

uploaded at http://www.rshanthini.com/PM3125.htm


Reference books used for ppts

1. C.J. Geankoplis Transport Processes and Separation Process Principles4th edition, Prentice-Hall India

2. J.D. Seader and E.J. HenleySeparation Process Principles2nd edition, John Wiley & Sons, Inc.

3. J.M. Coulson and J.F. RichardsonChemical Engineering, Volume 15th edition, Butterworth-Heinemann


Mass transfer could occur by the following three ways:

Diffusion is caused by concentration gradient.

Advection is caused by moving fluid. (It cannot therefore happen in solids.)

Convection is the net transport caused by both diffusion and advection. (It occurs only in fluids.)

Modes of mass transfer

Diffusion

Advection

Convection


Stirring the water with a spoon creates forced convection.

That helps the sugar molecules to transfer to the bulk water much faster.

Diffusion(slower)

Convection(faster)


Mass transfer by diffusion occurs when a component in a stationary solid or fluid goes from one point to another driven by a concentration gradient of the component.

Solute ASolvent B

concentration of Ais high

concentration of Ais low

Diffusion


At the surface of the lung:

Air Blood

Oxygen

Carbon dioxide

High oxygen concentrationLow carbon dioxide concentration

Low oxygen concentrationHigh carbon dioxide concentration

Example of diffusion mass transfer


CA

CA + dCA

dz

A & B

for mass transfer in z-direction only

(1)

Fick’s First Law of Diffusion

JA = - DAB

dCA

dz

JA


JA = - DAB

dCA

dz

diffusion coefficient (or diffusivity) of A in B

diffusion flux of A in relation to the bulk motion in z-direction

(mass/moles per area per time)

concentration gradient of A in z-direction

(mass/moles per volume per distance)

What is the unit of diffusivity?


For dissolved matter in water: D ≈ 10-5 cm2/s

For gases in air at 1 atm and at room temperature: D ≈ 0.1 to 0.01 cm2/s

Diffusivity depends on the type of solute, type of solvent, temperature, pressure, solution phase (gas, liquid or solid) and other characteristics.

Unit and Scale of Diffusivity


Molecular diffusion of Helium in Nitrogen: A mixture of He and N2 gas is contained in a pipe (0.2 m long) at 298 K and 1 atm total pressure which is constant throughout. The partial pressure of He is 0.60 atm at one end of the pipe, and it is 0.20 atm at the other end. Calculate the flux of He at steady state if DAB of He-N2 mixture is 0.687 x 10-4 m2/s.

Example 6.1.1 from Ref. 1

Solution:

Use Fick’s law of diffusion given by equation (1) as

JA = - DAB dzdCA

Rearranging equation (1) and integrating gives the following:


JA = - DAB dz dCA

DAB is given as 0.687 x 10-4 m2/s

(z2 – z1) is given as 0.2 m

(CA2 – CA1) = ?

⌠⌡z1

z2

⌠⌡ CA1

CA2

At steady state, diffusion flux is constant. Diffusivity is taken as constant.

(2)

Therefore, equation (2) gives

JA = - DAB(z2 – z1) (CA2 – CA1) (3)


Even though CA is not given at points 1 and 2, partial pressures are given. We could relate partial pressure to concentration as follows:

CA =nA

V

Number of moles of A

Total volume

pA V = nA RT

Partial pressure of A

Absolute temperature

Gas constant

Combining the above we get CA =pA

RT


Equation (3) can therefore be written as

JA = - DAB(z2 – z1)

(pA2 – pA1)

RT

which gives the flux as

JA = - DAB

(pA2 – pA1)

RT(z2 – z1)

JA = 5.63 x 10-6 kmol/m2.s

JA = - (0.687x10-4 m2/s)(0.6 – 0.2) x 1.01325 x 105 Pa

(8314 J/kmol.K) x (298 K) x (0.20–0) m


Diffusion is the net transport of substances in a stationary solid or fluid under a concentration gradient.

