Mass Transfer 2 Definitions and Diffusion v2

8/12/2019 Mass Transfer 2 Definitions and Diffusion v2

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2.2

S. Bickerton, 2001, R. Sharma 2007, S. Norris 2013 MechEng 713 Energy Technology

The definitions and equations listed above are necessary for the treatment of chemical mixtures. You should becomecomfortable with their use before we continue with the analysis of mass transfer processes.

For our course we will mostly consider binary mixtures, which are mixtures of two chemical species. For binary

mixtures where one species has a much higher mass (or molar) fraction than the other, the dominant species is called thesolvent, and the minor species is called the solute. For instance, for air dissolving into water, the water is the solvent,and the air is the solute.

2.1 Ideal Gas Mixtures

Most of the gasses we encounter in typical engineering applications can be considered to act as ideal gases. Here we

show that the molar fractions of such mixtures can be related to the partial pressures of the component gases.

Recall the Ideal Gas law relationships for a pure gas:

P RT= where R is the gas constant of the gas

UP CR T = where

UR is the universal gas constant

Note:U

R = 8.314 [kJ/kmol.K]

For the i

th

component of a mixture of ideal gases, the Partial Pressure Piis:

i i iP R T=

1

N

i

i

P P=

=

i i UP C R T = where

U i iR R M= = 8.314 [kJ/kmol.K]

We will prove the following useful relationship:

i

i

Px

P=

where Piis the partial pressure of component i, Pis the total pressure of the gas mixture, andxiis the molar fraction of

component i.

Proof: i i iP RT

P P

= using

i i iP R T=

i U

i

R T

M P

= using

U i iR R M=

U

i

R TC

P= using

i i iM C =

U

i

CR Tx

P= using i

i

Cx

C=

ix= using

UP CR T = Q.E.D.


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2.2 Concentrations at Interfaces

At the boundary between two phases of matter the concentrations will be discontinuous the concentration of achemical species in one phase will differ from that in the other. For instance, the salt in a solid block of pure salt will

have a concentration of 1. However, if the block is placed in some water, the concentration of the salt in the water willbe less than 1, since it is mixed with water. To solve mass transfer problems we need to be able to calculate theconcentrations at these phase interfaces.

At the phase interface the two phases can normally considered to be in thermodynamic equilibrium (unless there is achemical reaction occurring). Therefore the temperature will be the same across the interface. In contrast, the

concentrations in the two phases will normally be different. However, immediately adjacent to the interface themixtures can be considered to be saturated, they have absorbed as much mass from the other phase as they can, andcannot absorb any more.

In order to deal with discontinuities at interfaces when solving mass transfer problems, we will define two imaginarysurfaces, one either side of the physical interface. These are positioned an infinitesimally small distance away from the

real surface. Each surface is labelled as in the diagram below. You should get in the habit of defining these surfacescorrectly when setting out to solve a problem.

The surfaces are customarily labelled u and s, with u being the more solid of the two phases.

u for the solid side of solid-liquid interfaces

u for the solid side of solid-gas interfaces

u for liquid side of liquid-gas interfaces

For solid-solid and liquid-liquid interfaces the labelling is arbitrary, and we sometimes label them as uand v.

The subscripts and used together with the species label, to specify the mass and molar fractions of species at different

locations. For instance, n1,uis the mole fraction of species 1 at location u, and w2,sis the mass fraction of species 2 atthe surface s.

The data to determine the saturation condition that occurs at the interfaces is presented in different forms, depending onthe physical situation. For instance, the evaporation of liquid into a gas is presented in a different form to that for

solid/liquid interfaces. The different cases are discussed below.

Real Interface

Imaginary

Surfaces

u sSurface Labels


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2.2.1 Liquid / Gas interfaces (evaporation of liquid into gas)

The schematic below depicts the interface between a body of water, and an air stream into which water vapour isevaporating. For air/water mixtures it is customary to label the dry air as species a, and the water as species v.

Considering the concentration of H2O, what are the molar concentrations at the u and s surfaces? At surface u thewater is practically a pure liquid (very little air is absorbed by the water) and the molar fraction is approximately 1:

, 1v ux =

Adjacent to the water surface the air is fully saturated with water vapour . In order to calculate this concentrationwe will consult saturated steam tables to find the saturation pressure of the water vapour. We assume thermodynamic

equilibrium at the interface, and so the temperature of the water will equal the temperature of the gas.

s uT T=

We can then specify v,sx , if we know sT and P of the air/water mixture. From the ideal gas relations:

,

, , ,

v sat

v s v sat v s

PP P x

P= =

wherev,sat

P is the saturated vapour pressure ats

T which can be found in steam tables ([2] Table A.12a, [4] Table A.4).

xv,u=1

xv,s

xv

Water vapour / air

mixture

u s

Water

Pv,s

Entropy

Temperature

Ts

State of H2O at ssurface

State of H2O at usurface


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2.2.2 Liquid / Gas interface (absorption of gas by liquid)

The schematic below depicts the interface between pure CO2gas, and a body of water that is absorbing the gas.

