Biochemical Engineering: An Introductory Textbook [1 ed.] 9789814800433, 9780429031243, 9780429632044, 9780429630552, 9780429633539

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Biochemical Engineering

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Biochemical Engineering

Biochemical Engineering An Introductory Textbook

Debabrata Das Debayan Das

Published by Jenny Stanford Publishing Pte. Ltd. Level 34, Centennial Tower 3 Temasek Avenue Singapore 039190

Email: [email protected] Web: www.jennystanford.com British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Biochemical Engineering: An Introductory Textbook Copyright © 2019 by Jenny Stanford Publishing Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN 978-981-4800-43-3 (Hardcover) ISBN 978-0-429-03124-3 (eBook)

Contents

Foreword Preface List of Symbols 1. Introduction 1.1 Biological Sciences Microbiology 1.2 1.2.1 Microbial Culture 1.2.2 Microbial Growth Cycle Biochemistry 1.3 1.3.1 Biomolecules 1.3.2 Metabolism and Metabolic Pathways 1.4 Biochemical Engineering and Bioprocesses 1.4.1 Strain Selection and Improvement 1.4.2 Bioprocess Design and Optimization 1.5 Applications of Biochemical Engineering 1.5.1 Agricultural Bioprocesses 1.5.2 Single-Cell Proteins 1.5.3 Biopharmaceuticals 1.5.4 Food and Dairy Applications 1.5.5 Environmental Applications 1.5.6 Biofuels 1.5.7 Biosensors 1.6 Objectives of the Book 1.7 Scope and Relevance 1.8 Major Chapters 2. Stoichiometry of Bioprocesses 2.1 Law of Conservation of Mass 2.2 Mass Balance of Bioprocesses 2.3 Thermodynamic Efficiency

3. Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis 3.1 Chemical Reaction Thermodynamics

xiii xv xvii

1 2 4 5 8 9 10 12 14 16 17 19 19 20 20 20 21 21 21 22 22 22 25 25 26 30 45 45

vi

Contents

3.1.1 3.1.2 3.1.3 3.1.4 3.1.5

3.2

3.3 3.4 3.5 3.6

Scope and Introduction Reversible and Irreversible Processes Work and Heat Concept of Internal Energy First Law of Thermodynamics 3.1.5.1 Enthalpy 3.1.5.2 Constant-volume and constant-pressure process 3.1.5.3 Heat capacity 3.1.5.4 Isothermal and adiabatic process 3.1.5.5 Entropy 3.1.5.6 Gibbs free energy 3.1.5.7 Enthalpy of formation (Dhf) 3.1.5.8 Enthalpy of combustion (Dhcomb) 3.1.5.9 Hess’s law 3.1.5.10 Chemical equilibrium Chemical Reaction Kinetics 3.2.1 Rate of Reaction 3.2.2 Irreversible Second-Order Reaction 3.2.3 Irreversible Reactions in Parallel 3.2.4 Irreversible Series Reaction Reversible Reaction Dependency of Reaction Rate on Temperature Chemical Reactor Analysis 3.5.1 Batch Reactor 3.5.2 Continuous Stirred Tank Reactor 3.5.3 Plug Flow Reactor Multiple Reactor System

4. Enzymatic Reaction Kinetics 4.1 Characteristics of an Enzyme 4.2 Application of Enzymes 4.3 Enzyme Kinetics 4.3.1 Michaelis–Menten Approach 4.3.2 Briggs–Haldane Approach 4.4 Estimation of the Kinetic Parameters vmax and kM 4.5 Bioreactor Modeling for Enzymatic Reaction

45 48 48 49 50 51 52 53

54 55 57 59

59 59 60 61 62 66 70 72 74 74 76 77 79 80 82

115 118 118 119 123 125

126 128

Contents

4.6 4.7

4.5.1 Batch Bioreactor 4.5.2 Plug Flow Bioreactor 4.5.3 Continuous Stirred Tank Bioreactor Inhibition of Enzyme Reactions 4.6.1 Competitive Inhibition 4.6.2 Non-competitive Inhibition Factors Affecting Enzymatic Reactions 4.7.1 Effect of pH 4.7.2 Effect of Temperature 4.7.3 Effect of Shear

5. Immobilized Enzymes 5.1 Merits of Immobilized Enzymes 5.2 Demerits of Immobilized Enzymes 5.3 Classifications of Immobilization Techniques 5.4 Characterization of Immobilized Enzymes 5.4.1 Activity of Immobilized Enzymes 5.4.2 Bound Protein 5.4.3 Specific Activity of Bound Protein 5.4.4 Coupling Efficiency 5.4.5 Stability of Carrier–Enzyme Complexes 5.4.6 Materials, Methods, and Conditions of Binding Technique 5.4.7 Physicochemical Characteristics of Carrier 5.4.7.1 Solid matrix morphology and configuration 5.4.8 Application of Immobilized Enzymes 5.4.8.1 Industrial 5.4.8.2 Analytical 5.4.8.3 Medical 5.5 Kinetics of Immobilized Enzymes 5.5.1 External Mass-Transfer Resistance 5.5.1.1 Effectiveness factor and Damköhler number 5.5.1.2 Determination of the factor affecting overall rate of reaction 5.5.2 Effect of Internal Mass Transfer 5.5.2.1 Effectiveness factor and Thiele modulus

128 129 130 132 132 133 135 135 136 136

173 173 174 174 179 179 179 179 180 180 180 181 183 186 186 187 187 187 188 190 192 192 197

vii

viii

Contents

6. Microbial Growth Kinetics, Substrate Degradation, and Product Formation 6.1 Differences between Enzymatic and Microbial Reaction 6.2 Bacterial Growth Cycle 6.3 Ideal Reactor for Kinetics Measurements 6.3.1 The Ideal Batch Reactor 6.3.2 Cell Growth Models 6.3.2.1 Generalized growth model 6.3.3 Cell Growth Models with Cell Growth Inhibitors 6.3.3.1 Substrate inhibition 6.3.3.2 Product inhibition 6.3.3.3 Toxic compound inhibition 6.3.4 Logistic Equation 6.3.5 Cell Growth Characteristics of Multicellular Cells Like Mold 6.3.6 Product Formation Kinetics in Cell Culture 6.3.7 Determination of Maintenance Coefficient of Cells 6.4 Ideal Continuous Flow Stirred Tank Reactor 6.4.1 Determination of Kinetic Constants in Chemostat 6.4.2 Analysis of Chemostat with Cell Recycle 6.4.2.1 Chemostat with cell mass recycling 6.5 Continuous Operation Using Plug Flow Reactor 6.6 Kinetics of Fed Batch Cell Growth 6.6.1 Variable Volume Fed Batch 6.6.2 Constant Volume Fed Batch 7. Air Sterilizer 7.1 Airborne Microbes 7.2 Methods of Air Sterilization 7.3 Physical Implication of Single Fiber Collection Efficiency 7.3.1 Inertial Impaction 7.3.2 Interception 7.3.3 Diffusion

205 207 208 211 212 215 216 217 217 218 219 219 220

221

222 223

229 230 234 236 238 238 241 293 293 294 295 296 296 297

Contents

7.4

7.5 7.6 7.7

Types of Filters 7.4.1 Depth Filters 7.4.1.1 Pressure drop in air filter 7.4.2 Membrane Filters 7.4.2.1 PVA filter for air sterilization Experimental Setup for Air Sterilization Selection Criteria of Air Filter Economic Considerations

8. Medium Sterilizer 8.1 Filtration 8.1.1 Physical Characteristics of Microorganisms 8.1.2 Filter Type 8.1.2.1 Depth filters 8.1.2.2 Absolute filters 8.1.3 Filtration Strategy 8.2 Thermal Destruction 8.2.1 Thermal Death Kinetics 8.2.2 Effect of Temperature on Death Kinetics 8.2.3 Experimental Determination of Microbial Death Rate 8.2.4 Batch Sterilization of Medium 8.2.5 Temperature–Time Profile during Batch Sterilization 8.2.6 Scale-Up of Batch Sterilization 8.3 Continuous Sterilization 8.3.1 Equipment for Continuous Sterilization 8.3.1.1 Continuous steam injection 8.3.1.2 Continuous plate heat exchangers 8.4 Design of a Continuous Sterilizer 8.4.1 Fluid Flow 9. Transport Phenomena of Bioprocesses 9.1 Fluid Mechanics in Bioprocesses 9.2 Boundary Layer Theory 9.3 Incompressible Fluid Flow inside a Pipe

301 301 303 303 304 305 306 306 313 314

314 314 314 315 315 316 316 317 318 319 320 321 322

323 323 323 324 324 335 337 343 345

ix

x

Contents

9.4 9.5 9.6

9.7 9.8

9.9

9.10

Mass Transfer in Bioprocess Diffusion in Liquid and Gas Oxygen Diffusion in Fermentation Broth 9.6.1 Mechanism of Mass Transfer 9.6.2 Estimation of Mass-Transfer Coefficients via Correlations 9.6.3 Volumetric Mass-Transfer Coefficient Determination of Oxygen-Absorption Rate 9.7.1 Experimental Determination of kLa by Dynamic Gassing out Techniques Heat Transfer in Bioprocess 9.8.1 Conduction 9.8.2 Convection 9.8.2.1 Free convection 9.8.2.2 Forced convection 9.8.2.3 Forced convection across tube banks Evaluating the Overall Heat Transfer Coefficient (U) Power Consumption

10. Process Parameters Monitoring and Control 10.1 Operational Parameters for Bioprocess 10.2 Purpose of Monitoring Different Operational Parameters 10.3 Resistance Thermometer 10.4 Foam Control 10.5 Rotameter 10.6 Rotational Viscometer 10.7 Turbidity Meter 10.8 pH Probe 10.9 Dissolved O2 Probe 10.10 CO2 Probe 10.11 Oxidation–Reduction Potential Sensor 10.12 Automated Process-Control System 11. Downstream Processing 11.1 Solid–Liquid Separation 11.1.1 Pretreatment 11.1.2 Sedimentation

346 347 350 351 352 353 354

357 361 361 363 366 368 369 371 373 379 379

384 385 385 386 388 388 389 390 392 392 392 395 396 396 397

Contents

11.2 11.3 11.4

11.1.3 Centrifugal Settling 11.1.4 Filtration 11.1.4.1 Plate and Frame Filter Press 11.1.4.2 Rotary Vacuum Drum Filter 11.1.4.3 Membrane Filter Concentration Purification of the Product 11.3.1 Liquid–Liquid Extraction 11.3.2 Adsorption 11.3.3 Chromatography Formulation 11.4.1 Crystallizer

12. Industrial Fermentation Processes 12.1 Baker’s Yeast 12.1.1 History of Baker’s Yeast Production 12.2 Classification of Baker’s Yeast 12.3 Baker’s Yeast Fermentation Process 12.3.1 Raw Materials 12.3.1.1 Theoretical calculation of cane molasses requirement for Baker’s yeast production 12.3.1.2 Typical material analysis 12.3.2 Characteristics of Yeast 12.3.3 Other Physicochemical Parameters 12.4 Ethanol Fermentation Process 12.4.1 Classification of Ethanol 12.4.2 Microbes Used in Alcohol Fermentation 12.4.3 Factors that Affect Alcohol Fermentation 12.5 Citric Acid Fermentation Process 12.5.1 History of Citric Acid Production 12.5.2 Citric Acid Fermentation Process 12.5.3 Materials Analysis of Citric Acid Fermentation Process 12.5.4 Sample Calculation for Yield and Productivity

397 399 401 401 402 404 404 405 407 410 411 411 415 416 416 416 417 418 419 420 420 421 422 422

423

423 426 427 427 431

431

xi

xii

Contents

13. Biological Wastes Treatment Processes 13.1 Activated Sludge Process 13.1.1 Nature and Morphology of Mixed Microbes 13.1.2 Settling Characteristics of the Sludge 13.1.3 Process Parameters 13.1.4 Calculation of Oxygen Uptake Rate 13.1.5 Design of Activated Sludge Process 13.1.6 Kinetic Models 13.1.7 Effect of Temperature and pH 13.1.8 Sedimentation Tanks or Separator 13.2 Biological Wastewater Treatment Process by Immobilized Whole Cell Reactor 13.2.1 Rotating Disk Biological Reactor 13.3 Anaerobic Digestion Process 13.3.1 Process Design 13.4 Development of Overall Kinetic Model for AD of Organic Biomass (A) 13.4.1 Two-Step Anaerobic Digestion Process

Index

445 446 447 447 448 449 450 451 452 453 459 459 465 465 470 471 479

Foreword

Foreword

The importance of biotechnology has been increasing due to the manipulation of DNA for the production of several new bioproducts that are useful in chemical, pharmaceutical, agricultural, and environmental management. Through research work, it is possible to produce new products useful for mankind. A book on biochemical engineering would certainly help not only the undergraduate and postgraduate students but also researchers and process engineers involved in developing and applying bioprocesses to get all the necessary information for designing and operating bioreactors, which are the heart of any biochemical process. I congratulate the authors, Prof. Debabrata Das and Dr. Debayan Das, for seeing the need for such a book and bringing it out in a timely manner. The present book titled Biochemical Engineering: An Introductory Textbook comprehensively covers all aspects of the applications of bioprocesses. The objective of this book is to carry out the mathematical analysis of the processes in simplified forms in order to be understandable and utilizable by most of the biologists. Derivation of mathematical equations is explained in details so that the majority of the students can understand the meaning of the equations. I strongly recommend this excellent book to the undergraduate and postgraduate students in biotechnology, biochemical engineering, chemical engineering, food technology, biochemistry, and related fields and also to the researchers and process engineers involved in the production of bioproducts so that they can enrich their knowledge in the area of application of biosciences and benefit mankind.

Prof. Roberto De Philippis Department of Agrifood Production and Environmental Sciences University of Florence, Italy

xiii

Preface

Education is not the learning of the facts, but training of the mind to think. —Albert Einstein

All engineering disciplines have been developed from basic sciences. Science gives us the knowledge to develop new products whereas engineering applies science to scale up production for commercial purposes. Biological processes involve different biomolecules, which come from living sources. It is now possible to modify DNA to get desired changes in biochemical processes. Developments in gene expression, protein engineering, and cell fusion have significantly affected product development in biotechnology industries. Chemical processes deal with different reactions to get desired products. Biochemical reactions in nature are mostly reversible and chain reactions. To understand biochemical engineering, it is necessary to know the principles of chemical engineering, which involves not only mathematical modeling but also scaling up of processes for commercial applications. Any biochemical industry can be divided into three major steps. The first step is upstream processing, which involves medium preparation, medium and air sterilization, etc. The second step involves the bioreactor where biochemical reactions lead to the desired products. The third and crucial step is purification of the products, known as downstream processing. This book begins with the identification of the differences between conventional chemical reaction engineering and biochemical reaction engineering. It gradually makes readers conversant with the rate laws and their applications to help them understand reaction engineering behavior and give them the expertise to apply the acquired knowledge in designing bioreactors. It discusses the stoichiometry of bioprocesses for materials and energy analysis and the transport phenomena that are important for the operation of bioprocesses. The book also enables students to contribute their knowledge in various professional fields like bioprocess development, modeling and simulation, environmental engineering, etc.

xvi

Preface

The chapters are organized in broad engineering sub-disciplines such as mass and energy balances, reaction theory using chemical and enzymatic reactions, microbial cell growth kinetics, and transport phenomena. Other chapters such as different control systems used in the fermentation industry, case studies of some industrial fermentation processes, different downstream processes, and effluent treatment are also included. Each chapter begins with a fundamental explanation for general readers and ends with in-depth scientific details suitable for expert readers. The book also includes solutions of more than 100 problems. It is written in a manner so that it can be useful to senior and graduate students of biotechnology and those studying courses in food and environmental engineering. It is also appropriate for chemical engineering graduates, undergraduates, and industrial practitioners. We would like to acknowledge help of Ms. Jhansi L. Varanasi and Mr. Chandan Mahata at various stages of manuscript preparation. We are also thankful to Mr. Tapas Mohanty for his help in creating most of the book’s illustrations using computer graphics. We hope this book will be useful to our readers! Debabrata Das Debayan Das Summer 2019

List of Symbols

List of Symbols

Symbols a A Ab

Ap

am, av b

C

Constant Area

Frontal area of the float Area of the particle

Surface area per unit volume Constant

Heat capacity; Concentration, Concentration ratio, Cunningham’s correction factors for slip flow

CA*

Equilibrium concentration of solute in fluid phase

CASm

Maximum solute adsorption capacity of the solid

CAS*

CDm, CD CE CP

CR CS

CV D

Da

Des, DBM, DAB dP, df, dS Dz

Equilibrium concentration of solute on the solid surface Drag coefficient

Equilibrium solute concentration in extracting solvent

Product concentration; Heat capacity at constant pressure Equilibrium solute concentration in raffinate Substrate concentration

Heat capacity at constant volume

Dilution rate, Decimal reduction time Damköhler number

Effective diffusivity of substrate, Diffusivity of the particle, Diffusivity A into B Particle diameter, Fiber diameter, Sauter mean diameter Axial dispersion coefficient

xvii

xviii

List of Symbols

E, Ea, Eo or eo Activation energy, Total enzyme concentration, Potential difference EIME

Overall activity of immobilized enzyme

f

Friction factor

F

Fd Fb G

gc, g Gr

h

h hf

hcomb I

K, k ka

KA kG Ki

kL

kLa

kf

Keq Km kp

KS kr l

m, mS

Force, Volumetric flow rate, Faraday constant, Flow rate Drag force

Buoyancy force

Free energy, Substrate feed rate

Conversion factor, Acceleration due to gravity

Grashof number

Heat transfer coefficient Enthalpy

Enthalpy of formation

Enthalpy of combustion Inhibitor concentration

Constant, Thermal death rate constant, Rate constant Mass transfer coefficient Adsorption constant

Gas side mass transfer coefficient Inhibition constant

Mass transfer coefficient in the liquid phase

Volumetric mass-transfer coefficient Rate constant of forward reaction Equilibrium constant

Michaelis–Menten constant

Product inhibition constant Saturation constant

Rate constant of backward reaction Length

Mass, Maintenance coefficient, Empirical constant

List of Symbols

M N n NP Nr

NS

Nu P

Pe

Po Pr

q, Q qP qS r

R Ra

Re, NRe rO2 rx T

t, tgn, td, tb, thd S, S0, SSS Sc Sh u, U V, v

Amount of biomass

Number of moles, Concentration of the cells, Number of discs in the stack Order of reaction, Constant Power number

Reaction number

Rate of mass transfer Nusselt number

Pressure, Product concentration Peclet number

Power number

Prandtl number

Thermal energy, Rate of uptake of oxygen, Heat flux, Volumetric flow rate Specific product formation rate

Specific substrate consumption rate Rate of reaction, Radius

Gas constant, Radius, Pellet radius Rayleigh number

Reynolds number

Rate of oxygen consumption Specific oxygen uptake rate Temperature

Time, Generation time, Doubling time, Batch time, Holding time Entropy, Substrate concentration, Initial substrate concentration, Steady state substrate concentration Scmidth number

Sherwood number

Internal Energy, Overall heat transfer coefficient Volume, Air velocity, Rate of reaction

xix

xx

List of Symbols

vmax

Maximum velocity of reaction

W, w

Work done

VP, VR, VC, Vg X x90 Y

YC

YX/S, Y YP/S

YX/O Z

Greek Letters m mmax md h

xp

sb

sP sS gb

gp gs

t, tCSTR, tPFR e

h b j r

Volume of the particle, Volume of the reactor, Critical air velocity, Settling velocity Fraction of the substrate converted 90 % removal of cell mass Mole fraction

g C-atom biomass/g C-atom substrate Cell mass yield coefficient

Product yield coefficient

g biomass/g oxygen as O2 consumed Height of the column

Micron, Specific growth rate of the cell Maximum specific growth rate Specific death rate of the cell

Energetic growth yield, Efficiency Energetic product yield

Weight fraction of carbon in biomass Weight fraction of carbon in product

Weight fraction of carbon in substrate Degree of reduction of biomass Degree of reduction of product

Degree of reduction of substrate

Space time, Space time in CSTR, Space time in PFR Voidage, Void fraction Effectiveness factor

Saturation parameter, Non-growth associated coefficient Thiele modulus, Inertial parameter Density

List of Symbols

a αg,m αG

—total eM t

ω l θ θC

Abbreviations A

AC

Growth associated coefficient, Recycle ratio, Volume fraction of the filter

Fraction of carbon utilized for cell growth and maintenance Fraction of carbon converted to gaseous form Sterilization criterion

Eddy diffusivity of momentum Stress

Angular velocity

Extraction factor

Hydraulic retention time

Mean cell residence time Arrhenius constant Activated charcoal

Acetyl-CoA

Acetyl co-enzyme A

AMP

Adenosine monophosphate

AD

ADP

6-APA ASP

ATP

BOD CA

CAA

CAM CMC CoA

COD

CSTR

DEAE

Anaerobic digestion

Adenosine diphosphate

6-aminopenicillanic acid

Activated sludge process Adenosine triphosphate

Biochemical oxygen demand Citric acid

Citric acid anhydrous

Citric acid monohydrate Carboxymethylcellulose Coenzyme A

Chemical oxygen demand

Continuous stirred tank reactor

Diethylaminoethyl

xxi

xxii

List of Symbols

DHAP

Dihydroxy acetone phosphate

EMP

Embden–Meyerhof–Parnas

DNA DO

FAD

FADH FDA F6P

F1,6 P FTU GC

HAc

HEPA HFCS HLac HMP IE

IR

KE

LAB LC

MLSS

MLVSS NAD+

NADH

NADP+

NADPH NTU PE

Deoxyribonucleic acid Dissolved oxygen

Flavin adenine dinucleotide (oxidized form) Flavin adenine dinucleotide (reduced form) Food and Drug Administration Fructose-6-phosphate

Fructose 1,6 diphosphate Formazin turbidity unit Gas chromatograph Acetic acid

High efficiency particulate air High fructose corn syrup Lactic acid

Hexose monophosphate Immobilized enzyme Infrared

Kinetic energy

Lactic acid bacteria

Liquid chromatograph

Mixed liquor suspended solids

Mixed liquor volatile suspended solids Nicotinamide form)

adenine

dinucleotide

(oxidized

Nicotinamide adenine dinucleotide (reduced form) Nicotinamide adenine dinucleotide phosphate (oxidized form) Nicotinamide adenine dinucleotide phosphate (reduced form) Nephelometric turbidity unit Potential energy

List of Symbols

PEP

Phosphoenolpyruvate

RBC

Rotating disc biological contactor

PFFP PFR

RNA RQ

RVF SCP SVI

TCA TA

UQ UV

VFA

Plate and frame filter press Plug flow reactor Ribonucleic acid

Respiratory quotient Rotary vacuum filter Single cell protein

Sludge volume index Tricarboxylic acid Total acid

Ubiquinone Ultraviolet

Volatile fatty acid

xxiii

Chapter 1

Introduction

All engineering disciplines have been developed from the basic sciences. Biological processes involve different biomolecules that come from living sources. Chemical processes deal with the reactions of different reactants to get the desired products. The major bottlenecks of these processes are high temperature and pressure and use of different raw materials for different products. On the other hand, biochemical processes are mostly operated at the ambient temperature and atmospheric pressure and can produce different products from the substrate like sucrose using different microorganisms. Biochemical reactions are mostly chain and reversible reactions in nature. However, biochemical processes are complicated as compared to chemical processes. Biochemical engineering involves not only mathematical modeling but also scaling up of the process for commercial application. The purpose of this book is to initially distinguish the differences between conventional chemical reaction engineering and biochemical reaction engineering. Students will be gradually conversant with the rate laws and their applications in understanding the reaction engineering behavior. With this expertise, they will be able to apply the acquired knowledge in designing bioreactors. They will also learn the stoichiometry of the bioprocesses for material and energy analyses. The transport phenomena of bioprocesses are also very important for their operation. This textbook will provide students with the ability to contribute their knowledge in various professional Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

2

Introduction

fields such as bioprocess development, modeling and simulation, environmental engineering, etc.

1.1 Biological Sciences

Biological science or biology is the study of living organisms such as plants, animals, and other living organisms. The word biology comes from the Greek words bios meaning “life” and logos meaning “the study of” (knowledge). The subject of biology is divided into many separate fields such as behavior, human anatomy, botany, physiology, zoology, ecology, and genetics. Biology is a natural science concerned with the study of life and living organisms, including their structure, function, growth, evolution, distribution, identification, and taxonomy. Living organisms are characterized by their ability to grow, reproduce, metabolize, acclimatize, generate waste products, and display sensitivity to the surrounding environment. All living things are composed of one or more cells, which are the basic structural and functional units of life. Cells are considered to be the building blocks of life. There are two distinct types of cells: prokaryotic and eukaryotic cells. The difference lies in their structure, nature of nucleus, organelles, metabolism, and so on. Table 1.1 shows the major differences between prokaryotic and eukaryotic cells. About two million different kinds of organisms live on earth. It is impossible to study every living organism individually; therefore, they are classified into various groups based on their similarities. The science dealing with the description, identification, naming, and classification of organisms is known as taxonomy. It was first developed by Carl Linnaeus. He is known as the father of taxonomy. For the naming of organisms, he introduced a binomial nomenclature comprising a genus name and a species name. Microorganisms can be classified based on their cell type, phenotypic, genotypic, and analytical. The most widely accepted classification based on cell type is the three-domain system introduced by Carl Woese et al. in 1977. It divides cellular life forms into archaea, bacteria, and eukarya domains (Table 1.2). For each domain, the final scientific hierarchy for classification is as follows: Domain > Kingdom > Phylum > Class > Order > Family > Genus > Species.

Biological Sciences

Table 1.1

Comparison of prokaryotic and eukaryotic cells

Feature

Prokaryote

Eukaryote

Cell size (diameter)

Average 0.5–5 µm

> 10 µm

Cell wall

Rigid in plants (cellulose) Rigid, containing and fungi (chitin), absent polysaccharide plus protein. Murein is the main in animal cells strengthening compound.

Organization

Unicellular or filamentous

Unicellular or filamentous or multicellular

Cell membrane Present

Present

Chromosome

Multiple, containing DNA and protein

Nucleus

Ribosome

No nuclear envelope, no nucleolus

Single circular, containing only DNA 70S, 50S+30S subunits

80S, 60S+40S subunits

Asexual (binary fission)

Sexual (mitosis and meiosis)

Mitochondrion Absent Cell division Flagellum

Table 1.2

Nucleus bound by nuclear envelope, nucleolus present

Simple, lacking microtubule (not enclosed by cell surface membrane)

Present

Complex, with 9+2 microtubules, (surrounded by cell surface membrane)

Classification of living organisms based on the three-domain system

Group

Cell type

Typical organisms

Archaea (Archaebacteria)

Prokaryote

Methanogens, acidophiles, halophiles, extreme thermophiles

Eukarya

Eukaryote

Algae, fungi, protozoa, plants, and animals

Bacteria (Eubacteria)

Prokaryote

Enteric bacteria, cyanobacteria

3

4

Introduction

The study of living things has undergone tremendous expansion in the recent years, and topics such as cell biology, neuroscience, evolutionary biology, and ecology are advancing rapidly. This rapid expansion has been accompanied by a blurring of the distinctions between disciplines: A biologist with an interest in tropical plants may well use many of the tools and techniques that are indispensable to a molecular geneticist.

1.2 Microbiology

Microorganisms can be viewed only with the aid of a microscope. Microbes are found in a variety of shapes, and their sizes range from 0.01 mm to 20 mm. The first microbe was discovered by Antonie van Leeuwenhoek using a compound microscope. He is known as the father of microbiology. Microbes are diversified in nature and can be grouped into six classes, which include eubacteria, archaebacteria, viruses, fungi, algae, and protozoa. The specific characteristic features of each of these classes can be found in standard microbiology textbooks. Microorganisms are ubiquitous in nature. They are tenacious and can thrive in extreme environments. The growth conditions (such as nutrition, temperature, pH, etc.) of each of these organisms vary distinctly. Based on nutrition, they can either be autotrophic (can utilize sunlight and inorganic carbon as their energy and carbon source) or heterotrophic (dependent on organic carbon for energy and carbon source) in nature. Some microorganisms can grow at −20°C (psychrophiles), while others can grow at 20–40°C (mesophiles) and 50–120°C (thermophiles). Similarly, microbes that prefer pH values below 6 are known as acidophiles, while others that grow well above 9 are termed alkaliphiles. The organisms that grow in high salt regions are known as halophiles. Depending on the availability of oxygen in the surrounding environment, microbes have developed different metabolic regimes. The microbes that cannot survive without oxygen are known as aerobes, while others that are inhibited by the slightest presence of oxygen are obligate anaerobes. However, a few organisms that can switch their metabolic pathways to thrive in both conditions are called facultative microbes. Microorganisms play a crucial role in day-to-day life. The concept of using microbes for the production of industrially

Microbiology

relevant products is not new. From several decades, microbes were used for the production of wine, beer, vinegar, etc. without the acknowledgement of their presence. In addition, dairy industries widely rely on thermophilic bacteria for acquiring yoghurt from milk. The discovery of microbes has helped in the development of modern microbiology and its contribution to industrial biotechnology. Over the years, different microbial metabolites such as ethanol, butanol, acetic acid, lactic acid, riboflavin, and enzymes such as protease, amylase, invertase, etc. have been produced for commercial usage. Figure 1.1 shows the different microbial products that have been explored till date. The first antibiotic produced on a large scale was penicillin (a fungal product) during World War II. After that, other antibiotics have been discovered using microbial exploitation. By using genetic engineering approaches, many useful non-microbial products such as insulin, human growth hormone, vaccines, interferons, and other pharmaceutical products have been successfully commercialized since the 1980s. This great diversity provides a nexus of opportunities for microbial technologies in the coming centuries. Food products and Beverages

Hazardous waste management

Oxychemicals (ethanol, acetic acid etc.)

Vaccines

Microbes

Therapeutic agents

Single cell proteins

Antibiotics Hormones

Figure 1.1 Application of microbes for bioproduct development.

1.2.1 Microbial Culture Microorganisms require different nutrients for their growth and development (Table 1.3). The essential nutrients include

5

6

Introduction

macronutrients (required in large quantities) and micronutrients (required in trace amounts), vitamins (required as cofactors), and water. These nutrients perform several functions that aid in their growth. Water helps in the soaking up of nutrients and for the elimination of waste products. In the laboratory conditions, microorganisms are grown in a culture media that comprises all these essential nutrients. Specialized media are developed for isolating and identifying organisms, testing their antibiotic sensitivity, distinguishing one species from another, etc. (Table 1.4). The nutritional requirement of each organism depends on the natural habitat from which it is isolated. For example, organisms growing in deep-sea sediments require complete depletion of oxygen from its surrounding environment for their growth and survival, while the organisms growing in saline soils require high salt conditions. Table 1.3

Essential nutrients required for microbial growth

Nutrients

Functions Carbon (C)

Required for growth, product, and to provide energy

Nitrogen (N)

Essential for protein synthesis

Oxygen (O) Macronutrients

Hydrogen (H) Sulphur (S)

Phosphorus (P) Calcium (Ca2+)

Cations

Potassium (K+)

Magnesium (Mg2+)

Iron (Fe2+, Fe3+)

Adequate amount of oxygen is needed for the growth if the organism is an aerobe

Contribute to components of carbohydrates, lipids, proteins, and nucleic acids In cell activation, heat resistance of endospores, for activity of number of enzymes

Serves as cofactor for many enzymes, stabilizes membranes and ribosomes

A part of cytochromes, cofactor for enzymes, and electron-carrying proteins

Microbiology

Nutrients

Functions Manganese (Mn)

Trace elements

Zinc (Zn)

Molybdenum (Mo) Cobalt (Co) Nickel (Ni)

A part of enzyme and cofactors, help in catalysis of reactions, maintenance of protein structure, etc.

Copper (Cu) Table 1.4

Type of culture media

Type of media

Components/Purpose

Simple

A general media used to grow or cultivate normal microorganisms of environment, e.g. nutrient agar media

Complex

Defined/ Synthetic

The exact concentration of all the components are unknown.

All the components and concentrations are known.

Selective

Used for the growth of only selected microorganisms

Minimal

A few and necessary nutrients are supplied.

Differential Enrichment

Used to distinguish one microorganism type from another growing on the same medium Commonly used to harvest different types of microbes from the same source

In the natural habitat, a mixed population of bacteria coexists to perform synergistic functions and often compete for nutrients to survive. However, for an experimental and industrial point of view, isolation of a single pure culture (i.e., a population of cell originating from single species) is a prerequisite. Robert Koch developed techniques for isolating pure cultures to identify human pathogens. These techniques are now used to isolate a species of interest. There are three means of establishing a pure culture: streak plate, spread

7

Introduction

plate, and pour plate methods. In the streak plate technique, the inoculum (or the environmental sample) is streaked over the agar surface in such a way that it “thins out” the bacteria. In the spread plate technique, the inoculum is diluted and inoculated to a molten medium, which is spread on the solid agar media. In the pour plate method, the inoculum is poured over an empty plate and kept for incubation. These plate-based techniques are useful when the concentrations of microbes in a sample are high. For samples with low bacterial count, the isolation and enrichment can be performed using selective, differential, or enrichment media as described above.

1.2.2 Microbial Growth Cycle

Microbial growth can be defined as an increase in the cellular constituents, which in turn increase the microbial cell size, number, or both. When microorganisms are grown in a liquid medium in closed conditions, their growth follows a definite course with distinct phases. These phases can be observed by using a growth curve as shown in Fig. 1.2. The curve usually comprises four phases: lag phase, log phase, stationary phase, and decline phase. Stationary phase Log number of viable cells

8

Log phase (Exponential phase)

Decline phase

Lag phase Time

Figure 1.2 The different phases of microbial growth curve.

During the lag phase, there is no appreciable increase in cell number or size as during this initial phase of growth, the cells

Biochemistry

adapt to the new environment and the necessary enzymes and metabolic intermediates are built up in adequate quantities for the multiplication of cells to proceed. This phase is also known as acclimatization or adaptation phase. Depending on the lifespan of a microorganism, the duration of lag phase can vary from a few hours to days. The lag phase is immediately followed by the log (logarithmic) or exponential phase. In this phase, cells divide exponentially with time (Fig. 1.2). The cells are highly active in this phase, and the cell division occurs at its maximal rate. This is also known as the active phase. This phase plays an important role for the development of the inoculum. The inoculum for the fermentation medium is usually prepared in between the mid-log and the latelog phases. The growth rate remains constant, and the chemical and physiological properties of cell population are uniform. The duration of the exponential phase depends on the nature of microbe, carbon and nitrogen sources, culture conditions, etc. After a varying period of exponential growth, the cell enters the stationary phase where the cell division declines due to the depletion of nutrients and accumulation of toxic products. There exists an equilibrium between the newly growing cells and the cells dying, thereby making the overall count of viable cells constant. This is known as the starvation phase. It is observed that the growth of cells ceases when a critical population density is reached, i.e., the cell enters the decline or death phase. In this phase, the cells die due to nutritional exhaustion and toxic accumulation.

1.3 Biochemistry

Biochemistry is the study of the structure, composition, and chemical reactions of substances in living systems. It primarily focuses on understanding how biological molecules give rise to the processes that occur within living cells and in between cells. The chemistry of the cells depends on the reactions of different biological molecules that perform a specific function. Biochemistry mainly deals with the structure and function of different biomolecules and the various biochemical reactions devoted for the functioning of biological molecules.

9

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Introduction

1.3.1 Biomolecules Living cells are composed of different high-molecular-weight biomolecules such as carbohydrates, lipids, proteins, and nucleic acids. These biomolecules constitute the major structural elements of living cells. Each biomolecule is a part of intracellular organelle and functions in its unique microenvironment. Table 1.5 enlists the various properties of the different biomolecules. Carbohydrates are a key source of energy in living organisms. Its general formula is (CH2O)n, where n is greater than or equal to 3. Green plants synthesize carbohydrates by the photosynthesis process using sunlight. The functional groups present are either ketone or aldehyde. Carbohydrates exist in two stereo-isomers: “D” or “L” forms. Depending on the number of sugar molecules, carbohydrates can be classified as monosaccharides (single sugar unit; e.g., glucose), disaccharides (two sugar units; e.g., sucrose), oligosaccharides (2–10 sugar units; e.g., raffinose), and polysaccharides (>10 sugar units, e.g., starch). Table 1.5

Properties of different biomolecules

Biomolecule

Elements

Monomer

Function

Examples

Carbohydrate

Carbon, hydrogen, oxygen

Monosaccharide

Source of energy

Amino acid

∑ Structural component ∑ Transport/ storage ∑ Control biochemical reactions (enzymes) ∑ Protection against the toxins (antibodies)

Glucose, fructose, starch (potato)

Protein

Carbon, hydrogen, oxygen, nitrogen

Collagen, albumin, enzymes, eggs, soy beans

Biochemistry

Biomolecule

Elements

Monomer

Function

Lipid

Carbon, hydrogen

Nucleic acids

Fatty acid + ∑ Stores glycerol energy ∑ Membrane constituent

Nucleotide Carbon, hydrogen, oxygen, nitrogen, phosphorus

Stores and transmits genetic information

Examples Fats, oils, waxes DNA (deoxyribonucleic acid), RNA (ribonucleic acid)

Proteins are the most abundant organic molecules in living cells, constituting 40% to 70% of their dry weight. They are polymers of amino acids joined by peptide bonds (i.e., –CONH–). Proteins have diverse biological functions and can be classified into five major categories: (i) structural proteins (collagen, keratin), (ii) catalytic proteins (enzymes), (iii) transport proteins (hemoglobin, serum albumin), (iv) regulatory proteins (insulin), and (v) protective proteins (antibodies, thrombin). Fats or lipids are hydrophobic biological compounds that are formed by esterification of longchain fatty acids with glycerol. The general structure of fatty acids is (H3C-CH2-COOH)n. The value of n may vary from 12 to 20. Fatty acids may be saturated (having no double bonds, e.g., myristic acid, stearic acid, etc.) or unsaturated (at least one double bond, e.g., oleic acid, linoleic acid, etc.). Steroids can also be classified as lipids. They are essential for controlling development and metabolism at very low concentrations. A most common example of steroid is cholesterol, which is present in animal tissue. Nucleic acids play a central role in the reproduction of living cells. They store and preserve genetic information. They carry genetic code as well as are responsible for protein synthesis. Nucleotides are the major building block of DNA and RNA, which comprises phosphoric acid, pentose sugar, and nitrogenous bases. There are two types of nitrogenous bases: (i) purines (e.g., adenine and guanine) and (ii) pyrimidine (e.g., thymine, cytosine, and uracil). The base sequences carry genetic information, which is expressed and passed on from one generation to another.

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Introduction

1.3.2 Metabolism and Metabolic Pathways The mechanisms by which cells harness energy from their environment via different biochemical reactions are known as metabolism. It is the sum of all chemical reactions that take place within a cell providing energy for vital processes (catabolism) and for synthesizing new organic materials (anabolism). Catabolism involves break down of biomolecules to a more oxidized end product for generating energy and reducing power. On the other hand, anabolism is the biosynthesis of more complex compounds from simple precursor molecules. Anabolism involves consumption of energy and reducing power generated during the catabolic processes. The cell stores energy in the form of adenosine triphosphate (ATP), while the reducing power is stored in the form of nicotinamide adenine dinucleotide (NADH) and nicotinamide adenine dinucleotide phosphate (NADPH). Figure 1.3 shows the major metabolic pathways involved during anabolism and catabolism. Carbohydrates

Proteins

Nitrogen Pool

glycolysis

Lipogenesis

Fatty acid spiral

Pyruvic acid

Acetyl-CoA

Urea cycle Urea

glyconeogenesis

Lactic acid

CO2

Fatty acid, glycerol

Glucose-6phosphate

Glycogen

Tissue protein NH3

Fats and Lipids

Glucose, fructose, galactose.

Amino acids

CO2

Citric acid cycle

2H+

ADP

ADP

ADP

Electron Transport Chain 2e-

ATP

ATP

ATP

O2 H2O

Figure 1.3 Interrelationship between major metabolic pathways.

Glucose metabolism has been studied in much detail as glucose is the major carbon and energy source for many organisms. Depending on the availability of oxygen, glucose can be catabolized aerobically (in the presence of oxygen) or anaerobically (in the absence of oxygen). Aerobic glucose catabolism mainly involves three major

Biochemistry

metabolic pathways: (i) the Embden–Meyerhof–Parnas (EMP) pathway, or commonly known as glycolysis in which glucose is converted to pyruvate, (ii) the tricarboxylic acid (TCA) cycle in which pyruvate is oxidized to NADH and CO2, and (iii) the electron transport chain (ETC) in which electrons are transferred from NADH to the final electron acceptor (oxygen) to form ATP. The end product of glycolysis, i.e., pyruvate is the key metabolite in glucose metabolism. Under anaerobic conditions, the EMP pathway is followed by fermentation in which pyruvate is converted to lactic acid, ethanol, acetic acid, butanol, etc. depending on the enzymatic machinery available. In this process, substrate-level phosphorylation supplies the ATP. The net ATP yield of the anaerobic process is too low (2 mol ATP/glucose) as compared to aerobic glucose metabolism (38 mol ATP/glucose). Most of the intermediates in the TCA cycle are used for biosynthesis of amino acids, lipids, polysaccharides, and nucleic acids. The major anabolic pathway is the pentose phosphate pathway or the hexose monophosphate (HMP) pathway, which converts glucose-6-phosphate into a variety of carbon skeletons (small organic compounds with 3–7 carbon atoms) and glyceraldehyde3-phosphate as the end product. The carbon skeletons are used for the synthesis of nucleic acids, amino acids, and other biomolecules, while glyceraldehyde-3-phosphate can be converted to pyruvate to yield further energy. Acetyl-CoA is the precursor for the synthesis of lipids and polysaccharides (Fig. 1.3). Nitrogen metabolism in the cell usually occurs by the hydrolysis of proteins to peptides and amino acids by the action of proteases. The amino acids are further converted to organic acids by the deamination process. Depending on the enzyme systems involved, deamination can be oxidative, reductive, or dehydrative. Apart from deamination, transamination is another mechanism for the conversion of amino acids to organic acids. In this process, the amino group is exchanged for the keto group of a-keto acid. Like carbohydrates and proteins, nucleic acids can also be used for the carbon, nitrogen, and energy sources. Nucleic acids can be hydrolyzed by nucleases into its precursor molecules, i.e., sugar, nitrogenous base, and phosphorus. Sugar can be used by glycolysis and TCA cycle under aerobic conditions, and phosphorus can be used for the synthesis of ATP, phospholipid, and nucleic acid synthesis.

13

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Introduction

The nitrogenous bases are further degraded into urea and acetic acid, ultimately forming ammonia and CO2. Most of the metabolic pathways discussed above are observed in heterotrophic organisms, which are dependent on organic molecules as a source of carbon and energy. The autotrophic organisms on the other hand have alternate metabolic pathways to assimilate inorganic carbon in the form of CO2 to obtain energy. These organisms rely on the Calvin–Benson cycle to fix carbon from CO2 to form cellular biomolecules. The energy is derived from either sunlight (photoautotrophs) or other inorganic chemicals (chemo-autotrophs). These organisms have a pivotal role in fixing the atmospheric CO2 and maintaining the carbon balances.

1.4 Biochemical Engineering and Bioprocesses

Biochemical engineering (also termed bioprocess engineering) is a branch of chemical engineering that mainly deals with the design and construction of unit processes that involve biological organisms or molecules, such as bioreactors, upstream processing, and downstream processing (Fig. 1.4). Biochemical process is an essential part of several food, chemical, and pharmaceutical industries. These processes make use of microbial, animal, and plant cells and components of cells such as enzymes to manufacture new products and also for waste treatment. Biochemical engineering may be considered a multidisciplinary field that implements engineering principles to design and operate unit processes required to successfully produce high-quality bioproducts. Chemical engineering principles play a central role in producing biochemical products on a large scale for marketing in purified form. Though the principles behind both the disciplines are similar, the characteristic features of chemical processes and biochemical processes vary distinctly (Table 1.6). Biochemical engineers play an important role in the bioprocess design and metabolic engineering by using fundamental biological research. For the industrial production of bioproducts, the major tasks for a biochemical engineer include the following:

Biochemical Engineering and Bioprocesses

∑ To obtain the best possible biological catalyst (microbial cell, enzymes, plant cell, or animal cell) to drive the biochemical reaction efficiently at large, ∑ To provide a suitable physicochemical environment for the biocatalyst by designing a bioreactor, ∑ To separate the desired product in the most economical manner. B I O

Upstream

∑ Strain Selection ∑ Media Optimization ∑ Feed stock collection ∑ Sterilization ∑ Enzymatic Process

Immobilized enzyme Suspended Cell

P R

Free enzyme

Bioreactor

∑ Microbial Process

O C E S

Downstream

S

Whole Cells

Immobilized whole Cell

Recombinant Cell ∑ Cell Disruption ∑ Solid/Liquid separation ∑ Concentration ∑ Purification ∑ Formulation ∑ Effuent treatment

Figure 1.4 General scheme of the biochemical processes.

Table 1.6

Differences between chemical and biochemical engineering

Chemical engineering

Biochemical engineering

Deals with synthetic or chemical processes

Deals mostly with biological organisms and their biochemical pathways

Reactants form products in the presence of a chemical catalyst

Reactants form products in the presence of a biological catalyst

Involves the design and operation of industrial manufacturing plants Chemical reactions occur at high temperature and pressure

Involves studying the behavior of living cells in the bioreactor Biochemical reactions occur at ambient temperature and pressure

Most of the industrially relevant products are obtained from microorganisms due to their fast growth rates and simple metabolic

15

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Introduction

machinery. As explained in Section 1.2, microorganisms that grow under different environmental conditions develop different metabolic strategies to acclimatize to the surrounding environments. These metabolic pathways can be exploited for the production of desired end products by altering the growth conditions. The intermediates of major metabolic pathways such as glycolysis, TCA cycle, and fermentation are of immense industrial importance. These intermediates or metabolites are basically of two types: primary metabolites, which are produced during active period of growth (e.g., amino acids, vitamins, ethanol, lactic acid, etc.), and secondary metabolites, which are produced in response to specific environmental conditions during the stationery phase (where cell growth equals cell death) of growth (e.g., antibiotics, steroids, pigments, etc.). Once the desired product is identified (primary or secondary metabolite), a number of considerations are to be made for the realization of commercially viable industrial bioprocess as described below.

1.4.1 Strain Selection and Improvement

A key step in the development phase of a biochemical process is the identification of the most potent strain that can generate high yields of the desired product. To achieve this purpose, suitable microbes are isolated from the natural habitats and are screened based on their metabolic capability, productivity, robustness, and tolerance to toxic metabolites and substrates. For example, if the targeted product is ethanol, the fermentative microorganisms such as yeast or bacteria are isolated and grown in a culture media consisting a high dose of ethanol to assess their tolerance for high concentration of ethanol. After the screening process, the selected species are subjected to strain improvement, which helps in manipulating and improving the microbial strains to enhance their metabolic capacities for the production of valuable products. Different kinds of genetic and metabolic engineering tools are applied to improve the performance of the wild species (i.e., the natural species) with respect to the type and quantity of the product. By introducing new metabolic pathways in suitable hosts, it is possible to redirect the carbon fluxes such that the productivity or yield of the desired product is improved.

Biochemical Engineering and Bioprocesses

The most commonly used strain improvement techniques include the following:

∑ Random mutation, in which mutagenic agents such as X-rays, UV, and other chemical mutagens are employed to obtain a mutant with desired traits. For example, the penicillin yields have been significantly improved by using random mutations to the strain Penicillium chrysogenum, which is now commercially used for penicillin production. ∑ Heterologous protein production, in which a heterologous gene is inserted into the production hosts to initiate a de novo protein synthesis pathway. It has been applied for the production of various pharmaceutical proteins (hormones, antibiotics, vaccines, etc.) and novel enzymes. ∑ Protoplast fusion, which makes it possible to combine the genetic material of two different species in a single cell. It has been successfully used in yeasts and fungi. ∑ Transgenic technique, in which genes are transferred from one organism into another and are transcribed in the recipient organisms. ∑ Pathway engineering, in which new metabolic pathways are introduced in host organisms to improve substrate utilization efficiency, elimination of byproducts, synthesis of new products, etc.

1.4.2 Bioprocess Design and Optimization

The criteria for the design and optimization of a biochemical process depend on the type of product and biocatalyst used. For instance, during industrial microbial bioproduction, controlled conditions of aeration, agitation, temperature, pH, etc. can aid in the enhancement of the desired end product. Each microbe has a maximum production rate at its optimal physicochemical environment. The biochemical reactions occur in a specially designed vessel, called a bioreactor. In laboratory conditions, the different bioreactors used for microbial growth include culture tubes, shake flasks, stirrer fermenters, or other mass culture systems. These lab-scale bioreactor systems can be used to optimize the suitable physicochemical parameters

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Introduction

for maximizing the yield or productivity of the desired bioproduct. Bioreactor design and analysis will be discussed in Chapters 3, 4, and 6. Once the design criteria are met, the next challenge is to reduce the overall production costs of the process. Depending on the market value, bioproducts can be broadly classified into three major categories: (i) low value, high volume products; (ii) medium value, medium volume products; and (iii) high value, low volume products. The bioproducts that cost less than £6/kg and are usually required in large quantities (millions of kilograms per year) are termed low value, high volume products. This category includes mostly whole-cell products (baker’s yeast) and primary metabolites (citric acid). The cost of raw materials plays a crucial role for the production of low value products. Usually lower-cost crude materials (such as whey, corncobs, corn-steep liquor, etc.) are preferable raw materials for the production of low value products. The bioproducts that cost less than £60/kg and are needed in quantities less than a million kilogram per year are termed medium value, medium volume products. Most of the antibiotics are included in this category. For these products, the raw material cost cannot be compromised due to the need for good-quality product; however, the duration of the fermentation process, i.e., the productivity of the process is the major criteria for the selection of raw material. Productivity ensures an efficient utilization of the production capacity, i.e., the bioreactor. High value, low volume products are those bioproducts that cost more than £60/mg and are needed in very less quantities (about 1 kg per year). Human insulin and interferons are a few examples of high value, low volume products. For these products, nutrient and utility cost are not that critical, and major emphasis is given on improving the stability and the level of expression of the industrial strain. Moreover, the process time-to-market and product quality are of greater importance. Apart from production costs, purification costs are also very important for a commercially viable bioprocess. It is estimated that the separation costs are about 90% of the total production costs. Thus, the selection of a suitable downstream process is of utmost importance. Chapter 11 deals with various bioseparation technologies that are implemented at commercial scales.

Applications of Biochemical Engineering

1.5 Applications of Biochemical Engineering The history of bioprocesses dates back to 6000 BC when yeasts were used to produce wine and beer. Since then, various commercial bioproducts have been successfully developed and used at the global scales. Figure 1.5 shows the chronology of the development of bioprocesses over the years. Currently, biochemical engineering principles have found its niche in various industries, including food processing, agriculture, oxychemicals, energy, health care, and environment. With further research developments in the applied biosciences, the horizon of biochemical industries is ever expanding. This section illustrates some of the recent technological advancements made in the bioprocess industries. Yeasts used to make wine and beer

Sewage treatment systems using microbes developed

The use of monoclonal antibodies Large First for scale successful diagnosis production genetic of engineering approved penicillin experiments in U.S.A

About Before 4000 BC About 1912-14 1944 6000 BC 1910 BC

Yeasts used for making leavened bread

Large scale production of acetone, butanol and glycerol using bacteria

1962

Mining of uranium with the aid of microbes (Canada)

1973

1980

Animal interferons produced by GEM’s for the protection of cattle againat diseases

Food & Drug Administ ration (FDA) approves the first GM food

FDA issues draft rules for biosimilar drugs.

1981 1983 1984 1990 1994 2003 2012

Marketing of human food of fungal origin (U.K.)

Approval of insulin produced by GEMs (U.S.A & U.K.)

First federally approved gene therapy treatment

The Human Genome Project is completed (U.S.A)

Figure 1.5 The chronology of bioprocess development.

1.5.1 Agricultural Bioprocesses The unprecedented increase in world population has triggered the demands for food and energy crops. Bioprocess engineering has helped in accelerating the productivity of agricultural crops by utilizing the direct gene transfer techniques. In recent times, recombinant DNA technologies have been tremendously used for

19

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Introduction

the improvement of cultivable varieties, raising disease resistant crops, and production of certain transgenic plants. Moreover, several steps have been taken for the development of biofertilizers and biopesticides to reduce the harmful effects of chemical fertilizers and pesticides.

1.5.2 Single-Cell Proteins

Microbial biomass is rich in nutrients and is exploited for human consumption. For example, single-cell protein (SCP) technology uses bacteria, algae, yeasts, or fungi to make protein fit for human consumption. The single-cell biomasses of Spirulina (cyanobacteria) and Chlorella (microalgae) are grown commercially in ponds to produce food materials for human consumption. Single-cell protein develops when microbes ferment waste materials such as wood, straw, alcoholic residues, etc. The major challenge for this process is the product dilution and cost.

1.5.3 Biopharmaceuticals

The first bioprocess application was the production of antibodies (penicillin). Since then various biopharmaceutical applications have emerged apart from antibody production, which include antigens, endorphins, growth hormones, serum albumin, interferons, interleukins, immune regulators, monoclonal antibodies, and vaccines.

1.5.4 Food and Dairy Applications

Microbes are most commonly used for industrial food and dairy products. Lactic acid bacteria such as Lactobaccilus sp. are used for the production of various types of cheese. Other bacteria such as Brevibacterium are used for the development of flavor and aroma. The yeast Saccharomyces cerevisiae is used for the commercial production of fermented juices and distilled liquors. Similarly, Aspergillus niger has been used for commercial citric acid production. Other food products and supplements produced microbially include vinegar (Acetobacter aceti), vitamins (Propionibacterium shermanii), glutamic acid (Corynebacterium glutamicum), etc.

Applications of Biochemical Engineering

1.5.5 Environmental Applications The utilization of undefined mixed microbial cultures in wastetreatment processes constitutes one of the largest scale uses of bioprocesses. The biological waste-treatment processes can be aerobic (such as activated sludge process, trickling filter, etc.) or anaerobic (such as anaerobic digestion). Apart from waste treatment, several microbial systems have been used for the biodegradation of xenobiotic compounds, mineral leaching, pollution control, and enhanced oil recovery.

1.5.6 Biofuels

The concept of biofuel emerged from the discovery of biodiesel, which was produced by the transesterification of vegetable oils. With the rising concerns over depleting fossil fuels, the focus on biofuel has gained appreciable interest. Research on biodiesel now is primarily based on the search for the non-edible substitute for vegetable oils. The third-generation feedstock microalgae is a potential candidate for high-scale biodiesel production. Apart from biodiesel, ethanol made from cane sugar by fermentation and distillation processes is another form of biofuel. In many countries such as USA and Brazil, bioethanol is used in blended form with gasoline. Another major gaseous biofuel is biogas or methane produced as a byproduct of anaerobic digestion. Biogas can be burned to provide power and heat. Other potential emerging biofuels include biobutanol, biomethanol, and biohydrogen.

1.5.7 Biosensors

A biosensor is a device that uses living organisms or biological molecules to detect the presence of chemicals. Depending on the type of biological element, different classes of biosensors are developed, which include enzymatic biosensors (such as glucose oxidase biosensor, which detects glucose), immobilized cell biosensors (for the detection of toxic chemicals, carcinogens, biological oxygen demand, heavy metals, etc.), and antibody-based sensors such as immunosensors and optical biosensors (which are used for the detection of specific antigen in various lateral flow assays).

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Introduction

1.6 Objectives of the Book This book aims to provide a general understanding of the basic concepts of bioprocess engineering principles such as microbiology, biochemistry, chemical/biochemical reaction kinetics, and bioreactor design analysis, which are essential for the industrial application of bioprocesses. It will help the readers to get acquainted with the different unit operations that are involved in a biochemical process and help in evaluating how the process is to be operated more efficiently. The major focus is on the central unit of the bioprocess, i.e., the bioreactor and the processes that occur inside it. This book is targeted toward a wide audience, mainly undergraduates, postgraduates, researchers, scientists in industries and organizations, and others who deal with the study of biochemical and bioprocess developments. Students involved in biotechnology, biochemistry, chemical engineering, biochemical engineering, and bioenergy courses may find this book as a ready reference.

1.7 Scope and Relevance

Biochemical engineering mostly deals with the most complicated life systems as compared to chemical engineering. Describing the behavior of any life system in terms of mathematical form is the basic essence of biochemical engineering. It is an interdisciplinary subject where biologists, physicists, chemists, technologists, and mathematicians all join hands to develop realistic models. This specialized subject comprises transport processes, sophisticated advanced control system besides conventional biology, physics, and mathematics. This book is designed to cater to all the above aspects of the subject. The readers will be able to apply the acquired knowledge in various professional fields such as process development, design, simulation environmental science, etc.

1.8 Major Chapters

This textbook is structured into 13 chapters, which mark the journey from the basic natural sciences to the more advanced applied sciences

References

and engineering approaches involved in the bioprocess design. Beginning from this chapter (Chapter 1), readers are elucidated with the basic concepts of biological sciences such as microbiology and biochemistry and are familiarized with the different bioprocesses and bioproducts, their classification and application. Chapter 2 deals with the stoichiometry of the bioprocesses, which is helpful mostly for the mass and energy analysis of the bioprocesses. In Chapter 3, the fundamental concepts of chemical reaction thermodynamics and kinetics are revisited, which will be useful for differentiating between the chemical and biological processes. Chapter 3 will also provide tools used for reactor analysis and different types of bioreactors that have been developed for carrying out biochemical reactions at large scales. Chapters 4 and 5 focus entirely on free enzymes and immobilized enzymes, including their characteristic features, reaction kinetics, applications, and limitations. Chapter 6 will be centered on the kinetics of growth, substrate utilization, and product formation of whole cells. Chapters 7 and 8 include the air and medium sterilization, which are prerequisites for the operation of most of the bioprocesses. In Chapter 9, the various transport phenomena involved in bioprocesses are discussed in details. Process control and monitoring systems play an important role in the operation of the bioreactor as well as upstream and downstream processes. Different process monitoring equipment are included in Chapter 10. In Chapter 11, the downstream processes involved for the purification of products are discussed in details. Chapter 12 will enlighten the readers on how an industrial fermenter is operated and will briefly describe the various material analysis of the process. Industrial effluents of both chemical and biochemical processes pose severe environmental pollution problems. So in the last chapter, Chapter 13, industrial effluent treatment processes are discussed in details. It is hoped that the proper understanding of the topics discussed in the chapters will help in providing the readers some useful tips for the design and development of new industrialscale bioprocesses.

References

1. Levenspiel, O. Chemical Reaction Engineering, Third Edition, WileyIndia, 2010.

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Introduction

2. Shuler, M. L. and Kargi, F. Bioprocess Engineering: Basic Concepts, Second Edition, Prentice-Hall Inc., New Delhi, India, 2002.

3. Blanch, H. W. and Clark, D. S. Biochemical Engineering, Marcel Dekker Inc., New York, USA, 1997. 4. Bailey, J. E. and Ollis, D. F. Biochemical Engineering Fundamentals, McGraw-Hill Inc., New Delhi, India, 2010.

5. Doran, P. M. Bioprocess Engineering Principles, Second Edition, Academic Press, Waltham, USA, 2012.

6. Madigan, M. T., Martinko, J. M., Bender, K. S., Buckley, D. H., and Stahl, D. A. Brock Biology of Microorganisms, Fourteenth Edition, Pearson, Boston, 2015. 7. Liu, S. Bioprocess Engineering: Kinetics, Biosystems, Sustainability, and Reactor Design, Elsevier, Amsterdam, The Netherlands, 2013. 8. Lehninger, A. L. Biochemistry, Worth Publishers, Gloucester, UK, 1970.

9. Pelczar, M. J., Chan, E. C. S., and Kreig, N. R. Microbiology, Fifth Edition, Tata McGraw-Hill, New Delhi, India, 1998.

10. Prescott, S. C. and Dunn, C. G. Industrial Microbiology, McGraw Hill Book Co. Inc., and K O Gakusha Co. Ltd., Tokyo, 1959. 11. Katoh, S. and Yoshida, F. Biochemical Engineering: A Textbook for Engineers, Chemists and Biologists, Wiley-VCH, 2010.