Advection is the net transport of substances by the moving fluid, and so cannot happen in solids. It does not include transport of substances by simple diffusion.

Convection is the net transport of substances caused by both advective transport and diffusive transport in fluids.

Diffusion of gases A & B plus convection

JA is the diffusive flux described by Fick’s law, and we have already studied about it.

Let us use NA to denote the total flux by convection (which is diffusion plus advection.


JA = - DAB

dCA

dzMolar diffusive flux of A in B: (1)

The velocity of the above diffusive flux of A in B can be given by

vA,diffusion (m/s) = JA (mol/m2.s)

CA (mol/m3)

The velocity of the net flux of A in B can be given by

vA,convection (m/s) = NA (mol/m2.s)

CA (mol/m3)

The velocity of the bulk motion can be given by

vbulk (m/s) = (NA + NB) (mol/m2.s)

(CT) (mol/m3)

(4)

(5)

(6)

Total concentration Ignore the derivation, if you wish


vA,diffusion + vA,convection = vbulk

(NA + NB)

CT

CA vA,diffusion + CA vA,convection = CA vbulk

JA + NA = CA (7)

Substituting JA from equation (1) in (7), we get

(NA + NB)

CT

NA = + CA(8)dCA

dz-DAB

Multiplying the above by CA, we get

Using equations (4) to (6) in the above, we get

Ignore the derivation, if you wish


Using (9a) and (9b), equation (8) can be written as

(NA + NB)NA = - + (10)dpA

dz

DAB

Let us introduce partial pressure pA into (8) as follows:

CA =nA

V(9a)=

pA

RT

CT =nT

V(9b)=

P

RT

Total number of molesTotal pressure

RT

pA

P



Using (11), equation (8) can be written as

(NA + NB)NA = + xA(12)dxA

dz-CT DAB

Let us introduce molar fractions xA into (8) as follows:

xA = NA

(NA + NB) =

CA

CT

(11)



(NA + NB)NA = (12)dxA

dz-CT DAB

Diffusion of gases A & B plus convection:Summary equations for (one dimensional) flow in z direction

(NA + NB)NA = - + (10)dpA

dzDAB

RT

pA

P

NA = + (8)dCA

dz-DAB

In terms of concentration of A:

In terms of partial pressures (using pA = CART and P = CTRT):

In terms of molar fraction of A (using xA = CA /CT):

xA

(NA + NB)CT

CA

+

Total concentration

Total pressure

Molar fraction xA = NA

(NA + NB) =

CA

CT


Evaporation of a pure liquid (A) is at the bottom of a narrow tube.

Large amount of inert or non-diffusing air (B) is passed over the top.

Vapour A diffuses through B in the tube.

The boundary at the liquid surface (at point 1) is impermeable to B, since B is insoluble in liquid A.

Hence, B cannot diffuse into or away from the surface.

Therefore, NB = 0

A diffusing through stagnant, non-diffusing B

Liquid Benzene (A)

Air (B)

1

2

z2 – z1


Substituting NB = 0 in equation (10), we get

(NA + 0)NA = - + dpA

dz

DAB

RT

pA

P

Rearranging and integrating

NA (1 - pA/P) = - dpA

dz

DAB

RT

NA dz = - dpA

(1 - pA/P)

DAB

RT⌠⌡

z1

z2

⌠⌡ pA1

pA2

NA = ln P - pA2 DAB P

RT(z2 – z1) P – pA1

(13)


Introduce the log mean value of inert B as follows:

NA = (pA1 - pA2 )

DAB P

RT(z2 – z1) pB,LM

(14)

pB,LM = = (pB2 – pB1 )

ln(pB2 /pB1 )

(P – pA2 ) – (P – pA1 )

ln[(P - pA2 )/ (P - pA1 )]

Equation (13) is therefore written as follows:

Equation (14) is the most used form.

