Applications of Markov Chains in Chemical Engineering

Applications o f Markov Chains in Chemical . . t n g i neeri ng I

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Applications o f Markov Chains in Chemical 0

Engineering

bY

Abraham Tamir Department of Chemical Engineering Ben-Curion University of the Negev Beer-Sheva, Israel

1998 ELSEVIER AMSTERDAM- LAUSANNE-NEW YORK-OXFORD-SHANNON -SINCAPORE-TOKYO

ELSEVIER SCIENCE B.V. Sara Burgerhartstraat 25 P.O. Box 211,1000 AE Amsterdam, The Netherlands

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Markov chains enable one to predict the fiture state of a system from its present state ignoring its past history. Surprisingly, despite the widespread use of Markov chains in many areas of science and technology such as: Polymers, Biology, Physics, Astronomy, Astrophysics, Chemistry, Operations Research, Economics, Communications, Computer Networks etc., their applications in Chemical Engineering has been relatively meager. A possible reason for this phenomenon might be that books containing material on this subject have been written in such a way that the simplicity of Markov chains has been shadowed by the tedious mathematical derivations. This caused one to abandon the book, thus loosing a potential tool for handling his problems. There are many advantages, detailed in Chapter 1, of using the discrete Markov-chain model in Chemicai Engineering. Probably, the most important advantage is that physical models can be presented in a unified description via state vector and a one-step transition probability manh. Consequently, a process is demonstrated solely by the probability of a system to occupy a state or not to occupy it. William Shakespeare profoundly stated this in the following way: to be (in a state) or not to be (in a state), that is the question". I believe that Markov chains have not yet acquired their appropriate status in the Chemical Engineering textbooks although the method has proven very effective and simple for solving complex processes. Thus, the major objective of writing this book has been to try to change this situation. The book has been written in an easy and understandable form where complex mathematical derivations are abandoned. The demonstration of the fundamentals of Markov chains in Chapter 2 has been done with examples from the bible, art and real life problems. The majority of the book contains an extremely wide collection of examples viz.,

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reactions, reactors, reactions and reactors as well as combined processes, including their solution and a graphical presentation of it. All this, to my opinion, demonstrates the usefulness of applying Markov chains in Chemical Engineering. Bearing all the above in mind, leads me also to suggest this book as a useful textbook for a new course entitled Applications of Markov chains in Chemical Engineering. Abraham Tamir Beer Sheva, Israel May 1, 1998

ACKNOWLEDGMENTS A few persons have contributed either directly or indirectly to this book; I would like to mention them by name. Professor Arie Dubi, a great teacher and scientist, deserves special thanks. He was the one who skillfully polished my knowledge in Markov chains to such a level which made it possible for me to write this book. Mr. Moshe Golden, a personal friend and a talented programmer, deserves many thanks. He assisted me in all technical problems which developed in producing the book in a camera ready copy form. Professor T.Z.Fahidy, of Waterloo University, was extremely influential in the creation of this book, in reviewing part of it; I deeply thank him. To Ms. Stella Zak, an extremely talented artist, many thanks for helping design the book cover. The most significant impact, however, has been that of my graduate students who participated in my course related to Markov chains. Their proclivity to ask 'why'? has forced me to rethink, recognize and rewrite many parts of the book again and again. In particular, many thanks are due to my student Adi Wolfson, who reviewed Chapter 2. Also thanks are due to Ben Gurion University, which provided generous assistance and a pleasant atmosphere in which to write this book. Finally, since I have no co-authors, I must accept responsibility for all errors in this book. Abraham Tamir

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CONTENTS

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Chapter 0 Biblical Origins and Artistic Demonstrations of Markov Chains-a Humorous Introduction ................................. 1 Chapter 1.Why Write this Book? ..................................................

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Chapter 2 Fundamentals of Markov Chains ................................. 11 2.1 Markov Chains Discrete in Time and Space.................... 11 2.1-1 The conditional probability ...................................... 11 19 2.1-2 What are Markov chains?........................................ 2.1-3 Construction elements of Markov chains ...................... 27 The state space................................................... 28 The one-step transition probability matrix...................-29 The initial and the n-step state vectors ........................ 32 The relationship between the one-step probability matrix and the state vector .............................................. 32 The discrete Chapman-Kolmogorov equation ...............34 2.1-4 Examples........................................................... 39 39 Biblical examples............................................... Artistic examples ................................................ 48 Real life examples............................................... 71 2.1-5 Classification of states and their behavior.................... 116 2.2 Markov Chains Discrete in Space and Continuous in Time .........................................................................132 132 2.2- 1 Introduction...................................................... 2.2-2 The Kolmogorov differential equation ..............133 2.2-3 Some discontinuous models .......................... 138 2.3 Markov Chains Continuous in Space and Time .............170 2.3-1 Introduction...................................................... 170 2.3-2 Principles of the modeling ..................................... 172 174 2.3-3 Some continuous models ...................................... 2.4 Concluding Remarks ................................................. 180 2.5 Artistic Ending of the Chapter .................................... 180

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Chapter 3 Applications of Markov Chains in Chemical Reactions. 186 3.1 Modeling the Probabilities in Chemical Reactions.........187 3.2 Application and Verification of the Modeling ............... 193 3.2-1 Non-linear reversible reactions with all aj = 1............... 193 3.2-2 Non-linear reversible and irreversible reactions with a j * 1............................................................. 198 3.2-3 Linear reactions ................................................. 201 3.2-4 Linear-non linear reactions with aj * 1....................... 204 3.3 Major Conclusions and General Guidelines for Applying the Modeling .............................................. 204 3.4 Application of Kinetic Models to Artistic Paintings ......204 3.5 Introduction to Modeling of Chemical Reactions .........210 3.6 Single Step Irreversible Reaction ................................ 213 3.7 Single Step Reversible Reactions ................................ 219 3.8 Consecutive Irreversible Reactions ............................. 228 3.9 Consecutive Reversible Reactions............................... 237 3.10 Parallel Reactions Single and Consecutive Irreversible Reaction Steps ...................................... 250 3.11 Parallel Reactions Single and Consecutive Reversible Reaction Steps ........................................ 287 3.12 Chain Reactions ...................................................... 301 3.13 Oscillating Reactions [55-691 ................................... 305 3.14 Non-Existing Reactions with a Beautiful

Progression Route ...................................................

323

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Chapter 4 Applications of Markov Chains in Chemical Reactors ..334 4.1 Modeling The Probabilities in Flow Systems ...............335

4.1-1. Probabilities in an interacting configuration ................336 4.1-2. 'Deadstate' (absorbing) element............................. 348 4.1-3. Plug flow element.............................................. 348

4.2 Application of the Modeling and General Guidelines., ...349 4.3 Perfectly Mixed Reactor Systems................................ 353 4.4 Plug Flow-Perfectly Mixed Reactor Systems................406 4.5 Impinging-Stream Systems ........................................ 462

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Chapter 5 Applications of Markov Chains in Chemical Processes 498 5.1 Modeling of the Probabilities .....................................498 5.2 Application of the Modeling and General Guidelines..... 521 5.2-1 Reacting systems ................................................ .522 5.2-2 Absorption systems............................................. 542 5.2-3 Combined processes............................................ 563 Nomenclature ..........................................................................

590

References ..............................................................................

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Chapter 0

BIBLICAL ORIGINS AND ARTISTIC DEMONSTRATIONS OF MARKOV CHAINS A HUMOROUS INTRODUCTION The origin of Markov chains, a probabilistic model for predicting the future given the present of a process and ignoring its past, goes back to biblical times, i.e. to the Book of Books. This we know thanks to what has been said in Exodus 28, verse 13-14, "Make gold rosettes and two CHAINS of pure gold worked into a form of ropes, and fix them on the rosettes". A thorough investigation of this verse led to the conclusion that the word CHAINS is an abbreviation of MARKOV CHAINS. Thus, it turns out that Markov chains is a very old subject and, as said in Ecclesiastes 1 verse 9, "... And there is nothing new under the sun". It is also surprising that available books [2-8, 15-18] related to the subject matter do not refer at all to biblical Markov processes. Such a process, for example, can be generated on the basis of Genesis 1 and is related to the order of the duys ofthe week in the Creation. According to verse 27, man was created on Friday. The Bible describes this event very nicely as follows: " And God created mun in HIS image, in the image of God he created him ... And there was evening and there was morning, the sixth day." Independent of the past history, i.e. Sunday to Friday, the probability that mcut will occupy a Saturday on the next day is 100%. In other words, since the present state is known, namely, Friday, and the probability of moving to the next state is also known, 100%, it is possible to predict Saturday as the future state of man with respect to the days of the week The above example, elaborated later in example 2.11, demonstrates for the first time the essence of Markov chains proposed by Markov only in 1906 [l],much later than Biblical times.

2 An additional example of a Markov process is related to the states day and night in Genesis 1 verse 4-5. The creation of these complicated states is described simply as: " God saw that light was good, and God separated the light from the darkness. God called the light day, and the darkness He called night. And there was evening and there was morning, a first day." The occurrence of the state night (or day) depends only on the previous state unless something unexpected happens in the universe. The last Biblical example of a Markov process is concerned with the famous trial of king Solomon [ l Kings 31. The story develops as follows (verse 16-22): "Then came there two women, that were harlots, unto the king, and stood before him. And the one women said: "Oh, my lord, I and this woman dwell in one house; and I was delivered of a child with her in the house. And it came to pass the third day after I was delivered, that this woman was delivered also; and we were together; there was no stranger with us in the house, save we two in the house. And this woman's child died in the night; because she overlay it. And she arose at midnight, and took my son from beside me, while thy handmaid slept, and laid her dead child in my bosom. And when I arose in the morning to give my child suck, behold, it was dead; but when I had looked well at it in the morning, behold, it was not my son, whom I did bear". And the other woman said: "Nay; but the living is my son, and the dead is thy son". And this said: "No; but the dead is thy son, and the living is my son. Thus they spoke before the king." King Solomon was faced with an extremely hard human problem of how to find out to whom does the living child belong? In order to resolve the problem, king Solomon made a wise decision described in verse 24-25 as: And the king said: "Fetch me a sword." And they brought a sword before the king. And the king said: "Divide the living child in two, and give half to the one, and half to the other." The above example generates the following Markov process. There are two states here, namely, that of a living child and that of a divided child. By his brave decision, king Solomon fixed the probability of moving from the first state to the final one to be 100%. Consequently, if his verdict would have been materialized, an ultimate state known in Markov chains as dead state would have been reached. Fortunately, according to verse 26-27 the woman the child belonged to said, for her heart yearned upon her son: ... "Oh, my lord, give her the living child, and in no wise

3 slay it" while the other woman said: ... "It shall be neither mine nor thine; divide it." Then the king answered and said: "Give her the living child, and in no wise slay it: she is the mother thereof." In this way, the terrible result predicted by the Markov process was avoided. The following example of afrog in a lily pond was mentioned by Howard in 1960 in the opening of his book [3, p.31 as a graphic example of a Markov process.

Fig.0-1. Escher-Howard Markov process (M.C.Escher "Frog" @ 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved) Surprisingly, M.C.Escher, the greatest graphic artist (1898-1972), probably unfamiliar with Markov processes, has already demonstrated in 1931 the same situation in his woodcut Frog [lo, p.2311. This is reproduced in Fig.0-1. As time goes by, the frog, system, jumps from one lily pad, state, to another according to

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his whim of the moment. The latter assures a true Markov process since the effect of the past history is assumed to be negligible. The state of the system is the number of the pad currently occupied by the frog; the state transition is, of course, his leap. If the number of lily pad is finite, then we have a finite-state process. In 1955,Escher prepared the lithograph Convex and Concave [lo, p.3081 which is reproduced in Fig.0-2. It is interpreted below as a Markov process and elaborated in Chapter 2 in example 2.14.

Fig.0-2. Demonstration of Escher's Markov process ("Convex and Concave", 0 1955 M.C.Escher Foundation 8 - Baarn, the Netherlands. All rights RSeNed)

It is interesting, first of all, to explore some interesting phenomena in the lithograph, which is a visual shock. The columns appearing in the picture can be seen as either concave or convex. On the right-hand side, the solid floor underfoot

5 can become the ceiling overhead, and that one may once climb the selfsame staircase safely, and after some time, while climbing, suddenly be falling down because the stairs seem upside down. Similarly is the situation with the woman with the basket walking down the stairs. The upper floor in the middle of the lithograph, with the flute player, may be seen as convex or concave. Thus, when it looks concave, the flute player, when climbing out of the window, stands safely on the vaulting. However, if the appearance looks convex, and the flute player does not pay attention, he might land far down when leaving the window. In addition, the element on the floor above the two lizards, may be observed as shellshaped ceiling or shell-shaped basin. All above behaviors are phenomena related to the cognition of vision by the brain.The reader can see in Fig.0-2 six locations, designated 1 to 6, selected as possible states that a person, system, can occupy. In principle, the possible occupations depend on the original location of the person, namely, the initial state vector, and on the probabilities of moving from one state to the other, i.e. the single-step probability matrix. The above concepts are elaborated in Chapter 2.1-3. As demonstrated later, a person trying to walk along Escher's Convex Concave structure, will end up walking up and down along the staircase connecting states 2 and 3. This result solely depends on Eq.(2-42) which may be looked upon as the policy-making matrix of the person. The matrix depends on his mood, but for the sake of simplicity it has been assumed to remain unchanged. Such a Markov process, i.e. ending walking endlessly between states 2 and 3, is known as periodic chain. However, more interesting is the fact that the final situation the person has been trapped in, is independent of the initial state. This is known as without memory or ergodic process. In conclusion, the aforementioned examples indicate that the origin of Markov chains goes back to very ancient days and many wonderful examples can be found in the Book of Books to demonstrate this process. In addition, some interesting relationships may also arise between the subject matter and art, which are demonstrated in 2.1-3 of Chapter 2. However, from Chapter 3 on, applications of Markov chains in Chemical Engineering are demonstrated.

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Chapter 1

WHY WRITE THIS BOOK? Markov chains or processes are named after the Russian mathematician A.A.Markov (1852-1922) who introduced the concept of chain dependence and did basic pioneering work on this class of processes El]. A Markov process is a mathematical probabilistic model that is very useful in the study of complex systems. The essence of the model is that if the initial state of a system is known, i.e. its present state, and the probabilities to move forward to other states are also given, then it is possible to predict the future state of the system ignoring its past history. In other words, past history is immaterial for predicting the future; this is the key-element in Markov chains. Distinction is made between Markov processes discrete in time and space, processes discrete in space and continuous in time and processes continuous in space and time. This book is mainly concerned with processes discrete in time and space. Surprisingly, despite the widespread use of Markov chains in many areas of science and technology such as: Polymers, Biology, Physics, Astronomy, Astrophysics, Chemistry, Operations Research, Economics, Communications, Computer Networks etc., their applications in Chemical Engineering has been relatively meager. A possible reason for this phenomenon might be that books containing material on this subject have been written in such a way that the simplicity of discrete Markov chains has been shadowed by the tedious mathematical derivations. This caused one to abandon the book, thus loosing a potential tool for handling his problems. In a humorous way, this situation might be demonstrated as follows. Suppose that a Chemical Engineer wishes to study Markov processes and has been suggested several books on this subject. Since the mathematics is rather complex

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or looks complicated, the probability of moving to the next book is decreasing and diminishes towards the last books because the Chemical Engineer remembers always the difficulties he has encountered in studying the previous books. In other words, his long-range memory of the past has a significant and accumulative effect on the probability of moving to the next book. However, had he known Markov chains, he should have made efforts to forget the past or to remember only the effect of the last book which might be better than the previous ones. In this way, his chances of becoming familiar with Markov chains would have been significantly increased. M.C.Escher demonstrated the above situation very accurately in his woodcut Still Life and Street [ 10, p.2711, which is reproduced in Fig.1-1 .

Fig.1-1. Abandoned books on Markov chains according to Escher (MCEscher "Still Life and Street" 0 1998 Cordon Art B.V. - B a r n - Holland. All rights reserved)

8 The reader can observe on the right- and left-hand sides of the desk a total of twelve books on Markov processes among which are, probably, refs.[2-8, 15-18, 841. Some support to the fact that the books are abandoned is the prominent fact that the immediate continuation of the desk is the street ... and that the books are leaning on the buildings. There are many advantages of using the discrete Markov-chain model in Chemical Engineering, as follows. a) Physical models can be presented in a unified description via state vector and a one-step transition probability matrix. Consequently, a process is demonstrated solely by the probability of a system to occupy a state or not to occupy it. William Shakespeare profoundly stated this in the following way: " to be (in a state) or not to be (in a state), that is the question". It is shown later that this presentation coincides with the finite difference equations of the process obtained from the differential equations. In some cases the process is also of probabilistic nature, such as a chemical reaction, where the Markov-chain model presentation seems natural. b) Markov-chain equations provide a solution to complicated problems. The increase in the complexity of the problem increases the size of the one-step transition probability matrix on the one hand, however, it barely increases the difficulty in solving it, on the other. c) In some cases, the governing equations of the process are non-linear differential equations for which an analytical solution is extremely difficult or impossible. In order to solve the equations, simplifications, e.g. linearization of expressions and assumptions must be carried out. For these situations the discrete Markov-chain model may be advantageous. d) The application of an exact solution is sometimes more complicated in comparison to Markov-chain finite difference equations. For example, an analytical solution with one unknown where the equation has no explicit solution, or an equation with two unknowns where there is no analytical expression of one unknown versus the other. Both cases may be encountered in problems with chemical reactions where the solutions involve iterative means. e) It is extremely easy to obtain all distributions of the state vector versus time from the Markov-chain solution, However, it is not always easy or convenient to

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obtain these distributions from the analytical solution. f ) Elements of consecutive state vectors yield the transient response of the system, undergoing some process, to a pulse input. Thus, RTD of the fluid elements or particles are obtained, which gives an insight into the mixing properties of a single-or multiple- reactor systems. Such a solution, given by Eq.(2-24), is the product of the state vector by the one-step transition probability matrix. g) One can model various processes in Chemical Engineering via combination of flows, recycle streams, plug-flow and perfectly-mixed reactors. The processes may be also associated with heat and mass transfer as well as chemical reactions. The author believes that Markov Chains have not yet acquired their appropriate status in the Chemical Engineering textbooks and applications although the method has proven very effective and simple for solving complex processes. Surprisingly, correspondence of the author with eminent Professors in Chemical Engineering revealed they hardly have heard about Markov chains. Thus, the major objective of the proposed book is to try to change this situation. Additional objectives are: a) Present a comprehensive collection of various applications of Markov chains in Chemical Engineering, viz., reactions, reactors, reactions and reactors as well as other processes. This is materialized as from Chapter 3. b) Provide the university Professor with a textbook for a possible course on "Applications of Markov chains in Chemical Engineering". Alternatively, the book can be used as reference book in other courses such as Reactor Design where examples presented in the present book may be very useful. c) Provide the practical engineer with numerous models and their solutions in terms of the state vector and the one-step transition probability matrix, which might be useful in his work. In addition, to convince the engineer about the simplicity of applying Markov chains in solving complicated problems. d) Stimulate application of Markov chains so as to become a common tool in Chemical Engineering. e) Last, but not least, to demonstrate the application of Markov chains in art and biblical problems. The organization of the book is as follows. The fundamentals of Markov chains will be presented in Chapter 2 in an easy and understandable form where complex

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mathematical derivations are abandoned and numerous examples are presented including their solution. The chapter contains processes discrete in time and space, processes discrete in space and continuous in time and processes continuous in space and time. In Chapter 3, modeling of chemical reactions is presented as well as demonstrations of their transient behavior. In Chapter 4,modeling of chemical reactors is presented and their dynamic behavior with respect to a pulse input. The latter are important parameters for describing the RTD behavior of a system where graphical presentations follow the modeling. Chapter 5 presents modeling of a few processes encountered in Chemical Engineering where effects of heat and mass transfer as well as chemical reaction are also accounted for. A general presentation of the model and its solution by Markov chains is also provided.

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Chapter 2

FUNDAMENTALS OF MARKOV CHAINS Markov chains have extensively been dealt with in refs.[2-8, 15-18, 841, mainly by mathematicians. Based on the material of these articles and books, a coherent and a short "distillate" is presented in the following. The detailed mathematics is avoided and numerous examples are presented, demonstrating the potential of the Markov-chain method. Distinction has been made with respect to processes discrete in time and space, processes discrete in space and continuous in time as well as processes continuous in space and time. Demonstration of the fundamentals has been performed also on the basis of examples generated from unusual sources, art and the Bible. Surprisingly, biblical stories and paintings can be nicely analyzed by applying Markov chains discrete in time and space. For each example, a solution was obtained by applying Eq.(2-23) and the EXCEL software. The solution is presented graphically, which demonstrates the dynamical behavior of the system in occupying the various states under consideration. Such an information was missing in the above refs.[2-8, 15-18, 841, The latter contained only the one-step transition matrix, termed also as policymaking matrix.

2.1 MARKOV CHAINS DISCRETE IN TIME AND SPACE 2.1-1 The conditional probability The conception of conditional probability plays a fundamental role in Markov chains and is presented firstly.

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In observing the outcomes of random experiments, one is often interested how the outcome of one event Sk is influenced by that of a previous event Sj. For example, in one extreme Case the relation between sk and Sj may be such that Sk always occurs if Sj does, while in the other extreme case, Sk never occurs if Sj does. The first extreme may be demonstrated by the amazing lithograph WuterjiuZZ by Escher [10, p.3231 depicted in Fig.2-0.

Fig.2-0. Conditional probability demonstrated by Escher's Waterfall (M.C.Escher "Waterfall" 0 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved)

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If we follow the various parts of the construction in the figure one by one, we are unable to discover any mistake in it. Yet the combination is impossible as one may reveal. The basis for this phenomeon is a particular triangle, "the impossible triangle", 3-4-5 in Fig.2-0 with sum of angles of 2700, first introduced by Oscar Reutesverd in 1934. Two sides of such a triangle can exist in reality, however, the overall element can never be constructed but can easily be perceived by the brain. Therein lies the ingenuity of the artist who can give free play to the imagination, relying on optical illusions to make the fanciful look real. As observed, Fig.2-0 is based on three such triangles, i.e., 1-4-5, 3-4-5 and 2-3-4. Despite this fact, the conception of conditional probability may be clearly demonstrated in the following way. Assume that S1, S 2 , ..., S5 are five events occurring along the trajectory of the moving water, i.e. that water pass through points 1, 2, ..., 5 in Fig.2-0 along their perpetual motion uphill. According to the above conception, the passing of water at point 2 is completely dependent on their previous passing at point 1. In other words, given that water pass at point 1, ensures also their passing at point 2 with 100% probability. Similar are the relationships between events S3 and S2 and the other consecutive events. Finally, it should be noted that Fig.2-0 has also been widely used in Chapter 2.1-5 to demonstrate the application of the equations developed there in classifying the various states of Markov chains. To characterize the relation between events Sk and Sj, the conditional probability of Sk occurring under the condition that Sj is known to have occurred, is introduced. This quantity is defined [5, p.251 by the Bayes' rule which reads:

For the Escher's example, Eq.(2-1) reads, prob{S2 I S l } = prob{S2Sl}/prob{S1}. prob[ Sk I Sj} is the probability of observing an event Sk under the condition that event Sj has already been observed or occurred; prob[Sj} is the probability of observing event Sj. SkSj is the intersection of events Sk and Sj, i.e., that both Sk and Sj may be observed but not simultaneously. prob{SkSj} is the probability for the intersection of events Sk and Sj or the probability of observing, not

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simultaneously, both Sk and Sj. Simultaneous observation of events or, alternatively, simultaneous occupation of states which is impossible according to the aforementioned, is nicely demonstrated in Fig.2-1, an oil on canvas painting [ l l , p.921, The Empire of Lights (1955) of Magritte, the greatest surrealist philosopher.

Fig.2-1, The coexistence of two states, Day and Night, according to Magritte ("The Empire of Lights", 1954,O R.Magritte, 1998 c/o Beeldrecht Amstelveen)

15 The painting shows a house at night surrounded by trees. The only bewildering element about this peacefully idyllic scene is the surprising fact that it has been placed under the light blue clouds of a daylight sky. This is Magritte's amazing skill in combining seemingly disparate elements by simultaneously showing two states that are mutually exclusive in time. Thus, in mathematical terms, Magritte's painting depicted in Fig.2-1 contradicts the fact that Day and Night can not coexist, i.e. prOb(SkSj} = prob(Day Night} = 0. Possible explanations for this contradiction by Magritte are the following ones. The first is that Magritte was not familiar with Markov chains or probabilistic rules. The second one is based on Genesis 1, which Magritte was, probably, familiar with. In verse 1-2 is said: "When God began to create the heaven and earth-the earth being unformed and void...". Thus, recalling the philosophical character of Magritte, it may be assumed that in The Empire of Lights Magritte has described the last second before the Creation, i.e. when "...the earth being unformed ..." and Day and Night could live in harmony together. Additional support is provided by Rashi, the greatest Bible commentator. In Genesis 1 verse 4 it has been said: "God saw that light was good...". According to Rashi: "God saw that it is good, and that it is not appropriate that darkness and light should be mixed; he fixed the light for the day and the darkness for the night". In other words before God's action, Day and Night were mixed and the prob{Day Night} = 1. Following the above example, Eq.(2-1) gives:

prob{Day/Nightl =

rob{Day Night} prob{ Night}

(2-la)

and the question presented by the equation is: what is the probability that Day will come, knowing that Night has already occurred? Although the answer is trivial, i.e. prob{Day/Night} = 1, it will be demonstrated on the basis of the above equation as follows. For a time interval of 24 hours prob{Day Night} = 1, i.e. it is certain that both events will occur within that time interval. In addition, prob(Night} = 1, thus it follows from Eq.(2-la) that, indeed, prob(Day/Night} = 1. On the other hand, because prob{Day Day} = 0 it follows that:

16 rob{DayDay} - 0prob{ Daymay} = p prob{Day} - 1 -

(2-lb)

Let us now return to Eq.(2-1) for further derivations. If the two events s k and Sj are independent of one another, then the probability of observing Sk and Sj not simultaneously is given by the product of the probability of observing Sk and the probability of observing Sj; that is:

It follows then from Eq.(2-1) that prob{Sk I Sj} = prob{Sk}. If prob{SkSj} = 0, Eq.(2-2) yields that prOb{Sk} = 0. In general, the events S1, S 2 , ..., S , are said to be mutually independent if: prob{ S1S2 ... S , } = prob{ Sl}prob{S 2 }

... prob{S , }

(2-2a)

The following examples demonstrate the above concepts. a) Assume a single coin is tossed one time where the two sides of it are designated as event Sk and event Sj. For a single toss prob{ Sk} = prob{Sj} = 112. However, prob{ SkSj} = 0, since in a single toss either event Sk or event Sj may occur. In other words the two events are dependent according to prob { s k } + prob( Sj} = 1. b) Assume now that the coin is tossed twice. Again prOb{Sk} = prob{Sj} = 1/2. However, the two tosses are independent of each other because in the second toss one can obtain either Sk or Sj. Therefore, according to Eq.(2-2) one obtains that prob{SkSj} = 1/4. If the number of tosses is three, prob{SkSkSk} = PrOb{sksjsk}= ... = (1/2)( 1/2)(1/2) = 1/8. c) Let Sk be the event that a card picked from a full deck (52 cards) at random is a spade (13 cards in a deck), and Sj the event that it is a queen (4 cards in a deck). Considering the above information, we may write that prOb{Sk} = 13/52; prob{Sj} = 4/52; prob{SkSj} = 1/52 where the last equality designates that there is only one card on which both a spade and queen are marked. It follows from the above probabilities that Eq.(2-2) is satisfied, namely, the events Sk and Sj are independent. In other words, if it is known that a spade is withdrawn from a deck, no information is obtained regarding to the next withdrawn.

17 In calculating the probability of observing an event Sk, the conditional probability is applied in the following way. Suppose S1, S 2 , S3, ... is afull set of mutually exclusive events independent of each other. By a full set is meant that it includes all possible events and that the events S1, S 2 , S3, ... may always be observed, but not simultaneously. If Sk is known to be dependent on S1, S 2 , S3, ..., then we can find prob{ Sk} by using the total probability fornula [ 5 , p.271 where Z is the total number of events; it reads:

j=l

Alternative conceptions to event and observe, more suitable in Chemical Engineering, are, respectively, state and occupy. Thus, the prOb{ Sk} designates the probability of occupying state Sk at step n+l. The prob{Sk I Sj} designates the probability of occupying state Sk at step n+l under the condition that state Sj has been occupied at step n. The prob{Sj} is the probability that state Sj has been occupied at step n where Z is the total number of states. The application of Eq.(2-3) has been demonstrated on the basis of the painting The Lost Jockey [12, p.181 of Magritte (1898-1967) in Fig.2-2 which was slightly modified. The surreal element in the picture has been achieved by the trees which appear like sketched leaves, or nerve tracts. As seen the rider, defined as system, leaves point 0 towards the trees designates as states. Eight such states are observed in the figure, S1, S 2 , ..., S7 and Sk. From each state the system can also occupy other states, i.e., riding to other directions from each tree. The question is what is the probability of the rider to arrive at Sk noting that, at first, he must pass through one of the trees-states S1, S 2 , ..., S7? The solution is as follows. The trees S1, S 2 , ..., S7 are equiprobable, since, by hypothesis, the rider initially makes a completely random choice of one of these when leaving 0. Therefore: prob{Sj} = 117, j = 1, ..., 7 Once having arrived at S1, the rider can proceed to Sk only by making the proper choice of one of the five equiprobable roads demonstrated by the five arrows in

18 Fig.5-2. Hence, the conditional probability of arriving at Sk starting from S1 is 1/5. The latter may be designated by:

Fig.2-2. Application of Eq.(2-3) to Magritte's "Lost Jockey" ("The Lost Jockey", 1948, 0 R.Magritte, 1998 c/o Beeldrecht Amstelveen) prOb{Sk I S l } = 1/5. Similarly, for the other states S1, ..., S7: prob{ Sk I s2} = 1/3, prob{Sk I s3}= 1/4, prob{Sk I s4}= 1/4, prob{Sk I s 5 } = 1/3, prob{sk I sg} = 1/4, prob{Sk I s7) = 113 Similarly, for the other states S1, ..., S7: prOb{Sk I s2} = 1/3, prob{Sk I s3}= 1/4, prob{s k 1 s4} = 114, prob{ s k I s5} = 1/3, prob{Sk I s g } = 1/4, prob{Sk 1 s7} = 1/3 Thus, it follows from Eq.(2-3) that the probability of arriving at point Sk is: prOb{Sk}= (1/7)(1/5 + 1/3 + 1/4 + 114 + 1/3 + 1/4 + 1/3) = 341190 = 27.9%

19

2.1-2 What are Markov chains? Introduction A Markov chain is a probabilistic model applying to systems that exhibit a special type of dependence, that is, where the state of the system on the n+l observation depends only on the state of the system on the nth observation. In other words, once this type of a system is in a given state, future changes in the system depend only on this state and not on the manner the system arrived at this particular state. This emphasizes the fact that the past history is immaterial and is completely ignored for predicting the future. The basic concepts of Markov chains are: system, the state space, i.e., the set of all possible states a system can occupy and the state transition, namely, the transfer of the system from one state to the other. Alternative synonyms are event as well as observation of an event. It should be emphasized that the concepts system and state are of a wide meaning and must be specified for each case under consideration. This will be elaborated in the numerous examples demonstrated in the following. A state must be real, something that can be occupied by the system. Fig.2-3, probably the most famous of Magritte's pictures, showing a painting in front of a window, can nicely demonstrate the above. The painting is representing exactly that portion of the landscape covered by the painting. Assume the tree to be the state. Thus, the tree in the picture is an unreal state, hiding the tree behind it outside the room, which is the real state. The latter can be occupied by a system, for example, the Lost Jockey in Fig.2-2. Another example of the above concepts is presented by a drunkard, the system, living in a small town with many bars, the state space. As time goes by, the system undergoes a transition from one state to another according to the mood of the system at the moment. The drunkard is also staying in the bar for some time to drink beer; in other words, the system is occupying the state for some time. If the system transitions are governed by some probabilistic parameters, then we have a stochastic process.

20

Fig.2-3. The real and unreal state of Magritte ("The human condition", 1933, 0 R.Magritte, 1998 c/o Beeldrecht Amstelveen)

Another example is that of particles suspended in a fluid, and moving under the rapid, successive, random impacts of neighboring particles. This physical phenomenon is known as Brownian motion, after the Botanist Robert Brown who first noticed it in 1927. For this case the particle is the system, the position of the particle at a given time is its state, the movement of the particle from one position to the other is its transition and its staying in a certain position is the occupying of the state; all states comprise the state space. The difference between the above

21 examples is that the first one may be considered as discrete with respect to the states (bars) and the second one is continuous with respect to the states (position of the particle in the fluid). To summarize we may say the following about the applications of Markov chains. Markov chains provide a solution for the dynamical behavior of a system in occupying various states it can occupy, i.e., the variation of the probability of the system versus time (number of steps) in occupying the different states. Thus, possible applications of Markov chains in Chemical Engineering, where the transient behavior is of interest, might be in the study of chemical reactions, RTD of reactors and complex processes employing reactors. Markov chains aim, mainly, at answering the following questions: 1) What is the unconditional probability that at step n the system is occupying some state where the first occupation of this state occurred at n = O? The answer is given by E ~ s(2-23)-(2-25). . 2) What is the probability of going from state j to state k in n steps? The answer is given by Eqs.(2-30)-(2-32). 3) Is there a steady state behavior for a Markov chain? 4)If a Markov chain terminates when it reaches a state k, defined later as absorbing state or dead state, then what is the expected mean time to reach k (hence, terminate the chain) given that the chain has started in some particular state j ?

A few more examples In order to examine the characteristics of Markov chains and the application of the basic conceptions system,state, occupation ofstate as well as to elaborate the idea of the irrelevance of the past history on predicting the future, on the one hand, and the relevance of the present, on the other, the following examples are considered. Example 2.1 is the following irreversible first order consecutive reaction

where a molecule is considered as system . The type Aj of a molecule is regarded as the state of the system where the reaction from state Ai to state Aj is the

22 transition between the states. A molecule is occupying state i if it is in state Ai. The major characteristic of the above reaction is that the transition to the next state depends solely on the state a molecule occupies, and on the transition probability of moving to the next one. How the system arrived at the occupied state, i.e. the past history, is immaterial. For example, state & is governed by the following equation dC4 = k3C3 - k4C4 dt

(2-5)

where the finite difference equation between steps n and n+l (time t and t+At) reads C4(n+l) = C3(n)[k3Atl+ C4(n)[l - k4Atl

(2-6)

The quantities [k3Atl and 11 - k4Atl may be looked upon, respectively, as the probabilities to transit from state A3 to state and the probability to remain in state A4. Eq.(2-6) indicates that the condition of state A4 at step n+l depends solely on the conditions of this state prevailing at step n where the past history of the reaction prior to step n is irrelevant. Example 2.2 is also a Markov chain. It deals with a pulse input of some dye introduced into a perfectly-mixed continuous flow reactor. Here the system is a fluid element containing some of the dye-pulse. The state of the system is the concentration of the dye-pulse in the reactor, which is a continuous function of time. The change of system's concentration with time is the state transition given by C(t')/Co = exp(-t'/t,)

(2-7)

C, is the initial concentration of the pulse in the reactor, C is the concentration at each instant t' and t, is the mean residence time of the fluid inside the reactor. It may be concluded that once the state of the system is known at some step, the prediction of the state at the next step is independent of the past history.

23 Example 2.3 where the outcome of each step is independent of the past history, is that of tossing repeatedly a fair coin designated as the system. The possible states the system can occupy are heads or tails. In this case, the information that the first three tosses were tails on observing a head on the fourth toss is irrelevant. The probability of the latter is always 1/2, independent of the past history. Moreover, the future is also independent on the present, and from this point of view the above chain of tosses is a non-Markov one. Example 2.4 demonstrates a non-Markov process where the past history must be taken into account for prediction of the future. We consider the state of Israel as the system which has undergone many wars during the last fifty years. This situation is demostrated schematically as follows: Independence + Sinai + Six day war 1947 war1956 war1967

+

+ Attrition + Yom kipur + Lebanon war 1968

war 1973

war 1982

The Gulf + Intifadah uprising war 1991 1987-9 1/92

It is assumed that the system may occupy the following three states: w a r , no-war and peace. In 1995 the system was in a state of war with Lebanon and in a state of no-war with Syria. It may be concluded that the prediction of the future state of the system, if possible at all, depends not only on the present state, but also on the outcome of preceding wars. In other words such a situation is not without memory to the past and is affected by it. Example 2.5 concerns the tossing of a die, the system, numbered 1 to 6, the states. The probability of obtaining a 6 upward at the 6th toss, conditioned that previous tosses differed from 6, is 1/6. This is because the probability of obtaining any number at any toss is 1/6, since the outcome of any toss is independent of the outcome of a previous toss. Similarly, the probability of obtaining a 6 at the 3rd toss, conditioned that previous tosses differed from 6, is again 1/6. Thus, the future is independent neither on the past and on the present, and the tossing chain is a non-Markov one.

24

Mathematical formulation The formulation for the discrete processes may be presented as follows. Let the possible states that a system can occupy be a finite or countably infinite number. The states are denoted by S1, S 2 , S3, S4, Sg, S6, ..., Sj, ... where S stands for state. The subscript i designates the number of the state and if we write Si =, it means that after the equality sign must come a short description about the meaning of the state. A discrete random variable X(t) is defined, which describes the states of the system with respect to time. The quantity t designates generally time where in a discrete process it designates the number of steps from time zero. t is finite or countably infinite. X(t) designates the fact that the system has occupied some state at step t. X(t) can be assigned any of the values corresponding to the states S1, S2, S3, S4, ... However, at a certain occupation (observation) of a state by the system, only one value can be assigned. When the following equality is applied, i.e.,

this indicates that state i was occupied by the system on step t, or that the random variable X(t) has been realized by acquiring the value Si. Thus, Si may be looked upon as the realization of the random variable. In example 2.5, the number of states is six where the observations are: S1 = 1, S2 = 2, S3 = 3, S4 = 4, S5 = 5, s6 = 6. The figures corresponding to the states are those appearing on the die. If the die was tossed 5 times, a possible outcome while considering the upward observation in each toss as a result, may be presented as : [X(1), XG), W3), W4), X(5)I =

PI,Si, S3, S2, S6l

Thus, for example, on the fourth toss state S2 was obtained, where 2 was observed upward. In general, after n observations or steps of the system one has a sample X(n>l. [X(1), In the above examples, the observations are obtained sequentially. An *.*¶

important question that can be asked is: Does our knowledge of the past history of the system affect our statements for the probability of the future events? For example, does knowledge of the outcome on the first k-1 observations affect our

25 statements of the probability of observing some particular stare, say Si, on the kth observation ? In example 2.4 above, the answer was yes, in examples 2.1 and 2.2 the answer was no, whereas examples 2.3 and 2.5 are non-Markov chains, because neither the past and the present are relevant for predicting the future. Therefore, the general question may be presented in the following form: What is the probability of occupying state Si by the system on step k, knowing the particular states that were occupied at each of the k-lsteps? This probability may be expressed as: prob{X(k) = Si 1 X(1) = s1, X(2) = S2, ..., X(k-1) =

(2-9)

where Sn,for n = 1, 2, ..., k-1, is a symbol for the state that was observed on the nth observation or for the state that was occupied on the nth step. As shown later, the answer to this question will be given by Eqs.(2-23)-(2-26). Eq.(2-9) contains the conception of conditional probability previously elaborated, designated as prOb[Sk 1 Sj] which reads îthe probability of observing Sk given that Sj was observedî. The Escher’s Waterfall in Fig.2-0 is a nice demonstration of Eq.(2-9), i.e., prob(X(5) = S5 I X(l) = S1, X(2) = S2, X(3) = S3, X(4) = S4} In other words the probability of the system, a water element, to occupy S5, i.e. point 5 , is conditioned of previous occupation by the system of S1 to S4. If the outcomes of the observations of the states in a system are independent of one another, it can be shown [6, p.151 that the conditional probability in Eq.(29) is equal to the unconditional probability prob{X(k) = Si}; that is: prob{X(k) = Si I X(l) = S1, ..., X(k-1) = Sk-l} = prob{X(k) = Si}

(2-10)

Example 2.6 where observations are independent of one another, is that of tossing repeatedly a fair coin, the system. The possible states are S1 = head and S2 = tail. The probability of observing a head on the fourth toss of the coin, given the information that the first three tosses were tails, is still simply the probability of observing a head on the fourth toss, that is, 112. Thus, Eq.(2-10) becomes for this example:

26 prob{X(4) = S1 I X(l) = S2, X(2) = S2, X(3) = S2} = probIX(4) = S l } = 1/2 In general, however, physical systems show dependence, and the state that occurs on the kth observation is conditioned by the particular states through which the system has passed before reaching the kth state. For a probabilistic system this fact may be stated mathematically by saying that the probability of being in a particular state on the kth observation does depend on some or all of the k- 1 states which were observed. This has been demonstrated before in example 2.4. A Markov chain is a probabilistic model that applies to processes that exhibit a special type of dependence, that is, where the state of the system on the kth observation depends only on the state of the system on the (k-l)st observation. In other words, the processes are conditionally independent of their past provided that their present values are known. Thus, for a Markov chain: prob{X(k) = Si I X(1) = S1, ..., X(k-1) = Sk-I} = prob{X(k) = Si I X(k-1) = Sk-l}

(2-1 1)

or alternatively: prob{X(k+n) = Si I X(1) = S1, ..., X(k) = Sk} = prob{X(k+n) = Si I X(k) = Sk}

(2-1 la)

that is, the state of the system on the (k+n)th observation depends only on the state of the system on the kth observation. We, therefore, have a sequence of discrete random variables X( l), X(2), ... having the property that given the value of X(k) for any time instant k, then for any later time instant k+n the probability distribution of X(k+n) is completely determined and the values of X(k-1), X(k-2), ... at times earlier than k are irrelevant to its determination. In other words, if the present state of the system is known, we can determine the probability of any future state without reference to the past, or on the manner in which the system arrived at this particular present state. The theory of Markov chains is most highly developed for homogeneous chains and we shall mostly be concerned with these. A Markov chain is said to be time-homogeneous or to posses a stationary transition mechanism when the probability in Eq.(2-1la) depends on the time interval n but not on the time k.