12. Najafpour, G. Biochemical Engineering and Biotechnology, Elsevier, 2006.

13. Villadsen, J., Nielsen, J., and Liden, G. Bioreaction Engineering Principles, Springer, 2003. 14. Prasad, K. K. and Prasad, N. K. Downstream Process Technology: A New Horizon in Biotechnology, Prentice Hall of India, 2010.

15. Mansi, E. M. T., Bryce, C. F. A., Dahhou, B., Sancher, S., Demain, A. L., and Allman, A. R. Fermentation Microbiology and Biotechnology, CRC Press, 2011. 16. Asenjo, J. A. and Andrews, B. A. (Eds.). Recombinant DNA Biotechnology III: An Integration of Biological and Engineering Sciences, Vol. 782, Annals of the New York Academy of Sciences, 1996.

Chapter 2

Stoichiometry of Bioprocesses

Stoichiometry of the biochemical processes gives the following information:

∑ The study of quantitative relationships between the amounts of reactants used and the amounts of products formed by a chemical reaction or mutual relationships and internal limitations within the biochemical system ∑ Validity of the experimental result ∑ Heat evolved in the aerobic fermentation processes

2.1 Law of Conservation of Mass

Stoichiometry is based on the law of conservation of mass in a system, which is represented in Fig. 2.1. Rate of substrate outflow

Rate of substrate inflow

System

Figure 2.1 Substrate inflow and outflow of a system.

Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

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Stoichiometry of Bioprocesses

The mass balance of the system can be represented as follows (Fig. 2.2). Mass in through – system boundaries

Mass out through + system boundaries

Mass generated – within system

Mass consumed within system

=

Mass accumulated within system

Figure 2.2 Mass balance of a system.

A material balance on biological reactions can be easily written when the compositions of substrates, products, and cellular material are known. Usually, electron–proton balances are required in addition to elemental balances to determine the stoichiometric coefficients in biochemical reactions. Accurate determination of the composition of cellular material is a major problem. One mole of biomass is defined as the amount containing 1 g atom of carbon, such as CHpOnNq. A typical cellular composition can be represented as CH1.8O0.5N0.2 with the error usually smaller than 5%. A microbial system may be represented as shown in Fig. 2.3. Reactants

Products

Microbial cell

Figure 2.3 Microbial fermentation process.

2.2 Mass Balance of Bioprocesses The aerobic fermentation process may be represented as follows:

Cells + Medium + O2 Æ More cells + Product + CO2 + H2O (2.1)

The medium comprises a carbon source like sugars, a nitrogen source like (NH4)2SO4, metal ions and vitamins as cofactors. The macroscopic mass balance of the microbial system concerning biomass production and another product can be written in its original form: CHmOl + aNH3 + bO2 = YcCHpOnNq + YpCHrOsNt + (1 − Yc − Yp) CO2 + cH2O

(Substrate)

(Biomass)

(Product)

(2.2)

Mass Balance of Bioprocesses

Compositions of the substrate, biomass, and of the product in the equation are expressed by elemental chemical analysis. The degree of reduction (g ) is defined as the number of equivalents of available electrons in that quantity of material containing 1 g atom carbon. Free electrons of the different components of Eq. 2.2 are shown in Eqs. 2.3–2.5. gb = 4 + p − 2n − 3q

Biomass:

gp = 4 + r − 2s − 3t

Product:

gs = 4 + m − 2l

Substrate:

(2.3)

(2.4)

(2.5)

where the number of free electrons is taken as 4 for one atom of carbon, 1 for the atom of hydrogen, 5 for the atom of phosphorus, −2 for the atom of oxygen, and −3 for the atom of nitrogen. There is no free electron in the end products such as H2O, CO2, and NH3, while oxygen in the O2 form accepts four electrons. Degree of reduction of H2O = 2 × (+1) + 1 × (−2) = 0 Degree of reduction per 1 g C atom glucose (CH2O): 1 × (+4) + 2 × (+1) − 1 × (−2) = 4 (C6H12O6 = 6 CH2O) The free electron balance of Eq. 2.2 can be written as follows:   gs + b(−4) = Yc gb + Yp gp (2.6) The electron balance for oxygen demand is shown in Eq. 2.6, where Number of available electrons in the substrate + Number of available electrons in O2 = Number of available electrons in the biomass + Number of available electrons in the product b=

(g s - Ycg b - Ypg p )

= oxygen demand (2.7) 4 The requirement for oxygen is related directly to electrons available for transfer to oxygen. Respiratory quotient (RQ) provides the information on the metabolic rate of the microorganism. It is expressed as follows: RQ =

Moles of CO2 Mole of O2

(2.8)

A high RQ value indicates that metabolism in the microorganism is taking place at high efficiency. Equation 2.6 is divided by gs to get a relationship that reflects the free electron content of different components with respect to the organic substrate.

27

28

Stoichiometry of Bioprocesses

4b Ycg b Ypg p + + =1 gs gs gs

(2.9)

The left-hand side of Eq. 2.9 consists of three fractions as mentioned below:

4b : The fraction of available electrons transferred from the gs substrate to oxygen Ycg b : The fraction of available electrons transferred from the gs substrate to the biomass Ypg p

: The fraction of available electrons transferred from the gs substrate to the product The second term in this equation is the fraction of free electrons transferred to biomass from the utilized substrate, which is known as energetic growth yield (h). h=

Ycg b gs

(Free electron present in 1 g C atom biomass) (g C atom biomass) (g C atom biomass) = ¥ (g C atom substrate) (Free electron present in 1 g C atom substrate) (g C atom substrate)

=

(Free electron present in 1 g C atom biomass) (Free electron present in n 1 g C atom substrate)

(2.10)

The third term designates the fraction of total substrate internal energy that is transferred to the product. It is called the energetic product yield (xp). xp =

Ypg p gs

(Free electron present in 1 g C atom product) (g C atom product) (g C atom product) = ¥ (g C atom substrate) (Free electron present in 1 g C atom substrate) (g C atom substrate)

Mass Balance of Bioprocesses

=

(Free electron present in 1 g C atom product) (Free electron present in n 1 g C atom substrate)

Equation 2.9 may be written as

(2.11)

4b + h + xP = 1 gs

b=

(1 - h - xp )g s

(2.12) 4 Another way of characterizing compounds participating in the microbial process is to use the weight fraction of carbon in organic matter defined by the following relationship: Biomass: sb =

Atomic weight of carbon 12 = Molecular weight of 1 g C atom biomass 12 2 + p + 16m + 14q

Product: sp =

(2.13)

Atomic weight of carbon 12 = Molecular weight of 1 g C atom product 12 2 + r + 16s + 14t

Substrate: ss =

(2.14)

Atomic weight of carbon 12 2 = Molecular weight of 1 g C atom substrate 12 + m + 16l (2.15)

The process efficiency can be expressed with regards to the actual values of macroscopic yield coefficients Yx/s and Yp/s determined by experimental laboratory measurements:

g of biomass g of substrate (g C atom biomass) (mol. wt. of 1 g C atom substrate) = (g C atom substratee) (mol. wt. of 1 g C atom substrate) 12 s s g (mol. wt. of 1 g C atom substrate) = Yc = Yc s = h s s (2.16) 12 sb s bg b (mol. wt. of 1 g C atom biomass)

Yx/s =

29

30

Stoichiometry of Bioprocesses

Similarly, the product yield coefficient (Yp/s) may be mathematically represented as in Eq. 2.17. Yp/s =

g p s sg s s g g of product = Yp = xp s s g of substrate g s s pg p s pg p

and oxygen consumption yield may be written as Yx/o =

(2.17)

g of biomass g of oxygen

g of biomass ¥ M.W. of substrate g C atom of substrate ¥ M.W. of substrate b(mol. wt. of O2 )

(1 g C atom of substrate) =

s sg s ¥ M.W. of substrate s bg b (1 - h - xP )g s 32 4

(1 g C atom of substrate)h =

=

(1 g C atom of substrate) hs s ¥ M.W. of substrate 8(1 - h - xp )s bg b 12 ¥ M.W. of substrate (1 g C atom of substrate) ¥ M.W. of substrate 8(1 - h - xp )s bg b

(1 g C atom of substrate)h = =

3h 2(1 - h - xp )s bg b

(2.18)

The process yield coefficient with regards to nitrogen can be derived from the species mass balance for nitrogen of Eq. 2.2, which is given below: a = Yc q + Yp t

2.3 Thermodynamic Efficiency

(2.19)

To find out the validity of the experimental data, it is important to consider the thermodynamic efficiency, which is the summation of the energetic growth yield (h) and energetic product yield (xp). Thermodynamic efficiency = h + xp

(2.20)

Thermodynamic Efficiency

Thermodynamic efficiency of aerobic processes varies between 0.5 and 0.6. It is approximately 0.7 for anaerobic processes. This fact can assist in the estimation of yield coefficients for new fermentation process or for pointing out inadequate or erroneous measurements of the process studied. Again, the total amount of heat evolved for the unit organic substrate containing a gram atom carbon can be expressed according to Minkevich Q = 4Q0b [kJ/g atom of C substrate]

(2.21)

where Q0 is approximately 133 kJ/equivalent of free electron transferred from the substrate to CO2. The invariant Q0 directly links the mass balance of the process with its energy balance. The attractiveness of macro stoichiometry and theory of mass and energy balance is that all growth coefficients are interrelated and could be measured from any available compounds of the culture mass balance. The microbial growth cannot be determined by conventional routine as dry weight of biomass (because of the presence of solids in broth liquid), it is possible to calculate on the basis of N or O2 uptake evolution, pH titration rate, and so on. PROBLEM 2.1 Klebsiella aerogenes is produced from glycerol in aerobic culture with ammonia as the nitrogen source. The biomass contains 8 %w/w ash. Around 0.40 g biomass is produced per gram of glycerol consumed, and no major metabolic products are formed. What is the oxygen requirement for this culture? SOLUTION MW of glycerol = 92 MW of biomass (Klebsiella aerogenes), CH1.73O0.43N0.24 = 23.97/0.92 = 26.1

(since biomass contains 8 %w/w ash) Degree of reduction for substrate, i.e., glycerol is gs = 4.7. Degree of reduction for biomass is gB = 4 ¥ 1 + 1 ¥ 1.73 – 2 ¥ 0.43 – 3 ¥ 0.24 = 4.15 Molecular weight of glycerol is C3H8O3 = 92. Molecular weight of 1 g C atom glycerol is 92/3. Yield of biomass is (YX/S) = 0.40 g/g

31

32

Stoichiometry of Bioprocesses

YX/S =

g biomass produced MW of 1 g C atom substrate ¥ g substrate consumed MW of 1 g C atom biomass

Ê g mole of biomass ˆ = Yc Á Ë g mole of substrate ˜¯ Now, Yc = (0.4 ¥ 92/3)/26.1 = 0.47 g atom biomass per gram atom substrate Oxygen requirement (b) per one carbon atom substrate is (g s - Ycg b - Ypg p ) = 0.687 mol 4 Amount of oxygen required is (0.687 ¥ 3)(32/92) = 0.72 g oxygen per gram substrate

PROBLEM 2.2 Candida utilis utilize ethanol as a carbon source in place of hexose and pentose under aerobic condition. Determine substrate consumption for biomass production and O2 consumption, fermentation heat produced and stoichiometry of the process. Stoichiometric equation:

CH3O0.5 + bO2 + cNH3 Æ dCH1.79O0.5N0.2 + fCO2 + gH2O (Ethanol)

SOLUTION Calculation of gs, gb, ss, and sb:

 

(Biomass)

gs = 4 + 3 – 2 ¥ 0.5 = 6

 

gb = 4 + 1.79 – 2 ¥ 0.5 – 3 ¥ 0.2 = 4.19 ss =

sb =

12 12 = = 0.522 12 + 3 + 16 ¥ 0.5 23

12 12 = = 0.488 12 + 1.79 + 16 ¥ 0.5 + 14 ¥ 0.2 24.59

For aerobic process, the value of thermodynamic coefficient is in the range of 0.5–0.6. Assuming 0.6 is the value of thermodynamic coefficient, Yx/s = h

h = 0.6

s sg s g of biomass (0.523)(6) = 0.6 = 0.92 s bg b g of substrate (0.488)(4.19)

Thermodynamic Efficiency

O2 consumption: Yx/o =

Yx/o =

(assuming ep = 0)

3(0.6) 1.8 g of biomass = = 1.1 2(0.488)4.19(1 - 0.6) 1.63 g of O2 use

Oxygen demand as O2:

b=

h=

b=

3h 2s bg b (1 - h - e p )

(g s - g cg b - Ypg p ) 4

Ycg b hg 0.6 ¥ 6 = 0.859 , Yc = s = gs gb 4.19 Yc = d = 0.859

g mole of O2 6 - 0.859 ¥ 4.19 = 0.6 (Ypg p = 0) 4 g C atom substrate

Heat evolved during the process is

Q = 4 ¥ Qo ¥ b = 4 ¥ 133 ¥ 0.6 = 319.2

= 319.2 ¥

kJ g C atom substrate

1 kJ = 13.8 23 g of substrate

(MW of CH3O0.5 = 23)

Calculation of stoichiometric coefficients C:

O: fi fi fi H:

1 = d + f fi f = 1 – d = 1 – 0.859 = 0.141 0.5 + 2b = 0.5d + 2f + g

0.5 + 2(0.6) = 0.5(0.859) + 2(0.141) + g 0.5 + 1.2 = 0.4295 + 0.282 + g

1.7 – 0.4295 – 0.282 = g = 0.988 3 + 3c = 1.79d + 2g

33

34

Stoichiometry of Bioprocesses

fi

3 + 3c = 1.79(0.859) + 2(0.98) c = 0.171

fi N:

c = 0.2d = 0.2 (0.859) = 0.171

Stoichiometric equation, CH3O0.5 + 0.6O2 + 0.171NH3 Æ 0.859CH1.79O0.5N0.2 + 0.141CO2 + 0.989H2O

PROBLEM 2.3 In a batch fermentation process, ethanol is produced from glucose under anaerobic condition and experimental data are given in the table below: Time (h)

Biomass (g/L)

Sugar (g/L)

Ethanol (g/L)

0

1.0

50

0.04

39

10

2

4

6

8

10 12

1.2

2.0

3.2

4.5

5.8 6.4

47

43

24 8 0

Check the data reported are correct or not. SOLUTION

C6H12O6 Æ CH2O

Sugar:

C2H6O Æ CH3O0.5

Ethanol:

Biomass:

   

CH1.79O0.5N0.2

Yx/s =

( x - xo ) (6.3 - 1.0) = = 0.106 ( so - s ) (50 - 0)

Yp/s =

( p - po ) (27 - 0.04) = = 0.539 ( so - s ) (50 - 0) gs = 4

gp = 4 + 3 + 0.5(–2) = 6

1

4

18 25 27

Thermodynamic Efficiency

 

gb = 4 + 1.79 – 2(0.5) – 3CO2 = 4.19

12 = 0.4 12 + 2 + 16 12 sb = = 0.488 12 + 1.79 + 16 ¥ 0.5 + 14 ¥ 0.2 ss =

12 = 0.522 12 + 3 + 16 ¥ 0.5

sp =

Yx/s = h fi

0.106 = h

fi

s sg s s bg b

0.4 ¥ 4 0.488 ¥ 4.19

h = 0.135

Yp/s = e p ep =

Yp/ss pg p s sg s

=

s sg s s pg p

0.539 ¥ 0.522 ¥ 6 = 1.055 4 ¥ 0.4

Thermodynamic coefficient is h + ep = 0.135 + 1.055 = 1.19 >> 0.7 For the validation of data, the thermodynamic coefficient for anaerobic process should be around 0.7. To validate experimental data in this case, the ep and Yp/s values should be low. For lowering these two parameters, substrate concentration should be high. Therefore, substrate concentration should be multiplied by some factor. Now, let us see at which multiplication factor, the value of h + ep becomes 0.7 on multiplication, there is an effect on the h value. Multiplication factor

YP/s

Yx/s

h

ep

h + ep

1.2

0.449

0.088

0.069

0.879

0.948

2.0

0.270

0.053

0.068

0.528

0.596

1.5 1.8 1.6 1.7

0.359 0.299 0.337 0.317

0.071 0.059 0.066 0.062

0.091 0.075 0.084 0.079

0.703 0.585 0.660 0.621

0.794 0.660 0.744 0.700

35

36

Stoichiometry of Bioprocesses

Therefore, it can be concluded that the substrate concentration data are not correct. It can be corrected by multiplying with 1.7 to get a thermodynamic coefficient value of 0.7.

PROBLEM 2.4 Elemental analysis of refuse shows that it contains 76% (w/w) of organic matter of the composition CH2.1O0.9N0.15. The remaining 24% is ash. Calculate the minimum methane production by anaerobic digestion per kilogram of dry refuse. Also determine the stoichiometric equation of the process. SOLUTION

CH2.1O0.9N0.15 + aH2OÆ bCH4 + cCO2 + dNH3 + eH+ (Substrate) ss =

(Product)

12 = 0.392 12 + 2.1 + 16 ¥ 0.9 + 14 ¥ 0.15 sp =

12 = 0.75 12 + 4

gs = 4 + 2.1 – 2 ¥ 0.9 – 3 ¥ 0.15 = 3.85 Calculation of Yp/s For anaerobic process,  

Yp/s = e p

gp = 4 + 4 = 8 ep = 0.7

s sg s g of CH4 (0.392)(3.85) = 0.7 s pg p (0.75)(8) g of organic refuse

Yp/s = 0.176

g of CH4 g of CH4 = 0.176 ¥ 0.76 g of organic refuse g of refuse

= 0.134

g of CH4 g of refuse

Total methane produced is 0.134 Yp/S = 0.176 = 0.32

g of CH4 g of refuse

g atom methane = b g atom organic refuse

Calculation of stoichiometric coefficients: b+c=1

Thermodynamic Efficiency

c=1–b

fi fi

c = 1 – 0.32 = 0.68

Nitrogen balance indicates d = 0.15. Oxygen balance indicates

a = 2c – 0.9 = 2 ¥ 0.68 – 0.9 = 0.46

Hydrogen balance indicates fi fi

2.1 + 2a = 4b + 3d + c

2.1 + 2 ¥ 0.46 = 4 ¥ 0.32 + 3 ¥ 0.15 + c c = 3.02 – 1.73 = 1.29

Stoichiometric equation is

CH2.1O0.9N0.15 + 0.46H2O Æ 0.32CH4 + 0.68CO2 + 0.15NH3 + 1.29H+

PROBLEM 2.5 The microorganism Mycobacterium vaccae can grow with ethane as the sole source of carbon and energy. It can grow (at 30°C) in a CSTR reactor (chemostat) with a continuous supply of gas (5 vol% ethane and 95 vol% air) and of a sterile aqueous medium that contains various minerals and N source, (NH4)2SO4. The limiting substrate is ethane. YX/S = 23.7 g cell mass (C mol ethane)−1. 1. Except the small amount of S and P, an analysis of dry cell mass is C: 47.40 wt%, N: 8.30 wt%, H: 7.43 wt%, and ash: 4.00 wt%. The remainder is taken to be oxygen, which cannot be detected in the analysis. Determine the stoichiometric formula for the (ash-free) biomass, CHaObNc, and the formula weight per C atom. Determine YX/S in the unit of C atom biomass/C atom ethane. 2. Calculate the oxygen consumption YO/X(mol O2/C mol biomass) when it is assumed that CO2 is the only product besides the biomass. Write the full stoichiometric equation. Determine the total heat evolved per kilogram of dry cell mass.

SOLUTION 1. The stoichiometric formula of the “ash-free” biomass can be calculated in different ways (on the basis of ash-free or nonash-free biomass). The result should, of course, be the same if the calculations are consistent. Here we show the calculation on an “ash-free basis” in the following table:

37

38

Stoichiometry of Bioprocesses

Analysis

Wt.% of the total

“Ash-free basis”

Atoms per C atom

C

47.40

49.375 = 4.112 C atom

1

N

8.30

8.645

0.150

H

7.43

“Ash”

7.739

4.00

O

Formula: YX/S =

32.87

—

34.24

CH1.88O0.52N0.15

1.881 —

0.52

Mbiomass = 24.30 g (C atom )-1

(0.96)(23.7) 24.3 (2)

= 0.4681 (C atom biomass)(C atom ethane)-1 2. gb = 4.39, gs = 7, and suppose the stoichiometric equation be CH3 + aNH3 + bO2 Æ YcCH1.88O0.52N0.15 + (1 – Yc)CO2 + cH2O where Yc = YX/S We know

b = YO/S

C atom biomass C atom ethane

(g s - g b ¥ Yc ) 4 mol O2 (7 - 4.39 ¥ 0.4681) = = 1.236 C atom ethane 4

Oxygen demand = b =

YO/X =

YO/S YX/S

=

1.236 = 2.64 mol O2 (C atom biomass)-1 0.4681

CH3 + aNH3 + 1.236O2 Æ 0.4681CH1.88O0.52N0.15 + (1 – 0.4681)CO2 + cH2O Balancing the equation, we have CH3 + 0.0702NH3 + 1.236O2 Æ 0.4681CH1.88O0.52N0.15 + 0.5319CO2 + 1.165H2O Heat evolved during the process is

Thermodynamic Efficiency

4Qo b

kJ = 4 ¥ 133 ¥ 1.236 C atom ethane = 657.552

kJ C atom ethane

=

657.552 kJ 0.4681 g C atom ash free biomass

=

657.552 kJ 0.4681 ¥ 24.3 g ash free biomass

= 57.80

kJ g C atom ash free biomass

= 0.96 ¥ 57.80 = 55.49

kJ g biomass

kJ g biomass

PROBLEM 2.6 In case of Baker’s yeast fermentation process, glucose is converted to yeast according to the following stoichiometry: C6H12O6 + 0.48NH3 + 3O2 Æ 0.48CH1.8O0.5O0.2 + 3.12CO2 + 4.32H2O

For the production of 40 g/L of yeast in 10 m3 reactor, determine the following: (a) Amount of glucose and ammonia required (b) Amount of O2 required (c) Determine YX/S and YO/X

SOLUTION Given stoichiometry equation C6H12O6 + 0.48 NH3 + 3 O2 → 0.48 CH1.8O0.5O0.2 + 3.12 CO2 + 4.32 H2O 180 17 32 24.6 (biomass) Yeast concentration = 40 g/L = 40 kg/m3 Volume of the reactor = 10 m3 (a) Total amount of yeast produced = 40 kg/m3 ¥ 10 m3 = 400 kg From the above equation, 0.48 x 24.6 g of biomass is produced from 180 g of glucose So, amount of glucose required for the production of yeast =

180 ¥ 400 = 6,097 kg 0.48 ¥ 24.6

39

40

Stoichiometry of Bioprocesses

Similarly, total amount of ammonia required = 17 ¥ 400 = 276 kg 24.6

(b) Total amount of oxygen required = 3 ¥ 32 ¥ 400 = 3252 kg 0.48 ¥24.6 (c) Yx/S 0.48 ¥ 24.6 = 0.0656 180 3 ¥ 32 = 8.13 YO/x = 0.48 ¥ 24.6

PROBLEM 2.7 Aerobic microorganisms follow the following stoichiometry during respiration using glucose: C6H12O6 + 6O2 Æ 6CO2 + 6H2O

Pseudomonas putida converts glucose to CO2 and H2O during growth. The cell composition is CH1.84O0.55N0.2 plus 7% ash. The yield of biomass from substrate is 0.5 g/g. Ammonium sulfate is used as the nitrogen source. (a) Find out the oxygen demand during the growth of the cell? (b) P. putida can also grow with phenol as substrate and producing cells of the same composition as mentioned above. Find out the stoichiometry of the process.

SOLUTION Given data organic cell mass composition = CH1.84O0.55N0.2 Molecular weight (M.W.) of the cell mass = 25.44

g total biomass g organic biomass = 0.465 g glucose g glucose Stoichiometry of the process may be written as Yx/S = 0.5

CH2O + bO2 + aNH3 → dCH1.84O0.55N0.2 + eCO2 + fH2O

From the above equation, the degree of reduction may be calculated as follows  

gS = 4; gb = 4.14; sS = 0.4; sb = 0.47

g S sS = η 0.822 = 0.465 g b sb η = 0.565 3h 3 ¥ 0.565 = Again, Yx/O = 2(1 - h - xp )s bg b 2 (1 - 0.565)0.47 ¥ 4.14 Again, Yx/S = η

Thermodynamic Efficiency

= 1.00

g of cell mass g of oxygen

So, 1 g oxygen is required for the production of 1 g cell mass.

(b) Stoichiometry of the process using phenol may be written as  

CHO0.17 + bO2 + aNH3 → dCH1.84O0.55N0.2 + eCO2 + f H2O

Degree of reduction of the process

gS = 4.66; gb = 4.14; sS = 0.76; sb = 0.47 gb Again, η = d = 0.565 gS g C atom biomass g C atom substrate e = (1– 0.635) = 0.365

So, d = 0.635

a = 0.635 ¥ 0.2 = 1.27 1 - h - x p) g s b= ( = 0.506 4 f = 1.82 The stoichiometry of the process is CHO0.17 + 0.506 O2 + 1.27 NH3 → 0.635 CH1.84O0.55N0.2 + 0.365 CO2 + 1.82 H2O

PROBLEM 2.8 An analysis of dry cell mass is C: 47.40 wt%, N: 8.30 wt%, H: 7.43 wt%, and ash: 4.00 wt%. The remainder is taken to be oxygen, which cannot be detected in the analysis. Find out the stoichiometric formula for the (ash-free) biomass in the form CHaObNc.

SOLUTION

Composition of 100 g total Elements biomass C

47.40

Elemental Composition composition Stoichiometric of 100 g of Mol. ash free Elemental in 1 C atom formula of biomass Wt. biomass composition biomass 49.375

4.114

1

N

8.30

8.646

0.617

0.15

O

32.87

34.239

2.134

0.52

H Ash

7.43

4.0

7.739 -

7.739

1.88

CH1.88O0.52N0.15 24.3

41

42

Stoichiometry of Bioprocesses

PROBLEM 2.9 The empirical bacterial biomass formula, CH1.7O0.46N0.18 (molecular weight = 23.6 g) is grown aerobically on glucose (C6H12O6) as substrate. The following data are obtained: YX/S = 0.47 g biomass/g substrate YX/O2 = 1.22 g biomass/g O2

This organism produces no significant amount of product under these conditions. (a) Prove that the measured values of YX/S and YX/O2 are stoichiometrically consistent with each other, indicating all assumptions. (b) 0.01 g of biomass and 20 mmol glucose used as an inoculum and substrate , respectively in a batch culture. The culture is incubated overnight, and subsequent optical density measurements suggest that the cells are no longer growing. The total biomass in the culture is calculated to be 1.0 g. Estimate the final amount of glucose in the medium (mmol), and speculate on what caused the plateau in biomass amount.

SOLUTION (a) The stoichiometry equation of the process may be written as CH2O + bO2 + aNH3 → dCH1.7O0.46N0.18 + eCO2 + f H2O From the above equation

 

gS = 4; gb = 4.24; sS = 0.4; sb = 0.509 g S sS Again, Yx/S = η = η ¥ 0.74 = 0.47 g b sb Again, Yx/O =

η = 0.63

3h 3h = = 1.22 2(1 - h - xp )s bg b 2(1 - h )0.509 ¥ 4.24

η = 0.63 So, the values of YX/S and YX/O2 are stoichiometrically consistent. (b) Given data X0 = 0.01 g, S0 = 20 mmol = 20 ¥ 10-3 ¥ 180 = 3.6 g, X = 1.0 g (S0 – S) =

(X - X 0 ) Yx/S

(1.0 - 0.01) 2.106 = (3.6 – S) 0.47

3 S = 1.494 g = 1.494 ¥ 10 = 8.3 mmol = Final glucose concentration 180 in the medium

References

The plateau in biomass amount may be due to the exhaustion of nitrogen content in the medium.

PROBLEM 2.10 Consider an anaerobic fermentation by yeast. The empirical formula of the biomass is CH1.7O0.45N0.15 (Molecular weight = 23.0 g). The carbon and nitrogen sources are glucose (C6H12O6) and ammonium salts, respectively. The possible products are biomass and ethanol (C2H6O), along with carbon dioxide and water. Find out the maximum possible yield coefficient of ethanol (g ethanol/g glucose), and under what condition is it realized? SOLUTION The stoichiometry equation of the process may be written as

CH2O + aNH3 → bCH1.7O0.45N0.15 + dCH3O0.5 + eCO2 + f H2O Substrate(s) Biomass (b) Ethanol (p) From the above equation

 

gS = 4; gb = 4.24; gp = 6; sS = 0.4; sb = 0.52, sp = 0.52 s sg s Again Yp/S = ξp s pg p Ethanol production will be maximum when we assume the biomass formation is negligible. In anaerobic process the thermodynamic efficiency is 0.7. g of ethanol So, Yp/S = 0.7 4 ¥ 0.4 = 0.36 g of glucose 6 ¥ 0.52 So, the maximum ethanol production will be 0.36 g per g of glucose. In the real condition, theoretical ethanol yield is 0.51 g per g glucose.

References

1. Minkevich, I. A. and Eroshine, V. K. Folio Microbiologica, 18, 376, 1973. 2. Volesky, B. and Votruba, J. Modeling and Optimization of Fermentation Processes, Elsevier, Netherlands, 1992.

3. Flickinger, M. C. and Drew, S. W. (Eds.) Encyclopedia of Bioprocess Technology: Fermentation, Catalysis and Bioseparation (Vol. 3), John Wiley & Sons Inc, Canada, 1999. 4. Nielsen, J. and Villadsen, J. Bioreaction Engineering Principles, Plenum Press, New York, 1994.

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Stoichiometry of Bioprocesses

5. Shuler, M. L. and Kargi, F. Bioprocess Engineering: Basic Concepts, Second Edition, Prentice-Hall Inc., New Delhi, India, 2002.

6. Blanch, H. W. and Clark, D. S. Biochemical Engineering, Marcel Dekker Inc., New York, USA, 1997. 7. Bailey, J. E. and Ollis, D. F. Biochemical Engineering Fundamentals, McGraw-Hill Inc., New Delhi, India, 2010. 8. Doran, P. M. Bioprocess Engineering Principles, Second Edition, Academic Press, Waltham, USA, 2012.

9. Katoh, S. and Yoshida, F. Biochemical Engineering: A Textbook for Engineers, Chemists and Biologists, Wiley-VCH, 2010.

10. Najafpour, G. Biochemical Engineering and Biotechnology, Elsevier, 2006.

11. Bhatt, B. I. and Thakore, S. B. Stoichiometry, Tata McGraw-Hill Education, 2010.

12. Ghasem, N. and Henda, R. Principles of Chemical Engineering Processes, CRC Press, 2012.

13. Felder, R. M., Rousseau, R. W., and Bullard, L. G. Felder’s Elementary Principles of Chemical Processes, Wiley, 2016.

Chapter 3

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

3.1 Chemical Reaction Thermodynamics 3.1.1 Scope and Introduction Thermodynamics is a field in science and technology for a proper understanding of heat and temperature flow in any system and their relations with energy and work. In other words, the field of thermodynamics relates the flow of heat caused due to the temperature difference and the energy contained in the system. Also, thermodynamics gives us an idea as to how much heat can be converted to work. Every chemical engineering operation is bounded with the use of thermodynamics. Generally, a chemical process system consists of chemical reactors, flow devices, various heat transfer and separation equipment, etc. All these equipment can be primarily designed and analyzed only with the basic thermodynamics principle. However, thermodynamics may not be useful in evaluating the rate of physical or chemical processes. We already know that the rate of any process is directly dependent on the resistance (inversely proportional) and driving force (directly proportional). For example, the rate of conduction heat transfer in a rectangular slab of length L Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

and thermal conductivity k is dependent on the temperature gradient (driving force) and the ratio of length to thermal conductivity (resistance). Although thermodynamics is effective in explaining the driving force, elaboration of the resistance to energy or material flow is out of scope of this chapter. In other words, for any system, thermodynamics cannot predict the rate at which the system is approaching equilibrium. Despite its limitations, thermodynamics remains a vital tool in estimating the maximum work that any system can yield from a particular heat input, the maximum efficiency of various mechanical systems (such as refrigerators, turbines, etc.). In addition, for a particular chemical reaction, it is instrumental in evaluating the maximum yield. Overall, there are three important categories of thermodynamics: (1) classical thermodynamics, (2) statistical thermodynamics, and (3) irreversible or non-equilibrium thermodynamics. Classical thermodynamics involves analysis with macroscopic properties of matter such as pressure, temperature and presents its properties on the basis of a given mass of material. Statistical thermodynamics covers mostly the microscopic or molecular level, and it involves the calculation of thermodynamic properties based on the observed energy levels, statistical distribution of these levels and spectroscopic data. Irreversible or non-equilibrium thermodynamics is a comparatively new field, which attempts to directly treat systems that are not at equilibrium, such as heat, electric, or mass flow systems. The current book is restricted to the discussion of only classical thermodynamics. Before going into much detail, one should understand the important terminologies in the field of thermodynamics. The entire thermodynamics is explained based on either system or surroundings. System refers to a substance or group of substances of special interest, whereas the part of the universe outside and separated by the system is called surroundings. Let us consider the example of water heating on a pan. The water-filled pan is considered to be the system, whereas the heating aid and the air outside the system may be considered the surroundings. A system may be classified as (1) open, (2) closed, and (3) isolated. Any system that possibly involves exchange or transfer of energy and mass with the surroundings may be referred to as an open system.

Chemical Reaction Thermodynamics

In contrast, the system that can exchange only energy across the boundaries may be called a closed system. On the contrary, when a system is unable to transfer or exchange both energy and mass with the surroundings, it may be referred to as an isolated system. Hence, an isolated system is totally unaffected by the changes brought in the environment. A system may also be classified based on the phase. A system consisting of two or more distinct phases may be defined as a heterogeneous system. A typical example of a heterogeneous system is the simultaneous existence of water vapor and water in a closed container. Interestingly, an immiscible liquid mixture of water and benzene can also be referred to as a heterogeneous system. On the other hand, a system is said to be homogenous if the properties are same throughout. Water in a beaker and an air-filled glass column are typical examples of a homogenous system. Hence, in order to understand the importance of thermodynamics for a specific system, homogeneity or heterogeneity plays a pivotal role. In the field of thermodynamics, parameters such as pressure, volume, and temperature are extremely important in order to specify the system conditions. These variables or parameters depict the state of a particular system, and hence they are known as the state functions. Hence, pressure, volume, and temperature may be referred to as state functions. On the other hand, many properties such as work and heat are totally independent of the final and initial states of the system. However, these properties are dependent on the path via which the transformation or change has occurred. Such properties may be termed as path functions. Also it is important to mention that based on the quantity or the amount of matter in a particular system, properties are mainly of two types: (1) extensive and (2) intensive properties. The properties that are function of mass or the quantity of a particular system are known as extensive properties. Typical examples of extensive property involve volume and mass. On the other hand, if the property of a particular system is independent of the quantity of matter or the mass, then it may be referred to as an intensive property. Density, heat capacity, viscosity, thermal conductivity, and specific volume are typical examples of intensive properties.

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

3.1.2 Reversible and Irreversible Processes In order to understand the concept of reversibility or irreversibility, one must know the proper definition of a process. A process may be defined as a phenomenon in which the system in consideration has undergone a change or a transformation to the final state from its initial state. It may be noted that a thermodynamic process is accompanied with changes in the system properties such as volume, temperature, pressure, entropy, enthalpy, etc. The thermodynamic process is said to be reversible if the major thermodynamic entities of the universe (system + surroundings) remain unaffected or restored while the process undergoes a change from the final state to the initial state. As the gradient is extremely small or infinitesimal in a reversible process, it corresponds to a slow and gradual change. Also the accompanying changes occurring within the system are in equilibrium during the course of the reversible process. One must remember the above two conditions for a process undergoing reversible change.

3.1.3 Work and Heat

Work may be defined as the process that occurs due to the application of force over a particular distance. Let force F be applied over a small elemental length dl. Then the work done dW will be given as dW = Fdl

(3.1)

Now force F can be defined as the magnitude of pressure P and area A. Hence, Eq. 3.1 may be modified as dW = PAdl

(3.2)

This equation must be integrated if the work for a finite process is required. Most of the processes in the field of chemical thermodynamics involve volumetric change in the system. Consider the expansion or compression of a fluid in a cylinder caused by the movement of a piston. The displacement of piston may be defined as the ratio of the change in the volume of a fluid by the area of the piston. Hence, Eq. 3.2 may be further modified as ÊV ˆ dW = PAd Á ˜ Ë A¯

(3.3)

Chemical Reaction Thermodynamics

As the area is constant, Eq. 3.3 may be rewritten as dW = PdV

(3.4)

An extremely well-known fact of the physical world is the transfer of energy to a cold body from a hot body while the two bodies are in close contact with each other. This transfer of energy may be referred to as the heat transfer between the two objects. It must be remembered that heat always flows from a body of greater temperature to a body of relatively lesser temperature. The reverse situation is impossible in the physical world. Note that the transport process of heat transfer can only occur when there is a significant gradient or driving force for temperature. In other words, the difference in temperature between two bodies is directly proportional to the rate at which heat is transferred from a body to another one. If both the bodies are at similar temperature, the gradient in temperature corresponds to zero and that further leads to no heat transfer. One must remember that for a body gaining heat, the energy is stored in the molecules and atoms in the form of potential and kinetic energy. Not surprisingly, the energy theory of heat did not prevail until the atomic theory of matter was well established.

3.1.4 Concept of Internal Energy

Till now, we know that potential energy and kinetic energy are the two major forms of energy that depend on the position and motion of the substance, respectively. These two major forms of energy can be categorized under external energy. In addition to these, the substance also inherits energy within its molecules, which undergoes almost ceaseless motion and a minute amount of internal vibration. This energy possessed by the molecules of the system is known as internal energy. An external input of heat energy (by temperature variation) can result in the enhancement of the molecular interaction or activity and that further increases the internal energy of a particular system. In a similar manner, an external work done on the system (e.g., agitator, impeller, etc.) can also result in the enhancement of internal energy due to the increase in the molecular interaction. It is important to mention that the absolute internal energy value of the system at a given state is not known. However, from a thermodynamics perspective, the change

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

or difference of internal energy is only necessary and important for analysis.

3.1.5 First Law of Thermodynamics

In the recent past, a significant number of theories have suggested that a generalized law of conservation of energy is applicable to internal and heat energies. In addition, the energy conservation law is also applicable to the work done, potential and kinetic energies. Truly speaking, there are various other forms of energy—magnetic, surface, electrical energies—to which the energy conservation applies. In other words, the law of conservation of energy has assumed the stature of the first law of thermodynamics, which is as follows: Although energy assumes many forms, the total quantity of energy is constant, and when energy disappears in one form, it appears simultaneously in other forms.

The above law is applicable to the entire universe, which encompasses both the surroundings and the system. Note that the change in the energy of the system will be balanced with the simultaneous change in the energy of the surroundings. In other words, one can write DEsystem = -DEsurroundings

which can be rewritten as

DEsystem + DEsurroundings = 0

(3.5)

For a better understanding of the first law of thermodynamics, let us assume a random system that gains a particular amount of thermal energy q. The energy gained by the system may be through any mode of heat transfer process. Due to the gain in thermal energy, the system will also do a particular amount of work, say w. Hence, it may be imagined that the remaining or excess energy can be calculated by subtracting w from q. This remaining energy can be referred to as the energy pertained by the system or internal energy. Hence, one can write Du = q - w

(3.6)

Chemical Reaction Thermodynamics

The following points are important to understand and summarize the first law of thermodynamics:

∑ If heat is absorbed, then the value of q is positive and vice versa. ∑ If work is done by the system, then w assumes a negative value. On the other hand, if work is done on the system, then a positive sign will be assigned to w.

Hence, for a system absorbing heat and for the case involving work done by the system, Eq. 3.5 is applicable. In addition, Eq. 3.5 is applicable to the systems undergoing finite changes. The differential changes in Eq. 3.6 may be written as dU = dQ - dW

(3.7)

It is important to mention that the left-hand side of Eq. 3.7 can be further extended for the moving systems either undergoing motion or falling from some height. In such cases, the change in kinetic energy (KE) and potential energy (PE) should be added to the change in internal energy on the left-hand side of Eq. 3.7. Hence, Eq. 3.7 can be modified as dU + d(KE ) + d( PE ) = dQ - dW

(3.8)

However, in most of the cases, the potential and kinetic energies are found to be negligible; hence, Eq. 3.8 is reduced to Eq. 3.7. Note that Eq. 3.7 is useful when U, Q, and W are expressed as functions of process variables.

3.1.5.1 Enthalpy

In addition to internal energy, a number of other thermodynamic functions are commonly used for their practical importance. Enthalpy (H) is explicitly defined for any system by the mathematical expression H = U + PV where U denotes the internal energy, P represents the absolute pressure, and V denotes the volume. It is evident that the units of U and PV should be the same as those of H. If we consider the SI system, the internal energy U will have the units of joules. Hence, the units of pressure and volume should be ideally N/m2 and m3, respectively. Then the term PV will have the units of Nm or joules. In order to have homogeneity, the unit of H should also be joules in SI unit. It must be remembered that P, V, and H are state functions and not path functions. Therefore, the enthalpy H should

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

also effectively be a state function. The differential form of H = U + PV can be written as dH = dU + d( PV )

dH = dU + PdV + VdP

(3.9)

(3.10)

For a constant-pressure or isobaric process, Eq. 3.10 reduces to dH = dU + PdV

(3.11)

For a finite change occurring in the system, Eq. 3.9 can be modified as DH = DU + D( PV )

(3.12)

Like volume and internal energy, enthalpy can be categorized as an extensive property. However, as the molar or specific enthalpy is dependent on mass, it is an intensive property.

3.1.5.2 Constant-volume and constant-pressure process Based on the first law of thermodynamics, one can write dU = dQ - dW

Also the work done can be written as

dW = PdV

Hence, Eq. 3.13 can be modified as

dU = dQ - PdV

For a constant-volume process, Eq. 3.15 may be modified as dU = dQ

Putting Eq. 3.15 in Eq. 3.11, we get

dH = dQ - PdV + PdV

or

dH = dQ If we integrate Eq. 3.18, we get

(3.13) (3.14) (3.15) (3.16) (3.17) (3.18)

(3.19) Q = DH Equation 3.19 signifies that the quantity of heat transferred is equal to the enthalpy change of a process for a constant-pressure and mechanically reversible process. It is important to now define the two major forms of reactions that largely depend on the positive or negative value of the heat transferred, Q. A reaction that involves

Chemical Reaction Thermodynamics

heat liberation from the system is known as exothermic reaction. Heat liberation is associated with the negative value of DH, and hence for an exothermic process, one can write Q < 0. Combustion reactions are typically exothermic in nature. On the other hand, if heat is absorbed by a system during a reaction, then such a reaction is called an endothermic process. Clearly, an endothermic process involves a positive value of DH and that corresponds to Q > 0. When sodium hydroxide is added to water, the system undergoes an absorption of heat, and hence the nature of reaction is endothermic. Enthalpy changes are so far discussed essentially for a physical process. It must be remembered that chemical reactions are also accompanied by the transfer of heat, by temperature changes during the course of reaction, or by both. The amount of heat required to carry out a specific chemical reaction is largely dependent on the products and reactants. The difference in the enthalpy of the products and reactants equals to the enthalpy change, which can be given as DH = Hproducts - Hreactants

3.1.5.3 Heat capacity The heat capacity of a body (C) may be defined as C=

dQ dT

(3.20)

(3.21)

Equation 3.21 physically signifies that the quantity of heat capacity is greater if the change in temperature is relatively lesser in magnitude. In addition, Eq. 3.21 also clarifies that an increase in the difference of the amount of heat results in the enhancement of the heat capacity. Hence, similar to Q, it can be clearly verified that C is basically a path function. In practice, essentially two different types of heat capacities are defined. One heat capacity is defined at a constant volume: Ê ∂U ˆ Cv = Á ˜ Ë ∂T ¯ v

(3.22)

Ê ∂H ˆ Cp = Á ˜ Ë ∂T ¯ p

(3.23)

The other heat capacity is defined with respect to constant pressure:

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

Equation 3.22 can be rewritten for a constant-volume process as dU = C v dT

The integration of Eq. 3.24 yields

T2

Ú

DU = C v dT T1

We can also write Eq. 3.25 as

(3.24) (3.25)

T2

Ú

Q = DU = C v dT T1

In a similar manner, Eq. 3.23 can be written as dH = C pdT

(3.26) (3.27)

Integrating Eq. 3.27 for a constant-pressure process, one can get T2

Ú

Q = DH = C pdT T1

For an ideal gas, Eq. 3.9 can be written as dH = dU + d(RT )

dH = dU + RdT + TdR

(3.28) (3.29) (3.30)

As R is the universal gas constant, the differential form of R will be zero. Hence, Eq. 3.30 will be reduced to dH = dU + RdT

Putting Eqs. 3.24 and 3.27 in Eq. 3.31, we get C pdT = C v dT + RdT Cp = C v + R

3.1.5.4 Isothermal and adiabatic process

(3.31) (3.32) (3.33)

An isothermal process refers to a system involving a constant temperature profile. During the isothermal process, the internal energy is not supposed to change. Hence, the first law of thermodynamics as given by Eq. 3.13, can be re-written as 0 = dQ - dW dQ = dW

(3.34)

Chemical Reaction Thermodynamics

Integrating Eq. 3.34, one can get

Q =W

(3.35)

For a mechanically reversible non-flow process, one can write V2

Ú

W = PdV

(3.36)

V1

If we consider ideal gas, then we can replace P = RT/V. Hence Eq. 3.36 becomes: V2

Ú

V2

W = PdV = V1

RT

ÚV

dV = RT ln

V1

V2 V1

(3.37)

We can also write P1V1 = P2V2 and hence Eq. 3.37 can be modified as W = RT ln

P1 P2

(3.38)

An adiabatic process refers to a system that undergoes nil exchange of energy with the outer surroundings, and hence dQ corresponds to 0 for such systems. Therefore, Eq. 3.13 can be written as dU = 0 - dW

C v dT = - PdV

(3.39) (3.40)

Equation 3.40 can be further modified for the ideal gas situation by replacing P = RT/V and then it can be integrated. However, it is up to the reader to take that exercise and do the entire solution.

3.1.5.5 Entropy

Entropy is a property of the thermodynamic systems. Entropy measures the randomness or disorder of a system. A higher entropy denotes a highly disordered system, whereas a well-organized system is represented by a lower entropy system. It must be remembered that the quantity entropy cannot be negative. In order to define entropy, let us first discuss the Clausius inequality. It states that a reversible process can only occur within an enclosed homogenous system. It can be mathematically expressed as

Ú

dQ =0 T

(3.41)

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

Equation 3.41 signifies that the line integral of the infinitesimal change in heat transfer over temperature is equal to zero. In other dQ words, the quantity does not depend on the path or it may be T referred to as a path-independent function. Hence, a state function S (entropy) can be defined such that

Ú

d Q dQ = (3.42) T T Equation 3.42 can be further modified for an isothermal process: dS =

qrev (3.43) T In Eq. 3.43, qrev is the heat transferred for a reversible process at a constant temperature, T (K). For an ideal reversible system, one can write DS =

DS univ = DS system + DS surroundings = 0

(3.44)

In Eq. 3.44, DSuniv denotes the change in the entropy of the universe and that should be ideally equal to zero. Now if we consider that the two major parts of the universe are the system and the surroundings, then DSuniv can be written as the summation of the change in the entropy of the system (DSsystem) and the surroundings (DSsurroundings). Note that, for an irreversible or a real and spontaneous system, one can express that DS univ = DS system + DS surroundings > 0

(3.45)

Hence, Eq. 3.45 clearly demonstrates that the entropy change is an ever-increasing process and is always positive. The spontaneity of a reaction can be easily expressed based on the signs of DS. If a reaction is exothermic with positive DS, then the reaction is said to be spontaneous. However, for an endothermic reaction with negative DS, the reaction is mainly non-spontaneous. Let us analyze the various cases of spontaneity for an exothermic reaction (DH < 0). If both the components DSsystem and DSsurroundings are positive, the entropy change is positive for the universe (system + surroundings) based on Eq. 3.45, and hence the reaction will be spontaneous. However, if DSsystem is negative, then DSsurroundings should be greater than DSsystem in order to make DSuniv as positive and the reaction spontaneous. Otherwise, the reaction will be nonspontaneous.

Chemical Reaction Thermodynamics

Next we will analyze the cases related to the spontaneity of the endothermic reactions. Generally in an endothermic reaction, DSsurroundings should always be lesser than zero. If DSsystem > 0, then DSsurroundings must be smaller than DSsystem in order to make the reaction spontaneous or DSuniv as positive. Truly speaking, Eq. 3.45 is an alternative expression of the second law. It physically signifies that the universe is tending toward increasing randomness or disorder. It must be remembered that entropy is an ever-increasing property for the universe and also acts as the major driving force for a spontaneous process.

3.1.5.6 Gibbs free energy

Another interesting parameter in the field of thermodynamics is Gibbs free energy, depicted by G. It is defined as a thermodynamic potential that can be used to calculate the maximum reversible work at constant temperature and pressure. It must be remembered that the absolute value of G is difficult to measure. However, it is relatively easier to calculate DG or the change in Gibbs free energy. Mathematically speaking, Gibbs free energy can be expressed as DG = DH - T DS

For standard conditions, Eq. 3.46 can be written as DG∞ = DH ∞ - T DS ∞

(3.46) (3.47)

The spontaneity of the reaction can also be expressed based on the Gibbs free energy. Let us suppose the following reversible chemical reaction: A´ B

(3.48)

If DG is negative, then the forward reaction (A ´ B) is said to be spontaneous, whereas the reaction is spontaneous in the reverse direction (B ´ A) if DG is positive. Note that the system is said to be at equilibrium when DG = 0. Now let us analyze the conditions for which DG should be spontaneous or non-spontaneous. As discussed previously, DH is negative for an exothermic process. In order to make the reaction spontaneous, DS should be a positive quantity. Similarly for an endothermic reaction, DH is positive. Hence, the reaction will be spontaneous if DS is positive and much greater than DH.

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PROBLEM 3.1: If the value of DH is 158 kJ and DS is 411 J/K, then estimate the temperature at which the reaction will be spontaneous? SOLUTION Based on Eq. 3.46, we can write

DG = DH - T DS = 158, 000 - 411T < 0

Based on Eq. 3.49, we can write

T > 15, 800 / 411

(3.49)

T > 384 K

or

PROBLEM 3.2: Estimate the change in enthalpy for heating 100 kg of water from 27°C to steam at 100°C. It is given that the specific heat of water is 4180 J/°C kg and the latent heat of vaporization is 23 ¥ 105 J/kg.

SOLUTION We can modify Eq. 3.42 as dS = DS1 =

mC pdT T

T2

= mC p

dT

ÚT

= 100 ¥ 4180 ¥ ln

T1

373 = 91082.2 J/∞C 300 (3.50)

DS1 is the change in entropy for heating 100 kg of water from 27°C to 100°C. Similarly, the entropy change for vaporization, DS2, may be calculated as DS2 =

dq 100 ¥ 23 ¥ 105 = = 6.16 ¥ 105 J/∞C T 373

(3.51)

PROBLEM 3.3: Let us suppose the following set of chemical reactions: A + B Æ C DG = -20kJ

B +C Æ D

DG = -28 kJ

Calculate DG for the reaction A + 2B + C Æ D + C. SOLUTION In order to get the following reaction,

A + 2B + C Æ D + C

(3.52)

Chemical Reaction Thermodynamics

we need to add the reactions given in the question, which gives DG = -20 - 28 kJ = -48 kJ

3.1.5.7 Enthalpy of formation (Dhf)

The enthalpy change when 1 mol of a pure compound is formed from its elements in their most stable states is called the enthalpy of formation and is denoted by DHf. Let us consider the oxidation of carbon to produce carbon dioxide. C (solid) + O2 (gas) Æ CO2 (gas)

(3.53)

The heat of formation of carbon dioxide from carbon is almost DHf = −393.5 kJ. This clearly indicates that the reaction is exothermic in nature.

3.1.5.8 Enthalpy of combustion (Dhcomb)

The enthalpy of combustion is defined as the heat evolved or the enthalpy change accompanying the complete combustion of 1 mol of any combustible compound in the presence of oxygen at a given temperature and 1 bar pressure.   C2H5OH (liquid) + 3O2 (gas) Æ 2CO2 (gas) + 3H2O (liquid)

(3.54)

The heat or enthalpy of combustion (DHcomb) of C2H5OH is −1365.6 kJ/mol.

3.1.5.9 Hess’s law

We may have come across Hess’s law in our high schools. Hess’s law states that the summation of the standard enthalpy of each reaction equals to the standard enthalpy of the overall reaction. Let us consider an example in order to understand Hess’s law more in details. As we all know, the clean hydrogen gas can be generated by the hydrolysis of carbon and the reaction looks like C + 2H2O Æ CO2 + 2H2

(3.55)

The right-hand side of Eq. 3.55 has the compound CO2, which can be produced using the reaction given by Eq. 3.53. As already mentioned, the heat of formation of CO2 is −393.5 kJ. Now 2 mol of hydrogen gas can be burned in the presence of oxygen to produce water as given below: 2H2 + O2 Æ 2H2O

(3.56)

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The heat of formation of H2O is nearly −483.6 kJ. Now both Eqs. 3.53 and 3.56 can be combined to get the reaction given by Eq. 3.55. In other words, the difference between the heat evolved, −393.5 kJ, in the combustion of carbon and the heat evolved, −483.6 kJ, in the combustion of hydrogen will give the enthalpy of formation of hydrogen (DHf = 90.1 kJ). For solving further examples with Hess’s law, the readers are suggested to refer some basic books on thermodynamics.

3.1.5.10 Chemical equilibrium

When a reversible reaction is performed in a closed container, a state or a point occurs where the backward and forward reaction rates are identical. This state is said to be at chemical equilibrium as the reaction rates are similar. It can be alternatively defined as the state at which the concentrations of the reactants and products do not change with time. The equilibrium is dynamic in nature, i.e., the reaction does not stop at this stage but takes place in both the directions with equal speed. Equilibrium state is not influenced by the presence of a catalyst, and it only helps to attain the equilibrium state rapidly. The state of equilibrium is disturbed when the pressure and temperature of concentrations are changed. Note that the free energy change (DG) is zero at equilibrium. Let us assume that a reversible reaction aA + bB ´ cC + dD

Forward reaction rate is kf C aAC Bb

(3.57)

Backward reaction rate is kbCCc C Dd . At equilibrium,

K eq =

kf CCc C Dd = kb C aAC Bb

(3.58)

In Eqs. 3.57 and 3.58, the concentrations of the components A, B, C, and D are given by CA, CB, CC, and CD, respectively. For an endothermic reversible reaction, Keq increases with temperature, whereas for an exothermic reaction, Keq decreases with increasing temperature. The parameter Keq varies with temperature, which is further given by the van’t Hoff equation as follows:

Chemical Reaction Kinetics

log

K eq2 K eq1

=

DH Ê 1 1 ˆ Á 2.303R Ë T1 T2 ˜¯

(3.59)

For the reaction given by Eq. 3.57, the Gibbs free energy change can be written as DG = DG∞ + RT ln K eq

(3.60)

In Eq. 3.60, DG° is the standard Gibbs free energy. Note that at equilibrium, DG = 0, and hence Eq. 3.60 reduces to DG∞ = -RT ln K eq

(3.61)

One can deduce a very important significance of Eq. 3.61. For Keq > 1, DG° is negative, whereas DG° is positive for Keq < 1. PROBLEM 3.4: At 2000 K, DG° = 5.08 kJ/mol for a certain reaction. Evaluate the Keq for the same reaction. SOLUTION Using Eq. 3.61, one can write

DG∞ = - RT ln K eq

or or

5080 = -8.314 ¥ 2000 lnK eq Keq = 0.737

Therefore, the value of equilibrium constant is 0.737.

3.2 Chemical Reaction Kinetics

Most of the bioprocessing operations involve a large number of biochemical or chemical reactions occurring in various reactors. Hence, it is important to understand the changes in compositions of the products and reactants, as well as their rates of utilization and production under given conditions. This will ultimately lead in determining the size of a reactor. The mass balance across the reactors will also be important in designing and analyzing the reactors. This chapter is essential in understanding the fundamentals of reaction kinetics and reactor analysis. Later, this chapter will act as the stepping stone in understanding the biochemical reaction kinetics, which will involve the enzyme-catalyzed reactions associated with the growth of microorganisms.

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Also this chapter will be helpful in understanding the various mechanisms of biochemical reactions, and also for the modal operation of bioreactors.

3.2.1 Rate of Reaction

In order to define the rate of reaction, let us suppose that the following homogenous reaction occurs inside a closed system at uniform pressure, temperature, and composition: AÆ B

(3.62)

Equation 3.62 represents a stoichiometry where the product B is being formed from reactant A. For such a system, the rate at which reactant A disappears is equal to the rate at which the product B is formed. The reaction rate of A (rA) is defined as follows: rA = -

dC A dt

(3.63)

In Eq. 3.63, the “−” sign denotes the disappearance of reactant A. The right-hand side of Eq. 3.62 describes the depletion of the concentration of reactant A (CA) with time (t). In a similar manner, we can also define the rate of production of B (rB) as follows:

dC B (3.64) dt It may be noted that the “−” sign disappears from the right-hand side of Eq. 3.63 as the product B is being produced from reactant A. Now the term concentration is mainly written as the ratio of the number of moles per unit volume (V). Hence, Eqs. 3.63 and 3.64 may be rewritten as rB =

rA = -

and

rB =

d( N A /V ) dt

d( NB /V ) dt

(3.65)

(3.66)

In Eqs. 3.65 and 3.66, NA and NB represent the number of moles of A and B taking part in the reaction, respectively. Hence, the rate of reaction may be summed up as the rate of change in the number of moles of the components taking part in the reaction per unit volume.

