1 Introduction To Mass Transfer And Diffusion.pdf

EKC 217: MASS TRANSFER 1

TOPIC 1: DIFFUSION

COURSE OUTCOME 2

CO1: Perform calculations on diffusion and mass transfer problems. Course Strategies: Lectures covering: Laws on diffusion and their applications as related to mass transfer in practical applications such as diffusion with bulk flow and through stagnant media. The use of dimensionless co-relations of Sherwood number with Reynolds Number and Schmidt Number.

Course Activities: Lectures; in-class examples; tutorials and assignments EKC 217: Introduction to Mass Transfer and Diffusion

Course Learning Objectives 3

By end of this topic, student should be able to: 1) Understand the concepts of mass transfer and diffusion. 2) Carry out calculations on diffusion and mass transfer problems.

EKC 217: Introduction to Mass Transfer and Diffusion

What is MASS TRANSFER? 4

 A term used to indicate the transference/movement of a component (at molecular level) in a mixture from a region where its concentration is high to a region where the concentration is lower.  Among some of the familiar phenomena that involves mass transfer are: 1) Liquid in an open pail evaporates into still air because of the difference in concentration of water vapor at the water surface and the surrounding air. Liquid to gas EKC 217: Introduction to Mass Transfer and Diffusion

2) A piece of sugar added into a cup of coffee eventually dissolves by itself and diffuses to the surrounding solutions. Solid to liquid

3) A piece of solid CO2 (dry ice) also gets smaller and smaller in time as the CO2 molecules diffuse into the air. Solid to gas EKC 217: Introduction to Mass Transfer and Diffusion

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 Mass transfer occurs by two basic mechanisms: Molecular diffusion - random and spontaneous microscopic movement of individual molecules in a gas, liquid or solid. Eddy (turbulent) diffusion – due to turbulent flow of fluid.  In a binary mixture, molecular diffusion occurs because of one or more different potentials or driving force, including differences (gradients) of concentration, pressure (pressure diffusion), temperature (thermal diffusion) and external force fields (forced diffusion).  This chapter will only focus on molecular diffusion caused by CONCENTRATION GRADIENTS as it is the most common type of molecular diffusion in commercial separation process. EKC 217: Introduction to Mass Transfer and Diffusion

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The rate at which the process takes place is dependent both on the driving force (concentration difference) and on the mass transfer resistance. In most of these applications, mass transfer takes place across a phase boundary where the concentrations on either side of the interface are related by the phase equilibrium relationship. Where a chemical reaction takes place during the course of the mass transfer process, the overall transfer rate depends on both the chemical kinetics of the reaction and the mass transfer resistance.

EKC 217: Introduction to Mass Transfer and Diffusion

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Classification of Mass Transfer Operations 8

Based on the three various phases available, i.e.: solid, liquid and gas, six possibilities of contact are available: 1. 2. 3. 4. 5. 6.

Gas-Gas Gas-Liquid Gas-Solid Liquid-Liquid Liquid-Solid Solid-Solid

EKC 217: Introduction to Mass Transfer and Diffusion

1) Gas-Gas  The phase contact between gas-gas is not practically realized as most gasses are completely soluble in each other.

2) Gas-Liquid  This is on of the most common phase contact in the industries.  Among some of the industrial process that has this two phase contact are distillation, absorption, desorption/stripping.

EKC 217: Introduction to Mass Transfer and Diffusion

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a) Distillation  The main function is to separate a liquid mixture, of miscible and volatile substances into individual components by vaporization;  Utilizes the differences in volatility for separation while the vapor phase is created from the liquid by application of heat;  Example of industrial process: Separation of a mixture of alcohol and water into its component and the separation of a mixture of benzene and toluene into its component (Figure 1). EKC 217: Introduction to Mass Transfer and Diffusion

