Mass Transfer in Multicomponent Mixtures

Mass Transfer in Multicomponent Mixtures

Mass Transfer in Multicomponent Mixtures J.A. Wesselingh R. Krishna

DELFT UNIVERSITY PRESS

CIP-data Koninklijke Bibliotheek, Den Haag Wesselingh, J.A. Mass transfer in multicomponent mixtures - Delft : Delft University Press. - TIl. ISBN 90-407-2071-1 NUGI 812,813,831

©VSSD First edition 2000 Published by: Delft University Press P.O. Box 98, 2600 MG Delft, The Netherlands tel. +31 152783254, telefax +31152781661, e-mail [email protected] website: http://www.library.tudelft.nlldup On behalf of:: Vereniging voor Studie- en Studentenbelangen te Delft Poortlandplein 6, 2628 BM Delft, The Netherlands tel. +31 152782124, telefax +31 152787585, e-mail: [email protected] internet: http://www.oli.tudelft.nllvssdlhlflhandleidingen.html

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photo-copying, recording, or otherwise, without the prior written permission of the publisher. ISBN 90-407-2071-1

5

Foreword It was twenty years ago. A little before that, I had left the Equipment Engineering

Department of Shell Research in Amsterdam for a less hectic job at Delft University. At least, so I thought at that moment. In my former section at Shell we had worked on catalytic crackers, on polymerisation reactors, on cleaning of oil tankers and other exciting developments, but I had found managing this a bit too much. There I was, with a lower salary, teaching separation processes to second year students, and running the undergraduate laboratory with one hundredth of my Shell budget. I had written a little book on Separation Processes, and sent it to friends in Amsterdam. One of the pieces of equipment that we had in Shell Research was (what was then) the largest distillation test column in the world. It was two and a half metres in diameter and some twenty metres high. The column was so big that we could only run it in the summer: the reboiler used the complete capacity of our boiler house. The operating pressure could be varied between vacuum and fifteen atmospheres. We had a beautiful time trying out all kinds of trays and packings. In time we started to get interested in trying to understand not only distillation of binary mixtures, but also of mixtures with more components. We started to gather measurements and to try to understand them. However, much of what we saw was baffling, to say the least. Only gradually did we realise that our binary mass transfer tools were not adequate; that we needed to try something different. That something was a young graduate from Manchester who had picked up wild ideas on mass transfer doing his PhD. His name was Krishna. I left Shell just after he arrived. One day, Krishna came along at home to visit me. He had read my book and told me politely that my approach to mass transfer was not all that good. I was a little vexed because I was professor, he was not, and besides, I had copied my ideas from wellknown handbooks. Even so, I did try to listen, and three weeks later went back to him for more explanation. It was all about multicomponent mass transfer, it had connections with thermodynamics and was quite different from anything I knew .. I had difficulties in following what he was telling me, and I can remember: 'Hans, if you really want to understand this, we should give a course together.' That is where this book started. We are now twenty years, fifteen PhD students, twenty-three courses and some thousand participants further. The course has evolved and so has the book. It now has examples from membrane technology, reaction engineering, sorption processes, biotechnology; from mixtures of gases or liquids, but also porous media and

6

Mass Transfer in Multicomponent Mixtures

polymers. The basics have not changed: they are still almost the same as presented by James Clerk Maxwell in 1866 and (more clearly) by Josef Stefan in 1872. (Maxwell is the one of the theory of electrical and magnetic fields, and Stefan the one associated with the name of Boltzmann.) I am a little ashamed when I look in editions of the Encyclopaedia Britannica from around 1900 and see how well diffusion was then understood. It feels as if it has taken me more than a century to pick up the brilliant but simple ideas of these two long-dead scientists. Krish and I hope this book will help you to do that a bit more quickly. Hans Wesselingh Groningen, May 2000

7

Contents FOREWORD

5

1

BEGINNING ... 1.1 Who should read this Book? 1.2 What this Book covers 1.3 Structure of the Book 1.4 Guidelines to the Reader 1.5 Guidelines to the Teacher 1.6 Symbols 1.7 Conventions

13 13 14 15 16 17 18 21

2

Is SOMETHING WRONG? 2.1 The Starting Point 2.2 Three Gases 2.3 Two Cations 2.4 Two Gases and a Porous Plug 2.5 Summary 2.6 Further Reading 2.7 Exercises

24 24 26 28 28 30 30 30

PART 1. MAss TRANSFER IN GASES AND LIQUIDS

33

3

DRIVING FORCES 3.1 Potentials, Forces and Momentum 3.2 Momentum (Force) Balance of a Species 3.3 The Driving Force: a Potential Gradient 3.4 The Maxwell-Stefan Equation 3.5 Simplifying the Mathematics 3.6 The Film Model 3.7 Difference Form of the Driving Force 3.8 Summary 3.9 Further Reading 3.10 Exercises

34 34 36 38 39 40 41 42 44 44 45

4

FRICTION 4.1 Friction Coefficients and Diffusivities 4.2 Velocities and the Bootstrap 4.3 Velocities and Fluxes 4.4 The Difference Equation 4.5 Mass Transfer Coefficients 4.6 Summary 4.7 Further Reading 4.8 Exercises

46 46 48 50 50 52 53 53 53

5

BINARY EXAMPLES 5.1 Stripping 5.2 Polarisation

55 55 56

8

Mass Transfer in Multicomponent Mixtures

5.3 Vaporisation 5.4 Gasification of a Carbon Particle 5.5 Binary Distillation 5.6 Summary 5.7 Further Reading 5.8 Exercises

56 58 59 60 60 61

6

TERNARY EXAMPLES 6.1 From Binary to Ternary 6.2 A Condenser 6.3 A Ternary Distillation 6.4 A Ternary Reaction 6.5 Binary Approximation of a Ternary 6.6 Summary 6.7 Further Reading 6.8 Exercises

63 63 64 65 68 68 70 70 71

7

MASS AND HEAT TRANSFER 7.1 Temperature Gradients 7.2 Enthalpy 7.3 Mass Transfer Relation 7.4 Energy Transfer Relation 7.5 Condensation in Presence of an Inert Gas 7.6 Heterogeneous Reacting Systems 7.7 An Ammonia Absorber 7.8 Summary 7.9 Further Reading 7.10 Exercises

74 74 75 76 77 79 80 80 81 81 82

8

N ON-IDEALITIES 8.1 Chemical Potential and Activity 8.2 Non-ideal Binary Distillation 8.3 A Simple Model of Non-idealities 8.4 Large Non-Idealities: Demixing 8.5 Maxwell-Stefan versus Fick 8.6 When can we neglect Non-ideality? 8.7 Mass Transfer in Liquid-Liquid Extraction 8.8 Summary 8.9 Further Reading 8.10 Exercises

84 84 84 85 87 88 89 90 92 92 92

9

DIFFUSION COEFFICIENTS 9.1 Diffusivities in Gases 9.2 Diffusivities in Liquids 9.3 How do you measure diffusivities? 9.4 Summary 9.5 Further Reading 9.6 Exercises

10 TRANSFER COEFFICIENTS 10.1 Introduction 10.2 Dimensionless Groups

95 95 97 100 102 102 103 106 106 106

Contents

10.3 Tubes and Packed Beds 10.4 Packed Gas-Liquid Columns 10.5 Single Particles, Bubbles and Drops 10.6 Using 'Single' Coefficients for Swarms 10.7 Using Binary Coefficients in Multicomponent Calculations 10.8 Summary 10.9 Further Reading 10.10 Exercises

9

108 110 111 117 118 118 119 120

11 ELECTRICAL FORCES AND ELECTROLYTES 11.1 Electrolytes 11.2 The Electroneutrality Relation 11.3 Electrical Forces 11.4 Transport Relations 11.5 Diffusion of Hydrochloric Acid 11.6 Plus a Trace of Sodium Chloride 11.7 Diffusion of Proteins 11.8 Conduction and Friction between Ions 11.9 Diffusivities in Electrolyte Solutions 11.10 Summary 11.11 Further Reading 11.12 Exercises

122 122 123 124 124 125 126 126 128 130 132 133 134

12 CENTRIFUGAL AND PRESSURE FORCES 12.1 Volumetric Properties 12.2 The Pressure Gradient 12.3 Gravitational Force 12.4 Centrifuges 12.5 Gas and Protein Centrifugation 12.6 Difference Equation for the Pressur~ Force 12.7 The Maxwell-Stefan Equations (again) 12.8 Summary 12.9 Further Reading 12.10 Exercises

137 137 138 138 139 140 141 142 143 143 143

13 WHY WE USE THE MS-EQUATIONS 13.1 Three Ways 13.2 A Mixture of Three Gases 13.3 The Fick Description 13.4 Thermodynamics of Irreversible Processes 13.5 The Maxwell-Stefan Description 13.6 Units 13.7 Further Reading 13.8 Exercises

146 146 147 148 150 150 152 152 153

PART 2. MASS TRANSFER THROUGH A SOLID MATRIX

155

14 SOLID MATRICES 14.1 The Applications 14.2 Membrane Processes 14.3 Adsorption and Chromatography 14.4 Heterogeneous Catalysis

156 156 156 158 160

10

Mass Transfer in Multicomponent Mixtures

14.5 Structured and Non-structured Matrices 14.6 Effects of a Matrix on Mass Transfer 14.7 Compositions with a Matrix 14.8 How Further? 14.9 Further Reading 14.10 Exercises

161 162 163 164 165 165

15 PROPERTIES OF POLYMERS 15.1 A Few Words on Polymers 15.2 Thermodynamics of Mixtures in a Polymer 15.3 Summary 15.4 Further Reading_ 15.5 Exercises

167 167 171 177 177 177

16 DIFFUSION IN POLYMERS 16.1 Behaviour of Diffusivities 16.2 The Free Volume Theory 16.3 Summary 16.4 Further Reading 16.5 Exercises

180 180 182 188 189 189

17 DIALYSIS AND GAS SEPARATION 17.1 Dialysis 17.2 Gas Separation 17.3 Summary 17.4 Further Reading 17.5 Exercises

191 191 194 197 197 198

18 PERVAPORATION AND REVERSE OSMOSIS 18.1 Pervaporation 18.2 Reverse Osmosis 18.3 Summary 18.4 Further Reading 18.5 Exercises

200 200 204 208 208 209

19 ELECTROLYSIS AND ELECTRODIALYSIS 19.1 Introduction 19.2 Polarisation in Electrolysis 19.3 Electrodialysis 19.4 Summary 19.5 Further Reading 19.6 Exercises

211 211 212 214 220 220 221

20 ION EXCHANGE 20.1 Fixed-Bed Processes 20.2 Ion Exchange EquiIibria 20.3 Linear Driving Force Model 20.4 Ion Exchange (Film Limited) 20.5 Ion Exchange (Particle Limited) 20.6 Summary 20.7 Further Reading 20.8 Exercises

223 223 224 225 227 228 230 230 231

Contents

11

21 GAS PERMEATION 21.1 Transport in Cylindrical Pores 21.2 The Diffusion Coefficients 21.3 Looking Back 21.4 Transport in a Bed of Spheres 21.5 The Dusty Gas Model 21.6 Summary 21.7 Further Reading 21.8 Exercises

233 233 233 235 236 238 240 240 241

22 IN POROUS CATALYSTS 22.1 Introduction 22.2 Pressure gradients inside a particle 22.3 Separate Transport Equations 22.4 Single-variable Pressure and Rate Expressions 22.5 Solution for a Slab 22.6 Summary 22.7 Further Reading 22.8 Exercises

242 242 243 245 245 247 249 249 250

23 IN ADSORBENTS 23.1 Adsorption 23.2 Equilibria - Langmuir Isotherm 23.3 Maxwell-Stefan and Fick Diffusivities 23.4 Macropore Diffusion 23.5 Transport Equations 23.6 Transient Adsorption of a Binary 23.7 Membrane Applications 23.8 Summary 23.9 Further Reading 23.10 Exercises

252 252 253 254 257 258 259 260 261 261 262

24 ULTRAFILTRATION 24.1 The Module 24.2 Membrane and Permeants 24.3 Osmotic Pressure (No Ions, no Charge) 24.4 Size Exclusion 24.5 Polarisation 24.6 Transport Equations 24.7 Inside the Membrane 24.8 Electrical Effects 24.9 Summary 24.10 Further Reading 24.11 Exercises

264 264 265 266 266 267 269 272 273 275 276 277

25

......... ENDING 25.1 Looking Back 25.2 Thermodynamic Models - the Potentials 25.3 Driving Forces 25.4 Friction Terms 25.5 Friction Coefficients

279 279 279 281 281 282

12

Mass Transfer in Multicomponent Mixtures

25.6 Additional Relations (Bootstraps) 25.7 The Many Variants 25.8 Goodbye 25.9 Further Reading 25.10 Last Exercise

283 285 286 287 287

THANKS

289

ApPENDIX 1. READING MATHCAD

293

ApPENDIX 2. UNITS A2-1 Molar basis A2-2 Mass basis A2-3 Volume basis A2-4 Molar and mass diffusivities A2-5 Molar and volume diffusivities A2-6 Molar and mass driving force A2-7 Molar and volume driving forces A2-8 Difference equation for a 'film' A2-9 Difference equation, molar basis A2-10 Difference equation, mass basis A2-11 Difference equation, volume basis A2-12 Summary

296 296 297 297 298 299

ApPENDIX 3. PoRING OVER PORES A3-1 Introduction A3-2 The System A3-3 Forces and Velocities A3-4 The Two Transport equations A3-5 Comparing the Two Models A3-6 Summary A3-7 Further Reading A3-8 Exercises

304 304 305 305 308 310 313 314 314

ANSWERS

317

INDEX

326

CD-ROM FOR 'MASS TRANSFER IN MULTICOMPONENT MIXTURES'

329

300 300 301 301 302 302 303

13

1 Beginning ... 1.1 Who should read this Book? This book is about the diffusion and mass transfer processes that are really important, but which are neglected in most textbooks: • those with three or more species, the 'multicomponent' mixtures, • those with more than one driving force, including electrical or pressure gradients, and • those with a solid matrix such as a polymer or a porous medium. If you want to know more about these subjects, but find existing texts too difficult, then this is the book for you. Also, if you already understand the intricacies of multicomponent mass transfer, you may find it enjoyable to see how far you can get with simple means. We are assuming that you are interested in processes or products. This may be in an academic or industrial setting: in chemicals, water treatment, food, biotechnology, pharmaceuticals ... you name it. The book assumes that you have a working knowledge of: • thermodynamics and phase equilibria: chemical potentials, enthalpies, activity coefficients, partial molar volumes and distribution coefficients, • transport phenomena: simple mass balances, binary diffusion and mass transfer coefficients, and • fluid flow, especially around particles and in porous media. If you are not too sure, do not despair. We will repeat all important concepts in a leisurely manner. However, this is not a book for a complete beginner in mass transfer; you must have heard of the above concepts. Because there are many new ideas to get used to, we have tried to avoid mathematical complexity. For the greater part of the text you do not need more than the ability to solve three linear equations with three unknowns. You can even do a fair bit with pencil, paper and a calculator. Of course you will need a computer for larger problems, but not to obtain a first understanding.

14

Mass Transfer in Multicomponent Mixtures

1.2 What this Book covers This book takes motion in a mixture to be governed by forces on the individual species. There are two kinds of forces: • driving forces, which stem from the potential gradient of a species, and • friction forces between the species, which arise from their velocity differences. Maxwell and Stefan already used this method more than a century ago. This mechanical viewpoint is much more general than Fick's law, which is usually taken as the basis of diffusion theory. It has not caught on, probably because the mathematics is thought to be difficult. This is not really a problem however: • There are simple approximations to the solutions of the equations. • The computer and numerical techniques now make 'exact' calculations much easier. Using potential gradients allows the incorporation of different driving forces: • composition gradients (or more pr:ecisely: activity gradients), • electrical potential gradients, • pressure gradients, • centrifugal fields and others. The friction approach to interaction between the species allows a consistent handling of any number of components. Working with force balances makes it easy to link the subject of mass transfer to other parts of science. Thermodynamics and transport processes become neighbours: equilibrium is simply the situation where driving forces have disappeared. The use of forces on the species in a mixture fits in the way of thinking of engineers: it is a logical extension of mechanics of a single species. For friction coefficients we can make use of the many relations for hydrodynamics of particles or porous media. These show that flow and diffusion are two sides of the same coin. With our starting points we can describe almost any mass transfer process. Examples in this book cover: • multicomponent distillation, absorption and extraction, • mUlticomponent evaporation and condensation, • sedimentation and ultracentrifugation, • dialysis and gas separations, • pervaporation and reverse osmosis, • electrolysis and electrodialysis, • iop exchange and adsorption • heterogeneous catalysis and • ultrafiltration.

1. Beginning ...

15

The examples treat diffusion in gases, in liquids, in electrolyte solutions, in swollen polymers and in porous media. The book includes methods for estimating multicomponent diffusivities and mass transfer coefficients. A major limitation of the book is that it mainly covers examples with a single transfer resistance, not complete pieces of equipment. Such a resistance will be a building block for the simulation of separation columns, membrane modules or chemical reactors. The reader must be prepared to incorporate the equations into his own simulations. The approximations used should be sufficiently accurate for most engineering applications. With this book, we hope to make you feel at home in the equations of multicomponent mass transfer. However, we do not derive these equations. If you are inquisitive and have some perseverance, you will be able to retrace the fundamentals in the references that we give.

1.3 Structure of the Book This book has twenty-five chapters, covering a range of subjects. You may feel that it is a jumble of facts and problems, but there is an underlying structure. The theme is the development of the Maxwell-Stefan equations. There are two main parts: on transfer in gases and liquids (Chapters 3 ... 13), and on transfer through a solid matrix (Chapters 14 ... 24). Chapters 3 and 4 introduce the two sides of the Maxwell-Stefan equations: the driving forces for mass transfer and the frictional forces between moving species. Chapters 5 and 6 apply the equations to simple binary and ternary examples. Chapters 8, 11 and 12 complete the description of the driving forces by including the effects of non-ideality in a mixture (8), electrical forces (11) and centrifugal and pressure forces (12). Chapters 9 and 10 consider parameters in the friction terms: diffusivities or mass transfer coefficients. Chapter 13 discusses the relation between the MS-equations and other ways of describing mass transfer. You may wonder what has happened to Chapter 7. It introduces the effects of a temperature gradient, and does not quite fit into the structure. We could have put it almost anywhere. The second part of the book considers mass transfer through solid matrices. Chapter 14 gives a preview of the subject and discusses the two types of matrix: • polymer matrices (Chapters 15 ... 20) and • structures with defined pores (Chapters 21 ... 24). Chapters 15 and 16 give a brief description of polymers and of the behaviour of diffusion coefficients in polymers. These chapters are a sideline, introducing concepts that we need further on. We continue with a series of examples with

16

Mass Transfer in Multicomponent Mixtures

different driving forces: composition gradients (Chapter 17), pressure gradients (Chapters 17 and 18) and electrical gradients (Chapters 19 and 20). In the chapters on porous media, we mainly focus on the friction side of the MSequations. Chapter 21 covers transport of non-adsorbing gases and introduces the effects of viscous flow. Chapter 22 shows how the Maxwell-Stefan equations are applied when chemical reactions are taking place. Chapter 23 considers diffusion of species which do adsorb, such as in microporous adsorbents. In Chapter 24 we consider an example where viscous flow is very important: ultrafiltration. We finish by looking back at the many different aspects of the MS-equations in Chapter 25.

1.4 Guidelines to the Reader The text was written to accompany overhead transparencies in a full week's course on multi-component mass transfer. Most transparencies have found their way into the figures: they are important, not just illustrations. The figures contain all formul~e and much of the other information. Not all chapters are equally important. As a minimum, we recommend that you work through Chapters 3, 4, 5, 6, 14, 17 and 21. Together, these will give you a working idea of multicomponent mass transfer theory for about two day's work. Other comments: • If you are convinced that you know all about mass transfer (as we used to be!) you should read Chapter 2. It may contain a few surprises. • Chapters 7,8 and 12 cover subjects which, although important, can be omitted on first reading. • The Chapters 9, 10, 11 (second half), 15 and 16 are on the estimation of properties and model parameters such as diffusivities and mass transfer coefficients. You can skip these on first reading. • If you are not interested in ions, electrolytes and electrical fields you can skip Chapters 11, 19, 20 and parts of 24. (However, do note that electrical fields are much more prevalent than thought by most chemical engineers!) • If you never encounter polymers, you will not need Chapters 15 ... 20. • When porous media play no role in your life, you can omit Chapters 21 ... 24. It is all up to you. Chapters 2 ... 6 contain a number of questions and small sums in the text. We recommend that you try these. The answers are buried in the text or figures. Behind each chapter is a series of exercises. These are to help you to go through the material more thoroughly than you will with a single reading. There are short questions, discussions and additions to the material of the main text. The answers are given at the end of the book.

17

1. Beginning ...

Beginning with Chapter 5, there are assignments in Mathcad - a fairly accessible programming language. There are two kinds: short ones, which you are to program yourself and longer files, which are demonstrations of more complicated problems. Our students very much favour the first type. We have marked them with and we hope that you will try a fair number of these. The second kind of files is for you to use for your own problems, to look at, to play with, to modify or to criticise. We leave it to you. Many of these files work out examples given in the text. The Mathcad assignments are in the folder Exercises/Questions on the CD-ROM. Completed Mathcad files are in the folder Exercises/Answers on the CD-ROM. You can read these, and change their parameters, using the free program Mathcad Explorer which is also on the CD-ROM as a self extracting file in MathcadlExplorer. Appendix 1 in the book contains an introduction to Mathcad; enough to allow you to read the files. You can further improve your Mathcad skills with the tutorial in Mathcad Explorer. To make full use of the Mathcad exercises, you will need Mathcad 7.0, Student Edition or higher. Before our courses, we give students a short self-instruction course in Mathcad. You will find this in the MathcadlTutor7 folder. It consists of ten Mathcad files; they should help you to get a good start in Mathcad in less than half a day. The regular text of the book continues in Chapter 2. However, you should glance through the list of symbols, and the list of conventions at the end of this chapter. You may not understand all details on first reading, but you should know where they are so you can look back later.

*

1.5 Guidelines to the Teacher This book has evolved in a series of twenty-three courses that we have given at different universities since 1982. The participants were mainly PhD and Masters students, but we have also had many participants from industry and a fair number of our colleagues: together about nine hundred of them. Most have been from chemical engineering, but we have also had mathematicians, chemists, physicists, mechanical engineers, and the occasional pharmacy or biology student. Because we always have an audience coming from many different places, our courses have mostly been in five days consecutively. In such a course we have about 36 hours for lectures and computer assignments. We divide these into (roughly) 16 hours of lecturing and 20 hours of computer assignments in changing groups of two. Except for Chapters 1 and 25 (which require no lecturing time) and 2 and 14 (which take less than half an hour), all chapters need about an hour. This means that you will have to make a choice of about 15 chapters from the 19 others that we provide. The 'Guidelines to the Reader' above should help you in making a choice. As a minimum for a course we recommend Chapters 3, 4, 5, 6, 14, 17 and 21. Together, these will give participants a working idea of multicomponent mass

18

Mass Transfer in Multicomponent Mixtures

transfer theory for about two day's work. The CD-ROM that accompanies this book contains a complete set of PowerPoint 7.0 files of the colour transparencies that we use in the course. They are in the folder Transparencies. You can use and edit these freely for your own teaching, but you are not allowed to use them for commercial purposes. They are our property! On assessing the knowledge of students. You can of course do that in the traditional way. We also have good experience with giving each student some article on a mass transfer problem and asking him or her to construct a new Mathcad example. Our students find this difficult but instructive. Success!

1.6 Symbols Symbols used only at one or two points, are defined there and not listed here.

fl

Avogadro constant # mol- I

A

non-ideality p'arameter

a

activity-

c

molar concentration

cp

D

molar heat capacity J mor l Fick diffusivity m 2S-I

d

diameter

D

Maxwell-Stefan diffusivity m 2 S-I energy flux J m-2 S-I

E

-

mol m-3

m

F

activation energy J mol- I force per mole N mol- I

'.F

Faraday constant

g

acceleration of gravity enthalpy J mol- I

E

H

C mol-I m S-2

I

heat transfer coefficient W m- 2 KI electrical current density A m-2

J

flux with respect to the mixture mol m-2 S-I

k

mass transfer coefficient

K

equilibrium constant

M

molar mass

m

mass

N

flux with respect to an interface mol m-2 S-I

n

number of moles pressure N m-2

h

p

m S-I

-

kg mol-I

kg mol

1. Beginning ...

R

gas constant J mol-I K"I

R

retention

r

radius

r

reaction rate

mol m- 3 S-I

T

temperature

K

t

time

Tg u

glass transition temperature diffusive velocity ms-I

v

viscous velocity

m

s K

ms-I m3

mol-I

V

molar volume

w

whole (overall) velocity

x

mole fraction

y

mole fraction -

z

distance (position)

z

charge number

ms-I

m

-

Greek symbols a viscous selectivity

-

Ll

increase of...

E

volume (void) fraction



electrical potential

V

'Y

activity coefficient

-

11

A

viscosity Pa s thermal conductivity J m-I K"I

/l

chemical potential

V

stoichiometric coefficient

J mol-I

1t

osmotic pressure N m-

P

density

cr

interfacial tension

-

2

kg m- 3

't

stress

't

'\)

tortuositydiffusion volume

CO

angular speed

~

friction coefficient, structured

'If

J mol-I friction coefficient, unstructured N s mol-I m-I

S

N m-

N m-I

2

m 3 mol-I

rad S-I

potential of a species

N s mol-I m-I

19

20

Mass Transfer in Multicomponent Mixtures

Superscripts average of x in a film .: pressure of pure '1' (vapour pressure) PI*

11*

.: volume of pure' 1'

D I,2

Q in free space

o

d I,2

V

pr

in the dispersed phase boiling pressure

=1

DxI=I

V

pfq

pressure of '1' if a reaction were to go to s;guilibrium

T,.ej

ui

reference thermal velocity velocity of '1' at the average composition

X',X"

x in two different phases

1,2

v

T

1,2

at x,

Subscripts Cw

concentration of water

dp

diameter of particle or pore effective Fick diffusivity of '1' free volume

Dt,ejf

\j (I,M

VC xIa' xIf3

friction coefficient between '1' and the matrix molar volume of a chain element species '1' at different positions <X, ~, "t, 0

UI, U2

reference temperature species '1' and '2'

UI.

tracer with properties equal to species '1'

Ui

species i. under consideration

Uj

species i other than that under consideration viscous friction coefficient

T,.ej

(v

Abbreviations ED electrodialysis

PH MS RO

Flory-Huggins Maxwell-Stefan Reverse Osmosis

UP

Ultrafiltration

1. Beginning ...

21

1.7 Conventions Below are a few notes on the conventions used in this book. You may want to look back at this list occasionally while you are reading the book; do not expect to understand every detail on fIrst reading. (1) In the drawings and sketches, the positive direction is from left to right.

Velocities and fluxes in that direction are also positive. (2) A force is directed down a potential gradient. Examples are: 1)

= - dJ.ll dz

or

This convention holds both for differential and for difference equations. (3) When computing a difference, we begin with the value at the most positive position (the right hand value) and subtract the other: Llx2

= x2f3 -

x2a

(4) Examples may consider compositions in many positions; these are denoted by Greek subscripts. If the problem considers different phases, these are distinguished by accents (Figure 1.1). I I I

phase i

I

I I

phaseD

three different phases

mole fractions of species '2'

different positions positive direction Fig. 1.1 Three different phases

(5) A 'property' can mean several different things in a mixture. We illustrate this for the molar volumes in a ternary mixture. (Figure 1.2)

Mass Transfer in Multicomponent Mixtures

22

molar volumes of the pure species

V;*,

V;, ~*

molar volumes of the species as in the mixture (partial molar volumes) molar volume of the mixture V =

XIV;

+ X2V2 + X)l~

Fig.1.2 Component volumes: pure and in a mixture

In an ideal mixture, the species volumes are the same as those of the pure species,

but only then. Note that a property of the mixture has no separate subscript or superscript. (6) A mixture moving through a solid matrix can be described in several ways. As an example, we consider the movement of two permeants '1' and '2' through a membrane. We can regard the membrane material as a third component '3', or as something separate - the matrix 'M' - which is not part of the mixture (Figure 1.3). The friction forces exerted on component '1' in the two notations are:

(a) (b) friction exerted by '2' on '1'

friction exerted by the matrix on '1'

Fig.l.3 Two notations: (a) the matrix is part of the mixture, (b) the matrix is separate

Each of the terms separately has the same size in both notations. Also the velocities are identical (including u 3 = uM ). However, the numerical values of both the mole fractions and the friction coefficients differ. (7) We will be describing porous media in two ways. In the first, we consider the effects of the structure of the medium: effects of pore or particle sizes, or the effect of the void fraction. In this model it is also common to distinguish between two species velocities: diffusive and yiscous (or convective) velocities. Viscous flow is governed by hydrodynamics. The sum of the two velocities is the whole (or overall velocity) . Note how the symbols for the velocities are contained in the letters of their . names (Figure 1.4).

23

1. Beginning ...

Structured model

S (zeta)

I Non-structured model

,

........

•.. ~~ . :." ....

:::...

~

~

F; = X2S1,2(WI-W2)+SI,MWl F2

= X1S 2,I(W2 -W1)+S2,M W2

~ (ksi)

Fig. 1.4 Notation in the structured and non-structured models

In the non-structured model, the effects of the structure are built into the friction coefficients. This kind of model only considers the whole velocities. The two models give the same result when applied properly. However, the friction coefficients can behave quite differently. The relations between the two sets of coefficients are complicated.

24

Is Something Wrong? In this chapter, we look back at how we have learned mass transfer. We see that Fick's law is incomplete and that it leads to wrong predictions, even in simple multicomponent problems.

2.1 The Starting Point We expect your working knowledge of mass transfer to be something like that summarised in Figure 2.1. Once these were also the only tools we had. So let us have a look at them. The upper part of the figure shows two 'laws' which govern motion of a species i in a mixture, with respect to that mixture: • the flux of a species is proportional to its diffusivity and concentration gradient, and • the flux is proportional to a concentration difference times a mass transfer coefficient. The flux is defined relative to some 'frame of reference' , for example one that moves with the average molar velocity of the mixture.

IJ

i :

flux of i with respect to the mixture

dc. l. =-D - ' Fick's law , " dz !

.<1c

D

J. = -D - ' = -k..<1c. k. =-' , '& ",' , & diffusiVity

mass transfor coefficient

Fig. 2.1 Mass transfer as you have learned it

The first law (Fick's law) tells us that a species should move down its concentration gradient. You may regard the second 'law' as a difference form of the first, which is applicable to a thin film. A film is a rough model for a mass transfer resistance. The two laws are handy little formulae for describing simple mass transfer problems. In passing we note that the fluxes and concentrations used here are in molar units

2. Is Something Wrong?

(mol m-2 S-1 and mol m-3). This implies that the diffusivity has units of m2 mass transfer coefficient ofm S-1 (a velocity).

25

S-1

and the

We are usually interested not in the fluxes with respect to the mixture, but in the fluxes with respect to some boundary or interface. We give these the symbol Ni (Figure 2.2). Only if the flux N of the mixture (as a whole) is zero, are the fluxes Ji and Ni equal. Otherwise, we must add a drift flux to the diffusion flux. Also here, we can write the flux relation as either a differential equation or a difference equation. The last is simpler, but approximate. We can force the difference equation into a simple looking form by including a Stefan or drift correction. Unfortunately, this correction can have any value and sign; the result is not as useful as it looks.

INi : flux with respect to an interface I differential equation N

i=-Di

flux of mixture

ck+

dc.

N

Xi

N=LNi

difference equation Ni = -k/1Ci + NXi ~'-v-'

diffusion flux

drift flux

---. -kisi!:.Ci

Ste;;;; or drift correction

Fig. 2.2 The flux with respect to an interface

In most of the examples that we consider in this chapter, drift is not important. So you should expect both fluxes J i and Ni to be equal, and directed down the concentration gradient of i. The only exception will be in the very last example, where drift is important. Conventional mass transfer in a binary mixture of gases (Figure 2.3) is especially simple. If pressure and temperature in the gas are assumed constant, the total molar concentration is also constant. By definition the sum of the fluxes with respect to the mixture is zero. If you then write down the flux equations for the two components it is immediately clear that there is only one binary diffusion coefficient. Also experiment (and the kinetic theory of gases) tells us that this diffusivity is a constant (that is: independent of composition, not of pressure and temperature)_ There is more to be said about conventional mass transfer theory, but this should be enough for the moment.

26

Mass Transfer in Multicomponent Mixtures

gas: c1 + C z = c = constant

J 1 --D dC1 1 dz

fluxes with respect to mixture

dc z J 2 --D 2 dz

+-----J1 + J 2

_0 __Dd(C1+C2) -

-

dz

only one binary D, which is independent of composition

Fig. 2.3 A binary gas is simple

2.2 Three Gases Look at the experiment in Figure 2.4. There are two equal glass bulbs filled with mixtures of ideal gases. The left bulb consists of hydrogen and nitrogen, and the right bulb of carbon dioxide and nitrogen. The amounts of nitrogen in the two bulbs only differ slightly. Both bulbs are at the same pressure and temperature. At a certain moment we connect them by a capillary; this is fairly narrow, say with a diameter of one millimetre, but otherwise nothing special. Gases start diffusing from one bulb to the other. Before you read on we would like to ask you to think a moment about the questions given in the figure, and to make your own decision on which answer you choose. ideal gases, 100 kPa, 298 K

®8~==8® beginning:

= 0.46 xH2 =0.54

X N2

x N2 =0.52 XC02

=0.48

Question: Does N2 transfer (a) from A to B? (b) from B to A? (c) not at all? (d) or does it do (a), (b) and (c)? Fig. 2.4 A 'simple' experiment, with some questions

Have you answered the questions?

2. Is Something Wrong?

27

The results of the experiment are shown in Figure 2.5. The behaviours of hydrogen and carbon dioxide (bottom part of the figure) are as expected. Their compositions change monotonically in such a way that after a few days the amounts in the two bulbs will have become equal. Hydrogen (which is the smaller molecule) moves more rapidly than carbon dioxide. Nitrogen (note the difference in the composition scale) behaves quite differently. Initially it diffuses from the high concentration (bulb B) to the low concentration, and the two concentrations become equal after about one hour. However, nitrogen keeps on diffusing in the same direction, now against its concentration gradient. The gradient keeps increasing up to about eight hours after the start of the experiment. Only after that, do the two bulbs gradually go back to equal compositions.

: t«-=;:m;on mole fraction Xi

reverse

0.4

0.6 0.4

0.2

time h

0.0

o

10

20

Fig. 2.5 The strange behaviour of nitrogen

With the conventional mass transfer theory of the first two figures in mind you will probably find it difficult to understand what is happening. So just try to forget these for a moment and try a different viewpoint. It is fairly obvious why hydrogen is going from left to right. There is far more hydrogen in the left bulb than in the right one; the random thermal motions of the molecules will on average cause them to move to the right. The same mechanism causes carbon dioxide to move to the left. Now it looks as if nitrogen is being dragged along by the carbon dioxide. This is understandable: you would expect more friction between the heavy carbon dioxide molecules and nitrogen than between nitrogen and hydrogen. At least initially, the movement of nitrogen is mainly determined by the carbon dioxide and hydrogen gradients, and not by its own gradient (which is rather small). You can extend conventional mass transfer theory to explain the results, but the result is much more complicated and less convincing than that above.

28

Mass Transfer in Multicomponent Mixtures

2.3 Two Cations This experiment involves a membrane that is permeable to cations, but not to anions (Figure 2.6). We bring a dilute solution of sodium chloride in the right compartment and a much more concentrated solution of hydrochloric acid in the left compartment. You would expect the sodium ions to diffuse from the right to the left until the two concentrations have become equal. They do indeed go in that direction. However, they may go on until their concentration in the left compartment is many times higher than in the right one! The experiment has similarities with the previous one. The explanation, however, is rather different. Hydrogen ions diffuse through the membrane to the right. (Remember that the chloride ions cannot.) This causes a small positive excess charge in the right compartment. The resulting electrical gradient forces the sodium ions to the left and restricts the amount of hydrogen that can be transferred. There is no such mechanism in conventional mass transfer. cation penneable membrane

high concentration

H+;:

Na+"

CI-

Cl-

CD H+ moves rapidly so Na+ can move against its concentration gradient!

~

~

0 0

low concentration

~

excess +charge and electrical field

3

Fig. 2.6 Sodium moves against its gradient

2.4 Two Gases and a Porous Plug As a last example (Figure 2.7) we consider a porous plug. It is a plug with fine openings (you might think of compressed cotton wool), but otherwise inert. On one side there is helium, on the other argon, both at exactly the same pressure and temperature. Such a situation can be maintained by having flows of pure helium and pure argon past the ends of the plug, passing out into the open. The concentration gradients of the two gases are equal. A careless application of Figure 2.3 might let you expect that the fluxes should also be equal.

2. Is Something Wrong?

29

In reality, experiment shows that the helium flux is about three times highet than the argon flux! If you want to understand this you should realise that Figure 2.3 only tells you something about the movement of helium and argon with respect to each other. Figure 2.3 tells you nothing about the movement of the mixture with respect to the plug. This movement is balanced by friction between the two gases and the plug. To let the two friction terms cancel, helium must have the higher velocity. There is no interaction with any plug in the conventional analysis of mass transfer; clearly you should take the plug as a component of the mass transfer system. He

)

-

N He =-3NAr

t ~

298K

298 K+-= 100 kPa

Ar

100 kPa

, •

M He

friction (He / plug) < friction (Ar / plug) ---)~

the plug, matrix or membrane is a (Joseudo)component

Fig. 2.7 The fluxes need not be equal-but-opposite

By applying a pressure difference (Figure 2.8) you can equalise the two fluxes. The viscous flow due to the pressure gradient increases the argon flux and decreases the helium flux. This is analogous to the effect of the electrical gradient in the previous example. The pressure difference required depends on the structure of the plug: the finer the pores, the larger the pressure difference. He 298 K 11( 100 kPa

)

-

Ar N He =-NAr

298K 101 kPa for example

main reason: viscous flow retards He, accelerates Ar

Fig. 2.8 Equal fluxes with a small pressure difference

I

The ratio is equal to the square root of the ratio of the molar masses.

30

Mass Transfer in Multicomponent Mixtures

In the three-gases problem we saw earlier, the capillary is much wider. Even so, rapid diffusion of hydrogen does cause a minute pressure difference. The transport of nitrogen against its gradient is in small part due to the resulting viscous flow.

2.5 Summary In this second chapter we have briefly reviewed conventional mass transfer. We have then applied it to a few examples and seen that it does not work well. The examples suggest that a better approach to mass transfer would have to take the following phenomena into account: • friction between each pair of components, including any solid matrix, • the occurrence of other driving forces than only composition gradients. You can think of electrical and pressure gradients, and • viscous flow in heterogeneous media (solid matrices). We will be working these ideas out in the rest of this book.

2.6 Further Reading Cussler, E. L. (1997) Diffusion, Mass Transfer in Fluid Systems, 2nd edition, Cambridge University Press, Cambridge, 1997. A lively book and a good primer. If you decide not to proceed into the subject of multicomponent mass transfer, Cussler will give you a few excuses. Duncan, J.B. and Toor, H.L. (1962) An experimental study of three component gas diffusion. A.I.Ch.E.J. 8, 38-41. This describes the two-bulb diffusion experiment.

2.7 Exercises 2.1 The concentrations of oxygen and nitrogen in the atmosphere change with altitude. At the top of the Mount Everest their values are about one half of those at sea level. According to Fick's law, air should be diffusing away into outer space; rough estimates show that the half-life of the atmosphere should be about ten thousand years. This is incorrect; the atmosphere is several hundred times older. Which force counteracts the diffusion and is neglected by Fick's law? 2.2 Consider a glass of water. The concentration of water in the liquid is 55 000 mol m-3 ; that in the surrounding air 1 mol m-3 • So a large concentration difference exists across the water/air interface, which is thought to be a few molecules thick. It may not seem to be surprising that water is vaporising. Now we put a lid on the glass. The air in the glass becomes saturated; the water concentration might go up to 2 mol m-3• We still have the concentration gradient in the interface, yet vaporisation stops. (This is obviously at variance with Fick's law, although few people seem to notice.)

2. Is Something Wrong?

31

In the next chapter we will take the driving force for mass transfer to be the chemical potential gradient. What does this theory tell when we put the lid on? 2.3 We fill the two compartments of the cell in Figure 2.6 with the same aqueous solution of NaCl. There are no concentration differences, so there is no driving force for either water or for the ions. If we now apply an electrical potential difference across the membrane, sodium will be forced through. This is not surprising: there is a driving force on the sodium, be it one that is not incorporated in Fick's law. What is surprising, is that the electrical force leads to a considerable flux of water. Water is not charged, and there is no obvious driving force. How can this be? 2.4 Flow is quite important in gas and water pipelines, in the airing of clothes, in the wetting of toilet paper ... in just about everything in daily life. However, transport in media with extremely small pores, such as cellophane, seems to be dominated by diffusion. Why would this be? 2.5 Study the introduction to the review of Krishna and Wesselingh (1997), 'The Maxwell-Stefan approach to mass transfer' Chem.Eng.Sci. 73, 861-911. Here the limitations of conventional approaches to mass transfer are shown using several other examples.

33

rt 1 Mass Transfer in Gases and Liquids

34

Driving Forces In this chapter we introduce a new way at looking at mass transfer. We look at the individual species in a mixture, at the forces working on them, and at their resulting motion. There are two kinds of forces: driving forces and friction forces. Their balance is the Maxwell-Stefan equation, which is the basis of this book. We end with a simplified form of the MS-equation and an estimate of the driving force.

3.1 Potentials, Forces and Momentum Before we consider motion in mixtures, we briefly repeat a few concepts from mechanics and thermodynamics. Our first concept is that of a potential. Consider the mass in Figure 3.1 and suppose you raise it slowly (reversibly) in the earth's gravity field. The work you perform is equal to the increase of potential of the mass. If you increase the height by one metre, the potential increases by almost ten Joules. The mass could - potentially - return the same amount of work if its motion were to be reversed. In this book, we will mostly consider not the mass of a species, but the number of moles. We obtain the potential per mole by mUltiplying our fIrst result by the molar mass.

o

o

the potential difference is the work required to change the condition of the weight here:

Ll\jl=mgAz=9.81J (",,9.8lNm)

lm or, per mole

o

the driving force is the negative potential gradient: F =- d'JIi =-Mg

,

dz!'

the force is downwards Fig. 3.1 Gravity: a simple example of a potential gradient·

The gravitational force on the mass is the negative of the potential gradient. It has a value of about ten Newton per kilogram; for a simple chemical this might be one

3. Driving Forces

35

Newton per mole. We will see that this is a very small force when considering molecular motion. The negative sign shows that it is directed downwards. There are other more important forces, such as those due to electrical fields and pressure fields, those in centrifuges, and the support forces in solid matrices such as membranes. We will come back to these in Chapter 11 (electrical forces), Chapter 12 (centrifugal and pressure forces) and Chapter 14 (solid matrices). Our second concept is that of the chemical potential. Consider a separation of one mole of i from a large amount of a mixture (Figure 3.2). The work required to do this reversibly is the change of the chemical potential. It typically has a value of a few thousand Joules per mole. The chemical potential is related to composition; it usually increases with the concentration of the species. As you can see, the chemical potential is a logarithmic function of the activity of the species. This activity is the product of the activity coefficient and the mole fraction.

~i

chemical potential

'YI mixture

Ili = const(p,T)+ RTln ai ".ai = "fiXi

d;tivity work required: change in the chemical potential

"'·activity coefficient

/",.--'~

/

11; = const(p,T)

pure i (one mole) Fig. 3.2 An important potential: the chemical potential

Gases, and liquid mixtures of similar components, form ideal solutions (Figure 3.3). The activity is then the same as the mole fraction. This simplifies formulae, and we will often assume ideality in our examples. Note that the chemical potential of a gas can be written in terms of a partial pressure. chemical potential in an ideal solution

J.1i = const(p, T) + RTln(xi)

in an ideal gas

Ili = const(p, T) +

R~~ )

partial pressure Fig. 3.3 Chemical potentials in ideal solutions

36

Mass Transfer in Multicomponent Mixtures

In a moment we shall see that also the gradient of the chemical potential causes a force, in a manner analogous to gravity. Our last concept is that of momentum. The momentum of a particle - which is sometimes called its 'amount of motion' - is the product of the mass and velocity of the particle. The momentum of a system is the sum of that of all parts. The momentum of the system can change due to flows carrying momentum in, out, or by forces working on the system (Figure 3.4). The momentum balance shown is the onedimensional form of a more general relation. The symbol v in this equation stands for 'some velocity'; it does not have the special meaning that it has in the rest of the book. You can regard the momentum balance as a generalisation of the Newton law of mechanics. For a closed system with a single force it reads F = ma.

change of momentum

Fig. 3.4 The momentum balance

This ends our repetition of mechanics and thermodynamics: we now go back to mass transfer.

3.2 Momentum (Force) Balance of a Species The previous chapter has indicated that we need to look at mass transfer in a different way. As a start, we consider a simple experiment with two glass bulbs connected by a capillary (Figure 3.5). The bulb at the left contains hydrogen (species' 1'); that at the right carbon dioxide (species '2'). The whole system is at ambient pressure and temperature. Consider the mixture between two nearby points z and z + dz in the capillary. The two components are moving through each other with local velocities u j and uz. The differences in velocity cause friction between the two species. The velocities that we are talking about are the average 'diffusive' velocities of the species. These diffusive velocities should not be confused with the thermal velocities of individual molecules: they are much lower. Thermal velocities are hundreds of metres per second; the diffusive velocity in a gas might be one centimetre per second. However, the thermal velocities are for the greater part random, and the random part gives neither a contribution to the species velocity, nor to the momentum of the species.

3. Driving Forces

37

Fig. 3.5 Gases moving through each other

We now consider the momentum balance of hydrogen in the slice between z + dz (Figure 3.6) This contains the following terms: 1. a force due to the partial pressure at PIAlz

z

and

z:

2. a force due to the partial pressure at Z + dz: - PIAIz+dz 3. the friction force exerted by carbon dioxide on hydrogen.

z z+dz

area A

volumeAdz

Fig. 3.6 Forces on species (1)

We expect friction between carbon dioxide and hydrogen to be proportional to the amounts of the two gases in the slice and to their difference in velocity. The amounts are proportional to the partial pressures: (friction fo rce )

oc

PIP2 (u2 - uI)

There will also be friction between hydrogen and the wall. However, except in very narrow pores or at low pressures the interactions between the two gases are much larger!. Finally the velocity of hydrogen changes a little across the slice, but the resulting terms are also smalf. So we neglect these terms. In a steady state the three remaining terms must cancel: We consider this in Chapter 21. See exercise 3.1. Velocity effects are important in centrifuges (Chapter 12) and in shock waves from detonations. I

2

Mass Transfer in Multicomponent Mixtures

38

PIAi z - PIAi z+dz

cc

PIP2(~ - uI)

Taking the limit for dz ~ 0 yields: dPI

- - cc

dz

PIP2(uI -~)

These are forces per unit volume. We obtain the force per mole of hydrogen by dividing by its concentration:

This yields:

The left-hand side is the driving force on hydrogen:

Fi =- RT dPI PI dz

You will find that this is in Newton per mole of hydrogen. The right-hand side is the friction force exerted by carbon dioxide on hydrogen. Using x2 cc P2 we rewrite it as: SI,2 X 2(UI -U2)

Here

Sl2 is the friction coefficient between species (1) and (2).

3.3 The Driving Force: a Potential Gradient You can easily check that the driving force on hydrogen is the negative gradient of its chemical potential3:

Fi = - dill dz

In this form, the formula is not only valid for binary gases but for any fluid not too far from equilibrium at constant temperature and pressure 4 • A few forms of this driving force in fluids are given in Figure 3.7. The gradient of the chemical potential causes an internal force. Internal forces cause motion inside a mixture, but not of the mixture as a whole. The force we saw in the beginning of this chapter - gravity - is an external force. It does have an effect on the whole mixture. 3 In our example p and T are constants: their values in the chemical potential disappear in the differentiation 4 This is well founded in the theory of Thermodynamics of Irreversible Processes, but we do not cover that in this book.

39

3. Driving Forces

for a given T and p

in ideal solutions

Fig. 3.7 Driving force from the chemical potential gradient

There may be several contributions to the potential of a species. The driving force is then the negative gradient of the total potential: F I

= _ dlfli dz

This means that we are allowed to split up the force in different parts due to the different potentials.

3.4 The Maxwell-Stefan Equation The complete equation with driving force and friction force reads:

Ft =~12X2(UI-Uz) This is easily generalised to any component i surrounded by other components j (Figure 3.8). driving force on i friction coefficient between i and j

(diffusive) species velocities

Fig. 3.8 The Maxwell-Stefan equation

This equation is called the Maxwell-Stefan (MS) equation, after the two nineteenth century scientists who fIrst derived it. It is the basis of the rest of this book. We shall see that it is much more general than the Fick equation. The MS equation does yield the Fick equation as a limiting case for some simple, but important, diffusion problems. However, it can also lead to results that are quite different. Because the Maxwell-Stefan equation is so important, we repeat it once more below - in words (Figure 3.9).

40

Mass Transfer in Multicomponent Mixtures

-

the driving force on a species i in a mixture

the sum of the friction forces between i and the other species;

=i. ? the friction exerted by ; on i is proportional - to the fraction of; in the mixture and - to the difference in velocity between i and ;. Fig. 3.9 MS-equation in words

3.5 Simplifying the Mathematics The complete theory of multicomponent diffusion is extensive: it is too large for a short course. This has forced us to cut a corner. Our choice has been to avoid mathematical complexity as far as we can: 1. The book only considers diffusion in one direction: one-dimensional diffusion. 2. It uses a simple engineering model of the flow near interfaces: the 'film' model. 3. You will be doing your exercises using a difference form of the Maxwell-Stefan equation, such as _110/ A_

~

=~ r ..x.(u.I -u.) £..~I.J J J ji'i

The left side of the equation contains the ratio of the change of the potential across the film to the film thickness. The right side of the equation uses suitably chosen average values of composition and the velocities. We come back to these in Chapter 4. The difference form of the equation contains all the main features of the differential equation. Also its accuracy is adequate for most engineering applications. Using it allows us to cover a broad range of mass transfer problems in a short time (Figure 3.10). Finally: when you have mastered the approximate form, you will be quite far in understanding the behaviour of the complete equations. We will occasionally step outside the limitations of the difference equation. In Chapter 10 we show how to use the difference equation to allow for more realistic flow models than the film theory. In a number of examples we will also look at complete concentration profiles using the differential equation.

3. Driving Forces

41

great scientist

one month scientist dlfl· dZ

__ I

= L ~I" r .. x.(u.-u.) ,I ,

engineer !1lf1i -A_=

.#.

, l o n e week

L(. ·x·- (-u·-u·-) I"

.#.

L4. ,

I

,I

,

one day

Fig. 3.10 Why we use a difference equation

3.6 The Film Model Mass transfer through an interface is usually caused by interplay of diffusion and fluid flow. So we need to solve the Maxwell-Stefan equations simultaneously with a flow model. two thin, one dimensional '.films' next to the phase boundary

eddies and large scale convection

"---'0:,:' • ", :,

1 1 1 3':"'1

.,001 '\--'1, 0":.'1

"~ l

30

000

:

:

,

,

'film': .. ---------no eddies

'

I

-----}

---t

... !1z

phase boundary

Fig. 3.11 Film theory: a model of flow near an interface

A simple model - very popular with chemical engineers - that has many of the main properties of real systems, is the film model5 (Figure 3.11). The idea is that bulk fluids are turbulent: that they contain convective currents and eddies of all sizes. These cause rapid mixing of the fluid, so that concentration gradients in the bulk cannot develop. Near phase interfaces the eddies die out, and transport is only by 5 To avoid misunderstanding: the film model is not a part of the general theory of mass transfer that we are developing here. It is just a convenient starting point, and a useful model for engineering problems. The MS-equations can be combined with any flow model.

42

Mass Transfer in Multicomponent Mixtures

diffusion. The film theory assumes that the eddies disappear at a defined distance from the interface - at the film thickness. The film is usually very thin; Figure 3.12 shows a few orders of magnitude. We can of course also use the film model to describe a membrane, and we shall see that it is even (approximately) applicable to diffusion inside solid particles. gases &",,10-4

ID

&=10-7 ••• 1O-4 m membrane Fig. 3.12 Thickness of films

3.7 Difference Form of the Driving Force As noted, we will mostly be using a difference form of the Maxwell-Stefan equations. The difference form of the composition driving force is given in Figure 3.13, together with an approximation.

for a given T and p

in ideal solutions

Fig. 3.13 Difference form of the driving force

Figure 3.14 shows how the potential difference varies with the ratio of the activities on the two sides of the film. It also shows that the approximation holds over wide spans of compositions. The approximate difference formula gives better results in the MS-equations than the exact one, so we always use the approximate formula. Apparently there are compensating errors.

3. Driving Forces

+1

LlJll RT

43

exact

In( a

l

/3)

ala

approximate Lla1 =

04-----~----4-----~

al

al~

- ala. 0.5(al~ + ala.)

we do not use the 'exact' potential diffirence in our approximations Fig. 3.14 The difference approximation

To become accustomed to the previous formula, you might try the exercise in Figure 3.15. You are to calculate the force driving carbon dioxide out of beer (or cola, if you prefer) into bubbles. Also check the dimensions. The problem should only take you a minute or so to solve. Try it. growing bubble of CO2

mole fractions Xl~ = 0.001 of CO2 I

- RT dx _ RT Llxl Ft1 - - - l- - - - xl dz Xl & Fig. 3.15 A glass of beer

You should have obtained a value of about two hundred and fifty meganewton per mole or five hundred thousand ton force per kilogram of carbon dioxide. Every kilogram feels the weight of a supertanker! (Figure 3.16). This illustrates the enormous size of molecular forces in mass transfer. Although the forces are large, their effects are not dramatic. This is because the force has to propel huge numbers of molecules; the force per molecule is not extreme.

44

Mass Transfer in Multicomponent Mixtures

1 kg of CO2 RT Llx j F j "" - .&: Xj

= 2.4 x 108

=

8.314x300 -0.002 10-5 X +0.002

N "" 5 X 109 kgf kg CO2 mol CO2

Fig. 3.16 Each kilogram feels the weight of a supertanker

3.8 Summary In this chapter we have seen the following.

• The concepts of potential, potential gradient, chemical potential, momentum; and the momentum balance. • We have set up a mass transfer theory based on a momentum balance of each species in a mixture. • The important terms in the balance are the driving force on the species and friction forces with other species. • The driving force on a species is its potential gradient. • An important driving force is the gradient of the chemical potential. • We use difference equations and the film theory to keep our mathematics simple. • On a molecular scale, forces due to potential gradients can be extremely large. You should memorise the approximate form for the potential difference and driving force in Figures 3.13 and 3.14.

3.9 Further Reading Sherwood, T.K., Pigford, RL. and Wilke, C.R (1975) Mass Transfer. McGraw-Hill, New York. A good explanation of the film theory of mass transfer. Taylor, Rand Krishna, R (1993) Muiticomponent Mass Transfer. Wiley, New York. A rigorous development of the proper driving forces for mass transfer.

45

3. Driving Forces

3.10 Exercises 3.1 A species in a gas has a molar mass of M = 0.03 kg mol-I. The gas diffuses through a film with a thickness & = 10-4 m. The velocity of the gas changes from UfJ. = 1 cm S-I to UJ'> = 2 cm S-I. There are driving forces and friction forces on the species. The driving force during transport through a film has a value of 107 N mol-I. The momentum balance for the species reads: Fdr!ving onl

+ Ffri.ction = il(Mjuj ) On!

ilt

How does the change in momentum compare with the driving force? 3.2 Chemical potential gradients yield forces which are of the order of magnitude of 108 N mol- I (see for example the glass of beer in Figures 3.15 and 3.16). You will find similar values in different situations, independent of particle size. This is quite different from the behaviour of particles under influence of gravity (Figure 3.1) which is proportional to the mass of the species. Estimate for which particle size, gravity will become more important than the chemical potential gradient. Take the particles to be spheres with a density p=1000kgm-3 . You will also need the number of particles per mole: Jl = 6.02 x 1023 mol-I. 3.3 Consider a room with central heating. The air at the ceiling will be warmer than that at the floor; the density near the ceiling will be several percent lower than that at the floor. Fick's law tells us that we should expect a diffusion flux from the floor to the ceiling. This prediction is a little surprising: a cold floor producing air and a hot ceiling which annihilates the same amount (?!). What does the equation in Figure 3.13 say about the driving force? You may assume that there are no differences in the mole fractions of oxygen and nitrogen between the floor and the ceiling. 3.4 The correct form of the driving force for diffusion can be derived from the theory of Irreversible Thermodynamics. We have not followed that approach in this book. It is however instructive to follow the fundamental derivations. Study Chapter 2 of Taylor and Krishna (1993) and satisfy yourself that the relations used in this book are well founded.

46

Friction We continue with the right hand side of the Maxwell-Stefan equation: the friction between the diffusing species. We then see that the MS-equations themselves are not sufficient to define mass transfer problems: we need other 'bootstrap' relations to connect a problem with its surroundings. Finally we consider the calculation of fluxes using the MS-equations and how the calculations are implemented in difference equations.

4.1 Friction Coefficients and Diffusivities In the previous chapter we have introduced the Maxwell-Stefan equation. We have seen that the forces exerted on a species i are given by: Fj

= L~i.jX/Ui -Uj) i*)

So far we have focus sed on the 'driving force': the left hand side of the equation. We now look at the friction forces in the right hand side. We begin with a few remarks on the friction coefficients (and the closely related diffusivities), and on mole fractions. The velocity differences will come later. To get a first idea of the behaviour of friction coefficients, we consider a binary mixture. This consists of a dilute solute (1) in a solvent (2). The right hand side of the MS-equation then becomes:

If also the solvent is stagnant, we get a simple formula for the friction force:

~1,2Ul

Now suppose that species (1) consists of large spherical molecules with a diameter d l and that the solvent is a liquid with a viscosity 112 (Figure 4.1) The friction coefficient of a single sphere is then given by Stokes' law, and that for one mole of spheres by: ~1,2

= }l X 61ITJ2 dl

Here }l is the number of spheres in a mole: the Avogadro number. This estimate which goes back to Einstein - leads to friction coefficients of small molecules in ordinary liquids in the range from 1012 ... 1013 N mot l (m s-lyl. This is not far from

47

4. Friction

realityl. We can now understand why friction coefficients have such large values: it is because of the huge numbers of molecules that are transported. We have seen that driving forces are large, but so are the friction forces. spherical molecules' I '

in a stagnant liquid '2' with viscosity 112

Fig. 4.1 Binary mixture of spheres in a liquid

We have been developing our theory using a friction model, This leads naturally to the use of friction coefficients. However, mass transfer theory has also been developed along other. lines, and these have led to the widespread use of MaxwellStefan diffusivities. There are simple relations between the two: RT

RT

':>1, I

I, I

Vi,; = ~ and Si,; = D .. In most problems you can regard RT as a constant with a value of about 2500 J morl. Except for the constant, the one coefficient is then the inverse of the other. We will be using either, depending on which gives the simplest formulae. Note that we give the Maxwell-Stefan diffusivity its own symbol: D. In a binary mixture the MaxwellStefan diffusivity is a close relative of the Fick diffusivity D, but in general the Fick and Maxwell-Stefan diffusivities have a different meaning and a different behaviour. We consider this in Chapters 8 and 13. MS-diffusion coefficients in liquids are around 10.9 m2 s-\ in gases around 10.5 m 2 S·I. In concentrated solutions both friction and diffusion coefficients can depend on composition, but often the variation is small. We discuss the behaviour of the coefficients further in Chapter 9. Our last remark here concerns the local composition of the 'other component j'. Including this in the friction terms makes the friction coefficients much less dependent on composition than they would be otherwise. It also shows that the effects of composition can be asymmetric: the friction force exerted by the solvent per mole of solute is much larger than the converse. We have denoted the local 1 The Einstein-Stokes fonnula is not valid for friction coefficients of small molecules in gases. These are typically ten thousand times lower than in liquids, in the range of 108 •• 109 N mor 1 (m S·l)"l; we come back to them in Chapter 9.