27

Example 2.7 is concerned with the application of the above formulation for two jars, one red and one black. The red jar contains 10 red balls and 10 black balls. The black jar contains 3 red balls and 9 black balls. A ball is considered as system and there are two states viz. S1= red ball and S 2 = black ball. The process begins with the red jar where a ball is drawn, its color is noted, and it is then replaced. If the ball drawn was red, the second drawn is from the red jar; if the ball drawn was black, the second draw is from the black jar. This process is repeated with the jar chosen for a draw determined by the color of the ball on the previous draw. It is also assumed that when drawing from an urn, each ball in that urn has the same probability of being drawn. The probability that on the fifth drawing one obtains a red ball, given that the outcomes of the previous drawings were (black, black, red, black) = (S2, S2, S1,S,), is simply the probability of a red ball on the fifth draw, given that the fourth draw produced a black ball and that the first draw was from the red jar. That is: prob(X(5) = S1 I X(1) = S2, X(2) = S2, X(3) = S1, X(4) = S2} = prob(X(5) = S1 I X(4) = S2 } = 3/12 = 114 Note that: prob(X(5) = S1 I X(l) = S1, X(2) = S1,X(3) = S1, X(4) = S2} = prob(X(5) = S1 I X(4) = S2} = 3/12 = 114 while: prob(X(5) = S1 I X(l) = S1, X(2) = S2,X(3) = S1, X(4) = S l } = Prob(X(5) = S1 I X(4) = S l } = 10/20 = 1/2

2.1-3 Construction elements of Markov chains The basic elements of Markov-chain theory are: the state space, the one-step transition probability matrix or the policy-mukingmatrix and the initial state vector termed also the initial probabilityfunction In order to develop in the following a portion of the theory of Markov chains, some definitions are made and basic probability concepts are mentioned.

28

The state space Definition. The state space SS of a Markov chain is the set of all states a system can occupy. It is designated by:

In Si, S designates state where the subscript stands for the number of the state. The states are exclusive of one another, that is, no two states can occur or be occupied simultaneously. This point is clearly elaborated in example 2.9 in the following and its opposite in Fig.2-1 with the associated explanations. Markov chains are applicable only to systems where the number of states Z is finite or countably infinite. In the latter case, an infinite number of states can be arranged in a simple sequence S1, S 2 , S3, .... For the preceding example 2.1, the state space is SS = [Sl, S2, S3, S4, S51 = [Ai, A2, A3, &, As]. For example 2.3, SS = [Sl, S21 = [heads, tails]. For example 2.5, SS = [Sl, S2, S3, S4, S5, s6] = [l, 2, 3, 4, 5, 61 and for example 2.7 in the following, SS = [Sl, S2] = [red ball, black ball]. Some properties of the state space are [6, p.181: 1 2 prob{Si} 2 0

i = 1, 2, ..., Z

(2-12a)

where prob[ Si) reads the probability of occupying state Si. An alternative expression for Eq.(2-12a) for all Sj in the state space, in terms of the conditional probability defined in Eq.(2-l), reads: 1 2 prob{Sj I Si) 2 0

(2-12b)

Note that Si must be occupied before occupying each of the others. It should be noted that Eqs.(2-12a) and (2-12b) satisfy:

probi

Si\ = 1 or prob[

Iall i in ss 1

I all j in ss

An additional property of the state space is:

Sj I Si\ = 1

J

(2- 12c)

29

prob{

z i=l

Sit = J

prob[ Sil or prob{ i=l

2 Sj I S i } = 2 prob{Sj I Si> (2-124

Ij=1

1

j=1

The summation on the left-hand side designates a state (or event) comprised of Z fundamental states. The prob{summation} means the probability of occupying at least one of the states [Sl, S2,S3, ..., Szl. Definition. A Markov chain is said to be a finite Markov chain if the state space is finite.

The one-step transition probability matrix Definition. The one-step transition probability function pjk for a Markov chain is a function that gives the probability of going from state j to state k in one step (one time interval) for each j and k. It will be denoted by: Pjk = prOb{s k 1 Sj } = prob{k I j }

for all j and k

(2-13)

Note that the concept of conditional probability is imbedded in the definition of pjk. Considering Eq.(2-1la), we may write also: Pjk = prob{X(m+n) = Si I X(m) = Sj}

(2- 13a)

pjk is time-homogeneousor stationary transition probability function if it satisfies: pjk = function(time interval between j and k)

(2-14)

Considering Eq.(2- 13a), Eq.(2- 14) is expressed by: pjk = function(n) pjk f function(m) as well as pjk # function(m+n)

(2-14a)

Thus, for a time-homogeneouschain, the probability of a transition in a unit time or in a single step from one given state to another, depends only on the two states and not on the time. In general, the one-step transition probability function is given by:

30 Pjk(n, n+l) = prob{X(n+l) = Sk I X(n) = Sj}

(2-15)

which gives the probability of occupying state k at time n+l given that the system occupied state j at time n. This function is time-dependent, while the function given by Eq.(2-13) is independent of time, or homogeneous in time. Since the system must move to some state from any state j, Eq.(2-18) below is satisfied for all j The one-step transition probabilities can be arranged in a matrix form as follows: P11 P12 P21 P22

.*.

PlZ p2z (2-16)

p = (Pjk) =

PZl Pz2

.-. p z z

where pjk denotes the probability of a transition from state j (row suffix) to state k (column suffix) in one step. The matrix is time-homogeneous or stationary if the pjk's satisfy Eq.(2-14a). A finite Markov chain is one whose state space consists of a finite number of states, i.e., the matrix P will be a ZxZ square matrix. In general the state space may be finite or countably infinite. If the state space is countably infinite then the matrix P has an infinite number of columns and rows. Definition. The square matrix P is a stochastic matrix if it satisfies the following conditions:

1) Its elements obay:

otherwise the transition matrix loses meaning. 2) For every row j: Z

Pjk = k= 1

(2-18)

31

where Z is the number of states which can be finite or countable infinite. However, z

one may notice that Zpjk * 1, a fact that apparently violates the standard theory of b l

Markov chains, and encountered, for example, in non-linear chemical reactions. Other characteristicsof the square matrix P are: 3) The elements pii on the diagonal designate probabilities of remaining in same state j. 4) The elements pjk above the diagonal designate probabilities of entering state k by the system, from statej. 5) The elements Pjk under the diagonal designate probabilities of leaving state j by the system, to state k. 6) The sum of the products of the elements Pjk (over a certain column k) by the elements of the state vector (defined below) has the significance of the conditional probability defined by Eq.(2-3) and is also identical with Eq.(2-23). The latter equation is of utmost significance, giving the probability of occupying a certain state at step (n+l) knowing that this state was influenced by other states at step n. The transition matrix P is, thus, a complete description of the Markov process. Any homogeneous Markov chain has a stochastic matrix of transition probabilities and any stochastic matrix defines a homogeneous Markov chain. In the non-homogeneous case the transition probability: prob(X(r) = k I X(n) = j} (r > n) will depend on both n and r. In this case we write: pjk(n,r) = prob{X(r) = k I X(n) = j}

(2-19)

In particular the one-step transition probabilities pjk(n,n+l) will depend on the time n and we will have a sequence of the following stochastic matrices corresponding to Eq.(2-16) for n = 0, 1, 2, 3, ...:

32

(2-20)

I : The initial and the n-step state vectors Definition. The initial state vector, termed also as the initial probability function, is a function that gives the probability that the system is initially (at time zero or n = 0) in state i, for each i. The initial state vector will be denoted by: (2-21) si(0) = prob(X(0) = Si} i = 1, 2, ..., Z designating the probability of the system to occupy state i at time zero. The above quantities may be arrayed in a row vector form of the initial state vector, i.e.:

designating the initial occupation probability of the states [Sl, S2, S3, ..., SZ] by the system. Similarly:

is the state vector of the system at time n (step n). si(n) is the occupation probability by the system of state i at time n, where: si(n) = prob{X(n) = Si} i = 1, 2, ..., Z

(2-21a)

The relationship between the one-step probability matrix and the state vector This relationship takes advantage of the definition of the product of a row vector by a matrix [7, p.191. The product of S(n) defined in Eq.(2-22a), by the square matrix P defined in Eq.(2-16), yields the new row vector S(n+l). The sk(n+l) component of this vector, i.e., the (unconditional)probability of occupying s k at the (n + 1) step, reads:

33

(2-23)

It should be noted [15, p.3841 that intuitively it is felt that the influence of the initial state Sj(0) should gradually wear off so that for large n the distribution in Eq.(2-23) should be nearly independent of the initial state vector S(O), i.e., the state occupation is without memory to its initial history. Let us now consider the significance of Eq.(2-23) with respect to Eq.(2-3) given below, i.e.: Z U

j= 1

where prob{ Sj} is the probability that state Sj has been occupied at step n. Noting Eq.(2-21a), it may be concluded that prob{Sj} is identical to sj(n) in Eq.(2-23), where sj(n) is the occupation probability by the system of state j at time n. prob{ Sk I Sj} is the probability of occupying state Sk at step n+l under the condition that state Sj has been occupied at step n. Thus, it is identical to Pjk in Eq.(2-23) which designates the one-step transition probability from state j to k. Finally, prob{ Sk} in Eq.(2-3) designates the probability of occupying state s k at step n+l, which is identical to sk(n+l). Eq.(2-23), which is a recurrence relation, may be expressed in matrix notation as: S(n+l) = S(n)P

(2-24)

and on iteration we obtain: S(n+l) = S(O)Pn+l

(n = 0, 1,2, ... )

(2-25)

where P and S(0) are given by Eqs.(2-16) and (2-22). Alternatively, if S(n) denotes a column vector, then: S(n+l) = Pn+lS(o) since the choise of S(n) as a row vector is arbitrary.

(2-25a)

34 Eqs.(2-23)-(2-25)are the fundamental expressions of Markov chains because of the following reasons: a) The equations give an answer to question 1 (in the introduction of section 2.1-2), i.e., what is the unconditional probability that at time n+l (n+l steps after thefirst occupation) the system occupies state k? As can be seen, a Markov chain is completely described when the state space, initial state vector and the one-step transition probability matrix are given. For a physical system that is to be represented by a Markov chain, this means that first, the set of possible states of the system, SS, must be determined or defined. Second, the initial (at time zero) probabilities of occupying each of these states, Si(O), must be calculated or estimated. Finally, the probability of going from state j to state k in one time interval (one step), pjk, must be determined or estimated for all possiblej and k . Thus, the probabilities of future state of a system, namely, at step n+l, can be predicted from its present state at step n, and the transition probabilities in one step; the past has no influence at all in the predictions. b) On tha basis of Chapters 3 and 4,it may be concluded that Eqs.(2-23)-(225), or the state vector S(n) given by Eq.(2-22a) and the matrix P yielding the above equations, is an elegant way of writing the Euler integration algorithm for the differential equations which describe the mechanism of the process.

The discrete Chapman-Kolmogorov equation In deriving this equation, the following question is considered:What is the probability of transition of a systemfrom state Sj to state Sk in exactly n steps i.e.: pjk(n) = prob{X(n+t) = Sk I x(t) = Sj}

(2-26)

In other words, pjk(n), the n-step transition probability function, is the conditional probability of occupying Sk at the nth step, given that the system initially occupied Sj. pjk(n), termed also higher transition probability, extends the one-step transition probability pjk( 1) = pjk and gives an answer to question 2 in 2.1-2. Note also that the function given by Eq.(2-26) is independent of t, since we are concerned in homogeneous transition probabilities. In answering the above question, we refer again to The Lost Jockey depicted in Fig.2-2 defined as system. We designate now point 0, the initial state of the

35 system, by Sj. The other states are the trees S1 to S, where the final state is s k . It may be seen that the system can move from state Sj to Sk by a number of different paths. For example, if a system has Z possible states, then in two steps it may go from Sj to s k by:

(2-27)

where Z = 7 in Fig.2-2, excluding Sj and Sk. In order to compute the probability of the transition Sj + Si + Sk,assuming the states are independent of one another, one applies the concept of independence given by Eqs.(2-2) and (2-2a). Thus, for a Markov chain, the transition Sj + Si in one step is independent of the transition Si + sk, yielding:

prob{Sj + si and si + Sk} = prob{Sj + Si}prob{Si 4 sk} = PjiPik (2-28)

Noting that prob(Sj

+ Si + S},

= prob{Sj + Si and Si

+ sk},

we may now

have expressions for computing the probabilities of the transitions to the states listed in Eq.(2-27). Since the transitions to and from the Z states in Eq.(2-27) are mutually exclusive (that is, no pair of them can be occupied simultaneously),the probability of the transition from state j to state k in two steps, i.e. pjk(2), is equal to the sum of the probabilities over the Z different paths; it is given by: Pjk(2) = PjlPlk + Pj2P2k +

i-PjzPzk

(2-29)

It should be noted that the above result follows also from the concept of conditional probability given by Eq.(2-3). Assume now that the Jockey becomes tired and wants to rest along the paths i to k. Thus, the latter trajectories are performed in the two steps Si + Si + sk, i.e., he is resting at state Si for one time interval and then riding towards the state

36 Sk. The corresponding probability for this step is pik(2), i = 1, 2, ..., Z, where the total probability for Si to Sk, i.e., pik(3), is given recursively by:

The general case is where the Jockey is moving from Sj to Si in n steps and from Si to state Sk in m steps. Based on the above considerations, one can show that

(2-30) i=l

which is the discrete Chapman-Kolmogorov equation. In order to have Eq.(2-30) true for all n 2 0 we define pjk(0) by Pjj(O) = 1 and pjk(0) = 0 for j # k as is natural [15, p.3831. pji(n) and Pik(m) are the n and m-step transition probabilities, respectively. The latter quantities are arrayed in matrix form denoted by P(n) in the same way as Pjk form the matrix P in Eq.(2-16), i.e.:

(2-31)

The calculation of the components p,k(n) is as follows. In general: P(n) = Pn

(2-31a)

where Pn is the one-step transition probability matrix multiplied by itself n times. In a matrix form notation, we may write Eq.(2-30) as:

37 P(n + m) = P(n)P(m) = P(m)P(n)

(2-32)

where also: P(n

+ 1) = P(n)P = PP(n)

(2-32a)

The above equations require the multiplication of a matrix by a matrix yielding a new matrix. According to [7, p.191, we define the product: (2-33) to have the components: Z

j , k = l , 2 ,..., Z

qjk = x a j r b r k

(2-33a)

r=l

where A and B are both a ZxZ square matrices given by:

all

a12

a21

a22

* * a

a1z

311

b12

a2z

321

b22 * * . b2Z

B = (bjk) =

A = (ajk) =

2zl

a=

...

.

..* b1z

.

(2-34)

azz

Eq.(2-33a) states that to obtain the component qjk of Q in Eq.(2-33), we have to multiply the elements of the jth row of A by the corresponding components of the kth column of B and add all products. This operation is called row-into-column multiplication of the matrices A and B. S , corresponding to the following As an example, we consider the case S 2x2 matrix:

38

S1

P12

P11

P=

(2-35) s2

P21

P22

From Eq.(2-33), noting that A = B = P where Z = 2, follows from P2 that:

P12(2) = PllP12 + P12P22 for the paths

s1 + s1 + s2

s, + s2 + s2

where SI+S1 and S2+S2 indicate "resting" steps at states S1 and S2. P21(2) = P21P11 + P22P21 for the paths s2

+ s1 + s,

s2

+ s2 + s1

P22(2) = P21P12 + P22P22 for the paths s2 +

s, + s2

s2

+ s2 + s2

Similarly, from P3 = PP2 it is obtained that: P11(3) =P11P11P11 +PllP12P21+ PllP12P21+ P12P22P21 for the paths

39

2.1-4 Examples In the following examples, the application of Eqs.(2-23)-(2-25) is demonstrated to describe the dynamics of the occupation of the states by the system. The basic conceptions system and state are defined in each example selected from the Bible, art as well as real life problem.

Biblical examples Example 2.8 combines bible and art and refers to the oil on canvas painting Still Life with Open Bible (1885) by van Gogh [13, p.151, one of the greatest expressionists. It should also be noted that the Bible was a symbol of van Gogh's parents' home. To demonstrate Markov chains on this painting, the original painting was slightly modified by placing two identical books on the Bible located to the left-hand side of the candles as depicted in Fig.2-4. As a matter of fact, the two books are copies of the original one on the right-hand side to the Bible.

40

Fig.2-4. The modified 'Still Life with Open Bible' ("Still Life with Open Bible", 1885, V. van Gogh )

The following Markov chains is generated. It is clear that if one wants to study the Bible, the Bible has to rest on the top of the pile. For three books, designated as system, which are placed one on the top of the other, three states are possible according to the order of the books: S1 = Bible, book, book; S2 = book, Bible, book; S3 = book, book, Bible where the book on the left-hand side is placed on the top of the pile. A one-step transition from one state to the other is conducted by taking a book from the bottom of the pile and placing it on the top of it. Thus, the three states may be expressed by a 3x3 one-step transition matrix given by:

s

1 0

p = s 2

s

1

3

0 0

1

0

0 1 0

(2-36)

41 The matrix was applied for the following cases: 1) Assume the system is initially at state S 1. In this case p11 = 1, namely, the state is a trapping or a dead state which is impossible to escape from. Thus, one can study safely the Bible because the Book of books is always on the top of the other books. 2) The system is initially at state S2, i.e., ~ ( 0=) 1 where si(0) = s3(O) = 0. The question is how many steps are needed to move to S1 where the number of states Z = 3? The initial state vector reads:

S(0) = [ 0, 1, 0 1 Applying Eqs.(2-23), (2-24), i.e., multiplying the state vector by the matrix (236), yields: sl(n>, s2(n), S(1)= [O, 0, S(2) = 1, 0, S(3) = [ 1, 0, S(4)= [ 1, 0,

s3(n)

0

1 1 1

0

1

1

0

The above calculations indicate that at the second step, as expected, the system will be at S1. Once it reaches this state, it will remain there forever. 3) If the system is initially at S 3 , the initial state vector reads:

S(0) = [O, 0, Similar calculations yield:

1

1

slm9 s2(n), s3(n) [ 1,

0,

0

1

S(2) = 1, S(3) = [ 1,

0, 0,

0

1

0

1

i.e. a steady state at S1 will be reached after one step. It should be noted that S1 is always reached independent of S(O), namely, this state is without memory to the initial state.

42 Example 2.9 is related to the creation of the two states already mentioned in Chapter 0, vis., S1 = Day and S2 = Night in Genesis 1 verse 4-5. The system may be a man or a population who may occupy the above states. The two states are expressed by the following 2x2 matrix corresponding to 24 hours:

s1

0

1

P=

(2-37) s2

1

0

The significance of the above probabilities is the following. The probability of remaining in S1 is zero because after day comes always night, i.e., p11 = 0. The transition probability from S1 to S2 is, of course, p12 = 1. Similarly, p21 = 1, i.e., the transition probability from S2 to S1, as well as probability of remaining in S2, p22 = 0. It should be noted that above probabilities are independent of time. If the initial state vector reads:

application of Eqs.(2-23), (2-24) yields:

The behavior of the vector sl(n) indicates that if initially there was S1, the probability of remaining at this state after one step is sl( 1) = 0. After two steps (24 hours), s1(2) = 1, i.e., there will be again a state of Day, as expected. According to section 2.1-5 Table 2-2, the above chain is defined as a periodic chain.

43

Example 2.10 is a Markov chain representation of king Solomon's famous trial [ 1 Kings 31 discussed in Chapter 0. The child was selected here as system. The two states are: S1 = Living child and S2 = Divided child. These states are presented by the following 2x2 matrix where the probability of the system to remain in state S1 is, unfortunately, p11 = 0. This is because king Solomon said: Divide the living child in two. The transition probability from S1 to S2 is, therefore, loo%, i.e., pi2 = 1. Similarly, p21 = 0, which is the transition probability from S2 to S1. Finally, the probability of remaining in state S2, p22 = 1 and the matrix reads:

s1

s1

s2

0

1

P=

(2-38) s2

0

1

It is interesting to note that the matrix is characterized by the so-called dead state, i.e., once the system reaches this state, it will remain there for ever. The application of Eq.(2-24), assuming that initially sl(0) = 1, i.e., the system is at S1 = Living child and that:

The behavior of the vector sl(n) indicates that if initially there was a Living child, the probability of remaining at this state after one step is, unfortunately, sl( 1) = 0. Thus, the Markovian dead state will be also a real description of the actual situation.

Example 2.11 considers the order of the days of the week in the Creation mentioned in Chapter 0. The states are defined as S1 = Sunday, S2 = Monday, ..., and S7 = Saturday. The probability matrix reads:

44 s2

s3

s4

s5

s6

s7

0

1

0

0

0

0

0

0 0 0 0 0 1

0 0 0 0 0 0

1 0 0 0 0 0

0 1 0

0 0 1

0 0 0

0 0 0

0 0 0

0 0 0

1 0 0

0 1 0

s1

(2-39)

The system may be a man or the universe, which may occupy the states at some time according to matrix (2-39). Assuming that:

S ( O ) = [ 1, 0, 0, 0, 0, 0, 0 I i.e., that the system initially occupying S1, yields the following behavior: si(n) s2(n) s3(n) s4(n) sdn> sdn) s7(n) % I ) = [ 0, 1, 0, 0, 0, 0, 0 3 S @ ) = [ 0, 0, 1, 0, 0, 0, 0 I S(3)= [ 0, 0, 0, 1, 0, 0, 0 3 S(4)= [ 0, 0, 0, 0, 1, 0, 0 1 S ( 5 ) = [ 0, 0, 0, 0, 0, 1, 0 1 S(6) = 0, 0, 0, 0, 0, 0, 1 1 S(7)= 1, 0, 0, 0, 0, 0, 0 1

(2-39a)

The above behavior reveals that all states are periodic of a period of seven days.

Example 2.12 generates A Markov chain based on the biblical story about the Division of the Promised Land among the twelve tribes. In Book of Books [Joshua 13, verses 1 and 71 we read: "Now Joshua was old and advanced in years; and the Lord said to him: You are old and advanced in years, and there remains yet very much land to be possessed ... Now therefore divide this land..." In [25, p.521, a map depicted below, shows the results of this division, i.e. the boundaries of the inheritances of the tribes in the 12th century BC. corresponding to the above verses.

45

The Big Sea

s,=Asher s2=Naphtali

ACre

S l Simeon ~

A visitor, designated as system, wishes to visit the tribes. His transition between the states assumes that the probabilities of remaining in a state or moving to the other states is the same and that the following one-step transition matrix holds:

46 s 1 s 2 s3 s 4 s5 s6 s7 s8 s9 s10 s11 113 1/3 113 0 0 0 0 0 0 0 0 1/3 113 1/3 0 0 0 0 0 0 0 0 1/4 114 114 114 0 0 0 0 0 0 0 0 0 0 113 1/3 1/3 0 0 0 0 0 0 0 0 0 1/3 1/3 1/3 0 0 0 0 0 0 0 114 1/4 114 114 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0

0 0 0

0 0 0 0

0 0 0

0

0

0 1/5 0 0 0

0

0

0

0

1/4 1/4 0 1/5 0 115 1/6 1/6 1/6

0 0 0

0 1/4 1/4 114 0 0

s12 0 0 0 0

0

0 114 1/4 0 115 0 1/5 116 1/6 1/6 114 1/4 1/4 114 1/4 0 1 / 2 0 1/2

The application of Eq.(2-24) yields Fig.2-5(a,b) showing the probability distribution of visiting the twelve tribes, i.e., si(n), against the number of steps n. A step is equivalent to some time interval the visitor stays in a tribe after which he is leaving to the next one. Two cases were considered: a) The visitor begins at state S1 = Asher, where the initial state vector S(0) is given at the top of Fig.2-5(a). b) The visitor begins at state S12 = Simeon, where the initial state vector S(0)is given at the top of Fig.2-5(b).

47 1

0.8

0.6

-A

C

v)

0.4

0.2 0

0

5

10

15

20

15

20

n 1

0.8

c v A

a-

0.6

-

0.4

0.2

0

0

5

10 n

Fig.2-5(a, b). Probability of visiting the twelve tribes As observed, in both cases the visiting probability distribution of the tribes, which may be used to decide the policy of visiting of the tribes, reaches a steady state after several steps. In other words, he will start at the tribe of the highest probability and then move to the tribe of a lower probability, and so on. According

48 to the results, in case (a) he starts at S1 and moves according to the following order of states where he reaches the last tribe after 12 steps:

It should be noted that the results indicate that s2( 1) = s3( 1) = 0.333, s5(6) = Sg(6) = 0.044. Bearing this in mind this fact, the visitor has decided to move according to S2+S3 and S5 4 6 . In case (b), he starts at S12 and his transition was found to be according to the following trajectory:

where it has been observed that s9(2) = s11(2) = 0.125. However, the interesting result is that the visitor terminates his visits after 7 steps and will never reach states S1 to S5 because the values of the probabilities si(n) = 0, i = 1, ..., 5.

Artistic examples

Example 2.13 is a Markov-chain model for Magritte's painting 'The Castle in the Pyrenees' [14, p.1161 depicted in Fig.2-6. In 1958-1959 particularly, Magritte was obsessed by the volume and weight of enormous rocks, but he altered the laws of gravity and disregarded the weight of matter; for instance, he had a rock sink or rise. Similarly, in Fig.2-6, he visioned a castle on a rock floating above the sea. Considering the floating effect, two states may be visualized, i.e., S1 = the rock is floating at some height above sea level, as seen in Fig.2-6; S2 = the rock is floating at very small distance above sea level. The rock was chosen as system. Thus, one may assume the following one-step probability matrix given by Eq.(241). The reason for assuming p11 and p22 to be unity, i.e., the system remains in its state, is that since the rock is in a floating state, once it is "located" somewhere, it will remain there.

49

Fig.2-6. Magritte's gravityless world ("The Castle in the Pyrenees", 1959,O R.Magritte, 1998 c/o Beeldrecht Amstelveen)

50

(2-41)

If we assume the following matrix: S1

s2

1

0

s1

P=

(2-41a) 1

s2

0

where p21 = 1, this suggests that if the rock was placed very near the sea level at S 2 , it will move up to S1, a dead state, and remains there.

Example 2.14 demonstrates the situation depicted in Fig.0-2 by Escher as a Markov process in the following way. States Si ( i = 1, 2, ..., 6) are various locations which the system, a moving man, is occupying along its trajectory. The one-step transition of the system is according to the following matrix:

s

1

s 2 0

0

0

0

0 1

1

0

0

0

0

0

(2-42)

Some explanations are needed regarding to the underlying assumptions in the matrix. The probabilities pii on the diagonal are all zero, indicating that the system, never remain in these states. p45 has been assumed 1, namely, that climbing along the staircase from state 4 to 5 is possible. This is applicable if the reader covers with his palm the right half of the staircase. He then sees that the staircase is in the

51

upward direction. However, when he unveils the staircase, the latter seems to turn over. Therefore, it has been assumed that p55 = 0, i.e., it is impossible to remain at state 5 . Since gravitational forces are effective, the man will fall to state 2. Praying that he remains alive, it is assumed that p52 = 1. Assume the following initial state vector, namely, the starting position is at state S1:

(2-42a)

Inspection of the above state vectors behavior reveals that the system, at steady state from n = 4,will end up walking up and down along the staircase connecting states 2 and 3. The latter behavior was also independent of its initial state vector, i.e., this is an ergodic Markov chain which is without memory to the initial step.

Example 2.15, demonstrating the common situation of "dead state" (pii = 1) in Markov chains, is based on Escher's lithograph 'Reptiles' [ 10, p.2841 depicted in Fig.2-7. It demonstrates the life cycle of a little alligator. Amid all kinds of objects, a drawing book lies open. The drawing on view, is a mosaic of reptilian figures in three contrasting shades. Evidently one of them has tired of lying flat and rigid among his fellows, probably in a "dead state", so he puts one plastic-looking leg over the edge of the book, wrenches himself free and launches into real life. He climbs up the back of a book on zoology and works his laborious

52

way up a slippery slope of a set square to the highest point of his existence. At states 5 and 7 he might slip and fall on the book joining again the "dead state" situation. If this does not happen at this stage, after a quick snort, tired but fulfilled, he goes downhill again, via an ashtray, to the Ievel surface to that flat drawing paper, and meekly rejoins his previous friends, taking up once more his function as an element of surface division, i.e, the "dead state" situation.

Fig.2-7. Escher's reptiles demonstrating life cycle and "dead state" (M.C.Escher "Reptiles" 0 1998 Cordon Art B.V. - B a r n - Holland. All rights reserved) The above description can be framed by a Markov process in the following way. A reptile was defined as system and the following states shown in the figure were selected, i.e. Si = reptile at position i, i = 1, 2, ..., 11. On the basis of above states, the following matrix may be established. Some assumptions made were:

53 from S5, the reptile can move to states S6 and S7 with equal probabilities, i.e., 1/2. Similarly, from S7 it can move to S 8 and S9 with equal probabilities. Other transitions of the reptile are governed by the one-step probability matrix given by Eq.(2-43) which is the policy-making matrix of the reptile. SlO 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.5 0.5 0 0 0 1 0 0 0 0 0 0 0.5 0.5 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

s1 s2

s3 s4

s 5 s6 s7 s8 s9

0 0 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0

s1

0 0 0 0

0 0 0 0 0 1 1

(2-43)

To demonstrate the behavior of a reptile along his life cycle we assume that S(0) = 11,

0,

0,

0,

0,

0,

0,

0,

0,

0,

01

i.e., the system (reptile) in Fig.2-7 initially at state S 1, yields the following dynamical behavior: sdn) s 9 m s1o(n) sll(n) 0, 0, 0, 01 0, 0, 0, 01 0, 0, 0, 01 0, 0, 0, 01 0, 0, 0 01 0.25, 0.25, 0 , 01 0.25 , 01 0.25, 0, 0.25, 0, 0 , 0.251 9

54

s(9)= [o,

0,

0,

0,

0,

0.5,

0,

0.25, 0,

0 , 0.251

As seen, the system attains a steady state after seven steps where the reptile

has a probability of 50% to occupy state s6 and 25% probability to occupy Sg and S 11. Note, however, that in all these states the reptile is in a "dead state", pii = 1, as also demonstrated Escher's Fig.2-7. If the reptile is initially at Sg, i.e.

S(0) = LO, 0, 0, it is obtained that:

0,

0,

0,

0,

0,

si(n) s2(n) s3(n) s4(n) sdn) sdn) s7(n) s d n ) S(1) = [O, 0, 0, 0, 0, 0, 0, 0, S(2) = LO, 0, 0, 0, 0, 0, 0, 0, S(3) = LO, 0 , 0, 0, 0, 0, 0, 0,

1,

0,

01

sg(n) s d n ) sii(n) 1, 01 0, 0, 0, 11 0, 0, 11

i.e., the reptile is in a steady "dead state" at S11 already after two steps.

Example 2.16 models the movement of fish based on Escher's painting 'Fish [lo, p.3111 depicted in Fig.2-8. The selected states S1 to S12, are various locations of the fish as shown in Fig.2-8. The system is a fish. The underlying assumptions on the transition of the system are: a) The probability of remaining in some state pii = 0. b) The probability of occupying the state of an adjacent fish moving in counter current flow is also zero. c) A fish can not jump above another fish. d) There are equal probabilities to occupy two adjacent fish states. Bearing in mind the above assumptions yields the matrix below.

55

Fig.2-8. Movement of fish according to Escher (MCEscher "Fish" @ 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved)

s1

s2 s3 s 4

0

0

s5

s6 s7 s8 s9 SlO s11 s12

0.50.50 0 0 0 0 0 0 0 0 0.5 0 . 5 0 0 0 0 0 0 0 0 0.50 0 0 0 0 0 0 0.50 0.50 0 0 0 0 0 0 0 0 0.5 0 0 0 0 0 0.50.50 0 0 0 0 0.50 0 0.50 0 0 0 0 0 0 0 0 0 0 0 0 0.50.50 0 0 0 0 0 0 0.50.50 0 0 0 0 0 0 0 0 0 0 0 0 0.50 0

0

0

0

0

0

0

0.50.50

0

0

0

0

0

0

0

0

D.50

0

0

0

0

0

0

0 0

0

0

0

0 0 0 0 0 0 0.5

0

0.50 0.5 0 0.5 0

56 Results of the dynamical behavior of the system are presented in Fig.2-9 for the initial state vector S(0) = [ l , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 01, i.e., the fish is initially occupying state 1 in Fig.2-8. As seen, there are two groups of states attaining an identical steady state after ten steps where one group lags one step behind the other. In addition, the occupation probability distribution of the states at steady state, which equals 0.167, is periodic. One group lags by one step behind the other where the system has to decide at each step what state to occupy among six possibilities of an equal probability. It has also been observed that the steady state is independent of the initial state vectors S(0) which classifies this case as an ergodic Markov chain. l \

I

I

I

I

:Js,(n)

I

o.8

I I I

Examples 2.17-2.22 (and 2.41, 2.42) relate to what is normally called random walk [7, p.26; 4, p.891. In principle, we imagine a particle moving in a straight line in unit steps. Each step is one unit to the right with probability p or one unit to the left with probability q, where p + q = 1. The particle moves until it reaches one of the two extreme points called boundary points. The possibilities for its behavior at these points determine several different kinds of Markov chains

57

demonstrated in the following. An artistic demonstration of the different cases is based on Escher's painting Sun and Moon [lo, p.2951 depicted in Fig.2-10.

Fig.2-10. Random walk demonstrations according to Escher (M.C.Escher "Sun and Moon" 0 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved) We select nine states S1,S 2 , ..., Sg, which are various locations in the bird's state field that can be occupied by the system - the moving bird. The states are designated by 1, 2, ..., 9 in Fig.2-10. Although the movement is not in a straight line, the transition from one state to the other defined by the transition probability matrices below, ensures the random walk model. States S1 and Sg are the boundary states and S2, S3, ... , Sg the interior states.

58

Example 2.17 is a simple random walk between two absorbing barriers. It is characterized by the behavior of the moving bird, the system, that when it occupies states S1 and Sg, it remains there from that time on (p11 = pgg = 1). In this case the transition matrix is given by: s1

s2

s3

s4

s5

s6

sl

s8

s9

S1

1

0

0

0

0

0

0

0

0

s2

q o o

o q o

p o q

o o o

o o o

o o o

0 o

0 o

0 o

P

0 o

0 o

o 0 0

o 0 0

o 0 0

p o 0

o p 1

s3

P =

s4 s5

s6 s7 S8

S9 where p + q = 1.

o p o

o o p q

o o o

o

q

o

O p

o

o o 0

q o 0

o q 0

o 0

0

(2-45)

Fig.2-11 presents results for q = 0.5 and S(0) = [0, 0, 0, 0, 1, 0, 0, 0, 01, i.e., the bird is initially occupying S5 in Fig.2-10. It is observed that as time goes by, the probabilities of the bird to occupy states 1 and 9 is increasing as well as are identical. After 90 steps a steady state is achieved where S(90)= [0.5,0, 0, 0, 0, 0, 0, 0, 0, 0.51. In other words only the boundary states may be occupied with identical probabilities. The above behavior is evident recalling that S 1 and Sg are absorbing boundaries. It should be noted that the behavior of the other states is similar to S5 reported in the figure, i.e., their occupation probability diminishes versus time. For q = 0.2, the steady state state vector reads S(30) = [0.004, 0, 0, 0, 0, 0, 0, 0,0.996], i.e., the probability of occupying Sg is much higher than that of state 1.

59 1

0.8

0.6 0.4

0.2

0

0

30

20

10

n

40

Fig.2-11. Probability distribution of occupying various states for the two absorbing barriers S1 and S9 Example 2.18 is a simple random walk with reflecting barriers. Whenever the bird reaches states S1 and S9, it returns to the point from which it came, i.e. p11 = pgg = 0. The transition matrix reads

where p + q = 1.

s1

s2

s3

s4

0

1

0

0

q o o o o o o 0

o q o o o o o 0

p o q o o o o 0

o p o q o o o 0

SS

s6

s7

s8

s9

0

0

0

0

0

o o p o q o o 0

o o o p o q o 0

o o o o p o q 0

o o o o o p o 1

o o o o o o p 0

(2-46)

60

Fig.2-12presents results for q = 0.5 and S(0) = [0, 0, 0, 0, 1, 0, 0, 0, 01. The general trend observed is that after 20 steps, the probabilities of occupying the various states attain the following steady states corresponding to the groups designated clearly in Fig.2- 12: S(20) = [0.125,0,0.25,0,0.25,0,0.25,0,0.1251 S(21) = [0,0.25,0,0.25,0,0.25,0,0.25,01 S(22) = [0.125,0,0.25,0,0.25,0,0.25,0,0.1251 S(23) = [0,0.25,0,0.25,0,0.25,0,0.25,01

This behavior is plausible recalling that the boundaries are of reflecting barrier type. In other words, the moving bird will never be at rest. A final remark is that the limiting behavior is independent of S(O), characterizing an ergodic Markov chain. 1

I

1

1

0.8 0.6

sp(n) = sJn) = s,(n) = sJn)

A

I

C

Yv)

0.4

0.2 0

0

5

10

15

n

Fig.2-12. Probability distribution of occupying various states for the two reflecting barriers S1 and S9 Example 2.19 belongs also to the random walk model with reflecting barrier. However, it is assumed that whenever the moving bird, system, hits the

61 boundary states S1 or Sg, it goes directly to the central state S s . The corresponding transition matrix is given by Eq.(2-47) where p + q = 1. Fig.2-13 presents results for q = 0.5 and S(0) = [1,0,0,0,0,0,0,0,0],i.e., the system is initially at the S1 reflecting barrier shown in Fig.2-10. s1

s2

s3

s4

s5

s6

s7

s8

s9

0 q o o

0 o q o

0 p o q

0 o p o

1 o o p

0 o o o

0 o o o

0 o o o

0 o o o

o 0

o 0

o 0

q

o

p o

o p

o o

o o

o o 0

o o 0

o o 0

o o 0

q o 0

o q 0

p o 0

o p 0

o

q o o 1

(2-47)

As seen, the elements of the state vector oscillate towards a steady state distribution attained for n = 31, i.e., S(31) = [0.029, 0.059, 0.117, 0.177, 0.235, 0.177, 0.1 17, 0.059, 0.0291 where the maximum probability corresponds to Sg. As in previous cases, the steady state distribution is independent of S(0)and if the system has to occupy some state, it will be, probably, S 5 , of the highest probability.

62 1

I

I

5

10

i

i

i

i

15

20

25

30

0.8 0.6

0.4

0.2

0

0

n

Fig.2-13. Probability distribution of occupying the states for the two reflecting barriers S1 and S9 sending the bird directly to S5 Example 2.20 is a random walk with retaining barriers (partially reflecting). It has been assumed that the occupation probability by the system of the boundary state, or moving to the other boundary state is 0.5. Thus, the one-

P =

s5

s6

s7

s8

s9

0

0

0

0

0

0

0 0 0

0.5 0 0

S1

0.5 0

0

0

s s s s

9 0 0 0

0

P

0

9 0

0 q

P

0 0 0

0

p

o

o

o

o

0 0

0 0

q

o

p

o

o

o

0

9

O

P

O

O

0

0

0

0

0

0

0

0.5 0

0

0

0 0

9 0 0

0 P 9 0 0 0

2 3 4 5

s 6 0 s 7 0

S 8 0 sg

O P 0.5

(2-48)

63 where p + q = 1. Fig.2-14 presents results for q = 0.5 and S(0) = [0, 0, 0, 0,1, 0, 0, 0, 01, i.e., the system is initially occupying S 5 . As expected, the following steady state is obtained at n = 90, i.e., S(90) = [OSOO, 0, 0, 0, 0, 0, 0, 0, 0.5001. Under these conditions, the bird, has a 50% probability of occupying the boundary states S1 and Sg in Fig.2-10. It should also be noted that the limiting behavior for large n, is independent of the initial state, i.e. a situation without memory with respect to the far past.

'v 0.8

Ls5m a

0.6 h

C

.-

v

v)

0.4 0.2 0

0

5

10

15

20

25

30

35

40

n

Fig.2-14. Occupation probability distribution of some states for the partially reflecting barriers S1 and S9 Example 2.21 assumes that when the bird reaches one of the boundary states S1 or Sg, it moves directly to the other, like in aping-pong game (p1g = pgl = 1). Thus, the transition matrix reads:

64 s6

s7

s8

s9

0 0 o o o p o o o p o q o p 0 q o

0 o o o o p

0 o o o o o

1 o o o o o

o q

p o

o p

0

0

0

s1

s2

s3

s4

0 q o o

0 o q o

0 p o q

0

o 0 0

o 0 0 0 0

o 0 0

0

1

0 0

s5

o

o

0 0

0

q o

0

0

(2-49)

where p + q = 1. Fig.2-15 presents the variation against time of sl(n), s3(n) and sg(n) for q = 0.5 and S(0) = [0,0,1,0,0,0,0,0,0], i.e., the system is initially occupying S3 as shown in Fig.2-10.

0.8

0.6

0.4

0.2 0

0

5

10

15

20

n

25

30

35

40

Fig.2-15. 'Ping-pong' type probability distribution of the boundary states S1 and S9

65

As seen, s3(n) is oscillating and approaches zero at steady state. A similar behavior was observed also for the other si(n)'s excluding sl(n) and s9(n) which correspond to the boundary states S1 and S9. These oscillating quantities attain the following limiting behavior of the state vectors for S(0) = [0,0,1,0,0,0,0,0,0]: S(86) = [0.75, 0, 0, 0, 0, 0, 0, 0, 0.251 S(87) = [0.25, 0, 0, 0, 0, 0, 0, 0, 0.751

S(88) = [0.75, 0, 0, 0, 0, 0, 0, 0, 0.251 S(89) = [0.25, 0, 0, 0, 0, 0, 0, 0, 0.751 This behavior is plausible recalling the 'ping-pong' type behavior of the boundaries. It should be noted that the limiting values of S1 and S9 depend on

S(0). For example: a) S(0) = [l, 0, 0, O,O, 0, O,O, 01 yields

S(1) = [O, 0, 0, 0, 0, 0, 0, 0, 11 S(2) = [I, 0, 0, 0, 0, 0, 0, 0, 01

where for S(0) = [0, 0, 0, 0, 0, 0, 0, 0, 11 the values of S(2) replace these of b) S(0) = [0, 1, 0, 0, 0, 0, 0, 0, 01 yields S(79) = [0.875, 0, 0, 0, 0, 0, 0, 0, 0.1251 S(80)= [0.125, 0, 0, 0, 0, 0, 0, 0, 0.8751

where for S(0) = [0, 0, 0, 0, 0, 0, 0, 1, 01 the values of S(80) replace these of S(79). c) S(0) = [0, 0, 0, 1, 0, O,O, 0, 01 yields S(89) = [0.625, 0, 0, 0, 0, 0, 0, 0, 0.3751 S(90) = [0.375, 0, 0, 0, 0, 0, 0, 0, 0.6251

66 where for S(0)= [0, 0, 0, 0, 0, 1, 0, 0, 01 the values of S(90) replace these of S(89). d) S(0) = [0, 0, 0, 0, 1, 0, 0, 0, 01 yields S(90) = [OSOO, 0, 0, 0, 0, 0, 0, 0, 0.5001 S(91) = [OSOO, 0, 0, 0, 0, 0, 0, 0, 0.5001

Example 2.22 is a modified version of the random walk. If the system (bird) occupies one of the seven interior states S2 to S7, it has equal probability of moving to the right, moving to the left, or occupying its present state. This probability is 1/8. If the system occupies the boundaries S1 and Sg, it can not remain there, but has equal probability of moving to any of the other seven states. The one-step transition probability matrix, taking into account the above considerations, is given by: s1

s2

s3

s4

s5

0 1/8 118 1/3 1/3 1/3 0 1/3 1/3 0 0 1/3 0 0 0 0 0 0

1/8 1/8 0 0 1/3 0 1/3 113 1/3 1/3 1/3 0

0 0 0 0 1/8 1/8

0 0 0 0 1/8 1/8

0 0 1/8

s6

s7 s8

1/8 0 0 0 0 0 0 1/3 0 1/3 1/3 1/3 1/3 0 1/3 1/8

1/8 0 0 0 0 0 1/3 1/3

s9

1/8 0 0 0 0 0 0 1/3

(2-50)

1/8 1/8 1/8 0

Fig.2-16 presents results for S(0) = [O,O,O,O,l ,O,O,O,O]indicating that the probability distribution of approaching a steady state for n = 13 reads S(13) = t0.03, 0.078, 0.134, 0.168, 0.179, 0.168, 0.134, 0.078, 0.031. It should be noted that, generally, sl(n) = sg(n), s2(n) = ss(n), s3(n) = s7(n), s4(n) = sg(n), i.e., the curves are symmetrical. In addition, it was found that the limiting distribution is independent of S(O), i.e., the resulting Markov process generates an ergodic chain.