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Benzene

Mixture of benzene & toluene

Toluene

Figure 1

b) Gas Absorption  In gas absorption, a soluble vapor is absorbed by means of a liquid in which the solute gas is more or less soluble using selective absorbent.  For instance, if a mixture of air and ammonia is in contact with liquid water, a large portion of ammonia but essentially no air will dissolve in the liquid (Figure 2).  Through this, the air/ammonia mixture can be separated.  The process need more than one column in order to achieve 100% separation. EKC 217: Introduction to Mass Transfer and Diffusion

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Air (outlet)

Water (inlet)

Absorption column

Air & ammonia Water & ammonia (inlet) (outlet)

Figure 2

c) Gas Desorption/Stripping  Similar to gas absorption, but the only difference is purely in the direction of solute transfer;  For example, if air is brought into contact with an ammonia-water solution, some of the ammonia will leave the liquid phase and enters the gas phase. 3) Gas-Solid  If a solid is moistened with a volatile liquid and is exposed to a relatively dry gas, the liquid leaves the solid and diffuses into the gas, an operation generally known as DRYING.

EKC 217: Introduction to Mass Transfer and Diffusion

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4) Liquid-Liquid  Separations involving the contact of two insoluble liquid phase are known as liquid-liquid extraction.  A simple and familiar example is the separation of acetone from acetone-water solution using carbon tetrachloride. If the acetone-water solution is shaken in a separatory funnel with carbon tetrachloride and the liquids are allowed to settle, a large portion of the acetone will be found in the carbon tetrachloride-rich phase and will thus have been separated from the water. Water

 Acetone-water solution

Carbon tetrachloride-acetone solution

Carbon tetrachloride solution

EKC 217: Introduction to Mass Transfer and Diffusion

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5) Liquid-Solid  Crystallization is one of the process that involves liquid and solid phase. It is used to obtain materials in attractive and uniform crystals of good purity, separating a solute from the melt or a solution and leaving impurities behind .  Leaching is a process whereby liquid solvent is used to dilute selective component in a solid mixture. For example, the leaching of gold from its ore by cyanide solution and cotton seed oil from the seeds by hexane. 6) Solid-Solid  E.g.: diffusion of carbon into iron during case-hardening, doping of semiconductors for transistors, migration of doped molecules in semiconductors at high temperature. EKC 217: Introduction to Mass Transfer and Diffusion

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DIFFUSION 15

 Diffusion involves movement of a component (at molecular level) in a

mixture from a region where its concentration is high to a region where the concentration is lower.  The rate of diffusion is conveniently described in terms of a molar flux (mole per unit area per unit time), the area being measured in a direction normal to the diffusion.  In a binary mixture, molecular diffusion occurs because of one or more different potentials or driving force, including differences (gradients) of concentration, pressure (pressure diffusion), temperature (thermal diffusion) and external force fields (forced diffusion).  We will only focus on molecular diffusion caused by concentration gradients as it is the most common type of molecular diffusion in commercial separation process. EKC 217: Introduction to Mass Transfer and Diffusion

 In all mass transfer operations, diffusion occurs in at least one

phase and often in both phases.  For example, in gas absorption, solute diffuses through the gas phase to the interface between the phases and through the liquid phase from the interface.

GAS

LIQUID

 This chapter is only restricted to binary mixture and steady state

is assumed. EKC 217: Introduction to Mass Transfer and Diffusion

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MOLECULAR DIFFUSION

A container is filled with dye solution

Clear water is added on top, dye solution is undisturbed

Uniform mixture of dye and water is formed

Water molecules Dye molecules EKC 217: Introduction to Mass Transfer and Diffusion

Water and dye molecules move across the horizontal plane

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 A container is filled with dye solution.  Clear water is then carefully added on top, so that the dye

solution on the bottom is undisturbed.  At the first instance, a sharp boundary between the two layers. However, after a short while, the upper layer becomes colored and the lower layer becomes less colored.  The color change process is through diffusion of the dye molecules.  By taking the horizontal plane across the solution, on the average, a fraction of molecules in the solution below the plane will cross over to the region above, and the same fraction will cross in the opposite direction.