(pA1 – pA2 )

ln[(P - pA2 )/ (P - pA1 )]

=


NA = - (xA1 - xA2 )

DAB CT

(z2 – z1) xB,LM

(16)

NA = ln 1 - xA2 DAB CT

(z2 – z1) 1 – xA1

(15)

Using xA = CA /CT, pA = CART and P = CTRT, equation (13) can be converted to the following:

Introduce the log mean value of inert B as follows:

xB,LM = = (xB2 – xB1 )

ln(xB2 /xB1 )

(1 – xA2 ) – (1 – xA1 )

ln[(1 - xA2 )/ (1 - xA1 )]

(xA1 – xA2 )

ln[(1 - xA2 )/ (1 - xA1 )]

=

Therefore, equation (15) becomes the following:


Diffusion of water through stagnant, non-diffusing air: Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed to be dry) is 1 atm and the temperature is 293 K. Water evaporates and diffuses through the air in the tube, and the diffusion path is 0.1524 m long. Calculate the rate of evaporation at steady state. The diffusivity of water vapour at 1 atm and 293 K is 0.250 x 10-4 m2/s. Assume that the vapour pressure of water at 293 K is 0.0231 atm.

Answer: 1.595 x 10-7 kmol/m2.s



Solution: The set-up of Example 6.2.2 is shown in the figure. Assuming steady state, equation (14) applies.

Water (A)

Air (B)

1

2

z2 – z1

NA = (pA1 - pA2 )

DAB P

RT(z2 – z1) pB,LM

(14)

Data provided are the following: DAB = 0.250 x 10-4 m2/s;

P = 1 atm; T = 293 K; z2 – z1 = 0.1524 m;

pA1 = 0.0231 atm (saturated vapour pressure);

pA2 = 0 atm (water vapour is carried away by air at point 2)

(pA1 – pA2 )

ln[(P - pA2 )/ (P - pA1 )]

pB,LM =

where


Substituting the data provided in the equations given, we get the following:

NA = (0.0231 - 0) atm (0.250x10-4 m2/s)(1x1.01325x105 Pa)

(0.0231 – 0 )

ln[(1 - 0 )/ (1 – 0.0231 )]

pB,LM = = 0.988 atm

(8314 J/kmol.K) (293 K) (0.1524 m) (0.988 atm)

= 1.595 x 10-7 kmol/m2.s


Diffusivities for different systems could be estimated using the empirical equations provided in the following slides as well as those provided in other reference texts available in the library and other sources.

Estimating Diffusivity


Diffusivity of gases

An example at 1 atm and 298 K:

System Diffusivity (cm2/s)

H2-NH3 0.783

H2-CH4 0.726

Ar-CH4 0.202

He-CH4 0.675

He-N2 0.687

Air-H2O 0.260

Air-C2H5OH 0.135

Air-benzene 0.0962


DAB - diffusivity in cm2/s

P - absolute pressure in atmMi - molecular weight

T - temperature in KVi - sum of the diffusion volume for component i

DAB is proportional to 1/P and T1.75

Binary Gas Diffusivity


Binary Gas Diffusivity



T - temperature in Kμ - viscosity of solution in kg/m sVA - solute molar volume at its normal boiling point

in m3/kmol

DAB is proportional to 1/μ and T

DAB =9.96 x 10-12 T

μ VA1/3

For very large spherical molecules (A) of 1000 molecular weight or greater diffusing in a liquid solvent (B) of small molecules:

applicable for biological

solutes such as proteins

Diffusivity in Liquids



MB - molecular weight of solvent B

T - temperature in Kμ - viscosity of solvent B in kg/m sVA - solute molar volume at its normal boiling point in m3/kmol

Φ - association parameter of the solvent, which 2.6 for water, 1.9 for methanol, 1.5 for ethanol, and so on

DAB is proportional to 1/μB and T

DAB =1.173 x 10-12 (Φ MB)1/2 T

μB VA0.6

For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B):

applicable for biological solutes

Diffusivity in Liquids


DoAB is diffusivity in cm2/s

n+ is the valence of cation

n- is the valence of anion

λ+ and λ- are the limiting ionic conductances in very dilute

solutions T is 298.2 when using the above at 25oC

DAB is proportional to T

DoAB =

8.928 x 10-10 T (1/n+ + 1/n-)

For smaller molecules (A) diffusing in a dilute liquid solution of solvent (B):

(1/λ+ + 1/ λ-)

Diffusivity of Electrolytes in Liquids


Diffusion in solids are occurring at a very slow rate.