Chemical Reaction Kinetics

It can also be defined as the rate of change in concentration of the species taking part in the chemical reaction. The variation in concentration for the participating species in a chemical reaction mostly results in the variation in the rate of the reaction. An equation describing the rate of the reaction is termed as the rate equation. In order to measure the reaction rate, several simple ways exist. Let us suppose that gas comes out from a set up after reaction. The measurement will be in terms of the volume of gas that comes out per unit of time. One must remember that the rate of 1 cm3/s is twice as lower in magnitude compared to the rate of 2 cm3/s. For the reaction described through Eq. 3.62, one can write -rA µ C nA

(3.67)

After removing the proportionality sign, the rate equation describing the depletion of reactant A may be written as follows:

dC A = k C nA (3.68) dt In Eq. 3.68, k depicts the rate constant of the reaction, while the concentration of component A is given by the term CA. Also the order of the reaction is demonstrated by the letter n. Hence, the reaction rate rA refers to the rate at which the component A gets depleted in a system. The unit of the rate equation is mol/m3s, and k has dimensions of concentration1−n/time. Although generally we assume k to be constant throughout the reaction, k is a function of temperature T in many cases. The rate of the reaction is dependent on the concentration of the reactants via a proportional sign, which implies that the increase in the concentration of the reactant will result in the enhancement of the reaction rate. The extent of the reaction rate also depends on the order n to which the concentration is raised to. Note that order n only refers to the order of the specific substance that is in the reaction. For example in Eq. 3.68, n refers to the order of reactant A. However, since only component A is participating in the reaction, the overall order of the reaction is also n. This will surely change if two or more components are participating in the reaction. One can evaluate the order of the reaction only by doing experiments. -rA = -

63

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

A common confusion that mostly occurs among the readers is that whether the sum of the stoichiometric coefficient is equal to the overall order of the reaction! The answer is no, not always. Only for a rate-limiting step, the stoichiometric coefficients and exponents are equal. The exponents and the stoichiometric coefficients are not identical for a reaction that involves a single step. It is important to mention that the order of the reaction is estimated mostly from the experiments. One such example is demonstrated through Problem 3.5. PROBLEM 3.5: For the reaction and data given below, evaluate the rate law and rate constant at 298 K: A + B Æ C + D Trial 1 2 3

CA (mol/m3) 1 1 2

CB (mol/m3) 1 2 2

rinitial (mol/m3s) 2 8.1

15.9

SOLUTION Overall, the problem clearly shows that the experiment has been carried out for three trials. In order to find the order of the reactions, we have to consider two trials at a time. Let us first take into account trials 1 and 2. ∑ Trial 1: r1 = kC A x C B y = k(1.00)x (1.00) y ∑ Trial 2: r2 = kC A x C B y = k (1.00)x (2.00) y Dividing the second equation by the first equation, we get 8.1 ª 4 = (2.00) y 2 22 = 2y

So, y=2 Similarly for trials 2 and 3, we can write the following: ∑ Trial 2: r2 = kC A x C B y = k (1.00)x (2.00) y ∑ Trial 3: r3 = kC A x C B y = k (2.00)x (1.00) y Similar to the previous case, dividing the second equation by the first gives us 2 = (2.00)x

Chemical Reaction Kinetics

x=1

Since we got the orders corresponding to the components A and B, we can now write the rate law as r = 2.0CACB2

The overall order of the reaction is 1 + 2 = 3. Once we got the overall order of the reaction, it is easy to calculate the rate constant k. One has to just replace the values in any of the rate equations corresponding to trials 1–3. 2.0 mol/m3s = k (1.00 mol/m3) (1.00 mol/m3)2 k = 2.0 mol-2/m6 s

At this moment, it is extremely important to introduce the term “percentage conversion.” Percentage conversion is defined as the extent to which a reactant is converted to products. Hence, based on Eq. 3.63, the percentage conversion of reactant A may be defined as the fraction of A converted to products, or the fraction of A reacted away. In the present book, the symbol Xi will be used to denote the percentage conversion of any reactant i. Hence, the percentage conversion of A will be denoted by the symbol XA. Let us suppose that the initial concentration of A is CA0 at time t = 0, and that CA is the amount present at time t. Then the conversion of A in the constant-volume system is given by XA =

C A0 - C A C =1- A C A0 C A0

Equation 3.69 can be rewritten as

C A = C A0 (1 - X A )

(3.69)

(3.70)

PROBLEM 3.6: In an irreversible reaction A Æ products, 50% of A is converted into products. If the initial concentration of A is 10 mol/m3, then find the final or residual concentration of A.

SOLUTION The final concentration of A can be calculated using the Eq. 3.70 as follows: CA = 10(1 – 0.5) = 5 mol/m3

Irreversible first-order reaction Now let us consider the reaction

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

A Æ Products

(3.71)

Assuming this is a first-order reaction, Eq. 3.68 can be written as -rA = -

dC A = kC A dt

Separating and integrating, we obtain CA

Ú

-

C A0

t

dC A = k . dt CA

- ln

(3.72)

Ú 0

CA = k .t C A0

C A = C A0e - kt

(3.73) (3.74) (3.75)

Hence, Eq. 3.75 clearly demonstrates that the concentration of A decays exponentially with time t. In terms of conversion (see Eq. 3.72), the rate equation (Eq. 3.72) becomes -

d C A0 (1 - X A ) = kC A0 (1 - X A ) dt dX A = k(1 - X A ) dt XA

Ú 0

t

dX A = k dt 1- XA

Ú 0

Integrating Eq. 3.78, we get

- ln(1 - X A ) = kt

(3.76)

(3.77)

(3.78) (3.79)

PROBLEM 3.7: For an irreversible first-order reaction, 50% of A is converted to product after 3 min. What is the rate constant k? SOLUTION Using Eq. 3.79, k is deduced as k=

- ln(1 - 0.5) = 0.231 min -1 3

3.2.2 Irreversible Second-Order Reaction Let us consider the following second-order reaction: A + B Æ Products

(3.80)

Chemical Reaction Kinetics

Equation 3.80 demonstrates that the two reactants A and B react together and form products. The rate equation for the reactants can be written as follows: -rA = -rB = -

dC A dC = - B = k C AC B dt dt

(3.81)

Since, rA and rB have similar expressions, we express the later part of the rate equation with respect to reactant A. Hence, Eq. 3.81 can be simplified as -

dC A = k C AC B dt

(3.82)

If CAoXA is the amount of A reacted after any time t, one can express the final concentration of A and B after time t as follows: C A = C A0 - C A0 X A

C B = C B 0 - C A0 X A

and

(3.83) (3.84)

In Eq. 3.84, CB0 refers to the initial concentration of B. Using Eqs. 3.83 and 3.23, Eq. 3.84 can be modified as follows: -

ÊC ˆ d C A0 (1 - X A ) = kC A0 (1 - X A )C A0 Á B 0 - X A ˜ dt C Ë A0 ¯

(3.85)

In Eq. 3.85, the ratio CB0/CA0 may be denoted by N. Differentiating and simplifying Eq. 3.85, we get dX A = kC A0 (1 - X A )( N - X A ) dt

XA

Ú 0

t

dX A = kC A0dt (1 - X A )( N - X A )

Ú 0

Integrating Eq. 3.87, we get ln

C C N - XA 1 - XB = ln = ln B A0 N(1 - X A ) C B 0C A 1- XA C = ln B = C A0 ( N - 1) kt = (C B 0 - C A0 ) kt NC A

(3.86) (3.87)

(3.88)

It is noteworthy to mention that if CB0 >> CA0, the constant values of CB can be found at each time, and Eq. 3.88 approaches Eq. 3.74.

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

Therefore, that effectively gives rise to a pseudo first-order reaction. It may also happen that a single reactant results into product formation and the reaction is of second order. A typical example for such a reaction is as follows: 2A Æ Products

(3.89)

Let us assume that the stoichiometric coefficient of the reactant and the order of the reaction is identical. We can write the rate equation as follows: -rA = -

dC A = kC 2A dt

Putting Eq. 3.83 in Eq. 3.90 and integrating Eq. 3.90, we get 1 1 = kt C A0 C A

(3.90) (3.91)

In a similar manner, we can also integrate Eq. 3.68 and generate an expression for the nth-order reaction: -rA = -

dC A = k C nA dt

C 1A- n - C 1A-0n = (n - 1)kt

(3.92) (3.93)

Due to the nonlinearity in Eq. 3.93, it is difficult to estimate n explicitly. Consequently, a trial-and-error approach has to be followed in order to evaluate n. For example, let us assume a random value of n corresponding to which we will evaluate k. Now the variation in k will be minimized for a particular value of n and that is the desired order of the reaction. It is important to mention that the rate does not go to completion in finite time for n > 1. In contrast, the concentration of the reactant will fall to zero and then it further reduces below zero (becomes negative) when the order of the reaction (n) is lower than 1 (n < 1), such that C A = 0 at t ≥

C 1A-0n (1 - n)k

(3.94)

For the values of n lower than 1, the concentration does not dip below zero, and hence the integration should not be carried out beyond the particular time t as given by Eq. 3.94.

Chemical Reaction Kinetics

It is interesting to note from Eq. 3.92 that for n = 0 (a zeroorder reaction), the rate equation becomes independent of the concentration of the reactant. Hence, Eq. 3.92 reduces to -rA = -

dC A =k dt

Integrating and simplifying Eq. 3.95, we get

C A0 - C A = C A0 X A = k t for t < C A = 0 for t ≥

C A0 k

(3.95)

C A0 k

(3.96) (3.97)

It has been found that the zero-order reactions occur only for a specific concentration domain or limit. However, many special cases clearly display that the larger concentrations result in the zeroorder reaction. In contrast, the order of the reaction deviates from the zero order and tends to rise to higher order if the concentration is minimized significantly. Hence, such cases correspond to a high dependency on the reactant concentration. Truly speaking, as the zero-order reaction is concentration independent, there are various other adjoining factors that determine the rate for a zero-order reaction. These factors mainly involve physical phenomena such as available surface area for a solid phase or solid-catalyzed reactions, radiation or light intensity needed to carry out a photochemical reaction, etc. Hence, it must be noted that these factors must be correctly incorporated in the rate equation in order to achieve the correct expression. At this moment, it is important to define the reaction half-life (t1/2). Reaction half-life may be defined as the situation where the reactant concentration is reduced to one-half or 50% of the original reactant concentration participating in the reaction. In other words, we can mathematically express such situations as follows: CA =

C A0 2

Putting Eq. 3.98 in Eq. 3.93, we get t 1/2 =

0.51- n - 1 1- n C A0 k(n - 1)

(3.98)

(3.99)

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3.2.3 Irreversible Reactions in Parallel Irreversible reactions in parallel are very common in biochemical reaction pathways where a single reactant gives two or more parallel products. Let us consider the following simple example in order to illustrate the phenomenon of parallel reaction. k

1 A ææ ÆR

k

2 A ææ ÆS

(3.100)

(3.101)

In Eqs. 3.100 and 3.101, reactant A gives two parallel products R and S, separately. Since the reactions are occurring separately, the reaction equation given by Eq. 3.100 has a rate constant k1 and Eq. 3.101 is governed by the rate constant k2. The rates equations for reactant A and products R and S can be expressed as follows: -rA = -

dC A = k1C A + k2C A dt

rR = rS =

dC R = k1C A dt

dC S = k2C A dt

(3.102) (3.103) (3.104)

The values of rate constants k1 and k2 are evaluated using the differential equations given by Eqs. 3.102 and 3.103. One can first integrate Eq. 3.102 and get the following expressions: - ln

CA = (k1 + k2 )t C A0

Now, dividing Eqs. 3.103 and 3.104, we get rR dC R k1 = = rS dC S k2

(3.105) (3.106)

Integrating Eq. 3.106 while taking the limits from the initial concentration of R (CR0) and S (CS0) to the final concentration of R (CR) and S (CS), we get C R - C R0 k1 = C S - C S 0 k2

(3.107)

The ratio kl/k2 can be achieved if one plots CR versus CS. The slope of such a plot will give the ratio kl/k2. Also if we plot CA and t, we will get

Chemical Reaction Kinetics

the slope as (k1 + k2) based on Eq. 3.105. Hence, by knowing k1/k2 and (k1 + k2), one can easily calculate the values of the rate constants k1 and k2. This is the right time to illustrate the phenomenon of the homogenized catalytic reactions and autocatalytic reactions. Let us consider the examples of a catalytic reaction (see Eq. 3.109) along with the normal reaction given by Eq. 3.108. This is as follows: k

1 A ææ ÆR

k

2 A + C ææ ÆR + C

(3.108) (3.109)

The rate expressions for Eqs. 3.108 and 3.109 can be written as follows: -rA = rR = -

dC A dC R = = k1C A + k2C ACC dt dt

(3.110)

It may be noted that CC is unaffected, and it does not undergo any change along the course of the reaction. In other words, the entire reaction involves the constant concentration of the catalyst C (CC). Keeping this in mind and integrating Eq. 3.109, we get - ln

CA = (k1 + k2CC )t C A0

(3.111)

Next we will consider the example of an autocataytic reaction. In an autocatalytic reaction, the product of the reaction acts as the catalyst. A+R ÆR+R

The rate equation is given as follows: -rA = -

dC A = kC AC R dt

(3.112) (3.113)

One must remember that the total concentration is fixed throughout the course of reaction. In other words, the total concentration equals the sum of the reactant and catalyst concentrations (initial as well as final): C0 = C A + C R = C A0 + C R0 = constant

Replacing the expression of Eq. 3.114 into Eq. 3.113, we get -rA = -

dC A = kC A (C0 - C A ) dt

(3.114)

(3.115)

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

On integrating Eq. 3.115, we get ln

C A0 (C0 - C A ) C /C = ln R R0 = (C A0 + C R0 )kt C A (C0 - C A0 ) C A /C A0

(3.116)

Microbial fermentation process is an example of autocatalytic reaction.

3.2.4 Irreversible Series Reaction

A series reaction is defined as one where the product formed from the reactants reacts and forms further products. These types of reactions are very common in enzymatic systems where the enzyme combines with a substrate to form an enzyme substrate complex, which further results in the formation of product. In order to explain this phenomenon, let us consider the following first-order reaction in series where reactant A forms an intermediate product R, which further reacts and forms the final product S. k

k

1 2 A ææ Æ R ææ ÆS

The rate equations for A, R, and S can be written as follows: -rA = -

rR =

dC A = k1C A dt

dC R = k1C A - k2C R dt

(3.117) (3.118) (3.119)

dC S = k2C R (3.120) dt Equation 3.118 is a first-order homogenous ordinary differential equation, and it can be easily solved analytically. Integrating Eq. 3.118, we get rS =

C A = C A0e - kt

(3.121)

Equation 3.60 clearly demonstrates the exponential decay of CA with time. Putting Eq. 3.121 in Eq. 3.119, we get a non-homogenous firstorder ordinary differential equation. Further integrating Eq. 3.119, we get Ê e - k1t e - k2t ˆ C R = C A0 k1 Á + ˜ Ë k2 - k1 k1 - k2 ¯

(3.122)

Chemical Reaction Kinetics

Based on the stoichiometry and assuming that there is no loss in the number of moles, one can simply write C A0 = C A + C R + C S

(3.123)

Putting Eqs. 3.121 and 3.122 in Eq. 3.123 and further integrating the modified differential equation, we get Ê k e - k1t k1e - k2t ˆ C S = C A0 Á 1 + 2 + ˜ k1 - k2 k2 - k1 ¯ Ë

(3.124)

Now there may be cases when the reaction rate involving the formation of S is higher than R. In other words, if k2 >> k1, Eq. 3.124 reduces to C S = C A0 (1 - e - k1t )

The rate of the reaction is determined by the first step. If k1 >> k2, then C S = C A0 (1 - e - k2t )

(3.125) (3.126)

Hence, the rate is determined by the second step of the chemical reaction and k2 governs the first-order reaction. In a series reaction, the product R forms a maxima and then it steeply falls down as it further reacts to give the final product S. This can be easily found out from Eq. 3.122 as the double differentiation will give a negative value. In order to locate the maximum concentration of R, Eq. 3.122 can be differentiated once with respect to time (t) and equated to 0 (dCR/dt = 0). The maximum concentration of R is observed at time tmax, which can be written as t max =

Êk ˆ ln Á 2 ˜ Ë k1 ¯ k2 - k1

(3.127)

Now Eqs. 3.122 and 3.127 can be combined together, which will further yield the maximum concentration of the intermediate product R. C R ,max C A0

Êk ˆ =Á 1˜ Ë k2 ¯

k2 ( k2 - k1 )

(3.128)

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

3.3 Reversible Reaction Many enzymatic reactions are reversible; hence, it is important to learn the kinetics of reversible reactions. For example, enzyme substrate complex formation by the combination of enzyme and substrate is a typical example of a reversible first-order reaction. A typical reversible reaction can be expressed as follows: k

1 A ¨æ ÆR k 2

(3.129)

Let the symbol N denote the ratio of CR0/CA0; the rate equation can be written as dC A dC R = = k1C A - k2C R dt dt = k1 (C A0 - C A0 X A ) - k2 ( NC A0 + C A0 X A )

-rA = rR = -

(3.130)

The value of dCA/dt corresponds to 0 at the equilibrium. One will get the following expression: KC =

C Re N + X Ae = C Ae 1 - X Ae

(3.131)

In Eq. 3.131, KC may be defined as the ratio of k1/k2. Also, XAe is the percentage conversion at equilibrium. Putting these into Eq. 3.130 and integrating, we get Ê X ˆ N +1 - ln Á 1 - A ˜ = kt X Ae ¯ N + X Ae 1 Ë

(3.132)

Equation 3.132 clearly demonstrates that if ln(1 − XA/XAe) is plotted for various t, then a straight line will be obtained with a slope of (N + 1)k1/(N + XAe).

3.4 Dependency of Reaction Rate on Temperature

Till now, it has been assumed that the concentration of reactant and product has a significant effect on the reaction rate at a specified temperature. However, in order to achieve the rate expression, the role of temperature dependency cannot be neglected, which plays a

Dependency of Reaction Rate on Temperature

pivotal role on the reaction rate. A typical rate equation mostly looks like -r = k f (C )

(3.133)

The dependency of the two terms on the right-hand side of the above equation on temperature is a matter of major discussion. Although the term f(C) is independent of the temperature T, the rate constant is largely dependent on T and that varies with various values of temperature. The chemical theory states the following expression for relating the temperature dependency with the rate constant via the following expression: k µ T me - E /RT

(3.134)

k = Ae - E /RT

(3.135)

In Eq. 3.134, the unit of temperature is K and E has the units of J/mol. However, based on the mathematical approximations, one can easily imagine the higher sensitivity of the temperature in the exponential term in contrast to the power term. An Arrhenius-type relationship yields the following: where k is the reaction rate constant, A is the pre-exponential factor or the steric factor, E is the activation energy, R is the gas constant, and T is the temperature in K. Taking log on both sides, we get

E (3.136) RT If we plot ln k versus 1/T, we get a straight line with the intercept ln A and the slope −E/R. For two rate constants k1 and k2 at temperatures T1 and T2, one can easily get ln k = ln A -

k2 E È 1 1 ˘ (3.137) = Í - ˙ k1 R ÎÍ T1 T2 ˚ Equation 3.137 clearly states that the temperature and the corresponding rate constants for two particular states are enough to evaluate the activation energy E. The important points are listed as follows: ln



∑ Reactions associated with higher activation energy are largely dependent on temperature and vice versa.

75

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis



∑ Any given reaction is much more temperature sensitive at a low temperature than at a high temperature. ∑ From the Arrhenius law, the value of A does not affect the temperature sensitivity.

3.5 Chemical Reactor Analysis

The topic of chemical kinetics is important to understand the reaction mechanism. It may be noted that a chemical reaction occurs inside a reactor, and hence it is highly essential to analyze and design a chemical reactor. In this chapter, performance equations will be derived and developed in three ideal reactors. All the problems and reactions considered will be isothermal homogeneous in nature. The first ideal reactor to be discussed is a batch reactor. A batch reactor refers to a system where the components or, specifically, the reactants are introduced at a particular time. Consequently, they are nicely mixed with the aid of an agitator/impeller and left to react for a specific period of time. After the specified time, the mixture (products + unreacted reactants) is then taken out of the reactor. Hence, such systems correspond to an unsteady state. One must remember that if concentration is measured at a particular instance of time, it will always be found to be uniform. In a batch reactor, there is no continuous inflow and outflow of reactants or products. Such reactors involve only the unsteady state condition. However, if an inlet and outlet stream is added to a batch reactor such that there is constant addition of reactants along with simultaneous removal of products, the batch reactor turns into a continuous stirred tank reactor (CSTR). In contrast to a batch reactor, a CSTR always operates at steady state and the components are thoroughly mixed and stirred in the reactor. Hence, there is a uniform distribution of the contents throughout the reactor. It is important to note that the concentration of the outlet stream and the contents inside a reactor at any point of time is supposed to be identical. A CSTR is also known as a mixed flow reactor (MFR) since the contents are uniformly mixed throughout the reactor at any instant of time. The other ideal steady-state flow reactor is variously known as the plug flow, slug flow, piston flow, ideal tubular, and unmixed flow

Chemical Reactor Analysis

reactor. This reactor is tubular in configuration. We refer to it as the plug flow reactor, or PFR, and to this pattern of flow as plug flow. The main characteristic of such a reactor is that there is no backmixing of the fluid elements and the fluid flow in such a reactor is very orderly. Actually, there may be lateral mixing of fluid in a plug flow reactor; however, there must be no mixing or diffusion along the flow path. The necessary and sufficient condition for plug flow is that the residence time in the reactor must be the same for all elements of fluid. In addition, the velocity gradient is expected to be constant at each cross section of the reactor. These three ideal reactors are comparatively easy to treat. In practice, industries try to design real reactors so that their flows approach these ideals. In the following paragraphs, the symbol V will be used to designate the reactor volume or the total volume of fluid in the reactor. When this differs from the internal volume of the reactor, Vr designates the internal volume of the reactor while V designates the volume of the reacting fluid. The best example is that of a solid-catalyzed reactor. With voidage (e), we have V = eVr

(3.138)

For our simplicity and benefit, we will only consider homogenous reactions in the present chapter and, hence, the term V will be used alone. Let us suppose that the reaction given by Eq. 3.62 occurs in all the above-mentioned reactors. Keeping this in mind, we will first design the performance equation of a batch reactor.

3.5.1 Batch Reactor

Let us construct the material balance equation for reactant A in a batch reactor. For such an accounting, we usually select the limiting component.

Input – output + formation – disappearance = rate of accumulation (3.139) In a batch reactor, since the composition is uniform at any instance of time, the entire reactor is considered for the design. Note that fluid neither enters nor exits the reaction mixture during the reaction. In addition, component A is not formed since it reacts to give products. Hence, the first three terms on the left-hand side of Eq. 3.139 are cancelled. So Eq. 3.139 is modified to

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

−disappearance = rate of accumulation

The disappeared term can be written as

(3.140)

disappearance of A by reaction (moles/time) = –rAV (3.141) rate of accumulation of A (moles/time) =

dN A dt r

Hence, Eq. 3.140 can be written mathematically as follows: -( -rA )V =

-rA = Integrating Eq. 3.144, we get

dN A dt r

dC 1 dN A =- A dt r V dt r

tr = -

CA

Ú

C A0

dC A -rA

For a constant-volume batch reactor,

C A = C A0 (1 - X A )

Putting Eq. 3.146 back to Eq. 3.145, we get XA

t r = C A0

dX A

Ú -r 0

A

(3.142)

(3.143) (3.144) (3.145) (3.146) (3.147)

This is a general equation that shows the time required to achieve a conversion XA for either isothermal or non-isothermal operation. The volume of the reacting fluid and the reaction rate remain under the integral sign, for in general they both change as the reaction proceeds. It may be noted that tr in Eq. 3.147 denotes the reaction time in a batch reactor. It is important to mention at this point that a cycle in a batch reactor consists of fill, react, settle, decant, and idle phases. Hence, the total cycle time t is the sum of all these phases. t = tf + tr + ts + td + ti

(3.148)

In Eq. 3.148, tf is the fill time, tr is the reaction time, ts is the settle time, td is the decant time, and ti is the idle time.

PROBLEM 3.8: A homogenous first-order reaction A Æ B occurs inside a batch reactor. For a conversion of 50% and an initial

Chemical Reactor Analysis

concentration of 10 mol/L, calculate the overall reaction time in the batch reactor. The rate constant for the reaction is 5 min−1.

SOLUTION The reaction rate expression will be given by

-rA = kC A = kC A0 (1 - X A )

From Eq. 3.147, we get XA

tr =

Ú 0

dX A 1 = k (1 - X A ) 5

0.5

(3.149)

dX A

Ú 1- X 0

A

= 0.2 ¥ ( - ln 0.5) = 0.2 ¥ 0.693 = 0.1 1386 The total reaction time in the batch reactor is 0.1386 min.

3.5.2 Continuous Stirred Tank Reactor

A CSTR or an MFR involves continuous input and output along with uniform agitation in the reactor. The agitation maintains uniform concentration throughout the reactor. In such a system, the rate of accumulation of the components is equal to zero. Also component A reacts to give product R. Therefore, the rate of formation of A is also zero. Hence, Eq. 3.139 for such a system boils down to Input – output – disappearance = 0

(3.150)

FA0 - FA = ( -rA )V

(3.151)

Let the input to the reactor be FA0 moles/time and the corresponding output be FA moles/time. The disappearance of component A from the reactor is equal to –rAV. Hence, we can write For a constant-volume system, FA in Eq. 3.151 can be written as FA = FA0 (1 - X A )

Putting Eq. 3.152 in Eq. 3.150, we get

FA0 - FA0 + FA0 X A = ( -rA )V FA0 X A = ( -rA )V

Equation 3.153 on rearrangement give

t X V = A = CSTR C A0 FA0 -rA

(3.152) (3.153) (3.154)

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

In Eq. 3.154, τCSTR denotes the space–time or the hydraulic retention time. A space–time of 2 min means that every 2 min one reactor volume of feed at specified conditions is being treated by the reactor. If we further rearrange Eq. 3.154, we can get t CSTR =

V C A0 X A = v -rA

(3.155)

In Eq. 3.155, v refers to the volumetric flow rate. The four terms XA, −rA, V, and FA0 are related easily with each other through the expressions through Eq. 3.154. Thus, knowing any three allows the fourth to be found directly. In design, then, the size of the reactor needed for a given duty or the extent of conversion in a reactor of given size is found directly. In kinetic studies, each steady-state run gives, without integration, the reaction rate for the conditions within the reactor. The ease of interpretation of data from MFR, it’s better to use CSTR. It makes its use very attractive in kinetic studies, in particular with messy reactions (e.g., multiple reactions and solidcatalyzed reactions).

3.5.3 Plug Flow Reactor

In a PFR, the composition of the fluid varies from point to point along the flow path. Hence, the material balance for a reaction component A must be made for a differential element of volume dV. Similar to a CSTR, the rate of accumulation is zero and reactant A undergoes disappearance rather than formation. Therefore, Eq. 3.150 is also valid for designing a PFR. At each small volume dV, the input of reactant A is FA moles/time, whereas the output of the reactant A from that small volume is FA + dFA moles/time. The disappearance of component A can be written as (−rA)dV. Putting the above information in the material balance Eq. 3.155, we get FA - FA - dFA = ( -rA )dV -dFA = ( -rA )dV -d ÎÈFA0 (1 - X A )˘˚ = ( -rA )dV FA0dX A = ( -rA )dV

(3.156) (3.157) (3.158) (3.159)

Chemical Reactor Analysis

This, then, is the equation that accounts for A in the differential section of reactor of volume dV. For the reactor as a whole, the expression must be integrated. Now FA, the feed rate, is constant, but rA is certainly dependent on the concentration or conversion of materials. Grouping the terms accordingly, we obtain V

Ú 0

dV = FA0

XA

dX A

Ú -r

(3.160)

A

0

t V = PFR = FA0 C A0

XA

dX A

Ú -r

A

0

Further Eq. 3.161 can be modified as follows: XA

t PFR = C A0

0

CA

t PFR = -

dX A

Ú -r

Ú

C A0

A

dC A -rA

(3.161) (3.162) (3.163)

These performance equations, Eqs. 3.160 to Eq. 3.163, can be written either in terms of concentrations or conversions. For systems of changing density, it is more convenient to use conversions; however, there is no particular preference for constant density systems. Whatever its form, the performance equations interrelate the rate of reaction, the extent of reaction, the reactor volume, and the feed rate, and if any one of these quantities is unknown, it can be found from the other three. PROBLEM 3.9: Consider the following first-order reaction: AÆB

The initial concentration of reactant A is 20 mol/L. For a maximum conversion of 60%, determine the best steady-state reactor to process the reaction. The rate constant of the reaction is 10 min−1. Assume the inlet volumetric flow rate to be 10 L/min.

Hint: The best reactor will be decided based on the minimum volume.

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SOLUTION Let us first evaluate the volume of a CSTR. Initial concentration (CA0) = 20 mol/L Volumetric flow rate (F) = 10 L/min Molar flow rate (FA0) = CA0 ¥ F = 20 ¥ 10 mol/min = 200 mol/min Now, the design equation of the CSTR is VCSTR X = A FA0 -rA

VCSTR = FA0 ¥ = 200 ¥

XA kC A0 (1 - X A ) 0.6 0.6 = = 1.5 L 10 ¥ 20 ¥ (1 - 0.6) 0.4

Therefore, the volume of the CSTR is evaluated as 1.5 L. Let us next evaluate the volume of PFR. VPFR = FA0

XA

VPFR = FA0

Ú kC 0

0.6

= 200 ¥

XA

dX A

Ú -r 0

A

dX A A0 (1 - X A ) dX A

Ú 10 ¥ 20 ¥ (1 - X 0

A)

= 0.916 L

Hence, the volume of the PFR is calculated as 0.916 L. Therefore, we can conclude that for the given reaction and conditions, a PFR is a better reactor as it gives the minimum volume as compared to a CSTR.

3.6 Multiple Reactor System

In biochemical industries, we generally come across multiple reactors that are connected in series. First, let us consider N PFRs connected in series. Let X1, X2, …, Xn be the fractional conversion of component A leaving reactor 1, 2, ..., N. Based on the material balance on the feed rate of A to the first reactor, we find for the ith reactor

Multiple Reactor System

Vi = F0

Xi

Ú

X i -1

dX -r

(3.164)

For the N reactors in series, V = F0 =

N

Vi

ÂF i =1 X1

Ú

X0

=

0

dX + -r

V1 + V2 +  + VN N F0

X2

Ú

X1

dX + + -r

XN

Ú

X N -1

dX = -r

XN

Ú

X0

dX -r

(3.165)

Hence, N PFRs in series with a total volume V give the same conversion as a single PFR of volume V. Next we will discuss N CSTRs in series. Though the concentration is uniform in each reactor, there is, nevertheless, a change in concentration as the fluid moves from reactor to reactor. This stepwise drop in concentration suggests that the larger the number of CSTR units in series, the closer should the behavior of the system approach plug flow. This will be demonstrated mathematically. Let C0 be the concentration of the reactant entering the first CSTR and C1 be the concentration of the reactant at the outlet of the CSTR. Again C1 be the inlet concentration of the second reactor, and the corresponding outlet concentration be C2. In general, we can say that Ci−1 is the concentration of the reactant entering the ith reactor and Ci is the exit concentration of the reactant from the ith reactor. Therefore, the percentage conversions Xi and Xi−1 may be computed as Xi = 1 -

Ci C  and  X i -1 = 1 - i -1 C0 C0

The design equation of the CSTR can be written as follows: ti =

C0Vi C0 ( X i - X i -1 ) = F0 F0

(3.166)

(3.167)

Since we consider a constant-volume reactor, Xi and Xi−1 in Eq. 3.162 can be replaced by the expressions of Eq. 3.161. Hence, we get ti =

ÊÊ C ˆ Ê C ˆˆ C0 Á Á 1 - i ˜ - Á 1 - i -1 ˜ ˜ C0 ¯ Ë C0 ¯ ¯ ËË F0

=

C i -1 - C i kCi

(3.168)

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

C i -1 = 1 + kt i Ci

(3.169)

Remember, we are considering only first-order reactions taking place in the reactors. The space–time t is similar in all reactors as all the reactor volumes are considered to be identical. Hence, we can modify Eq. 3.169 as follows: C0 C C C 1 = = 0 1 ◊◊◊◊ N -1 = (1 + kt i )N C N 1 - X N C1 C2 CN

(3.170)

Rearranging Eq. 3.165 and making N Æ •, Eq. 3.170 reduces to a plug flow equation t PFR =

1 C0 ln k C

Hence, N CSTRs will behave as a single PFR.

(3.171)

PROBLEM 3.10: For the reaction 3A + 2B Æ C + 6D, find out the correlation among rA, rB, rC, and rD. SOLUTION Suppose a single phase reaction

aA + bB Æ rR + sS

The rates of reaction of all materials are related to -rA -rB rR rS = = = a b r s

For the reaction 3A + 2B Æ C + 6D, -

r rA r r =- B = C = D 3 2 1 6

Multiplying by 6 gives

−2rA = −3rB = 6rC = rD

PROBLEM 3.11: Which of the following statements on the Arrhenius model of the rate constant k = Ae–E/RT is correct?

a. A is always dimensionless. b. For two reactions 1 and 2, if A1 = A2 and E1 > E2, then k1(T) > k2(T). c. For a given reaction, the percentage change of k with respect to temperature is higher at low temperature.

Multiple Reactor System

d. The percentage change of k with respect to temperature is higher for higher A.

SOLUTION The rate constant from the Arrhenius model k = Ae–E/RT

Case (a) In the Arrhenius model, e–E/RT is dimensionless. Therefore, units of k and A are same. So A is not dimensionless.

Case (b) If

A1 = A2 1

( E1 - E2 ) k1 = e RT k2

E1 > E2, k1 < k2

If

Case (c)

ln(k ) = ln( A) -

E RT

Differentiate with respect to T

d ln(k ) E = dT RT 2

For a lower temperature,

d ln(k ) will be higher. dT

The percentage change in k with respect to temperature is higher at low temperature. Case (d) The percentage change in k with respect to temperature is not related to A as can be seen in case (c). PROBLEM 3.12: For an isothermal second-order aqueous phase reaction A Æ B, find out the ratio of the time required for 90% conversion to the time required for 45% conversion? SOLUTION

A æææææ ÆB second order

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

For the second-order reaction

dC A = kC A2 dt If CA0 Æ Feed concentration, XA Æ Fraction of A converted ( -rA ) = -

Then \

CA = CA0(1 − XA).

dX A = kC A02 (1 - X A )2 dt

C A0 XA

t

dX A

Ú (1 - X 0

2 A)

Ú

= kC A0 dt 0

X

È 1 ˘ A Í ˙ = kC A0t Î 1 - X A ˚0 1 È XA ˘ Í ˙ k C A0 Î 1 - X A ˚ Time required for 90% conversion is t=

t 90 =

1 È 0.9 ˘ 9 = Í ˙ kC A0 Î 1 - 0.9 ˚ kC A0

Time required for 45% conversion is t 45 =

Thus,

1 kC A0

È 0.45 ˘ 45 1 Í 1 - 0.45 ˙ = 55 kC Î ˚ A0

t 90 55 = 9¥ = 11 t 45 45

PROBLEM 3.13: What is the order of the reaction in case the halflife period of the reaction is independent of the initial concentration of the reactant? SOLUTION The half-life of the nth order reaction is t 1/ 2 =

(0.51- n - 1) 1- n C A0 k(n - 1)

where CA0 is the initial concentration and k is the rate constant.

Multiple Reactor System

If n = 1, t1/2 will be independent of CA0.

PROBLEM 3.14: The rate equation for an autocatalytic reaction k

A + RÆR + R

dC A = k C AC R dt When is the rate of disappearance of reactant A maximum? is

-rA = -

SOLUTION For an autocatalytic reaction

k

A + RÆR + R

-rA = -

dC A = kC AC R dt

C A + C R = C A0 + C R0 = C0 = constant Therefore,

CR = C0 – CA

dC A = kC A (C0 - C A ) dt The rate of disappearance of reactant A is maximum where -rA = -

d( -rA ) =0 dC A

d( -rA ) = k(C0 - C A ) - kC A = 0 dC A 2CA = C0 = CA + CR CA = CR

PROBLEM 3.15: What is the order of reaction if the time required to complete a definite fraction of reaction varies inversely as the concentration of the reactants? SOLUTION Using the fractional life method (from Eq. 3.93), tF =

(F 1- n - 1) 1- n C A0 k(n - 1)

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

F=

CA = fraction of reactant in time t F C A0

where n is the order of the reaction and k is the rate constant of the reaction. If n = 2, then only tF µ

1 C A0

PROBLEM 3.16: For a chemical reaction A Æ B, the rate of disappearance of A is given by -rA = -

dC A kC = 1 A dt 1 + k2C A

At high CA, the reaction is of a. b. c. d.

First order with rate constant k1 First order with rate constant k1/k2 Zero order with rate constant k1 Zero order with rate constant k1/k2

SOLUTION The rate equation is

-rA = -

dC A kC = 1 A dt 1 + k2C A

For a high CA value, k2CA >> 1. Then the rate equation can be modified as -rA = -

dC A k1C A k1 0 = = C dt k2C A k2 A

Therefore, at a high CA value, the reaction is of zero order with rate constant k1/k2.

PROBLEM 3.17: Liquid A decomposes by second-order kinetics, and in a batch reactor, 50% of A is converted in a 5-min run. How much longer would it take to reach 75% conversion? SOLUTION Using the fractional life method (Eq. 3.93), tF =

(F 1- n - 1) 1- n C A0 k(n - 1)

Multiple Reactor System

where F = CA/CA0 is the fraction of reactant in time tF, n is the order of the reaction, and k is the reaction rate constant. F1 =

C A C A0 (1 - 0.5) = 0.50 at tF1 = 5 min = C A0 C A0

F2 =

C A C A0 (1 - 0.75) = 0.25 at tF2 = ? = C A0 C A0

In the given problem, n = 2. Hence,

t F 2 (F2-1 - 1) (0.25-1 - 1) = = =3 t F 1 (F1-1 - 1) (0.50-1 - 1) tF2 = 3tF1 = 3 ¥ 5 = 15 min

PROBLEM 3.18: The rate of a chemical reaction doubles for every 10°C rise in temperature. If the temperature is raised by 50°C, what is the increase in the rate of reaction? SOLUTION The rate of a chemical reaction doubles for every 10°C rise in temperature: +10∞C

+10∞C

+10∞C

+10∞C

+10∞C

k Æ 2k Æ 4k Æ 8k Æ 16k Æ 32k

Therefore, if the temperature is raised by 50°C, the rate of the reaction increases by about 32 times.

PROBLEM 3.19: The half-life of a first-order chemical reaction is 6.93 min. What is the time required for the completion of 99% of the reactant? SOLUTION For the first-order reaction,

For a half-life time t1/2,

C A = C A0e - kt CA =

C A0 2

0.693 k where k is the reaction rate constant. t 1/ 2 =

89

90

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

Given data:

So

t1/2 = 6.93 min.

Now for 99% conversion,

k = 0.1 min–1

C A = C A0 (1 - 0.99) = 0.01C A0 = C A0e -0.1t 0.01 = e–10t

t = 46.05 min

PROBLEM 3.20: Find the mathematical expression for the activation energy Ea if the reaction rate constants at two different temperatures T1 and T2 are k1 and k2, respectively? SOLUTION From the Arrhenius model,

k = Ae - Ea /RT

where A is the frequency factor, Ea is the activation energy, and R is the universal gas constant. For two different temperatures, k1 = Ae - Ea /RT1

k2 = Ae - Ea /RT2 Thus,

Ea È 1

1˘

- Í - ˙ k2 R T T =e Î 2 1˚ k1

Êk ˆ E ln Á 2 ˜ = - a R Ë k1 ¯

È1 1˘ Í - ˙ Î T2 T1 ˚

Êk ˆ E È1 1 ˘ ln Á 2 ˜ = a Í - ˙ Ë k1 ¯ R Î T1 T2 ˚ Rearranging

Ea =

Êk ˆ RT1T2 ln Á 2 ˜ (T2 - T1 ) Ë k1 ¯

PROBLEM 3.21: The following reaction rate curve is shown for the reaction A Æ B. Here, −rA and XA represent the reaction rate and

Multiple Reactor System

fractional conversion, respectively. The feed is pure A, and 90% conversion of A is desired.

1 –rA

0

0.2

0.4

XA

0.6

0.8

1

Figure 3.1 Correlation between −1/rA versus XA.

What will be the design of reactor(s) configuration for the reaction?

SOLUTION From Fig. 3.1, we can determine the minimum time required for the conversion of 90% of the substrate (A) as shown Fig. 3.2. 1 –rA

t PFR VPFR = C A0 FA0

t CSTR VCSTR = C A0 FA0

0 0.2

0.4

XA

0.6

0.8

1

Figure 3.2 Determination of the values of ƬCSTR and ƬPFR. C Ao PFR 0.1 C Ao CSTR

Figure 3.3 Schematic diagram of the process for determining the minimum time required for 90% conversion of the substrate (A).

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

We know that space time α volume of the reactor So Fig. 3.3 (CSTR followed by PFR) gives the lowest total volume of the reactor. PROBLEM 3.22: A homogeneous liquid-phase reaction AÆB

-rA = kC A2

takes place with 50% conversion in an isothermal CSTR. What will be the conversion if the reactor is replaced by a PFR of equal size all else remaining the same?

SOLUTION Let us assume that CA0 is the initial concentration of A, FA0 is the molar flow rate, k is the reaction rate constant, XA is the fractional conversion, and V is the volume of the reactor. In the case of CSTR, X V = A FA0 -rA XA = 0.5

-rA = kC A2 = kC A02 (1 - X A )2 Now,

XA V = 2 FA0 kC A0 (1 - X A )2 V 0.5 = 2 FA0 kC A0 (1 - 0.5)2 V k C A02 = 2 FA0

In the case of PFR,

V = FA0

XA

V = FA0

dX A

Ú kC 0

2 A

=

XA

dX A

Ú -r 0

A

1

XA

dX A

Ú (1 - X

kC A02 0

2 A)

Multiple Reactor System

V kC A02 = FA0

XA

Ú 0

X

È 1 ˘ A XA = Í ˙ = 1- XA (1 - X A )2 Î (1 - X A ) ˚0 dX A

as the volumes of CSTR and PFR are equal. Also CA0 and FA0 remain V kC 2 = 2 for PFR also. unchanged. Hence, FA0 A0 Now put the value

XA =2 1- XA XA = 0.67

PROBLEM 3.23: In an aqueous solution, the reaction A Æ B occurs under isothermal conditions following first-order kinetics. The feed rate is 500 mL/min, and the concentration of A in the feed is 1.5 × 10−4 mol/mL. The reaction is carried out in a 5 L CSTR. At steady state, 60% conversion of A is observed. Find the rate constant (min−1) of the reaction? SOLUTION Given data:

F = 500 cm3/min, CA0 = 1.5 × 10–4 mol/cm3,

Now,

V = 5 L = 5000 cm3, XA = 0.6

X X XA V = A = A = FA0 -rA kC A kC A0 (1 - X A )

Again

Therefore,

FA0 = CA0 × F

XA V V = = FA0 C A0 ¥ F kC A0 (1 - X A ) k= k=

XA ¥ F V ¥ (1 - X A )

0.6 ¥ 500 = 0.15 min -1 5000 ¥ (1 - 0.6)

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

PROBLEM 3.24: A liquid-phase reaction A Æ B is conducted isothermally in a CSTR having residence time 2 s. The inlet concentration of species A is 2 mol/L, and the outlet concentration is kC 1 mol/L. The rate equation for the reaction is ( -rA ) = 1 A , where k2 + C A k1 = 5 mol/Ls. Determine the value of k2 in mol/L. SOLUTION Given data:

tCSTR = 2 s, CA0 = 2 mol/L, CA = 1 mol/L

We know that ( -rA ) =

k1C A k2 + C A

k1 = 5 mol/Ls

From the design equation of CSTR,

C A - C A0 ( -rA )

t CSTR =

Hence,

2=

2-1 5 k2 + 1

By solving, we get k2 = 9 mol/L.

PROBLEM 3.25: The time required for 50% decomposition of reactant in a liquid-phase isothermal batch reactor following first-order kinetics is 2 min. What is the time required for 75% decomposition? SOLUTION For first-order reaction, we can write CA = CA0e−kt

[we know that CA = CA0(1 – XA) ]

ln (1 − XA) = −kt

(3.172)

where CA is the concentration at time t, CA0 is the initial concentration, k is the reaction rate constant, and XA is fractional conversion. Using Eq. 3.172 and comparing for two reactions, we get t2 ln(1 - X A2 ) = t1 ln(1 - X A1 )

Multiple Reactor System

XA1 = 0.50 at t1 = 2 min

Given data:

XA2 = 075 at t2 = ?

So

Putting the known values, we get

t2 =2 t1

t2 = 4 min

or

PROBLEM 3.26: A pollutant A degrades according to first-order kinetics. An aqueous stream containing A at 2 kmol/m3 and volumetric flow rate 1 m3 /h requires a mixed flow reactor of volume V to bring down the pollutant level to 0.5 kmol/m3. The inlet concentration of the pollutant is now doubled, and the volumetric flow rate is tripled. If the pollutant level is to be brought down to the same level of 0.5 kmol/m3, find out the volume of the mixed flow reactor? SOLUTION Case 1 Given data:

C A10 = 2

kmol m3

Case 2

,  F1 = 1 C A20 = 4 F2 = 3 C A = 0.5

We know that ƬCSTR =

m3 kmol ,  P1 = 0.5 3 h m kmol m3 m3 h kmol m3

C - CA V = A0 -rA F

Assuming ƬCSTR is constant, we can write

V2 F2 (C A20 - C A ) 3 (4 - 0.5) = ¥ = ¥ =7 V1 F1 (C A10 - C A ) 1 (2 - 0.5)

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

So the volume of the second CSTR will be seven times more than the first CSTR.

PROBLEM 3.27: The conversion of A for a first-order liquid-phase reaction A Æ B in a CSTR is 50%. If another CSTR of the same volume is connected in series, then what is the overall conversion?

SOLUTION Assume the initial concentration of the reactant to be CA0. For 50% conversion in the first reactor (CSTR), the outlet concentration is CA1 = CA0(1 − 0.50) = 0.50CA0

If a similar CSTR is connected in series, the outlet concentration from the second reactor is CA2 = CA1(1 − 0.50) = 0.25CA0

Then the overall conversion is

(C A0 - C A2 ) = 0.75 C A0

PROBLEM 3.28: The liquid-phase reaction A Æ B is governed by the 1

kinetics ( -rA ) = kC A 2 . If the reaction undergoes 75% conversion of A in 10 min in an isothermal batch reactor, what is the time (in min) required for the complete conversion of A? SOLUTION Using the fractional life method, tF =

(F 1- n - 1) 1- n C A0 k(n - 1)

where F = CA/CA0 is the fraction of reactant in time tF, n is the order of the reaction, and k is the reaction rate constant.

Here,

tF1 = 10 min for XA = 0.75 n=

1 2

tF 2 (01/2 - 1) = =2 t F 1 ((1 - 0.75)1/2 - 1) tF2 = 20 min

Multiple Reactor System

PROBLEM 3.29: The liquid-phase reaction A Æ B + C is conducted isothermally at 50°C in a continuous stirred tank reactor (CSTR). The inlet concentration of A is 8.0 gmol/L. At a space–time of 5 min, the concentration of A at the exit of CSTR is 4.0 gmol/L. The kinetics of the reaction is −rA = kCA. Find out the rate constant (k) for this reaction at 50°C. SOLUTION For a CSTR,

C A0 - C A C - CA = A0 ( -rA ) k CA

t CSTR =

where CA0 is the inlet concentration of A, CA is the outlet concentration of A, and tCSTR is the space–time for CSTR. Given data: CA0 = 8 gmol/L



CA = 4 gmol/L tCSTR = 5 min

Putting all the known values in the above equation (design equation), 5=

or

Now what is the unit of k? Therefore,

8-4 k¥4

k = 0.2

unit of k = (unit of CA)1–1 (unit of time)–1 unit of k = (min)–1 k = 0.2 (min)–1

PROBLEM 3.30: The first-order reversible liquid reaction A  B ,  C A0 = 0.5

mol , CBO = 0 L

takes place in a batch reactor. After 8 min, the conversion of A is 33.3%, while the equilibrium conversion is 66.7%. Find the rate equation for this reaction. SOLUTION Given data:

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Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

mol L CB0 = 0

C A0 = 0.5

XA = 0.33 at t = 8 min The rate equation is

-

-

XAe = 0.667

dC A dC B = = k1C A - k2C B dt dt

dC A = k1 (C A0 - C A0 X A ) - k2 (C B 0 + C A0 X A ) dt -

dC A = k1 (C A0 - C A0 X A ) - k2 (C A0 X A ) dt

C A0dX A = k1 (C A0 - C A0 X A ) - k2 (C A0 X A ) = C A0 ÈÎk1 (1 - X A ) - k2 ( X A )˘˚ dt dX A = ÈÎk1 (1 - X A ) - k2 ( X A )˘˚ dt Now at equilibrium,

-

Therefore,

dX dC A = 0  or  A = 0 dt dt k1 (1 - X Ae ) = k2 ( X Ae ) k2 1 - X Ae = 0.50 = k1 X Ae

From Eq. 3.173,

k2 = 0.50 k1 dX A = ÈÎk1 (1 - X A ) - 0.5 k1 ( X A )˘˚ dt dX A = k1 ÈÎ(1 - X A ) - 0.50( X A )˘˚ dt

(3.173)

Multiple Reactor System

dX A = k1dt ÎÈ(1 - X A ) - 0.50( X A )˘˚ 0.333

Ú 0

From Eq. 3.173,

8

dX A = k1dt [(1 - 1.50 X A )]

Ú 0

1 ln (1 - 1.5 ¥ 0.333) = k1 ¥ 8 -1.5 k1 = 0.057 min−1

k2 = 0.50 × 0.057 = 0.0285 min−1

Therefore, the rate of reaction may be written as -

dC A dC B = = (0.057 min -1 ) C A - (0.0285 min -1 ) C B dt dt

PROBLEM 3.31: The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 650°C than at 500°C? SOLUTION Given data:

Ea = 300

We know that

kJ mol

Ê ( -rA )2 ˆ Ea Ê 1 1 ˆ ln Á = Ë ( -rA )1 ˜¯ R ÁË T1 T2 ˜¯ T1 = 500°C = 773 K

We know that

T2 = 650°C = 923 K

Ê ( -rA )2 ˆ 300, 000 Ê 1 1 ˆ ln Á = = 7.58 Á ˜ 8.314 Ë 773 923 ˜¯ Ë ( -rA )1 ¯ ( -rA )2 = e7.58 = 1958.6 ( -rA )1

99

100

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

Therefore, the decomposition will be 1958.6 times faster at 650°C than at 500°C. PROBLEM 3.32: In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and CA alone determines this rate: CA (mol/L)

−rA (mol/L h)

1

0.06

2

0.1

4

0.25

6

1.0

7

2.0

9

1.0

12

0.5

We plan to run this reaction in a batch reactor at the same catalyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from CA0 = 10 mol/L to CA = 2 mol/L.

SOLUTION Given data: CA0 = 10 mol/L; CA = 2 mol/L We know that CA

t batch = -

Ú

C A0

dC A -rA

The integration can be evaluated by graphical method (area under the curve of CA versus 1/–rA between CA0 and CA) or by numerical method CA (mol/L)

(−rA) (mol/L h)

1/(−rA) (L h/mol)

1

0.0600

16.6667

4

0.2500

4.0000

2 6 7 9

12

0.1000 1.0000 2.0000 1.0000

0.5000

10.0000

1.0000 0.5000 1.0000

2.0000

Multiple Reactor System batch time calculation

18 16

1/(-rA) (liter.hr/mol)

14 12 10 8 6 4 2 0

0

2

4

6 CA (mol/liter)

8

10

12

Area under the curve (using trapezoidal rule) tbatch = 22.42 h

PROBLEM 3.33: An aqueous feed of A and B (400 L/min, 100 mmol A per liter, 200 mmol B per liter) is to be converted to product in a plug flow reactor. The kinetics of the reaction is represented by A + B Æ C,  -rA = 200C AC B

mol Lmin

Find the volume of the reactor needed for 99.9% conversion of A to product. SOLUTION Given data:

C A0 = 100

mmol mol = 0.1 L L

C B0 = 200

mmol mol = 0.2 L L

v = 400

L mim

XA = 0.999 k = 200

101

102

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

Thus,

N=

C B 0 200 = =2 C A0 100

For bi-molecular type second-order reaction,

È N - XA ˘ k t PFR C A0 ( N - 1) = ln Í ˙ Î N(1 - X A ) ˚ Putting all the known values, tPER = 0.31 min We know

VPFR = tPFR ¥ v = 0.31 ¥ 400 = 124.3 L

PROBLEM 3.34: A plug flow reactor (2 m3) processes an aqueous feed (100 L/min) containing reactant A (CA0 = 100 mmol/L). This reaction is reversible and represented by A ⇌ R −rA = 0.04 min–1CA – 0.01 min–1CR

First find the equilibrium conversion and then find the actual conversion of A in the reactor. SOLUTION Given data:

k1 = 0.04 min–1, k2 = 0.01 min–1 C A0 = 100

mmol mol = 0.1 L L

CB0 = 0

v = 100

Thus,

L min

VPFR = 2 m3 = 2000 L N=

C B0 =0 C A0

For first-order reversible reaction,

k1 N + X Ae = k2 1 - X Ae

Putting all the known values,

XAe = 0.80

Multiple Reactor System

So equilibrium conversion is 80%. We know that

Ê X ˆ N +1 kt - ln Á 1 - A ˜ = X Ae ¯ N + X Ae 1 PFR Ë

where

t PFR =

VPFR 2000 = = 20 min v 100

Putting all the known values in the above equation and applying trial and error method, XA = 0.50

So actual conversion is 50%.

PROBLEM 3.35: A mixed flow reactor (2 m3) processes an aqueous feed (100 L/min) containing reactant A (CA0 = 100 mmol/L). This reaction is reversible and represented by A ⇌ R −rA = 0.04 min–1CA – 0.01 min–1CR

First find out the equilibrium conversion and then find out the actual conversion of A in the reactor. SOLUTION Given data:

k1 = 0.04 min–1, k2 = 0.01 min–1 mmol mol = 0.1 L L CB0 = 0

C A0 = 100

v = 100

Thus,

L min

VMER = 2 m3 = 2000 L N=

C B0 =0 C A0

For first-order reversible reaction,

k1 N + X Ae = k2 1 - X Ae

103

104

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

Putting all the known values,

XAe = 080

So equilibrium conversion is 80%. It is known that t CSTR =

VMFR 2000 = min v 100

When conversion is XA,

(−rA) = k1CA0(1 − XA) − k2CA0XA t CSTR =

C A0 X A ( -rA )

Putting all the known values in the above equations and solving by trial-and-error method, we get XA = 0.40. Therefore, actual conversion is 40%. PROBLEM 3.36: An aqueous feed of A and B (400 L/min, 100 mmol A/L, 200 mmol B/L) is to be converted to product in a mixed flow reactor. The kinetics of the reaction is represented by

mol Lmin Find the volume of the reactor needed for 99.9% conversion of A to product. A + B Æ R,  -rA = 200C AC B

SOLUTION Given data:

C A0 = 100

mmol mol = 0.1 L L

C B0 = 200

mmol mol = 0.2 L L

v = 400

L min

XA = 0.999 Thus,

k = 200

N=

C B 0 200 = =2 C A0 100

Multiple Reactor System

It is known that where

t MFR =

VMFR C A0 X A = v -rA

-rA = 200(C A0 - C A0 X A )(C B 0 - C A0 X A )

Putting all the known values in the above equations, we get VMFR = 19960 L.

PROBLEM 3.37: We plan to operate a batch reactor to convert A into B. This is a liquid reaction. The stoichiometry is A Æ B, and the rate of reaction is given in the following table. How long must we react each batch for the concentration to drop from CA0 = 1.3 mol/L to CA = 0.3 mol/L? CA mol/L

−rA mol/L min

0.1

0.1

0.4

0.6

0.2 0.3 0.5 0.6 0.7

0.80 1.0 1.3

SOLUTION Given data:

2.0

0.3 0.5 0.5

0.25 0.10 0.06 0.05

0.045 0.042

C A0 = 1.3 C A = 0.3

mol L

mol L

105

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

We know that

CA

t batch = -

Ú

C A0

dC A -rA

The integration can be evaluated by graphical method or by numerical method (area under the curve of CA versus 1/–rA between CA0 and CA). CA mol/L

−rA mol/L min

1/(−rA) L min/mol

0.1

0.1

10.00

0.4

0.6

1.67

0.2

0.3

0.3

3.33

0.5

0.5

2.00

0.5

0.6

0.25

0.7

0.10

0.80

0.06

1.0

0.05

1.3

0.045

2.0

0.042

2.00 4.00

10.00 16.67 20.00 22.22 23.81

Batch time calculation

25

1/(-rA) (liter.min/mol)

106

20

15

10

5

0

0

0.2

0.4

0.6

0.8 1 1.2 CA (mol/liter)

1.4

1.6

1.8

2

Multiple Reactor System

Area under the curve (using trapezoidal rule) is tbatch = 12.70 min.

PROBLEM 3.38: For the reaction of previous problem (Problem 3.37), what size of plug flow reactor would be needed for 80% conversion of a feed stream of 1000 mol A/h at CA0 = 1.5 mol/L? SOLUTION Given data:

C A0 = 1.5

mol L

XA = 0.80

FA0 = 1000 Thus,

mol 1000 mol = h 60 min

C Af = C A0 (1 - X A ) = 1.5(1 - 0.80) = 0.3

We know that

CA

t PFR = -

Ú

C A0

mol L

dC A -rA

The integration can be evaluated by graphical method (area under the curve of CA versus 1/–rA between CA0 and CAf). 25

1/(-rA) (liter.min/mol)

20

15

10

5

0

0

0.2

0.4

0.6

0.8 1 1.2 CA (mol/liter)

1.4

1.6

1.8

2

107

108

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

Area under the curve (using trapezoidal rule) is tPER = 17.19 min. We know that VPFR t PFR = FA0 C A0

VPFR = FA0

t PFR 1000 17.19 L = 191 L = ¥ 60 1.5 C A0

PROBLEM 3.39: For the reaction of Problem 3.37,

(a) What size of mixed flow reactor is needed for 75% conversion of a feed stream of 1000 mol A/h at CA0 = 1.2 mol/L? (b) Repeat part (a) with the modification that the feed rate is doubled, thus 2000 mol A/h at CA0 = 1.2 mol/L are to be treated. (c) Repeat part (a) with the modification that CA0 = 2.4 mol/L; however, 1000 mol A/h are still to be treated down to CAf = 0.3 mol/L.

SOLUTION

(a) Given data:

C A0 = 1.2 FA0 =

Thus,

1000 mol 60 min

XA = 0.75 C Af = C A0 (1 - X A ) = 0.30

Now at

C Af = 0.30 -rA = 0.50

Now,

mol L

t MFR =

mol L

mol , L

mol Lmin

C A0 X A 1.2 ¥ 0.75 = = 1.8 min 0.5 -rA

Multiple Reactor System

VMFR

1000 t MFR ¥ FA0 1.8 ¥ 60 = = = 25 L 1.2 C A0

(b) It is seen that VMFR µ FA0 . Therefore, on doubling the feed rate, VMER will be double. (c) Given data:

VMFR = 50 L

C A0 = 2.40 FA0 =

mol L

1000 mol 60 min

XA = 0.75

Thus,

C Af = 0.30

mol L

Now at

C Af = 0.30

mol , L

-rA = 0.50 XA = 1t MFR =

VMFR

mol Lmin

CA 0.30 =1= 0.875 C A0 2.40

C A0 X A 2.4 ¥ 0.875 = = 4.2 min 0.5 -rA

1000 t MFR ¥ FA0 4.2 ¥ 60 = 29.12 L = = 2.4 C A0

PROBLEM 3.40: At an initial concentration of CA0 = 1 mol/L, reactant A is introduced into a batch reactor where it reacts to form product B according to the following stoichiometry: A Æ B. The concentration of A in the reactor is monitored at different times, as given below:

109

110

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

t (min) CA

0

(mmol/m3)

1000

100

500

200

333

300

250

400

200

If CA0 = 500 mol/m3, find the conversion of the reactant after 5 h in the batch reactor. SOLUTION From the data table, At

At

t = 0, CA0 = 1000 mol/m3

t = 100 min + 5 h = 400 min, CA = 200 mol/m3

So the conversion after 5 h is (1000 − 200)/1000 = 0.80 PROBLEM 3.41: For the elementary reactions in series k

k

C = C A0 R 0 = CS 0 = 0

1 2 A ææ Æ R ææ Æ S , k1 = k2 , at t = 0 {C A

determine the maximum concentration of R and when it is reached. SOLUTION The chain reaction is AÆ RÆ S. The rate equations are rA =

rR =

dC A = -kC A dt

dC R = kC A - kC R dt dC S = kC R dt k1 = k2 = k

rS = When

Let us start with a concentration CA0 of A, no R and no S present. So at any time, CA0 = CA + CR + CS Integrating the first rate equation (rA), we get ÊC ˆ - ln Á A ˜ = kt Ë C A0 ¯ C A = C A0e - kt

To find the CR, substitute CA in the second rate equation (rR) dC R + kC R = kC A0e - kt dt

Multiple Reactor System

Multiplying both side by ekt e kt

dC R + kC R e kt = kC A0 dt d(C R e kt ) = kC A0 dt

Úd(C e R

kt

Ú

) = C A0tdt

C R e kt = kC A0t + constant Applying the initial condition, CR = CR0 = 0 at t = 0 constant = 0 C R = kC A0e - kt t

Therefore,

If the desire product is R, have to achieve CR,max At

C R,max , 

dC R = 0 when t = t max dt

{

}

dC R = kC A0 e - kt - e - kt kt = 0 dt e - kt - e - kt kt = 0 [ e - kt π 0 ] t max =

1 k

C R ,max = kC A0e - kt max ¥ t max C R ,max = kC A0e C R ,max =

-k ¥

1 k

¥

1 k

C A0 e

PROBLEM 3.42: Find the conversion after 1 h in a batch reactor for A Æ B,  -rA = 3C A0.5

SOLUTION Given data:

mol mol , C A0 = 1 Lh L

CA0 = 1 mol/L

111

112

Chemical Reaction Thermodynamics, Kinetics, and Reactor Analysis

The rate of reaction of the above reaction is ( -rA ) = -

dC A = 3C A0.5 dt

Rearranging and integrating, we get CA

dC A

ÚC

C A0

0.5 A

t

Ú

= -3 dt 0

C A0.5 - C A00.5 = ( -0.5) ¥ 3 ¥ t t=1h

When

C A0.5 - C A00.5 = ( -0.5) ¥ 3 = -1.5 C A0.5 - 10.5 = -1.5 C A0.5 = -0.5 C A = 0.25

mol L

Therefore, the conversion is

C A0 - C A 1 - 0.25 ¥ 100% = ¥ 100% = 75% C A0 1

References 1. Levenspiel, O. Chemical Reaction Engineering, Third Edition, WileyIndia, 2010. 2. Bailey, J. E. and Ollis, D. F. Biochemical Engineering Fundamentals, McGraw-Hill Inc., New Delhi, India, 2010.