Considering the concentration of CO2 on a molar basis, what are the molar concentrations at surfaces u and s?Within the pure CO2:

2,1

CO sx =

For gases that are only sparingly soluble in a liquid, the solubility data can be conveniently represented by Henrys law,which states that the solubility of species i is proportional to the pressure and a function of temperature:

( )

( )

,

, ,

,

,

i u

i s He i

i

i u

He i

x P

x C T

Px

C T

=

=

The Henry constant for a given species, CHe,i, is a function only of temperature, and can be found in [1] Table A.21.Note that this data is indexed to temperatures in K and pressures in bar (105Pa). If you use other units your answer will

be wrong.

Note that the concentration of the gas in the liquid is proportional to the partial pressure of the gas, and inverselyproportional to the Henry constant (which increases with temperature). Therefore, you can add a gas to a liquid byincreasing its partial pressure, and remove it by heating the liquid.

With the solubility of gases being very low, the molecular weight of the solution at the usurface will be approximatelythat of the pure liquid, and so the mass fraction can be approximated by:

1, 1 2, 2 2u u uM x M x M M= +

1

1, 1,

2

u u

Mm x

M

xCO2,s=1

xCO2,u

xCO2

Aqueous CO2solution

u s

Pure CO2


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2.2.3 Solid / Liquid interface (dissolving of solid into liquid)

The schematic below depicts the interface between solid salt (NaCl), and water into which the salt is dissolving.

Considering the concentration of salt on a mass basis, what are the mass concentrations at surfaces u and s? Withinthe solid salt:

1, 1um =

At the liquid side of the phase interface (surface s) the salt / water mixture will be fully saturated with salt. Assuming

thermodynamic equilibrium, the concentration at this point can then be found from published solubility data.

The data in [1] Table A.24 is presented in the form of kg of the solid species that can be dissolved in 100 kg of water.Therefore the mass fraction at the ssurface can be calculated as

1,100

m

s

m

Sm

S=

+

where Smis the solubility from the table.

u s

mNaCl,u=1

mNaCl,s

mNaCl

Salt / water

mixtureSolid salt


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2.7


2.2.4 Solid / Gas interface (absorption of gas by solid)

Finally we consider the case of a gas being absorbed by a solid (something you might not have encountered before). Inthis example we will consider Hydrogen gas being absorbed into Iron.

There are several ways of presenting solubility data. The simplest is a dimensionless solubility coefficient, where the

concentration of the gas in the solid is proportional to the concentration in the surroundings.

,

,

gas u

gas s

C

C= S

If we can approximate the gas phase by the ideal gas equation, the concentration inside the solid is a function of thepartial pressure of the gas.

,

, ,

gas s

gas u gas s

U

PC C

R T = =S S

Sometimes the data is presented as a solubility S, where the dependence on temperature is ignored. Note that this form

is no longer dimensionless but has units of kmol/m3.Pa or kmol/m3.bar, so it is important to specify pressure in the

same units as the solubility data.

, ,gas u gas sC P= S

For gas-solid mixtures the mass or molar fraction of the gas species in the solid is typically very small. Therefore the

mixture density and molecular weight can be approximated by that of the solid component, and the molar fraction canbe calculated from

,

, ,

,

, , ,

gas u solid

gas u gas u

solid u solid

gas gas

gas u gas u gas u

solid

C Mx C

C

M Mm x C

M

=

=

x2,s=1

xH2,u

xH2

Solid Fe

u s

Pure H2


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2.3 Air / Water Vapour Mixtures

Air / water vapour mixtures are commonly encountered in engineering applications. It is possible to develop simplerelationships specifically for this mixture. These are used extensively in the air-conditioning industry.

Let: Air = species a (dry air)

H2O = species v (water vapour)

Given: aM = 28.84 [kg/kmol]

vM = 18.02 [kg/kmol]

Utilising these molecular weights, we can define relations for the mass fractions of the water vapour, based upon the

mole fractions:

1.6 0.6

v v v v

v

a a v v

v

v

v

x M x Mm

M x M x M

xm

x

= =+

=

The mole fraction of the water vapour can be determined from the mass fraction by:

1.6

1 0.6

v

v

v

mx

m=

+

A commonly encountered measure of the water vapour content of air is the relative humidity , often presented as apercentage. This is the ratio of the molar fraction of water vapour to the maximum molar fraction that will evaporate at

the mixtures temperature, the saturationmolar fraction. To find the saturation molar fraction we can use the saturatedvapour pressure data from steam tables, as was discussed in 2.2.1 above.