Diffusion in solids

In gas: DAB = 0.1 cm2/s

Time taken 2.09 h

In liquid: DAB = 10-5 cm2/s

Time taken 2.39 year

In solid: DAB = 10-9 cm2/s

Time taken 239 centuries


Diffusion in solids are occurring at a very slow rate.

However, mass transfer in solids are very important.

Examples:

Leaching of metal oresDrying of timber, and foodsDiffusion and catalytic reaction in solid catalystsSeparation of fluids by membranes

Treatment of metal at high temperature by gases.

Diffusion in solids


Diffusion in solids

Diffusion in solids occur in two different ways:

- Diffusion following Fick’s law (does not depend on the structure of the solid)

- Diffusion in porous solids where the actual structure and void channels are important


Diffusion in solids following Fick’s Law

NA = + (8)dCA

dz-DAB (NA + NB)

CT

CA

Start with equation (8):

Bulk term is set to zero in solids

NA = dCA

dz-DAB

Therefore, the following equation will be used to describe the process:

(17)


Diffusion through a slab

NA = (CA1 - CA2)

Applying equation (17) for steady-state diffusion through a solid slab, we get

(18)

z2-z1

CA2

CA1

z2 - z1

where NA and DAB are taken as constants.

DAB

Similar to heat conduction.


Relating the concentration and solubilityThe solubility of a solute gas in a solid is usually expressed by the notation S.

m3 solute at STP

m3 solid . atm partial pressure of solute

CA = kmol solute /m3 solid S pA

22.414

STP of 0oC and 1 atm

Unit used in general is the following:

Relationship between concentration and solubility:

where pA is in atm


Relating the concentration and permeabilityThe permeability of a solute gas (A) in a solid is usually expressed by the notation PM. in m3 solute at STP (0oC and 1 atm) diffusing per second per m2 cross-sectional area through a solid 1 m thick under a pressure difference of 1 atm.

m3 solute at STP . 1 m thick solid

s . m2 cross-sectional area . atm pressure difference

PM = DAB S

Unit used in general is the following:

Relationship between concentration and permeability:

where DAB is in m2/s and S is in m3/m3.atm


Diffusion of H2 through Neoprene membrane: The gas hydrogen at 17oC and 0.010 atm partial pressure is diffusing through a membrane on vulcanized neoprene rubber 0.5 mm thick. The pressure of H2 on the other side of neoprene is zero. Calculate the steady-state flux, assuming that the only resistance to diffusion is in the membrane. The solubility S of H2 gas in neoprene at 17oC is 0.051 m3 (at STP of 0oC and 1 atm)/m3 solid. atm and the diffusivity DAB is 1.03 x 10-10 m2/s at 17oC.

Answer: 4.69 x 10-12 kmol H2/m2.s



Diffusion through a packging film using permeability: A polythene film 0.00015 m (0.15 mm) thick is being considered for use in packaging a pharmaceutical product at 30oC. If the partial pressure of O2 outside the package is 0.21 atm and inside it is 0.01 atm, calculate the diffusion flux of O2 at steady state. Assume that the resistances to diffusion outside the film and inside are negligible compared to the resistance of the film. Permeability of O2 in polythene at 303 K is 4.17 x 10-12 m3 solute (STP)/(s.m2.atm.m).