EKC 217: Introduction to Mass Transfer and Diffusion

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 Since the concentration of dye molecules is higher in

the lower region than the upper, there will be a net rate of mass transfer of dye molecules will take place from the lower region to the upper region.  After a longer period, the concentration of dye will be uniform throughout the solution.

EKC 217: Introduction to Mass Transfer and Diffusion

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 Based on these observations, we can conclude that:

Mass transfer by ordinary molecular diffusion occurs because of a concentration difference or gradient, i.e.: a species diffuses in the direction of decreasing concentration.  The mass transfer rate is proportional to the area normal to the direction of mass transfer and not the volume of the mixture. Therefore, the rate of diffusion can be expressed as a FLUX.  Mass transfer stops when the concentration is uniform. 

 These observations were then quantified by Fick in

1855, and it is known as FICK’S LAW.

EKC 217: Introduction to Mass Transfer and Diffusion

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FICK’S LAW 21

 Fick’s Law for steady state diffusion of a binary mixture of

may be written as:

J A   DAB

dc A dz

(for component A)

A and B ------ (1.1)

where:

JA DAB cA z

= molar flux of component A in the z direction due to molecular diffusion (mole per unit area per unit time, mole A/m2.s) = Diffusivity/Diffusion coefficient of molecule A in B (m2/s) = Concentration of component A (mole per unit volume, mole A/m3) = Distance in the direction of diffusion (m)

EKC 217: Introduction to Mass Transfer and Diffusion

 Similarly, for component B :

dcB (for component B) ------ (1.2) J B   DBA dz  Fick’s Law for steady state diffusion of a binary mixture of A and B may also be written in many other ways, such as: 1) Gradients of mole fraction:

J A   cDAB

------ (1.3)

dx A dz

where: c = Total molar concentration of component A and B xA = Molar fraction of component A NOTE: cxA = cA EKC 217: Introduction to Mass Transfer and Diffusion

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2) In term of partial pressure: cA 

(since yA = pA/pT)

p A p A pT p    yA  T RT pT RT RT

where: pA = partial pressure of A pT = total pressure of the system Therefore:

J A   DAB

dc A  p  dy    DAB  T  A dz  RT  dz

EKC 217: Introduction to Mass Transfer and Diffusion

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------ (1.4)

REVIEW: Different expressions of the concentrations in a binary mixtures of component A and B.

EKC 217: Introduction to Mass Transfer and Diffusion

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Exercise: The composition of air is often given in terms of only the two principal species in the gas mixture: Oxygen, O2 Nitrogen, N2

yO2 = 0.21 yN2 = 0.79

Determine the mass fraction of both oxygen and nitrogen and the mean molecular weight of the air when it is maintained at 25C (298 K) and 1 atm (1.013 x 105 Pa). The molecular weight of oxygen is 0.032 kg/mol and of nitrogen is 0.028 kg/mol. [Ans: O2 = 0.23, N2 = 0.77, mean molecular weight of mixture = 0.0288 kg/mol] EKC 217: Introduction to Mass Transfer and Diffusion

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Example 1: Molecular diffusion of He in N2 A mixture of He and N2 gas is26contained in a pipe at 298 K and 1 atm total pressure, which is constant throughout. At one end of the pipe at point 1 the partial pressure, pA1 of He is 0.6 atm and the other end 0.2 m pA2 = 0.20 atm. Calculate the flux of He at steady state if DAB of the He-N2 is 0.687 x 10-4 m2/s. The universal gas constant, R is given as 82.057 cm3.atm/gmol.K Solution: Beginning from the Fick’s Law:

J He   DHeN 2

dcHe dz

EKC 217: Introduction to Mass Transfer and Diffusion

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Rearranging Eq. (1.1) and integrating: z2

cHe 2

z1

cHe1

J He  dz   DHeN 2 

J He 

dcHe

DHeN 2 cHe1  cHe 2  z2  z1

From the perfect gas law, pHeV = nHeRT, thus:

pHe1 nHe cHe1   RT V

EKC 217: Introduction to Mass Transfer and Diffusion

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Substituting into the earlier equation: J He 