67

1

I

I

I

5

10

15

-

0.8

c

h

0.6

v

vi-

0.4

0.2

0

0

n

20

25

30

Fig.2-16. Probability distribution of occupying various states Example 2.23 is the last one on artistic examples. In Fig.2-1, Magritte has demonstrated an impossible situation on the coexistence of D a y and Night, two s t a t e s which can not coexist. It is also interesting to show how Escher demonstrated the above situation in his woodcut Day and Night [ 10, p.2731 and to model this behavior. The situation is depicted in Fig.2-17. One sees gray rectangular fields develop upwards into silhouettes of white and black birds. The black ones are flying towards the left and the white ones towards the right, in two opposing formations. To the left of the picture, the white birds flow together and merge to form a daylight sky and landscape. To the right, the black birds melt together into night. The day and night landscapes are mirror images of each other, united by means of the gray fields out of which, once again, the birds merge. The difference between the two demonstrations of day and night is therefore the following. In Magritte's picture, 'half' of the picture, i.e., clouds and sky, are at daylight. The other 'half', house surrounded by trees, are at night. In Escher's woodcut, however, the right 'half is at night, where the other one, on the left (mirror image of the right), is at day light.

68

Fig.2-17. The "coexistence" of Day and Night according to Escher (MCEscher "Day and Night" 0 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved) In Fig.2-17 sixteen states S1 to S16, i.e., sixteen possible locations of birds in the sky along the flying route, are shown; the states are designated by 1, 2, ..., 16. The system is a bird. The underlying assumptions for the system are: a) A bird occupying states S11 to S16 moves only to the right. b) A bird occupying S5 to S1 moves only to the left. c) A bird occupying S6 to Slo can move to the left and to the right. d) A bird occupying S1 and s16 remains there, Le., pi1 and p16,16 = 1. Other assumptions can be concluded from the 16x16 one-step transition probability matrix given by Eq.(2-51).

69

s1 S1 s2 s3

s4 s5

P

s6 = s7

s8 s9

SlO

s11 s12 s13

s14 s15 s16

s2 s3 s4 SS s6 s7 s8 s9 SlO s11 s12 s13 s14 s15

1 0 0 0 0 0 0 0 0 0 0 0 0 1 / 2 0 1 / 2 0 0 0 0 0 0 0 0 0 0 1/21/20 0 0 0 0 0 0 0 0 0 0 0 1/20 01/20 0 0 0 0 0 0 0 0 0 1/21/20 0 0 0 0 0 0 0 0 0 0 0 1 / 3 0 0 01/31/3 0 0 0 0 0 0 0 0 1 / 3 0 0 01/31/3 0 0 0 0 0 0 0 0 1 / 3 0 0 1 / 3 0 1/3 0 0 0 0 0 0 0 1/5 1/5 1/5 0 1/5 0 1/5 0 0 0 0 0 0 0 1 / 3 0 1 / 3 0 0 0 1/3 0 0 0 0 0 0 0 0 0 0 0 1 / 2 0 1 0 0 0 0 0 0 0 0 0 0 1 / 3 0 1/3 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 / 0 0 0 0 0 0 0 0 0 0 0 0 0

s1

0 0 0

0 0 0

0 0 0

0

0

0

0

0

0

0

0

0

0 0

0 0

0 0

0 0 0 0 / 2 0

0

12-51)

0 0 0 1/30 0 0 0 01/21fi K 2 O 0 0 1

Fig.2-18 presents results for S(0) = [0,0,0,0,0,1/3,1/3,0,1/3,0,0,0,0,0,0,0] indicating equal initial probabilities, 1/3, of occupying Sg, S7 and S9 (6, 7, 9 in Fig.2-17). Since this initial condition enables the bird to fly to the left or to the right, it is observed in Fig.2-18 that at steady state the bird has a probability of 47.2% to occupy Sl(1eft) and 52.8% to occupy S16(right). This behavior is explained by the fact that S1 and s16 are absorbing (dead) states, thus, the occupation probability of other states must diminish versus time.

70 0.6

I

1

1

0.5 0.4 n

5 vi-

0.3 0.2 0.1

0

0

5

10

15

20

n Fig.2-18. Occupation probability of states by the bird Table 2-1 shows limiting occupation probabilities (for n = -) for the case where the initial occupation of the bird is of a certain state, i.e., si(0) = 1, i = 1, 2, ...., 16. Table 2-1. Limiting occupation probabilities of states S1 and S16 for S i ( 0 ) = 1

i

sl(Oo) 1 1-5 0.542 6 0.542 7 0.292 8 213 9 0.292 10 0 11-16

S16(Oo)

0 0.458 0.458 0.708 113 0.708 1

71 Generally, the results comply with the assumptions spelled out above. For example, if the bird is initially at states S1 to S5, it will eventually occupy state S1. If it is initially at states S11 to s16, it will finally occupy state s16. If it is initially at states Sg to S l o , it will at the end occupy state S1 or s16 depending on the magnitude of the relative probability.

Real life examples Examples 2.24 and 2.25 were inspired by the assassination of the Prime Minister of the state of Israel Itzhak Rabin (1922-1995) on Saturday, November 4th, 1995. During his life, Rabin was a soldier, the Chief of Staff, participating in all Israeli wars and the greatest motivating force for peace in the Middle East. Thus, the examples deal with Life and Death as well as with Peace, War and No peace-No war situations.

Example 2.24 assumes two states a man, designated as system, can occupy, viz., S1 = Life, S2 = Death. The two states may be expressed by the following 2x2 matrix:

s1

q

1-q

0

1

P= s2

(2-5 1)

where q is a parameter depending, among others, upon the age of the man, his profession (soldier, university professor, worker), health, etc. For q = 0, the matrix is of the "dead-state" type, i.e., once the system occupies S2, it will remain there for ever. In other words, if the man is initially alive, after one step he will die. Fig.2-19 presents results for S(0) = [l, 01 and q = 0.5, i.e., the man is initially alive and his probability to occupy this state is 50%. His probability to die is also 50%. It is observed that after ten years the system will occupy state S 2 , namely, "dead state". It should be noted that sl(n) = 1/2n where S 2 is occupied only as n j - . Of course sl(lO>= 1/210 is nearly zero for d l practical purposes.

72 1

0.8

0.6

c

n W

0.4

vi-

0.2 0

2

0

4

n

6

8

10

Fig.2-19. Probability of remaining alive Example 2.25 presents a three-state model for which S1 = War-No war, S 2 = War and S3 = Peace. The system is at some state. The matrix for this case is the following one:

s1

p = S2 s3

9 r

P t

1-q-p 1-r-t

u

v

1-u-v

(2-52)

Eq.(2-52) is a multi-parameter matrix which depends on time and other factors not easy to evaluate. This is because we deal with complicated states. Fig.2-20 presents results for S(0) = [0, 1,0], i.e., the system (some country) is initially at a state of war. The following values were also assumed for the transition probabilities: q = p = 0.1, r = 0.5, t = 0, u = 0.2 and v = 0.1. It is observed that after five steps the system approaches a steady state for which S(5) = [0.206, 0.09 1,0.704]. The state vector indicates that the chances for peace are quite high, 70.4%, promising a bright future.

73

Fig.2-20. Dynamical probability of S1 = War-No war, S2 = War and S3 = Peace A very interesting behavior may be obtained by varying the initial state vector S(0). It is observed that the steady state behavior is independent of S(O), where such a Markov chain, later discussed, is defined as ergodic.

Example 2.26 is a simulation of a tennis game. The system is a tennis ball for which the following states are defined and schematically depicted in Fig.221a. S1 = the ball is down on the ground on the right-hand side, briefly designated, DR (down right); S 2 = the ball is up in the air on the right-hand side, UR (up right); S3 = the ball is up in the air on the left-hand side, UL (up left); S4 = the ball is down on the ground on the left-hand side, DL (down left).

74

S3=m=UpLeft

0 Right

Fig.2-21a. Scheme of the states in a tennis game The corresponding matrix reads: S1=DR 1

P2 1 P3 1 0

S2=UR S3=UL

S4=DL

0 0

0

O

P23

P24

P32

0

P34

0

0

1

I

(2-53)

where the one-step transition probabilities are: p21 = UR + DR, p23 = UR j UL, p24 = UR j DL, p31 = UL + DR, p32 = UL + UR, p34 = UL + DL. p11= p u = 1 means that once the ball is on the ground, it will remain there until the game starts again. p22 = p33 = 0 indicates that if the ball is in the air the game must go on. It should also be noted that the various probabilities in the matrix depend on the characteristics of the players which depend on time and their talent. In the following demonstrations the probabilities remain unchanged. Assume the following values: p21 = 0.01, p23 = 0.99, p24 = 0, p31 = 0, p32 = 0.8, p34 = 0.2

75

The above data indicate that the tennis player on the right-hand side of the tennis court is a better one. This is because p21 is significantly lower than p34, i.e., the probability to hit the ball to his court is lower. On the other hand, both players will always hit the ball up to the air to the other court, i.e., p31 = p24 = 0. Fig.2-21b presents results for the dynamics of the tennis game corresponding to S(0) = [0, 1, 0, 01 indicating that the ball is initially at the right-hand side court. It is observed that as time goes by, the probability for state S4 to be occupied is increasing, and the player on the right-hand side is going to win the game. The steady state value of the state vector reads S(68) = [0.048,0,0, 0.9521, indicating the above trend. 1

0.8 0.6

c

h

v

vi-

0.4 0.2

0

0

5

10

15

20

25

30

35

40

n Fig.2-21b. Dynamics of a tennis game (better player on the right-hand side) Assume now the following values for the one-step transition probabilities: p21

= 0.01, p23 = 0.99, p24 = 0, p31 = 0.12, p32 = 0.8, p34 = 0.08 Considering the values of pzl, p23 and p24, indicates that the performance of the tennis player on the right-hand side is unchanged. However, the player on the lefthand side improved his performance; there are good chances he can now hit the ball from his court down to the ground on the right-hand side, i.e., p31 = 0.12 instead of nil before. This situation is reflected in Fig.2-22 where the probability for occupying S1 is increased, which is opposite to the behavior in Fig.2-21b. At

76 steady state S(68) = [0.619, 0, 0, 0.3811, indicating the above trend. It is interesting to note that if p31 is increased, i.e., p31 = 0.2 ( p32 = 0.8, p34 = 0), S(68) = [l, 0, 0, 01, i.e., the left-hand player has excellent chances to win the game at the end.

c

h

ui-

W

0

10

20

n

... - - . -_

30

40

Fig.2-22. Tennis game dynamics of the better player (on left-hand side) Example 2.27 is a simulation of an ideal pendulum, taken as system, where drag force is ignored. The four states assumed for the system, demonstrated in Fig.2-23a, are: S1= maximum height of the pendulum on left-hand side; S2jeft = minimum height of the pendulum while reaching this point from left-hand side; S2sight = minimum height of the pendulum while reaching this point from righthand side; S3 = maximum height of the pendulum on the right-hand side. The

above states are schematically depicted in Fig.2-23a

77

Fig.2-23a. Scheme of the states of an ideal pendulum The governing one-step transition matrix is given by: S1

s2,left

1

S2,right

s3

0

0

(2-54) 0

1

0

If the following initial state vector is assumed S(0) = [l, 0, 0, 01, i.e., the pendulum is initially at S1, the behavior depicted in Fig.2-23b is obtained. As seen, S1 is occupied after four steps whereas states S2,right and S2,left, which are practically the same point, are occupied after two steps, once from left-hand side and once from right-hand side. As observed also, the behavior of the system is un-damped oscillating. The more complicated case of a damped oscillations may be obtained by varying the transition matrix versus the number of steps.

78 1

0.8 0.6 c a’-

n Y

0.4

0.2

0

0

2

6

4

8

10

n Fig.2-23b. The behavior of an ideal pendulum

Example 2.28, a student progress scheme at a university, is a slightly modified version of the example appearing in [7, p.301. In the faculty of Engineering at Ben-Gurion university of the Negev in Israel, a student, the system, is studying for four years. The following states are assumed: S1-first year; S2second year; S3-third year and &-fourth year. Additional states are: Sg-the student has flunked out; Sg- the student has graduated. Let r be the probability of flunking out, p the probability of repeating a year and q the probability of passing on to next year where r + p + q = 1. The six states are governed by the following matrix:

S 1 p q

0

0

S 2 0

q

0

p

r r

0 0 (2-55)

0

::lo

0 0

0 0

0 0

1 0

O1 I

The first situation considered is a student having some probability of repeating a year, p = 0.1, and good chances for passing to next year, i.e., q = 0.9;

79 thus, his flunk out probability r = 0. In addition, the student is a first year student, thus S ( 0 ) = [l, 0, 0, 0, 0, 01. Fig.2-24 demonstrates the progress of the student at the university, where after eight steps (years) the probability of graduating is loo%, sg(8) = 1. The reason for not graduating after four years are his slight chances, lo%, for repeating a year. If p = 0, the student will graduate exactly after four years. 1

0.8

n

C

0.6

v

vi-

0.4

0.2 0

0

1

2

3

4

5

6

7

8

n Fig.2-24. Student's progress dynamics at the university in the absence of flunk out probability (r = 0) Fig.2-25 demonstrates the progress of the student with good chances to pass to next year, however, there are chances of 16% to flunk out. The prominent result is that at his fourth year, he has only 50% chances for graduating, i.e., he has equal probabilities to occupy S5 or Sg. The probability of occupying the other states has diminished.

80 1

0.8

c

n

0.6

W

Vi-

0.4 0.2 0

0

1

2

3

4

5

6

7

8

n Fig.2-25. Student's progress dynamics at the university in the presence of flunk out probability (r = 0.16)

Examples 2.29-2.31 relate to weather forecast, whereas example 2.3 1 demonstrates a way to improve the forecast by considering previous days information. Example 2.29 [4, p.781 predicts the weather in Tel Aviv (Israel) by a twostate Markov chain during the rainy period December, January and February. The system is the city of Tel Aviv where the states it can occupy are: S1 = D, Dry day and S2 = W, Wet day. Using relative frequencies from data over 27 years it was found that during this season the probability of wet day following a dry day is 0.250 (= pl2) and the probability of a dry day following a wet day is 0.338 (= ~ 2 1 ) .These data resulted in the following transition matrix:

P=

s1

0.750

0.250 (2-56)

S2

0.338

0.662

81 Results of the calculation of the state vectors S(n), are depicted in Fig.2-26 for different initial state vectors S(0) spelled out at the top of the figure. Thus for example, given that January 1st is a dry day (right-hand side of Fig.2-26), yields a probability of 0.580 that January 6th will be a dry. If January 1st is a wet day (middle of Fig.2-26), then the probability that January 6th is a dry day is 0.586. However, after ten days, the equilibrium conditions has for all practical purposes been reached. Thus, for example, if we call December 31st day 0 and January 10th day 10, then whatever distribution S[O] we take for day 0, we find that S(10) = [0.575,0.425]. Such a Markov chain is defined as ergodic and is without memory to the initial state. 1

S(0) = [0.5, 0.51

F j S(0) = [O,11

S(0) = [ I t 01

S -) : : E ] v)

0.4

0.2 0 0

q n r 2

4

n

6

8 1 0 0

2

4

n

6

8

1 0 0

--

2

4 n 6

8

10

Fig.2-26. Weather in Tel Aviv for different initial weather conditions Example 2.30 is related to the weather in the land of Oz [7, p.291 where they never have two nice days one after the other. If they have a nice day, they are just as likely to have snow as rain the next day. If they have snow (or rain), they have an even chance of having the same the next day. If there is a change from snow or rain, only half of the time is this a change to a nice day. The system is the land of Oz whereas the states it can occupy are: S1 = Rain; S2 = Nice; S3 = Snow. The transition matrix reads:

s2

112 114 114 112 0 112

s3

114

S1

P =

(2-57)

114 112

Three cases, differing by the initial state vector S(0) given above each graph, were studied. The results of the computations are depicted in Fig.2-27, indicating that

82

after two steps (days) the weather will always be snowy; it will continue like t h s independent of the weather two days before, i.e., on S(0). The limiting behavior is similar to that obtained in example 2.29, indicating an ergodic Markov chain.

Fig.2-27. Weather forecast Example 2.31. In examples 2.29 and 2.30 the Markov property clearly held, i.e., the new step solely depends on the previous step. Thus, the forecast of

the weather could only be regarded as an approximation since the knowledge of the weather of the last two days, for example, might lead us to different predictions than knowing the weather only on the previous day. One way of improving this approximation is to take as states the weather of two successive days. This approach is known as e x p d i n g a Murkov chain [7, p.30, 1401, which may be summarized as follows. Consider a Markov chain with states S1, S2. ..., S,. The states correspond to the land of Oz defined as system. We form a new Markov chain, called the eqandedprocess where a state is a pair of states (Si, Sj) in the original chain, for which Pij > 0. We denote these states by Sij. Assume now that in the original chain the transition from si to Sj and from Sj to s k occurs on two successive steps. We shall interpret this as a single step in the expanded process from the state Sij to the sjk. With this convention, transition from State Sij to state Su in the expanded process is possible only if j = k. Transition probabilities are given by

*

P(ij)(jl) = pil; P(ij)(kl) = 0 for j k Consider now example 2.30 applying the expanded process approach. Making the following designations, S = Snowy day, N = Nice day and R = Rainy day, yields the following states: S 1 = NR, S2 = NS, S3 = RN, S4 = RR, S5 = RS, s6 = SN, S7 = SR, s8 = SS. Note that NN is not a state, since p" = 0 in the

83

original process in example 2.30. The transition matrix for the expanded process is:

P=

RR RN RS NR NS SR SN

0

0

0

ss

0

0

0

112 112 0 0 0 0 0 114 1/4 112

The results of the computations are depicted in Fig.2-27a for two initial state vectors given at the top of the graphs. The one on the left-hand side gives 9% chances for the two previous days to be either Rainy-Nice or Snowy-Nice. The one on the right-hand side gives 100% chances for the two previous days to be Snowy-Rainy. The prominent observation is that the probability distribution of the states becomes unchanged from the 7th day on and is independent of S(0). It is given by S (7) = [0.2, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.21, indicating that the weather on the 7th day has 20% chances to be either Rainy or Snowy. If it was Snowy on the 7th it will remain like this; if it was Rainy, it will continue Rainy. Recalling the previous example, where only the information on the previous day was taken into account, it was obtained that the weather in the winter season will be 100% Snowy from the 3rd day on. S(0)= [0,0.5,0,0,0,0,0.5,0]

1

I

0.8

I

- s&n)=sJn)

I

I

I

I

sin) = sJn) = .. = sJn)

0

1

2

3

n

4

5

6

S(0) = ~0,0,0,0,0,1,0,01 I

-

7

I

I

I

I

I

sJn) = sJn) = ... = s,(n)

0

1

2

3

n

4

5

6

Fig.2-27a. Weather forecast by the expanded Markov process

-

7

84

Examples 2.32-2.33 below treat the behavior of a drunkard. In the first example the new step solely depends on the previous step. In example 2.33, the behavior depends on the last two steps. Example 2.32. A drunkard, the system, is living in a small town with * . four bars, states Sj = 1, 1 = 1, ..., 4. As time goes by, the drunkard jumps from one bar to the other according to the following one-step transition matrix. As seen, the probabilities of moving from one bar to the other are equal; also, the drunkard eventually leaves the bar to the next one, i.e. pii = 0.

P=

S1 S2 S3 S3

0 1/3 1/3 0

1/3 1/3

1/3 1/3

1/3 1/3 1/3 1/3

0 1/3

1/3 0

(2-59)

Fig.2-28 shows the behavior of the drunkard corresponding to two initial state vectors S(0) shown above the graphs. It is observed that after six steps the behavior of the drunkard reaches a steady state where his chances to visit any of the bars on the next step are 25%. The steady state is independent of S(O), thus the chain is an ergodic one.

-

l

S(0) = [I ,0,0,01

I

S(0) = [0.5,0,0,0.5] I

I

I

I

I

Example 2.33 applies the expanded process approach (elaborated in example 2.31) for the drunkard, the system, in example 2.32. The states are designated in the present example by Sij indicating that before moving to state s j k at step n+l, the drunkard spent in state Si some time at step n-1, and at step n he

visited state Sj. In this way, the effect of the last two steps, rather than one step, on his next visits can be studied. Considering the assumptions made in example 2.32, yields the following matrix where q = 113: s12 s13 s14 s21 s 2 3 s 2 4 s31 s32 s34 s41 s 4 2 s 4 3

S S S s s s

1 2 0 0 o q ~ 3 0 0 0 1 4 0 0 0 o 2 1 q q q 0 2 3 0 0 0 o 2 4 0 0 0 o

P = S 3 1 q q q S 3 2 0 0 0 S 3 4 0 0 0

0 q 0

q

q

o

o

0 o 0 o o

0 o 0 o o

0 o 0 q o

q o 0 q o

0 0 0 q q O 0 0 0

o q o 0 q o

o 0 0 q 0 0 q q q 0 0 0 o o o q q q

0 0 0 0 O O 0 0 0 0 q q

0 0 q

0

(2-60)

S 4 i q q q 0 0 0 0 0 0 0 0 0 S 4 2 0 0 0 q q q O O O 0 0 0 S 4 3 0 0 0 o o o q q q o o o The results of the computation for two initial state vectors given at the top of the graphs, each comprising of twelve states, are depicted in Fig.2-29. They indicate that after 7 steps, the drunkard is always at the same situation, i.e., his chances to occupy the next state, depending on his past deeds of two steps before, are 8.3%. This situation is independent of his initial step, which is plausible recalling that we deal with a drunkard.

Fig.2-29. The drunkard's behavior according to the expanded Markov process

86 Example 2.34 deals with actuarial considerations needed for premium calculations. The problem may be presented by assuming the following states that a system, the customer of an insurance company, may occupy: S1 = Healthy customer, S2 = Handicapped customer, and S3 = Dead customer. The following matrix clarifies the interaction between the states S1

s2

s3

Healthy Handicapped Dead S1

P=

s2 s3

P 0 0

9 r 0

1-P-9

1-r

(2-6 1)

1

For example, the matrix indicates that a healthy man can remain healthy, can become handicapped or die, i.e., p11 = p, p12 = q and pi3 = 1- p - q. If he is handicapped, he can never become healthy again, thus p21 = 0, and if he is dead, he is in the so-called dead state. It should be noted that the parameters p, q and r depend strongly on age but are taken here as constants. Fig.2-30 demonstrates the situation of a young man and an old man after five steps (years); both are initially healthy, i.e., S(0) = [I, 0, 01. It has been assumed that the young man has 95% chances to remain healthy (p = pi1 = 0.95) whereas the old man has 50% chances to remain in S1 (p = pi 1 = 0.5); other quantities shown on the graphs, r and q, are identical. It is observed in the figure that the probability of the young man to remain healthy after five years are quite good, Sl(5) = 0.774; for the old man si(5) = 0.0313, quite small to remain healthy. An interesting observation is related to the effect of r, i.e. to remain handicapped, on the state vector S(n); for example: r =0 Young man: S ( 5 ) = [ 0.774,0.0613,0.210] r=1 S ( 5 ) = [ 0.774,0.0905,0.136] Old man: S ( 5 ) = [ 0.0313,0.0380,0.930] r =0 r=1 S ( 5 ) = [ 0.0313, 0.0013,0.968] It may be observed that sl(5) is not affected by r where the other quantities are influenced by varying r.

87

Old man

Young man

~C

o.61 0.4

p = 0.50 q = 0.02

p=0.95 q = 0.02 r=0.97

0

1

2

n

3

4

5

0

1

2

n

3

4

5

Fig.2-30. The future dynamics of a young man and an old man Example 2.35 is the Ehrenfest diffusion model [6, p.211 for a simple random walk with reflecting barrier presented in example 2.18. The model assumes two containers A and B containing Z molecules. The containers are separated by a permeable membrane so that the molecules may move freely back and forth between the containers. It is assumed that at each instant of time t, one of the Z molecules chosen at random, is moving from one container to the other. The system are molecules in container A and the state Sj of the system is the number of molecules in container A which equals j - 1. Thus, the following states are assumed: S1 = 0, S2 = 1, S3 = 2, S4 = 3, ..., Sz+l = Z molecules. In the Ehrenfest model, if A has j molecules, i.e., it is in state Sj+l, it can on the next step move to Sj or to Sj+2 with probabilities Pj+lj

= YZ; pj+l,j+2= (Z - j)/Z; j = 1, ..., z

Pj+l,j+l = 0; j = 0, 1, ..., Z The transition probability matrix is then given by:

(2-62)

88 s1

0

1/z 0 0

...

s,

1 0 0 0 ..* 0 (Z-I)/z 0 0 ... 2$2 0 (Z-2)E 0 ... 0 3E 0 (2-3)/Z ...

0 0 0 0

s2

s3

s4

s5

0

0

0

0

0

0

0

0

0

0

.

.

.

.

.

.

... 0 ... 1

Sz+l 0 0

0 0

(2-63)

1LZ 0

Fig.2-3 1 presents results for total number of molecules Z = 3 and S(0)= [ 1, 0, 0, 01. The initial state vector indicates that the system (molecules in container A) is initially at S1, i.e., does not contain any molecule. In other words the three molecules are in vessel B. It is observed in the figure that the system attains the following two sets of constant values:

S(9) = [1/4,0, 314, 01, S(l0) = [0, 314, 0, 1/41 S(11) = [1/4, 0, 3/4, 01, S(12) = [0, 3/4, 0, 1/41 The results indicate that states S2 and S3 have the same occupation probability which oscillate against time. Note that state S2 corresponds to one molecule in container A where in state S3 two molecules should occupy container A. Thus, if at step n = 10 there is one molecule in A, two molecules will occupy container B. If at step n = 11 there are two molecules in A, then one molecule will occupy B and this process repeats its self ad infnitum.

89 1

0.8

c ui-

h

0.6

v

0.4

0.2 0

0

2

4

n

6

8

10

Fig.2-31. The approach towards equilibrium for a total number of molecules in the containers Z = 3 Fig.2-32 presents results for an even number of molecules in the containers, i.e., Z = 8 and an identical initial state vector as before, S(0) = [ l , 0, 0, 01. It is observed that the system attains the following behavior after about 20 steps, which is similar to that before, i.e.: S(25) = [0, 0.063, 0, 0.438, 0, 0.437, 0, 0.062, 01 S(26) = [0.008, 0, 0.219, 0, 0.547, 0, 0.218, 0, 0.0081 S(27) = [0, 0.063, 0, 0.438, 0, 0.437, 0, 0.062, 01 S(28) = [0.008, 0, 0.219, 0, 0.547, 0, 0.218, 0, 0.008] The results indicate that s4(25) = s6(25) and that states S4, S5 and S6 have the highest occupation probability which oscillate against time; other state probabilities are lower. Note that S4 corresponds to three molecules in container A, S6 to five molecules where in S5 four molecules should occupy container A. Thus, if at step n = 25 there are three or five molecules in A, because the states are of equal probability, the mean value is four. Thus, since the total number of molecules is m

= 8, four molecules will occupy container B. If at step n = 26 there are four molecules in A corresponding to S5 with the highest probability, then also four molecules will occupy container B. Therefore, at steady state the eight molecules will be equally distributed between the two containers.

90 1

0.8

c

0.6

h

ui-

W

0.4

0.2

0

0

5

10

15

20

n Fig.2-32. The approach towards equilibrium for a total number of molecules in the containers Z = 8 The general conclusion drawn is that in both cases, i.e., with odd and even number of molecules in the containers, the tendency of the system, molecules in container A, is to shift towards an equilibrium state of half molecules in each container. This trend is also expected on physical grounds.

Example 2.36 is the Bernoulli-Laplace model of diffusion [ 15, p.3781, similar to the one suggested by Ehrenfest. It is a probabilistic analog to the flow of two incompressible liquids between two containers A and B. This time we have a total of 22 particles among which Z are black and Z white. Since these particles are supposed to represent incompressible liquids, the densities must not change, and so the number Z of particles in each container remains constant. The system are particles in container A of a certain color and the state Si of the system, is the number of these particles in container A where Si = i - 1 (i = 1, 2,

..., Z+l)

(2-64)

If we say that the system is in state Sk = k - 1, i.e., the container contains k - 1 black particles, this implies that it contains Z - (k -1) white particles. The transition probabilities for the system are:

Pj+l,k+l

(2-65)

= 0 whenever Ij - k I >1; j = 0, 1 , ..., Z

The transition probability matrix is then given by:

sz

Sz+l

0

0

... ...

0 0

0

...

0

*.*

0

s1

s2

s3

s4

s5

0

1

0

0

0

0

P34 P44

P45

P21 P22 P23 0 P32 P33 0 0 P43

* a .

0

0 0 (2-66)

0

0

0

0

0

* ’ *

0

0

0

0

0

...

Pzz 1

Pz,z+l 0

where

pzz = 2(Z -1)/z2; pz,z+1= 1/z2 Fig.2-33 presents results for 2 2 = 8, i.e., 4 black and 4 white particles. Two initial state vectors S(0) were considered, given at the top of the figure. On the lefthand side, it is assumed that initially container A contains 2 black particles. On the right-hand side, there are 50% chances that container A will contain one black particle and 50% to contain four black particles. The results indicate that after ten steps, the system attains a steady state, independent of the initial conditions, where S(l0) = [0.014, 0.229, 0.514, 0.229, 0.0141. Such a process is known as

92 ergodic Markov chain later discussed. As seen, s3( lo) = 0.514 corresponding to two black particles in each container at steady state, is the highest probability with respect to the other states (Sl = 0, S2 = 1, S3 = 2, S4 = 3 and S 5 = 4). In other words, the state of the highest probability will exist at steady state; this is also the expected physical result. An interesting behavior observed in Fig.2-33 on the lefthand side is the following. In S(O), it was assumed that S3 = 2 has 100% probability. As seen, the probability of this state, i.e., s3(n), remains always the highest along the path towards equilibrium until it reaches its ultimate value. On the other hand, on the right-hand side, the state S3 = 2 had initially a probability of 0%. As seen, along its approach towards equilibrium, s3(n) is continuously increasing and attains the value of s3( 10) = 0.5 14, which remains constant from thereon.

0.8

-1

c\

0

2

4

n

6

8

10

,

,

,

S ( 0 ) =,[0,0.5,~,0,0.51

,

10

2

4

n

6

8

10

Fig.2-33. The dynamics of approach towards equilibrium for a total number of particles in the containers 22 = 8 (4 black and 4 white) Example 2.37 considers the random placement of balls [ 15, p.3791 where a sequence of independent trials, each consisting in placing a ball at random in one of Z given cells, is performed. The system are balls and the state of the system is the number of cells occupied by the balls. The state Si of the system is given by: Sj = i - 1 (i = 1, 2,

..., Z+l)

(2-67)

Thus, if we say that the system is in state sk, this implies that k-1 cells are occupied and Z - (k -1) cells are still free. For the placing process of the balls in the cells, the following transition probabilities apply:

93

Pj+l,j+l = jlz; pj+l,j+2= (Z-j)/Z; j = 0, 1 , ..., Z

(2-68)

yielding the following matrix: s1

s2

s3

s4

s5

0

1

0

0

0

0

0

P34

0

p44

p45

0

p22 p23

0 0

0 0

p33 0

**.

... ...

sz 0 0 0

- I f

0

Sz+l 0

0 0 0 (2-69)

0 0

0 0

0

0

0

0

0

0

... ...

Pzz 0

Pz,z+l 1

where P22 = 11Z; p23 = (Z - 1)E; P33 = 212; p34 = (Z - 2 ) E PU = 3E; p45 = (Z - 3 ) l z ; Pzz = (Z - 1)E; pz,z+1= (Z - l ) E Fig.2-34 presents results for three cells, i.e., Z = 3. Thus, the four states are: S1 = 0, S2 = 1, S3 = 2 and S4 = 3 cells occupied. S2 = 1 indicates that one cell has already been occupied by the ball. It may be observed that in both cases depicted in the figure, differing by the initial state vectors S(O), the ultimate situation is identical, i.e., all four cells are occupied, as expected. This occurs after 15 steps. The maximum in the si(n) curves, i = 2, 3 is interesting and clear. For example, s2(n) corresponding to S2 = 1 is attaining a maximum after one step, n = 1, because after S1 = 0, S2 must come.

94

1

v,

, s,(N

-

0.6

-

C Y-

**--+----

I

0.8

-

I

s,(n)

,*

0

-

0.4

0.2 0

n n Fig.2-34. The dynamics of cell occupation by balls for Z = 3

Example 2.38 is concerned with cell genetics [15, p.3791 where a Markov chain occurs in a biological problem which may be described roughly as follows. Each cell of a certain organism contains N particles, some of which are of type A, others of type B. The system is a cell and the state of the cell is the number of particles of type A it contains; there are N + 1 states. A cell is said to be in state i if it contains exactly i - 1 particles of type A, where

Si = i - 1 (i = 1, 2, ..., N

+ 1)

(2-70)

The cell is undergoing the following process. Daughter cell are formed by cell division, but prior to its division each particle replicates itself. The daughter cell inherits N particles chosen at random from 2i particles of type A and 2N - 2i particles of type B present in the parental cell. The probability that a daughter cell occupies state k is given by the following hypergeometric distribution

Pj+l,k+l =

(N!I2 (2j)! (2N - 2j)! k!(2j - k)! (N - k)!(N - 2j + k)! (2N)!

-

(2-7 1)

j , k = O , 1 , 2 ,..., N Note that pi1 and P N + I , N + ~= 1; pj+l,k+l = 0 if the expressions in the parenthesis in the denominator < 0.

95

The behavior predicted by the model is demonstrated for N = 4, yielding the following states from Eq.(2-70): S1 = 0, S 2 = 1, S3 = 2, S4 = 3 and S5 = 4 particles of type A. The corresponding matrix obtained from Eq.(2-71) reads: s2 0

S1

s4

s3

s5

0 0 0 1 0.2143 0 0 0.2143 0.5714 s2 0.0143 (2-72) 0.2286 0.2286 0.5143 P = s3 0.0143 0 0.2143 0.5714 0.2143 s4 0 0 0 1 0 0 s5 Results for the formation dynamics of particles of type A from their cells is demonstrated in Fig.2-35. Three initial state vectors, given at the top of each figure, were considered. The calculation indicate that a steady state is reached after thirty steps (generations), always at S1 and S5, which are dead or absorbing states. ) where for that For the case on the left-hand side, sl(30) = 0.75 and ~ ~ ( 3=00.25 on the right-hand side sl(30) = 0.25 and ~ ~ ( 3=00.75. ) The state with the highest probability is always the one nearest to the state of the highest probability in the initial state vector. The case in the middle is symmetrical, hence, sl(50) = ~ ~ ( 5=0 ) 0.50. It is interesting to mention that the results after sufficiently many generations, n +OO, comply with the following theoretical predictions. The entire population will be (and remain) in one of the pure states S1 and & + I ; the probability of these two contingencies at steady state are sl(-) = 1 - i/N, sN+*(=)= i/N where i is the number of particles of A at the initial S1

state. For example, i = s3(O) = 2 and N = 4, for the case on the middle in Fig.2.35. Thus, sl(-) = s&-) = 0.5. 1

S(0) = [0,1,0,0,01

S(0) = [0,0,~,0,01

S(0) = [0,0,0,1,01

0.6

-C

0.4

0

5

10

n

15

20 0

5

10

n

15

20 0

5

10

n

15

20

Fig.2-35. The formation dynamics of A-type particles for N = 4

96

Example 2.39 is taken from population genetics [ 15, p.3801. Consider the successive generation of a population which is kept constant in size by the selection of N individuals in each generation. A particular gene, assuming the forms A and a, has 2N representatives. If in the nth generation A occurs i times, then a occurs 2N - i times. The system is the population (such as plants in a corn field). The state of the system is the the number of times that the A-gene occurs after some generations. A population is said to be in state i if A occurs 2i times, i.e., Si = i - 1 (i = 1, 2, ..., 2N + 1)

(2-73)

where the number of states is 2N + 1. Assuming random mating, the composition of the following generation is determined by 2N Bernoulli trials in which the probability of the occurrence of the A-gene is i/2N. We have, therefore, a Markov chain with (2-74) j, k = 0, 1 , ..., 2N

Note that pi1 and P2N+1,2N+1 = 1; Pj+l,k+l = 0 if the expressions (2N - k) in the denominator < 0. The above indicates that at states S1 and S2N+1, called homozygous, all genes are of the same type, and no exit from these states is possible. The behavior predicted by the model is demonstrated for N = 2, yielding the following states from Eq.(2-73):S1 = 0, S 2 = 1, S3 = 2, S4 = 3 and S5 = 4 A-gene occurrence. The corresponding matrix reads: S1 1 0.3164 0.0625 0.0039 0

s2

s3

s4

S5

0 0.4219 0.2500 0.0469 0

0 0.2109 0.3750 0.2109

0 0.0469 0.2500 0.4219

0 0.0039 0.0625 0.3164

0

0

1

(2-75)

97 Fig 2.35 demonstrates the dynamical behavior of the A-gene Occurrence for two initial state vectors S(0). The behavior is, in general, similar to example 2.38 where the process is terminated at one of the dead states for which pii = 1. In other words, generally S1 and s2N+1 designate the homozygous states at one of which the ultimate population will be fixed, depending on the magnitude of the corresponding probabilities. It should be emphasized that the ultimate results are dependent only on the initial step where the steady state probabilities are given by 1 - i/(2N), S * ~ + ~ ( C O=) i/(2N). I is the number of A-genes at the initial

sl(oo) =

step. For example, for the case on the right-hand side in, I = q(0) = 1 and N = 2. Thus, S ~ ( O O ) 0.75 and s,(w) = 0.25.

-

7 S(0) = [0,1/3,1/3,1/3,0]

t

15 5 n 10 Fig.2-35a. The dynamics of A-gene occurrence for N = 2

0

Example 2.40 is a breeding problem [15, p.3801 where in the so-called brother-sister mating, two individuals are mated. Among their direct descendants, two individuals of opposite sex are selected at random. These are again mated, and the process continues indefinitely. With three genotypes AA, Aa and aa for each parent, we have to distinguish six combinations of parents, designated as states, which we label as follows: S1= AA*AA, S2 = AA*Aa, S3 = Aa*Aa, S4 = Aa*aa, Ss = aa*aa, S6 = AA*aa. The system is two individuals of opposite sex which are mating. Based on above reference, the matrix of transition probabilities reads:

98 s2

s1

s3

s4

s.5

s6

1 0 114 112 1116 114

0 114 114

0 0 0

114 1/2 114 0 0 0 1 0 0 0 0 1

0 0 0

0 0 0 0 0 0 114 1116 118 (2-76)

Fig.2-36 demonstrates the results of brother-sister mating against time. As expected from the matrix given by Eq.(2-76), containing two dead states S1 and S5 (p11 and pss = l), the system will eventually occupy one of these states, depending on the magnitude of the relative probabilities sl(n) and ss(n). The latter depends on S(0) given in Fig.2-36 at the top of the graphs.

F-7 S(0 = [0,0,0,0,0,1]

& sg(n)

0

5

n

10

15

0

5

n

10

15

Fig.2-36. Breeding dynamics Examples 2.41-2.42 are examples of random walk type on real life problems, in addition to examples 2.17-2.22. Example 2.41 demonstrates a game [6, p.211 related to a simple random walk between two absorbing bam'ers. considered also in example 2.17. Assume that Jacob and Moses have five shekels (Israeli currency) divided between them. On one side there is a symbol of a leaf and on the other side appears the number one. The shekels are assumed fair ones, so that the probability of a leaf on a toss equals the probability of a one on a toss, equals 112. Jacob tosses a coin first and records the outcome, Zeaf or one. Then Moses tosses a coin. If Moses matches

99

Jacob (obtains the same outcome as Jacob), then Moses wins the shekel, otherwise Jacob wins. Note that Moses or Jacob win with probability 1/2. Let the states Si = i - 1 (i = 1, 2, ..., 6) represent the number of shekels that Moses, the system, has won. The game ends when Moses has 0 or 5 shekels, i.e., states S1 and S6 are then dead or an absorbing states. Given that Moses is in state k (has k - 1 shekels), he goes to state k + 1 (wins) with probability 1/2 or goes to state k - 1 (loses) with probability 1/2. The above rules of the game may be summarized in the following matrix:

s s2

P=

1

1 0 0 112 0 112 1/2 0 0 1/2

::I :

0 0 0 0 112 0 0

0 0 0

1/2 0

(2-77)

where, for example, p45 = 1/2 is the probability that Moses wins a fourth coin given that he already has three coins. Fig.2-36 demonstrates the dynamics of the game for two cases. On the lefthand side is observed that initially Moses is in S2, he has one shekel. As time goes by, his chances to win are decreasing where after 20 steps his situation is given by S(20) = [0.8, 0, 0, 0, 0, 0.21. This means that there are 80% chances he will be left without money. It is also observed that after one step, there are 50% chances he will be left without money at all (be in Sl) or win one shekel (be in S3) and from thereon his chances to loose are increasing. So if he is smart, on the one hand, and knows Markov chains, on the other, he should better stop the game after one step. On the right-hand side, Moses begins with three shekels, he is in S4, and as time goes by, his chances to win are increasing where S(25) = [0.4,0,0,0,0,0.6].

100

h

C

Y-

I ' r J

I

v)

0

5

n

10

15 0

5

n

10

15

Fig.2-36. Dynamics of the matching shekels game Example 2.42 deals with a political issue of establishing a coalition in Israel in the eighties. Generally, the treatment of the problem is based on the random walk model incorporating the reflecting-absorbingbarrier effects, i.e., p11 = t and pgg =1, respectively, assuming that the maximum size of the coalition comprise nine parties. In Fig.2-37, the Israeli caricaturist Moshik [ 191 demonstrates the efforts made by Mr. Menahem Begin, the Prime Minister these days, to establish a coalition. He is seen trying to attract to the coalition additional parties in order to establish a stable government. As observed in the figure, so far five parties have joined the coalition. The system is the coalition headed by Mr. Begin, where a state is the number of parties in the coalition, i.e., Si = 1,2, ..., 9. The underlying assumptions of the "political game" are: a) The probability to increase the coalition from one state to the next one is p, independent of Si. b) There is a probability, r, that the coalition will remain unchanged in its size. c) Similarly, there is a probability, q, that the size of the coalition will decrease due to unsuccessful negotiations. Note that q + p + r = 1.