48

Mass Transfer in Multicomponent Mixtures

composition of the mixture by mole fractions. This is arbitrary: you can set up the Maxwell-Stefan equations using any measure of composition. For example it is possible to use mass fractions, volume fractions, but also mass or molar concentrations. A change in the measure requires an appropriate change in the size (and sometimes also the dimension) of the friction or diffusion coefficient. We have worked out a few cases in Appendix 2.

4.2 Velocities and the Bootstrap The species velocities in the equations can differ greatly. In gases they might be around 10-2 m s- \ in liquids around 10- 4 m s-I, and in a solid matrix such as a membrane they might be 10-6 m S-l or even lower. You may have noticed that the binary MS-equation only contains a velocity difference. There is no single velocity in the equation. The value of the velocity difference does not depend on how you look at a mixture - whether you move along with it or not. The Maxwell-Stefan equations are 'independent of the frame of reference'. Also the value of the velocity difference does not depend on the measure of composition used in the Maxwell-Stefan equation. The velocity differences are absolute values. For a binary you can write the equations of both species, but they contain the same difference: you cannot calculate the velocities separately. The equations only say something about motion inside in the mixture: nothing about the mixture as a whole. With more components something similar applies: there is always one independent equation less than the number of velocities (Figure 4.2) 2 components: ~

3 components: ~

n components: ~

1 relative velocity 1 independent equation 2 relative velocities 2 independent equations

n - 1 relative velocities n - 1 independent equations

Fig. 4.2 Not enough transport equations

So our equations do not determine the absolute values of the velocities: they are 'floating' relations, which have to 'tied' to the surroundings in some way or other (Figure 4.3). To do this, you need one or more relations of a completely different kind. We will call these bootstraps. There are good examples in Chapter 2: they are repeated in Figure 4.4. In the two bulb experiment the bootstrap is that there can be no volume flow of the gases. In the experiment with Argon and Helium diffusing through a plug the problem is fully defined by stating that the plug is stagnant.

4. Friction

49

'floating' transport relations: have to be 'tied' to surroundings Fig. 4.3 Equations floating in thin air

The experiment with hydrogen and sodium ions diffusing through a membrane is a little more complicated. Here we have as unknowns not only the velocities of the three species, but also the value of the electrical potential difference between the two compartments. So we need two bootstrap relations. Here also the membrane is stagnant, but in addition we use the fact that the amount of charge transferred to cause the electrical potential difference is almost zero.

no net volume flow

plug does not move membrane does not move (almost) no charge transfer

Fig. 4.4 Examples of 'bootstraps'

To obtain a good set of bootstraps, you need to understand the problem that you are dealing with. In the coming two chapters we only consider problems where the bootstrap is fairly obvious: more complicated cases will be covered later in the book (especially in the exercises).

50

Mass Transfer in Multicomponent Mixtures

4.3 Velocities and Fluxes The process engineer is not usually interested in the velocity of a species, but in its flux. The flux is the product of the species velocity and the species concentration:

Here c is the total concentration of the mixture. A simple way to obtain a relation between the fluxes is to multiply both sides of the Maxwell-Stefan equation with the concentration of species i (Figure 4.5):

force on i per unit volume of mixture Fig. 4.5 Flux form of the MS-equation

Here!; is the driving force on species i per volume of mixture. This form of the MSequation is often more practical than the form with velocities. However, it does not show the mechanism of the equation quite so clearly. In this book we use both the velocity and the flux forms.

4.4 The Difference Equation As noted before, we will be using the difference form of the Maxwell-Stefan equation in most of our exercises. Writing out the difference terms is straightforward, but you have to do it carefully. Also it is handy to rearrange the equation a little. We illustrate this for a binary mixture with only an activity gradient as driving force. Figure 4.6 shows a film with its left position a and its right position ~, and a flux going down its gradient. The first equation is the flux velocity form of the MSequation: here it is simplest to write it using a diffusivity. In the second equation we have cancelled the RTs and rearranged the two differentials. In the third we have replaced the differential by a difference. The activities, mole fractions and total concentration are replaced by average values. Also we have introduced our old friend: the mass transfer coefficient. Note that a difference is equal to the value at the right hand side ~ minus that at the left hand side a and not the other way around. For example Aal = alp - ala' The average values are as you might expect: X2 = O.5(x2a + x2p), Here you must realise that velocities in a film usually vary with position. They will be highest where the concentrations are lowest (Figure 4.7). The velocities that we are estimating with the difference equation are those at the average concentration.

4. Friction

L1a

1 _ (_

l --=-=kX2

positive direction

a1

51

D k 1,2 =~ L1z

_)

U 1 -U

2

1,2

t

mass transfor coefficient Fig, 4.6 Deriving the difference equation

ctl-film

,-A--.,

o

a

avemge concentration

c,

~

species velocity (depends on position in film)

U;

"'{V o --+-------1f-)

species velocity at the average composition

positive velocity Fig. 4.7 Velocities in a film

Also here you can obtain the flux form via a simple multiplication (Figure 4.8):

L1a1 1 _ _ _ -- =k x2 (u1 -u) 2 al

1,2

Fig, 4.8 Difference equation in flux form

Both the velocity and the flux equations are easily written for any number of components (Figure 4.9). The forms for ideal solutions are quite simple: these are the equations that we will be using in the coming three chapters.

52

Mass Transfer in Multicomponent Mixtures

using velocities

using fluxes

'---.r---'

for ideal solutions - Ax; Fig. 4.9 Multicomponent difference equations

4.5 Mass Transfer Coefficients Our difference equations may contain several mass transfer coefficients: one for each pair of interactions. These form a natural extension of the single mass transfer coefficients in binary mixtures. Figure 4.10 gives an idea of the ranges of mass transfer coefficients encountered. For gases they are usually between one centimetre and one metre per second. For liquids, they are a factor of a thousand lower. Transport coefficients in porous media are typically an order of magnitude lower than in free solution. In tight polymer matrices they may be two or three orders of magnitude lower. The coefficients in solids used for packaging are far lower, but these are not of interest in mass transfer operations. We come back to estimating mass transfer coefficients in Chapter 10. D kc=_'_·1 ',I

&

10-2

10-4

~ ~c

+-- gases

c..

'" 10

-1

-1

m pores

~

c +--

CH

cc .. '" ~

10-6

ms

liquids 10-4 m s-1

mpores

Fig. 4.10 Mass transfer coefficients

In many problems, one of the friction terms will dominate in a given equation. You will usually find that the velocity of the species considered is then of the same order of magnitude as that of the mass transfer coefficient in the dominating term. This ends our introduction of the basic theory of multicomponent mass transfer. You should now be ready for the applications.

4. Friction

53

4.6 Summary • In this chapter we have looked at the friction forces in the right hand side of the MS-equations. • These contain friction coefficients that are related to a new kind of diffusivity. • The MS-equations contain velocity differences: they do not determine absolute velocities or fluxes on their own. • To obtain velocities with respect to a phase interface you must therefore have one or more 'bootstrap' relations to tie down the problem. • Once you know the velocities, you calculate fluxes as the product of the species velocity and the species concentration. • You can also write the MS-equations directly in terms of fluxes. • We will do our calculations using difference forms of the MS-equation. These calculate average velocities and fluxes using mass transfer coefficients. • You should memorise the difference forms of the MS-equation in Figure 4.9.

4.7 Further Reading A. Einstein, Investigations on the Theory of Brownian Movement, Dover Publications Incorporated, New York, 1956. The sphere-in-liquid model is from A. Einstein, who uses the gradient of the osmotic pressure as the driving force. Otherwise his treatment is quite in line with the M-S approach. Taylor, R. and Krishna, R. (1993) Multicomponent Mass Transfer. Wiley, New York. A rigorous development of the Maxwell-Stefan formulation.

4.8 Exercises 4.1 Look at Figure 4.5. We consider a solution dilute in component '1' (Xl ~ 0). The mass transfer coefficient in the final equation has a value kl2 =10-4 ms-I. (a) What is the largest value that the dimensionless driving force -f1xdxI can attain? (b) What is then the value of the velocity difference (UI - Uz)? 4.2 Texts on the classic theories of diffusion often begin with deliberations on the many different kinds of velocities in mixtures. Examples are the mass average velocity, the molar average velocity, the volume average velocity and different species velocities taken with respect to these averages, or with respect to different frames of reference. All these velocities greatly complicate the subject. In this book we use the Maxwell-Stefan equations from the beginning and you will see only one diffusive velocity U i for each component i. You do not need all the others! Can you

54

Mass Transfer in Multicomponent Mixtures

see why the Maxwell-Stefan system is so much simpler in this respect? 4.3 We have a mixture of four species. How many independent velocity differences are there? 4.4 In a ternary system, consisting of species 1, 2 and 3, the expression for the friction experienced by species 1 (say) does not have a contribution from the term (u 2 - u3 ). Why is this? Hint. Have a look at Example 4.3. 4.5 What would the bootstrap relation be for the glass of beer problem in Figure 3.15? Remember that there are two major components in this diffusion problem: carbon dioxide and water. What is the motion of water with respect to the bubble interface?

55

Binary Examples This chapter covers examples with ideal binary mixtures having only a composition gradient as a driving force. Here the advantages of our new approach to mass transfer are not large. However, the examples allow you to become acquainted with the method on small sets of small equations.

5.1 Stripping In the first example (Figure 5.1) ammonia is being stripped from drops of water, into an atmosphere of nitrogen. We assume that water evaporation is negligible, and that the bulk concentration of ammonia is very small. Also, we focus on the gas film. There are only two components. So there is one independent transport equation. We choose the equation for ammonia. (If you choose that for nitrogen you will see that you get the same equation after a little rearranging.) Nitrogen does not transfer through the interface: the nitrogen flux is zero. The nitrogen mole fraction is unity. Inserting all this into the transport relation gives the flux. This is equal to the product of the mass transfer coefficient and the concentration difference. In this simple example the new method gives the same result as the classic method, albeit in a roundabout manner.

•r l

mtl

drops on a tray gas: trace ofNH3 (1) transport relation

-Ax

]I x2

bulk ofN2 (2) x2N] -x]N2 \ k],2

~l

C

Nz

()~ =0

bootstrap

Fig. 5.1 Stripping of a trace of ammonia

Mass Transfer in Multicomponent Mixtures

56

Above, we have considered gas containing only a trace of ammonia; we now consider a concentrated gas. In Figure 5.2 the average mole fraction ammonia in the gas film has a value of one half. Set up the transport relation and bootstrap equation and solve the flux yourself. The answer is shown in the figure, but try yourself.

---"-----'-----'------- 0

a

(3

Fig. 5.2 Stripping of concentrated ammonia

As you can see, the flux is now equal to twice the product of mass transfer and concentration difference. This may seem a little surprising, but classical mass transfer yields the same result. There, a drift correction or Stefan correction is required for this example. The value of the correction factor is indeed two. The new theory does not require such a correction; drift forms an integral part of the equations.

5.2 Polarisation A phenomenon that can be important in membrane processes is polarisation (Figure 5.3). Here, water is permeating through a membrane that retains dissolved salt. The water transport causes an increase in salt concentration close to the membrane interface. How large is this increase? The transport relation for the salt should be obvious. The bootstrap relation is that the salt velocity is zero: salt cannot pass through the membrane. Working out the equations is trivial. You see that the salt concentration starts rising very sharply as the value of the water velocity approaches the value of the mass transfer coefficient. An 'exact' solution of the steady-state film model for this case is not difficult to obtain. As you can see, the 'exact' and approximate solutions are similar.

5.3 Vaporisation This (Figure 5.4) is a case that has puzzled many engineers. A drop consisting of a binary mixture of benzene and toluene is evaporating in a flash vessel. Benzene is the more volatile, and it evaporates much more rapidly. It has the expected concentration gradient downward towards the interface. However, this implies that toluene must

57

5. Binary Examples

have a gradient upwards. Yet toluene is evaporating. So it must be diffusing against its concentration gradient. How can this be? water (l) permeates salt (2) does not: X2

&2

transport:

x2 small, u2

-1 O-u -&2 --I

also

3

=0 .

x2

k l ,2

-

= X2 +2

X2

&2

o;

uI Ax2=~ 2

2 so I

exact solution: 2

1

2-~ kl ,2

lli kl ,2

00

X2 «

Llx2 = X2a

ex p(

ul )

kl ,2

Fig. 5.3 'Polarisation' upstream of a desalination membrane

heat benzene (l), volatile toluene (2) Yl =K1x1B

l--"';"

x'B

::!_!.2 =

~ K 2 x2B

vapour removed by convection

bootstrap:

Fig. 5.4 Vaporisation of a binary drop

With your knowledge of the Maxwell-Stefan equation, this should be clear. The mixture moves towards the interface. Toluene diffuses in the opposite direction, but the diffusive velocity is smaller than the mixture velocity. The bootstrap relation here is found in the vapour phase. We assume that vapour moves away from the interface by convection (which is a good approximation). The two fluxes must then have the same ratio as the equilibrium compositions in the vapour at the interface. The bootstrap and the flux relations yield the result in Figure 5.5. The MS-relations allow

Mass Transfer in Multicomponent Mixtures

58

us to calculate the drift corrections; in the example shown they have values of plus four and minus two for benzene and toluene.

x,N2 -x2 N,

k,,i: N

-'=v N2

example

v =2

x, = x2 =0.5

Fig. 5.5 Fluxes from the vaporising drop

Modelling this problem with a single steady-state film resistance is rather an oversimplification. However, even this simple model contains most of the essentials of the problem.

5.4 Gasification of a Carbon Particle At sufficiently high temperatures, carbon particles in oxygen react to form carbon monoxide (Figure 5.6). Oxygen diffuses to the particle surface. Its concentration there is very low: it is almost immediately consumed by the reaction. Twice as much carbon monoxide is formed as oxygen is consumed. This diffuses away from the surface into the bulk of the gas. The bootstrap relation is given by the reaction stoichiometry: the carbon monoxide flux is equal to twice the oxygen flux and opposite in direction. both components are moving and have a high concentration c =10 mol m- 3

1.0

bootstrap:

0.6 0.4

0.0

calculate NI and N2

Fig. 5.6 Gasification of carbon; the problem

All data required to calculate the fluxes are given in the figure. You might try it as a last binary exercise. As before, the solution is given (Figure 5.7). Note how good the

5. Binary Examples

59

agreement is between the approximate difference method and the exact solution of the differential equation. N z =-2N1 ~

-Llxl = xlNj -xjNl = (Xl +2Xj)Nj kj,i: kj,lC 2

NI = -

kI •2c LlxI = _ 10- X 10 (-0.6) x2 +2xi 0.7+2x0.3

=0.046 (exact:0.047)molm- l

NI

S-I

Nl =-0.092 (exact :-0.094)molm-l

S-I

Fig. 5.7 Gasification of carbon: the solution

5.5 Binary Distillation As a last example, we consider a tray in a distillation column. The column separates a mixture of hexane (1) and heptane (2). On the tray, the more volatile (1) will be vaporising, while (2) will be condensing into the liquid (Figure 5.8). This problem looks similar to the stripping problem with which we started this chapter. However, it is quite different because both species are transferring. 1

1

X1o. :

heptane (2)

1

~----

/1 1 /

1

1

x l~ XI~

xzo. : hexane (1)

W

\N2;1

---'--'--'!"'---

et

~

transport relation _

A ..

_

LUj-

xlNI-xlNl kl,lC

bootstrap N = - Nl (equimolar exchange) j

~

_

A .. _

LUI -

(X, +xl)N, -'--'--'=":"'-"':"

kl,lC

0

Fig. 5.8 Equimolar exchange in binary distillation

The bootstrap follows from the mass and energy balances on the tray. On the tray, the temperature changes are usually small. Also, the two components have almost equal molar heat capacities and molar enthalpies of vaporisation. As a result, the fluxes of the two components are (almost) equal but opposite - we have equimolar exchange. We will not derive this here, but we discuss similar problems in Chapter 7. The fluxes are just the negative of 'mass transfer coefficient times concentration difference'. This is one of the few examples, which obeys the simple law in Figure 2.1 without a drift correction. Here, this is so for any set of compositions, not only for dilute mixtures.

Mass Transfer in Multicomponent Mixtures

60

5.6 Summary In this chapter we have seen the following. • The theory used here gives the same results as classic binary mass transfer. (This holds for ideal solutions.) • Our new theory requires no drift corrections. • The velocity and flux forms of the MS-equations are identical; the flux forms are often handier. • We have seen a number of different bootstrap relations. Figure 5.9 summarises them.

CD membrane stagnant W bulk stagnant (absorption) Cl) @

equimolar exchange (distillation)

0

interface determined (vaporisation)

@

UM

N2 =0

a; =0

trace stagnant (polarisation)

reaction stoichiometry

=0

NI

+N2 =0

NI

=2:'L

N2

Y2

NI

=N2

VI

V2

Fig. 5.9 Some bootstrap relations

5.7 Further Reading Cussler, E. L. (1997) Diffusion, Mass Transfer in Fluid Systems, 2nd edition, Cambridge University Press, Cambridge. A good introductory text on binary mass transfer. Bird, RB., Stewart, W.E. and Lightfoot, E.N. (1960) Transport Phenomena. Wiley, New York. This book contains the definitive treatment of binary mass transfer. Drift phenomena are covered very clearly. Taylor, Rand Krishna, R (1993) Multicomponent Mass Transfer. Wiley, New York. Nice examples to demonstrate the influence of drift on binary mass transfer. Also bootstrap relations. Many fuZZy worked numerical examples.

5. Binary Examples

61

5.8 Exercises In this chapter we begin with our first computer exercises. We will be doing these in Mathcad, a computer language that is briefly described in Appendix 1. You will find the files in the directory Exercises/Questions on the CD-ROM. The short exercises, that you are to program yourself, are marked with We suggest that you at least try a number of these.

*.

5.1* Stripping of Ammonia (Mathcad). This is the problem given in Figure 5.1. We investigate the differences between the simple approach to mass transfer of Figure 2.1, the difference approximation of the MS-equation in Figure 5.1 and the 'exact' solution of Stefan for diffusion through a stagnant film. 5.2 What is WRONG with this Mathcad file? (Mathcad) A little warning: this problem has nothing to do with the Maxwell-Stefan equations. It is a problem that you may encounter in any programming language (also in Mathcad). You may see the answer immediately. If you do not, try a few things in the Mathcad file.

*

5.3 Binary Distillation of Methanol - Ethanol (Mathcad). Here you will have to set up the transport equation and bootstrap relation in Mathcad yourself. The equations are those in Figure 5.8. We advise you to let Mathcad solve these equations with the Newton-Raphson method (in a Given ... Find block). You will need to understand this method anyhow as we go on. 5.4 Polarisation in Reverse Osmosis and Ultrafiltration (Mathcad, but also possible without). The pressure driven membrane processes of reverse osmosis (RO) and ultrafiltration (UF) are similar. The difference is in the size of the molecules rejected. A reverse osmosis membrane rejects salt ions with a diameter of around 0.3 nm; ultrafiltration rejects proteins which have a diameter of around 3 nm. So the proteins diffuse much more slowly than the ions. The mass transfer coefficients in the polarisation layer (Figure 5.3) might have values of 10.4 m S·l for RO and 10-5 m sol for UFo The volume fluxes of water for the two processes are roughly equal: 10-5 m sol. For which of the two processes is polarisation the most important? How much difference does it make? 5.5 Exact Solution of the Polarisation Problem. The exact solution of the polarisation problem in Figure 5.3 is not difficult to derive. This at least, if you restrict yourself to dilute solutions. Regard the polarisation layer as a stagnant film running from z = 0 to z = &. If you assume that water is ideal, then the Maxwell-Stefan equation in the film is:

62

Mass Transfer in Multicomponent Mixtures

RTdxl

RT

dz

DI ,2

- - - = - - X Z ( U I - UZ ) Xl

With the additional relations u2 = O,x2 = 1- Xl and Xl ~ 1 and assuming that u l is constant, you can arrange this in such a way that Xl only shows on the left side of the equation and dz on the right. You can then integrate the equation. Check whether you get the same result as in the figure. 5.6* Evaporation of a Binary Mixture (Mathcad). This is the problem in Figures 5.4 and 5.5. It is more difficult than the previous ones because you only know the compositions in the bulk of the liquid. Again you can best let Mathcad do the solving. 5.7 In this book we mostly use the difference approximation of the Maxwell-Stefan diffusion equations. This is good enough for most engineering problems because of the uncertainties in the mass transfer coefficients in practical situations. It is, however, instructive to compare the approximate solutions with exact analytical solutions. As an example, take gasification of a carbon particle; see Figure 5.6. Develop the exact analytical solutions and compare the final answers with those of the difference approximation. Hint. The exact analytic expressions can be found in Chapter 8 of Taylor and Krishna (1993).

63

Ternary Examples This is an important chapter. After finishing it you should have done your first multicomponent calculations.

6.1 From Binary to Ternary The extension of the transport relations from binary to three components is straightforward (Figure 6.1). There are now two independent transport relations. In each relation there are two friction terms: for component 1: • that between (l and 2), and • that between (l and 3). The forces in these equations are forces per mole of the component considered.

F; = SI,2 X 2(UI - u2 )+ SI,3 X3(UI -u3 )

forces per mole of' l'

= S2,I X I(U 2 -UI )+S2,3 X 3(U 2 -u3 )

forces per mole of'2'

F2

Fig. 6.1 From binary to ternary ...

If we multiply the first equation with XI' and the second with x 2' we obtain the forces per mole of mixture (Figure 6.2). In the mixture, friction between (1 and 2) must balance the friction between (2 and 1). A look at the equations then shows that the friction coefficients between (l and 2) and between (2 and 1) must be equal. (This a form of the Onsager relation). So, although there are four friction terms in the two equations, there are only three friction coefficients. The same holds for MaxwellStefan diffusivities and mass transfer coefficients. - Xl

Fi l

r--------------SI,2 XI X2(UI - U2y:+ S1,3 XIX3(UI - U3) I

I

}forces per mole

f

'

0

mIxture

~

-x2F2 =!S2,IXIX2(~ -UI):+S2,3X2X3(U2 -u3) ___________ J

these should cancel:

S2,1

=SI,2

D2 ,l

=DI,2

10. I =kl 2

Fig. 6.2 Forces per rnole of rnixture and the equality of two friction coefficients

Also the difference form of the Maxwell-Stefan equations is just an extension of the binary case. Figure 6.3 shows the flux form that you will need many times. Further extensions, to quaternary and so on, only require more equations with more terms.

64

Mass Transfer in Multicomponent Mixtures

binary

ternary quaternary

Fig. 6.3 Difference equations, binary, ternary, quaternary ...

6.2 A Condenser Your first multicomponent exercise concerns a condenser (Figure 6.4). The tubes of the condenser are cooled internally, and water and ammonia condense from a vapour mixture of ammonia, water and hydrogen. The hydrogen is insoluble and does not condense. You are to find the fluxes through the gas film.

NH3 (1) and Hp (2)

condense on a tube

H2 (3) does not condense k12

=lxlO-3

0.6

ms- 1

k1,3 = k2,3 = 3 X 10-3 m

s-1

c=30molm-3

0.4 0.2

find the f/uxes through the gas film

0.0 Fig. 6.4 Condensation of a ternary mixture

It is important that you try this exercise yourself. Further on in this book there will be

many more examples of multicomponent transfer (usually more complicated). Once you have the idea you will be able to glance over them and you will recognise the same things coming back, again and again. But you must first try yourself. All data for the exercise are given in the figure. So go ahead. Write down the transport equations and the bootstrap relations and solve the fluxes. It may cost you a quarter of an hour, but once you have done it you will see how simple it is. The exercise is

6. Ternary Examples

65

worked out in Figure 6.5. If you write out the equations you have a set of three linear equations in the three fluxes. In this example the bootstrap relation is so simple that you can immediately reduce the system to two equations and solve them. transport (MS) relations: -0.2= Oo4N) -0.3N2 + 0.3N) - 004N3 (1 x 10-3 )30 (3 x 10-3 )30 004= 004N2 -0.3N1 + 0.3N2 -0.3N3

(1 x 10-3 )30

(3 x 10-3 )30

bootstrap three linear equations, three unknowns ~

NI =0.015

N2 =0.045 molm- 2 s-1

exact solutions: NI =0.013

N2 =0.049 molm- 2 s-1

mixture velocity H20 moves down its gradient NH3 drifts against its gradient H2 does not move at all

H

2

I I

Fig. 6.5 The solution to the condenser

The mixture moves towards the liquid. The driving force on water is in the same direction and it moves more rapidly. Ammonia is retarded by a driving force in the opposite direction. Hydrogen does not move at all even though it has a concentration gradient. Its driving force cancels against friction with the other components. If you know the driving forces, the theory here always gives a linear set of equations for the fluxes or velocities. The problems arise when you do not know the driving forces. You then usually have to solve the equations by trial and error. Even then, the fact that the MS-equations are linear in the fluxes or velocities is a distinct advantage.

6.3 A Ternary Distillation We now turn to another mUlticomponent problem: the prediction of tray efficiencies in distillation. Doing this completely would require setting up a complete model of a tray and this is outside the scope of this book. We can, however, obtain a qualitative idea of the problems by looking at our transport relations.

Mass Transfer in Multicomponent Mixtures

66

The (Murphree) efficiency of a tray is defined as the ratio of two quantities (Figure 6;6). The first is the change of the vapour composition on a real tray for a certain component. The second is the change that would occur in an equilibrium stage. Such a stage is an ideal concept: you may regard it as a tray with both phases well mixed and with no mass transfer resistances. In binary distillation, efficiencies are very useful. They are always positive, and usually have a value somewhat below one. They tend to be fairly constant throughout a column, and are easily incorporated in distillation design. Can analogous concepts be used in multicomponent distillation? We think not and will try to explain this. equilibrium line

Y

n

y~

operating line

x = Yn - Yn-J = _--..:c~h~a~n~g>.=::e....!;in~re~a~l-=s'""'ta"",g:>::e,-----

E mv

y~q -

Yn-J

change in equilibrium stage

Fig. 6.6 The Murphree efficiency of a tray

Look at the vapour film of the distillation column given in Figure 6.7. (We will neglect any mass transfer resistance in the liquid). There are three components: ethanol, water and a trace of butanol. Ethanol is vaporising; it has a gradient from the liquid down to the gas. Water has an opposite gradient and is condensing. The butanol compositions have been chosen equal at both sides of the film: butanol has no gradient. The transfer coefficients involving water are much higher than those involving the alcohols are. The bootstrap relation is that of equimolar exchange, as already discussed.The friction between butanol and ethanol is much larger than that between butanol and water. Now the usual question: will butanol be moving, and in which direction? Of course butanol will be dragged along by the ethanol: it will be vaporising. However, butanol has no driving force: its vapour compositions on the two sides of the film are in equilibrium. So the butanol efficiency must be unity. We look at a few small variations of the butanol gradient (Figure 6.8). Suppose that we increase the bulk concentration of butanol a little to give it a gradient towards the interface. At a certain point the driving force and the drag on butanol will cancel. There will be no flux and no change in the butanol mole fractions on the tray. The efficiency will be zero. In between these two points, the denominator in the Murphree efficiency goes to zero. (This is not obvious; you have to believe this for

67

6. Ternary Examples

the moment). The Murhpree efficiency of butanol is then undetermined: plus or minus infinity.

CD CV

ethanol G) water

vapour

a trace of butanol

large friction between CD and k1,2 =8xlO-2 m s-l k 1,3

=10.;3 =20 X 10-2

I

liquid

CV

m S-l

bootstrap: equimolar exchange ul.Yl

+ u2.Y2 + 113.Y3 =0 0.02

in which direction does(j) move?

Fig. 6.7 The vapour film on a temary distillation tray

Y2a: : Y2l> ( Ho.022 0.018 I I I

0.020

~ =-3.75 X 10-2

I

P

M

Q0.020

0.02l4~

0.022

no motion

0.0186

:

I

I

I

0.018

("

Murphree efficiency undetermined

lr----------+~'--------~ I

Y2a

= Y2j3

I I I I

I I

zero

=0.020 o~-----------+~----~

Fig. 6.8 Murphree efficiency of butanol versus composition

More complete calculations confirm the picture found here. Murphree efficiencies in multi-component mixtures are not the simple well-behaved functions of binary systems. The binary concept is useless. Please note that the problems are not due to non-idealities. In this case they are due to interactions between molecules in ideal gas

68

Mass Transfer in Multicomponent Mixtures

mixtures! We believe efficiency approaches to multicomponent distillation to be a dead-end alley. They only increase confusion. We prefer models that compute the mass transfer fluxes directly. It may be remarked at this point that the three-gases problem from Chapter 2 is formally identical to the problem we have just considered. It can be described almost

quantitatively with the same equations.

6.4 A Ternary Reaction As a third example, we consider a reaction of gases on a catalytic surface. The gases are nitrogen and hydrogen, and they react to form ammonia (Figure 6.9). The reaction causes differences in composition between the bulk and the surface. As a result hydrogen and nitrogen diffuse to the right, and ammonia diffuses to the left. You must expect the mass transfer coefficients involving hydrogen to be larger than the nitrogen-ammonia coefficient. transport relations: A __

- LiAl -

x2 N l- x l N 2

N

A _

_

- LiA2 -

x N

x3 l - l 3 + ----"-----"---"---"'-

k1,2C

k1;3c

x 1N 2 - x 2N l

+

x3 N 2- x 2 N 3 ---"------"'----=---"'-

k1,2C

NH3 (3):

A _

_

- LiA3 -

I

I I

/

Xl N3 - X3 N I

/

bootstrap: NI I

N

N

X2 3 - X3 2 + ----"' ------"----"'-----"'-

k 2,3c

kIf

a/13 catalytic surface

k2,3C

= Nz = N3 3

-2

Fig. 6.9 Ammonia reaction on a catalytic surface

The reaction stoichiometry provides the bootstrap relation. Because much hydrogen is consumed, there is a large drift flux towards the catalytic surface. You will notice that we have written all equations in the flux form; this is always handiest when dealing with chemical reactions. In the following figure, it is easy to get the symbols for nitrogen (Nz) and the flux of hydrogen (Nz) mixed up. So look carefully. With these complications, you might expect considerable multicomponent effects. We come back to this in a moment.

6.5 Binary Approximation of a Ternary Engineers often use binary mass transfer relations for ternary mixtures. From the previous examples you might obtain the impression that this should lead to disaster. This is not so: in most cases the problems are not spectacular.

69

6. Ternary Examples

When can we approximate ternaries as 'effective' binaries? That is when the two friction terms can be taken together. Figure 6.10 shows three cases. a ternary can be approximated as a binary when

Axl _ ul -u2 _ ul -u3 - - = x 2 - - + x3 - xl kl ,2 kl ,3

CD

~

_ ul -ueft "'=Xtt---=e kl,eft

one friction term dominates:

= X3 kl,eft = kl ,3 Xeff

examples: mobile species in many membranes solutes in a solvent

(2)

equal velocity of two species: (U2

=u3

b(

x2

+

x3 )(Ul-U2)

kl ,2 kl ,3 example: Na+ and Cl- in water

®

.

equal diffusivities: (.

=k

k \,2

) ~ (x2 + x3 )ul - (x2 u2 + x3 u3) k

\,3 .

1,2

example: '1' in 0- and m-xylene Fig. 6.10 When a ternary behaves as a binary

• In the ftrst case, one friction term disappears in both equations. This might happen

when friction with a membrane or a solvent dominates. • When two components have the same velocity, they behave as a single component. This happens when hydrogen diffuses through stagnant air (nitrogen and oxygen). The two ions of a salt may also behave as a single component. (However, with three ions, the situation is completely different, as we shall see in Chapter 11.) • In the third case, two species are similar and two mass transfer coefficients are identical. We can now group the species into an effective binary. Many problems are approximately described by these situations. Their behaviour is then similar to that of a binary. You should not, however, take this as an excuse for not learning the multicomponent approach. If you use the binary viewpoint you cannot see when mUlticomponent effects do become important. There is another important case where multicomponent mixtures behave as binaries. This happens when all fluxes are coupled by the stoichiometry of a single chemical reaction. An example is the ammonia reaction from the previous section. If we take

70

Mass Transfer in Multicomponent Mixtures

the transport equation for nitrogen, we can eliminate the hydrogen and ammonia fluxes with the stoichiometry relations (Figure 6.11). The result is an effective binary relation and we can obtain similar ones for the other components. So - possibly to your surprise - this reacting multicomponent mixture shows a simple behaviour! We will make use of this when we discuss heterogeneous catalysis in Chapter 22. simplifying the transport equation ofN2 in Fig 6.9: eliminate N2 and N3 with

N2

=3Nl

N3

=-2NI

1 x2 -3xl +---"-----"x3 +2xl --=

k1,eff

k1,2

k1,3

similarly for H2 and NH3 Fig. 6.11 Effective binary relation in the ammonia reaction

6.6 Summary • In the driving force and friction description of mass transfer, multicomponent systems are a straightforward extension of binary systems. • You need one mass transfer coefficient for each pair of interactions. So one for a binary, three for a ternary, six for a quaternary, and ten for a five-component mixture. • For given boundary conditions of the film the transport and bootstrap relations are linear equations in the velocity. You can solve these easily. • In multicomponent systems, friction can well cause a component to move against its gradient. This is why certain binary concepts, such as that of a tray efficiency, are not always transferable to multi component systems. • Binary theory can be used to approximate multicomponent systems when all mass transfer coefficients are equal, and when all components are dilute except for the solvent. • When all fluxes are coupled by the stoichiometry of a single chemical reaction, we can describe a mixture as a set of pseudo binaries.

6.7 Further Reading Krishna, R. and Standart, G. (1976) A multicomponent film model incorporating an exact matrix method of solution to the Maxwell-Stefan equations. A.l.Ch.E.J., 22, 383-389. Presents an exact analytic solution to the Maxwell-Stefan equations for ideal gas mixtures. This approach is often used as a benchmark to test simpler approximate solutions. /"

6. Ternary Examples

71

Taylor, R. and Krishna, R. (1993) Multicomponent Mass Transfer. Wiley, New York. Gives several worked examples to show the differences between binary and ternary mass transfer. Comprehensive treatment of mass transfer in ternary distillation and condensation of mixed vapours. Pseudo-binary treatment of multicomponent mixtures using the effective diffusivity approach is discussed critically. Toor, H.L. (1957) Diffusion in three component gas mixtures. A.I.Ch.E.J., 3, 198207. A classic paper which emphasises the peculiarities of ternary mass transfer.

6.8 Exercises 6.1 In Figure 6.1 we have written the forces per mole of each of the species. How do you modify the driving force terms to obtain forces per unit mass? In Figure 6.2 we have written the forces per mole of mixture. How do you obtain the driving forces per unit mass of the mixture? Using different units is explored systematically in the Appendix 2 on Units. 6.2 We consider the ternary mixture of Figure 6.1. Although there are only two independent transport relations, we can write out all three relations. We have put the friction coefficients for these equations in a matrix below:

~1.2 ~2,1

hi

~1,3 ~2,3

~3,2

The coefficients in the first two rows are the same as those in the figure. Check that you understand those of the last row for component '3'. Note that there are no diagonal terms in the matrix: a component does not exert any friction on itself. The Onsager relation tells us that all three coefficients in the lower left triangle are equal to the corresponding values in the upper right triangle. So there are three different friction coefficients, not six. Extend this reasoning to obtain the number of coefficients in a mixture of four components.

*

6.3 Ternary Diffusion in a Stefan Tube (Mathcad). Your first real multicomponent exercise. You are to set up the transport relations for two gases diffusing through stagnant air in a tube. You can use the same Given ... Find blocks as before, but now with more equations (two or three; you can solve this problem in several ways). 6.4* Diffusional Distillation (Mathcad). An exercise on a surprisingly simple way of breaking an azeotrope. As in the previous example, you know the compositions on

72

Mass Transfer in Multicomponent Mixtures

both sides of a film beforehand. This makes the problem relatively easy. 6.3 and 6.4 are important exercises; they show the structure of most future calculations. 6.5 Accuracy of the Difference Equation (a Condenser) (Mathcad). This is a large file (which took several days to construct). It considers the problem of the condenser in Figures 6.4 and 6.5. It first solves the simple difference equations that we use (the 'one step' approximation). It then subdivides the film into four thin layers; this should approximate the differential equation quite accurately. Comparing the two solutions allows you to see what the accuracy is of the one-step method. The fourstep method also gives composition profiles in the film. You will spend most of your time reading the file and only little in making a few modifications. 6.6 Efficiency on a Distillation Tray (Mathcad). We have built a model of a distillation tray for you to play with. Have a look at it and check that the behaviour of the efficiencies of a trace component is like that shown in Figure 6.8. 6.7 The Maxwell-Stefan equations for ternary diffusion can been solved analytically without making the difference approximation; the solution is given by Krishna and Standart (1976). This exact analytic solution can also be programmed without great difficulty using MathCad. For the diffusional distillation example 6.4, try to compare these results with those obtained using the difference approximation. Hint. You may look at Chapter 8 of Taylor and Krishna (1993) for help, if needed. 6.8 In the distillation tray efficiency example 6.6 we took the vapour phase to be well mixed. In actual distillation tray operations, a better model is to assume that the vapour phase is in plug flow. Try to work out the component efficiencies for the plug flow vapour situation. Can you explain the differences in the efficiency values for the two sets of assumptions? Hint. You may consult Chapter 13 of Taylor and Krishna (1993) for help. 6.9 Air consists of a mole fraction of 0.21 of oxygen; the rest is nitrogen. In many problems the two components do not move with respect to each other and so behave as a single component. The diffusivities of hydrogen-in-oxygen and hydrogen-innitrogen are estimated as 3.22 x 10-5 and 3.41 x 10-5 m2 s-'. Write down the Maxwell-Stefan equation for hydrogen, and from this determine the effective diffusivity of hydrogen in air.

6. Ternary Examples

73

6.10 A species '1' is diffusing through a mixture where all other species have the same velocity. The friction side of the MS-equation then has the form: XZSl,Z(Ul - Urn) + X3S1,3(Ul - Urn) + X4S1,4(Ul - urn) + ...

=Sl,eff(Ul -

urn)

(a) Write the general formula for the effective friction coefficient of a species i moving through other species j which all have the same velocity. Do this also for Maxwell-Stefan diffusivities and mass transfer coefficients. (b) Could you also use an effective friction coefficient when the velocities of the other species are not equal? (c) Can you think of any special cases, where the velocities of the other components are not equal, but where an effective coefficient is still handy.

74

Mass and Heat Transfer In the previous chapters we have considered problems which were isothermal (or almost so). Here we look at processes with notable temperature gradients. We will see that mass and heat transfer processes can influence each other strongly.

7.1 Temperature Gradients In many mass transfer processes, such as condensation, evaporation and drying, we encounter temperature gradients within the diffusional film. Figure 7.1 shows this qualitatively. evaporation

drying

condensation

® x

liquid vapour

x

x

Fig. 7.1 Conductive heat and mass fluxes, temperatures and mole fractions in three mass transfer processes

Systems where chemical reactions with large heat effects occur also show such gradients. An example is the gasification of coal, which is an endothermic reaction (Figure 7.2).

H 20+CO

it

Fig. 7.2 Conductive heat flux, mass fluxes, and some of the reactions and equilibria in gasification of coal

Temperature gradients have several effects on mass transfer: • they are the direct cause of molecular motion,

7. Mass and Heat Transfer

75

• they may cause bulk motion of the fluid, and • they influence phase equilibria and reaction rates. At the same time, concentration gradients have effects on heat transfer: • they are the direct cause of temperature gradients, • they may cause bulk motion of the fluid, and • they influence compositions and so phase equilibria and reaction rates. The direct cause or 'thermal diffusion' effects play a role in the separation of isotopes and in thermal deposition of dust from gases. They are also important in chemical vapour deposition. Otherwise the effects of thermal diffusion are small, and we will not take them into account. Temperature changes alter the bulk and interfacial properties of fluids. Such variations can give rise to free convection and interfacial motion (Benard effect). We can - in principle - take both effects into account via the mass transfer coefficients in our models. Finally, temperature levels and compositions influence phase equilibria and reaction rates. In that way they influence heat and mass transfer rates. Mass transfer fluxes carry enthalpy, and in turn they affect temperature gradients. These simultaneous effects are the subject ofthis chapter.

7.2 Enthalpy We begin with a brief repetition of a few aspects of energy transfer. We take one mole of a substance and add heat, keeping the pressure constant (Figure 7.3). The heat taken up by the substance is its change of enthalpy. This enthalpy is a property of the substance. It depends on temperature and pressure. Except in the vicinity of the critical point, the effect of pressure is small however.

CD Cl) Cl) ~

I mol of i at T, p

H

add heat samep enthalpy change H + dH

Fig. 7.3 Definition of a change in enthalpy

For a liquid, enthalpy increases more or less linearly with temperature (Figure 7.4). The proportionality constant is the molar heat capacity. In transport processes the absolute value of the enthalpy is unimportant: we only need enthalpy differences. The enthalpy can then be put equal to zero at any convenient reference temperature.

76

Mass Transfer in Multicomponent Mixtures

H H

=C pL (T - T;.ef)

o+-~"-------

I

T

/

/

molar heat capacity

H = 0 at the reference temperature

H

H=cpG(T -Tref ) + Ml;:r

entharp;;;;~;:~~sation,

(depends on choice ofTrej)

Fig. 7.4 Enthalpy of a single liquid or gas

When the liquid is vaporised, the enthalpy increase is the enthalpy of vaporisation. The gas also has an enthalpy, which is again a function of temperature. As the molar heats of gas and liquid usually differ, the enthalpy of vaporisation does depend on the temperature. However, it does not vary rapidly. In a mixture (Figure 7.5), the enthalpy is a summation of contributions from the different species. These species or partial molar enthalpies depend on composition. In the examples in this chapter, we will neglect such variations. mixture species enthalpies

effect of composition often small:

Hi

"".H*-i-pure ,., 1

Fig. 7.5 Enthalpy of a mixture

If chemical reactions occur between species, we cannot assign independent zero points to the enthalpies of all components. As an example, we look at the nitrogenhydrogen-ammonia system (Figure 7.6). We assign a value of zero to both reactants at the same reference temperature. Then the enthalpy of ammonia is equal to the reference enthalpy of reaction. Note that the enthalpy of reaction is a weak function of temperature because the heat capacities of reactants and products are not equal.

7.3 Mass Transfer Relation Temperature gradients do require extra terms in the MS-equations (Figure 7.7). However, we have already noted that these are usually small, and that we are going

7. Mass and Heat Transfer

77

to neglect them. So our mass transfer relation retains the same form, irrespective of whether there is heat transfer or not. reacting system

1N2

+1 H 2 ~ NH3

(1)

(2)

(3)

c pI = c p2 = 29 J morI K- I T ref

H

=

i

298

c p3 =4lJmorI K-I

reactants ~T

---- ---------

_-----

Mlreaction ref

= -46 • 1kJ (mol NH 3 )-1

------_ Mlreaction (T)

NH3(3) L_-------·----

H

A

(T

ureaction

3 =unref

+Cp3

T -

) ref

Fig. 7.6 Enthalpy in a reacting system

non-isothermal MS-equation small

Ft = Ili,jX/Ui -U j ) + (thermal diffusion terms) Fig. 7.7 No important change in the MS-equation

There are, however, a few details of which you should be aware. The driving force remains the potential gradient, but taken at constant temperature l • This means that you should not differentiate the temperature in the formula of the chemical potential (Figure 7.8). In the difference form of the equation, we need the average temperature in the film. This cancels in the dimensionless form of the equation, so here there is no change.

7.4 Energy Transfer Relation The energy flux through a film is given in Figure 7.9. There are three contributions: • heat conduction, • enthalpy carried by the mass transfer fluxes, and • thermal diffusion terms. Also here, we neglect the thermal diffusion terms.

1

This is derived in the theory ofThennodynamics of Irreversible Processes.

Mass Transfer in Multicomponent Mixtures

78

driving force in non-isothermal mixtures (ideal solutions) F: - _(dJi i ) 1dz T

_

RT dxi X. dz

at constant

~ temperature

difference form:

________ _~ -----average film temperature F =_ RT Llxi Xi Llz I

Fig. 7.8 Taking gradients at constant temperature

~

1

i.!!..-../

differential equation:

i-----:[1-AT

:iT: CV :N : 1

1 1

1 •

small

conduction thennal drift

1

E

=? + L;;;~ +

/dz thermal conductivity

(thermal diffusion terms)

species enthaipies

difference equation:

heat transfer coefficient

at the average T

Fig. 7.9 The energy flux relations

We approximate the transport relation by a difference equation2 involving a heat transfer coefficient. This coefficient usually has a value around 10 W m·2 Kl in gases, and around 1000 W m· 2 Kl in liquids. Its behaviour is similar to that of the mass transfer coefficient; it can often be predicted from the same relations (see Chapter 10). Most heat transfer problems involve at least two phases. To describe these problems, you will need another relation: the continuity of the energy flux across the phase boundary (Figure 7.10). You may regard this equation as a 'thermal bootstrap'. It is similar to the continuity of the molar fluxes at an interface that we will be using further on.

2

For examples of how to use the differential equation, see the references.

79

7. Mass and Heat Transfer

the film other phase

.

,,

calculated with the energy transport relation

means

Fig.7.10 Continuity of the energy flux

7.5 Condensation in Presence of an Inert Gas To illustrate combined heat and mass transfer we consider condensation of methanol (2) in the presence of nitrogen (1). Nitrogen does not condense, and is swept against the cooling tube by methanol vapour (Figure 7.11). So methanol has to diffuse through a nitrogen 'blanket'. This can greatly reduce the flux and the heat transfer rate. The bulk vapour composition and temperature are given, and also the bulk coolant temperature. The questions are: • what is the flux of methanol, and • what is the energy flux? Here we only analyse the gas film. The relation between the energy flux in the liquid and the interfacial temperature is given in the figure. condensate,

~ I

~\

wall coolant 1// /

inert (1) accumulates at interface

methanol Y2 : , ,, ,,

nitrogen Y\

-----~j ,"

I I I

I I .

I I I

~,

vapour

1 = exp[ -Mfva R - ( Tb

1 T/3

Jl

#-_ .

"I" ,\

I

p

Y2/3

,

\

interfacial equilibrium of (2)

boiling -;;~-int

2.___ ~\\_-- low temperature for condensation

L. E' -- N 2 HL(T. 2 J3 )+h'(T.J3 -T.) y

Fig. 7.11 Condensation in the presence of an inert gas

Let us have a look at the interface and assume that we know its temperature. Also, we assume that vapour and liquid are in equilibrium at the interface. The vapour

80

Mass Transfer in Multicomponent Mixtures

pressure relation of methanol then gives the interfacial composition of the vapour. We can now calculate the methanol flux in the same way as in Chapter 5. Calculating the energy fluxes, to and from the interface, is also straightforward. The methanol fluxes from the two equations should of course be equal; this allows us to find the interfacial temperature.

7.6 Heterogeneous Reacting Systems As a second example we have a look at the synthesis of ammonia (Figure 7.12). This is a steady state system, so there is no net energy transport to the particle. The reaction is exothermic, so the particle has a high temperature and heat is transferred by conduction to the gas. This is compensated by a thermal drift towards the interface. In the following figure, it is easy to get the symbols for nitrogen (N2) and the flux of hydrogen (N2 ) mixed up. So look carefully. 1N2 +~H2 ~NH3 (1) (2) (3)

N2 = 3N1 N3 = -2N1

bootstraps:

no net energy flux through film:

-h(Tcp -Ta)+N1H1(T)+N2H2(T)+ N3H3(T) =0 '----v----'

conduction

'

v

thermal drift

'"

T

= O.S(Tcp + T,J

Fig. 7.12 Convective heat transfer in the synthesis of ammonia

7.7 An Ammonia Absorber Finally we consider absorption of ammonia. Ammonia transfers from air into water, in a packed column (Figure 7.13). This problem is not as simple as it looks. We assume counter-current operation, with fresh water entering at the top. The rich ammonia and air mixture enters at the bottom where the ammonia is absorbed. The enthalpy of absorption3 is released; this causes a rise in temperature of the liquid. As a result, water vaporises. The mass transfer process in the vapour therefore involves three species: ammonia, water and (stagnant) air. Ammonia and water vapour diffuse against each other at the bottom of the column. Towards the top of the column the vapour encounters cold entering water. Therefore water vapour condenses near the top of the column, and we now have co-diffusion of ammonia and water through air. We should not ignore water vaporisation at the bottom and condensation at the top in

3

This is the negative of the enthalpy of vaporisation.

7. Mass and Heat Transfer

81

the analysis. The resulting temperature profiles along the column show a pronounced bulge towards the bottom water (1)

condensation of '1 ' co-diffusion absorption of '2 ' evaporation of '1' counter difjitsion 10

air (2) + ammonia (3) Fig. 7.13 Absorption and evaporation in an ammonia absorber

7.8 Summary • Temperature gradients influence mass transfer in several ways. Most important, they influence phase equilibria and chemical reaction rates. In that way, they influence the driving force for mass transfer. The direct influence (via thermal diffusion) is small. • Mass fluxes in turn carry enthalpy fluxes with them. The contribution of this thermal drift is significant in several applications such as condensation and heterogeneous catalytic reaction. • If energy and mass transfer occur together, you must solve the energy and transport relations simultaneously. The energy relation to remember is the difference equation in Figure 7.9. • You also need the continuity of the energy flux through the interface (Figure 7.10).

7.9 Further Reading Bird, RB., Stew art, W.E. and Lightfoot, E.N., 1960, Transport Phenomena. Wiley, New York. A good treatment of simultaneous heat and mass transfer. Taylor, Rand Krishna, R (1993) MUlticomponent Mass Transfer. Wiley, New York.

82

Mass Transfer in Multicomponent Mixtures

A comprehensive treatment of simultaneous heat and mass transfer with emphasis on application to the design of mixed vapour condensers. Many fully worked numerical examples. Raal, J.D. and Khurana, M.K. (1973) Gas absorption with large heat effects in packed columns. Canad. J. Chem. Eng., 51,162-167. This discusses the ammonia absorber example along with experimental details. Katti, S. (1995) Gas-liquid-solid systems: An industrial perspective. Chem. Eng. Res. Design. Trans. l. Chem. E., Part A, 73, 595-607. This emphasises the importance of rigorous modelling of heat and mass transfer in sour gas absorption processes. Kohl, A.L. and Riesenfeld, F.C. (1985) Gas purification. 4th Edition, Gulf Publishing Co., Texas. An industrially oriented book, which emphasises the importance of heat effects in sour gas absorbers. Von Stockar, U. and Wilke, C.R. (1977) Rigorous and short-cut design calculations for gas absorption involving large heat effects. Ind. Eng.Chem. Fundam., 16, 88103. Design procedures for absorbers including heat effects.

7.10 Exercises 7.1 We consider the vaporisation of water from a vertical film. The film (phase') is at the left, the gas into which the water is vaporising is to the right. The thermal bootstrap relation (Figure 7.10) reads:

E'=E

(7.1)

We can replace the energy fluxes to obtain: NHL =-hI1T+NHG

(7.2)

Here HL is the liquid enthalpy at the interface and Hv the enthalpy of the vapour half way through the vapour film (Figure 7.11). For a given temperature difference, heat transfer coefficient and values of the enthalpies, this equation yields a value of the flux. For the system water-steam the reference point for the enthalpy is commonly chosen as the liquid at 0 QC, so the enthalpies (Figure 7.4) are given by: (7.3)

7. Mass and Heat Transfer

83

A different way is to choose as reference point the gas at 0 DC:

HL _ = cpL(T- Trej )-Ml::J HG_ = cpG(T- T,.ej)

(7.4)

Check that we get the same flux from the above equations irrespective of our choice of the reference state. 7.2* A Drop in a Spray Dryer (Mathcad). Here heat is transferred from a hot gas to a drop. This causes vaporisation of water, which keeps the temperature of the drop down. This exercise is governed by three main equations: the equilibrium relation between water and water vapour, the diffusional transport relation and the heat transfer relation. In addition, we need a diffusional bootstrap and a thermal bootstrap. 7.3 * Condensation of Methanol (Mathcad). This is the problem in Figure 7.11. It is essentially the same as the previous example, except that heat is transferred in the same direction as methanol. 7.4* Synthesis of Ammonia (Mathcad). This is the problem shown in Figure 7.12. 7.5 For adiabatic distillation the bootstrap relation that is commonly used is that the sum of the molar fluxes adds to zero; see Figure 6.7. Why is this so? Try to derive the general bootstrap relation for adiabatic distillation using the continuity of energy fluxes across the vapour-liquid interface as a starting point (see Figure 7.10). Hint. In distillation operations the temperatures of the vapour and liquid phases are usually close to each other and the conductive contribution to the energy flux vanishes. See Chapter 11 of Taylor and Krishna (1993) for a rigorous development of heat and mass transfer. 7.6 In the classic, but now outdated book of Treybal (Mass Transfer Operations, McGraw-Hill, New York, p. 317, 1980) there is an analysis of the problem of absorption of ammonia from air into water in a packed column. Study this problem and try to reconcile the mass and heat transfer analysis with that outlined in this book.

84

Non-idealities So far we have only considered ideal solutions (mainly gases). This was to keep our equations simple; in our new mass transfer theory it is not difficult to take the effects of non-ideality into account.

8.1 Chemical Potential and Activity The driving force on each species is its potential gradient; here the chemical potential gradient (Figure 8.1). It is usually handier to replace the chemical potential by activities, as shown in the figure. chemical potential ideal solution:

l1i F I

non-ideal solution:

F I

=constant + RTlnai

=_ dPi =_RTdlnxi ~_ RT axi dz

dz

&; xi

=_ dPi =_RTdlnai ~_ RT l1ai dz

dz

&;lli

+

difference equation Fig. 8.1 Driving force in non-ideal solutions

Activities are mostly calculated via activity coefficients: the activity of a species is the product of its activity coefficient and mole fraction.

8.2 Non-ideal Binary Distillation As an example of a calculation on a non-ideal mixture we look at the liquid phase in the distillation of water and ethanol (Figure 8.2). The ethanol fractions at the two sides of the film are given in the left-hand figure, as is the mass transfer coefficient. The activity coefficient of ethanol and the total concentration are shown as a function of composition in the right-hand figure. They both vary considerably with composition: this is typical for non-ideal systems. Calculations in non-ideal systems tend to be a little messy. We have worked out the driving force including non-idealities for ethanol in Figure 8.3. We advise you to check whether you can follow the details.

85

8. Non-idea/ities

ethanol (1) kl2

water (2)

=4.3 X 10-4 m s-1

:N 6

2

o

c

3 83 . X1Q4

4

~1--'Y_Ii>",,"~:'~~ L-~~-L~____L -__- L__~

o

2

o

1

Fig. 8.2 Liquid phase in the distillation of ethanol - water

bootstrap: 'equimolar exchange' transport:

NI = -N2

_ Llal _ x2 NI - x l N2 - --"'----"---"---"-

- Xl -

~al

kl ,2e

_xI(Ll(rIXI))=_O.2 3.04xO.l-1.72xO.3 = ... (rlxl) O.5x(3.04xO.l+1.72xO.3)

working out yields:

NI =8.8molm- 2 s-1

'exact' value:

NI =9.3molm- 2 s-1

non-ideality neglected:

NI =l7molm- 2 s-1

Fig. 8.3 Solution to the ethanol - water distillation

In this example the driving force is very large. Also the non-ideality is considerable. Even so, the difference equation is not far from the 'exact' result including varying concentrations and activity coefficients in the film. You get a poor result if you neglect non-ideality.

8.3 A Simple Model of Non-idealities Many thermodynamic models are available to calculate activity coefficients. Wellknown ones are the Margules, Van Laar, Wilson, NRTL and UNIQUAC models. These models lead to quite complicated relations for activity coefficients, especially in mUlticomponent mixtures. However, their details are not important for understanding mass transfer in non-ideal systems. So the greater part of this discussion will use the simple model of Figure 8.4.

Mass Transfer in Multicomponent Mixtures

86

4 simple binary model:

= A(1- X\)2 In 'Y2 = A(l- X 2 )2 In 'Y\

~~

-----

non-ideality parameter

ideal: A = 0 Fig. 8.4 Activity coefficients in our simple model

This is a model for a binary system. It contains a parameter A that is a measure of the non-ideality of the system. If A is zero, the system is ideal. Increasingly positive values of A correspond to increasing non-idealities. The corresponding activities are shown in Figure 8.5 as a function of composition. For an ideal mixture, the activity is equal to the mole fraction of the species considered. Large values of A give a more complicated behaviour. The behaviour does remain simple for either small or large values of the mole fraction. For small values, the activity is proportional to the mole fraction; for large values it goes to unity.

t------------- -- non-ideality increasing 4

important efficts of nonideality

o

L..-_ _--'--_ _----'

o

Fig. 8.5 Effect of non-ideality on activity

87

8. Non-idea/ities

8.4 Large Non-Idealities: Demixing In the upper part of Figure 8.5 we observe a number of interesting things. For A = 1 (a moderately non-ideal system) the activity still rises monotonically with composition. For A = 2 the curve has one point with a horizontal tangent (at Xl = 0.5). This is a critical point. We shall see what this means in a moment. For A > 2 there is a domain where the component activity decreases with increasing mole fraction (see also Figure 8.6). This domain is bounded by the two spinodal points. Within the spinodal domain the mixture is completely unstable: a component tends to diffuse to regions where it has a higher concentration. The mixture then splits in two parts: one with a high, and one with a low mole fraction of that component. spinodal points

,I ~ ~ 'i' moves to regions with ..,.."."" .....

/

,

W

higher concentrations ~

demixing

Fig. 8.6 Instability between the spinodal points

A complete analysis of what happens is outside the scope of this book. We have, however, summarised the main results in Figure 8.7. For A < 2 there is no demixing. The critical point (Xl = 0.5, A = 2) is at the boundary of the demixing zone. For A > 2 we see the spinodal zone where demixing occurs spontaneously. It is bounded by two supersaturation zones. Fluids with a composition in such a zone can exist indefinitely, but they are metastable. Any disturbance tends split them into two liquids with compositions on the binodal curve or phase boundary.

i

A, non-ideality

unstable 4

supersaturation spinodal curve binodal curve (phase boundary)

2 critical point

-j one liquid f-0 0

Xl

1

Fig. 8.7 Demixing due to non-ideality

88

Mass Transfer in Multicomponent Mixtures

8.5 Maxwell-Stefan versus Fick This book uses the Maxwell-Stefan relations to describe mass transfer. Most other texts use the classical Fick description, and most diffusivity data you will encounter in literature are Fick coefficients. So it is important to have an idea of the relations between the two kinds of coefficients. In the conventional theory the effects of nonideality are incorporated in the Fick diffusivity. In the generalised Maxwell-Stefan (MS) approach they are part of the driving force (Figure 8.8). Fick: non-ideality in the diffusivity:

Maxwell Stefan: non-ideality in the driving force:

t

d(ln(Ylxl)) _ dz

-----'-----'-'-'----'-'-'---x

Fick MS

2

uI -u 2 D I,2

\

dx

J I =-CDI2 - 1 , dz

D1,2 =f=l+xl d(lnYl) -Dt,2/, dx 1 ",

thermodynamic correction factor

D example: InYI =A(l-XI)2~ ~=1-2Axl(l-Xl) DI,2 Fig. 8.8 Difference between the Maxwell-Stefan and the Fick descriptions of diffusion

For a binary mixture the relation between Fick and MS diffusivities is given in the figure. The relation is not difficult to derive: you need to replace the velocities in the MS equation by fluxes with respect to the mixture, and then to arrange the equations in the Fick form. For our simple model the ratio of the two diffusivities is given at the bottom of the figure. Proponents of the Fick theory often call this ratio 'the thermodynamic correction factor' . Figure 8.9 summarises the behaviour of the Fick diffusivities; it shows the ratio of the Fick to the MS-diffusivities. The MS diffusivity is usually a positive, monotonic and well-behaved function. For an ideal binary solution the Fick and MS diffusivities are equal. In non-ideal solutions the Fick coefficient shows a minimum at intermediate concentrations. This minimum becomes progressively deeper as the non-ideality increases. At the critical point the Fick diffusivity becomes zero. A further increase in the non-ideality gives a spinodal zone where the Fick coefficients are negative and supersaturation zones where they rapidly fall to zero. Clearly the behaviour of Fick diffusivities is considerably more complicated than that of MS diffusivities. This is even more so in multicomponent mixtures.

