mass transfer in multicomponent mixtures (j.a. wesselingh,r. krishna) completo

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 well-known 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 ... 13 1.1 Who should read this Book? 13 1.2 What this Book covers 14 1.3 Structure of the Book 15 1.4 Guidelines to the Reader 16 1.5 Guidelines to the Teacher 17 1.6 Symbols 18 1.7 Conventions 21

2 Is SOMETHING WRONG? 24 2.1 The Starting Point 24 2.2 Three Gases 26 2.3 Two Cations 28 2.4 Two Gases and a Porous Plug 28 2.5 Summary 30 2.6 Further Reading 30 2.7 Exercises 30

PART 1. MAss TRANSFER IN GASES AND LIQUIDS 33 3 DRIVING FORCES 34

3.1 Potentials, Forces and Momentum 34 3.2 Momentum (Force) Balance of a Species 36 3.3 The Driving Force: a Potential Gradient 38 3.4 The Maxwell-Stefan Equation 39 3.5 Simplifying the Mathematics 40 3.6 The Film Model 41 3.7 Difference Form of the Driving Force 42 3.8 Summary 44 3.9 Further Reading 44 3.10 Exercises 45

4 FRICTION 46 4.1 Friction Coefficients and Diffusivities 46 4.2 Velocities and the Bootstrap 48 4.3 Velocities and Fluxes 50 4.4 The Difference Equation 50 4.5 Mass Transfer Coefficients 52 4.6 Summary 53 4.7 Further Reading 53 4.8 Exercises 53

5 BINARY EXAMPLES 55 5.1 Stripping 55 5.2 Polarisation 56


5.3 Vaporisation 56 5.4 Gasification of a Carbon Particle 58 5.5 Binary Distillation 59 5.6 Summary 60 5.7 Further Reading 60 5.8 Exercises 61

6 TERNARY EXAMPLES 63 6.1 From Binary to Ternary 63 6.2 A Condenser 64 6.3 A Ternary Distillation 65 6.4 A Ternary Reaction 68 6.5 Binary Approximation of a Ternary 68 6.6 Summary 70 6.7 Further Reading 70 6.8 Exercises 71

7 MASS AND HEAT TRANSFER 74 7.1 Temperature Gradients 74 7.2 Enthalpy 75 7.3 Mass Transfer Relation 76 7.4 Energy Transfer Relation 77 7.5 Condensation in Presence of an Inert Gas 79 7.6 Heterogeneous Reacting Systems 80 7.7 An Ammonia Absorber 80 7.8 Summary 81 7.9 Further Reading 81 7.10 Exercises 82

8 N ON-IDEALITIES 84 8.1 Chemical Potential and Activity 84 8.2 Non-ideal Binary Distillation 84 8.3 A Simple Model of Non-idealities 85 8.4 Large Non-Idealities: Demixing 87 8.5 Maxwell-Stefan versus Fick 88 8.6 When can we neglect Non-ideality? 89 8.7 Mass Transfer in Liquid-Liquid Extraction 90 8.8 Summary 92 8.9 Further Reading 92 8.10 Exercises 92

9 DIFFUSION COEFFICIENTS 95 9.1 Diffusivities in Gases 95 9.2 Diffusivities in Liquids 97 9.3 How do you measure diffusivities? 100 9.4 Summary 102 9.5 Further Reading 102 9.6 Exercises 103

10 TRANSFER COEFFICIENTS 106 10.1 Introduction 106 10.2 Dimensionless Groups 106

Contents 9

10.3 Tubes and Packed Beds 108 10.4 Packed Gas-Liquid Columns 110 10.5 Single Particles, Bubbles and Drops 111 10.6 Using 'Single' Coefficients for Swarms 117 10.7 Using Binary Coefficients in Multicomponent Calculations 118 10.8 Summary 118 10.9 Further Reading 119 10.10 Exercises 120

11 ELECTRICAL FORCES AND ELECTROLYTES 122 11.1 Electrolytes 122 11.2 The Electroneutrality Relation 123 11.3 Electrical Forces 124 11.4 Transport Relations 124 11.5 Diffusion of Hydrochloric Acid 125 11.6 Plus a Trace of Sodium Chloride 126 11.7 Diffusion of Proteins 126 11.8 Conduction and Friction between Ions 128 11.9 Diffusivities in Electrolyte Solutions 130 11.10 Summary 132 11.11 Further Reading 133 11.12 Exercises 134

12 CENTRIFUGAL AND PRESSURE FORCES 137 12.1 Volumetric Properties 137 12.2 The Pressure Gradient 138 12.3 Gravitational Force 138 12.4 Centrifuges 139 12.5 Gas and Protein Centrifugation 140 12.6 Difference Equation for the Pressur~ Force 141 12.7 The Maxwell-Stefan Equations (again) 142 12.8 Summary 143 12.9 Further Reading 143 12.10 Exercises 143