101

Fig.2-37. Establishment of a coalition in Israel The following matrix summarizes the above considerations: s3

s.5

s1

s2

s8

s9

t q O O O O

1-to 0 0 0 0 0 r p O O O O O q r p O O 0 0 O q r p O O O

0 O 0 o

O O

O O

q O

r q

p r

O O 0

O O 0

O O 0

O O 0

O O 0

q O 0

s4

s6

s7

O p r q 0

O O

O O

p r 0

O p 1

(2-78)

102 Fig.2-38, containing the input data and S(0) above the graphs, demonstrates results of the calculations with respect to the following points. Note that S(0) corresponds in all cases to a coalition already with five parties, i.e. S5, as seen in Fig.2-37. In general, depending on time, a stable coalition consisting of nine parties will be established because Sg = 9 will, eventually, acquire the highest probability. In case c, the approach towards a stable coalition is very fast because the 'negotiation-success factor' p = 0.8 is a relatively high value, in comparison to p = 0.25 in cases a and b. It is also observed in case c that at each step, after intensive rounds of talks, the highest probability corresponds to the state of a higher number of parties, i.e., a larger coalition. In cases a and b, S5 remains of the highest probability until n = 15; from thereon the probability of Sg becomes the highest. However, this behavior depends on the reflecting barrier effect governed by the factor t in the matrix given by Eq.(2-78). If t = 0 (case a), i.e., an ideal reflector, sg(n) > sl(n). If t = 1 (case b), S1 becomes an absorbing or dead state like Sg, sg(n) = q(n) and the chances to establish a stable coalition or to fail are the same. It should be noted that for high p's, the effect of t is negligible, as observed in case c.

103

1

0.8 C

0.6

v,

0.4

v-

r = 0.1

0.2 0

20

n

(c> Fig.2-38. The dynamics of establishing the coalition

Examples 2.1-3-2.45 treat a few models of periodic chains, also referrec to as recurrent events. Generally speaking, a system undergoes some process as a result of which it occupies the states sequentially. The latter repeats its self ad in3nitum or attains some steady state. Example 2.43, recurrent events 1 [15, p.3811, obeys the following transition probabilities matrix:

s1

P11

s 2 1

P12

0

P13 0

P14 . . 0 . . . (2-79)

s s o

0

0

1

. . . . . . . . .

To visualize the process which generates from above matrix, suppose that initially the system occupies S1. If the first step leads to Sk-1, the system is bound to pass successively through states Sk-2, Sk-3, ..., and at the k th step it

104 returns to S1, whence the process starts from scratch passing, in principle, the above steps again and again. Practical examples conforming with the above model are the following ones. A drunkard, the system, treated also in examples 2.32 and 2.33, is acting now according to different rules dictated by the transition matrix, Eq.(2-79). The states S1 to S4 are four bars in the small town the drunkard is occupying. Another example is concerned with a dancer, the system, acting as follows. There are four nice dancers, states S 1 to S4,standing on a circle at equal distances. The dancer is moving from one nice dancer to the other, performing with her a dance, and moving to the next one. His occupation ofstutes are according to Eq.(2-79). The interesting question is what happens versus time with the system? Fig.2-39a,b demonstrate the behavior of the system as a function of the Pij's in the matrix, Eq.(2-79), and S(0). The characteristic behaviors observed in the figure are: a) In each state the system oscillates until a steady state is achieved. b) The magnitude of the steady state is independent of S ( O ) , i.e., the chain is ergodic. c) The magnitude of the steady state depends on the policy-making matrix, Eq.(2-79), i.e., on the Pij'S. Regarding to the behavior of the system, the drunkard or the dancer, it is observed in Fig.2-39a,b that at steady state, it will remain in S1, the state of the highest probability. The ultimate values of the state vectors are: Fig.2-39aYS(24) = [ 0.333, 0.300, 0.234, 0.1331 Fig.2-39b, S(l0)= [ 0.500, 0.300, 0.150, 0.0501

n

n

a) p11 = 0.1, p12 = 0.2, pi3 = 0.3, pi4 = 0.4

105

1

S(0) = t1,0,0,01 I

I

I

0.8 .

5

0

10

15

20

n

b) pi1 = 0.4, pi2 = 0.3, pi3 = 0.2, pi4 = 0.1 Fig.2-39 a,b. The dynamical behavior of the system Example 2.44, recurrent events 2 [15, p.3821, obeys the following transition-probability matrix:

P=

0 0 0

. . .

0

0 0 p3

0

0

p4

. . .

.

. . .

. .

. . . . . .

0

q2

p1 0

s3

q3

0

s4

q4

0

ss

*

s1

q1

s2

p2

. . . . . .

(2-80)

The matrix indicates that the system moves from one state to the other, and upon reaching the new state it has always some probability of returning to the initial state. Fig.2-40 demonstrates results of the calculations for the case of four states, i.e., S1, ..., S4. The corresponding matrix reads:

(2-80a)

106

The input data are reported on the figure. On the left-hand side the values of the qi's are relatively high in comparison to the right-hand side, thus, the probability of remaining in or returning to S 1 is high, as a result of which the approach towards a steady state is after one step. By increasing qi, the occupation of the states by the system is of a recurrent type, where eventually a steady state is achieved of the state-probability distribution. In both cases, the Markov chain is ergodic.

y

n

0

,'

I

I

5

10

n

I

15

20

Fig.2-40. The dynamical behavior of the system Example 2.45, recurrent event 3, demonstrates a periodic chain [4,p.1021 by considering the behavior of a Saudi sheik. Fig.2-41 shows the Saudi sheik, defined as system, opening the door for an amazing beauty symbolizing the West at the entrance of his harem. The Israeli caricaturist Moshik [ 191 demonstrates in the figure the approaching process of the West towards Saudi Arabia at the beginning of the eighties. The following model is developed for the caricature, which is applicable for cases 1 and 2 below. The states are seven Saudi beauties, Si = Saudi beauty , i = 1, ..., 7 and sg = Western beauty. The caricature may be understood in several ways, dictating the construction of the 8x8 one-step transition probability matrix.

L A -

Fig.2-41. The harem of the Saudi sheik The matrix demonstrates several interesting characteristics, for example: from S1 the sheik never goes to S2; similarly, from S;! he never goes to S1; however, from these states he always moves to S3. From S3 the sheik has to decide where to go, because p34 = p35 = p36. As seen, by analyzing the various pij's, which are

108

assumed to remain constant, a complete understanding of the sheik's behavior can be obtained. Case 1. It has been assumed that in the matrix p = 0 and q = 1/2, indicating that s8 has no preference over the other states. sg has been located on the third floor of the harem, right to S7 as seen in Fig.41. It has also been assumed that the sheik is initially occupying S1, i.e., S(0)= [l, 0, 0, 0, 0, 0, 0, 01. The occupation dynamics of the sheik is obtained by applying Eq.(2-24), i.e., multiplying the state vector by the matrix, Eq.(2-81), yielding Fig.2-42. 1

0.8

0.6 t

h

W

ui-

0.4

0.2

0

0

2

4

6

n

8

10

12

14

Fig.2-42. States occupation dynamics of the Saudi sheik

The prominent observation in Fig.2-42 is the periodic behavior of the system (sheik), where each state is reoccupied after four steps. However, in general, the sheik i s occupying at each step another state, thus acting very very hard, sooner or later might affect his health. The only way to change the dangerous results predicted by the model, is by modifying the matrix, Eq.(2-81); the latter depends on the habits of the sheik. Thus, he should be advised by his doctor accordingly. Additional interesting observations are: a) Whenever the sheik is moving to S7, he should consider to occupy instead s8 because the OccUpQtion probability of

109 these states is 50%. He has the same problem with S1 and S2. b) The occupation problem of S4, Ss and s6 is more complex since he has to decide among three beauties. c) No problem with Sg; the occupation probability of this state, whenever reached, is 100%. Case 2. It has been assumed here that in matrix, Eq.(2-81), p = 1, q = 0 and, as in previous case, S(0) = [I, 0, 0, 0, 0, 0, 0, 01. The result in Fig.2-43 indicate clearly a significant preference for $43 over the other states. After 43 steps it is obtained that S(43) = [O,0, 0, 0, 0, 0, 0, 11, i.e., the sheik will be 'absorbed' at sg. Noting the amazing beauty in Fig.2-41, his behavior is not surprising at all. 1

0.8 0.6 K

n

ui-

Y

0.4

0.2

0

0

5

10

15

20

25

30

n Fig.2-43. States occupation dynamics of the Saudi sheik Case 3. In cases 1 and 2 the Saudi sheik was the system and the beauties were the states. Now we look at the caricature in Fig.2-41 from a different point of view. The system is the Western beauty where the stares Si , i = 1, ..., 8, are the eight rooms occupied by the Saudi beauties. Note that number 8 in the figure, representing the western beauty in the above cases, designates now room number 8

located on the third floor of the harem, right to S7. The figure shows also that initially the West is invited by the sheik to join the harem, hoping it will occupy

110

only one of the rooms, i.e. number 8 which is the only available one. The behavior of the system may be deduced from the following single-step transition matrix: s6

s1

s2

s3

s4

1/5 1/4

115 1/4

1/5 0

1/5 115 0 1/4 1/4 0

0 0

1/5

0

1/5

1/5 0

1/5 0

118 1/6

118 1/6

1/8 0

1/8 1/8 1/8 1/8 1/8 1/6 1/6 0 1/6 1/6

0

0

1/4

1/4

0

0

0

0

1/6 0

1/6 1/6 1/4 1/4

s5

0

1/5

s7 s8

0 0

(2-82)

1/4 1/4 0 1/6 1/6 1/6 0 1/4 1/4

Two cases were explored for the dynamical behavior of the West. In the first case, the West occupies initially S1; this is expressed by the initial state vector S(0) given on the top of Fig.2-44, left-hand side. On the right-hand side, the state vector corresponds to the case where the West has equal probabilities to occupy states S2, S4 and S6, i.e., Si(0) = 1/3, i = 2, 4, 6. The results depicted in the figure are very interesting, indicating that after ten steps the system has a certain probability to be found in every state, i.e., the domineering process of Saudi Arabia by the West is very effective. Moreover, this process after some time, becomes independent of the initial step, i.e., an ergodic Markov chain which is without memory to the past. The probability distribution at steady state after ten steps is given by S(10) = [0.119, 0.095, 0.119, 0.190, 0.143, 0.095, 0.143, 0.0951 noting that the probabilities are not too different from each other.

111

it

0.8 A

C

Y-

ic

0.6

S(0) = [0,1/3,0,1/3,0,1/3,0,0]

0.4

0.2 0

0

5

n

10

15

0

5

n

10

15

Fig.2-44. The domineering dynamics of Saudi Arabia by the West Example 2.46 is of fundamental importance and completes our real life examples. It considers the imbedded Markov chain of a single-server queuing process [4, pp.8, 891. Queuing is encountered in every comer of our life such as queues at servers, telephone trunk lines, traffic in public transportation - bus, trains as well as airports, queues at surgery rooms in hospitals and governmental offices, department stores, supermarkets and in a variety of industrial and service systems. One of the simplest models of queuing is the following one. Let customers arrive at a service point in a Poisson process [see 2.2-31 of rate h [customers arriving per unit time]. Suppose that customers can be served only one at a time and that customers arriving to find the server busy queue up in the order of arrival until their turn for service comes. Such a queuing policy is called First In First Out (FIFO).Further, suppose that the length of time taken to serve a customer is a random variable with the exponential p.d.f. = (probability density function) given by

f(t) [Mime] is the probability density that the length of time it takes to complete the service is exactly t, f(t)dt is the probability that the length of time to complete the service is between t and t + dt where

6 6

prob{z 2 t> = f(t)dt =

pe-Ptdt = 1 - e-p7

(2-83a)

112 is the probability to complete the service between 0 - 7. For z -j 00 the probability equals unity, i.e. the probability to complete the service at infinite time is unity. The constant l / p [time/customer] is the expected mean service time per customer where p is the mean number of customers being serviced per unit time. The exponential distribution is employed to describe the service probability distribution function in many queuing systems. It should also be noted that for any other distribution the problem becomes intractable. It can also be shown that if the service of a customer is in progress at time t, and the p.d.f. of the service time is given by Eq.(2-83), then the probability that the service time is completed in the time (t, t + At) is:

pAt + O(At2)

(2-84)

where O(At2) denotes a function tending to zero at the same rate as At2. The above queuing process, and more precisely between consecutive times of two customers who completed to receive their service, can be described also by the so-called discrete imbedded Markov chain of a single server. In the queuing literature this process is noted by W / 1 . The first M denotes a Poisson arrival time, the second M is an identical independent exponential probability distribution service time and the 1 represents a single server. We define the system as the queue and the state of the system is the number of customers waiting in the queue. The state space is the possible number of states, i.e., S1 = 0, S2 = 1, S3 = 2, ... If the state space is finite, S1 = 0, S 2 = 1, S3 = 2 , ..., Si = i - 1 customers. It should be noted that the state of the system is evaluated at the moment the nth customer has completed to receive herihis service. By observing the state of the system at these points (completion of service time) the queuing process of the system is described by the Markov property of absence of memory, i.e., ignoring the past history of the customers which have already been served. Accordingly, let Xn denote the number of customers in the queue immediately after the nth customer has completed his service. Thus, X,, includes the customer, if any, whose service is just commencing. Then we can write down the following equations:

113

(2-85a) (2-85b) where Yn+l is the number of customers arriving during the service time of the (n+l)th customer. Eq.(2-85a) expresses the fact that if the nth customer does not leave an empty queue behind him (Xn 2 l), then during the service time of the (n+l)th customer, Yn+l new customers arrive and his own departure diminishes the queue size by 1. If the nth customer leaves an empty queue, then the (n+l)th customer arrives and depart after the completion of his service during which Yn+l new customers arrive. We can distinguish by a simple argument between two types of behavior of the system. The rate of arrival of customers is h and within a long time to the average number of customers arriving is At,. As indicated before, the mean service time per customer is l/p. If the service of the customers were to go on continuously, the average number of customers served during time to would be Pto. Hence, if h > p, we can expect the queue of unserved customers to increase indefinitely; in a practical application when this occurs, customers would be deterred from joining the queue and h will therefore decrease. If, however, h c p the server needs to work only for a total time of about htO/pin order to serve the customers: that is the server will be idle, i.e., the process will be in state S1 = 0, for about a proportion 1- Up of the time. In the following presentation we assume that customers arrive in a Poisson process of rate h and that their service times are independently distributed. The distribution function B(t) (0 I t < 00) is the probability that the service time t, satisfies B(t) = prob{t, 5 t}

(2-86)

where dB(t) = B'(t)dt

(2-87)

114 needed later is the probability that the service time lies between t and t + dt. B'(t) is the p.d.f. defined in Eq.(2-83). A useful quantity based on the above concepts, needed for evaluating the probabilities in the one-step transition matrix, is the following one bi = prob{Y, = i}

(i = 0, 1, 2, ...)

(2-88)

i.e., bi is the probability that the number of customers arriving during the service of the nth customer is equal to i. The above quantity may be calculated in the following way. For the Poisson process Pi(t) = prob{N(t) = i} = e-ht(ht)i/i! (i = 0, 1, 2, ...)

(2-89)

where Pi(t) is the probability that the number N(t) of events occurred (customers arrived) is equal to i, given that the service time is t. From the above definitions, Eqs.(2-86) and (2-89), it follows that the product Pi(t)dB(t) designates the probability that i additional customers will be added to the queue (see Eq.2-85a) if the service time is between t and t + dt. Since the service time ts varies as 0 I ts 5 00, the probability that the number of customers arriving during the service time is equal to i, is given by [4, pp.881 bi =

J,

Pi (t)dB(t) (i = 0, 1, 2,

...)

(2-90)

From Eqs.(2-85a) and (2-85b) the elements of the transition matrix of the process Xnare given by the following matrix: s3

s4

s5 . . .

b2 b2 bl

b3 b3 b2

b4 b4 b3

bo 0

bl bo

b2 bl. . .

. .

. . . .

. . . . . . . .

. . .

. . .

(2-91)

115 It should be emphasized that the transition matrix, Eq.(2-91), applies to the time interval between two consecutive service completion where the process between the two completions is of a Markov-chain type discrete in time. The transition matrix is of a random walk type, since apart from the first row, the elements on any one diagonal are the same. The matrix indicates also that there is no restriction on the size of the queue which leads to a denumerable infinite chain. If, however, the size

of the queue is limited, say N - 1 customers (including the one being served), in such a way that arriving customers who find the queue full are turned away, then the resulting Markov chain is finite with N states. Immediately after a service completion there can be at most N - 1 customers in the queue, so that the imbedded Markov chain has the state space SS = [0, 1,2, ..., N - 1 customers] and the transition matrix: s1

s2

bo bo

bl bl

0 0 0

bo 0 0

SN-1

0

0

SN

0

0

(2-92)

where dN-1 = 1 - (bo + bl + ... + bN-2). An example demonstrating the above concepts was calculated for N = 4 states; thus, SS = [0, 1,2, 3 customers]. The corresponding matrix is the following one where the probabilities were selected arbitrarily. s2

s4

0.2

s2

0.5 0.2

s3 S4

0

0.5

0.2

0.3

0

0

0.5

0.5

I

0.5 0.2

s3

0.1 0.1

S1

P=

I

s1

0.2

(2-93)

116 Fig.45 shows results for two initial state vectors S(0). On the left hand-side the system is initially at S1, i.e., there are no customer in the queue. On the right hand-side the system is initially at S4,i.e., there are three customers in the queue. The propagation of the probability distribution indicates that the system (queue) reaches a steady state independent of S(O), i.e., an ergodic Markov chain. The calculation indicates that the steady state is achieved after nine steps and the appropriate state vector reads S(9) = [0.211, 0.212, 0.254, 0.3221, i.e., S4 with three customers is the state of the highest probability. Certainly, if the possible number of states N (with maximum number of customers N - 1) is changed, a new steady state would have been achieved, which depends also on the matrix, Eq.(293).

-

-\

-

\ \

-

\

--___ 4

n

5

0

1

2

n

3

4

5

Fig.2-45. The dynamics of a queuing process

2.1-5 Classification of states and their behavior The examples presented in section 2.1-4 indicate that the states of a Markov chain fall into distinct types according to their limiting behavior. In addition, it is also possible to identify from the transition matrix some characteristic behavior of the states. Suppose that the system is initially at a given state. If the ultimate occupation to this state is certain, the state is called recurrent; in this case the time of first return will be called the recurrence time. If the ultimate return to the state has probability less than unity, the state is called transient. On the basis of the examples presented above, the following classification of states may be suggested [4, p.91; 6, p.28; 15, p.3871.

117

Ephemeral state. A state j is called ephemeral if pij = 0 for every i. Thus, an ephemeral state can never be reached from any other state. For example, S1 in the matrix given by Eq.(2-43) is such a state since pi1 = 0 for every i. Accessible state. A state k is said to be accessible from state j if there exists a positive integer n such that pjk(n) > 0 where pjk(n) is defined in Eq.(2-26). For example, according to Eq.(2-49), S4 is accessible from S2 since it may be obtained from Eqs.(2-32) to (2-34) that p24(2) = p2 > 0. However, in the same matrix S2 is not accessible from S1 since p12(n) = 0. In Eq.(2-51), S2, ..., s16 are not accessible from S 1 since it is an absorbing state. Inter-communicating states. Two states j and k are said to communicate if k is accessible from j in a finite number of transitions, i.e., if there is an integer n such that pjk(n) > 0. We define njk to be the smallest integer n for which this is true. If j is accessible from k and k is accessible from j, then j and k are said to be inter-communicating. As indicated before, in Eq.(2-49), state S4 is accessible from S2 since p24(2) = p2 > 0. In fact, S 2 is also accessible from S4 since it may be shown that p42(2) = q2 > 0. Thus, S2 and S4 are inter-communicating. It may also be shown that other states in Eq.(2-49) demonstrate the same behavior. Irreducible chain. Perhaps the most important class of Markov chains is the class of irreducible chains. An irreducible chain is one in which all pairs of states communicate, i.e., pjk(n) > 0 for some integer n where j, k = 1, 2, ..., Z; the latter is the number of states. In other words, every state can be reached from every other state. The following matrix, corresponding to example 2.32 s1

S1

10

s2

s3

s4

1/3

1/3

1/3

P=

satisfies the above condition since it follows from Eqs.(2-32) to (2-34) that p11(2), P22(2), P33(2), P44(2) > 0 as well as the other Pjk'S. Pjj(n), defined in Eq.(2-26),

118 is the probability of occupying Sj after n steps (or at time n) while initially occupying also this state. Absorbing state. State j is said to be an absorbing (dead or a trapping) state if the occupation probability of the state by the system satisfies pjj = 1. In other words, once a system occupies this state, it remains there forever. Examples of such a state are: 2.8, 2.10, 2.13, 2.15, 2.17, 2.23, 2.24, 2.28, 2.34, 2.37, 2.38, 2.39, 2.40, 2.41, 2.42. Prior to further definition of states, additional notation is established in the following[ 15, p.3871. Throughout the following chapter we will use fj(n) as the probability that in a process starting from Sj, the next occupation of Sj occurs exactly at the nth step. In other words, we may say that conditional on Sj being occupied initially, fjj(n) is the probability that Sj is avoided at steps (times) 1, 2, ..., n - 1 and re-occupied at step n. fjj( 1) = pjj and for n = 2, 3, ..., fjj(n) = prob{X(r) # j, r = 1, ..., n - 1; X(n) = j I X(0) = j } (2-94) It is interesting to demonstrate the above concepts considering Fig.2-0 by Escher. Assume that X(r) designates locations along the water trajectory after r steps. Thus, X( 1) = 1 is location of the system at point 1, the top of the waterfall, after one step. Similarly, X(5) = 2 designates location at point 2 after 5 steps. On the basis of Eq.(2-94), the following probabilities may be established, i.e.: f11(5) = prob{X(r) # 1, r = 1, ..., 4;X(5) = 1 I X(0) = 1) = 1 f11(3) = prob{X(r) # 1, r = 1, 2; X(3) = 1 I X(0) = l } = 0 On the basis of fj(n) the following quantities, later applied, are defined: Probability fjj that, starting from Sj, the system will ever pass through Sj, reads: (2- 95) n= 1

In other words, fjj designates the probability that Sj is eventually re-entered. The application of Eq.(2-95) can be demonstrated on the basis of Fig.2-0 in the following way. Instead of the perpetual motion of the water, the loop is opened at

119 S3, point 3. A certain amount of water is then poured at S1 which flows through S2 and is leaving away at point 3, state S3. For a water element, the system, Eq.(2-95) yields that f l l = f11(1) + f11(2) + ... = 0 + 0 + ... = 0, i.e., the probability of the water element to re-enter at point 1 is nil. Similarly, f22 = f33 = 0. Mean recurrence time is given by

(2- 96) n= 1

where for the above example, it gives pi = p2 = p3 = 0.

Probability fjk(n) that in a process starting from Sj, the first occupation of Sk (or entry to Sk) occurs at the nth step, is defined in Eq.(2-97). In other words, fjk(n) indicates that Sk is avoided at steps (times) 1, ..., n - 1 and occupied exactly at step n, given that state Sj is occupied initially. Thus, fjk( 1) = Pjk and for n = 2, 3, ..., fjk(n) iS give n by: fjk(n) = prob{X(r)

#

k, r = 1, ..., n - 1; X(n) = k I X(0) = j }

(2-97)

The application of Eq.(2-97) to Fig.2-0 yields that:

f15(4) = prob{X(r) # 5, r = 1, ..., 3; X(4) = 5 I X(0) = I } = 1 fls(7) = prob{X(r) # 5 , r = 1, ..., 6; X(7) = 5 I X(0) = l } = 0 whereas: f13(7) = prob{X(r) f 3, r = 1, ..., 6; X(7) = 3 I X(0) = 1) = 1 The calculation of fjk(n) is detailed in [15, p.3881. We put fjk(0) = 0 and on the basis of fjk(n) we define

120 (2-98)

fjk is the probability that, starting from Sj, the system will ever pass through Sk. Thus, fjk 5 1. When fjk = 1, the {fjk(n), n = 1, 2, ...) is a proper probability distribution and we will refer to it as the first-passage distribution for s k . In particular, {fjj(n),n = 1, 2, ...} represents the distribution of the recurrence time for Sj. It should be noted that the definition in Eq.(2-96) is meaningful only when fjj = 1, that is, when a return to Sj is certain. In this case pj 5 00 is the mean recurrence time for Sj. Considering again the case of an open loop between points 1 and 3, states S2 and S3, in Fig.2-0, and applying Eq.(2-98) for a fluid element initially at point 1, yields: f12 = f12( 1) + f12(2) + f12(3) + ... = 1 4- 0 + 0 + ... = 1 f13 = f13( 1) + f13(2) + f13(3) + ... = 0 + 1 + 0 + ... = 1 i.e., the probability of the water element to pass S2, after one step and S3 after two steps is 100%. Transient or non-recurrent state. State j is said to be trunsient if the conditional probability of occupying (or returning to) Sj, given that the system initially occupies Sj, is less than one. Thus, the eventual return to the state is uncertain. According to [15, pp.3891, Sj is transient if, and only if

n= 1

(2-99)

In this case:

n= 1

(2-100)

for all j in the state space. On the basis of the quantity defined in Eq.(2-95), Sj is transient if:

121 fjj < 1

(2- 101)

For the open loop between states S1 and S3 corresponding to points 1 and 3 in Fig.2-0, where a certain amount of water is poured at S1, Eqs.(2-99) and (2100) yield

i.e., S2 and S3 are transient states. By applying Eqs.(2-95), it is obtained that f22 (= 0) c 1 and f33 (= 0) c 1 results which are in agreement with Eq.(2-101) for characterizing transient states. Recurrent or persistent state. State j is said to be recurrent if the conditional probability of occupying Sj or returning to it, given that the system initially occupied Sj or started in it, is one. Thus, the eventual return to the state is certain. Taking into account Eq.(2-95),it follows that a state Sj is recurrent if (2-102)

fjj= 1

Considering Eq.(2-96), the mean recurrence time pj = 00 and Sj is called nullrecurrent state or null state. If Fj is finite, Sj is defined as positive-recurrent. The application of Eq.(2-102) can be demonstrated in the following way on the basis of Fig.2-0, now for a closed water loop in its perpetual motion uphill. Considering a fluid element, and applying Eq.(2-95), yields:

f22 = f22( 1) + f22(2)

and similarly:

+ ... + f22(5) + ... = 0 + 0 + ... + 1 + ... = 1

122 f33 = f44 = f55 = 1

Thus, on the basis of Eq.(2-102), the states Sj (points i, i = 1, 2, recurrent states. From Eq.(2-96) it follows that:

..., 5) are

p1 = p2 = ... = p5 = 5 i.e., each state is reoccupied after 5 steps, where the states are also positiverecurrent considering the definition following Eq.(2- 102). Periodic state. Suppose that a chain starts in state Sj. Subsequent occupations of Sj can only occur at steps (times) lv, 2v, 3v, 4v ... where v is an integer.

Ifv>1

(2-103)

and the chain is fj ite, Sj is periodic. The peri d of Sj is the greatest common divisor of the set l v , 2v, 3v, 4v ... for which pjj(n) > 0 where n is an integral multiple of V. In the absence of the latter, pjj(n) = 0. If v = l

(2-104)

and the chain is finite, Sj is called an aperiodic Markov chain. The above concepts are demonstrated on the basis of Fig.2-0 by considering point 2, state S 2 . Occupation of this state occurs at steps 5, 10, 15, 20, ... , hence, lv = 5, 2v = 10, 3v = 15, 4v = 20, ..., yielding that v = 5. From Eq.(2-103) S 2 is periodic with a period of 5. It should be noted that the same results applies also for the other states S1, S3, S4 and S5. An additional example is 2.9 for which the transition probability reads:

s1

0

1

s2

1

0

P=

123 From Eq.(2-33a), it follows that:

s.

s,

Hence, lv = 2,2v = 4,3v = 6,4v = 8,5v = 10, thus v = 2, i.e., S1 and S2 have a period of 2. Other examples showing periodic behavior are: Example 2.11 with a 7x7 matrix showing a period of v = 7 in Eq.(2-39a); Example 2.14 with a 6x6 matrix and a period of v = 2 for states S2 and S3 in Eq.(2-42a) where the probability of the other states vanishes; Example 2.16 with a 12x12 matrix and a period of v = 2 for all states demonstrated in Fig.2-9; Example 2.18 with a 9x9 matrix and a period of v = 2 for all states depicted in Fig.2-12; Example 2.21 with a 9x9 matrix and a period of v = 2 shown in Fig.2-15; Example 2.27 with a 4x4 matrix and a period of v = 2 for all states demonstrated in Fig.2-23; Example 2.35 with a 4x4 matrix and a period of v = 2 for all states depicted in Figs.2-31, 32; Example 2.45 case 1 with a 8x8 matrix and a period of v = 4 as demonstrated in Fig.2-42. The following may be shown: a) A Markov chain is aperiodic if for a state Sj there exist pjj = pjj( 1). Thus, if any of the diagonal elements in P is non zero, the chain is aperiodic. b) A Markov chain is aperiodic if there exists an integer n such that pjk(n) > 0 for all j and k. The following examples demonstrate the above behaviors: If

P =

s 1 0 1 0 112 112 s2 0 s

3

1

0

0

the chain is aperiodic, since P22 = 112 > 0.

124 If

s2

0 112 112 112 0 112

s3

112

s1

P =

112 0

the chain is aperiodic, since

P2 = s2

112 114 114 114 112 114

s3

114 114 112

S1

and pjk(2) > 0 for all j and k. Other examples exhibiting aperiodic behavior may be found among examples 2.8-2.46.

Ergodic state. A finite Markov chain is ergodic if there exist probabilities nk such that [6, p.41; 4, p.1011: lim pjk(n) = nk

for all j and k

(2- 105)

n+= These limiting probabilities nk are the probabilities of being in a state after equilibrium has been achieved. As can be seen from Eq.(2-105), the are independent of the initial state j, i.e., they are without memory to the past history. Similarly, from Eq.(2-25), we may write that:

where .n is the stationary distribution of the limiting state vector. Thus:

nP = n

(2- 105a)

Hence, if the system starts with the distribution n over states, the distribution over states for all subsequent times is 71. This is the defining property of a stationary

125 distribution. Ergodic systems retain the property of having unique equilibrium distribution and these are also unique stationary distributions. The limiting probabilities nk may be found by solving the following system

of equations: Z -

for k = 1, 2, ..., Z

ICk = x n j p j k j= 1

(2- 106)

subject to the conditions: n k 2 0 for all k Z

&k=

(2-107) 1

k= 1

As indicated before, t,.e probability distribution (nk defined in Eqs.(2-106)

and (2-107) is called a stationary distribution. If a Markov chain is ergodic, it can be shown [ 17, pp.247-2551 that it possesses a unique stationary distribution; that is, there exist nk that satisfy Eqs.(2-105), (2-106) and (2-107). There are Markov chains, however, that possess distributions that satisfy Eqs.(2-106) and (2-107), i.e., they have stationary distributions, which are not ergodic. For example, if the probability transition matrix is given by:

P=

s1

s2

S1

0

1

s2

1

0

then: Pll(n>= {

1 ifniseven 0

ifnisodd

126

so the chain is not ergodic. However, one may solve Eqs.(2-106) and (2-107) to obtain the stationary probabilities nl and 7c2 = 1/2. Recall that this P is the transition matrix for an irreducible periodic Markov chain. Some sufficient conditions for a finite Markov chain to be ergodic are based on the following theorems, given without proof [17, pp.247-2551. The first one states that: Afinite irreducible aperiodic Markov chain is ergodic. Let:

P =

S1 S2 s3

114 1/4 112 0 2/3 1/3 314 114 0

This chain is irreducible (all pairs of states communicate) since pjk(2) > 0 for all j, k. It is aperiodic since pi1 = 1/4 > 0. Hence by above theorem the chain is ergodic. To find the limiting probabilities, solve Eqs.(2-106) for Z = 3 to obtain the following equations: 7c1 = (1/4)nl + (3/4)7c3; 7c2 = (1/4)nl + (2/3)7c2 + (1/4)?~3; 7c3 = (1/2)7c1 + (1/3)7c2 The solution of these equations is 7c1 = 2/7, 7c2 = 3/7, 7c3 = 2/7. Thus, the asymptotic probability of being in state S1 is 2/7, in S2 is 3/7 and in S3 is 2/7. The following Table summarizes part of the classification of the above states: Table 2-2. Classification of states (4, p.93) Type of state Periodic Aperiodic Recurrent Transient Positive-recurrent Null-recurrent Ergodic

Definition of state (assuming it is initially occupied) Return to state is possible only at times lv, 2v, 3v, ..., where the period v > 1 Not periodic. Essentially it has a period of v = 1 Eventual return to state is certain Eventual return to state is uncertain Recurrent, finite mean recurrence time Recurrent, infinite mean recurrence time, pj = 00 Aperiodic, positive-recurrent, pj < 00

127

Doubly stochastic matrix. A transition probability matrix is said to be doubly stochastic if each column sums to 1, that is, if

(2-108)

pjk = 1 for each k j=l

On the basis of the following theorem, i.e., if the transition matrix P for u finite irreducible aperiodic Murkov chain with Z states is doubly stochastic, then the stationary probabilities are given by

it follows that the transition matrix s4

1/2 1/8

P= S3

I

0

114

1/2

1/8 1I4

is irreducible, aperiodic, and doubly stochastic. Hence nk = 1/4 fork = 1,2, 3,4.

Definition. Closed sets of states. A non empty set C of states is called a closed set if each state in C communicates only with other states in C. In other words, no state outside C can be reached from any state in C. A single state Sk forming a closed set is an absorbing state. Once a closed set is entered it is never vacated. If j belongs to a closed set C then pjk = 0 for all k outside C. Hence, if all rows and columns of P corresponding to states outside C are deleted from P, we are still left with a stochastic matrix obeying Eqs.(2-17) and (2-18). A Markov chain is irreducible if there exists no closed set other than the set of all states. According to [17, p.2101, a closed communicating class C of states essentially constitutes a Markov chain which can be extracted and studied independently. If one writes the transition probability matrix P of a Markov chain so that the states in C are written first, and P can be written as:

128

P= where

pc

0

*

*

(2- 109)

* denotes a possibly non-zero matrix entry and pc is the sub matrix of P

giving the transition probabilities for the states in C, then:

(2- 109a)

Thus, if a Markov chain consists of one or more closed sets, then these sets are sub-Markov chains and'can be studied independently. The following example demonstrates the above ideas, i.e.: s2 O

s3 O

s4 a

s5

s6

s7

s8

s9

O

O

O

O

O

p

O 0 1 0 0

y 0 0 0 1

6 0 0 0 0

0 0 0 0 0

& 0 0 1 0

0 0 0 0 0

0 0 0 0 0

0 1 0 0 0

h 0 0 0 0

0

0

s1

0

0

0

0

0

~

T

0

1

0

0

0

0

0

0

o

v

o

o

o

o

p

(2-110) 0

In order to find the closed sets, it suffices to know which pjk vanishes and which are positive. In the fifth row in matrix (2-110) p55 = 1, thus S5 is absorbing. S3 and Sg form a closed set since p3g = p83 = 1. From S1, passages are possible into S4 and Sg, and from there only to S1, S4,Sg. Accordingly the latter states form another closed set. The appearance of the matrix and the determination of the closed sets can be simplified by renumbering the states in the order S5 S3 Sg S1 S4 Sg S 2 s6 S7 so that the modified matrix reads:

129

s5

O

O

O

O

0

0

0

0

o

1

1

0

o

0

0

0

p

o

o

o

0

0

0

0

(2-11Oa)

o

o

v

0 0

~

o

0 0

0 0

0 0

0

0 0

0 0

0 0

0

s 1 0 (p)"= S4

0

0 0

sg

0

1

s2

E

6

s6

0 0

0

s7

1

o

"

s8

a

0

0

0

0

0

0 0

0

0

a P " 0 1 0 0 0 p v 0 y h

0 0

0 0

0 0

0 0

0

0

0

1

0

0

0

g

z

(2-11Ob)

Examples To conclude section 2.1-5 on classification of states, we consider the following matrix corresponding to five states which belongs to the random walk model considered in examples 2.17-2.22, 2.41 and 2.42.

130

S1

P =

s2 s3 s4 s5

By changing the parameters p, q and r, many behaviors are revealed depicted in Fig.2-46. It should be noted that examination of the various behaviors has been done with respect to the state vector S(n) rather than exploration of the quantities pij(n) referred to in part of the above definitions. Eqs.(2-23) and (2-25) were used in the calculation of S(n) and an initial state vector S(0) = [0, 0, 1, 0, 01 was assumed, i.e., s3(O) = 1. Case a in Fig.2-46, for which q = 0, p = 1 and r = 0, demonstrates that S 5 is an absorbing state which is occupied after two steps. Thus, the probability of occupying the other states vanishes. S1, which is also an absorbing state, can never be occupied unless it is initially occupied. Case b where q = 1, p = 0 and r = 0, has a reflecting barrier at the origin. Fig.2-46 demonstrates a periodic behavior of states S 1 and S2. For the other states sl(n) = s2(n) = 0, n = 1, 2, ...; s3(n > 1) = 0. The period equals v = 2.

> q (q = 0.3, p = 0.7 and r = 0) reveals a transient behavior of the states as well as their being ergodic, i.e., independent of S(0). The Case c for which p

stationary distribution reads S(25) = [0.045, 0.045, 0.104, 0.242, 0.5641. Case d , also for p > q (q = 0.1, p = 0.7 and r = 0.2), reveals a similar behavior as in the previous case. However, the approach towards equilibrium is less oscillatory by comparison to case c due to the damping parameter r > 0, i.e., the probability of remaining in the state. The limiting distribution reads S(24) = [0.001, 0.003, 0.018, 0.122, 0.8561. Case e, for p = q = 0.5 is transient with damped oscillations towards an equilibrium of S(32) = [0.2, 0.2, 0.2, 0.2, 0.21, i.e., the occupation probability of all states is the same. The states are also ergodic.

131

Cusef with q = 0.2, p = 0.2 and r = 0.6 is transient, free of oscillations as well as ergodic. The equilibrium state vector reads S(46) = [0.058,0.236,0.236, 0.235, 0.2351. Cases g and h with q = 0.8, p = 0.2, r = 0 and q = 0.4, p = 0.2, r = 0.4, respectively, behave similar to cases e and f, i.e., they are transient and ergodic.

Note that p < q. The ultimate distributions are: S(65) = [0.430, 0.430, 0.107, 0.027, 0.0061 for case g and S(20) = C0.210, 0.420, 0.211, 0.106, 0.0531 for case h. q=l,p=O,r=O

0.2

-

f I

01 0

a)

1

2

n

3

4

0

5

a = 0.3.D = 0.7.r = 0

1

b)

2

3 4 n q = 0.1, p = 0.7,r = 0.2

1

5

0.8 h

C

0.6

I

i

I.

6- 0.4

v

0.2 0

0

5

I

1

10

15

0

5

10

15

132 a = 0.2.P = 0.2, r = 0.6

q = 0 . 5 , p = 0 . 5 ,r = O

1

I

' l~

I

S p )

0.8 1

I

n

a = 0.8, D = 0.2,r = 0

9)

0

5

n

10

15

q = 0.4, p = 0.2,r = 0.4 I

I

n

Fig.2-46. Demonstration of states behavior for S(0) = [0, 0, 1, 0, 01

2.2 MARKOV CHAINS DISCRETE IN SPACE AND CONTINUOUS IN TIME 2.2-1 Introduction In the preceding chapter Markov chains has been dealt with as processes discrete in space and time. These processes involve countably many states S1, S2, ... and depend on a discrete time parameter, that is, changes occur at fixed steps n =o, 1 , 2 , .... Although this book is mainly concerned with the above mode of Markov chains, a rather concise presentation will be given in the following on Markov chains discrete in state space and continuous in time [2, p.57; 4,p.146; 5, p.102; 15, p.4441. This is because many processes are associated with this mode, such as telephone calls, radioactive disintegration and chromosome breakages where changes, discrete in nature, may occur at any time. In other words, we shall be concerned with stochastic processes involving only countably many states but

133 depending on a continuous time parameter. Such processes are also referred to as discontinuous processes. Unfortunately, application of the basic models presented in 2.2-3 to chemical reactions, behavior of chemical reactors with respect to RTD, as well as to chemical processes, is very limited or even inapplicable. While the distinction between discrete time and continuous time is mathematically clear-cut, we may in applied work use discrete time approximation to a continuous time phenomena and vice versa. However, discrete time models are usually easier for numerical analysis, whereas simple analytical solutions are more likely to emerge in continuous time. In the following, we derive the Kolmogorov differential equation on the basis of a simple model and report its various versions. In principle, this equation gives the rate at which a certain state is occupied by the system at a certain time. This equation is of a fundamental importance to obtain models discrete in space and continuous in time. The models, later discussed, are: the Poisson Process, the Pure Birth Process, the Polya Process, the Simple Death Process and the Birthand-Death Process. In section 2.1-3 this equation, i.e. Eq.2-30, has been derived for Markov chains discrete in space and time.

2.2-2 The Kolmogorov differential equation The following model represents a simplified version of the Kolmogorov equation. A big state of a total population No consists of many cities. The cities are arranged in two circles, internal and external. The external circle consists of J cities, j = 1, 2, ..., J, where each city is a state designated as Sj = j = jth city in the external circle. The internal circle consists of K cities, k = 1, 2, ..., K. A state in this circle is designated by Sk = k = kth city in the internal circle. Inhabitants, designated as system, are moving from cities in the external circle to cities in the internal circle, i.e., along the trajectory Sj+ Sk+ out of kth city, where from each external city, inhabitants can move only to each internal city. In other words inhabitants can not move from one external city to another external city. The movement of inhabitants through an internal city was convincingly demonstrated by Magritte [20] in his painting Golconda depicted in Fig.2-47.