3. Sinclair, C. G. and Kristiansen, B. Fermentation Kinetics and Modelling, Open University Press, 1987. 4. Shuler, M. L. and Kargi, F. Bioprocess Engineering: Basic Concepts, Second Edition, Prentice-Hall Inc., New Delhi, India, 2002.

5. Blanch, H. W. and Clark, D. S. Biochemical Engineering, Marcel Dekker Inc., New York, USA, 1997. 6. Doran, P. M. Bioprocess Engineering Principles, Second Edition, Academic Press, Waltham, USA, 2012.

References

7. Gavhane, K. A. Introduction to Process Calculations Stoichiometry, Nirali Prakashan, 2012.

8. Bhatt, B. I. and Thakore, S. B. Stoichiometry, Tata McGraw-Hill Education, 2010. 9. Ghasem, N. and Henda, R. Principles of Chemical Engineering Processes, CRC Press, 2012.

10. Katoh, S. and Yoshida, F. Biochemical Engineering: A Textbook for Engineers, Chemists and Biologists, Wiley-VCH, 2010.

11. Nauman, E. B. Chemical Reactor Design, John Wiley & Sons, 1987.

12. Himmelblau, D. M. Basic Principles and Calculations in Chemical Engineering, Sixth Edition, Prentice Hall of India, 1997. 13. Kelly, R. M., Wittrup, K. D., and Karkare, S. (Eds.). Biochemical Engineering VIII, Vol. 745, Annals of the New York Academy of Sciences, 1994.

113

Chapter 4

Enzymatic Reaction Kinetics

Enzymes are globular proteins with an active site and act as a biological catalyst. Chapter 3 involved lots of discussion on catalysts and various types of reactions involving catalysts (such as autocatalytic reaction and various types of catalytic reactions). In order to gain a proper understanding of the functioning of enzymes, we will again briefly discuss catalysts since enzymes are a special type of catalysts. As discussed previously, a catalyst may be defined as a substance that has the ability to enhance the reaction rate by not altering its own concentration and composition throughout the course of the reaction. In other words, the concentration of the catalyst remains identical before and after reaction. Technically speaking, a catalyst basically lowers or diminishes the activation energy of the associated molecules. In addition, the reaction equilibrium also remains unaffected (Fig. 4.1). Enzymes are important as catalysts from the perspective of biochemical/biological reactions. A large number of living cells in various plants (including fruits, vegetables), microorganisms, and animals generate a significant amount of enzymes. For example, papain (an enzyme produced by papaya), cellulolytic enzymes (produced by microorganisms), and amylase (produced by animals) are typical examples of enzymes. It may be noted that a specific enzyme is needed for a specific reaction. Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

116

Enzymatic Reaction Kinetics

An enzymatic reaction is associated with the breaking of chemical bonds. In addition, enzymes are also responsible for making or building of chemical bonds. Similar to synthetic catalysts used in chemical reactions, the concentration and composition of an enzyme remain typically unchanged after the completion of any biochemical reaction. Uncatalyzed

Catalyzed

Ea is high Activation energy (Ea) is low Glucose 1-phosphate

Standard free energy change of reaction

G-1-P G° = –1.75 kcal Glucose 6-phosphate

G-6-P

Figure 4.1 Change of activation energy in catalyzed and uncatalyzed states.

Enzyme catalysts are categorized under six major types: (1) oxidoreductases (mostly used in transport chains of bacteria, certain chloroplasts, mitochondria), (2) transferases (used for biotechnological purposes and in producing plasmid vectors), (3) hydrolases (found in liver and helps in secreting bile salts), (4) lyases (found in biological membranes and used in breaking of bonds), (5) isomerases (used in various biological processes and helpful in carbohydrate metabolism along with glycolysis), and (6) ligases (present as membrane proteins across various biological membranes). Oxidoreductases catalyze oxidoreduction reactions by transferring electron(s); for example, Alcohol dehydrogenase C2H5OH + NAD Æ CH3CHO + NADH

The shifting of a particular functional group from a molecule to another is catalyzed by the enzyme transferase; for example, Methylamine-glutamate N-methyl transferase Methyamine + l-glutamate Æ NH3 + N-methyl-l-glutamate

Hydrolases catalyze to cleave chemical bond by using water.

Enzymatic Reaction Kinetics

Invertase

Sugar + H2O Æ Glucose + Fructose

A carbon–carbon, carbon–nitrogen, or carbon–oxygen bond is broken by the aid of the enzyme lyase. The breaking of the bonds are executed by processes other than oxidation and hydrolysis. The product contains mostly a ring or a double-bond structure; for example, Decarboxylase CH3COCOOH Æ CH3CHO + CO2

Isomerases catalyze a molecule to convert one isomer to another; for example, Glucose isomerase  Glucose Æ Fructose

Ligase is an enzyme that initiates the formation of a chemical bond by combining two large molecules together. The best example of a ligase enzyme is DNA ligase. The three-dimensional configuration of a protein molecule is responsible for the development of an active site, which is further useful for various catalytic activities. In addition, a cofactor is extremely necessary for enzyme activity since cofactors largely assist enzymes during catalytic reactions (Fig. 4.2). Cofactors are basically non-protein components that combine with an inactive protein component to generate the final compound, which is active catalytically. In general, the inactive enzymes are known as apoenzymes, and the final compound is basically called a holoenzyme. Metal ions

Cofactor ------Coenzymes

Figure 4.2 Classification of cofactors.

Although many enzymes have similar names, they are acquired from different organisms. In addition, these enzymes have different catalytic activity, e.g., glucose isomerase. It has been found that glucose isomerase obtained from B. coagulans requires Co2+,

117

118

Enzymatic Reaction Kinetics

whereas glucose isomerase from a mutant of B. coagulans does not require Co2+ at a pH greater than 8. A specific enzymatic activity indicates the purity of an enzyme and is expressed as micromoles of substrate converted per minute per milligram of protein. Enzymatic activity is also expressed as turnover number. It is defined as the number of substrate molecules reacted per catalyst site per unit time. A comparative study on enzymes and solid-catalyzed reactions reveals that some enzymatic reactions are faster than non-biological catalytic reactions, e.g., papain (proteolytic enzyme). The protein structure of an enzyme is responsible for the catalytic ability. One must understand the significance of active sites as they are important in explaining the science behind enzyme kinetics. An active site refers to a small portion on the surface of an enzyme where a specific chemical reaction is catalyzed.

4.1 Characteristics of an Enzyme The characteristics of enzymes are as follows:

∑ Higher specificity ∑ Catalyzes only a single or smaller number of chemical reactions ∑ Required in smaller amounts to produce the desired effect ∑ Conditions associated with enzymatic reactions, e.g., pH, temperature, etc. ∑ Highly sensitive and unstable molecules in contrast to chemical catalysts

4.2 Application of Enzymes

The actual working of enzymes was unknown to people although they used enzymes for preparing different food products for a long time. Enzymes have been used for making cheese from milk (chymosin enzymes), extracting and clarifying fruit juices (pectinase enzymes), tendering meat or animal protein (papain enzymes), processing raw plants (microbial protease enzymes), reducing the maturation time for wine making (acetolactate decarboxylase enzymes), etc. Commercially, enzymes have been used since the late 19th century.

Enzyme Kinetics

Enzymes were used for the first time for making sugar from starch. In addition, history also reveals that in the United States, digestive enzymes were available in the form of takadiastase. As already discussed, enzymes are a special kind of protein. The function of this protein largely depends on the amino acid sequence. Since proteins have an extremely complicated structure, synthesizing a large amount of enzyme becomes truly difficult. Then the question arises: how do we acquire enzymes? In general, various microorganisms are capable of producing enzymes in pure culture. Enzymes can be classified into three types: 1. Medical enzymes: These enzymes are used in the medical field. Typical examples are asparaginase, proteases, lipases, and streptokinase. 2. Industrial enzymes: These enzymes are used in various industrial processes, e.g., glucose isomerase, lipase, proteases, etc. 3. Analytical enzymes: These enzymes are largely used for various laboratory and clinical purposes, e.g., glucose oxidase, alcohol dehydrogenase, etc.

4.3 Enzyme Kinetics

The previous chapter clearly illustrated that chemical kinetics involves the rate of chemical reaction and various conditions on which the chemical reaction depends. In a similar manner, enzyme kinetics is associated with the reaction rate of an enzymatic reaction and the corresponding factors associated with it. Enzyme kinetics is extremely important to understand the mechanism of various enzymatic reactions. Similar to the study of chemical kinetics, the reaction rate associated with an enzymatic reaction is developed to estimate the product yield, selectivity, reaction time along with various other parameters. This will, in turn, help in designing a bioreactor. A typical enzymatic reaction involves the conversion of the substrate into products via enzymatic pathways. The following example gives a generalized picture of an enzymatic reaction: E(Enzyme) + S(Substrate) Æ P(Product)

(4.1)

119

120

Enzymatic Reaction Kinetics

Let the rate constant of the above reaction be denoted by the symbol k. From Eq. 4.1, one can clearly make out that the concentration of the substrate will deplete and reach a minima. On the other hand, the product concentration will keep on increasing till it reaches a maxima. Note that the time taken to reach the minima and maxima should be identical. Based on the fundamentals of chemical kinetics, one can easily write the expression for the rate law associated with the diminishing substrate concentration and increasing product concentration as follows:

dCS dC and rP = P (4.2) dt dt For a better understanding of the sensitivity and effectivity of enzyme kinetics, one must deeply understand the effect of the enzyme, product, and substrate concentration on the reaction rate. In general, a plot of different substrate concentrations versus reaction rates yields a profile depicting a parabolic increase in the reaction rate till a particular substrate concentration. From this profile, one can conclude the following: rS = -

1. A linear variation in the reaction rate and substrate concentration is observed especially in the lower range of substrate concentration. Hence, we can infer that the substrate concentration and the reaction rate are directly proportional to each other for the lower ranges in the substrate concentration. Generally, the order of the reaction in such cases is found to be almost 1. 2. For the cases involving a greater range of substrate concentration, the rate of reaction becomes independent of the substrate concentration. Hence, in such cases the order of the reaction tends to zero. 3. The reaction rate reaches a maximum (vmax) after a particular substrate concentration CS,opt. Beyond CS,opt, the reaction rate remains constant irrespective of the substrate concentration.

In 1902, Henri observed this behavior and proposed a rate equation as follows: v=

vmax CS kM + C S

(4.3)

Enzyme Kinetics

Equation 4.3 is the most important equation in the field of enzyme kinetics. The observations made by all the three aforementioned points have been covered effectively by the rate equation given by Eq. 4.3. The kinetic parameters denoted by kM and vmax are determined experimentally. Let us analyze Eq. 4.3 in brief. One can easily make out that the lower the substrate concentrations (CS > kM), Eq. 4.3 reduces to the case where v = vmax. Various kinetic mechanisms have been proposed over the years. But the noteworthy proposal is that of Brown, who proposed that a complex is formed between the enzyme and the substrate. Free enzyme is generated after the complex breaks down into products. Hence, the enzyme is used only for binding and building the complex. The mechanism of the enzyme–substrate reaction is given by k

1 E + S ¨æ Æ ES k 2

k

3 ES ææ ÆE + P

(4.4)

(4.5)

In Eqs. 4.4 and 4.5, ES depicts the enzyme–substrate or the intermediate complex. Note that Eq. 4.4 is a reversible reaction, whereas the product-formation step, as given by Eq. 4.5, is an irreversible reaction. Since Eq. 4.4 is reversible, one can imagine that E and S combine to give ES. Also ES can separate individually into E and S. This type of mechanism can be clearly imagined based on the “lock and key” theory, which says that a similarity in the structural and topological compatibility between the enzyme and the substrate favors the affinity of the substrate (key) toward the enzyme (lock). The following assumptions are made based on the mechanisms that have been discussed via Eqs. 4.4 and 4.5. In addition, the rate equation is also constructed based on the relation between the reaction rate and the substrate concentration. 1. Throughout the course of the reaction, the total concentration of the enzyme remains constant. The initial enzyme concentration is given by CE0, whereas the concentration of the ES (enzyme–substrate) complex is denoted by CES, and CE

121

122

Enzymatic Reaction Kinetics

refers to the concentration of the enzyme taking part in the reaction. Then CE0 = CE + CES. 2. It must be remembered that the substrate concentration is significantly higher compared to the enzyme concentration. Hence, one can surely assume that the substrate concentration does not undergo a significant depletion although the ES complex is generated. 3. The product formed may be insignificant in concentration, and hence the inhibition caused by the product formation may be assumed to be negligible.

Considering the above assumptions in mind, two approaches are highly established in order to derive the rate equation for an enzymatic reaction process:

1. Michaelis–Menten approach: The Michaelis–Menten approach assumed that the product-formation step given by Eq. 4.5 is a slower step and, hence, a rate-determining step in contrast to Eq. 4.4. In chemical engineering, heterogeneous catalytic reactions are typical examples where these types of assumptions are employed. As the enzyme composition is similar to the protein composition, the chemical structures of enzyme molecules are pretty large and complicated. However, these enzyme molecules are very much soluble in water. As a result, enzymes are almost analogous to solid catalysts, which are used in surface catalytic reactions in chemical engineering. In addition, it has been also found that the interaction between the enzyme and the substrate in the ES complex is weak and, hence, the ES complex is less stable. Taking this fact, one can confidently conclude that the ES-complex-formation step is reasonably faster, while the product-formation step is relatively slower and more stable. 2. Briggs–Haldane approach: This approach was introduced by assuming the pseudo-steady-state model. This model clearly states that the rate of change in the concentration of the intermediate product (ES complex) is close to zero. We can express that mathematically as (dCES/dt) = 0. This type of assumptions is normally taken into account for homogeneous catalytic reactions in the field of chemical kinetics.

Enzyme Kinetics

4.3.1 Michaelis–Menten Approach Equation 4.5 represents the rate-determining step as it demonstrates a slower reaction in contrast to Eq. 4.4. Hence, based on Eq. 4.5, the concentration of the ES complex and the rate of product formation are directly proportional to each other. If the rate of product formation is represented by v, the rate equation may be written as

dCP = k3CES (4.6) dt For maintaining homogeneity and simplicity, let us consider the unit of concentration as mol/L in all cases. Note that based on the reversible reaction represented by Eq. 4.4, CES is related to the enzyme concentration CE and the substrate concentration CS. As the forward and backward reactions of Eq. 4.4 are in equilibrium, we can write v=

k1CECS = k2CES CES =

k1CECS k2

(4.7)

(4.8)

We can also express Eq. 4.8 with respect to CE. Then Eq. 4.8 will be modified as CE =

k2CES k1CS

(4.9)

As mentioned previously, the total enzyme contents are conserved. Hence, we can write CE0 = CE + CES

Putting Eq. 4.9 in Eq. 4.10, we get CE0 =

Rearranging Eq. 4.11, we get

k2CES + CES k1CS

CES =

CE0CS k2 + CS k1

(4.10)

(4.11) (4.12)

Now, replacing Eq. 4.12 in Eq. 4.6 we get the expression of the rate equation as follows:

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Enzymatic Reaction Kinetics

v=

C C v C dCP = k3 E0 S = max S k dt kM + C S 2 + CS k1

(4.13)

Equation 4.13 is also referred to as the Michaelis–Menten equation. One can surely visualize that the above equation is identical to Eq. 4.3 as given by Henri. The term vmax in the numerator is obtained by combining the terms k3 and CE0. This is mainly done as the concentration of the enzyme cannot be expressed in molar units. In the denominator of Eq. 4.13, the term kM is referred to as the constant associated with the Michaelis–Menten equation, or simply the Michaelis–Menten constant. It may be noted that the Michaelis–Menten constant (kM) and the equilibrium constant (Keq) are inversely related to each other, which is expressed as follows: kM =

k2 CSCE 1 = = k1 CES K eq

(4.14)

It is quite clear from Eq. 4.14 that the units of kM and CS are identical. Note that if kM and CS are equal, then v = (vmax/2). Hence, it can be concluded that the actual reaction rate is 50% or half of the maximum reaction rate when the Michaelis–Menten constant and the substrate concentrations are identical in magnitude. Note that the kinetic parameter kM is an important quantity as it denotes the extent of the interaction between the substrate and the enzyme. Although the substrate and product concentrations can be easily expressed in molar units or moles per unit volume, the expression of the enzyme concentration in molar units may be slightly confusing. It must be understood that molar or mass unit is not the exact or correct way of expressing enzyme contents. This may be due to the fact that there is a wide variation in enzyme contents based on the enzyme purity. Hence, it is better to express the concentration of an enzyme according to its catalytic ability. Let us illustrate an example to have a better idea on this. Let us suppose than an enzyme E is hydrolyzed to produce C. Then we can define one unit of the enzyme E as the enzyme quantity required to hydrolyze E to synthesize 1 mmol of C per minute. Hence, the units of enzyme concentration should be chosen very carefully such that the unit of k3CE0 is same as that of vmax or moles/volume (time).

Enzyme Kinetics

4.3.2 Briggs–Haldane Approach In order to illustrate the Briggs–Haldane approach, let us first write the rate equations with respect to the product formation, substrate depletion, and ES complex formation based on Eqs. 4.4 and 4.5: dCP = k3CES dt

-

dCS = -k2CES + k1CSCE dt

dCES = k1CECS - k2CES - k3CES = 0 dt If we further rearrange Eq. 4.17, we get

k1CECS – k2CES = k3CES

Putting Eq. 4.18 in Eq. 4.16, we get

(4.15) (4.16)

(4.17)

(4.18)

dCS = k3CES (4.19) dt Assuming that the total enzyme contents are conserved, we can write -

CE0 = CE + CES

or

CE = CE0 - CES

Putting Eq. 4.20 in Eq. 4.17, we get CES =

CE0CS k2 + k3 + CS k1

Replacing Eq. 4.21 in Eq. 4.19 and Eq. 4.15, we get v=

v C dC C E 0C S dCP = - S = k3CES = k3 = max S k + k dt dt kM + C S 2 3 + CS k1

(4.20) (4.21)

(4.22)

Note that Eq. 4.22 is the same as Eq. 4.13, which is the Michaelis– Menten equation. However, the meaning of the constant kM for the Briggs–Halden approach is different from the Michaelis–Menten approach. As already discussed, kM is equal to the ratio k2/k1, while in the Briggs–Haldane approach, it is equal to (k2 + k3)/k1.

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Enzymatic Reaction Kinetics

Equation 4.22 can be simplified to Eq. 4.13 if k2 >> k3, which means that the product-releasing step is much slower than the ES-complexformation step.

4.4 Estimation of the Kinetic Parameters vmax and kM

The rate expressions given by Eqs. 4.13 and 4.22 clearly demonstrate the presence of two kinetic parameters vmax and kM. In other words, two variables have to be solved using one equation, which is impossible to solve algebraically. Consequently, we need to stick to the graphical approach. However, for a graphical approach, a series of batch runs for various concentrations of the substrate is necessary. This will help in obtaining a relation between the reaction rate (v) and the various concentrations of the initial substrate (CS). Hence, if one plots v versus CS, the maximum reaction rate (vmax) is the asymptote of v, and when kM equals CS, the reaction rate is halved compared to the maximum reaction rate. But the question arises: how to test the validity of the kinetic model? Also how can one accurately estimate the asymptotes? In order to avoid all the above questions, one can simply invert Eqs. 4.13 and 4.22. The rearrangement results in the following form: 1 kM + C S = v vmax CS

kM 1 1 = + v vmax CS vmax

(4.23) (4.24)

Equation 4.24 resembles the straight-line equation y = mx + c with a positive slope m and a positive intercept c. In Eq. 4.24, the slope m is equal to kM/vmax and the intercept c equals 1/vmax when 1/CS is plotted against 1/v. This plot is known as the Lineweaver–Burk plot. Equations 4.13 and 4.22 can also be rewritten in the following form: v = vmax - kM

v CS

(4.25)

It can be clearly deduced from Eq. 4.25 that a negative slope of kM with a positive intercept vmax can be obtained if one plots v versus

Estimation of the Kinetic Parameters vmax and kM

v/CS. This plot is known as the Eadie–Hofstee plot. Similar to the various other methods, the Eadie–Hofstee plot can be effectively used to determine the different kinetic parameters (kM and vmax) quickly. In addition, the robustness of the Eadie–Hofstee plot is found to be much higher compared to the Lineweaver–Burk plot as it provides similar weightage to the different data points for any substrate concentration range or reaction rate. Equations 4.13 and 4.22 can also be expressed in the following form: v=

or or

vmax CS kM + C S

C S kM + C S = v vmax CS C k = M + S v vmax vmax

(4.26) (4.27)

Equation 4.27 is also known as the Langmuir plot, and such plots are essential for the data associated with the gas adsorption on the solid. Equation 4.27 clearly displays that if we plot CS/v versus CS, a straight line will be obtained with a slope of 1/vmax and an intercept of kM/vmax. In general, the Lineweaver–Burk plot is frequently implemented in contrast to the Eadie–Hofstee and Langmuir plots since it demonstrates the relationship between the dependent variable v and the independent variable CS. Equation 4.26 nicely depicts that the decrease in CS makes l/v approach infinity. Hence, there is an imbalance in the weight distribution to Eq. 4.26 for both low and high concentrations of the substrate (CS). In contrast, Eq. 4.25 (Eadie–Hofstee plot) displays a relatively better balance of data on both sides. Both the Eadie–Hofstee and Lineweaver–Burk plots incorporate minute or minimal variation in the kinetic parameter values in contrast to the Langmuir plot. This results in inappropriate weight distribution for the cases involving lower concentration of the substrate for the Eadie–Hofstee and Lineweaver–Burk plots. In addition, for the cases with higher substrate concentration, the

127

128

Enzymatic Reaction Kinetics

data deviation is lesser and mostly ignored for the Eadie–Hofstee and Lineweaver–Burk plots in contrast to the Langmuir plot. Hence, in contrast to the Eadie–Hofstee and Lineweaver–Burk plots, the Langmuir plot is relatively better because the points are spaced equally. The kinetic parameters kM and vmax can be evaluated based on the following points:





∑ For the various levels of substrate concentration, we need to carry out an extensive number of batch runs involving a constant initial enzyme concentration. Based on these data, one can surely evaluate the rate of change in the final product concentration or the rate of change in the substrate concentration. ∑ The initial reaction rate can be estimated from the plots of CS (concentration of substrate) or CP (concentration of product) versus time (t) involving the various concentrations of the substrate. ∑ Thereafter, kM and vmax (kinetic parameters) are estimated by plotting any one of the three plots discussed previously (Lineweaver–Burke plot, Eadie–Hofstee plot, and Langmuir plot).

4.5 Bioreactor Modeling for Enzymatic Reaction

Biochemical transformations are caused by the action of living cells or enzymes in bioreactors. A bioreactor is also called a fermenter whether the transformation is carried out by living cells or in vivo cellular components (i.e., enzymes). However, we call a bioreactor employing enzymes an enzyme reactor to distinguish it from a bioreactor that employs living cells, which we call the fermenter.

4.5.1 Batch Bioreactor

A batch bioreactor is the simplest reactor configuration for enzymatic reactions. It consists of an agitator or mixer, which is used to uniformly mix the reactants. It is important to mention that the pH of the reactant mixture largely affects the reaction process. Hence,

Bioreactor Modeling for Enzymatic Reaction

the pH of the reactant mixture is balanced by adding a pH controller or buffer solution. An ideal batch reactor is supposed to be well and uniformly mixed so that the contents have uniform composition at all times. Let us suppose that a substrate is first charged in a batch reactor. At time t = 0, an enzyme is added to initiate the reaction and the entire reaction mechanism can be expressed as v=-

dCS vmax CS = dt kM + C S

(4.28)

By integrating Eq. 4.28, an expression can be obtained, which will relate the change in substrate concentration with time: CS

Ú

CS 0

t

Ê k + CS ˆ -Á M ˜ dCS = vmax dt Ë CS ¯ 0

kM ln

Ú

C S0 + (CS0 - CS ) = rmax t CS

(4.29) (4.30)

Note that at time t = 0, the substrate concentration CS is equal to the initial substrate concentration CS0, whereas at some time t = t, the substrate concentration is CS = CS. Equation 4.30 demonstrates the rate of change in the substrate concentration. With known values of vmax and kM, the change in CS with time in a batch reactor can be predicted from Eq. 4.30.

4.5.2 Plug Flow Bioreactor

In a plug flow bioreactor, the immobilized enzymes are packed inside a tubular reactor and the substrate is introduced from one end of the reactor such that the substrate reacts with the enzymes at each cross section of the reactor. Consequently, the product stream leaves through the other end of the reactor. Hence, based on the flow pattern, one can imagine that mixing is absent in the long tube due to the lack of a stirring device. It is important to mention that there will be radial and longitudinal variation in the properties of the flowing stream. As the structure of the reactor is tubular, the variation in the physical properties of the flow stream in the radial direction is smaller compared to that in the longitudinal direction. If a plug flow reactor is operated at steady state, the properties will

129

130

Enzymatic Reaction Kinetics

be constant with respect to time. An ideal plug flow enzyme reactor can approximate the long tube, packed-bed, and hollow fiber or multistaged reactor. Equation 4.30 is applicable for an unsteady-state plug flow reactor only if the time t in Eq. 4.30 is replaced by the space–time t (refer previous chapter for details). Hence, Eq. 4.30 can be modified as kM ln

C S0 + (CS0 - CS ) = vmaxt CS

(CS0 - CS ) vmaxt = - kM C S0 C S0 ln ln CS CS In Eq. 4.32, if

is plotted versus

(4.31)

(4.32)

(CS0 - CS ) C ln S0 CS t , C ln S0 CS

then the slope is vmax and the intercept is −kM.

4.5.3 Continuous Stirred Tank Bioreactor As already discussed in the previous chapter, a continuous stirred tank reactor/bioreactor (CSTR) is largely used for the uniform mixing of reactants. Therefore, the concentrations of the various components of the outlet stream are assumed to be the same as the concentrations of these components in the reactor. Continuous operation of the reactor is supposed to decrease the downtime and increase the productivity of the reactor. In order to reduce the overall labor costs, it is also easy to automate. Let the initial concentration of the substrate be CS0 and the final concentration be CS. If the volumetric flow rate is F and the volume of the reactor is V, then we write the material balance for the substrate as follows:

Bioreactor Modeling for Enzymatic Reaction

Input – output + formation – disappearance = Rate of accumulation FCS0 - FCS + 0 - ( -v )V = V

dCS dt

(4.33)

Equation 4.33 represents the unsteady-state equation for the continuous stirred tank fermenter. Note that −v is the rate of substrate consumption for the enzymatic reaction, whereas dCS/dt is the rate of change of the substrate concentration in the reactor. For a steady-state CSTR, CS should be constant and uniform throughout the reactor at any point of time. Hence, the rate of accumulation for such cases would be zero. Thus, Eq. 4.33 boils down to FCS0 – FCS + 0 – (–v)V = 0

vmax CS F 1 =D= = V t CSTR (CS0 - CS )(kM + CS )

(4.34)

(4.35)

In Eq. 4.35, D is the dilution rate, and it can be defined as the reciprocal or reverse of the space–time tCSTR. Equation 4.35 can be rearranged as follows: C S = - kM +

vmax CS t CSTR C S0 - C S

(4.36)

Based on Eq. 4.36, if one plots CS versus CStCSTR/(CS0 − CS), the slope will be vmax and the intercept will be −kM. The Michaelis–Menten kinetic parameters can also be estimated by running a series of steady-state CSTR runs with various volumetric flow rates. Equation 4.34 can also be modified as follows: -v =

F (C - CS ) V S0

(4.37)

Hence, we can conclude that the rate of the reaction can be calculated if the initial and final substrate concentrations, volumetric flow rate, and volume of the reactor are known. Now Eq. 4.35 can be modified and rearranged based on the concentration of the product. The general product balance may be written as Input – Output + Formation – Disappearence = Accumulation (4.38) 0 – FCP + vV – 0 = 0

vmax CS F v = = V C P C P ( kM + C S )

(4.39)

(4.40)

131

132

Enzymatic Reaction Kinetics

Hence, we can also write

D=

vmax CS 1 = t C P ( kM + C S )

(4.41)

Equation 4.41 is solvable if the final substrate and product concentrations are known.

4.6 Inhibition of Enzyme Reactions

Some substances combine with enzymes and either increase or decrease catalytic activity. A substance that decreases the catalytic activity is known as an inhibitor. An inhibitor can decrease the reaction rate competitively, noncompetitively, or partially competitively.

4.6.1 Competitive Inhibition

A competitive inhibitor has strong structural resemblance to the substrate, and the substrate and the inhibitor compete for the same active site of the enzyme. The formation of an enzyme–inhibitor complex reduces the available surface area of the enzyme for interaction with the substrate and that largely decreases the reaction rate. A competitive inhibitor normally combines reversibly with an enzyme. Therefore, the effect of the inhibitor can be minimized by increasing the substrate concentration, unless the substrate concentration is greater than the concentration at which the substrate itself inhibits the reaction. The mechanism of competitive inhibition can be expressed as follows: k

1 E + S ¨æ Æ ES k 2

k

3 E + I ¨æ Æ EI k 4

k

5 ES ææ ÆE+ P

(4.42)

(4.42a)

(4.42b)

The product-formation step is the slower reaction step; hence, it is the rate-determining step. The rate of the reaction can be expressed as vP =

dCP = k5CES dt

(4.43)

Inhibition of Enzyme Reactions

The entire enzyme balance can be given as CE0 = CE + CES + CEl

(4.44)

The two equilibrium reactions represented by Eqs. 4.42 and 4.42a can be written as CES k1 = = kS CECS k2 + k5

and

k CE I = 3 = kI C EC I k 4

(4.45) (4.46)

where kS and kI are the dissociation constants, and they are the reciprocal of the equilibrium constants. Combining Eqs. 4.43–4.46 along with Eq. 4.13 (Michaelis–Menten equation), we get vP =

Vmax CS

Ê C ˆ CS - kS Á 1 + I ˜ KI ¯ Ë

rP =

Vmax CS C S - kM I

(4.47)

(4.48)

Note that the second term in the denominator can be written as follows: Ê C ˆ kM I = kS Á 1 + I ˜ KI ¯ Ë

(4.49)

Since kMI is significantly greater than kS, the reaction rate decreases due to the presence of the inhibitor. It is interesting to note that the maximum reaction rate is not affected by the presence of a competitive inhibitor. However, a larger amount of substrate is required to reach the maximum rate.

4.6.2 Non-competitive Inhibition

In this type of inhibition process, the inhibitor and the substrate may not compete for the same active site. However, an irreversible or reversible binding of the inhibitor with the enzyme at the active site or any other region of the enzyme results in the enzyme–inhibitor

133

134

Enzymatic Reaction Kinetics

complex, which is extremely inactive. The mechanism of the noncompetitive inhibition process is described as follows: k

1 E + S ¨æ Æ ES k 2

k

3 E+ I ¨æ Æ EI k 4

k5 EI+ S ¨æ Æ EIS k6 k

7 ES+ I ¨æ Æ E IS k 8

k9

ES ææ Æ E+ P

(4.50)

Since substrates and inhibitors do not compete for the same site for the formation of enzyme–substrate or enzyme–inhibitor complex, we can assume that the dissociation constant for the first equilibrium reaction is the same as that of the third equilibrium reaction, as CECS k2 C C k = = kS = ES I = 6 = kIS CES k1 CEIS k5

and

C C k C EC I k 4 = = kI = EI S = 8 = kSI CEI k3 CEIS k7

The rate of product formation is given by v=

dCS dt

(4.51)

(4.52) (4.53)

Similar to the previous section, the total enzyme concentration is constant, and hence one can write CE0 = CE + CES + CEI + CEIS + CESI

(4.54)

Combining Eqs. 4.51–4.54 and rearranging the rate equations, one can get vP =

vI,max CS CS + kS

In Eq. 4.55, the term vI,max can be expanded as follows: vI,max =

vmax C 1+ I kI

(4.55)

(4.56)

Factors Affecting Enzymatic Reactions

Equation 4.56 clearly demonstrates that the maximum reaction rate will be decreased by the presence of a non-competitive inhibitor, while the Michaelis constant kS will not be affected by the inhibitor. Several variations of the mechanism for non-competitive inhibition are possible. One case is when the enzyme–inhibitor–substrate complex can be decomposed to produce a product and the enzyme– inhibitor complex. This mechanism can be described by adding the following slow reaction to Eq. 4.50, and it is called the partially competitive inhibition: k

10 EIS ææ Æ EI+ P

4.7 Factors Affecting Enzymatic Reactions

(4.57)

As previously discussed, enzymatic reactions are often carried out in various reactors. Various chemical and physical conditions within the reactor may affect the rate of enzymatic reactions. Concentrations of the substrate and enzymes play a major role in the enzymatic reaction process. They have been discussed in details in the preceding sections. Other than those, the most important factors are listed in the following sections:

4.7.1 Effect of pH

The pH of the reaction mixture largely affects the enzymatic reaction. It may be noted that there is an optimum pH for each enzyme. For example, stomach has the enzyme pepsin, which has an optimum pH between 2 and 3.3, while the optimum pH of amylase, from saliva, is 6.8. Chymotrypsin, from the pancreas, has an optimum pH in the mildly alkaline region between 7 and 8. Enzymes are protein consisting of amino acid residues. The pH of such solutions may be acidic, basic, or neutral based on the negative or positive charge. For example, glutamic acid is acidic at a lower pH, and the increase in pH ionizes glutamic acid. On the other hand, an amino acid lysine is basic in the range of higher pH value. As the pH is decreased, lysine is ionized. An enzyme is catalytically active when each of the amino acid residues at the active site possesses a particular charge. Therefore, the fraction of the catalytically active enzyme depends on the pH.

135

136

Enzymatic Reaction Kinetics

4.7.2 Effect of Temperature The effect of temperature on the enzymatic reaction can be explained via the Arrhenius equation as follows: k = Ae - E A

RT

(4.58)

where k is the rate constant, A is the frequency factor, EA is the activation energy, and R is the gas constant. Taking log on both sides of Eq. 4.58, we get ln k = ln A -

EA RT

(4.59)

Based on Eq. 4.59, if ln k is plotted against 1/T, a straight line is obtained with slope −EA/R and intercept ln A. Note that the reaction rate increases with the increase in temperature since the atoms get excited and move randomly at a greater rate. However, the rise in temperature beyond a certain limit may lead to the denaturation process, which further deactivates enzyme molecules. This may be attributed to protein unfolding after the breakage of weak bonds. For many proteins, denaturation begins to occur at 45–50°C. Some enzymes are very resistant to denaturation by high temperature, especially the enzymes isolated from thermophilic organisms found in certain hot environments.

4.7.3 Effect of Shear

Shearing has tremendous effect on the stability of enzymes. The structure of an enzyme is extremely vulnerable to mechanical agitation. Note that the shape of an enzyme molecule changes, resulting in significant denaturation. The mechanical force that an enzyme solution normally encounters is fluid shear, generated either by flowing fluid, the shaking of a vessel, or by stirring with an agitator. However, the contents of the reactor need to be agitated or shook to minimize mass-transfer resistance.

Examples

PROBLEM 4.1: Urease catalyzes hydrolysis of urea. At T = 21°C, the activation energy of an uncatalyzed reaction is 3.98 kJ/mol. In the

Factors Affecting Enzymatic Reactions

presence of urease, the activation energy is 1.54 kJ/mol. By what factor does urease increases the velocity of this reaction at the same temperature? SOLUTION The Arrhenius equation is as follows:

k = Ae - Ea /RT

Let k1 be the rate constant for the uncatalyzed reaction and k2 the rate constant for the catalyzed reaction. Then, k2 Ae - Ea 2/RT = k1 Ae - Ea 1/RT

k2 = e( Ea 1 - Ea 2 )/RT k1 Given Ea1 = 3.98 kJ/mol = 3980 J/mol and Ea2 = 1.54 kJ/mol = 1540 J/mol. So (3980 J/mol - 1540 J/mol ) k2 =e k1

( 8.314

J )(294 K ) mol K

k2 = 2.71 k1

or

k2 = 2.71k1

or

PROBLEM 4.2: Consider an industrially important enzyme, which catalyzes the conversion of a protein substrate to form a much more valuable product. The enzyme follows the Briggs–Haldane mechanism: k

k

3 1 E + S ¨æ Æ ES ;  ES ææ ÆE + P k 2

At the initial rate, analysis for the reaction in solution, with E0 = 0.10 μM and substrate concentrations S0, yields the following Michaelis– Menten parameters: vmax = 0.60 μM/s and kM = 80 μM. A different type of experiment indicates that the dissociation rate constant k1 = 2.0 × 10−6 M−1s−1 (2.0 μM−1s−1). Find out the values of k2 and k3.

137

138

Enzymatic Reaction Kinetics

SOLUTION We know that Therefore, Again,

k3 = kM =

vmax = k3E0

0.6 mM/s = 6 s-1 0.1 mM

k2 + k3 k +6 = 2 = 80 2 k1 k2 = 156 μM

PROBLEM 4.3: The enzyme-catalyzed conversion of a substrate at 25°C has kM = 0.035 M. The rate of the reaction is 1.15 × 10−3 M/s when the substrate concentration is 0.110 M. What is the maximum velocity of this reaction? SOLUTION Given v = 1.15 × 10−3 M/s; kM = 0.035 M and [S] = 0.110 M The Michaelis–Menten equation may be written as v=

vmax [S] kM + [S]

Therefore, vmax =

v (kM + [S]) 1.15 ¥ 10-3 (0.035 + 0.110) = 1.52 × 10−3 M/s = [S] [0.110]

PROBLEM 4.4: The kM value of lysozyme is 6.0 × 10−6 M with hexaN-acetyl glucosamine as the substrate. The initial rate measured at 0.061 M substrate concentration was 3.2 mM/min. What would be the rate at a substrate concentration of 6.8 × 10−5 M? SOLUTION

Given kM = 6 × 10−6 M; v = 3.2 mM/min and [S] = 0.061 M

Now,

v = 3.2 mM/min ×

1 min 1M × = 5.33 × 10−5 M/s 60 s 1000 m M v=

vmax [S] kM + [S]

Factors Affecting Enzymatic Reactions

Therefore,

vmax =

v (kM + [S]) 5.33 ¥ 10-5 (6 ¥ 10-6 + 0.061) = [S] [0.061]

Now at [S] = 6.8 × 10−5 M, v=

= 5.33 × 10−5 M/s

5.33 ¥ 10-5[6.8 ¥ 10-5 ] 6 ¥ 10-6 + [6.8 ¥ 10-5 ]

= 4.8 × 10−5 M/s

PROBLEM 4.5: An enzyme follows M–M kinetics with vmax = 2.5 mM/m3s and kM = 5 mM. It was used to carry out the reaction in a batch stirred tank reactor. Starting with an initial substrate concentration of 0.1 M, what is the time required for 50% conversion of the substrate? SOLUTION Given Now,

S0 = 0.1 M, vmax = 2.5 mM/m3s, kM = 5 mM S = 0.5 (S0) = 0.05 M

vmax = 2.5 mM/m3s

= 2.5 mM/m3s ×

1M 1000 m M

= 0.0025 M/m3s = 0.0025 = 2.5 × 10−6 M/L s kM = 5 mM ×

For a batch reactor, =

M m3 s

s−1 ×

1 m3 1000 L

1M = 0.005 M 1000 m M

tbatch = 1/vmax [kM ln (S0/S) + (S0 – S)]

1 0.1 È ˘ Í0.005 ln 0.05 + (0.1 - 0.05)˙ = 6 h 2.5 ¥ 10-6 Î ˚

PROBLEM 4.6: An enzyme with a kM of 1 × 10−3 M was assayed using an initial substrate concentration of 3 × 10−5 M. After 2 min, 5% of the substrate was converted. How much substrate will be converted after 10, 30, and 60 min?

139

140

Enzymatic Reaction Kinetics

SOLUTION Given data:

S0 = 3 × 10−5 M; kM = 1 × 10−3 M

There is 5% conversion in 2 min. Therefore, S = 0.95 S0. It is known that ÊS ˆ vmax t = S0 - S + kM ln Á 0 ˜ Ë S¯

Putting all the known values in the above equation, we get vmax = 2.64 × 10−5 M/min

Case I: For t = 10 min,

2.64 ¥ 10-5 ¥ 10 = 3 ¥ 10-5 (1 - x ) + 1 ¥ 10-3 ln

or

3 ¥ 10-5 x ¥ 3 ¥ 10-5

Ê 1ˆ 26.4 ¥ 10-5 = 3 ¥ 10-5 (1 - x ) + 10-3 ln Á ˜ Ë x¯

By trial-and-error method, it has been found that when x = 0.77, LHS = RHS. Therefore, conversion will be 23%. Cases II and III: Similarly, percentage conversions after 30 and 60 min are 54% and 80%, respectively.

PROBLEM 4.7: The data in Table 4.1 have been obtained for two different initial enzyme concentrations for an enzyme-catalyzed reaction.

Table 4.1 Data of enzyme-catalyzed reactions v ([E0] = 0.015 g/L) (g/L min)

[S] (g/L)

v ([E0] = 0.00875 g/L) (g/L min)

1.14

20.0

0.67

0.59

5.0

0.34

0.87 0.70 0.50 0.44 0.39 0.35

10.0 6.7 4.0 3.3 2.9 2.5

0.51

0.41 0.29

Factors Affecting Enzymatic Reactions

(a) Find kM. (b) Find vmax for [E0] = 0.015 g/L (c) Find vmax for [E0] = 0.00875 g/L

SOLUTION The Lineweaver–Burk plot can be used to determine vmax and kM. 1 kM 1 1 = + v vmax [S ] vmax

Table 4.2 Calculated Lineweaver–Burk plot data 1 (L/g) ÎÈ S ˘˚

1 (E0 = 0.015) v

1 (E0 = 0.00875) v

0.05

0.877

1.50

0.20

1.70

2.94

0.10 0.15 0.25 0.30 0.34 0.40

1.15

1.96

1.43

2.44

2.00

3.45

2.27 2.56 2.86

From the Lineweaver–Burk plot for [E0] = 0.015 g/L, we have Intercept =

1 = 0.58 vmax

vmax = 1.72

Slope =

g L min

kM = 5.7 vmax

kM = vmax ¥ 5.7 = 1.72 ¥ 5.7 = 9.8

g L

From the Lineweaver–Burk plot (Fig. 4.3) for [E0] = 0.00875 g/L, we have Intercept =

1

vmax

= 0.99

141

142

Enzymatic Reaction Kinetics

vmax = 1.01

g L min

Slope =

1/v

Km vmax

1 vmax

1/[S]

–1/Km

Figure 4.3 The Lineweaver–Burk plot for enzyme-catalyzed reactions.

PROBLEM 4.8: Find out the volume required to produce 1 kg of product per day by using the enzyme X (vmax = 1.5 × 10−4 kg/s m3 biocatalyst; kM = 5 × 10−3 kg/m3; YP/S = 1 kg/kg; degree of conversion = 99%; initial concentration of substrate is 5 kg/m3; downtime = 12 h).

SOLUTION The M–M reaction kinetics are given as vmax = 1.5 × 10−4 kg/s m3 biocatalyst; kM = 5 × 10−3 kg/m3; YP/S = 1 kg/kg; degree of conversion = 99%; downtime = 12 h; initial concentration of substrate 5 kg/m3. The activity of the biocatalyst can be assumed to be constant with time. tb =

1 { -kM ln(1 - X s ) + S0 X s } (since S = S0 (1 − Xs)) vmax

Putting all the values from the given data, we get tb =

1

1.5 ¥ 10-4

{-5 ¥ 10

-3

}

ln(1 - 0.99) + 5 (0.99) = 33153.5 s

Therefore, tb = 9.2 h

tb(total) = tb + tdowntime = 9.2 + 12 = 21.2 h

Thus, the number of batch per day is 24/21.2 = 1.13. Basis: 1 kg of product per day

Factors Affecting Enzymatic Reactions

For 1 kg product per day, product to be provided per batch is 1/1.13 = 0.885 kg. As YP/S = 1 kg/kg, substrate required per batch is 0.885 kg. For Xs of 0.99, actual substrate required per batch is 0.885/0.99 = 0.893 kg. Volume of the reactor (batch) required is Actual substrate required Initial substrate concentration Vbatch =

0.893 kg 3

5 kg/m

= 0.176 m3

PROBLEM 4.9 Glucose is to be converted to fructose by using glucose isomerase enzyme. Initial glucose concentration in the reaction mixture is 200 g/L and the degree of conversion is 60% w/w. Calculate the volume needed to produce 10 kg of fructose per day by glucose isomerase in a i. Batch reactor ii. CSTR iii. Plug flow reactor

Data:

∑ Michaelis–Menten reaction kinetics with vmax = 1.5 × 10–2 kg/s m3 biocatalyst and kM = 5 kg/m3, YP/S = 1 kg/kg ∑ Downtime for batch reactor is 4 h. ∑ The activity of the biocatalyst can be assume to be constant in time.

SOLUTION Given stoichiometry: glucose Æ fructose (P) (S) From the given data, we have the initial substrate concentration g kg S0 = 200 = 200 3 L m Degree of conversion is XS = 0.60

vmax = 1.5 × 10–2 kg/s m3

143

144

Enzymatic Reaction Kinetics

kM = 5 kg/m3

YP/S = 1 kg/kg td = 4 h

Downtime,

i. Batch reactor It is known that

S

t batch = -

dS

Ú ( -r )

S0

S

as it follows the M–M kinetics ( -rS ) =

vmax S kM + S

Integration yields t batch =

1 È-kM ln(1 - X S ) + X S S0 ˘˚ vmax Î

Putting all the known values, we get

[-5ln(1 - 0.60) + 0.60 ¥ 200] 1.5 ¥ 10-2 = 8305.43 s = 2.307 h ttotal = tbatch + tdowntime

t batch =

1

ttotal = 2.307 + 4 = 6.307 h

Now, number of batches per day is 24/6.307 = 3.80. Base: 10 kg of fructose (P) per day Product to be produced per batch is 10/3.80 = 2.63 kg. As YP/S = 1 kg/kg, the substrate required per batch is 2.63 kg. For Xs = 0.60, the actual substrate required per batch is 2.63/0.60 = 4.38 kg. The required volume of the reactor is

Actual substrate required 4.38 kg = = 0.022 m3 = 22 L kg Initial substrate concentration 200 3 m ii. CSTR t CSTR =

S0 - S ( S0 - S )(kM + S ) S0 X S {kM + S0 (1 - X S )} = = ( -rS ) (vmax S ) vmax S0 (1 - X S )

Factors Affecting Enzymatic Reactions

t CSTR =

0.60 ¥ {5 + 200(1 - 0.60)} 1.5 ¥ 10-2 ¥ (1 - 0.60)

tCSTR = 8500 s = 2.36 h

Base: 10 kg of fructose (P) per day For YP/S = 1, the substrate required per day is 10 kg per day. As Xs = 0.60, the actual substrate required per day is 10/0.60 = 16.67 kg per day. The actual substrate required per hour is 16.67/24 = 0.70 kg/h. Volumetric flow rate is kg 3 Actual substrate required h = 0.70 m F= Initial substrate concentration kgg 200 h m3 = 3.5 ¥ 10-3

m3 h

Now we know that

t CSTR =

VCSTR F

VCSTR = F ¥ t CSTR = 3.5 ¥ 10-3 ¥ 2.36 m3 = 8.26 ¥ 10-3 m3 = 8.26 L iii. Plug flow reactor (PFR) It is known that

S

t PFR = -

dS

Ú ( -r )

S0

S

As it follows the M–M kinetics, -rS =

Integration yields t PFR =

1

vmax S kM + S

È-k ln(1 - X S ) + X S S0 ˘˚ vmax Î M

145

146

Enzymatic Reaction Kinetics

Putting all the known values gives

[-5ln(1 - 0.60) + 0.60 ¥ 200] 1.5 ¥ 10-2 = 8305.43 s = 2.307 h

t PFR =

1

Base: 10 kg of fructose (P) per day For YP/S = 1, the substrate required per day is 10 kg per day. As Xs = 0.60, the actual substrate required per day is 10/0.60 = 16.67 kg per day. The actual substrate required per hour is 16.67/24 = 0.70 kg/h. Volumetric flow rate is kg 3 Actual substrate required h = 0.70 m F= Initial substrate concentration kgg 200 h m3 = 3.5 ¥ 10-3 Now we know that

t PFR =

m3 h

VPFR F

VPFR = F ¥ t PFR = 3.5 ¥ 10-3 ¥ 2.307 m3 = 8.07 ¥ 10-3 m3 = 8.07 L Table 4.3

Volume (L)

Volume of the reactor in different processes Batch reactor

CSTR

PFR

22

8.26

8.07

PROBLEM 4.10: At room temperature, sucrose is hydrolyzed by the catalytic action of the enzyme invertase as follows: Sucrose ææææ Æ Fuctrose + Glucose Invertase

Starting with a sucrose concentration CA0 = 1.0 mmol/L and enzyme concentration CE0 = 0.01 mmol/L, the following kinetics data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):

Factors Affecting Enzymatic Reactions

Table 4.4

Concentration of sugar at different time in a batch process

CA 0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 (mmol/L) t (h)

1

2

3

4

5

6

7

8

9

10

Determine whether these data can be reasonably fitted by a kinetic equation of the Michaelis–Menten type, or -rA =

k2C A CE0 C A + kM

where kM is the Michaelis–Menten constant. If the fit is reasonable, evaluate the constants k2 and kM. Solve by the integral method. SOLUTION Given data:

CA0 = 1.0 mmol/L

CE0 = 0.01 mmol/L

The Michaelis–Menten equation is -rA = -

dC A k2C A CE0 = dt C A + kM

(C A + kM ) dC A = -k2CE0dt CA CA

Ú

CA0

t

(C A + kM ) dC A = - k2CE0dt CA

Ú 0

ÊC ˆ (C A0 - C A ) + kM ln Á A0 ˜ = k2CE0t Ë CA ¯

k t = M (C A0 - C A ) k2CE0

ÊC ˆ ln Á A0 ˜ Ë CA ¯ 1 + (C A0 - C A ) k2CE0

y = mx + c

147

148

Enzymatic Reaction Kinetics

Table 4.5

Calculated data

t CA C A0 (h) (mmol/L) C A

Ê C ˆ (CA0 – CA), ln Á A0 ˜ (mmol/L) Ë CA ¯

t ÊC ˆ ln Á A0 ˜ Ë C A ¯ (C A0 - C A ) (C A0 - C A ) (h L/ mmol) (L/mmol)

1

0.84

1.190476 0.174353 0.16

1.089709

6.25

4

0.38

2.631579 0.967584 0.62

1.560619

6.451613

2 3 5 6 7 8 9

10 11

0.68 0.53 0.27 0.16 0.09 0.04

0.018 0.006

0.0025

1.470588 0.385662 0.32 1.886792 0.634878 0.47 3.703704 1.309333 0.73 6.25

1.832581 0.84

11.11111 2.407946 0.91 25

3.218876 0.96

55.55556 4.017384 0.982 166.6667 5.115996 0.994 400

5.991465 0.9975

From the plot (Fig. 4.4) of

we have

1.205195 1.350805 1.793607

2.181645

2.646094

3.352996

4.091022 5.146877

6.006481

ÊC ˆ ln Á A0 ˜ Ë CA ¯ t versus (C A0 - C A ) (C A0 - C A ) Slope =

Intercept =

By solving, we get

kM = 0.9879 h k2CE0

1 = 5.0497 h L/mmol k2CE0 mmol L k2 = 19.80 h–1

kM = 0.195

Therefore, the rate equation can be written as -rA =

19.80C A CE0 C A + 0.195

6.25

6.382979 6.849315

7.142857

7.692308

8.333333

9.164969 10.06036

11.02757

Factors Affecting Enzymatic Reactions

12 y = 0.987x + 5.049 R2 = 0.998

t/(CA0–CA)

10 8 6 4 2 0

0

1

2

3

4

5

6

7

In (CA0/CA)/(CA0–CA)

Figure 4.4 Plot of t/(CA0 – CA) versus ln (CA0/CA)/CA0 – CA).

PROBLEM 4.11: Enzyme E catalyzes the transformation of reaction A to product R as follows: Enzyme E

A ææææÆ R , -rA =

200 C A CE0 mol 2 + C A L min

In the case of introducing the enzyme (CE0 = 0.001 mol/L) and the reactant (CA0 = 10 mol/L) into a batch reactor and letting the reaction proceed, find the time required for the reactant concentration to drop to 0.025 mol/L. Assume the enzyme concentration remains unchanged during the reaction. SOLUTION Given data:

CE0 = 0.001 mol/L, CA0 = 10 mol/L, CA = 0.025 mol/L

The rate equation of the following enzymatic reaction can be expressed as -rA = -

dC A 200C A CE0 = dt CA + 2

Integrating the above (for a batch reactor) gives

ÊC ˆ (C A0 - C A ) + 2 ln Á A0 ˜ = 200CE0t batch Ë CA ¯ ÊC ˆ 2 ln Á A0 ˜ Ë CA ¯ (C A0 - C A ) = t batch + 200CE0 200CE0

149

150

Enzymatic Reaction Kinetics

t batch

Ê 2 ˆ 2 ln Á Ë 0.025 ˜¯ (10 - 0.025) = + = 93.69 min 200 ¥ 0.001 200 ¥ 0.001

PROBLEM 4.12: An enzyme acts as a catalyst in the fermentation of reactant A. At a given enzyme concentration in the aqueous feed stream (25 L/min), find the volume of the plug flow reactor needed for 95% conversion of reactant A (CA0 = 2 mol/L). The kinetics of the fermentation at this enzyme concentration is given by A ææææ Æ B , -rA = Enzyme

SOLUTION Given data:

0.1 C A mol 1 + 0.5C A L min

CA0 = 2 mol/L

Volumetric feed flow rate is F = 25 L/min XA = 0.95

CA = CA0 (1 − XA) = 2(1 − 0.95) = 0.1 mol/L

The rate equation is

-rA =

0.1C A 0.2C A = 1 + 0.5C A 2 + C A

Integrating for a plug flow reactor gives

ÊC ˆ (C A0 - C A ) + 2 ln Á A0 ˜ = 0.2t PFR Ë CA ¯ Ê 2 ˆ (2 - 0.1) + 2 ln Á = 0.2t PFR Ë 0.1 ˜¯

Again,

tPFR = 39.46 min t PFR =

VPFR F

VPFR = t PFR ¥ F = 39.46 min ¥ 25

L = 986.43 L min

Factors Affecting Enzymatic Reactions

PROBLEM 4.13: Enzyme E catalyzes the fermentation of a substrate A to produce B. Find the size of the CSTR needed for 95% conversion of the reactant in a feed stream (25 L/min) of the reactant (2 mol/L) and the enzyme. The kinetics of the fermentation at this enzyme concentration are given by A ææææ Æ B , -rA = Enzyme

SOLUTION Given data:

0.1 C A mol 1 + 0.5 C A L min

CA0 = 2 mol/L

Volumetric feed flow rate, F = 25 L/min XA = 0.95

CA = CA0 (1 − XA) = 2 (1 − 0.95) = 0.1 mol/L

The rate equation is

-rA =

0.1C A 0.2C A = 1 + 0.5C A 2 + C A

For a mixed flow reactor or CSTR,

VCSTR C A0 - C A 2 - 0.1 = = = 199.5 min 0.2 ¥ 0.1 F ( -rA ) 2 + 0.1

VCSTR = 199.5 ¥ F = 199.5 ¥ 25 L = 4987.5 L ª 5 m3 PROBLEM 4.14: Substrate A and enzyme E flow through a CSTR (V = 6 L). From the entering and leaving concentrations and flow rate (Table 4.6), find a rate equation to represent the enzymatic reaction. Table 4.6

Data of enzyme-catalyzed reactions in a CSTR

CE0 (mol/L)

CA0 (mol/L)

CA (mol/L)

F (L/h)

0.02

0.2

0.04

3.0

0.01

0.001

SOLUTION

0.3

0.15

0.69

0.60 k

k

1 2   E + S ÆE + P   ES ææ

k-1

4.0

1.2

151

152

Enzymatic Reaction Kinetics

We know that Therefore,

t CSTR =

C A0 - C A VM FR = ( -rA ) F

-rA =

(C A0 - C A )F VCSTR

Again from the Michaelis–Menten rate equation -rA =

k2CE0C A C A + kM

By rearranging the above equation, we get CE0 1 kM 1 = + ( -rA ) k2 k2 C A

CE0 k 1 1 = M + ( -rA ) k2 C A k2

Table 4.7

y = mx + c

Calculated data of enzyme-catalyzed reactions

- rA =

(C A0 - C A )F (mol/L h) VCSTR

0.080

0.100

C E0 (h) ( - rA )

1 (L/mol) CA

0.250

25.00

0.100

0.018

0.055

6.67

1.67

The plot of CE0/(–rA) versus 1/CA gives the slope of kM/k2 and intercept of 1/k2. From Fig. 4.5, we have 1 = 0.0426 h k2

kM h mol = 0.0083 k2 L By solving, we get k2 = 23.47 h–1.

Factors Affecting Enzymatic Reactions

mol L Therefore, the rate equation can be written as kM = 0.195

-rA =

23.47CE0C A C A + 0.195

0.3

CE0 /(-rA)/(h)

0.25

y = 0.008x + 0.042 R2 = 0.999

0.2 0.15 0.1 0.05 0

0

5

10

15

20

30

25

1/CA (L/mol)

Figure 4.5 Plot of CE0/(–rA) versus 1/CA.

PROBLEM 4.15: In a number of separate runs, different concentrations of substrate and enzyme are introduced in a batch reactor and allowed to react. After a certain time, the reaction is quenched and the vessel contents are analyzed. From the results found below, find a rate equation to represent the action of the enzyme on the substrate. Table 4.8

Data of the enzyme-catalyzed reactions in different runs

Run

CE0 (mol/m3)

CA0 (mol/m3)

CA (mol/m3)

1

3

400

10

2

3

2

1

200 20

SOLUTION The Michaelis–Menten equation is

5

1

t (h) 1

1

1

153

Enzymatic Reaction Kinetics

-rA = -

dC A k2C A CE0 = dt C A + kM

(C A + kM ) dC A = -k2CE0dt CA CA

Ú

CA0

t

(C A + kM ) dC A = - k2CE0dt CA

Ú 0

ÊC ˆ (C A0 - C A ) + kM ln Á A0 ˜ = k2CE0t Ë CA ¯ ÊC ˆ ln Á A0 ˜ Ë CA ¯ tCE0 k 1 + = M (C A0 - C A ) k2 (C A0 - C A ) k2 y = mx + c

0.06 0.05 tCE0/(CA0-CA) (h)

154

y = 0.304x + 0.004 R2 = 1

0.04 0.03 0.02 0.01 0

0

0.05

0.1

0.15

In(CA0/CA)/(CA0-CA) (m3/mol)

ÊC ˆ ln Á A 0 ˜ Ë CA ¯ tCE0 . versus Figure 4.6 Plot of (C A 0 - C A ) (C A 0 - C A )

From the plot of

ÊC ˆ ln Á A0 ˜ Ë CA ¯ tCE0 versus (C A0 - C A ) (C A0 - C A )

0.2

1

20

200

1

2

400

3

1

1

CA0 (mol/m3)

Calculated data

t CE0 (h) (mol/m3)

Table 4.9

1

5

10

CA (mol/m3)

20

40

40

C A0 CA

2.995732

3.688879

3.688879

ÊC ˆ ln Á A0 ˜ Ë CA ¯

19

195

390

(CA0 –CA) (mol/ m3)

1

2

3

tCE0 (h mol/m3)

0.15767

0.018917

0.009459

(m3/mol)

ÊC ˆ ln Á A0 ˜ Ë CA ¯ (C A0 - C A )

0.052632

0.010256

0.007692

tC E0 (h) (C A0 - C A )

Factors Affecting Enzymatic Reactions 155

156

Enzymatic Reaction Kinetics

we get

Slope =

kM = 0.3042 h mol/m3 k2

Intercept =

By solving kM = 64.72

mol

1 = 0.0047 h k2

, we get m3 k2 = 212.76 h–1

Therefore, the rate equation can be written as -rA =

212.76C A CE0 C A + 64.72

PROBLEM 4.16: Carbohydrate A decomposes in the presence of enzyme E. We also suspect that carbohydrate B, in some way, influences this decomposition. To study this phenomenon, various concentrations of A, B, and E flow in and out of a CSTR (V = 240 cm3). a. From the following data, find a rate equation for the decomposition. b. What can you say about the role of B in the decomposition? c. Can you suggest a mechanism for this reaction?