13 WHY WE USE THE MS-EQUATIONS 146 13.1 Three Ways 146 13.2 A Mixture of Three Gases 147 13.3 The Fick Description 148 13.4 Thermodynamics of Irreversible Processes 150 13.5 The Maxwell-Stefan Description 150 13.6 Units 152 13.7 Further Reading 152 13.8 Exercises 153

PART 2. MASS TRANSFER THROUGH A SOLID MATRIX 155

14 SOLID MATRICES 156 14.1 The Applications 156 14.2 Membrane Processes 156 14.3 Adsorption and Chromatography 158 14.4 Heterogeneous Catalysis 160


14.5 Structured and Non-structured Matrices 161 14.6 Effects of a Matrix on Mass Transfer 162 14.7 Compositions with a Matrix 163 14.8 How Further? 164 14.9 Further Reading 165 14.10 Exercises 165

15 PROPERTIES OF POLYMERS 167 15.1 A Few Words on Polymers 167 15.2 Thermodynamics of Mixtures in a Polymer 171 15.3 Summary 177 15.4 Further Reading_ 177 15.5 Exercises 177

16 DIFFUSION IN POLYMERS 180 16.1 Behaviour of Diffusivities 180 16.2 The Free Volume Theory 182 16.3 Summary 188 16.4 Further Reading 189 16.5 Exercises 189

17 DIALYSIS AND GAS SEPARATION 191 17.1 Dialysis 191 17.2 Gas Separation 194 17.3 Summary 197 17.4 Further Reading 197 17.5 Exercises 198

18 PERV APORA TION AND REVERSE OSMOSIS 200 18.1 Pervaporation 200 18.2 Reverse Osmosis 204 18.3 Summary 208 18.4 Further Reading 208 18.5 Exercises 209

19 ELECTROLYSIS AND ELECTRODIALYSIS 211 19.1 Introduction 211 19.2 Polarisation in Electrolysis 212 19.3 Electrodialysis 214 19.4 Summary 220 19.5 Further Reading 220 19.6 Exercises 221

20 ION EXCHANGE 223 20.1 Fixed-Bed Processes 223 20.2 Ion Exchange EquiIibria 224 20.3 Linear Driving Force Model 225 20.4 Ion Exchange (Film Limited) 227 20.5 Ion Exchange (Particle Limited) 228 20.6 Summary 230 20.7 Further Reading 230 20.8 Exercises 231

Contents 11

21 GAS PERMEATION 233 21.1 Transport in Cylindrical Pores 233 21.2 The Diffusion Coefficients 233 21.3 Looking Back 235 21.4 Transport in a Bed of Spheres 236 21.5 The Dusty Gas Model 238 21.6 Summary 240 21.7 Further Reading 240 21.8 Exercises 241

22 IN POROUS CATALYSTS 242 22.1 Introduction 242 22.2 Pressure gradients inside a particle 243 22.3 Separate Transport Equations 245 22.4 Single-variable Pressure and Rate Expressions 245 22.5 Solution for a Slab 247 22.6 Summary 249 22.7 Further Reading 249 22.8 Exercises 250

23 IN ADSORBENTS 252 23.1 Adsorption 252 23.2 Equilibria - Langmuir Isotherm 253 23.3 Maxwell-Stefan and Fick Diffusivities 254 23.4 Macropore Diffusion 257 23.5 Transport Equations 258 23.6 Transient Adsorption of a Binary 259 23.7 Membrane Applications 260 23.8 Summary 261 23.9 Further Reading 261 23.10 Exercises 262

24 ULTRAFILTRATION 264 24.1 The Module 264 24.2 Membrane and Permeants 265 24.3 Osmotic Pressure (No Ions, no Charge) 266 24.4 Size Exclusion 266 24.5 Polarisation 267 24.6 Transport Equations 269 24.7 Inside the Membrane 272 24.8 Electrical Effects 273 24.9 Summary 275 24.10 Further Reading 276 24.11 Exercises 277

25 ......... ENDING 279 25.1 Looking Back 279 25.2 Thermodynamic Models - the Potentials 279 25.3 Driving Forces 281 25.4 Friction Terms 281 25.5 Friction Coefficients 282


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

THANKS

ApPENDIX 1. READING MATHCAD

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

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

ANSWERS

INDEX

CD-ROM FOR 'MASS TRANSFER IN MULTICOMPONENT MIXTURES'

283 285 286 287 287

289

293

296 296 297 297 298 299 300 300 301 301 302 302 303

304 304 305 305 308 310 313 314 314

317

326

329

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 multi-component 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.