134

Fig.2-47. Motion of inhabitants through an internal city k according to Magritte ("Golconda ", 1953, 0 R.Magritte, 1998 c/o Beeldrecht Amstelveen) The figure has slightly been modified by adding a circle on it, demonstrating a city, as well as arrows indicating 'in' and 'out' from city k. The painting i s one of Magritte's most scrupulous displays of reordering where he allows his figures no occupation, no purpose, and despite of their rigid formation, no fixed point. They could be moving up or down or not moving at all. In setting up the model, we designate by Nj(t) the number of inhabitants occupying sfufej(an external city j) at time t. Similarly, Nk(t) corresponds to the number of inhabitants occupying state k (an internal city k) at time t. The change in the number of inhabitants in state Sk during time interval At is given by:

135 mk(t) = number of inhabitants entering the city in the time interval (t, t + At) - number of inhabitants leaving the city in the interval (t, t + At) (2-111) The quantities on the right-hand side may be expressed by the following probabilities [2, p.591: qj(t) - a rate (Mime) or an intensity function indicating the rate at which inhabitants leave state Sj (external city j). This function has the following interpretation, i.e., qj(t)At - stands for the probability of the system (inhabitants)to leave state Sj in the time interval (t, t + At), This quantity is the probability for a change to occur at Sj not indicating the 'direction' of the change. The following quantity gives the 'direction', i.e., the probability that inhabitants leaving Sj will occupy exactly Sk, namely, they will occupy city k in the internal circle. Thus, Qjk(t) - gives the direction of the change at the interval (t, t + At), i.e., the transition probability of the inhabitants from Sj to occupy Sk at time t [ 15, p.4731. Note that 0 IQjk(t), Qb(t) I1. Other quantities pertaining to the cities in the internal circle are: qk(t) - the rate (Mime) at which inhabitants (system)leave state Sk (internal city k). Thus, qk(t)At - is the probability of the inhabitants to leave Sk at the interval (t, t t At). Accounting for the above quantities and applying Eq.(2-111) yields: 1

Nj(t)qj(t)AtQjk(t)- Nk(t)qk(t)At

A&(t) =

(2-11 la)

j= 1

Defining the following state probabilities, i.e., Pk(t) - probability of the system to occupy state Sk at time t, Pj(t) - probability of the system to occupy Sj at time t and assuming that the total number of inhabitants No in the big state remains constant, gives Nj(t) = Pj(t)N,;

Nk(t) = Pk(t)No

(2-112)

Substitution of Eq.(2-112) into Eq.(2-11 la) and approaching At to zero, yields a simplified version of the forward KoZmogorov differential equation for the transition Sj + s k + out of kth city. This equation is continuous in time and discrete in space; it reads:

(2-1 13) j=l

It should be noted that Eq.(2-113) may be looked upon as an "unsteady state probability balance" on city k in the internal circle; J is the total number of cities in the external circle and K in the internal circle. In other words, the equation gives the rate of change of the probability of occupying state k at a certain time. Note that the origin of the equation was an unsteady state "mass balance" on the transition of inhabitants through city k. Another version of the Kolmogorov equation is obtained by considering the transition Sj + Si + s k . Eq.(2-30) gives the Chapman-Kolmogorov equation, discrete in time and space. For a continuous time and discrete space, this equation is generally written as [2, p.611: (2-1 14) i

and is valid for z c s c t. This relation expresses the fact that a transition from state Sj at time 7 to Sk at time t occurs via some state Si at the intermediate time s, and for a Markov processes the probability Pik(S,t) of the transition from Si to Sk is independent of the previous state Sj. Pjk(7,t) is the transition probability of a system to occupy state k at time t subjected to the fact that the system occupied state j at time T. Another way of expressing the above is that we shall write Pjk(7,t) for the conditional probability of finding the system at time t in state Sk, given that at a previous time z the system occupied Sj. As indicated above, the symbol Pjk(7,t) is meaningless unless 7 5 t. Differentiation of Eq.(2-114) [ 15, p.4721 yields the forward Kolmogorov differential equation for the transition Sj + Si + s k . It is continuous in time and discrete in space and reads:

137 (2-115) Here j and z are fixed so that we have, despite of the formal appearance of the partial derivative, a system of ordinary differential equations for the function Pjk(7,t). The parametersj and z appear only in the initial conditions, i.e.: 1 forj=k (2- 115a)

pjk(z,z) = 0 otherwise

Thus, the system of ordinary differential equations reads: (2-1 16)

The above equation becomes identical to Eq.(2-113) by replacing the notation of the state transitions, i.e. i with j. As Pjk(7,t) stands in Eq.(2-113, it is not time-homogeneous, since it depends explicitly on t and 7. However, if a restriction is made to the timehomogeneous or stationary case, then:

i.e., the transition probability Pjk(7,t) depends only on the duration of the time interval (t - z) and not on the initial time z. Hence, in the time-homogeneous case, Eq.(2-114) reads: (2-1 18) i

The significance of Eq.(2-118) is as follows. Given that at time = 0 a system is at state Sj = j. If we ask about its state at time (T + t), then the probability that the system will occupy state k at time (z + t), i.e., Pjk(T + t), may be computed from the above summation based on the fact that the system will occupy an intermediate

138

state Si = i before a transition to occupy Sk takes place. Pji(Z) is the probability of

the system to occupy Si at time Z and Pik(t) is the probability of occupying Sk at time (z + t).

2.2-3 Some discontinuous models The following models for the transition Sj

+ S , are considered because they

find applications in many corners of our life, later elaborated. It should be noted that the above transition indicates the basic property of Markov chains, i.e. that occupation of Sk is conditioned on a prior occupation of Sj and that the past history is irrelevant. Unfortunately, the application of the models in Chemical Engineering is very limited, however, some applications are mentioned. For each case we derive the difference equation describing the probability law of the process and report the final solution of the differential equation accompanied by its graphical presentation. In some cases we derive the differential equation from the Kolmogorov equation.

The Poisson Process. This process is the simplest of the discontinuous processes which occupies a unique position in the theory of probability and has found many applications in biology, physics, and telephone engineering. In physics, the random emission of electrons from the filament of a vacuum tube, or from a photosensitive substance under influence of light, and the spontaneous decomposition of radioactive atomic nuclei, lead to phenomena obeying the Poisson probability law. This law arises frequently in the field of operations research and management science, since demands for service, whether upon the cashiers or salesmen of a department store, the stock clerk of a factory, the runways of an airport, the cargo-handling facilities of a port, the maintenance man of a machine shop, and the trunk lines of a telephone exchange, and also the rate at which service is rendered, often lead to random phenomena either exactly or approximately obeying the Poisson probability law. Part of the above examples are elaborated in the light of the governing equation describing the law. Finally, it should also be noted that the Poisson model obeys the Markov chains fundamental property, i.e., that future development depends only on the present state, but not on the past history of the process or the manner in which the present state has been reached.

139

In deriving the Poisson model, a rather general description arising from the above applications can be established. We assume that events of a given kind occur randomly in the course of time. For example, we can think on "service calls" as (requests for service) arriving randomly at some "server" (service facility) as events, like inquiries at an information desk, arriving of motorists at a gas station, telephone calls at an exchange, or emission of electrons from the filament of a vacuum tube. Let X(t) be a random variable designating the number of events occurring during the time interval (0, t). An interesting question regarding to the random variable X(t) may be presented as what is the probability that the number of events occurring during the time interval (0, t) = t will be equal to some prescribed value x. Mathematically it is presented by: Px(t) = prob{X(t) = x); x =. 0, 1, 2,

...

(2-119)

where the exact relationship is derived in the following. The above equation indicates also the realization of the random variable X(t) by acquiring the value x. An expression for Px(t) for the transition Sj -+ S,, where the two discrete states are Sj = x - 1 and Sk = x, may be derived bearing in mind the following assumptions: a) The events are independent of one another; more exactly, the random variables X(tl), X(t2), ... are independent of each other if the intervals tl, t2, ... are non-overlapping. In other words, if for example t l = t2 then X(t1) = X(t2). X(ti) designates the number of events occurring during the time interval (0, ti). b) The flow of events is stationary, i.e., the distribution of the random variable X(t) depends only on the length of the interval t and not on the time of its occurrence. c) The probability of a change in the time interval (t, t + At), or of a transition from Sj to Sk in the time interval (t, t + At), or the probability that at least one event occurs in a small time interval At, is given by: prob{X(t, t + At) = 1) = hAt + o(At) = pjk

(2- 120)

140

where h(events/time) is a positive parameter characterizing the rate (or density or intensity) of occurrence of the events. A possible interpretation of the above definition, later elaborated, makes use of the conception birth, i.e., the occurrence of an event in the time interval (t, t + At) may be looked upon as a single birth. Thus, the parameter h is the birth rate. If we approach At to zero, no change occurs. Here o(At) is an infinitesimal of a higher order than At, i.e.:

At + 0

o(At) emerges from the expansion Pjk(At) = ho + hAt + h2At2 +

The second

term on the right-hand side is responsible for the probability of the occurrence of at least one event where the third term accounts for more than one event to occur during At, i.e., "twin birth". should be omitted in the expansion noting that the probability of the number of events per unit time, i.e., pjk(At)/At approaches infinity as At is approaching to zero. d) The probability of no change in (t, t + At), or of remaining in Sj, or that no events occur during At, is given by: prob{X(t, t + At) = 0 ) = 1 - Ut + O(At) = pjj

(2-121)

e) The probability of more than one change in the interval (t, t + At) is o(At), thus it is negligible as At is approaching zero. In other words, this assumption excludes the possibility of a "twin birth". It should be noted that the above probabilities are independent of the state of the system. Having established the one-step transition probabilities pjk and pjj, the differential equation for Px(t) will be derived by setting up an appropriate expression for Px(t + At). If the system occupies state Sj = x - 1 at time t, then the probability of occupying Sk = x, i.e., making the transition Sj to sk, is equal to the product &AtPx-l(t). If the system already occupies state Sk = x at time t, then the probability of remaining in this state at (t, t + At) is equal to (1 - U t )Px(t). Thus, since the above transition probabilities are independent of each other, and following Eq.(2-3), we may write that:

141 Px(t + At) = (1 - hAt)Px(t) + k i t Px-1(t) + o(At)

(2-122)

If we transpose the term Px(t) on the right-hand side, divide by At, and approach At to zero, we obtain the following differential equation: (2- 123) When x = 0, Px-l(t) = 0; hence: (2-124) Eqs.(2-122) and (2-123) characterize the Poisson process and are to be solved with the initial conditions: Po(0) = 1 Px(0) = 0

for x = 1, 2, ...

(2-125)

Having obtained Po(t), and using Eq.(2-123), it is possible to obtain by induction [2, p.741 that: Px(t) = -e-hf

X.

; x = 0, 1, 2,

...

(2-126)

which is called the Poisson distribution. Some interpretations of Px(t) are: a) It gives the probability that at time interval (0, t) = t (20) the system occupies state x (x = 0, 1, 2, ...). b) It is also the probability of exactly x changes or events occurring during the time interval of length t. c) Px(t) indicates the probability to remain at a prescribed state x during the time interval t. Let us analyze in more details some examples, assuming they obey the Poisson model: 1) Requests for service arriving randomly at some single service facility, arrive at a mean rate h calls per time. If the calls have not yet been answered by the

142 service facility, Eq.(2- 126) gives an answer to the following possible questions which might be of some interest: a) What is the probability that during a prescribed time interval (0, t), the number of the arriving calls will be equal to a certain value x? or, b) How long will it take until a call is answered if the probability of remaining at a certain number x of "waiting calls" is prescribed? 2) A new cemetery of a known size of x graves has been opened at some town. The mean death rate h in this town is known and the following questions may arise: a) What is the probability that during the next ten years, i.e., time interval (0, t) = 10, the cemetery will be full? b) If the probability of occupying 50% of the cemetery is known, how long will it take to reach this state? c) If the probability to reach a certain time in the future is known, Eq.(2-126) gives an answer to the occupation state of the cemetery. 3) It has been announced that some urban area became polluted and consequently men might become infertile. If the process of becoming infertile is random, thus, Poissonic, a mean parameter h may be defined as the rate of infertility per day (say!), and Eq.(2- 126) becomes applicable yielding the following information: a) What is the probability that all men in town become infertile during a week, a month, a year etc.? b) If we prescribe the probability of remaining at some state x (= number of men which be came infertile), then we may calculate the time interval of remaining in this state. Certainly, the most important data needed is

4)Electrons are emitted randomly from the filament of a vacuum tube. If the emission rate, h electrodunit time, is known, then Eq.(2-126) gives the probability that within the time interval (0, t) the total number of electrons emitted is a prescribed value x. 5 ) A mailbox, designated as system, has a finite capacity for letters. The number of letters is the state of the system x = 0, 1, .... The rate of filling of the mailbox is h letters per day and Eq.(2-126) gives an answer to several questions of the type demonstrated above. Some characteristic properties of the Poisson equation are:

143 a) Fig.248 demonstrates the Px(ht) - At, P&t) - x relationships computed from Eq.(2-126). 1

1

1

1

I

I

I

l

0.6-

l

-

1

-

o,8

-

1

1

1

1

ht = 0, x = 0

1

-

0.6 0.4

- ht = 1

1

1

1

1

-

-

-

Fig.2-48. The Poisson model Characteristic behavior observed is: Px=o(O)= l,Pxso(0)=O and Px , o ( ~) = O

(2-126a)

m

(2- 127) x-0

c) The mean number ofevents, m(t), occurring in the time interval of length t is given by:

(2-128)

It should be noted that for the Poisson distribution, the variance is equal to the mean and as ht + 00 the distribution tends to normality.

d) The meun time until the Occurrence of the first event, < t >, reads: < t > = llh where h is the mean rate of Occurence of the events.

(2-129)

144 It is interesting to demonstrate the derivation of Eq.(2-123) from the Kolmogorov Eq.(2-113). In this equation J = K = 1 as well as: k = Sk = s k = l = X = x-1 j = Sj = qj(t) = qj=l(t) = h, Pj(t) = Px-l(t) and Qjk(t) = 1 qk(t) = qk=l(t) = h and Pk(t) = Px(t) Substitution of the above quantities into Eq.(2-113) gives Eq.(2-123). Few applications of the Poisson Distribution in Chemical Engineering are: For x = 0, Eq.(2-126) reduces to: Po(t) = e-ht

(2- 130)

where Po(t) stands for the probability of remaining at the state x = 0, i.e., no change (or event) is expected to occur in the system during the time interval of length t. The above equation is applicable in the following cases: a) 1st order chemical reaction, A + B, for which -dCA/dt = kCA occurring in a batch reactor. The solution for the concentration distribution reads: (2-131) where CA(t) and CAOare, respectively, the concentration of A at time t and t = 0. k is the reaction rate constant. Similarly, NA(t) and NAOare the number of moles of A at time t and t = 0. The system in this case is a fluid element containing the species A and B. The states of the system in Eq.(2-126) are: x = 0 designating species A and x = 1 for species B. As observed, Eqs.(2-130) and (2-131) are similar, i.e.: Po(t) = e-kt

(2-132)

therefore, the ratio NA(t)/NAO = Po(t) may be looked upon as the probability that the system remains at x = 0 until time t. If t = 0, it is obtained that Po(0) = 1, indicating that the probability of remaining at the initial state at t = 0 is 100%.

145 If k = 0, Eq.(2-132) yields Po(t) = 1, namely, in the absence of a chemical reaction, the probability that the system remains in its initial state x = 0 for all t is 100%. If k + 00, it is obtained from Eq.(2-132) that Po(t) = 0, namely, the probability of remaining at the initial state for all t is zero. In other words, in the presence of an intensive reaction, the initial number of moles will immediately diminish, i.e., the system will immediately occupy the state x = 1, i.e. in the state of species B. If Po(t) = 0.5, this means that the probability of remaining in x = 0 until time t is 50%. In other words, 50% of A will decompose to B because the probability of remaining in the initial state until time t is 50%. b) An additional example is concerned with the introduction of a pulse of an initial concentration CAOinto a single continuous perfectly-mixed reactor, the system. The states of the system are x = 0, designating the initial concentration CAO,and x = 1, the concentration of species A at time t, i.e. CA(t). The relationship CA(t) is given by: (2-133) where tm is the mean residence time of the fluid in the reactor. Thus, If tm + 00, Eq.(2-133) yields Po(t) = 1. This means that the probability of the pulse to remain at the initial state x = 0 along the time interval t is 100%because the residence time of the fluid in the reactor is infinity. If tm + 0, Po(t) = 0. This means that the probability of remaining at x = 0 at t > tm is zero because of the extremely short mean residence time of the fluid.

c) The following example is concerned with a closed recirculation system consisting of N perfectly-mixed reactors of identical volumes. If we introduce a pulse input into the first reactor, then the output signal at the Nth reactors is given by [21, p.2941:

(2-134 )

146 where C is a dimensionless concentration at the Nth reactor. tm is the mean residence time of the fluid in a reactor which is equal for all reactors. n is the number of passes of the fluid through the system. For short times where only one term in the series expansion is taken into account, n = 1, Eq.(2-134) is reduced to:

(2-134a) x = N - 1, h = l/tm where the above equation is exactly the Poisson distribution for x = N - 1. The description about the behavior of the pulse in the previous example, is also applicable here. d) The final example is related to a step change in concentration at the inlet to a single continuous perfectly-mixed reactor from CA,inlet = 0 to CA,inlet = CAO. The response at the exit of the reactor, designated as system, is given by: (2- 135) where Po(t) is given by Eq.(2-133). CA(t) is the concentration of A inside the reactor. The states of the system are: x = 0, i.e. CA(O) = 0 and x = 1 for CA(t) > 0. The above result can be interpreted as the probability of not remaining at the initial state along the time interval t. For example: If tm + 0, Po(t) = 0 and Eq.(2-135) gives CA(t)/CAO = 1, indicating that the probability of not remaining at the initial state x = 0 along t is 100%. Indeed, this is plausible because the residence time of the fluid in the reactor is extremely short and the reactor will acquire instantaneously the final concentration CAO,i.e, the inlet concentration. If tm+ 00, Eq.(2-135) gives CA(t)/CAO = 0, i.e, the probability of remaining at the initial state x = 0 along t is 100% because of the extremely long residence time. The Pure Birth Process. The simplest generalization of the Poisson process is obtained by permitting the transition probabilities to depend on the actual state of the system. Thus, if at time t the system occupies state Sj = x (x = 0, 1, 2,

147 ...), and at time (t, t

+ At) the system occupies state Sk = x + 1 (a single birth), then

the following probabilities may be defined: a) The probability of the transition Sj to Sk is: Pjk = hxAt

+ o(At)

(2- 136)

b) The probability of no change reads: pjj = 1 - &At

+ O(At)

(2-137)

where & is the mean occurrence rate of the events which is a function of the actual state x. The dependence of h on x avoids, in the context of birth, the phenomenon of infertility unless h, is a constant. c) The probability of a transition from x to a state different from x + 1 is o(At), i.e., twin or multiple birth is impossible. In view of the above assumptions and following Eq.(2-3), we may write that: P,(t

+ At) = (1 - hxAt)Px(t)+ h,-,P,-,(t)At + o(At)

(2-138)

The reason that x - 1 occurs in the coefficient of P,-1 (t) is that the probability of an event has to be taken conditional on X(t) = x - 1. If we transpose the term Px(t) on the right-hand side, divide by At, and approach At to zero, we obtain the following system of differential equations: (2- 139)

(2- 139a) subjected to the following initial conditions: Px(0) = 1 for X = X O P,(O) = O for X > X O

(2-139b)

148 Depending on the h, - x relationship, two cases will be considered, viz, the linear birth process and the consecutive-irreversible z-states process.

The linear birth process. This process, sometimes known as the Yule-Furry process, assumes [2, p.771 that for a constant h: h,=hx x>l,h>O (2-140) In this case Eq.(2-138) becomes: P,(t

+ At) = (1 - hxAt)P,(t) + h(x - 1)At P,-l(t) + o(At)

(2- 141)

where in terms of the conception of birth, it may be interpreted as prob{X(t + At) = x} = prob{X(t) = x and no birth occurs in (t, t + At)} + prob{X(t) = x - 1 and one birth occurs in (t, t + At)} The following definitions are applicable: Px(t) = prob{X(t ) = x}; 1 - hxAt = prob{X(t, t + At) = 0 } Px-l(t) = prob(X(t ) = x - I}; (x - 1)hAt = prob{X(t, t +At) = I } where X(t) is a random variable designating the population size at time t ; x or x - 1 designate the actual population size The corresponding physical picture which may be visualized in the light of the above, is the following one [4,p.156; 15, p.4501. Consider a population of members which can, by splitting or otherwise, give birth to new members but can not die. Assume that during any short time interval of length At, each member has probability hAt + o(At) to create a new one; the constant h births/(timexmember) determines the rate of increase of the population. If their is no interactions among the members and at time t the population size is x, then the probability that an increase takes place at some time between t and t + At equals hxAt + o(At). It is interesting at this stage to compare the "birth characteristics" of the above process with the Poisson one. In the Pure Birth Process, each member at each time interval is capable of giving birth. Also each new born member continues to give birth and this process repeats its self ad infiniturn. Thus, the birth rate is hx. In the Poisson process, once a member gave birth, he is becoming impotent but remains alive, and only the new born member is giving birth and then becoming

149

also impotent. In this case, the birth rate is h. Table 2-3 in the following is a numerical comparison between the two process for h = lbirth per unit time and an initial population size of x = 1. Table 2-3. Comparison between Pure Birth and Poisson Processes Poisson Process Pure Birth Process BR= hx At* t* B R = h At* t* X ~irthsper population births per unit time unit time at time t 1 1 / o / I 0 2 2 1 1 1 1 1 3 3 1/2 1.50 1 2 1 4 4 113 1.83 I 1 3 5 5 114 2.08 1 1 4 6 6 115 2.28 1 1 5 7 7 116 2.45 1 1 6 8 8 V7 2.59 1 1 7 9 9 118 2.72 1 1 8 10 10 119 2.83 1 I 9 7

BR=birthrate; At=-= AX

birthrate

* number of time units

't+At-'t -=(BRIt

1 (BR),

The major conclusion drawn from Table 2-3 is that the increase of the population in the Pure Birth Process is significantly greater because the members don't become infertile after their first birth giving. For example, in the Pure Birth Process the population size becomes 10 after 2.83 time units where in the Poisson Process it takes 9 time units. The results in the Table can also be demonstrated in Fig.2-49 by Escher's painting Metamorphose [10, p.3261 modified by the author of this book. On the left-hand side of the original upper picture hexagons can be seen which make one think of the cells in a honeycomb, and so in every cell there appears a bee larva. The fully grown larvae turn into bees which fly off into space. But they are not

150 vouchsafed a long life of freedom, for soon their black silhouettesjoin together to form a background for white fish seen on the right-hand side. The modified painting below the original one demonstrates the Poisson Process of rate h = 1 birth-metamorphosis per unit time corresponding to x = 1, 2, 3,4. An interesting question is concerned with the birth mechanism in the figure, which is beyond the scope of this book.

Fig.2-49. Poisson process by the modified painting "Metamorphose" (M.C.Escher "Metamorphose" 0 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved)

Returning to Eq.(2- 141), the following basic system of differential equations are obtained: dP,(t)

--dt - -hxP,(t) + h(x - 1)Px-1(t)

x21

(2-142) (2- 142a)

151 where the solution reads [2, p.781: px(t) = e-ht(1- e-ktIx-l =O

for x = 1,2, ...

(2-143)

for x = 0

If the initial population size is denoted by xo and the initial conditions are PxO(O)= 1, P,(O) = 0 for x > XO, the solution reads [ 16, p.4501: PJt) =

(x- l)! e-htx,( 1 (x - xo)!(xo - l)!

e-ht)x-xfJ

xlxo>O

(2- 143a)

The above type of process was first studied by Yule [15, p.4501 in connection with the mathematical theory of evolution. The population consists of the species within a genus, and the creation of a new element is due to mutations. The assumption that each species has the same probability of throwing out a new species neglects the difference in species sizes. Since we have also neglected the possibility that species may die out, Eq.(2-143) can be expected to give only a crude approximation for a population with initial size of xo = 1. Thus, if the mutation rate h is known, the above equation gives the probability that within the time interval (0, t) the population will remain at some prescribed state x > xo. Some characteristic properties of the above distribution are: a) Fig.2-50 demonstrates the Px(ht) - ht, P,(ht) - x relationships computed from Eq.(2-143) with ht as parameter. It is observed that for x > 1 and for a constant ht, the probability of remaining in a certain state decreases by increasing x. For x > 1, and at a constant ht, the probability of remaining in the state decreases as time increases, where in general, Eq.(2- 126a) is satisfied.

152

0

*

At

4

6

0

2

4

6

8

10

X

Fig.2-50. The Pure Birth model b) Px(t) obeys Eq.(2-127). There is, however, some problem with the summation in the equation. Eq.(2-139) indicates that the solution for Px(t) depends generally on the value of &. Hence, it is possible for a rapid increase in the A, to lead to the condition where:

i.e., is a dishonest process. However, in order to comply with Eq.(2-127) for all t, it is necessary that [2, p.811:

c$=00

(2-144)

x=o

c) The mean number ofevents, m(t), occurring in the time interval of length t, is defined by Eq.(2-128) where for Px(t)given in Eq.(2-143) it reads: m

(2-145) x=o

The only application of Eq.(2-143)in Chemical Engineering is for x = 1, i.e., that during the time interval t only one event occurs with the probability Pl(t) given by:

153

This is the same equation obtained by the Poisson model, i.e., Eq.(2-130).

The consecutive-irreversible z-state process. In this process, encountered in chemical engineering reactions, the system undergoes the following succession of transitions: (2-146) In this context, the system is defined as a fluid element containing chemical species. The above scheme may be looked upon as a birth process where the born member gives birth to only one new member. Once the new member gives birth, it becomes infertile for any reason. At a certain time interval, it may happen that all members are alive at different ages. However, at the end of the birth process, only one member remains and all previous born ones disappeared. The present birth scheme is similar to the Poisson process, however, no disappearance of members versus time occurs in the latter. In addition, the magnitude of the birth rate hi in Eq.(2146) depends on the state Si, where in the linear birth process, according to which mankind growth is conducted (in the absence of death), the birth rate is proportional to the state size x according to Eq.(2-140). In view of the above assumptions, we may conclude that Eqs.(2-138) to (2-139a) are applicable. Let us now apply the above birth model to a well-known process, i.e., a consecutive-irreversible z-stage first order chemical reaction, with a single initial substance, the "first member of the family". The various states are Si i Ai (i = 0, 1, ..., Z) where A designates concentration of a chemical species i acquiring some chemical formula; from Px(t) it follows that i = x. Considering Eq.(2-146), the system occupies the various states at different times, i.e. a fluid element contains different species along its transitions among the states. Px(t) is the probability of occupying state x, i.e., occupying the chemical state of species i = x. x = 0 is the initial chemical substance, x = 1 is the second chemical species, etc., where x = Z is the "last born member" of the family which remains alive for ever, according to Eq.(2-146). Another interpretation of Px(t) is the probability that at time interval

154 (0, t), the system will still occupy the state of chemical species x of concentration Px(t) = Ax/(t)Ao(0)where &(O) is the initial concentration of species x = 0. On the

basis of Eqs.(2-138) to (2-139a) we may write for the consecutive-irreversible reactions the following equations, designating hi = ki where the latter are the chemical reaction rate constants (Mime): (2- 147a) for species 0, (2- 147b) for species 1. (2-147~) for specie 2. .

.

.

.

.

.

.

.

.

.

a

.

I

.

(2-147d) for specie 2-1. For species z the probability or mass balance reads (2- 147e) The following are the initial conditions: P,(O) = 1 for x = 0 P,(O) = O for x > O

(2-1470

A general solution for the above set is available [22, p. 113. For Z = 2 considered in the following, it reads:

Po(t) = e-kot

(2- 148a)

155 (2- 148b)

P2(t) = 1 - Po(t) - Pl(t)

(2-148~)

where K = kl/ko. Some characteristic properties of the above distributions are demonstrated in Fig.2-51 where the relationship P,(kot ) - kot is plotted for K = 0.25. 1

0.8

P c)

x=o

2/

K = k l k =0.25 1

0

0.6

Y

0.2

0

k,t Fig.2-51. The Pure Birth model for consecutive reactions Generally, it is observed that the system, eventually, occupies state 3, i.e., only species 3 is present; the other "members of the family" have died during the time. On the basis of curves in the figure, we may ask also the following questions which are of some interest. If the occupation probability of some state is, say, 0.368, then what is this state, what is the situation of the other states and during what time interval are the above occupation probabilities valid? According to Eq.(2-148c) the summation of all probabilities at each time must be unity. Therefore, for the above probability, the following is the probability distribution of the system among the states: PO(1) = 0.368, PI(1) = 0.548 and P2( 1) = 0.084. In other words, at time kot = 1, 36.8% of the system still occupies the state x = 0 (SO = &), 54.8% underwent a transition to state x = 1 (Sl = Al) and 8.4% to state x =

156 2 ( S 2 = A2). However, at time bt = 20, the whole system, practically, occupies state x = 2 ( S 2 = A2). To conclude birth models, it is interesting to present in Fig.2-52 the painting Development ZI by Escher [ 10, p.2761 which is originally a woodcut in three colors. The painting demonstrates the development of reptiles and at first glance it seems that their number is increasing along the radius. Although their birth origin is not so clear from the figure, it was possible, by counting their number aiong a certain circumference, to find out that it contains exactly eight reptiles of the same size. This number is independent of the distance from the center. Thus, it may be concluded according to Eq.(2-142) that dP,(t)/dt = 0, namely, Px=g(t)= 1. In other words, the probability of remaining at the state x = 8 reptiles during time interval (0, tj is 100% since no birth takes place along the radius.

Fig.2-52. Escher's demonstration for reptiles' birth rate of h = 0 (M.C.Escher "Development 11" 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved)

157

The P6lya Process. In the above models the rate of occurrence of the events h was independent of time, i.e., homogeneous. In the present process, it has been assumed that the probability of a transition from Sj to Sk is time dependent. Two cases are considered in the following.

The inversely proportional time dependence. It has been assumed [2, p.821 that in the time interval (t, t + At) the transition probability is given by: prob{X(t, t + At ) = 1) = h

1 + ax

nAt + o(At) = pjk

(2- 149)

where a and h are non negative constants. The above probability designates an occurrence of one event or one occupation during the time interval At. By proceeding as before, we obtain the following differential equations: dP,(t) -= dt

1 +ax -h 1 + a h t PX(t) +

+

a(x- ‘)P

1 +aht

x-I

(t)

x = 1, 2,

...

(2-150)

(2- 150a) subjected to the following initial conditions: P,(O) = 1 for x = 0 P,(O) = O for x > O

(2- 15 Ob)

Eq.(2- 150a) yields that: po(t) = (1 + aht)-”a

(2-151)

and the solution of Eq.(2-150) reads:

(2- 152) i= 1

158 For a = 0 the equation reduces to Poisson model. For a = 1 and by introducing the new time parameter (1 + At) = exp(hz>, one obtains the Yule-Fumy process given by Eq.(2-143a). Some characteristicproperties of the above distribution are: a) Fig.2-53 demonstrates the Px(t) - ht, Px(t) - x relationships computed from Eq.(2-152) with ht as parameter and for a = 1, 10. It is observed that generally Eq.(2-126a) is satisfied. 1

1

1

1

1

1

1

1

1

ht=l-2 a =10

A

z x

Q

h 1.1.l’j 2 5 ht=O,x>

ht=l

......

I

0

2

4

X

6

10

8

Fig.2-53. The Pblya model for inversly proportional time dependence b) Px(t) obeys Eq.(2-127). c) The mean number of events, m(t), occurring in the time interval of length t, is defined by Eq.(2-128) where for Px(t) given by Eq.(2-152) (2-153)

The exponential time dependence. Another dependence on time of the transition probability from Sj to Sk, leading to a closed form solution of the differential equation, assumes that the process rate parameter h is given by:

-

h hoe-kt

(2-154)

where ho and k are adjustable parameters. Let us relate h to the following

humorous example noting that numerous cases fall in this category. A girls’

159

dormitory accommodates N girls in every academic year. The mean rate of a girl acquiring a boy-friend throughout the year varies with time according to Eq.(2-154) where t denotes the number of months lapsed from the beginning of the academic year. Assume that the same girls stay in the dormitory, there is no departure or arrival of new girls during the academic year, a girl can have only one boy-friend throughout the year, and that there are no break-ups during the year. The decrease of h with time is plausible because as time goes by, the girls become more engaged in their studies and don't have too much time for other activities. If the system is the N girls in the dormitory and x is the state of the system, i.e., the number of girls already acquired a boy-friend, where Px(t) is the probability that at time interval (0, t) x girls are already occupied, then the set of Eqs.(2-122)-(2-124) is applicable under the initial conditions Po(0) = 1, P,(O) = 0 for x = 1, 2, ..., N. The solution of Eq.(2-124) with h given by Eq.(2-154) is: p,(t) = e[h&l[exp(-kt)-ll The rest of the solution is obtained as follows [23]:

Thus, Eq.(2-123) is expressed in terms of the variable h instead oft by:

where P,&)

= 0 for x 2 1. If the following solution is assumed:

Px(h) = Po(h)F,(h) = e(h-no)kF,(h) it is possible to obtain for F,(h) that:

(2- 155)

160 where Fx(ho) = 0. Applying the solutions for Fx(h) and the above relationships, yields the following expression for Px(t) which is a modified Poisson model: P,(t)

(-OX

x = 0, 1, 2, ... (2-156)

- l]}xe[ho’kl[exp(-kt)-l];

= -{[hok][exp(-kt)

X!

If k = 0, the original Poisson model is obtained, i.e., Eq.(2-126). Some characteristic properties of the probability distributions computed from Eq.(2-156) are demonstrated in Fig.2-54. The figure on the left-hand side indicates that the distributions versus time attain a steady state depending on ho and k. For example, for x = 1 and ho/k = 1, namely, one girl has already been occupied, it is observed that the probability of remaining at this state during time interval (0, a)is 0.368 where the probability of remaining in state x = 0 is zero at t > 0. Thus, the probability of occupying the other states in the above time interval is 1 - 0.368 = 0.632 , i.e., each girl will acquire a boy-friend throughout the year with some probability. The figure on the right-hand side indicates that increasing the number x of the occupied girls, other parameters remain constant, reduces the probability of remaining in this state. A possible explanation for this behavior is that an increased number of occupied girls by a boy-friend, indicates that this is a good situation. Thus, more girls would like to acquire boy-friends which is nicely predicted by the model. l

h

E

0.6

l

l

l

l

l

l

l

2 2

x=1

-

--

kt= 0, x = O

-

l

1

k

I

I

1

1

1

kt=O,x=O

1

1

1

1

-

1

kt=l

nx

0

2

4

k-t

6

8

10 X

Fig.2-54. The P6lya model for exponential time dependence

161

The Simple Death Process. In the above birth processes, the transition of the system is always from state Sj = x - 1 to Sk = x where x > 0. This is clearly reflected in Eqs.(2-122) and (2-138, 2-141). In the death process, the transition of the system is in the opposite direction, i.e., from Sj = x + 1 to Sk = x, and the derivation of the governing equations follows the assumptions: a) At time zero the system is in a state x = XO, i.e., the random variable X(0) = xo 2 1.

b) If at time t the system is in state x (x = 1, 2, ...), then the probability of the transition x j x - 1 at the interval (t, t + At) is: Pjk = pxAt

+ O(At)

(2- 156)

c) The probability of remaining in state x reads: pjj = 1 - pxAt + O(At)

(2- 157)

where CL, is the mean occurrence rate ofthe events which is a function of the actual state. d) The probability of the transition x j x - i where i > 1 is o(At), i.e., twin or

multiple death is impossible. In view of the above assumptions and following Eq.(2-3), we may write that: (2-158) The reason that x + 1 occurs in the coefficient of Px+l(t)is that the probability of an event has to be taken conditional on X(t) = x + 1, The differential equation which follows reads: (2-159) where pxis the death rate which is a function of the state the system occupies. If px= p = constant, we have a constant rate death process. The case of p = 1 death per unit time may be demonstrated in Fig.2-49 as follows. If one is observing the figure, while beginning from the middle and continuing to left or to

162 the right, he can see that the number of fish or beens is reducing up to one, due to their death for some reason. If it is assumed that the death rate follows a linear relationship, i.e., px= px where p > 0, we obtain the so-called simple death process or linear death process satisfying: (2-160) subjected to the following initial conditions: P,(O) = 1 for x = xo > 1 Px(0) = 0 for x # xo

(2-161)

The solution of the differential equation is [2, p.851: (2-162)

with the following characteristics: a) Fig.2-55 demonstrates the Px(ht) - x, P,(ht) - ht relationships computed from Eq.(2-162) with pt as parameter for two initial values of xo = 5, 10. A typical behavior revealed in the present model indicates that: P, = xo(o) = P, = o(-)

=1

(2- 163)

which is due to xo > x. On the other hand, in the previous models: (2-164)

163

o.:

3

0.6

3

pt = 0,x = x

xo= 10

lj

ax 0.4

pt = 0,x c xo

0.2

0

4

2

0

4

6

8

1

0

0

2

4

6

0.4

1

1

1

1

1

8

1

0

X

X 1

1

1

1

1

1

1

1

1

1

1

xo=5

-

-

-

I

1

Fig.2-55. The Simple Death model b) Since xo > x and depending on the magnitude of XO, the probability of remaining in a certain state increases versus time. C) Px(t) obeys Eq.(2-127). d) The mean number of events, m(t), occurring in the time interval of length t, is defined by Eq.(2-128)where for Px(t) given in Eq.(2-162):

(2-165)

The Birth-and-Death Process. An example of such a process are cells undergoing division, i.e., birth, where simultaneously they also undergo apoptosis, i.e., programmed cell death. The latter might be due to signals coming from exterior of the cells. The existence of human being may also be looked upon, approximately, as a simultaneous birth-death process although population in the

164 world is generally increasing. One should imagine what would have happened if death phenomenon would not have existed. Finally, it should be noted that birthand-death processes are of considerable theoretical interest and are also encountered in many fields of application. The derivation of the governing equations is based on the following assumptions : a) If at time t the system is in state x (x = 1, 2, ...), the probability of the transition x x + 1, i.e. Sj to sk, in the time interval (t, t + At) is given by:

(2-166) where hx is the mean birth rate which is an arbitrary functions of x. b) If at time t the system is in the state x (x = 1 , 2, ...), the probability of the transition x -$ x - 1, i.e. Sj to Sk,in the time interval (t, t + At) is given by: pjk = pXAt+ o(At)

(2-167)

where p, is the death rate. c) The probability of a transition to a state other than the neighboring state is o(At). d) If at time t the system is in the state x (x = 1, 2, ...), the probability of remaining in the state is given by: pjj = 1 - (hx+ px)At + o(At)

(2-168)

e) The state x = 0 is an absorbing or deud state, i.e., poo =l. In view of the above assumptions and following Eq.(2-3), we may write that: P,(t

+ At) = hx-lPx-,(t)At + p,+lPx+l(t)At+ El - (h,+

px)At)IP,(t) + dAt)

(2-169) leading to the following differential equation:

(2-170)

165 In Eq.(2-169), the first two terms on the right-hand side describe the transitions from states x-1 and x+l to state x; the third term designates the probability of remaining in state x. Eq.(2-170) holds for x = 1, 2, ... where for x = 0, noting assumption e above, we have: (2-171) since h-1 = ho = po = 0. If at t = 0 the system is in the state x = XO, 0 < xo < 00, the initial conditions are:

P,(O) = 1 for x = xo Px(0) = 0 for x z xo

(2-172)

Depending on the h, - x and px - x relationships, two cases will be considered, viz., the linear birth-death process and the z-stage reversible-consecutive process.

The linear birth-death process. It is assumed here that p, = px and h, = hx where p and h are constant values. The solution obtained for xo = 1[2, p.881, which satisfies Eqs.(2-172), reads: (2-173a) (2-173b) where (2-174a)

(2- 174b)

166 Some characteristics of the solution are: a) Fig.2-56 demonstrates the Px(t) - t relationship computed by Eqs.(2-173a,b) for different values of p and h designated on the graph.

0

1

2

t

3

4

5

'1

0.8

c A

0.6

-

px 0.4

0.2 0

0

1 0 2 0 3 0 4 0 5 0 t

0

10

20

t

30

40

50

Fig.2-56. The Birth-Death model

The following trends, also demonstrated in the figure, are observed regarding to the extinction probability Po(t), i.e., the probability that the population will eventually die out by time t. The trends are obtained from Eq(2-173b) by letting t + 03. Hence:

(2-175) From the above results we can see that the population dies out with a probability of unity when the death rate is greater than the birth rate, but when the birth rate is greater than the death rate, the probability of eventual extinction is equal to the ratio

167 of the rates. Additional computations indicate that if h > p (left-hand side graphs), increasing h at constant p decreases Px(t). For example: for p = 1 and h = 2, Po(t) = 1 if t > 3 for p = 1 and h = 10, Po(t) = 0.1 if t > 1 for p = 1 and h = 50, Po(t) = 0.02 if t > 1. If p > h, increasing p causes Po(t) to approach unity and the others to faster approach zero. b) Px(t) obeys Eq.(2-127). c) The mean number of events, m(t), occurring in the time interval of length t, is defined by Eq.(2-128), where for Px(t) given in Eq.(2-173) m(t) = e(Wt

(2-175)

The following asymptotic behavior is obtained: lim m(t) = 0 for h c p t+= = 1 for h = p =-

for h > p

(2-176)

From the above we can see that when h = p, the expected rate of growth is zero and the mean population size is stationary.

The z-state reversible-consecutive process. In this process, widely encountered in chemical engineering reactions, the system, defined below, undergoes the following succession of simultaneous birth-death transitions:

where the magnitude of the birth rate 3\1 and the death rate pi depend on the state Si. Let us now apply the above simultaneous birth-death model to a well-known process, i.e., a z-stage irreversible-consecutivefirst order chemical reaction, with a single initial substance. The various states are Si Ai (i = 0, 1, ..., Z) where Ai designates concentration of a chemical species i acquiring some chemical formula;

168 considering the quantity Px(t), it follows that i I x. In this context, the system is defined as a fluid element containing chemical species which undergo a chemical reaction or not. Considering Eq.(2-177), the system occupies the various states at different times. Px(t) is the probability of occupying state x, i.e., occupying the chemical state of species i = x. x = 0 is the initial chemical substance, x = 1 is the second chemical species, etc., where x = Z is the last born member of the family which remains alive for ever. Another interpretation of Px(t) is the probability that at time interval (0, t), the system still occupies the state of chemical species x of concentration Px(t) = Ax(t)/Ai(O);Ai(0) is the initial concentration of species x = i. On the basis of Eq.(2-170) we may write the following equations where hi and pi are the chemical reaction rate constant (Mime): (2- 178a) for species 0. (2- 178b) for species 1.

(2-1 7 8 ~ ) for specie 2.

.............. (2- 178d) for specie z-1 where for specie z (2- 178e)

169 For Z = 2, a solution is available [22, p.421 if we take species 0 with initial concentration Ao(0) to be the starting substance. Then, P,(t) = A,(t)/&(O) and the solution reads:

(2-179a)

(2-179b)

(2-179~)

where yl and y2 are roots of the following quadratic equation taken with the reverse signs:

Some characteristic properties of the above probability distributions, or dimensionless concentration of the species, are demonstrated in Fig.2-57 where the relationship Px(t ) - t is plotted.