89

8. Non-idealities

A~ 1 c--...;:O---'r--~~

D's positive, well behaved

'-

in ideal binaries D,,2 =

in non-ideal binaries, D,,2 varies considerably

2 \

o ~

D,,2 = 0 at a critical point

\\............../---- ---- @ forIn y\

£),,2

= A(l- x}

D, 2 falls rapidly in super-saturation zones, is zero at spinodal points and negative in between.

-1 '-----'------'

o

Fig. 8.9 Relation between the MS and the Fick diffusivities

8.6 When can we neglect Non-ideality? Even in this book, you will see that we often neglect the effect of non-ideality - there are often no activity coefficients in our equations. When is this allowable? To put it another way: how large is the error that we make? Figure 8.10 gives the ratio of the real driving force to the force when the solution is considered ideal. It is the same formula that we encountered when comparing the binary Fick and Maxwell-Stefan diffusivities (Figures 8.8 and 8.9). F

_ dln(xi ) dz

ideal -

F'.xact

F;deal

= dln(ri x ) d In( X)

1+ dln(r) d In( X)

Fig. 8.10 The effect of neglecting non-ideality in the driving force

A look at Figure 8.9 tells us that we can neglect non-idealities (for the system considered here): • for any sufficiently dilute solution, and • over the whole concentration range if the parameter Ais small (say IAI < 0.1). It will also be clear that neglecting non-idealities leads to large errors in concentrated non-ideal solutions. It is unacceptable near a critical point. In multicomponent mixtures, we must further specify the derivative in Figure 8.10. For example, if we consider component '1' in a ternary mixture, then we must specify how the other compositions change as we take the derivative. We might

90

Mass Transfer in Multicomponent Mixtures

specify x 2 = constant, X3 = constant or something in between. What we do can make a difference to the effect ofnon-ideality.

8.7 Mass Transfer in Liquid-Liquid Extraction Several separation processes make use of strong non-idealities in liquids. Wellknown examples are liquid-liquid extraction, and extractive and azeotropic distillation. In these processes there are at least three components. It is outside the scope of this book to treat them properly, but we will discuss a few aspects of their mass transfer characteristics. The example used is a liquid-liquid system that extracts water from acetone, using glycerol as a solvent. Figure 8.11 shows the composition diagram of the system. Acetone and glycerol are poorly miscible: they have a demixing zone along the bottom line of the triangle. Water is miscible with both acetone and glycerol. For water concentrations that are not too high there are two phases: one rich in acetone and one rich in glycerol. Points on the phase boundary that are in equilibrium are connected by tie lines. These slope upward to the right, showing that water has a preference for the glycerol phase.

t water (3) 0.4

0.2

0.0 0.0 acetone (2)

0.4

I

0.8 Xl glycerol (l)

Fig. 8.11 Ternary diagram for liquid-liquid extraction, showing the demixing zone, tie lines, and a possible set of compositions in the two phases and at the phase interface

In computer calculations, a thermodynamic model provides infonnation on the activity coefficients of the components. This model can also calculate the phase boundaries and tie lines in the diagram. The extraction is done in a series of countercurrent stages. Each of these consists of a mixer for mass transfer and a settler for separating the phases. We will only consider a single stage and neglect any mass transfer in the settler (Fig 8.12). In the mixer, both phases are considered to be well

8. Non-idea/ities

91

mixed. There is a mass transfer resistance on either side of the interface. In such a strongly non-ideal system, we cannot assume a bootstrap relation beforehand; we have to solve all mass balance and transport relations simultaneously. The variables in this model are shown in the figure. There are thirteen unknowns! This complexity, by the way, has nothing to do with our using the MS equations. The thirteen equations can be chosen as follows: • three overall mass balances for the three components, • three equilibrium relations at the interface, • two transport relations in the glycerol film, • two transport relations in the acetone film and • three partial mass balances relating the fluxes through the interface to the flow changes in one of the phases.

XI

••

, , , ,

Xla XI~ X 2a X2~

X2

L'

X3

,,

,,{

/

/

XI~

XI'Y

X2~

X2'Y N2

L

NI N3

.//jlows

13 unknowns, but the transport relations remain simple: _ Llal

- X 1 llj

X3 N I -~N3 = x2 N I -xI N 2 + -"---'---'---"-

k1,2 C

k1,3 C

et cetera...

Fig. 8.12 Even a single, well mixed stage, has many unknowns

As you can see, the bootstrapping of this problem is not a trivial task. On the other hand, the transport relations of each film separately are nothing special. We have written one of them down for the glycerol phase in Figure 8.12. It is a simple extension of the binary case: hardly anything to be remarked upon. You should not have the illusion that you can solve all problems in liquid-liquid extraction by solving the MS-equations. Extraction calculations are still unreliable. Some reasons are: • poor accuracy of thermodynamic data, • lack of good data on multicomponent diffusivities and mass transfer coefficients, and • the occurrence of interfacial flows and emulsion formation due to interfacial tension gradients.

92

Mass Transfer in Multicomponent Mixtures

8.8 Summary • For non-ideal mixtures you need a thermodynamic model that predicts activity coefficients. With such a model, including the non-ideality effects in our mass transfer theory is trivial. Figure 8.3 shows how that is done. • Dilute systems behave as ideal systems. • Non-idealities become particularly important in the vicinity of a critical point. There is, however, nothing special in the behaviour of the MS diffusivities around such a point. The same is true in the supersaturation zones. This is because the effect of non-idealities is included in the driving forces. • In contrast to the MS diffusivities the Fick diffusivities of classic mass transfer do show a complicated behaviour in non-ideal mixtures. • In liquid-liquid extraction (a fairly typical non-ideal process) bootstrapping requires the simultaneous solution of many equations. The transport relations, however, are nothing special.

8.9 Further Reading Reid, RC. Prausnitz, J.M. and Poling, B., 1987, The properties of gases and liquids. Fourth edition, McGraw-Hill, New York. A good introduction to the estimation of activity coefficients in non-ideal liquid mixtures. Krishna, R, Low, C.Y., Newsham, D.M.T., Olivera-Fuentas, C.G. and Standart, G.L. (1985) Ternary mass transfer in liquid-liquid extraction. Chem. Eng. Sci., 40, 893903. Experimental data on diffusion in the system glycerol-water-acetone to demonstrate the strong influence of thermodynamic non-idealities near the critical point. Pertler, M., Blass, E. and Stevens, G.W. (1996) Fickian diffusion in binary mixtures that form two liquid phases. A.l.Ch.E.1., 42, 910-920. Taylor, Rand Krishna, R. (1993) Multicomponent Mass Transfer. Wiley, New York. A good compilation and analysis of literature data.

8.10 Exercises 8.1 Distillation of Methanol-Water (Mathcad). In this file we have taken the liquid phase to be ideal. You are to include the effect of non-ideality and to note how it affects the fluxes. This problem is similar to that in Figure 8.3. The model for the activity coefficients has already been programmed for you.

93

8. Non-idea/ities

8.2 Non-ideal Binary Distillation (Mathcad). This calculation shows that the difference approximation is quite good, even in strongly non-ideal mixtures. You must, however, include the effect of non-ideality in the driving force. In the calculation we compare the one-step difference approximation with an eight-step calculation (which is supposed to be accurate). The part that you are to program is quite small; the assignment takes some time because you need to read a rather large file. 8.3 Composition Profiles in a Non-ideal Binary Mixture. Below you see a simple diffusion cell, with a porous plate between the two compartments. The left-hand compartment contains only component '1' of a binary mixture: the right-hand compartment only component '2'. The two components have equal molar volumes, so the volume and mole fluxes must add up to zero. Diffusivities in the mixture are given in Figure 8.8. On the right side of the figure, you see the concentration profiles for steady diffusion for four values of the non-ideality parameter A. You are to answer the following questions: a) What happens to the tlux with increasing non-ideality? You can easily see this qualitatively when you realise that all solutions are ideal in the dilute limits, so you can use Fick's law there. b) For A = 2 and Xl = 0.5 the Fick diffusivity becomes zero. Does the flux go to zero? How can this be? c) What happens when A > 2? d) For A > 2, Figure 8.9 shows negative Fick diffusivities in the central part of the composition range. What would these mean? Why do they not show up in the figure above? The calculation of the profiles above is given in a separate Mathcad file. You do not need to read this file to answer the questions. A=O

A=l

A=3

A=2

increasing non-ideality

)

8.4 A· Single Stage Liquid / Liquid Extractor (Mathcad). This file calculates the compositions and fluxes in a single stage such as in Figures 8.11 and 8.12. It uses the

94

Mass Transfer in Multicomponent Mixtures

ternary Margules equation for the solution thermodynamics. The result is a large and complicated file: it took us a week to get it running. You will need time to read it, and get an idea of what it does. You will see that the fluxes in liquid / liquid extraction can behave in ways which are counter-intuitive. 8.5 Myerson and Senol (A. 1. Ch. E. J., 30, 1004-1006 (1984)) have measured Fick diffusion coefficients near the spinodal curve for urea in aqueous solutions. Using their data try to estimate the Maxwell-Stefan diffusivities and check the validity of the trends shown in Figure 8.9.

95

Diffusion Coefficients Our transport relations require Maxwell-Stefan diffusion coefficients or diffusivities. This chapter summarises our knowledge of these for gases and non-electrolyte liquids.

9.1 Diffusivities in Gases Gas diffusivities can be estimated quite well from the kinetic theory of gases. In the simplest form of this theory, molecules are hard spheres. Friction between species is caused by momentum transfer in binary collisions of unlike molecules. The theory is not extremely difficult, but too long to give here. Important results (Figure 9.1) are: • friction is proportional to the velocity difference between the species, • the diffusion coefficients do not depend on the gas composition, • diffusivities increase rapidly with temperature and are inversely proportional to the pressure, • diffusivities are lower for larger and heavier molecules. molecules are little (hard) spheres moving around with their thermal velocity and occasionally bumping into each other d 1,2

= d l +d2 2

Fig. 9.1 Diffusion coefficients in gases

Real molecules are not hard spheres. At higher temperatures they undergo harder collisions and then show a lower effective diameter. The empirical modification of the theory in Figure 9.2 takes this into account. To use the empirical equation you need molar diffusion volumes. These are tabulated in handbooks; their values are roughly two thirds of the molar volume of the liquid at its normal boiling point. The equation is easy to use. Errors are less than 10% up to pressures of 1 MPa. At higher pressures the assumptions of the kinetic theory, • only binary collisions, and • a 'free path' much larger than the molecule diameters, are not correct.

Mass Transfer in Multicomponent Mixtures

96

empirical equation:

..' diffusion volume::; liquid volume, nf mo[-l I H2 I 7.07

I

NH3 I H2O 14.9 I 12.7

CO2

N2 17.9

26.9

i I

Fig. 9.2 Modification of the kinetic gas theory

Figure 9.3 gives an example of the use of the empirical formula. Diffusivities of gases at ambient conditions are around 10-5 m2 s-'. example: N2 (1) CO2 (2) DJ2

= 3.16xlO-8

,

p=10 5 Pa

T=300K

3001.75 105 [(17.9 X 10-6)113 + (26.9 x 10-6)113]

Fig. 9.3 Calculation of the diffusivity in

N2 -

2

CO2 mixtures

Friction is determined by binary collisions. For mixtures of more than two components you can therefore simply use the binary coefficients (Fig. 9.4). This is a distinct advantage of the Maxwell-Stefan approach. However, you can only do this with ideal gases at low to moderate pressures. binary:

muIticomponent:

D . = f)bmary I,'

1,1

also does not depend on composition

o D,,2 does not depend on composition

mole

1

2

Fig. 9.4 Using binary gas diffusivities in multicomponent calculations

9. Diffusion Coefficients

97

9.2 Diffusivities in Liquids For liquids the situation is less satisfactory. There is no good, accurate theory. The 'sphere in liquid' theory that we have seen earlier gives the right orders of magnitude for dilute binary solutions: around 10-9 m2 S-I (Figure 9.5). The 'constant' in the equation depends on the ratio of the diameters of the two species molecules. When the solute is much larger than the solvent, the constant has a value of 31t; for molecules of similar size it is nearer 21t.

f)

_

RT

1')1t)l112dl

/

\.

d l »d2

this 'constant' varies with --~. 31t -----;z:::--,/~-----the size ratio of the species: ' , - - - - - - - - - - - - - - . . 2 1 t ---, --- d l = d2 RT -10-9 2 -1 J D12 -2/ m s I , 21t)llh d1 d2 I1t

:J'

"'--

------o

+--~--~-1_-~

2

4

Fig. 9.5 Diffusivity of a dilute spherical species in a liquid

For the sphere-in-liquid model, we need the diameters of both molecules. We can estimate these from the cubic root of the diffusion volumes. For the latter, we can also take two thirds of the molar volume of the liquid at its boiling point. These rules only apply to simple liquids such as hydrocarbons, where there are no special interactions between the different molecules. In many mixtures, the solute binds to solvent molecules. This is especially so, if the two can form a hydrogen bond. Molecules that can do so, usually contain positively charged hydrogen (such as in hydroxyl or ammonia) and negatively charged oxygen (also in hydroxyl groups). Bonds can also occur between solvent molecules. When such bonds occur, you should calculate diameters from the volume of the bonded molecule. Hydrogen bonding always occurs in mixtures involving water - this is one of the reasons why diffusivities in aqueous solutions are difficult to estimate. If the binary diffusivities at infinite dilution are known, values at intermediate compositions can be estimated by logarithmic interpolation (Figure 9.6). The example shown in the figure concerns two simple hydrocarbons, hexane and hexadecane. Their mixtures are almost ideal, and the Maxwell-Stefan and Fick

Mass Transfer in Multicomponent Mixtures

98

diffusivities almost coincide. The two species differ in size, and the diffusivity varies by a factor of three. It is highest when the more mobile species (hexane) dominates. hexane (1)hexadecane (2)

D I ,2,DI ,2

1O-9 m 2 s-1 2

o

MS and Fick diffusivities almost equal ~ 2 "" DI 2

o

logarithmic interpolation:

o

D larger in the mobile fluid

0.5 '-----_ _ _ _---'-_ _ _ _----l

o Fig. 9.6 Diffusivity in an ideal liquid mixture

Ethanol-water (Figure 9.7) is a non-ideal mixture. Here, the Fick coefficient goes through a minimum, where its value is one third of the Maxwell-Stefan diffusivity. The MS-diffusivity does not vary much with composition; again logarithmic interpolation is not bad. 5

ethanol (1) - water (2)

DI ,2,DI ,2 1O-9 m 2s-1 25°C

o o

2

MS and Fick diffusivities differ -DI 2 ~ Dt 2 only at

XI

ro

0

DD DD

D I ,2

=1 and x2 =1 is

r.

D 1,2 = D1,2

o

JL..il"",

LS

0

0

interpolation roughly logarithmic for D

--v

D12

u~'o

0.5

~

o

Fig. 9.7 Diffusivities in a non-ideal liquid mixture

OCT

0

9. Diffusion Coefficients

99

Many non-ideal mixtures behave like ethanol-water, but there are also systems where the Fick diffusivity is larger than the MS-diffusivity. This can be caused by dimerisation or oligomerisation of one of the species. Our last binary example is a strongly non-ideal mixture: methanol-hexane (Figure 9.8). Here, the two sets of diffusivities differ by as much as a factor of thirty. If the temperature is a few degrees lower, this system demixes and the Fick coefficient goes through zero. The Maxwell-Stefan coefficients do nothing special; you can still estimate them roughly using logarithmic interpolation. 5

methanol (l) hexane (2) 2

I:;--00

0

-~

0

[:g

.of

u

0

D12

0

u

D[,2,D[,2

0

40°C

1O-9 m 2 s-[ 0

~

0,5

0 0

~2 ~

0,2

Pr> v

0

x

0.1

?'---___--'I_l___----'1

Fig. 9.8 Diffusivities in a strongly non-ideal liquid mixture

In multicomponent liquids the situation is worse again. Experimental data exist only for a small number of systems. Also they usually only encompass a few measurements. For lack of something better you might use the pseudo-binary logarithmic interpolation of Figure 9.9. In this ternary interpolation, we need three diffusivities at infinite dilution. We consider the diffusivity for species '1' and '2'. We then need:

100

Mass Transfer in Multicomponent Mixtures

the diffusivity of a trace of '1' in '2', the diffusivity of a trace of '2' in '1' and the diffusivity of traces of '1' and '2' in '3'. The first two are binary diffusivities as we have already discussed. The third parameter is very difficult to measure. The few data that we have suggest a value close to the geometrical average of the '13' and '23' diffusivities in pure '3'. Interpolation in mixtures with three or more components is still guesswork. We can say one thing in its favour. The friction terms counting most heavily are those involving the highest mole fractions. These are the terms in which the compositions are relatively close to the corresponding binary compositions. for lack of better...

logarithmic scale

i

mole fraction composition triangle

3

2

Fig. 9.9 Using logarithmic interpolation in multicomponent liquids

We finish by showing a single example of data on a ternary mixture (Figure 9.10). The system is toluene (l)-chlorobenzene (2)-bromobenzene (3). The species are numbered in order of decreasing mobility and increasing molar mass. We see that the toluene-chlorobenzene '12' diffusivities are highest, and also that they have the highest values in the mobile liquid. Conversely, the lowest diffusivity is that of chlorobenzene-bromobenzene '23' in bromobenzene. For this simple, almost ideal mixture, logarithmic interpolation does a good job; the measured points (not shown, but six in each diagram) differ less than 10% from the values drawn.

9.3 How do you measure diffusivities? We could spend a whole book on that; here you only get a brief description of two common methods. They are the diffusion cell with a porous membrane and the Taylor or dispersion capillary Figure 9.11).

9. Diffusion Coefficients

101

2

[!]3 /

2

mole traction composition triangle Fig. 9.10 Three diffusivities in the mixture toluene (1 )-chlorobenzene (2)-bromobenzene (3) at 30°C. The triangles are the best-fit logarithmic interpolations

CD diffusion cell

follow concentration changes ~ 105 s

(2)

Taylor capillary residence time ~ 1000 s

sharp peak in.

broad peak out

peak smeared out by velocity profile smearing reduced by radial diffusion Fig.9.11 Two techniques for measuring diffusivities

102

Mass Transfer in Multicomponent Mixtures

The diffusion cell is filled with two solutions with different compositions. You then follow the compositions in the two compartments as a function of time. From the rate of equilibration you can deduce the binary diffusivity. Diffusion cell experiments are simple, but they take a long time and have to be done carefully. A modern technique uses a 'Taylor capillary'. Here the bulk fluid is pumped continuously through a narrow capillary. At a certain moment a pulse is injected in the inlet of fluid with a different composition. The pulse is dispersed by the combined action of the flow profile (which pulls the pulse apart) and radial diffusion, which narrows the pulse. From the pulse broadening you obtain the binary diffusivity. Measurements with a capillary take much less time than with a diffusion cell. However, they do require a very stable and sensitive detector. Both techniques yield the Fick diffusivity 'with respect to the volume averaged velocity of the mixture'. You need species volumes and an accurate thermodynamic model to transform these into Maxwell-Stefan diffusivities, using formula from Chapter 8 and the Appendix 2 on Units.

9.4 Summary • For ideal gases, the kinetic theory of gases predicts binary diffusivities. Multicomponent gas diffusivities in the Maxwell-Stefan theory are identical to these binary diffusivities. • For liquids, you should try at least to get experimental data for dilute solutions. If you cannot get anything, use the modified sphere-in-liquid model or one of the empirical relations in the literature. • For non-dilute binary solutions, use dilute diffusivities and logarithmic interpolation. • For non-dilute multicomponent mixtures you have little choice but to use pseudobinary logarithmic interpolation. The method is not reliable, but there is nothing better. • Except for gases, our knowledge of multicomponent diffusivities is still very incomplete. • Measuring multicomponent diffusivities is not easy.

9.5 Further Reading Clark, W.M. and Rowley, RL. (1986) The mutual diffusion coefficient of methanol-n-hexane near the consolute point. A.!. Ch.E.]., 32, 1125-1131. Source of data in Fig. 9.7. Fuller, E.N., Schettler, P.D., and Giddings, J.C. (1966). A new method for prediction

9. Diffusion Coefficients

103

of binary gas-phase diffusion coefficients, Ind. Eng. Chem., 58, 19-27. The empirical modification of the kinetic gas theory for estimation of binary dif.fusivities is from these authors. Reid, RC. Prausnitz, J.M. and Poling, B. (1987) The properties of gases and liquids. Fourth edition, McGraw-Bill, New York. This text gives all standard methods to estimate dif.fusivities in binary mixtures of gases and liquids. Taylor, R. and Krishna, R (1993) Multicomponent Mass Transfer. Wiley, New York. Discusses estimation of dif.fusivities in binary and ternary mixtures along with worked numerical examples. Extensive literature references. Tyn, M.T. and Calus, W.F. (1975). Temperature and concentration dependence of mutual diffusion coefficients of some binary liquid systems, J. Chem. Eng. Data, 20, 310-316 Source of data in Fig. 9.6. Vignes, A. (1966) Diffusion in binary solutions. Ind. Eng. Chem. Fundam.,5, 189199. The logarithmic interpolation formula for binary liquid dif.fusivities used in Fig. 9.5 was first suggested by Vignes. Kooijman, B.A. and Taylor, R (1991) Estimation of Diffusion Coefficients in Multicomponent Liquid Systems, Ind. Eng. Chem. Res., 30,1217-1222. This develops the logarithmic interpolation formula for multicomponent liquids in Fig.9.B. Rutten, Ph.W.M, Diffusion in Liquids (1992) PhD thesis, University of Delft. Delft University Press, Delft, ISBN 90-6275-838-X / CIP

9.6 Exercises 9.1 Spheres in a Gas. We consider small spheres (2) moving in a gas (1). When the size of the spheres reduces to molecular dimensions, the friction coefficient will be given by the kinetic gas theory:

104

Mass Transfer in Multicomponent Mixtures

(This is the fonnula in Figure 9.1, rearranged.) Larger spheres will move according to Stokes' equation. For such spheres, we expect the friction coefficient to follow:

Sl,2 = 3njl1hd2 Sketch the behaviour of the friction coefficient as a function of the diameter of the sphere. Use a figure with logarithmic scales. At which diameter is the crossover between the two fonnulae? Is the friction coefficient of the smallest spheres smaller or larger than predicted by Stokes' law? Use the following parameters:

jl = 6.02 X 1023 mol- 1 R = 8.31 J mol- 1 K- 1 T = 300 K p = 105 N m-2

Ml = 0.03 kg mol- 1

9.2 Mutual and Self-Diffusivities (Mathcad). The diffusivities and friction coefficients that we use in this book are mostly mutual coefficients. They are characterised by two different subscripts such as in DI2 or 1;12' In literature you will often find self-diffusivities. These are diffusivities of a species with the same physical properties as the solvent. They can be measured fairly easily by using radioactive isotopes, or by tagging molecules with NMR-techniques. The selfcoefficients have two identical subscripts. It is often assumed that the mutual friction coefficients can be calculated from the self-friction coefficients as the geometric average:

Sl,2 = ~SI,1 .S2,2 The mutual coefficients in a binary gas are found by rearranging the fonnula in Figure 9.1:

The self-coefficients are given by the same fonnulae:

_ (;3 jlp( 2dd2 ( SI,1 -

fn

(RT)o.5

2 Ml

)-0.5

You are to investigate how well the geometric average fonnula agrees for ideal gases with different species diameters. You may assume that the molar mass is proportional to the cube of the diameter. 9.3 Diffusivity of Gaseous NH 3 -H 20 (Mathcad). This file uses the empirical fonnula in Figure 9.2 to calculate the diffusivity in a mixture of two gases. You can use it for other gases than the example by modifying the constants.

9. Diffusion Coefficients

105

9.4* Diffusivity of Dilute Spheres in a Liquid (Mathcad). Calculate the diffusivities of hexane dilute in hexadecane, and of hexadecane dilute in hexane. You are given viscosity and volume data; you are to use the method in Figure 9.5. 9.5* Fick and MS Diffusivities of Acetone-Benzene (Mathcad). This file uses the NRTL model for solution thermodynamics and logarithmic interpolation (the Vignes equation) to calculate the two diffusivities in a non-ideal binary. If you have NRTL parameters and diffusivities in the dilute solutions, you can also use it for other mixtures. 9.6* Plotting the Diffusivities of Acetone-Water (Mathcad). From a list of experimental Fick diffusivities and thermodynamic correction factors from the UNIQUAC equation, we obtain MS diffusivities. We then compare the two in a plot of diffusivity versus composition. 9.7* Diffusion near the Critical Point (Mathcad). In this file you use experimental Fick diffusivities on the system methanol-n-hexane. The system is just above the temperature below which demixing occurs, and the Fick diffusivities show a deep minimum. (The result is shown in Figure 9.8.) 9.8 Three-dimensional Plotting of Ternary Diffusivities (Mathcad). This file allows you to explore in three dimensions the behaviour of the MS-diffusivities in a ternary liquid mixture. 9.9 The six infinite dilution MS diffusivities in the system acetone (1) - benzene (2) - carbon tetrachloride (3) at 298 K are: f){r~l = 2.75 x 10-9 m 2 Is; D;r~l =4.15x1O-9 m 2 /s f){r~l = 1.70 x 10-9 m 2 / s; D;r-71 = 3.57 x 10-9 m 2 / s

D;~--71 = 1.42 X 10-9 m 2 / s; D;,~--71 = 1.91 X 10-9 m 2 / s

Estimate the values of Dij at a composition of Xl = 0.7, x 2 = 0.15, X3= 0.15. Use the formula shown in Figure 9.9. Also use the data given here as inputs to Example 9.8.

106

Transfer Coefficients This chapter contains relations to estimate mass transfer coefficients, and shows how they are used in multicomponent calculations.

10.1 Introduction The engineering form of the Maxwell-Stefan equation contains mass transfer coefficients: one coefficient for each pair of species. At this moment, we can usually only estimate transfer coefficients for solutes in binary mixtures. The MS-equations allow us to interpolate between these binary extremes. So far, we have used the film theory, where the transfer resistance is due to a stagnant layer next to the interface. This theory is only strictly applicable to thin homogeneous membranes. For other geometries, and for flowing mixtures, we do have better models. Here, we will also see how these can be used in the MaxwellStefan equations. Even the dilute binary transfer coefficients are often not known accurately. They depend on the pattern of flow past the interface, and the flow often shows transitions that are not completely predictable. Also, with deforming bubbles and drops, the geometry of the interface may not be known. So predictability varies. It is • fairly good for tubes and packed beds, • less for single drops, bubbles and particles, and • poor for swarms of bubbles, drops and fluidised particles. 'Fairly good' means within thirty percent, 'less' is usually within a factor of two, and 'poor' within a factor of ten. For the 'poor' cases, you can sometimes find better data for specific situations. Below, we give a collection of our favourite relations for dilute binary mixtures. In these systems, the Fick and the Maxwell-Stefan approaches are identical, so we can replace the Fick diffusivities by MS diffusivities. Even though we have spent much time on this collection, we must warn that you use it at your own risk. This part of the subject of mass transfer is still far from reliable.

10.2 Dimensionless Groups Relations for mass transfer coefficients are traditionally given in terms of dimensionless groups. Figure 10.1 shows the important groups: the Sherwood, Reynolds and Schrnidt numbers.

10. Transfer Coefficients

107

You see one difference with other texts: in our problems there are often many Sherwood numbers and many Schmidt numbers. There is one set of such numbers for each pair of components. There is, however, only one Reynolds number for a given problem. Both the Sherwood number and the Reynolds number contain a diameter. Which diameter depends on the problem considered. For transfer to a tube wall, it will be the tube diameter; for transfer to a spherical particle, the particle diameter. For relations describing more complicated geometries, you may have to check how the diameter has been defined. Sherwood number mass transfor

kd Sh . =-'-"-

Reynolds number fluid flow

Re. =pvd I,' 11

Schmidt number mixture property

Sci j ,

'"

fJ.

'"

_

11 pDi,j

---

diameter film thickness

Re < 1 laminar Re» 1 turbulent Sc "'" 1 in gases Sc "'" 10 3 in liquids

Fig. 10.1 Dimensionless groups used in mass transfer

The Sherwood numbers contain the mass transfer coefficients. You may regard the Sherwood number as the ratio of system diameter to film thickness. However, do not take this too seriously. The 'film' concept is not a very good one and we will often find that the 'film thickness' is different for every pair of components. The Reynolds number describes the character of the flow. For small values, the flow is laminar; for large values it is turbulent. In flow through tubes there is a sharp transition at Re "'" 2000. This, however, is an exception, not the rule. In other flows the transition is gradual and more complicated. Many flows show whirls and instabilities when Re;::: 100. The Reynolds number contains a velocity. This is usually the 'superficial velocity': the volume flow divided by the whole cross section of the equipment. The Schmidt number only contains fluid properties. It tells how quickly velocity fluctuations are evened out by viscosity, compared to the smoothing of concentration differences by diffusion. For gases the Schmidt number has a value close to one. In ordinary liquids this goes up to around one thousand. For diffusion of small molecules in viscous liquids, the Schmidt number might have a value of one million. Heat Transfer Coefficients

We end this paragraph with a sideline. As we have seen in Chapter 7, mass transfer and heat transfer often occur simultaneously. To describe this, we also need a heat transfer coefficient. Diffusion in a dilute stagnant mixture follows Fick's law; heat conduction in a stagnant mixture follows Fourier's law. These have the same form.

108

Mass Transfer in Multicomponent Mixtures

Also the mechanisms of convective transport of matter and heat are similar. The result is that you can use the same relations for obtaining heat transfer coefficients as for mass transfer coefficients. You only need to change a few variables. In all relations, the Sherwood number is replaced by a Nusselt number, and the Schmidt number by a Prandtl number (Figure 10.2). This is the same as replacing the ratio of the mass transfer coefficient to the diffusivity by the group shown in the figure, and the diffusivity by the thermal diffusivity.

» heat transfer

mass transfer Sh= f(Re,Sc)

same as

Nu= f(Re,Pr)

replace kif) by hit.. 1/ thermal diffusivity f) by a / a = t../(pc p ) Fig. 10.2 Heat transfer coefficients from mass transfer relations

This analogy between mass and heat transfer is only valid for a restricted set of conditions. However, these are just the conditions where mass transfer coefficients are usually determined: in dilute solutions with only concentration gradients as a driving force. So the analogy is a truly useful tool.

10.3 Tubes and Packed Beds Tubes, capillaries and hollow fibres, are used for mass transfer in membrane modules. Figure 10.3 shows such a tube, with fluid flowing from left to right. The flow is laminar. The fluid contains a species that is being removed through the wall. At first, the concentration is high across the whole diameter. However, material near the wall is rapidly removed, and a concentration profile forms. After a certain distance, the concentration in the middle of the tube begins to decline as well. From that point onward, the profile retains its shape, but becomes lower and lower. So we have two zones: the inlet zone and a zone with a fully developed declining profile. The behaviour of the mass transfer coefficient between fluid and wall is different in the two zones. In the inlet region, it decreases with increasing distance; in the fully developed zone it has a constant value. The relation for the inlet zone gives the average values up to a distance z. To obtain the local transfer coefficients you should replace the constant 1.62 by 1.08. In liquids, where diffusivities are low, the inlet zone is often quite long; in gases it is usually negligible. In a large tube (Figure lOA) flow will be turbulent. The mass transfer coefficient to the wall can be much higher in this case. Also here, transfer is faster in the inlet.

109

10. Transfer Coefficients

However, the inlet zone is short (of the order of ten tube diameters) and the effect of the inlet correction is usually small. laminar