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 multi-component 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 multi-component 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


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 MS-equations. 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.

1. Beginning ... 17

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 Math-cad 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


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 mol m-3

cp molar heat capacity J morl

D Fick diffusivity m2 S-I d diameter m D Maxwell-Stefan diffusivity m2 S-I E energy flux J m-2 S-I E activation energy J mol-I

F force per mole N mol-I

'.F Faraday constant C mol-I g acceleration of gravity m S-2 H enthalpy J mol-I

h heat transfer coefficient W m-2 KI I electrical current density A m-2 J flux with respect to the mixture mol m-2 S-I k mass transfer coefficient m S-I K equilibrium constant -M molar mass kg mol-I m mass kg N flux with respect to an interface mol m-2 S-I n number of moles mol p pressure N m-2

1. Beginning ...

R gas constant J mol-I K"I

R retention r radius m r reaction rate mol m-3 S-I T temperature K t time s Tg glass transition temperature K u diffusive velocity ms-I

v viscous velocity ms-I

V molar volume m3 mol-I w whole (overall) velocity ms-I x mole fraction -y mole fraction -z distance (position) z charge number -

Greek symbols a viscous selectivity -Ll increase of ...

m

E volume (void) fraction electrical potential V 'Y activity coefficient -11 viscosity Pa s A thermal conductivity J m-I K"I /l chemical potential J mol-I V stoichiometric coefficient -1t osmotic pressure N m-2

P density kg m-3

cr interfacial tension N m-I 't stress N m-2

't tortuosity-'\) diffusion volume m3 mol-I CO angular speed rad S-I ~ friction coefficient, structured N s mol-I m-I 'If potential of a species J mol-I S friction coefficient, unstructured N s mol-I m-I

19


Superscripts

* PI

11* o D I,2 d V I,2

pr DxI=I

1,2

pfq T,.ej vT

ui X',X"

Subscripts Cw dp Dt,ejf \j (I,M VC xIa' xIf3 T,.ej UI, U2

UI.

Ui

Uj (v

average of x in a film .: pressure of pure '1' (vapour pressure) .: volume of pure' 1 ' Q in free space in the dispersed phase

boiling pressure V 1,2 at x, = 1

pressure of '1' if a reaction were to go to s;guilibrium reference thermal velocity velocity of '1' at the average composition x in two different phases

concentration of water diameter of particle or pore effective Fick diffusivity of '1' free volume friction coefficient between '1' and the matrix molar volume of a chain element species '1' at different positions

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)


molar volumes of the pure species

molar volumes of the species as in the mixture (partial molar volumes)

V;*, V;, ~*

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 u3 = 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).

1. Beginning ...

Structured model

I Non-structured model

, .. :::... . ........ :." .... ~~

~ ~

F; = X2S1,2(WI-W2)+SI,MWl F2 = X1S 2,I(W2 -W1)+S2,M W2

S (zeta)

~ (ksi)

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

23

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.

I Ji : flux of i with respect to the mixture dc.

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

! .

2. Is Something Wrong? 25

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

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.

I Ni : flux with respect to an interface I differential equation

N dc. N i=-Di ck+ Xi difference equation

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

diffusion drift flux flux

flux of mixture N=LNi

---. -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.


J --D dC1 gas: c1 + Cz = c = constant 1 - 1 dz

J --D dcz fluxes with respect to mixture 2 - 2 dz

+------

J J _0 __ Dd(C1+C2) 1 + 2 - -

only one binary D, which is independent of

composition

Fig. 2.3 A binary gas is simple

dz

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: X N2 = 0.46

xH2 =0.54 xN2 =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?


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.

mole fraction Xi reverse

:: t-=;:m;on 0.4

0.6

0.4

0.2

time 0.0 h

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.


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+;:

CI-

CD H+ moves rapidly

Na+"

Cl-

0 so Na+ can move against ~ 0 its concentration gradient! ~ 3

low concentration

~ excess +charge and electrical field

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.


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 ) Ar NHe =-3NAr 298 K+-= 298K t

100 kPa

-

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

298K NHe =-N Ar

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.


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.)


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 lm

o the potential difference is the work required to change the condition of the weight

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

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 'YI mixture

work required: change in the chemical potential

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

d;tivity "'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

in an ideal gas

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

Ili = const(p, T) + R~~ ) partial pressure

Fig. 3.3 Chemical potentials in ideal solutions


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 one-dimensional 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.


Fig. 3.5 Gases moving through each other

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

I We consider this in Chapter 21. 2 See exercise 3.1. Velocity effects are important in centrifuges (Chapter 12) and in shock waves from detonations.