Answer: 2.480 x 10-12 kmol O2/m2.s

Would you prefer nylon to polythene? Permeability of O2 in nylon at 303 K is 0.029 x 10-12 m3 solute (STP)/(s.m2.atm.m). Support your answer.



Diffusion through a cylinder wall

CA2

Applying equation (17) for steady-state diffusion through a cylinder wall of inner radius r1 and outer radius r2 and length L in the radial direction outward, we get

r1r2

r

NA = = dCA

dr-DAB

(19)nA

2 π r L

Area of mass transfer

Mass transfer per time

CA1

Mass transfer per area per time

nA = 2πL DAB(CA1 - CA2) (20)ln(r2 / r1)



Diffusion through a spherical shell

CA2

Applying equation (17) for steady-state diffusion through a spherical shell of inner radius r1 and outer radius r2 in the radial direction outward, we get

r1r2

r

NA = = dCA

dr-DAB

(21)nA

4 π r2

Area of mass transfer

Mass transfer per time

CA1

Mass transfer per area per time

nA = 4πr1r2 DAB(CA1 - CA2) (22)(r2 - r1)



Microscopic (or Fick’s Law) approach:

Macroscopic (or mass transfer coefficient) approach:

NA = - k ΔCA

where k is known as the mass transfer coefficient

(1)JA = - DAB

dCA

dz

(50)

good for diffusion dominated problems

good for convection dominated problems


NA

Mass Transfer Coefficient Approach

kc ΔCA=

NA

CA1

CA2

A & B

kc (CA1 – CA2 )= (51)

kc is the liquid-phase mass-transfer coefficient based on a concentration driving force.

What is the unit of kc?


CA1 = pA1 / RT; CA2 = pA2 / RT

Using the following relationships between concentrations and partial pressures:

Equation (51) can be written as

NA kc(pA1 – pA2) / RT= (52)kp

(pA1 – pA2)=

where kp = kc / RT (53)

kp is a gas-phase mass-transfer coefficient based on a partial-pressure driving force.

What is the unit of kp?

NA kc ΔCA= kc (CA1 – CA2 )= (51)

Mass Transfer Coefficient Approach


Mass transfer between phases across the following interfaces are of great interest in separation processes:

- gas/liquid interface

- liquid/liquid interface

Such interfaces are found in the following separation processes:

- absorption

- distillation

- extraction

- stripping

Models for mass transfer between phases


Models for mass transfer at a fluid-fluid interface

Theoretical models used to describe mass transfer between a fluid and such an interface:

- Film Theory

- Penetration Theory

- Surface-Renewal Theory

- Film Penetration Theory


Film Theory

Entire resistance to mass transfer in a given turbulent phase is in a thin, stagnant region of that phase at the interface, called a film.

Bulk liquid

Liquid film

Gas

pA

z=0 z=δL

CAi

CAb

For the system shown, gas is taken as pure component A, which diffuses into nonvolatile liquid B.

Mass transport

In reality, there may be mass transfer resistances in both liquid and gas phases. So we need to add a gas film in which gas is stagnant.


Two Film Theory

Liquid phase

Liquid film

Gas phase

pAb

CAi

CAb

pAi

Gas film

Mass transport

There are two stagnant films (on either side of the fluid-fluid interface).

Each film presents a resistance to mass transfer.

Concentrations in the two fluid at the interface are assumed to be in phase equilibrium.


Liquid phase

Gas phase

pAb

CAi

CAb

pAi

Mass transport

Concentration gradients for the film theory

More realistic concentration gradients

Liquid phase

Liquid film

Gas phase

pAb

CAi

CAb

pAi

Gas film

Mass transport

Interface Interface

Two Film Theory


Liquid phase

Liquid film

Gas phase

pAb

CAi

CAb

pAi

Gas film

Mass transport, NA

Two Film Theory applied at steady-state

NA kc (CAi – CAb )= (51)

NA (52)kp(pAb – pAi)=

Mass transfer in the gas phase:

Mass transfer in the liquid phase:

Phase equilibrium is assumed at the gas-liquid interface.