DHeN 2  pHe1  pHe 2  RT ( z2  z1 )

Putting in all the known values: J He

(0.687 x 10 4 )(0.6  0.2)  (82.06 x 10-3 )(298)(0.2  0)  5.63 x 10-6 kgmole He/m 2  s

EKC 217: Introduction to Mass Transfer and Diffusion

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Correlation between Diffusivity 29

For diffusion of A and B in a gas at constant temperature and pressure:

P c A  cB  c  RT

------ (1.5)

The relationship between DAB and DBA is easily determined for ideal gases, since the molar density (concentration) does not depend on the composition:

dcA  dcB  dc  0 EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.6)

Choosing the reference plane for which there is zero volume flow, the sum of the molar diffusion fluxes of A and B can be set to zero:

J Az  J Bz  0

------ (1.7)

The subscript z is often dropped when the direction is obvious. Writing Fick’s Law for A and B for constant total concentration, c :

dc A J A   DAB dz

and

dcB J B   DBA dz

EKC 217: Introduction to Mass Transfer and Diffusion

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------ (1.8)

Replacing Eq. (1.8) into (1.7):  DAB

dc A dcB  DBA 0 dz dz

------ (1.9)

Since dcA = -dcB , therefore:

DAB  DBA

EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.10)

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Diffusion process together with convection 32

Up to now, we have considered Fick’s Law for diffusion in a stationary fluid; i.e.: there has been no net movement or convection flow of the entire phase of the binary mixture A and B. Many practical problems such as the evaporation of water from a lake under the influence of the wind or the mixing of two fluids as they flow in a pipe, involve diffusion in moving medium, i.e. bulk motion is cause by the external force. EKC 217: Introduction to Mass Transfer and Diffusion

What is CONVECTIVE MASS TRANSFER? -Mass transfer between a moving fluid and a surface or between immiscible moving fluids separated by a mobile interface (gas/liquid or liquid/liquid contactor)  Mass transfer = diffusion + bulk motion of medium (convective) The diffusion flux, JA occurred because of the concentration gradient. The rate at which moles of A passed a fixed point to the right, which will be taken as a positive flux, is JA kg mole A/m2·s. This flux can be converted to a velocity of diffusion of A to the right by:

 m kg mole  J A (kg mole A/m 2  s)   Ad c A   3 m s  where Ad is the diffusion velocity of A in m/s. EKC 217: Introduction to Mass Transfer and Diffusion

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------ (1.11)

Now let’s consider what happens when the whole fluid is moving in bulk or convective flow to the right. The molar average velocity of the whole fluid relative to a stationary point is M (m/s). Component A is still diffusing to the right, but now its diffusion velocity, Ad , is measured relative to the moving fluid. To a stationary observer, A is moving faster than the bulk of the phase, since its diffusion velocity, Ad , is added to that of the bulk phase, M . Expressed mathematically, the velocity of A relative to the stationary point is the sum of the diffusion velocity and the average or convective velocity: ------ (1.12)  A   Ad  M

where A is the velocity of A relative to a stationary point. EKC 217: Introduction to Mass Transfer and Diffusion

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Expressed pictorially: A Ad

A = actual velocity of A Ad = diffusional velocity of A M = velocity of bulk

M

observer sees actual movement as A

Multiplying Eq. (1.12) by cA:

c A A  c A Ad  c AM

------ (1.13)

Each of the three terms represents a flux. The first term, cAA, can be represented by the flux NA kg mole A/m2·s. This is the total flux of A relative to the stationary point. The second term is JA, the diffusion flux relative to the moving fluid. The third term is the convective flux of A relative to the stationary point. EKC 217: Introduction to Mass Transfer and Diffusion

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Hence, Eq. (1.13) becomes:

N A  J A  cAM

------ (1.14)