170

Fig.2-57. The Birth-Death model for reversible-consecutive reactions Generally, it is observed that the system, i.e., a fluid element containing some chemical species, attains a steady state distribution of its occupation probability among the various states. The distribution depends on the rate constants designated in the figure and are characteristic of reversible reactions. Considering the figure on the left-hand side, it may be concluded that the probability of remaining in states 0, 1 and 2 at time interval (0, w) is 1/3 for each state, i.e., in a form of the initial component, component 1 and component 2. The above probabilities indicate also that in the above time interval, exactly three changes have occurred, i.e., the formation of states 2 and 3 due to the reaction and also the remaining in state 0. Certainly, the quantities Px(=)= Ax(=)/Ao(0), x = 0, 1, 2, may be looked upon as dimensionless concentrations encountered in reaction engineering. On the basis of the figure on the right-hand side, it may be concluded that if the rate constants pi and p2 will approach zero, i.e., the effect of the irreversibility is diminished, the probability of remaining at states 0 and 1 will also diminish at time interval (0, w), and the only state to be occupied by the system is 3, i.e., the state of solely component 3.

2.3 MARKOV CHAINS CONTINUOUS IN SPACE AND TIME 2.3-1 Introduction

The above topic is presented in the following because the equations developed and the basic underlying mechanism, i.e. diflusion, are of fundamental

171 importance in Chemical Engineering. However, the major problem of applying the equations is that not always an analytical solution is available. Let us consider a dancing hall where people are arriving at and leaving from after they have enjoyed the evening. If the hall is defined as system and the number x of people present in the hall at some time during the evening is the state of the system, x = 1, 2, 3, ..., then this situation may be looked upon as a Markov chain continuous in time and discrete in space, elaborated in the preceding section 2.2. More specifically the situation may be categorized as a birth-death model where the entering people designate birth and the leaving ones symbolize death. The mean rates of birth and death can easily be determined by counting the people arriving and leaving the hall. In the above example, one may be interested in the probability that at a certain time of the evening, the number of dancers in the hall is some prescribed value. Certainly, early in the evening, x = 0 and P,(O) = 1. The main characteristic of processes similar to the above one, as well as the ones discussed in section 2.2, is that in a small time interval there is either no change of state or a radical change of state. Therefore in a finite interval there is either no change of state or a finite or possibly denumerably infinite number of discontinuous changes. Assume now that some time in the evening the dancing hall becomes extremely congested and the doors are suddenly shut. Under these conditions, the movement of the dancers in the hall may be governed by random impacts of neighboring dancers. A typical physical situation, similar to the above one, is that of particles suspended in a fluid, and moving under the rapid, successive, random impacts of neighboring particles. If for such a particle the displacement in a given direction was plotted against time, we would expect to obtain a continuous or somewhat erratic graph which would, in fact, be a realization of a stochastic process in continuous time with continuous state space. The characteristic of such processes is that in a small time interval, displacement or change of state are small. The physical phenomenon related to the above behavior is known as Brownian motion, first noticed by the botanist Robert Brown in 1827. The modeling of such motions is based on the theory of diffusion and kinetic theory of matter. Markov processes associated with above motions, in which only continuous changes of state occur versus time, are therefore called difusion processes. The above points will be elaborated in the following.

172

2.3-2 Principles of the modeling Brownian motion of particles is the governing phenomenon associated with transitions between states in the above examples as well as in the mathematical derivations in the following [4,p.2031. If we consider a particle as system and the states are various locations in the fluid which the particle occupies versus time, then the transition from one state to the other is treated by the well-known random walk model. In the latter, the particle is moving one step up or down (or, alternatively, right and left) in each time interval. Such an approach gives considerable insight into the continuous process and in many cases we can obtain a complete probabilistic description of the continuous process. The essence of the above model is demonstrated for the simple random walk in the following. Let X(n) designate the position at time or step n of the moving particle (n = 0, 1, 2, ... ). Initially the particle is at the origin X(0) = 0. At n = 1, there is a jump of one step, upward to position 1 with probability 1/2, and downwards to position -1 with probability 1/2. At n = 2 there is a further jump, again of one step, upwards or downwards with equal probability. Note that the jumps at times n = 1, 2, 3, ... are independent of each other. The results of this fundamental behavior are demonstrated in Fig.2-58 where two trajectories 1 and 2 for a single particle, out of many possible ones, are shown. I

I

I

I

I

I

I

I

/-

3 -

I

I

I

I

I

I

I

I

173

In general, the trajectory of the particle is given by: X(n) = X(n - 1) + Z(n)

(2-181)

where Z(n), the jump at the nth step, is such that the random variables Z(1), Z(2;, ... are mutually inaependent and ali have the same distribution whch reais, prob{Z(n) = I } = prob(Z(n) = - 1] = L’2

(ri

= I, 2, ... ;

(2-182)

Finally, it should be noted that Eq.(2-181), taken with the initiai condition X(0j = 0, is equivalent to: X(n) = Z(l) + Z(2) + Z(3) + ... + Z(nj

(2-183)

The mathematicai elements used in the following derivations ars X(t) - a random variabie designating the position of a particle, system, in the fluid at time t. X(t) = x, indicates realization of the random variable, i.e., that at time t the random variable acquired a value x, or, that the system occupied state x. p(y, z; x, t) - a probability densityfunction, i s . probability per unit length, wherc, p(y, z; x, t)dx - is the probability of finding a Brownian particle at time t in the interval (x, x + dx) when it is known that the particle occupied state y at an earlier time z. It follows from above definitions that: prob(a c X(t) c b 1 X(z) = y) =

1 pCy, z; x, t)dx rb

(t > z)

(2-184)

i.e., the probability of finding a particle at time t between states a and b, subjected to the fact that the particle occupied state y earlier at time z, is given by the integral on the right-hand side. Assume now that the lower limit is a = 0, then we define a new function P(y, T; x, t), the transition probability function by:

174 P(y,z; b, t) = jobp(y,7; x, t)dx = prob{ X(t) < b 1 X(z) = y} (t > 2 ) (2-185) where (2- 186a)

P(y, 2; x, t), the probability of finding a particle at time t in the interval (0, b) while recalling it occupied state y at time z, also satisfies the following conditions: lim P(y, 2 ; x, t) = 0 x+o lim P(y, z; x, t) = 1

(2- 186b)

X+-

2.3-3 Some continuous models In the following, two basic models are presented [2, p.129; 4, p.2031; the Wiener process and the Kolmogorov equation. Applications of the resulting equations in Chemical Engineering are also elaborated. The common to all models concerned is that they are one-dimensional and, certainly, obey the fundamental Markov concept - that past is not relevant and that f i t w e may be predicted from the present and the transition probabilities to the future. The Wiener process. We consider a particle governed by the transition probabilities of the simple random walk. The steps of the particle are Z( l), Z(2), ... each having for n = 1, 2, ... the distribution: prob{Z(n) = 1) = p, prob{Z(n) = - 1) = q = 1 - p

(2-187)

where p and q are constant one-step transition probabilities. Assume that a particle is in the time coordinate n - 1. It undergoes a transition to a state coordinate k from the state coordinates k - 1 and k + 1. Thus, we may write the following forward equation for a fixed j:

175 (2-188) where Pjk(n) is the probability that a particle occupies state k at time n, if it occupied state j at n = 0. Considering the definition of p(y, 2;x, t)dx above, let:

be the conditional probability that the particle is at x at time t, given that it started at yo = xo at time 2 = 0; x is a continuous state coordinate. We have also xo = jAx,

x = kAx and t = nAt, thus, Eq.(2-188) in terms of the new scale reads: P(xo; X, t) = P(XO;x - AX,t - At)P + ~ ( x ox; + AX, t - At)q

(2- 190)

the factor Ax canceling throughout. Suppose that p(x0; x, t) can be expanded in a Taylor series and the first and second derivatives exist, we obtain that:

a + Ax& + 0.5(Ax)2??! ax2

p(xo; x + Ax, t - At) = p(xo; x, t) - At-$

(2-191)

If we expand Eq.(2-190) according to Eq.(2-191) and apply the following expressions for p and q [4,p.2061: p = 0.5[ l +

5 ai 1 -,

q = 0.5[ l-

L] a t O

(2-192)

where p is the mean (lengthhime) and 0 2 i s the variance (length%ime) of the random variable X(t) designating the position of the particle at time t. If At + 0, we obtain the forward equation:

ap(xo; x, t) 2a2p(xo;x, t) ap(xo; x, t) = 0.50 -1 dx dt

ax2

(2-193)

176 which is a partial differential equation of the second order with respect to x and is first order in t. It is an equation in the state variable x at time t for a given initial state XO. The solution of the above equation conditional on X(0) = xo reads:

(2- 194) with the two parameters p and 6. Eq.(2-194) satis--;s Eqs.( 18342-186b). Eq.(2-193) is familiar in Chemical Engineering in the following cases [24]: a) Diffusive-convective heat transfer. In this case, the equation of energy for constant properties, ignoring viscous dissipation, taking into account axial heat diffusion only in the x-direction which is superimposed on the molecular diffusion of heat, reads: dT(x, t)

7

=

a

a2T(x, t)

ax2

dT(x, t)

- Uax

(2- 195a)

Comparing the above equation with Eq.(2-193) demonstrates the similarity between the two for xo = 0. The parameters appearing in Eq.(2-195a),i.e., the heat thermal diffusivity a which accounts for the molecular diffusion, and the constant axial velocity U accounting for the axial diffusion, are characteristic to the heat transfer process. The underlying model of particle's motion associated with the derivation of Eq.(2-193) may be applied also to heat transfer. In this case, a fluid element (or a molecule) at some temperature moves due to both, diffusion as well as axial diffusion generated by the fluid velocity, transferring heat to the other fluid elements. The motion of the fluid elements may be looked upon also from probabilistic arguments which led to Eq.(2-193). Finally, it should be mentioned that in particular problems, the solution of the equations will depend on initial and boundary conditions. b) Diffusive-convective mass transfer. The equation of continuity for species A in a binary solution of A+B, assuming constant density of the fluid mixture as well as the diffusion coefficient DAB,taking into account axial mass diffusion only in the x-direction which is superimposed on the molecular diffusion of mass, reads:

177 (2- 195b)

CA is the local concentration of species A in the mixture. Other remarks made in (a) are also applicable here. c) Heat or mass transfer into semi-infinite bodies by pure conduction or diffusion. The governing equations may be obtained from the above equations by ignoring the axial component; thus: aT(x, t>

a2T(x, t)

at=a ax2 (2-195~) The above equations are similar to Eq.(2-193) for p = 0 which assumes that a fluid element of some concentration, or at some temperature, is moving by pure molecular diffusion, thus generating mass or heat transfer in the system. An additional equation of a similar form, belongs to momentum transfer. For a semi-infinite body of liquid with constant density and viscosity, bounded on one side by a flat surface which is suddenly set to motion, the equation of motion reads [24, p.1251: (2-195d) ux is the velocity of the liquid in the x-direction, and v is the kinematics viscosity of the liquid governing the momentum transfer by pure molecular diffusion.

The Kolmogorov equation. In the preceding model, the major assumption made was that the one-step transition probabilities p and q remain constant. In the following it is assumed that these probabilities are state depending, namely, designated as Pk and qk. The following forward equation for the present model reads:

178

The equation may be interpreted in the following way: the probability of being at state k after n steps = the probability of being at state k-1 after n-1 steps times the one-step transition probability (which depends on state k-1) plus the probability of being at state k+l after n-1 steps times the one-step transition probability of moving from state k+l to k (which depends on state k+l) plus the probability of being at state k after n-1 steps times the probability of remaining in this state, (1- Pk - qk); d l transitions take place in the time interval n and n+l . Again we consider the process as a particle taking small steps -Ax, 0, or +Ax in the small time interval of length At. From Eq.(2-189) the probability density p(x0; x, t)Ax may be looked upon as the conditional probability that a particle is at location x at time t, given that initially it occupied XO. Thus, Eq.(2-196) becomes: p(xo; X, t) = ~ ( x ox; - AX, t - At)p(x- AX) + p(x0; x + AX, t - At)q(x+ A ~ )

the factor Ax canceling throughout. It has been shown elsewhere [4, p.2141 that:

(2-198) Ax,At+O

where p(x) and a(x) are the instantaneous mean (lengthhime) and variance (length%me), respectively. Applying Eqs.(2-197) and (2-198), resulted in the following forward equation:

179 which is a parabolic partial differential equation. A more general equation can be obtained by allowing the transition mechanism to depend not only on the state variable x but also on time t. In this case we are led to define p(x, t) and a(x, t), both depending on x and t, i.e., the instantaneous mean (lengthhime) and variance (length*/time), respectively. Thus, if xo denotes the state variable at time to and x that at a later time t, then the transition probability density p(x0, to;x, t) satisfies the followingforward Kolmogorov equation also called the Fokker-Plank equation:

where it has been assumed in the derivations that the above partial derivatives exist. The functions p(x, t) and a(x,t) are sometimes called the infinitesimal mean and

variance of the process. If these functions are assumed as constant values, the above equation reduces to Eq.(2-193) where a(x, t) = o2and p(x, t) = p. Eq.(2-200) is familiar in Chemical Engineering in turbulent flow. For example, the energy equation for one-dimensional flow [24, p.3771 for a fluid of constant properties, in the absence of viscous dissipation effects and for xo = to = 0, reads: (2-201)

where T = ? + T'; the thermal diffusivity a has been assumed constant. Note that in turbulent flow the temperature T is a widely oscillating function of time,

-

fluctuating about the time-smoothed value of the temperature T, where T' is the temperature fluctuations. The major problem of Markov chains continuous in time and space is that the availability of analytical solutions of the governing equations, which depend also on the boundary conditions, is limited to simplified situations and for more complicated cases, numerical solutions are called for.

180

2.4 CONCLUDING REMARKS In the present chapter, Markov processes discrete in time and space, processes discrete in space and continuous in time as well as processes continuous in space and time, have been presented. The major aim of the presentation has beer; to provide the reader with a concise summary of the above topics which should give the reaaer an overview of the subject. The fundamentals of Markov chains have been presented in an easy and understandable form where compiex mathematical derivations are abandoned on the one hand, and numerous examples are presented on the other. Despite of the simplifications made, the author believes that the needed tools have been provided to the reader so that he can solve complicated problems in reactors, reactions, reactor plus reactions and other processes encountered in Chemical Engineering, where Markov chains may provide a useful tool. The models discrete in space and continuous in time as well as those continuous in space and time, led many times to non-linear differential equations for which an analytical solution is extremely difficult or impossible. In order to solve the equations, simplifications, e.g. linearization of expressions and assumptions must be carried out. However, if this is not sufficient, one must apply numerical solutions. This led the author to a major conclusion that there are many advantages of using Markov chains which are discrete in time and space. The major reason is that physical models can be presented in a unified description via state vector and a one-step transition probability matrix. Additionai reasons are detailed in Chapter 1. It will be shown later that this presentation coincides also with the fact that it yields the finite difference equations of the process under consideration on the basis of which the differential equations have been derived.

2.5 ARTISTIC ENDING OF THE CHAPTER The present chapter begun with Fig.2-0, the painting Wategall by Escher, which made it possible to present in various places of this chapter the basic concept of conditional probability. We end this chapter by 'Markovization', discrete in time and space, of the amazing oil on canvas painting by Magritte Carte blancheSignature in blank [12, p.451 depicted in Fig.2-59. This demonstrates again a way of entertaining the combination of art and science. A few words about this

181

painting. Magritte, in an extremely subtle and deceptive way, demonstrated the simultaneous movement in two planes of the horsewoman. Between two trunks the normal backdrop of foliage is visible, and this conceals a portion of the horse and rein, the horse appearing to be passing between the same two trunks. Spatially, the rider and the woods become an absurdity, due to this section of the intruding background, the position of one of the horse's hind legs, and another tree trunk in the background, part of which passes in front of the horse and rider. Magritte expressed his thoughts on his painting as follows: "Visible things can be invisible. If somebody rides a horse through a wood, at first one sees them, and then not, yet one knows that they are there. In Carte blunche, the rider is hiding the trees, and the trees are hiding her. However, our powers of thoughts grasp both the visible and the invisible."

182

Fig.2-59. The impossible state S5 ("Carte blanche", 1965,O KMagritte. 1998 c/o Beeldrecht Amstelveen)

Let us now apply Markov chains to investigate the trajectory of the syslem, the horsewoman, riding in the foresl. The possible states that the system can occupy on the basis of Fig.2-59 are defined as follows: S1-the passage between trees 1 and 2 through which the system can move, &the passage between trees 4 and 3, Sythe passage between trees 4 and 5, Sq-the passage between trees 7 and 6,

183 and S5-the impossible situation depicted in Fig.2-59. It is assumed that once the horsewoman abandons S5, she never returns to it and continues to ride in the forest according to her mood at the moment. Thus, S5 is an emepheral state. The policy-making matrix distinguishes among all cases analyzed below where the common to all of them is that S(0) = [0, 0, 0, 0, 11, i.e., the rider is initially at S5. The results of the computations are depicted in Fig.2-60; on the top of each figure, the corresponding one-step transition probability matrix is presented. Generally, in cases a to c the states attain a steady state whereas the stationary distribution in each case obeys Eq.(2-105a), i.e. the states reveal Ergodic characteristics. In cases d to f the states exhibit periodic behavior whereas S5 is eventually abandoned in all cases. In cases a, c, d and fit is abandoned after the first step whereas in cases b and e, abandoning occurs after a few steps. This is because in the corresponding matrix, there is also a probability of remaining in the state, pii # 0. In the following we consider each case separately. In case a the steady state occupation probability is different for each state; the corresponding state vector reads, S(11) = [0.300, 0.400, 0.200, 0.100, 0.0001. As seen, S 2 is of the highest probability, i.e. s2( 11) = 0.400. Case b differs from a only by the values of psi; thus, causing vanishing of s5(n) to occur after 6 steps rather than after 1 step. From thereon, the values of S(n ) are the same as in a. Case c causes a problem to the horsewoman because S(2) = [0.25, 0.25, 0.25, 0.25, 0.001; thus she has a problem what state to occupy because all states acquire an identical probability. Case d reveals periodic characteristics of the states with period of v = 2 as follows: S(21) = [0, 113, 0, 213, 01 S(22) = [2/3, 0, 113, 0, 01 S(23) = [0, 1/3, 0, 2/3, 01 S(24) = [2/3, 0, 1/3, 0, 01

In other words, in each step two states, S 2 and S4 or S1 and S3, acquire a certain occupation probability. Case e demonstrates the following periodic results at steady state

184 S(12) = [0.133, 0.267, 0.533, 0.067, 01 S(13) = [0.067, 0.133, 0.267, 0.533, 01 S(14) = [0.533, 0.067, 0.133, 0.267, 01 S(15) = [0.267, 0.533, 0.067, 0.133, 01

where the period of each state is v = 4. Case f is similar to e whereas the occupation probability of each state at each step is unity. In conclusion we may say that the major problem of the horsewoman, once abandoning the strange state Sg, is as William Shakespeare profoundly stated:" to be (in a state) or not to be (in a state), that is the question". Case b

s1

s2

s3

s4

s5

0

1

0

0

0 0

0

1

0

0

0

S1

112

0

112 0

0

112 0

112 0

0

112 0

112 0

s2 P = s3

0

112

0

112 0

1

0

0

0

0

s4

1

0

0

0

0

112

0

0

112 0

s5

113

0

0

113

113

t

185 Case c s1

s2

s3 s4

112 0

0 P = s3

112 0

112 0

112 0

0

ss 0

1

0

0

0

0

0

1

0

0

s 4 1

0

0

0

0

112 0

s 5 0

0

0

1

0

112 0

s4

112 0

112 0

SS

112 0

0

P=S3

0

2

4

l

l

s1

s2

s3 s4

s5

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

l

l

l

6

n

8 1 0 1 2 1 4

Case f

Case e

l

112 0

0

s 2 0

n

1

112 0

0

s1

l

l

l

s1

s2

s3

s4

ss

S1

0

1

0

0

0

s2 P = s3

0

0

1

0

0

0

0

0

1

0

s4

1

0

0

0

0

SS

0

0

0

1

0

l

0.8 C

vv)

0.6

0.4 0.2 0

n n Fig.2-60. Policy-making matrices and state occupation probabilities of the horsewoman

186

Chapter 3

APPLICATION OF MARKOV CHAINS IN CHEMICAL REACTIONS Chemical reactions occur due to collisions between similar or dissimilar species. The collision process requires, first of all, the coming together of molecules, which is of probabilistic or stochastic nature. Thus, chemical reactions may be looked upon as probabilistic or stochastic processes and more specifically as a Markov chain. According to this model, concentration of the species at time (n+l) depends only on their concentration at time n and is independent on times prior to n. Therefore, the governing equations for treating chemical reactions, as elaborated later, are Eqs.(2-23) and (2-24) below, derived in Chapter 2:

(2-23)

S(n+l) = S(n)P)

(2-24)

The application of the equations to chemical reactions requires the proper definition of the above quantities as well as correctly defining the transition probabilities pjj and pjk; this is established in the following. It should also be noted that the models derived below for numerous chemical reactions, are applicable to chemical reaction occurring in a perfectly-mixed batch reactor or in a single continuous plug-flow reactor. Other flow systems accompanied with a chemical reaction will be considered in next chapters.

187 The attractiveness of Markov chains in chemical reactions is particularly due to the following reasons: a) Simplicity, elegance and the didactic value. b) Demonstrating the power of probability theory in handling a priori deterministic problems. c) Applying fundamentals of linear algebra to chemical reaction problems of practical importance. d) The attractiveness of the method increases with the number of reacting components, if the reactions are higher-order or if they are non-isothermal. In such cases the governing differential equations are nonlinear, and an analytical solution is extremely difficult, if not impossible; finally, e) The solution of Eq.(2-24) yields the transient response of the reaction towards attainment of a steady state, which is important in practice. The above points are demonstrated by the numerous examples presented in this book. Throughout this chapter it has been decided to apply Markov chains which are discrete in time and space. By this approach, reactions can be presented in a unified description via state vector and a one-step transition probability matrix. Consequently, a process is demonstrated solely by the probability of a system to occupy or not to occupy a state. In addition, complicated cases for which analytical solutions are impossible are avoided.

3.1 MODELING THE PROBABILITIES IN CHEMICAL REACTIONS Definitions. The basic elements of Markov chains associated with Eq.(224) are: the system, the state space, the initial state vector and the one-step transition probability matrix. Considering refs.[26-30], each of the elements will be defined in the following with special emphasize to chemical reactions occurring in a batch perfectly-mixed reactor or in a single continuous plug-flow reactor. In the latter case, which may simulated by perfectly-mixed reactors in series, all species reside in the reactor the same time. The system is simply a molecule. The state of the system is the specific chemical formula of the molecule, or what kind of species is the molecule. The state space is the set of all states that a molecule can occupy, where a molecule is occupying a state if it is in the state. For example, in the following irreversible consecutive reaction:

188 kl

A,

k2

k3

k4

+ A, + A, + A, + A,

type Ai = i of a molecule is regarded as the

st

te of th system, i.e.

specific

chemical formula. The state space SS, which is the set of all states a system can occupy, is designated by:

Finally, the reaction from state Ai to state Aj is the transition between the states. The initial state vector is given by Eq.(2-22), i.e.:

Si(0)

designates the probability of the system to occupy state Ai = i at time zero,

whereas S(0)designates the initial occupation probability of the states [Al, A2, A3, ..., Az] by the system. Z designates the number of states, i.e. the number of chemical species involved in the chemical reaction. In the context of chemical reactions, as shown later, the probabilities S j ( 0 ) may be replaced by the initial concentration of the molecular species and S(0) will contain the initial concentration of all species. The one-step transition probability matrix is generally given by Eqs.(2-16) and (2-20) whereas pjk represent the probability that a molecule Aj will change into a molecule Ak in one step. p~ represent the probability that a molecule Aj will remain unchanged within one step. In the following, general expressions are derived for the determination of pjk and pjj for the model below. The reacting system. Consider a chemically reacting system containing the species Al, A2, A3, ...., AZ. A chemical reaction among the species induces the change in the state of the mixture. It is also assumed that a certain species Aj can react simultaneously in several reactions, designated in the following by superscripts (i), i = 1, 2, ..., R where R is the total number of reactions in which Aj is involved. The following scheme of irreversible reactions by which Aj is converted to products is assumed, where each set of reactions involving reversible reactions can always be written according to the scheme below in order to apply the following derivations.

189 (1) J‘

1 s t reaction:

... + a j(1)A, + ... + products

2nd reaction:

... + aj

Rth reaction:

... + aj(R)A, + ...

(2)

(2)

A,

I‘

+ ... +

products

(i)

J

4

products

(3-1)

where af) is the stoichiometric coefficients of speciesj in the ith reaction. The rate

of conversion of species j in the ith reaction based on volume of fluid V, i.e. ry), is defined by:

where N, are the number of moles of species j and C, is its concentration, moles of j/m3. kf), in consistent units, is the reaction rate constant with respect to the conversion of speciesj in the ith reaction. In the case of a plus sign before kp), this means moles of j formed/(sm3). The discrete form of Eq.(3-2) reads:

where the reaction rate and the concentrations Cj(n) refer to step n. In addition, the conservation of the molar rates for all reacting species in the ith reaction in Eq.(3-1) is governed by:

(3-3) Eq.(3-3), corresponding to the ith reaction, makes it possible to compute the reaction rates of all species on the basis of ry) given by Eq.(3-2).

190 Definition of p ~ If. species j is converted simultaneously in R reactions, the overall rate of conversion of species j is the following sum:

i

where the summation is over all reactions in which species j is involved, i.e. i = 1, 2, ..., R. The integration of the equation between t and t + At, taking into account that rf) is negative according to Eq.(3-2),yields that: Cj(d - Cj (n+l) = xrii)(n)At i

(3-5)

where Cj(n) and Cj(n+l) are, respectively, the concentration of species j in the mixture at time interval t and t + At or at step n and n+l. rii)(n) is the reaction rate corresponding to step n, i.e. the concentrations in Eq.(3-2a) correspond to this step. Here it is assumed that the variation of rii)(n)between step n and n+l is not significant. If we consider the following quantity: VCj(n+l) amount of species j present at time t = (n+l)At I - amount of species j present at time t = nAt vCj(d

(3-5a)

where V is the volume of fluid, we may look on it as the probability pjj that a molecule Aj will remain unchanged within one step. Thus, from Eq.(3-5) it follows that:

The summation is over all reactions in which species j is converted to products, i.e. i = 1, 2, ..., R. Eq.(3-6) indicates that if At + 0, pjj = 1. Indeed Aj will remain unchanged under such conditions. However, if At is large enough, the probability that Aj will remain unchanged approaches zero, as expected. Thus, the above expression may, indeed, serve as a probability term, provided that 0 5 pjj I 1.

191

Definition of pjk. In deriving an equation for the probability pjk corresponding to the transition Aj + A,, we consider the ith reaction in Eq.(3-1). In this case, dCf)/dt = r:), and an integration of it yields that:

amount of speciesj converted in iaaetime interval At in reactm i amount of j available at time t = nAt

(3-7)

Since the formation of the products in the ith reaction is associated with reaction among all reactants, and each species contributes according to its stoichiometric coefficient, we account for this effect in the determination of Pjk by introducing the ratio between the stoichoimetric coefficients. In addition, the transition probability for Aj+ A, depends also on the stoichiometric coefficient of Ak, Thus, on the basis of Eq.(3-7), the probability of the transition Aj + A, for the ith reaction in (i)

Eq.(3-l), i.e. P,k, reads:

1=1

/

where N(i) is the number of the reacting species in reaction i. Indeed, the properties of the above expression are appropriate for describing a transition probability, namely, as At = 0, pfi = 0 while for relatively large values of At, pj(2 is increasing with a limiting value not exceeding unity. If the ith reaction in Eq.(3-1) is reduced to:

then in order to comply with the results obtained by direct integration of the rate equations, the transition probability for Aj 4 A, must read:

192

(3-8a)

If species j is involved in several reactions, the total of which is R as in Eq.(3-l), the following expression may be obtained on the basis of Eq.(3-5) for the overall transition probability of Aj + A,, i.e.:

The summation is over all reactions in which species j is involved, i.e. i = 1,2, ..., R. The justification for the above expressions for pjj and Pjk is confirmed later by the agreement with determinations made on the basis of integration of the rate equations. If all R reactions are of the type: 6)

‘1

q ) A j -+

... + ak(i) A, + ...

the transition probability is given by:

(3-9a)

Finally, if species j undergoes a change solely by a single irreversible reaction of the type: r.J

... + ajAj + ... + ... + akAk + ... we may ignore, for the sake of simplicity, some of the subscripts and superscripts in Eq.(3-8), which is reduced to:

193 (3-10)

N is the number of the reacting species on the left-hand side of the above reaction. If the above reaction is reduced to:

the transition probability for Aj + Ak will read: (3-10a)

3.2 APPLICATION AND VERIFICATION OF THE MODELING The modeling in section 3.1 will be applied to non-linear and linear reactions. In section 3.2-1, all the stoichiometric coefficients of the species equal unity, i.e. aj = 1. In section 3.2-2 part of the stoichiometric coefficients are different from unity, aj # 1. In section 3.2-3 linear reactions are dealt with whereas in section 3.2-4, linear-non linear reactions with aj # 1 are demonstrated in detail by case 3.13-6. The validity of the results for Cj(n+l) computed by Eq.(3-20) on the basis of pjj and pjk predicted by Eqs.(3-6), (3-8) to (3-lo), will be compared to those obtained by direct integration of the kinetic equations.

3.2-1 Non-linear reversible reactions with all aj = 1 A procedure for determination of the transition probabilities is demonstrated by a detailed treatment of the following non-linear irreversible reactions:

194 kl

A1 +A2

t

A3 + A4

(3-lla)

k- 1 k2

A 2 + A 3 - + A4

(3-1 lb)

t k-2

The following rate equations are assumed satisfying Eq.(3-3): rl = dCl/dt = - klC1C2 + k-1C3C4

(3- 12a)

+ k-1C3C4 + k-2C4 r3 = dCs/dt = - k2C2C3 - k-iC3C4 + klC1C2 + k-2C4 r4 = dC4/dt = - k2C4 - k-lC3C4 + klClC2 + k2C2C3

(3-12b)

1-2 = dC2/dt

= - klCiC2 - k2C2C3

(3-12~) (3- 12d)

1st step: Transformation of Eqs.(3-11) into the following set of irreversible reactions which best demonstrate the transition between the states: kl

i = 1:

A1 + A 2 4 Ag+A4

i=2:

Af+Aq+.Al+A2

i = 3:

A2

k- I

k2

+ A3 + &

(3- 13a)

(3-13b) (3-13~) (3- 13d)

2nd step: Determination of the reaction rates for Eqs.(3-13) on the basis of Eqs.(3-12). The definition in Eq.(3-2) yields for species j = 1 reacting in the first reaction, i = 1, that:

where kill = kl according Eq.(3- 13) for i = 1. Similarly, for species j = 2 reacting in the first and third reactions, i = 1, 3:

195

(3-14b) where ky) = k, according Eq.(3-13) for i = 1 and k2(3) = k2 for i = 3. For species j = 3 reacting in the second and third reactions, i = 2,3:

(3-14~) For speciesj = 4 reacting in the fourth reaction, i = 4:

(3-14d)

3rd step: Determination of the probabilities pjj. Applying Eqs.(3-6) and (314a) yields for species j = 1 where i = 1 that: pi1 = 1 - klC~(n)At

(3-15a)

For species j = 2, converted according to Eqs.(3-13),i = 1, 3, Eqs.(3-6) and (314b) yield:

For species j = 3, converted according to Eqs.(3-13), i = 2, 3, Eqs.(3-6) and (314c) yield:

For species j = 4, converted according to Eqs.(3-13), i = 2,4, Eqs.(3-6) and (314d) yield: p44 = 1 - [klC3(n) + k-21At

(3- 15d)

196 4th step: Determination of the probabilities Pjk. As observed in the reactions given by Eqs.(3-13): pi2 = 0 and p21 = 0 because A1 is not converted to A2 and vice versa. Applying Eqs.(3-9) and (3-14a) to species j = 1, converted to species j = 3 and j = 4 according to Eqs.(3-13), i = 1, noting that N(1) = 2, yields: (3-16a) Applying Eqs.(3-9) and (3-14b) to species j = 2 which is converted to species j = 3 according to Eq.(3-13), i = 1, and to species j = 4 according to Eq.(3-13), i = 1,3, noting that N(1) = N(3) = 2, yields: P23 = (1/2)klCl(n)At p24 = (1/2)[klCi(n) + k2C3(n)]At (3-16b) Applying Eqs.(3-9) and (3-14c) to species j = 3 which is converted to species j = 1 according to Eq.(3-13), i = 2, to species j = 2 according to Eq.(3-13), i = 2, and to species j = 4 according to Eq.(3-13), i = 3, noting that N(2) = N(3) = 2, yields:

Finally, applying Eqs.(3-9) and (3-14d) to species j = 4 which is converted to species j = 1 according to Eq.(3-13), i = 2, to species j = 2 according to Eq.(3-13), i = 2,4, and to species j = 3 according to Eq.(3-13), i = 4, noting that N(2) = 2 and N(4) = 1, yields:

The above probabilities may be grouped in the matrix given by Eq.(3-17). It should be noted that Eq.(2-18) is not satisfied along each row because the one-step ftaasitian probabilities pjk depend QEI time n. This is known as the nonhomogeneous case defined in Eqs.(2-19) and (2-20), due to non-linear rate equations, i.e. Eqs.(3-12).

197

1 2

P= 3 4

(3-17)

Verification of the model. Several assumptions were made in section 3.1 which led to Eqs.(3-6), (3-8) to (3-10) for the determination of p,, and pjk. For the reactions given by Eqs.(3-13) the results are summarized in the matrix given by Eq.(3-17). The validity of the results will be tested by writing the Euler integration algorithm for the differential equations, Eqs.(3-12), which describe the reaction mechanisms. Integration of Eq.(3-12a) yields after a few manipulations that

where p21 = 0. The other nj's,as well as those for the results below, are given in Eqs.(3-15a to d) and (3-16a to d) which are summarized in the matrix given by Eiq~(3-17).Integration of Eq.(3-12b) yields: C2(n+l) = Cl(n)P12 + CZ(n)p22 + C3(n)P32 + C4(n>P42

(3-18b)

where p12 = 0. Integration of Eqs.(3-12c) and (3-12d) yields:

(3- 18d) Equations (3-18) reveals the following characteristics:

198 a) The equations are a function of the transition probabilities Pij and Pik detailed in the matrix given by Eq.(3-17). b) Each of the Eqs.(3-18a to d) obey Eq.(2-23) for a number of states Z = 4 as well as Eq.(2-24). Thus, the following equalities may be obtained: sj(n) = Cj(n); sj(n+l) = Cj(n+l) S(n) = C(n>= [C,(n), C&n>,C3(n), ..., Cz(n>l

(3-19)

where C(n) may be looked upon as the state vector of the system at time nAt (step n). In addition, the initial state vector reads:

No) = C(0) = [Cl(0), C2(O>,C3(O>,..., Cz(0)l

(3-19a)

c) Eqs.(3-18a to d) indicate that each Cj(n+l) is a result of the product of the row vector C(n), defined in Eq.(3-19), by the square matrix P defined in Eq.(3171, i.e.: j= 1

(3-20)

C(n+l) = C(n)P where Z is the number of states.

3.2-2 Non-linear reversible and irreversible reactions with ajz 1 kl

Example a: A,

+ 2A2

(3-21)

for which: rl = dCl/dt = - klC1

(3-22a)

It follows from Eq.(3-3) that: r2 = dC2/dt = 2klC1

(3-22b)

Integration of the above equations yields: Cl(n+l) = Cl(n)[l - klAt1

(3-23a)

Cz(n+l) = Cl(n)[2k1At] + C2(n)

(3-23b)

199 which can be arranged on the basis of Eq.(3-20) in the following matrix form:

1 1 -klAt

2klAt

P= 2

0

1

(3-24)

Applying Eqs.(3-6) and (3-10) yields identical probabilities. k,

Example b: A1

* 2A2 t

(3-25)

k2

for which: rl = dCl/dt = - klCl + k2C;

(3-26a)

It follows from Eq.(3-3) that: r2 = dC2/dt = 2klC1 - 2k2C,2

(3-26b)

Integration of the above equations gives: Cl(n+l) = Cl(n)[l - klAt] + C2(n)[k2C2(n)At]

(3-27a)

C2(n+l) = Cl(n)[2klAt] + C2(n)[1 - 2k2C2(n)At]

(3-27b)

yielding the following matrix:

P=

1 2

1 -k,At k2C2(n)At

2klAt 1-2k2C2(n)At

(3-28)

Applying Eqs.(3-6), (3-10) and for p2 1 Eq.(3-lOa), yields identical probabilities while Eq.(3-25) was expressed as the following two irreversible reactions:

200 kl

Example c: 2A,

--+

A2

(3-29)

for which: rl = dC,/dt

=

- k,C:

(3-3oa)

gives from Eq.(3-3) that: r2 = dC2/dt = 0. 5kl C:

(3-30b)

Cl(n+l) = Cl(n)[l - klCl(n)At]

(3-3la)

C2(n+l) = Cl(n)[O.SklCl(n)At] + C2(n)

(3-31b)

Thus,by integration it is obtained that

hence:

P=

A1 = 1 1 1 - klCl(n)At

A2=2 O.SklCl(n)At 1

0

2

(3-32)

Identical probabilities are obtained from Ekp(3-6) and (3-1Oa) for p12. kl

Example d 2A1

t

(3-33)

A2

k2

for which ri = dC,/dt

- k,C: + k2C2

r2 = dC2/dt = O.Sk:Cl

(3-34)

- 0.5k2C2

(3-34b)

yields by integration that Cl(n+l) = Cl(n)[l - klCl(n)At] + C2(n)[k2At] C2(n+l) = Cl(n)[OSklCl(n)At]

P=

A1 = 1 1 1 - klCl(n)At 2

k2At

+ C2(n)[l - 0.5k2AtI

(3-35) (3-35b)

A2=2 O.SkiCl(n)At 1 - 0.5k2At

(3-36)

201 The above probabilities are identical to those computed from Eqs.(3-6), (310) and Eq(3-1Oa) for pi2 considering Eq.(3-33) as the irreversible set kl

2A1+ A2 and A2

k2 --+

2A1

.

3.2-3 Linear reactions Consider the following reactions:

(3-37) +

k-1

C

k-2

The kinetics of the reactions satisfying Eq.(3-3) is given by the following expressions: rl = dCl/dt = - k l C l + klC2

(3-38a)

r2 = dC2ldt = - k2C2 - klC2

(3-38b)

+ k l Cl + k2C3 r3 = dC3/dt = - k3C3 - k2C3 + k2C2

(3-384

(3-38d) r4 = dCLJdt = k3C3 1st step: Transformation of Eqs.(3-21) into a set of irreversible reactions as follows: kl

i = 1:

A1 + A2

i=2

A 2 4 A1

i=3:

A2+A3

i =4:

A3 + A2

i=5

A3+A4

(3-39a)

k-1

(3-39b)

k2

(3-39c)

k-2

(3-39d)

k3

(3-3%)

2nd step: Determination of the reaction rates for Eqs.(3-39) on the basis of Eqs.(3-38). Following the definition in Eq.(3-2) yields:

202 rl(1) (n) = - kl(1) Cl(n) = - klC,(n)

(3-40a)

Similarly: (2)

(2)

r2 (n) = - k2 C2(n>= - klC2(n>

(3-40b)

h3)(n) = - k2(3)C2(n) = - k2C2(n)

(3-40c)

r3(4) (n) = - k3(4)C3(n>= - k&(n)

(3-40d)

$'(n) = - k3( 5 )C3(n)= - k3C3(n)

(3-40e)

3rd step: Determination of the probabilities pjj. Applying Eqs.(3-6) and (340a) yield for species j = 1 where i = 1 that: p11= 1 - klAt

(3-41a)

For species j = 2 which is converted according to Eqs.(3-39b, c), i = 2,3, Eqs.(36 ) and (3-40b, c) yield: p22 = 1 - [ k l + k21At

(3-41b)

For species j = 3 which is converted according to Eqs.(3-39d, e), i = 4, 5, Eqs.(36) and (3-40d, e) yield: p33 = 1 - [k2+ k31At

(3-41C)

For species j = 4, formed according to Eq.(3-39e), i = 5 , and for remaining at this state: P44 = 1

(3-4 1d)

4th step: Determination of the probabilities pjk. Applying Eqs.(3-9) or (310) and (3-40a) to species, j = 1 which is converted to species j = 2 according to in Eq.(3-39a), i = 1, noting that N(l) = 1 and that A1 is not converted to A3 and one step, yields:

203 P12 = klAt

P13 = P14 = 0

(3-42a)

Applying Eqs.(3-9) or (3-10) and (3-40b) to species j = 2 which is converted to species 1 according to Eq.(3-39b), i = 2, to species 3 according to Eq.(3-39c), i = 3, noting that N(2)= 1 and N(3) = 1, yields: p21= k l A t

p23 = k2At

p24 = 0

(3-42b)

Applying Eqs.(3-9) or (3-10) and (3-4Oc) to species j = 3 which is converted to species j = 2 according to Eq.(3-39d), i = 4, to species j = 4 according to Eq.(339e), i = 5, noting that N(4) = N(5) = 1, yields: p31= 0

p32 = k2At

p34 = kgAt

(3-42~)

Species j = 4, formed according to Eq.(3-39e), i = 5, remains in its state, thus: P41 = 0

P42 = 0

p43 = 0

(3-42d)

The above probabilities may be grouped in the matrix given by Eq.(3-43). It should be noted that Eq.(2-18) is satisfied along each row because the one-step transition probabilities pjk are independent of the time n. This is known as the time-homogeneous case defined in Eqs.(2-14a) and (2-16), due to the fact that the rate equations (3-23) are linear.

P =

1

1 -klAt

2 3 4

k-1 At 0

0

klAt 1 - [k-1

+ k2]At

k-2 At

0

0

0

k2At

0

1 - [k-2

+ k3]At 0

kgAt 1

(3-43)

Verification of the model. Integration of Eq.(3-38a to d) yields after a few manipulations Eqs.(3-18a to d). The pij's are given by Eqs.(3-41a to d) and (3-42a to d), which are summarized in the matrix governed by Eq.(3-43).

204

3.2-4 Linear-non linear reactions with aj # 1 Example 3.13-6 (chapter 3.13) is presented in details demonstrating also the derivation of the kinetic equations satisfying Eq.(3-3). This example should be studied thoroughly since it contains important aspects of applying the equations for dcdating pjj and Pjk.

3.3 MAJOR CONCLUSIONS AND GENERAL GUIDELINES FOR APPLYING THE MODELING The major conclusions drawn from treating the above reactions, and many others reported in the following without detailed derivations, are: a) The results obtained by the Euler integration are in complete agreement with the results obtained by the model presented in section 3.1 yielding Eqs.(3-6) and (3-8) to (3-1Oa) for predicting the probabilities pll and pjk. Thus, one may apply each of the methods, depending on his conveniece. However, by gaining enough experience, one starts to 'feel' that the method based on the Markov chains is easier and becomes 'automatic' to apply. In addition, chemical reactions are presented in unified description via state vector and a one-step transition probability matrix. b) Reversible reactions should be transformed into a set of irreversible reactions. c) The above reactions, treated in detail, provide the reader with a good introduction for applying the probabilities pjj and Pjk to chemical reactions.