( ) ( )

v v

v sat v sat

x P

x P = =

( )

( )

v sat

v satPx

P=

Therefore, if a mixture is described in terms of relative humidity, the molar fraction of the mixture can be calculated as,

( )v sat

v

Px

P

=

By using the relationship between mass and mole fractions for air / water vapour mixtures we get

( )

( )1.6 0.6

v sat

v

v sat

Pm

P P

=

Another commonly used measure of water vapour content is the specific humidity, , which is the pass of water vapour

per unit mass of air,

Mass

Mass

v

a

=

The following relationships are useful for converting between the various measures of humidity,

( )

( )

0.6250.625

1

0.625

v satv

v v sat

v

Px

x P P

x

= =

=+

( ) ( )0.625

1

v sat

v

P

P

m

=+

=+


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When calculating convective mass transfer you may need to know the density, viscosity and thermal conductivity of themixture. As a first approximation the viscosity and thermal conductivity may be approximated by that of dry air at thesame temperature. The density of the mixture varies with temperature and composition. Assuming that the mixture isan ideal gas,

( )

10.82 28.84

v av v

v a

U U

v

U

P P MP M

R T R T

P PR T

= + = +

+=

2.4 Example Questions

1. Assuming air to be composed exclusively of O2and N2, with their partial pressures in the ratio 0.21 to 0.79,what are their mass fractions?

2. A mixture of CO2 and N2 is in a container at 25C, with each species having a partial pressure of 1 bar.Calculate the molar concentration, the mass density, the mole fraction, and the mass fraction of each species.

3. Consider a gas mixture of helium, argon and xenon. The mixture contains equal concentrations of eachcomponent on a molar basis. What is the composition of this mixture on a mass basis?4. Consider again a gas mixture of helium, argon and xenon. The mixture now contains equal concentrations of

each component on a mass basis. What is the composition of this mixture on a molar basis?

5. Methane is burned with 20% excess air. At 1250 K the equilibrium composition of the product is:Species i: CO2 H2O O2 N2 NO

xi: 0.0803 0.160 0.0325 0.727 0.000118

Determine the mean molecular weightMof this mixture.

6. Consider an air-water vapour mixture at 100 kPa and 303 K, having a relative humidity of 50%. What arethe mass and mole fractions of each component? What is the specific humidity?

7. Consider an air/water vapour mixture adjacent to a pool of water at 310 K, and 1 atmosphere. What are themass and molar fractions for the water vapour in the mixture just above the surface of the pool?

8. Pure CO2is dissolving into a pool of water. The CO2is at a pressure of 3 bar, and the water is at a temperatureof 27

oC (300 K). What is the concentration of CO2at the surface of the water?

9. Consider a salt / water solution adjacent to a piece of solid salt. The entire system is at a temperature of 30C.What is the mass fraction of the salt in the solution at the interface?

10. Helium at a pressure of 1 bar and 470 K (molar concentration 2.5610-1kmol/m3) is held in a glass container.What is the concentration of He at the inner surface of the glass wall if the container is made of,

a. Pyrex glass?b. 7740 glass?

11. Other solubility relationships for gasses in solids also occur. For instance, the solubility of hydrogen H2gas insteel can be given by:

4 3950 /

1, 1,2.09 10

T

u sm e P

where T is in Kelvin and P is in bar. For hydrogen gas stored at 400C and 500 kPa in a steel vessel,determine the mass and molar fraction of hydrogen on the steel side of the steel-hydrogen interface.


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3.1

S. Bickerton, 2001, S. Norris 2013 MechEng 713 Energy Technology

3 Ordinary Mass DiffusionMass diffusion is analogous to heat conduction. Both heat conduction and mass diffusion are transport processes thatoriginate from molecular activity. For a gas, the random motion of the molecules tends to evenly distribute thetemperature. Similarly, the random motions of molecules of a particular species in a gas mixture results in the speciesbeing evenly distributed through the mixture. Ordinary mass diffusion, which we will study here, is driven solely bygradients in the concentration of a species through a body, with molecules of a species on average migrating to regions

of lower concentration. Under conditions typical for engineering applications, ordinary mass diffusion is governed by asimple and useful equation, Ficks law.

In addition to Ordinary diffusion which is driven by concentration gradients, diffusion can be driven by temperaturegradients, a process known as Thermal diffusion. This is dependent on the presence of large temperature gradients, and

so is often negligible compared with ordinary diffusion. Similarly, Pressure diffusion is driven by very large pressuregradients, such as occur in centrifuges, but is normally negligible elsewhere. A fourth type of diffusion, forced

diffusion, where the diffusion is driven by an external force (such as an electric field) also occurs under specialcircumstances. See [1] 9.2.4 or [2] 1.2.4 for further discussion of these mechanisms. Whilst for simplicity theseforms of diffusion are not covered in this course, they do have many useful applications.

We will also restrict ourselves to low mass transfer rate diffusion, where mass transfer across interfaces is due todiffusion only, and the medium through which the diffusion takes place will be assumed to be either stationary, or themomentum of the diffusion flux is small compared to the momentum of the mixture. This is appropriate forevaporation of water from a pond, but not for boiling where there is a large velocity normal to the interface. For a fullerdefinition see [1] 9.7 and [2] Chapter 2.