PIAiz - PIAiz+dz cc PIP2(~ - uI) Taking the limit for dz ~ 0 yields:

dPI -- cc PIP2(uI -~) dz

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,2X2(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 pressure4 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.


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 = _ dlfli I 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).


the driving force the sum of the friction on a species i in a - forces between i and the

-

mixture 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/ = ~ r .. x.(u. -u.) A_ ..~I.J J I 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.


great scientist

one month

scientist dlfl L

__ I = r .. x.(u.-u.) d ~I" ,I , Z .#.

, lone week

engineer !1lf1i L( - (- -)

---= . x u-u A _ I" ,I , L4. .#.

, 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

1 1 1

3':"'1

eddies and large "---'0:,:' .,001 scale convection ", :, "~ l

000

30

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

, -----} ... 'film': .. ---------- ---t

no eddies !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.


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.

+1

LlJll RT

3. Driving Forces

exact In( a l /3) ala

approximate 04-----~----4-----~ Lla1 = al~ - ala.

a l 0.5(al~ + ala.)

we do not use the 'exact' potential diffirence in our approximations

Fig. 3.14 The difference approximation

43

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

I mole fractions Xl~ = 0.001 of CO2

Ft - RT dx l _ RT Llxl 1--------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.


1 kg of CO2

RT Llxj 8.314x300 -0.002 Fj "" - .&: Xj = 10-5 X +0.002

= 2.4 x 108 N "" 5 X 109 kgf mol CO2 kg 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.


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 + Ffri.ction = il( Mjuj ) onl 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 dl 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 61ITJ2dl 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 motl (m s-lyl. This is not far from

4. Friction 47

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 Maxwell-Stefan diffusivities. There are simple relations between the two:

RT RT Vi,; = ~ and Si,; = D ..

':>1, I I, I

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 Maxwell-Stefan 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 m2 SI. 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 mor1 (m Sl)"l; we come back to them in Chapter 9.


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

'floating' transport relations: have to be 'tied' to surroundings

Fig. 4.3 Equations floating in thin air

49

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).


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 MS-equation 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 MS-equation: 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

L1al 1 _ (_ _) --=-=kX2 U1-U2 D k =~ 1,2 L1z positive

direction a1 1,2

t mass transfor coefficient

Fig, 4.6 Deriving the difference equation

film ,-A--.,

o ctl-- avemge concentration c, a ~

U; species velocity (depends on position in film)

"'{V species velocity at the average o --+-------1f-- 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 _ _ _ --=-x (u -u)

- k 2 1 2 al 1,2

Fig, 4.8 Difference equation in flux form

51

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.


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 ~ +-- gases -1 -1 ~c c.. '" 10 m s m pores 10-4 ~c +-- liquids CH cc .. '" 10-4 m s-1 ~ mpores 10-6

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


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 (u2 - 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 .

rl mtl

drops on a tray

gas: trace ofNH3 (1) bulk ofN2 (2)

transport relation

-Ax x2N] -x]N2 \ ] I k],2C ()~ x2 ~l Nz = 0

bootstrap

Fig. 5.1 Stripping of a trace of ammonia


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

5. Binary Examples 57

have a gradient upwards. Yet toluene is evaporating. So it must be diffusing against its concentration gradient. How can this be?

X2

water (l) permeates salt (2) does not: x2 small, u2 = 0

&2 transport: &2 -1 O-uI .

-----

x2 k l,2

3 also - &2 X2 = X2 o; +2

2 2 uI

so Ax2=~ X2 2-~

I kl,2 lli kl ,2

exact solution: Llx2 = exp( ul ) 00 1 2 X2a kl ,2

Fig. 5.3 'Polarisation' upstream of a desalination membrane

heat

benzene (l), volatile toluene (2)

Yl =K1x1B l--"';"

x'B ~ ::! _!.2 = K 2x2B

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


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

N -'=v N2

x,N2 -x2N, k,,i:

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.

1.0

0.6 0.4

0.0

both components are moving and have a high concentration

c =10 mol m-3

bootstrap:

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 NI = - kI 2c LlxI = _ 10-

2 X 10 (-0.6)

x2 +2xi 0.7+2x0.3

NI =0.046 (exact:0.047)molm-l 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 ~---- x

/1 l~ 1

/ 1 XI~ 1

xzo. : hexane (1)

W \N2;1

---'--'--'!"'--- 0 et ~

transport relation _ A .. _ xlNI-xlNl

LUj-kl,lC

bootstrap Nj = - Nl (equimolar exchange)

_ A .. _ (X, +xl)N, ~ LUI - -'--'--'=":"'-"':"

kl,lC

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 (j.a. wesselingh,r. krishna) completo

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