Re < 2000

-v

I.~

+-----41 fully developed 11----Sh .. = 1.62 Re 1/3 ScY3 ( ~ ) I" I" d

-1/3 .

~~~vd d

20D

J:3.66 1,/

Fig. 10.3 Transfer coefficients in a tube from bulk to wall-Iaminar flow

turbulent d

Re> 2000 (see text)

Sh i ,; = 0.027(1+ 0.7~)Re4/5 SC~:7 '--v----'

inlet correction

1--J··
If-------

Fig. 10.4 Transfer coefficients in a tube from bulk to wall - turbulent flow

The transition between the laminar and turbulent flow patterns occurs at a value of the Reynolds number of a few thousand. The change in the Sherwood number is sharp, but not very predictable; turbulence is only fully developed and predictable at Reynolds numbers above 10 000. Unfortunately, the transition region is often important. In several membrane processes, we use flat membranes, with slit shaped channels in between. In the channels we need a spacer, to keep the membranes apart. The spacer is also a turbulence generator, and it markedly increases the mass transfer from bulk to membrane. The mass transfer coefficients for such a system are given in Figure 10.5. There is no sharp transition between laminar and turbulent conditions in this system. Another piece of equipment with a (fairly) well-defined interface is a packed bed (Figure 10.6). This is used for adsorption and chromatography and as a chemical reactor. The mass transfer coefficient for the 'film' outside the particles is given by the formula in the figure. Again there is no sharp transition from laminar to turbulent conditions.

110

Mass Transfer in Multicomponent Mixtures

Sh 1,1 .. = 1.9 Re

) )

1/2 sc1/3(~) 1,1

Z

300 < Re < 2000

) )

spacer wires

)

,, v

)

~

00

8

;rz ~

/1 membranes 0

~

L

8

~

8 It d

Fig. 10.5 Sherwood numbers for the space between membranes

\ \ \

superficial velocity v

void fraction e particle diameter d

external (film) coefficient

Sh j

,.

,

23 = 0.34 e Re / SC .,I/3

minimum value (chromatography)

j

Sh i ,;

'"

4

Fig. 10.6 Transfer coefficients outside particles in a packed bed

At low Reynolds numbers, the first formula from Figure 10.6 predicts that the Sherwood number can drop below unity, so that the diffusional distance is larger than the particle diameter. Although such results have often been reported they are thought to be due to poor packing and channelling. In chromatography, where much effort is spent on good packing, the lower limit of the Sherwood number is often taken to have a value of four. The mass transfer coefficient is then independent of the flow velocity, proportional to the diffusivity, and inversely proportional to the particle diameter.

10.4 Packed Gas-Liquid Columns For absorption and distillation we can use a packing. There are now two phases. The liquid flows downwards over the packing, and the gas upwards through the channels between the packing. The flow in both phases is usually rather turbulent. Figure 10.7 shows formulae for estimating the transfer coefficients in the liquid and in the gas. In such two-phase systems, our equations are not accurate: they might be in error by as much as a factor of two.

10. Transfer Coefficients

liquid, superficial velocity vL

I

€

111

gas through and around the packing

void fraction

liquid as a film over the packing

I /

/

/

/

liquid: gas:

gas, superficial velocity vG Fig. 10.7 Coefficients in packed gas/liquid columns

10.5 Single Particles, Bubbles and Drops The greater part of all mass transfer in engineering occurs in two-phase equipment between swarms of particles, bubbles, drops, and their surroundings. Unfortunately, mass transfer coefficients in these two-phase swarms cannot (yet?) be predicted well. The situation is a little better for single bubbles, drops and particles. The Problem

Even for these single particles, bubbles or drops, the subject is a little frustrating. There are several reasons for this: • The subject has challenged several generations of engineers, and this has led to many models and empirical correlations. For the beginner it is difficult to distinguish between the trees and the forest. • The velocity difference between the particle, bubble or drop, and the surrounding fluid, is not an independent variable. It is determined by the diameter, gravitational acceleration and the physical properties of the system. So the Reynolds number is an awkward parameter, and the traditional (Sh, Re, Sc) description of mass transfer is not handy. • Bubbles and drops can have an interface which is rigid, mobile, or in between. The transition is only partly understood. To avoid the first two problems, we have chosen to show only a few important models, and to give equations from which the velocities have been eliminated. These equations look different from those that you will find in literature, but they give similar predictions.

112

Mass Transfer in Multicomponent Mixtures

In our problems, we usually require mass transfer coefficients in two phases: • the phase inside the particle, drop or bubble (also known as the dispersed phase) with its particle coefficient, and • the phase outside the particle, drop or bubble (also known as the continuous phase) with its film coefficient. We will consider three regimes: • the transient regime, which is important during and just after formation of the interface between the two phases, • the rigid interface regime, which is always encountered with rigid particles, but also with small drops and bubbles, and • the mobile interface regime, which can be important for large bubbles and drops. Inside Rigid Particles

We begin with a spherical solid particle that is immersed in a fluid (Figure 10.8). A trace of a certain component i is leached out of the particle into the fluid. We consider the case where the diffusion resistance outside the particle is negligible, so the composition of the solid at its interface immediately becomes equal to that of the bulk fluid. In the very first moment, potential gradients at the interface are infinite, and so is the transfer rate. However, the rate in this transient regime goes down

interface formation aft =

0

transient regime (penetration theory)

time

1

1

k .= 1,/

1.13~Di'it

l

semi-stationary regime Dd k. =10_'_'/ 1,/ d

1

Fig. 10.8 Leaching of a spherical particle

10. Transfer Coefficients

113

rapidly as concentration changes penetrate into the particle, and diffusion occurs over longer and longer distances. After a short period (which might be as small as one millisecond, but could be as large as a thousand seconds), the concentration profiles from the two sides of the particle meet in the centre. From then on transfer is described by a different semi-stationary mechanism, to which we come back in a moment. In the initial period, transfer is described by the penetration theory. The transfer coefficient is proportional to the square root of the diffusivity and to the inverse of the square root of time. This only applies strictly to a surface that is flat and rigid, and infinitely thick, but the formula is good enough for our purposes. The transfer coefficient given in the formula is the value averaged over time. For drops and bubbles forming at a nozzle, you may assume that the initial transfer begins at the moment that the drop or bubble detaches. This gives a conservative estimate of the transient mass transfer. When the concentration changes have penetrated to the middle of the particle, a new transfer regime begins. Now the shape of the concentration profile no longer changes, but its amplitude goes down exponentially in time until the particle and its surroundings are in equilibrium. In this regime the mass transfer coefficient inside the particle is proportional to the diffusivity and inversely proportional to the particle diameter. The 'constant' 10 in the lower equation in Figure 10.8 is not really a constant. It depends on the concentration history of the particle; it usually has a value between 7 and 15. For large particles this semi-stationary transfer coefficient can be low. To find whether you need the transient or the semi-stationary formula: calculate both coefficients and take the highest. . Outside Rigid Particles

Figure 10.9 summarises our knowledge on mass transfer coefficients outside single, freely falling solid particles (or small bubbles and drops). for short times for small particles

k. =1.13 I,'

t

D. d

k=2~ I"

for large particles boundary layer theory

·IF

~

Ll . k . = 0.3 g P D~· 1,1 1,/ [ 1]

P,11 and Di,i are values in the fluid outside the particle Fig. 10.8 Coefficients outside rigid interfaces

]1/3

114

Mass Transfer in Multicomponent Mixtures

When a particle is brought into contact with a fluid, there is again a transient regime. The transient period tends to be shorter than that inside the particle. For small particles (usually below one tenth of a millimetre) we then get a regime similar to the semi-stationary regime inside the particle. For larger particles, transfer is governed by the fluid sweeping and swirling past the interface. It roughly follows the boundary layer theory where the transfer coefficient is proportional to the diffusivity to a power of two thirds. In this regime the coefficient is a weak function of particle diameter; we have neglected this effect to keep our formula simple. Also here, you find which formula to use by calculating all three transfer coefficients and taking the highest. Rigid and Mobile Interfaces

Of course a rigid particle has a rigid interface. However, the interface of bubbles and drops can deform. Also, fluid can move in the interface; the interface can be mobile. Deformation and mobility only occur when the bubbles or drops are large enough: roughly above one millimetre. The reason is that impurities accumulate at interfaces, and lower the interfacial energy and interfacial tension. In process liquids there may be many of such suiface-active agents. If a bubble rises in a liquid, shear forces will move these towards the back of the bubble. This causes a gradient of the interfacial tension, which opposes motion of the interface. Figure 10.10 shows the idea and gives a simple formula to estimate whether a bubble (or drop) will have a mobile interface. This depends on how clean the liquid is, and that is rarely known beforehand. The formula shows that the transition diameter increases • in dirty liquids, with a high value of the 'fouling coefficient', • when the interfacial tension of the clean interface is high (for aqueous solutions), and • when the density difference between the two phases is small (in liquidlliquid extraction) .

D surface active agents

!

swept to the b a c k - - - - . .

D cause gradients of the •. (tensIOn ; . cr surJace

E ilcr

the transition is at

- gilp d "" 1 mm

fis the fouling (fudge) factor

f"" 0.2

f "" 1

d

t

D these immobilise the d < [ fa ]112 drop or bubble when

t

in clean liquids in dirty liquids

Fig. 10.9 How surface active agents immobilise an interface

10. Transfer Coefficients

115

Bubbles and drops, with a diameter less than one half of the transition value, behave as solid particles. Those above twice the transition value are largely mobile. The transition is not very predictable; this is one of the worst uncertainties in mass transfer. Free Rising Bubbles - Mobile Surface Regime

In bubbles with a diameter of more than two millimetres, the interface is usually mobile. Here we can roughly predict the behaviour with one of the suiface renewal theories. The idea is that we continually form fresh interface, for example at the top of a rising bubble. This disappears after a certain lifetime at the bottom of the bubble. The theories lead to a mass transfer coefficient that depends on the square root of the diffusivity (just as in the penetration theory). The empirical formula for the mass transfer coefficient is given in Figure 10.11. Also here, we have neglected a weak influence of the size of the bubble. The mobile values are typically three to five times the rigid values. There is little data on transfer inside bubbles, but you may assume that the same formula applies. bubbles: outside

2 2n3 .]116

k . . =0.4 g !1p l.l

[

11P

I"

p and 11 are for the liquid (outside) Fig. 10.10 Coefficients for bubbles

Note that the density and viscosity in the formula are those of the outer phase; the diffusivity is that of the phase considered. Drops in Liquids -Intermediate

Large drops (> 2 mm) are often applied in liquid/liquid extraction. They also can have mobile interfaces. However, the interface is slowed down by the effect of the inner phase (which has an appreciable density and viscosity). You might expect a drop with a viscous inner fluid to behave like a rigid sphere. Most of existing data is indeed described by the formula in Figure 10.12, which interpolates between the mobile bubble and the rigid particle. The transfer coefficients inside mobile drops are not well known. We suggest that you use the same interpolation as for the outside, but using the inner coefficients. Note that the interpolation formula also works for bubbles. There, the viscosity ratio is so small that only the mobile term is important.

Mass Transfer in Multicomponent Mixtures

116

mobile drops in a liquid

o

o

calculate the rigid and mobile (bubble) coefficients separately (these differ for the outer and inner coefficients) interpolate using k .= I"

K

f).d

1,1

lld

=lld /11

1 kmobile + O.3K krigid 1+0.3K I,' 1+0.3K I"

Fig. 10.11 Coefficients for mobile drops in a liquid

Drops Falling through a Gas - Rigid or Intermediate

Also small drops falling through a gas behave as rigid spheres. There is only a small range of drop sizes where the interface is mobile; large drops in a gas are not stable and they shatter (Figure 10.13).

rigid sphere

mobile oscillating drop

drop shatters

Fig. 10.12 Drops falling in a gas

The transfer coefficient outside the drop is equal to that of a rigid particle; this is because the inner density and viscosity are much higher than the outer values. This effect is correctly predicted by the interpolation formula in Figure 10.12. Mobility of the interface does increase the transfer coefficient inside a drop. The semi-stationary coefficient in a large drop is so small that even a little mobility has a large effect. The transfer coefficients for the few 'inside' data points that we have, are about one and a half times larger than the predictions from the interpolation formula. Choosing the Mass Transfer Coefficient

How do you choose the mass transfer coefficient from the many possibilities? Follow this recipe: 1. First decide whether you need the inner or the outer coefficient. 2. Then you must know how long the particle, bubble or drop will exist. Calculate the transient coefficient (Figures 10.8 and 10.9). Use the diffusivity in the phase

10. Transfer Coefficients

117

considered. 3. If you are dealing with a solid particle, the interface will be rigid. Calculate the rigid coefficient. For the inner coefficient this is the semi-stationary value from Figure 10.8. For the outer coefficient it is largest of the semi-stationary and boundary layer values from Figure 10.9. 4. Take the largest of the transient and the rigid interface coefficients. For a rigid particle this ends the calculation. 5. If you have a bubble or drop, the interface may be mobile. Calculate the transition diameter (Figure 10.10). If the diameter is less than about one half of this value, you may assume that the bubble or drop is rigid and nothing changes. 6. If the diameter is larger than twice the transition value, you may assume that a bubble or drop is mobile. Calculate the mobile coefficient (Figure 10.11) using the diffusivity of the phase considered, but always the viscosity and density of the outer phase. 7. Interpolate the coefficients with the formula of Figure 10.12. This also works for bubbles and for drops in a gas, where usually one of the two terms dominates. 8. For the final coefficient choose the largest of the transient and interpolated values. If the diameter is near the value for the transition between rigid and mobile, then your guess is as good as ours ...

10.6 Using 'Single' Coefficients for Swarms Transfer coefficients for single particles, bubbles and drops are often used for calculations involving swarms. It seems probable that the coefficients for the individual entities in swarms will not differ greatly from those of the single ones. However, there are serious complications, of which we note a few. An important problem is that different parts of a fluid may flow through a two-phase mixture in different ways. A few examples: 1. In a bubble column, both small and large bubbles are formed. The large ones have a short residence time and a low interfacial area. So they have a poor contact with the liquid. If we use average values, we get a too optimistic idea of the mass transfer. This effect may be compensated by an exchange between large and small bubbles due to break up and coalescence. 2. In a spray on a distillation tray, parts of the gas pass through the spray bed quickly as jets. The contacting of the jet will be less than that of the gas passing through the spray bed. Also here, there is compensation by exchange between the spray bed and the jets. 3. Similar effects occur in beds of fine particles fluidised by a gas. Here, most of the gas can pass through the bed as large bubbles. Mass transfer between the bubbles and the bed is mainly due to convective exchange of gas between the particles and

118

Mass Transfer in Multicomponent Mixtures

that in the bubbles. Break up and coalescence of bubbles and drops may renew the interfaces in a swarm of bubbles or drops quite often. This enhances mass transfer, but is not (yet) very predictable. Sometimes you can make a reasonable guess. For example in sieve tray columns for liquid/liquid extraction, drops are formed anew at every tray passage. Their lifetime is then the rising time between the trays. These phenomena are the subject of active research. You may expect the next edition of 'Mass Transfer in Multicomponent Mixtures' to say more about them.

10.7 Using Binary Coefficients in Multicomponent Calculations All current methods for calculating mass transfer coefficients are for dilute binaries only. How do we obtain estimates for concentrated multicomponent mixtures? The procedure is simple: 1. We estimate the properties (density, viscosity and Maxwell-Stefan diffusivities) of the mixture. 2. We then use these properties in the binary relations to obtain ij mass transfer coefficients. 3. We use these in the difference form of the Maxwell-Stefan equations. There is no general theory behind this procedure. However: • the procedure gives the correct result (including drift corrections) for concentrated binaries, • it gives a plausible interpolation for more components, and • we have been able to check that it is seldom far off in the multicomponent film and penetration theories. So, we believe that it is good enough for engineering calculations. Besides, we have nothing better!

10.8 Summary In this chapter, we have seen how to estimate mass transfer coefficients for dilute binary mixtures: • For mass transfer in tubes, we must distinguish between laminar and turbulent flow. Transfer coefficients in the transition between these two are not very predictable. For both cases we need separate relations for the inlet region and for the fully developed region downstream. Keep in mind that, for liquids in laminar flow, the inlet region can be long. • In slits between flat membranes, we use wire spacers, which promote turbulence. Here, there is no sharp transition between laminar and turbulent flow. These

10. Transfer Coefficients

119

spacers work well in the transition region between laminar and turbulent. • Mass transfer outside particles in a packed bed, is described by relations similar to those in tubes. Again, there is no sharp transition between laminar and turbulent flow. • In a two phase packed bed, the behaviour of the liquid (which flows over the packing) differs from that of the gas (which flows through and around the packing). • For particles, bubbles and drops, transient phenomena can be important. We describe these with the penetration theory. • Bubbles and drops can have mobile interfaces. The interface will only be mobile for sufficiently large diameters; the transition diameter also depends on how clean the interface is, and on the surface tension of the clean interface. The transition is not very predictable. • For bubbles, drops and particles with rigid interfaces, we determine mass transfer coefficients from an empirical modification of the boundary layer theory. • For bubbles and drops with mobile interfaces, we use empirical modifications of the surface renewal theory. We calculate multicomponent mass transfer coefficients, using the binary relations, but with the appropriate properties of the multicomponent mixture.

10.9 Further Reading Scriven, L.E. (1968-1969). Flow and transfer at fluid interfaces, Chemical Engineering. Education, pp. 150-155, (Fall 1968); pp. 26-29, (Winter 1969); pp. 9498, (Spring 1969). A masterly discussion on theories of mass transfer. Clift, R., Grace, J.R. and Weber, M.E. (1978). Bubbles, drops and particles. Academic Press Inc., San Diego. A comprehensive analysis of fluid dynamics, heat and mass transfer within and outside bubbles, drops and particles. Extensive literature references. Perry, R.H. and Green, D.W. Perry's Chemical Engineers' Handbook, 7th edition, McGraw-Hill, New York 1997 Contains an extensive list of mass transfer correlations in Chapter 5. See also under the different types of equipment. J.A. Wesselingh and A.M. Bollen, (1999), Single particles, bubbles and drops: their velocities and mass transfer coefficients, Chem Eng Res Des 77 (1999) A2 89-96.

120

Mass Transfer in Multicomponent Mixtures

This reference contains the background to the recommended correlations for mass transfer inside and to single particles, bubbles and drops. Kister, H.Z. (1992). Distillation design. McGraw-Hill, New York. Encyclopedic compilation of distillation hydrodynamics and mass transfer data in tray and packed columns. Lockett, M.J., (1986). Distillation tray fundamentals, Cambridge University Press, Cambridge. Estimation procedures for mass transfer coefficients and efficiencies in tray columns. Godfrey, J.C. and Slater, M.J. (1994). Liquid-liquid extraction equipment. Estimation procedures for hydrodynamics and mass transfer in liquid-liquid dispersions in the context of equipment design. J.D. Thomton (1992) Science and practice of liquid-liquid extraction, Volume 1., Clarendon Press, Oxford. Estimation procedures for hydrodynamics and mass transfer in liquid-liquid dispersions in the context of equipment design.

10.10 Exercises 10.1 (Mathcad). The following are examples of the use of the relations for mass transfer coefficients to solid surfaces given in Figures 10.3 .. 10.7. You can modify them for your own use; do not spend time on them unless you need them. 1O.1a Capillary (Dialysis Module) 1O.1b Capillary (Gas Separation) 1O.1c Tube (Ultrafiltration Module) 1O.1d Slit (Electrodialysis Module) 1O.1e Packing (Gas Adsorption) 1O.lf Packing (Ion Exchange) 1O.1g Packing (Chromatography) 1O.1h Gas-Liquid Packing 10.2 It is important that you start obtaining an idea of the orders of magnitude of mass transfer coefficients. Make an enlarged copy of Figure 4.10 and plot and name all values calculated in exercises 1O.1x and 1O.3x. 10.3 (Mathcad). The next are examples of the use of relations for mass transfer coefficients inside and outside particles, bubbles and drops. We have set each of them up using the full recipe given in this chapter. (With a little experience you

10. Transfer Coefficients

121

should be able to skip many of the steps of this recipe, but as a beginner you will not know.) We suggest that you reconstruct one of these examples yourself, after you have deleted everything except the data block at the beginning of the file. 1O.3a Particles (Liquid Fluidised Bed) 1O.3b Bubbles (High Pressure Distillation) 1O.3c Drops (Atmospheric Distillation) 1O.3d Drops (Extraction Mixer) 1O.3e Drops (Extraction Column) 10.4 Discuss the reasoning behind the recipe 'Choosing the Mass Transfer Coefficient' . Could you see ways of improving it? 10.5 In Chapter 10 of this book we recommend that you insert the Maxwell-Stefan diffusivity Di,j in available theories and correlations for mass transfer to estimate the ki,j coefficients for multicomponent mixtures. To understand the background and justification of this recommendation, study Chapters 8, 9 and 10 of Taylor and Krishna (1993).

122

Electrical Forces and Electrolytes If we have charged species, electrical forces are always important. In this chapter, we consider such forces in electrolyte solutions. They can give rise to many unusual effects.

11.1 Electrolytes The most important applications of electrical forces are in electrolytes. Electrolytes (Figure 11.1) consist of a solvent - usually water - and dissolved positive and negative ions (cations and anions). Each ion has a charge number: the number of unit charges that it carries. For simple electrolytes this number is small; proteins and charged microscopic particles can have large charge numbers. electrolyte:

with

water

Zi charge number (Na+:

EEl cations, Na+, Ca2+ anions, CI-, SOi-...

+ 1; SOi-: -2 ... )

Fig. 11.1 An electrolyte with ions and their charge numbers

In electrochemistry, compositions are usually given in molalities (moles per kilogram of solvent). To stay consistent with the rest of this book we will, however, use mole fractions. (We require these anyhow in concentrated electrolytes, such as molten salts.) We regard each ion as a separate species and define mole fractions in electrolytes as shown in Figure 11.2. 1 mol CaCl2 ~ 1 molCa2+ 2 mol Cl54 mol H2 0 n H20 = 54 nCa2+ = 1 n Cl _ = 2

mixture composition:

n

mole fractions: xH20

54 57

_ n _ H20 -----

n

x

= n H20 + nCa2+ + n C1- = 57 - nCa2+_ 1 -n- - 57

Ca2+ -

x

= Cl-

dilute solutions in water Fig. 11.2 Mole fractions in an electrolyte

nCl- =2n 57

123

11. Electrical Forces and Electrolytes

Classical electrochemical engineering deals mainly with dilute aqueous solutions. For these, there is a simple relation between molalities and our mole fractions: the molality is fifty-five times the mole fraction. The interfaces of electrolyte solutions can differ sharply from the bulk solution. They nearly always carry a charge. This is due to several mechanisms: an important one is the preferred adsorption of certain ions at the interface. The surface charge attracts opposite charges, which form a more or less diffuse counter-layer. This 'double layer' is very thin - of the order of a few molecule diameters (Figure 11.3). The electrical field in the double layer is extremely high, and we can easily alter it dramatically by imposing a small potential difference. Such a change can give rise to the most interesting phenomena in electrochemistry: surface reactions. These however, are not the subject of this book. electrical double layer, thin A. "" 3 nm large potential gradient ~

dif> "" 107 V dz m

charge separation

Fig. 11.3 The electrical double layer

11.2 The Electroneutrality Relation In mass transfer theory we deal with much thicker layers. A diffusion film (Figure 11.4) has a thickness of the order of ten micrometres or thirty thousand molecule diameters. The electrical potential gradient in this film is small (however, it is important for the transport of ions). As a consequence the charge separation in the diffusion film (and also in the bulk liquid) is negligible.

.& "" 10000 nm

diffusion film, 'thick'

H ~

I

I II

almost no charge separation

d

~ "" 102

V

-

dz

electroneutrality: "£.,.ZiXi = 0

m

.

i

Fig. 11.4 The diffusion film with the electroneutrality relation

For all practical purposes the concentrations of positive and negative charges are equal l • This yields the electroneutrality equation. The electroneutrality equation is essential in describing mass transfer in a way similar to that of the bootstrap relation. Because of electroneutrality, there is no net electrical force on a bulk solution; the forces on the two kinds of ion cancel. 1

The difference in the mole fraction of the two ions is ofthe order of 10- 10, see exercise 11.3.

124

Mass Transfer in Multicomponent Mixtures

11.3 Electrical Forces An electrical field causes forces on charged species. These are proportional to the negative of the gradient of the electrical potential (Figure 11.5). This potential is conventionally expressed in volts. To obtain forces in molar units we multiply the gradient by the Faraday constant. This constant has a large value; even small electrical potential differences can give rise to enormous forces. The force is also proportional to the charge number of the species.

96.5kCmor1

q; Faraday constant
electrical potential, V

Fig. 11.5 The electrical force in the Maxwell-Stefan equations

11.4 Transport Relations Figure 11.6 shows the transport relations of ions in electrolytes. The most important difference with previous examples is the occurrence of the electrical driving force. You can of course write similar equations using velocities. differential equation difference equation

dl/» i -cx· (dlna RT--+PZ·I dId Z Z - (l1a. PZ· ) -Xi ~+-1111/> ai RT

RT =I. -.-(x.N--x.N.) D..' i'lc-i

I

I,

I"

=I. (XjNi - _xiN j ) ,'of-'I

ki, jC

Fig. 11.6 Transport relations in electrolytes

The Debye-Huckel theory gives the activity coefficients in the equations in dilute solutions. Figure 11.7 shows this theory for single and double-charged ions (1-1 and 2-2 electrolytes). The parameter along the horizontal axis is the mole fraction analogue of the ionic strength used in classic electrolyte theory. We can take the activity coefficients into account in the same way as in Chapter 8. As you see, the changes of the coefficients with composition are not very large. The theory has its limitations. It predicts the average of the activity coefficients of the two ions and not the individual values. Also, it does not predict the sharp rise seen in concentrated solutions (these are not shown). In very dilute solutions the activity coefficients become equal to unity. In such solutions, friction between the ions and the solvent dominates. If the solvent velocity is also zero, we obtain the equations shown in Figure 11.8. Here it is possible to write

125

11. Electrical Forces and Electrolytes

flux explicitly. This yields the Nernst-Planck equation, which is a special case of the Maxwell-Stefan equation. In the following examples, we will be using the MSdifference equation, but we could also have used the Nernst-Planck equation. i=~2£.. ~ z~x. I I

i

~

i

10-3

10-4 1.0

'Yi 0.8 DebyeHuckel

0.6

in dilute solutions, the nonideality variation is not large

~'Y «~ 'Y

x

Fig. 11.7 Non-idealities in dilute electrolytes

in dilute solutions of stagnant water _ RT d In Xi dz

_

pz dep = RT u I

dz

Di,w

I

~

difference equation Llx

_~/

RT

I

I,W

[dxdzi + PZixi difJ] RT dz

Nernst-Planck equation

PZLlep U __ '_=_'

Xi

N = -cD.

~

N

ki,w

I

=-k·

[ Pzx]

.c Llx. + __ I _I Lln.

1,1

I

RT

'I'

Fig. 11.8 The equations for dilute electrolytes

11.5 Diffusion of Hydrochloric Acid We consider diffusion of hydrochloric acid through a film (Figure 11.9). The ions in this example have greatly differing diffusivities. Even so, hydrochloric acid diffuses in water as a single substance. How can this be? D1,3

I

I

I I I

I

I

I

I

I I I

Cl

ep

/3

V1

1 I I

I I I I

Cr(2)

=9.3xlO-9 m 2 s- 1

D 2,3 = 2.0x 10-9 m 2 s-l

why do they move at the same rate? ~ electrical gradient retards H+,

• »

accelerates CI-

Fig. 11.9 Diffusion of hydrochloric acid in water

Mass Transfer in Multicomponent Mixtures

126

The reason is not difficult to understand qualitatively. The hydrogen ion has a tendency to move more rapidly. This causes a minute charge imbalance (immeasurably small) and a potential gradient. This gradient slows down the hydrogen and speeds up the chloride so they go along together. To describe this system we use the electroneutrality and no-current relations (Fig 11.10). Solving the transport equations shows us that there is a potential difference across the film: the diffusion potential. The diffusion potential is very small - in the order of millivolts. It is, however, essential in the transport relations as it causes the two ions to move with the same velocity. As a result, the transport of hydrogen chloride can be described by a single mass transfer or diffusion coefficient, whose value is in between those of the two ions separately. diffusion potential

electroneutrality: Xl

=x2 =x

iltjJ= RT(_I_ _ _ 1

)u

no current:

2P ~,3 velocity

transport equations: u X RT k l ,3

u = _2(_1_ + _1_)-1 Ax k1,3 k2 ,3 X diffusivity

•

_ Ax + Pil = u X RT k 2 ,3

k1,3

~CL = 2(_1_ + _1_)-1 Dl,3

D2 ,3

Fig. 11.10 Equations governing diffusion of hydrogen chloride

11.6 Plus a Trace of Sodium Chloride A salt, base or acid on its own, diffuses in water as a single substance. You may be tempted to assume that mixed electrolytes also diffuse as separate substances. This can, however, lead to completely erroneous conclusions. In Figure 11.11 we have added a trace of sodium chloride to the previous example. The diffusivity of NaCI according to the previous formula is given in the figure. You might expect NaCI to diffuse down its gradient with a velocity corresponding to this diffusivity. This idea is wrong. In reality sodium will diffuse in the opposite direction! The movement of the sodium ion is determined by the electrical gradient caused by the hydrogen ion.

11.7 Diffusion of Proteins Proteins are large polyelectrolytes. Their outer surfaces are studded with weak acid and basic groups. Their degree of ionisation depends on the pH, and so does the charge of the protein. In sufficiently acid solutions, the protein is strongly positive: it

11. Electrical Forces and Electrolytes

127

might have a charge number between + 10 and +20. As the pH increases, the charge goes down and becomes equal to zero at the isoelectric point. Here, the positive and negative charges canceL In more basic solutions, the protein becomes negative; the charge number might go down to -10 or -20. Because of these charges, electrical effects are important in the diffusion of proteins. concentrations in mol m-3

18

you might think that NaCl would diffuse to the right with a diffusivity: )

I

I

I

j

3

~l~

I



I I

~! I

I

=1.3 X 10-9 m 2 s-I 9 2 DC1,W = 2.0 X 10- m s-l ~ ~aCI

NaCl 2

~a,W

I I

=1.6 X 10-9 m 2 s-1 this is wrong!

the electrical gradient (due to HCI) forces Na+ against its composition gradient

!

Fig. 11.11 In a mixed electrolyte solution, a trace (here Nael) can diffuse against its concentration gradient.

Lysozyme is a small, roughly spherical protein. It can be bought in a form with chloride as counter ions; when the protein dissolves in water it yields a slightly acid solution with about 13 chloride ions to each lysozyme molecule (Figure 11.12). In a solution with an excess of an electrolyte such as KC1, the diffusivity of the protein has a value near 1.2 x 10. 10 m 2 S·l. This is what the Stokes-Einstein equation predicts for a sphere with the diameter of hydrated lysozyme. However, in solutions with no extra electrolyte, the diffusivity appears to be almost five times higher. What is going on? What happens is that the protein does not diffuse alone, but accompanied by thirteen chloride ions. To maintain electroneutrality, they must move together, and we measure some average of the diffusivities of all fourteen ions. The chloride ions 'pull' the protein. When we add potassium chloride, the role of the protein is taken over by the more mobile potassium ions, and the effect disappears. The effective diffusivity is measured in a cell with two liquid layers. The bottom layer contains a dilute solution of the protein; there is no protein in the top layer. Both layers contain an equal concentration of KCL The effective diffusivity follows from the transfer rate across the interface. We model this system by assuming that diffusion occurs at a steady state through a film (Figure 11.13). Solving the four transport relations and the no-current relation yields all velocities and the potential difference. From the velocity of the protein, we

Mass Transfer in Multicomponent Mixtures

128

compute its effective diffusivity. The results are the drawn line in Figure 11.12. We see that they describe the experiments quite well. 10

Ly12+ 't:-C KCl

=0

e e e 13 CIe e ee e e e ee

e

E9

H+

Q lysozyme is pulled by \ /heory

its counter ions

~ r--- excess electrolyte I )-

1

o

0

Q effect disappears at high ionic strengths C Kel

100

200

molm-3

Fig. 11.12 Effect of electrolyte on the diffusivity of a protein

transport equations i =1... 4

toplt,lyer

_ Axi

_

xr

PZi Ll= RT

U;

Dr, w

/&-

'no current' LX/liZi =0 i

(4) mole fractions

these yield a; and

~

Dt,eff

=-Ut &-

Fig. 11.13 Calculating the effective diffusivity of lysozyme

11.8 Conduction and Friction between Ions So far, we have considered very dilute solutions where we could neglect the interactions between ions. However, friction between ions is already noticeable for ionic mole fractions of well below 1%. Ordinary molecules interact mainly with their nearest neighbours. Not so for ions, whose electrical fields can extend over several molecule diameters, and so cause large friction with other charged species. A simple example where friction between ions shows up is in an electrical conductivity cell (Figure 11.14). The cell has the same cross-section throughout and is fed by an alternating voltage. If the frequency is sufficiently high you may assume that no concentration gradients develop in the cell.

11. Electrical Forces and Electrolytes

(1) water

129

,1.l/J "" 1 V f\;

current

+

•

(2) cation (3) anion

no concentration gradients ..i

Az =10-2 m

small water velocity

dJJ,1a

d _ RT

... RT

d _ RT dz D\,3

RT D2 ,3

2 -RT. ,.'j.;'-PZ 2 -d --D x\(u2 -!6)+-D X 3 (U 2 -u 3 ) '!Z, z 1,2 2,3

\d~a3

rD

-RT-,-\-~"L·Z3 ---x\(u3 -~)+--X2(U3 -u2 )

.' . dz

~N2,N3

Fig. 11.14 Electrical conductivity cell

In conductivity experiments, the velocity of water is small. There are two reasons for this: • the mole fraction of water is usually high, so only large water fluxes yield an appreciable velocity, and • the volume fluxes of the two ions tend to cancel, so there is only a small volume flux of water. If we neglect the water velocity, we obtain the equations in Figure 11.14. These yield the fluxes of the two ions. From these, we can calculate the current density and the electrical conductivity (Figure 11.15). The conductivity is the ratio of the current density and the (negative) potential gradient. In dilute solutions, it is proportional to the electrolyte concentration. In concentrated solutions, however, the rise is less than proportional because friction between the ions reduces the fluxes. current density using electroneutrality

I K=

_d

dz conductivity

effect offriction between ions Fig. 11.15 Electrical conductivity

Mass Transfer in Multicomponent Mixtures

130

A concept related to conductivity, is the transport numbe,-l. The transport number of an ion is the fraction of the total current that it carries, under conditions where there are no concentration gradients. For the conductivity cell of the previous example, the ratio of the two transport numbers is easily calculated as in Figure 11.16. In contrast to conductivity, the transport numbers hardly depend on concentration because the two ion-water diffusivities vary in about the same way. t z = Nzzz = _ Z2 DI,Z t3 N3 Z 3 Z3 DI,3 Fig. 11.16 Transport numbers

11.9 Diffusivities in Electrolyte Solutions There is a reasonable amount of information on ion-water diffusivities, and on the interactions between oppositely charged ions. Figure 11.17 gives a complete set for aqueous solutions of sodium chloride. 10-8

------- ----- D+,wandD_.w I-----t-DW,Na --l--''-'U~---l almost constant

-------- ------- D+,_ ~ C°.5 10-10 1 - - - - - + -

10-1 Fig. 11.17 Diffusivities in sodium chloride solutions

We see that the diffusivities of sodium and chloride in water both have values of around 10-9 m 2 S-I; they depend little on concentration. The sodium-chloride diffusivity is much smaller (there is a lot of friction between the ions). Its value increases roughly proportionally to the square root of the salt concentration. More ion-water diffusivities are given in Figure 11.18. The values drop slowly at mole fractions in excess of a few percent, mainly because of the increasing viscosity of the solutions. The diffusivities of hydrogen and hydroxyl ions are considerably 2

Also known as the transference number.

11. Electrical Forces and Electrolytes

131

higher than those of other species. These ions propagate through water by 'bumping off' new ions from the other side of a water molecule. 10-7

much higher } than the others

La3+

increasing charge and size

1

Csalt

molL I

10-1 Fig. 11.18 Ion-Water diffusivities in aqueous chloride solutions

A series of cation-anion diffusivities is plotted in Figure 11.19. For all components the diffusivity increases approximately with the square root of the salt concentration. The diffusivities decrease strongly with increasing charge numbers. Interactions between pairs of ions with the same charge numbers are almost identical. HCI NaCI K2S04 LaCl3

10-11

CUS04

empirical relation: D

13

10-

csaJt molL-I

10-4

D+ w +D_ W iO. 55 , +,- 2 Iz+z_12.3 -'

i=O.5LZ;xi

10- 2 Fig. 11.19 Cation-anion diffusivities

All this indicates that ion-ion interactions are mainly due to electrostatic forces, as might be expected. Note that the diffusivities of hydrogen ions are again higher than the others. The empirical relation in the figure for ion-ion diffusivities is not very

132

Mass Transfer in Multicomponent Mixtures

accurate. We have encountered deviations of up to 30%. Even then it can be useful: ion-ion friction is often a small part of the total friction because of low ionic concentrations, so we need not evaluate it accurately. Figure 11.20 contains a few diffusivities of ions of like charge. This is one of the few cases where Maxwell-Stefan diffusivities are negative. This means that the friction of an ion with its surroundings tends to be reduced by adding ions of like charge. (The net effect may not be a reduction, because we must also add ions of unlike charge to retain electroneutrality.) -10-8

,----,----,----,----,----,

-10-10

1-1 _10-12

1-2 1-3

-::iI"~--_+__--+---___l Csalt

molL l

10-4 Fig. 11.20 Diffusivities of ions of like charge

Also here, we see that the absolute value of the friction is largest for the largest electrical charges (for the 1-3 electrolyte).

11.10 Summary • The electrical driving force in electrolyte solutions gives rise to many new effects. • It requires the use of a new relation: that for electroneutrality. • Sometimes, we need multiple bootstrap relations. A new type is the no-current relation. • The electrical field causes a strong coupling of the transport of ionic species. It causes the two constituents of a single salt to move as one component. In mixed salt solutions it may cause some species to move against their composition gradients. • The friction between ions of unlike charge is quite high, and already noticeable at relatively low concentrations. • When we apply an electrical field, we get a current. The current density depends on the conductivity of the solution. This is governed by the diffusivities of the

11. Electrical Forces and Electrolytes

133

ions: it falls at high concentrations because of friction between the ions. • The behaviour of ion-water diffusivities is nothing special. However, ion-ion diffusivities depend strongly on concentration and on the charge numbers of the ions. They are usually negative for ions of like charge. • You should remember the formulae with the electrical driving force in Figure 11.5.

11.11 Further Reading Atkins, P.W., 1998, Physical Chemistry. Sixth Edition, Oxford University Press, Oxford. Lovely book. Excellent to brush up your electrochemistry. Especially Chapters 10 and 29. Newman, J., 1991, Electrochemical Systems. Second Edition, Prentice-Hall, Englewood Cliffs, New Jersey. An excellent book, with an extensive treatment of multicomponent mass transfer in electrolytes using the M-S approach. Lightfoot, E.N. (1974) Transport Phenomena and Living Systems. McGraw-Hill, New York. One of our favourites. It contains a clear introduction to the Maxwell-Stefan approach to multicomponent mass transfer and also nice examples on electrochemical systems. Cadman, A. D. et aI, Diffusion of Lysozyme Chloride in Water and Aqueous Potassium Chloride Solutions, Biophys. J. 37 (1981), 569..574 Data on lysozyme chloride in Figure 11-13. Wesselingh, J.A., Kraaijeveld, G. and Vonk, P. (1995) Exploring the Maxwell-Stefan description of ion exchange. Chem. Eng. Jl, 57, 75-89. Estimation of diffusivities in electrolyte solutions. Also discusses how these are determined. Kraaijeveld, G. , Kuiken, G.D.C. and Wesselingh, J.A. (1994) Comments on "negative diffusion coefficients". Ind.Eng.Chem. Res., 33, 750-851. A discussion on the negative diffusivities shown in Figure 11.16. Muralidhara, H.S. (1994) Enhance separations with electricity. Chemtech, 24 (May), 36-41. A survey of separations using electrical fields.

Mass Transfer in Multicomponent Mixtures

134

Vinograd, J.R. and McBain, J.W. (1941) Diffusion of electrolytes and ions in their mixtures. J. Am. Chem. Soc., 63, 2008-2015. One of the first papers to highlight mixed ion effects.

11 .12 Exercises 11.1 Use Figure 11.5 to get an idea of the size of electrical forces. Consider a film with a thickness Az = 10-5 m, a monovalent ion Zi = 1 and a potential difference flt/J = 0.1 V. Compare the electrical force per mole of ions with the force calculated in the glass-of-beer (Figure 3.15). 11.2 A single salt diffuses as a single component through water (see Figure 11.9). So the salt-water mixture behaves as a binary. In not too concentrated solutions, the activity coefficient of the ions is given by the Debye-Hlickel equation. In the mole fraction notation that we use here, this equation reads:

We often neglect the effect of non-ideality in electrolyte solutions. Use the formula in Figure 8.10 to determine the ratio between the 'exact' driving force and the value assuming an ideal solution. Do this for a 1-1 electrolyte such as NaCl. 11.3 Concentration of charges in a non-equilibrium electrolyte. We consider the film of water in Figure 11.9. The film consists of water (3) through which HCI is diffusing as hydrogen ions H+ (1) and chloride ions CI- (2). Their diffusivities are given in the figure. The film has a thickness Az = 10-5 m. The mole fractions of the ions on the two sides of the film are small: xl a = x2a = 2 X 10-4 and xlf3 = x2f3 = 1 x 10-4

Due to unequal diffusion rates of ions, there is a potential difference flt/J across the film, which you can calculate using Figure 11.10. You may assume that the potential varies linearly across the film:

The charge density in an electrical field is given by the Poisson equation: 2 d (J'el --2=dz eel

t/J

Here (J'el is the electrical charge density in C m-3; eel z7xlO- lO CV- 1 m- 1 is the dielectric constant of water. Integrate this equation to obtain the charge density and from that the mole fraction of unbalanced charges xub:

11. Electrical Forces and Electrolytes

x

ub -

135

(Jel

Pc

Here P =96500Cmol- 1 is the Faraday constant and c =55000 mol m -3 the (total) concentration. 11.4 How accurate are the Nemst-Planck equations? (Mathcad) In the Nemst-Planck equations (Figure 11.8) ion-ion interactions are neglected. Investigate how accurate this assumption is for conduction of electricity in the cell of Figure 11.14. The electrolyte is a solution of 2 mole L- 1 of NaCI in water at 20 °C. Use the data in Figure 11.17 to support your arguments.

*

11.5 Diffusion of Three Ions (Mathcad). Calculate the fluxes of the three ions in Figure 11.11. Number the species as H+ (1), Na+ (2), Cl- (3) and water (4). You may consider water as stagnant. The concentrations are given in the figure. The film has a thickness & = 10-5 m; the hydrogen / water diffusivity has a value £)1.4 = 9.3 m2 S-I. In which direction does sodium diffuse according to your calculations?

X

10-9

11.6 Diffusion of Lysozyme Chloride in Aqueous Potassium Chloride (Mathcad). The full calculations behind the example in Figure 11.13. 11.7 Electrical conduction through a turbulent liquid (Mathcad). The two ions in a salt solution in an electrodialysis cell must be removed at the same rate. The bulk liquid in the cell is turbulent, so it is often assumed that there can be no concentration gradients. Can these two considerations be reconciled if the two ions (such as Na+ and Cn have different mobilities? Do you see a solution? 11.8 Measuring Maxwell-Stefan Diffusivities. How do you measure Maxwell-Stefan diffusivities in electrolytes? Answering this question fully is outside the scope of this book, but we can give you an idea. Consider a solution of water, Na+ and Cl- ions. There are four different transport experiments that you can do with such a solution; they are shown below. We need three diffusivities to describe such a solution: D+ w ' D -w andD+_. In addition, you might like to have the interaction between like ions such as Cl- and CI-(isotope). The clue is that you get different information out of each experiment; you get the diffusivities by combining information from several experiments. This is not easy; that is the main reason why our knowledge of MS-diffusivities is still so incomplete. In the following table, you are to indicate which diffusivity plays an important (++) role, some role (+) or no role (0) in the experiment:

136

Mass Transfer in Multicomponent Mixtures

diffusivity

9 water

+

+

conductivity transport numbers tracer diffusivity

diffusivity conductivity transport numbers tracer diffusivity

0(

+ t

137

1 Centrifugal and Pressure Forces So far, we have considered driving forces due to activity gradients and electrical gradients. These are forces internal to the mixture: the forces, on all species together, balance. In this chapter, we add two driving forces: the centrifugal force and gravity. (The two are similar, but the former can be much stronger.) Both are known as body forces. In most diffusion problems, the body forces cause counteracting pressure gradients, so that fluid as a whole hardly accelerates. These three forces: centrifugal, gravity and pressure forces, all lead to new terms on the left-hand side of the Maxwell-Stefan equation.

12.1 Volumetric Properties In the investigation of other forces we shall need two new properties of a species in a mixture: • the species volume, and • the species density. These properties are not necessarily equal to those of the pure species. We can determine the species volume (usually called the 'partial molar volume') by adding one mole of the pure species to a large amount of mixture. The resulting increase in volume (Figure 12.1) is the species volume. The species density is simply the molar mass of the species divided by its volume. species volume

1 mole of pure 'i'

mixture (» 1 mole)

= Mi

p 1

species

V; density

Fig. 12.1 Volume and density of species in a mixture

138

Mass Transfer in Multicomponent Mixtures

In some cases the species properties are identical to those of the pure species. This is so: • in ideal gases, • in mixtures of similar molecules forming ideal solutions, and • for small suspended solid particles, such as in the following example. In non-ideal solutions, the volume in the mixture usually differs from that of the pure species. Adding a species to a mixture sometimes even causes the mixture to contract: the species volume is then negative.

12.2 The Pressure Gradient In our first example, we consider a fluid under influence of gravity (Figure 12.2). We assume that there is no acceleration of the fluid as a whole. There are two forces working on the mixture: gravity and the resulting pressure gradient. These two must balance; this yields the hydrostatic pressure formula. Note that this formula contains the average density of the mixture, and that the pressure increases in the negative (downward) direction. average mixture properties:

density molar mass

p M

molar volume V

one mole .. of mixture :-

Fig. 12.2 Pressure gradient as caused by gravity

12.3 Gravitational Force Figure 12.3 shows small particles settling in a fluid. Gravity drives the particles downwards. However, unless the fluid has no density, gravity will also cause a pressure gradient. This gradient opposes the effect of gravity. So again, there are two forces working on the particles. (There can also be forces due to activity gradients, but we neglect these for the moment.) Note that the effect of gravity is proportional to the mass of the particles, and that the effect of the pressure gradient is proportional to their volume. Note also, that both forces have the proper unit of N mol-I.

12. Centrifugal and Pressure Forces

139

We see that the combined effect of gravity and pressure gradient is a force proportional to the difference in density between particle and mixture. There is no driving force if the particle and the mixture have an equal density. gravitational plus pressure forces on i:

mixture density Fig. 12.3 The force on settling particles due to gravity and the pressure gradient

Gravity and pressure gradients are important for large particles. They largely determine the behaviour of a human body in a swimming pool; the molar mass and volume of such bodies are several times the mass and volume of the earth. However, gravity and pressure gradients are less important on a molecular scale.

12.4 Centrifuges The forces due to gravity are quite insignificant in mass transfer operations. Centrifugal devices can provide forces which are more than five orders of magnitude larger. In its simplest form a centrifuge is a cylindrical container, rotating rapidly around its axis. A fluid in the centrifuge is flung outward against the wall. Species denser than the fluid then move through the fluid and collect at the wall (Figure 12.4).

>r

liquid

~

repIace g by - m2 r ---+ pressure grad·lent -dp = pm2r dr Fig. 12.4 The centrifugal force is a body force analogous to gravity

140

Mass Transfer in Multicomponent Mixtures

The centrifugal force causing this movement is similar to gravityl. Formally, we only need to replace the gravitational acceleration by a centrifugal acceleration. This is proportional to the angular velocity squared and to the radius. There are differences: a centrifuge has a cylindrical symmetry and the force depends on the radial position. The pressure difference over a centrifuge containing a liquid can be over one hundred megapascals! The equipment has to be very sturdy to withstand this. Also, in centrifuges containing gases, the pressure varies strongly with the radius. When we start a centrifuge (Fig 12.5, situation a) there are no concentration or activity gradients. The difference between the centrifugal and pressure forces, causes the species to move with respect to each other and activity gradients build up (Fig 12.5, b). Eventually, the activity, centrifugal and pressure forces balance, and the system reaches equilibrium (Fig 12.5, c). The heavier species concentrates near the outside of the rotor. Centrifugal forces are not extremely large and transport distances in a centrifuge are of the order of centimetres. So it can take a long time to reach equilibrium. +M·olrVI dp I d

r

= £..J ~ x ,':JI,J f. ·(u· -u·)

dln~ 2 dp -RT--+M·ro r-V-= d I I d

...

"11·"

r

r

Lxf. ;

_RTdlnai +Mro2r-V dp =0 dr

1

1

dr

I

;

,

·(u·-u·) I ,

,':JI,J

(equilibrium)

Fig. 12.5 How the terms in the MS-equation change in time

12.5 Gas and Protein Centrifugation Centrifugal forces can be large enough to concentrate molecular species. There are currently two kinds of application: • for the separation of gases (uranium hexafluoride) in the enrichment of nuclear fuel, and • for the separation of proteins in biotechnology. For the gases, only a modest separation is obtained (Figure 12.6). Centrifugal gas separators consist of huge series of continuous centrifuges; we will not discuss them here. Ultracentrifuges for proteins are widely used in the laboratory, but they are losing ground to other techniques.

I

For the origin of the centrifugal force, see exercise 12.2.

141

12. Centrifugal and Pressure Forces

small molecules

~

large molecules

' •.'-:;. e... :-..•.... • :. -.-1 t.. ~i ~~. .Ia

o

.'

•

~r

~r

gas separations

U235 F6 I U238F 6

protein centrifugation

Fig. 12.6 Centrifugal equilibrium. Small molecules (left) have a small gradient. Large (heavy) molecules form a diffuse layer near the rim.

12.6 Difference Equation for the Pressure Force Centrifugal and gravity forces are seldom important in mass transfer processes. We will not consider them any further after this chapter. However, the pressure force does occur often in membrane processes, so we will devote a few words to it. There is a large variation in the size of species volumes (Figure 12.7). The volume of water, for example, is very small. So only very large pressure gradients influence the transport of water. (We use these in the membrane process known as reverse osmosis: see Chapter 18.) Pressure gradients are more important for gas separations (Chapter 17) and for macromolecules (in ultrafiltration: Chapter 24). species volumes water: most liquids: macromolecules:

Vi "",1.8xlO-5

m 3 moI- 1

VI "'" 10-4 m3 mol- 1

10-2 ... 10° m3 mol- 1

for gases: pressure force: in general [ -V; dp dz gases

[- RT dp pdz

1~ -v'&~-x RT & ~

[- V; RT

~1

1~ -RTp&-~-x RT & ~ I-¥l

in differential equations

~

in difference equations

Fig. 12.7 Species volumes and the pressure force

As before, we can introduce difference equations of the different forces (Figure 12.7). We make these dimensionless by mUltiplying with a film thickness and dividing by RT.

142

Mass Transfer in Multicomponent Mixtures

12.7 The Maxwell-Stefan Equations (again) We are approaching the end of the first part of this book, where we have treated multicomponent diffusion in gases and liquids. This is a good point to have a look at the complete Maxwell-Stefan equations for bulk fluids. Below, we have included the activity, electrical, pressure and centrifugal forces (Figure 12.8). You can add more forces (such as gravity or magnetic forces) just by adding new terms to the left side of the equation. driving force I

\

_RTdlnai -pz d -v dp +Molz= dz 'dz'dz'

I,

,

i '" .. (u. -u.) = RT['" X. u -u;] £... x.!" !,~l.' ~ D .. rtl

joti \

,

I" I

friction with other species 'j' Fig. 12.8 MS-equations including electrical, pressure and centrifugal forces

For gases, it is often handy to combine the activity and pressure forces into a single force containing the gradient of the partial pressure of the species (Figure 12.9). for gases: _RTdlnxi _ V dp dz I dz

=-RT[~ dxi +! dP]=_ RT dPi Xi dz

P dz

Pi dz ~~~

/

partial pressure Fig. 12.9 Using partial pressures in the MS-equation for gases

The difference form of the equations is analogous (Figure 12.10). However, you will probably seldom use it for centrifugal separations, because conditions change too much across a centrifuge. centrifugal friction forces

activity electrical

for gases

_ !}.pi Pi

Fig. 12.10 Difference equation including the electrical, pressure and centrifugal forces

12. Centrifugal and Pressure Forces

143

12.8 Summary In this chapter we have seen a number of driving forces for mass transfer: • gravity (a very weak force, except for large species) and • centrifugal forces. These external forces usually give rise to pressure gradients. The pressure gradient is the sum of all external forces per volume of mixture. It is the cause of a pressure force, which is proportional to the species volume (partial molar volume ). We have seen how all the different driving forces combine in the left-hand side of the Maxwell-Stefan equations. You should memorise the pressure force and its difference equation; we will be using it many times in the coming chapters.

12.9 Further Reading Lightfoot, E.N. (1974) Transport Phenomena and Living Systems. McGraw-Hill, New York. Some nice examples. Rickwood, D. et aI, Centrifugation, Essential Data (1994) John Wiley & Sons, Chichester / BIOS Scientific Publishers Ltd.

12.10 Exercises 12.1 Small Bubbles and Footballs. Determine the species volumes of small bubbles with a diameter d1 = 10-8 m and of footballs with a diameter d 2 = 0.2 m. Use these to determine the pressure force per mole on bubbles and footballs submerged in a large amount of water (see Figures 12.2 and 12.3). Why would it be that pressure forces are quite important in daily life, but relatively unimportant in molecular systems? 12.2 Centrifugal Force on a Pure Liquid. Consider a cylindrical centrifuge containing a single incompressible liquid and running under equilibrium conditions. Because the system is at equilibrium, the liquid must have the same potential everywhere. (Remember that the potential change is the amount of work that must be provided reversibly to change the system). There are two contributions to the potential of the fluid (taken here per mole): 2

= M ((f)r )2 and 2 2 2. the pressure energy V{p - Prej). 1. the kinetic energy M v

Forces are the negative of the gradient of the potential. Take the derivative of the potential with respect to the radius and check that you get the same centrifugal and pressure terms as in the Maxwell-Stefan equation in Figure 12.8.

144

Mass Transfer in Multicomponent Mixtures

12.3 Friction Coefficient of a Swarm of Spheres. (Mathcad) If you wish to apply the Maxwell-Stefan equations to centrifugation of concentrated protein solutions, you will need the MS-friction coefficients between protein and fluid. You can obtain this from a relation for the settling velocities of a swarm of spheres, which is known under the names of Richardson and Zaki. 12.4* Separation of Protein in an Ultracentrifuge (Mathcad). Estimate the time required to separate the protein using the formulae in Figure 12.5. 12.5 Separation of Uranium Isotopes in a Gas Centrifuge (Mathcad). Derive an analytical solution for the equilibrium distribution of the two forms of uranium hexafluoride UZ35 F 6 and U Z38 F 6 in an ultracentrifuge. 12.6 Electrical Gradients in an Ultracentrifuge. We consider a solution of water 0), a charged protein (2) and a co-ion Na+ (3). The solution is dilute, and has the density PI of water. The protein has a negative charge number Zz and a density pz. The ions have a charge number Z3 = 1; you may assume that they have the same density as water. In an ultracentrifuge, the protein will collect as a thin diffuse layer at the rim (r = R) of the centrifuge. The acceleration in this layer is practically constant and equal to aiR. We consider the forces (Figure 12.8) on the three species when the centrifuge has been spinning for some time, so that it is in equilibrium: dlnxl udp M 2 - RT - - - vl-+ 1(0 R - 0 dz dz

(1)

dlnx2 IV dlP dp 2 -RT---TZ2-- \'2-+M2(O R=O dz dz dz

(2)

dlnx3 IV dlP dp 2 -RT---TZ3--\'3-+M3(O R=O dz dz dz

(3)

The fIrst term in the fIrst equation is very small, so dp = Ml (02 R . dz "1 The densities of '1' and '3' are equal; equation (3) becomes _RTdlnx3 -PZ:3 dlP = 0 dz dz (a) Use electroneutrality to obtain a relation between X z and x 3 • Then eliminate the mole fraction gradients in equations (2) and (3) to obtain an equation for the electrical potential gradient. (b) Calculate the value of the gradient for a spherical protein with a diameter d2 =5xlO-9 m and density P2=1400kgm-3 in an ultracentrifuge with 2 6 (02 R = 10 ms- .

12. Centrifugal and Pressure Forces

145

12.7 So far we have considered the following driving forces: the activity gradient (composition part of the chemical potential gradient), the electrical gradient, the gravitational gradient, the centrifugal gradient and pressure gradients. Then we have also considered frictional forces between the species. Can you think of any other forces that might be important for motion in mixtures? 12.8 In this book we have ignored the contribution of thermal diffusion to mass transfer. This thermal diffusion contribution can be easily incorporated into the Maxwell-Stefan equation given in Figure 12.8. In order to see how this is to be done see the review of Krishna and Wesselingh (1997). Thermal diffusion effects are of importance in chemical vapour deposition reactors where we have large temperature gradients across diffusion layers and the species being transferred have large molar masses.

146

Why we use the MS-equations A complicated subject such as multi component mass transfer can be described in many ways. We have focused here on one type of description: the 'driving force = friction' or Maxwell-Stefan-type of approach. You may wonder why. We will tell you using an example. The example is diffusion in an ideal ternary mixture, where diffusion is caused by concentration gradients only.

13.1 Three Ways Figure 13.1 shows three common ways of describing this system. The first should by now be familiar to you. The second shows the Fick equations, which make the process engineer feel at home. The third is the form mostly used in the thermodynamics of irreversible processes (TIP). There are many variations possible on these equations, for example in the units used. All this is fairly obvious.

CD Maxwell-Stefan three coefficients

WFick four coefficients

CD Thermodynamics of Irreversible Processes (TIP) three coefficients

Fig. 13.1 Three descriptions of multicomponent diffusion

The three descriptions - although they look very different - can be transformed into each other. They give the same results and are all equally correct. So you choose the one you find most handy or the easiest to understand. Also in all three systems you must choose any two out of the three transport relations for the three components.

13. Why we use the MS-equations

147

13.2 A Mixture of Three Gases As an example we have taken the diffusivities in a ternary mixture of ideal gases. (This is the only kind of system in which we can calculate ternary diffusivities with any accuracy.) The pressure is 100 kPa and the temperature 298 K. The components are hydrogen, nitrogen and di-chloro-di-fluoro-methane (a Freon). We have chosen the molar masses of the three far apart to accentuate a few things, but otherwise the system is nothing special (Figure l3.2).

CD H ®N Cl)

2,

hydrogen, a light gas;

2,

nitrogen, an intermediate gas

CC1 2F2, a 'Freon'; a heavy gas

At T = 298 K and p = 100 kPa the gas theory gives the MS-diffusivities (in m2 S·I): 5 5 5 D D12 =7.7xlO- D13 =3.4xlO23 = 0.81 X 10,

~

'0

,

Fig. 13.2 Our mixture of three gases

In the coming paragraphs, we present a series of graphs that need a little explanation (Figure l3.3). We plot diffusivities and phenomenological coefficients in a prism (with three dimensions).

....•................. +10

o

-10

Transport coefficients are projected in a 3-D pIjsm . -................ -. Compositions are given by position in the triangular diagram (bottom) The main components are '1' and '2'; they are always on the front axis ...... Values positive above the central plain, negative below Fig. 13.3 How we plot the transport coefficients

Ternary compositions are given by points on the bottom triangle of the prism. Each corner represents a pure component. The vertical distance above the bottom plane

148

Mass Transfer in Multicomponent Mixtures

gives values of the coefficients. They can be positive (above the mid-plane) or negative (below). We must choose two independent components. These are always on the front axis. We can choose them in three ways (Figure 13.4) 'dependent' -----+

H2

~

1

'independent' ----.. N 2

212 CC1 2F2 CCl 2F2 H2

CCl 2F2

L

Fig. 13.4 The three choices of the independent components

13.3 The Fick Description In the first instance the Fick equations look attractive. They seem a logical extension of mass transfer theory, and look the right way for a process engineer. Also the fluxes are explicit and concentrations are handy in calculations involving mass balances. No thermodynamic model is required in mass transfer calculations. There are, however, disadvantages. The driving forces in the Fick equations are concentration gradients. If you try to include the effects of non-idealities and electrical and pressure gradients, the equations become very messy and almost incomprehensible. Another problem is the behaviour of the diffusivities. In a binary gas mixture (see Chapter 2) the diffusivity is independent of composition. You might expect such simple behaviour also to prevail in ternary mixtures. This is not so. You must choose two independent components and this can be done in three ways (Figure 13.5). In the first we have chosen nitrogen and the Freon. You see that all four coefficients are positive. However, their behaviour is far from simple: they form strongly curved surfaces in space. Only two straight lines are visible along the binary boundaries. If you choose the Freon and hydrogen as the independent components (second set) the behaviour is completely different. In this case one of the cross-coefficients is always negative. Clearly the size of the diffusivities depends not only on pressure, temperature and composition, but also on the choice of the sequence of the components! The result of the third choice is even wilder. The two cross-coefficients are now both negative; see how beautifully curved in space the two main coefficients are. If you try this on a strongly non-ideal system, you will find that it yields a scientific edifice of great and lasting proportions. This is nice for 3D-graphics, but not if you want to understand what is going on.

13. Why we use the MS-equations

149

• H2

3

1

0

N2

2 CCl 2

o I-f------i--

Ol-f-----+_~

Fig. 13.5 The three sets of Fick diffusivities

2 N

150

Mass Transfer in Multicomponent Mixtures

13.4 Thermodynamics of Irreversible Processes The TIP representation can be derived from fairly simple assumptions on entropy production. It is explicit in the fluxes (which remains an advantage in numerical calculations). Also the form is such that you can immediately see what the effects of changing the driving forces are. The driving force contains the effect of non-idealities. You can extend the equations to include forces such as electrical and pressure gradients, but these other forces do change the values of the L coefficients. In TIP, the cross-coefficients are equal, so only three coefficients are required. However the complicated behaviour of the coefficients remains. Also the behaviour depends on which components are chosen as independent (Figure 13.7; next page). The difference equation in the TIP description usually has a smaller range of validity than that of the following Maxwell-Stefan description.

13.5 The Maxwell-Stefan Description You already know the behaviour of the Maxwell-Stefan diffusivities (Figure 13.6). There are just three of them. They are independent of composition and the same for the three choices of independent components. Many people believe that the Maxwell-Stefan equations are difficult to use. We hope to have shown you that this is not at all so if you accept a few approximations.

o 1-1'----+--+--'1

o three diffusivities o positive constants o independent of sequence

-10 Fig. 13.6 One single set of Maxwell-Stefan diffusivities

13. Why we use the MS-equations

H2

+10

o

1

C62F21 -10

"'--_ _ _ _----"

+10

Ao 1

H2121 N2 -10

Fig. 13.7 The three sets of TIP phenomenological coefficients

151

Mass Transfer in Multicomponent Mixtures

152

The pros and cons of the three descriptions are summarised in Figure 13.8. The Fick and TIP systems are occasionally simpler, but mostly we prefer the Maxwell-Stefan system. Fick

MS

TIP

easily extended to other driving forces

0 0 0

0

number of ternary coefficients

3

3

4

coeff s independent of driving forces

0 0 0 0

0

simple behaviour of coefficients independence of reference frame

coefficients independent of sequence fluxes explicit integration with thermodynamics looks like 'chemical engineering'

>

0

0

Fig. 13.8 Advantages and disadvantages of the three descriptions

13.6 Units Not only can we choose different systems to describe diffusion; there are also many different ways of choosing compositions. Three of these are covered in Appendix 2 on Units. They use: 1. mole fractions (as in the greater part of this book), 2. mass fractions (when mass transfer is combined with hydrodynamics) and 3. volume fractions (which often show diffusion theory most simply).

13.7 Further Reading Kuiken, G.D.C., 1994, Thermodynamics of irreversible processes: Applications to diffusion and rheology. John Wiley, Chichester. This book presents a rigorous treatment of irreversible thermodynamics and provides a theoretical foundation for the Maxwell-Stefan approach. Taylor, R. and Krishna, R. (1993) Multicomponent Mass Transfer. Wiley, New York. Interrelationship between the Fick, Maxwell-Stefan and TIP approaches.

13. Why we use the MS-equations

153