As we study diffusion we will discover many useful analogies between heat conduction and mass diffusion. Astemperature gradients drive heat conduction, concentration gradients drive mass diffusion. We will discuss Ficks law

which governs ordinary mass diffusion, as Fouriers law governs heat conduction. We will then use this physical law todevelop solutions for steady state, and then transient diffusion problems.

Species ASpecies B

Concentration

species A

Concentration

species B


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3.2


3.1 Notation

The following notation will be used in our studies of mass diffusion, as well as subsequent topics. You will need to feelcomfortable with this notation in order to apply the concepts we will learn.

T = Temperature of mixture

,i jP = Pressure of species i, at surface j

ij = diffusive mass flux of species i [kg/m2s]

iJ = diffusive molar flux of species i [kmol/m2s]

im = mass flow of species i [kg/s]

iN = molar flow of species i [kmol/s]

12D = binary diffusion coefficient, or mass diffusivity [m

2/s]

3.2 Ficks Law of Mass Diffusion

Recall Fouriers law of heat conduction, which states that the local heat flux in a body is proportional to the localtemperature gradient.

Tq k

z

This statement is the building block for the study of heat conduction, for steady and transient applications, in multipledimensions.

Ficks law of mass diffusion states that the local mass flux in a body is proportional to the local concentration gradient.

This statement is true for:

binary gas mixtures at normal pressures

dilute liquid solutions

dilute solid solutions

This list encompasses many mixtures encountered in engineering applications. In a binary mixture of species 1 and 2,the equation governing diffusion of species 1 is;

1

1 12

dmj

dz D

where D12is the binary mass diffusion coefficient (units [m2/s]), and is the density of the mixture.

1j is the mass flow

rate of species 1 per unit area, a mass flux. We can also write an equation governing the diffusion of species 2 in themixture;

2

2 21

dm

j dz D

The relations stated above provide the diffusive mass flux [kg/m2s]. A similar relations can be stated for the diffusive

molar flux[kmol/m2s], and we will make use of both in our treatment of mass transfer. The following is a molar flux

statement of Ficks law for one-dimensional diffusion problems:

1

1 12

dxJ C

dz D

where Cis the mixture molar concentration, and 1J is the mole flux of species 1 (the molar flow rate per unit area).

Note that the same diffusivity constant, D12, is used for both the mass and molar forms of the law, and that D12= D21.


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3.3 Steady Mass Diffusion

In this section we develop steady state diffusion of mass, particularly for one-dimensional applications. For theseproblems, the system in consideration has reached a steady state, with concentrations at any point in a body remainingconstant with time. We will use Ficks law, remembering the mixtures for which it is applicable:

binary gas mixtures at normal pressures

dilute liquid solutions

dilute solid solutions

3.3.1 Steady Mass Diffusion Through a Plane Wall

We start by considering the simple case of one-dimensional mass diffusion across a plane wall. This wall mightrepresent a solid partition, a stationary liquid film, or a stationary gas layer. The binary mixture is made up of species 1

and 2, and we will consider the diffusion of species 1 through the wall. The geometry is shown below, the wallhaving a thickness ofL, and surface areaA. Given boundary conditions for the concentration of species 1 at either sideof the wall, we will calculate the flow of species 1 through it.

The principal of conservation of mass speciesstates (for a steady system) that flow of species 1 across the plane atz,must equal the flow of species 1 across the plane atz+ z. This will hold for any value ofzwithin the body. Therefore,

1 1 Constantm j A

Substituting Ficks law for1

j ,

11 12 Constant

dmm A

dz D

integrating across the wall from 0 toL, and applying the appropriate mass fraction boundary conditions, gives

1,

1,

1 12 1

0

v

u

wL

w

m dz A dm D

The areaAis constant through the thickness of the wall. If we now assume there are no significant variations of and

D12withz, we arrive at the following equation.

121 1, 1,u vA

m m mL

D

This is the rate that species 1 diffuses through the wall in kg/s. An identical analysis can be completed applying molefraction boundary conditions, providing the rate that species 1 diffused through the wall in kmol/s:

121 1, 1,u vC A

x xL

D

N

m1,u

z=0 z z+z z=L

m1,v

vu

AreaA

L

im , or iN


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3.4


Compare these two equations with the equivalent equation for heat conduction through a plane wall. The followingexpression gives heat flow rate through a plane wall for steady state heat conduction:

u vkA

Q T TL

Note: The above equations allow you to solve a diffusion problem in terms of either the mass concentration or themolar concentration. When deciding which method to use, it is often best to stick to using the concentration that your

boundary conditions are stated in.

How does concentration of species 1 vary through the wall?