Let N be the total convective flux of the whole stream relative to the stationary point. Then,

------ (1.15)

N  cM  N A  N B Solving for M (bulk velocity) gives:

M 

N A  NB c

------ (1.16)

Substituting Eq. (1.16) into (1.14):

cA N A  J A  (N A  NB ) c EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.17)

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Since JA is Fick’s law:

N A  cDAB

dx A c A  (N A  NB ) dz c

------ (1.18)

Equation (1.18) is the final general equation for diffusion plus convection to use when the flux NA is used, which is relative to the stationary point. A similar equation can be written for NB:

N B  cDBA

dxB cB  (N A  NB ) dz c

------ (1.19)

To solve Eq. (1.18) or (1.19), the relation between the flux NA and NB must be known. Eq. (1.18) and (1.19) hold for diffusion in gas, liquid or solid.

For equimolar counter diffusion, NA = -NB and the convective term in Eq. (1.18) becomes zero. Then, NA = JA = -NB = -JB . EKC 217: Introduction to Mass Transfer and Diffusion

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Eq. (1.18) and (1.19) can also be used in different forms. For example, since N = cM and cxA = cA, thus:

N A  c A M  DAB

dc A dz

------ (1.20)

Therefore, the appropriate equation used to solve a problem would entirely depends on the information given in the problem. Equation (1.18) or (1.20) is the basic equation for mass transfer in a non-turbulent fluid phase. It accounts for the amount of component A carried by the convective bulk flow of the fluid and the amount of A being transferred by molecular diffusion. EKC 217: Introduction to Mass Transfer and Diffusion

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There are several types of situation covered by Eq. (1.18) or (1.20). Among two of them that will be discussed in the syllabus are: 1) Equimolar counterdiffusion

2) Unimolecular diffusion (diffusion of a single component through stationary second component).

EKC 217: Introduction to Mass Transfer and Diffusion

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Equimolar counterdiffusion 40

In equimolar counterdiffusion, the molar fluxes of A and B is equal, but in opposite direction or the net volumetric and molar flows are zero. A typical example of this case is the diffusion of A and B in the vapor phase for distillation that have constant molar overflow.

EKC 217: Introduction to Mass Transfer and Diffusion

Since the net volumetric and molar flows are zero, thus Eq. (1.18) can be used with the convective term is set to zero, as shown below: ------ (1.21)

dx A J A  cDAB dz

Eq. (1.21) is then integrated over a film thickness of zT, assuming a constant flux, JA: xA

zT

 DAB c  dx A  J A  dz x Ai

------ (1.22)

0

EKC 217: Introduction to Mass Transfer and Diffusion

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Integrating Eq. (1.22) and rearranging gives: DAB c DAB JA  ( x Ai  x A ) or J A  (c Ai  c A ) zT zT

------ (1.23)

The concentration gradient for A is linear in the film, and the gradient for B has the same magnitude but the opposite sign.

EKC 217: Introduction to Mass Transfer and Diffusion

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Example 2: Ammonia gas (A) is diffusing through a uniform tube 0.10 m long containing N2 gas (B) at 1.0132 x 105 Pa pressure and 298 K. At a point 1, pA1 = 1.013 x 104 Pa and at a point 2, pA2 = 0.507 x 104 Pa. The diffusivity DAB = 0.230 x 10-4 m2/s. Calculate the flux JA at steady state and repeat for JB. Solution: Given: P = 1.0132 x 105 Pa z2 – z1 = 0.10 m

T = 298 K

Substitute known values into the following equation: DAB  p A1  p A2  JA  RT ( z2  z1 ) EKC 217: Introduction to Mass Transfer and Diffusion

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0.23 x 10 1.013 x 10  4

JA

 0.507 x 10 4 8314(298)(0.10  0) 4



 4.70 x 10-7 kgmole A/m2  s For component B: pB1 = P - pA1 = 1.0132 x 105 – 1.013 x 104 = 9.119 x 104 Pa pB2 = P – pA2 = 1.0132 x 105 – 0.507 x 104 = 9.625 x 104 Pa Hence,

 0.23 x 10 9.119 x 10  4

JB

 9.625 x 10 4 8314(298)(0.10  0) 4



 - 4.70 x 10-7 kgmole B/m 2  s The negative value for JB means the flux goes from point 2 to point 1.