Applying Henry’s law,

pAi = HA CAi (53)


Liquid film

pAb

CAb

Gas film

Henry’s Law

pAi = HA CAi at equilibrium,

where HA is Henry’s constant for A

Unit of H:

[Pressure]/[concentration] = [ bar / (kg.m3) ]

CAi

pAi

Note that pAi is the gas phase pressure and CAi is the liquid phase concentration.


Liquid phase

Liquid film

Gas phase

pAb

CAi

CAb

pAi

Gas film

Mass transport, NA

Two Film Theory applied at steady-state

We know the bulk concentration and partial pressure.

We do not know the interface concentration and partial pressure.

Therefore, we eliminate pAi and CAi from (51), (52) and (53) by combining them appropriately.


Two Film Theory applied at steady-stateNA (54)From (52):

Substituting the above in (53) and rearranging:

(56)

pAi = pAb - kp

NA (55)From (51): CAi = CAb + kc

NA =pAb - HA CAb

HA / kc + 1 / kp

The above expression is based on gas-phase and liquid-phase mass transfer coefficients.

Let us now introduce overall gas-phase and overall liquid-phase mass transfer coefficients.


Introducing overall gas-phase mass transfer coefficient:Let’s start from (56).

Introduce the following imaginary gas-phase partial pressure:

pA* ≡ HA CAb

where pA* is a partial pressure that would have been in

equilibrium with the concentration of A in the bulk liquid.

Introduce an overall gas-phase mass-transfer coefficient (KG) as

(57)

(58)1

kp

HA

kc

≡ +1

KG

Combining (56), (57) and (58):

NA = KG (pAb - pA* ) (59)


Introducing overall liquid-phase mass transfer coefficient:Once again, let’s start from (56).

Introduce the following imaginary liquid-phase concentration:

pAb ≡ HA CA

*

where CA* is a concentration that would have been in

equilibrium with the partial pressure of A in the bulk gas.

Introduce an overall liquid-phase mass-transfer coefficient (KL) as

(60)

(61)1

HAkp

1

kc

≡ +1

KL

Combining (56), (60) and (61):

NA = KL (CA* - CAb) (62)


CAi

pAi

pAb

CAb

pA*

CA* CA

pA

pAi = HA CAi

pA* = HA CAb

pAb = HACA*

Gas-Liquid Equilibrium Partitioning Curve showing the locations of p*

A and C*A


Summary:

where

NA = KL (CA* - CAb)

= KG (pAb - pA*)

1

kp

HA

kc

+1

KG

pA* = HA CAb

CA* = pAb

/ HA

KL

= =HA

(62)

(59)

(60)

(57)

(58 and 61)


Example 3.20 from Ref. 2 (modified)

Sulfur dioxide (A) is absorbed into water in a packed column. At a certain location, the bulk conditions are 50oC, 2 atm, yAb = 0.085, and xAb = 0.001. Equilibrium data for SO2 between air and water at 50oC are the following:

pA (atm) 0.0382 0.0606 0.1092 0.1700

CA (kmol/m3) 0.03126 0.04697 0.07823 0.10949

Experimental values of the mass transfer coefficients are kc = 0.18 m/h and kp = 0.040 kmol/h.m2.kPa.

Compute the mass-transfer flux by assuming an average Henry’s law constant and a negligible bulk flow.