For steady flow, the mass flow rate through the wall is constant. At each point in the wall,

1

1 12Constant

dmm A

dz D

If and D12are constant, then the gradient dm1/dzwill also be constant. Therefore the concentration of species 1 varies

linearly between the boundary conditions at the surfaces of the wall:

Expressing m1as a function ofzgives a linear profile for mass fraction:

1, 1,

1 1,( )

v u

u

m mm z z m

L

You may recall that the temperature profile through a plane wall was also linear.

3.3.2 Steady Diffusion Through Cylindrical and Spherical Geometries

The analysis completed above can be repeated for cylindrical and spherical geometries. This must be completed usingthe appropriate coordinate systems, integrating over the radial coordinate. Expressions are given below for mass flowrates, but it is straightforward to convert these expressions to given molar flow rates.

Cylindrical Geometry:

121 1, 1,

2

ln u v

v u

Lm m m

r r

D

m1,u

z=0z

z=L

m1,v

vu

u

v

m1,v

m1,u

L

Inner radius, ru

Outer radius, rv


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3.5


Spherical Geometry:

121 1, 1,4

1 1u v

u v

m m m

r r

D

3.3.3 Permeability of a Wall

The molar flow rate for steady diffusion of a gas through a plane wall is given by

12 121 1, 1, 1, 1,u v u vC A A

x x C CL L

D D

N

C1,u and C1,v are the concentrations of the gas inside the surfaces of the solid wall. These concentrations can bedetermined from the partial pressure of the gas outside the wall, by use of one of the solubility relationships.

1, 1,u sC S P or 1, 1,u U sC S R T P

ThepermeabilityP,is defined as

12 12SP D or 12 12US R TP D

and has units kmol/m.s.Pa, kmol/m.s.bar or kmol/m.s.atm. By use of the permeability, the rate of mass diffusionthrough a wall can be expressed directly in terms of the partial pressures of the gas adjacent to each side of the wall.

12

1 1, 1,

12 1

1 1, 1,

kmol/s

kg/s

s t

s t

AP P

LM A

m P PL

PN

P

For water vapour in air, the partial pressures may be determined from the saturation pressure Psat and the relative

humidity . Using these, the mass flow rate of water vapour through a wall can be calculated.

12 , ,vv s sat s t sat t M A

m P PL

P

For building materials of a given thickness L, the permeability may be defined in terms of the permanence M(units

kg/s.m2.Pa or ng/ s.m

2.Pa. ng = 10

-12kg) allowing a further simplification

212

12 kg/m .s.Pa

M

L

PM

1 12 1, 1,

12 , ,

s t

s sat s t sat t

m A P P

A P P

M

M

m1,v

m1,u

Inner radius, ru

Outer radius, rv u

v


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3.6


3.3.4 Obtaining the Binary Diffusion Coefficient 12

To solve a problem using the diffusion equations above the binary diffusion coefficient D12needs to be known. Note

that the diffusion coefficient is different for every combination of species and phase (ie: you cant use D12for hydrogen

in solid iron to model hydrogen in aluminium or hydrogen in molten iron). The diffusion coefficient varies with

temperature, and for gasses it varies with pressure as well.

The Schmidt number is the mass transfer equivalent of the Prandtl number, and is given by,

12 12

Sc

D D.

If you know the Schmidt number of a mixture, and the kinematic viscosity (or the dynamic viscosity and the mixturedensity), then you can calculate the mass diffusivity using,

12Sc Sc

D .

Sometimes the density the mass diffusivity is needed, and so you can use,

12Sc

D .

Diffusion in Gasses:

From the kinetic theory of gasses it can be shown that the dynamic viscosity and thermal conductivity vary only withtemperature (and not pressure), and vary as

and .T k T

From the ideal gas equation the density of the gas is proportional to P/T(ie: P/T), and so the kinematic viscositythermal diffusivity, and mass diffusivity vary as,

3 3 32 2 2

12, , .

p

T k T T

P c P P

D

The ratio of the viscosity and thermal and mass diffusivities (the Prandtl and Schmidt numbers) are therefore constant.

To calculate the diffusivity of a dilute mixture of a gas in air, you need the Schmidt number for the mixture and thekinematic viscosity of the mixture, , which may be approximated as that of air at the same temperature and pressure.Tables of Schmidt numbers can be found, as can tables for the properties of air for a range of temperatures atatmospheric pressure.

If the problem is at atmospheric pressure, use the Schmidt number of the mixture, and the kinematic viscosity taken atthe same temperature as the problem.

If the problem is not at atmospheric pressure, use the Schmidt number, the dynamic viscosity of air at the sametemperature as the problem, and the density of the mixture calculated using the ideal gas law,

U

PM

R T .

This requires the molecular weight of the mixture, which may be approximated by that of dry air. However, sometimes

the combination of D12is needed, in which case you just need the dynamic viscosity of air and the Schmidt number.

Diffusion in Liquids:

For liquids the density is relatively constant. The viscosity, thermal conductivity, and mass diffusivity all vary with

temperature, but not pressure. However, the Prandtl and Schmidt numbers vary with temperature.