EKC 217: Introduction to Mass Transfer and Diffusion

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Unimolecular Diffusion 45

In unimolecular diffusion, mass transfer of component A occurs through stagnant component B, NB = 0. Therefore, the total flux to or away from the interface, N is the same as NA.

EKC 217: Introduction to Mass Transfer and Diffusion

A typical example of this case is the evaporation of a liquid with the diffusion of the vapor from the interface into a gas stream. Based on the definition of unimolecular diffusion, Eq. (1.18) becomes:

N A  cDAB

dx A  xA N A dz

EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.24)

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Rearranging: dx A N A (1  x A )  cDAB dz

NA 1 or dz   dx DAB c 1  xA

------ (1.25)

Integrating: x A dx N A zT 1  xA dz     ln  x Ai 1  x DAB c 0 1  x Ai A

------ (1.26)

Or:

DAB c 1  x A NA  ln zT 1  x Ai

EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.27)

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MOLECULAR DIFFUSION IN GASES 48

Special case for A diffusing through stagnant B: In the gas phase, e.g.: if ammonia (A) were being absorbed from air (B) into water, only ammonia diffuses since air does not dissolve appreciably in water.

EKC 217: Introduction to Mass Transfer and Diffusion

Thus, NB = 0 and NA = constant. From Eq. (1.18): N A  cDAB

dx A c A  (N A  NB ) dz c

------ (1.28)

Since NB = 0, therefore: N A  cDAB

dx A c A  (N A ) dz c

EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.29)

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PT c , RT

------ (1.30)

p A  x A PT

Hence: DAB dp A p A NA    NA RT dz PT

------ (1.31)

Rearranging:  pA  DAB dp A   N A 1    RT dz  PT 

------ (1.32)

Integrating:

NA

z2

z1

DAB p A 2 dp A dz   RT p A1 1  p A / PT

EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.33) 50

After integration, Eq. (1.33) becomes:

DAB PT PT  p A2 NA  ln RT ( z2  z1 ) PT  p A1

------ (1.34)

Since: PT – pA2 = pB2 , PT – pA1 = pB1 , pB2 – pB1 = pA1 – pA2 , then:

DAB PT p A1  p A2 pB 2 NA  ln RT ( z2  z1 ) pB 2  pB1 pB1

------ (1.35)

The logarithmic mean of pB1 and pB2 is given by:

pBM

pB 2  pB1  ln( pB 2 / pB1 )

EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.36) 51

Substituting Eq. (1.36) into Eq. (1.35) gives:

DAB PT NA  ( p A1  p A2 ) RT ( z2  z1 ) pBM

------ (1.37)

Compare with the earlier equation for equimolar counterdiffusion: DAB  p A1  p A2  JA  RT ( z2  z1 )

Therefore, in the present case, PT/pBM can be regarded as correction factor.

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In addition, for gases, Eq. (1.14): N A  J A  c AM can also be expressed using mole fraction in vapor phase (yA), since:

cA  M y A

and

N M  M

where: M = molar density (kgmole/m3) = 1/22.41 kgmole/m3 (at standard conditions, 0C & 1 atm) yA = mole fraction of component A in vapor phase N = total convective flux of the whole stream relative to the stationary point (kgmole/m2·s) M = molar average velocity (ms-1) cA = molar concentration of component A (kgmole/m3) EKC 217: Introduction to Mass Transfer and Diffusion

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Eq. (1.14) becomes:

dy A N A  y A N  DAB  M dz

------ (1.38)