Table 4.10 Data of enzyme-catalyzed reaction in a CSTR CA0 (mol/m3)

CA (mol/m3)

CB0 (mol/m3)

CE0 (mol/m3)

v (cm3/min)

200

50

0

12.5

80

700

33.3

33.3

33.3

24

900

1200 200 900

SOLUTION Given data:

300

0

80

33.3

800 500

0

33.3

5 5

10

20

volume of CSTR = 240 cm3

24 48 80

120

Factors Affecting Enzymatic Reactions

In the absence of B (CB = CB0 = 0) CA0 (mol/m3)

CA (mol/m3)

CE0 (mol/m3)

F (cm3/min)

200

50

12.5

80

1200

800

5

48

900

300

5

In the presence of B (CB = CB0 = 33.3 mol/m3)

24

CA0 (mol/m3)

CA (mol/m3)

CE0 (mol/m3)

F (cm3/min)

700

33.3

33.3

24

200

900

80

10

500

(a) For a CSTR, we know that t CSTR =

Therefore,

20

80

120

C A0 - C A VCSTR = ( -rA ) F

-rA =

(C A0 - C A )F VCSTR

In the absence of B, from the Michaelis–Menten rate equation -rA =

k2CE0C A C A + kM

By rearranging the above equation, we get CE0 1 kM 1 = + ( -rA ) k2 k2 C A

CE0 k 1 1 = M + ( -rA ) k2 C A k2 y = mx + c

157

Enzymatic Reaction Kinetics

Table 4.11 Calculated data C E0 ( - rA )

1 CA

(mol/m3 min)

(h)

(m3/mol)

50

0.25

0.020

- rA =

(C A0 - C A )F VMFR

60

0.0833

80

0.00333

0.062

0.00125

Plot of CE0/(–rA) versus 1/CA results in the slope of kM/k2 and the intercept of 1/k2. 0.3

y = 10.01x + 0.049 R2 = 1

0.25 CE0/(-rA) (h)

158

0.2 0.15 0.1 0.05 0

0

0.005

0.01

0.015

1/CA (m3/mol)

Figure 4.7 Plot of

CE0 1 versus . (- rA ) CA

From the graph, we have

1 = 0.0497 h k2

and

By solving

kM h mol = 10.016 k2 m3 k2 = 20.12 h–1

kM = 201.53

mol m3

0.02

0.025

Factors Affecting Enzymatic Reactions

Therefore, the rate equation can be written as -rA =

20.12CE0C A C A + 201.53

In the presence of B B will inhibit the enzymatic reaction. We can write the Michaelis–Menten rate equation as -rA =

k2¢CE0C A C A + kM¢

By rearranging the above equation, we get CE0 1 kM¢ 1 = + ( -rA ) k2¢ k2¢ C A

CE0 k¢ 1 1 = M + ( -rA ) k2¢ C A k2¢

Table 4.12 Calculated data - rA =

(C A0 - C A ) F VMFR

y = mx + c

(mol/m3 min)

66.67 40

200

C E0 (h) ( - rA )

1 (m3/mol) CA

0.49

0.03

0.25 0.1

0.0125 0.002

Plot of CE0/(–rA) versus 1/CA results in the slope of kM¢ k2¢ and the intercept of 1 k2¢ .

From the graph,

1 = 0.0737 h k2¢ kM¢ h mol = 13.907 k2¢ m3

By solving,

k2¢ = 13.56 h -1

159

Enzymatic Reaction Kinetics

kM¢ = 188.70

mol

m3 Therefore, the rate equation can be written as -rA =

13.56CE0C A C A + 188.70

0.6 y = 13.90x + 0.073 R2 = 0.999

0.5 CE0/(-rA) (h)

160

No inhibition

0.4 0.3

With inhibition

0.2 y = 10.01x + 0.049 R2 = 1

0.1 0

0

0.01 1/CA

Figure 4.8 Plot of

0.02

0.03

0.04

Linear (No inhibition) Linear (With inhibition)

(m3/mol)

CE0 1 versus . (- rA ) CA

(b) Both kM and k2 are decreased in the presence of B. Thus, B reduces the rate of reaction although a lower concentration of A is favorable for the reaction. (c) Mechanism: Since both kM and k2 are decreased in the presence of B, it is uncompetitive inhibition. Inhibitors do not bind to the free enzyme but affect enzyme reaction by binding to the EA complex at a location away from the active side. B binds to the EA complex only; increasing [A] favors the inhibition.

Factors Affecting Enzymatic Reactions

PROBLEM 4.17: Enzyme E catalyzes the decomposition of substrate A. To see whether substance B acts as an inhibitor, we make two kinetic runs in a batch reactor, one with B present and the other without B from the data recorded below, a. Find a rate equation to represent the decomposition of A. b. What is the role of B in this decomposition? c. Suggest a mechanism for the reaction.

Run 1: CA0 = 600 mol/m3, CE0 = 8 g/m3, B not present CA

350

160

t (h) 1

40

2

10

3

4

Run 2: CA0 = 800 mol/m3, CE0 = 8 g/m3, CB = CB0 = 100 mol/m3 CA

t (h)

SOLUTION Given data

560

340

1

180

2

80

3

4

30 5

Run 1: CA0 = 600 mol/m3, CE0 = 8 mol/m3, CB = CB0 = 0 CA (mol/m3) 350 t (h)

160

1

40

2

3

10 4

Run 2: CA0 = 800 mol/m3, CE0 = 8 mol/m3, CB = CB0 = 100 mol/m3 CA (mol/m3) t (h)

560 1

The Michaelis–Menten equation is -rA = -

340 2

180 3

dC A k2C A CE0 = dt C A + kM

(C A + kM ) dC A = -k2CE0dt CA

80 4

30 5

161

162

Enzymatic Reaction Kinetics

CA

Ú

CA0

t

(C A + kM ) dC A = - k2CE0dt CA

Ú 0

ÊC ˆ (C A0 - C A ) + kM ln Á A0 ˜ = k2CE0t Ë CA ¯ k t = M (C A0 - C A ) k2CE0

ÊC ˆ ln Á A0 ˜ Ë CA ¯ 1 + (C A0 - C A ) k2CE0

y = mx + c

(a) Run 1 (in the absence of B)

Table 4.13 Calculated data

CA t (mol/ (h) m3)

C A0 CA

1

1.714286 0.538997 250

2

3

4

350

160

40

10

3.75

15

60

ÊC ˆ ln Á A0 ˜ Ë CA ¯

(CA0 – CA) (mol/m3)

1.321756 440

2.70805

560

4.094345 590

ÊC ˆ ln Á A0 ˜ Ë CA ¯ (C A0 - C A )

t (C A0 - C A )

(m3/mol) (h m3/mol) 0.002156

0.003004

0.004836

0.00694

0.004

0.004545

0.005357

0.00678

ÊC ˆ ln Á A0 ˜ t Ë CA ¯ results in the slope of versus Plot of (C A0 - C A ) (C A0 - C A ) kM 1 and the intercept of . k2CE0 k2CE0

From the graph, we have

1 h m3 = 0.0028 k2CE0 mol kM = 0.5674 h k2CE0

Factors Affecting Enzymatic Reactions

t/(CA0-CA)(h.m3/mol)

0.008 0.007

y = 0.567x + 0.002 R2 = 0.992

0.006 0.005 0.004 0.003 0.002 0.001 0

0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008

0

In(CA0/CA)/(CA0-CA) (m3/mol)

ÊC ˆ ln Á A 0 ˜ Ë CA ¯ t . versus Figure 4.9 Plot of (C A 0 - C A ) (C A 0 - C A )

By solving,

k2 = 44.63 h–1

mol L Therefore, the rate equation can be written as kM = 202.64

-rA =

44.63CE0C A 357.04C A = C A + 202.64 C A + 202.64

Run 2 (in the presence of B) Table 4.14 Calculated data

CA (mol/ t (h) m3) 1

2

3

4

5

560

340

180

80

30

C A0 CA

ÊC ˆ ln Á A0 ˜ Ë CA ¯ (C A0 - C A )

ÊC ˆ ln Á A0 ˜ (CA0 – CA) Ë CA ¯ (mol/ m3) (m3/mol)

1.428571 0.356675 240

2.352941 0.855666 460

4.444444 1.491655 620

10

2.302585 720

26.66667 3.283414 770

0.001486

0.00186

0.002406

0.003198

0.004264

t (C A0 - C A ) (h m3/mol) 0.004167

0.004348

0.004839

0.005556

0.006494

163

Enzymatic Reaction Kinetics

ÊC ˆ ln Á A0 ˜ Ë CA ¯ t versus Plot of results in the slope of (C A0 - C A ) (C A0 - C A ) kM 1 and the intercept of . k2CE0 k2CE0 0.008 y = 0.858x + 0.002 R2 = 0.996

0.007 t/(CA0-CA)(h.m3/mol)

164

No inhibition

0.006

With inhibition

0.005 0.004

y = 0.567x + 0.002 R2 = 0.992

0.003 0.002 0.001 0

0

0.002

0.004

0.006

0.008

In(CA0/CA)/(CA0-CA) (m3/mol)

ÊC ˆ ln Á A 0 ˜ Ë CA ¯ t . versus Figure 4.10 Plot of (C A 0 - C A ) (C A 0 - C A )

From the graph, we have

1 h m3 = 0.0028 k2CE0 mol kM = 0.8582 h k2CE0

By solving,

k2 = 44.63 h–1

kM = 306.5

mol L

Factors Affecting Enzymatic Reactions

Therefore, the rate equation can be written as -rA =

44.63CE0C A 357.04C A = C A + 306.5 C A + 306.5

(b) In this case, kM increases and k2 remains same. B inhibits the enzymatic reaction. The type of inhibition is competitive inhibition. B has molecular similarity with A. (c) Mechanism: B binds to free E only and competes with A. Increasing A can overcome inhibition.

PROBLEM 4.18: Cellulose can be converted to sugar by the following enzymatic attack: Cellulose ææææÆ Sugar Cellulases

and both cellobiose and glucose act to inhibit the breakdown. To study the kinetics of this reaction, a number of runs are made in a CSTR kept at 50°C and using a feed of finely shredded cellulose (CA0 = 25 kg/m3), enzyme (CE0 = 0.01 kg/m3, same for all runs), and various inhibitors. The results are given in Table 4.15.

Table 4.15 Data of enzyme-catalyzed reactions

Run

Exit stream CA (kg/m3)

Series 2 with Series 1 no cellobiose CB0 = 5 kg/m3 inhibitor τ (min) τ (min)

1

1.5

587

940

1020

4

21.0

36

40

50

2 3

4.5 9.0

279 171

387 213

Series 3 with glucose CG0 = 10 kg/m3 τ (min) 433 250

Find a rate equation to represent the breakdown of cellulose in the absence of inhibitor. SOLUTION Given data: For CSTR,

CA0 = 25 kg/m3, CE0 = 0.01 kg/m3 t CSTR =

C A0 - C A ( -rA )

165

166

Enzymatic Reaction Kinetics

Therefore,

-rA =

(C A0 - C A ) t CSTR

In the absence of inhibitor, from the Michaelis–Menten rate equation -rA =

k2CE0C A C A + kM

By rearranging the above equation, we get CE0 1 kM 1 = + ( -rA ) k2 k2 C A

CE0 k 1 1 = M + ( -rA ) k2 C A k2 y = mx + c

Table 4.16 Calculated data CA (kg/m3)

1.5

4.5

9.0

21.0

tCSTR (min)

587

279

171

36

C E0 ( - rA )

1 CA

(kg/m3 min)

(h)

(m3/kg)

0.04

0.25

0.667

- rA =

(C A0 - C A ) t MFR

0.073

0.14

0.093

0.107

0.111

0.090

0.222

0.111

0.0476

Plot of CE0/(–rA) versus 1/CA results in the slope of kM/k2 and the intercept of 1/k2. From the graph, we have 1 = 0.0794 h k2

kM h kg = 0.2571 3 k2 m By solving,

k2 = 12.60 h–1 kM = 3.24

kg

m3

Factors Affecting Enzymatic Reactions

Therefore, the rate equation can be written as -rA =

12.6CE0C A C A + 3.24

0.3 y = 0.257x + 0.079 R2 = 0.998

CE0/(-rA) (h)

0.25 0.2 0.15 0.1 0.05 0

0

0.2

0.4

0.6

0.8

1/CA (m3/kg)

Figure 4.11 Plot of

CE0 1 versus . (- rA ) CA

PROBLEM 4.19: Continuation of PROBLEM 4.18. What is the role of cellobiose in the breakdown of cellulose (find the type of inhibition and the rate equation)?

SOLUTION In the presence of cellobiose (CB0 = 5 kg/m3) From the Michaelis–Menten rate equation -rA =

k2¢CE0C A C A + kM¢

By rearranging the above equation, we get CE0 1 kM¢ 1 = + ( -rA ) k2¢ k2¢ C A

CE0 k¢ 1 1 = M + ( -rA ) k2¢ C A k2¢ y = mx + c

167

Enzymatic Reaction Kinetics

Table 4.17 Calculated data CA tCSTR (kg/m3) (min)

1.5

940

4.5

387

9.0

C E0 ( - rA )

1 CA

(kg/m3 min)

(h)

(m3/kg)

0.025

0.40

0.667

- rA =

(C A0 - C A ) t MFR

0.053

213

21.0

0.190

0.075

40

0.133

0.100

0.100

0.222

0.111

0.0476

Plot of CE0/(–rA) versus 1/CA results in the slope of kM¢ k2¢ and

the intercept of 1 k2¢ . 0.45

y = 0.4819x + 0.0794 R2 = 0.9996

0.4 0.35 CE0/(-rA) (h)

168

No inhibition

0.3

Inhibition with cellobiose

0.25 0.2 0.15 0.1

y = 0.257x + 0.079 R2 = 0.998

0.05 0

0

0.4

0.2 1/CA

Figure 4.12 Plot of

From the graph,

0.6

(m3/kg)

CE0 1 versus . CA (- rA )

1 = 0.0794 h k2¢ kM¢ h kg = 0.4819 3 k2¢ m

0.8

Factors Affecting Enzymatic Reactions

By solving,

k2¢ = 12.60 h -1 kM¢ = 6.072

kg

m3 Therefore, the rate equation can be written as -rA =

12.6CE0C A C A + 6.072

It is observed that the kM value increases and the k2 value remains unchanged. Thus, it is competitive inhibition. PROBLEM 4.20: Continuation of PROBLEM 4.18. What is the role of glucose in the breakdown of cellulose (find the type of inhibition and the rate equation).

SOLUTION In the presence of glucose (CG0 = 10 kg/m3) From the Michaelis–Menten rate equation, -rA =

k2¢¢CE0C A C A + kM¢

By rearranging the above equation, we get CE0 1 kM¢¢ 1 = + ( -rA ) k2¢¢ k2¢¢ C A

CE0 k ¢¢ 1 1 = M + ( -rA ) k2¢¢ C A k2¢¢

Table 4.18 Calculated data CA (kg/m3)

tCSTR (min)

y = mx + c

- rA =

(C A0 - C A ) t CSTR

(kg/m3 min) 1.5

4.5

9.0

21.0

1020

433

250

50

0.023

0.047

0.064

0.080

C E0 ( - rA )

1 CA

(h)

(m3/kg)

0.430

0.212

0.156

0.125

0.667

0.222

0.111

0.0476

169

Enzymatic Reaction Kinetics

Plot of CE0/(–rA) versus 1/CA gives the slope of kM¢¢ k2¢¢ and the

intercept of 1 k2¢¢ . 0.5

CE0/(-rA) (h)

170

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

y = 0.492x + 0.101 R2 = 1

No inhibition Inhibition with glucose

y = 0.257x + 0.079 R2 = 0.998

0

0.2

0.4

0.6

0.8

1/CA (m3/kg)

Figure 4.13 Plot of

From the graph,

CE0 1 versus . (- rA ) CA

1 = 0.1018 h k2¢¢ kM¢¢ h kg = 0.4924 3 k2¢¢ m

By solving,

k2¢¢= 9.82 h -1 kM¢¢ = 4.83

kg m3

Therefore, the rate equation can be written as -rA =

9.82CE0C A C A + 4.83

It is observed that kM increases and k2 decreases. So the inhibition is a mixed-type inhibition.

References

References 1. Bailey, J. E. and Ollis, D. F. Biochemical Engineering Fundamentals, McGraw-Hill Inc., New Delhi, India, 2010. 2. Doran, P. M. Bioprocess Engineering Principles, Second Edition, Academic Press, Waltham, USA, 2012.

3. Shuler, M. L. and Kargi, F. Bioprocess Engineering: Basic Concepts, Second Edition, Prentice-Hall Inc., New Delhi, India, 2002.

4. Levenspiel, O. Chemical Reaction Engineering, Third Edition, WileyIndia, 2010. 5. Sinclair, C. G. and Kristiansen, B. Fermentation Kinetics and Modeling, Open University Press, 1987.

6. Blanch, H. W. and Clark, D. S. Biochemical Engineering, Marcel Dekker Inc., New York, USA, 1997.

7. Ghose, T. K. and Das, K. A. A simplified kinetic approach to cellulose– cellulase system, In: Advances in Biochemical Engineering (T. K. Ghose and A. Fiechter, Eds.), volume 1, Springer, Berlin, Heidelberg, 1971. 8. Underkofler, L. A. Manufacture and use of industrial enzyme, CEP Symp. Bioeng. Food Process, 69, 1966.

9. Faith, W. T., Neubeck, C. E., and Reese, E. T. Production and applications of enzymes, In: Advances in Biochemical Engineering (T. K. Ghose and A. Fiechter, Eds.), volume 1, Springer, Berlin, Heidelberg, 1971

10. Guilbault, G. G. Enzymatic Methods of Analysis, Pergamon Press, New York, 1970. 11. Katoh, S. and Yoshida, F. Biochemical Engineering: A Textbook for Engineers, Chemists and Biologists, Wiley-VCH, 2010.

12. Najafpour, G. Biochemical Engineering and Biotechnology, Elsevier, 2006.

13. Yoo, Y. J., Feng, Y., Kim, Y. H., and Yagonia, C. F. J. Fundamentals of Enzyme Engineering, Springer, 2017. 14. Yong, G. Enzyme Engineering, Science Press, 2014.

15. Kartan, P. Enzyme Engineering, Arcler Education Incorporated, 2017.

171

Chapter 5

Immobilized Enzymes

Immobilization of an enzyme confines or localizes it to an inert, insoluble material so that it can be reused continuously and its stability is greatly increased. Although an immobilized enzyme is held in a solid matrix throughout the reaction, the product can be easily separated from the matrix. An immobilized enzyme may be reused until and unless the activity of the enzyme decreases drastically. So immobilized enzymes are widely used in the industry for enzymecatalyzed reactions. An important application of immobilized enzymes is the production of high fructose corn syrup (HFCS) from corn starch. The merits and demerits of the immobilized enzyme system are discussed here.

5.1 Merits of Immobilized Enzymes

The main advantages of using immobilized enzymes are as follows:

∑ Overcome the problems of enzyme removal from the end products. ∑ Reuse of the enzyme, greater turnover number. ∑ Impart increased stability to the enzyme (reduction in contamination and destruction).

Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

174

Immobilized Enzymes



∑ Deactivation due to higher temperature, and range of pH is reduced. ∑ Prolongation of the half-life of the activity of the enzyme and improved storage qualities.

5.2 Demerits of Immobilized Enzymes

However, immobilized enzymes have some disadvantages:

∑ Substrate and product diffusivity may become a problem, particularly in the case of insoluble substrates. ∑ Partial steric hindrance of the active site of the enzyme may also occur. ∑ Due to insolubility imparted, stirring may become essential. ∑ All types of immobilization principles may not be applicable to all enzymes.

5.3 Classifications of Immobilization Techniques

Enzyme immobilization techniques are classified based on the following parameters:

∑ Mode of interaction between enzymes and the solid matrix for immobilization. ∑ Characteristics of the solid matrix. ∑ Nature of the resulting complex.

Different techniques of enzyme immobilization are shown in Fig. 5.1. In the case of insoluble enzymes, immobilization may be done by

∑ Binding ∑ Entrapment

Binding with the solid matrix can be done by either crosslinking or carrier binding. A crosslink is a bond that links one polymer chain to another, e.g., glutaraldehyde. The structure of glutaraldehyde is shown in Fig. 5.2. The chemical formula of glutaraldehyde is as follows: OHC-(CH2)3-CHO (empirical formula C5H8O2)

Classifications of Immobilization Techniques

Immobilization techniques

Using soluble enzymes

Using insoluble enzymes

Carrier binding

Crosslinking

Physical adsorption

Covalent binding

Hollow fiber devices

Entrapping

Binding

Gel entrapping

Metal binding

Fiber entrapping

Ultrafiltration membrane

Microencapsulation

Ionic binding

Figure 5.1 Classification of immobilization techniques.

Figure 5.2 Structure of glutaraldehyde.

It has two aldehyde groups, which can hold two enzyme molecules as shown below: Enzyme-NH2 + OHC-(CH2)3-CHO + H2N-Enzyme



Æ Enzyme-N=HC(CH2)3CH=N-Enzyme

In this method, covalent bonds between the functional group of polyfunctional reagents like glutaraldehyde and that of the enzyme are formed. As compared to other methods, there is no need of the solid matrix. This method is cheap but rarely used for the immobilization of pure enzymes. Binding can be classified as follows:

∑ ∑ ∑ ∑

Physical adsorption Ionic binding Metal binding Covalent binding

Physical adsorption is a very simple technique. The characteristics of this process are as follows:

∑ This is a physical or surface phenomenon. The enzymes are attached to the solid matrix with the help of weak van der Waals forces.

175

176

Immobilized Enzymes





∑ Solid matrix usually binds with other than the active site of the enzyme and most of the enzymatic activity is retained after the immobilization by adsorption. ∑ The common problem of this technique is desorption or detachment of enzymes from the solid matrix due to the presence of strong hydrodynamic forces, since binding forces are weak.

The examples of solid matrices used for the adsorption technique are porous glass, diatomaceous earth, silica, alumina, ceramics, clay, and bentonite, or organic materials such as diethylaminoethylcellulose (DEAE-C), carboxymethyl cellulose (CMC), starch, activated carbon, and ion-exchange resins such as sephadex, amberlite, and dowex. Immobilization by ionic binding occurs due to the ionic charges of the solid matrix and the enzyme molecules. They should be opposite in nature. Several polymeric materials acquire ionic charges. Metal binding is largely used for the development of biosensors, such as platinum (Pt) covalent binding is mainly due to electron sharing between the solid matrix and the enzyme. This is the strongest bond between the enzymes and the solid matrix. Enzymes lose some of their activities during the process of immobilization mainly due to the chemical treatment. The enzymes attach to the solid matrix by covalent bond formation between functional groups on the enzyme and those on the solid matrix. The functional groups are as follows: ∑ Carboxyl (–COOH) ∑ Hydroxyl (–OH) ∑ Amino (–NH2) ∑ Sulfhydryl (–SH) The functional groups present at the active site of the enzyme do not take part in this immobilization, e.g., covalent binding using CMC: 3OH ,HCl 1. -CH2COOH æCH æææ Æ -CH2COOCH3

Hydrazide

NH2 - NH2

ææææÆ -CH2 - CO - NH - NH2 NaNO

2 2. -CH2 - CO - NH - NH2 æææ Æ -CH2 - CON3

Acid azide

: 8-9 3. -CH2 - CON3 + NH2 - protein æpH æææ Æ -CH2 - CONH - protein

Classifications of Immobilization Techniques

The entrapment of the insoluble enzymes may be done by three techniques: 1. Gel 2. Fiber 3. Microencapsulation

In the case of gel entrapment, the enzyme solution is mixed with a polymetric fluid, which is followed by solidification (Fig. 5.3). This may solidify into different forms depending on application. The polymetric material is semi-permeable. In the gel, the largemolecular-weight enzyme cannot diffuse out, but smaller substrate and product molecules can. Examples of gel entrapment include k-carrageenan, polyacrylamide, Ca-alginate, agar, collagen, etc. Gel

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Figure 5.3 Enzyme immobilization by gel entrapment.

In the case of fiber entrapment, fibers are used as the solid matrix for the immobilization of enzymes. Hollow fibers are considered for entrapping an enzyme. Examples of membrane materials used for the entrapment of enzymes are polyacrylate, cellulose, nylon, polysulfone, etc. In the case of microencapsulation, a membrane capsule is used for the immobilization of enzymes (Fig. 5.4). An example of a capsule is nitrocellulose. Effectiveness of this technique depends on the stability of enzymes inside the membrane.

177

178

Immobilized Enzymes Semi-permeable membrane

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Enzyme

Figure 5.4 Enzyme immobilization by encapsulation.

Soluble enzymes may be immobilized by an ultrafiltration membrane. The selection of an immobilization technique for a particular enzyme depends on the desirable characteristics of certain parameters such as preparation, cost, and binding force, as shown in Table 5.1. Table 5.1

Characteristics of different immobilization techniques Covalent binding

Entrapment

Membrane encapsulation

Binding force Variable

Strong

Weak

Strong

Enzyme leakage

No

Yes

No

Properties

Adsorption

Cost

Low

Applicability

Wide

Preparation

Operational problems

Effect of the solid matrix Diffusional problem Microbial protection

Simple Yes

High Yes No

No

High

Difficult

Moderate Difficult

Selective

Wide

Yes

Yes

Low No

No

High Yes

Yes

High

Simple Very wide High No

Yes

Yes

Characterization of Immobilized Enzymes

5.4 Characterization of Immobilized Enzymes The characteristics of immobilized enzymes play an important role for the operation of the process.

5.4.1 Activity of Immobilized Enzymes

The activities of immobilized enzymes are generally expressed in international units, defined as micromoles of substrate converted per minute per gram of immobilized enzyme. The activities of immobilized enzymes depend on different parameters such as

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

Initial substrate concentration, Concentration of immobilized enzymes, pH, Temperature, Reaction time, Agitation or flow rate, Buffer concentration, Ionic strength, and Physical dimensions of the solid matrix.

These parameters are usually mentioned with the activity data.

5.4.2 Bound Protein

Bound protein is the amount of protein bound to the solid matrix. This is also important for studying the kinetic behavior of immobilized enzymes. The unit of bound protein is expressed as milligram of protein per gram of carrier.

5.4.3 Specific Activity of Bound Protein

The specific activity of immobilized enzymes is defined as micromoles of substrates converted per minute per milligram of bound protein. The difference between the specific activity of immobilized enzymes and that of the soluble enzyme indicates the effectiveness of the immobilization procedure.

179

180

Immobilized Enzymes

5.4.4 Coupling Efficiency Coupling efficiency is an indicator of the activity of immobilized enzymes as compared to the amount of enzyme that initially binds with the solid matrix. Coupling efficiency may be expressed as follows: Coupling yield =

Overall activity of immobilized enzyme ¥ 100 Overall activity of the initial enzyme solution immoblized

=

EIME ¥ 100 E initial - E recovered

where EIME is the activity of immobilized enzyme, Einitial is the initial activity of free enzyme solution before immobilization, and Erecovered is the activity of the free enzyme solution recovered after immobilization.

5.4.5 Stability of Carrier–Enzyme Complexes

The stability of an immobilized enzyme is expressed in two ways: shelf life or storage stability and operational stability. The shelf life of the enzyme is determined by measuring its activity during regular intervals of time when stored at specific environmental conditions. The operational stability of the enzyme is determined by monitoring the activity of the immobilized enzyme during continuous use under the environmental conditions. The primary cause of decreasing enzymatic activity may be due to the denaturation of enzyme. It may be due to other reasons such as inactivation by bound inhibitors, enzyme elution, and fouling of the reactor.

5.4.6 Materials, Methods, and Conditions of Binding Technique

The cost incurred in producing immobilized enzymes with a given amount of stable enzymatic activity plays an important role. So it is necessary to find out the expenditure involved for reagents, specialized equipment, energy, labor, and maintenance for the operation of the process. Regenerability of the solid matrix influences the process to some extent.

Characterization of Immobilized Enzymes

For example, some enzymes may be irreversibly inactivated in the presence of certain crosslinking agents, while others can tolerate the conditions extremely well. The mode of attachment of the enzyme and the solid matrix is also important. However, the overall efficiency of the process depends on the amount of protein immobilized.

5.4.7 Physicochemical Characteristics of Carrier

The suitability of carriers in a continuous system can be determined on the basis of the following parameters:

∑ Surface area should be high for enzyme immobilization. ∑ Tolerance to chemicals present in the reactor. ∑ Microbial resistance to safeguard both enzyme and the solid matrix. ∑ Mechanical strength and dimensional stability to protect the structure of enzyme. ∑ Thermal stability of the solid matrix. ∑ Charge of the pore as compared to the substrate. ∑ Solid matrix regeneration. ∑ Size and shape of the solid matrix.

The characteristics of the solid matrix have significant influence on the performance of an immobilized enzyme for product formation. The overall economy of the process is influenced by the cost and availability of the solid matrix. An important characteristic of the solid matrix is microbial resistance. If the solid matrix has a significant amount of carbon, such as starch or protein, it may undergo degradation. Due to the degradation of the organic compound, an enzyme can be released into solution. This problem may be overcome by using an inorganic oxide such as silica, titania, etc. There are different organic materials that are resistant to microbes, e.g. polypropylene, fluorocarbons, etc. The thermal stability of a solid matrix also plays an important role. Enzymatic reactions are generally carried out in the temperature range of 0°C to 80°C. If the solid matrix has a large coefficient of expansion, it may be responsible for the distortion of the active site of the enzyme.

181

182

Immobilized Enzymes

The size or shape of the particles of a solid matrix can affect the performance of an immobilized enzyme in a continuous reactor. Less pressure drop takes place when particle size is large. This helps in maintaining a uniform high flow rate. However, in this case, the bound enzyme concentration is reduced. Table 5.2 shows the uses of solid matrices by considering the specific properties of carriers with respect to different applications. The life of a carrier has a significant influence on the total economics of an immobilized enzyme system. The conditions include the initial shipment of the solid matrix to the plant, the reuse or regeneration of the solid matrix, and the disposal problem of unregenerable solid matrices. Organic carriers cause the disposal problem. Normally, the regeneration of an organic solid matrix is very expensive. Table 5.2

Properties suggesting carrier applications

Carrier

General application

Collagen, cellulose acetate

Membrane to separate Polymer hydrolysis and collection of product free substrate and products (protein hydrolysis) of substrate

Controlled pore alumina

Continuous plug flow reactors at pH 5–11

Controlled pore glass or silica

Controlled pore titania

Conditions suggesting type of application

Continuous plug flow reactors at pH below 7.0

Dimensional stability, low pressure drops, durable below 7.0

Continuous plug flow reactors at pH 3–9

Dimensional stability, low pressure drops, durable at pH 3–9

Dimensional stability, low pressure drop, durable above pH 5.0

Inorganic solid matrices can be mostly regenerated by a simple pyrolysis process. Different types of supporting matrices are available for immobilization processes. Examples of some carriers are as follows:

∑ Carriers used for the adsorption of enzymes: alumina, bentonite, calcium carbonate gel, carbon, cellulose, clay, collagen, sepharose, glass (porous), hydroxyapatite, ionexchange resins, kaolin, phenolic polymers, silica gel, PVC.

Characterization of Immobilized Enzymes





∑ Carriers used for the covalent attachment of enzymes: agarose (sepharose), cellulose, dextran (sepharose), glass, polyacrylamide co-polymer, polyaminostyrene.

∑ Carriers used for the entrapment of enzymes/cells: polyacrylamide gels, calcium alginate gels, kappa-carrageenan, agar gel.

5.4.7.1 Solid matrix morphology and configuration

Solid matrices are either organic or inorganic in nature. The morphology of a solid matrix plays an important role with respect to surface area and pore diameters. This will affect the loading of enzymes. The carrier may be classified as follows: nonporous and porous. Nonporous solid matrices: Disadvantages of nonporous solid matrices are as follows:







∑ These materials have low surface area. So enzymes get limited surface area for immobilization. This problem can be partially overcome by using fine particles. ∑ It is very difficult to remove fine particles from the solution. This may lead to high pressure drops and limited flow rates.

Advantages of nonporous carriers are as follows:

∑ Nonporous materials such as nylon may be used in the blood system. They do not cause a vigorous clotting reaction as compared to inorganic materials having a high surface area, such as porous glass. ∑ The enzyme is immobilized to an external surface. So the diffusion effects are minimal. The substrate in solution can be easily reacted.

Porous carriers: Disadvantages of porous solid matrices are as follows:

∑ Enzymes are mostly immobilized in the internal structures of the solid matrix. Enzyme molecules must penetrate into the internal surface of the solid matrix for immobilization. In the case of a substrate larger than the enzyme, molecules may not penetrate the pores or may have access only to larger pores.

183

184

Immobilized Enzymes







So the purpose of immobilized enzymes may never be fully realized. ∑ The internal pore environment has an important role to play. If the pore surface and the substrate have the same charge, repulsion occurs. In this case, the substrate may never come in contact with the enzyme or the number of contacts between the enzymes and the substrate is very less. ∑ The presence of different sizes of pores (broad pores) may also reduce the substrate–enzyme interaction. ∑ Diffusion problem is significant in entrapped or encapsulated enzymes. Use of these solid matrices is difficult in plug flow reactors since they are readily compacted and, as a result, high pressure drops occur.

Advantages of porous pore solid matrices are as follows:

∑ High surface areas available for immobilization. ∑ Protection of the enzyme due to internal surface bonding from the turbulent external environment. ∑ Opposite surface charge of the solid matrix with respect to substrate may enhance the enzyme substrate reaction. ∑ In the case of encapsulation or entrapment, a large amount of enzymes can be loaded without the need for surface attachment.

Pore diameter plays an important role in the immobilization of enzymes. If the diameter of the enzyme is greater than that of the substrate particles, then the pore dimension of the carrier should be chosen with respect to the enzyme. In the reverse case, the pore dimension of the solid matrix should be chosen with respect to the substrate. Table 5.3 shows that the amount of enzyme adsorbed is a function not only of the pore diameter but also of the surface area. In the case of some solid matrices, the availability of surface for the attachment of enzyme is limited by the restrictive openings but not on the surface area. For example, bentonite has a very high surface area, but the available internal surface is less for enzyme attachment. It has been reported that the optimum pore diameter of a solid matrix for enzyme immobilization is approximately two times the

Characterization of Immobilized Enzymes

major axis of the enzyme. Glucose–glucose oxidase reaction is shown below: Glucose ææææææ Æ Glucono 1,5 lactone + H2O2 Glucose oxidase

So in the case of immobilized glucose oxidase, hydrogen peroxide accumulates within the pores. Hydrogen peroxide can be degraded by catalase as shown below: 2H2O2 ææææ Æ 2H2O + O2 Catalase

So in the absence of catalase, glucose oxidase is destroyed by hydrogen peroxide. However, in the presence of catalase, this enzyme does not lose its activity. The maximum diameter of glucose oxidase is 84 Å. So this enzyme has substantial access to a pore diameter of 168 Å. On the other hand, catalase has the largest dimension of 182 Å, which has access to a pore diameter of 364 Å. It has been observed that glucose oxidase is unstable in the case of a solid matrix such as alumina, which has a pore size of 175 Å. The instability of the enzyme is due to the exclusion catalase. On the other hand, a very stable glucose oxidase–catalase system has been achieved with a titania having an average pore of 350 Å and a maximum pore of 400 Å. The relationship between the stability of glucose oxidase activity and pore diameter is shown in Fig. 5.5. 80 Stable activity (GOU/g) 60 40 20 0

0

200

400

600

800

Pore diameter (A°)

Figure 5.5 Plot of glucose oxidase activity versus pore diameter.

185

186

Immobilized Enzymes

Table 5.3 Relation of pore size and surface area of glass with the quantity of adsorbed stabilized enzyme Pore diameter (Å)

Surface area (m2/g)

7740 (test tube)

Not porous

Very low Papain (21,000)

0

Controlled pore glass

900

20

0.33

Glass corning

7740 (F-font) 7930 7930

Controlled pore glass 7930

Enzyme (mol. wt.)

50,000

0.2

Papain (21,000)

92

60

Papain (21,000)

68

118

900

20

68

118

Note: Taken from Ref. [3].

Papain (21,000) Papain (21,000)

Glucose oxidase (150,000) Glucose Oxidase (150,000)

Milligram of active bound enzyme per gram of glass

0.01 0.08 0.07 0.12 0.0001

5.4.8 Application of Immobilized Enzymes 5.4.8.1 Industrial Immobilized enzymes have very good potential in industries. However, only a few immobilized enzyme processes have been commercialized. One of the major industrial applications in biochemical industries is the production of HFCS using immobilized glucose isomerase. Glucose ææææææææææÆ Fructose Immobilized glucose isomerase

Fructose is sweeter than glucose. In 1976, USA produced more than a billion pounds of HFCS (containing approximately 42% fructose). However, relatively smaller amounts of HFCS is also produced in Belgium, Holland, Spain, Italy, and Japan. Commercialized immobilized glucose isomerase processes are marketed by the

Kinetics of Immobilized Enzymes

following companies: Clinton Corn Processing Company (USA), Imperial Chemical Industries (USA), Nova Enzyme Corporation (Denmark), and Gist Brocades, N.U. (The Netherlands). Immobilized penicillin acylase is used in the hydrolysis of penicillin to 6-aminopenicillanic acid (6-APA). This is commercially produced in Europe. Examples of other applications of immobilized enzymes are as follows:

∑ Steroid transformation ∑ Chill proofing of beer ∑ Use of immobilized glucoamylase for the saccharification of starch ∑ In various food products for the removal of residual oxygen ∑ Conversion of dl-α-amino-ε-caprolactam to l-lysine

5.4.8.2 Analytical

Immobilized enzymes have a great potential in the manufacture of biosensors. This can be used for analytical purpose. Immobilized enzymes are used to determine glucose, urea, uric acid, amino acids, cholesterol, etc. at the micro level. A penicillin-specific enzyme electrode is used for penicillin assay in the fermentation broth. Cholesterol oxidase has been used successfully for the measurement of serum cholesterol.

5.4.8.3 Medical

In the pharmaceutical industries, immobilized enzymes have a wide range of applications, such as clinical analysis, therapy, preventative and environmental medicine. Enzyme-sensitive disorders can be treated by using immobilized enzymes. These can also be used in several types of cancer. Utilization of immobilized enzymes has a greater promise as compared to using soluble enzyme therapy. Collagen is found suitable as a solid matrix for the immobilization of therapeutic enzymes because it is a very good biomaterial.

5.5 Kinetics of Immobilized Enzymes

The diffusion problem of the substrate used in the immobilization of an enzyme can be classified as

∑ External mass-transfer resistance ∑ Internal mass-transfer resistance

187

188

Immobilized Enzymes

5.5.1 External Mass-Transfer Resistance In case the enzyme is immobilized only on the external surface of a solid matrix, then mass transport is to be considered from the bulk solution to the surface of the solid matrix and reaction occurs at that position. This is known as external mass-transfer resistance. The rate of mass transfer from the bulk solution to the surface is given by NS = kaam (Sb – Ss)

(5.1)

where ka is the mass-transfer coefficient, am is the surface area per unit volume, Sb is the bulk-phase substrate concentration, and Ss is the substrate concentration at the surface of the solid matrices. The Michaelis–Menten equation may be written as v=

vmax S s km + S s

(5.2)

At steady-state (SS) conditions, Mass-transfer rate = Rate of reaction at the surface of the solid matrix where the internal diffusion limitations are not significant. So at steady-state condition, we can write from Eqs. 5.1 and 5.2, ka am ( S b - S s ) =

\

Ss = Sb -

vmax S s =v km + S s

(S k a - v ) v = b a m ka am ka am

Equation 5.2 may be written as

k 1 1 1 = + m v vmax vmax S s

Putting the value of Ss from Eq. 5.4 in Eq. 5.2, we get ka am k 1 1 = + m v vmax vmax ( S b ka am - v )

ka am k 1 1 1 = + m v vmax vmax S b Ê v ˆ ÁË ka am - S ˜¯ b 1 1 = ( S b ka am - v + km am ka ) v vmax ( S b ka am - v )

(5.3)

(5.4) (5.5) (5.6) (5.7)

Kinetics of Immobilized Enzymes

S b ka am v 1 - = ( S b ka am + km am ka - v ) v v vmax S b ka am vmax = S b ka am + km am ka - v + vmax v 1 1 1 = ÈS b ka am + km am ka - v + vmax ˘˚ v Sb Î ka am vmax k 1 1 È Sb v 1 ˘ = + m + Í ˙ v S b Î vmax vmax ka am vmax ka am ˚ The plot of 1/v versus 1/Sb is shown in Fig. 5.5.

(5.8)

Ê k 1 ˆ ¨ Slope = Á m + Ë vmax k 2 a m ˜¯

1 v

km Slope = v max Intercept =

1 vmax 1 Sb

Figure 5.6 Plot of 1/v versus 1/Sb using immobilized enzyme.

Again Eq. 5.7 may be written as

˘ È Í ˙ 1 Í 1 = 1 + Ê km ˆ Ê 1 ˆ ˙ ÁË v ˜¯ ÁË S ˜¯ Ê Ív v ˙ ˆ v max max b Í ˙ 1 ÁË S b ka am ˜¯ ˙˚ ÍÎ

(5.9)

189

190

Immobilized Enzymes

When Sb Æ •, i.e., at =

1 . vmax

k 1 Æ 0 , in Fig. 5.6, slope = M and intercept Sb vmax

Again, when Sb Æ 0, Eq. 5.8 may be written as 1 1 Ê km 1 ˆ = + Á v S b Ë vmax ka am ˜¯

In Fig. 5.6, when

1 Æ •, Sb

(5.10)

Ê k 1 ˆ The slope of the straight line is Á m + . Ë vmax ka am ˜¯

So the values of km, vmax, and kaam can be determined from Fig. 5.6, which are the reaction kinetics parameters of the immobilized enzyme system.

5.5.1.1 Effectiveness factor and Damköhler number

The influence of mass transfer on the overall reaction process is expressed using the effectiveness factor (η) as follows: h=

Observed rate of reaction Rate of reaction in case of no mass-transfer resistance, i.e., S s = S b

The Michaelis–Menten equation may be written as

Observed reaction rate: vobs =

vmax S s km + S s

The reaction rate with no mass-transfer limitation is v SS Æ S b = h= \

h=

vmax S b km + S b

vobs

(5.11) (5.12) (5.13)

v SS Æ S b

vmax S s (km + S b ) (km + S s )vmax S b

h=

S s ( km + S b ) S b ( km + S s )

(5.14)

Kinetics of Immobilized Enzymes

Putting x =

Ss k , K = m , we get Sb Sb h=

x/(k + x ) x(k + 1) = 1/(k + 1) (k + x )

(5.15)

The Damköhler number (Da) is a dimensionless number and may be represented as Da =

v Maximum rate of reaction = max Maximum rate of mass transfer km am S b

(5.16)

Here η < 1 indicates the effect of increasing mass-transfer resistance and reduction in observed activity. Under the steady-state condition, from Eq. 5.3 we have ka am ( S b + S s ) =

vmax S s km + S s

S vmax s Ê ˆ S Sb ka am S b Á 1 - s ˜ = S b ¯ km S s Ë + Sb Sb vmax v (1 - x ) = max Da k+x x 1- x = Da K +x Ss Sb

h=

vobs

vSs Æ S b

vmax S s km S s x + Sb Sb km + S s x(k + 1) = = = k+x = Sb vmax S b 1 k+x k +1 Sb K m + Sb km S b + Sb Sb

From Eqs. 5.17 and 5.18, we get

1- x h = Da k +1

(5.17)

(5.18)

(5.19)

191

192

Immobilized Enzymes

h=

x (1 - x )(K + 1) (k + 1) = (K + x ) Da

Equation 5.20 gives the correlation between Da and h.

(5.20)

5.5.1.2 Determination of the factor affecting overall rate of reaction In case Da Æ 0 (very slow rate of reaction as compared to maximum mass transfer) and x Æ 1, then from Eq. 5.15, η = 1. Here the observed reaction kinetics is the same as the true intrinsic kinetics at the fluid–solid interface. So the reaction kinetics may be represented as v=

vmax S b km + S b

Again from Eq. 5.17, the substrate mass balance becomes

(5.21)

1- x x = Da K +x

where 0 ≤ x ≤ 1.0 The diffusion-limited regime of the system with coupling between the chemical reaction and the mass transfer arises when vmax is much larger than kaamSb. When Da >> 1 and x approaches zero, Eq. 5.19 becomes 1+k Da So the reaction kinetics may be represented as h=

v = km am S b

(5.22)

As long as Da is very large, the observed rate of reaction, v , is first order with respect to the bulk substrate concentration and is independent of the intrinsic rate parameters vmax and km.

5.5.2 Effect of Internal Mass Transfer

In the case of enzymes immobilized in a porous spherical particle, internal mass transfer is considered (Fig. 5.7). The following assumptions are made:

∑ The particle is isothermal. ∑ Mass transfer takes place by diffusion only.

Kinetics of Immobilized Enzymes



∑ Fick’s law can be used to describe diffusion using constant effective diffusivity. ∑ The particle is homogeneous. ∑ The particle is at steady state. ∑ Variation of substrate concentration with a single spatial. Bulk liquid 1

Immobilized enzyme 2

Sb

3

Sb

S

Catalyst surface

S

Figure 5.7 Internal mass transfer in a spherical particle.

Dr

r Diffusion of substrate

R

Figure 5.8 Diffusion of substrate inside the spherical particle.

Assuming a thin spherical shell of thickness ∆r located at radius r from the center, the steady-state shell mass balance for a shell on substrate S may be written as dS Ê ˆ Rate of diffusion in input = Á Des 4p r 2 ˜ Ë ¯ r + Dr dr

(5.23)

193

194

Immobilized Enzymes

dS Ê ˆ Rate of diffusion in output = Á Des 4p r 2 ˜ Ë ¯r dr

(5.24)

Rate of substrate degradation = (–rS)4pr2Dr

(5.25)

Rate of generation = 0

At steady state, rate of accumulation is 0. Here Des is the effective diffusivity of substrate, S is the concentration of the substrate in the particle, and r is the distance. The substrate balance may be written as follows: Rate of input + Rate of generation = Rate of output

  + Rate of consumption + Rate of accumulation

For steady-state shell mass balance

(5.26)

dS dS Ê Ê ˆ 2ˆ 4p r 2 ˜ + ( -rS )4p r 2 Dr + 0 (5.27) + 0 = Á Des ÁË Des dr 4p r ˜¯ Ë ¯r dr r +Dr

Rearranging the above equation, we get

dS dS Ê Ê ˆ 2ˆ 4p r 2 ˜ = ( -rS )4p r 2 Dr - Á Des ÁË Des dr 4p r ˜¯ Ë ¯r dr r +Dr

Steady-state shell mass balance Dividing both terms by 4pDr gives

dS 2 ˆ dS 2 ˆ Ê Ê r - Á Des ÁË Des dr r ˜¯ Ë dr ˜¯ r r +Dr Dr

= ( -rS )r 2

Taking limit Dr Æ 0 lim

dS 2 ˆ dS 2 ˆ Ê Ê r - Á Des ÁË Des dr r ˜¯ Ë dr ˜¯ r r + Dr

Dr At steady-state shell mass balance, Dr Æ0

= ( -rS )r 2

d Ê dS 2 ˆ D r = ( -rS )r 2 dr ÁË es dr ˜¯

Ê d2S dS ˆ Des Á 2 r 2 + 2r ˜ = ( -rS )r 2 dr ¯ Ë dr

(5.28)

(5.29)

(5.30) (5.31) (5.32)

Kinetics of Immobilized Enzymes

Ê d 2 S 2 dS ˆ = ( -rS ) Des Á 2 + r dr ˜¯ Ë dr

(5.33)

Equation 5.33 represents diffusion and reaction in a spherical biocatalyst. Steady-state shell mass balance Ê d 2 S 2 dS ˆ = ( -rS ) Des Á 2 + r dr ˜¯ Ë dr Ê d 2 S 2 dS ˆ ( -rS ) Á 2 + r dr ˜ = D Ë dr ¯ es

(5.34) (5.35)

Boundary condition for the above equation is as follows: At r = R, S = Sb, where Sb is the bulk substrate concentration.

dS = 0. dr Assuming non-dimensional parameters as follows r = 0,

At

S=

S Sb

(5.36)

r (5.37) R Replacing S and r with non-dimensional parameters in Eq. 5.35, we get r=

Sb d2S

R dr 2

d2S dr 2

+

2

+

2S b dS ( -rS ) = Des R2r dr

2 dS R2 ( -rS ) = r dr Des S b

Boundary condition for this equation is r = 0,

At

r = 1,

At

(5.38) (5.39)

dS =0. dr S =1.

If the reaction follows the M–M kinetics, then d S 2

dr

2

+

2 dS = r dr

vmax S km + S Des S b

R2

(5.40)

195

196

Immobilized Enzymes

vmax S km + S Des S b

R2 RHS =

=

RHS =

R vmax Des km 2

R vmax Des km 2

È S Í S b Í Í1 + S Í km Î

È ˘ Í ˙ 2 S Í ˙ = R vmax Í Ï S S b ¸ ˙ Des km Í1 + Ì ˝˙ ÍÎ ÓÔ km S b ˛Ô ˙˚

˘ ˙ ˙ ˙ ˙ ˚

(5.41) (5.42)

È ˘ È ˘ Í ˙ Í ˙ Í S ˙ = 9j 2 Í S ˙ Í ÊSˆ˙ Í ÊSˆ˙ Í1 + Á ˜ ˙ Í1 + Á ˜ ˙ ÍÎ Ë b ¯ ˙˚ ÍÎ Ë b ¯ ˙˚ (5.43)

where b = km/Sb is the saturation parameter; when b = 0, the reaction is zero order, and when b = •, the reaction is first order. Here j is called the Thiele modulus: 2

Ê Vp ˆ vmax Maximum reaction rate j =Á ˜ = Ë Ap ¯ Des km Maximum mass-transfer rate (intra particulate) 2

(5.44)

where Vp and Ap are the volume and area of the particle, respectively. For spherical geometry, 2

Ê 4 3ˆ pR Á ˜ vmax R2 vmax = j2 = Á 3 2 ˜ 9 Des km Á 4p R ˜ Des km Ë ¯ 9j 2 =

R2vmax Des km

(5.45) (5.46)

To determine the effect of mass-transfer resistance on the rate of reaction, the effectiveness factor of an immobilized enzyme (h) can be defined as h=

Observable reaction rate Reaction rate in case of no mass-transferr resistance

Kinetics of Immobilized Enzymes

=

Performance of heterogeneous system ystem Performance of homogeneous sy

5.5.2.1 Effectiveness factor and Thiele modulus In the case of zero-order reaction rate (β Æ 0), η ≈ 1. For a first-order reaction rate (β Æ ∞), η = f(φ, β) and η is approximated to the following equation for high values of φ. h=

3È 1 1˘ - ˙ Í j Î sinh j j ˚

(5.47)

Effectiveness fractor, h

1.0 0.8 0.6 Zero order reaction 0.4

1st order reaction

0.2

2nd order reaction 0.1 1

2

4

6 8 10

20

40

Thiele modulus, f

Figure 5.9 Correlation between Thiele modulus and Effectiveness factor.

It is difficult to determine the kinetic parameters such as km and vmax in many practical cases. To overcome this problem, the observable Thiele modulus (Φ) is considered, which is known as the Weisz criteria (Table 5.4). 2

Ê Vp ˆ ( -rS )obs F=Á ˜ Ë Ap ¯ Des S b

It is an intrinsic kinetic parameter.

(5.48)

197

198

Immobilized Enzymes

Table 5.4

The Weisz criteria

Φ

η

Limiting regime

Mass-transfer significance

3

µ 1/Φ

Examples

Diffusion

Mass resistance significance

PROBLEM 5.1: The rate of degradation kinetics of lactose by β-galactosidase includes a product inhibition terms as follows: v=

vmax S +S km (1 + p/ki )

To allow easy recovery of β-galactosidase from milk, the enzyme solution was encapsulated in thin microcapsular cellulose nitrate membranes of diameter d ≈ 30 µm (microcapsules). If vmax = kpe0 , kp = 0.57

mmol , k = 0.54 mM, ki = 1.5 mM, mg enzyme min m

construct the Lineweaver–Burk plots for p = 0, 0.5, 1.5, 5, and 10 mM (e0 = 75 mg per 100 mL). SOLUTION The given equation can be written in the Lineweaver–Burk plot: 1 1 = v vmax

Ê Ê p ˆ ˆ 1 km Á km Á 1 + ˜ + s ˜ = ki ¯ ¯ s vmax Ë Ë

Ê pˆ1 1 ÁË 1 + k ˜¯ s + v i

max

Putting p = 0 and assuming different values of S, the corresponding v values can be calculated. Similarly at p = 0.5, 1.5, 5, 10 mM, different values of v can be calculated at different S. Given data: kp = 0.57

mmol = 0.57 mmol/g min, km = 0.54 mM mg enzyme min

ki = 1.5 mM, e0 = 75 mg per 100 mL = 0.75 g/L vmax = kpe0 = 0.57 mmol/g min ¥ 0.75 g/L = 0.427 mM/min Intercept will be same for all the cases. Intercept =

1 1 = min/mM = 2.34 min/mM vmax 0.427

Kinetics of Immobilized Enzymes

For

km vmax

Slope = For

km vmax

Slope = For

km vmax

Slope =

For

km vmax

Slope =

For

Slope =

km vmax

p = 0,

Ê Ê pˆ 0.54 mM 0 mM ˆ ÁË 1 + k ˜¯ = 0.427 mM/min ÁË 1 + 1.5 mM ˜¯ = 1.264 min i p = 0.5,

Ê Ê pˆ 0.54 mM 0.5 mM ˆ ÁË 1 + k ˜¯ = 0.427 mM/min ÁË 1 + 1.5 mM ˜¯ = 1.685 min i p = 1.5,

Ê Ê pˆ 0.54 mM 1.5 mM ˆ ÁË 1 + k ˜¯ = 0.427 mM/min ÁË 1 + 1.5 mM ˜¯ = 2.528 min i p = 5,

Ê Ê pˆ 0.54 mM 5 mM ˆ ÁË 1 + k ˜¯ = 0.427 mM/min ÁË 1+ 1.5 mM ˜¯ = 5.477 min i p = 10,

Ê Ê pˆ 0.54 mM 10 mM ˆ ÁË 1 + k ˜¯ = 0.427 mM/min ÁË 1 + 1.5 mM ˜¯ = 8.93 min i

The Lineweaver–Burk plots at different values of p are shown in Fig. 5.10. Lineweaver–Burk plots for p = 0, 0.5, 1.5, 5, and 10 mM

50 45

1/v (min M–1)

40 35

p = 0 mM

30

p = 0.5 mM

25

p = 1.5 mM

20

p = 5 mM

15

p = 10 mM

10 5 0

0

1

2

3 1/S (mM–1)

Figure 5.10 Plot of 1/v versus 1/S.

4

5

6

199

200

Immobilized Enzymes

PROBLEM 5.2: Immobilization of baby hamster kidney cells is done in alginate beads. The average particle diameter is 5 mm. The rate of oxygen consumption at a bulk concentration of 8 × 10−3 kg O2/m3 is 8.4 × 10−5 kg/s m3 catalyst. The effective diffusivity of oxygen in the beads is 1.88 × 10−9 m2/s. The oxygen concentration at the surface of the catalyst is assumed to be equal to the bulk concentration. It is assumed that oxygen uptake follows zero-order kinetics. Do you think internal mass-transfer effects are significant? SOLUTION The observable Thiele modulus may be written as 2 Ê R ˆ ( -rO2 )obs F=Á ˜ Ë 3 ¯ Des S b

with

R=

5 ¥ 10-3 m = 2.5 ¥ 10-3 m 2 2

Ê 2.5 ¥ 10-3 m ˆ 8.4 ¥ 10-5 kg/s m3 Then F = Á = 3.88 ˜ 3 Ë ¯ (1.88 ¥ 10-9 m2/s)(8 ¥ 10-3 kg/m3 )

From the Weisz criteria (Φ > 3), the internal mass-transfer effects are significant.

PROBLEM 5.3: The diameter of the agarose beads is 8 mm. It contents 0.018 kg protein per m3 gel. Ten beads are immersed in a stirred tank reactor containing 3.2 × 10−3 kg/m3 substrate. The effective diffusivity of substrate in agarose gel is 2.1 × 10−9 m2/s. The kinetics of the enzymatic reaction is assumed to be first order with specific rate constant 3.11 × 10 5 s−1 kg−1 protein. Find out the type of the limiting regime. SOLUTION Given data:

R = 4 × 10−3 m

Des = 2.1 × 10−9 m2/s

In the absence of external mass-transfer effects, Sb = 3.2 × 10−3 kg/m3. 4 Volume per bead = ¥ p ¥ (4 ¥ 10-3 ) m3 = 2.68 × 10−7 m3 3

Kinetics of Immobilized Enzymes

Therefore, volume of 10 beads is 2.68 × 10−6 m3. The amount of enzyme present is 2.68 × 10−6 m3 (0.018 kg/m3) = 4.83 × 10−8 kg. Therefore, K = 3.11 × 105 s−1 kg−1 (4.83 × 10−8 kg) = 0.015 s−1 The kinetics is first order: (−rS)obs = kSb

The observable Thiele modulus is 2

Ê Vp ˆ ( -rS )obs F=Á ˜ Ë Ap ¯ Des S b

For spherical beads,

Ê R ˆ ( -rS )obs F=Á ˜ Ë 3 ¯ Des S b 2

2

2 Ê 4 ¥ 10-3 ˆ È 0.015 ˘ Ê R ˆ kS b F=Á ˜ =Á ˙ = 12.69 ˜ Í Ë 3 ¯ Des S b Ë 3 ¯ ÍÎ 2.1 ¥ 10-9 ˙˚

From the Weisz criteria (Φ > 3), diffusion is the limiting regime.

PROBLEM 5.4: Glucose is converted to fructose by using immobilized glucose isomerase. Find the height of the immobilized enzymes column? Following data are given: Diameter of the column DT = 5 cm Particle size 30/40 mesh (average diameter (dp) = 0.71 mm) Feed rate F = 500 mL/h Glucose concentration in feed at 60°C is 500 g/L. Glucose conversion (C.E.) efficiency is 60%. Feed viscosity µ = 3.6 c.p. at 60°C Feed density r = 1.23 g/mL at 60°C Substrate diffusivity D = 0.21 × 10−5 cm2/s at 60°C Void fraction e = 0.35

SOLUTION We assume that Z is the height of the column, ε is the void fraction, av is the ratio of the particle surface area to volume, Y2 is the mole fraction of the substrate in the product, Y1 is the mole fraction of the substrate in the feed, and DT is the diameter of the column.