For a dilute mixture in water, the simplest option is to assume a constant Schmidt number and calculate the massdiffusivity using the kinematic viscosity of water at the same temperature as your problem.

Diffusion in Solids:

Obviously the Schmidt number is not defined for diffusion through solids. [1] Table A.19 gives the diffusion

coefficients as functions of temperature in the form

/12 0

a UE R TeD D


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3.7


3.3.5 Heterogeneous Catalysis

Heterogeneous catalysis describes a catalytic reaction occurring at the surface of a body, due to reactants applied to thatsurface. A chemical reaction is promoted by contact of the reactants with a suitable surface, termed the catalyst. The

inner surfaces of an automobile catalytic converter are one example, providing a site for the oxidation of carbonmonoxide in exhaust gases. Platinum oxide is one example of a catalyst, promoting the following reaction:

2 22CO+O 2CO

CO and O2are absorbed at the surface, and CO2is given off.

We will analyse the mass transfer occurring in the binary mixture adjacent to a catalytic surface as a steady state

diffusion problem. The surface catalytic reaction will be treated as an alternative boundary condition, as it is effectivelya sink or source for a particular chemical species.

The Catalytic Boundary Condition

In the following analysis we will consider the diffusion of species 1, as it diffuses across the gas mixture, and isconsumed at the u surface. The rate of consumption at the surface is assumed proportional to the concentration ofspecies 1 at the u surface, m1,u, and proportional to the rate of reaction at the surface:

1 1 1,''

um j A m k A

where1m is the mass flow rate to the surface, and ''k is the rate constant for the reaction.

Since the system is assumed to be at a steady state, the rate that the species diffuses to the wall is given by,

121 1, 1,e uA

m m mL

D

As the mass that diffuses through the gas mixture must equal that consumed at the catalytic surface, both of the

equations above are valid expressions for the mass flow through the system. An expression for the mass flow rate1

m

can be found if we first eliminate1,um , and solve:

1,

1

12

1

''

eAm

mL

k

D

This result provides the mass flow,1m , given the mass fraction of species 1 at surface e, and various parameters that

define the problem. A very similar result could be generated for the rate that species 1 is consumed on a molar basis.

1,

1

12

1

''

eC Ax

L

k

N

D

1,uj

m1,e

eu

Catalyst for

reaction

z=0 z=L

z

m1,u

CO2

O2, CO


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3.8


Note that the sign convention for the above expression is that a positive mass flow is one that flows towards the wall(with a species being destroyed at the wall), and a negative mass flow is one with mass flowing away from the wall (thespecies is being created at the wall).

The behaviour of this type of system is now considered for two extreme cases, a very slow catalytic reaction, and a veryfast reaction.

The Rate-Controlled Limit

If the surface catalysis reaction is very slow, and therefore 12''k L D , it is said to be rate-controlled,

1 1, ''em m k A

In this limit, 1, 1,u em m , and the mass flow through the system is independent of the diffusion coefficient within the

mixture.

The Diffusion-Controlled Limit

If the catalysis reaction is very fast, and therefore 12''k L D , it is said to be a diffusion-controlled reaction,

121 1,e

Am m

L

D

In this limit, 1, 0um . Any of species 1 diffused through the mixture is consumed quickly at the catalytic surface.

The variation of 1m withzis plotted below, for various reaction rates:

m1,eeu

Catalyst

z = 0 z = L

m1,u

Diffusion

Controlled

Rate

Controlled

IncreasingReaction

Rate


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3.9


3.4 Transient Mass Diffusion

We now extend our analysis of ordinary mass diffusion to include time dependant behaviour. As for the steady state

analysis, the tools developed here are basically identical to those for transient heat conduction problems. We will focuson a one-dimensional analysis, which proves to be suitable for many engineering applications. Diffusion will beassumed to be described by Ficks law.

For transient problems species concentrations will be functions of time and position. In order to solve these problems

we will need to establish appropriate boundary conditions, and one initial condition.

3.4.1 Governing Equation

Consider one-dimensional diffusion through a mixture in Cartesian coordinates (excluding any chemical reactions).The principal of conservation of mass species will be applied to an elemental control volume having a volume of Vand thickness z.

1 zm 1 z zm

Over a time interval of tthe conservation of species requires that the change of species 1 in the element is equal to thenet flow of species 1 across the element boundaries.

Change of species 1 in volume = Net inflow of species 1 across boundaries

Stating this mathematically,

1 1 1 1t t t z z z V m m t

Substituting V A z , and 1 1m m A , and dividing throughout by A z t ,

1 1 1 1t t t z z z j jt z

z z+z

V=Az

z

Sudden increase inconcentration at

surface

Increasing time


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3.10


Now we take the limit, letting , 0z t ,

1 1j

t z

At this point we introduce Ficks law, substituting for the diffusive mass flux, 1j :

1 1

12

m

t z z

D (Note that1 1m )

If we now assume there are no significant variations of and D12, withzand t, we have the following equation, which

governs the mass fraction, m1.