Solution:

T = 273oC + 50oC = 323 K; PT = 2 atm;

yAb = 0.085; xAb = 0.001;

kc = 0.18 m/h;kp = 0.040 kmol/h.m2.kPa

y = 1.4652x

R2 = 0.9759

0

0.04

0.08

0.12

0.16

0.2

0.02 0.04 0.06 0.08 0.1 0.12

CA (kmol/m3)

pA (

atm

)

HA = 1.4652 atm.m3/kmol

Data provided:

slope of the curve

HA = 161.61 kPa.m3/kmol


where

NA = KL (CA* - CAb)

= KG (pAb - pA*)

1

kp

HA

kc

+1

KG

pA* = HA CAb

CA* = pAb

/ HA

KL

= =HA

(62)

(59)

(60)

(57)

(58 and 61)

Equations to be used:


1

kp

HA

kc

+1

KG KL

= =HA (58 and 61)

1

kp

=0.040

1h.m2.kPa/kmol = 25 h.m2.kPa/kmol

HA

kc

=0.18 m/h

161.61 kPa.m3/kmol= 897 h.m2.kPa/kmol

KG = 1/(25 + 897) = 1/922 = 0.001085 kmol/h.m2.kPa

KL = HA KG = 161.61/922 = 0.175 m/h

Calculation of overall mass transfer coefficients:


NA = KL (CA* - CAb)

CA* = pAb

/ HA = yAb PT / HA

(62) is used to calculate NA

= 0.085 x 2 atm / 1.4652 atm.m3/kmol = 0.1160 kmol/m3

CAb = xAb

CT = 0.001 CT

CT = concentration of water (assumed) = 1000 kg/m3 = 1000/18 kmol/m3 = 55.56 kmol/m3

CAb = 0.001 x 55.56 kmol/m3 = 0.05556 kmol/m3

NA = (0.175 m/h) (0.1160 - 0.05556) kmol/m3

= 0.01058 kmol/m2.h


NA = KG (pAb - pA*)

pA* = CAb

HA = xAb CT HA

(59) is used to calculate NA

= 0.001 x 55.56 x 161.61 kPa = 8.978 kPa

pAb = yAb

PT = 0.085 x 2 x 1.013 x 100 kPa = 17.221 kPa

NA = (1/922 h.m2.kPa/kmol) (17.221 - 8.978) kPa

= 0.00894 kmol/m2.h

Alternatively,


Liquid phase

Liquid film

Gas phase

pAb

CAiCAb

pAi

Gas film

Mass transport, NA

NA = kp (pAb – pAi) = kc (CAi – CAb )

pAi = HA CAi (53)

Summary: Two Film Theory applied at steady-state

(52) (59)(51) (62)

1

kp

HA

kc

+1

KG

pA* = HA CAb

pAb = HA CA*

KL

= =HA

(60)

(57)

(58 and 61)

= KG (pAb - pA*) = KL (CA

* - CAb)


Liquid phase

Liquid film

Gas phase

yAb

xAi

xAb

yAi

Gas film

Mass transport, NA

NA = ky (yAb – yAi) = kx (xAi – xAb )

yAi = KA xAi

Summary equations with mole fractions

yA* = KA xAb

yAb = KA xA*

= Ky (yAb - yA*) = Kx (xA

* - xAb)

1

ky

KA

kx

+1

Ky Kx

= =KA

(63)

(65)

(66)

(64)

(67)


Notations used:

xAb : liquid-phase mole fraction of A in the bulk liquid

yAb : gas-phase mole fraction of A in the bulk gas

xAi : liquid-phase mole fraction of A at the interface

yAi : gas-phase mole fraction of A at the interface

xA* : liquid-phase mole fraction of A which would have been in

equilibrium with yAb

yA* : gas-phase mole fraction of A which would have been in

equilibrium with xAb

kx : liquid-phase mass-transfer coefficient

ky : gas-phase mass-transfer coefficient

Kx : overall liquid-phase mass-transfer coefficient

Ky : overall gas-phase mass-transfer coefficient

KA : vapour-liquid equilibrium ratio (or equilibrium distribution coefficient)


xAi

yAi

yAb

xAb

yA*

xA* xA

yA

yAi = KA xAi

yA* = KA xAb

yAb = KAxA*

Gas-liquid equilibrium ratio (KA) curve


Gas & Liquid-side Resistances in Interfacial Mass Transfer

1

KL

1

H kp

= +1

kc

1

KG

1

kp

= +H

kc

fG = fraction of gas-side resistance

=1/KG

1/kp

1/kp

1/kp=+ H/kc kc

kc=+ H kp

fL = fraction of liquid-side resistance

=1/KL

1/kc

1/Hkp

1/kc=+ 1/kc + kc/H

kp=kp


If fG > fL, use the overall gas-side mass transfer coefficient and the overall gas-side driving force.