13.8 Exercises The first three files are not really exercises. They just demonstrate that the different descriptions of mass transfer, described in this chapter, give the same results. You can skip them if you believe us! 13.1 The Effect of the Component Numbering on the Fick Diffusivities (Mathcad). Shows that just changing the sequence of the species changes the signs of the Fick cross coefficients (as in Figure 13.5). 13.2 MS and Fick are the same (Mathcad). Shows that you get the same fluxes with the two sets of equations, independent of the input values. You will see that the transformation between the two descriptions is quite complicated, even for this ternary mixture of ideal gases. We do not derive the transformation; we only use it. 13.3 MS and TIP are the same (Mathcad). Shows that you get the same fluxes with the two sets of equations, independent of the input values, just as in the previous example. 13.4 Molar, Mass and Volume Based Diffusivities (Mathcad). The Maxwell-Stefan equations can be written in different units, as explained in the Appendix on 'Units'. In this exercise you investigate how the choice of units influences the dependence on composition of the Maxwell-Stefan diffusivity in a binary mixture.

Mass Transfer through a Solid Matrix

156

1 Solid Matrices In the second part of this book, we consider mass transfer through mixtures which contain a solid matrix. This chapter introduces the subject and paves the way.

14.1 The Applications Permeable matrices have a great number of applications in technology. The most important ones are: • as membranes in separation technology, • as sorbents in adsorption and chromatography columns, and • as heterogeneous catalysts for chemical reactions. This book is on mass transfer, and not on the separate applications. So we will not be discussing these technologies systematically: only as examples of mass transfer problems. However, to put the examples in some perspective, we give a brief description of the three technologies and their classification. We begin with membrane processes.

14.2 Membrane Processes A membrane module consists of (at least) two compartments, separated by the membrane (Figure 14.1). The feed enters the upstream side and splits into a retentate and a penneate flow. These flows should have different compositions. How different, depends on the membrane, the driving forces used to transfer matter through the membrane, and the properties of the components in the feed. The components moving through the membrane are called penneants. feed

drivingforce

retentate

membrane

permeate

Fig. 14.1 A simple membrane module

Membrane processes can be roughly classified according to the main driving force in the process (Figure 14.2). The figure has three rows. Processes in the top row are mainly driven by composition gradients, in the second by electrical gradients, and in the third by pressure gradients.

14. Solid Matrices

157

Of the composition-driven processes, dialysis is the oldest and most important. In dialysis the solvent (usually water) is the same on both sides of the membrane. Dialysis has found considerable application in the laboratory, in medical techniques and in the food industry. In pertraction, two different solvents are used. Pertraction has much in common with the process of liquid-liquid extraction. It is not (yet) of great significance. Of the electrically driven processes we will only discuss electrolysis, membrane electrolysis and electrodialysis. Electrodialysis uses two kinds of membranes: • positively charged membranes which only transfer negative ions, and • negatively charged membranes which transfer positive ions. Electrodialysis is used on a fairly large scale, both for purifying and for concentrating electrolyte solutions. driving forces

C;<'YXD

pertraction

dialysis

(ed) electrodialysis

salts

~ -~1IIIII1IIII1I •.:'.-,~-,.I11III"'1

(gs) gas separation

(pv) pervaporation

(ro) reverse osmosis

proteins

microorganisms <=> 0

i> i. Q,o ••• 1I . . . . (.' , L,

(ut) ultrafiltration

(mt) microfiltration

Fig. 14.2 The three groups of membrane processes

Nowadays the most important membrane processes are those driven by pressure differences. These can be subdivided according to the size of the species that they separate. Small molecules (sizes smaller than about one nanometre) can be separated with tight (but very thin) polymer membranes. An important example is gas separation. Pervaporation is similar to evaporation. The process is operated with a pressure difference over the membrane. This difference is, however, smaller than that in the other pressure-driven processes. At the pressure side is a liquid, at the other side a vapour. The membrane can greatly modify the selectivity of the evaporation process. Pervaporation is used for breaking azeotropes which cannot be separated by conventional distillation.

158

Mass Transfer in Multicomponent Mixtures

We can make fresh water from salt water using reverse osmosis. The membranes here are somewhat swollen, but still sufficiently tight to avoid passage of the hydrated salt ions. Molecules in the colloidal range (roughly between one and one hundred nanometres) are separated by ultrafiltration. Ultrafiltration membranes are sieves having fairly well defined pores. Ultrafiltration has found many applications in biotechnology and water treatment. Microfiltration retains again larger particles. The distinction between ultrafiltration, microfiltration and conventional filtration is a little arbitrary. They all operate via the sieving action of: • the membrane itself, or • a cake or gel layer formed on the membrane.

14.3 Adsorption and Chromatography These processes are popular for removing or separating traces of materials. They both use a column filled with sorbent particles; the mixture flows through this 'bed' of particles (Figure 14.3). Different species have different interactions with the bed, and so move with different velocities. feed

not to scale Fig. 14.3 Bed of particles for adsorption

In all sorption processes it is important that species can move rapidly in and out of the particles. Because of this, sorbents can be swollen polymers or they may contain both large macro-pores and small micro-pores. The large ones look after transport; the small ones provide a large surface area for molecular interactions. The particles used mostly have a size of around one millimetre; they might be spherical or cylindrical (extrudates). The different interactions of the components with the sorbent are caused by four mechanisms: differences in size, differences in charge, differences in polarity, and differences in shape (Figure 14.4). Often, one of these mechanisms dominates in a certain separation, but usually several of them play a role simultaneously.

159

14. Solid Matrices

In size exclusion chromatography, the pores are small enough to exclude certain species. This mechanism is also important in molecular sieves, such as zeolites. Ion exchange uses a charged matrix, which has a different affinity for different ions. Classical adsorption on activated carbon is mainly governed by polarity differences; shape differences are important in separations using molecular sieves and in affinity chromatography .

•

• size

charge 0

Q

0

QO

0

e¥

0

• polarity •

0

o shape ..

0

Fig. 14.4 Four separation mechanisms in sorption

A chromatography column uses a constant flow of a carrier fluid (Figure 14.5). At a certain moment, we inject a small sample of the feed into the carrier; the different components migrate with different velocities and after a sufficient length of column we can withdraw them separately. Adsorption (Figure 14.6) is similar except that we load the column for a longer time, and then separate the components using special washing and elution streams. Adsorption makes better use of the capacity of the column material than chromatography, but is not as flexible. Sorption processes are non-steady; the composition of the fluid around the particles changes during the separation process.

Mass Transfer in Multicomponent Mixtures

160

~I feed (as a 'pulse')

WB

different velocities

1 B

concentration

A

11

carrier liquid

~time

Fig. 14.5 A simple chromatography column

~

~

feed ---+

B

B

A

regenerant

A

A

~

.

-----------------~)t1me

Fig. 14.6 Separating A and B with adsorption followed by washing and regeneration

14.4 Heterogeneous Catalysis Porous particles are also used as the support for industrial catalysts. Two groups of reactors in which they are used are packed columns, and fluidised or slurry reactors (Figure 14.7). The packed columns are similar to those used in adsorption, but the particles may be larger: up to about one centimetre. In fluidised beds and slurry reactors, the particles are suspended. Particles used here are much smaller; they might be around one tenth of a millimetre in diameter. In heterogeneous catalysis, we have mass transfer combined with a chemical reaction. So the fluxes in a particle depend gready on position. However, they are all coupled via the reaction stoichiometry.

14. Solid Matrices

161

I I

.' •• ·• ; •· .',. • ·• ••·•·•·• • •• ., • •• •• • • • · .•'•. • ••·•• • ,0.

;.

!,-,; •

',#I O'~:}',.t ~

".'1'~

packed bed

slurry reactor

",~.

·

fluidised bed

.':

.:

'~.:

':,' ,:e:",. .,> {'~~~

,,~'.

,~

I

Fig. 14.7 Catalyst particles in reactors

14.5 Structured and Non-structured Matrices From our applications, two main groups of matrices come forward (Figure 14.8). The first are the non-structured or homogeneous matrices, usually polymers, which are used in many membrane processes, but also in gel ion exchangers and other applications. The second group is that of the structured or porous matrices, such as used in most catalytic reactors and many sorption processes. Both are used in the form of membranes and of particles.

7::'::'::::

~ neutral

particle

membrane

structured (porous)

~.

.. • •

••

charged

~.t;.: :: ~~~~

• :: ®:: • • • :: ::

with macro and micropores

Fig. 14.8 Non-structured and structured matrices

Non-porous polymers are homogeneous (more-or-Iess) down to scales near those of ordinary molecules. In some ways they behave like viscous liquids. Polymers can be dense or swollen; the swollen ones are much more permeable than dense ones. Polymers may carry fixed charges on their matrix: these are opposed by mobile counter ions. The porous matrices have larger openings. These can be anywhere from a few nanometres up to one millimetre. The pores can have all kinds of shape; we will only consider cylindrical pores and the pores formed in a bed of spherical particles.

162

Mass Transfer in Multicomponent Mixtures

14.6 Effects of a Matrix on Mass Transfer Mass transfer through a matrix is not fundamentally different from that through a fluid. Also here, transfer is driven by the potential gradient of a species and limited by friction with the surroundings, but now including the matrix. However, there are a few complications: 1. The matrix, being a solid, can transfer forces to a support. As a result there can be pressure gradients, independent of body forces. These gradients can be large: they are the prime driving force in processes such as gas separation by membranes and reverse osmosis. As we have discussed in Chapter 4 (Figure 4.2), there are (n - 1) independent transport equations, so we leave one of them out. We usually omit the equation for the matrix. Then the support forces do not show up in our calculations (although we can always calculate them). 2. Pressure gradients (and also other unbalanced forces on the permeants) may cause viscous flow. The viscous velocities of the components are often about the same. This is not necessarily so; for example, a species adsorbing on pore walls may have a lower viscous velocity than species in the centre. ~

There are two common ways of modelling transport through a matrix (Figure 14.9). You should be aware of which you are using. Do not get the two mixed up! 1. In the first non-structured or 'homogeneous' model we regard the matrix with permeants as a single phase. All components (including the matrix) are thought to form a molecular mixture. 2. In the second structured model the matrix is regarded as a heterogeneous structure. A simple example is the model of parallel cylindrical pores through a plate of an impermeable solid. non-structured

N

j

= S,membrane(~ +V;! \ \

\

\ \

viscous velocities

structured (taking structure into account)

,,/"

Ni = fC;, pore (if; + il;)

"

/

velocities are the same Fig. 14.9 The two kinds of model of a matrix

Structured models are the natural choice for structured matrices. They allow some understanding of the influence of structure on mass transfer properties. Non-

163

14. Solid Matrices

structured models are more natural for polymers and gels. We shall see, however, that the two models are closely related, and give the same results i.

14.7 Compositions with a Matrix A mixture moving through a solid matrix can be described in several ways. As an example, we consider the movement of two permeants '1' and '2' through a matrix: (a) we can regard the matrix as a third component '3', or (b) we can regard it as separate 'M' (it is not part of the mixture). The friction forces exerted on species '1' in the two notations are: (a)

(b)

r-----------, r-----------, :X2Si,2(U1-Uz}tfjX3S1,3(U1 I I

:X S 2(U -

I I I I

-u3 ): I I

U2)~ SI M (U -U ) :

2 1 1 1 M 1 ,----:..-______ 1 ,__ .:.. __________

friction exerted friction exerted by by '2' on' l' the matrix on' l' Fig. 14.10 The two ways of describing the matrix

Each of the terms separately has the same size in both notations. Also the velocities are identical (including u3 =uM ). However, the numerical values of both the mole fractions and the friction coefficients differ. You can write the friction side of the multicomponent MS-equations in many ways. Figure 14.11 shows eight of them for the common situation that the matrix is stagnant. The first four are for the differential equation; they have the dimension of N mol,i. As above, the numerical values of transport coefficients, mole fractions and the total concentration depend on how you define the mixture: the values in the left and right-hand equations are different. The last four equations give four different ways of writing the friction terms in the difference equation. They are dimensionless, but otherwise similar remarks apply to them. As before, the velocity forms refer to one mole of species i; the flux forms to one mole of mixture. Which form of the equation that you use depends on the problem. In some mixtures it is difficult to define 'one mole of matrix'. You then regard the matrix as separate. The equations with velocities are easy to 'understand'; those with fluxes are often more practical in applications. You will find examples of most of these forms of the equations in the coming chapters.

1 We can already see an indication of this in Figure 14.9. The species velocities in the two models are, of course, the same. The concentration of a permeating species in a pore (used in the pore model) is higher than the concentration averaged along the membrane (as used in the non-structured model). However, in the first model, transport only occurs in the pores; in the second it occurs over the whole membrane and these two effects cancel.

164

Mass Transfer in Multicomponent Mixtures

matrix is a component differential equations

IJi,;X/U; -ui)

matrix is separate ~ ~I,l r .. x.(u· MU'1 £.. I I -u·)+(;· I I,

i'#

IJi,/x;Ni -XiNj)

;*i

difference equations

for a stagnant matrix

UM

= 0

Fig. 14.11 Some forms of the friction terms

14.8 How Further? We have subdivided the rest of the book in two parts: • Part 2a, mass transfer through polymers, and • Part 2b, mass transfer through porous structures. The first will deal with non-structured models; in the second we look at structured models. The part on polymers begins with a brief description of their properties and of the thermodynamics of mixtures containing a polymer (Chapter 15). This is followed by some information on diffusion through polymers (Chapter 16). Then we look at a series of applications: • Dialysis and Gas Separation (Chapter 17), • Pervaporation and Reverse Osmosis (Chapter 18), • (Membrane) Electrolysis and Electrodialysis (Chapter 19), and • Ion Exchange (Chapter 20). In each application, we look at the process and the mass transfer equipment; then at the equilibria; at which terms are important in the transport relations, and at some complications. Part 2b on transport through porous structures, starts with diffusion of gases in porous structures (Chapter 21). We apply the results to: • transport in heterogeneous ~atalysts (Chapter 22), and • transport in microporous adsorbents (Chapter 23).

14. Solid Matrices

165

In transport of liquids through porous media, viscous flow is often important, as we will illustrate with ultrafiltration (Chapter 24).

14.9 Further Reading Lightfoot, E.N. (1974) Transport Phenomena and Living Systems. McGraw-HiU, New York. Modelling of membrane transport. Mulder, M, 1991, Basic principles of membrane technology. Kluwer Academic Press, Dordrecht. A good primer. Ruthven, D.M. (1984) Principles of Adsorption and Adsorption Processes. John Wiley, New York. A thorough introduction to adsorption processes. Ruthven, D.M., Farooq, S. and Knaebel, K.S. (1994) Pressure Swing Adsorption, VCH Publishers, New York. A detailed account of the design ofpressure swing adsorption processes. Spiegler, K.S. (1983) Principles of Energetics, Springer-Verlag, Berlin, 1983. Nice examples on membranes, by one of the fathers of the frictional model of mass transfer. Yang, R.T. (1987) Gas Separation by Adsorption Processes. Butterworth, Boston. A good text on gas adsorption. Wijmans, J.G. and Baker, R.W., 1995, The solution-diffusion model: a review. J. Membrane Sci., 107, 1-21. A good review.

14.10 Exercises 14.1 The Two Descriptions. We are often asked why there are two different descriptions for solid matrices. Would it not be simpler to have only one? In this dialogue, you are to sort that out for yourself. Before we begin: the difference between these two descriptions is: (A) in the first, we regard the matrix as a species which is part of the mixture, and (B) in the second we regard the matrix as a separate entity. We begin by looking at two extreme cases:

166

Mass Transfer in Multicomponent Mixtures

(1) a solution of disaccharides (sugar) in water and (2) a bed of sand, with water in the voids. How would you classify these? Probably (1) = (A) and (2) = (B). However, do realise that the driver trucking wet sand will regard it as a mixture (and not worry about the structure). On the other hand, when you are estimating the friction factor between sugar and water with the Einstein-Stokes equation, you are (in a way) regarding the mixture as two separate phases. Now two less extreme examples: (3) a cross linked gel of polysaccharides in water, and (4) an adsorbent of agglomerated zeolite particles. What now? Here you may well have problems deciding. It might depend on what you want to do: (I) to model a separation process using the gel or adsorbent as an auxiliary phase, or (IT) to try to improve the structure of the gel or adsorbent. In the first case, you might choose for (A): in the second almost certainly for (B). You see that there are valid reasons for both descriptions to exist. A few more practical points: • Many membrane processes use dense membranes, where the volume fraction of the matrix is close to one. If you use a sensible definition of mole fractions, then this also applies to the matrix mole fraction: X3 ---7 1. What can you then say about the importance of the '12' friction terms? And what can you say about the relation between S1,3 and SI,Min Figure 14.1O? • It is sometimes difficult to assign a concentration to the matrix. You might regard polyethene as consisting of ethene units. But what is the molar concentration of soil? A final remark: in this book we consider four different descriptions of matrices. They are the structured and the non-structured models, in which we can either regard the matrix as a component or not. In practice only three models are used: in the structured models, the matrix is never regarded as a component.

167

Properties of Polymers In this chapter we have a brief look at some properties of polymers, and at the thermodynamics of permeants in polymers.

15.1 A Few Words on Polymers In any university library, you will find many metres of books and journals on polymers. We cannot cover these, but we wi11look at a few things that you need to get into the subject of diffusion of two or more permeants (small molecules) through a polymer. Terminology and Properties

Polymers are extremely long molecules. They consist of chain units, with dimensions similar to those of other 'normal' molecules. However, polymer molecules may contain thousands of such units (Figure 15.1). If they were to be fully stretched, they would have lengths in the order of micrometres. Actually, they are usually strongly coiled and intertwined, a bit like a bag of worms. The dimensions of a coil might be around ten nanometres.

*

",,10 nm

""1J1m

.(----------------------------------~)

( ) ~

0.3 nm

104 chain units

thin, flexible polyethene

H H H H H H H H I -C-C-C-C-C-C-C-C~oar H H H H H H H H hydrophobic

thick, rigid cellulose

polar hydrophylic

~

long molecules

coned~

Fig. 15.1 Properties of polymer molecules

168

Mass Transfer in Multicomponent Mixtures

The number of possible variations in polymers is almost unlimited. The chain units can be small, such as in polyethene. These produce thin flexible polymers. If the units are bulkier, such as in cellulose, the polymers can be much stiffer. Apolar chain units (such as ethene and other hydrocarbons and fluorocarbons) produce hydrophobic polymers, which hardly interact with water, but often swell strongly in apolar solvents. Polar polymers such as cellulose, which contain large numbers of hydroxyl groups, swell strongly in water, and may even dissolve. We can have polymers consisting of different chain units (copolymers) and also mixtures of polymers. A cross-linked polymer forms a three-dimensional network with a molar mass that is practically infinite (Figure 15.2). Cross-linked polymers can swell in a solvent, but they are not soluble. This is important for many applications, for example in the membrane processes that we will be looking at in Chapters 17 ... 19. Cross-links need not be chemical; entanglements and crystallites (to be discussed in a moment) can also provide cross-links. entanglement

H H H H H -C-C-C-C-CH H H I ~~;.H HCH

H I H H H H

-C-C-C-

Fig. 15.2 Cross-links in a polymer. (Polymers are much denser than shown in this drawing.)

All polymers are at least partly amorphous; in the amorphous regions the molecules show little ordering. However, polymers often contain substantial 'crystalline' parts, where the polymer chains are more or less aligned (Figure 15.3). The crystallinity varies with the type of polymer, but it also depends on the history of the polymer, and it often changes in time. Also the size and shape of the crystalline areas can depend on the history of the polymer, and not only on local composition, temperature and pressure. These inhomogeneities have dimensions that are small compared to that of a polymer film or particle; the dimensions might be around ten nanometres. The crystalline areas are typically a tenth denser than the amorphous parts; for many species they are practically impermeable. So diffusion occurs mainly in the amorphous regions in a polymer, where there is some space between the polymer chains and also more chain mobility. These amorphous zones are not homogeneous themselves; they contain open and dense regions. Diffusion occurs primarily through the most open parts.

169

15. Properties of Polymers

amorphous

with some space open, that allows motion of permeants

o< crystallinity < 1 depends on the polymer and its history

dimensions"" 10 nm

Fig. 15.3 Crystalline and amorphous regions in a polymer

Phase Transitions

Figure 15.4 shows the behaviour of one of the many properties of a polymer: the modulus of elasticity. There are two sharp breaks indicating phase transitions. At low temperatures the polymer is rigid and brittle: it forms a glass. At the glass transition temperature Tg the modulus drops dramatically, here by a factor of ten thousand. Many of the properties of the polymer change a little at this temperature. Above Tg the polymer becomes soft and elastic; it forms a rubber. At high temperatures, the polymer may melt, to form a viscous liquid. This does not occur when the polymer is cross-linked. (A cross-linked polymer will break down thermally if it is heated too far.) The polymers that we know as glasses, such as polystyrene or window glass, have a glass transition temperature above ambient temperature. Substances such as silicone rubber and the polymer used in an eraser have a Tg well below ambient temperature. It is important to keep the distinction between glasses and rubbers in mind; the behaviour of diffusivities in these two phases is very different. 1010

/glass transition

E

Nm-

2

108

/

glass

modulus of elasticity

rubber

106

Iliquid T ----------- melting

T

m

temperature with cross

....:_--- .....

104

rigid

creeping flow

102

200

300

linking

400

temperature T 500 K

Fig. 15.4 Modulus of elasticity against temperature, showing the glass transition and melting temperatures

Mass Transfer in Multicomponent Mixtures

170

The glass transition temperature of a polymer is not as well defined as phase transitions in simpler substances. This is not surprising: polymers seldom consist of a single species. They are mixtures of species with different molar masses. However, there is more to this. If we start with a rubber, and cool it, we find that the glass transition temperature depends on the rate of cooling (Figure 15.5). The slower the cooling, the lower the glass point. Fast cooling yields a structure with fine crystals that freeze in below the glass transition temperature; slow cooling gives a coarser and more crystalline structure. Only with extremely slow cooling can we expect to obtain an equilibrium structure. There are two lessons to be gleaned from this: • polymers can have a non-equilibrium structure frozen in, so that • their properties can depend on their history (and not only on composition, pressure and temperature). At temperatures above the glass transition point, polymer chains equilibrate rapidly (although not as quickly as simple liquids). Here, properties depend little on the history of the polymer. (At least the local properties do not. Large inhomogeneities will not disappear in a short time.) Near Tg the time needed to attain equilibrium is similar to the time of many diffusion processes, and diffusivities vary with time. Well below Tg , the properties do depend on the thermal and mechanical history of the polymer, but otherwise hardly change in time. 10 10 modulus of elasticity 108

slow cooling: lower glass point

E

Nm- 2

106 temperature 104 200 relaxation times

~

long

300 •

I I

III

:

between

T 500 K

I I

:400

short

Fig. 15.5 Properties are 'frozen in' at the glass transition temperature and depend on the rate of cooling (and the mechanical history of the polymer).

Polymers often have irregular internal stresses, and regions where the polymer is more or less aligned. In these regions you must expect diffusivities to depend not only on position, but also on the direction of diffusion.

15. Properties of Polymers

171

We hope that this overview gives you some idea of the difficulties in developing the theory of both multi component equilibrium thermodynamics and diffusion in polymers: • the number of polymers and polymer combinations is almost unlimited, • polymers do not consist of one (or a few) well defined species, • the properties of a polymer can depend on its history, • the properties can vary with position, • they can depend on the direction of diffusion, and • the properties of a polymer may vary in time. The last point is one of the reasons why membrane processes take some time to settle down. Also, if conditions cycle, you may find hysteresis: the process does not return to the same condition in each cycle. All this is not quite as bad as it might seem. Many technical applications only use a few standardised polymers under fairly constant and well-defined conditions. We do our best to avoid inhomogeneities or to control them. At temperatures far above or far below Tg , properties hardly vary in time. While many membrane processes use polymers near their Tg , they work at a steady state; after some time each part of the membrane then goes to equilibrium. In the rest of this book, we will regard polymers as materials with properties that do not change in time.

15.2 Thermodynamics of Mixtures in a Polymer Because a polymer is often not at equilibrium, you might wonder whether thermodynamics would be applicable. (We also wonder occasionally ... ) However, it turns out that thermodynamic models do yield useful insights in the behaviour of polymers and permeants (certainly for liquid and rubbery polymers). We will need a model anyhow for the driving forces in the Maxwell-Stefan equations. Figure 15.6 shows one of them: that of Flory and Huggins (FH). Flory-Huggins equation

The FH equation in its simplest form deals with molecules that are similar chemically, but differ greatly in length. An example might be polyethene with the solvent propane. The model is based on the idea that the chain elements arrange themselves randomly (but with the molecules remaining connected) on a threedimensional lattice. It does not take effects of crystallisation or other inhomogeneities into account. The resulting equation for the activity of the solvent is a simple function of the volume fraction! of the solvent. For low values, it is

I Books on polymers use potential.



for the volume fraction, but this clashes with our symbol for the electrical

172

Mass Transfer in Multicomponent Mixtures

proportional to this volume fraction. The activity of the solvent is independent of the length (or molar mass) of the polymer, as we might have expected. (This only applies if the length of the polymer chains is large, but this is usually the case.) In this graph of activity versus volume fraction, the mixture behaves as non-ideal, but not strongly. The activity in dilute solutions is e (2.71) times higher than it would have been if activity had been proportional to the volume fraction. The FIory-Huggins equation does not consider volume changes on mixing. You can take the volumes in the formulae equal to those of the pure components .

I1

...

"..

. ...

i-

r=

-

r=

11 i-

/

polymer (1)

........

I I:: F"

.., In

_.... v

solvent activity 10°

III

solvent (2)

polymer and solvent fit on a 3D lattice

10-2 10- 2

E2~

le

volume fraction Fig. 15.6 Permeant activity according to the Flory-Huggins equation

Mole or Volume Fractions?

We can use the FH equation to calculate the activity of the propane in polyethene as a function of its mole fraction. The mole fraction is much higher than the volume fraction; how much depends on the ratio of the molar volumes of polymer and solvent. We show the results in the lower part of Figure 15.7, for polymers with a molar volume that is one hundred, one thousand and ten thousand times that of propane. The curves show a very strong non-ideality. Activity coefficients are a few hundredths, a few thousandths, or a few ten thousandth for the three curves. We see that non-ideality depends on the way that we represent composition. Clearly, the mole fraction is not a good measure of composition for mixtures of molecules of greatly differing sizes. The volume fraction is a more natural variable. To get around these problems in our diffusion equations, we can do several things: • write the equations in terms of volume fractions, • write the equations in terms of mass fractions, or

15. Properties of Polymers

173

• define mole fractions using the mole numbers of the chain units instead of those of the polymer. The three methods yield similar results. (If the solvent and chain unit have the same size and density, they are identical.) We show how to set up different methods and to find the relation between the resulting diffusivities in the Appendix 2 on 'Units'. There are more methods for dealing with mixtures containing a solid matrix. We will see many examples in the coming chapters. Here, we simply go on using volume fractions. volume fraction

10-2 £2 ~ 10° 10-0 , - - . , . - - - , . - , ------ ---- activity is a simple function of the volume fraction

solvent activity 10-2

strong volume and nonideality efficts if we use mole fractions

mole fraction

10-6 ' - _ . . . L . - _ - - ' 10-2 X2~ 10°

Fig. 15.7 Mole fractions are not a good choice for composition

Non-ideality

Polyethene swells in propane, but hardly in water. The reason is that the energy of a water-water bond is much lower than that of a water-methylene group. So water will hardly dissolve in polyethene. The complete FH-equation contains a term which takes this into account approximately (Figure 15.8). This equation contains a nonideality parameter X, which is similar to the parameter A that we introduced in Chapter 8. The effect of this term is shown at the right side ofthe figure. We see that the behaviour of the activity of the solvent is similar to that in non-ideal liquid mixtures (be it, that we use volume fractions here and mole fractions there). If X is positive, the solution can split into two phases, one rich in polymer and one rich in solvent. The equation always yields some solubility of the polymer in the solvent; it is not applicable to a cross-linked (non-soluble) polymer. We can extend

174

Mass Transfer in Multicomponent Mixtures

the equation to three or more species (but we do then need more xparameters, one for each pair of species).

~=

101

c2

ex p[

(I-\'\(~,}l + xel J \1, solvent /~

//(' // ///

activity

i

a2

demi~ing!

"':::'"

//

non-ideality parameter

X>o

2

10°

I

solvent and polymer dislike each other

X=o

solvent and polymer are similar

X
solvent and polymer attract each other

0

-I i

10-1

X 10-2 10-2

£2~

10°

volume fraction Fig. 15.8 Non-ideality in the FH equation

Swelling

If we contact polyethene (with a not-too-high molar mass) with propane vapour, it takes up propane until the activities of propane are equal in the two phases. As the propane fraction rises, the polymer transforms from a glass into a rubber, from a rubber into viscous liquid and finally we end up with propane liquid at the propane vapour pressure. We see that addition of a solvent has an effect on the polymer similar to that of raising temperature. The solvent lowers the glass transition temperature. This is known as plasticising. If the polymer had been cross-linked, the polymer would not have dissolved. A simple extension of the FH equation takes cross-linking into account (approximately). The coiled polymer strands pull the matrix together, causing an internal 'swelling pressure'. The value is high: often of the order of 10 MPa. Solvent entering the polymer has to overcome this pressure. This yields the swelling term in the exponent in the extended equation (Figure 15.9). We see that a large molar volume of the solvent, and a high cross-link density increase the activity of the solvent. In this equation, we need the molar volume of the chain unit Vc For a cross-linked polymer, there is a maximum degree of swelling in a solvent. At this point, the activity of the solvent inside the polymer is equal to that outside. Figure 15.10 shows the swelling for different values of B. We can use this figure to estimate the value of B from swelling data.

15. Properties of Polymers

175

1

B""'-nc

7

number ofchain units between each cross link Fig. 15.9 Extension of the FH equation to allow for cross linking

gels, protein separation media (electro) dialysis, ion exchange reverse osmosis

pervaporation

gas separations (high pressure)

Fig. 15.10 Volume fraction of solvent in a swollen cross-linked polymer

In the figure, we have indicated the range of solvent volume fractions for different

applications. These range from extremely high in gels down to low in gas separation membranes. Distribution of Two Components

With the FH equation, we can calculate the equilibrium distribution of a species between a liquid or gas, and the polymer. We first do this for a vapour at low pressure (Figure 15.11). The activity of the vapour is the ratio of the partial pressure to the vapour pressure of the species. Only a little vapour dissolves in the polymer and the activity there is given by the second formula. Equating the two activities shows that the volume fraction of the vapour in the polymer is the ratio of partial to vapour pressure, multiplied by a constant. This constant - the Henry coefficient - is small if the non-ideality parameter is large. It is also strongly influenced by cross-linking and the size of the permeant. If we have two gases, both at low pressures, they dissolve independently. This is the situation in

Mass Transfer in Multicomponent Mixtures

176

most gas separation membranes. Only when their volume fractions start to exceed a few percent, do the solubilities begin to influence each other. activity of vapour '2'

activity of '2' in polymer

~ =C2 exp[l + X+B~;] (Cl -t 1) equilibrium

Henry coefficient Fig. 15.11 Distribution of solute between a vapour and a polymer

In our last example, we consider a polymer strongly swollen in a solvent. To keep the formulae short, we only consider the case where the non-ideality parameters are zero. The polymer might be a dialysis membrane. The swelling is given by Figure 15.10 as we have already discussed. Now we add a trace of a solute '3' to the liquid. The activities of '3' in the two phases are then as shown in Figure 15.12. The exponents are constants for a given system. Equating the two activities tells us that the volume fractions of the solutes in the two phases are proportional to each other. Their ratio the Nernst coefficient - depends on the sizes of the three species and the crosslinking. In the general case it also depends on the sizes of the non-ideality parameters. ;'''

if~O L;, solvent (2) .e;large·· .. .

.

solute (3)

e;small

a3

(1- ~)c2 ]

11'.,11 11.1

'"

e1 polymer(l)

~

....

- . ' ... -e

2,·

".,.~

.

.

--

3

(

·1\

swelling: Fig 15.10

equilibrium calculation:

a] = C] ex p[

l1';;'" "

a; =a3 ---e--------...

~23

~c3 c:/3] ~ et = constant

Y'

y,

= c3 ex p[c1 + (1- )C2 + B

volume distribution (Nernst) coefficient

Fig. 15.12 Distribution of a solute between swollen polymer and solvent

15. Properties of Polymers

177

15.3 Summary • In the first part of this chapter we have had a look at a number of properties of polymers. We have seen that they are mainly amorphous, but can contain crystalline regions. They often show a phase transition from a rubbery to a glassy phase. • In the glassy phase, very little motion is possible; the structure of the polymer is frozen in. Its properties depend on history, but hardly on time. • In the rubber phase, motion is fairly rapid. Rubbers can be locally in equilibrium during non-steady diffusion processes. • In the region around the glass transition temperature, relaxation times of polymer chains are of the same order as times in non-steady diffusion. Here, diffusivities will depend on the history of the polymer, but they also can vary in time. • In the second part, we have looked at thermodynamics of polymer solutions. We have seen that mole fractions based on polymer molecules lead to a very distorted picture of composition. Volume fractions, mass fractions or mole fractions based on polymer chain units are better. • The Flory-Huggins thermodynamic model can easily incorporate non-idealities and the effects of cross-linking. The modified forms allow us to calculate swelling and the distribution of permeants between liquid (or gas) and polymer.

15.4 Further Reading Young, R.I and Lovell, P.A. (1991) Introduction to Polymers. 2nd edition, Chapman & Hall, London. A good overview. The binary form of the Flory-Huggins theory starts on page 130. A method for estimating the X parameter is given on page 149. Ugelstad, I.et al. Thermodynamics of Swelling, Makromol.Chem. Suppl. 10111 (1985), 215-234. Covers the multicomponent version of the Flory-Huggins equation. Gusler, G.M. and Cohen, Y. Equilibrium Swelling of Highly Cross-Linked Polymeric Resins, Ind Eng Chem Res (1994), 33, 2345-2357. A good overview of the subject of swelling. It includes methods for estimating our parameter B.

15.5 Exercises 15.1 Activity and Activity Coefficient. We begin with the equation for the chemical potential in Figure 3.2. We see that the activity is equal to aj = YjXj, or for an ideal solution aj = Xj. Now we look at the activity formula for the Flory-Huggins equation in Figure 15.6.

178

Mass Transfer in Multicomponent Mixtures

(a) If the two molar volumes are equal, what kind of behaviour does the FH equation predict? (b) The mole fractions and volume fractions in a binary are related by equations such as: X2

Work out the activity coefficient as a function of composition for component '2' when "1/V2 = 2. (c) Compare your result with the activity coefficients in Figure 8.4. What do you notice? 15.2 Mole Fractions and Volume Fractions. An oligomer has a molar volume nine times larger than that of a solvent: "1/V2 = 9. In a mixture of the two, the volume fraction of the solvent is £2 = 0.5. (a) What is the mole fraction of '2'? (b) And for "1/V2 = 99? (c) For "1/V2 = 999? 15.3 Non-ideality. We again consider a mixture of two species with the same molar volumes "1 = Vz, so £2 = x2' What does the Flory-Huggins equation with a nonideality term in Figure 15.8 tell us about the activity coefficient as a function of composition? Compare the resulting equation with that in Figure 8.4. Hint: look at Figure 3.2. 15.4 Swelling. A polymer (1) swells in its monomer (2). You may expect the nonideality parameter %1,2 to be close to zero. The volume fraction of monomer in the polymer at equilibrium is £2 = 0.2. We then contact the pure polymer with another solvent '3'. The polymer now swells to a volume fraction £3 = 0.03. Finally we bring the pure polymer into contact with a vapour mixture. This consists of an inert gas '4' which does not dissolve, and the vapour '3'. The partial pressure of '3' is one half of its vapour pressure. All the above experiments are done at the same temperature. (a) What is the value ofthe swelling parameter B(V2/VC)? (b) What is the value of the non-ideality parameter %1,3? (c) What will the volume fraction of solvent '3' be in the polymer in equilibrium with the vapour mixture? Hint: use Figures 15.10 and 15.11. 15.5 Distribution of a Solute between Swollen Polymer and Solvent (Mathcad). A polymer' I' is swollen in a solvent '2'. This contains a trace of a solute '3'. All three

15. Properties of Polymers

179

species have a similar chemical structure, so you may take the X parameters equal to zero. This is the situation of Figure 15.12. The swelling parameter has a value: B(V2/VC) = 1 (note that the first volume has a subscript '2', and not '3' as in the figure). Plot the volume distribution coefficient of '3' as a function of the ratio l-3/Vz. Which species are taken up preferentially by the polymer: the small or the large ones?

180

1 Diffusion in Polymers The subject of multicomponent diffusion in polymers is not well developed. We can only paint a rough picture using (too) simple models and some speculation. Some day, we will have to rewrite this chapter.

16.1 Behaviour of Diffusivities There is much on diffusivities of a single permeant in a polymer. (There is much less on simultaneous diffusion of two or more permeants, and what is available contains insufficient information to assess Maxwell-Stefan diffusivities.) We present a small selection, to show some of the main trends. To interpret the measurements below, we have used mole fractions with the polymer consisting of chain units. (In the examples here, mole, mass or volume fractions give almost the same results.) Glassy and Rubbery Polymers

Figure 16.1 shows diffusivities of a trace of benzene. These are in a series of polymers with different glass transition temperatures. The figure suggests a strong correlation, but we warn you to have a good look at the vertical axis. The diffusivities vary by a factor of 10 10 , so over ten decades! The variation of diffusivities in polymers is very, very large. You cannot expect any general correlation to predict such variation with much accuracy. What we plot along the bottom axis is the difference between the temperature at the measurement (300 K) and the glass transition temperature of the polymer. To the left of the zero point we have glassy polymers, to the right rubbery polymers. We see that diffusivities tend to increase very rapidly above the glass transition point. The highest diffusivity shown (that in silicone rubber) is similar to that of benzene in normal liquids. Except in extremely thin layers, glassy polymers are impermeable for all but the smallest molecules; here we only consider them briefly. A good look at the figure shows that some diffusivities lie substantially outside the band drawn. The diffusivities also appear to depend not only on the glass transition temperature, but also on other details of the structure of the polymer. Poly(isobutene) is an extreme example; a little thought will tell you why it is such a popular material for bicycle tires.

181

16. Diffusion in Polymers

glassy polymer

----.) rubbery polymer

<1110(--I I I

10-9

.0 --- -

silicone rubber

0------------.

poly-isobutene

I

T=Tg

-£)1,2 m 2 S-I

···· ...·0 <00

: I

10-12

0

I

:

p

0

o.

: I I

10-15

I

:' 0

j

I I

:;

r

10-18

9--------------------------

poly-vinyl-acetate

I

o

100

200

T-T --g

K

temperature difference

Fig. 16.1 Diffusivity of a trace of benzene in different polymers at 300 K. The effect of the glass transition temperature

Effect of the Size of the Permeant

The effect of the size of the permeant is also extreme (Figure 16.2) at least near and below the glass transition temperature. The diffusivities of the smallest permeants are much bigger than those of the larger ones. (Almost) glassy polymers have a low permeability, but an attractive size selectivity. The size selectivity is much less pronounced in the rubbery polymer (although it is still important). hexane octane o ...0 hexadeaane ,0

10-9

o

methanol

---------lO-15

rubbery polymer: swollen pp at 328 K

o propanol

o propanon

almost glassy polymer: PVAat313 K

o benzene 18

10-

molar volume of permeant

11;

Fig. 16.2 Effect of permeant volume in a rubbery and an almost glassy polymer

Mass Transfer in Multicomponent Mixtures

182

There may appear to be a correlation not only with size, but also with polarity of the permeant. However, the effects are mainly size effects, as other data show. Effect of Swelling Swelling usually increases diffusivities in a polymer. Figure 16.3 shows two examples. Again the effect is much larger in the more glassy polymer. (This is largely because poly(vinyl acetate) becomes more rubbery as it swells in the solvent.) The change in the diffusivity is often exponential in the volume fraction of the permeant, at least over not-too-Iarge variations. This means that we can use logarithmic interpolation, as with simple liquids, but not over the whole concentration range. 10-9 polymer-permeant diffusivity

0

DI,2 m

2 S-I

0

00

0

rubbery

10-12

00 0

10-15

0 0 0

almost glassy

10-18 0.0

0.1

volume fraction of permeant

0.2 £2

Upper curve: polyethene with i-octane at 300 K. Lower curve: poly(vinyl acetate) with benzene at 318 K. Fig. 16.3 The effect of swelling

16.2 The Free Volume Theory We end this chapter with a theory that predicts the phenomena that we have seen, also for mixtures with more than two components. This free volume theory is based on loose arguments, but it yields formulae with the right shape and (more than) enough adjustable constants. When 'Holes' occur The central idea in the free volume theory is that a molecular mixture contains 'holes'. These are openings with a volume of the order of that of a molecule. A molecule can only move if there is a large enough hole next to it. Both diffusion and viscous flow require such holes.

183

16. Diffusion in Polymers

We begin with a pure fluid (Figure 16.4). At a low temperature, or high pressure, the fluid will assume a closely packed structure, which contains no free volume. (For >imple fluids, you can estimate this minimum volume from the liquid branch of their ~q~ation of state. It is usually about one quarter of the critical volume.) At higher :emperatures (or lower pressures), the fluid expands and we get free volume. The Tee volume is an almost linear function of temperature. As we shall see, only a small lart of the free volume is actually accessible to diffusing molecules. )tatistical models with spherical particles tell us that the free volume distributes to ~ive holes with dimensions similar to those of the particles. The chance of finding a lole larger than a certain volume is a simple function of the free volume per particle. IVe are interested in the fraction of the volume consisting of holes larger than our nolecules. This is the volume available for diffusional motion. You should have a ~ood look at the formula for the available volume in Figure 16.4 as it plays an mportant role in the rest of this discussion. Within the exponent, we see a minus sign nd a fraction. The molecular volume is on top and the free volume per molecule dow. If we plot this available volume against the ratio of free volume to molecular olume, we see that it stays near zero until the free volume exceeds about one quarter f the molecular volume. It then begins to rise rapidly. closely packed minimum volume Vo (per molecule) this distributes itself as 'holes ':

actual volume

v

........

free volume vf

o

T

chance of a hole> Vo

-I-_--.-''---,_ _ V f

o

0.25

Vo

Fig. 16.4 Free volume and holes in a pure fluid

lass Transition and Self Diffusion

"e can also plot the available volume against temperature (Figure 16.5). The mperature at which it begins to rise rapidly is the glass transition temperature. ~low this temperature there is insufficient available volume to allow much motion. le fluid will behave as a solid, and there will be little diffusion. Above the glass msition point this changes rapidly. Simple liquids usually crystallise on cooling

Mass Transfer in Multicomponent Mixtures

184

and do not show a glass transition temperature. However, with precautions, many fluids can be sub-cooled to form viscous liquids and glasses. These glasses have an amorphous structure, like that of the liquid frozen in. The free volume theory predicts the glass transition temperature fairly well. fraction of volume 'available'

little internal motion

-+-~-f----

0/1

T

internal motion possible

Tg

glass transition temperature Fig. 16.5 The glass transition temperature

Our ideas are useful in predicting the self-diffusivity of our fluid (Figure 16.6). This is a measure of the local rate of diffusional mixing of molecules. You can determine it by following a radioactive tracer '1-' with properties identical to those of the fluid '1' .

A slightly artificial, but simple model of diffusion assumes that molecules are arranged on a lattice. After a certain period (extremely short in practice), they can move one position. This happens randomly in one of six directions: up, down, forward, backward, left or right. This happens many, many times per second and causes diffusion. The theory tells us that the diffusivity on the lattice is proportional to the frequency of the steps and to the square of the lattice distance. We can also write this as the product of displacement distance and the velocity. For molecules in a fluid, we would expect the distance to be roughly the molecule diameter. The velocity should roughly equal the thermal velocity. In the free volume theory, we assume that a displacement on the lattice will only take place if there is a free position (a hole of sufficient size) next to the molecule. The combination of these two ideas yields the formula in Figure 16.6. This formula gives reasonable estimates of self-diffusivities. Note that the formula contains the ratio of two volumes. This allows us to replace the molecular volume and free volume per molecule by molar volumes. In a Ternary Mixture

We will work out the theory for a ternary mixture; the ideas can be extended to any number of components. For mixtures, we need two more assumptions. The first is that all species in the mixture contribute to the free volume; it is the sum of the free

16. Diffusion in Polymers

185

volumes per mole multiplied by their mole fractions (Figure 16.7). With this assumption, we can calculate the diffusivity of any tracer species '1·', '2·' or '3·' through a stagnant mixture of '1', '2' and '3'. These effective diffusivities are a combination of the different Maxwell-Stefan diffusivities in the system, including the self-diffusivities. The further calculations are more easily done with friction coefficients. 'self diffusivity' of tracer identical to '1':

'1.J /'~ DI exp[- Vfl.J,

£1,1' / / -'

' '\molar volumes

pre-expon~~tial factor from lattice theory: T

.n _ VI d l -10-8 m 2-1 '-1 - - .- , s

,/ 6 \

thermal velocity' 300 m S·l

:"'olecular diameter 3 x 10- 10 m

Fig. 16.6 Self diffusivity of a tracer in a pure fluid

CD

free volumes are additive:

@

tracer friction coefficients (3x): RT

_I'

_1'0

- - - ~l'.eff - ~l D1'. eff

®

Vf

-

I'

Xl~l',1

I' I' + x2~1'.2 + x3~1'.3

relation between tracer and normal coefficients (3x): Sl.2

@

["1]-

exp -

=~ Sl',lS2.,2 ~1'.2 = Sl.2

~1 •.3 = ~1,3

solving for Sl.2'~1,3,S2.3 Fig. 16.7 Derivation of the ternary free volume theory

The second assumption is the relation between the self-friction coefficients and the Maxwell-Stefan friction coefficients. We assume that the mutual coefficient is the geometrical average of the self-friction coefficients. This sounds plausible and has some theoretical justification. ('Some' means weak, but that we have nothing better.) We now have three equations for the three MS-friction coefficients. Solving the equations for the three friction coefficients yields the formulae and calculation scheme of Figure 16.8.

Mass Transfer in Multicomponent Mixtures

186

input ~,V2' ~,vfl,vf2,v(3' D1 , D2 , D3

Vf

=

x 1Vf1

I

~

......

".

+ x 2 V f2 + X3V f3

s=RT 1 D

~l.,eff =~1 exp[ ~~ 1

S =RT

~ p[ ~;1 ~2.,eff = ~2 ex r+ r+

X1~1.,eff + X2~2.,eff + X3~3.,eff

~3.,eff =~3 exp[ ~ 1

X1~1.,eff + X2~2.,eff + X3~3.,eff

1

2

D2

S =RT 3

D3

output

~

_ 12 ' _ 13 '

~1.,eff~2.,eff X1~1.,eff + x2~2.,eff + X3~3.,eff

~1.,eff~3.,eff

~

_ 23 '

Dr,2

~2.,eff~3.,eff

RT =r1,2

Dr,3

•

RT D =r1,3

23 '

RT

=~2,3

Fig. 16.8 Calculation scheme for the free volume theory

Some Results

The free volume theory is really a simple theory for liquid mixtures. However, it turns out that we can use it to describe the movement of small permeants through a polymer. We do that by regarding the polymer as a stagnant component consisting of its chain units. Apparently, the permeating species only 'see' local parts of the polymer, roughly of the size of a chain unit. (The chain units do have a lower free volume than the other components, but are otherwise similar). Figure 16.9 shows three examples of what the free volume theory can do. It is typical for what we would expect for the diftusivities in a system consisting of a polymer (1), a solvent (2) and a larger solute (3). The solute has a very low concentration. Figure (a) shows the behaviour of a strongly swollen, rubbery membrane, such as might be used in dialysis. The diffusivities increase with the swelling (or mole fraction of the solvent), but not strongly. With the parameters as chosen, the solventsolute (2,3) diffusivity is highest, followed by the polymer-solvent (1,2) diffusivity. Figure (b) shows the same membrane, but now with much less swelling. The diffusivities are much lower, especially the polymer-solute diffusivity. If we decrease the free volume of the polymer (Figure c) we obtain a system with low diffusivities. These are a very strong function of swelling and of the size of the permeant. This simulates the behaviour of a glassy polymer. As far as we know, nobody has yet completely measured such diagrams for a real ternary mixture with a polymer. ..

16. Diffusion in Polymers

I r---

D1,2-

r--r---

D1. 3

~

I----

£)2.3

f----::-

- ---

D2 ,3

--

I"" f-------

£)1,2

i::==

D1,3

187

-

,,--~

,--------

D23

I---

.- '"

=== /

£)1,2

10-

12

./

7

7'

7 13

10-

(ab

(bb

-

-

Vf1 ~

10-14

0.4

0.5

Vf1

01= . --

x2

0.1

I

01= . --

~

0.6 0.0

.L:icb

r===== D1,3'r== --

x2

0.2 0.0

-

Vf1 )~ '-==O.OJ

/

-~

0,1

x2

0.2

(a) a strongly swollen rubbery polymer (b) a slightly swollen rubbery polymer ( c) a slightly swollen glassy polymer

DiO

/m 2 s-1

3

v,/m mor

1

Vfi

IV;

polymer (1)

2xlO-8

Vc =lOxlO-5

see above

solvent (2)

2xlO-8

V2 =6xlO-5

0.4

solute (3)

lxlO-8

l'3 == 8x 10-5

(0.3)

Fig. 16.9 Three results from the free volume theory

Remarks on the Model Coefficients You may be surprised that the loose set of arguments in the free volume theory leads anywhere. One reason for the success of the theory is the large number of adjustable constants. You will have to fit these in some way to experimental data; in its current form the theory is not accurate enough for real predictions. In the example above, we have nine constants: • three pre-exponential factors, • three minimum molar volumes, and • three free volumes.

188

Mass Transfer in Multicomponent Mixtures

These parameters usually have physically realistic values. Even so, you can probably better regard them as fitting parameters and not expect them to have much meaning. Before searching for parameters, you should have an idea of the properties of each component: the molar mass, molar volume and viscosity. For the mass and volume of the polymer, you may start with that of a chain unit. The pre-exponential factors usually have values in the range 0.5 ... 2 x 10-8 m 2 S-I. The values are roughly what you would find from lattice theory. You may expect the smallest or lightest component to have the highest value. The diffusivities vary in a roughly linear fashion with these constants. You can get an idea of the minimum volume from the molar mass and molar volume of a pure species. You will probably have to adjust the minimum volumes to .obtain good results. The form of the free volume theory presented here overestimates the effect of volume. So you may have to choose compressed volume parameters nearer to each other than the pure component data would suggest. The results are very sensitive to the values of the free volumes of the dominant components. Rough starting values for the pure components are: ordinary liquids viscous liquids and rubbery polymers glasses and glassy polymers

Vfi "'" 0.3 V; Vfi "'" 0.1 V; Vfi "'" 0.05 V;

Usually, there will be other constraints, simplifying the search of good constants. For instance, in the example of the previous paragraph, the mole fraction of the solute is zero, so its free volume plays no role. Even so, you will find that you hardly ever have enough data to fit all parameters properly. Such is life.

16.3 Summary • In the first part of this chapter we have had a look at the diffusivities of a small permeant through a polymer. We have seen - that the diffusivity increases rapidly with temperature, especially near the glass transition, - that large permeants have very low diffusivities in glassy polymers and - that swelling of the polymer by the permeant (plasticising) greatly increases the diffusivity . • In the second part, we have derived the free volume theory. This theory gives a fairly good description of the behaviour of diffusivities of both liquids and mixtures containing a polymer. We have extended it to ternary mixtures and shown that it gives results which agree (at least qualitatively) with experience. The theory does require a large number of fitting parameters.

16. Diffusion in Polvmers

189

16.4 Further Reading Wesselingh, J.A. and BoIlen, A.M., (1997) Multicomponent Diffusivities from the Free Volume Theory Trans IChemE, 75, Part A, 590-601. A more detailed version of the theory in this chapter, with many examples. This theory uses surface fractions instead of the volume fractions used here; these are better when the species differ in size, but also a little more complicated. Curtiss, C.F. and Bird, R.B. (1996) Multicomponent diffusion in polymeric liquids. Proc. Nat. Acad. Sei., 93, 7440-7445. The Maxwell-Stefan approach applied to diffusion in polymers. Cussler, E.L. and Lightfoot, E.N. (1965) Multicomponent diffusion involving high polymers. m. Ternary diffusion in the system polystyrene 1 - polystyrene 2 - toluene. J.Phys. Chem., 69, 2875-2879. Data showing the strong coupling effects in polymeric systems. Peppas, N.A. and Meadows, D.L., 1983, Macromolecular structure and solute diffusion in membranes: An overview of recent theories. J. Membrane Sci., 16,361377. A review. Vieth, W.R. Diffusion In and Through Polymers, (1991) Carl Hanser Verlag, MUnchen / Oxford University Press One of the more recent books on this topic. Hardly concerned with multicomponent diffusion. Bitter, J.G.A. Transport Mechanisms in Membrane Separation Processes (1991) Plenum Press, New York This book contains many interesting and novel ideas, and also valuable data on diffusion of binaries in polymers. We advise you to be critical on the form of the Maxwell-Stefan equations used.

16.5 Exercises 16.1 Binary Free Volume Theory. Consider a mixture of a pplymer (1) and a permeant (2). a) Write out the free volume theory equation for SI.2' using Figure 16.8. b) Determine the value of SI,2 when x 2 ~ O. c) Now consider the case that the permeant has a mole fraction small, but not zero. Remember that the free volume of the permeant may be as much as ten times higher than that of the polymer. Derive an approximation for SI,2 in this case.

190

Mass Transfer in Multicomponent Mixtures

16.2 Diffusion in a Polymer (Mathcad). This is a short exercise to let you explore the sharp effects of varying the parameters in the free volume theory. It uses the approximation derived in the previous exercise. 16.3 3D Plots of Free Volume Diffusivities (Mathcad). This file demonstrates the free volume theory calculations for a ternary mixture (Figure 16.8). It plots results such as in Figure 16.9 in a three dimensional fashion. Nice colours ... but you cannot do much except vary the parameters.

191

Dialysis and Gas Separation Dialysis and membrane gas separation are two of the simpler membrane processes. Both use polymer membranes, which we will regard as homogeneous. These are our first examples where we have a solid matrix; the matrix gives a few complications.

17.1 Dialysis Moneywise, dialysis is the most important membrane process; it is used on a large scale for treating of kidney patients (Figure 17 .1). The dialysis module removes substances such as urea from blood. The module contains a large number of hollow cellulose fibres. Blood flows in the fibres, the aqueous rinsing solution outside. The dimensions shown are typical for this moment, but smaller modules are being developed.

rinsing solution blood - -----+-~

membrane module

./

t 50 mm

~

rinsing solution + urea

t ~:.Llz=lOllm

200 Jlffi

~

J

,

•..

m

/

hollow cellulose fibres Fig. 17.1 Treating of blood with dialysis

In a moment, we will be setting up the Maxwell-Stefan transport equations for species passing through the membrane (the permeants). Before doing so, we will first have a look at their equilibrium solubilities in the membrane (Figure 17.2). The membrane is strongly swollen - the liquid occupies a volume fraction of 0.6. The polymer matrix occupies the other 0.4. In this example, we take the matrix to be a

192

Mass Transfer in Multicomponent Mixtures

separate phase, not part of the mixture. The two permeants, water and urea do not adsorb specifically, so their concentrations inside the membrane are 0.6 times those outside.

at equilibrium: c' = 55 kmolm- 3 (1) water

(2) urea

c= 33kmolm-3

0.6 x 0.6 x

Fig. 17.2 Compositions in and around the membrane

Under transport conditions (Fig 17.3), the compositions on the two sides of the membrane differ. However, at the two membrane-liquid interfaces, we still assume equilibrium. (The interface is only a few molecules thick, and potential changes across this zone are usually very small.) This again means that both components have concentrations just inside the membrane, which are 0.6 times the external values. Because of this, the mole fractions do not change across the interface. Please note that the accents in the figure now denote a different phase. in this example: same mole fractions on both sides of an interface

blood

0.03

rinsing solution

0.01

Fig. 17.3 Compositions at the membrane interfaces

Figure 17.4 gives the transport equations. In dialysis, the only driving forces are due to activity (composition) gradients. For simplicity we take the solutions to be ideal. The friction sides of the two equations contain permeant-permeant '12' terms and permeant-matrix 'lM' and '2M' terms. The velocity of the matrix is zero. In the swollen membrane used here, all diffusivities are of the same order of magnitude. (Most other membrane processes use denser membranes where the permeant-matrix diffusivities are small. Then friction with the matrix dominates.)

17. Dialysis and Gas Separation

difference equations

-Ax

193

= x2N I -xI N 2 +~ k l ,2 C

I

kl,M C

-Ax =xIN2-X2NI+~ 2 kl ,2 C k 2,M C

k 1,2

= D!lzI ,2

k I,M

= DI,M !lz

k2,M

D !lz

=~

Fig. 17.4 Transport equations for a dialysis membrane

For our exercise we use the rounded values for diffusivities given in Figure 17.5. Calculating the velocities and fluxes of the two components is simple. The interesting thing to note is the value of the water flux; it goes into the blood and has a higher value than the flux of urea. transport coefficients:

!lz =10-5 m f)12 =5 X 10 -10 m 2 s-I~ k12 =50 X 10-6 m s -I , -10 ' -6 Dt.M=IOX1~1Q ~ kl ,M=100xlO D 2 ,M=2xlO ~ k2 ,M = 20 X 10-6

MS-equations (1) 002 - 0.02N1 - 0.98N2 + NI - . - (50xlO-6 )(33xI03 ) (l00xlO-6 )(33xI0 3 ) (2)

+0.02 =

0.98N2 -0.02NI

(50 x 10-6)(33 x 103 )

+

N2

(20 x 10-6 )(33 x 103 )

Fig. 17.5 Calculation of fluxes

Unless we do something (by adding other solutes to the rinsing solution or applying a pressure difference) this will have disastrous effects (Figure 17.6). As usual, we have only needed transport equations of two of the three species. However, in this example, the force balance on the matrix also tells us something interesting (Figure 17.7). If there is no pressure difference across the membrane, there will be no support force, and the two friction terms must cancel. We see that the permeant-matrix transfer coefficients govern the ratio of the two fluxes. This remains so even in extremely open matrices; friction with the matrix is never totally unimportant. Here, we have only looked at transport through the membrane on a local scale. We have neither considered transport through the boundary layers on both sides of the membrane, nor the effects of the flow pattern in the module. We come back to these points in other examples and in the exercises.

194

Mass Transfer in Multicomponent Mixtures

the patient will swell and burst (not shown)

Fig. 17.6 The effect of water entering the patient

no support/orce on the membrane:

O=~+J'!.L kl,M C

k 2,M C

ratio o/the two fluxes: NI

=_ kl,M

N2

k2,M

Fig. 17.7 Force balance on the matrix and flux ratio

17.2 Gas Separation Gas separation is one of the fastest growing parts of membrane technology. There are already many units in operation for hydrogen recovery from process gases and for recovery of organics from nitrogen and air. Purification of natural gas looks promising and large-scale enrichment of air is in sight. In gas separation processes, we apply a pressure difference. This causes gases to diffuse through the membrane. The rates depend mainly on: • the applied pressure difference, • the solubility of the gas in the membrane, and • the diffusivity of the gas in the membrane. There are two groups of processes (Figure 17.8): those with high upstream pressures, and those with (roughly) atmospheric upstream pressures. High-pressure systems mostly use glassy polymer membranes; low-pressure systems use rubbery membranes which have a higher permeability, but lower selectivities than glassy membranes. In both types of process, compression costs are substantial, and we use extremely thin membranes. These may be as thin as 0.1 /lm; they can only be

17. Dialysis and Gas Separation

195

handled if they have a thicker (porous) support layer. In the example that we discuss here, we neglect the effect of the support layer. p""O.1 MPa

N 2' organics

N2

~------~--~

1

pS; 10 MPa

~----'"

ps; 0.01 MPa

N,

organicq

Fig. 17.8 High-pressure and low-pressure gas separations

The modules can use hollow fibres (usually with the high pressure on the outside) or flat membranes that are wound up. The hollow fibre modules are similar to those discussed under dialysis, but much larger. We discuss the spiral wound modules in Chapter 18 under reverse osmosis. To obtain the desired flows we may need very large membrane areas. These may be hundreds or thousands of square metres. Our example considers the separation of carbon dioxide from natural gas (methane). The gas is at a pressure of 10 MPa; it contains ten percent of carbon dioxide. The scheme shown in Figure 17.9 is to reduce this to five percent. We are to calculate the required membrane area and the methane losses. I

F=10mol

S-1

p=10MPa

F-P

YCH4

=0.90

YCH4

= 0.95

YC02

=

0.10

YC02

=0.05

Fig. 17.9 A simple separation scheme

To simplify the problem, we take the upstream compartment of the membrane system to be well mixed. It then has the same composition throughout. We now have a look at the membrane (Figure 17.10). The active layer is one micrometre thick and we take it to be a homogeneous polymer. The solubility of both gases in the polymer is proportional to their partial pressures. That of carbon dioxide is higher than that of methane; we can see this from the two Henry constants. Even so, both gases have a low solubility in the membrane material; even at these high pressures it is not more than a few mass percent. The driving forces in the Maxwell-Stefan equations are the partial pressure differences (as discussed in Chapter 12). They are multiplied by the respective mole

196

Mass Transfer in Multicomponent Mixtures

fractions because we are using the flux form ofthe equations (Figure 17.11). equilibrium solubilities: porous support layer

C

,

2

CH 4 (l)

P2 RT

=-

Cl

= HelPI

Hel =lxlO-5 ///

£:.------

CO/2) c2 = He 2P2 He2 =2xlO-3

------------

'Henry constants'

mol m -3Pa-1

Fig. 17.10 The membrane and the solubilities of CH 4 and CO 2

_X

= J(2 N r;-" ~N2 + _NI

/).pI

k'

I

PI., , ,

i,2C"

N

k

.,'

=-k

I,MC" "

,_N2 -'-N N 2 Y N 2 -"'-'f/2 _ XI X I -x: - - " " z + - z PZ./ ' >kl 2~ k2 ,MC An

/

/

C; /1p1

I,M I

I

-

Pi

k2 MC"2/1p2 , P 2

/

small ('no coupling ')

DtM =2x1O-11

D2M =2X1O- l2 m 2 s- l

kl,M = 2 X 10-5

~,M = 2 X 10-6 m s-l

Fig. 17.11 The transport equations

The friction side of the MS-equations contains permeant-permeant and permeantmatrix friction terms. Because of the low permeant concentrations, the permeantpermeant terms are not important. This leads to the simple difference equations in Figure 17 .11. We take the diffusi vities to be constant; here, that of methane is higher than that of carbon dioxide. If the downstream pressure is much lower than the upstream pressure then we can calculate the fluxes directly. (Check that you do know all parameters in the flux equations). A mass balance for carbon dioxide then gives the required membrane area (here, about twenty-five hundred square metres). With this area, we then calculate the methane losses (here, about five percent of the throughput). If the downstream pressure is not negligible, you will have to solve a number of equations simultaneously, using the results with zero downstream pressure as a starting point.

In a real design, we would do several things more accurately: • We would make a better model of the flow through the system. This would probably be a plug flow model, or some multi-module configuration . • We would use a better thermodynamic model for the solubilities, one that takes the influence of the two components on each other into account.

17. Dialysis and Gas Separation

197

• We might take the effects of pressure and composition in the membrane on the diffusivities into account. • We might take the effects of mass transfer resistances on the two sides of the membrane into account. (These effects are small in gas separation because of the high diffusivities in gases.) • Finally, for these very high pressures we might need a better model for the potentials in the gases, using fugacities instead of partial pressures. The first point is the most important, but in a multi million euro plant also the effects of the other refinements could be worthwhile. We will explore a few of these points in the exercises.

17.3 Summary In this chapter we have had a first look at membrane processes.