201

202

Immobilized Enzymes

Product (C.E. 60%)

Immobilized glucose isomerase

Z

Glucose, 500 g/L

Satterfield has suggested an expression for the column height as follows: 2

2

e(Re )3 ( Sc )3 Ê Y1 ˆ Z= ln Á ˜ 1.09av Ë Y2 ¯

where the Reynolds number Re = Velocity of the fluid (u)=

DTur m

Volumetric feed flow rate Cross-sectional area of the column

Volumetric feed flow rate = 500

mL mL = 0.139 h s 2

ÊD ˆ Cross-sectional area of the column = p Á T ˜ = p (2.5)2 cm2 Ë 2¯ mL s = 7.079 ¥ 10-3 cm u= 2 s p (2.5) cm2 0.139

Re =

(5)(7.079 ¥ 10-3 )(1.23) = 0.0121 3.6

Scmidth number = Sc =

m 3.6 = = 1.3937 ¥ 106 Dr 0.21 ¥ 10-5 ¥ 1.23

References

av =

=

Ê dp ˆ 4p Á ˜ Ë 2¯

2

6 Surface area of particle = = 3 Volume of particle dp 4 Ê dp ˆ pÁ ˜ 3 Ë 2¯ 6 6 = = 84.50 cm -1 0.71 mm 0.071 cm

500 Mole of glucose Y1 = = 180 = 1 Total moles of all constituents 500 180 Mole of glucose Y2 = = Total moles of all constituents

500 ¥ (1 - 0.60) 180 500 180

500 ¥ 0.40 180 = = 0.40 500 180 Putting all the known values in the proposed equation, we get 2

Z=

2

0.35(0.0121)3 (1.3937 ¥ 106 )3 1.09 ¥ 84.50 cm

-1

Ê 1 ˆ ln Á = 2.29 cm Ë 0.40 ˜¯

References 1. Bernath, F. R., Venkat Subramanian, K., and Veith, W. L. Immobilized enzymes, In: Annual Reports on Fermentation Processes, D. Perlman (Ed.), Vol. 1, Academic Press, 2014. 2. Bailey, J. E. and Ollis, D. F. Biochemical Engineering Fundamentals, McGraw-Hill Inc., New Delhi, India, 2010.

3. Messing, R. A. (Ed.). Immobilized Enzymes for Industrial Reactors, Academic Press, New York, 1975.

4. Pitcher, W. H. (Ed.) Immobilized Enzymes for Food Processing, CRC Press, 1980.

203

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Immobilized Enzymes

5. Kennedy, J. F. and Roig, M. G. Principles of immobilized enzymes, In: Handbook of Enzyme Biotechnology, A. Wiseman (Ed.), Halsted Press, 1985.

6. Woodward, J. Immobilized enzymes: Adsorption and covalent coupling, In: Immobilized Cells and Enzymes: A Practical Approach, J. Woodward (Ed.), IRL, Oxford, U.K., pp. 3–17, 1985. 7. Messing, R. A. Carriers for immobilized biologically active systems (enzymes), In: Advances in Biochemical Engineering, T. K. Ghose, A. Fiechter, and N. Blakebrough (Eds.), Vol. 10, Springer-Verlag Berlin Heidelberg, 1978.

8. Das, D. and Biswas, A. K. Short-term course on “Application of immobilized techniques in biotechnology,” Indian Institute of Technology Kharagpur, India, 1992. 9. Chibata, I. (Ed.). Immobilized enzymes, In: Methods in Enzymology, Halsted Press, New York, 1978.

10. Hashem, M. S. and Abd El-Ghaffar, M. A. Immobilized Enzymes, Lambert, 2016.

11. Guisan, J. M. (Ed.). Immobilization of Enzymes and Cells, Humana Press, 2013. 12. Torchilin, V. P. Immobilized Enzymes in Medicine, Springer-Verlag, 1991. 13. Yoo, Y. J., Feng, Y., Kim, Y. H., and Yagonia, C. F. J. Fundamentals of Enzyme Engineering, Springer, 2017. 14. Yong, G. Enzyme Engineering, Science Press, 2014.

15. Kartan, P. Enzyme Engineering, Arcler Education Incorporated, 2017.

16. Aris, R. The Mathematical Theory of Diffusion and Reaction in Permeable Catalysts, vol. 1, Oxford University Press, London,1975.

Chapter 6

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

In enzymatic reactions, enzymes require a suitable substrate and operating conditions such as temperature, pH, etc. Living cells require a proper medium and different operational parameters for growth and product formation. The liquid medium contains the following ingredients: carbon source, nitrogen source, minerals, and vitamins. The carbon source may be utilized for cell growth, product formation, energy source, and maintenance. The nitrogen source mostly contributes to the growth of cells, whereas minerals and vitamins are utilized mostly as cofactors in the metabolic pathways of living cells. The growth of cells is monitored in two ways. In unicellular organisms, the mass of living cells is proportional to the number of cells. So the concentration of the cells may be expressed in either the number of viable cells/volume or the mass of cells/ volume. But in the case of mold growth, the size and density of the cells are not necessarily proportional to their numbers. So the concentration of mold-like cells is expressed only in mass/volume.

Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

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Microbial Growth Kinetics, Substrate Degradation, and Product Formation

Two processes are associated with cell growth: (i) utilization of materials from the medium by the cells and (ii) generation of metabolic end products in the medium. Figure 6.1 depicts the kinetics of the biochemical processes involved for cell growth and metabolism. It is impractical to consider a kinetic model that has all the features mentioned in Fig. 6.1. So it is necessary to simplify the model with some approximation and develop a useful representation of cell growth kinetics. First, it is assumed that the concentration of all components present in the medium is sufficiently high and the rate of reaction depends on the concentration of that component only. This is termed as the rate-limiting substrate. The process parameters are regulated by controlling the environmental parameters such as pH, temperature, and dissolved oxygen concentration. Arnold Fredrickson and Henry Tsuchiya classified the microbial systems according to the number of components used in the cellular representation (Fig. 6.2). The structured model usually deals with the kinetics of the change in individual components present in the cells, such as RNA, DNA, proteins, etc. The unstructured model is considered a singlecomponent system assuming the kinetics of change of all components is same. In the segregated model, the kinetics of individual cells is taken into consideration separately. In the unsegregated model, the growth characteristics of individual cells are assumed to be same. The real condition of the living system is a structured, segregated one. In ideal condition, the cell growth kinetics is assumed to be in the unsegregated, unstructured mode. Medium environment

• • • • • •

Multicomponent Reactions in solution Acid-base equilibria Variable pH, temp. etc. Changing rheological properties (density, viscosity etc.) Multiphase

Liquid medium

Products Heat Mechanical

Cell population

• Cell-to-cell heterogeneity • Chain reactions • Internal control • Adaptibility • Stochastic • Genetic change

interactions

Figure 6.1 Summary of some of the important parameters, phenomena, and interactions that determine cell population kinetics. Adapted from Ref. [1].

Unsegregated

Differences between Enzymatic and Microbial Reaction

Unstructured Cell population is treated as one component solute

Structured Cells population treated as multicomponent like DNA, RNA Balanced growth

Real case

Segregated

Ideal case

Single component, heterogeneous cells Balanced growth

Multicomponent, heterogeneous cells

Figure 6.2 Different growth characteristics of cells. Adapted from Ref. [1].

6.1 Differences between Enzymatic and Microbial Reaction In an enzymatic reaction, the reaction mixture usually consists of a substrate, an enzyme, and a cofactor (if any). The reaction normally takes place at a particular temperature and pH. Here the substrate is specific with respect to the enzyme; e.g., glucose isomerase reacts only on glucose to produce fructose. On the other hand, in a microbial reaction, one substrate (carbon source) present in the medium can produce different products with the help of different microorganisms. In special cases, the same microorganism can produce different products from the same substrate; e.g., Aspergillus niger produces citric acid when the pH is allowed to change during the fermentation process. However, it produces gluconic acid when the pH is controlled at 6.0. The differences between enzymatic and microbial fermentation processes are shown in Table 6.1.

207

208

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

Table 6.1

Differences between enzymatic fermentation processes

Enzymatic reactions

reactions

and

microbial

Microbial fermentation processes

Usually globular proteins with an Different types of reactions carried active site that catalyzes a specific out by living organisms reaction Each enzyme requires a specific Can act on a variety of substrates substrate

Works only at a particular pH and Requires an optimal range of pH and temperature temperature

Does not have the characteristics Has acclimatization properties of acclimatization with respect to the environmental conditions or substrate Every reaction substrate.

requires

a A cell growth medium consists of a carbon source, a nitrogen source, vitamins, minerals, etc.

Does not have the mechanism of Has reproduction characteristics reproduction

6.2 Bacterial Growth Cycle

During growth, the bacteria pass through different phases: lag phase, log phase, stationery phase, and death phase, as shown in Fig. 6.3. Different growth phases have different significances. During the lag phase, bacteria acclimatize themselves to the environment. In this period, the individual bacteria become mature and cannot divide. In this phase, RNA, enzymes, and other molecules undergo synthesis. The lag phase can last for 1 h to several days. It is also known as the acclimatization phase. The log phase is also termed as the logarithmic phase or the exponential phase. Cells are active in this phase. The age of inoculums plays an important role in the fermentation process. Usually microbial cells in between the mid-log phase and the latelog phase are most suitable for fermentation. The number of new bacteria produced per unit time is proportional to the present population. A straight line will be obtained in this phase of growth if the logarithm of cell mass is plotted against time (Fig. 6.3). The

Bacterial Growth Cycle

slope of this line gives the specific growth rate of the organism. This is a measure of the number of divisions per cell per unit time in the case of unicellular cells. The specific growth rates of different microorganisms are not uniform. The rate of microbial growth is characterized by the specific growth rate (mg): mg =

1 dX 1 dN = X dt N dt

(6.1)

where X is the cell mass concentration, N is the number of cells, and t is time. Here mnet is the difference between the gross specific growth rate (mg) and the rate of cell death (md)

mnet = mg – md

In the case of no cell death, Eq. 6.2 may be modified as mnet = mg =

d ln X dt

(6.2)

(6.3)

Plot ln X versus t is shown in Fig. 6.3. The slope of the straight line is equal to µg. Slope = mg

ln X

t

Figure 6.3 Plot of ln X versus t.

PROBLEM 6.1: At each cell division, a new microorganism produced three daughters. The cell growth rate data is given below. Estimate the generation time of the microorganism. Time (h) 0 0.5 1.0 1.5 2.0

Dry cell weight (g/L) 0.10 0.15 0.23 0.34 0.51

209

210

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

SOLUTION From the given data, the following table can be obtained. t

X

ln (X/X0)

0

0.1

0.00

1

0.23

0.83

0.5

0.15

1.5

0.34

2

0.51

0.41

1.22

1.63

Figure 6.4 The plot of ln (X/X0) versus t gives the slope µ.

From the graph, µ = 0.82 h−1 We know that generation time t gn Given n = 3.

Therefore, t gn =

ÊX ˆ ln Á n ˜ Ë X ¯ = m

ln(3) = 1.34 h 0.82

(6.4)

The stationary phase (Fig. 6.5) is considered the starvation phase due to the depletion of an essential nutrient, and/or the formation of an inhibitory product such as organic acids. In the stationary phase, the rate of cell growth is equal to the rate of cell death. Mutations

Ideal Reactor for Kinetics Measurements

can occur during the stationary phase. Secondary metabolites are formed in this phase. The death phase is also known as the decline phase where the bacteria die. This is mostly due to the lack of nutrients, environmental temperature above or below the tolerance limits for the species, or other injurious conditions. Log of numbers of bacteria

Stationary phase

Exponential phase

Death phase

Lag phase Time

Figure 6.5 Growth cycle of the bacteria.

6.3 Ideal Reactor for Kinetics Measurements Three different processes are mostly used for microbial fermentation, as shown in Fig. 6.6. Microbial cell growth

Batch

Fed batch

Continuous

Figure 6.6 Different microbial fermentation processes.

It is difficult to obtain useful kinetic information on microbial populations from reactors in nonuniform condition. So it is necessary to study kinetics in well-mixed reactors. A detailed analysis of cell growth kinetics is performed by using batch stirred-tank reactors and continuous flow stirred-tank reactors.

211

212

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

6.3.1 The Ideal Batch Reactor Several biochemical processes are carried out to study cell growth in batch process. Inoculum is generally prepared by using seed culture in a liquid medium. In a batch reactor, the concentrations of nutrients, cells, and products vary with time as cell growth takes place. Batch process is depicted in Fig. 6.7. Seed culture

Medium

Figure 6.7 Batch stirred-tank reactor.

The microbial biomass and product formation can be represented as follows: Substrate + Microbial cells Æ Extracellular products + More                microbial cells, i.e.,

ÂS + X Æ ÂP + nX

(6.5)

where S is the substrate concentration (mass/vol), X is the cell mass concentration (mass/vol), P is the product concentration (mass/ vol), and n is the number of cell divisions. From the material balance of any component “a,” we get

d( S aVR ) dS = VR a = VR ra (6.6) dt dt where VR is the working volume of the reactor, which is usually constant; Sa is the moles of “a” per volume; ra is the rate of formation of “a” (in moles of “a” per volume per time). If no liquid is added to or removed from the reactor, and if the gas stripping of the culture liquid is negligible, then

Ideal Reactor for Kinetics Measurements

dS a = ra dt

(6.7)

In general, the rate of formation, ra, depends on the characteristics of the cell population (composition, morphology, and age distribution) and all environmental parameters that influence the rate of reactions in the cells and the medium. In the case of batch process, bacterial cell growth cycle can be developed. The roles of different phases are shown in Table 6.2. Table 6.2

Functions of different phases of cell growth

Phase

Description

Specific growth rate

Lag

Cells acclimatize to the new environment, no growth

mnet ~ 0

Log

Stationary Death

Cells are active and growth achieves its maximum rate

Cells are under starvation and growth ceases due to starvation Cell losses viability and lyse

mnet ~ mmax

mnet ~ 0 mnet < 0

Batch process is an unsteady state operation where the concentration of cell mass, substrate changes with time. The cell mass (X) balance can be written as: Input + Cell generation = Output + Accumulation + Cell death

d( XV ) + mdXV (6.8) dt where mg and md are the rate of specific cell growth and specific cell death, respectively. For constant volume V, Eq. 6.8 can be written as 0 + mgXV = 0 +

dX = ( m g - m d )X (6.9) dt If specific cell death rate (µd) is negligible as compared to cell growth (mg = mnet = m),

dX = mX (6.10) dt In batch growth in the log phase, mg remains approximately constant and approaches mmax. Rearranging and integrating Eq. 6.10, we get

213

214

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

X

Ú

X0

t

dX = m dt X

Ú 0

Ê X ˆ ln Á ˜ = mt Ë X0 ¯ Ê X ˆ ln Á ˜ Ë X0 ¯ t= m

(6.11) (6.12) (6.13)

The doubling time of the cell (td) is defined as the time required for doubling the microbial mass (X = 2X0): ln2 µ

td =

The minimum doubling time is

td min =

ln 2 µmax

Xn

t gn

(6.14)

(6.15)

The generation time of a cell is defined as the time required between cell divisions. If Xn is the number of cells produced during the cell division, the generation time (tgn) may be calculated as follows:

Ú

X0

dX = X

Ú µdt 0

ÊX ˆ ln Á n ˜ = µ(t gn ) Ë X0 ¯

t gn

ÊX ˆ ln Á n ˜ Ë X0 ¯ = µ

(6.16) (6.17) (6.18)

1 dX ˆ Ê ÁË µ = X dt ˜¯ , specific substrate consumption rate and specific product formation rate may be expressed as follows: The specific substrate consumption rate is Like

specific

growth

rate

qS =

1 dS X dt

(6.19)

Ideal Reactor for Kinetics Measurements

The specific product formation rate is qP =

1 dP X dt

(6.20)

6.3.2 Cell Growth Models Different mathematical models have been proposed to describe the growth kinetics of microbial cells. Jacques Monod described a cell growth model like the Michaelis–Menten equation, which describes enzymatic reaction. This is considered an ideal equation, i.e., an unstructured, unsegregated model (Eq. 6.21): m=

mmax S KS + S

(6.21)

where m is the specific growth rate (h−1), mmax is the maximum specific growth rate (h−1), KS is the saturation constant (g/L), and S is the growth-limiting substrate concentration (g/L). The correlation between m and S is shown in Fig. 6.8. In the case of enzymatic reaction, the substrate is very specific with respect to enzyme. In the case of microbial growth kinetics, the growth-limiting substrate might be either a carbon source or a nitrogen source or any component present in the growth medium, provided the variation in the concentration of the same affects the specific growth rate of the cell. mmax

Specific growth rate (m)

S = Ks Limiting substrate concentration (S)

Figure 6.8 Correlation between specific growth rate (m) and limiting substrate (S).

215

216

Microbial Growth Kinetics, Substrate Degradation, and Product Formation



The limitations of the Monod model are as follows: ∑ ∑ ∑ ∑ ∑

When S →∞, m → mmax. m is finite when S is finite. Does not explain what will happen when S → 0. Does not take care of cell death. Not applicable in the case of substrate and product inhibitions.

Other substrate-limited growth models proposed by different researchers as an alternative to the Monod equation are as follows: Blackman equation:

mg = mmax if S ≥ 2Ks

(6.22)

mg = mmax (1 - e - K S )

(6.24)

mg =

mmax S if S < K S 2K S

Tessier equation:

Moser equation: mg =

mmax S n

KS + S n

= mmax (1 + K S S - n )-1

Contois equation: mg =

mmax S K SX X + S

(6.23)

(6.25) (6.26)

The Contois model very much resembles with the Monod equation, where the saturation constant depends on the cell mass concentration.

6.3.2.1 Generalized growth model

The different growth equations can be described by a single differential equation as follows: du = K u a (1 - u )b dS

(6.27)

where u = mg/mmax ; S is the rate-limiting substrate concentration; and K, a, and b are constants.

Ideal Reactor for Kinetics Measurements

Table 6.3

Values of a, b, and K for different growth equations a

b

K

Monod

0

2

1/Ks

Tessier

0

1

1/K

Moser

Contois

Note: Taken from Ref. [3].

1 + (1/n) n/Ks1/n

1 − (1/n) 0

2

1/KSX

6.3.3 Cell Growth Models with Cell Growth Inhibitors Three types of inhibitions may take place in microbial growth: 1. Substrate inhibition 2. Product inhibition 3. Inhibition by toxic compounds

The inhibition pattern of microbial growth is analogous to the inhibition of enzymatic reaction. Often the underlying mechanisms are complicated. The kinetic constants are obtained from the experimental data by curve fitting.

6.3.3.1 Substrate inhibition

At high substrate concentrations, the microbial growth rate is inhibited by the higher substrate concentration, as shown in Fig. 6.9. No inhibition

Substrate inhibition m≠

S

Figure 6.9 Correlation between µ and S in the case of substrate inhibition.

217

218

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

The substrate inhibition for cell growth may be either competitive or non-competitive. In non-competitive substrate inhibition, the specific growth rate of the cell may be expressed as m=

If KI >> KS, then

mmax

KS ˆ Ê Ê Sˆ ÁË 1 + S ˜¯ Á 1 + K ˜ Ë I¯

m=

mmax S Ê S2 ˆ KS + S + Á ˜ Ë KI ¯

(6.28)

(6.29)

In the case of competitive substrate inhibition, the cell growth equation may be written as m=

mmax S

Ê Sˆ KS Á 1 + ˜ + S KI ¯ Ë

(6.30)

Substrate inhibition may be alleviated by a slow, intermittent addition of substrate to the growth medium.

6.3.3.2 Product inhibition

High product concentration can be inhibitory for microbial growth. The product inhibition of growth may also be competitive or noncompetitive. The non-competitive product inhibition may be expressed as m=

mmax

KS ˆ Ê Ê P ˆ ÁË 1 + S ˜¯ Á 1 + K ˜ Ë P¯

where KP is the product inhibition constant. The competitive product inhibition is given as follows: m=

mmax S

Ê P ˆ +S KS Á 1 + K P ˜¯ Ë

(6.31)

(6.32)

Other cell growth rate expressions for the product inhibition are shown in Eqs. 6.33 and 6.34.

Ideal Reactor for Kinetics Measurements

m=

mmax S Ê P ˆ 1Á KS ˆ Ë Pm ˜¯ Ê ÁË 1 + S ˜¯

n

(6.33)

where Pm is the product concentration at which cell growth stops. m=

mmax e - P/K P KS ˆ Ê KS Á 1 + Ë S ˜¯

(6.34)

6.3.3.3 Toxic compound inhibition Inhibition by toxic compounds is analogous to enzyme inhibition as follows: Non-competitive inhibition: m=

mmax KS ˆ Ê Ê I ˆ ÁË 1 + S ˜¯ Á 1 + K ˜ Ë I¯

(6.35)

Competitive inhibition: m=

mmax S Ê I ˆ KS Á 1 + ˜ + S KI ¯ Ë

(6.36)

Uncompetitive inhibition: m=

mmax S Ê KS ˆÊ I ˆ Á I+ I K + S ˜ Á I+ K ˜ Ë ( I) ¯Ë I¯

(6.37)

6.3.4 Logistic Equation The logistic equation characterizes cell growth in terms of carrying capacity, i.e., the maximum cell mass that can be obtained (Xm). It does not consist of substrate concentration. The expression for cell growth rate can be represented as: Ê dX X ˆ = mmax X Á 1 dt X max ˜¯ Ë

(6.38)

219

220

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

where mmax is the maximum specific growth rate constant (h−1), Xmax is the maximum biomass concentration at the end (g/L), and X is the biomass concentration at any time t (g/L). The integral form of the above equation can be written as: X=

X 0e mmaxt

(

)

Ï ¸ X0 1 - e mmaxt ˝ Ì1 X Ô˛ max ÓÔ

(6.39)

The differences between the logistic equation and the Monod equation are given in Table 6.4. Table 6.4

Comparison of logistic equation and Monod equation

Logistic equation

Monod equation

Microbial growth is related to Independent of substrate concentration and is only related to biomass concentration and limiting substrate concentration. biomass concentration. Cell growth rate is directly proportional to biomass concentration and the carrying capacity (Xm – X).

It does not consider more than one growth-limiting substrate concentration.

6.3.5 Cell Growth Characteristics of Multicellular Cells Like Mold Filamentous organisms such as mold form microbial pellets at high cell densities in suspension culture. Cells growing inside pellets are subjected to diffusional limitations. In the absence of mass transfer limitations, the radius of pellet in the submerged culture increases linearly with time such as: dR = kP dt

where R is the pellet radius. The biomass M can be expressed as follows: 4 M = r p R3 3

(6.40)

(6.41)

Ideal Reactor for Kinetics Measurements

From Eqs. 6.40 and 6.41, the growth rate can be expresses as dM dR = r 4p R2 = kp r 4p R2 dt dt dM = g M2/3 dt where g = kp(36pr)1/3 Integrating Eq. 6.43 with an initial biomass of M0, we get gtˆ Ê M = Á M01 3 + ˜ Ë 3¯

(6.43)

3

Since M0 > θ. But in case of chemostat θC = θ. So the purpose of cell recycling is to increase the cell residence time.

235

236

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

On the other hand, if x is known, Eq. 6.118 can be used to find the reactor volume (V) as follows: V=

F0 ( S0 - S )YX/S q c X (1 + q c md )

6.5 Continuous Operation Using Plug Flow Reactor

(6.119)

Analysis of a plug flow reactor (Fig. 6.14) for cell culture follows the same procedure as that for an enzymatic reaction. The material balance for cell mass in a small section (DZ) can be represented as follows: Input + Generation = Output + Consumption + Accumulation L

F

F, Xc

F

Xz

Z

X z+∆z

F, X

∆Z

Figure 6.17 Cell mass balance across a plug flow reactor.

FXz + mXADz = FXz+Dz + 0 + 0

(6.120)

F(Xz+Dz – Xz) = mXADz

(6.121)

where A is the cross-sectional area of the reactor, F is the volumetric flow rate, and dV = ADz. ( X z + Dz - X z ) = mX Dz where u is the velocity of the liquid. Applying the limit z Æ 0 to the above equation, we get u

- Xz )˘ È( X u Í z + Dz ˙ = mX Dz Î ˚ Ê dX ˆ uÁ ˜ = m X Ë dz ¯

The drawbacks of a plug flow reactor are as follows:

(6.122) (6.123) (6.124)

Continuous Operation Using Plug Flow Reactor





∑ Not suitable for suspended cell growth. ∑ Can be used for cell recycling or immobilized cell reactions, but difficult to operate. ∑ Difficult to control due to temperature and composition variations. ∑ Maintenance is more expensive in PFR. ∑ Industrial fermentations are very less.

Comparison between major modes of cultivation is as follows: ∑ Kinetic characteristics of PFRs are the same as those of the batch reactors. ∑ When many CSTRs are connected in series, the conversion characteristics approach those of batch and PFRs. ∑ A decrease in substrate concentration in PFR or batch is observed. ∑ In a single CSTR, the outlet substrate concentration is the same under steady-state condition at a particular flow rate and feed substrate concentration. ∑ In the case of cascade (CSTRs are connected in series), there is step-wise drop in substrate concentration between each stage mimicking PFR or batch, as shown in Fig. 6.18. s Large number of CSTRs

Four CSTRs of equal size PFTR (or batch reactor) Single CSTR

s

Figure 6.18 CSTR (chemostat) in series is equivalent to PFR or batch process.

237

238

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

The rates of conversion of substrate in a chemostat operated at Dmax are usually 10–20 times greater than those of batch. A CSTR has significant theoretical advantages over PFR. Despite the benefits of CSTRs, most commercial fermentations are conducted in batch. Batch fermentations have lower risk of contamination compared to CSTRs. Equipment and control failures during long-term operation are the associated problems with CSTRs.

6.6 Kinetics of Fed Batch Cell Growth

A major problem of the fermentation industry is substrate inhibition. The fed batch process is usually used when a substrate acts as an inhibitor of the fermentation process. Examples are Baker’s yeast production, penicillin fermentation, etc.

6.6.1 Variable Volume Fed Batch

If the substrate is added continuously at a constant flow rate F, the rate of change in volume (V) can be given as dV =F dt Rearranging and integrating the above equation, we get V

t

V0

0

ÚdV = ÚFdt V = V0 + Ft

(6.125)

(6.126)

(6.127)

where V is the volume of the reactor at time t and V0 is the initial volume of the reactor (at t = 0)(Fig. 6.19). Figure 6.20 shows the profiles of cell mass concentration, substrate concentration, specific cell growth rate, and volume of a fed batch reactor. At a quasi-steady state, Sadded → Sconsumed and X is constant. The cell mass balance can be given as Input + Cell generation = Output + Accumulation + Cell death FX 0 + m XV = 0 +

(assuming cell death is negligible)

dXV +0 dt

(6.128)

Kinetics of Fed Batch Cell Growth

FX 0 + m XV = X

dV dX +V dt dt

d db da (ab) = a + b ) dx dx dx dX At steady state, X0 = 0; =0 dt

(6.129)

(Since

mXV = X

Therefore,

m=

dV dt

(6.130)

1 dV F = V dt V

(6.131)

F, So

V0

Figure 6.19 Fed batch process with variable volume of feed.

The ratio of flow rate (F) and volume (V) is called dilution rate (D), which can be represented as D=

Thus, from Eq. 6.127

m=D=

F V

F V0 + Ft

Applying the Monod kinetics gives m=D=

mmax S F = K s + S V0 + Ft

(6.132) (6.133) (6.134)

239

240

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

By rearranging, we get

S=

KsD mmax - D

(6.135)

V

m

So X S Time

Figure 6.20 Profiles of cell mass concentration, substrate concentration, specific cell growth rate, and volume of a fed batch reactor.

Now, the biomass concentration at time t can be given as

Xt V where Xt is the total biomass concentration. At quasi-steady state, X=

ÊX ˆ dÁ t ˜ dX ËV ¯ = 0 i.e., =0 dt dt

Ê dX ˆ Ê dV ˆ V Á t ˜ - Xt Á Ë dt ˜¯ Ë dt ¯ V2 da db b -a d dx dx ) (Since (a/b) = 2 dx b

=0

dX t X t dV = = FX dt V dt

(6.136)

(6.137)

Kinetics of Fed Batch Cell Growth

The total biomass concentration (Xt) can be expressed as Xt = X0 + (Xt – X0)

(Since YX / S = as

Xt = X0 + YX/S(S0 – S)

Xt - X0 ) S0 - S

(6.138)

(6.139)

Now, when S = 0 and X0 > μmax, there will be no cell present in the bioreactor due to washout of the cell. So X = 0 g/L. D=

247

248

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

PROBLEM 6.9: Lactococcus lactis has a maximum specific growth rate (μm) of 1.23 h−1 in a glucose–yeast extract medium. Find out the specific growth rate of this organism at steady state in a 4 L reactor at a flow rate of 2 L/h? SOLUTION Given

μmax = 1.23 h−1, V = 4 L, and F = 2 L/h D=

F 2 = = 0.5 V 4

At steady state, μ = D. Therefore, μ = 0.5 h−1

PROBLEM 6.10: Recombinant porcine growth hormone is produced by Escherichia coli. The bacteria are grown aerobically in a batch culture with glucose as the growth-limiting substrate. Cell and substrate concentrations are measured as a function of culture time as shown in Table 6.5: Table 6.5

Cell and substrate concentration with respect to time in a batch process

Time (h)

Cell concentration X (kg/m3)

Substrate concentration S (kg/m)

0.0

0.20

25.0

0.75

0.32

24.6

0.33 0.5 1.0 1.5 2.0 2.5 2.8 3.0 3.1 3.2 3.5 3.7

0.21 0.22 0.47 1.00 2.10 4.42 6.9 9.4

10.9 11.6 11.7 11.7

24.8 24.8 24.3 23.3 20.7 15.7 10.2 5.2

1.65 0.2 0.0 0.0

Kinetics of Fed Batch Cell Growth

(a) What are the values of mmax and KS? (b) What is the observed cell yield coefficient?

SOLUTION Calculated data obtained from Table 6.5 are shown in Table 6.6. Table 6.6

Calculated data

Time X

1/X

0.00

0.20

5.00 25.00 0.04

0.75

0.32

3.13 24.60 0.04 0.50

0.33 0.50 1.00 1.50 2.00 2.50 2.80 3.00 3.10 3.20 3.50

3.70

0.21 0.22 0.47 1.00 2.10 4.42 6.90 9.40

S

1/S

dX/dt dS/dt µ(1/X dX/dt) 1/µ

4.76 24.80 0.04 0.04

0.40

0.19

5.25

2.13 24.30 0.04 0.91

1.73

1.93

0.52

4.55 24.80 0.04 0.26 1.00 23.30 0.04 1.63 0.48 20.70 0.05 3.42 0.23 15.70 0.06 6.00 0.14 10.20 0.10 9.96 0.11 5.20

10.90 0.09 1.65 11.60 0.09 0.20 11.70 0.09 0.00

11.70 0.09 0.00

0.19 13.33 0.61 11.00 5.00 2.00 0.20

0.48 1.00 3.60 7.60

13.13 21.00

1.19 1.56 1.63 1.63 1.36 1.44

0.84 0.64 0.61 0.61 0.74 0.69

28.50

1.42

0.71

0.40

0.02

58.50

25.00 4.13

1.01

0.17

0.99

5.80

(a) From Table 6.6, a plot of 1/m versus 1/S can be used to determine mmax and KS values. From the Lineweaver–Burk plot, we have K 1 1 1 = + S m mmax mmax [S ]

From the graph (neglecting deviating points), 1

mmax

= 0.59 ; mmax = 1.69 h–1

KS = 0.65 ; KS = 1.10 kg/m3 mmax

249

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

1.50

Lineweaver-Burk plot

1.00 1/m

250

y = 0.6546x + 0.5909 R2 = 0.9926

0.50 0.00 0.00

0.20

0.40 1/S

0.60

0.80

Figure 6.22 Plot of 1/m versus 1/[S].

(b) The overall biomass yield (YX/S) is the ratio of cell mass ( X - X0 ) production and substrate consumption, i.e., . S0 - S

From Table 6.5, taking X and S values at t = 3.7 h, we get YX/S = 0.46 kg/kg.

PROBLEM 6.11: A genetically engineered strain of E. coli is used to produce human protein. A batch culture is started by inoculating 12 g cells into 100 L fermenter containing 10 g/L glucose. The maximum specific growth rate (µmax) of the culture is 0.9 h−1. The biomass yield from glucose is 0.575 g/g. (a) Find the time required to reach the stationary phase. (b) What will be the final cell density if the fermentation is stopped after only 70% of the substrate is consumed?

SOLUTION

(a) Given X0 = 12 g, S0 = 10 g/L, V = 100 L, µmax = 0.9 h−1, and YX/S = 0.575 g/g. For a batch culture in the log phase, we know that tb =

(assuming m = mmax) Now,

ln( X X 0 ) mmax

YX / S =

X - X0 S0 - S

Kinetics of Fed Batch Cell Growth

So X = X0 + YX/S (S0 – S). Putting in the above equation, we get tb =

1

mmax

È (YX/S ( S0 - S ) + X 0 ) ˘ ln Í ˙ X0 ÍÎ ˙˚

È (YX/S ( S0 - S )) ˘ ln Í1 + (6.151) ˙ mmax ÍÎ X0 ˙˚ Assuming S = 0 at the stationary phase, putting the given values in the above equation, we get =

tb =

1

1 È (0.575 (10 - 0)) ˘ ln 1 + ˙ 0.9 ÍÎ 0.12 ˚

Therefore, tb = 4.32 h.

(b) Given S = 0.3, S0 = 0.3 (10) = 3 g/L. Like the previous problem, putting the given values in Eq. 6.151, we get tb =

1 È (0.575 (10 - 3)) ˘ ln 1 + ˙ 0.9 ÍÎ 0.12 ˚

Therefore, tb = 3.93 h. The initial cell mass concentration is X0 = 12 g in 100 mL, i.e., 0.012 g/L. Now cell density at this time can be given as X = X 0e mt b = 0.012 e0.9(3.93) X = 0.412 g/L

PROBLEM 6.12: Nicotiana tabacum cells are cultured to high density for producing polysaccharide gum. A stirred-tank reactor was used containing initially 100 L medium. The maximum specific growth rate of the culture is 0.18 d−1, and the yield of biomass from the substrate is 0.5 g/g. The concentration of the growth-limiting substrate in the medium is 3%(w/v). The reactor is inoculated with 1.5 g/L cells and operated in batch until the substrate is totally exhausted. Medium flow is then started at a rate of 4 L/d. Fed batch operation occurs under quasi-steady-state conditions. (a) Estimate the batch culture time and final biomass concentration.

251

252

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

(b) Fed batch operation is carried out for 40 d. What is the final mass of cells in the reactor? (c) The fermenter is available 275 d per year with a downtime between runs of 24 h. How much biomass is produced annually?

SOLUTION

(a) Given µmax = 0.18 d−1, S0 = 3% (w/v) = 30 g/L, X0 = 1.5 g/L, YX/S = 0.5 g/g We know that for a batch reactor, tb is the time required to achieve S = 0, which can be expressed as (assuming m = mmax) tb =

1 È (YX/S ( S0 - S ) + X 0 ) ˘ ln Í ˙ m ÍÎ X0 ˙˚

Putting the given values in the above equation, we get tb =

È (0.5 (30 - 0)) ˘ 1 ln Í1 + ˙ = 13.32 d 0.18 ÍÎ 1.5 ˙˚

Now, cell density at this time can be given as

X = X 0e mt b = 1.5 e0.18(13.32) = 16.5 g/L

(b) The cell mass at the start of the fed batch operation is the product of the final batch cell concentration and the initial medium volume, i.e., X0 = Xbatch × V = 16.5 × 100 = 1650 g

For a fed batch system at quasi-steady state (i.e., variable volume fed batch), cell mass at time t can be given as Xt = X0 + FYX/SS0t

Given F = 4 L/d; t = 40 d Assuming yield is constant, we get

Xt = 1650 + (4 ¥ 0.5 ¥ 30 ¥ 40) = 4050 g = 4.05 kg

(c) The total biomass produced per run can be given as Xper run = Xt − Xi

where Xt is the fed batch cell concentration and Xi is the initial biomass concentration used for inoculation, i.e., Xper run = 4050 – (1.5)(100) (Since Xi = X0V)

Kinetics of Fed Batch Cell Growth

Therefore, Xper run = 3900 = 3.9 kg Now, the total reaction time (tt) can be given as tt = tb + tfb + tdn (batch time + fed batch time + downtime)   =13.32 + 40 + 1 = 54.32 d Given that the fermenter is available for 275 d per year. So the number of runs in a year is 275/54.32 = 5.06 runs. Now the total biomass produced annually can be given as Xannual = Xper run × total number of runs per year = 3.9 × 5.06 = 19.73 kg

PROBLEM 6.13: For the manufacture of Swiss cheese, Lactobaccilus casei is cultivated under anaerobic conditions as a starter culture. The culture produces lactic acid as by-product of energy metabolism. The organism has the following characteristics: YX/S = 0.23 Kg/Kg, Ks = 0.15 Kg/m3, µmax = 0.35 h−1, and ms = 0.135 kg/kg h A stirred-tank reactor is operated in fed batch mode at quasisteady state with a feed flow rate of 4 m3/h and feed substrate concentration of 80 kg/m3. After 6 h, the liquid volume is 40 m3.

(a) What was the initial culture volume? (b) What is the concentration of the substrate at the quasi-steady state? (c) What is the cell mass concentration at the quasi-steady state? (d) How much biomass is produced after 6 h of fed batch operation?

SOLUTION

(a) Given F = 4 m3/ h, V = 40 m3, and t = 6 h. For a fed batch operation at quasi-steady state (variable volume fed batch), the volume of the reactor at time t and initial culture volume V0 can be given as So

V = V0 + Ft

V0 = V – Ft = (40) – (4)(6) = 16 m3

(b) Given F = 4 m3/h, V = 40 m3, Ks = 0.15 kg/m3, µmax = 0.35 h−1. We know that the dilution rate is

253

254

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

D=

F 4 = 0.10 h–1 = V 40

At quasi-steady state,

m=D=

mmax S Ks + S

So the substrate concentration at quasi-steady state can be given as S=

KsD mmax - D

Putting the given values in the above equation, we get S=

0.15(0.10) = 0.06 kg/m3 0.35 - 0.10

(c) At quasi-steady state, the substrate balance (considering substrate requirements for growth, product formation, and maintenance) across the fed batch reactor can be given as Input + Generation = Output + Consumption + Accumulation FS0 + 0 ÈÊ dS ˆ ˘ d( SV ) Ê dS ˆ Ê dS ˆ = 0 + ÍÁ ˜ ˙V + +Á ˜ +Á ˜ Ë ¯ Ë ¯ Ë ¯ dt dt dt dt ÍÎ cell growth product maintain nence ˙ ˚ Ê m ˆ qp d( SV ) FS0 – Á + + ms ˜ X V = dt Ë YX/S YP/S ¯

(6.152)

Assuming the product formation is coupled to energy metabolism, qp = 0. The above equation becomes Ê m ˆ dV dS FS0 – Á + ms ˜ XV = S +V dt dt Ë YX/S ¯ Ê m ˆ dS FS0 – Á + ms ˜ XV = V Y dt Ë X /S ¯

Dividing both sides by V, we get

Ê m ˆ dS DS0 – Á + ms ˜ X = dt Ë YX/S ¯

(Since S = 0) (Since

F = D) V

Kinetics of Fed Batch Cell Growth

At steady state, dS/dt = 0 and m = D, so from the above equation, X can be estimated as X=

D( S0 )

Ê D ˆ + ms ˜ Á Ë YX / S ¯

(6.153)

Given YX/S = 0.23 Kg/Kg, ms = 0.135 kg/kg h, S0 = 80 kg/m3, and from part (b) D = 0.10 h–1 Putting the values in the above equation, we get X=

0.10(80) = 14.04 kg/m3 ÈÊ 0.10 ˆ ˘ ÍÁË 0.23 ˜¯ + 0.135˙ Î ˚

(d) The total biomass produced from the fed batch reactor can be given as Xt = (XVf – XVi)

where Vf and Vi are the final and initial volumes of the reactor during fed batch operation. After 6 h of fed batch operation, Vf = 40 m3 (given) and X = 14.04 kg/m3 (from part c). Therefore, XVf = 14.04 (40) = 561.6 kg. At the start of fed batch operation, Vi = 16 m3 (from part a) and X = 14.04 kg/m3 (from part c). Therefore, XVi = 14.04 (16) = 224.64 kg. So the total biomass produced from the fed batch operation is (561.6 − 224.64) = 336.96 kg.

PROBLEM 6.14: A baker’s yeast factory produces 1 MT compressed yeast (Saccharomyces cerevisiae) per day using cane molasses as a raw material in a chemostat. Compressed yeast contains 70% w/w moisture. The µmax, Ks, and Yx/s values of the yeast are 0.5 h−1, 2 g/L, and 0.5 g/g, respectively. Cane molasses contain 45% (w/w) of sucrose. The initial substrate concentration of the fermentation broth is 200 g/L. Compute the following: (a) (b) (c) (d)

Minimum doubling time of the cell Total amount of cane molasses required Volume of the fermenter Maximum cell mass productivity

255

256

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

SOLUTION Basis: 1 MT compressed yeast per day ≡ 1000 kg compressed yeast per day (70 %w/w moister) Total amount of dry yeast is 1000 (1 − 0.7) = 300 kg per day. (a) We know, minimum doubling time is td,min =

ln(2) 0.693 = = 1.386 h ª 1.4 h µmax 0.5

(b) Amount of sugar required per day is 300 300 = = 600 kg YX S 0.5

We know that

Ê ˆ Ê KS 2 ˆ Dmax = mmax Á 1 = 0.45 h -1 ˜ = 0.5Á 1 (K S + S0 ) ¯ 2 + 200 ˜¯ Ë Ë Steady-state substrate concentration is S=

K S Dmax 2 ¥ 0.45 = = 18 g/L mmax - Dmax 0.5 - 0.45

Therefore, substrate conversion efficiency is 200 - 18 = 0.91 200

So the actual amount of sugar required is 600 = 659 kg/d 0.91

The amount of cane molasses required is (c) t CSTR =

659 = 1464 kg/d = 1.464 MT/d 0.45

S0 - S ( -rS )

( -rS ) =

1 YX

mX = S

1

YX

DX S

(6.154) (6.155)

(under steady-state conditions and sterile feed) Assuming chemostat operate at Dmax to get maximum cell mass production,

Kinetics of Fed Batch Cell Growth

( -rS ) =

1 YX

S

Dmax X

For sterile media, X0 = 0. Steady-state biomass concentration is

(6.156)

X = X0 + YX/S(S0 – S)

Substituting respective values, we get Now

X = 0.5(200 – 18) = 91 g/L

( -rS ) =

1 1 g g Dmax X = ¥ 0.45 ¥ 91 = 81.9 0.5 Lh Lh YX S t CSTR =

S0 - S V = ( -rS ) F

V (200 - 18) = h = 2.22 h F 81.9 Volumetric feed flow rate is

Substrate required Initial substrate concentration 659 kg/d 659 kg/d = = 0 g/L 200 kg/m3 200

F=

= 3.295

m3 m3 = 0.1373 d h

m3 ¥ 2.22 h = 0.305 m3 = 305 L h (d) Maximum cell mass productivity is V = 0.1373

Dmax X = 0.45 ¥ 91

g g = 41 Lh Lh

PROBLEM 6.15: The growth of a microorganism on glucose is described by the Monod model. During the growth of the cell, glucose concentration is 10 g/L and the μmax and Ks values are 0.5 h−1 and 0.1 g/L, respectively. Find the time required to triple the biomass concentration? SOLUTION Given μmax = 0.5 h−1, Ks = 0.1 g/L, and S = 10 g/L.

257

258

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

The Monod equation for the cell growth is µ =

µmax S 0.5 ¥10 = = 0.495 h -1 0.1 + 10 KS + S

If cell death is negligible as compared to growth (mg = mnet), dX = mX dt

Rearranging and integrating the above equation, we get X

Ú

X0

t

dX = m dt X

Ú 0

Ê X ˆ ln Á ˜ = mt Ë X0 ¯ t=

ln X /X 0 m

Time required to triple cell mass (X = 3X0) is t triple =

ln 3 ln 3 = m 0.495

t triple =

ln 3 = 2.22 h m

PROBLEM 6.16: A distillery factory produces 100 m3 rectified sprit (containing 90% v/v ethanol) in a chemostat from cane molasses (containing 50% w/w sugar) using S. cerevisiae. The characteristics of the yeast are as follows: mmax = 0.05 h–1, KS = 2 g/L, YX/S = 0.05, YP/S = 0.5, and S0 = 300 g/L

Find the volume of the bioreactor and the amount of cane molasses required per day. SOLUTION For sterile media, X0 = 0. Steady-state biomass concentration is

Now

X = X0 + YX/S (S0 – S)

Ê Ê ˆ KS Dmax = mmax Á 1 ˜ = 0.05Á 1 (K S + S0 ) ¯ Ë Ë

(6.157)

ˆ 2 = 0.046 h -1 (2 + 300) ˜¯

Kinetics of Fed Batch Cell Growth

Also steady-state substrate concentration is S=

K S Dmax 2 ¥ 0.046 = = 23 g/L mmax - Dmax 0.05 - 0.046

Substituting in Eq. 6.157, we get

X = 0.05(300 – 23) = 13.85 g/L

Basis: 100 m3 spirit ≡ 90 m3 ethanol production per day (90% v/v) = 90,000 L ethanol per day

Density of ethanol is 780 g/L. Amount of ethanol is

780 × 90,000 g = 70.2 × 106 g = 70.2 × 103 kg ethanol per day

Substrate required is

70.2 ¥ 103 kg/d 70.2 ¥ 103 kg/d = = 140.4 ¥ 103 kg/d YP / S 0.5

Actual amount of sugar required is

140.4 ¥ 103 ≈ 143 ¥ 103 kg/d 0.98

(Assuming sugar conversion efficiency 98%) Volumetric feed flow rate is F=

Substrate required Initial substrate concentration

=

143×103kg/d 143×103kg/d = 300 g/L 300 kg/m3

= 476 Now

m3 m3 = 19.86 d h t CSTR = ( -rs ) =

S0 - S ( -rS ) 1

YX

mX S

We substitute Dmax for µ in the case of sterile feed and under steadystate condition,

259

260

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

( -rS ) =

1 YX

S

Dmax X =

1 g g ¥ 0.046 ¥ 13.85 = 12.74 0.05 Lh Lh

t CSTR =

S0 - S V = ( -rS ) F

V (300 - 23) = h = 21.74 h F 12.74 V = 19.86

m3 ¥ 21.74 ª 432 m3 h

Volume of the reactor is 432 m3. Substrate required per day is 143 ¥ 103 kg sugar per day. Cane molasses contains 50% w/w sugar. Therefore, cane molasses required is 286 ¥

103 kg MT = 286 d d

PROBLEM 6.17: Pseudomonas sp. has a minimum doubling time of 2.4 h when grown on acetate in a chemostat that follows the Monod model. g cell g g , S = 38 . Given K S = 1.3 , YX S = 0.46 g acetate 0 L L 1 (a) Find out steady-state S and X when D = Dmax . 2

(b) Find the cell mass productivity at 0.8 Dmax. (c) Determine the value of Dwashout?

SOLUTION It is known that

t dmin = mmax =

ln(2) mmax

ln(2) = 0.288 h -1 t dmin

Ê Ê KS ˆ 1.3 ˆ Dmax = mmax Á 1 ˜ = 0.288 Á 1 K S + S0 ¯ 1.3 + 38 ˜¯ Ë Ë = 0.288 (1 - 0.1818) = 0.235 h -1

Kinetics of Fed Batch Cell Growth

1 (a) D = Dmax = 0.1178 2 S=

KSD 1.3(0.1178) g = = 0.899 mmax - D 0.288 - 0.1178 L

X = YX S ( S0 - S ) = 0.46 (38 - 0.899) = 17

g L

(b) 0.8Dmax = 0.8 ¥ 0.235 h–1 = 0.188 h–1 Cell mass productivity is XD. S=

X = YX/S (S0 – S)

1.3(0.188) g = 2.44 0.28 - 0.188 L

X = 0.46(38 - 2.4) = 16.35 Cell mass productivity (at 0.8Dmax) is XD = 16.35 ¥ 0.188 = 3.07

(c) Dwashout =

g L

g Lh

mmax S0 0.288 ¥ 38 = = 0.278 h -1 K S + S0 1.3 + 38

PROBLEM 6.18: In a chemostat, the steady-state substrate and biomass concentration are given in Table 6.7. The initial substrate concentration (S0) is 700 mg/L. Find out mmax and Ks, the growth yield coefficient, Y ¢X/S(growth), and maintenance coefficient m. Table 6.7

Steady-state cell mass and substrate concentration D (h–1)

S (mg/L)

X (mg/L)

0.3

45

326

0.12

8

342

0.25 0.20 0.08

SOLUTION The Monod equation is

41 16

3.8

328 340 344

261

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

µ=

µmax S KS + S

Under steady-state chemostat operation and sterile feed m=D



D=

So

µmax S KS + S

K 1 1 1 = S + D µmax S µmax

Plotting 1/D versus 1/S yields the value of KS and mmax. Table 6.8

Steady values of 1/S and 1/D 1 (L/mg) S

1 (h) D

0.0222

3.33

0.1250

8.33

0.0244

4.00

0.0625

5.00

0.2632

12.50

12 11 10

KS = 37.53 Slope = m max

9 1/D

262

8 1 = 2.9 Intercept = m max

7 6 5 4 3

0.05

0.15

0.1 1/S

Figure 6.23 Plot of 1/D versus 1/S.

0.2

0.25

Kinetics of Fed Batch Cell Growth

Therefore,

µmax =

1 = 0.34 h -1 2.9

K S = 37.53 ¥ µmax = 12.76

mg L

According to the Pirt model,

1 1 m = + YX/S (overall ) YX¢ /S ( growth ) µ

Under steady-state CSTR operation and sterile feed, m = D. 1 1 m = + YX/S (overall ) YX¢ /S ( growth ) D

So

Plotting 1/YX/S(overall) versus 1/D of Table 6.9 yields the values of m and 1 YX¢ /S ( growth ) . Table 6.9

Calculated values

D (h–1)

1/D (h)

S (mg/L) X (mg/L)

YX/S(overall)

1/YX/S(overall)

0.3

3.33

45

326

0.498

2.008

0.20

5.00

16

340

0.4970

2.012

0.25

0.12

0.08

4.00

8.33

12.50

41 8

3.8

328

342

344

YX / S (overall ) = -

0.4977

0.4942

0.4941

dX X n - X 0 = dS S0 - S n

2.009

2.023

2.024

For the first sampling point, n = 1 YX / S (overall ) = -

dX X1 - X 0 326 - 0 = = 0.498 = dS S0 - S1 700 - 45

From Fig. 6.24, we get m = 0.0019 h–1 (from slope) and YX/S(growth) = 0.50 (from intercept of the Y-axis).

263

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

2.024 2.022 1/Yx/s(overall)

264

2.02 2.018 2.016 2.014 2.012 2.01 2.008 3

4

5

6

7

8 1/D

9

10

11

12

Figure 6.24 Plot of 1/YX/S(overall) versus 1/D.

PROBLEM 6.19: Zymomonas mobilis is used to convert glucose to ethanol in a batch fermenter under anaerobic conditions. The yield of biomass from substrate is 0.06 g/g and the product yield (YP/X) is 7.7 g/g. The maintenance coefficient is 2.2 g/gh. The specific rate of product formation for maintenance is 1.1 h−1. The maximum specific growth rate of Zymomonas mobilis is 0.3 h−1. Five gram of bacteria is inoculated in 50 L media containing glucose 12 g/L. Determine the batch culture time required to produce (a) 15 g of biomass (b) Achieve 90% substrate conversion (c) Produce 100 g ethanol

SOLUTION Given YX/S = 0.06 g/g, YP/X = 7.7 g/g, ms = 2.2 g/gh, µmax = 0.3 h−1, mp = 1.1 h−1, X0 = 5 g/50 L, and S0 = 12 g/L. (a) We know that for a batch culture, Ê X ˆ ln Á ˜ Ë X0 ¯ tb = µmax

Now, X0 = 5 g/ 50 L = 0.1 g/L Similarly, X = 15 g/ 50 L = 0.3 g/L

Kinetics of Fed Batch Cell Growth

Putting in the above equation, we get tb =

ln(0.3 0.1) = 3.66 h 0.3

(b) The substrate balance (considering substrate requirements for growth, product formation, and maintenance) across the batch reactor can be given as Input + Generation = Output + Consumption + Accumulation ÈÊ dS ˆ ˘ Ê dS ˆ Ê dS ˆ 0 + 0 = 0 + ÍÁ ˜ ˙V +Á ˜ +Á ˜ ÍÎË dt ¯ cell growth Ë dt ¯ product Ë dt ¯ maintenancce ˙˚ +

d( SV ) dt

Ê m ˆ qp d( SV ) + + ms ˜ XV = Á Y Y dt Ë X /S ¯ P /S

For a batch reactor, V is constant and when µ = µmax, X = X 0e mmaxt . Putting in above equation, we get Ê mmax ˆ qp dS + + ms ˜ X 0e mmaxt = Á dt Ë YX / S YP / S ¯

Rearranging and integrating the above equation, we get t S Ê mmax ˆ qp mmax t m X e dt + + = dS Á s˜ 0 Ë YX / S YP / S ¯ 0 S

Ú

Ú

0

Ê mmax ˆ Ê e mmaxtb - 1 ˆ qp + + ms ˜ X 0 Á Á ˜ = ( S - S0 ) Ë YX / S YP / S ¯ Ë mmax ¯ Ê 1 qp m ˆ + + s ˜ X 0 e mmaxt b - 1 = ( S0 - S ) Á Ë YX / S YP / S mmax mmax ¯

(

e mmaxt b - 1 =

e mmaxt b =

)

( S0 - S ) Ê 1 qp m ˆ + + s ˜ X0 Á Ë YX / S YP / S mmax mmax ¯ ( S0 - S )

Ê 1 qp m ˆ + + s ˜ X0 Á Ë YX / S YP / S mmax mmax ¯

+1

265

266

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

˘ È Í ˙ Í ˙ ( S0 - S ) 1 tb = ln Í + 1˙ mmax Í Ê 1 qp m ˆ ˙ + + s ˜ X0 ÍÁ Y ˙ m m Y P S max max ¯ ÎË X S ˚

(6.158)

Now, after 90% substrate conversion,

S = (1 − 0.9) S0 = 1.2 g/L

Since ethanol production is coupled to energy metabolism, qp = 0. Putting in Eq. 6.158, we get È ˘ ˙ 12 - 1.2) ( 1 ÍÍ ln tb = + 1˙ = 5.68 h ˙ 2.2 ˆ 0.3 Í Ê 1 Í ÁË 0.06 + 0 + 0.3 ˜¯ 0.1 ˙ Î ˚

(c) The rate of product formation in the batch process can be given as d( PV ) = qp XV dt

Since V is constant,

dP = qp X 0e mmaxt dt

Rearranging and integrating the above equation, we get P

Ú

tb

Ú

dP = qp X 0 e mmaxt

P0

0

Ê e mmaxt b - 1 ˆ P - P0 = qp X 0 Á ˜ Ë mmax ¯ e mmaxt b - 1 = tb =

1 mmax

mmax ( P - P0 ) qp X 0

Èm ˘ ln Í max ( P - P0 ) + 1˙ ÍÎ qp X 0 ˙˚

(6.159)

Kinetics of Fed Batch Cell Growth

Now, the specific rate of product formation can be given as qp = YP/Xm + mP

Since m = mmax, qp = YP/Xmmax + mP Putting the given values in Eq. 6.160, we get qp = (7.7 ¥ 0.3) + 1.1 = 3.41 h

(6.160)

(6.161)

To achieve 100 g ethanol, P=

100 g = 2 g/L and P0 = 0 g/L 50 L

Putting all the values in Eq. 6.159, we get tb =

˘ 1 È 0.3 ln Í (2 - 0) + 1˙ = 3.38 h 0.3 Î 3.41 (0.1) ˚

PROBLEM 6.20: To produce secondary metabolite, a two-stage chemostat system is used. The volume of each reactor is 0.5 m3. The flow rate of feed is 50 L/h. Mycelial growth occurs in the first reactor. The second reactor is used for product synthesis. The concentration of substrate in the feed is 10 g/L. Kinetic and yield parameters for the organism are YX/S = 0.5 kg/kg, KS = 1.0 kg/m3, µmax = 0.12 h−1, qp = 0.16 kg/kg h, and YP/S = 0.85 kg/kg. Assume that product synthesis is negligible in the first reactor and growth is negligible in the second reactor. (a) Find out the cell and substrate concentration entering the second reactor. (b) What is the overall substrate conversion? (c) What is the final concentration of product?

SOLUTION Given V = 0.5 m3, F = 50 L/h, S0 = 10 g/L = 10 kg/m3 YX/S = 0.5 kg/kg, KS = 1.0 kg/m3, µmax = 0.12 h−1, qp = 0.16 kg/kg h, and YP/S = 0.85 kg/kg. (a) We know that

D=

F 50 = = 0.1 h -1 V 500

S1 leaves the first reactor, and S1 enters the second reactor.

267

268

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

Therefore,

S1 =

KSD 1.0(0.1) = = 5 kg/m3 mmax - D 0.12 - 0.1

Similarly, X1 leaving the first reactor = X1 entering the second reactor X1 = X0 + YX/S (S0 – S1)

For a sterile feed, X0 = 0. Therefore, X1 = 0.5(10 –5) = 2.5 kg/m3 F X0 S0

F X2

F X1 S1

S2

P0

P Liquid Volume = Working Volume =V

Figure 6.25 Two chemostat are in series.