2

1 112 2

m m

t z

D

A similar equation can be derived for analysis done on a molar basis.

2

1 112 2

x x

t z

D

At this point we have derived an equation for transient mass diffusion in one-dimensional geometries. We can comparethis result with the governing equation for transient heat conduction. Recall,

2

2

T T

t z

where is the thermal diffusivityof a medium.

Note that these equations have the same exact mathematical form, and note the identical roles played by the binary mass

diffusivity, D12, and the thermal diffusivity, . This mathematical equivalence allows us to directly apply solutions

derived for heat conduction problems to equivalent mass diffusion problems.

The mass and thermal diffusivities are properties of a substance or mixture, and provide a feel of how easy mass, orthermal energy, is diffused into that substance or mixture.

3.4.2 Transient Diffusion in a Semi-Infinite Domain

The semi-infinite domain proves to be the most useful for application to transient diffusion processes, due to theirslow nature. We will now apply the governing equation for mass fraction to transient diffusion in the semi-infinitedomain depicted below:

A particular problem is defined by the geometry, boundary conditions, and initial condition. The following boundaryconditions will be applied:

1 1,

1 1,0

at 0

as

um m z

m m z

The mass fraction of species 1 at surface u is known, and the mass fraction is known as we move far away fromsurface u. The following initial condition is applied:

1 1,0at 0m m t

Now compare this problem to an analogous heat transfer problem:

z

Solving for m1(z,t), >0

u


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3.11


2

2

T T

t z

0

0

at 0Boundary Conditions

as

at 0 Initial Condition

uT T z

T T z

T T t

This problem has the following solution (see [1] 3.4.2):

0

0

,erfc

2u

T z t T z

T T t

where erfc is the complimentary error function (tabulated on p3.13). This may be non-dimensionalised as

1erfc

2 Fo

where is the non-dimensional temperature, and Fo is the Fourier number.

0

2

0

, Fo

u

T T t

T T L

The heat transfer and mass transfer problems are mathematically equivalent, and so we can write the solution for themass transfer problem we have defined as:

1erfc

2 Fom

m

where Fomis the mass transfer Fourier number, Fom= D12t/L2, and mis a non-dimensionalised concentration, which

can take many forms;

1 1,0 1 1,0 1 1,0 1 1,0

1, 1,0 1, 1,0 1, 1,0 1, 1,0

m

u u u u

m m x x C C

m m x x C C

The behaviour of 1w with position and time is depicted schematically below:

In this case the concentration was increased at the surface (ie:1, 1,0u

m m ). However, if the surface concentration was

decreased the solution would look like:

z

m1(z,t)

u

m1,u

m1,0

increasing t

z

m1(z,t)

u

m1,u

m1,0

increasing t


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3.12


Penetration Depth

The penetration depth, which can be defined in a variety of ways, is a measure of how deep the diffusing species haspenetrated at time t. We will define the penetration depth of species 1, 1, as the location where a tangent to theconcentration profile atz= 0 intercepts the line m1= m1,0.

Taking the spatial derivative of the concentration profile and evaluating it atz= 0,

1, 1,0 1, 1,01

0 112

u u

z

m m m mmz t

D

Therefore, our measure of the penetration depth is,

1 12t D

or the Fourier number based on the penetration depth is equal to .

Fo= .

The important thing to note is the dependency of 1 to the square root of time. Initially the diffusing substancepenetrates quickly, the rate of penetration continuing to decrease with time.

Rate of Diffusion per unit surface area

The mass flux at surface u is given by,

1

1, 12

0

u

z

mj

z

D

121, 1, 1,0u uj m mt

D

or as a function of the penetration depth,

12 1, 1,01,

1

u

u

m mj

D

z

m1(z,t)

m1,u

m1,0

PenetrationDepth

Time

1 t


22/25


23/25

3.14


3.4.3 Transient Diffusion in a Plane Wall

Due to the similarity of the governing equations, solutions of transient mass diffusion problems in plane walls, cylindersand spheres can be quickly generated by inspection of existing heat conduction solutions. However, it should be notedthat these solutions find relatively few practical applications, as we will often have significant variations in the diffusion

coefficient with position, due to dependence on temperature, or concentrations in the mixture.

However, to further demonstrate the application of existing heat conduction solutions, we will look briefly at transient

diffusion in a plane wall. Consider the problem described below:

Defining the required boundary conditions:

1 1,um m at ,z L L

and initial condition:

1 1,0m m at 0t

This problem can be compared to an analogous heat conduction problem, having the following conditions:

uT T at ,z L L Boundary.Conditions

0

T T at 0t Initial Condition

This has the following solution, which is developed in [1] pp138 -141:

2

0

1

2

0

12 exp Fo cos

,

n

n

u

u

z

L

T Tn

T T

By careful inspection of the governing equations, geometry, boundary conditions and initial conditions, we find anequivalent equation for the mass diffusion problem:

20

1 1,

1,0 1,

12 exp Fo cos

n

m m

n

u

m

u

z

L

m m

m m

Again, the non-dimensionalised concentration mcould be expressedin mole fraction, partial density or species concentration.