• Our first example was on dialysis, a process driven by activity gradients. • We have seen that the two components are driven in opposite directions. • The rates are determined by friction between the permeants and (usually more important) by friction between the permeants and the matrix. • Diffusion coefficients between permeants and matrix are very small in tight (nonswollen) membranes, but they increase rapidly when the matrix swells. They also tend to be low for large molecules. • Permeant-permeant diffusivities are lower than in free solution. They are important in strongly swollen membranes. • Our second example was on the separation of two gases. This process uses a tight membrane. • Here, the driving forces contain the gradients of the partial pressures of the components (at least for ideal gases). • The fluxes depend on the driving forces, but also on solubilities and diffusivities of the components. • Finding the membrane area and the flows requires a simultaneous solution of the transport equations and mass balances (and several auxiliary equations).

17.4 Further Reading Wankat, P.C. Rate Controlled Separations, Elsevier 1990 Keurentjes, I.T.F., Ianssen, A.E.M., Broek, A.P., van der Padt, A., Wesselingh, I.A. and van 't Riet, K. (1992) Multicomponent diffusion in dialysis membrane. Chem.Eng.Sci.,47,1963-1971. Illustrates the Maxwell-Stefan approach

198

Mass Transfer in Multicomponent Mixtures

Rautenbach, R. and Albrecht, R. Membrane Processes (1989) John Wiley & Sons, New York

17.5 Exercises

*

17.1 Driving Forces in Membranes (Mathcad). The composition of matrices can be described in different ways. The description should not have an influence on the values of driving forces and fluxes. This exercise shows the most important problems. 17.2* A Dialysis Module (Mathcad). This problem introduces calculations on membrane processes. It also shows how to use mass balances instead of bootstrap relations. This is an important exercise. 17.3 Avoiding Water Flow. In dialysis, the flow of water can be a problem (Figure 17.6). Can you think of two ways to avoid this? 17.4 Purifying of CH4 with a Membrane (Mathcad). This file explores the separation scheme for CH4 and CO 2 in Figures 17.8 and 17.9. It uses the free volume theory to describe diffusion in the membrane and to obtain the complicated effects of change of pressure. 17.S Forces on a Membrane Matrix. The Maxwell-Stefan equations are momentum or force balances on the separate species. They should also apply to a matrix! Here we study this in a simple model of a gas separation membrane. Consider the membrane to consist of narrow parallel pores. (In reality the pores are more complicated and have molecular dimensions, but this does not change the idea.) There are two gases passing through the membrane; their partial pressures and also the total pressures are shown below. The matrix takes up a volume fraction EM' the pores a fraction (1 - EM). The average concentrations of the two gases in the pores are: C I

=

O.S(Pla + PI{3)

RT

C 2

O.S(P2a + P2{3)

= --'------'-'RT

The numbers of moles per unit area of membrane are: nl

= (1- eM )c1Llz

n2

= (1- eM )C2Llz

We can distinguish five forces on the matrix (all per unit of area): 1. the pressure force upstream on the solid cross section, 2. the pressure force downstream on the solid cross section,

17. Dialysis and Gas Separation

199

3. the friction force between '1' and the matrix, 4. the friction force between '2' and the matrix, and 5. the support force on the matrix. The friction forces on the matrix per mole of gas are equal to the driving forces (Figure 17.11): !1J.lI RT !1PI !1J.l2 RT !1p2 ---=--- ---=----

&

PI &

&

P2 &

Calculate the support force per unit area. You should get a very simple answer. upstream pores (1- eM ) membrane eM support

downstream

200

Pervaporation and Reverse Osmosis Here, we consider two pressure-driven membrane processes: • breaking the ethanol-water azeotrope with pervaporation, and • desalination of sea water by reverse osmosis.

18.1 Pervaporation Pervaporation (permeation followed by vaporisation) is a membrane process with a liquid on the feed side of the membrane and a vapour (vacuum) on the other side (Figure 18.1). It resembles distillation and is often coupled with distillation processes. The membrane is a slightly swollen polymer; the average volume fraction of permeants is a few percent. The membrane has a large influence on the transport rates of components through the liquid-vapour interface, mainly via the different solubilities of the components in the membrane. p=200kPa

==cw1 tHinl@., 1-(-~~kPa vacuum

heating (usually)

membrane unit

. condensor

permeate

Fig. 18.1 A pervaporation system (simplified)

We use pervaporation to remove traces (solutes) from a bulk liquid (the solvent). The solute should dissolve well in the membrane; the solvent should not. There are currently two important applications: • removal oftraces of organics from water, and • removal of traces of water from organics. An example of the ftrst is the removal of tri-chloro-ethene (TeE) from contaminated underground water. Here we use an apolar membrane, in which TeE is very soluble, but water is not (Figure 18.2). TeE has a high activity coefficient in water, so that the TeE concentration can be much higher inside the membrane than outside. We also note that the TeE concentration in the membrane is highest on the water side, so that the swelling varies across the membrane.

201

18. Pervaporation and Reverse Osmosis

water (l), saturated with TCE(2h \

'.

small

vacuum = low pressure ofTCE

apolar membrane

Fig. 18.2 Using a membrane with a high solubility for TCE

An example of the second group of processes is the removal of water from the ethanol-water azeotrope that is obtained by distillation (Figure 18.3). Here we use a polar membrane in which water dissolves preferentially. This was the first pervaporation system of commercial importance, and we will say a few more words about it.

distillation

o azeotrope

(1) water

o membrane unit

(2) ethanol

)

X 2d

X 2b

=0.00

> 0.89

water removal with a pervaporation unit; requires a polar membrane

Fig. 18.3 Breaking the water-ethanol azeotrope with pervaporation

In pervaporation we transfer a component from a liquid to a vapour under vacuum. This means that we have to supply the enthalpy of vaporisation. We also condense the vapour. So pervaporation equipment often contains many heat exchangers (Figure 18.4). The volume flows of the vapour on the vacuum side are large, and this leads to voluminous and expensive equipment. Even so, pervaporation often competes with conventional schemes for separating azeotropes, such as azeotropic and extractive distillation, and liquid-liquid extraction.

202

Mass Transfer in Multicomponent Mixtures

vacuum vessel reed ____-+-.c===r1 kg S-1 membranes 1000 m2

heater

10m (ethanol) condenser permeate (water)

Fig. 18.4 Construction of a pervaporation unit

The driving force on a component i is (as always) the chemical potential gradient, or in a difference model, the chemical potential difference (Figure 18.5). We take the liquid and lower the pressure until it starts to boil (at the 'boiling pressure'). This lowers the chemical potential by a small amount. (Remember that the molar volume in the formula is the small liquid volume.) decreasing the pressure

liquid

CD

vapour

boiling point

LlJ..li = ~ (ph - p) = -1... -10 J mor1

o Llll =RTln(' i

__~)

)=-2000 ... -10000 J mol-

Lllli _ In(l!i..) ", RT -

negligible

pr

1

pr pr)

Pi O.5(Pi +

Fig. 18.5 Contributions to the chemical potential difference

203

18. Pervaporation and Reverse Osmosis

Only when we further reduce the pressure (and obtain a vapour with a large volume) does the potential go down rapidly. So the chemical potential difference is governed by the difference between: • the partial pressure of i at the 'boiling pressure' of the system, and • the partial pressure of i at the vacuum side of the system. We can approximate this by the difference formula in the figure. As already noted, the swelling of the polymer varies across the membrane. The membrane swells most on the liquid side; there is little permeant in the membrane on the vacuum side. In the Maxwell-Stefan equations, the dominating friction terms are those with the matrix, and we usually neglect permeant-permeant interactions (Figure 18.6). However, because of the variations in swelling, the permeant-matrix diffusivities may vary considerably across the membrane. This leads to non-linear concentration profiles in the membrane (Figure 18.7). It also yields a non-linear behaviour of the fluxes with pressure. small because of low permeant and large _1,2

--- ---'

10-10 D m 2 s-1

-'-"-

__I_dJ..l1 = RT dz __I_dJ..l2 RT dz

.-------varies through the membrane

10-14

-'---'------,----,--,----,

o

0.1

0.2

Fig. 18.6 Friction with the matrix dominates

considerable changes in the gradients

Fig. 18.7 Swelling and diffusivities vary across the membrane

Suppose we begin with ambient pressure on both sides of the membrane. When we lower the pressure on the vacuum side, to below the boiling pressure of the liquid, the flux starts to increase (Figure 18.8). At this point, the two phases are still close to equilibrium and the resistance of the membrane is unimportant. The partial pressures of the components govern the ratio of the two fluxes. On a further decrease, the

204

Mass Transfer in Multicomponent Mixtures

membrane becomes more important, and we can obtain quite high selectivities with sufficiently low downstream pressures. Here, we have only considered the mass transfer resistance in the membrane. This is not always allowable. Especially when the solute has a low solubility in the solvent, polarisation may be important. selectivity highest at low pressures

ethanol (1) water (2)

3

10-

ratio = p;

N

p;

~1 IV 2..................

o

zero at the boiling pressure

,,~

,

--------------~... -----.. ..........,

o

Pdownstream

Pa

Fig. 18.8 The fluxes as a function of downstream pressure

18.2 Reverse Osmosis' The production of potable water from seawater has been one of the most successful applications of membrane processes. The principle is to force water through a membrane, which is almost impermeable to salt. Figure 18.9 shows a simple single stage process. (Real processes apply multi-module schemes, but this does not change the principle.) A feed pump brings pre-filtered seawater into a high-pressure loop. We need a fairly high velocity along the membrane to avoid polarisation and to minimise fouling. This is provided by a circulation pump. Part of the feed permeates through the membrane; this is the product. The concentrated brine goes back to sea via a power-recovery turbine. We need large membrane areas. These use a few flat membranes, which are kept apart by gauze spacers and wound up in a module as shown in Figure 18.10. Permeate is removed in cross flow via a central channel.

1 Osmosis is diffusion of water through a membrane into a salt solution. When swimming in fresh water, you take up water by osmosis. In reverse osmosis, we force water out of a salt solution.

205

18. Pervaporation and Reverse Osmosis

circulation membrane module

fresh water

brine power recovery turbine

Fig. 18.9 A single stage reverse osmosis process

feed (sea water)

~ Wcross flow""

1m s

-1

retentate (brine) permeate (fresh water) 2 membranes

Fig. 18.10 A 'spiral wound' reverse osmosis module

In reverse osmosis, we use an extremely thin membrane (Figure 18.11). This is supported by a much thicker, open, spongy structure. Here, we neglect the effect of the support structure, and only consider the actual membrane or 'skin'. We regard the skin as a homogeneous medium. Figure 18.12 shows concentrations in and around a typical reverse osmosis membrane. The volume fraction of water is about ten percent. Because water is the dominant component on both sides, the water fraction varies little across the membrane. Salt is only slightly soluble in the membrane; here we have taken it to have a distribution coefficient of one hundredth. The salt concentration varies strongly across the membrane. In water, salt dissociates in two ions; we regard these

206

Mass Transfer in Multicomponent Mixtures

as separate molecules when computing mole fractions. Inside the membrane, salt probably moves as pairs of ions. support structure ", 200 I.1m

(neglected here) polymer skin & '" 0.1 l.1m

the actual membrane regarded as homogeneous Fig. 18.11 A reverse osmosis membrane

concentrations in mol L-i

NaCl(2) 0.5 Na+ { 0.5 Cl-

C

Fig. 18.12 Concentrations in and around the membrane

Before looking at the transport equations, we first analyse one special case (Figure 18.13). There is a pressure difference - the osmotic pressure difference - at which the driving force on water is zero. For seawater against fresh water, the osmotic pressure difference has a value of 2.6 MPa. A reverse osmosis membrane only works well if the pressure difference is larger than the osmotic pressure difference. (This is a serious limitation for many processes, but not for desalination of seawater.) the driving force on water (1) is zero when: ~l Vi _ _ osmotic pressure - xl - RT!1rc - ~ !1rc - 2.6 MPa difference

°

only transport of water to the right when !1p > flIt Fig. 18.13 The osmotic pressure difference of seawater

18. Pervaporation and Reverse Osmosis

207

In the transport relation for water (Figure 18.14) we only take friction with the membrane into account. The main driving force is the pressure gradient. We can replace the composition term by an osmotic pressure term: the water velocity is proportional to the difference between the pressure-difference and the osmoticpressure-difference.

ilx;

~ An ----~ =--(~p-~1t)

water:

V;

x;

RT -0.02 +0.08

ilx;

salt:

~ ~_ N2 - k2,MC2 +0.33

- x; - RT +2

~=-10MPa

k I,M

RT

= NI -kl,M~

= 10

-3

ms

~1t=2.6MPa

-I

k

=10- m s-I 5

2,M

c = 5000 mol m- 3

Fig. 18.14 Transport equations for a reverse osmosis membrane

In the salt transport equation the major driving force is the composition term. The driving force on the salt is much larger than that on the water. Even so, the salt velocity is lower. This is because of the very much lower diffusivity of the hydrated salt ions. The flux relations are summarised in Figure 18.15. The water flux is proportional to the pressure difference; the salt flux is barely dependent on it .. So an increasing pressure differential improves the salt retention. water:

~=-!i(~p-~1t) kl,MCI

salt:

RT

~=_ilx;

salt retention: c

R=l-A

c2a

O+---~-------------

Fig. 18.15 Flux relations and retention in reverse osmosis

208

Mass Transfer in Multicomponent Mixtures

In this example we have kept the equations very simple. Details we have neglected in the calculations are: • non-idealities, • water-salt friction in the membrane, • viscous flow, and • the effect of the salt concentration downstream on the osmotic pressure difference. We shall include some of these in the exercises. If we use realistic values of the parameters, these new effects modify the behaviour of the system only slightly, except for pressure differences near the osmotic pressure.

18.3 Summary In this chapter we have looked at the peculiarities of two pressure driven processes: pervaporation and reverse osmosis. • In pervaporation the potential difference in the driving force is proportional to the difference in boiling pressure of the solute upstream and its partial pressure downstream. • In reverse osmosis the potential difference for water is proportional to the difference in pressure, from which we subtract the difference in osmotic pressure. • Pervaporation only works well if the downstream pressure is much lower than the boiling pressure of the feed. Reverse osmosis only works well if the downstream pressure is much lower than the osmotic pressure difference. • In both processes, transport is dominated by friction between permeants and the matrix. • In pervaporation, swelling may vary considerably across the membrane. This may cause a large variation of the diffusivities. This effect is much less pronounced in reverse osmosis. • In both processes, selectivity increases with increasing values of the driving forces.

18.4 Further Reading Heintz, A. and Stephan, W., 1994, A generalized solution-diffusion model of the pervaporation process through composite membranes, J. Membrane Sci., 89, 143169. Nice illustration of the Maxwell-Stefan approach for pervaporation. Rautenbach, R. and Albrecht, R. Membrane Processes (1989) John Wiley & Sons, New York Spiegler, K.S. and EI-Sayed, Y.M., A Desalination Primer (1994) Balaban Desalination Publications Santa Maria Imbaro, Italy, ISBN 086689-034-3

209

18. Pervaporation and Reverse Osmosis

18.5 Exercises 18.1 Driving Force for Pervaporation. The potential changes of the permeants in pervaporation are analysed in Figure 18.5. Check whether you find similar results and also how good or bad the difference form for the potential change is. Data: upstream pressure

p=l.O

boiling pressures

Pb

downstream pressures molar volumes of liquids

= 0.03

MPa PbI

=0.02

= 0.001 "l = 8 X 10-5

PI

PbZ

MPa

=0.01

pz=lx1O

3

Vz =1Ox1O-

MPa 5

m3 mor 1

When you have finished, read the remarks on Figure 3.14. 18.2 Film Limitation in Pervaporation. In pervaporation, the main resistance to mass transfer is usually thought to be inside the membrane. This is, however, not always so. For example, in the process for removing tri-chloro-ethene (TCE) from water in Figure 18.2 the solubility of TCE is so low that diffusion through water can become limiting. In the following, you will have to look forward in the book, to Figure 20.7. There you will find a rough estimate of the ratio of the two friction losses: \

friction in the film friction in the matrix

ki MC(

--------::::: -,-

kj,SCi

Here ki,M and ki ,s are transfer coefficients of the species i in the matrix and in the solvent. The concentrations c(andci are average values in the matrix and in the solvent film. For a pervaporation membrane, the solubility of the transferring species should be quite high, say c(= 10 3 mol m -3. Transfer parameters might have values 5 11 Z Bi,M = 10- m s-I andLlz = 10- m. For the film transfer coefficient you may expect a value of ki,s::::: 10-4 ms-I. Below which bulk concentration of the solute, would you expect the film limitation to start playing a role? 18.3 Pervaporation of Hexane-Heptane (Mathcad). This exercise explores properties of a pervaporation membrane. The feed side of the membrane is in contact with a liquid. The downstream side is in contact with a vapour under vacuum. So the membrane is more swollen and permeable at the feed than at the downstream side. This is an example where the Maxwell-Stefan diffusivities can vary greatly with composition, and where we have to subdivide the membrane into a number of layers. You are to calculate the fluxes and selectivities for a series of downstream pressures and to see whether you can understand the behaviour of this process.

210

Mass Transfer in Multicomponent Mixtures

18.4 Osmotic Pressure. This is a misnomer: osmotic pressure is not a pressure. It is a property of a solution that is closely related to the composition. From Figure 18.13:

RT &1

RT.

Iln = - - - z --&1 orwlth &2

"1

XI"!

= -&1 ~

RT

Iln z - & 2

"1

(a) Why does the osmotic pressure of a solution contain the molar volume of the solvent, and not that ofthe solute(s)? Hint: have a look at Figure 24.3. (b) Reverse osmosis (RO) only works well when the pressure difference across the membrane is more than (say) two times the osmotic pressure of the feed. (We study this in the next exercise). The maximum pressure difference in membrane processes is about 10 MPa. How far can we concentrate in an RO process with water as a solvent. Vwater = 1.8 X 10-5 m 3 mol-I. (c) And how far can we concentrate with a non-aqueous solvent with 3 Y;.olvent = 10-4 m mol-I? 18.5* Retention of a Reverse Osmosis Membrane (Mathcad). You are to investigate the effect of the pressure difference across the membrane on retention (see Figures 18.14 and 18.15). You are to look at what happens when the pressure is lowered to below the osmotic pressure of the feed. The problem here is to get Mathcad to converge: you are provided with a solve block with all equations and a converging set of data. You are to get Mathcad to use earlier results as a starting point for new calculations. When you finish you should understand the two different regimes under which an osmosis membrane can operate. 18.6 Change of pH in Reverse Osmosis (Mathcad). It is said that there is a measurable change in the pH of water as it passes through an RO membrane: something like half a pH unit. In this assignment we consider a slightly acid feed, containing Na+, CI- and H+ ions. You are provided with a model for transport of water and ions through the membrane. You are first to have a good look at the model (answering a few questions). Then you are to investigate possible causes of the change in pH.

211

Electrolysis and Electrodialysis In this chapter, we consider the transport of ions in membranes and near electrodes. Here, electrical forces are important. We assume that you have already read Chapter 11, 'Electrical Forces and Electrolytes' .

19.1 Introduction A simple example of an electrochemical process is the electrolysis cell shown in Figure 19.1. The cell contains a solution of silver nitrate (that may be contaminated with less noble metals). It has two electrodes. If we pass a current through the cell it will cause pure silver to deposit on the negative electrode. Reactions happening at the electrodes largely govern the behaviour of the cell. These reactions depend on the concentrations in the cell, the surface composition and structure of the electrodes, and the applied current (or potential differences). The behaviour also depends to some extent on mass transfer near the electrodes, and that is what we will be looking at in a moment. We will not consider the electrode reactions any further, even though they are important.

°t c>W-... 2H.20,4.NO.3 ~·4H+ ·4Ag+ 2

silver solution with impurities

/pure silver

Fig. 19.1 An electrolysis cell for separating silver

Many electrochemical processes use membranes. An example is the chlor-alkali electrolysis shown in Figure 19.2. This process splits a sodium chloride solution into chlorine gas, a concentrated alkali solution and hydrogen (a by-product). Here a negatively charge membrane is placed between the two compartments. This membrane only transports (positively charged) cations, so that the alkali produced in the right hand compartment is not contaminated with chloride. We use electrodialysis (Figure 19.3) to split a saline solution into a concentrate and a diluate (desalinated) stream. This process uses a stack with a large number of membranes. These are alternating cation and anion transporting membranes, with the solution in between. When we pass a current through the stack, the ions can only travel in such a way that they concentrate in one half of the compartments. In

212

Mass Transfer in Multicomponent Mixtures

electrodialysis we use many diluate and many concentrate compartments, so the electrode compartments are relatively unimportant. 2 NaOH

t

t2H2 20H: ::le:::> 2Hp current

////

cation permeable membrane

Fig. 19.2 Chlor-alkali membrane electrolysis

two electrode compartments "r-___________ _

t "cr e:::>' . . . g ::Jo~·;c·o °2

Na +

.

t

----------

.~f

···0-------,,·.

Q

i e:::> 2 H2

.

current

" """'" many diluate and concentrate streams /

//

Fig. 19.3 An electrodialysis stack (simplified)

19.2 Polarisation in Electrolysis ill electrolysis we use an electrical field to separate ions from a solution. There is a

big difference compared with most examples in Chapter 11. There we applied a concentration difference and obtained an electrical potential difference. Here we apply an electrical field and obtain concentration differences. In our first example, we analyse the situation near an electrode. The electrolyte is a dilute solution of silver nitrate. We only consider the negative electrode (Figure 19.4). This is a fairly simple system: only silver deposits on the electrode. Neither nitrate nor water partakes in the reaction at this electrode; the bootstrap relations should be clear. The electroneutra1ity relation leads to the conclusion that the silver and nitrate concentrations must be the same everywhere. The transport relations for the two ions should be obvious. From the transport relation of the nitrate we see that the ionic concentration at the electrode becomes zero when the potential difference is one twentieth of a volt. At this condition the silver transport rate reaches a maximum value. The corresponding charge transfer rate is known as the limiting current. Note that the electrical and composition forces on the silver are both in the same direction. (For small forces you will find that this yields a flux equal to twice the product of the mass transfer

213

19. Electrolysis and Electrodialysis

coefficient and the concentration gradient.) The two forces on the nitrate ion just balance. Ag+ (1)

X

N2 =N3 =0

XJ3 --'-+-'"

.

transport equations

(1) - D.x _!£.;1.$ = .!!.L.

F:lectrical

EB===*

~

H20 (3)

bootstraps:

----7-~

FeleC/fical

N0 3- (2)

~om'Position

x

kl,3~

RT

P (2) --+-;1.$=0 D.x

)

x-

FcompOSilion

RT

limiting current when: xf3 =0 ~

D.x

x

2RT 1 =-2 ~ ;1.t/>=-p~- 20 V

Fig. 19.4 Polarisation in electrolysis

Higher potential differences do not increase the current above the limiting value until another reaction - the electrolysis of water - takes over. Figure 19.5 gives formulae for the current density and limiting current. Here we also see that the complete differential equation ('exact solution') predicts a less sharp transition to the limiting current than our difference equation. (This is one of the few cases where more sophistication is worthwhile.)

1

I

;1.t/> 1/40

o

o

2

4

Fig. 19.5 The limiting current density in electrolysis

The effect of adding a considerable amount of an inert support electrolyte such as NaN03 (Figure 19.6) is to almost eliminate the electrical potential drop over the electrode film. It also halves the limiting current (the contribution of the electrical

214

Mass Transfer in Multicomponent Mixtures

driving force disappears). Qualitatively the reasoning should be clear; you can easily follow the details numerically.

NO~(2)

-"'--

Na + (3)

---"~-

consider the situation at the limiting current Llx3

Llx2 +_L1$=O p x2 RT

c_ _

- Llx3 _ ~ L1$ = 0 X3 RT

\

small, so

L1$ small

Ag+ is now only driven by Fig. 19.6 The effect of an inert electrolyte on electrolysis

19.3 Electrodialysis Electrodialysis is used for removing electrolytes from solution and for concentrating electrolytes. It is slightly more complicated than other membrane processes in that it uses two kinds of membrane. (Figure 19.7). Both consist of a polymer backbonepolystyrene or a fluorinated hydrocarbon. One type has fixed negative charges such as sulphonic acid groups; the other has fixed positive charges such as quaternary ammonium groups. The negative membrane only transfers positive ions and vice versa.

~ ~

cation exchange membrane, negative

Q

transports +ions

-RSO;

charged, swollen polymers

Q :+:

:+:

anion exchange membrane, positive transports -ions

-RN+

Fig. 19.7 The two kinds of membranes in electrodialysis

215

19. Electrolysis and Electrodialysis

We arrange the membranes in stacks of several hundred (Figure 19.8). These alternate: positive-negative-positive and so on. An electrical potential is applied across the stack; both the stack potential and the electrical current can be large. The electrical field forces ions through the membranes, alternately diluting and concentrating the electrolyte between the membranes. A pump forces liquid through the thin slits between the membranes; the flow is turbulent. These slits are alternately connected to form two sets of compartments (Figure 19.9). feed velocity 0.1..1 m S-1

400 ... 800 membrane pairs

turbulent flow (spacers!)

current 10 .. .1 000 A m-2

114> "" 1 V per membrane pair distance between membranes 1 mm concentrate streams Fig. 19.8 Arrangement of the membranes in the stack

electrical field feed ions can only pass through membranes ofopposite charge

concentrate

" ,, ,, , diluate

Fig. 19.9 The ion fluxes and forming of concentrate and diluate

216

Mass Transfer in Multicomponent Mixtures

There are several resistances to mass and charge transfer in an electrodialysis stack. Figure 19.10 shows the important ones.

CD ® CD

o

electrodes (not considered here) polarization layers bulk fluid layers membranes Fig. 19.10 Important resistances in the stack

There are two contributions to the resistance of the fluid layers (Figure 19.11). Polarisation in the boundary layers causes limiting current effects similar to those in electrolysis. The bulk of the fluid is mixed by turbulence, so there is little mass transfer resistance. However, the electrical conductivity is not increased by turbulence: the effects of forward and backward transport of charge by the eddies cancel. The resistances in the diluate compartments are more important than those in the concentrate compartments.

polarization: similar to electrolysis x f3

~

0

limiting current

bulk turbulent, well mixed diffusion not important but electrical conduction is!

Fig. 19.11 The two resistances in the fluid layers

Before looking at transport in the membranes we first look at the compositions in and around the membrane (Figure 19.12). We only consider the negative membrane - the

19. Electrolysis and Electrodialysis

217

principles are equal for the other positive one. Water has a molar concentration of 55 mol per litre. It contains only small amounts of the electrolyte (here, sodium chloride). We characterise the membrane composition in a manner different from that in the previous chapter. Here we choose the fixed charge concentration as the membrane concentration. Because the fixed charges are polar, the membrane swells in water. The swelling is, however, limited by cross-linking of the polymer. Overall, membrane material occupies about three-quarters of the volume. The chloride co-ion is effectively excluded from the membrane by the fixed charges. (We shall look at this in a moment.) So the sodium concentration must be equal to the fixed charge concentration. Together these yield a concentrated electrolyte solution in the membrane, quite different from the external solution.

water

co. @

negative membrane

component H2O 1

cimolL-1

,

Xi

or Xi

=1

55 1

2 x 10-4.. 2 X 10-3

2

Na+

3

cr

10-2 ••• 10-1

2xl0-4.. 2x1O-3

1

H 2O

12

0.8

2

Na+

1.5

0.1

3

cr

small

small

1.5

0.1

4

-HSO~

polymer

10-

2

••• 10-

cp =0.75

Fig. 19.12 Concentrations in and outside the membrane

Figure 19.13 gives a brief summary of the ionic equilibria of the membrane. We assume here that the membrane is homogeneous and that the liquids are ideal. The sodium in the membrane tends to diffuse outward. This causes a potential difference of the order of a tenth of a volt. (You obtain the equilibrium relation at the top of Figure 19.13 by setting the driving force in our transport relations equal to zero.) The potential difference excludes the chloride ions in a similar manner. Combining the two relations shows that the chloride concentration in the membrane varies with the square of the external concentration. This is the Donnan relation. At low external concentration it predicts that chloride exclusion will be very effective. In reality, chloride is not excluded so well; experiment gives a roughly linear isotherm (Figure 19.14). This is thought to be due to inhomogeneities in the membrane: 'clusters' of liquid may contain electrolyte concentrations similar to those in the external liquid. Even so, the chloride concentration is much lower than the sodium concentration. This, as long as the external concentration is much lower than the internal electrolyte concentration.

218

Mass Transfer in Multicomponent Mixtures

counter ion attracted by membrane

RTlnx2 + 0""<1>= RTln x; +0""<1>'

co-ion repelled by membrane

RTlnx3 -0""<1> = RTlnx; -0""<1>'

effective co-ion exclusion for low X3

Fig. 19.13 Donnan equilibrium for the two ions

: membrane

~ high charge

experiment Donnan theory

co-ion exclusion not as good as predicted; isotherm linear

liquid

~lowcharge

~-l

cZowcharge "" CUquid

I

I I

Fig. 19.14 Effect of inhomogeneity on co-ion exclusion

So there are three species in the membrane: water, sodium and the membrane material. The water-sodium and water-membrane diffusivities are about one order of magnitude lower than in free solution (Figure 19.15). This is mainly due to geometrical effects that we will discuss in Chapter 21. The sodium-membrane diffusivity is very small: just as in free solutions the friction between ions is high. Note the extremely low value of the sodium-membrane mass transfer coefficient. Figure 19.16 gives a set of transport calculations. We have chosen the concentration to the right of the membrane ten times higher than to the left. The applied potential difference is two tenths of a Volt.

219

19. Electrolysis and Electrodialysis

1

H 2O

2

Na+

3

Cl"

= 1.4 x 10-10 D14 = 2.0 X 10-10 D12

ten times lower than in free solution

-HSO~

4

very low!

polymer membrane and species inside

&=2xlO-4 m D

k.=_',_1 1,1 &

neglected

Fig. 19.15 Transport properties of an electrodialysis membrane

X la

X 2a

=0.998 conditions just outside the double layer

=0.001 "------' d<j> = -0.2 V dxl _, (u -u ) - , U! = Xz I 2 +X4 -

--=XC

"'" 0

kl2

kl4

large flux ofwater, dragged along by sodium

_ dx 2 _ ~ d<j> = x' (u2 - uI ) + x' fl2 x2 RT I kl ,2 4 k 2 ,4 "",-2

"",8

Fig. 19.16 Transport equations of an electrodialysis membrane

In this example we can calculate the species potential difference most easily from the changes between points just outside the membrane. As you can see, the driving force on the water is almost zero. Water is dragged along by friction with the moving sodium ions: its velocity is between that of sodium and the membrane. This is called electro-osmosis. The main driving force on sodium is the electrical force. The composition driving force is, however, not negligible. Before leaving the subject of electrodialysis, a few additional remarks are in order. • Polarisation is very important. • At high external electrolyte concentrations the chloride concentration in the negative membrane does start to have effect.

220

Mass Transfer in Multicomponent Mixtures

Our model can take both effects into account. This does not imply that building a complete model of an electrodialysis stack is a trivial matter: we know better from experience.

19.4 Summary In this chapter, we have had a look at two major electrically driven processes: electrolysis and electrodialysis. • We have seen that electrical forces cause changes of the total electrolyte concentration near electrodes and membrane surfaces. • If the electrolyte concentration goes to zero near an interface, we approach a limiting current. The value of the limiting current depends on bulk electrolyte concentration and on mass transfer parameters. • We have studied the use of two kinds of membrane in electrodialysis: cation and anion exchange membranes. The fIrst is negatively charged. It transports positive ions (cations) preferentially. The opposite holds for anion exchange membranes. • Ions transported through a membrane, drag liquid with them (electro-osmosis). So liquid can be transported even when there is no driving force on the liquid.

19.5 Further Reading Prentice, G., (1991) Electrochemical Engineering Principles, Prentice-Hall Eaglewood Cliffs, New Jersey Spiegler, K.S. (1958) Transport processes in ionic membranes. Trans. Farad. Soc., 54,1408-1428. A classic paper on membrane transport where the friction formulation is laid out. Spiegler, K. S. (1983) Principles of Energetics, Springer-Verlag, Berlin. An introductory text, containing many examples of different approaches to mass transfer processes. Nice examples on membranes. Scattergood, E.M. and Lightfoot, E.N. (1968) Diffusional interaction in an ionexchange membrane. Trans. Faraday Soc., 64,1135-1143. A classic paper. Wills, G.B. and Lightfoot, E.N., 1966, Transport phenomena in ion exchange membranes. Ind. Eng. Chem. Fundam., 5, 114-120. Another classic paper from Wisconsin. Kraaijeveld, G., Sumberova, V., Kuindersma, S. and Wesselingh, J. A. (1995) Modelling electrodialysis using the Maxwell-Stefan description. Chem. Eng. JI,57,

19. Electrolysis and Electrodialysis

221

163-176. Process modeling.

Rautenbach, R. and Albrecht, R. Membrane Processes (1989) John Wiley & Sons, New York

19.6 Exercises 19.1 Why is the membrane important in the chlor-alkali process (Figure 19.2)? For which components should it have a low permeability? 19.2 In the chlor-alkali process (Figure 19.2), there are two electrodes and a membrane, so four polarisation films. Suppose we have flat electrodes separated from the membrane by spacers as in Figure 10.5. Assume also that the flows are the same in the two compartments, and that both have a bulk electrolyte concentration of 1 mol L- 1• Get the ion-water diffusivities from Figure 11.17 and mass transfer coefficients from the file 'Slit in an Electrodialysis Module' in Chapter 10. (a) Which film will limit the current? (b) Calculate the limiting current density. (Actually the geometry described here is not suitable for industrial electrolysis modules. One of the reasons is that we have to remove gas at the electrodes.) 19.3 Why would the current densities in electrodialysis be much lower than in chloralkali electrolysis? 19.4 The membranes used in electrolysis and electrodialysis are fairly tight. The volume fraction of pores is about 0.25 (Figure 19.12, which also contains other data that we need in a moment). The pores are thought to have a diameter of about 2 nm. Assume that they are cylindrical. (a) What is the surface area of the pores per unit volume of membrane? (b) Calculate the surface charge density and the distance between the sulfonic acid groups. Could manufacturers increase the charge density much? (c) Regard the solution in the pores as one of water, Na+ and -HS04-. Assume that the sulfonic acid groups behave like CL Estimate the electrolyte concentration in the pore liquid and use Figure 11.17 to obtain D Na+,Cl- =D Na+,-HS04-' Multiply this value by 0.1 to allow for tortuosity and constrictions in the pores. Do you get anywhere near the value of D 2,4 in Figure 19.15? 19.5 Driving Force for Electrodialysis. Across a negative electrodialysis membrane, we apply an electrical potential difference A<j>. We consider the driving force on a positive ionic species with a charge number z. The mole fractions of this species

222

Mass Transfer in Multicomponent Mixtures

differ in the solutions at the two sides of the membrane. At the interfaces ex and ~ of the membrane we have equilibrium between the two phases. So we can calculate the potential difference of the species in two ways: (1) from ex Gust outside) to ~ Gust outside), and (2) from ex Gust inside) to ~ Gust inside). The first method yields a total potential difference: .t1lf1i

LU = RT-' + ".!"zi .t1f/J rr::'

Xi

Now the second method. For low external concentrations of electrolyte the concentration of counter ions in the membrane is more or less constant and equal to that of the fixed charges. So the .t1x term in the membrane disappears and the potential difference according to the second method becomes: .t1lf1i

= PZi .t1f/J

The two calculations should give the same answer. What have we forgotten? (This happened to me (JAW) in my first publication on electrodialysis. I still feel a little ashamed.) 19.6* Electrodialysis Membrane with Co-ion Transport (Mathcad). In this exercise, you are provided with the model of the electrodialysis membrane in Figure 19.16. You are to extend the model to include transport of traces of the co-ion CL This is a major source of losses in electrodialysis. 19.7 A Seawater Driven Desalination Unit (Mathcad). Unbelievable, but true. It is possible to desalinate brackish river water by using the potential differences between the river water and seawater. And this with no external power, and no moving parts. In this file you analyse how such a system works. Whether it is economical is a different matter.

223

Ion Exchange Ion exchange is a sorption process that uses a charged polymer matrix. We use it as an example to introduce fixed bed sorption processes and equilibria. Ion exchange also shows a few surprising mass transfer effects; the Maxwell-Stefan equations describe these well.

20.1 Fixed-Bed Processes Sorption processes are mostly carried out in a fixed bed of the sorbent partid:~s (Figure 20.1). Their prime application is for the removal of trace amounts of solutes from gas and liquid streams. We pump the solvent and solute into a fresh column, which contains little or no solute. As the solvent passes through the column, the solute is adsorbed. First the area near the feed point becomes saturated; this saturation zone gradually fills up the column. The sorbent usually stores the solute in a much higher concentration than that of the feed. So the column only fills up slowly. solvent + solute

II

CD

external (film) resistance

_-------'''-r'-..''--''''

>e---" "\

_-----fr

\

I I

I

-------------_~~,,-

I

,//

......... _---""'/

sorbent solvent

1

CD

internal (particle) resistance

Fig. 20.1 Sorption in a fixed bed

After some time the front reaches the end of the column and we stop the sorption. Then we regenerate the column. We can do this by changing conditions in the column (pressure, temperature, electrical field, composition, pH ... ) so that the solute is no longer strongly adsorbed. The solute is then removed in a concentrated form by

224

Mass Transfer in Multicomponent Mixtures

passing a regenerant through the column. The front is the actual mass transfer zone. Its width is quite an important design criterion. The front width increases when we increase the flow rate through the column. It also depends on the mass transfer processes in and around the particles. Before looking at these processes, we fIrst look at an ion exchanger and its equilibria. (We shall not look at the development of the front or breakthrough behaviour. That would require too much space.)

20.2 Ion Exchange Equilibria An ion exchanger consists of a charged, swollen polymer, similar to that in an electrodialysis membrane. We consider a cation exchanger with negative sulphonic groups attached to the polymer (Figure 20.2). The exchanger is in eqUilibrium with a dilute solution of sodium chloride and hydrochloric acid in water. The negative chloride ions (the co-ions) are largely excluded by the negative charges on the matrix; here we will neglect them.

EB ~ matrix with HS03charges

• •

C EB

m •

CI- co-ions (excluded by the matrix) Na+) counter ions (can exchange) H+

total cations in solution: c Fig. 20.2 Ion exchanger with fixed charges, co- and counter ions

The exchanger contains the two counter ions, with a concentration equal to that of the sulphonic acid groups (Figure 20.3). This capacity is a convenient reference to defIne mole fractions. We show concentrations inside the exchanger with an accent to distinguish them from those in the liquid, and note that the total concentration of cations in the liquids is usually much smaller than that in the solid. Also its value can be chosen freely; this is in contrast to that in the exchanger. The mole fractions of sodium in the two phases are not equal. In this example, the exchanger has a preference for the sodium ion. A simple model that describes much of ion exchange is the 'law of mass action'. This assumes that the exchange of the two cations is a fast chemical eqUilibrium reaction (Figure 20.4). The law gives a

20. fon Exchange

225

hyperbolic relation between the mole fractions of sodium in the liquid and in the exchanger.

• • •

•

concentration of counter ions in the particle giveu by concentration of _------ fixed charges

•

,.R>-~~~~----J I)

•

I

capacity

C' I

c' "" 4 mol (L of solid phase usually c << c'

r

external concentration of counter ions can be chosen freely Fig. 20.3 Cation concentrations in and outside the particle

Na+ ions in the particle

in the particle

~,

,

I

Na++(H+) {::}(Na+) +H+

x

,

equilibrium constant

_ x' (I-x)

/~:~ (I-x')

cati;;~;~~~~ns (Na+)

o

x

o

I

Na+ ions in the liquid

Fig. 20.4 'Law of Mass Action' Isotherm

The relation between the two compositions is non-linear. So, if we take the dilute solution in the liquid to be ideal, then sodium in the exchanger must be non-ideal. Figure 20.5 gives its activity coefficient. It varies with composition, and we have to take this into account in our diffusion equations.

20.3 Linear Driving Force Model Just as in the other examples, we will restrict ourselves to mass transfer to and inside a single particle. Also we will regard the transfer to be steady (at least for short periods). There are two mass transfer resistances: • the external (or film) resistance, and

226

Mass Transfer in Multicomponent Mixtures

• the internal (or particle) resistance. Both of these resistances usually play some role. isotherm rearranged:

x

,

x=~~~~-

K +(1-K)x'

we take Na+ in water to be ideal:

two extremes:

x , ~O

rNa+~

K- 1

x' ~1

rNa+ ~1

Fig. 20.5 Activity coefficient of Na+ in the particle

The external film is similar to the films we have seen earlier. Transport inside the particles is a very complicated process (Figure 20.6). The solute first penetrates rapidly into the outer layers of the particle. After that, a concentration profile develops of which the shape only changes a little. The largest part of the gradients is in the outer layers of the particle: we model this with a resistance located in the particle-liquid interface. This resistance has a value corresponding to a 'film' thickness of about one tenth of the particle diameter. This is approximate and you have some reason to be dissatisfied. However, a complete treatment of transient multicomponent problems is too difficult for this text, and you can learn quite a lot from the Linear Driving Force (LDF) model.

reality

model

'm I I I

I I

Xl

o

I I I

corresponding to a film with

position and time

d &""10

o large gradients near the o

t>1

I I I

o resistance in the interface,

o velocities depend on interface average diffusion distance:

I

d

~z""-

lO

o

average concentration depends on time

Fig. 20.6 Modelling the resistance inside the particle

20. Ion Exchange

227

As the diffusivities in the solid matrix are usually lower than those outside, you might expect the particle resistance to dominate. However, this is not necessarily so (Figure 20.7). For a given flux of a species through the interface, the species velocity is highest in the phase with the low concentration. The solute concentration in the solvent is often orders of magnitude lower than in the sorbent, so the resistance in the solvent can be important. film

particle (accented)

solvent

I I I

(s)-~~

I

/--1:heir ratio is: friction in film friction in matrix

I I I I I I I

1,...---------------.

major friction terms for i are usually:

I

I

I

and

_/

I

I I I

friction in the 'film' becomes important for low external concentrations

Fig. 20.7 Which resistance dominates?

20.4 Ion Exchange (Film Limited) Ion exchange removes an ionic species from an aqueous solution and replaces it by another. In the following examples the two ions are hydrogen and sodium. They are both positive. The ion exchanger particles are made of the same type of swollen polymer as electrodialysis membranes; here the particles have a negative charge. Again co-ions such as chloride are excluded from the matrix (at not too high external concentrations). Mass transfer in ion exchange is complicated by the transient nature of the process. This can cause swelling or shrinking of the particles. We can take the resulting water flux through the interface into account with our models, but we shall not do so here. We look at the film limitation outside the particle (Figure 20.8). This tends to be the overruling resistance at low external solute concentrations (in the sorption step). In the example shown, we consider two situations. In the first, we remove sodium ions from the electrolyte and replace them by hydrogen ions from the exchanger. In the second case the reverse occurs. In both situations the exchange has just begun and the interface and bulk compositions have not yet changed appreciably. Quite surprisingly these two processes are not at all symmetrical. The second one is much

228

Mass Transfer in Multicomponent Mixtures

more rapid than the first.

I I I I

I I I I I I

(in this example)

I I I

•• I

: CI...l ............. . •••••• I Na+ ..... : H+

\.: Cll ............ H+ Na+

the rate depends on the direction! Fig. 20.8 Two seemingly symmetrical processes

The culprit is the hydrogen ion, in combination with the chloride ion. The chloride does not participate in the exchange, but does influence the mass transfer process. The fast-moving hydrogen ion causes an electrical potential difference. In the lefthand example this pushes chloride ions away from the interface. In the right-hand example the reverse occurs. So in the left case there are low salt concentrations and low gradients near the interface. In the right-hand example it is merely the other way around. We need a fair number of equations to describe these examples (Figure 20.9). There are three transport relations for the three ions. Additionally there are two bootstrap relations. One states that chloride does not partake in the exchange. The other requires that the net current be zero. Finally we need the electroneutrality relation. The ionic mole fractions at the interface are not known beforehand. In the partly worked-out example we have taken the mass transfer coefficient of hydrogen to be nine times that of sodium. (In reality the difference is not quite as large, but this gives nice round figures.) You can easily check that the electrical terms in the left- and right hand equations are then plus one and minus one respectively. The fluxes in the right-hand case turn out to be three times higher than in the left-hand case.

20.5 Ion Exchange (Particle Limited) With high external concentrations, the external mass transfer resistance may become unimportant. Transport is then limited by phenomena inside the particle (Figure 20.10). We will regard the particle as homogeneous and again neglect swelling. We will also neglect any co-ion transport, although this is not necessarily correct at high external concentrations.

20. Ion Exchange /;'

229

.... -- .........

(~Na+

\~H+ ,~--_/

transport relations

_~I_PZl~=~

RT

Xl _

~2 X2

~3

_

kl,w

Na+

PZ2~ = U2 RT k 2,w

H+

PZ3~_0

Cl-

---::::::::=------

X3 X3

RT

=Xl +X2

~Xl +U2 X2

~l PZl~_ ~ --=-----RT Xl kl,w _

~2

_

x2 _

~3

PZ2~ = U2 RT k 2,w

_

PZ3~ =0

RT

X3

= Xl +X2 U;Xl +U2 X2 = 0

electroneutrality

=0

X3

bootstrap

Fig. 20.9 Same equations, different fluxes!

for high external concentrations, or a tight matrix matrix (3) no current:

Na+ (2) H+(l) (trace)

the important terms in the transport equations are: -

~('fJXl) 'fJXj

PZl A", _ - - L l ' l ' - X2

RT

~ U; -+X3 kl2 k l3

_~_PZ2~=~+~ ~

RT

~ ~

transport is determined by the trace component Fig. 20.10 Transport of a dilute species in a particle

We consider the case that the exchanger has initially been in the hydrogen form. Only a trace of hydrogen is left; it is diffusing outward. From the no-current relation we see that the sodium velocity is much lower than the hydrogen velocity. All terms in the sodium transport relation are then very small. This also applies to the electrical gradient. A look at the hydrogen relation shows that friction of the trace component determines the transport rate. In this case the rate is determined by the faster of the two ions! For you, this may be obvious by now. It was not so for earlier investigators who were used to thinking in terms of the Fick description of diffusion.

230

Mass Transfer in Multicomponent Mixtures

From this example you see that also here, ion exchange rates depend on the direction of transfer. The exchange is faster for the situation shown here than for the opposite case.

20.6 Summary • Ion exchange is one of the sorption processes. Sorption is always transient, and rather complicated. • Ion exchange equilibria are non-ideal. They are dominated by electrical effects and can be roughly modelled with the 'law of mass action' . • There are mass transfer resistances both inside and outside the particle. The external resistance can be important, especially with low external concentrations. • In ion exchange, mass transfer rates depend on the direction of transfer. We have analysed the situation both for external and for internal resistances.

20.7 Further Reading F. Helfferich (1962) Ion Exchange, McGraw-Hill, New York, 1962. Also in a reprint from Dover Books Inc. 1997. Probably still the best introduction to mass transfer in charged matrices and in ion exchange. Helfferich, F. (1983) Ion-exchange kinetics - Evolution of a theory. In Mass Transfer and Kinetics of Ion Exchange, L. Liberti and F. G. Helfferich (Editors), NATO Advanced Study Institute Series E: Applied Sciences - No. 71, Martinus Nijhoff, The Hague. A more recent review by the master. Helfferich, F. (1962) Ion-exchange kinetics. Ill. Experimental test of the theory of particle-diffusion controlled ion exchange. J.Phys.Chem., 66, 39-44.. Demonstrates asymmetry in transport for particle-limited ion exchange (Figure 20.10) Kraaijeveld, G. and Wesselingh, J.A. (1993), The kinetics of film diffusion limited ion exchange. Chem.Eng.Sci., 48, 467-473. Demonstrates asymmetry in transport for film-limited ion exchange (Figure 20-8,9). Pinto, N.G. and Graham, E.E. (1987), Characterisation of ionic diffusivities in ion exchange resins. Ind. Eng. Chem. Research., 26, 2331-2336. Wankat, P.c. (1990) Rate-controlled separations. Elsevier Applied Science, London.

20. Ion Exchange

231

A good text dealing with chromatography, sorption and ion exchange processes. Numerical examples.

Wesselingh, J.A., Kraaijeveld, G. and Vonk, P., 1995, Exploring the Maxwell-Stefan description of ion exchange. Chem. Eng. Jl, 57, 75-89. A review.

20.8 Exercises 20.1 Divalent-Monovalent Equilibrium. The simplest kind of ion exchange equilibrium is the exchange of two monovalent ions (' 1-1' exchange) shown in Figure 20.4. Here, we consider a slightly more complicated example: the ion exchange equilibrium between calcium (divalent) and hydrogen (monovalent). This is known as a '2-1' exchange:

,

,

Ca2 + +2(H+) <=>(Ca 2 +) +2H+ The equilibrium relation is given by:

The total concentrations of positive charges in the two phases are: in the liquid c' = 2cCa +cH

in the particle

The liquid concentration can be varied over wide ranges; the capacity is fixed by the manufacturer of the exchanger. The fractions of the charges of the two ions are: Calcium

Hydrogen

2cCa , _ 2cCa x=-- x - - c c' 1- x

= cH c

1- x' = cH

c'

Write out the equilibrium relation in terms of the fractions as for the example in Figure 20.4. There are several differences. One is that you cannot get rid of the total concentrations in your final result: the equilibria depend on the total concentration in the liquid. Do you get a high affinity for calcium at low or at high concentrations? Can you think of practical applications of this effect? 20.2 Dominating Resistance in Ion Exchange. Which mass transfer resistance dominates, depends on the ratio given in Figure 20.7. At low concentrations of ions

232

Mass Transfer in Multicomponent Mixtures

in the liquid, you may expect the external 'film' to dominate, at higher concentrations diffusion inside the particle. For exchange of monovalent ions the parameters are roughly: film transfer coefficient k =10-5 ms-I; particle diameter d p =0.6 X 10-3 m; internal diffusivities

f)

= 10-10 m 2 s-I; internal counterion concentration

c' = 4000molm-3 .

At which electrolyte concentration in the liquid do you expect the changeover between the two regimes? Use the LDF model for the internal resistance.

*

20.3 Effect of the Direction of Mass Transfer on Ion Exchange (Mathcad). You are to redo the calculations in Figures 20.8 and 20.9 using the equations given there. There are two cases: (1) Na+ being transferred from left to right, and (2) Na+ transferring from right to left. There is a large difference in the transport rates and you are to check whether you understand Why. 20.4 Particle Limited Ion Exchange (Mathcad). You are to investigate the situation in Figure 20.10. Take a constant value of the composition change LU inside the particle, and see how the fluxes vary with the average composition.

233

1 Gas Permeation The movement of gases through porous structures is fairly well understood. The theory is used in heterogeneous catalysis and adsorption. In this chapter we consider the transport coefficients of gases in structures consisting of either parallel cylindrical pores, or a random packing of equal-sized spheres.

21.1 Transport in Cylindrical Pores We begin with a matrix consisting of parallel cylindrical pores. The matrix is regarded as a separate phase. There are two permeating gases, but we can easily extend the reasoning to more components. Figure 21.1 shows the transport equations. First of all, if there is a pressure gradient, there will be viscous flow. We can calculate the viscous velocity with the Poiseuille equation; in not-too-narrow pores, with no adsorption, the viscous velocities are equal for all species. Viscous flow is small in narrow pores, but it increases rapidly with increasing pore sizes. Secondly, there are two diffusion equations. Each contains two friction terms: one covering the interaction between the two gases, and the other friction between the gas and the matrix. If we omit either of the two, we obtain one of the two classical diffusion equations: • If we omit friction with the matrix, we recover the equation for free binary diffusion. • If we omit the free diffusion term the result is the Knudsen equation. This describes diffusion in solids where the pores are smaller than the mean free path of the gas. The Maxwell-Stefan equations used here describe both phenomena and the transition from one mechanism to the other. Of course, the total or whole velocity is the sum of the viscous and diffusive velocities. We use the whole velocity to calculate fluxes of the species.

21.2 The Diffusion Coefficients Before looking at the diffusivities in the matrix, we repeat the formula for the diffusivity in free space, according to the kinetic gas theory (Figure 21.2). We will be referring to this several times. For molecules of greatly different size, the larger of the two diameters dominates. However, the smaller of the two molar masses is the most important.

Mass Transfer in Multicomponent Mixtures

234

membrane with cylindrical pores

Pa

x1a x2a

CD viscous flow v=

ii

P~ XI~

X2~

:E~

1 _flpd 2 P

32 11&

--&~

Q)

diffusive transport

gas-matrix friction (Knudsen diffusion)

--In(p,) --

:-;;--------1 r-T---:

d

dz

=r---(u\-u2),-It--u\'

iD, 2

Pre!

::

Dl M

:

-~In[ P2 )~~(U2 -ulj~_i-U2i dz Pre! : D D 1,2

: :

2,M

:

L - - - - - - T . - - - ... L_______ I

gas-gas friction (free diffusion)

CD

whole (overall) velocity

Fig. 21.1 Transport of gases in a porous matrix

DO 1,2

=

I;; .,.3

'"

(RT)312 t;f pd 2 /1

1,2

(_1_ + _1_)"2 M

1

M

d 1,2

=d1 +d2 2

2

Fig. 21.2 (Free space) binary diffusivity from the kinetic gas theory

Figure 21.3 gives the formulae for the three diffusivities in the matrix. These formulae are valid when • there is no adsorption at the pore walls, and • the pores have a diameter considerably larger than the gas molecules. The gas-gas diffusivity is then equal to the value in free space. The gas-wall diffusivity can be derived from momentum transfer arguments; here we only show the final result (Figure 21.3). Gas-wall friction becomes more important for smaller pores and heavier gases. As usual the mass transfer coefficients are equal to the ratio of the diffusivity and the thickness of the porous layer. We will discuss the effects of adsorption and 'surface diffusion' in Chapter 23.

21. Gas Permeation

235

-

cylindrical pores with: void fraction £. length and diameter

gas-gas diffusivity same as in free space

[

gas-matrix diffusivities (Knudsen)

1 =d

f) i,M

p

~91t 8 RT Mi

diffusion larger (friction smaller) in large pores Fig. 21.3 Diffusivities in cylindrical pores

Figure 21.4 shows the results of experiments on a membrane with cylindrical pores. There is only diffusive transport in these experiments. Can you distinguish the two regimes of our diffusion equation? Ne Ar

_He __~)t

mol m- 2

100

S-1

r----,----~----.---_,

1000!lm 40!lm no fitting parameters!

10-3~---+----~----+---~

10 1 Fig. 21.4 Diffusion of three gases in cylindrical pores

21.3 Looking Back Playing a bit with the cylindrical pore model will give you more understanding of two examples in Chapter 2: the 'three gases' (Figure 2.4) and the 'two gases with a porous plug' (Figure 2.7). A fairly fundamental way of describing the 'three gases' example is to consider it a problem with four components: • the three gases, and

236

Mass Transfer in Multicomponent Mixtures

• the capillary wall. If the volume of the bulbs is large compared with that of the capillary, you may regard the flow at any instant as stationary . You can then set up the transport equations for the three gases, including wall friction and a pressure gradient. This of course causes viscous flow. The geometry also indicates that the net flow must be close to zero. If you solve the set of equations you will find an extremely small pressure difference because the hydrogen migrates more rapidly than the other gases. It is this difference which balances the flows. A simpler way of describing the problem is to regard it as free diffusion of three gases with a 'no net flow' bootstrap relation. For not-too-narrow capillaries you will find that the results of the two approaches are almost identical. The cylindrical pore model also gives a good description of the experiment with 'two gases and a porous plug'. If there is no pressure difference over the plug, the mole fraction differences of the two gases are equal but opposite. It is then easy to show that the gas-matrix friction terms determine the ratio of the fluxes. These in turn contain the square root of the molar masses of the two gases, as we had already noted in Chapter 2.

21.4 Transport in a Bed of Spheres Many porous media are more realistically described as aggregates of spherical particles. Models for such systems have also been developed. Also here, we consider a matrix with two gases (Figure 21.5)

The mixture consists of three 'gaseous' components: an ordinary gas 'I ' an ordinary gas '2' spherical 'dust' particles '3' Fig. 21.5 A bed of spheres: large 'dust' molecules

We ftrst look at the gas-matrix (Knudsen) diffusivity. Possibly to your surprise there is a simple theory which describes the behaviour of this coefficient quite well. In the following you may think your leg is being pulled. But really it works. The idea is that the matrix particles are giant gas molecules ('dust') in an ideal gas mixture (!). Because they are so large their thermal motion is negligible and that is just what happens in a sintered matrix. We can now use the kinetic gas theory to obtain the gas-'dust' diffusivity (step (1) in Figure 21.6). However, we are not so much interested in this parameter as in the gas-matrix diffusivity. This is the ratio of the

237

21. Gas Permeation

gas-dust diffusivity over the mole fraction of dust. We obtain this mole fraction using the ideal gas law (step (2)). The gas-matrix diffusivity, which is derived in step (3) is seen to increase • linearly with the coarseness (particle size) of the membrane, • with the square root of the absolute temperature, and • inversely with the square root of the molar mass of the diffusing species.

CD gas-dust diffusivity @ mole fraction of dust: 1-£

(1-£)RT

n =-3

"d3

"6

3

DI3

Cl) matrix diffusivity:

D1,M =

X3

x~

=

~d{p)l

23/2 (RT)1/2 d d ~ = 31t1l2 (1- £ )M 172 = 1~£ ~ 91t MI 1

D2,M similar

this is only valid for a 'gas': Fig. 21.6 Deriving an expression for the gas-matrix diffusivity

The derivation above is only valid for a 'gas', so for low concentrations of the dust particles. We come back to this in a moment. Figure 21.7 shows a few aspects of the gas-gas diffusivity. The idea is that we are dealing with tortuous pores with wide parts and narrow parts (constrictions). The tortuosity has two effects: it increases the pore length and reduces the effective pore width. So the effect on the diffusivities is quadratic. The constrictions also reduce the effective diffusivity: the extra friction in the narrow parts is not balanced by the reduced friction in the wide parts. Together the effect is to reduce the gas-gas diffusivity by roughly one order of magnitude (less in very open structures, more in compressed structures). In many references you will find the ratio of tortuosity squared over constriction factor designated as the 'tortuosity'. We have kept the two things separate. Note that the gas-gas coefficient does not depend on the size of the channels or the particles of the matrix. For dense structures, we need to apply a tortuosity-constrictivity correction to the gas-matrix diffusivity just as we have done for the gas-gas diffusivity. This correction depends on details of the structure of the matrix. However, the formula in

238

Mass Transfer in Multicomponent Mixtures

Figure 21.8 roughly describes the correction for many structures. This yields our final formulae for the diffusivities in a bed of spherical particles. the matrix contains tortuous pores with constrictions

~ .....•••....••. ' . :; ~ 't& CD these cause longer paths .

.

'tu

Cl) and higher velocities

. ......

.

(for a given flux)

·V·.·

the friction

.••. . : ; ' .

/"-.....~ . ./"-.. .

I ~-

I

constriction 1,2 -:2 ,actor 't, ", tortuosity . .'!J...... ~

D -Dol 1,2 - /

/

./ / diffusivity in free solution

-- .{;

0.03 ... 0.3

Fig. 21,7 Gas-gas diffusivity in a porous medium - tortuosity and constriction factor

for many structures we find roughly:

1

o o

L...-_ _- - L_ _ _- '

€

1

Fig. 21.8 Diffusivities in a bed of spheres

We calculate the viscous flow in a matrix of spherical particles with relations for packed beds (such as the Carman-Kozeny equation, Figure 21.9). The viscous flow increases with the pressure gradient and the square of the diameter of the matrix particles. The velocity we need is not the superficial velocity, but the average velocity in the pores. That is why the exponent in the void fraction is one less than in packed bed relations.

21.5 The Dusty Gas Model The results of the bed-of-spheres model are very similar to those of the model with cylindrical pores. You will note that the effects of temperature, pressure, molar mass,

21. Gas Permeation

239

diameters of the permeants and dimensions of the matrix are identical. The two models are examples of a more general model known as the 'dusty gas model' (Figure 21.10). The constants in the transport coefficients are usually determined experimentally. VsuperfiCial

=V£

Fig. 21.9 Viscous flow in a bed of spheres

constants determined by experiment for each material

Fig. 21.10 Practical Dusty Gas Model

The Dusty Gas Model is one' of those parts of science that has been discovered many times (Figure 21.11).

The Dusty Gas Model has been discovered, independently, by at least five different scientists in five different periods ... (We only read our own publications.)

Fig. 21.11 Science

240

Mass Transfer in Multicomponent Mixtures

21.6 Summary This chapter has introduced two heterogeneous models for transport of gases through a porous medium: • a model with parallel cylindrical pores, and • a model consisting of a packed bed of equal spheres. Both models are similar. They contain terms for • gas-gas interactions, • gas-wall interactions and • viscous flow. They require a dimension: a pore size or particle diameter. In addition the bed-ofspheres model requires the void fraction, and tortuosity and constriction factor of the membrane. The two models can be regarded as special cases of a more general model. This is known as the 'dusty gas model'. If we know the structure of the matrix, we can estimate the coefficients of the model. However, they are usually determined via diffusion and permeation experiments.

21.7 Further Reading Remick, RR. and Geankoplis, C.J. (1974) Ternary diffusion of gases in capillaries in the transition region between Knudsen and molecular diffusion. Chem. Eng. Sci., 29, 1447-1455. The source of the experimental data used in Figure 21.4 to verify the dusty gas model. Jackson, R, Transport in Porous Catalysts. Elsevier, Amsterdam, 1977 An excellent book about the dusty gas model. Written for the chemical engineer. Cunningham, RE. and Williams, RJ.J., 1980, Diffusion in gases and porous media. Plenum Press, New York. Provides a very good historical perspective of diffusion inside porous media. Mason, E.A. and Malinauskas, A.P., 1983, Gas transport in porous media: The dusty gas model. Elsevier, Amsterdam. The first author is considered to be one of the fathers of the dusty gas model. The book is very readable. Epstein, N. (1989) On the tortuosity and the tortuosity factor in flow and diffusion through porous media. Chem. Eng. Sci., 44, 777-779. A clear exposition of the proper interpretation of the tortuosity factor.

21. Gas Permeation

241

21.8 Exercises 21.1 Modifying the Knudsen Relation. A diffusivity can often be regarded as the product of a jump velocity, a jump distance and a coefficient which is a little smaller than unity. For gases, the velocity is obviously the thermal velocity ~ RTf M; .

vi "'"

(a) What would the jump distance be in the Knudsen regime in the parallel pores of Figure 21.3? (b) You would expect the jump distance to actually be smaller by one molecule diameter, because the molecule can only approach the wall up to its radius. Modify the Knudsen relation for the gas-solid diffusivity to take this into account. (c) For which pore diameter does the gas-solid diffusivity now become zero? 21.2* Three Gases with the Wall (Mathcad). We go back to the three gases problem of Figures 2.4 and 2.5. You are to include the effect of friction with the tube wall and to find how small it is. This is a good exercise for getting hold of all main effects of the Dusty Gas Model. 21.3 Helium, Neon and Argon in a Capillary (Mathcad). In this file you are to analyse the experimental data of Remick and Geankoplis leading to Figure 21.4. This is quite a challenge. 21.4 Different Models for Tortuosity (Mathcad). If you want to know how we came to choosing the tortuosity - constriction factor in Figure 21.8, have a look at this file. If you still find this a baffling subject, do have a look in the article of Epstein. 21.5 Diffusivities of Nitrogen, Hydrogen and Ammonia in a Catalyst (Mathcad). This is a four component system (the fourth is the matrix). So you need the following set of diffusivities:

D1,z

DI,3

D1,M

DZ,3

DZ,M D 3 ,M

This file shows how to calculate these for non-adsorbing species in an array of spherical particles. This is lengthy but straightforward; only use this file if you need it. 21.6 A Novel Process for Gas Separation (Mathcad). A student having followed a course by Wesselingh and Krishna on the Maxwell-Stefan approach to mass transfer has come up with an idea for selective removal of styrene (1) from a mixture with ethyl benzene (2) and hydrogen (3). You are required to examine this invention and comment on its feasibility. This requires carefully reading and thinking about a file.

242

In Porous Catalysts When a single chemical reaction occurs in a porous catalyst, reaction stoichiometry governs the ratios of all fluxes. This has interesting consequences.

22.1 Introduction In this chapter, we consider diffusion and reaction in porous particles that are used in the chemical industry as catalysts. The subject is not an easy one, and we will have to accept a few limitations. • We only consider ideal gases. • We only consider a single reaction (because two simultaneous reactions greatly complicate the problem). • We neglect viscous flow in the particle. (Viscous flow is usually small in catalysts; we will see how it can be simply taken into account in Appendix 3). • We assume a uniform temperature in the particle. (This is usually so, although there can be large temperature gradients in the gas outside the particle.) • We neglect diffusion of adsorbed species along the pore surfaces (surface diffusion). (We discuss surface diffusion in Chapter 23, but without chemical reactions. It is not important here.) In a catalyst, there is both diffusion and chemical reaction. To obtain compositions and pressures, we must solve the Maxwell-Stefan differential equations; difference approximations are not useful here. Figure 22.1 shows the form of the equations used here. As in the previous chapter, we regard the matrix as a separate phase. Note that the fluxes are averaged over all space (as everywhere in this book): they are not the fluxes in the pores that are often used in literature on catalysis. The equations contain partial pressures as driving forces, fluxes instead of velocities and friction coefficients instead of diffusivities. The concentrations in the equations are those in the gas; they do not contain any adsorbed species. Catalysts have complicated three-dimensional (3-D) shapes. Solving of 3-D equations is outside the scope of this text. Fortunately, most 3-D structures behave much like an infinite slab with the same volume to surface ratio (Figure 22.2). We will solve the one-dimensional equations for the slab, to obtain estimates of the fluxes through the catalyst surface. This exercise also gives us an idea of concentration and pressure profiles in the real particles.