Since N = NA + NB , and when only component A is being transferred (i.e.: NB = 0), the total flux to or away from the interface N is the same as NA, then Eq. (1.38) becomes:

dy A N A  y A N A  DAB  M dz

EKC 217: Introduction to Mass Transfer and Diffusion

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------ (1.39)

Rearranging and integrating:

dy A N A (1  y A )   DAB  M dz z2

yA2

z1

y A1

N A  dz   DAB  M 

------ (1.40)

dy A (1  y A )

DAB  M 1  y A2 NA  ln z2  z1 1  y A1

EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.41)

------ (1.42)

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Similarly,

yB1 = 1 – yA1 yB2 = 1 – yA2 yB2 – yB1 = yA1 – yA2 Then,

DAB  M y A1  y A2 yB 2 NA   ln z2  z1 yB 2  yB1 yB1 The logarithmic mean of yB1 and yB2 is given by:

yBM

yB 2  yB1  ln( yB 2 / yB1 )

------ (1.43)

------ (1.44)

Finally, by substituting Eq. (1.44) into Eq. (1.43) gives:

DAB  M NA  ( y A1  y A2 ) ( z2  z1 ) yBM EKC 217: Introduction to Mass Transfer and Diffusion

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------ (1.45)

Example 3: Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assume dry) is 1.01325 x 105 Pa (1.0 atm) and the temperature is 293 K (20C). Water evaporates and diffuses through the air in the tube and the diffusion path z2 –z1 is 0.1524 m (0.5 ft) long. The diagram is similar to the shown figure. Calculate the rate of evaporation at steady state in lb mol/ft2· h and kgmole/m2· s. The diffusivity of water vapor at 293 K and 1 atm pressure is 0.250 x 10-4 m2/s. Assume that the system is isothermal. Use SI and English units. EKC 217: Introduction to Mass Transfer and Diffusion

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pA2 = 0

Air (B)

0.5 ft

NA pA1 = 0.0231 atm Water (A)

T = 293 K

Solution: The diffusivity is converted to ft2/h by using the conversion factor (refer Appendix 1, McCabe, Smith and Harriott).

DAB  (0.250 x 10-4 )(3.875 x 104 )  0.969 ft 2 /h Using Appendix 7 (McCabe, Smith and Harriott), the vapor pressure of water at 20C (68 F) is 0.3402 lbf/in2 or 2345.6 N/m2.

2345.6 p A1   0.0231 atm 5 1.01325 x 10

and

p A2  0 (pure air)

T = 460 +68 = 528R = 293 K

R = 82.057 cm3· atm/gmole· K = 0.730 ft3 · atm/lbmole· R EKC 217: Introduction to Mass Transfer and Diffusion

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To calculate the value of pBM (from Eq. 1.36):

pB1 = PT – pA1 = 1.00 – 0.0231 = 0.9769 atm pB2 = PT – pA2 = 1.00 – 0 = 1.00 atm Therefore: pBM 

pB 2  pB1 1.00  0.9769   0.988 atm  1.001 x 105 Pa ln( pB 2 / pB1 ) ln(1.00 / 0.9769)

Since pB1 is close to pB2 , the linear mean (pB1 +pB2)/2 could be used and would be very close to pBM .

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Substituting in Eq. (1.37) with z2 – z1 = 0.5 ft (0.1524 m), thus: NA  

DAB PT ( p A1  p A 2 ) RT ( z 2  z1 ) pBM 0.969(1.0)(0.0231  0) 0.730(528)(0.5)(0.988)

 1.175 x 10  4 lbmole/ft 2  h

NA 

DAB PT ( p A1  p A2 ) RT ( z2  z1 ) pBM

(0.250 x 10  4 )(1.01325 x 105 )(2.341 x 103  0)  8314(293)(0.1524)(1.001 x 105 )  1.595 x 10 7 kgmole/m 2  s

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Exercise: a) For the diffusion of solute A through a layer of gas to an absorbing liquid with yAi = 0.20 and yA = 0.10, calculate the rate transfer for unimolecular diffusion compared to that for equimolar counter diffusion. b) What is the value of yA halfway through the layer for unimolecular diffusion? [Ans: yA = 0.1515]