If fL > fG use the overall liquid-side mass transfer coefficient and the overall liquid-side driving force.

Gas & Liquid-side Resistances in Interfacial Mass Transfer


1

kp

HA

kc

+1

KG KL

= =HA (58 and 61)

The above is also written with the following notations:

1

KOG

1

KG

= +H

KL

=H

KOL


Other Driving Forces

Mass transfer is driven by concentration gradient as well as by pressure gradient as we have just seen.

In pharmaceutical sciences, we also must consider mass transfer driven by electric potential gradient (as in the transport of ions) and temperature gradient.

Transport Processes in Pharmaceutical Systems (Drugs and the Pharmaceutical Sciences, vol. 102), edited by G.L.

Amidon, P.I. Lee, and E.M. Topp (Nov 1999)

Encyclopedia of Pharmaceutical Technology (Hardcover)by James Swarbrick (Author)


Example 1

An exhaust stream from a containing 3 mole% acetone and 90 mole% air is fed to a mass transfer column in which the acetone is stripped by a countercurrent, falling 293 K water stream. The tower is operated at a total pressure of 1.013x105 Pa. If a combination of Raoult-Dalton equilibrium relation may be used to determine the distribution of acetone between the air and aqueous phases, determine

(a) The mole fraction of acetone within the aqueous phase which would be in equilibrium with the 3 mole% acetone gas mixture, and

(b) The mole fraction of acetone in the gas phase which would be in equilibrium with 20 ppm acetone in the aqueous phase.

Example 1 worked out



Example 2

The Henry’s law constant for oxygen dissolved in water is 4.06x109 Pa/(mole of oxygen per total mole of solution) at 293 K. Determine the solution concentration of oxygen in water which is exposed to dry air at 1.013x105 Pa and 293 K.



Henry’s law can be expressed in terms of the mole fraction units by

pA = H’ xA

where H’ is 4.06x109 Pa/(mol of oxygen/total mol of solution).

Dry air contains 21 mole percent oxygen. By Dalton’s law

pA = yA P = (0.21)(1.013x105 Pa)

= 2.13 x 104 Pa



Example 3

In an experiment study of the absorption of ammonia by water in a wetted-wall column, the overall mass-transfer coefficient, KG, was found to be 2.74 x 10-9 kgmol/m2.s.Pa.

At one point in the column, the gas phase contained 8 mole% ammonia and the liquid-phase concentration was 0.064 kgmole ammonia/m3 of solution. The tower operated at 293 K and 1.013x105 Pa. At that temperautre, the Henry’s law constant is 1.358x103 Pa/(kgmol/m3).

If 85% of the total resistance to mass transfer is encountered in the gas phase, determine the individual film mass-transfer coefficients and the interfacial compositions.







Example 4

A wastewater stream is introduced to the top of a mass-transfer tower where it flows countercurrent to an air stream. At one point in the tower, the wastewater stream contains 10-3 mole A/m3 and the air is essentially free of any A. At the operation conditions within the tower, the film mass-transfer coefficients are KL = 5x10-4 kmole/m2.s.(kmole/m3) and KG = 0.01 kmole/m2.s.atm. The concentrations are in the henry’s law region where pA,i = H CA,i with H =1 0 atm/(kmole/m3). Determine the following:

(a) The overall mass flux of A

(b) The overall mass-transfer coefficients, KOL and KOG.






Documents

Prof. R. Shanthini 27 Feb & 05 Mar 2012 Content: Mass transfer: concept and theory PM3125: Lectures 1 to 5 uploaded at