Typical behaviour of 1 ,m z t is shown schematically to the right.

We might also be interested in mass flux at the surface u.This is provided by Ficks Law:

1

1, 12

0

u

z

mj

z

D

m1,u

z = 0zz=L

m1,u

uu

z=-L

At t=0, m1=m1,0

z=0

zz=L

m1,u

uu

z=-L

m1,0

Increasing

time


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3.15


Utilising the solution for m1(z,t) above:

21 1, 1,00

21 exp Fo sinn

u m

nz L

dmm m

dz L

Simplifying further:

21

1, 1,00

2

exp Fou mnz L

dm

m mdz L

And finally;

2121, 1,0 1,0

2exp Fo

u u m

n

j m mL

D

.

Example:

Consider a mild steel plate of total thickness 20 mm. It has an initial carbon concentration 0.2% by weight, and is to

have the carbon concentration increased to at least 0.8%. An equilibrium concentration of 1.5% carbon is established atthe surface of the plate. Calculate the time required for the centreline of the plate to reach a carbon mass concentration

of 0.8%. The diffusion coefficient of carbon in steel at the process temperature is approximately 5.610-10

m2/s.

Solution:The series solution for m1(z,t) has been implemented using a simple Fortran code. The solution was stepped forward intime, until m1atz= 0 exceeded 0.008. This required 62400 s, or 17.33 hrs.

We might also be interested in the rate at which mass is diffused through the surface of the plate (@ z=L). The seriessolution for surface mass flux has been applied in the same code, producing the following result:

Increasing

Time


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3.16

3.5 Example Questions

1. A thin plastic membrane is used to separate helium from a gas stream. The concentrations of helium at theinner and outer surfaces of the membrane are 0.02 and 0.005 kmol/m3, respectively. If the membrane is 1 mmthick, and the binary diffusion coefficient of helium with respect to the plastic is 10

-9m

2/s, what is the molar

diffusion flux? What is the mass diffusion flux (the molecular weight of helium is 4.0 kg/kmol?

2. A steel tank of 1 litre capacity and 2 mm wall thickness is used to store hydrogen at 400C. The initial pressure

is 9 bars, and the tank is located in an evacuated chamber. Estimate the initial rate of hydrogen loss per unitsurface area on both mass and molar bases. For hydrogen in steel you can use

6 4630/ 2

12

4 3950/ 1/2

1, 1,

1.65 10 [m /s]

2.09 10 [ in K, in bar]

T

T

u s

e

m e P T P

D

3. A helium-cadmium (blue) laser used in a photocopying machine printer contains He at a nominal pressure of460 Pa. The glass outer shell has a volume of 150 cm

3, an area of 550 cm

2, and a wall thickness of 1.52 mm.

At normal operating conditions, the average temperature of the gas in the tube is 225C and the shell outertemperature is 115C. Estimate the (i) initial helium leak rate; and (ii) time required to lose 1% of the originalhelium. You can use the following.

1,8 3280/ ( ) 2 5

121,

1.40 10 m /s ' 0.007 3.0 10 ( )uT K

s

Ce S T C

C

D

4. Helium gas at 27C and 4 bars is contained in a Pyrex glass cylinder of 100 mm inside diameter and 5 mmthickness. What is the rate of mass loss per unit length of the cylinder?

5. A tube of length 10 cm and cross-sectional area 0.785 cm2contains an Air/CO mixture with a mixture densityof 0.442 kg/m3, with D12= 97.8 10

-6m2/s. When the mass fraction of CO in the gas mixture flowing over

the tube is 0.001, the rate of removal of CO is measured to be 0.01855 g/s. Determine the rate constant of thereaction on the particular catalyst used. If the rate constant can be increased by some means, what is themaximum rate that CO could be consumed in the rig?

6. A mild steel rod with an initial carbon concentration 0.2% by weight is to be case-hardened. It is packed in acarbonaceous material (eg: charcoal) and maintained at a high temperature. Thermodynamic data for thechemical system indicates an equilibrium concentration of carbon in iron at the interface of 1.5%.

a. What is the time required for the location 1 mm below the surface to have a carbon massconcentration of 0.8%?

b. What is the penetration depth for this time?c. What mass of carbon per unit surface area has diffused into the steel up to this time?d. What is the carbon molar flux through the surface at this time?e. Plot penetration depth - time, and mass diffused - time curves for this process.

You may use D12= 5.610-10

m2/s for carbon in steel at the process temperature.

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Mass Transfer 2 Definitions and Diffusion v2