3.4 APPLICATION OF KINETIC MODELS TO ARTISTIC PAINTINGS Prior to modeling of chemical reactions in the next sections, it is interesting to demonstrate how simple kinetic models can also be applied to artistic paintings.

No reaction

The first example applies to Fig.2-52. The painting in the figure, Development ZZ by Escher [lo, p.2761, demonstrates the development of reptiles, and at first glance it seems that their number is increasing along the radius.

205

Although their birth origin is not so clear from the figure, it was possible, by counting their number along a certain circumference, to conclude that it contains exactly eight reptiles of the same size. This number is independent of the distance from the center. Thus, if we designate by C1 the number of reptiles at some distance from the origrn, it follows that r l =- -dC1 dt

-0

or alternatively in a discrete form

Cl(n+l) = Cl(n)

This result indicates the absence of a chemical reaction. Although the reptiles become fatter versus the number of steps (time), their number is unchanged.

zero order reaction The second example refers to Fig.3-la showing various kinds of "winged creatures' in a drawing of a ceiling decoration designed by &her in 1951 [lo, p.791 for the Philips company in Eindhoven. If the number of the "winged creatures", designated as the "concentration" C1, is counted along the lines corresponding to steps (time) n = 0, 1, 2, ... shown schematically in Fig.3-lb for cases 1 to 3, the results summarized in Table 3-1 are obtained. The general trend observed is a decrease versus time of the 'concentration'.

206

Fig.3-la.

Winged creatures” demonstrating zero order reaction (M.C.Escher “Ceiling Decoration for philips” 0 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved)

207 n=o

, Case 1

Case 3

Case 2

Fig.3-lb. Three configurations for determination of the creatures concentration" C1

I

winged

Table 3-1. "Concentration" C1 of "winged creatures" versus time n

4

5

9

10

casel*,Cl(n) 12 11 10 9 8 7 6 5 4 case2*, Cl(n) 25 23 21 19 17 15 13 11 9 case3*, Cl(n) 48 40 32 24 16 8

3

2 5

10

1

2

3

6

7

8

timen

7

* see Fig.3-lb It should be noted that in case 1 of Fig.3-lb, the lines for n = 0 and 11 correspond to twelve butterflies and a single fly, respectively, in Fig.3-la. In case 2 these lines correspond to twenty five and three "winged creatures", respectively, whereas in

case 3 the lines correspond to forty eight and eight "winged creatures", respectively. In order to fit the concentration data C1 (number of "winged creatures" along a line) versus the time n reported in Table 3-1, the common approach of fitting experimental data to a hnetic model is applied. Thus, the simplest model of a zero k

order reaction is tested which corresponds A1 --+ A2. The rate equation reads

208 (3-44) Eqs.(3-6) and (3-8) are applied for determination of the probabilities which yield the following matrix:

P=

1 1 - [WCl(n)]At

2

0

[k/C~(n)]At 1

(3-45)

Cl(n+l) = C,(n) - kAt C2(n+1) = kAt + C2(n)

(3-46)

Fitting the data in Table 3-1 by Eq.(3-46) for At = 1, yields the following equations corresponding to Fig.3-1b: Case 1: Cl(n+l) = Cl(n) - 1; C2(n+l) = 1 + C2(n) where Cl(0) = 12; C2(0) = 0 Case 2: Cl(n+l) = Cl(n) - 2; CZ(n+l) = 2 + C2(n) where Cl(0) = 25; C2(0) = 0 Case 3: Cl(n+l) = Cl(n) - 8; C2(n+l) = 8 + C2(n) where Cl(0) = 48; C2(0) = 0 The excellent fit to Eqs.(3-30) of the data given in Table 3-1, indicates that the concentration-time dependence of the "winged creatures" in Fig.3- 1a according to the configurations depicted in Fig.3-lb, obeys a model of zero order reaction. The significance of the quantities C2(n) is as follows. Since Cl(n) is decreasing versus time, i.e. the number of the "winged creatures", the conservation of mass requires that they are found in state A2 according to the reaction A, + A,.

mth order reaction The third example refers to Fig.3-2 which is a woodcut by Escher [lo, p.118, 3251 showing moving fish of changing size. Here Escher demonstrated an infinite number by a gradual reduction in size of the figures, until reaching the limit of infinite smallness on the straight side of the square. If the number of fish along the square perimeter, designated as "concentration" C 1, is counted, the obtained results are summarized in Table 3-2. Fig.3-lb, case 3, shows schematically the

209 fish orientation along a squre which was counted, where each circle symbolizes a fish. Also, along a certain square, each fish is located exactly behind (or before) the other, and all are of the same size. The case of n = 0 corresponds to the the square located almost at the sides of the square where n = 6 corresponds to the most inner square comprising of four fish.

Fig.3-2. Fish orientation for demonstrating an rn th order reaction (MCEscher "Square limit" 0 1998 Cordon Art B.V. - Baarn - Holland. All rights reserved)

Table 3-2. "Concentration" C1 versus time of moving fish along the square perimeter time n

1 0

1

2

3

4

5

6

376

184

88

40

16

4

399

185

84

37

15

5

t

C 1(n> C1,calc(n)*

760

* rounded values computed by Eq.(3-49)

210 In order to fit the "concentration" data C1 (number of moving fish along a square) versus the time n reported in Table 3-2, an mth order reaction is tested k

correspondng to A1 +A2. The rate equation reads:

where the application of Eqs.(3-6) and (3-8) yields the following transition matrix:

P=

1 1 - kClm-l(n)At 2

0

kClm-l(n)At 1

(3-48)

Thus, Cl(n+l) = Cl(n)[l - kAtC?-'(n)] C2(n+l) = kAtC?(n) + C2(n)

(3-49)

Fitting the data in Table 3-2 by Eq(3-49) for At = 1, which was modified to the following equation [Cl(n) - Cl(n+l)] = kCf(n), yields m = 0.904 and k = 0.8% with a mean deviation of 8.3% between calculated data with respect to counted values in Fig.3-2. The above examples indcate that the application of kinetic models to artistic paintings has been successful.

3.5 INTRODUCTION TO MODELING OF CHEMICAL REACTIONS In the following, a solution generated by the discrete Markov chains is presented graphically for a large number of chemical reactions of various types. The solution demonstrates the transient response Cj-nAt and emphasizes some characteristic behavior of the reaction. The solution is based on the transition probability matrix P obtained on the basis of the reaction kinetics by applying

21 1 Eqs.(3-6), (3-8) to (3-10a) for computing the probabilities p i and pjk. It should be emphasized that the rate equations for the kinetics were tested to satisfy Eq.(3-3). In order to obtain the transient response, Eq.(3-20) is applied where the initial state vector S(0) is given by Eq.(3-19a). In each case, the magnitude of S(0) are the quantities on the Ci axis of the response curve corresponding to t = 0. An important parameter in the computations is the magnitude of the interval At. This parameter has been chosen recalling that Pjj and pjk should satisfy 0 5 pjj and pjk I1 on the one hand, and that Cj versus nAt should remain unchanged under a certain magnitude of At, on the other. In addition, a comparison with the exact solution has been conducted in many cases, which made it possible to evaluate the accuracy of the solution obtained by Markov chains. The quantities reported in the comparison are the maximum deviation, Dmax, and the mean deviation, Dmean. On the basis of these comparisons, a representative value of At = 0.01 is recommended, which is the parameter of Markov chains solution. Finally, it should be emphasized that by equating the reaction rate constant (one or a few) to zero in a certain case, it is possible to generate numerous interesting situations. The reactions are presented according to the following categories: 1) Single step irreversible reaction. 2) Single step reversible reaction. 3) Consecutive-irreversiblereactions. 4)Consecutive-reversible reactions. 5 ) Parallel reactions: single and consecutive-irreversible reaction steps. 6) Parallel reactions: single and consecutive-reversible reaction steps. 7) Chain reactions. 8) Oscillating reactions. The following definitions are applicable [31,321: Consecutive chemical reactions are those in which the initial substance and all the intermediates products can react in one direction only, i.e.:

Parallel chemical reactions are those in which the initial substance reacts to produce two different substances simultaneously, i.e.:

212

Reversible reactions are those in which two substances entering a single simple consecutive chain reaction interact in both forward and backward directions, 1.e.:

Conjugated reactions are two simultaneous reactions in which only one substance A1 is common to both, i.e.:

A,

+

A,+

+

%-As

A,

All three substances Al, A2 and A3 must be present in the reaction mass in order for both reactions to take place concurrently. Consecutive-reversible reactions are those in which two or more reactions, each of different type, occur simultaneously,for example:

Parallel-consecutive reactions belong to the mixed type which have the characteristics of both parallel and consecutive reactions. The following example comprises two parallel chains, each composed of three simple reactions:

A parallel-consecutive reaction becomes complex when species that belong to different chains interact as shown below:

213

Chain reactions. If the initial substance and each intermediate reaction product interact simultaneously with different substances and in different

directions, such processes are known as chain reactions. For example, the following scheme is a chain reaction in two stages. Other types of chain reactions are described in [22].

3.6 SINGLE STEP IRREVERSIBLE RE-CTIO k

3.6-1

(3.6-1)

A1 -) A3

where

rl = - kC7

(3.6- 1a)

By applying Eqs.(3-6) and (3-lo), the following one step transition probability matrix is obtained:

P=

1 1 - kCim-l(n)At kCim-l(n)At 2

0

1

(3.6- 1b)

214 where 1, 2 stand for states (chemical species) A1 and A2, respectively. From Eqs.(3-19a), (3-20),one obtains that:

(3.6-Ic)

The variation of Ci against t = nAt for the initial state vector C(0) = [Cl(O), C2(0)] = [ l , 01 is depicted in Figs.3.6-1 (a to d) for different values of the parameter m = 0,0.5, 1 and 3 where also the effect of the reaction rate constant k is demonstrated. 1

0.8

u

.-

0.6 0.4

0.2 0

0

0.2

0.4

t

0.6

0.8

10

0.2

0.4

0.6

t

0.8

1

Fig.3.6-la. Ci versus t demonstrating the effect of k for m = 0 in Eq.(3.6-la)

0

0.5

1 t

1.5

2 t

Fig.3.6-lb. Ci versus t demonstrating the effect of k for m = 0.5 in Eq.( 3.6- l a )

215

k = 1-

-

t

t

Fig.3.6-lc. Ci versus t demonstrating the effect of k for m = 1 in Eq.( 3.6- 1a) 1

0.8 0.6 .-

u

0.4 0.2

0 t

t

Fig.3.6-ld. Ci versus t demonstrating the effect of k for m = 3 in Eq. (3.6-l a ) It should be noted that exact solutions for the above models are available in refs.[32, vol.1, p.361; 34, pp.4-5, 4-61. For At = 0.01, the agreement between the Markov chain solution and the exact solution is Dmax= 0.4% and Dmean= 0.2%. k

3.6-2

a l A l + a2A2+ a3A3

(3.6-2)

where rl = - kC',C';I

r2 = - rkC1C;

r = a2/al

The following one step transition probability matrix is obtained:

(3.6-2a)

216

1 l-kC'F'(n)Cy(n)At

kRC'r'(n)C:(n)At

0

P = 2

0

l-rkC\(n)Cy-'(n)At

r2kRC\(n)Cy-'(n)At

3

0

0

1

(3.6-2b)

The following cases were explored:

3.6-2a

For the initial state vector C(0) = [Cl(O), C2(0), C3(0)] = [l, 0.5,0]:

1

0.8

u

.-

0.6 0.4

0.2

0

1 i=l

t

0

0.5

1 t

1.5

1

2

t

Fig.3.6-2a. Ci versus t demonstrating the effect of r for 1 = 0, m = 3/2 and k = 5 in Eq.(3.6-2a) For At = 0.04, the agreement between the Markov chain solution and the exact solution [32, vol.1, p.3611 is Dmax = 1.9% and Dmean = 0.3%.

217

3.6-2b

For an initial state vector C(0) = [Cl(O), C2(0), C3(0)] = [2, 3,Ol:

3 2.5

2

6-1.5 1

0.5 0

0

0.5

1

1.5

t

t

Fig.3.6-2b. Ci versus t according to Eq.(3.6-2a). r l = kC11’2C2 (left), rl = kC1C21’2 (right) for k = 1, r = 1

-

-

For At = 0.005, the agreement between the Markov chain solution and the exact solution [32, vol.1, p.3611 is Dmax= 1.2% and Dmean = 0.6%.

3.6-3

A , + A,+A,

where

k

+ A4

(3.6-3)

rl = r2 = r3 = - r4 = - kClC2C3

(3.6-3a)

yields the following transition probability matrix

1-

1

0

0

-kC2(n)C3(n)At 1 3

0

;kCl(n)C3(n)At

kC2(n)Cg(n)At

2

0

3

0

0

0

0

P=

1kC, (n)C3(n)At

11 -kC ,(n)C,(n)At kCl(n)C2(n)At 3

I

4 I

0

1

(3.6-3b)

218 Ci(ni-1) is obtained by applying Eq.(3-20) where the effect of the initial state vectors C(0) = [Cl(O), C2(0), C3(O), C4(0)] = [l, 2, 3, 01 and [3, 2, 1, 01 is demonstrated in Fig.3.6-3a. 3

2.5

I

I

I

I

I

,

. I

2 0'- 1.5

1

0.5

______------

4------

0

0

I

I

I

I

0.2

0.4

0.6

0.8

t

1

t

Fig.3.6-3. Ci versus t demonstrating the effect of Ci(0) for k = 1 For At = 0.01, the agreement between the Markov chain solution and the exact solution [31, p.201 is Dmax = 2.3% and Dmean = 1.5%. For At = 0.005, Dmax = 1.4% and Dmean = 1%. kl

3.6-4

*I

where

--j

(3.6-4)

A2

rl = - klC1/(l + k2C1)

(3.6-4a)

yields the following transition probability matrix:

1 1 - [kl/(l+ k2Cl(n)lAt [k,/(l+ k2CI(n)lAt

P= 2

0

1

(3.6-4b)

Fig.3.6-4 demonstrates the effect of the reaction rate constants kl and k2 on the species concentration distribution for the initial state vector C(0) = [C1(0),

C2(0)1 = E l , 01.

219 1

0.8 .-

u

-

0.6

-

0.4

-

0.2

.-

\\

u

OO

u

I

I

-

0.5

1

'

k 1 = 1 , k 2 =0.1

I /

/

/

\,

\

-

, '

2

-

/ I

-

I

I

2

1.5

t

t

0.6 0.4

0.2

0

Fig.3.6-4. Ci versus t demonstrating the effect of k l and k2

3.7 SINGLE STEP REVERSIBLE REACTIONS

3.7-1

kl

(3.7-1)

A1 = A 2 k2

where rl = - r2 = - (klC1- k,C2)

(3.7- 1a)

By applying the approach detailed in section 3.2-1, i.e., treating the reversible reactions as two irreversible ones demonstrated in Eqs.(3-lOa) and (3lob), the following transition probability matrix is obtained: 1 1 1 - klAt

2 k,At

P= 2

k,At

1 - k,At

(3.7-1b)

220 Thus, from Eqs.(3-19a), (3-20), one obtains that: Cl(n+l) = Cl(n)[l - klAtl + C2(n)[k2Atl C2(n+l) = Cl(n)[klAtl + C2(n)[l - k2Atl

(3.7-lc)

yielding the following curves for the initial state vector C(0) = [Cl(O), C2(0)] = [l, 01: 1h

I

I

I

I

t

0.8

u

.-

0.6 0.4

-1

t

1

t t

Fig.3.7-1. Ci versus t demonstrating the effect of k2 for k l = 1 At steady state, the results verified the relationship which follows from Eq.(3.7-la), i.e., ( C Z / C ~ ) ~klk2. ~ . = For At = 0.01, the agreement between the Markov chain solution and the exact solution [34, p.4-7; 48, p.20; 49, p.851 is Dmax= 0.3% and Dmem= 0.1%.

3.7-2

kl

A1 j 2 A 2 t

(3.7-2)

k2

The transition matrix based on Eqs.(3-26a,b) was developed before and is given by Eq.(3-28). The transient response of C1 and C2 for the initial state vector C(0) = [Cl(O), C2(0)] = [ l , 01 is demonstrated in Fig.3.7-2. At equilibrium, the results verified the relationship which follows from Eqs.(3-26a,b), i.e., ( C ~ / C I ~ ) ~kik2 , . = where no analytical solution is available for comparison.

22 1 1

0.8 0.6 0;-

0.4

i

t7

L2-J

o.2 OO

0.5

1

1.5

2

0

0.5

t

1

1.5

2

t

Fig.3.7-2. Ci versus t demonstrating the effect of k2 for k l = 1 according to Eqs.(3-26a,b)

3.7-2.1

kl

(3.7-2.1a)

A1 + A 2 + A 3 t k2

where rl = - r2 = - r3 = - k,Cl

+ k2C2C3

(3.7-2.1b)

yields the following transition probability matrix:

The transient response of C1, C2 and C3 for the initial state vector C(0) = [C1(0),C2(0), C3(0)] = [ l , 0, 01 is demonstrated in Fig.3.7-2.1.

222 I

-

-

c2 = c3

0.2 -

I

I

I

t

Fig.3.7-2.1. Ci versus t demonstrating the effect of k2 for kl = 1 For At = 0.01, the agreement between the Markov chain solution and the = 0.36% and Dmem = exact solution [32, p.79; 44;48, p.20; 49, p.851 is 0.23%.

(3.7-3) The transition matrix based on Eqs.(3-34a,b) was developed before and is given by Eq.(3-36). The transient response of C1 and C2 for the initial state vector C(0) = [C1(0), C2(0)] = [I, 01 is demonstrated in Fig.3.7-3. The remarks made in 3.7-2 are also applicable here. I

G-

0.2 -

0.4

O

2

A

H I

/---=I

I

I

k =5

I

-

- -

-

- -

2 - - - - I

I

I

223 kl

3.7-4

2A1

t

(3.7-4)

A2+ A3

k2

where rl = - kC:+ k2C2C3 r2 = r3 = 0.5kC:

(3.7-4a)

- 0.5k2C2C3

(3.7-4b)

yields the following transition probability matrix:

(3.74) where the computation of pi2 = pi3 was made by Eqa(3-10a). The transient response of C1, C2 and C3 for the initial state vectors C(0) = [Cl(O), C2(0), C3(0)] = [ l , 0.1, 01 and [l, 0, 01 is demonstrated in Fig.3.7-4. 1,

V.U

--

0.2

t

t

Fig.3.7-4. Ci versus t demonstrating the effect of k l for k2 = 1

For At = 0.005, the agreement between the Markov chain solution and the = 0.7%. For At = exact solution [32, p.35; 49, p.861 is Dmax = 3.1% and D, 0.01, Dmax= 7.2% and Dmean = 2.6%.

224 kl

3.7-5

2A1+A2

(3.7-5)

t k2

where

rl = - r3 =

- 2klC:C2 + 2k2C:

(3.7-5a)

r2 = - klCfC2 + k2C:

(3.7-5b)

yields the following transition probability matrix: 1 1 1-2kl C (n)C2(n)At

2 0

3

4 3klC1(n)C2(n)At

P= 2 (3.7-5c)

3

The transient response of C1, C2 and C3 for the initial state vector C(0) = [Cl(O), C2(0), C3(0)] = [l, 1, 01 is demonstrated in Fig.3.7-5.

6 '

0.4

Oe2

----

I

-

_-

_.

'I

3, -,/

I I

I

k =1 1

- -

I

t

t

Fig.3.7-5. Ci and the ratio Ci/(C?C,

)

versus t demonstrating the

effect of k2 for k l = 1 No exact solution is available for this reaction [32, v01.2, p.761. However, it should be noted that the ratio C:/(CfC2 ) approaches at steady state the ratio k l k 2

as predicted from Eqs.(3.7-5a,b).

225 kl

3.7-6

(3.7-6)

A1+A2 + A 3 c k2

where rl = r2 = - r3 = - klClC2+ k2C3

(3.7-6a)

yields the following transition probability matrix:

(3.7-6b) The transient response of C1, C2, C3 and C3/(CiC2)for the initial state vector C(0)= [Cl(O), C2(0), C3(0)] = [ l , 0.5,0] is demonstrated in Fig.3.7-6.

,

I

0.2 - '

' 3 / -

I

t

-,----=

/ I

t

Fig.3.7-6. Ci and the ratio C3/(CiC2)versus t demonstrating the effect of k2 for kl = 1 For At = 0.01, the agreement between the Markov chain solution and the exact solution [33, p.43; 48, p.201 is Dmax = 6.5% and Dmean = 1.5%. In addition, the ratio C3/(C1C2) approaches at steady state the ratio kl/k2 as predicted from Eqs.(3.7-6a,b).

226

3.7-7

kl

A1+A2 -)A3+A4

(3.7-7)

t k2

where rl = r2 = - r3 = - r4 = - k1ClC2+ k2C3C4

(3.7-7a)

yields the following transition probability matrix:

0

7klC2(n)At 1

7klC2(n)At 1

l-kiCi(n)At

T1k l C l (n)At

71k , C l (n)At

1 l-klC2(n)At

P=

0

2

3 .4

1 7k2C4(n)At 1 l-k2C4(n)At 7k2C4(n)At 7k2C3(n)At 1 7k2C3(n)At 1

0

0

1-k2C3(n)At

(3.7-7b)

The transient response of C1 to C4 and C3C4/(ClC2) for the initial state vector C(0) = [Cl(O), C2(0), C3(O), C4(0)] = [I, 0.5, 0.25, 01 is demonstrated in Fig.3.7-7. k =l!k=l

I

I

\

... '

-_..___.

W

0.6 .-

0.4

C3C4/(C,C2)-,

A'1 ' . -

0.2 -

d

-

'

---

-

'

-----__k I = l i k 2 = 4 1 -

-

-

I

_-

I

I

+

i= 1

- -

.

.

,

- . . . ..

f

I

For At = 0.01, the agreement between the Markov chain solution and the exact solution, not existing for kl = k2, [3 1, p. 187; 44; 48, p.20; 49, p.861 is Dmax

227 = 8.1% and Dmean = 5.3%. The ratio C3C4/(ClC2) approaches at steady state the

ratio k l k 2 as predicted from Eqs.(3.7-7a). kl

3.7-8

(3.7-8)

A 1 + A 2 + A 3 :A4 k2

where rl = r2 = r3 = - r4 = - klC1C2C4+ k2C4

(3.7-8a)

yields the following transition probability matrix:

1klC2(n)C3(n)At

P=

0

7klC2(n)C3(n)At 1

0

2

0

3

0

0

4

k2At

k2At

1k,Cl(n)C3(in)At

-$k,Cl (n)C3(n)At

0

1I k , C , (n)C2(n)At klC (n)C2(n)At 3

k,At

l-k?At

(3.7-8b)

C ~the ) initial state The transient response of C1 to C4 and C ~ / ( C I C ~ for vector C(0) = [Cl(O), C2(0), C3(O), C4(0)] = [ l , 0.9, 0.8, 01 is demonstrated in Fig.3.7-8. I

U V

0

I

I

I

0.5

1 t

1.5

I

V

2 0

1

I

I

I

I

I

I

0.5

1 t

1.5

2

Fig.3.7-8. Ci and the ratio C4/(ClC2C3) versus t demonstrating the effect of k2 for k l = 1

228 The present reaction appears in [35, p.1481 with no exact solution. As seen in Fig.3.7-8, the ratio Cd(C1C2C3) approaches at steady state the ratio kl/k2 as predicted from Eqs.(3.7-8a).

3.8 CONSECUTIVE IRREVERSIBLE REACTIONS

(3.8-1) where rl = - klC;l

r2 = - (l/al)dCl/dt - k2C9

r3 = - (l/a2)dC2/dt

(3.8- 1a)

yields the following transition probability matrix:

P = 2 3

(3.8- 1b)

where pi2 and p23 were computed by Eq.(3-lOa). The transient response of C1, C2 and C3 for the initial state vector C(0) = [Cl(O), C2(0), C3(0)] = [l, 0, 01 is demonstrated in Figs.3.8-la to d for various combinations of a1 and a2 in Eq.(3.81).

229

3.8-la

kl

k2

A1 + A2 + A3

t

t

Fig.3.8-la. Ci versus t for a1 = a2 = 1 in Eq.(3.8-1) demonstrating the effect of k2 for kl = 1 For At = 0.015, the agreement between the Markov chain solution and the exact solution [31, p.166; 49, p.90; 511 is Dmax = 2.2%and Dmean= 1.1%.

I

0 t

1

3

2

4

5

t

Fig.3.8-lb. Ci versus t for a1 = 2, a2 = 1 in Eq.(3.8-1) demonstrating the effect of k2 for k l = 1 For At = 0.01, the agreement between the Markov chain solution and the exact solution [33, p.951, which is very complicated, is Dmax = 0.4% and D m e m =

230 0.3%. It should be noted that for large values oft, C3 should approach 1 whereas fort = 150, C3 = 0.497.

1,

I

I

I

I

I

I

I

C2(0)= 0, k2= 1

I

\

OO

2

4

6

8

I

I

t

1

0.8

I

I

C$O) = 0, k2 = 5

10

-

i=l

-

6‘

-

0.2

_----

-

---”---.--

3’

I

2

4

6 t

8

10 t

Fig.3.8-lc. Ci versus t for a1 = 2, a2 = 2 in Eq.(3.8-1) demonstrating the effect of k2 and C2(0) for kl = 1

For At = 0.02, the agreement between the Markov chain solution and the exact solution [36; 511 is Dmax = 0.7% and Dm,,, = 0.5%.

23 1

t

t

Fig.3.8-ld. Ci versus t for a1 = 1, a2 = 2 in Eq.(3.8-1) demonstrating the effect of Ci(0) for k l = 1 and k2 = 5 For At = 0.01, the agreement between the Markov chain solution and the exact solution [36; 32, ~01.2,p.511 is Dmax = 2.4% and Dmean = 4.9%. (3.8-2)

where rl = - klC1 + k,C,

r2 = - k2C2+ klCl

r3 = - k3C3+ k2C2

(3.8-2a)

yields the following transition probability matrix:

1

P= 2

3

l-kiAt 0 k3At

k1At l-k,At 0

0 k,At l-k,At

(3.8-2b)

232 The transient response of C1, C2 and C3 for the initial state vector C(0) = [C1(0), C2(0), C3(0)] = [l, 0, 01 is demonstrated in Fig.3.8-2 for various combinations of the reaction rate constants. No exact solution is available for comparison.

3/

5.

I

I

I

I

I 1

-

-

-

- - _. -

1 -

I

I

I

I

I

k =3,k = 2 2

-

\I

\

-

-Ib

- - - - -

-i‘v[l - k2Atl C3(n+l)

-

C2(n)[k2Atl + C,(n)[l

- k3Atl

(3.8-7~)

An exact solution for the present case appears in [22, p.9; 33, p.521. Particular solutions by Markov chains appear above in cases 3.8-la (n = 3) and 3.8-6 (n = 4).

3.9 CONSECUTIVE REVERSIBLE REACTIONS

3.9- 1

k,

k,

(3.9-1)

A1 + A 2 2 A 3 k-2

where

rl

=

- klCl r2 = - k2C2 + k-2C3 + klCl

r3 = - k-2C3

yields the following transition probability matrix:

+ k2C2

(3.9-1a)

238 1 l-klAt

k1At

0 0

P = 2 3

l-k,At k-2At

0 k2Aht l-k-2At

(3.9-1b)

The transient response of C1, C2, C3 and the ratio C3K2 for the initial state vectors

C(0) = [Cl(O), C2(0), C3(0)] = [l, 0, 01 and [l, 0, 11 is depicted in Fig.3.9-1

where the effect of C3(O) is demonstrated. 1

d

-

0.8

0.6

u"

-- ---

//2-

0.4

0.2

0

-

-1-

C3(0)= 1 -

-

I

0

0.2

0.4

0.6

0.8

1

t

t

Fig.3.9-1. Ci and the ratio C3/C2 versus t demonstrating the effect of C3(O) for kl = k2 = k.2 = 5 For At = 0.005, the agreement between the Markov chain solution and the exact solution [32, v01.2, p.261 is Dmax= 6.2% and DmW = 3.1%. As seen also in Fig.3.9-1, the ratio C 3 Q approaches at steady state the value k2k-2 as predicted from Eqs.(3.9-la).

(3.9-2)

rl

--

klCl + k-,C2

r2 = - (k-l + k2)C2 + klC,

yields the following transition probability matrix:

r3 = k2C2

(3.9-2a)

239

1 1 l-k, At P = 2 k-1At 3 0

2 k1At l-(k-l+ k2)At

3 0 k2At

0

1

(3.9-2b)

The transient response of C1, C2 and C3 for the initial state vector C(0) = [Cl(O), C2(0), C3(0)] = [l, 0, 01 is demonstrated in Fig.3.9-2. 1

1

I

I

k =O-

0.2

-

-

3 I

I

1

t

t

Fig.3.9-2. Ci versus t demonstrating the effect of k2 fork1 = k l = 5 For At = 0.01, the agreement between the Markov chain solution and the exact solution [22, p.24; 511 is Dmax= 3.6% and D ,, = 0.8%.

(3.9-3)

yields the following transition probability matrix:

240 1 2 3 1 1-k1 At 0 k1At k2At P = 2 k-1At l-(k-,+ k,)At 0 k-2At 1-k-2A t 3 (3.9-3b) The transient response of C1, C2, C3 and the ratios C21C1, C3/C2 for the initial state vector C(0) = [Cl(O), C2(0), C3(0)] = [l, 0, 01 is demonstrated in Fi 8.3.9-3. 1.5-

I

I

,

t

.-

_

_

I

-

___------

t

Fig.3.9-3. Ci, C2/C1 and C3/C2 versus t demonstrating the effect of ki and k i f o r kl = 5 For At = 0.01, the agreement between the Markov chain solution and the exact solution [22, p.42; 31, p.175; 421 is D- = 1.4%and Dmem = 0.5%. As observed in Fig.3.9-3, the ratios C21C1 and C3/C2 approach at steady state the limits k l k l and k2k2, respectively.

(3.9-4)

A3

241 where

From Eq.(3.9-4),the steady state conditions for the system follows from rl = r2 = r3 = 0, yielding

(3.9-4b)

(3.9-4) Eqs.(3.9-4a) yield the following transition probability matrix:

(3.9-4)

The transient response of C1, C2, C3 and the ratios C1/C3, C2/C1 and C3K2 for the initial state vector C(0) = [Cl(O), C2(0), C3(0)] = [l, 0, 01 are demonstrated in Fig.3.9-4.

242

0.2 t,2 0

3-

0

--I

0.2

-

I

I

I

0.8

0.6

0.4

i

__

--

1

0

I

4 -1

I

k as in (a)

0.8

1

t

1 (C)

,

-

-

3 . V” . V 2

I:,

I

I

k I as in (b)

I

i

1

0

0.6

0.4

0.2

t

0

0.2

0.4

0.6

0.8

0

0.2

t

0.4

---0.8

0.6

t

Fig.3.9-4. Ci and Ci/Cj versus t demonstrating the effect of k.i for k l = k2 = k3 = =5 For At = 0.01, the agreement between the Markov chain solution and the exact solution [32, v01.2, p.31; 35, p.921 is Dmax= 2.0% and Dmean= 0.3%. As observed in Fig.3.9-4, the ratios Ci/Cj approach at steady state values predicted by Eq.(3.9-4b).

3.9-5

(3.9-5)

where rl = - klCl + k l C 2

r2 = - (k-l

r3 = - k3C3 + k2C2

r4 = k3C3

yields the following transition probability matrix:

+ k2)C2+ klC1 (3.9-5a)

243

1 2 3 1-k1 At k1At 0 1 p = 2 k l A t l-(k-l+ k2)A.t k,At l-k,At 0 0 3 4

0

0

4

0 0 k3At

1

0

(3.9-5b)

The transient response of C1 to C4 for the initial state vector C(0) = [Cl(O), C2(0), C3(O), C4(0)] = [l, 0, 0, 01 is depicted in Fig.3.9-5 where the effect of k3 is demonstrated.

-

1.5

1

2

2.5 0

0.5

t

1.5

1

2

2.5

t

Fig.3.9-5. Ci versus t demonstrating the effect of k3 f o r k l = k.1 = k2 = 5 and At = 0.01 A complicated exact solution is available [22, p.271.

(3.9-6) where rl = - klCl + k l C 2

r2 = - (k-l+ k2)C2 + k,C,

r3 = k2C2 + k-3C4 - (k-2 + k3)C3 yields the following transition probability matrix:

+k2C3

r4 = k3C3- k-3C4

(3.9-6a)

244

(3.9-6b) The transient response of C1 to C4 for an initial state vector C(0) = [Cl(O), C2(0), C3(O), C4(0)] = [l, 0, 0, 01 is depicted in Fig.3.9-6 where the effect of ki is demonstrated.

t

t

Fig.3.9-6. Ci versus t demonstrating the effect of ki for kl = kz,= k3 = 5 and At = 0.01

A complicated exact solution is available [22, p.441.

(3.9-7)

--

-

+ k-,C2 r2 - (kl + k2)C2 + klCl + k-2C3 r3 = k2C2 - (k-2 + k3)C3 r4 = k3C3 - k4C4 rs = k4C4 rl

klCl

yields the following transition probability matrix:

(3.9-7a)

245

1 2

P= 3

1 2 l-k, At k1At k-1At l-(k-l+ kJAt 0 k-zAt

3

4

5

0

0 0

0 0 0

k4At 1

k2At l-(k-2+ k3)At

4

0

0

0

k3At l-k4At

5

0

0

0

0

(3.9-7b)

The transient response of C1 to Cs for the initial state vector C(0) = [Cl(O), C2(0), C3(O), C4(O), C5(0)] = [l, 0, 0, 0, 01 and [l, 0, 0.5, 01 is depicted in Fig.3.9-7 where the effect of l~ is demonstrated. 1 0.8

0.6

6'

0.4

0.2 0 t

t

Fig.3.9-7. Ci versus t demonstrating the effect of f o r k l = k z = k j = l O , k l = k 2 = 1 a n d A t = 0.01 A complicated exact solution is available [22, p.291.

(3.9-8)

(3.9-8a)

246 yields the following transition probability matrix:

(3.9-8b) The transient response of C1 to C4 for the initial state vector C(0) = [Cl(O), C2(0), C3(O), C4(0)] = [ l , 0.9, 0, 01 is depicted in Fig.3.9-8where the effect of ki, k i is demonstrated.

t

t

Fig.3.9-8. Ci versus t demonstrating the effect of ki and k.i for At = 0.01 The present reaction is considered in [35, p.1491 without an analytical solution.

3.9-9 where

(3.9-9)

247

- kIC1C2+ k-1C3 - k3C1C5 + k3C4 r3 = - (k-l + k2)C3 + k,ClC2 + k-2C4 r4 = - (k-2 + k3)C4 + k2C3+ k-3C1C5 rl =

1

1-

r2 =

- k,ClC2 + k l C 3

+ k3C4

r5 = - k-3C1Cs

(3.9-9a)

0

i k C2(n)At ik-3C5(n)At

0

[klC2(n)+k_,C5(n>lAt

2

1TklCl(n)At 1 klCl(n)At

0

P= 3

k- 1At

k-lAt

4

k3At

0

k,At

5

0

0

0

1

I

0.8 .-

0.6

-

5

I

k, = k, = k3 = 10

1-

(k_,+k,)At

I

. - . - _ _ .. . . . . ~

. .

U

0.4 -

-

0 , 2 - - - - - -4 - - - - / I

-

-

0

k3At

i k - 3 C 1 (n)At

I

1k-,Cl(n)At

1

k, = k,= k,= 1

-L/
+ C3(n) + C2(n+l) = C2(n)[l - k,C,(n)Atl

+ Cz(n))[klAtl + {C,(n> + C4(n>+ + Cz(n)}[k-lAtl

C3(n+l) = Cl(n)[0.5klC2(n)Atl+ C2(n>[0.5klCl(n)Atl + C3(n)[l - (2k-l + k2)Atl C4(n+l) = C3(n)[k2Atl + C4(n)[l - (2k-,

+ k2)Atl

(3.9-10d) For Z = 5 the transient response of C1 to C5 for the initial state vector C(0) = 1, 0, 0, 01 is depicted in Fig.3.9-10

l h

I

n o l \ : -
[l- PlAtI + Cz(n"1AtI i = 2, ..., Z-1 Ci(n+l) = Ci-l(n)[piAt] + Ci(n)[l - piAt] Cz(n+l>= Cz-l(n>[PzAtl+ Cz(n" - PZAtl A particular solution is obtained for Z = 5 and a constant Pi, i.e. all reactors have the same volume. For the pulse input C(0) = [Cl(O), C2(0), C3(O), C4(O), C5(0)] = [ l , 0, 0, 0, 01 the response curve for C1 to C5 is depicted in Fig.4.3-8a.

,,'

/ ,:-4

5

For At = 0.0005, the agreement between the Markov chain solution and the exact solution for C5 [21, p.2951 is as follows: for C5,exact= 0.001, D = 8.9%;

369 = 0.01, D = 3.3%; for C5,exact = 0.1, D = 0.074% and for C5,exact = 0.2, D = 0.012%. C5,exact

4.3-9 The following scheme is an open recirculating system with a recycle of magnitude Qzl. Q1+ Qzl

Q,+ Qzl

Ql+ Qzl

...

Q*l

Fig.4.3-9. An open recirculation system with recycle The above system can be deduced from Fig.4-1 as follows: j = 1, a = 2, etc. where E, = Z+1; in addition Q2 = 4 3 =, ..., Q, = 0 or a2 = a3 = ,..., a, = 0. Eq.(4-12a) gives 1 = PZ,,+l. Eq.(4-12b) gives 1+ a,l = a12. If we define recycle by R = Qzl/Q1 and consider Eq.(4-2) for kj = z l , it follows that a,l = R and that also a12 = 1 + R. Eq.(4-12c) gives for i = a = 2 that a12 = a23, for i = 3, 4. ..., Z-1, it gives that ai-l,i= ai,i+l= 1 + R, and for i = Z it is obtained that az-l,z = PZ,,+l(=l) + a,l = R + 1. The following probabilities were obtained: Eq.(4-14) gives p11 = 1- al2plAt = 1- (1 + R)plAt i = 1, ..., Z-1 Eq.(4-16) gives pii = 1 - ai,i+lpiAt = 1- (1 + R)piAt where p,, = 1 - (p,,,+l + azl)pzAt = 1 - (1 + R)p,At . Eq.(4-15) gives p12 = a12p2At = (1 + R)p2At Eq.(4-18a) gives pi,i+l = ai,i+1pi+lAt= (1 + R)pi+lAt i = 1, ..., Z - 1 i=Z Eq44-17b) gives Pz,z+l = Pz,z+lpz+lAt= lPZ,lAt Eq.(4-17a) gives pzl = azlplAt = RplAt For Z = 5 , while assuming that pi = p and considering the above probabilities, gives that:

370 2

3

4

(1+R)pAt

0

1

1 1- (1+R)pAt

6

0

5 0

0

2

0

1- (1+R)pAt

(l+R)pAt

0

0

0

3

0

0

1-(l+R)pAt

(l+R)pAt

0

0

0

0

0

l-(l+R)pAt

(l+R)pAt

0

5

RpAt

0

0

0

6

0

0

0

0

P= 4

l-(l+R)pAt pAt

1

0

For the pulse input C(0) = [l, 0, 0, 0, 0, 01, the response curve for C1 to Cg, computed from C(n+l) = C(n)P, is depicted in Fig.4.3-9a. 0.81-

0.6

9

i=l

--

-

0-

R=2

-

t

t

-

Fig.4.3-9a. Ci versus t demonstrating the effect of R for p 10 and At = 0.01

4.3-10

Fig.4.3-10. Perfectly mixed reactors with back flow

The equations for the above configumtion [21, p.2981 can be obtained from Fig.4-1 by designating j = 1, a = 2, b = 3, ..., 2 = 5 and = 6. In addition, Qji = Q j = O or a i j = a j i = 0, i = 3, ..., Z; Q = ai = 0, i = 2, ..., Z.

37 1 Eq.(4-12a) gives 1 = p56; Eq.(4-12b) gives 1 + a21 = a12; Eq.(4-l2c) for i = a = 2 gives a 1 2 + a32 = a 2 1 + a23, hence 1 + a 3 2 = a23; for i = 3, a 2 3 + a 4 3 = a 3 2 + a34, hence 1 + a 4 3 = a34; for i = 4,a 3 4 + a 5 4 = a 4 3 + a45, hence, 1 + a 5 4 = a45; for i = 5 , a45 = p56 + a54. The following probabilities were obtained considering the above results: p11 = 1 - a12plAt from Eq.(4-14). pii = 1 - (aij+ ai,i-l + ai,i+l)piAtfrom Eq.(4-16), i = 2, ..., Z - 1, i # j, 5. For i > 2, aij = 0. In addition, pzz = 1 - (pzs + az,z-l)pZAt. p12 = 1 - a12p2At from Eq.(4-15). pi,i+l = ai,i+lpi+lAt from Eq.(4-18a), i = 2, ..., Z - 1 where from Eq.(417b) p g = PzgpkAt. Finally, pi+l,i = ai+l,ipiAtfrom Eq.(4-18b), i = 1, ..., Z - 1. Assuming that pi = p, a21 = a 3 2 = 0143 = a54 = R, thus, a12 = a 2 3 = a34 = a45 = 1 + R where the rest values of a are zero. Considering the above, yields the following transition probability matrix which can be easily extended to a higher number of states:

1 2

1-

1 2 (l+R)pAt (l+R)pAt

3 0

4 0

5 0

6 0

RpAt

1- (1+2R)pAt

(l+R)@t

0

0

0

0

1- (1+2R)pAt

(l+R)pAt

0

0

RpAt

1- (1+2R)pAt

(l+R)jdt

0

3 P= 4

0

RpAt 0

5

0

0

0

RpAt

1- (1+R)at

pAt

6

0

0

0

0

0

1

For the pulse input C(0) = [l, 0, 0, 0, 0, 01 the response curve for C1 to C6, computed from C(n+l) = C(n)P according to above matrix, is depicted in Fig.4.3-10a. It may be observed that for large value of the back flow R, the concentration of the tracer becomes uniform after some time, i.e. all reactors act as a single reactor.

372 1.

0.8 -

I

I

I

I

1I

-

6

6

I

-

:1

0.6 -

"-0.4

I

R = 10

li=1

1

\

-

\,

R=O

I

4.3-11 The following configuration simulates a flow pattern in a tubular reactor [21, p.334, case d] in the presence of side-leaving streams.