22. In Porous Catalysts

243

basic equation 'h WIt C

N u=-',

€CX

~

---isthe total concentration of permeants in the pores

also

j

RT

I

p

C

Fig. 22.1 The Maxwell-Stefan equations as used in this chapter

~

these shapes behave like the same slab with a thickness: d=2 volume surface the fluxes through their surfaces are also equal

partial pressure profile

o Fig. 22.2 Approximating real catalyst shapes by a slab

22.2 Pressure gradients inside a particle If a reaction causes a change in the number of moles, then it will cause a pressure gradient. Figure 22.3 illustrates this for the simple irreversible gas reaction A ~ vB. This can represent a cracking reaction (where the number of moles increases) or a polymerisation reaction (where it decreases). In the first case, there is a net molar flow of gas out of the particle, and the pressure inside is higher than outside. With polymerisation, the pressure inside is lower than outside. To illustrate the effects, we consider a catalyst with small pores, where friction is mainly between gases and the matrix ('Knudsen diffusion'). From the transport equations, we obtain the ratio of the pressure gradients of the two components in

244

Mass Transfer in Multicomponent Mixtures

Figure 22.4. The stoichiometry relation and the ratio of the gas-matrix diffusivities cause the ratio to have an absolute value equal to the square root of the stoichiometric coefficient. a simple reaction

A~vB

Fig. 22.3 Pressure profiles inside a catalyst

transport equations

dpI

_I'

-e--~IM

dz

stoichiometry and Knudsen relations

N

'

_2=-V

NI

N

I

dP2 dz

-I'

-e--~2M

'

N2

_~2_~ -- -

S2,M -S],M

M]

V

Fig. 22.4 Ratio of the two pressure gradients

Figure 22.5 shows a simple situation where the bulk gas consists of the pure reactant, and the reaction goes to completion inside the particle. The pressure differences are then at their largest. We see that the reaction, producing four moles of gas per mole of reactant, doubles the pressure inside the particle. On the other hand, a tetramerisation reaction reduces it to one half.

A~4B

1

A~-B

4

Fig. 22.5 Pressure differences with reactions going to completion

22. In Porous Catalysts

245

In many reactors, the effects of internal pressure gradients in the catalyst are not large. So they are often neglected. However, neglecting pressure gradients leads to inconsistencies in the flux calculations. The Maxwell-Stefan equations include the pressure gradients automatically.

22.3 Separate Transport Equations Again consider the reaction A ~ vB. We first write down the transport relation of A (Figure 22.6). Then we eliminate the flux of B using the stoichiometry relation. We end with an equation containing only the flux of species A and an effective friction coefficient. N2 =-vN\

binary transport equation of A (l)

dp _e \ dz

=t;\2(X2N~2)+t;\MN\ ' ,

-e dPl = [( x 2 + vxIK12 + SI M]NI dz

"

effective transport equation

Fig. 22.6 Effective friction coefficient for a binary mixture

For the case that v = 1 (but only then), we obtain the commonly used Bosanquet formula (Figure 22.7). when v= I Bosanquet formula

D I,eft --

[

1

I

1,2

I,M

D +D-

-1 ]

Fig. 22.7 Bosanquet formula for a binary mixture

Also in mUlticomponent mixtures, we can obtain separate flux equations for each component. The formula for the effective friction coefficients is a generalisation of the binary case (Figure 22.8). As an example, we have worked out the friction coefficients for the synthesis of ammonia. Here, effective friction coefficients are only a weak function of composition: you can use values calculated at the bulk composition. What is more important: their ratios are almost independent of composition.

22.4 Single-variable Pressure and Rate Expressions The following paragraphs are a little formal and you may wonder where we are heading. The whole exercise is to obtain the transport and reaction equations in terms of a single variable. We will then be able to integrate these equations to obtain fluxes

246

Mass Transfer in Multicomponent Mixtures

and composition and pressure profiles in a slab. As an example, we will use the synthesis of ammonia. We arbitrarily choose nitrogen (1) as the key component and will express all pressures in terms of its partial pressure PI' 1'1 ,fF '::> ,eJJ

multi component formula

= ~(x. ~ I

+ v·',1' .x.)!'. . + 1'. M '::>',1 '::>"

j#

NI =N2 =_N3

132 SI,eff =(x2 -3XI)~1,2 +(X3

i )Sl,2 + ( + i )SI,3 + (

+2XI)~1,3 +~I,M

+ ~ x2 )S2,3 + S2,M

S2,e/f

= (Xl -

X3

S3,e/f

=( Xl

x2 +

~ x3 )S2,3 + S3,M

Fig. 22,8 Effective multicomponent friction coefficients

When we are dealing with a single reaction, the fluxes of all species are coupled by the reaction stoichiometry. Again, the effective friction coefficients have constant ratios. As a result, the partial pressure gradients of the different species also have fixed ratios (Figure 22.9). This means that any partial pressure can be expressed in terms of the partial pressure of the key component

(1) (2)

- e dPl dz

=

~1 'effNl

(3)

key component

-e dP2 = ~2 effN2 = 3~2 effN1

dP2

dz'

,

dPl

- e dP3 = ~3 effN3 dz'

= -2~3 ,",e«NI

dP3 dpI

= 3 ~2,eff ~1,eff = -2 ~3,eff ~I,eff

at an arbitrary point a~(--)-~-----~~-s~,:;-(----=----)-! : P2 PI - P2,a ~ PI,a PI: :

r

I I

(

) _

: P3 PI - P3,a -

l,eff

2 ~3,eff

: (

-~- Pl,a -

)

I I

PI :

L________________ __________: ~~~

Fig. 22.9 All partial pressures expressed in the pressure of the key component

22. In Porous Catalysts

247

The kinetics of heterogeneous reactions are usually complicated. Figure 22.10 shows an expression for the production of nitrogen in the synthesis of ammonia. It contains partial pressures of all three participating components. We note that there are two. terms with opposite signs. When these are equal, the reaction rate is zero and we have equilibrium. If we let the reaction go to equilibrium locally each of the reactants has its own equilibrium pressure. net rate offonnation ofN2 : ,-,-0 KI-7 -

if the reaction were to go to equilibrium locally:

( eq)2 P3 eq eq eq 3 -7 PI PI (P2 )

Fig. 22.10 Rate of formation of nitrogen and equilibrium relation

Because the pressures are inter-related, we can write the rate equation in tenns of a single pressure. (In the example, this is again that of nitrogen.) The resulting equation will look complicated, but we can easily plot it. The range of interest is usually between the pressure of the key component outside the particles and its equilibrium pressure. Over this range, the reaction rate is usually linear in the partial pressure of the key component (Figure 22.11). So we may regard the reaction as a pseudo first-order equilibrium reaction.

range of interest ~

I

I I I I I I

Fig. 22.11 Rate of formation of nitrogen as a function of its partial pressure

If you go too far outside the 'range of interest', one of the partial pressures may

become negative. This leads to a wild asymptotic behaviour of the rate function. Another point to notice is, that the constant kl is not a true chemical rate constant. For example, it depends on the ratios of the friction coefficients of the components.

22.5 Solution for a Slab We can now set up the equations for diffusion and reaction in a slab. A balance over a layer with thickness dz gives the differential equation in the flux NI shown in

Mass Transfer in Multicomponent Mixtures

248

Figure 22.12. By inserting the flux and the reaction rate relations, we get our final differential equation in the partial pressure PI of ammonia. The boundary conditions should be clear. I

I

NI"NI ~ ~z+dz :.+-. rldz I

o

z

I I

z+dz

boundary conditions

z = 0 ~ PI =

PI,a

d 2

dp dz

Z=_~_I =0

Fig. 22.12 Differential equation governing the nitrogen pressure

The equation is second order and linear; we can solve it quite simply (Figure 22.13). make the equation dimensionless with: P=

eq

PI-PI eq PI Iz=O - PI

boundary conditions the solution is:

A=

d 2P -=KP dZ 2

Z=2z d Z=O~P=1

Z=I~ dP =0 dZ

P(Z)= Aexp(Z.JK) + Bexp(-Z.JK)

exp(-.JK) exp(.JK) + exp(-.JK)

B=

exp(JK) exp(JK) + exp(-JK)

Fig. 22.13 Solving the differential equation

Figure 22.14 shows a typical set of partial pressure profiles in the catalyst. The total pressure inside the catalyst is lower than outside, but only very little. We see that the nitrogen pressure drops rapidly to its equilibrium value, and the central part of the catalyst is hardly utilised. A measure for the utilisation of the catalyst, is its effectiveness factor. This is the ratio of the average reaction rate in the catalyst, compared with the value at the outer surface (which in this case is also the maximum rate). It has a low value when fast chemical reactions occur in large particles. You can calculate the effectiveness factor as the average value of P (Figure 22.15).

22. In Porous Catalysts

249

20,----,----,

-

Pi MPa

10

NH3 ~

0

0

N2 I

dj2

Fig. 22.14 Pressure profiles in a catalyst

effectiveness factor

1J=

average rate . maXImum rate

I

f P(Z)dZ o

Fig. 22.15 Calculating the effectiveness factor of the catalyst

22.6 Summary • In a heterogeneous catalyst with a single reaction, all fluxes have fixed ratios because of stoichiometry. • As a result, there is only one independent Maxwell-Stefan relation. • To solve this equation we need effective friction coefficients of the separate species. These are almost constant; we can use values calculated at the bulk composition. • We can express the reaction rate in the partial pressure of any single component. It is usually possible to find a component for which the rate is linear in its partial pressure. • This 'linear first-order equilibrium' reaction can easily be handled in a slab to yield fluxes, pressures and composition profiles.

22.7 Further Reading Haynes, H.W., 1978, Calculation of gas phase diffusion and reaction in heterogeneous catalysts. The importance of viscous flow. Canad. J. Chem. Eng., 56, 582-587.

250

Mass Transfer in Multicomponent Mixtures

Hite, R.H. and Jackson, R., 1977, Pressure gradients in porous catalyst pellets in the intermediate diffusion regime. Chem.Eng.Sci., 32, 703-709. Jackson, R., Transport in Porous Catalysts. Elsevier, Amsterdam, 1977 Kaza, K.R. and Jackson, R., 1980, Diffusion and reaction of multicomponent gas mixtures in isothermal porous catalysts. Chem. Eng. Sci. 35, 1179-1197. Schnitzlein, K. and Hofmann, H., 1988. Solving the pellet problem for multicomponent mass transport and complex reactions. Comput. Chem. Eng, 12, 1157-1161.

22.8 Exercises 22.1 The Chapter-22 MS-equation. Have a look at the equations in Figure 22.1. The basic equation is that for a gas diffusing together with another through a solid matrix. The driving force is the chemical potential gradient; this only contains the partial pressure. The matrix is taken to be separate. The velocities in this MS-equation are velocities averaged over all space, not those in the pores. With a reaction occurring, we use fluxes, not velocities. To get the fluxes into the equations, we need a gas cOhcentration also averaged over all space. Two questions: a) Why is it easier to use fluxes than velocities when dealing with a chemical reaction? b) In the final equation we see an e before the pressure gradient. Where does this come from? 22.2 Pressure Gradients inside a Particle. Just after Figure 22.3 we state ' ... we consider a catalyst with small pores, where friction is mainly between the gas and the matrix ... '. Is this statement in agreement with the equations for diffusivities in Figure 21.8? Do gas-gas interactions become unimportant in small pores? 22.3 Pressure Gradients within a Catalyst Particle (Mathcad). See what happens to the pressure gradients when the reactions shift from cracking to dimerisation. 22.4* Reaction in a Diffusion Cell (Mathcad). This is really a preparation for the next exercise. It considers a cell with two volumes separated by a porous membrane. Gas flows through the left volume; it can react on a catalyst in the other volume. You are to calculate the diffusion rates through the membrane and to check whether it is allowable to neglect viscous flow.

22. In Porous Catalysts

251

*

22.5 Reaction in a Diffusion Cell using Fluxes (Mathcad). This concerns the same cell as in the previous example. You are to derive explicit approximations for the fluxes in a diffusion problem with a single chemical reaction, and to compare this approximation with the previous example. 22.6 Pseudo First-order Reaction. Under which conditions do you expect a reaction· to behave as pseudo first-order? 22.7 Water Gas Shift Reaction (Mathcad). This fIle considers the reaction: CO + Hp {:=} CO 2 + H2 The reaction occurs in a porous cylindrical catalyst. You are to investigate what happens to the effectiveness of the catalyst when the cylinders are replaced by cylinders with a hollow core. This is a large and instructive fIle: it uses the greater part of Chapter 22 and can be easily adapted to study other reactions.

252

In Adsorbents In adsorbents, permeants concentrate on the internal surfaces of porous solids. They diffuse mainly by slipping along the pore walls - by 'surface diffusion'!.

23.1 Adsorption Adsorption is usually done in a fixed bed, like ion exchange (Chapter 20). However, the mechanism that binds a solute to the matrix differs from that in ion exchange. Here, binding is mainly due to Van der Waals forces between solutes and the pore walls and not due to electrical effects. Adsorbents require large surfaces for a reasonable capacity. This implies very narrow (micro) pores, in which diffusion is slow. To speed things up, adsorbents are made as agglomerates of fine sub-particles. The diffusion to the sub-particles occurs through the macropores between them (Figure 23.1). In such agglomerates we can distinguish three main diffusional resistances: • the film resistance outside the particle, • the macropore resistance, and • the micropore resistance. In a rough kind of way you can regard these three as resistances in series. Here, we only regard resistances inside the particles (which are often the most important). macropores between sub-particles

sub-particles with

real

Fig. 23.1 Particle consisting of sub-particles with macropores in between

The diameters of micropores are around one nanometre; macropores might have a diameter around one hundred nanometres. Micropores are at most a few molecules I

Diffusion over the surface, not diffusion o/the surface.

23. In Adsorbents

253

wide. The pore shape and the distribution of sizes depend very much on the kind of adsorbent. Some, such as activated carbon, have a wide range of pore sizes. The micropores in activated carbon can be considered as slits left by randomly removing rings of six carbon molecules from a graphite structure (Figure 23.2). Zeolites are crystalline substances; they have micropores with a very regular structure. Such adsorbents are also known as molecular sieves because they can separate species sharply by differences in size and shape. For adsorbing species in micropores, slipping past the surface is the dominant transport mechanism. In the Maxwell-Stefan equations, a potential gradient drives the slipping. Before we can describe it quantitatively, we must first look at the thermodynamics of solid-fluid equilibria.

23.2 Equilibria - Langmuir Isotherm A simple model for solid-fluid equilibria, often used in theoretical studies, is that of Langmuir. It assumes that the surface consists of sites, which are all equal and equally accessible to the adsorbing species2• The idea of an 'open' surface is a little misleading; you can better think in terms of narrow slits or interconnected pores (Figure 23.2), but this makes no difference in the reasoning. surface with sites

activated carbon: slits

surface

zeolites: pores

Fig. 23.2 Surface of the adsorbent with 'sites'

You can understand the Langmuir model most easily by considering adsorption to be an equilibrium reaction between a free solute (A) and free site (S) to form an occupied site (AS) in Figure 23.3. The concentration of free sites is equal to the concentration of all sites c, minus that of the occupied sites. This leads to the

Solid surfaces are nearly always heterogeneous - their properties differ from point to point. Also the accessibility of different sites depends on the size and shape of molecules. Finally, adsorption is not necessarily restricted to a monolayer. As a result, solid-fluid equilibria can be very complicated; such complications are not taken into account by the Langmuir isotherm.

2

254

Mass Transfer in Multicomponent Mixtures

equation in the first line. When we are dealing with gases, it is simpler to replace the free solute concentration by its partial pressure, and the concentration of occupied sites by a mole fraction 3. This yields the pressure form of the equation. It contains two adjustable constants: the concentration of sites C and the parameter p~, which is a measure of the tendency of the adsorbed species to escape from the adsorbent. The resulting isotherm shows a rapid rise at low pressures; it levels off as all sites become occupied. If we set the activity coefficient of the adsorbed species equal to one at low pressures, then the value at higher pressures is given by the last formula in Figure 23.3. The activity coefficient rises sharply as we approach saturation of the surface. equilibrium A + S {:::} AS

~/

free A

free site

~

occupied site

using mole fractions activity coefficient

o

K=

>

=

CAS CA(C-C AS )

./

total concentration of sites

_ CAS

>

X1 - C

"fJ

CAS CACS

= (1- x1rl

L -_ _ _ _ _ _~_ _ _ _ _ _~_ _ _ _ _ _ _ _~_ _ _ _ _ _~_ _ _ _ _ _~

o

2

4

6

8

10

Fig. 23.3 The Langmuir isotherm

We can easily extend the Langmuir model to include more adsorbing species. Figure 23.4 shows the result for two species. The important thing to note is, that the two species strongly influence each other (at least for high surface coverage). A site occupied by one species cannot be occupied by the other. Also note that the activity coefficients of both species are equal.

23.3 Maxwell-Stefan and Fick Diffusivities We can regard the surface of the adsorbent as consisting of 'sites', in which the adsorbing species has an energy -E lower than in the surroundings. The distance between the sites is d. Molecules are jumping around with a (varying) thermal

3

In literature on adsorption, this mole fraction is called the surface occupancy and it has the symbol

8.

23. In Adsorbents

255

velocity vi; only if their thermal energy exceeds E can they jump to an adjacent site. The resulting formula for the diffusion coefficient is shown in Figure 23.5. It shows an Arrhenius type of dependency on the temperature.

Xl

v::---------------------------component 1 I

," ,,'

,-

--- ---- -------

capacity occupied by'] , cannot be occupied by '2'

component 2

I

activity coefficients:

:: k?------~--~-:~~-;---------

Yl =Y2 =(1-x,- xzy l

",----------------!

I

I

I

Fig. 23.4 Two component Langmuir isotherm

thennal jump velocity

~

~--~~

-~;~-~ energy of particle

activation energy E ~

jump distance d;ump

chance of a jump succeeding

diffusion coefficient

I\

vid

jump

V",M '"

(E )

--4- exp - RT

----

Fig. 23.5 Diffusion by jumping between sites

256

Mass Transfer in Multicomponent Mixtures

The diffusion coefficient of a single species usually shows the temperature dependency predicted. An example is the behaviour of the diffusivity of butane in silicalite-l shown in Figure 23.6. We see that the surface diffusivity can vary over a few orders of magnitude.

400

300

500 600

~

T K

butane in silicalite-l

10-

9

10-10

10-11

3.5

1.5

2.5

T- I ~ 1O-3 K- I

Fig. 23.6 Effect of temperature on surface diffusivity

In many cases we find that the Maxwell-Stefan diffusivity is independent of the coverage of the surface. We can understand this if a jumping molecule has such a high energy that it dislodges any molecule occupying a neighbouring site. When the Maxwell-Stefan diffusivity is constant, the corresponding Fick diffusivity must vary because the activity coefficient varies. Using the relation between the two diffusivities that we derived in Chapter 8 yields the result in Figure 23.7. We see that the Fick diffusivity rises sharply with increasing surface coverage of the adsorbent. often (not always)

D1,M '"

constant

YI =(l-xlr

Fick MS

DI,M DI,M

l

5

=l+x d(lnYI) =_1_ I

dxl

I-XI

the Fick diffusivity often rises sharply with surface coverage

1

o

Fig. 23.7 Relation between the Fick and Maxwell-Stefan diffusivity

Figure 23.8 shows an example of this kind of behaviour: the Fick diffusivities of oxygen and nitrogen both increase rapidly with increasing surface coverage. We see

23. In Adsorbents

257

that the surface diffusivities can differ greatly (even for similar molecules). That of oxygen is thirty times higher than that of nitrogen. (These are results from uptake experiments in which the size of the crystallites cannot be determined accurately. They are around one micrometre. From the experimental data, you can only determine the ratio of the diffusivity to the crystallite radius squared, as is plotted.) You should not have the illusion that Maxwell-Stefan surface diffusivities are always independent of composition. Indeed, there are examples where the Fick diffusivity varies less than the MS-diffusivity. The models that we have used here are too simple to cover all of surface diffusion. 1 X 10-4

T=300K D"M / rC;ystallite s-'

D1,M / rc~stal/ite s-'

lxlO-3

o Fig. 23.8 Diffusivities of oxygen and nitrogen on a carbon molecular sieve

23.4 Macropore Diffusion In Chapter 21 we have looked at non-adsorbing gas~s in pores. There, friction between gas and matrix was caused by molecules bumping against walls. Here, we have been looking at friction caused by adsorbed molecules slipping past a surface. In macropores, these two mechanisms may occur in parallel (Figure 23.9). We see that the permeant-matrix coefficient now consists of two parts: a surface and a Knudsen contribution. The'S' diffusivities are those from this chapter; the 'K' diffusivities are those from Chapter 21. Which of the two parts dominates, depends both on the fraction adsorbed and on the magnitudes of the'S' and 'K' diffusivities. Note that the concentrations used above are in mole per m 3 of (adsorbent including pores). In macropores, permeant-permeant diffusivities can also be important. They can be averaged in the same way.

Mass Transfer in Multicomponent Mixtures

258

_ _ NW ~

~

~

~UK ~ et

>

ut

Surface contribution }

Knudsen contribution

MM ElM *'"»

Fig. 23.9 Surface and Knudsen diffusion in parallel

23.5 Transport Equations The transport relations in microporous adsorbents are as usual. The driving force on a species is the gradient of its chemical potential, and this is counteracted by friction. We regard the matrix as a separate phase. Friction is mostly between the permeants and the matrix; we will neglect permeant-permeant interactions (Fig 23.10).

- dll 2 = r U N _ D

dZ

~2,M 2 -----;.

2 --

dlna2 _ D d(ln(Y2 x2» 2M CX2 - - - 2M CX2 -----'------'-'--"----"-'-"'

dz

'

dz

Fig. 23.10 The diffusion equations used here

We write out the chemical potential gradients in terms of the mole fractions of the two permeants (Figure 23.11). We see that both gradients depend on both mole fractions, and that the coefficients before the gradients can be of a similar size. This means that the two permeants influence each other; they are strongly 'coupled'. _ dJ11 dz

= _RTd(ln(Ylxl» = -RT[ dz

(1- x2) dx l XI (1- XI - x2) dz

+

1 dx2] 1- XI - x2 dz

_dJ12=_RTd(ln(Y2X2»=_RT[ 1 dx,+ (I-x,) dx2] dz dz 1- XI - x2 dz x2(1- XI - x2) dz Fig. 23.11 Driving forces written in mole fractions using 'Langmuir'

As usual, we will be doing our exercises, not with the complete differential equations, but with difference equations. We use the same linear driving force or LDF model that we have already introduced for ion exchange in Chapter 20, but taking the strong non-idealities into account. Figure 23.12 repeats the main features.

259

23. In Adsorbents

D

o resistance in the interface, corresponding to a film with d DM Llz""~ k =10-1, 10 I,M d

j · · . · · .. ,

I.. I'

1 1

1

1

1

1

k--d~

:1

o

t> :1

bulk concentration depends on time

VCdxi~ =NA ~ dxi~ =~N dt

[

o

~-_x 1 1 1_1 1

.

la

dt

I

cd

I

flux relation

- -kiMCXi~ - /:).a i Ni , a

X. I

i

Xi~-

Fig. 23.12 The linear driving force model

23.6 Transient Adsorption of a Binary The strong non-idealities in the phase equilibria and the large differences in the diffusivities of the permeants in adsorbents lead to interesting effects. As an example, we look at the simultaneous uptake of n-heptane and benzene in NaX zeolite crystals (Figure 23.13). To our surprise, we see that the uptake of n-heptane goes through a pronounced maximum. (The solid curves are the result of LDF modelling ofthe problem.) 1

benzene (2)

fraction of adsorbent occupied by species

n-heptane (1)

o o

2.104 ~ s

Fig. 23.13 Uptake of n-heptane and benzene in NaX zeolite crystals

There are two reasons for this behaviour. First of all, benzene adsorbs more strongly in the crystal than does n-heptane. On the other hand, n-heptane is much more mobile: its MS-diffusivity is about thirty times higher than that of benzene. This is in

260

Mass Transfer in Multicomponent Mixtures

part due to the lesser bonding, but also to differences in shape. N-heptane is a snakelike molecule that can sliver through the pores of the crystal; benzene is a rigid molecule that only just fits in the pores. As a result, n-heptane first occupies the crystal and is then slowly displaced by benzene. You will understand that there are two ways of operating separation processes using such an adsorbent. The classical way is to utilise the difference in binding strength: we fill up the column with benzene. This takes a long time. As an alternative, we can operate the column with short cycles in which it is mainly filled with n-heptane. The column is then much smaller. Designing such a dynamic process does require a good understanding of mass transfer processes.

23.7 Membrane Applications Figure 23.14 shows another interesting effect. This is a series of experiments with hydrogen and n-butane passing through a zeolite membrane. The left side shows experiments with the pure gases, which are fed through the membrane starting at zero time. We see that hydrogen comes through quickly, and with a high flux. Butane stays longer in the membrane, and also its final flux is much lower. If we feed a mixture (the right side of the figure) the flux ratios reverse. The flux of butane is almost the same (the two figures have different scales), but that of hydrogen is greatly reduced. The reason is that all sites are occupied by butane: this hardly allows any passage of hydrogen. Butane is transported by surface diffusion.

binary

single component hydrogen

20 10

o o

100 t /s

o

200 t /s

Fig. 23.14 Selectivity reversal in a zeolite membrane

A commercial process uses this phenomenon for hydrogen recovery from refinery waste gases (Figure 23.15). Also here, hydrocarbons permeate, while hydrogen is retained. The advantage over a conventional gas separation membrane is that the hydrogen does not have to be compressed after separation

23. In Adsorbents

o

hydrogen

•

n-butane

~.

Q

refinery waste gases

surface

·o~~~Sion highp 0.0

.~

'--J

261

hydrocarbons

lowp

LJ

molecular sieve membrane

hydrogen recovery

Fig. 23.15 Hydrogen retention in a waste-gas membrane unit

23.8 Summary • In adsorbents, the adsorbing species can have a high surface concentration on the pore walls. So diffusion can be dominated by slipping of this adsorbed material past the pore walls ('surface diffusion'). • Because of the strongly non-linear isotherms in adsorbents, the different species influence each other greatly. The Maxwell-Stefan equations take this into account properly. • MS-diffusivities in adsorbents are often (roughly) independent of surface coverage. This leads to a sharp increase of the Fick diffusivity with increasing coverage. • In simultaneous adsorption of two components, one component often has a high binding strength, but a low diffusivity. This occurs even for similar components. This effect allows for interesting dynamic separations. • The same effect may cause reversal of the selectivity in a microporous membrane fed with a binary mixture.

23.9 Further Reading Krishna, R. and Wesselingh, I.A. (1997) The Maxwell-Stefan approach to mass transfer, Chem.Eng.Sci. 73, 861-911. This discusses the Maxwell-Stefan theory for transport inside a particle in detail. Extensive literature references. Karger, I. and Ruthven, D.M., 1992, Diffusion in Zeolites. Wiley, New York. A good reference book

262

Mass Transfer in Multicomponent Mixtures

Ruthven, D.M., 1984, Principles of Adsorption and Adsorption Processes. John WHey, New York. The principles of adsorption are clearly laid out. Still the best book on the subject. Ruthven, D.M., Farooq, S. and Knaebel, K.S.,1994, Pressure Swing Adsorption, VCH Publishers, New York. Covers the design aspects ofpressure swing adsorption equipment. Yang, R.T., 1987, Gas Separation by Adsorption Processes. Butterworth, Boston. A good text.

23.10 Exercises 23.1 Macro and Micropores. This is a good point to look at the differences between the macropores of Chapter 21 and the micropores considered here. Take an ordinary sheet of paper and draw a circle with a diameter of 10 centimetres (this is a little over one half of the page width). Now draw a dot inside the circle of half a millimetre. This is the thickness of the (thin) leads used in most pencils nowadays. If you could reduce the scale of this drawing by one million times, you would have a macropore with a diameter of 100 nm, containing a molecule with a diameter of 0.5 nm. There is a lot of empty space in the macropore. Even if there is a bit of adsorption on the walls most molecules will still be in the bulk. Transport is primarily through the bulk - not along the surface. Draw a second circle with a diameter of one millimetre and place again a dot inside of half a millimetre. You now have a micropore, which scales down to one nanometre. The molecule is never more than one diameter away from the wall. There is no bulk fluid - transport can only occur by slipping along the walls of the pore. 23.2 We have an adsorbent with slit shaped pores with a thickness of 1 nm. The volume fraction of the pores is 0.3. A 'site' occupies an area of 0.3 (nm)2. What is the concentration of sites in mol (m3 of adsorbent)"!? This is the value required to define the mole fractions in the Langmuir isotherm (Figure 23.3). 23.3 Indicate on the line below the position of a diffusivity in (1) an ordinary gas, (2) a macroporous material containing a gas, and (3) the two extremes (3) and (4) shown in Figure 23.6. By which factor do the extreme values in your figure differ?

23. In Adsorbents

10- 15

10-12

10-9

263

10-6

10-3

23.4 The pre-exponential factor in Figure 23.5 can be estimated as v T djump / 4"" 10-8 m 2 S-I • Look at the value of the diffusivity of butane in silicalite at 300 K in Figure 23.6. Which fraction of the attempted jumps is effective? What is the corresponding value of the activation energy in Figure 23.5? 23.5 Diffusion of benzene in silicalite (MathCad). In this problem you are given data for the Pick diffusivity of benzene in silicalite. You are asked to calculate the Maxwell-Stefan diffusivity using the relations given in Fig. 23.7. The Langmuir isotherm parameter has a value PI $ = 1100 Pa. 23.6 Permeation of hydrogen and butane across a silicalite membrane (MathCad). In this problem you are to estimate the ratio of the flux of hydrogen to that of butane in Figure 23.14. The permeation through the silicalite membrane is at steady state. The data provided are the Maxwell-Stefan diffusivities, DIM = 2 X 10- 10 m 2/s; D 2M = 1 X 10- 11 m 2/s, and the Langmuir isotherm parameters (see Fig. 23.4) PI$ = 10000 Pa, P2$ = 250 Pa. 23.7 Adsorption of n-Heptane and Benzene on NaX Zeolite (Mathcad). This program simulates the simultaneous uptake of n-heptane and benzene in NaX zeolite crystals shown in Figure 23.13. Values of the constants have been chosen such as to simulate experiments by J. Km-ger and M. Bulow, 'Theoretical prediction of uptake behaviour in adsorption of binary gas mixtures using irreversible thermodynamics', Chem Eng Sci 30 (1990), 1779-1791. This is a nice example of a non-steady state problem; it shows how to do that in Mathcad. You cannot do much with the file you,rself.

264

U Itrafi Itration In this chapter, we analyse mass transfer in ultrafiltration. This is one of the membrane processes that uses porous membranes. It separates large molecules, such as proteins, from water. Ultrafiltration is largely governed by what happens in the boundary layer before the membrane. These 'polarisation' effects are complicated by electrical interactions between the charged proteins and other ions, and by high volume concentrations of the proteins. In the membrane, transport is mainly by viscous flow - this in contrast to the processes we have discussed so far. Proteins are excluded from the narrow pores in the membrane: this is the main separation mechanism. There is one other difference with previous examples. Because proteins and water molecules differ greatly in size, ultrafiltration is best described using the Maxwell-Stefan equations in a volume form. These use forces per volume, friction coefficients per volume, and volume fluxes.

24.1 The Module Figure 24.1 shows typical dimensions of an ultrafiltration module. It consists of tubular membranes with a diameter of about one centimetre. The liquid feed flows through the tubes with a velocity of one metre per second or higher; this yields a highly turbulent flow. The protein always accumulates near the membrane; we need the turbulence to minimise this effect.

t

W cross

flow

=

I 1...3 m s-1

turbulent core

r 1

boundary layer important!

0.5m

0.01 m

membrane 0.001 m

0(

t)

-----c~f-,-

porous tube wall

0.1 m

Fig. 24.1 Dimensions and flow in an ultrafiltration module

24. Ultrafiltration

265

24.2 Membrane and Permeants We consider four permeating species (Figure 24.2): 1. water, a very small molecule, 2. the protein, which we take to be spherical, with a diameter of a few nanometres, 3. a cation such as Na+, which has a diameter similar to that of water, and 4. an anion such as Cl", again with similar dimensions. We take the active layer of the membrane to be one micrometre thick. It consists of parallel cylindrical pores with a diameter of the same size as the protein (or smaller). These exclude the protein completely. (We will have a separate look at the behaviour of wider pores.) Note that the pores are a hundred times longer than wide: this could not be drawn in the figure. The pores take up one half of the volume of the membrane. The active layer is so thin that it can not be handled in its own. So the membrane has a spongy support layer that is much thicker; however, the mass transfer resistance of this layer is small and we will neglect it here. The dimensions in and around the membrane vary over an enormous range. Water molecules are less than a nanometre; protein and pore diameters are in the nanometre range. Active layer and pore length are around a micrometre; the boundary layer thickness is around ten micrometres and the thickness of the whole membrane with support layer is a few hundred micrometres. You cannot draw that to scale! (1) water, no charge

(2) protein, Z2 =-20 ... +20 (3) cation (Na+) (4) anion (CI-)

7S'F

7$

the active layer of the membrane, consisting of cylindrical pores

/ boundary layer

open support structure (neglected here)

"" 10 pm Fig. 24.2 Dimensions of the permeants an·d the membrane

266

Mass Transfer in Multicomponent Mixtures

We first consider the situation where the protein has no charge and the ions are absent. The effects of charges will be discussed later, but only in a qualitative manner.

24.3 Osmotic Pressure (No Ions, no Charge) Before looking at transport through the membrane, we first consider the equilibria involved. Figure 24.3 shows a membrane separating a solution of a protein in water, and pure water. The pores of the membrane are smaller than the protein, so protein does not pass through the membrane. If both compartments are at the same pressure, water will try to dilute the protein solution; it will flow from right to left. We can stop this flow with a pressure difference - the osmotic pressure difference. In ultrafiltration, osmotic pressure differences are of the order of 10 ... 100 kPa - much smaller than in reverse osmosis, but still important. Every solution has an osmotic pressure. This is a property of the solution; it is directly related to the chemical potential of water as we see in the formulae at the bottom of Figure 24.3.

membrane-,not permeable for the protein "

protein in water

P/~n

~'_;J__~_:_.~_0- i:._," -_.-_-.I I E:L-_- -p ure

//

0

°O~

I'

water

I

/

osmotic pressure equilibrium between water on the two sides: R Tln(Yl Xl ) + ~ (p + n) = RTln(1) + ~ p ~

RTln(Ylxl)=-~n

Fig. 24.3 Defining the osmotic pressure

We will use the 'excluded volume' formula for the osmotic pressure that is given in Figure 24.4. This contains volume fractions and the molar volume of the protein, even though osmotic pressure is best regarded as a property of the water. We see the osmotic pressure rising more than linearly with the volume fraction of the protein.

24.4 Size Exclusion The volume fractions of water and protein are not equal inside and outside the pores. An important reason for this is size exclusion. The centre of a protein molecule can only occupy the central part of a pore: the part that is at least one protein diameter away from the wall. If we assume that the concentration in this central part is equal

24. Ultrafiltration

267

to that in the surrounding solution, we obtain the expression for the distribution coefficient in Figure 24.5. A good ultrafiltration membrane wi111argely exclude the protein. Proteins larger than the pores are totally excluded. excluded volume theory

1t

-

RT 1t = c2 (1 + 4cJ

/11;

-- _-----

V2 =2xlO-2 m 3 mor1

40 kPa

20

T=300K

-properties of the protein ,//

/ ,~-

....

........

c2

0 0.0

0.1

0.2

Fig. 24.4 Osmotic pressure of a solution of neutral spheres

lr---'---'---'---~

protein centre can only occupy this volume

K2

size exclusion

K2

=£2 =

£2

(

1-~

2

d pore )

~

2 d pore

Fig. 24.5 Difference in protein concentration in the bulk fluid and the pore due to size exclusion

The protein concentration is lower in the pores than in the bulk liquid. So the water concentration must be higher in the pore than in the bulk. This causes a pressure reduction going from bulk liquid to the pore (Figure 24.6). When the pore is sufficiently narrow, it fully excludes the protein. The pressure jump at the interface is then equal to the osmotic pressure of the solution next to the membrane.

24.5 Polarisation If we feed a protein solution and apply a small pressure across an ultrafiltration membrane, we get a low flux. Increasing the pressure increases the flux, but only to a point. It rapidly levels off to a maximum of about ten micrometres per second (Figure 24.7). This is perhaps ten times smaller than would be obtained with clean

268

Mass Transfer in Multicomponent Mixtures

water and the same pressure. Surprisingly, the maximum flux depends on the liquid flow outside and along the membrane; it increases more-or-Iess linearly with the outside velocity. This shows that polarisation is very important.

concentration of water higher in the pore

dp inteiface

p

I:!.pinterface

=It -It'

z Fig. 24.6 The 'osmotic' pressure jump at the pore mouth

j},p

~ __ -----------increasing

... precipitate (gel) (not always)

plateau (height

- f),pinteiface

volume flux, NI

. W cross flow

t

-j},p

Fig. 24.7 The flux versus pressure difference

Polarisation causes the protein concentration near the membrane to rise sharply. The effect is much larger than for reverse osmosis, because of the low diffusivity of the protein. A high concentration near the membrane has two effects, both of which limit the flux to a maximum: • the 'osmotic' pressure jump at the pore mouth becomes large and may limit the flux, and • the concentration of the protein may become so high that it precipitates as a thin gel layer, impeding any further increase of the flux. The two effects are difficult to separate by experiment. Here, we only consider the first effect.

269

24. Ultrafiltration

24.6 Transport Equations We have discussed the equations describing polarisation in reverse osmosis in Chapter 5 (Figure 5.3). Here, the volume fraction of protein in the boundary layer can be quite high. This causes several problems. l. The difference approximation is not accurate (this is one of the few cases where it is not). 2. The effects of non-ideality are important. 3. The friction coefficient between protein and water is not constant. To allow for these effects we will integrate the Maxwell-Stefan equation over the boundary layer. This leads to somewhat complicated equations, which we will be writing in volume terms. These contain a volume flux of water, a force per unit volume of water, a friction coefficient in volume terms and so on. The behaviour of these is similar to that of the molar parameters with which you are by now familiar, but they have different units. We will distinguish them from the molar parameters by using bold symbols N, F and Z (We have no bold Greek font for the friction factor.) A few important bits of our equation are given in Figure 24.8. This shows how the volume fraction of protein rises rapidly and non-linearly in the boundary layer. We note that the volume flux of water in the boundary layer is the product of water velocity and water volume fraction. The protein flux through the membrane is zero; this yields the bootstrap relation. The driving force per volume of water turns out to be equal to the gradient of the osmotic pressure. (You see this by comparing Figures 24.3 and 24.8). volume flux N) =e)u

protein flux zero; bootstrap N2 = 0

driving force on '} ': \

~

a \ \

I

'.

strongly non-linear

o per mole of'}' o per m of '1' 3

F=F;=dn 1 V; dz

Fig. 24.8 Fluxes, volume fractions and driving force

We obtain the friction coefficient between water and the protein spheres from the settling velocity of a swarm of spheres (Figure 24.9). This velocity can be predicted by a relation from Richardson and Zaki. This tells us that the velocity is proportional to the volume fraction of the liquid to a power of 4.7. (To avoid very messy equations, we take a value of 5).

270

Mass Transfer in Multicomponent Mixtures

settling velocity of a swarm of particles u2 _ ,,4.7 _ ,,5 o - Cl - cl U2

see text

Z)?f 37r1Jd2

~ 12 .• ////'

•. '

=

-4

cl

V2

//

vo!um-: based friction coefficient

Fig. 24.9 Friction coefficient between spheres and liquid

The driving force on the spheres is due to a combination of gravity and the pressure gradient as we have seen in Chapter 12. The velocity of the liquid follows from the fact that the volume fluxes of spheres and liquid must cancel in a batch settling vessel. With the driving force and velocity difference we then obtain the friction coefficient between water and spheres. It turns out to be equal to that of spheres in a dilute solution, multiplied by the liquid volume fraction to a power of minus four. The Maxwell-Stefan equation in volume terms has the same structure as that in molar terms (Figure 24.10). Inserting the relations for the driving force, the friction coefficient and the flux yields a differential equation for the volume fraction of protein as a function of position. The solution is implicit and a little complicated, but not difficult to use.

---+ solution f(C2)

= In(c2) +3c2 -

with 2 70 3 75 4 39 5 4 6 15c2 +-c2 --c2 +-c2 --c2 3 4 5 3

Fig. 24.10 Solving the MS-equation of the boundary layer.

Figure 24.11 shows how the volume fraction of protein next to the membrane varies with the flux. For a zero flux it is equal to the bulk value, but it rises sharply with increasing flux. When the volume fraction goes up to about 0.6 we would expect the protein to start behaving like a fixed bed or a precipitate. This yields the highest possible flux.

24. Ultrafiltration

271

highest flux

0.6

.-----,------r--.---...-f--,

0.4 0.2 bulk e2,a

= 0.01 -----» 0.0

b:==:::r::=_---1.-_----lL-..--!--.J

o

2

4

Fig. 24.11 Volume fraction of protein next to the membrane

We note that the highest flux is inversely proportional to the thickness of the boundary layer (so it should go up with an increasing cross flow). It goes down with increasing size of the protein. The concentration profile inside the boundary layer or film has the same shape as the curve in Figure 24.11. You only need to replace the & by the distance z into the film to obtain the local volume fraction. This model allows us to calculate many other aspects of ultrafiltration. We end with two examples. Figure 24.12 shows the flux versus the pressure drop across the pore mouth. (This is usually the largest part of the pressure drop across the membrane.) We see the sharp rise up to a plateau as is seen in practice. 40

11 =10-3 N s m-2 d 2 =4nm T=300K Llz =10-5 m

20

o

e2,a

o

0.2

0.4

=0.01

- dp ;nteryace MPa 0.6

Fig. 24.12 Flux versus pressure drop across the interface

Figure 24.13 shows the effect of the bulk concentration of protein on the limiting flux. With increasing bulk concentrations the flux goes down dramatically. In the plot the horizontal axis is logarithmic; we then get an almost straight line. This is also commonly found in experiments.

272

Mass Transfer in Multicomponent Mixtures

20

'" "

same parameters as in previous figure except for e2,a

"

o 0.1

0.01

'"

"- ..... 1

Fig. 24.13 Limiting flux versus bulk composition

24.7 Inside the Membrane Ultrafiltration is largely governed by what happens in the boundary layer. Even so, what happens inside the membrane is sometimes of interest and we spend a few words on it. As we have seen in Chapters 2, 14 and 21, there are two transport mechanisms in a porous membrane: • transport by viscous flow, and • transport by diffusion. In the narrow pores, no protein is present. In the wide pores, transport occurs by both mechanisms. Here, we neglect the effect of diffusion in the membrane; we come back to this in a moment. Viscous flow increases with the square of the pore diameter (Figure 24.14). Viscous flow is small in reverse osmosis, of some importance in nanofiltration (with pores of about one nanometre) and dominant in ultrafiltration. If a membrane has a range of pore sizes, the largest part of the flow tends to go through the large pores. yiscous flow

water:

1 d~ore J.p v =-----1 32 11 &

the dominant terms in protein: ultrafiltration Fig. 24.14 The viscous flow equations in the membrane

If the concentration of protein in the membrane is low, then we calculate the viscous

flow of water with the Poiseuille equation in the figure. (Higher protein concentrations give a notable increase of the pressure drop, but we will not consider them here.) For a complete description of the ultrafiltration process, we should add this pressure drop to the interface drop that we have calculated earlier.

24. Ultrafiltration

273

The protein tends to move along near the centre of the pore. So its velocity is a little higher than that of water. The viscous selectivity of the protein is then larger than one. Figure 24.15 shows an approximate formula to calculate the viscous selectivity. When the protein is much smaller than the pore, it travels with the same velocity as water. This also happens in the other extreme, where the protein just fits into the pore, and protein and water must travel as a 'plug'. In between, the protein velocity has a maximum of about one and a half times the water velocity.

o

3.L

o

d pore

1

Fig. 24.15 Viscous selectivity of the protein

We have already seen that the exclusion of proteins goes down as pore size increases. All these factors cause the retention of proteins to be determined by the largest pores. Even a few large pores may have a considerable effect. The result is that the retention goes down with increasing pressure difference across the membrane. With a higher pressure difference there will be a higher protein concentration in the boundary layer and more protein transport through the large pores. We can reduce the concentration in the boundary layer by increasing the cross flow velocity: this improves the retention. Figure 24.16 shows some measurements on the retention of poly(ethene glycol), an uncharged, water-soluble polymer, by an ultrafiltration membrane. We see all the effects that are predicted, and one which is not: near zero pressure difference, the retention falls to zero. There, diffusion causes an equalisation of concenttations on both sides of the membrane. You will remember that we had neglected diffusion. (The retention of this membrane would be too small for most practical purposes.)

24.8 Electrical Effects So far we have only dealt with neutral species. However, proteins are usually charged and accompanied by smaller ions. Also, the membrane itself is often charged. It has only recently been realised that the resulting electrical effects are substantial. It appears that electrical effects can be taken into account quite well by

Mass Transfer in Multicomponent Mixtures

274

the Maxwell-Stefan equations. However it would take too much space to discuss these problems in detail. (You will find a few of them in the exercises). So we finish with some qualitative remarks.

CPHG. permeate

increasing W cross flow C PEG, retentate

-j},p

o

CD

diffusion controls

Ci)

proper operation

CD

MPa

o

CD

increasing polarisation

0.4

Fig. 24.16 Measured retention of PEG by an ultrafiltration membrane

When charged proteins accumulate in the polarisation layer, they repel each other (Figure 24.17). As a result, the protein concentration does not rise as sharply as predicted by the theory above: diffusion back to the bulk solution can be enhanced by as much as ten times. The charged layer behaves a bit like an electrodialysis membrane: it concentrates co-ions and repels counter ions. The rejection of the small ions can be influenced substantially. Quite surprisingly the rejections can be negative: salt concentrations in the permeate can be higher than in the retentate. This is found experimentally and also predicted by theory. charged polarisation layer El:!

~

~

backdiffusion of protein enhanced by electrical field concentrations

----e----

--

"

\

\ \ \

Fig, 24,17 Effects of charge accumulation in the boundary layer

24. Ultrafiltration

275

As remarked earlier, the charge of a protein depends on the pH of the solution. Electrical effects are smallest near the isoelectric point of the protein and there the flux through the membrane often goes through a minimum (Figure 24.18). isoelectric point

protein charge

+t o

flux

pH~

Fig. 24.18 Effect of pH (protein charge) on the flux

Membranes often have an electrical charge themselves. When the charge has the same sign as that of the proteins, then the proteins are repelled. This enhances their rejection (Figure 24.19a). When the membrane and protein have charges of opposite sign, they attract each other. This can lead to pore blocking (Figure 24.19b). The charge density of ultrafiltration membranes is usually small, so these effects disappear with sufficiently high electrolyte concentrations. c::>

~ ~

€I'

~

GEB c::> ~

c::> ~

c::>fI ~

G) charges with same sign ~

enhanced retention

C0 charges with opposite sign ~

pore blocking

Fig. 24.19 Effects of charge of the membrane

24.9 Summary In this chapter we have analysed ultrafiltration. In ultrafiltration, we transport liquid

through a porous medium. We have found a few important effects.

276

Mass Transfer in Multicomponent Mixtures

• The concentrations of species inside and outside the pores are unequal due to size exclusion. The unequal concentrations of water cause a pressure jump at the pore inlet. • Polarisation in the boundary layer is important because of the low diffusivity of the proteins. It may cause a large increase of the protein concentration near the membrane. This leads to a large pressure jump at the pore mouth, which results in a limiting value of the water flux. A higher turbulence in the boundary layer increases the flux; a higher pressure difference does not. • Transport in the membrane is primarily by viscous flow. Diffusion of the protein is only important for small pressure differences across the membrane. • Accumulation of electrical charge in the boundary layer enhances diffusion of proteins back to the bulk. It reduces the effect of polarisation and allows for higher fluxes than are obtained with neutral species. • The membrane itself may also have a charge. The effects can be substantial, especially for low electrolyte concentrations.

24.10 Further Reading Van den Berg, G.B. and Smolders, c.A., 1992, Diffusional phenomena in membrane separation processes. J. Membrane Sci., 73, 103-118. A good review. Deen, W.M., 1987, Hindered transport of large molecules in liquid-filled pores. A.I.Ch.E. J., 33, 1409-1425. A good review, worth reading. Robertson, B.c. and Zydney, A.L., 1988, A Stefan-Maxwell analysis of protein transport in porous membranes. Sep. Sci. and Technology, 23, 1799 - 1811. A nice example. Wesselingh, J. A. and Vonk, P., Ultrafiltration of a large polyelectrolyte, J. Membrane Sci., 99 (1995), 21-27 Analysis of electrical effects in the boundary layer of an ultrafiltration membrane. Rautenbach, R. and Albrecht, R., Membrane Processes (1989) John Wiley & Sons, New York The subject of ultra-filtration is covered also in the following two theses:

• 24. Ultrafiltration

277

Van Oers, C.W., 1994, Solute rejection in multicomponent systems during ultrafiltration. Ph.D Thesis in Chemical Engineering, Eindhoven University of Technology, Eindhoven, The Netherlands. Vonk, P., 1994, Diffusion of large molecules in porous structures. Ph. D. thesis in Chemical Engineering, University of Groningen, Groningen, The Netherlands.

24.11 Exercises 24.1 Osmotic Pressure. Why is the osmotic pressure in ultrafiltration (Figure 24.4) so much lower than in reverse osmosis (Figure 18.13)? 24.2 Osmotic Pressure again. Consider the formulae in Figures 24.3 and 24.4. Take the limits for low fractions of protein (where the solutions become ideal) and see whether you can transform the two formulae into each other. Where does the Vz in the second figure enter (there is no Vz in the first figure)? 24.3 Effect of Counter Ions. In this chapter we consider proteins to be uncharged spheres. However, proteins usually do have a charge, and often quite a large one. This means that they will be accompanied by a large number of counter ions. These will also contribute to the osmotic pressure of the solution. Consider water (1) with protein (2) with a charge number Zz and counter ions (3) with a charge number Z3' The solution is dilute, so we can neglect non-ideality. We place the solution on the left side of the membrane in Figure 24.3. The protein is larger than the pores of the membrane and cannot pass to the other side. On the right side we have pure water. The electroneutrality relation also prevents the counter ions from moving through the membrane. Derive a formula for the osmotic pressure in terms of the mole fraction of the protein and the two charge numbers. 24.4 Several Pore Diameters. Real membranes usually have a range of pore diameters. The exclusion of protein will be different in the different pores. So according to Figure 24.6 you must expect different pressures in different pores, even at equilibrium. How can this be? Why don't the fluids in the pores diffuse from high to low pressure? 24.5* Symbolic Integration (Mathcad). Look at the third line in Figure 24.10. You need the left-hand side of the equation in the third line. Type this expression in Mathcad. Then use the symbolic processor to expand it to a seventh order series. Integrate these term-by-term to check that you get the same result as in the figure.

278

Mass Transfer in Multicomponent Mixtures

24.6 The Boundary Layer in Ultrafiltration (Mathcad). This is a large file, which gives all calculations for Figures 24.10 to 24.13, and more. It contains the complete model that we have developed in this chapter. You can use it to tryout any variations. 24.7 Retention of a Polyelectrolyte in Ultrafiltration (Mathcad). This model is the basis of the publication of Wesselingh and Vonk in the Journal of Membrane Science and Figures 24.17 and 24.18. You can use it to investigate the effect of pH on the flux of an ultrafiltration membrane separating a protein in an electrolyte solution. 24.8 Transport Parameters from Data Sheets (Math~ad). You will not find membrane parameters, in the form needed for M-S calculations, in the data sheets of membrane manufacturers. Even so, engineers with a Sherlock Holmes attitude can get a lot of information out of product and application descriptions. Here we do this for a dialysis membrane.

279

......... Ending We end with an overview and a few remarks on 'how further?' .

25.1 Looking Back This book has presented you a complete theory of multicomponent mass transfer. You have seen a large number of greatly differing examples and you have dealt with many, many details. Because of this, you may have lost sight of the underlying structure, which is not all that complicated. To bring that back, we finish with an overview. Our starting point in Chapter 3 was the Maxwell Stefan equation in words: the driving force on a species i in a mixture

the sum of the friction forces between i and the other species i

Fig. 25.1 Our starting point

We will be looking back at how we have developed this equation. The discussion is divided in six parts: 1. Thermodynamic models for calculating driving forces and solubilities. 2. The different driving forces (the potential gradients). 3. Different ways of writing the friction terms. 4. The behaviour of the friction coefficients. 5. Additional relations (bootstraps). 6. The many variants of the diffusion equations.

In the greater part of this overview we use friction coefficients and velocities, because they give the simplest formulae. Only at the end do we consider the many changes needed to solve practical problems.

25.2 Thermodynamic Models - the Potentials In this book we have used many thermodynamic models, all of them relatively simple. They yield the potentials that we need. We need these both to calculate the driving forces in the left side of the Maxwell-Stefan equations, and for the concentrations (solubilities) in the right side. The potential of a species can be

280

Mass Transfer in Multicomponent Mixtures

divided into several parts. The way we do this is a bit arbitrary. Here we have chosen to use separate pressure, electrical, centrifugal, gravity and chemical potentials. (We could, for example, have incorporated the effects of pressure and electrical fields in the chemical potential). Because the chemical potentials are so important, their descriptions are summarised separately in Figure 25.2. Chemical Potential

at constant T, P and ifJ

ideal liquid and gases

l:!:L = const + In(x)

non-ideal mixtures

l:!:L RT = const+ 1n('Vx) I, ,

simple non-ideality, binary

l:!:L =const + In( Xi) + In( Ax~ )

solvent in polymer

l:!:L = const + In(l\) + In<Xc~)

RT

counter ions in an ion exchanger adsorbed species (Langmuir) ultrafiltration membrane

'

RT

RT

:~ = const+ In(xj) + In{{(K +(1- K~/}.)_I}

:d =

cons1 + In(x,) +

In (1- t 1 X;

J:!L = const + In( Xi) + In RT

(1

dj d pore

)2

Fig. 25.2 The different models of the chemical potential

Other Potentials l1/ P

pressure swelling pressure electrical centrifugal gravity

V

-'1'-'

=_' P

RT

RT

'1': = B V, £1/3 = ~ P

RT

Vc

M

RT

S

'1'; = tF.z; RT

RT

'I'~ = MiOl Z2 RT 2RT 'l'g Mgz

-'=-'RT

RT

Fig. 25.3 The other potentia Is that we have used

25......... Ending

281

At equilibrium, the total potential of any species is the same anywhere in a system. This is also the case at the two sides of an interface between two phases. If the 'other potentials' (Figure 25.3) are important, you will often find jumps at the interface, for example in the electrical potential and pressure.

25.3 Driving Forces This is the shortest and simplest part of this overview. The driving force on a species is the negative of its potential gradient. We have distinguished several contributions to the potential; each of these can contribute to the driving force (Figure 25.4). Driving Forces

derivatives at constant T and p

chemical potential gradient

Ffl

=_ dJii =_RTdlnai =

'dz

dz _RTd(ln(Yixi))

dz

electrical force pressure force swelling pressure gradient centrifugal force gravity Fig. 25.4 The different driving forces

We take chemical potential gradients at constant temperature and pressure. Effects of temperature and pressure gradients are taken into account by other terms. When we deal with external forces, or with porous matrices, we sometimes need the force on the mixture as a whole. This is the sum of the forces on all species. In this summation, many internal forces cancel. For example, the chemical potential gradients sum up to zero. In a bulk solution or a non-charged matrix, also the electrical forces cancel.

25.4 Friction Terms We can write the friction terms in the right hand side of the MS equations in several ways. First we must decide whether to use a structured or a non-structured description of the mixture. For gases, liquids and tight matrices, this distinction is not important (Figure 25.5). There is no viscous flow in such mixtures and the structured and non-structured equations are identical. For tight matrices, we have the choice of taking matrix elements as one of the species

282

Mass Transfer in Multicomponent Mixtures

in the mixture, or of regarding the matrix as a separate entity. The matrix can of course always exert friction on the permeants. Matrix elements might be the monomer or chain repeat units in a polymer. The choice does influence the values of the mole fractions and friction coefficients. It does not influence the final results of the calculations. structured

Friction Terms gases and liquids

permeants in tight matrix

permeants in a porous matrix

non-structured

for gases, liquids and tight matrices, structured and nonstructured are identical

you can al,o lake a mal,;x element as a species '3'

SI,2 X2(U1 -U2)+SI,3 x3(U1 -Lt:» S2,I XI(U2 -U1 )+S2,3 X3(U2 -Lt:»

7f SI,2 ;: S2,1 Onsager

L

SI,2 X2(U1 -U2)+SI,M(U1 -uM )

~ S2,1X t (U 2 -U1)+S2,M(U2 -uM)

parameters differ from above

SI,2 = S2,1

S12 , X2(U1-U 2)+Sl ,M(U1-uM )

;l,2x2(w j -w2 )+;j,M(Wj -W M )

S21, Xl (u 2 -u1)+ S2 ,M (U 2 -U M )

;2,1X t (W2 -W1)+;2,M(W2 -WM

F=SvV vj=ay Wj =uj+Vj

;1,2"* ;2,1

diffllsive, ]!.iscous, '!£.hole

see Appendix 3

Fig. 25.5 Different ways of writing the friction terms.

In porous matrices it makes rather a difference whether we use a structured or a nonstructured model. (The final results are the same.) In the structured models, we consider diffusion and viscous flow separately. Compared to the non-structured model, this leads to a larger set of equations (Appendix 3). However, the coefficients of the structured model show a simpler behaviour. They can often be obtained (at least partly) from hydrodynamic models The 'cross coefficients' for gases, liquids, tight matrices and the matrix in structured models, obey the Onsager relation. Those in non-structured models do not (so you need two separate cross coefficients).

25.5 Friction Coefficients The real problems in multi component diffusion are not the basic theory and mathematics. The real problem is the determination of the transport coefficients. We have seen quite a number of ways of describing these. They are summarised in Figure 25.6 for gases, liquids and polymers. Formulae for friction coefficients in porous media are given in Figure 25.7.

25......... Ending

283

Friction Coefficients in gases, liquids and polymers

n " 1 1 ~ flpd ( !;i,j = 2 (RT)~/i Mi + M 2

ideal gases, binary ideal gases, multi component

I; i,j -- I; i,jbinary

liquid, dilute, smail solute

!;i~i-->o = 2n}lTJ i d? / d j

j

f2

reckon with association in polar liquids

liquid, dilute, large solute

!;i~i-->O = 3n}lTJ A

liquid, concentrated, binary

I; 1,/. = (I;X;=lf' (I;Xi=lfi 1,/ 1,/

liquid, concentrated, ternary

( .. = (!;Xi=I)X; (Ci=I)Xi «(:"k=I)Xk (?) ',/

"/

',/

',/

!;:"k=1 = [(!;Xrl)(!;xrl)]1/2(??)

"

',/

/,

electrolytes ion-solventrwater) t' andt'_ ",,3xlOI2 (Nmol- I)(ms- I)-1 ~ ~+,w ~·.w tabulated as conductivities and mobilities

ion-ion

i = 0.5" Z?Xi L.J

!;+,- "" 2[_1_+_1_j-Ilz+z_1 t' t' ·0.55 ~+,w

~-,w

polymers single solute (2), dilute

t'XI-->O ~1,2

t' exp[ - "1 = ~l V

f2

multicomponent

free volume theory

1

1 (?)

Fig. 25.6 Friction coefficients in gases, liquids and polymers.

Many of the above relations are not reliable. Only in gases and dilute liquids (including electrolytes) can we predict friction coefficients and diffusivities with an accuracy of (say) ten percent. Many of the relations shown are too simple, or not well tested. However, at this moment we do not have anything better. Many relations contain parameters that have to be determined by experiment.

25.6 Additional Relations (Bootstraps) The Maxwell-Stefan equations describe the relative motion of species in a mixture. They do not describe the motion of the mixture as a whole. So to obtain absolute velocities of the species we also need other relations. The usual ones are balances of

284

Mass Transfer in Multicomponent Mixtures

mass, momentum, energy and electrical charge. In many cases we can replace the balances by simpler relations. These are obtained by inspection or from experience with the complete set of balances. Figure 25.8 summarises the bootstrap relations that we have used in this book. Friction Coefficients in porous media

effect of tortuosity and constrictions (general)

2

't t,i,i = t,i,i° y

~z~ =- --M;RT

gas-matrix in cylindrical pores

t;;,M

gas-gas, bed of spheres

t,

gas-matrix, bed of spheres

31Cl/Z (1- e) (RTM //z t; , ;,M- 23/Z ~

solute-matrix, adsorbents

t;;,M -

i,j

d pore

8

= e-LS t,0t,;

d;

(E)

4RT ---ex vT d. P -RT I

I

Fig. 25.7 Friction coefficients in porous media.

Bootstrap Relations

membrane stagnant

UM=O

bulk stagnant (absorption)

U

trace stagnant (polarisation)

UI

z =0 =0

NI +Nz =0

equimolar exchange (distillation)

NI

interface determined (vaporisation)

=.h

Nz

Y2 NI N2 -=-

reaction stoichiometry systems with heat transfer continuity of the energy flux

dT

E=-A,-+" dz £..J NB, I E=E'

electrochemical systems electroneutrality (always) current density given (may be zero) Fig. 25.8 Bootstrap relations

25......... Ending

285

25.7 The Many Variants With friction coefficients and velocities, the Maxwell-Stefan equations are simple to understand. However, this form of the equation is seldom the most practical and we have seen many variants. It is far more common to use diffusivities than friction coefficients. This is mainly because of historical reasons: mass transfer theory started with the Fick equations using diffusivities. We have found that friction coefficients often lead to simpler formulae and simpler physical pictures of what is going on. However, this is not always so, and the relation between friction coefficients and diffusivities is simple enough to allow us to use both simultaneously (Figure 25.9).

RT

H.=:1,1

r ..

~",

RT Si,j

=: H . 1,1

Fig. 25.9 Diffusivities or friction coefficients?

In practical problems we are seldom interested in the velocity of species, but in their fluxes (Figure 25.10). •....., d Nj=wjcj=(Uj-t1v;)cXj -E.1l=~ (x N -xN)+~ N ~"/" dz 1,2 2 I I 2 I,M 1 important in an open dp2 ~ (N N) ~ -E-= xI 2 -x2 I + 2,M N2 porous medium dz 1,2 but neglected here Fig. 25.10 Fluxes instead of velocities

We have often replaced the velocities in the MS-equations by fluxes. This is almost essential in problems involving chemical reactions. Also in problems with two or more phases, fluxes can be much simpler than velocities. Fluxes across an interface are continuous; velocities are not! A final point. The MS-equations are coupled non-linear partial differential equations. Even with modern software solving the complete equations is quite a job (Figure 25.11). In this book, we have restricted ourselves to the one-dimensional form of the equations. In most cases, we have even gone one step further. We have used a onestep central-difference approximation. This makes the mathematics almost trivial. We hope to have shown you, that this is good enough for many practical applications. It only breaks down in strongly asymmetrical applications with extremely large driving forces. Examples are in the boundary layers of electrodialysis and ultrafiltration membranes.

286

Mass Transfer in Multicomponent Mixtures

one step

one day

one dimensional

one week

three dimensional

one month

-'VJ.1i

= I.si,jX/Ui -Uj) j*i

Fig. 25.11 Engineering, scientific and greatly scientific forms of the MS-equations

25.8 Goodbye This has been a short - and possibly unconventional - tour through the vast subject of mass transfer. In contrast with most other engineering texts we have considered: • more than two components, • driving forces other than concentration gradients, and • the effect of solid matrices. We hope that you feel that this is a worthwhile extension. You have seen that all mass transfer processes can be understood from a single viewpoint. And you now have a basis for describing and understanding the new separation processes that are the challenge for this generation of engineers. Now you have had a start we hope you would like to learn a bit more on this subject. There are lists of further readings in the different chapters. You will find an enormous amount of information there. There is also much that you will not find . .One hundred and thirty years after the works of Maxwell and Stefan, multicomponent mass transfer is still in its infancy. It is your turn to do something about that. We wish you many good nights as shown below.

25......... Ending

287

Fig. 25.12 A good night.

25.9 Further Reading Krishna, R. and Wesselingh, J.A. (1997) The Maxwell-Stefan approach to mass transfer, Chem.Eng.Sci. 73, 861-91l. An overview of most of the recent literature on the Maxwell-Stefan equations.

25.10 Last Exercise 25.1 A Spray Dryer (Mathcad). Our very last example puts this book in perspective. It shows how the Maxwell-Stefan equations can be used in a model of a complete piece of equipment. You will see that the MS-equations are only a small part of the whole model; the equilibria, mass and energy balances and the many details of flow and mixing are just as important ... The model is that of a spray dryer, such as might be used for making milk powder. It uses the Maxwell-Stefan equations that you have developed in the file 'A Drop in a Spray Dryer': equations for heat transfer to, and mass transfer from an evaporating drop. You can use it to see the large effect on design of changing particle size, or the temperature level.

289

Thanks This book has had a predecessor: J.A. Wesselingh and R. Krishna 'Mass Transfer', which was published by Ellis Horwood in 1990. Below we first repeat the acknowledgements from that book. Obviously, some of the references here are no longer valid. From 'Mass Transfer'

Many people have contributed to this book in some way or other. We would like to thank them. The first ones are the students on two graduate courses on multicomponent mass transfer that we gave at Delft University of Technology in 1983 and 1984. The problems we encountered trying to teach them led to a series of eight short articles 'Beweging in Mengsels' (movement in mixtures). These appeared in our Dutch chemical engineering magazine, ]2-Procestechnologie in 1986-87. They may be regarded as the forerunners of this book. Of the material used, the data on ion-ion interactions and the external film model in ion exchange were sorted out by Margot Baerken. The TIP coefficients in Chapter 18 were calculated by Pieter Vonk. Both of them were students in Delft. The pervaporation example is from Wim Groot, of the Laboratory of Biotechnology of Delft University. The idea of the activated carbon-benzene-water example model originated in discussions with Wim van Lier of NORIT N.V. (activated carbon), Amersfoort. On the membrane examples we had many discussions with Joop Bitter of Shell Research Amsterdam and Jan-Henk Hanemaayer of TNO, Zeist. And we must thank Ross Taylor of Clarkson University, Potsdam, in the United States. He carefully checked all figures in the original version of this book. Ross also played a prominent role in our third course in 1988. Daniel Tondeur of the CNRS in Nancy, France, gave extensive and very helpful comments. A man who may not realise that he had an impact on this book is Alirio Rodrigues from Porto, Portugal. He was the main organiser of a series of NATO summer schools in Portugal. Courses on reaction engineering, ion exchange and adsorption brought us in contact with people such as Friedrich Helfferich, Gerhard Klein, Douglas Ruthven and Patrick Meares. They all contributed to our thinking about the subject. The summer course in 1988 also gave Hans the opportunity to start writing this book. The first version was finished in the home of Theresa and Leo Mudde in Lissabon, during the hot summer afternoons, in a large dusky room, shuttered to keep the sun out.