As growth is negligible in the second reactor, X1 = X2 = 2.5 kg/m3

The substrate balance in the second reactor can be given as Ê dS ˆ FS1 = FS2 + Á - ˜ V (since substrate is only used for Ë dt ¯ product product formation in the second reactor) = FS2 +

qP X 2V YP/S

Rearranging the above equation for S2, we get S 2 = S1 = 5-

(6.162)

qP X 2V YP/S F

(0.16 ¥ 2.5 ¥ 500) = 0.3 kg/m3 (0.85 ¥ 50)

Kinetics of Fed Batch Cell Growth

Now the overall substrate conversion from both the reactors can be given as Ê S0 - S2 ˆ Ê 10 - 0.3 ˆ ÁË S ˜¯ ¥ 100% = ÁË 10 ˜¯ ¥ 100% = 97% 0

(c) The product balance in the second reactor can be given as P0 = 0,

Since P=

qp X 2V F

Ê dP ˆ FP0 + Á ˜ V = FP Ë dt ¯

=

qpX2V = FP

(0.16 ¥ 2.5 ¥ 500) = 4 kg/m3 50

PROBLEM 6.21: A generalized form of the logistic equation for cell growth is 1

N•a + b

a

dN N ˆ Ê N ˆ Ê 1= kÁ ˜ Á Ë ¯ Ë dt N• N• ˜¯

b

where N• is the number of cells in the stationary phase; a, b are constants; and N is the initial number of cells. N a = Find the maximum growth rate occurring at and N • (a + b) kN•a + baa bb Ê dN ˆ that its value is given by Á . = ˜ Ë dt ¯ max (a + b)a + b SOLUTION Given that

1

N•a + b

a

dN N ˆ Ê N ˆ Ê 1= kÁ Ë N• ˜¯ ÁË dt N• ˜¯

Rearranging the above equation gives a

dN N ˆ Ê N ˆ Ê = kN•a + b Á 1˜ Á Ë N• ¯ Ë dt N• ˜¯

b

b

(6.163)

269

270

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

Differentiating Eq. 6.163, we get

b ˘ È Ê N ˆ a -1 1 dN Ê N ˆ Í ˙ a 1 ÁË N• ˜¯ Á ˜ 2 N• dt Ë N• ¯ d N Í ˙ = kN•a + b Í 2 b -1 a˙ dt Í+b Ê 1 - N ˆ Ê - 1 ˆ dN Ê N ˆ ˙ Í ÁË N• ˜¯ ÁË N• ˜¯ dt ÁË N• ˜¯ ˙˚ Î

Now for (dN/dt)max, d2N/dt2 = 0. Therefore, Eq. 6.164 becomes Ê N ˆ aÁ Ë N• ˜¯

a -1

1 dN Ê N ˆ N ˆ Ê 1- bÁ 1 Ë N• dt ÁË N• ˜¯ N• ˜¯

Ê N ˆ aÁ Ë N• ˜¯

a -1

1 dN Ê N ˆ N ˆ Ê 1= bÁ1 Ë N• dt ÁË N• ˜¯ N• ˜¯ Ê N ˆ aÁ Ë N• ˜¯

b

b -1

Ê 1 ˆ dN Ê N ˆ ÁË N• ˜¯ dt ÁË N• ˜¯ = 0

b

b -1

Ê 1 ˆ dN Ê N ˆ ÁË N• ˜¯ dt ÁË N• ˜¯

a -1

Ê N ˆ ÁË N• ˜¯

a

=

N ˆ Ê bÁ 1 Ë N• ˜¯

N ˆ Ê ÁË 1 - N• ˜¯ b = a Ê N ˆ ÁË N• ˜¯ N• b -1 = N a

N• b = +1 N a

Taking reciprocal on both sides, we get N a = N• a + b N a = N• a + b

a

b -1

N ˆ Ê ÁË 1 - N• ˜¯

N ˆ Ê Ê N ˆ aÁ1 = bÁ Ë Ë N• ˜¯ N• ˜¯

Now, for (dN/dt)max,

(6.164)

b

a

Kinetics of Fed Batch Cell Growth

Putting in Eq. 6.160, we get

a

a ˆ Ê dN ˆ Ê a ˆ Ê 1= kN•a + b Á ÁË dt ˜¯ ˜ Á Ë a + b¯ Ë a + b ˜¯ max

b

kN•a + b (a)a (b)b Ê dN ˆ = ÁË dt ˜¯ (a + b)a + b max PROBLEM 6.22: The Pirt model deals with the maintenance energy of the microbial cells as follows: -

1 1 mX = - ¢ mX - m X YX S YX S

where YX¢ S is the growth yield, grams of cells produced per gram of substrate consumed for growth; YX/S is the overall yield coefficient, grams of substrate consumed per gram of cell; m is the grams of substrate consumed for maintenance energy per gram of cell. (a) Prove that in a CSTR, X and S are related by X=

DYX¢ / S

mYX¢ / S + D

I/Yx/S , g glycerol/g cells

7

0 0

50

100

150

200

250

1/D,h

Figure 6.26 Plot of 1/YX/S versus 1/D in the case of A. aerogenes grown on glycerol.

271

272

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

(b) What relation must hold between YX¢ S and D in this circumstance? (c) Does the figure below for A. aerogenes growing on glycerol satisfy this model? Be quantitative. SOLUTION Given equation (a) -

1

YX

mX = -

S

1

YX¢

mX - m X

S

In a CSTR, at steady state and sterile feed m = D. Putting in Eq. 6.165 and rearranging, we get D

Therefore,

YX / S

YX / S =

Also, we know that

D

=

YX¢ / S D

YX / S =

+m

D

YX¢ / S

+m

X - X0 S0 - S

From Eqs. 6.166 and 6.167, we can write D

D

YX¢ / S

+m

=

(6.166)

(6.167)

X - X0 S0 - S

For sterile feed, X0 = 0, so X can be given as X=

(6.165)

YX¢ / S D ( S0 - S ) ¢ È ˘ ÎD + mYX / S ˚

(b) We know that practically Y < 1. From Eq. 6.166, we get D

˘ È D Í ¢ + m˙ ÍÎ YX / S ˙˚

1 D So from the above equation, 1

YX¢ / S

+

m >1 D

(6.169)

(c) Figure 6.27 shows hyperbolic nature of the curve; however, for the model given 1 1 m = + YX / S YX¢ / S D

The plot of 1/YX/S versus 1/D should be a straight line. So the plot of A. aerogenes growing on glycerol does not satisfy this model.

I/Yx/S , g glycerol/g cells

7

Plot of A. aerogenes Plot of the model Ê 1 1 Ko ˆ = + ËÁ V V D ¯˜

0 0

50

100

150 1/D,h

Figure 6.27 Plot of 1/YX/S versus 1/D of A. aerogenes.

200

250

273

274

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

PROBLEM 6.23: If a microbial species is inhibited by a volatile product (such as ethanol), the growth rate can be increased by the removal of inhibiting product via the continuous evacuation of the vapor space above the fermenter. For a batch fermentation, the following equation is given: m = mmax

KP S KS + S KP + P

Show that the time course of the substrate level for the overall reaction is

S Æ 0.3P + cell mass

KI dS - ÈÎ X 0 + YX / S ( S0 - S )˘˚ S mmax = dt YX / S K S + S K I + 0.3 ( S0 - S )

SOLUTION S Æ 0.3P + X

Given

We know that

1 dS =dt YX / S Now

dS dS dX = dt dX dt

-

dS 1 = mX dt YX / S È KP ˘ S Í mmax ˙X K S K + S P +P˚ Î

YX / S =

Therefore,

(6.170)

(6.171)

X - X0 S0 - S

X = YX/S(S0 – S) + X0

Putting in Eq. 6.171, we get

KP ˘ dS 1 È S =Í mmax ˙ YX S ( S 0 - S ) + X 0 dt YX S Î KS + S KP + P ˚

(

)

Kinetics of Fed Batch Cell Growth

Now

YP / S =

Therefore,

P - P0 S0 - S

P = YP/S(S0 – S) + P0

Assuming P0 = 0, the above equation can be written as P = 0.3(S0 – S)

Putting in Eq. 6.171, we get

(YP/S = 0.3)

KI dS -[ X 0 + YX / S ( S0 - S )] S mmax = (6.172) dt YX / S K S + S K I + 0.3 ( S0 - S )

where KP = KI (since product is the inhibitor).

PROBLEM 6.24: The mass balance equations for component C in a batch (or plug flow) reactor and a chemostat are Batch (or plug flow): dc = f (c ) dt

Steady-state chemostat:

D(c0 – c) = –f(c)

(6.173)

(6.174)

Since f(c) is available from Eq. 6.173 by differentiation, a plot of f(c) versus c can be constructed from batch data. Similarly, Eq. 6.174 indicates that a plot of D(c0 – c) versus c will intersect f(c) versus c at c* corresponding to the solution to Eq. 6.174. (a) Consider the logistic equation

Ê dX X ˆ = mX Á1 , dt X max ˜¯ Ë

with µ = 1.0 h−1, Xmax = 10 g/L, and X0 (feed) = 0. Solve for X* graphically for the cases D = 1.5 h−1, 0.75 h−1, and 0.25 h−1 using the method discussed above. Verify your results by direct solution analytically. (b) For fermenters in series, the outlet from one chemostat is the inlet of condition to the next. Show graphically how you would evaluate X3*, the biomass concentration in the third CSTR in a three-reactor cascade with an overall dilution rate of D = 0.75 h−1.

275

276

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

(c) Inspect the data for diauxic growth carefully. Differentiate it graphically and plot dX/dt versus X. How does the plot differ from the shape of simple logistic form? Sketch using the graphical design procedure above a solution for 5 tanks in series, which will consume all of the glucose and most of the second carbohydrate as well.

SOLUTION

(a) Given Logistic equation is

Ê dX X ˆ = mX Á1 ˜ dt X Ë max ¯

(6.175)

Also µ = 1.0 h−1, Xmax = 10 g/L, and X0 = 0. By assuming different values of X, the following table can be obtained.

Table 6.10 Calculated values of rate of cell mass growth X

dX/dt

0

0

3

2.1

1 2 4 5 6 7 8 9

10

0.9 1.6 2.4 2.5 2.4 2.1 1.6 0.9 0

By plotting dX/dt versus X, the X* at different D can be obtained as follows: At D = 1.5, X is negligible (no interaction); at D = 0.75, X = 2.47 g/L; at D = 0.25, X = 7.51 g/L.

Kinetics of Fed Batch Cell Growth

4 D = 1.5

(dx/dt)

3

2

D = 0.75

D = 0.25

1

0

0

2.5 X* @ D = 0.75

5 X g/L

7.5 X* @ D = 0.25

10

12.5

Figure 6.28 Plot of dX/dt versus X in a batch process.

Analytical solution:

Ê dX X ˆ = mX Á1 dt X max ˜¯ Ë

By rearranging , we get

Ê 1 dX X ˆ = m Á1 X dt X max ˜¯ Ë

We know that, at steady state

equation, we get

Solving for X, we get

1 dX = m = D . Putting in the above X dt

Ê X ˆ D = m Á1 X max ˜¯ Ë

Ê Dˆ X = X max Á 1 - ˜ m¯ Ë

Putting the given values in Eq. 6.175, we get Dˆ Ê X = 10 Á 1 Ë 1.0 ˜¯

At D = 1.5, X = −5 (negligible)

277

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

At D = 0.75, X = 2.5 g/L At D = 0.25, X = 7.5 g/L So the results for graphical and analytical solutions can be given in Table 6.11. Table 6.11 Comparative results of graphical and analytical analysis D

X (graphical)

X (analytical)

1.5

—

−5

0.75 0.25

2.47

2.5

7.51

7.5

(b) By plotting the curve b/w dX/dt versus X at D = 0.75, X3 is estimated to be 7.9 g/L. 4

3.5 3 (dx/dt)

278

2.5 2 1.5 1 0.5 0

0

2.5 X1

5

X2 X (g/L)

7.5 X3

10

12.5

Figure 6.29 Determination of steady-state cell mass concentration as any dilution rate (D).

(c) The plots of logistic growth versus diauxic growth are shown in Figs. 6.29 and 6.30. In the case of logistic growth, one carbon source is used by the cell, and in the case of diauxic growth, the cell will utilize the second carbon source after the exhaustion of the first carbon source.

Kinetics of Fed Batch Cell Growth

Logistic Growth 3.5

X

3 2.5 2 1.5 1 0.5 0

0

5

10

15

20

t

Figure 6.30 Logistic growth curve of the microorganism.

Diauxic growth 6 5

X

4 3 2 1 0

0

5

10

15

20

25

30

35

t

Figure 6.31 Diauxic growth curve of the microorganism.

The solution for five tanks in series for the consumption of all of the glucose and most of the second carbohydrate using the differential plot of diauxic growth can be given by Fig. 6.33.

279

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

3

Diauxic Growth

(dx/dt)

2.5 2

1.5 1

0.5 0

0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

X (g/L)

Figure 6.32 Plot of dX/dt versus X in the case of diauxic growth of the microorganism.

3

Diauxic Growth

2.5 2 (dx/dt)

280

1.5 1 0.5 0

0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

X (g/L)

Figure 6.33 Logical consideration to find out the cell mass concentration in diauxic growth.

PROBLEM 6.25: In a chemostat, Zymomonas mobilis cells are used in a 60 m3 fermenter. The yield of biomass from the substrate is 0.06 g/g, and the product yield YP/X is 7.7 g/g. The maintenance coefficient is 2.2 g/gh. The specific rate of product formation (qp) is 3.41 h−1. The maximum specific growth rate of Zymomonas mobilis is 0.3 h−1. The feed contains 12 g/L glucose; KS for the organism is 0.2 g/L. (a) What is the flow rate required for a steady-state substrate concentration of 1.5 g/L.

Kinetics of Fed Batch Cell Growth

(b) Find out the cell density at the flow rate in part (a)? (c) Determine the ethanol concentration at the flow rate in part (a)?

SOLUTION Given Ks = 0.2 g/L, YX/S = 0.06 g/g, YX/S = 7.7, ms = 2.2 g/gh, µmax = 0.3 h−1, ms = 2.2 g/gh, qp = 3.41 h–1, Si = 12 g/L, V= 60 m3 (a) At S = 1.5 g/L We know that at steady state m=D=

mmax S KS + S

Putting all the values in the above equation, we get D=

Now

D=

0.3 (1.5) = 0.26 h -1 0.2 + (1.5)

F Æ F = DV = (0.26)(60) V F = 15.6 m3/h

(b) Cell density when F = 15.6 m3/h The substrate balance (considering substrate requirements for growth, product formation, and maintenance) across the chemostat can be given as Input + Generation = Output + Consumption + Accumulation

˘ ÈÊ dS ˆ Ê dS ˆ Ê dS ˆ FS i + 0 = FS + ÍÁ ˜ ˙V + 0 +Á ˜ +Á ˜ ÍÎË dt ¯ cell growth Ë dt ¯ product Ë dt ¯ maintenance ˙˚ Ê m ˆ qp F(Si - S ) = Á + + ms ˜ XV Ë YX / S YP / S ¯ Dividing throughout by V and rearranging, we get X=

D( S i - S ) (At steady state, m = D) qp D + + ms YX / S YP / S

(6.176) (6.177)

Since the product is directly linked to energy metabolism, the above equation can be written as

281

282

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

X=

D( S i - S ) 0.26(12 - 1.5) = = 0.42 g/L D 0.26 + ms + 2.2 YX / S 0.06

(c) The product balance in chemostat can be written as Input + Generation = Output + Consumption + Accumulation FPi +

dP V = FP + 0 + 0 dt

Assuming Pi = 0, diving throughout by V, the above equation can be written as P=

qp X D

=

3.41(0.42) = 5.5 g/L 0.26

PROBLEM 6.26: For a chemostat culture, the kinetics of microbial growth, substrate consumption, and mixed growth associated product formation are given by the following equations: mmax S dX mmax S dS dP dX X; X; = = =a + b X = (am + b )X dt K S + S dt (K S + S )YX/S dt dt

The following data are given:

mmax = 0.7 h−1, Ks = 20 mg/L, YX/S = 0.5 g dw/g substrate, YP/X = 0.15 g P/g dw, a = 0.1, b = 0.02 h–1, and S0 = 1 g/L (a) Find the optimal dilution rate maximizing the productivity of product formation (PD). (b) Determine the optimal dilution rate maximizing the cell (biomass) formation (DX).

SOLUTION The optimal dilution rate can be found by plotting a differential curve between DP and D. By assuming different values of D, Table 6.12 can be obtained. Table 6.12 Calculated data

D

S

X

qp (αD + b)

DP (qpX)

0.00

0.00

0.15

0.02

0.00

0.10

0.00

0.50

0.03

0.01

Kinetics of Fed Batch Cell Growth

D

S

X

qp (αD + b)

DP (qpX)

0.15

0.01

0.50

0.04

0.02

0.25

0.01

0.49

0.05

0.02

0.20 0.30 0.35 0.40 0.45 0.60 0.65

0.01 0.02 0.02 0.03 0.04 0.12 0.26

0.50 0.49 0.49 0.49 0.48 0.44 0.37

0.04

0.02

0.05

0.02

0.06

0.03

0.06

0.03

0.07

0.03

0.08

0.04

0.09

0.03

From Fig. 6.34, Doptimal for maximum productivity of product formation (DP) is 0.6 h−1. 0.04

0.035 0.03 0.03 DP

0.025 0.02 0.015 0.01 0.005 0

0

0.1

0.2

0.4

0.3 D

0.5

0.6

0.7

0.8

(h-1)

Figure 6.34 Plot of DP versus D.

(c) Similar to the previous part, the optimal dilution rate can be found by plotting a differential curve between DX versus D. By assuming different values of D, the following table can be obtained:

283

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

Table 6.13 Calculated data D

S

X

DX

0.00

0.00

0.15

0.00

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.60

0.65

0.00

0.01

0.01

0.01

0.02

0.02

0.03

0.04

0.12

0.50

0.50

0.50

0.49

0.49

0.49

0.49

0.48

0.44

0.05

0.07

0.10

0.12

0.15

0.17

0.19

0.22

0.26

0.26

0.37

0.24

0.3

0.4

0.5

0.3 0.25 0.2 DX

284

0.15 0.1 0.05 0

0

0.1

0.2

0.6

0.7

0.8

D (h-1)

Figure 6.35 Plot of DX versus D.

From the above graph, Doptimal for maximum productivity of biomass formation (DX) is 0.6 h−1. Again, we know that Ê Ê KS ˆ 0.02 ˆ Dmax = µmax Á 1 ª 0.6 h -1 ˜ = 0.7 Á 1 K S + S0 ¯ 0.02 + 1 ˜¯ Ë Ë

Kinetics of Fed Batch Cell Growth

PROBLEM 6.27: A 1 m3 mixed flow reactor, at the initial substrate concentration of S0 = 500 g/m3 in the feed, produces 100 g/h of yeast cells in the exit stream at two different flow rates mentioned below: (i) At 0.5 m3/h of feed for which the steady state S = 100 g/m3 (ii) At 1 m3/h of feed for which the steady state S = 300 g/m3

The Monod cell growth kinetics follows yeast formation. Compute the following: (a) (b) (c) (d)

The fractional yield of yeast The kinetic equation for yeast formation The flow rate for maximum yeast production The maximum production rate of yeast

SOLUTION

(a) We know that Flow rate × Concentration of cells = Rate of cell mass production Therefore, concentration of yeast is 100 g/h

0.5 m3/h

= 200 g/m3

Case (i): In the case of sterile feed, XSS = YX/S (S0 – SSS).

Therefore, YX/S =

200 g/m3

(500 - 100) g/m3

= 0.5 = yield of yeast cells

(b) From the Monod model under steady state and sterile conditions, m=

Case (i): D =

µmax S =D KS + S

F 0.5 m3/h = 0.51/h V 1 m3

So from the Monod equation, we can write 0.5 =

Similarly, in Case (ii),

1=

µmax 100 K S + 100

µmax 300 K S + 300

(6.178) (6.179)

285

286

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

Then,

1 µmax = 2 and K S = 300 g/m3 h

So the kinetic equation is m =

2S 300 + S

(c) We know that

Ê ˆ KS Dmax = mmax Á 1 ˜ (K S + S0 ) ¯ Ë Dmax = 0.776 1/h =

F V

Flow rate for the maximum yeast production is 0.776 × 1 = 0.776 m3/h.

(d) S =

KSS 300 ¥ 0.776 = = 190 g/m3 ( mmax - D) (2 - 0.776)

X = YX/S (S0 – S) = 0.5 (500 – 190) = 155 g/m3

Maximum cell mass productivity is 0.776 × 155 = 120.28 g/h m3

PROBLEM 6.28: The effect of temperature on the lactic acid production by Lactobacillus delbrueckii is determined. From the data given in Table 6.14, determine the activation energy for this process. Table 6.14 Effect of temperature on the rate constant in the batch process Temperature (°C)

Rate constant (mol/L h)

40.4

0.0140

30.0

0.0051

36.8 33.1 25.1

0.0112 0.0074 0.0036

SOLUTION Temperature-dependent term rate constant (k) can be expressed using the Arrhenius equation as follows:

Kinetics of Fed Batch Cell Growth

k = Ae - Ea

RT

where A is the frequency factor, Ea is the activation energy of the reaction, R is the universal gas constant (8.314 J/mol K). Taking exponential of both sides, we get ln(k ) = ln( A) -

ln(k ) = -

Ea 1 R T

Ea 1 + ln( A) R T

Ea 1 - ln( A) R T y = mx + c

- ln(k ) =

T (°C)

T (K)

k (mol/L h)

1/T (K−1)

−ln(k)

40.4

313.4

0.014

0.003191

4.268698

30

303

0.0051

0.0033

5.278515

36.8

33.1

25.1

309.8

306.1

298.1

0.0112

0.0074

0.0036

0.003228

0.003267

0.003355

From the plot of –ln (k) versus 1/T,

4.491842

4.906275

5.626821

6

5

-In(k)

4 y = 8706.2x - 23.538 R2 = 0.9882

3 2 1

0 0.00315

0.0032

0.00325 0.0033 1/T (K-1)

Figure 6.36 Plot of ln k versus 1/T.

Slope =

Ea = 8706.2 K R

0.00335

0.0034

287

288

Microbial Growth Kinetics, Substrate Degradation, and Product Formation

Therefore, Ea = 72.38

kJ mol

PROBLEM 6.29: Prove that the Luedeking–Piret model related to cell growth and product formation and the Pirt model related to the maintenance energy are equivalent. SOLUTION We know that according to the Pirt model, 1

YX/S

=

1

YX¢ /S

+

m m

Multiplying the above equation by mX, we get mX mX dS = + mX = YX/S YX¢ /S dt

The above equation can be written as

(6.180)

dS = (Constant) m X + (Constant )X dt Now we know that -

dS Ê -dS ˆ Ê dS ˆ = dt ÁË dt ˜¯ growth ÁË dt ˜¯ product

The above equation can be written as -

dS 1 1 dP dS 1 1 = - ¢ mX = - ¢ mX (am + b ) X dt Y dt dt Y YX/S YX / S P/S P /S

(From the Leudeking–Piret model) By rearranging the above equation, we get -

Ê 1 dS X a ˆ = -m X Á ¢ + b ˜Á ˜ dt Ë YX / S YP / S ¯ YP / S

The above equation can be written as

dS = (Constant) m X + (Constant )X (6.181) dt From Eq. 6.180 and Eq. 6.181, we can conclude that both models are equivalent.

Kinetics of Fed Batch Cell Growth

PROBLEM 6.30: Two chemostats are connected in a series. The operational volumes of Reactor 1 and Reactor 2 are 120 L and 60 L, respectively. The feed to the first contains 5000 mg/L of substrate, and there is no cell mass, the flow rate being 20 L/h. The microbial growth follows the Monod kinetic model with μmax, KS, and YX/S of 0.25 h–1, 120 mg/L, and 0.3 g cell/g substrate, respectively. Calculate the steady-state substrate concentration S2 and the concentration of cell mass (X2) from the second bioreactor. SOLUTION The process can be schematically represented in Fig. 6.37.

S0, X0

S1, X1

S2, X2

120 L

60 L

Reactor 1

Reactor 2

Figure 6.37 Two-stage chemostat process.



The initial substrate concentration is 5000 mg/L = 5 g/L. Volumetric flow rate is F = 20 L/h. Also μmax = 0.25 h–1

KS = 120 mg/L = 0.12 g/L

YX/S = 0.3 g cell/g substrate

Two different volume reactors are in series for microbial cultivation. Dilution rate in Reactor 1 is D1 = F/V1 = 20/120 h−1 = 0.167 h−1

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Microbial Growth Kinetics, Substrate Degradation, and Product Formation

Dilution rate in Reactor 2 is

D2 = F/V2 = 20/60 h−1 = 0.33 h−1

Initial substrate concentration is

S0 = 5 g/L

Steady-state substrate concentration in Reactor 1 is S1. Steadystate substrate concentration in Reactor 2 is S2. Initial biomass concentration is X0 = 0 (for sterile feed). Steady-state biomass concentration in Reactor 1 is X1. Steady-state biomass concentration in Reactor 2 is X2. From the established formula S1 =

K S D1 0.12 ¥ 0.167 = = 0.24 g/L mmax - D1 0.25 - 0.167

X1 = X0 + YX/S (S0 – S1) = 0 + 0.3 (5 – 0.24) = 1.4 g/L

For the second reactor, dilution rate is

D2 = 0.33 h−1 < μmax = 0.25 h–1

Therefore, there will be cell mass washout in Reactor 2. Hence, S2 will be equal to S1 and X2 will be equal to X1. S2 = 0.24 g/L

References

X2 = 1.4 g/L

1. Bailey, J. E. and Ollis, D. F. Biochemical Engineering Fundamentals, McGraw-Hill Inc., New Delhi, India, 2010. 2. Sinclair, C. G. and Kristiansen, B. Fermentation Kinetics and Modelling, Open University Press, 1987. 3. Shuler, M. L. and Kargi, F. Bioprocess Engineering: Basic Concepts, Second Edition, Prentice-Hall Inc., New Delhi, India, 2002.

4. Blanch, H. W. and Clark, D. S. Biochemical Engineering, Marcel Dekker Inc., New York, USA, 1997. 5. Doran, P. M. Bioprocess Engineering Principles, Second Edition, Academic Press, Waltham, USA, 2012. 6. Metcalf and Eddy, Wastewater Engineering: Treatment, Disposal and Reuse (Revised by Tchobamoglous, G and Burton, F. L.), TATA McGrawHill Pub. Co. Ltd., 1991.

References

7. Lydersen, B. K., D’elia, N. A., and Nelseon, K. L. Bioprocess Engineering: Systems, Equipment and Facilities, Wiley India Pvt. Ltd., New Delhi, 1994. 8. Liu, S. Bioprocess Engineering: Kinetics, Biosystems, Sustainability, and Reactor Design, Elsevier, Amsterdam, 2013.

9. Aiba, S., Humphrey, A. E., and Millis, N. F. Biochemical Engineering, Academic Press, Inc. 1973.

10. Villadsen, J., Nielsen, J., and Lida, G. Bioreaction Engineering Principles, Springer, 2009. 11. Fredrickson A. Formulation of structured growth models, Biotechnology Bioengineering, 18: 1481, 1976.

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Chapter 7

Air Sterilizer

Airborne microorganisms present in the air are removed by air sterilizers and subsequently used for aerobic fermentation. There is a wide variation in the quantity of suspended particles and microbes in the outdoor air. The microorganisms present may vary from 10 to 2000/m3, while the suspended particles may be from 20 to 10,000/m3. The microorganisms present in the air usually contain fungal spores (50%) and Gram-negative bacteria (40%). A 100% collection efficiency of airborne microorganisms cannot be expected in a filter bed packed with fiber or granular particles. The purpose of such air filters is to prolong the interval between the passage of one airborne organism and the next. The length of the interval is governed by the type of fermentation but should be sufficient to protect the fermentation at least for the initial critical period of growth.

7.1 Airborne Microbes

Airborne microorganisms mainly consist of species of bacteria, bacterial spores, yeast, fungi, and viruses. The size of these organisms varies from 0.5 µm to several hundred micrometers. The airborne particles destroyed or collected during air sterilization are usually of the size of small bacteria, namely, 0.5–1 µm. However, in some typical cases, the size of the microorganisms may vary. For instance, Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

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in citric acid fermentation, the major interfering microorganism is yeast, such as Saccharomyces cerevisiae, whose size is much higher than that of the bacteria.

7.2 Methods of Air Sterilization Air may be used in two different forms:

1. Stagnant air, e.g., laminar flow, inoculum preparation room, etc. 2. Continuous flow of air, e.g., aerobic fermentation processes for air sparging.

The methods generally considered for the sterilization of stagnant air are as follows:

∑ Ultraviolet rays and other electromagnetic waves ∑ Chemicals or germicidal sprays

There are different methods available for the air sterilization of continuous air flow:

∑ Heat ∑ Mechanical filtration

However, sterilization of air by heat is ineffective because air has low conduction property as compared to liquids. The mechanical filtration method is widely used by the fermentation industry, e.g., fibrous air filters, polyvinyl alcohol (PVA) plate-type filters, etc. Filter sterilization has the following characteristics:



∑ Filtration results in the exclusion of microorganisms based on size. ∑ It does not kill the microorganisms but removes them physically. ∑ Sterilization by filtration depends on the content of air, such as the size, type, and moisture content of suspended matter. ∑ High-efficiency particulate air (HEPA) filters (used in the laminar flow) can remove up to 99.97% of particles greater than 0.3 µm in diameter. ∑ Air is first passed through prefilters to remove larger particles and then passed through HEPA filters.

Physical Implication of Single Fiber Collection Efficiency

7.3 Physical Implication of Single Fiber Collection Efficiency The mechanisms of collection of aerosol particles by fibrous media may be classified as follows:

∑ ∑ ∑ ∑ ∑

Inertial impaction Interception Diffusion Settling by gravitational force Electrostatic force

Settling by gravitational force in the case of fibrous filters may be neglected because the diameter of the particles to be collected is of the order of 1 µm. Again, it would be expected that charged organisms would be collected more effectively than neutral ones. It has been found that Bacillus subtilis possesses 70% positive, 15% negative, and 15% neutral charge. However, there is little quantitative data available to find out the contribution of the electrostatic charge of the microorganisms to the overall collection efficiency of fibrous air sterilization filter. So the first three mechanisms—inertial impaction, interception, and diffusion—are considered in detail. It is assumed that



∑ Single cylindrical fibers are placed perpendicularly to the aerosol flow in an infinite space, and that the air flow around the cylinder is laminar with no vortices. ∑ The following analyses are two dimensional, as shown in Fig. 7.1. df (diameter of the fiber)

Particle

b (width of the upstream)

Figure 7.1 Air flow is perpendicular against the cross section of the fiber.

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296

Air Sterilizer

7.3.1 Inertial Impaction The flow pattern of the particles deviates from those of air flow due to the inertia of the particles as they approach the cylindrical surface. In Fig. 7.1, the width of the upstream air flow is denoted by b. Particles that move in the streamlines of air beyond b will not touch the cylinder surface even after they deviate from the air stream line near the cylinder. Then, the collection efficiency of a single fiber due to the inertial effect of the particles is h0¢ =

b df

h0¢ = 0, when j = where

1 16

j = Inertial parameter =

C rpdp2V0 18µdf

At the critical air velocity Vc, j is equal to 1/16. Vc = (1.125)

m df

C rp dp2

(7.1)

(7.2) (7.3)

where C is Cunningham’s correction factor for slip flow, rp is the density of particle (g/cm3), dp is the particle diameter (µm, cm), df is the fiber diameter (µm, cm), and µ is the viscosity of air (g/cm s). The variation in Vc with dp is shown in Fig. 7.2. At the air velocity below the value of Vc, the inertial impaction of particle may be neglected.

7.3.2 Interception

The particles entrained in the streamline of air are collected by contact with the fibers. They are said to be intercepted. The streamline of air flow is at a distance of dp/2 from the fiber surface. This is a limited condition for the deposition of the entrained particles as they pass a cylindrical fiber. Then the collection efficiency due to interception may be written as follows: h0¢¢ =

¸Ô ÏÔ 1 1 ˝ Ì2(1 + NR )ln(1 + NR ) - (1 + NR ) + 2(2 - ln NRe ) ÓÔ (1 + NR ) ˛Ô

(7.4)

Physical Implication of Single Fiber Collection Efficiency

where

NR =

dp df

= Geometrical ratio

NRe = Reynolds number =

dur µ

7.3.3 Diffusion Small particles display Brownian motion and thus may be collected on the surface of the fibers as the particles are displaced from their median center of location. If the displacement of the particle is 2X0, the collection efficiency due to diffusion may be written as h0¢¢¢ =

È Ê ˘ 2X 0 ˆ Ê 2X ˆ Ê 2X ˆ 1 1 ln Á 1 + 0 ˜ - Á 1 + 0 ˜ + Í2Á 1 + ˙ ˜ 2(2 - ln NRe ) ÍÎ Ë df ¯ Ë df ¯ Ë df ¯ 1 + 2X 0 /df ˚

(7.5)

2X 0 2(2 - ln NRe )DBM = 1.12 df Vdf

(7.6)

where DBM = CKT/3pmdP is the diffusivity of the particle. 10

1 dr =19 mm

Vc (cm/sec)

13 mm

10-1

8 mm

1 mm 10-2

2 O

4 dp

6

mm

Figure 7.2 Effect of fiber diameter and particle diameter on the critical velocity for inertial impaction.

297

298

Air Sterilizer

Therefore, the overall collection efficiency is h0 = h0¢ + h0¢¢ + h0¢¢¢

Again, in Eqs. 7.4 and 7.5, when NRe varies from 1 µ NRe1/6 2 - ln NRe

10−4

to

10−1

(7.7) (7.8)

Assuming the values of NR and 2X0/df are small compared to unity and that the second and higher orders of each term can be neglected in Eqs. 7.4 and 7.5, we get h0¢¢ µ NR2 NRe1/6

(7.9)

h0¢¢¢ µ Nsc -2/3 NRe -11/18 where

Nsc = Schmidt number =

m r DBM

(7.10)

The collection efficiency due to the inertial impact is generally predominant at higher air velocity (Fig. 7.3), so the collection efficiency η0¢ can be disregarded in calculating η0 at low velocity (less than 10 cm/s by using glass wool fiber). The overall collection efficiency of a fibrous filter can be calculated as shown in Eq. 7.11.

h(%)

Diffusion Inertial impaction Interception

0.5 1.0

10

Ê cm ˆ Æ Ë s ˜¯

vs Á

Figure 7.3 Correlation between collection efficiency and superficial velocity of air using glass wool filter.

Physical Implication of Single Fiber Collection Efficiency

Figure 7.4 Determination of overall collection efficiency of glass wool filter. Reprinted with permission from Ref. [1], Copyright 2004, John Wiley and Sons.

where

h=

( N1 - N2 ) N1

(7.11)

N1 is the number of organisms in the incoming air, and N2 is the number of organisms in the outgoing air.

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Air Sterilizer

Again, single fiber efficiency is hµ = p df = p df

(1 - a ) N1 ln 4 La N2

(1 - a ) 1 ln 4 La 1-h

(7.12)

(7.13)

where L is the thickness of filter bed, and α is the volume fraction of the filter. Equation 7.11 indicates that the fraction of particles collected in any section within L is constant, which is known as the logpenetration relation. The L should be less than 4 cm. The correlation between hµ and h0 may be written as

hµ = h0 (1 + 4.5a), 0 < a < 0.10

(7.14)

The plot h0NRNpe (= h0NRNscNRe) versus NRNpe1/3NRe1/18 is shown

in Fig. 7.4. Figure 7.4 is usually used to estimate h0 for a specific set of operating conditions. From the h0 value, hµ for glass wool fiber can be calculated with the help of Eq. 7.14 because the a value of the glass wool fiber is 0.033. Equation 7.13 can be used to find out the value of the thickness of the air filter (L). The typical correlation between L and h is shown in Fig. 7.5. 100

L (cm)

10

90

98

99.5 h (%)

Figure 7.5 Relationship between the length of the air filter (L) and the overall collection efficiency ( h ).

Types of Filters

7.4 Types of Filters Two types of filters are used for air sterilization: 1. Depth filters: A depth filter consists of either a single layer or multiple layers of a medium having depth. This can remove the contaminants within its structure. In depth filters, particles penetrate the structure of the filter and form a filter cake on the surface. 2. Membrane filters: A membrane filter usually traps contaminants larger than the pore size of the membrane. In membrane filters, particles are collected on the surface of the membrane.

7.4.1 Depth Filters

It is assumed that if a particle touches a fiber, it remains attached to it, and if there is a uniform concentration of particles at any given depth of the filter, then each layer of a unit thickness of the filter should reduce the population entering it by the same proportion. This may be expressed mathematically as dN = -kN dx

(7.15)

where N is the concentration of particles in the air at a depth x in the filter and k is a constant. On integrating, the above equation may be written as N = e - kx N0

(7.16)

where N0 is the number of particles entering the filter and N is the number of particles leaving the filter. On taking exponential logarithms of Eq. 7.16, we get

Ê Nˆ ln Á ˜ = - k x (7.17) Ë N0 ¯ Equation 7.17 is known as the log-penetration relationship. A plot of log (N/N0) versus x filter length (x) (Eq. 7.17) will yield a straight line of slope k. The efficiency of the filter (E) is expressed by the ratio of the number of particles removed to the original number present:

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Air Sterilizer

E=

Now,

( N0 - N ) N0

(7.18)

( N0 - N ) N =1– = 1 – e–kx = E N0 N0

(7.19)

The log-penetration relationship can be used in filter design by using the concept X90, which is defined as the depth of the filter required to remove 90% of the total number of particles entering the filter. If N0 is 10 and x is X90, then N would be 1. From Eq. 7.17, we get ln (1/10) = −kX90

or

2.303 log10 (1/10) = −kX90 2.303(−1) = −kX90

Therefore,

X90 = 2.303/k

(7.20)

The value of k depends on the nature of the filter material and also on the linear velocity of the air passing through the filter. The k value increases to an optimum with increasing air velocity, after which any further increase in air velocity results in a decrease in k. k k

X90

X90

Linear air velocity

Figure 7.6 Effect of linear air velocity on the values of k and X90. Adapted from Ref. [4].

The increase in k with increasing air velocity is probably due to the increased impaction. This indicates the important contribution of this mechanism in the removal of contaminants. The decrease in k values at high air velocities is probably due to the disruption of the

Types of Filters

filter, allowing channels to develop and fibers to vibrate, resulting in the release of previously captured organisms.

7.4.1.1 Pressure drop in air filter

According to Kimura et al., the modified drag coefficient (CDm) can be related with the pressure drop and other variables present in the air filter as follows: CDm =

p gc df DP

2rLV 2 (1 - e )m

(7.21)

where gc is the conversion factor, DP is the pressure drop of air flow, r is the density of air, V is the air velocity, e is the void fraction (= 1 − α), and m is the empirical component. The variation in the empirical component (m) versus the depth of the filter is shown in Fig. 7.7. This indicates that m increases with the length of the filter, which also depends on the fiber diameter. The correlation between drag coefficient and Reynolds number by using cotton and glass wool filter has been experimentally found, which is shown in Fig. 7.8. For an air flow, the pressure drop in the case of cotton fiber is more in comparison to glass wool fiber. This is responsible for the high heat development in case of cotton fiber as compared to glass wool fiber. In addition, glass wool fiber can withstand high temperature and can be reused again. So, this is usually considered for the air sterilization in the industry. df = 8 mm

13 mm

19 mm

m

L

Figure 7.7 Plot of m versus depth of the filter.

7.4.2 Membrane Filters Membrane filters are made in two ways: The capillary pore membranes have pores produced by radiation, while the

303

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Air Sterilizer

labyrinthine pore membranes are produced by forced evaporation of solvents from cellulose esters. These are fixed pore filters (which have an absolute rating), which are very widely used in the fermentation industry. They usually require prefilter for trapping large particles such as dust, oil, carbon (from the compressor), pipe scale, and rust (from the pipework). These filters are made from a variety of polymeric materials such as cellulose nitrate, cellulose diacetate, polycarbonate, and polyester. These membranes have a pore diameter ranging from 0.015 μm to 12 μm. These filters are sterilized by autoclaving. CDm

Cotton

Glass wool NRe

Figure 7.8 Drag coefficient for air flowing through cotton and glass wool fiber.

7.4.2.1 PVA filter for air sterilization A plate filter can be made by acetylating polyvinyl alcohol and by coating it with a heat-resistant resin, such as silicon resin, to enable it to withstand repeated steam sterilization. This filter is widely used in Japan. The main advantage is that the space requirement is less. Figure 7.9 shows the effect of the superficial velocity of air (vs) on the value of collection efficiency ( h ). The value of h was controlled by inertial impaction when the value of vs exceeded 10 cm/s similar to the fibrous filter, but as vs fell below 1 cm/s, this removal of particle is probably due to diffusion effect, while in the intermediate region, interception is the controlling influence. It is interesting to note that the value of h increased appreciably with an increase in vs; however, beyond a critical value of vs, extending roughly from 150 to 200 cm/s, the value of h deteriorated sharply. This is because the force that retains the bacterial spores captured

Experimental Setup for Air Sterilization

by the fibrous network is overwhelmed by another force imposed from the air flow and this caused some of the spores to re-enter the stream. The drop-in pressure in the air flowing through a PVA filter significantly affects the value of h . Membrane air filters provide a low pressure drop at high air flow rates because the filter medium is thin and is pleated to maximize the filter area. Thickness of the plate-type filter is varied from 0.2 to 0.3 cm. Since the thickness is low, log-penetration law probably applies, as shown in Eq. 7.19. h = 1 - e - kL = 1 - e - s

(7.22)

where S is the stopping criterion, k is the stopping factor, and L is the thickness of the filter. 99.9

99 h(%)

90

0 0.59

1

10 vs, cm/s

100

Critical air velocity

Figure 7.9 Plot of collection efficiency versus superficial air velocity using PVA filter.

The value of k is proportional to the collection efficiency of a single fiber for fibrous filters with constant values of df and α.

7.5 Experimental Setup for Air Sterilization

The experimental setup for bench-scale studies is shown in Fig. 7.10. A definite volume of the air is passed through 0.85 %w/v sterile saline water under aseptic condition both at the entrance

305

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Air Sterilizer

and the exit. Then, these are used for viable cell count by plating technique. The air flow rate is monitored by using a calibrated rotameter. The diameter of the fiber is found out by using an ocular microscope. Glass wool fibers

Rotameter

Air inlet

Sampling Plating

Figure 7.10 Bench scale experimental setup for air sterilization.

7.6 Selection Criteria of Air Filter The criteria involved in selecting a fermentation air filter system for inlet or exhaust gas filtration are as follows:

∑ ∑ ∑ ∑

Filter retention efficiency Economy of operation Ease of filter use Service provided by the manufacturer

The most important selection criteria are filter efficiency and reliability of organism retention. In this regard, fixed submicron pore size membrane filters provide the highest level of filtration efficiency. The use of a hydrophobic filtering material minimizes or eliminates concerns of filter wetting due to air moisture content.

7.7 Economic Considerations

In calculating air filter system economics, one must consider costs associated with the loss of fermentation batches and production downtime if the air filters fail to provide reliable organism retention.

Economic Considerations

Absolute-rated, fixed pore membrane air filters provide the most reliable organism retention, minimizing or eliminating the risk of fermenter batch due to airborne contamination. A hydrophilic air filter medium can be wet by the moisture present in the inlet air supply resulting in decreased filtration efficiency and increased pressure drop. This can be overcome by using a hydrophobic membrane and increasing the temperature of air above the dew point, which is a costly affair. A coalescing filter reduces air moisture content and acts as a prefilter to lengthen the final filter life by removing debris such as rust particles present in piping. A second consideration affecting air filter system economics is the cost associated with pressure drop across the filter assembly. This factor can be readily calculated by determining the cost of producing the compressed air that is lost due to the pressure drop (Fig. 7.11). In this connection, a membrane air filter is found most suitable. When sizing a fermentation air system assembly, consideration should be given to minimizing the pressure loss and the number of filter elements required to obtain a low pressure drop. Figure 7.12 shows the plot of pressure drop versus total system cost (power consumption due to pressure drop + filter assembly). The point of maximum air filter system economy will be realized where the cost savings due to low pressure drop and the filter assembly size are optimum. 3

15

Annual

20 kwh × 104

2

1

6 7

8

4 3 2 1 Average annual pressure drop (psi)

5

Figure 7.11 Annual power consumption due to pressure drop.

Total air pressure

10 psig

2

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Air Sterilizer

Annual cost

Pressure drop

Figure 7.12 Plot of air system cost versus pressure drop.

Examples PROBLEM 7.1: It is desired to provide a 20 m3 fermenter with an air flow rate of 10 m3/min for a fermentation lasting 100 h. The air contains approximately 200 microorganisms per m3. From an investigation of the filter material to be used, the optimum linear air velocity was shown to be 0.15 m/s, at which the value of k was 1.535 cm−1. Calculate the dimensions of the filter if the acceptable degree of contamination is one in a thousand. SOLUTION From Eq. 7.17, we have

ln

N = -kx N0

The air in the fermentation plant contains approximately 200 microorganisms per m3 (given). Therefore, N0 = Total amount of air provided × 200 N0 = 10 × 100 × 60 × 200 = 12 × 106 organisms The acceptable degree of contamination is one in a thousand (given). Therefore, N = 10−3. Putting in Eq. 6.17, we get Ê 10-3 ˆ ln Á ˜ = -1.535x Ë 12 ¥ 106 ¯

Economic Considerations

x=

-23.21 = 15.12 cm -1.535

Therefore, the filter to be used should be 15.12 cm long. The cross-sectional area of the filter is given by the volumetric air flow rate divided by the linear air velocity: pr 2 =

10 m3min -1 = 1.11 m2 0.15 ¥ 60 m/min

(Given: Linear air velocity is 0.15 m/s.) Here, r is the radius of the filter. Solving the above equation for r, we get r = 0.59 m. Thus, the filter to be employed should be 15.12 cm long and have 0.59 m radius.

PROBLEM 7.2: The efficient operation of the filter is dependent on the supply of air at the optimum linear velocity. If the air velocity increases or decreases, the value of k decreases, resulting in a loss of filtration efficiency. Consider in the previous problem, if the linear air velocity drops to 0.03 m/s, the value of k declines to 0.2 cm−1. Calculate the number of organisms that would enter the fermentation in 1 min at this reduced air flow rate (operational conditions are same as mentioned in the previous problem).

SOLUTION At a linear velocity of 0.03 m/s, the amount of air entering the filter (in 1 min) would be 0.03 × 60 × πr2 (cross-sectional area of filter). Taking r = 0.59 m (from previous problem), the amount of air entering the filter is 1.98 m3. Now, since the air contains 200 microorganisms per m3, the total number of microbes entering the filter would be We know that Therefore, ln

N0 = 200 × 1.98 = 396 ln

N = -kx N0

N = -0.2 ¥ 15.12 (taking x from the previous problem) 396 N = 19.24

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Air Sterilizer

Thus, 19.24 microbes would have entered the fermenter in 1 min at the decreased air flow rate

PROBLEM 7.3: Determine the length of a glass wool filter required to reduce the concentration to 10−8 of its previous values. Following data are given: df = 19 mm, filter diameter = 1 m, a = 0.033, dp = 0.5 mm, Vs = 10 cm

Air viscosity = 2 ¥ 10–4 g/cm rr =

1g

cm3

particular diffusivity DBM = 2.78 ¥ 10-7 cm2/s

SOLUTION Here, from the given data, we can write NRe =

(19 ¥ 10-4 )(1.2 ¥ 10-3 )(10) (2 ¥ 10-4 )(1 - 0.033)

= 2.357 ¥ 10–1 (Assuming density of air = 1.2 ¥ 10–3 mL)

Nac =

(2 ¥ 10-4 ) m = = 6 ¥ 105 rDBM (1.2 ¥ 10-3 ) ¥ (2.73 ¥ 10-7 )

Again NPe = Nac NRe = (2.357 ¥ 10–1) ¥ (6 ¥ 105) = 1.4 ¥ 105 Therefore,

Ê 0.5 ¥ 10-4 ˆ 5 1/3 1/18 1/3 1/18 NR NPe NRe = Á = 2.0474 ˜ ¥ (6 ¥ 10 ) ¥ (0.2357) Ë 19 ¥ 10-4 ¯

From Fig. 7.4, we get

h0 =

We know that

Therefore,

h0NRNR = 2 ¥ 10

2 ¥ 10 = 0.0054 Ê 0.5 ¥ 10-4 ˆ 5 ˜ ¥ (6 ¥ 10 ) Á Ë 19 ¥ 10-4 ¯ hµ = h0 (1 + 4.5a)

hµ = 0.0054 (1 + 4.5 ¥ 0.033) = 0.0062

References

Therefore,

L= =

p df (1 - a ) 4hµa 3.14 ¥ 19 ¥ 10-4 ¥ (1 - 0.033) N1 ln 4 ¥ 0.0062 ¥ 0.033 N2

= 7.049 cm ¥ ln 8 = 14.65 cm

Therefore, the length or depth of the glass wool filter will be 14.65 cm.

References

1. Aiba, S., Humphrey, A. E., and Millis, N. F. Biochemical Engineering, Academic Press, Inc. 1973.

2. Conway, R. S. Selection Criteria of Fermentation Air Filter, Comprehensive Biotechnology-2, M. Moo-Young (Editor-in-chief), Pergamon Press, 1985. 3. Shuler, M. L. and Kargi, F. Bioprocess Engineering: Basic Concepts, Second Edition, Prentice-Hall Inc., New Delhi, India, 2002. 4. Richards, J. W. Air sterilization with fibrous filters, Process Biochemistry, 2(9), 21–25, 1967. 5. Bailey, J. E. and Ollis, D. F. Biochemical Engineering Fundamentals, McGraw-Hill Inc., New Delhi, India, 2010.

6. Doran, P. M. Bioprocess Engineering Principles, Second Edition, Academic Press, Waltham, USA, 2012.

7. Das, D. and Bhattacharya, B. C. Short term course on “Analysis and Design of Novel Bioreactors”, Indian Institute of Technology Kharagpur, India, 1989. 8. Block, S. S. Disinfection, Sterilization and Preservation, Fifth Edition, Lippincott Williams & Wilkins, 2001.

9. Whittet, T. D., Hugo, W. B., and Wilkinson, G. R. Sterilization and Disinfection, William Heinemann Medical Books Ltd., London, 1965.

10. Kimura, N. and Iinoya, G. Experimental studies on the pressure drop characteristics of fiber mats, Chemical Engineering, 23, 792, 1959.

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There are three processes available for the destruction of microbial contaminants in food products:

1. Thermization: It is used for partial removal of pathogens, i.e., to reduce the bacterial population. 2. Pasteurization: It aims to reduce all the viable pathogens, so they are unlikely to cause disease. 3. Sterilization: It is the process of killing or removal of all the contaminants (pathogenic and non-pathogenic, including spores) present in the medium.

Medium sterilization plays a critical role in the biochemical industry for a successful fermentation. The objective is to prevent the growth of undesired microorganisms during the fermentation, enzyme-catalyzed reaction, or medium storage. Sterilization may be achieved through the removal or destruction of any organism that will adversely affect the process or product. It is difficult, however, to remove selectively those organisms that may adversely affect a specific process. Sterilization can be done through two processes: 1. Filtration for the removal of microorganisms 2. Thermal treatment for killing microorganisms

Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

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The second process is generally considered in processing most media used in biochemical processes. The choice of the sterilization method depends on the following factors:

∑ ∑ ∑ ∑

Effectiveness in acquiring an acceptable level of sterility Reliability Effect (positive or negative) on medium quality Cost, including operating and capital expense

8.1 Filtration

8.1.1 Physical Characteristics of Microorganisms The physical properties of microorganisms are required for their removal from the liquid by filtration. The diameter of filamentous, mycelial organisms (such as molds) is several micrometers. So the filtration technique can be easily used for the removal of these organisms. The sizes of yeast, bacteria, and phages are 3.5 µm, 0.5–2 µm, and 0.04–0.1 µm, respectively. Phage contamination is not a major issue in most fermentation processes. For the design of filtration processes, it is desirable to remove the smallest organisms, namely, the bacteria.

8.1.2 Filter Type

8.1.2.1 Depth filters Depth filters are prepared from porous or fibrous materials. The diameter of the pore is usually more than the minimum size of the materials to be removed. Particle removal efficiency is calculated from the amount of particles retained by the filter. The principles of retention of particles by the filter are interception, inertial impaction, diffusion, gravitation, and electrostatic interaction. The first three principles play an important role during the particle separation by the filter, which are shown graphically in Fig. 8.1 (also see Chapter 7). The examples of depth filters are porous glass wool, sintered metals, ceramics, diatomaceous earth, and cellulose fiber.

Filtration

Particle trajectory Streamline

Collector A. Interception B. Inertial Impaction C. Diffusion

Figure 8.1 Schematic diagram of particle removal in a fibrous depth filter.

8.1.2.2 Absolute filters Total removal of microorganisms can be achieved in an absolute filter, which is based on the pore size of the membrane, e.g., ultrafiltration, microporous membrane, etc. The size of the pore should be less than the minimum size of the particles to be removed. So the principle of this filtration is nothing but absolute size exclusion.

8.1.3 Filtration Strategy

In general, media filtration is more expensive than thermal methods. However, it has the primary advantage because it is applicable to retain the nutritional properties of the heat-labile materials present in the medium. Sterilization by filtration is generally considered for the removal of heat-labile nutrient present in the medium. Pyrogen-free water can be prepared by means of microporous and ultrafiltration. Pressure drop in the filtration technique is high. The operational cost of the process is directly proportional to the pressure drop. In addition, the life of the filtration material also influences the cost of the process. So this technique is usually preferred in the case of high-value, low-volume products.

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8.2 Thermal Destruction 8.2.1 Thermal Death Kinetics Microbial death is a probabilistic phenomenon that follows the first-order kinetic, as shown in Fig. 8.2. Microbial death by heat takes place when the organism loses its ability to replicate itself in a given environment. Moreover, cell death does not mean destruction of all enzymatic activity. Enzymes may still have the capability for catalyzing one or more reactions. Cell death kinetics can be described by a first-order rate equation, as shown in Eq. 8.1. dN = -kN dt

(8.1)

where N is the number of viable cells, t is the time, and k is the death rate constant. 1.0

In

N No

0.1 Spores 0.01 0.001

0

Bacterial cells

0.5

1.0

Time, min

Figure 8.2 Thermal death kinetics of bacterial vegetative cells and bacterial spores at 121°C.

Equation 8.1 may be integrated with respect to time to reduce the initial viable cell number N0 to N (Eq. 8.2). The resulting equation is Ê Nˆ ln Á ˜ = -kt = — total Ë N0 ¯

(8.2)

Thermal Destruction

where —total is the extent of death that occurs during some time interval. (a)

1

1

(b)

54°C 10 ln

–2

10–2

N No

10

N No

56°C

–4

60°C 0

2

4

58°C 6

10–4 8

10

Time, min

110°C 121°C 116°C

0

5

105°C 108°C

20 15 10 Time, min

25

Figure 8.3 (a) Using E. coli and (b) using spores of Bacillus stearothermophilus.

Decimal reduction time D is the time of exposure to heat during which the original number of viable microbes is reduced by onetenth. Equation 8.2 may be written as Ê 1ˆ ln Á ˜ = -kD Ë 10 ¯

(8.3)

2.303 (8.4) k The death rate constant k is a function of temperature. This is varied in the case of vegetative bacterial cells and bacterial spores, which is depicted in Fig. 8.3a and Fig. 8.3b. The k values for vegetative cells are much greater than those for spores. It has been found that the dipicolinic acid present in spores may be responsible for their increased resistance to heat. D=

8.2.2 Effect of Temperature on Death Kinetics

The effect of temperature on the thermal destruction of cells may be expressed by the Arrhenius equation (Eq. 8.5) k = Ae - Ea /RT

(8.5)

where k is the thermal death constant, A is the Arrhenius constant, Ea is the activation energy, R is the universal gas constant, and T is the absolute temperature.

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Medium Sterilizer

The Ea values for the microorganisms vary from 250 to 290 kJ/mol. In the case of vitamins and amino acids, the typical activation energy varies from 84 to 92 kJ/mol. This indicates that high temperature has relatively more destructive effect on the cell than on nutrient. This is the basis of selecting high-temperature shorttime (HTST) technique for medium sterilization in the biochemical industry.

8.2.3 Experimental Determination of Microbial Death Rate

The main assumptions in determining the microbial death rate are as follows:

∑ The cell or spore suspension must be free of aggregates ∑ No lag of heating or cooling ∑ The composition and pH of the suspending medium must be such as to have the minimum inhibitory effect on the test organism Flipper

Hot

Cold

C.D. = 1.5 mm I.D. = 0.8 mm Spore

Figure 8.4 Manually operated apparatus for the determination of thermal death rates at relatively low temperatures.

The test cell suspension is sealed in a capillary tube, and a number of capillaries were hung over a cage on a “flipper” arm, as shown in Fig. 8.4. The “flipper” allows the capillaries to be dipped

Thermal Destruction

into either a hot or a cold bath for a given time. Biologically, plate count is used to find out the number of viable organisms before and after exposure to the higher temperatures. The values of reaction rate constant (k) and the decimal reduction time (D) of spores of different bacteria are given in Table 8.1.

8.2.4 Batch Sterilization of Medium

Batch sterilization of the fermentation medium is usually carried out by using either steam or electricity (Fig. 8.5).

Electrical Heating

Steam Sparging

Figure 8.5 Common equipment for batch sterilization of media.

Figure 8.5 shows different types of equipment for the batch sterilization of media. Table 8.1

Reaction rate constants (k) and decimal reduction time (D) of spores of different bacterial suspension in buffer at 121°C

Species

k at 121°C (min−1)

D at 121°C (min)

Bacillus subtilis FS 5230

~3.5–2.6~

~0.6–0.9~

Bacillus stearothermophilus FS 1518 Bacillus stearothermophilus FS 617 Clostridium sporogenes PA 3679

Note: Taken from Ref. [1].

~0.07~

~2.9~

~1.8~

~3.0~

~0.8~ ~1.3~

Equation 8.2 is used to find out the desired sterilization criterion for the reduction in the initial number of spores, Ni (spores/ml), present in the reactor volume of VF to N.

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Medium Sterilizer

— total = ln

N NiVF

(8.6)

The final level of contamination, N, represents the acceptable risk in sterilization. The sterilization criterion is the basis of the design of the sterilizer in the biochemical industry. For example, contamination risk depends on the particular biochemical industry: It may be one organism surviving every 10 million organisms or one organism surviving every 100 million organisms. It depends on the level of sterility requires in the biochemical industry.

8.2.5 Temperature–Time Profile during Batch Sterilization

Figure 8.6 shows the temperature profile of a batch sterilization process. The heating and cooling periods are dependent on the volume of the medium to be sterilized. It consists of heating, cooling, the desired temperature of the medium. The overall sterilization criterion (—total) may be divided into three parts: heating, holding, and cooling period, as mentioned below: — total = — heating + — holding + —cooling

(8.7)

The temperature–time profiles of heating and cooling depends on the volume of the medium (Fig. 8.6). The extent of cell death can be calculated from this profile using the Arrhenius equation. It is possible to plot the thermal death constant k as a function of time. The area under the curve is considered the sterilization criterion (—total). tf

Ú

ln( N/N0 ) = — total = kdt o

(8.8)

In the case of holding period, the temperature is constant. So at a particular holding temperature, k is fixed, and sterilization criterion is calculated as — holding = k(t2 - t1 )

(8.9)

Table 8.2 depicts the change in the values of k with respect to temperature using Bacillus stearothermophilus. The sterilization time (holding time) of the fermentation process is changed accordingly. In the laboratory, the sterilization temperature of 121°C for 15 min

Thermal Destruction

is found suitable. This is usually carried out in a batch process. But the sterilization temperature of 140°C is usually considered by the industries using the HTST technique. 140

100 Temp (°C)

60 40 0

t1

2

4

t2

6

Time (h)

Figure 8.6 Temperature–time profile of batch sterilization of medium.

Table 8.2

Typical values of k for Bacillus stearothermophilus spores (N/N0 = 10−15) Temperature (°C)

K (min−1 )

Holding time* (min)

100

0.02

1730

110

120

130

140

150

0.21

2.0

17.5

136

956

164

17

2

0.25

0.04

8.2.6 Scale-Up of Batch Sterilization

In batch sterilization, the destruction of nutrients in the medium will be increased if the volume of the fermenter is increased from V1 to V2 because the medium will be heated for a longer period of time (Fig. 8.7). If the nutrient content in the medium is critical for the

321

322

Medium Sterilizer

product formation, the product concentration may be reduced due to over-sterilization during scale-up. So medium sterilization for large-scale fermentation is usually done in the continuous mode. NV2 NV1 Increasing volume

Viable cells/ fermenter t1

t2

Co

Nutrient Conc.

C1

Time

t1

t2

Figure 8.7 Effect of scale-up on the nutrient quality of the medium in batch sterilization.

8.3 Continuous Sterilization The HTST treatment is the basis of the continuous medium sterilization. It has already been mentioned that an increase in temperature has relatively greater effect on the thermal destruction of cells than on nutrients. Around 20–25% steam consumption takes place in the case of continuous sterilization as compared to batch process. The total time required to sterilize media in the case of continuous sterilization varies between 2–3 h as compared to 5–6 h in the case of batch process.

Equipment for Continuous Sterilization

8.3.1 Equipment for Continuous Sterilization 8.3.1.1 Continuous steam injection This process is shown in Fig. 8.8a. In continuous steam injection (CSI), a rapid increase in the temperature of the medium will take place without the use of heat exchanger due to direct steam injection. The temperature–time profile is shown in Fig. 8.8b. This process is suitable with media that tend to foul heat exchanger. The disadvantage of this process is the dilution of the medium with condensed steam, and this causes difficulty in controlling pressure and temperature due to variation in the viscosity of the medium. Steam

2~3 min

140°C

Raw medium

140°C Flash Cooling 80°C

Temp. (°C) Expansion valve

37°C

Holding Section

(a)

(b)

Time (min)

Figure 8.8 (a) Schematic diagram of CSI process for medium sterilization and (b) the temperature–time profile.