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MOLECULAR DIFFUSION IN LIQUIDS 62

 Diffusion of solutes in liquid is very important in many industrial processes especially in such separation operations such as: 1) Gas absorption 2) Distillation 3) Liquid-liquid extraction or solvent extraction  Rate of molecular diffusion in liquids is considerably slower than in gases.  The molecules in a liquid are very close together compared to a gas. Therefore, the molecules of the diffusing solute A will collide with molecules of liquid B more often and diffuse more slowly than in gases. EKC 217: Introduction to Mass Transfer and Diffusion

 In diffusion in liquids, an important difference from diffusion in gases is that the diffusivities are often dependent on the concentration of the diffusing components.  Similar to those for gases, equations for diffusion in liquids can be classified in two cases: 1) Steady-state equimolar counterdiffusion: Starting from Eq. (1.18): N A  cDAB dx A  c A ( N A  N B ) dz c and knowing NA = -NB , then: DAB cav DAB JA  ( x Ai  x A )  (c Ai  c A ) zT zT

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------ (1.46)

cav is defined as follows:

cav

      M  av

 1 2     M1 M 2    2

------ (1.47)

where:

cav M1 1

= average total concentration of A + B (kgmole/m3) = average molecular weight of the solution at point 1 (kg mass/kgmole) = average density of the solution at point 1 (kg/m3)

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2) Steady-state diffusion of A through non-diffusing B:

NA = constant, NB = 0

DAB NA  ( z2  z1 ) xBM where:

   ( x A1  x A2 )  M  av

xBM 

------ (1.48)

xB 2  xB1 ln( xB 2 / xB1 )

Note that xA1 + xA2 = xB1 + xB2 = 1.0 For dilute solution, xBM is close to 1.0 and c is essentially constant. Then, Eq. (1.48) simplifies to:

DAB NA  (c A1  c A2 ) ( z2  z1 ) EKC 217: Introduction to Mass Transfer and Diffusion

------ (1.49) 65

Example 4: Calculate the rate of diffusion of acetic acid (A) across a film of nondiffusing water (B) solution 1 mm thick at 17C when the concentrations on opposite sides of the film are 9 and 3 wt %, respectively. The diffusivity of acetic acid in the solution is 0.95 x 10-9 m2/s. Solution:

Given: (z2 – z1) = 0.001 m MA = 60.03 kg/kmole MB = 18.02 kg/kmole At 17 C: Density of the 9% solution = 1012 kg/m3 Density of the 3% solution = 1003.2 kg/m3 EKC 217: Introduction to Mass Transfer and Diffusion

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Consider basis of solution = 1 kg, At point 1: x A1 

0.09 / 60.03 0.0015   0.0288 mole fraction acetic acid 0.09 / 60.03  0.91 / 18.02 0.0520

xB1  1  0.0288  0.9712 mole fraction water Molecular weight of the solution, M 1 

1

1012   52.7 kmole/m 3 M 1 19.21

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1  19.21 kg/kmole 0.0520

Similarly, at point 2: x A2 

0.03 / 60.03 0.0005   0.0092 mole fraction acetic acid 0.03 / 60.03  0.97 / 18.02 0.0543

xB 2  1  0.0092  0.9908 mole fraction water Molecular weight of the solution, M 2 

1  18.42 kg/kmole 0.0543

2

1003.2   54.5 kmole/m 3 M2 18.42

Then,  

    M  av

52.7  54.5  53.6 kmole/m 3 2

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xBM

xB 2  xB1 0.9908  0.9712    0.980 ln( xB 2 / xB1 ) ln(0.9908 / 0.9712)

Finally, substitute all known values in Eq. (1.48):

0.95 x 10 9 53.6(0.0288  0.0092) NA  (0.001)(0.980)  1.018 x 10-6 kmole/m 2  s

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