290

Mass Transfer in Multicomponent Mixtures

The final writing took place during 1988-89. The Koninklijke Nederlandse Academie voor Wetenschappen and the Shell group of companies provided financial support for a sabbatical year for Krish. He spent this with Professor Ton Beenackers at the University of Groningen, whom he would very much like to thank for the congenial atmosphere. This allowed us to prepare and give another one-week's intensive course in 1989. There we tried out the book on forty-four participants, mainly research students from the whole of the Netherlands. They discovered a fair number of the errors in the text, and got us to set the exercises accompanying this text. The one who probably suffered most from the writing of the book is Trudi Wesselingh, but she did not complain for a moment, We must have forgotten others, but hope they will forgive us. Hans Wesselingh/Rajamani Krishna GroningenlAmsterdam, July 1990

Added for 'Mass Transfer in Multicomponent Mixtures' It is almost ten years ago that we finished 'Mass Transfer'. In the words above we

mention a 'third course' on multicomponent diffusion. A lot has changed since then. We have not counted our courses carefully, but we think the next one is going to be number twenty-four. (We often give these courses several times a year). That implies that almost a thousand people have followed these courses. Many, many of them have contributed to this book, via assignments, giving us comments, asking questions and correcting our errors and inconsistencies. There must be hundreds of contributors: we do not even have a list of them. But if they read this, we hope that they will understand that we are grateful for what they have done. Several of our PhD students have made substantial contributions to our :.lllderstanding of multicomponent diffusion. We would like to mention Philippe Rutten who summarised all that is known on multi component diffusion coefficients, Arne Bollen who sorted out how to measure them and showed us how far we have to go to understand liquid/liquid extraction. Then Andries van der Meer, who first got us going in the field of electrolytes and ion exchange, later followed by Gerrit Kraaijeveld and Cor Visser who also covered electrodialysis. (In that area, we also learned a lot from the people from Akzo Nobel with their experience in chlor-alkali electrolysis). Pieter Vonk and Jaap Bosma showed us the way into separation of proteins, diffusion of polyelectrolytes, diffusion in gels and in gel ion-exchangers. Pieter also began work on ultrafiltration, which was later finished by Reinoud

Thanks

291

Noordman. This last lot was also involved in our first exercises on understanding diffusion in polymers. The greater part of this work was financed by the Netherlands Science Foundation (SONINWO), but also by the universities of Delft and Groningen and by Unilever Research. For our courses on multicomponent mass transfer, we have needed help from many colleagues. Two of them, Ross Taylor an Harry Kooijman from Clarkson University, helped us in the early years, and with them we developed our understanding of mass transfer in the classical separation processes such as distillation. In the past years we have had the pleasure of working with Jan-Willem Veldsink, and especially with Luuk van der Wielen. Our courses have been held at several different universities, in Amsterdam, Delft, Eindhoven, Wageningen, Twente, Oldenburg, Groningen, Lausanne, and La Plata. We would also like to thank those who received us and did all the organisational work around the courses. For the organisation of our courses we received substantial support from the Research School for Process Engineering in the Netherlands (OSPT). Krish is extremely grateful to the Netherlands Organisation for Scientific Research (NWO) for providing substantial financial support for his activities in the area of reactive separations. Then we must thank Jacques Schievink from Delft University Press, who guided us through the problems of document handling on different computer platforms, and who has managed to get our files into the form of a pleasant looking book. And yes, Trudi Wesselingh had to stand all of the book writing again, but she seems to be getting used to it. We nowadays listen carefully when she tells us that she could not follow what we were talking about. We then know that we have to improve our explanations. Hans Wesselingh/Rajamani Krishna Groningen/Amsterdam, February 2000

293

e

ix 1

Reading Mathcad In this course on mass transfer, we need to solve sets of non-linear equations. We have chosen to use Mathcad, a general-purpose mathematical programme of which you can learn the basics in a few hours. Even if you do not want to use Mathcad, you can still read its files. To do that, you must know a few details of the Mathcad syntax. We show the most important ones in the Mathcad file below. Mathcad is most easily demonstrated using Mathcad. You may not have noticed, but this is a Mathcad file. The text is just .that. The problems begin when we go to calculations. Equality Signs Mathcad has four different equality signs. They are best understood via examples. (1) the assignment command:

x := 3

(2) the calculate command:

x+y=5

~=1.5 y

z=2.718

(3) the global assign command: a",-1 (4) the 'malre equaf command

i - 7 = '1z

sinew)

=a - b

You use the assignment command to define a variable. You can only use a variable in a calculation after you have defined it. 'After' means to the right or below the point where the variable is defined. The calculate command uses variables which have already been defined. There is an exception: 'h' has not yet been defined, even so x= 3

x:= 3·b

X=

21

This is because we have used a global assignment for b: b '" 7 Mathcad runs through a file and reads global assignmeotsbefore it starts regular calculations. The malre equal command only works inside a solve block, to which we come back further on. Outside a solve block we can use it as a dummy equal sign, which does oot result in any calculation.

Mass Transfer in Multicomponent Mixtures

294

Functions You can also use variables which have not been defined in a/unction: 2

gee) := 3·e - 4·c

The variable inside the function is assigned at the moment that you use it: g(2) = 4

g(1) = -1

Note that the variable in a function is local. Its value is independent of what happens outside the function. Here, 'c' is not defined outside the function.

Range Variables These are also known as counting variables in other languages. They usually have letters such as i, j, k and so on. In Mathcad they are defined as follows: ;:=0 .. 3

j:= 1,-1 .. -5

You check what these mean with a calculate command (i = ...): The result is an output table.

i=

j =

1 ~~

~ 2

-3

3

-5

You need range variables to do a series of calculations, or for plotting a function. There you use them as subscripts.

Subscripted Variables The most important of these, are the variables with a single SUbscript: the vectors. As an example we construct two vectors, 'x' and 'y': Xi:= i + 1

You can output the separate elements if you wish: Yo= -1

Note that subscript counting begins at zero (this is the default).

Plotting You can plot anything based on subscripted variables: 20 , . . - - - - , - - - - , - - - ,

Yi

-10 L-_--L._ _...J...._--'

1

234

Appendix 1. Reading Mathcad

There is any number of variations in the plots, but you can easily read them.

Solve Blocks The most difficult concept in our use ofMathcad is that of the solve or Given.. Find block. Such blocks are used to find approximate solutions to sets of equations. They are normally preceded by the assignment of guess or initial values: guess values x+ y

Given

x:= 0

y:= 1

=3

y=i Find(x, y) = (1.303) 1.697

The equations between the operators Given and Find all have make equal assignments (the bold equal sign). Only between these operators does this sign have a meaning. The variables in the solve block are again local. They can have a completely different meaning from the same variables outside. If you want to make use of these variables you must export them, by assigning them to external variables. You can do this in several ways. The most connnon one is to assign the answer to the elements of a vector: guess values Given

x := 0

y:= 1

x+ y= 3

y=i

C)

:= Find(x. y)

Now X= 1.303 y= 1.697 x+ Y= 3 i= 1.697 This should be enough to let you read a Mathcad file. To do that, you will either need to install Mathcad Explorer from the CD-ROM accompanying this book, or to buy and install Mathcad itself. Mathcad Explorer allows you to read, modify and play with the files, but you can neither save your editing nor print the results. In the USA Mathcad is available from Mathsoft Inc., 101 Main Street, Cambridge MA 02142, USA Outside the USA contact Mathsoft International, Knightway House, Park Street, Bagshot, Surrey, GU19 5AQ, United Kingdom On the Internet: http://www.mathsoft.com

295

296

e

III

IX

Units The Maxwell-Stefan equations describe the movement (diffusion) of species in a mixture. In these equations the amounts of the species can be given in different ways. Three common choices are: • as the number of moles of a species, • as the mass of a species, and • as the volume of a species. Here we investigate the relations between the different forms of the equations. We want to show that the three descriptions are similar, and do this by only using different fonts for the corresponding variables: • small (lower case) letters for the molar system (the molar unit is small), • large (upper case) letters for the MASS system (mass is heavy), and • bold italic (voluminous) letters for the VOLUME system. The symbols in this chapter differ a little from those elsewhere. In particular, we use the symbols r ij , Rij and Rij for friction coefficients and dij , Dij and D;.j for diffusivities. (This is in contrast to the Sij and Dij used in the rest of this book). The gas constant R also appears in our equations. However, it does not have any subscript and always appears together with temperature: RT. The derivations in this appendix are elementary, lengthy and not exciting. If you only need to use the results to translate your models from one system to the other you can go directly to the summary.

A2-' Molar basis In the MS-equations each species has its own driving force. This is counteracted by friction with the other species. The friction is assumed to be proportional to the velocity difference between the species. For the first two components of a ternary mixture:

it = Ii2 X2(UI - Uz) + 'i3 X3(UI -

U3)

fz =r2I x I(u2 -uI)+r23 x 3(u2 -U3) The forces h in these equations are per mole of the components '1' and '2'. To obtain them per mole of mixture, we multiply by the appropriate mole fractions:

Appendix 2. Units

297

The first terms on the right hand side of the two equations are the frictional interactions: • between '1' and '2': 'i2 X IX 2(Ul -Uz) • between '2' and '1': r21x2xl(Uz -ud The two should have an opposite sign, but an equal magnitude, so:

A2-2 Mass basis We set up the equations on a mass basis in exactly the same manner. The only changes are that we use capital letters to designate forces per unit mass and mass fractions.

Fi = Rl2 X2(Ul -U2)+ R13 X3(Ul -U3) F2

= R21 X1(Uz - Ul)+ R23 X3(U2 -U3)

Note that the velocities are the same as in the equations before: the velocities are independent of the basis chosen. The friction per kg of mixture can be obtained in a manner similar to that above:

X1Fi =R12XIX2(UI-Uz)+R13XIX3(UI-U3)

x2F'z = R21 X2X1(U2 -Ud+ R23 X2X3(Uz -U3) Again we see the frictional interactions

Uz)

• between '1' and '2':

R12 X1X2(Ul -

• between '2' and '1':

R21 X2X 1(Uz -ud

Also these should be of opposite sign, but of equal magnitude. This is so if:

Rl2

= R21

A2-3 Volume basis Here we use bold italic letters to designate forces per unit volume of a species and volume friction coefficients and fractions.

298

Mass Transfer in Multicomponent Mixtures

The velocities are again the same. The friction per cubic metre of mixture can be obtained in a manner similar to that above:

Xl~ =R12XIX2(Ul-~)+R13XlX3(Ul-U3) X 2F2 =R21X2Xl(~ -ul)+R23 X 2X 3(U2 -U3)

Again we see the frictional interactions • between 'I' and '2':

R 12 X 1X 2(Ul -U2)

• between '2' and 'I':

R 21 X 2X 1(U2 -Ul)

Also these should be of opposite sign, but of equal magnitude. This is so if: R12 =R21

A2-4 Molar and mass diffusivities Now the relations between the molar and mass friction coefficients. To obtain the molar relations per kg of mixture we divide the molar relations by the molar mass of the mixture: M-1Xdl

= M- 1'12 XIX2(Ul -

1 U2) + M- 'i3 XIX3(Ul - U3)

M-1x2h =M- 1r21x2xl(U2 -ud+M-lr23x2x3(~ -U3) Each term in these equations must be identical to that in the equations on a mass basis. For the friction coefficients this yields:

M-1'i2XIX2

= R12 X1X2

1 M- 'i3 XIX3 = R13 X1X3

M- 1r21x2xl = R21 X2X1 M- 1r23x2x3 = R23 X2X3 It is clear that 'i2

= r21

yields R12

= R21 .

To obtain the diffusion coefficients we note that,

RT

• on a molar basis: d··=I] ~j

• on a mass basis:

RT DI].. =MR. -I]

This yields

299

Appendix 2. Units ~3 X X3 - = -I XI X2

d13

D21

X X

D23

= X2X3

d 2I

X2 XI

d 23

X2 X3

2 I --=-

Again we find that d12

XI X3

d 12

= d2I

yields

~2

= D2I . Furthennore so

X.

=_1

xi

M

_I

M·

This yields

A2-5 Molar and volume diffusivities To obtain the molar relations per m 3 of mixture we divide the molar relations by the molar volume of the mixture: V-IXIfi

= V- I 'i2 XI X2(UI -U2)+ V- I 'i3 XI X3(UI -U3)

v-Ixd2

= V- Ir2I x 2x I(U2 -UI)+ V-Ir23x2x3(~ -U3)

The terms in this equation must also be identical to those in the equations on a volume basis. For the friction coefficients this yields: I V- 'i2 XI X2 V-IW2IX2XI

It is clear that 'i2

= r2I

= RI2 XI X2

I V- 'i3 XI X3

=

R2I X2 XI

V- Ir23x 2x 3

= R23 X 2 X 3

=

yields

R12

=

R13 X I X 3

R2I . To obtain the diffusion coefficients we

note that

RT

• on a molar basis: dij='ij

RT

• on a mass basis: 0··=lJ VR IJ.. This yields:

0 12 = XI X2

013 = XI X3

d I2

xI x 2

d13

0 21

= X 2X I

0 23

d 2I

X2 XI

d23

XI X3

= X2X3 X2 X3

300

Mass Transfer in Multicomponent Mixtures

d12

Again we find that that

= d21

0 12

yields

= 0 21 • Furthermore, it is easily derived

D·· V: V. so -.!l. = -.!.,f dij V

A2-6 Molar and mass driving force The driving forces per unit mass of mixture derived from the two different bases must be equal:

or

Fi = M-I !.L fi. Xi

This can easily be rearranged to give the rather obvious result:

F=A M. I

I

The driving forces are potential gradients. As an example we consider a system in which only composition and electrical gradients play a role. With the molar basis the potentials are given by:

The potentials per unit mass are: AI

= RT In(rX). + r.FZi M. M.'!'

If>

I

I

Note that

(JX \

I

=(rX)i. The driving forces become: dJ1i d [ ] -r.Fz -dl{> fi=--=-RT-In(JX). i dz

dz

dz

I

and F I

= _ dAi = _ RT ~[ln(rx).]- r.FZi dl{> dz

Mi dz

I

Mi dz

A2-7 Molar and volume driving forces The sets of equations for these two bases must also be identical. So:

Xl~ Also

= V- 1Xlfi

X 2F2 = V- 1xd2 or Xi~

= V- 1Xifi

Appendix 2. Units

301

so

F=.hV; I

I

Again we consider a system in which only composition and electrical gradients play a role. With the molar basis the potentials are given by: f.1i

= RTln(rixi) + '.FziifJ

The potentials per unit volume are: RT ( GX ) .+-ifJ tp Zi I;=-ln

"i

Note that

I

"i

(r x)i =(GX)r The driving forces become: df.1i d [In(y.t). ] -'.Fz -difJ .h =--=-RTi dz

dz

dz

I

and

~ = _ dl; = _ RT ~[ln(GX).]- tp Zi dz "i dz "i

difJ dz

I

A2-8 Difference equation for a 'film' We now consider steady state transport through a 'film' with a thickness ilz. The potentials of the species are taken to be of the form: f.1i

= RTln(r x)i

or Ai

RT

= M. In(rX)i

or I;

RT

=V

I

Potential differences can be approximated by: RT. il(rX)i M j (rX)j

A2-9 Difference equation, molar basis Equating the potential gradient and the friction terms yields:

I

In( GX)i

302

Mass Transfer in Multicomponent Mixtures

The bars denote values at the average composition in the film. A similar equation can be derived for the second component. Dividing both sides by (RTf&.) yields:

'i2&' - (Ul - -U2 -) +--x3 'i3&' -(-) 3 - Ll(Y x)l =--x2 UI- U (yx)1 RT RT or

where

k

- d12 &.

12-

and k13

= d13 &.

A2-10 Difference equation, mass basis Now the same equation on a mass basis. The transport equation is: _ Ml =RI2X2(UI-U2)+R13X3(UI-U3)

&.

Dividing both sides by (RTf&.) yields:

M . Ll(rX)1 =Rl2&. - (ul- -) R13&. - (ul- -) u2 +--X3 u3 -_ --X 2 Ml (rX)1 RT RT or

.h WIt

K 12 - Dl2 Dr3- and K 13 =

&.

&.

The right side of the equation retains the same shape as with the molar basis. The left side (the driving force) needs to be modified in a similar way as found for the 'exact' equation.

A2-11 Difference equation, volume basis Here an almost identical derivation yields:

Appendix 2. Units

with K12

= 0Llz12

303

and K - 013 l3 -

L1z

A2-12 Summary In the three different bases considered for the Maxwell-Stefan equations (molar, MASS, and VOLUME), the driving forces and diffusivities are related via: driving forces: diffusivities:

D-. -.!L

dij

fi = MiFi M·M·

= --4M

0··1J

_

Vv. I J

---2dij V

Changes in the base of the difference equation require the following modifications:

mass transfer coefficients:

d··

D-.

D··

Llz

Llz

Llz

or -..!l. -7 -.!L -7 .....!L

304 II1II

I Poring over Pores In this appendix, we compare two different ways of looking at porous structures. This will help us to understand how diffusion and viscous flow interact. The theory has two practical consequences: • It allows us to estimate friction coefficients and diffusivities from hydrodynamic models. • It shows a simple way to include viscous flow in the Maxwell-Stefan equations by modifying the friction coefficients. The theory is rather complicated and parts of it are still under discussion. You do not need it to follow the rest of the book.

A3-1 Introduction There are two different ways of looking at fine porous structures (Figure P-l): • the 'non-structured' (N) approach, where we take the medium to be a pseudohomogeneous mixture, and • the 'structured' (S) approach, where we do take the pores and structure into account. We expect that the two approaches will be equal when the scale of the inhomogeneities is small enough. Two ways of looking at porous media:

®

Non-structured

®

Structured aggregated porous fibrous etcetera

expected to be equal when:

scale of < < scale of heterogeneity the phase

Fig. P-1 Two ways of looking at porous media

We shall see that comparing the two tells us a lot about the behaviour ofthe transport coefficients in porous media. It even allows us to make (partial) predictions of

Appendix 3. Poring over Pores

305

Maxwell-Stefan coefficients using hydrodynamic models of heterogeneous media such as the Cannan-Kozeny equation for a packed bed.

A3-2 The System We consider a system (Figure P-2) with two permeants, a small one '1' such as water, and another '2' which is at least as large, and might be a protein. The matrix 'M' is a porous solid, which can resist stresses and transmit support forces. We will regard the matrix as a separate phase, and not as part of the mixture. You can easily extend our reasoning to other cases; this system should be general enough for an introduction. Species:

'1' a small one (water) .. --___ ... '2' a large one (protein)

e<----_-::iO

The matrix 'M' is solid pha~e_ t h a t .

O :.;

o can resist stresses, and ----:n--.... .' o can transmit support forces / membrane support

··00. o·

>0

.. ~

Fig. P-2 System with two permeants, a porous membrane and a support

A3-3 Forces and Velocities We repeat the formulae for the driving forces on the two components (Figure P-3). Here, we will only consider compositionlnon-ideality, electrical and pressure forces, but you can easily include other forces such as gravity. In the transport equation of each species, it is simplest to specify the driving force per mole of the species. To describe viscous flow, we also need the force on the mixture as a whole. This is the sum of all forces on all components of one mole of mixture. When we sum these forces, the internal forces cancel. Compositionlnon-ideality forces are internal forces; they always cancel. If you try out, you will find that this cancelling is nothing else but the Gibbs-Duhem equation from multicomponent thermodynamics. If the matrix does not have an electrical charge, then the mixture of permeants must also be electrically neutral. In this case, the electrical forces will also be internal forces. An important external force is due to the pressure gradient.

306

Mass Transfer in Multicomponent Mixtures

Driving Forces on the components separately F. =_RTdlnaj_pz d -V, dp 1 dz ldz ldz

F =_RTdlna2 -pz d -V dp 2 dz 2 dz 2 dz

N

mol of i

'-v------'

internal forces cancel on the mixture as a whole

\

d dp F=xjF; +x2F2 =-Pz--Vdz dz

N

mol of mixture

with Fig. P-3 Driving forces on the permeants and on the mixture

In porous media, we need to distinguish between three different velocities (Figure P4):

• the diffysive velocity u, which you may regard as the velocity with which a permeant slips past the matrix, • the yiscous velocity v, which is caused by the force on the mixture as a whole, and • the total or whole velocity w, which is simply the sum of the two previous ones. Because these velocities play an important role in our reasoning, we have given them symbols, which we hope you will remember. Velocities

~Ui Vi

wi

=u, + Vi

diffgsive velocity slipping past the matrix yiscous velocity transport by convection

whole (overall) velocity

Fig. P-4 Three different velocities

We have already encountered diffusive velocities in many examples; we will not discuss them further for the moment. However, we do need to say a few things about the viscous velocity. Viscous flow is driven by the force on the mixture of permeants (Figure P-3). We write the viscous transport equation in the same way as the diffusive equations (Figure P-5), with the force left, and the friction term right. The

Appendix 3. Poring over Pores

307

viscous friction coefficient is directly related to the hydraulic permeability of the structure. The figure shows a few formulae for coarse (super molecular) structures. In all three cases shown, the permeability increases with the square of the channel diameter. F

-........ viscous friction coefficient

--

Sv =l cB

= Svv

hydraulic permeaJ.2ility 2 = d sphere _c__ 2

B

180 (I-Et

B= B=

d~ore

n

/

void fraction

d2

2 fibre_E_ _

80 (I-E)2 Fig. P-5 Viscous velocity and friction, and hydraulic permeability

We often apply these formulae to channels of colloidal dimensions. However, you should be aware of dangers in doing this. Figure P-6 shows two extra mechanisms that may play a role in small channels: the effect of adsorption of permeants on walls and the increased friction caused by particles in narrow pores. complications in channels of colloidal dimensions: surface adsorption changes diameters

particles cause increased friction

Fig. P-6 Viscous flow complications in narrow channels

It is tempting to assume that the viscous velocity will be the same for all permeants. This is often a good approximation, and it would greatly simplify our reasoning. However, • the permeants need not be well distributed across the diameter of a pore, and • the viscous velocity depends on position in the pore. This can lead to different viscous velocities (Figure P-7). For example, a nonadsorbing protein in an ultrafiltration membrane moves near the centre of a pore, so it has a velocity higher than average. Any species fully adsorbed on the pore wall has a viscous velocity of zero; partial adsorption lowers the viscous velocity of a species. We take these differences into account with viscous selectivities (J.,i for each

308

Mass Transfer in Multicomponent Mixtures

component. These selectivities are not independent. For example, if we assume that the mixture has a constant volume, then the volume fluxes of the components must add up to the velocity of the mixture. For a two-component mixture, one of the viscous selectivities will be larger than one, the other smaller. viscous .s~!ectivity

----v:-~<Xv I

I

viscous velocity of a centerline species <Xi > I higher than average viscous velocity of adsorbing species <Xi < I lower than average volume balance v = LCjVj = LCjajv ~ LCjaj = 1 Fig. P-7 Viscous selectivities and their relation

A3-4 The Two Transport equations We now write down the transport equations of our two models (Figure P-8). For the S-model, we have equations for diffusive velocities, viscous velocities, and the summations giving the whole velocities. In the N-model we can only look at the whole velocities; the separate contributions are not defmed. Note that we use two different sets of friction coefficients in the two models. The friction coefficients in the S-model are denoted by the Greek letter S(zeta), those in the N-model by ~ (ksi). These coefficients can be totally different. Structured model

F;

diffusive viscous

= X 2S t ,2(Ut -u 2 )+St,M Ut

F; = X t S I ,2(U2 -UI )+S2,M U2 F=Sv v

VI = <XIV

S (zeta)

V2 =<X2V

whole

Non-structured model

F;

= X2~1,2(WI -W2)+~I,MWI

F2

= X,S2"

(W2 - w,) + S2,M W2

whole velocities in one go Fig, P-8 Transport equations for the two models

~

(ksi)

Appendix 3. Poring over Pores

309

Now the important part: do these two sets of equations give the same result? The answer is by no means obvious, but it is yes! The results are identical if the two sets of friction coefficients are related as shown in Figure P-9. Two approaches are identical when:

r--------------

r-------------------l Non-structured

:

: Structured

I---------------..J

:

r--------------------------------------; ! =!S + a ISI,MS2,M + X2SI,2S2,M(al - a 2 ) 1,2j 1 1,2 r +a r + r : : ~v IXI~I,M a2x2~2,M ; ! -Is + a 2SI,MS2,M + X1S1,2SI,M( a 2 - a l ) 2,~ -! 1,2 Sv + aIX1S1,M + a 2X2S2,M 1

~

I_l

(sv + [XiSI,2 S2,M j(;I,M + X,X2S,,2 + X2S2,M ](a2 - a,))

:1 :1

~v+a,x'~IM+a2x2~2M

'::>',MI-'=>I,M

~ .j Jr

r

r

r

"

(sv + [XrS,,2 S"M j(;2,M + X,X2S,,2 +X,S"M ](al -a2)) 1~2,M r + r r : : ~v alxl~"M+a2x2~2,M ____ J L_____________________________________________________ _ '::>2,MI

Fig. P-9 Relations between the friction coefficients of the two models

RA. Mason derived these equations (be it in a different notation). We shall call them the Mason equations. We note three points that are currently under discussion. 1. In the S-model, the two permeant-permeant coefficients are equal (as we have seen in Figure 6.2). We shall see in a moment that those in the N-model usually differ. So the N-model does not obey the Onsager relation l . 2. Our S-model requires seven friction coefficients: the four diffusive coefficients, the viscous friction coefficient and the two viscous selectivities. However, the two permeant-permeant coefficients are equal, and we have seen that there is a relation between the two viscous selectivities. So there are five free parameters. The Nmodel has only four independent parameters. So there is some freedom in the choice of the parameters in the S-model. 1 Not obeying the Onsager relation, is a Mortal Sin in Thermodynamics of Irreversible Processes. One of the consequences is that we cannot regard the terms in the non-structured equation as forces between the species. However, the effect may not be very important. There are several special cases where it plays no role: • in tight matrices, where viscous flow is unimportant, • in very open matrices, where the viscous selectivities are equal, • for similar permeants, where the viscous selectivities are also equal, and • for solutions of dilute solutes in a solvent, where only the friction of the solvent on the solutes is important. It may be quite difficult to find examples where a departure from the Onsager relations is measurable.

310

Mass Transfer in Multicomponent Mixtures

3. An even more fundamental objection against the S-model is the fact that we can only measure the whole velocities. There appears to be no way to measure which part of the velocity is viscous, and which part diffusive. Clearly, there may be some loose ends to this subject. Even so, we have found comparing the S-and the N-models illuminating, as we shall see in the next section.

A3-5 Comparing the Two Models Pore with a single permeant

We begin with a simple case: a single liquid permeant. There can be no gradients in the mole fraction, and no electrical charge, so the only driving force is the pressure gradient. The only friction is that between the permeant and the matrix. Figure P-lO shows the two sets of transport relations. Structured

~d

Non-structured

dp -V-=SIMuI dz . p

dp -V-=SVVI dz w\ =u\+v\

~~I,M=[Svl+SI,M-Irl

Fig, P-10 A single liquid permeant in a pore

SI M ~ 00, the diffusive velocity is zero, and we have a no-slip condition at the pore wall. The flow is solely viscous and ~l M = Sv. The opposite happens when Sv ~ 00. There will usually be contributions of both diffusive and viscous transport. Slipping is caused by the shear stress in the liquid at the wall (Figure P-ll). The stress also moves the first layer of liquid molecules. (These are located at a distance of d/2 from the wall.) If we assume that this layer has a certain viscosity (which may be different from the bulk viscosity), then we can estimate this velocity and from that the S-liquid-matrix coefficient. The important thing to note is that the friction coefficient is inversely proportional to the pore diameter. If

U

1

=_'t_d\ ll atwall 2

)~

_Vdp dz

=[

4V 2TJatwall]Ul d pore d1

d pore dp

2

't=----

d\~k­

~dp~

4

dz

r

_ 8 VTJat wall - 4

~ ~l,M -

dd

1 pore

-

~

TJatwall d

Fig. P-11 A rough estimate of the S-permeant-matrix coefficient

pore

Appendix 3. Poring over Pores

311

With our estimates of the viscous and permeant-matrix coefficients, we can use the equations in Figure P-lO to find out when viscous and when diffusive flow dominates. For reasonable values of the parameters, the turnover point is for pores of about one nanometre diameter. In smaller pores diffusion dominates, in larger ones viscous flow is the important mechanism. Remember that we are discussing liquids, not gases. We can now speculate a little on the effect of the pore diameter on the different friction coefficients. We already know that the viscous friction coefficient is inversely proportional to the square of the pore diameter. Here, we have seen that the permeant-matrix coefficient is inversely proportional to the diameter. For not-toosmall pores, we would expect the friction between permeants to be independent of the pore diameter (Figure P-12). We will be using these ideas in further examples. In~ 2 d-pore

viscous

Y

Permeant-matrix

Y cc ~1,M

1 d-pore

permeant-permeant

~1,2

d~ore

~v

cc

cc

Fig. P-12 Effect of pore size on the different coefficients

Narrow pores

In narrow pores, the viscous friction coeffici~nt will be very large: Sv -7 00. In a tight matrix, there is no viscous flow. Here, transport is determined by the permeantmatrix coefficients. A glance at Figure P-9 shows that these coefficients become identical for the N- and the S-models. A matrix with narrow pores behaves like a homogeneous mixture. Wide pores

In wide pores, the situation may not be immediately clear. What happens, is that the whole velocities become equal to the viscous velocities, and that the permeantpermeant terms become unimportant (Figure P-13). Note that the viscous friction coefficient is equal to the molar average of the permeant-wall coefficients in the S-model. In a wide pore, the friction coefficients of the N-model and the S-model can differ by many orders of magnitude; their behaviours are totally different.

Mass Transfer in Multicomponent Mixtures

312

Wide pores:

o write N- equations per mole of mixture: =Xl X2~I,2 f~=-v~ + XI~I,MV x 2 F'z = XIX2~2,1 (v - v~+ X2~2,MV xJ'i

1____ _

o summing the equations: F =(Xl~l,M + X2~2,M)V == SvV identical to 'viscous flow through a tube' Fig. P-13 The N-equations for wide pores

Identical permeants If the two permeants are closely similar, then their viscous selectivities must be equal

to unity (Figure P-14). Also their permeant-matrix friction coefficients will be equal. The relations between the two sets of friction coefficients show here that: 1. The N-permeant-permeant coefficients are larger than the S-ones. 2. The N-permeant-matrix coefficients are smaller than the S-ones. Both of these effects are because non-selective viscous flow tends to equalise the whole velocities. Permeants similar, with slipping

~

~1,2

~

~2,1

+

_ I' -

':>1,2

_ I'

~2.1

-

~

_I'

~I,M

-

~

-I'

~2.M

+

-

~M

~M

SI,M

S~ Sv + SM S~

Sv + SM

0:1

= 0: 2 =1

=

S2,M

=

SM

viscous flow makes the N-coefficient larger than the S-coefficient

SV

Sv +SM SV

Sv +SM

viscous flow makes the N-coefficient smaller than the S-coefficient

Fig. P-14 Friction coefficients when the two permeants are identical

No slipping of solvent, dilute solute

This is roughly the situation in ultrafiltration membranes. Here, flow of water (l) through the pores is mainly viscous; the slipping of water is not important. The protein (2) tends to stay near the middle of the pores and has a viscous selectivity

Appendix 3. Poring over Pores

313

slightly larger than one. The protein has a low concentration. The relations between the friction coefficients reduce to those shown in Figure P-15. We can see several interesting things here: • The N-friction coefficient of water is equal to the viscous coefficient. Flow of water is completely viscous. • The two N-permeant-permeant coefficients are unequal. (The differences between the two coefficients are not large, however, and the coefficient ~1 2 has no meaning anyhow because the system is dilute.) • The N-protein-matrix friction coefficient is negative. (Indeed, the protein has a higher velocity than the surrounding liquid!) Protein (2) dilute, no slipping of water (1) SIM=oo

~2,l

= S2,1 + U 2 S 2,M ~1,M = SV ~2,M

=(S2,M + SI,Z)(1-U z )

Fig. P-15 Friction coefficients for the case solute (2) dilute and no slipping of solvent (1)

A3-6 Summary • We have compared the non-structured and the structured models of solid matrices; we have seen that they give the same results when their coefficients obey Figure P-9. • The non-structured model only calculates the whole velocities of all species. • The structured model calculates two velocities separately: - the diffusive or slip velocity and - the viscous velocity. • The viscous velocities of the species need not be equal; we take this into account using viscous selectivities. • The structured model shows that viscous flow should disappear in narrow pores. The S and N models then coincide. • In wide pores the results of the structured model are identical to those from hydrodynamic models of porous media. • Via the structured model, we can use hydrodynamic models to predict part of the behaviour of the friction coefficients in the non-structured modeL

314

Mass Transfer in Multicomponent Mixtures

A3-7 Further Reading The (S) structured way of looking at pores is discussed in the following papers by E.A. Mason and others: Mason, E.A. and Viehland, L.A., 1978, Statistical mechanical theory of membrane transport for multicomponent systems: Passive transport through open membranes. J. Chem. Phys., 68, 3562-3573. Mason, E.A. and Malinauskas, A.P., 1983, Gas transport in porous media: The dusty gas model. Elsevier, Amsterdam. Mason, E.A. and del Castillo, L.F., 1985, The role of viscous flow in theories of membrane transport. J. Membrane Science, 23, 199-220. Mason, E.A. and Lonsdale, H.K., 1990, Statistical mechanical theory of membrane transport. J. Membrane Science, 51,1-81. Mehta, G.D., Morse, T.F., Mason, E.A. and Daneshpajooh, M.H., 1976, Generalized Nernst-Planck and Stefan-Maxwell equations for membrane transport. J. Chem. Phys., 64, 3917-3923. The following book uses the (N) non-structured way of looking at pores. Lightfoot, E.N. (1974) Transport Phenomena and Living Systems. McGraw-Hill, New York.

A3-8 Exercises P-l We are sometimes asked why we think that a heterogeneous medium will behave as homogeneous when the scale of the heterogeneity is small enough (Figure P-l). We have no complete answer to this, but the following remarks may help. 1. Mechanical and civil engineers know from experience that they can do all kinds of things without worrying about small-scale inhomogeneities. They even build bridges and dams from materials like steel and concrete. These usually do not fall down. They only need macroscopic properties such as allowable stress, or modulus of elasticity. 2. As chemical engineers we can do nearly the whole of fluid flow without worrying about individual molecules (and even the greater part of the theory of diffusion). We only need macroscopic properties such as viscosity or diffusivities ... So it appears that we usually do not need to worry about inhomogeneity, as long as the scale is small enough and we have taken its effects into account in our macroscopic properties.

Appendix 3. Poring over Pores

315

Do you have any ideas on this your!>elf? P-2 Consider the electrodialysis membrane in Figure 19.16. The membrane contains 1 mol (m2 of membraner l of fixed negative charged groups -HS04-, and the same amount of positively charged mobile Na+ ions. There is no pressure difference and the electrical potential difference is -0.2 V. How large is the support force for a square metre of membrane? P-3 People working in the area of fluid flow, generally believe that there is no 'slip' at the interface between a fluid and a solid. Can you understand why they think this is so? Do you agree? Can you think of examples where there must be slip? P-4 Consider the situation in the bottom part of Figure P-7. The pore contains two species: '1' which is completely adsorbed and stagnant, and '2' which is completely mobile. The mole fraction and molar volume of '1' are Xl = 0.2 and l-l = 4 X 10-5 m 3 mol-I; those of '2' are x2 = 0.8 and V2 = 2 X 10-5 m 3 mol-I. a) What is the viscous selectivity of 'I'? b) And that of '2'? P-5 Checking the Mason Relations (Mathcad). Deriving the Mason relations in Figure P-9 is not simple. However, it is easy to check that they are correct. In this little file, we calculate the 'whole' velocities using both the structured and the nonstructured models. You will find that the two models give the same results, irrespective of your input. You do not need to look at this file if you already accept that the two models give the same result. .. P-6 Viscosity-at-Wall. The (very rough) estimate of the permeant-matrix coefficient in Figure P-l1 contains a 'viscosity at the wall' coefficient. You would expect this coefficient to depend on the interactions between the liquid and the wall. You might expect three cases: 1. the wall structure is such that the liquid adsorbs strongly, 2. the wall and the liquid have a similar structure, 3. the wall and the liquid 'do not like each other'. As examples, we consider (1) water on metal, (2) water on ice and (3) water on Teflon. How would you expect the value of Tl.t wall to compare with the bulk value of the viscosity for the three cases? P-7 Effect of Pore Size. a) How do the effects shown in Figure P-12 compare with the predictions from the Dusty Gas Model in Figures 21.8 and 21.9?

316

Mass Transfer in Multicomponent Mixtures

b) What can we learn about this diagram from Exercise 21.1? c) How does the behaviour of diffusion in micropores from Chapter 23 fit in? P-8 The Dusty Gas Model (Mathcad). In this file we work out all friction coefficients in both the structured and non-structured models, using the Dusty Gas Model from Chapter 21 and the Mason relations. This for a mixture of two gases and a porous matrix. The calculations are lengthy (ten pages!), but you get a detailed picture of the behaviour of all coefficients. P-9 The Sphere in Tube Model (Mathcad). This file shows the complexities of obtaining the Maxwell-Stefan friction coefficients of a sphere in a tube from an existing hydrodynamic model.

317

Answers There are several kinds of exercises in the book: 1. defined questions in the main text, 2. defined questions requiring only a bit of thinking or pencil and paper, 3. exercises using Mathcad and 4. open ended exercises, often referring to external sources.

Have you tried yourself?

The answers to the first group are in the main text or figures following the question. The answers to the second group are given here. To see the results of the Mathcad exercises, you will have to install Mathcad or Mathcad Explorer and read the files in the Answer folder on the CD-ROM. There are no answers to the fourth group.

Chapter 2 2.1 We have neglected gravity. 2.2 After we have put on the lid, the chemical potentials of water become equal in both phases and vaporisation stops. 2.3 Water is dragged along by the sodium ions. 2.4 For a given pressure difference, the velocity of flow is proportional to the diameter of a tube (or pore) squared. So flow disappears in small enough pores.

Chapter 3 3.1 The residence time in the film is u=1.5xlO-z ms-l~ I1t=&/u=7xlO-3 s resulting in a change in the momentum: I1(Mj uJ

---'---=--:.!.

I1t

=

(3 x lO-Z)X(1 x lO-Z)kgmol-1 ms- l 3

7x 10- s

'"

-Z -1 5 x 10 N mol

318

Mass Transfer in Multicomponent Mixtures

This is negligible compared to the driving force. 3.2 The two forces are of the same order of magnitude for particles of around 0.3 x 10-6 m. For particles of around this size, both sedimentation (gravity) and diffusion (chemical potential) play a role. 3.3 There is no mole fraction difference and no driving force, so the M-S equation predicts that there will be no flux. Actually there is no flux, but there will be minute mole fraction differences due to 'thermal diffusion', a phenomenon that we neglect in this book.

Chapter 4

ILlxdxll = 2. IUl - il21 = 2kl,dx2 = 2 X 10--4 ms- l .

4.1 When the value of XI is zero on either side ofthe film, then the ratio The velocity difference is then equal to

4.2 The Maxwell-Stefan equations use differences of velocities between the species. The differences are independent of any reference frame. 4.3 There are six velocity differences: Ui -u2,ul -u:"ul -u4,il2 -u3,u2 -u4,u3 -u4' However, only three of these are independent because

u2 -u3 = (Ul -U3)-(Ul -il2) u2 - u4 = (Ul - U4)- (Ul - il2) u3 - U4 = (Ul -U4)-(Ul - U3) 4.4. You can always remove any term like (u 2 answer of 4.3.

-

u 3 ) using the first equation in the

4.5 The bootstrap is that there is no transport of water through the film, so u2

=0

Chapter 5 5.4 The mole fraction of the rejected species increases by a factor: Llx2/ x2a = exp(udkl,2)' For reverse osmosis this has a value of around exp(O.I) "" 1.1; for ultrafiltration the value is around exp(l) "" 3 (and it can be much higher). Ultrafiltration is dominated by polarisation; reverse osmosis is not.

Chapter 6 6.1 dfl·I . (a) Instead of _dfl· _I use -Mi _ dz

dz

319

Answers

(b) Multiply the previous result by the mass fraction

00:

-COiMi dJ1i . dz

6.2 For a four component mixture the matrix becomes: SI,2 hI hI

S3,2

S4,I

S4,2

SI,3

SI,4

S2,3

S2,4 S3,4

S4,3

There are six different friction coefficients. 6.9 We write down the Maxwell-Stefan equation for hydrogen, taking hydrogen as component '1', oxygen as '2' and nitrogen as '3':

Pi =

~T X2(UI - U2) + ;T X3(UI - U3) = RTl~2 + ~3 J(UI -Uair) 1,2

1,3

1,2

1,3

The effective diffusivity is seen to have the value: D

I,ejf

= ~+~ [

n~1,2

n ) ~1,3

-1

I

=(0.21 + 0.79)3.22 3.41

X 10-5

m 2 s- I

6.10 (a) Si,ejf = .LXjSi,j Di,ejf = j#

.LxjD~J ki,ejf = .LXjki~J j#i

j#i

(b) In general, effective coefficients are complicated and incomprehensible functions of composition - and worse - of the velocities of all the species. Except in special cases, they are best avoided. (c) If the ratios between the fluxes or velocities have fixed values, you will also find effective coefficients to be handy. This is the case, for example, in the diffusion of the species for a single chemical reaction.

Chapter 7 7.1 The two choices yield the equations: N CpL( T - T',.ej) = -hilT + N[cpG( T - Trej ) + N[CpL(T - T',.ej These are indeed identical ...

Ml;:r]

)-Ml;:r] = -hilT + NCpG(T- Trej )

Mass Transfer in Multicomponent Mixtures

320

Chapter 8 8.3 (a) The initial slopes of the concentration profiles go down with increasing nonideality, and so do the fluxes. However, the fluxes do not become zero. (b) The flux does not become zero even though the Fick diffusivity does. At the point where this happens, the slope of the concentration profile is infinite, but only in that point. The flux is one half of that in the corresponding ideal solution. (c) When A > 2 the mixture fonns two phases. Because the system is symmetric, the phase boundary is half way down the pores. At the boundary, the two phases are in equilibrium, but have different compositions. (d) If a binary system has a negative diffusivity, this means that species will diffuse towards a higher concentration. So 'like will go to like' and the mixture will demix and fonn two phases.

Chapter 9 9.1 The turnover is at about 10-8 m; this is roughly at a diameter equal to the mean

free path of molecules in the gas. Below this value, the friction coefficient increases with the square of the diameter of the sphere; above that it increases linearly. The friction coefficients of the smallest spheres are much lower than predicted by Stokes' law. The hydrodynamicist will say that they 'slip' with respect to the surrounding fluid.

Chapter 11 11.1 Efl/! =-PZj

D.t/J ::::109 Nmol- 1 Llz

Electrical forces can be larger than those due to activity gradients. 11.2 The ratio of the exact to the approximate force is equal to: 1- 4.3S{X; 11.3 We integrate the Poisson equation to obtain:

dt/J = Gel Z or D.t/J = Gel Llz dz

eel

Llz

eel

(The integration constant is zero because of the given potential profile). A little manipulation then yields: - 2 eel D.t/J

_ 4 10-11

xub - FCLlz2 -

X

The mole fraction of unbalanced charges is seen to be very small. We can nearly always apply the electroneutrality relation in diffusion problems.

321

Answers

11.8 diffusivity

D+w ++

D_w

D+_

D++orD __

++

0 ++ ++ ++

0

conductivity

++

transport numbers

++

++

tracer diffusivity

0

++

0 0 ++

In the diffusivity experiment, there is no velocity difference between the + and - ions, so only the +w and -w diffusivities play a role. In both conductivity and transport number measurements, the difference in velocity between the ions is important. Transport number measurements give separate information on the two ions, and so are more useful than conductivity measurements. They are also more difficult to perform. Measurements with radioactive isotopes (or ions labelled otherwise) give information on ++ or -- interactions.

Chapter 12

tdl;

12.1 The species volume follows from \'i = A the two values are l't = 0.3 and V2 = 2.5 X 1021 m 3 morl. These lead to pressure forces of FiP "" 3000 and F! "" 2.5 x 1025 N mol-I. Pressure forces are proportional to molar volume: that is why they can be so large in macroscopic particles.

dif>

12.6 - =

dz

12.7 Obvious ones are magnetic gradients and capillary gradients. A little less obvious, but also important are the forces in solid matrices: elastic and support forces. Shear forces (for example between a gas and an adsorbed film) are frictional forces. All these can be brought into the Maxwell-Stefan formulation. There are other forces in society, where this is not so easy, such as the Air Force.

Chapter 15 15.1 (a) We get an ideal solution. (b) Y2

=exp(-O.5ed=exp[-~] l+xl

(c) The activity coefficient is smaller than unity; molecules of unequal size have a (slightly) greater tendency to mix than equal molecules. 15.2 (a) 0.9 (b) 0.99 (c) 0.999

322

15.3 Y2

Mass Transfer in Multicomponent Mixtures

= exp(xxf) or In(Y2) = X(l- xi).

This is identical to the formula in Figure

3.8. This is of course only so for species of an equal size. 15.4 (a) B

Vz

",,1

(b) XI,3 ",,1.5

(c) e3 ",,0.015. (You cannot read the graphs

Vc

accurately). 15.5 The polymer rejects large molecules.

Chapter 16 16.1

Chapter 17 17.3 1. Add a non-transferring solute to the rinsing side, so that the driving force on water is reduced to zero. 2. Apply a pressure difference so that the water flow is reduced by viscous flow. 17.5 The matrix is stagnant, so the net force on it must be zero: eMPa - eMPp + nlFi

+ nzEi + Fsupport

=0

Writing out the two friction terms yields:

Inserting these and rearranging gives the expected result: F support -_ -

(p a -

Pp

)

Chapter 18 18.2 The friction in the film will be more than 10% of the overall friction at a concentration of Cj :s; 10 mol m -3.

Answers

323

18.4 (a) Osmotic pressure is related to the equilibrium of the solvent on two sides of a membrane. It is the concentration difference of the solvent that has to be overcome by a pressure difference; not that of the solutes. (b) We can concentrate till Ax1 "" 0.04. (c) Ax1 ",,0.2.

Chapter 19 19.1 The membrane should pass sodium, but no chloride, no hydroxyl and as little water as possible. (Water dilutes the sodium hydroxide solution, which is evaporated further on in the process). 19.2 (a) There are two films where ions are removed and where their concentration could drop to zero. They are both in the left hand compartment. Because sodium has a lower diffusivity than chloride, you may expect the film on the left side of the membrane to be limiting. (b) The limiting current density can be calculated using Figure 19.5. (The flux of water through the membrane is not large enough to have much effect). Its value: 4 IUm "" 10 Am -2. Electrolysis units are operated at current densities well below this value. 19.3 You want a low salt concentration in the diluate compartments electrodialysis. This means a low limiting current density.

In

19.4 (a) 5x10 8 m 2 m-3 (b) The distance between charges is about 10-9 m. This hardly allows any further increase of the charge density. (c) You get surprisingly close (this may be fortuitous). 19.5 There are jumps in the electrical potential at the membrane interface. The jumps on the two sides are unequal (due to the Donnan equilibrium; see Figure 19.13). So there should be an extra electrical term in the second equation. This equals the Ax term in the upper equation.

Chapter 20 20.1 Answer:

x'

(I-x')

c'

2

=K -

x

c (I-x)

2

324

Mass Transfer in Multicomponent Mixtures

For low concentration c in the liquid, the affinity for the divalent ion becomes very large. You can use this to remove divalent ions from dilute solutions; you can regenerate the exchanger with a monovalent ion at a high concentration. 20.2 The changeover is at about 700 mol m-3; the changeover is gradual.

Chapter 21 21.1 (a) the pore diameter (b)D.M=(d I.

p

-d.)~ 9n 8 RT M. 1

I

(c) the value becomes zero when the molecule just fits in the pore.

Chapter 22 22.1 a) The fluxes have simple ratios, which are fixed by the reaction stoichiometry. Because of this we can eliminate all fluxes but one. b) The E is a consequence of needing a concentration averaged over all space. This concentration has a value that is E times that of the ideal gas in the pores. 22.2 Yes. Yes. 22.6 When the reaction rate is limited by one species only. This is usually the species with the lowest pressure.

Chapter 23 23.2 About 3000 mol m-3 23.4 Answer: about one of five hundred jumps is effective. The activation energy is about 15 kJ mol-I.

Chapter 24 24.1 Osmotic pressure is mainly determined by the mole fraction of the solvent. In ultrafiltration, the solute (protein) has a high molar mass, so the mole fraction of the solvent is always close to unity, and the osmotic pressure is small. 24.2 The first formula becomes:

Answers

325

RT RT RT RTln(xd= -'11f ~ 1f = --In(xd''''' --(I-Xl) = -x2

'1

'1

-"I.

Introducing the volume fraction of protein yields: E2 "'" x2 V2 "'" x2 V2, so Xl

'1

'1

1f

= RT E2' \t2

The V 2 is the result of introducing a volume fraction. 24.3 Answer:

1f

= ~(I+I~~1}2

24.4 The motion of the species is not governed by their pressure gradient, but by their potential gradient. At eqUilibrium all species have the same potential in the different parts of the membrane.

Appendix 3 P-2 The support force is zero. The effects of the electrical field on the two species cancel. P-3 In macroscopic fluid flow problems, velocities are much higher than slip velocities. However, they tend to go down quadratically with channel diameter. In sufficiently small channels, transport is by 'diffusion', which is another way of describing the slipping of species past each other. P-4

=0 (b) Here we need the volume fraction of '2': E2

(a) al

With

El al

= x;V2/{xi'1 + x;\t2) = 2/3.

+ E2a2 = 1 we obtain a2 = 1/ E2 = 1.5.

P-6

We don't know either. Our guess: (1) 11at wan» 11 (2) 11at wall"'" 11 (3) 11at wall::; 11 P-7

(a) The Dusty Gas Model and Figure P-I2 are in complete agreement. (b) Exercise 21.1 tells us that friction will rise sharply as the pore diameter is reduced towards the permeant diameter. This 'constrained Knudsen effect' is not shown in Figure P-I2. (c) Friction in micropores rises sharply due to direct interactions between permeant and pore wall. The reasoning used in the dusty gas model is not valid for micropores.

326

Index a absorption with heat effects, 80-81 activity, 35, 84, 86 in a polymer, 171 activity coefficient, 35, 85, 89 in electrolytes, 124 adsorption, 158-60, 252-60 eqnilibria, 253 micropore diffusion, 254 surface diffusion, 257 transient effects, 259 affinity chromatography, 159

b binary approximation of a ternary, 68-70 binodal curve, 87 bootstrap relation, 48-49, 60, 283 electroneutrality relation, 123 for chemical reactions, 58, 68 in distillation, 59 in electrolysis, 212 in ion exchange, 228 in liquid/liquid extraction, 91 in polarisation, 56 in stripping, 55 in vaporisation, 57 no current (no charge transfer), 126 thermal, 78 boundary layer theory, 114 C catalysis, 160, 242-49 Bosanquet formula, 245 concentration profiles, 247 effective binary, 70 effective friction coefficient, 245 effectiveness factor, 248 pseudo-first order kinetics, 247 reaction pressure gradients, 243 centrifugation, 139-41 charge number, 122 effect on ion-ion diffusivity, 131 of protein, 127

chemical potential, 35 ideal solution, 35 in gases, 202 non-ideal solution, 84 overview, 279-81 chemical reactions ammonia inside a catalyst, 245-49 ammonia, on a surface, 68 ammonia, with heat effect, 80 cracking and polymerisation, 243 gasification, 58 chemical reactors, 160 chromatography, 159 co-ion, 217, 224 condensation ternary, 64-65 with inert gas, 80 conduction, thermal, 77 conductive heat flux, 77 conductivity, electrical, 128 constriction factor, 237 constrictivity, 237 conventions, 21-23 counter ion, 217, 224 critical point, 87, 88, 89 current density, 129 Debye-Hiickel theory, 124

time dependence, 170 diffusion potential, 126 diffusivity. See diffusion coefficients dimensionless groups, 106 distillation effect of non-ideality, 8485 ternary, 65-67 with pervaporation, 200204 distillation efficiency, 66 Donnan equilibrium, 217 driving forces, 14, 34-44 centrifugal forces, 139 difference form, 42 effect ofnon-ideality, 8490 electrical forces, 124 gravity, 34, 138 in pervaporation, 202 on the whole mixture, 305 osmotic pressure gradient, 269 overview, 281 partial pressure gradient, 142 pressure gradient, 138, 141 sign convention, 21 support forces, 162

d

e

demixing, 87, 90 dialysis, 157, 191-94 difference equation, 40, 50-52 diffusion coefficients binary gas, 25 Fick coefficient, 25 Fick, ternary gas, 148 gas in bed of spheres, 23638 gas in cylindrical pores, 234-35 in electrodialysis membrane, 218 in electrolytes, 130-32 in gases, 95-96 in macropores, 257 in mass units, 299 in polymers, 180-88 in volume units, 300 lysozyme, 127 Maxwell-Stefan, ternary gas, 150

effective diffusivity, 69,127, 185,237 effective friction coefficient, 245 effective mass transfer coefficient, 69 efficiency, distillation, 66 electrical conduction, 128 electrical conductivity, 128 electrical double layer, 123 electrodialysis, 157,211,21420 Donnan equilibrium, 217 limiting current, 216 polarisation, 216 transport in membrane, 219 electrolysis, 211-14 effect of inert electrolyte, 213 limiting current, 214 electrolytes, 122 electroneutrality relation, 123

327

Index electro-osmosis, 219 energy flux, 77 enthalpy, 75-77 evaporation, 58 extraction, liquid-liquid, 90

f Fick equation, 24-26 in non-ideal mixtures, 88 ternary, 146 film model, 41, 106 fixed bed processes, 158,223, 252 Flory-Huggins equation, 172 flux, 24,50, 285 diffusion, 25 drift,25 energy, 77 free volume theory, 182-88 friction coefficient, 38, 4~7 overview, 282-84 spheres in a liquid, 46 swarm of spheres, 269 viscous, 307

g gas separation, 157, 194-97, 260 Gibbs-Duhem equation, 305 glass transition temperature, 169

h heat transfer coefficient, 78 heat transfer coefficients, 107 heat transfer with mass transfer, 74-81 heterogeneous catalysis. See catalysis

i interface, mobile, 114 ion exchange, 159,223-30 equilibria, 224 film limited, 227 particle limited, 228

k Knudsen diffusion, 233, 257

I Langmuir equation, 253 lattice model of diffusion, 184 law of mass action, 224 LDF model. See linear driving force model limiting resistance to mass transfer, 227 linear driving force model, 226, 258 liquid-liquid extraction, 90

m

n

macropore, 158,252 macropore diffusion, 257 mass transfer coefficient, 24, 106-18 choosing the right mechanism, 116 drops in a gas, 116 in and outside large bubbles, 115 in and outside liquid drops, 115 in packed bed (gas/liquid), 110 in packed bed (single phase),109 in space between membranes, 109 in swarms, 117 in tubes, 108 inside rigid particle, 112 limiting resistance, 227 outside rigid particle, 113 range, 52 mass transfer with heat transfer, 74-81 Mathcad location of files, 17 notes on exercises, 61 reading files, 293 Maxwell-Stefan equation, 39 comparing with Fick, 88, 152 comparing with TIP, 150, 152 difference form, 40 for gases and liquids, 142 in a matrix, 196 in a porous medium, 233 structured vs. nonstructured, 161,304-13 ternary, 63 membrane processes, 158 dialysis, 191-94 electrodialysis, 214-20 gas separation, 194-97 pervaporation,200-204 reverse osmosis, 204-7 ultrafiltration, 264-75 zeolite membranes, 260-61 micropore, 158,252 mobile interface, 114 momentum, 36 momentum balance, 36 in a gas, 36 Murphree efficiency, 66

Nernst-Planck equation, 125 non-ideality, 84-90 non-structured model, 23, 161, 162, 281, 282, 304-13 comparing with structured model,308 Mason equations, 308

o Onsager relation, 63, 282, 309 osmotic pressure in ultrafiltration, 266 seawater, 206

p partial molar volume. See species volume penetration theory, 113 pervaporation, 200-204 polarisation, 56, 204, 212, 216,267 polymers, 176 amorphous, 168 crystalline, 168 potential, 34 potential gradient centrifugal, 139 electrical, 124 gravitational, 34 protein, 140 proteins, 126

r rejection. See retention retention (of membrane), 207, 273 reverse osmosis, 158, 204-7 polarisation, 56-57

S sedimentation, 138-39 self-diffusivity, 184 settling, 138-39 species volume, 137 stripping, 55-56 structure of the book, 15 structured model. See nonstructured model supersaturation, 87, 88 support forces, 162 surface diffusion, 257 swelling of a polymer, 174

t thermal conduction, 77 thermal drift, 80 thermodynamic correction factor, 88 thermodynamics

328

Mass Transfer in Multicomponent Mixtures

overview, 279-81 solutions with a polymer, 176 thermodynamics of irreversible processes, 150 three gases problem, 26, 235 TIP, 150 tortuosity, 237 transference number, 130 transport number, 130 two cations problem, 28 two gases and porous plug, 28 two-bulb diffusion cell. See three gases problem U ultracentrifugation, 139-41

ultrafiltration, 158,264-75 driving force, 269 friction coefficient, 269 osmotic pressure, 266 polarisation, 267 size exclusion, 266 tubular module, 264 viscous flow, 272 viscous selectivity, 273 units in the MS-equations, 296 mass fractions, 297 mole fractions, 296 volume fractions, 269, 297

diffusive, 22, 306 viscous, 22, 273, 306 whole, 22, 306 viscous flow, 29, 162 Carman-Kozenyequation, 238 complications in narrow pores, 307 driving force, 305 Mason equations, 309 Poiseuille equation, 233 viscous selectivity, 273, 307, 312

V vaporisation, 57-58 velocity

Z zeolite, 159, 253

329

CD-ROM for 'Mass Transfer in Multicomponent Mixtures' The CD-ROM contains material associated with this book. You are allowed to use this material for any purpose directly related to learning or teaching the subject of the book. The first two directories contain all files (per chapter) of the Mathcad exercises in the book. The first set (Exercises) contains the files in Mathcad 2000 format. The second set (Exercises7) contains the same files in Mathcad 7 format. In both directories you will find subdirectories 'Questions' and 'Answers'. The Question files contain data and questions; you are expected to complete these yourself. You can view our anwers in 'Answers'. The third directory contains a self extracting ftle with the program 'Mathcad Explorer' . This program allows you to read Mathcad ftles, to modify the input and see the effects on output. You can use it to try out all of Mathcad, with two (important) exceptions: you cannot save or print your results. If you want to do that you will need a full version of Mathcad. (If you have a full version, we advise you to use that, and not to install Mathcad Explorer.) To install 'Mathcad Explorer', search the directory 'Mathcad Explorer' on your CD-ROM from inside Windows Explorer and double click on the icon. At a minimum you do need Windows 95 and 16 MB of memory to run this program. The following two directories contain exercises for learning those parts of Mathcad that you need in this book. They are best tried out after you have worked through the demonstration and the tutorial in the Mathcad Help system. Again one set is for Mathcad 2000, the other for Mathcad 7. The last directory contains the transparencies that we use in our courses. These are in PowerPoint 97 format. They are largely the same as the figures in the book, but in colour and in a bolder font. Also these are arranged per chapter. Exercises Answers Questions Exercises7 Answers7 Questions7 Mathcad Explorer Mathcad Tutor Mathcad Tutor7 Transparencies

Some other English language text books published by Delft University Press

Electronics. physics Electromagnetic waves - An introductory course prof.dr. P.M. van den Berg and prof.dr.ir. H. Blok 1999/ X+244 p. / ISBN 90-407-1836-9 includes (though not separately available) Electromagnetic waves - A repetitive guide prof.dr. P.M. van den Berg and dr.ir. M.D. Verweij 1999/ VI+30 p. / ISBN 90-407-1837-7 Electromagnetic waves appear in many forms and their applications are extremely widespread. Without exaggeration it may be said that our ability to employ and manipulate electromagnetic waves forms one of the reasons that communication plays such an important role in society. The macroscopic theory of electromagnetic waves has been formulated by Maxwell in 1864. But the mathematical-physical nature of the subject makes it difficult for students to master even today. The continuous stream of new college textbooks shows that many teachers encounter this problem and attempt to resolve it by presenting the theory in some suitable form. In the Electrical Engineering curriculum of the Delft University of Technology, the teaching of electromagnetic waves has been divided into three stages: 1) a basic course on Electricity and Magnetism, 2) an introductory course on Electromagnetic Waves, and 3) advanced courses on the application and computation of electromagnetic waves. The current book is written to facilitate the introductory course on Electromagnetic Waves. It is assumed that students are already acquainted with the basic phenomena and notions of the electric and the magnetic field, and that they know in which way Maxwell's equations describe the electromagnetic field. Starting from Maxwell's equations, this book deals with the derivation of plane wave propagation, plane wave reflection and transmission, electromagnetic rays, waves in two-wire transmission lines, waves in planar waveguides, and the excitation of electromagnetic waves. As such, the aim of the book is to provide a solid understanding of how the basic ingredients that make up the more sophisticated applications follow from Maxwell's equations. The aim of the introductory course on Electromagnetic Waves is to teach students to manipulate the fundamental formulas in order to

solve a problem at hand. To focus on this skill and to overcome the problem of having to learn many formulas by heart, an outline of this book is presented in the accompanying booklet entitled 'Electromagnetic Waves - A Repetitive Guide'. Chemical engineering Transport phenomena data companion prof.dr.ir. L.P.B.M. Janssen en dr.ir. M.M.C.G. Warmoeskerken 1997/ 168 p. / ISBN 90-407-1302-2/ hardback In dit boek is ernaar gestreefd een zodanige selectie van formules en tabellen op het terrein van de fysische technologie te presenteren, dat men hieruit voor het dagelijks gebruik kan putten. Het boek bestaat uit vier delen. In het eerste, algeme ne deel zijn gegevens ondergebracht varierend van het Griekse alfabet tot thermokoppels. Het tweede deel bevat veel gebruikte wiskunde. Algemene balansvergelijkingen, differentieren in vector- en tensornotatie, Besselfuncties, Laplace transformaties, elementaire differentiaalvergelijkingen en nog vele andere behoren hiertoe. Het derde deel is een compendium over fysische transportverschijnselen en bevat veel gegevens die in de vorm van grafieken zijn opgenomen. In het vierde deel wordt een keuze geboden uit de stofeigenschappen, uiteenlopend van die van water en lucht tot die van een aantal voedingsstoffen. Een uitvoerige index sluit het (Engelstalige) boek af. Mathematical Geodesy

Series on Mathematical Geodesy and Positioning Adjustment Theory prof.dr.ir. P.J.G. Teunissen 2000/ 201 p. / ISBN 90-407-1974-8 Adjustment theory can be regarded as the part of the mathematical geodesy that deals with the optimal combination of redundant measurements together with the estimation of unknown parameters. It is essential for a geodesist, its meaning comparable to what mechanics means to a civil-engineer or mechanical engineer. Historically, the first methods of combining redundant measurements originate from the study of three problems in geodesy and astronomy, namely to determine the size and shape of the Earth, to explain the long-term inequality in the motions of Jupiter and Saturn, and to find a mathematical representation of the motions of the Moon. Nowadays, the methods of adjustment are used for a much greater variety of geodetic applications, ranging from, for instance, surveying and navigation to remote sensing and global positioning. The two main reasons for performing redundant

measurements are the wish to increase the accuracy of the results computed and the requirement to be able to check for errors. Due to the intrinsic uncertainty in measurements, measurements redundancy generally leads to an inconsistent system of equations. Without additional criteria, such a system of equations is not uniquely solvable. In this introductory course on adjustment theory, methods are developed and presented for solving inconsistent systems of equations. The leading principle is that of leastsquares adjustment together with its statistical properties. The inconsistent systems of equations can come in many different guises. They could be given in parametric form, in implicit form, or as a combination of these two forms. In each case the same principle of least- squares applies. The algorithmic realizations of the solution will differ however. Depending on the application at hand, one could also wish to choose between obtaining the solution in one single step or in a step-bystep manner. This leads to the need of formulating the system of equations in partitioned form. Different partitions exist, measurement partitioning, parameter partitioning, or a partitioning of both measurements and parameters. The choice of partitioning also affects the algorithmic realization of the solution. In this introductory text the methodology of adjustment is emphasized, although various samples are given to illustrate the theory. The methods discussed form the basis for solving different adjustment problems in geodesy. Other text books in the Series on Mathematical Geodesy and Positioning Testing theory, ISBN 90-407-1975-6 Dynamic data processing, ISBN 90-407-1976-4

Civil engineering Introduction to bed, bank and shore protection Engineering the interface of soil and water ir. G.J. Schiereck 2000/ approx. 400 p. / ISBN 90-407-1683-8 The interface of land and water has always played an important role in human activities. Settlements are often located at coasts, riverbanks or deltas. Harbours, waterways, dikes, dunes and beaches, structures for water-control and water-resources management etc. are examples of hydraulic engineering on a macroscale. In this book, the interface is studied on a micro-scale. The occurring phenomena are

importClnt in all branches of hydraulic engineering. In a natural situation the interface moves freely with the forces of erosion and sedimentation. Actually, nothing is wrong with erosion, until some interest is threatened. Erosion is somewhat like weed: as long as it is in nobodies way, no action is needed or even wanted. There should always be a balance between the efforts of protection against erosion and the damage that would occur otherwise. Moreover, it should be realised that once a location is protected along a coast or riverbank that is eroded on a large scale, the protected part can induce extra erosion and in the end the whole coast or bank has to be protected. So, look before you leap, should be the motto. In many cases however, a protection is necessary: bottom protection behind outlet structures or around objects, revetments in rivers and canals: dike protection, coastal defence works etc. Contents: Introduction • Flow - Loads. FlowStability Flow - Erosion • Porous Flow - General • Porous Flow - Filters • Waves - Loads • Waves Erosion and stability • Ships. Proteetions· Construction and maintenance • Design • Appendix A: Materials • Appendix B: Environmental aspects • Appendix C: Cases. Principles of river engineering The non-tidal alluvial river P.Ph. Jansen (ed.) 1994/ XVI+509 p. / A4 / ISBN 90-407-1280-8 This book is intended both for the practising river engineer who, with its help, will be able to tackle problems giving all aspects the correct weights without overlooking any of these aspects - and for the post-graduate student. The book is divided in five parts. The first part presents a general introduction to river engineering. The next three parts deal with the basic subjects of river hydraulics, river surveys and river models. The final part deals with their applications. It should be stressed that most benefit will be obtained by studying the book as a whole, not just turning a particular section. Extensive use has been made of research work published by experts in many countries. System theory Mathematical systems theory prof.dr. G.J. Olsder 1998/VIII+192 p. / ISBN 90-407-1272-7 A system is part of reality which we think to be a separated unit within this reality. The reality outside the system is called the surroundings. The interaction between system and surroundings is realized via quantities, which are called input and

output. Quite often one wants, through a proper choice of the input, the system to behave in a desired way. Mathematical Systems Theory is concemed with the study and control of inpuVoutput phenomena. The emphasis is on the dynamic behaviour of these phenomena, i.e. how do characteristic features change in time and what are the relationships.

These course notes are intended for use at the undergraduate level and form the basis for other courses such as optimal control and filter theory. Contents: Introduction • Modelling principles • Linear differential systems • System properties • State and output feedback • InpuVoutput representations • Polynomial representations • Linear difference systems • Extensions and some related topics.