8.3.1.2 Continuous plate heat exchangers These exchangers are used to avoid direct steam injection. The principles of operations and the temperature–time profile are shown in Figs. 8.9a and 8.9b, respectively. The disadvantages of this process are as follows:

∑ Increased capital cost over use of steam ∑ Hot heat exchanger surfaces fouling ∑ Leakage of the exchanger seals and gaskets

This process is particularly effective for higher volume of the medium. The essential components of the continuous sterilizer

323

324

Medium Sterilizer

system include a heat exchanger for recovery of heat from the sterilized fluid and preheating fresh medium, known as economizer. This is followed by a heat exchanger or steam injector for heating the medium to sterilizing temperature, a holding section for sterilization, and a heat exchanger for cooling the medium to the fermentation temperature. The system is connected to the medium mixing tank. These components are typically arranged as shown in Fig. 8.10. The time required to achieve medium sterilization for the use of batch fermentation is typically 2–3 h. The time required to reach 99% of the final temperature of the particle present in the medium in the suspension form depends on its capacity. Sterile medium

20 S

Steam

2 ~ 3 min

20 S

Temp. (°C) 37°C

Cooling raw water medium

Holding section

Time

Figure 8.9 (a) Continuous plate heat exchanger (CPHE) for medium sterilization and (b) the temperature–time profile.

8.4 Design of a Continuous Sterilizer 8.4.1 Fluid Flow The residence time of the medium in the sterilizer depends on the type of fluid flow to assure adequate sterilization. Plug or piston flow is considered ideal (Fig. 8.11). In practice, the flow lies between viscous and fully turbulent flow so that the average velocity of the fluid varies from 0.5 to 0.82 times the maximum velocity (umax). To reduce the overheating of the medium, a fully turbulent flow is necessary. The Reynolds number (NRe) is at least 2.5 × 103 and preferably above 2 × 104. Moreover, this will reduce the extent of

Design of a Continuous Sterilizer

axial dispersion in the sterilizer. The Peclet number (Pe) deals with the degree of axial dispersion as described below: Pe =

uL Ez

(8.10)

where u is the average fluid velocity, L is the length of the sterilizer, and Ez is the axial dispersion coefficient. The Reynolds number in a tube is expressed as NRe =

Du r m

(8.11)

where D is the inner diameter of the tube, r is the density of the medium, and µ is the medium viscosity. Cooler Steam

Holding section

Economizer Raw material Heater

Fermenter

Mixing Tank

Figure 8.10 Continuous sterilization system for batch fermenter.

m = mmax

Piston flow

Figure 8.11 Types of fluid flow.

m = (0.5) mmax

Viscous flow

325

326

Medium Sterilizer

10-11 10-12 10-13

N2 N1

10-14 10-15 10-16 10-17 10-18 10-19

20

40

60

80 Nr =

100

120

140

kdL u

Figure 8.12 Removal of the contaminants present in the medium by heat as a function of the Peclet number (Pe) and the reaction number (Nr). Reprinted with permission from Ref. [1], Copyright 2004, John Wiley and Sons.

Figure 8.12 shows the dispersion effect on the residence time in a continuous sterilizer. At perfect plug flow, the Peclet number is infinite. Peclet numbers from 3 to 600 are maintained, and the cell concentration begins to decline the initial value N0 with a normal residence time 0.9. In the design of a sterilizer, the deviation from the plug flow is to be taken into account. So for designing the sterilizer, the first step in continuous sterilizer design is to determine the type flow. The NRe should be greater than 2 × 104. The liquid velocity is calculated after assuming a suitable inner diameter of the pipe. Then Peclet number is to be calculated or determined. The sterilization criterion N/N0 is to be chosen. The value of N reflects the desired

Design of a Continuous Sterilizer

tolerance for contamination, e.g., one in 10 million. The reaction number Nr can be found out from Fig. 8.13, from the values of N/N0 and Peclet number. Nr is written as Nr =

kL u

(8.12)

where k is the thermal death constant. The value of k is calculated at the turbulent flow and the holding temperature. Thus, from the reaction number (Nr), one can calculate the length of the sterilizer (L). Laminar

Turbulent

100

10 Dz

uD

Experimental 1

0.1 103

Theoretical 105

104

106

Dur Re = m

Figure 8.13 Correlation between axial dispersion coefficient and Reynolds number in the case of turbulent flow. Reprinted with permission from Ref. [4], Copyright 1958, American Chemical Society.

Examples PROBLEM 8.1: A 15 m3 chemostat is operated at a dilution rate of 0.1 h−1. A continuous direct steam injection sterilizer with flash cooling is utilized for medium sterilization. The temperature of the holding section for the medium sterilizer is maintained at 130°C.

327

328

Medium Sterilizer

The contaminants concentration in the raw medium is 105 mL−1. An acceptable contamination risk is one organism every 3 months. The activation energy for thermal death and Arrhenius constant are estimated to be 288.5 kJ/gmol and 7.5 × 1039 h−1, respectively. The inner diameter of the pipe of the sterilizer is 12 cm. Determine the length of the holding section assuming perfect plug flow. SOLUTION Given V = 15 m3, D = 0.1 h−1, A = 7.5 × 1039 h−1, Ed = 288.5 kJ/gmol, T = 130°C, and d = 12 cm. For a chemostat, we know that D = F/V

Therefore, F = DV = (0.1)(15) = 1.5 m3/h The linear velocity (u) in the holding section of the sterilizer can be determined by dividing F by the pipe cross-sectional area (SA). Thus, u=

F F 1.5 = 132.62 m/h = 2= A pr p 0.062

The specific death rate constant kd can be calculated as kd = Ae

-

Ed RT

= (7.5 ¥ 10 )e 39

-

2.885 ¥ 105 ( 8.314 )(130+273)

= 313.1 h−1

3 months = 90 days, So the number of cells entering the sterilizer is

N1 = 1.5 m3/h × (105 × 106 m−3) × (90 × 24) h = 3.24 × 1014

The acceptable number of cells remaining at the end of the 3-month sterilization (N2) is 1. For a perfect plug flow, with no axial dispersion, the holding time can be given as N1 3.24 ¥ 1014 N2 ln 1 = = = 0.107 h 313.1 kd ln

t hd

The temperature, fluid velocity, and diameter of the tube of the holding section are assumed to be constant. So the length of the holding section may be written as L = ut hd = 132.62 m/h (0.107 h) = 14.2 m

Thus, the length of the holding section is 14.2 m.

Design of a Continuous Sterilizer

PROBLEM 8.2: A fermentation medium is to be sterilized in a continuous sterilizer by a steam heat exchanger with a flow rate of 2 m3/h. The concentration of bacterial spores in the liquid is 5 × 1012 m−3. The Arrhenius constant and activation energy for thermal destruction of these contaminants are 5.7 × 1039 h−1 and 283 kJ/gmol, respectively. One organism surviving every 60 days’ operation is considered an acceptable contamination risk. The inner diameter of the sterilizer pipe is 0.1 m, and the length of the holding section is 36 m. The viscosity and density of the medium are 3.6 kg/m h and 1000 kg/m3, respectively. Find out the sterilizing temperature? SOLUTION The basis of the sterilizer is the one organism surviving every 60 days’ operation. Assuming no cell death in the heating and cooling sections, the number of cells entering the holding section over 60 days is Ê 24 h ˆ N0 = 2 m3/h(5 ¥ 1012 m -3 )Á (60 d) = 1.44 ¥ 1016 Ë 1 d ˜¯

Now, N = 1, the acceptable number of cells remaining after the sterilization. Therefore, N 1 = = 6.9 ¥ 10-17 N0 (1.44 ¥ 1016 )

The linear velocity (u) in the sterilizer can be calculated as follows: u=

2 m3h -1

Ê 0.1 m ˆ pÁ Ë 2 ˜¯

2

= 177 m/h

To calculate Peclet number (Pe), the value of Dz is determined by the correlation as shown in Fig. 8.13. Re =

Dv r (0.1 m)(177 m/h)(1000 kg/m) = = 4.91 ¥ 103 m 3.6 kg/mh

For Re = 4.91 ¥ 103, Dz can be determined from Fig. 8.13 using either the experimental or theoretical curve. The experimental curve is considered because this gives a larger value of Dz and a smaller value of Pe. Therefore,

329

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Medium Sterilizer

Dz = 0.65 uD Dz = (0.65)(254 m/h)(0.1 m) = 16.6 m2/h uL (177 m/h)(24 m) = = 256 Dz 16.6 m2/h

Pe =

Using Fig. 8.12, we can determine the value of k for the desired level of N cell destruction. The value of kl/µ corresponding to 2 = 6.9 ¥ 10-17 N1 and Pe = 256 is about 42. Therefore, k=

(177 m/h)(42) = 309.8 h -1 24 m

According to the Arrhenius equation, k = Ae - Ea

RT

Rearranging the above equation, we get Ê -Ea ˆ ÁË R ˜¯ T= Ê kˆ ln Á ˜ Ë A¯

Known data, Ea = 283 kJ/gmol = 283 × 103 J/kgmol, A = 5.7 × 1039 h−1, R = 8.314 J/Kgmol, and k = 309.8 h−1. T=

Ê -283 ¥ 103 J/kgmol ˆ Á 8.314 J/kgmol ˜ ¯ Ë Ê 309.8 h -1 ˆ ln Á ˜ Ë 5.7 ¥ 1039 h -1 ¯

= 396.5 K

So the temperature of sterilization is 123.5°C.

PROBLEM 8.3: A 60,000 kg fermentation medium is to be sterilized within 40 min using two continuous sterilizers in parallel. The pipes of the holding section of the sterilizer have an inner diameter d = 0.155 m and length L = 50 m. The steam injected into the sterilizer elevates the temperature of the raw medium almost instantly. Estimate the temperature of sterilization that will satisfy the following requirements:

Design of a Continuous Sterilizer

The levels of contamination and the sterility desired are: Contamination level of the medium: N0 = 105/mL Sterility level of the medium: N = 10−3 The characteristics of the fermentation medium at the temperature of sterilization are Cp = 1 kcal/kg°C, r = 103 kg/m3, and m = 3.6 kg/m h. Assume the Peclet number to be 1610. (Figures 8.12 and 8.14 are given.) 34 30

26

22 K 18

14

10 125

126

127

128 T(°C)

129

130

Figure 8.14 Values of the thermal death rate constant, k, at different sterilization temperatures

SOLUTION Given data: Amount of medium to be sterilized is 60,000 kg; diameter of the pipe is 0.155 m; length of the pipe is 50 m; Cp = 1 kcal/kg°C; r = 103 kg/m3; m = 3.6 kg/m h; Peclet number = 1610. Schematic diagram of the process may be represented as follows: Sterilizer I

Sterilized medium (N2)

Raw medium (N1)

Sterilizer II

331

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Medium Sterilizer

Volume of the medium to be sterilized is

60, 000 kg = 60 m3 103 kg/m3 Volume of the medium to be sterilized in each sterilizer is 60 m3 = 30 m3 2

Flow rate of the medium is

30 m3 = 0.0125 m3/s 40 min

Velocity of the medium is u=

Again From Fig. 8.12, So

0.0125 m3/s

0.1552 2 m 3.14 ¥ 4

= 0.66 m/s

N1 10-3 = 5 = 0.33 ¥ 10-15 N2 10 ¥ 106 ¥ 30

Nr =

kL = 33 u

k = 0.4356 s−1 = 26.135 min−1

From Fig. 8.14, we get the corresponding temperature of k as 129°C. So the temperature of the sterilizer is 129°C.

References

1. Aiba, S., Humphrey, A. E., and Millis, N. F. Biochemical Engineering, University of Tokyo Press, Tokyo, 1973. 2. Bailey, J. E. and Ollis, D. F. Biochemical Engineering Fundamentals, McGraw-Hill, New York, 1977. 3. Cooney, C. L. Media Sterilization, Comprehensive Biotechnology, Vol 2, Murray MooYoung (Editor-in-Chief), Pergamon Press, 1985.

4. Levenspiel, O. Longitudinal mixing of fluids flowing in circular pipes, Ind. Eng. Chem., 50, 343, 1958.

References

5. Das, D. and Bhattacharyya, B. C. Short term course on “Analysis and Design of Novel Bioreactor,” IIT Kharagpur, 1989.

6. Prescott, S. C. and Dunn, C. G. Industrial Microbiology, McGraw Hill Book Co. Inc., and K O Gakusha Co. Ltd., Tokyo, 1959. 7. Block, S. S. Disinfection, Sterilization and Preservation, Fifth Edition, Lippincott Williams & Wilkins, 2001.

8. Whittet, T. D., Hugo, W. B., and Wilkinson, G. R. Sterilization and Disinfection, William Heinemann Medical Books Ltd., London, 1965. 9. Doran, P. M. Bioprocess Engineering Principles, Second Edition, Academic Press, Waltham, USA, 2012.

10. Shuler, M. L. and Kargi, F. Bioprocess Engineering: Basic Concepts, Second Edition, Prentice Hall Inc., New Delhi, India, 2002.

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Chapter 9

Transport Phenomena of Bioprocesses

Transport phenomena is the study of the movement of different physical quantities in any chemical or biochemical process and consists of the basic principles and laws of transport. It also describes the relations and similarities among different types of transport that may occur in any system. A proper knowledge of transport phenomenon is necessary in order to effectively design and efficiently operate a bioprocess/biochemical plant. The current chapter lists the fundamentals of transport processes that one comes across in the bioprocessing industries. Hence, this topic is important for people from the chemical as well as non-chemical backgrounds. The word “transport phenomenon” in chemical engineering encompasses the subjects of momentum transfer or fluid mechanics, mass and energy, or heat transfer processes:





∑ Momentum transport deals with the transport of momentum in fluids and is also known as fluid dynamics, e.g., blood circulation in body, mixing phenomena in bioreactor.

∑ Energy or heat transport deals with the transport of different forms of energy in a system and is also known as heat transfer, e.g., sterilization of reactor, temperature control in bioreactor. ∑ Mass transport deals with the transport of various chemical species themselves and is also known as mass transfer, e.g., oxygen transport from bubbles to aerobic microorganism.

Biochemical Engineering: An Introductory Textbook Debabrata Das and Debayan Das Copyright © 2019 Jenny Stanford Publishing Pte. Ltd. ISBN 978-981-4800-43-3 (Hardcover), 978-0-429-03124-3 (eBook) www.jennystanford.com

336

Transport Phenomena of Bioprocesses

At the beginning, the important and fundamental topics of fluid mechanics will be covered, followed by the heat transfer principles and mass-transfer phenomenon. A typical example of fluid mechanics may be the flow of various fluids in conduit and pipes in the biochemical industry. Note that such flows result in significant pressure drop due to friction, and the mathematical expression of such situations is incorporated in the subject of fluid mechanics. Similarly, the cooling or heating of fluids in a fermenter involves the transfer of heat from the system to the fluid within the fermenter or vice versa. Such cases are typical examples of the phenomenon of heat transfer in biochemical industries. An example of masstransfer process in biochemical process plants or industries is that of the oxygen transport to the media broth (culture) from the atmosphere (air) within the aerobic fermenter. Three different types of physical quantities are used in transport phenomena: scalar (e.g., temperature, pressure, and concentration), vector (e.g., velocity, momentum, and force), and second-order tensor (e.g., stress or momentum flux and velocity gradient). Transport phenomena can be studied in three levels (Fig. 9.1): 1. Macroscopic level 2. Microscopic level 3. Molecular level

Macroscopic

Microscopic

Molecular

Figure 9.1 Different levels of transport phenomena. Adapted from Ref. [11].

The basic equations of transport phenomena are derived based on the following three axioms: 1. The equation of continuity based on the conservation of mass.

Fluid Mechanics in Bioprocesses

2. The equation of motion based on the conservation of momentum (Newton’s second law). 3. The equation of thermal energy based on the conservation of energy.

These equations are collectively called as the governing equations. The general statement for any transport process is Rate of transport process =

Driving force Resistance

where driving force means the factor that will make the transfer occur and resistance means the factor that will slow down the transport process. The driving force for momentum transport includes velocity gradient (microscopic/molecular) or, velocity difference (macroscopic). The driving force for energy or heat transport includes temperature gradient (microscopic/molecular) or, temperature difference (macroscopic). The driving force for mass transport includes concentration gradient (microscopic/molecular) or, concentration difference (macroscopic). There may be special cases where the driving force becomes negligible. In those cases, a significant resistance arises in the transport phenomenon and the system gradually reaches equilibrium. It may be noted that the transport phenomena may occur within a single phase or between multiple phases. Typical examples will be discussed and covered in the consecutive sections. Let us start our discussion with the phenomenon of fluid mechanics or momentum transfer.

9.1 Fluid Mechanics in Bioprocesses

In the bioprocessing industry, the flow of fluid occurs mostly through the piping. It is important to mention that the word “fluid” incorporates both gas and liquid. Fluid flow involves two distinct regimes, which largely depend on the physical properties of the fluids. The first regime involves the flow of the fluid particles only in a single direction. Hence, the components of velocity are either negligible or not present in the other directions. This type of flow regime is called the laminar flow and is also known by other names

337

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Transport Phenomena of Bioprocesses

such as viscous flow, streamline, etc. The second regime involves turbulence in the fluid flow, and this is mainly due to the motion of the fluid particle clusters. In general, the fluid flow is in one direction even in the second regime. In order to describe the laminar flow regime, let us take into consideration that fluid is present in the space between two parallel laminar plates (Fig. 9.2). One plate is separated from the other by a distance, say, y. Let us suppose that a force is applied to the top plate due to which it moves at a constant velocity u, while the bottom parallel plate is maintained stationary (velocity is zero). Once the top plate moves, the liquid layer adhering to the top plate, it also tends to move. Consequently, due to the associated inertia, the layers of fluid right under the top layer also gain momentum and move accordingly. However, the velocity will be surely lesser than the velocity of the topmost layer of fluid. At steady state, the bottommost layer of fluid adhering to the bottom stationary plate will have zero velocity. On the other hand, let us suppose the topmost layer of fluid (adhering to the plate) has a velocity u0. Then the question arises: how will the velocity distribution/profile look like across the layers? Figure 9.2 clearly shows that the velocity distribution is linear as the velocity will go on decreasing from the top plate along each subsequent layer, and the velocity of the fluid near the bottom plate will be almost zero. Plate A

Y

y X

Plate B

Figure 9.2 The velocity profile of laminar flow between parallel plates moving at different velocities.

Hence, we can very well establish that the nature of the velocity distribution or the velocity gradient is dependent on the force with which the top plate is moved. In physics, this applied force is called shear stress (t), and the resulting velocity gradient is known as the shear rate (du/dy). Note that the shear stress and shear rate are proportional to each other. In order to remove the sign of

Fluid Mechanics in Bioprocesses

proportionality, a constant must be introduced: viscosity. The overall equation looks like the following: t = -m

du dy

(9.1)

In Eq. 9.1, µ denotes the viscosity and the unit is Pa s. The dimension of viscosity is ML–1T–1. As the force or shear stress is applied from a domain of higher velocity to that of lower velocity, the sign of the velocity gradient is negative. The above equation is referred to as the Newton’s law of viscosity. Alternatively we can also express Eq. 9.1 as follows: t=

F du = -m A dy

(9.2)

du (9.3) dy In Eqs. 9.2 and 9.3, F denotes the force and A depicts the surface area on which force is applied. We can also express force F as follows: F = - Am

mass ¥ velocity momentum = (9.4) time time Hence, we see that force may be expressed as the rate of momentum change. Let the term momentum be denoted by mv and the time be depicted by t. Putting Eq. 9.4 in Eq. 9.2, one can get F = mass ¥ acceleration =

mv du = -m (9.5) At dy In the transport phenomenon, any transport rate per unit area is called the flux. In Eq. 9.5, the term mv/A is defined as the momentum flux (mf). Eq. 9.5 demonstrates that the rate of change in momentum flux is directly proportional to the negative value of velocity gradient. Now, if we divide and multiply the second term of Eq. 9.2 by the fluid density (r), we obtain t = -m

du d( ru) m d( ru) == -J dy dy r dy

(9.6)

Equation 9.6 clearly demonstrates that the negative of mass flux gradient and the momentum flux are directly proportional to each other.

339

Transport Phenomena of Bioprocesses

A new term (J) has been introduced in Eq. 9.6, which displays the ratio of the viscosity to the density of the fluid (μ/r) and is known as the kinematic viscosity of a fluid. The SI unit of kinematic viscosity is cm2/s (Stokes), and hence the dimension is L2T–1. Technically, the term kinematic viscosity can also be replaced by momentum diffusivity (analogous to thermal diffusivity in heat transfer and mass diffusivity in mass transfer) as it refers to the ability or extent of the fluid for momentum transport. In a similar manner, concentration and heat content are the driving forces for mass transfer and heat transfer, respectively. This will be discussed in the later part of this chapter. Based on its property, a fluid may be classified as Newtonian and non-Newtonian. The fluid viscosity and shear rates are independent of each other. If one plots shear rate versus shear strain, a straight line passing through the origin will be achieved for the Newtonian fluids. This is clearly shown in Fig. 9.3. Note that based on Eq. 9.1, the slope of such a straight line will give the viscosity (Fig. 9.3). Newtonian fluids mostly include liquids with low molecular weight and almost all gases. It is important to mention that increasing temperature results in a decrease in liquid viscosity and an increase in gas viscosity. An increase in pressure enhances both the liquid and gas viscosities. Bingham fluid Pseudoplastic fluid Shear stress, Pa

340

Newtonian fluid

Dilatent fluid

Shear rate (du/dy), s-1

Figure 9.3 Relationships between shear rate and shear stress for Newtonian and non-Newtonian fluids.

Fluid Mechanics in Bioprocesses

It may be noted that most of the fluids that exist in nature are non-Newtonian fluids. These fluids display a significant viscosity variation with shear rates. Based on the extent of variation of the shear rate with shear stress, the non-Newtonian fluids may be classified as pseudoplastic, Bingham, and dilatant (Fig. 9.3). This clearly demonstrates that the viscosity decreases with an increase in the shear rate for pseudoplastic fluids, whereas the viscosity increases with the shear rate for dilatant fluids. The Bingham plastic fluids are those that need a certain amount of shear stress till a threshold is reached. Beyond that threshold, the fluid flows and hence a linear relation is obtained between the shear stress and shear rate. The shear stress (t) can be represented by Ê du ˆ t =KÁ ˜ Ë dy ¯

n

(9.7)

where the flow behavior index is given by n, and K is called the consistency index. Note that n > 1 corresponds to dilatant fluids, whereas pseudoplastic fluids correspond to n < 1. In biochemical engineering, fermentation media containing microorganisms or fermentation broths mostly behave as nonNewtonian liquids. There exists a different category of fluids that display some amount of elastic behavior. These fluids are known as viscoelastic fluids. Typical examples include polymer solutions. The phenomenon of elasticity may be defined as the tendency to which an object comes back to its original form after the applied stress is released. The ratio of the applied stress (Pa) to strain (s−1) is termed the elastic modulus. As already mentioned, fluid flow can be categorized into two distinct regimes: laminar and turbulent. The regime of fluid transport in between the turbulent and laminar regimes is referred to as the transition flow regime. The behavior of fluid as turbulent or laminar depends on the Reynolds number, which is a dimensionless number. Re is defined as Re =

dv r m

(9.8)

In Eq. 9.8, d corresponds to the inner diameter of a circular pipe through which the fluid is flowing; v depicts linear fluid velocity; fluid density and viscosity are represented by r and m, respectively.

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If the Reynolds number is lesser than 2300, then the flow regime is laminar, whereas the flow regime is termed turbulent if Re is higher than 4000. Figure 9.4b represents the laminar regime in a tube (having circular cross section) for the cases with Re £ 2300. Similarly, Fig. 9.4a demonstrates the situation where Re = 10000. One can easily mark the difference between the two flow regimes. It is clearly seen from Fig. 9.4 that the laminar flow regime is much sharper and smoother compared to the turbulent flow regime, which demonstrates relatively flat profile. An instability in the flow transport occurs in the fluid regime where the flow starts changing from laminar to turbulent (2300 < Re < 4000) inside a tube (of circular cross section). However, it has been noticed that the turbulence is initiated in the fluid flow on the external portion of the tube at a much smaller Re. r/ri

M

(a) Turbulent flow (Re =10,000)

(b) Laminar flow (Re ≤ 2000)

Figure 9.4 Velocity profiles of laminar and turbulent flows through a tube.

Turbulent flow is significant in many applications of biochemical engineering. Various methods of measurement have been implemented to incorporate the velocity fluctuations of the eddies during the turbulent flow regime. Turbulence arises due to the greater contact of the flowing fluid with the solid boundary, and in such cases, it is known as wall turbulence. It may also arise when the fluid layers of two varying velocities are in contact with each other. This type of turbulence is known as the free turbulence. Note that free turbulence is extremely important during the phenomenon of mixing. It is important to remember that the turbulent flow consists of various sizes of eddies, which coexist in the flow stream. The smaller eddies have the tendency to combine and form a larger eddy.

Boundary Layer Theory

However, the larger eddies are extremely unstable, and they keep breaking into smaller eddies. Consequently, the smallest eddies disappear, and one can observe a wide spectrum of eddy size existing in the flow regime. It may be noted that the flow within this eddy is laminar and all eddies are of macroscopic size. By analogy with Eq. 9.1, the relationship between velocity gradient and shear stress in a turbulent stream is used to define the eddy viscosity Ev:

du (9.9) dy In Eq. 9.9, the quantity Ev is analogous to the absolute viscosity m. Also in analogy with the kinematic viscosity J, the quantity eM is also defined, which is called the eddy diffusivity of momentum Ê Ev ˆ Á e M = r ˜ . Hence, the total shear stress in a turbulent fluid is the ¯ Ë summation of viscous stress and turbulent stress: t t = Ev

t = (m + Ev )

t = (J + e M )

du dy

d( ru) dy

(9.10)

(9.11)

It is important to mention that Ev and eM depend on the fluid velocity and the geometry of the system. On the other hand, m and J are the true properties of the fluid.

9.2 Boundary Layer Theory

We will first discuss the boundary layer formation along a rectangular plate of length L. It has been found that a moving fluid region close to the solid boundary has a significant effect on the solid boundary or the wall. This region is known as the boundary layer. In the domain of the boundary layer, there is a variation in fluid velocity from zero to u. This velocity u is not exactly the free stream velocity. Rather, u corresponds to nearly 99% of the free stream velocity. The thickness of the boundary layer, i.e., δ is largely dependent on the forces of friction, which tend to decrease in the vertical direction away from the solid boundary. This concept is important since it is necessary to understand the development of fluid flow along the solid boundary.

343

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One must understand that the value of the fluid velocity corresponds to nearly zero near the solid–fluid interface at the boundary. This is true from the assumption that the fluid particles near the boundary will have the same velocity as the boundary. As we have assumed a stagnant solid boundary, the fluid particles adhering to the solid boundary will have zero velocity. In other words, the fluid particles are not slipping away from the solid boundary or the no-slip boundary conditions are maintained. The fluid flow near the boundary is essentially laminar. Farther away from the surface, the velocity of the fluid slowly enhances as the boundary effect decreases upward. Gradually, the flow becomes turbulent, and the eddies play a major role in making the flow pattern distorted. The onset of turbulence is characterized by an increase in the boundary layer thickness. Note that a transition layer or zone exists between the turbulence and laminar regions. Overall, one may infer that the turbulent boundary layer consists of three different zones, namely, the turbulent zone, viscous sublayer, and the buffer layer. The extent of turbulence occurring within the laminar boundary layer is governed by the Reynolds number definition: ReL =

Lu• r m

(9.12)

In Eq. 9.12, L is the distance from the leading edge of the plate, u• is the bulk fluid velocity, r is density, and m is the viscosity. Note that turbulent flow appears at a critical Reynolds number between 105 and 3 ¥ 106. The transition occurs at the lower Reynolds number when the plate is rugged or rough, and the intensity of turbulence in the approaching stream is higher. On the other hand, for a smooth surface and lower intensity of approaching turbulence, the value of Reynold’s number is relatively lesser. Let us next discuss the boundary layer formation inside a circular tube or pipe (Fig. 9.2). In contrast to a rectangular plate, we have two solid boundaries inside a tube. As expected, the boundary layer will start developing from the entrance and the boundary layer thickens as the fluid moves further down the channel. Near the entrance and almost 30% down the channel, the boundary layer occupies only part of the cross section and the total stream consists of a core of fluid flowing in a rod-like manner at a constant velocity. Further down the channel, the boundary layer reaches the center of the tube and

Incompressible Fluid Flow inside a Pipe

it occupies almost the entire cross section. The velocity distribution at this point resembles a parabolic profile. One can observe that the velocity distribution remains unchanged after a certain distance, and hence the flow can be termed fully developed.

9.3 Incompressible Fluid Flow inside a Pipe

Figure 9.4b displays the profile of velocity of a Newtonian and incompressible fluid inside a straight tube of circular cross section assuming the flow to be laminar and isothermal. The profile has been found to be parabolic, and the profile can be mathematically expressed as follows: u

uavg

È Ê ˆ2˘ r = 2 Í1 - Á ˜ ˙ Í Ë ri ¯ ˙ ˙˚ Î

(9.13)

Note that Eq. 9.13 represents a parabolic equation, and hence the velocity distribution inside the tube is also parabolic. In Eq. 9.13, the average velocity of the fluid inside the pipe is represented by uavg, ri denotes the inner radius of the pipe; local velocity (at a distance r with reference to pipe axis) is denoted by u. The maximum velocity of the fluid flow occurs near the center, i.e., at r = 0. Hence, Eq. 9.13 will be modified as umax = 2uavg

(9.14)

At r = ri, the velocity of fluid flow (u) is zero. This is also true based on our assumptions as discussed earlier. It has been mentioned that the velocity of fluid near the boundary (i.e., at r = 0) is zero based on the no-slip boundary conditions. Note that Eq. 9.13 is achieved after doing force balance across a circular cross section within a pipe. Two types of forces are taken into consideration: (a) pressure force acting on the circular cross section due to the motion of fluid inside the pipe and (b) forces due to the inner friction. Using the force balance, the mathematical expression for the pressure drop may be given by the Hagen–Poiseuille law: DP = 32

m uavg L D2

(9.15)

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In Eq. 9.15, the average velocity is denoted by the symbol uavg. The definitions of the other symbols are same as discussed previously in various sections and chapters. Thus, from Eq. 9.15, it can be clearly illustrated that DP varies inversely to the square of the diameter of the pipe or tube and DP is directly proportional to the length of the tube. Hence, in order to increase the pressure drop across a pipe or a tube, one can decrease the inner diameter or increase the length of the pipe. It has been observed that there is a significant variation in the distribution of the velocity profile with the Reynolds number or flow rate when the fluid flows through a tube of circular cross section. A flat profile of velocity and relatively thinner laminar sublayers are observed when the flow rates are increased, and this contrasts the cases with lower flow rates where thicker laminar sublayer and a parabolic velocity distribution are observed. In general, the ratio, uavg/umax corresponds to nearly 0.7–0.85 along the axis of the tube and this ratio enhances with Re. Fanning equation nicely provides an important relation between the pressure drop DP and the kinetic (average velocity uavg) along with the physical properties of the fluid (length L and diameter D) and the geometrical properties (density r), which is as follows: DP = 2

f ruavg2 L

(9.16) D In Eq. 9.16, the fanning friction factor is denoted by f, which is related to Re via many relations such as f =

0.079

(9.17) Re0.25 for the range of Re from 3000 to 100,000. Comparing and equating Eqs. 9.15 and 9.16, one can get the expression of friction factor f as a function of Reynolds number (Re). f =

16 Re

9.4 Mass Transfer in Bioprocess

(9.18)

In biochemical processes, the rate of the reactions are largely governed by gas–liquid (gas absorption, humidification, or de-

Diffusion in Liquid and Gas

humidification reactions), solid–liquid (leaching, adsorption, and immobilization process), and liquid–liquid (liquid extraction process) flow characteristics. A typical example is that of fermentation broth. The overall rates of aerobic fermentation are controlled by the rate of absorption of oxygen inside the broth. It is important to mention that mass transport may occur sometimes against the concentration gradient in various living systems. During the phenomenon of masstransfer processes, one must understand and distinguish between passive and active transport. Passive transport is also referred to as the downhill transport, whereas active transport is basically known as the uphill mass transport. The phenomenon of active transport mostly occurs across biological membranes. The most important application of mass transfer in biochemical engineering is that of aeration of the fermentation broth. Note that maintaining the dissolved oxygen concentration to a certain limit is a major challenge as that has the vulnerability to be consumed in a shorter time. In aerobic fermentation process, adequate supply of oxygen to cells is highly essential as slight oxygen depletion in the media may result in an irreparable cell damage or cell death. As a result, the oxygen demand of the microorganisms must be satisfied by the regular and constant supply of oxygen through the method of aeration. Note that in need for the perfect metabolism and growth of cells, the transfer of oxygen is a major limiting step. In a microorganism, the way or path of substrate transfer from the gas bubble to an organelle involves thorough understanding of the diffusion phenomenon, which can be categorized under the field of mass transfer. Hence, it is important to discuss the basic concepts of mass transfer before proceeding further.

9.5 Diffusion in Liquid and Gas

Analogous to Fourier’s law in heat transfer and Newton’s law of viscosity in fluid mechanics, Fick’s law governs the process of diffusion mass transfer. It states that the molar flux of any component and the gradient in concentration (dC/dz) are directly proportional to each other. Hence, for any component A, Fick’s law along the z-direction may be written as

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dC A (9.19) dz In order to remove the proportionality sign, a constant is multiplied on the right-hand side of Eq. 9.19. Hence, actual Fick’s law equation is given as follows: JA µ

J A = - DAB

dC A dz

(9.20)

In Eq. 9.20, DAB is the diffusivity of A into B. It indirectly refers to the diffusive mobility. The molar flux (assuming component A is not in motion) can also be expressed as

CA dC ( N A + NB ) - DAB A (9.21) C dz where the cumulative concentration of the components A and B is represented by C, and NB is the molar flux of B. It may be noted that the first term (on the right-hand side) denotes the flux due to the bulk flow, while the second term (on the right-hand side) represents diffusion. This is the right point of the chapter to discuss, the most important entity associated with the mass transfer and that is mass transfer coefficient. This will be illustrated based on Fick’s second law. It states that the mass-transfer rate/area or the mass flux is directly proportional to the concentration difference. For a solute undergoing gas–liquid diffusion, one can write the mass flux of solute on the gas side as NA =

qG µ (CG - CGi ) (9.22) A In Eq. 9.22, CG refers to the bulk phase gas concentration, whereas the concentration at the interface is denoted by CGi. In order to remove the proportionality sign, a constant is introduced in Eq. 9.22. Hence, Eq. 9.22 may be modified as NG =

qG = kG (CG - CGi ) (9.23) A The term kG in Eq. 9.23 is known as the gas-side mass-transfer coefficient. In a similar manner, the liquid-side mass flux may be written as follows: NG =

Diffusion in Liquid and Gas

qL = kL (CLi - CL ) (9.24) A In Eq. 9.24, kL is known as the mass-transfer coefficient for the liquid phase. It is evident that NG = NL. Substituting Eqs. 9.23 and 9.24, we get NL =

kG(CG – CGi) = kL(CLi – CL) kL (CG - CGi ) = kG (CLi - CL )

Li G C

CGi

C

C

C

CG

KL

C

(9.25)

C

Cc CL

C CL

CLi

Ci

Figure 9.5 Concentration profile near a gas–liquid interface and an equilibrium curve.

Using Eq. 9.25, one can get the slope as kG/kL if a graph is drawn between CG versus CL (Fig. 9.5). It is extremely difficult to evaluate CLi and CGi; therefore, one cannot easily determine the mass-transfer coefficients on either side. The equation looks as follows: NG = NL = K G (CG - CG* ) = K L (CL* - CL )

(9.26)

where CG* and CL* are the gas-side concentration and liquid phase concentration, respectively. Due to the identical mass-transfer rates, we can write K L (CL* - CL ) = kL (CLi - CL )

(9.27)

1 1 CL* - CL 1 (CL* - CLi ) + (CLi - CL ) 1 1 CL* - CLi = = = + (9.28) K L kL CLi - CL kL CLi - CL kL kL CLi - CL

The entire denominator of the second term in Eq. 9.28 can be replaced as follows: kL(CLi – CL) = kG(CG – CGi)

(9.29)

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Transport Phenomena of Bioprocesses

Hence, Eq. 9.28 can be modified as

1 1 1 CL* - CLi 1 1 = + = + K L kL kG CG - CGi kL kG M

(9.30)

Equation 9.30, thus, relates KL, kG, and kL. M is the slope (Fig. 9.5). In a similar manner, one can prove 1 1 m = + K G kG kL

(9.31)

Here, m can be obtained as a slope joining (CG*, CL) and (CLi, CGi).

9.6 Oxygen Diffusion in Fermentation Broth

Through interphase mass transfer, one must understand the mechanism governing the diffusion of solute in the gas phase to the liquid in contact. In the field of biochemical engineering, interphase mass transfer plays an effective role especially during the aeration process in the fermentation broth. In other words, the entire method governing the interaction of the gaseous substrate (present in the gas bubble) with the microorganism within the fermentation broth can be sub-divided into six different steps:

Step 1: From a gas bubble, bulk gas is transferred to the gas layers, which is relatively weakly mixed inside the fermentation broth. Step 2: In this weakly mixed gas layer, the process of diffusion must take place.

Step 3: Next, diffusion must also take place to the zone of bulk liquid in the fermentation broth from the unmixed or weakly mixed liquid layers.

Step 4: The diffusion must next take place from the bulk liquid to the region surrounding the microorganism, which is again supposed to be relatively unmixed. Step 5: The diffusion must then occur through the weakly mixed layer of liquid surrounding the microorganism. Step 6: The last step involves the diffusion from the microorganism surface to the organelle, and this is the point where the oxygen is consumed.

Oxygen Diffusion in Fermentation Broth

9.6.1 Mechanism of Mass Transfer In order to understand the interphase mass transfer, many mechanisms are available, and the best-known mechanisms are as follows:

∑ Two-film theory: This theory states that two fictitious films are present on either side of the interface that causes the overall resistance to mass transfer and that takes place due to molecular diffusion. Mathematically, the two-film theory states that kL (mass-transfer coefficient) varies inversely as the film thickness zf, whereas kL and DAB (diffusivity) are directly proportional to each other.

kL =



DAB zf

(9.32)

∑ Penetration theory: Higbie came up with the penetration theory in 1935. The penetration theory assumes that turbulent eddies travel from the bulk to the interface and at the interface, they remain for an exposure time equal to t0. In addition, the process of unsteady-state molecular diffusion is assumed by which the solute is assumed to penetrate a given eddy during its stay at the interface. In contrast to the two-film theory, this model predicts that kL (mass-transfer coefficient) and the square root of DAB (molecular diffusivity) are directly proportional, whereas kL varies inversely as the exposure time t0. kL = 2

DAB p t0

(9.33)

In Eq. 9.33, the term p comes from the definition of error function. In general, the error function erf(x) can be defined as erf( x ) =

2

x

exp( - Z p Ú 0



2

)dZ

(9.34)

∑ Surface renewal theory: In 1951, Danckwerts modified Higbie’s penetration theory for the liquid phase mass transfer. The modified theory is widely known as the surface renewal theory. The surface renewal theory states that a portion of the

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Transport Phenomena of Bioprocesses

mass-transfer surface is replaced with a new surface by the motion of eddies near the surface assuming that the liquid elements at the interface are being randomly swapped by fresh elements from bulk and each of the liquid elements at the surface has the same probability of being substituted by a fresh element. Similar to the penetration theory, the masstransfer coefficient is directly proportional to the square root of diffusivity. The mathematical expression of the surface renewal theory is as follows: kL = sDAB

In Eq. 9.35, s is the surface renewal rate.

(9.35)

9.6.2 Estimation of Mass-Transfer Coefficients via Correlations The mass-transfer coefficient kL is highly dependent on the vessel’s geometrical configuration and the physical properties of the fluids. In general, one can correctly fit the experimental data and obtain the relations corresponding to kL. Using various methods such as the Buckingham Pi theory, the correlations can be obtained and expressed using the dimensionless groups since they are dimensionally consistent. In the process of mass transfer, three dimensionless numbers are extremely relevant: 1. Sherwood number: This basically denotes the dominance of mass diffusivity over the rate of mass transfer by convection or vice versa. It is expressed as Sh =

kL l DAB

(9.36)

In the dimensionless Eq. 9.36, l represents the characteristic length. In bioprocessing applications especially during aeration, l will be equivalent to the diameter of the air bubbles.

2. Schmidt number: The dominance of mass diffusivity over momentum diffusivity or vice versa is given by the Schmidt number. Hence, this dimensionless number is most applicable to cases that simultaneously involve mass and momentum diffusion and can be expressed as

Oxygen Diffusion in Fermentation Broth

Sc =

m rDAB

(9.37)

The definitions of the various symbols in the above dimensionless equation are already discussed in various sections of the present chapter and in a few of the previous chapters.

3. Grashof number: This number is mostly associated with the cases involving free (natural) convection. It gives the dominance of the buoyancy force over the viscous forces and can be mathematically given as Gr =

gb(TS - T• )L3 J2

(9.38)

In the dimensionless Eq. 9.38, b and J are the volume expansion coefficient and kinematic viscosity, respectively; L is the characteristic length; T∞ and Ts are the temperature in the bulk phase and on surface, respectively. Experimental works have inferred that the Sherwood number is a function of the Schmidt and Grashof numbers. A few correlations are presented for the various diameters of air bubbles: For bubble diameter < 2.5 mm, the correlation derived is as follows: Sh = 0.31 Sc1/3Gr1/3

(9.39)

Sh = 2 + 0.31 Sc1/3Gr1/3

(9.40)

For solid suspended substances and rigid spherical bubbles of smaller size, the correlation given by Eq. 9.39 can be modified as

For bubble diameter > 2.5 mm,

Sh = 0.42 Sc1/2Gr1/3

9.6.3 Volumetric Mass-Transfer Coefficient

(9.41)

Once a correlation between the above-mentioned dimensionless numbers has been established, the mass-transfer coefficient kL can be easily found out. The discussion on volumetric masstransfer coefficient kLa is also essential as it is largely applicable in bioprocessing applications such as in a fermenter. Two parameters

353

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Transport Phenomena of Bioprocesses

make up kLa (a is the interfacial area, and the mass-transfer coefficient is kL). Hence, there may be a bit of confusion while varying the conditions in a fermenter since there is no effective way of identifying which parameter among kL and a may vary. Note that there are techniques such as laser optography or photography to estimate and evaluate a. Mathematically speaking, a is largely dependent on the gas-phase volume fraction (Vg) and dS (Sauter mean diameter): a=

6Vg ds

In Eq. 9.42, ds is mathematically expressed as ds

 = Â

(9.42)

i =n

n d3 i =1 i i i =n n d2 i =1 i i

(9.43)

Alternatively, the light transmission technique can be implemented to measure the drop size distribution. The variation in transmittance inverse (1/T) with an interfacial area is found to be linear, and the straight-line equation is given as 1 = m1 + m2a T

(9.44)

In Eq. 9.44, m1 is unity and the constant m2 does not depend on the distribution of drop size.

9.7 Determination of Oxygen-Absorption Rate

For accurate estimation of the necessary parameters for oxygen uptake in a fermenter, the calculation procedure using the correlations presented previously may be rigorous and the predicted value from those correlations can vary widely. In addition, one may not find a suitable correlation for their systems. In such cases, one can measure the oxygen-transfer rate or use correlations based on those experiments. The oxygen-absorption rate per unit volume q/v can be estimated by q = kL a(CL* - CL ) v

(9.45)

Determination of Oxygen-Absorption Rate

As oxygen is categorized as a sparingly soluble gas, the overall masstransfer coefficient KL is equal to the liquid-side individual masstransfer coefficient. Hence Eq. 9.45 can be rewritten as follows: q = K L a(CL* - CL ) v

(9.46)

It must be remembered that the oxygen-transfer rate can be maximized by minimizing the power consumption and air flow rate. The oxygen absorption can be enhanced by increasing the interfacial area a, mass-transfer coefficient kL, and the concentration difference C*L – CL. It may be noted that the concentration difference is quite limited and low. Therefore, the main parameters of interest in design are the mass-transfer coefficient and the interfacial area. In general, the rate of oxygen transfer should be measured within the fermenter, which consists of the microorganisms and nutrient broth. However, the process is extremely rigorous and complicated due to the changing nature and rheology of the fermentation broth. A common strategy is to use a synthetic system that approximates fermentation conditions. A few of the processes are listed as follows:

∑ Dynamic gassing out technique: Van’t Riet came up with the dynamic gassing out technique in 1979. This technique monitors the change in the oxygen concentration, while an oxygen-rich liquid is deoxygenated by passing nitrogen through it. Polarographic electrode is usually used to measure the concentration. The mass balance in a vessel gives dCL = kL a(CL* - CL ) dt

CL ( t2 )

Ú

C L ( t1

dCL

C * - CL ) L

t2

Ú

= kL a dt t1

(9.47)

(9.48)

Integrating Eq. 9.48, one can get the expression of kLa as È C * - C (t ) ˘ L 1 ln Í L* ˙ ÍÎ CL - CL (t2 ) ˙˚ kL a = t 2 - t1

(9.49)

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If one knows the equilibrium concentration and the bulk liquid concentration at two different intervals, then kLa can be measured directly from Eq. 9.49.

∑ Direct measurement: In this technique, a gaseous oxygen analyzer is used to estimate the oxygen concentration in the gas stream entering and leaving the fermenter. If the inlet flow rate of oxygen is Qin and the outlet flow rate of oxygen is Qout, then the residual rate of uptake of oxygen (q) can be estimated as follows: q = Qin CO2in - Qout CO2out

(9.50)

In Eq. 9.50, the inlet and outlet concentrations of oxygen are depicted by CO2in and CO2out , respectively. Once the oxygen uptake q is measured, then from Eq. 9.46, one can evaluate the mass-transfer coefficient kL. An oxygen sensor is generally used to measure and evaluate the oxygen concentration of the liquid in a fermenter. The difference in the concentration term (CL* − CL) largely depends on the size of the fermenter. Note that the difference in concentration is significant for the lesser volume of the reactor, whereas the concentration difference varies extensively for the greater volume of the reactor. Let us suppose that (CL* − CL) be denoted by DCL. In this case, the logmean value of CL* − CL or DCL of the inlet and outlet of the gas stream can be used as (CL* - CL )log M =



DCL,in - DCL,out

ln( DCL,in /DCL,out )

(9.51)

∑ Dynamic technique: Taguchi and Humphrey used the dynamic technique to estimate the kLa value for the oxygen transfer during an actual fermentation process with real culture medium and microorganisms. The present method is implemented in a microorganism-rich aerated batch fermenter over which the oxygen material balance is applied. The material balance can be mathematically expressed as dCL = kL a(CL* - CL ) - rO2 C X dt

(9.52)

where the cell respiration rate is given by rO2 (g O2/ g cell h) and CX denotes the microorganism concentration inside the fermenter.

Determination of Oxygen-Absorption Rate

Despite a steady dissolved oxygen level in the fermenter, the oxygen concentration will be decreased sharply if the air supply is turned off. Note that the mass-transfer coefficient will be equal to zero for such cases. Hence, the rate of decrease in the oxygen concentration will follow the following rate: dCL = -rO2 C X dt

(9.53)

Hence, it is evident from Eq. 9.53 that if one plots CL versus t, then the slope will give the magnitude rO2 C X . If the airflow is turned on again, the dissolved oxygen concentration will be increased according to Eq. 9.52 and that can be rearranged to result in a linear relationship as CL = CL* -

1 Ê dCL ˆ + rO2 C X ˜ Á ¯ kL a Ë dt

(9.54)

Ê dC ˆ Based on Eq. 9.54, if one plots CL versus Á L + rO2 C X ˜ , the slope ¯ Ë dt will be obtained as −1/kLa and the intercept will be CL*.

9.7.1 Experimental Determination of kLa by Dynamic Gassing out Techniques

Experiments were carried out using Saccharomyces cerevisiae in a 2 L BioEngineering AG fermenter. Microorganism: Saccharomyces cerevisiae Optimum temperature: 30–35°C

Theory: When air is sparged through a fermentation medium, one fraction of it remains in the broth, a certain amount is utilized by the cell for its growth, and the rest comes out of the fermenter. Rate of oxygen transferred is KLa(C* – CL) where C* is the saturated dissolved oxygen concentration (mg/L) (assume C* = 9 mg/L), CL is the dissolved oxygen concentration of the fermentation broth at any time t (mg/L), and KLa is the volumetric mass-transfer coefficient (time−1). Rate of oxygen uptake/consumed by the cell is rxX where rx is specific oxygen (O2) uptake rate (time−1), X is cell mass concentration (mg/L).

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Ê Rate of O2 ˆ Rate of O2 ˆ Rate of oxygen dissolution = Ê - Á consumed˜ Ë transferred¯ Á by cells ˜ Ë ¯

dCL = K L a(C * - CL ) - rx X dt Equation 9.54 can be written as fi

(C * - CL ) = CL = C * -

1 Ê dCL ˆ + rx X ˜ ¯ K L a ÁË dt

1 Ê dCL ˆ + rx X ˜ Á ¯ K L a Ë dt

Ê dC ˆ Plotting Á L + rx X ˜ versus CL yields the slope of Ë dt ¯

Ê 1 ˆ ÁË - K a ˜¯ . L

Limiting O2 concentration for yeast cells is 20–30%. Critical O2 concentration is 10–50% of air saturation. Dissolved oxygen concentration (CL) profile in the yeast fermentation process is shown in Fig. 9.6. Observable and calculated data are shown in Table 9.1. The specific oxygen uptake rate of the organism is determined from the dissolved oxygen profile when air is turned off (Fig. 9.7). Air on

Air off

CAL CAL2

CAL1

CAL

Ccrit

t0

t1

t2

Time

Figure 9.6 Dissolved oxygen concentration (CL) profile.

Determination of Oxygen-Absorption Rate

Table 9.1 Observation and calculation table Time (s)

%DO at 350 rpm

CL (mg/L)

10

53

4.77

30

45.8

4.122

37.8

3.402

31.4

2.826

20

Air off

40 50 60 70 80 90

43

34.8 26.6 22.2

100

18

[dCL/dt + rx]

4.59

−0.0324

0.0031

3.87

−0.036

−0.0005

3.132 2.394 1.998

−0.036

−0.0005

−0.0369

−0.0014

−0.0369

−0.0014

−0.0387

−0.0032

−0.0288 −0.0414

0.0067

−0.0059

17.5

1.62

1.575

0.02115

0.01435

120

14.6

1.314

0.01125

0.02425

140

21.2

1.908

0.0576

0.0931

33.3

2.997

0.05175

110

Air on

51

dCL/dt

130 150 160

15

27.8

1.35

2.502

−0.0153

0.0297

0.0202

0.0652

0.05445

0.08995 0.08725

170

39.3

3.537

0.05355

0.08905

190

48.8

4.392

0.0324

0.0679

180 200 210 220 230

45.2 52.4 53.2 53.2 53.2

4.068 4.716 4.788 4.788 4.788

0.04275 0.0198 0.0036 0

0.07825 0.0553 0.0391 0.0355

Cell mass concentration (X) at 300 rpm is 950 mg/L.

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6

CL (mg/L)

5 y =-0.0355x + 5.2236

4 3 2 1 0

0

20

40

60 Time (s)

80

100

120

Figure 9.7 CL versus time at a stirrer rotational speed of 300 rpm when air is off.

6 5 CL(mg/L)

360

4 3

y = -41.087x + 6.6448

2 1 0

0

0.02

0.04

0.06

0.08

0.1

[dCL/dt + rxX]

Figure 9.8 Dissolved oxygen (CL) versus [dCL/dt + rxX] plot.

From the slope of Fig. 9.7,

KLa = (1/41.08) s−1 = 0.0243 s−1

So the volumetric mass-transfer coefficient of the fermentation process is 0.0243 s−1.

Heat Transfer in Bioprocess

9.8 Heat Transfer in Bioprocess As already discussed, heat transfer occurs due to the temperature variation between the systems. The heat transfer process has tremendous importance in many bioprocessing applications such as in the implementation of heating or cooling jacket around the fermentation tank, sterilization processes, etc. Heat transfer in the bioprocessing industry occurs mainly due to conduction and convection. It may be noted that the heat flow due to radiation is not considered in the bioprocessing applications as the fermenters are operated at low temperatures. Let us start our discussions with the process of conduction and convection. Thereafter, we will proceed to the designing of the heat exchanger as it is extensively used in sterilization processes.

9.8.1 Conduction

The phenomenon of conduction heat transfer mostly occurs within a stationary liquid or solid medium or across a stagnant medium owing to molecular interaction. It may be noted that the phenomenon of conduction does not include any bulk heat transfer motion in the media. Conduction is associated with the molecular or atomic activity due to which heat or thermal energy is transferred from a zone of greater temperature to a zone of lower temperature. It is important to mention that substances differ in the mechanism of heat conduction. For an electrically non-conducting solid, waves associated with the vibration in the lattice mostly result in conduction. In contrast, both electron translational motion and lattice vibrations are essentially responsible for the conduction phenomenon in conducting solid. Note that molecular interaction along with random molecular motion causes conduction in gases or liquids. As heat conduction mostly occurs in incompressible fluid medium and the velocity of the bulk fluid is zero, the energy balance equation for a medium undergoing heat conduction can be written as Rate of heat accumulation = Heat input – Heat output +   Heat generated – Head dissipated (9.55)

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rC p

or

∂T = qin - qout + q¢¢ ∂t

(9.56)

In Eq. 9.56, q¢¢ denotes the volumetric internal heat generation (W/m3) and that is caused by the heating jacket, biochemical reaction, etc. Note that qin and qout denote the heat flux. Heat dissipated is neglected. In general, the gradient in temperature results in the rate of heat transfer and is expressed via Fourier’s law as follows: q = -k

∂T ∂x

(9.57)

∂T denotes the temperature gradient along the ∂x x-direction and the thermal conductivity is denotes by k. The negative sign on the right-hand side depicts that the temperature gradient and the heat flux vector vary in the opposite direction.

In Eq. 9.57,

T

T

T1

T

T(X)

T

qx¢ qx¢

T

T2

L

L

Figure 9.9 Conduction heat transfer in one dimension (along the x-axis).

Let us consider Fig. 9.9 for the illustration of conduction heat transfer along one direction (x-axis). In order to evaluate the heat flux q, Eq. 9.57 is integrated along the length L of the slab. Note that T varies only in the x-direction, and hence the partial differential form

Heat Transfer in Bioprocess

of Eq. 9.57 is replaced by the total differential form. Hence, Eq. 9.57 is modified as T =T2

Ú

dT = -

T =T1

q k

T1 - T2 =

x =L

Ú dx

x =0

qL k

(9.58)

(9.59)

Hence, the temperature profile along the slab of length L is linear. Equation 9.59 can also be written as k q = (T1 - T2 ) L

(9.60)

It may be noted that Eq. 9.60 denotes the heat transfer rate per unit area or the heat flux. The heat rate by conduction, qx, through a plane wall of area A is then the product of the flux: qx = q ¥ A

Putting Eq. 9.61 in Eq. 9.60, we get qx =

kA (T - T ) L 1 2

(9.61) (9.62)

It is important to mention that the expression of the heat conduction equation will be different for different geometries.

9.8.2 Convection

Convection heat transfer takes place in a fluid undergoing bulk motion. In other words, the multiple effects of both conduction and fluid motion result in the convection heat transfer process. Hence, convection is a bit complicated and non-trivial as compared to conduction. When a fluid is in motion, the intensity of heat transfer is largely enhanced as hot and cold fluids come into contact. This finally results in the greater conduction rate at various domains of the fluid. In contrast to conduction, convection dominates the heat transfer process in fluid especially when it is in motion. Note that convection is largely enhanced when the motion of the fluid or fluid velocity increases. This process can be better explained based on Fig. 9.10, which shows that a steady heat transfer occurs within a fluid occupying the

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zones within two plates parallel with each other. Let us imagine a situation where there is no fluid motion, and conduction is dominant over convection. In other words, the hot plate will emit energy that will be transferred to the next layer of fluid particles. These fluid particles will then transfer heat energy to the fluid particles in the next layer and this process will go on till the heat energy is transferred to the cold plate. Alternatively, if we draw some amount of fluid (by a syringe or a pipette) from the region near the relatively hotter plate and inject that near the cold plate at the bottom, the rate of heat transfer will be rapid due to the motion of fluid as some amount of energy will be transferred to the other side. Hence, in such cases, the process of heat transfer is called convection dominant. Hot plate 110°C

Fluid Q

Heat transfer through the fluid

Cold plate 30°C

Figure 9.10 Fluid flowing between a hot plate and a cold plate.

It is important to mention that the convection heat transfer largely depends on the properties of the fluid such as the density, viscosity, heat capacity, thermal conductivity, etc. Also the velocity of the fluid is expected to play a pivotal role in the convection process. In addition, the roughness of the geometry and also the size and structure of the geometry play a significant role in the convection heat transfer process. Hence, it is clear that the convection heat transfer process is comparatively complex and non-trivial as it depends on many factors. Similar to conduction, the convection heat flux expressed as watts/area and the temperature difference are found to be directly proportional to each other. Newton’s law of cooling is ideal to explain such a situation and that is as follows:

Heat Transfer in Bioprocess

Q µ (TS - T• ) (9.63) A Equation 9.63 is written based on the fact that the surface temperature (TS) is higher than the free stream temperature (T•), and hence heat is being transferred from the surface to the fluid. In order to remove the proportionality, a constant will be multiplied on the right-hand side, and hence Eq. 9.63 is modified as qconv =

qconv =

Q = h(TS - T• ) A

(9.64)

In Eq. 9.64, the heat transfer coefficient is denoted by the symbol h. If one re-arranges Eq. 9.64, h may be defined as the heat transfer rate per unit difference in temperature and per unit heat transfer area. The unit of h will be W/m2 K. Let us recap a few of the important points of heat transfer process. Till now, it is quite clear to us when there is contact among two or more bodies, heat will tend to move from the body at higher temperature to the body at lower temperature. This transfer process will keep on continuing till the bodies have similar kind of temperature near the point of contact. In addition, there is no jump in the temperature at those contact regions. This condition is also known as no-temperature-jump condition. Hence, at the solid surface, the rates of heat transfer through conduction and convection are supposed to be identical. We can write qconv = qcond

h(TS - T• ) = -k

dT dx

(9.65)

(9.66)

Note that in Eq. 9.66, k denotes the thermal conductivity of the fluid and the term dT/dx represents the temperature gradient along the x-direction. If we assume a finite temperature difference (DT) along the length L, then the rate of convection can be written as qconv = hDT

(9.67)

qcond =

(9.68)

And the rate of conduction can be written as k DT L

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The ratio of the rate of convection/rate of conduction will give the quantity qconv hDT hL = = qcond k DT /L k

(9.69)

The term hL/k is referred to as the Nusselt number, which plays an extremely critical role in convection. Note that if the Nusselt number is higher, the rate of convection becomes more and more effective. It may also be observed that Nu = 1 represents the situation when the rates of conduction and convection are identical. Another important dimensionless number is the Prandtl number, which is important in explaining the development of the thermal boundary layer along a solid surface. Similar to a velocity boundary layer, a thermal boundary layer develops when a fluid at a specified temperature flows over a surface that is at a different temperature. The thermal boundary layer thickness at any location along the surface may be defined as the distance from the surface at which the temperature difference (T – Ts) equals 99% of (T• – Ts). The relative thickness of the velocity and the thermal boundary layers can be best described by the dimensionless parameter Prandtl number. It is defined as the ratio of momentum diffusivity and thermal diffusivity. Pr =

n CP m = k a

(9.70)

It may be noted that fluids such as liquid metals correspond to a Prandtl number of almost 0.01, whereas higher viscous fluids such as oils have Prandtl number of almost 10,000–100,000. Notably the Prandtl number of gases corresponds to 1, which effectively means that both momentum and heat dissipate through the fluid at about the same rate. Heat diffuses very quickly in liquid metals (Pr < 1) and very slowly in oils (Pr > 1) relative to momentum. Consequently, the thermal boundary layer is much thicker for liquid metals and much thinner for oils relative to the velocity boundary layer.

9.8.2.1 Free convection

The mode of convection in which the fluid motion occurs by natural means such as buoyancy is called natural or free convection. The phenomenon of natural convection involves lower ranges of velocities (