Fig.4.3-11. Perfectly mixed reactors with side-leaving streams The equations for this configuration can be obtained from Fig.4-1 as follows, designatingj = 1, a = 2, b = 3, ..., Z = 4 and 5 = 5 while noting that ai =

0, Qil = 0 for i = 2, ..., 4. Eq.(4-12a) gives: 1 = p i 5 + p25 + p35 + p45. Eq.(4-12b) gives: 1 = p i 5 + a12. Eq.(4-12c) gives: for i = a = 2, a 1 2 = p25 + a23; for i = 3, a 2 3 = p35 + a34; for i = 4, a 3 4 = p45. Thus, if three values of p are given, the rest of the coefficients are known. The following probabilities were obtained considering the above results: pi1 = 1 - (pi5 + a12)plAt from Eq.(4-14) pii = 1 - (pi5 + ai,i+l)piAt from Eq.(4-16), i = 2, 3 where p44 = 1 - P45Wt In addition: pi2 = a12p2At from Eq.(4-15),

373 pi,i+l = ai, i+lpi+lAt from Eq.(4-18a), i = 2, 3 pi5 = P12p5At from Eq.(4-15b), and pi5 = PiSpSAt from Eq.(4-19), i = 2, 3 , 4 Defining pi = QlNi, p = QNwhere V = V1+ V2 +V3 +V4 and V1= mV, V2 = nV, V3 = pV, V4 = (1 - m - n - p)V for 0 < m + n + p c 1, while considering the above results, gives the following transition probability matrix:

1 2

P= 3 4 5

A specific example was obtained for V5 = V1, i.e. pg = p1 = p/m. Other parameters were: pi5 = 0.06 and 0.6, p25 = 0.2, p35 = 0.1; m = 0.5, n = 0.3, p = 0.1, and p = 2. For the pulse input C(0) = [ l , 0, 0, 0, 01, the response curve for C1 to C5, computed from C(n+l) = C(n)P, is depicted in Fig.4.3-lla for At = 0.005. 1

I

I

I

\

I

I

I

\

-

a = 0.06 15

-

I

I

5

\,

-

T1

-

-

1,

-,k /.>

-....

a = 0.6 I5

.-

.c4 . . . ....-.._ - . -....*

-

_'

. .

374

4.3-12

The following configuration is in some aspects similar to the previous one. It simulates a flow pattern in a tubular reactor [21, p.334, case b] in the presence of side feedings.

Fig.4.3-12. Perfectly mixed reactors with side feeding The equations for this configuration are obtained from Fig.4- 1 by designating a = 1, b = 2, ..., Z = 4 and 5 = 5 where reactor j is not active and the reference flow is Q = Q1 + 4 2 + 4 3 + 4 4 . As seen, the state space comprises of five reactors. Eq.(4-12a) gives: a1 + a 2 + a3 + a 4 = 1 = p45; Eqs.(4-12c) gives: a1 = a 1 2 for i = 1, a2 + a 1 2 = a 2 3 for i = 2, a3 + a 2 3 = a34 for i = 3 and a4 + a34 = p45. From the above equations it may be concluded that if a l , a2 and a3 are specified, the rest of the coefficients are known. Thus, the following probabilities are obtained from Eq.(4-16): P11 = 1 - al2plAt P22 = 1 - a23p2At P33 = 1 - a34p3At P44 = 1 - P45p4At The following probabilities are obtained from Eq.(4-18a): pi2 = aizp2At p23 = a23p3At p34 = a34p4At and from Eq.(4-19) P45 = P45p2At From Eq.(4-7) pi = QNi where p = QNand Q = Q1 + 4 2 + 4 3 + 4 4 . In addition, V = V1+ V2 +V3 +V4 where V1= mV, V2 = nV, V3 = pV and V4 = (1 - m - n p)V for 0 < m + n + p < 1. Thus, pi = l/m, p2 = pln, p3 = p/p and p4 = p/(1m-n-p). Considering the above results, gives the following transition probability matrix:

375 1

2

3

4

1 P11 p12 0 0 2 0 P22 P23 0 P= 3 0 0 P33 P34 4

5

0 0

0 0

0

0

5 0 0

0

p44 p45 0 1

A specific example was obtained for V5 = V1, i.e. p5 = p1 = p/m. Other parameters were: a1 = 0.2 and 0.6, a2 = 0.2 and a3 = 0.1; ai is defined by Eq(4-12d) where Qj = Q1 + Q2 + 4 3 + 4 4 . m = 0.5, n = 0.3, p = 0.1, and p = 2. For the pulse input C(0) = [ l , 0, 0, 0, 01, the response curve for C1 to C5, computed from C(n+l) = C(n)P, is depicted in Fig.4.3-12a for At = 0.005.

1

-

1

Fig.4.3-12a. Cj versus t demonstrating the effect of

4.3-13

ctl

The following configuration demonstrates an interacting system whereby it is possible to increase the mean residence time by applying operating different streams in the flow system. Reactor 5 is the tracer collector.

376

Fig.4.3-13. A model for demonstrating the increasing of the mean residence time The above configuration is obtained from the general scheme in Fig.4-1 by designating j = 1, a = 2, b = 3, ..., Z = 4 and 5 = 5 which is the 5th state. Eq.(4-12a) gives: 1 = p25 + p35 + p45 Eq.(4-12b) gives: 1+ a 2 1 + a31 = a12 + 0113 Eq.(4-12c) gives: a 1 2 + a 4 2 = p25 + a21 + a24 for i = 2, a13 + a 4 3 = p35 + a 3 1 + a34 for i = 3 and a 2 4 + a34 = p45 + a 4 2 + a 4 3 for i = 4. (4.3-13a) The above five equations contain eleven coefficients; thus, six of which must be prescribed in order to solve for the others. The following probabilities of remaining in the state are obtained from Eqs.(4-14)and (4-16): P11 = 1 - (a12 + a13)plAt P22 = 1 - (p25 + a 2 1 + a24)p2At P33 = 1 - (p35 + a31 + a341P3.6.t (4.3-13b) P44 = 1 - (p45 + a 4 2 + @43)p4At The following probabilities of leaving the state are obtained from Eqs.(4-15): P13 = al3p3At P12 = a12p2At From Eq.(4-18b): P21 = a21plAt P31 = a3lplAt P42 = a42p2At p43 = a43p3At From Eq.(4-18a): p24 = a24p4At p34 = a34p4At From Eq44-17b): P35 = P35FSAt P45 = P45p5At (4.3-13~) P25 = P25pSAt The above probabilities are arranged in the following matrix:

377

1

2

P= 3 4 (4.3-13d)

5

The above results are applied for the two cases below, 4.3-13(1) and 4.3-13(2).

4.3-13(1) The following assumption were made in the present case referring to Fig.4.3-13: 4 1 2 = 4 2 1 = 4 3 4 = 443,413 = 935 where 431 = 4 2 4 = 4 4 2 = q25 = q45 = 0. Thus, a13 = p35 = 1, a 1 2 = a 2 1 = a 3 4 = a 4 3 = a where the rest of the coefficients are zero. The above scheme reduces to the following on:

Also, all reactors are of the same volume, i.e. Vj = Vi = V, thus, it fc,,aws from Eqs.(4-3) and (4-7) that pj = pi = p = Q1N. The matrix given by Eq.(4.3-13d) is reduced to the following one where Eqs.(4.3-13a) to (4.3-13c) above, are applicable, considering the coefficients which are zero.

PAt

4 0

5 0

l-apAt

0

0

0

0

0

l-(l+a)pAt

apAt

pAt

4

0

0

aFAt

l-apdt

O

5

0

0

0

0

1

2

3

a@t

a@t

P= 3

1 1 1-( l+a)pAt 2

378 For pulse inputs C(0) = [ l , 0, 0, 0, 01 and [0, 1, 0, 0, 01, the response curves for C1 to C5 were computed from C(n+l) = C(n)P. A specific example was obtained for p = 1 where the effects of a = 0.5 and 5 as well as Cl(0) and C2(0) are demonstrated in Fig.4.3-13(1) for At = 0.005. As seen in cases (a) and (b) for the same Cl(0) = 1, by increasing a,i.e. the fluid exchange rate between

the reactors, the difference in the concentration between reactors 1-2 and 3-4, diminishes faster. A similar behavior is observed in cases (c) and (d) for the same C2(0)= 1. The effect of Ci(O), i.e. the initial location of the pulse, is demonstrated in cases (a) and (c) for a = 0.5 as well as (b) and (d) for a = 5. As seen, for a = 0.5, the concentration profiles are different, whereas for a = 5 they are identical with respect to reactors 3 and 4 where rector 1 replaces 2 because of the initial location of the pulse. The mean residence time tm in the flow configuration demonstrated above was computed by Eq.(4-26) for the response in reactor 3. The effect of a = a12 = a21 = a34 = a43 on tm is as follows. For a = 0, only reactors 1 and 3 are active and tm = 2; for a > 1, i.e. all four reactors are active and tm = 4, demonstrating the interaction between the vessels on tm. 1

0.8 0.6

u"

0.4

0.2 0 t

379

t

-

t

Fig.4.3-13(1). Ci versus t demonstrating the effect of Cl(O), C2(0) and a for p = 1

4.3-13(2)

Referring to Fig.4.3-13, the following assumption were made in the present case: 4 1 2 = 4 3 4 = 925 = q45. Thus, a12 = a 3 4 = p25 = p45 = a where the rest of

the coefficients are zero. The above scheme reduces to the following one:

Also, all reactors are of the same volume, i.e. Vj = Vi = V, thus, it follows from Eqs.(4-3) and (4-7) that pj = pi = p = Q1N.The matrix given by Eq.(4.3-13d) in case 4.3-13 above reduces to the following one where also Eqs.(4.3-13a) to (4.313c) there are applicable, considering the coefficients which are zero.

3 80 1 1-@t 0

2 a@t l-apAt

3 (1-a)pAt 0

P= 3

0

0

l-(l-a)pAt

4

0

0

0

5

0

0

0

1 2

4 0

5 0

0

apAt

apAt (1-2a)pAt 1-apAt apAt 0

1

For the pulse input C(0) = [ l , 0, 0, 0, 01, the response curve for C1 to C5 is computed from C(n+l) = C(n)P. A specific example was obtained for p = 1 where the effects of a = 0, 0.4 and 1 is demonstrated in Fig.4.3-13(2) for At = 0.02. As seen in cases a and c, a = 0 and 1, the concentration profiles are identical whereas rector 2 replaces 3. The mean residence time tm in the reactors' configuration demonstrated above was computed by Eq.(4-26) for the mean concentration of streams q25, q35 and q45. The effect of a = a 1 2 = a34 = p25 = p45 on tm is as follows. For a = 0, only reactors 1 and 3 are active and tm = 2; for 0 c a < 1, all four reactors are active and tm = 4 whereas for a = 1, again two reactors are active, 1 and 2, demonstrating the flow interaction effect between the vessels on tm.

38 1 1

0.8 0.6

6-

0.4 0.2

-

n 0

0.5

1 t

1.5

Fig.4.3-13(2). Ci versus t demonstrating the effect a for p = 1

4.3-14 The following configuration is a generalized scheme for demonstrating non ideal mixing vessel [77].

Fig.4.3-14. A generalized model for demonstrating non ideal mixing vessel The above configuration is obtained from the general scheme in Fig.4-1 by designating j = 1, a = 2, b = 3, ..., Z = 5 and the collector 5 = 6 which is the 6th state.

Eq.(4-12a) gives: a2 + a3 +.cq + a5 = p56 Eq.(4-12b) gives: a21 + a31 + a41 + a51 = a12 + a13 + a14 + a15

382 Eq.(4-12c) gives: a2 + a 1 2 + a32 + 0152 = a 2 1 + a 2 3 + a 2 5 for i = 2, a3 + a 1 3 + a 2 3 + a 4 3 = a31 + a 3 2 + a 3 4 for i = 3, a4 + a 1 4 + a 3 4 + a 5 4 = 0141 + a 4 3 + a 4 5 for i = 4, a5 + 0115 + a25 + a45 = a 5 1 + a 5 2 + a 5 4 + p56 for i = 5 (4.3-14a) The six equations contain twenty one coefficients; thus, fifteen of which must be prescribed in order to solve for the others. The following probabilities of remaining in the state are obtained from Eqs.(4-14) and (4-16): PI1 = 1 - (0112 + a13 a 1 4 + al5)plAt P22 = 1 (a21 a 2 3 a25)p2At P33 = 1 - (a31 + a32 + a34)p3At P44 = 1 - (a41 + a 4 3 + a451P4At (4.3-14b) P55 = 1 - (a51 + a 5 2 + a54 + P56)p5At The following probabilities of leaving the state are obtained: From Eqs.(4-15): p12 = al2p2At P13 = al3p3At PI4 = al4p4At pl5 = a15p5At (4.3-14~) From Eqs.(4-19aY19b): P21 = a21plAt P23 = a23p3 At P25 = a25p5At P31 = a3lplAt P32 = a32p2At P34 = a34p4At P41 = a41plAt P43 = a43p3At P45 = a45k5At p54 = a54p4At and from Eqs.(4-20) p52 = a52p2At p51 = a51plAt P56 = P56pdt (4.3- 14d) The above probabilities are arranged in the following matrix:

1 2

P= 3 4 5

6

(4.3-14e)

383 A simplified case appears in ref.[77], treated below in 4.3-14(1).

4.3-14(1) The following assumption were made in the present case while referring to Fig.4.3-14: 4 2 3 = 4 1 2 = 4 3 1 = 4 3 4 = 4 4 3 = 4 5 4 = 425 = 4 5 2 = 0, or a 2 3 = a 1 2 = a 3 1 = a 3 4 = a 4 3 = a 5 4 = a 2 5 = a52 = 0. Also, 42 = 4 3 = Q5 = 0, or a2 = a 3 = a5 = 0 as well as P16 = P26 = P36 = P46 = 0. However, a 4 = 1 since 4 4 is the reference flow rate noting that Qj = Q1 = 0. Additional assumptions made were: a 1 4 = a 4 1 = a;a 1 5 = a 5 1 = y; a 1 3 = a 3 2 = a21 = P. From Eq.(4.3-14a) for i = 4 and 5, it follows that a 4 5 = P56 = 1. All reactors are of the same volume, i.e. Vj = Vi = V, thus, it follows from Eq~.(4-3)and (4-7) that pj = pi = p = Q4/V. Consequently, the above scheme reduces to the following one.

'Q4

The matrix given by Eq.(4.3-14e) in case 4.3-14 above is reduced to the following one where also EqsS(4.3-14a) to (4.3-13d) there are applicable, considering the coefficients which are zero. 1 1 1-(a+P+y)pAt

2 0

5 w t

6

PW

4 apAt 0

0

0

0 WAt

0

3

2

P W

1-PW

0

3

0

PPAt

1-P+

awt

0

0

0 1-( 1+a)pAt

w t 0

0

0

0

0

0

0

P= 4

5 6

0

0

1-(l+y)pAt pAt 0 1

384 For a pulse input creating a unit concentration in the reactor, which is introduced into reactors 1 or 3 or 4 or 5 , the response curves given below for C1 to c6 is computed from C(n+l) = C(n)P. 1

I

,

,

,

,

,

I

CJO) = 1

'2

\ \

-

,". I

5

/'

10 0

5 t I

I

---

i .'I-

0 1

6-

I

I

I

"

"

J

,

2

_I ._..F1,._..__... _. _ .. , ' ,....... .. , _.., _ , . $_

5

%

~

p:

10

t

I

C,(O) = 1

0.8 t4

6

0.6

6-

0.4 0.2 0

t

t

Fig.4.3-14(1). Ci versus t demonstrating the effect of the introduction location of the tracer Fig.4.3-14(1) demonstrates the effect of the introduction location of the pulse on the response curves for a = a14 = a 4 1 = 0.05, p = a13 = a 3 2 = a 2 1 = 0.1, y = a 1 5 = a51, p = 1 and At = 0.004.

4.3-15 The following configuration is a generalization the example in [78, p.741 which aimed at describing various conditions of mixing.

385

Fig.4.3-15. Parallel reactors model The above configuration is obtained from the general scheme in Fig.4-1 by designating a = 1, b = 2, ...,Z and the collector 5 which is the Z + 1 state. Eq.(4-12a) gives: (4.3-15a) a1 + a 2 = 1 = Pz-1,t + P Z , t where the reference flow is Q1+ 4 2 and (4.3-15b) ai = Qj/(Q1+ 4 2 ) i = 1,2 Eq.(4-12c) gives: for i = 1 a1 + a 2 1 + a31 = a 1 2 + a 1 3 for i = 2 a 2 a12 + a 3 2 + a 4 2 = a21 + a 2 3 + a24 for i = 3 a13 + a 2 3 + a 4 3 -k a 5 3 = a31 + a 3 2 + a 3 4 + a35 for i = 4 a 2 4 + a34 + a54 a64 = a 4 2 + a 4 3 + a 4 5 + a 4 6 for i = 5 a 3 5 + a 4 5 + a65 + a 7 5 = a 5 3 + a54 + a56 + a57 for i = 6 a 4 6 + a 5 6 a76 a 8 6 = a 6 4 + a65 + a67+ a 6 8 ai-2,i

+ ai-1,i + ai+l,i + ai+2,i = ai,i-2 + aij-1+ ai,i+l + ai,i+2 3 Ii I Z - 2

az-4,z-2 + az-1.2-2

az-3,z-i

az,z-2 = az-2,z-4 + az-2,z-1 + az-2,z

+ az-2,~-1+ az,z-i = az-1,~-3+ az-1,z-2 + az-i,z + Pz-i,t

az-2,z + az-1,z = az,z-2 + az,z-1 + Pz&

where Z = 5 , 7, 9, ... and 5 = 6, 8,9,..., respectively.

for i = 2-2 for i = Z-1 for i = Z (4.3- 1%)

The following probabilities of remaining in the state are obtained from Eq.(4-16): P11 = 1 - (a12 -I-al3)PlAt P22 = 1 - (a21 + a 2 3 + a24)P2At P33 = 1 - (a31 + a32 + a34 + a35b3At P44 = 1 - (a42 + a 4 3 + a45 + a46)P4At

386 pii = 1 - ( a i j - 2 + aij-1 + ai,i+l+ ai,i+dviAt

3 I i I2-2

Pz-1,z-1= 1 (az-l,z-3 az-l,z-2 + a z - l , z + Pz-I,&z-lAt Pzz = 1 - (az,z-2 + az,z-1 + Pz,$PzAt The following probabilities of leaving the state are obtained: From Eqs.(4- 17b):

(4.3-15e)

Pz-1,k = Pz- 1,SP# Pzk = PzqqAt From Eqs.(4-18) for k, i = 1,2, ..., 2-2; k # i: p h = akipiAt The above probabilities are arranged in the following matrix:

P=

1

pi1 pi2 pi3

2

P21 P22 p23 p24 0

3

P31 P32 P33 P34 P35 0

4

0

5

0

0

0

0

...

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

...

0

0

0

0

i ... pii ... i ... 0 ...

0

0

0

0

0

0

0

0

0 ..*

P53 p54 P55 P56

i i

(4.3-150

0

P42 p43 p44 p45 p46 . * . 0

...

(4.3-15d)

. .. .

. .. .

: ...

Z-

0

0

0

0

0

0

Z-

0

0

0

0

0

0

Z

0

0

0

0

0

5

1 0

0

0

0

0

0

0

Pz-2,z-l pz-2,z

0

0

*.*

PZ-2J-2

* * *

0

* * *

Pz-1,z-2 Pz-1,z-1 pz-l,z Pz-1,t

0

-..

0

*.*

0

...

0

...

Pz,z-2 Pz,z-1 0 0

pzz 0

Pg 1

4.3-15(1) The above model was demonstrated graphically for the following configuration consisting of six states, i.e. Z = 5 and 5 = 6.

387

Considering Eqs.(4.3-15a) and (4.3-1%) above gives:

1 = p46 + p56 a1 + a21 + a 3 1 = a12 + a 1 3 a 2 + a 1 2 a32 + a42 = a21 + a23

a24

for i = 1 for i = 2

+ a 3 2 + a 3 4 -t a 3 5 a 2 4 + a 3 4 + a 5 4 = a 4 2 + a 4 3 + a 4 5 p46 a 3 5 a 4 5 = a 5 3 + a54 + p56

for i = 3 for i = 4 for i = 5 As observed, there are six equations and eighteen unknowns. It has been assumed that: a12=a21=b a23=a32=d a43=a34=e a45=a54=f a13 + a 2 3 + a 4 3

a53 = a31

a 4 2 = a53 = a31 = R Designating a1 and a2, defined in Eq.(4.3-15b), by:

(4.3-15( 1)a)

al=a a2=1-a (4.3-15(1)b) where from the above equations it follows that: a1 = p56 a2 = p46 a24 = R + 1 - a a13 = a 3 5 = R + a Thus, the solution is reduced to six parameter: a, b, d, e, f, R and pi = p. The following probabilities are obtained from Eqs.(4.3-15dYe, f ) and the above parameters: pi1 = 1 - (a + b + R)pAt pi2 = bpAt pi3 = (a + R)pAt p22 = 1 - (b + d + R + 1 - a)pAt p21 = bpAt p23 = dpAt p24 = (R + 1 - a )pAt p33 = 1 - (a + d + e + 2R)pAt p31= RpAt p32 = dpAt p34 = epAt P35 = (a + R W t p u = 1 - ( 1 - a + e + f + R ) p A t p42=RpAt p43=epAt p45=fpAt P46 = (1 - a)pAt p55 = 1 - (a + f + R)pAt p53 = RpAt p54 = fpAt p56 = apbt The above probabilities are presented in the following transition matrix:

388

1 2

P= 3 4 5

6 For the pulse input C(0) = [0, 1,0,0,0,0], the response curve for C1 to Cg was obtained from C(n+l) = C(n)P. A specific example was computed for p = 5, a = 0.5, b = d = e = f = 0.1, At = 0.001 where the effect of the recycle R = 0,5 is demonstrated in Fig.4.3-15( 1). As observed, by increasing the recycle, reactors 2, 4 and 1,3,5 acquire sooner the same concentration.

t

t

Fig.4.3-15(1). Ci versus t demonstrating the effect of the recycle R The following configurations, 4.3-16 and 4.3-16( l), are multiloop circulation models [74-761 for fitting experimental residence time distribution data of continuous stirred vessels.

4.3-16 The following scheme is a three loop model consisting of six perfectly mixer reactors which simulates a mixer [75].

389

Fig.4.3-16. Three loop model with inflow to impeller The above configuration is obtained from the general scheme in Fig.4-la by designating a = 1, b = 2, ..., Z = 6 and 5 = 7 which is the 7th state; the junction is j. The concentration at this point, Cj, is given by Eq.(4-la). Eq.(4-12a) gives: 1 = p57 where the reference flow is Qj in Fig.4.3-16. Eq.(4-12b) gives: (4.3-16a) 1 + a 2 j + a 4 j + a 6 j = ajl $. a j 3 + ajs Eq.(4- 12c) gives: ajl=a12 for i = 1; a 1 2 = 012j for i = 2; a j 3 = a34 for i = 3; a 3 4 = a 4 j for i = 4; ajs = P57 + a56 = 1 + a56 for i = 5 ; (356 = a6j for i = 6 (4.3-16b) The following probabilities of remaining in the state are obtained from Eq.(416) where pi = QjNi: p11= 1 - a12Cl1At ~ 2 =2 1 - a2j~2At P33 = 1 - a34p3At ~ 4 =4 1 - a4j~4At (4.3-16~) P55 = 1 - (p57 + a56)p5At P66 = 1 - a 6 j p d t The following probabilities of leaving the state are obtained: From Eq.(4-13~):p'kj = akj/xkajk k = 2,4, 6 From Eq.(4-15): Pji = ajipiAt k = 1 , 3 , 5 From Eq.(4-17b): p57 = P57p7At where p12, p34 and p56 are obtained From Eq.(4-18). The above probabilities are arranged in a transition matrix given below which is of the kind demonstrated by Eq.(4-13) with pjj = 0, noting those probabilities among p~ and Pkj which are zero according to the scheme in Fig.4.3-16.

390

(4.3- 16d) Eqs.(4.3-16a,b) is a set of seven equations with nine unknowns, two of which must be specified. Thus, it has been assumed that: 4 1 2 = 434 = r, or a12 = a34 = r/Qj = R, i.e. the circulatory flow in the loop 1-2-j. Therefore, Eq~(4.316a,b) yield: ajl = a 2 j = a j 3 = a 4 j = R and a6j = a56 and CXj5 = 1 + a 6 j where it is further assumed that: a6j = a56 = R thus, a j 5 = 1 + R If all reactors have the same volume V, i.e. pi = p = QjN, the probabilities in Eqs.(4.3-16c,d) satisfying the above matrix, read: pii = 1 - RpAt i = 1 , 2 , 3 , 4 , 6 p55 = 1 - (1 + R)pAt P12 = P34 = P56 = RpAt pj5 = [(R + 1)/(3R+ 1)]pAt pjl= pj3 = [(R/(3R + l)]pAt p'2j = p'4j = p'6j = 1/3 As seen, the solution is a function of the parameters p, R and At. For the pulse input given by Eq.(4-13a), i.e. C(0) = [Cj(O), 1, 0, 0, 0, 0, 0, 01, where Cj(0) is given by Eq.(4-1a), the response curve for C1 to C7 was obtained from C(n+l) = C(n)P. A specific example was computed for p = 1, At = 0.005 where the effect of the circulation R = 0.2, 1 and 5 is demonstrated in Fig.4.3-16a. As observed, by increasing R, reactors 1-6 acquire sooner an identical concentration and behave as a single reactor.

39 1 I

0.8 - i = 1

I

R = 0.2

I

-

t

t

t

Fig.4.3-16a. Ci versus t demonstrating the effect of the recycle R

4.3- 1 6 m

The following scheme is a simplified version of case 4.3-16for Qj = 0, i.e. a closed three loop system.

Assuming all reactors are of the same volume, that the reference flow is one of the internal flows, and that all flows in the loops are identical, thus all aij = 1, yields the following transition matrix:

392 j

j

0

1 @t

2 0

ClAt

4 0

pAt

6 0

0

0

0

0

0

0

0

0

3

5

1

O

1-@t

2

1/3

0

pAt 1-@t

0

0

0

1-@t

PAt

0

0

4

1/3

0

0

0

1-@t

0

0

5

0

0

0

0

0

l-PAt

pAt

6

1/3

0

0

0

0

0

1-@t

P= 3

For the pulse input C(0) = [Cj(O),1, 0,0,0, 0, 01, where Cj(0)is given by Eq.(4-la), the response curve for C1 to c6 was obtained from C(n+l) = C(n)P. A specific example was computed for At = 0.01 where the effect of p = 0.1, 1 is demonstrated in Fig.4.3-16( 1). As observed, by increasing p,reactors 1-6 acquire sooner an identical concentration which is equal to 1/6. 1 0.8

u

0.6

.-

0.4 0.2

n -

0

2

6

4 t

8

1 0 0

2

4 t

6

8

10

Fig.4.3-16(1). Ci versus t demonstrating the effect of p

4.3-17

This is another configuration simulating a continuous mixer with an equally divided flow between the upper two circulation loops [75].

393

Fig.4.3-17. Three loop model with inflow divided The derivation of the probabilities is quite similar to the former case. For the transition matrix given by Eq.(4.3-16d), assuming that all reactors are of the same volume and that the reference flow is Q in Fig.4.3-17, the following expressions were obtained: pii = 1 - (1/2 + R)@t i = 1,2,3,4 p55 = 1 - (1+ R)pAt p66 = 1 - RpAt P77 = 1 ~ j =5 (1+R)@t pjl = pj3 = RpAt p'2j = p'4j = (1/2 + R)/( 1 + 3R) P'6j = (1 + R)/( 1 + 3R) pi2 = p34 = (112 + R)pAt P56 = R@t P57 = @t where the rest of the probabilities are zero. The solution is a function of the parameters p,R and At, and for the pulse input C(0) = [Cj(O), 1, 0, 0, 0, 0, 0, 01, where Cj(n) is given by Eq.(4-la), the response curve for C1 to C7 was obtained by C(n+l) = C(n)P. A specific example was computed for p = 2, At = 0.01 where the effect of the circulation R = 0, 1 and 5 is demonstrated in Fig.4.3-17a. As observed, by increasing R, reactors 1-6 acquire sooner an identical concentration and behave as a single reactor. 1

0.8 0.6 W'

0.4

0.2

0

0

1

2

t

3

4

5

t

394 1

o.8 0.6

I

t

I

R=5

I

I

t

Fig.4.3-17a. Ci versus t demonstrating the effect of the recycle R

4.3-18 Another configuration simulating a continuous mixer with non divided inflow [74] is depicted below.

& QL

4

Q2*

j

21

5

Fig.4.3-18. Three loop model with non divided inflow Assuming that all reactors are of the same volume, that the reference flow in Fig.4.3-18 is 42, yields the following results by applying Eqs.(4-12a) to (4-12c): ajl=a j 3 = a3j = a5j = a12 = R where

a2j = aj4 = a45

= R + 1.

The transition matrix reads:

R = Qsj/Q2 = Qjk/Q2 k = 2,3

395

j 1

2

P= 3 4 5 6

where the probabilities obtained from Eqs.(4-14)to (4-19) are: pii = 1 - (1+ R)pAt i = 2 , 4 , 5 p11= p33 = 1 - RPAt pj1= pj3 = RpAt p'2j = (1 + R)/( 1 + 3R)

pj4 = (1 +R)pAt p'3j = p'5j = W (1 + 3R) P12 = RPAt p45 = (1 + R)@t p56 = @t Cases a and b in Fig.4.3-18a, demonstrate the effect of the recycle R for p = 1 and At = 0.004. Cases b, c and d demonstrate the effect of the introduction location of the tracer, reactors 1, 3 or 5, on the response curves for p = 1 and R = 5.

\

0.2 -,

-

-.

3 .

5

I

I

-

-1 -6 -

4-------+

;L, -T -I

t

t

I

(b) R = 5, Tracer in 1 -

3%

t

t

Fig.4.3-18a. Ci versus t demonstrating the effect of the recycle R and the introduction location of the tracer in reactors 1, 3 and 5

4.3-18(1) The following configuration is a simplified version of case 4.3-18 where Qj = 0, i.e. a closed three loop model.

Assuming that all reactors are of the same volume, that the reference flow is

one of the internal flows, and that all flows in the loops are identical, thus all aij =

1,yields the following transition matrix:

397

j

0

3 pAt

4 pAt

0

1-pht

pAt

0

0

0

1/3

O

l-pAt

0

0

0

3

113

0

0

l-pAt

O

0

4

0

0

0

0

1-pAt

FAt

5

113

0

0

0

0

1-pbt

0

1 PAt

1

0

P= 2

j

2

5

Cases a and b in Fig.4.3-18(1), demonstrate the effect of p = 0.1 and 1 for

At = 0.01. Cases b, c and d demonstrate the effect of the introduction location of the tracer, reactors 1, 3 or 5, on the response curves for p = 1. As observed, by increasing p, reactors 1-5 acquire sooner an identical concentration which is equal to 1/5. 1 p

0.8

-

0.1, Tracer at 1

U" 0.4 -

-

0.6

/

.-

/,--

0.2

0

0

2

4

t

0.8

p

0

2

4

t

6

8

-

1, Tracer at 3

6

8

10

0

2

10

0

2

4

4

t

t

6

8

10

6

8

10

Fig.4.3-18(1). Ci versus t demonstrating the effect of p and the introduction location of the tracer in reactors 1, 3 and 5

398

4.3-19 An extension of case 4.3- 18 where the central loop contains a varying number of reactors designated by 3,4, ..., 2-2, Z > 5 , is depicted in Fig.4.3-19.

Fig.4.3-19. Three loop system with the central loop of variable number of reactors From Eq.(4-12a) it is obtained that pzk = 1 where the reference flow is 4 2 . From Eq.(4-12b) it follows that: a 2 j + az-2,j + a z j = CXjl + aj3 + aj,z-1 where from Eq.(4-12c) it is obtained that: For i = 1: a j l = 0112; i = 2: 1 + a 1 2 = a2j; i = 3: a j 3 = ~134; i = 4: a 3 4 = ~ ~ 4 5...; i = i: ai-l,i = ai,i+I;... i = 2-2: az-3,z-2 = az-2,j; i = Z-1: aj,z-1 = az-1,z; i = Z: a z - l , z = a z j + 1 The above set yields Z (Z > 5) independent equations and Z+3 unknowns; thus three unknowns must be fixed. For example, if Z = 6 and 6 = 7, the process is as fOllOWS: Fix a 1 2 = R, thus, a j l = R and a 2 j = 1 + R; Fix a j 3 = R, thus, a 3 4 = a 4 j = R; Fix a56 = R + 1, thus, Olj5 = R + 1 and a 6 j = R; For the general case, the following transition matrix is applicable:

399

... ...

i-1

i

i+l

0

0

0

... 0 0 ... P33 P34 ...

0

0

0

0

0

0

...

0

j

1

2

3

4

j

0

Pjl

0

Pj3

0

1

0

PI1 P I 2 0

2

~'2j 0

p22

3

0

0

0

4

0

0

0

P = i-1

0

i+l

0

. ..

p44

0

0

. .. ...

...

0

.

.

0

.

0

21

Z

5

0

Pj,z-I

0

0

0

0

0

0

0

0

0

0

0

0

0

...

0

0

0

0

0

0

...

0

0

0

0

Pi-I,i-IPi-I,i

0

0

22

.

... 0 0 0 0 ... .. .. . .. ... 0 0 0 0 ... 0

i

0

.. .. . .

0

... ... ... ...

PiiPi,i+I

0

0

0

0

0

Z

P'zj

0

0

0

0

... ...

5

0

0

0

0

0

...

Z-1

0

0

0

0

0

0

0

0

0

0

.

0

0 :

0

. 0

0

... Pz-2,z-21

0 :

0

0

0

0

0

0

0

0

0

0

0

... ... ...

0

0

0 Pi+I,i+I

..

.

0

... ...

0

a

*

2 2 P'z-z,~

0

... ... ...

.. a

0

.

:I

Pz-1,z-IPZ-1,z 0

Pzz PZS 0

1

Specific expressions for the probabilities, obtained from Eqs.(4-15) to (4-19) are: p11= 1 - a12plAt = 1- RplAt ~ 2 =2 1 - a2jp2At = 1- (l+R)p2At p33 = 1 - a 3 4 ~ 3 A =t 1- Rp3At p44 = 1 - a 4 5 ~ 4 A =t 1- RMAt Pii = 1 ai,i+lpiAt = 1- RpiAt pZ-2,z-2= 1 - az-2j~Z-2At= 1- ( 1/3)pz-2At pz-1,z-l= 1 - az-l,zjLz-lAt = 1- (l+R)pZ-lAt pzz = 1 - (l+Ctzj)pzAt = 1- (l+R)PzAt pjl= CtjlplAt = RplAt pj3 = aj3~3At= Rp3At Pj,z-l= aj,z-lFz-lAt = (l+R)pz-lAt Pl2 = a12p2At = Rp2At P34 = a34p4At = Rp4At P45 = a45p5At = Rp5At Pi,i +I = ai,i+lPi+lAt = Rpi+lAt Pz-l,z = aZ-l,zpZ-lAt = (l+R)pz-lAt pzc = P@ z@ = pgAt p'2j = Ol2j/(ajl+aj~+aj,z-1)= (l+R)/( 1+3R) p'Z-2,j= a Z - 2 ~/(~tjl+~~j3+~~j,z-l) = R/(1+3R) p'zj = a z j l(ajl+aj~+aj,z-l) = R/(1+3R) Particular solutions were obtained for a total number of states of 7 , 9 and 13, in the following. The increase in the number of states was in the central loop. For a total number of states of 7, where the central loop contains the two states 3 and 4, Fig 4.3-19 is reduced to:

400

The following matrix is obtained for reactors of an identical volume V, i.e. pi =p = QN: 1

2

3

4

5

6

7

0

RFAt

0

(l+R)pAt

0

0

0

RpAt

0

I-RpAt

RpAt

0

0

0

0

0

0

1-(1+R)FAt

0

0

0

0

0

0

0

0

0

0

0

0

0

I-RpAt

0

0

0

0

0

0

0

1- (I+R)FAt

(l+R)pAt

0

61+3R

0

0

0

0

0

l-(l+R)j.tAt

FAt

7

0

0

0

0

0

j 1

1 +R

21+3R

P=3

0 R

41+3R 5

0 R

0

1-RpAt RpAt

0

1

For a total number of states of 9, where the central loop contains the four states 3 to 6, Fig 4.3-19 is reduced to:

The following matrix is obtained:

40 1

j 1

j

1

2

3

4

5

6

7

8

0

RpAt

0

RpAt

0

0

0

(l+R)pAt

0

0

I-RpAt

RpAt

0

0

0

0

0

0

11 +R)KAt

0

0

0

0

0

0

0

0

0

0

0

0

0

1-RpAt RpAt

0

0

0

0

2

1+R 1+3R

3

0

0

0

4

0

0

0

0

0

0

0

0

0

0

0

0

0

I-RpAt

0

0

0

0

0

P=5 6

R 1+3R

7

0

8

R 1+3R

9 I

0

l-RpAt RpAt

I-RpAt RpAt

1-

(l+R)pAt

l+R)PAt

0

0

0

0

0

0

0

0

1l+R)PAt

0

0

0

0

0

0

0

1

For a total number of states of 13, where the central loop contains the eight states 3 to 10, Fig 4.3-19 is reduced to

The following matrix is obtained while designating p = RpAt, q = 1 - RpAt, r = (l+R)pAt and s = 1 - (l+R)pAt.

402

j

j 1

0 0

1cR

21+3R 3 4 5 P=6 7 8 9

0 0 0 0 0 0 0

1 p q

2 O p

3 4 p O o o

5 O o

6 O o

7 O o

8 O o

9 O o

10111213 O r O O o o o o

o s o o o o o o o o o o o

R

o o o o o o o o

o o o o o o o o

q o o o o o o o

p q o o o o o o

o p q o o o o o

o o p q o o o o

o o o p q o o o

R

O 0

O 0

O 0

O

O

O

O

O

0

0

0

0

0

0

0

0

r O spAt

0

0

0

0

0

0

0

0

0

0

0

0

l01+3R 11 0 121+3R 13 0

o o o o p q o o

o o o o o p q o O

o o o o o o p q O

o o o o o o o o s

o o o o o o o o

o o o o o o o o

1

Fig.4-3.19 demonstrates response curves of various states (reactors) for a pulse introduced in state 1 raising its concentration to unity. The parameters of the graphs are the recycle rate R (= a12 = ajl = aj3 = a34 = a4j), 0.5 and 5, and the number of states in the central loop, 2,4 and 8 corresponding to a total number of states 7, 9 and 13, respectively. Common data were p = 1 and At = 0.01, The general trends observed were that the approach towards equilibrium becomes slower by increasing the number of states and that the streams attain faster a uniform concentration by increasing R. The effect of the number of states is reflected by curves 4 (case a), 6 (case c) and 10 (case e) corresponding to the exit reactors of the central loop, as well as the overall effect reflected in curves 7 (case b), 9 (case d) and 13 (case f) corresponding to the final collector of the tracer. The effect of R is demonstrated in cases a and b, c and d as well as e and f.

403 1

I

I

I

I

I

1

I

I

(b)

Number of states = 2, R = 5

t

t

I\

o.8

I

I

I

I

of states = 4, R = 0.5

tp2 The following transition matrices apply for which pii = 1 - pAt i = 1, 2; p12 = p21 = pat where p = p 1 = p2: O < t <tp1:

tP1 I t

1

2tp1+tp2 5 t

I

2

< 2tp1+2tp2: 1

2

1 P11 P12 P= 2 0 1

I

< tp1+tp2: 1 2

tp1+tp2 I t

< 2tp1+tp2: 1

2

< 2tp1+3tp2: 2tp1+3tp2 I t < 3tp1+3tp2: 1 2

2tp1+2tp2 I t

p=

:lp:l

i2I

p=

I 1 :I

1 P11 P12 2 0

43 1

The parameters of the solution are: tpl, tp2, p1 and p.2. A numerical solution was obtained for p = pi = p2, 1, 10 and 10, where p = QN,i.e. for reactors of an identical volume V. In addition p1= 1,2, 100 and p2 = 1,2,5, 10. The residence

time in the reactors was tpl = 1 and tp2 = 0,0.05,0.5 and 1 corresponding to At = 0.0005, 0.005 and 0.01, respectively. The tracer was introduced into reactor 1. The following behaviors are demonstrated in Fig.4.4-5(1b):the effect of tp2 is demonstrated in cases a to d; the effect of p in cases b, e and f; the effect of p2 in cases c, g and h; the effect of p.1 in cases c, i, j and k.

1

0.8

6 '

0.6 0.4

0.2

0

I

I

I

I

2

4

6

8

t t = 1 t =0.5 P'

p=1

t =1 D1

t t =1

p=1

DZ

-

0 -..:2

0

I

2

I

4

t

I

1

6

8

-

-...'

100

2

I

2

I

4

t

I

I

6

8

10

432

1

0.8 0.6

6 '

0.4 0.2

0

I

0

I

I

I

2

4

6

t

0

2

t

4 t

t =1 PI

I

8

6

8

10

6

4

8

10

t

P l = 2 P2 = 1

t =0.5 P2

2

0

I

10

t

PI

= I t =0.5 p =10 p =1 2

P2

(d

1 i=l

0.6

0.2

- ; 0

I

I

I

I

2

4

6

8

t

-

-

10

0

I

I

2

4

I

t

6

I

8

10

433

1k

(Q

l

0.8 -

-

-

-

0.6

u" 0.4

4.4-6 The fluid flow Q1 is divided into flows 4 1 2 and 4 1 3 as shown in Fig.4.4-6. The tracer, in a form of a pulse input, is introduced into reactor 1; reactor 4 is the collector of it.

*--$-q

5=4 5 1 .

434

Q3

Fig.4.4-6. An open system with divided flow between two plug flow reactors Noting the residence times in each plug flow reactor, i.e. tpl and tp2, and assuming that tppl > tP2, yields the following matrices: tp21 t < tpl:

P= 2

3

PI1 0 0

p12 1 0

p13 0 1

1

2

P= 3

4

Referring to Fig.4-1, yields for the configuration in Fig.4.4-6 the following probabilities considering reactor j (hence j = 1, a = 2, ... and 6 = 4) as well as Eqs.(4-12a) to (4-12c) and (4-14) to (4-17b) and Q1 as reference flow:

P11 = 1 - PlAt P12 = a12p2At P13 = a13p3At where a12 + a 1 3 = 1 p24 = a12 p34 = a13 P22 = 1 - a12p2A-t P24 = a 12p4At P33 = 1 - a13p3At P34 = (1 - a12)p4At . As seen, the parameters of the solutions are: pi = QlNi (i = 1, ..., 4), a12 = Q12/Q1, $1 and tp2. In the numerical solution it has been was assumed that tpl = 0.1, tp2 = 0.05, At = 0.001 as well as C(0) = [ l , 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.4-6. The effect of a12 = Q12/Q1, 0 and 0.5, is demonstrated in cases a and c in Fig.4.4-6a, as well as in cases a to d and e, f, for different values of pi. The effect of pi is demonstrated in cases a to c, c to e and d to f. In case d, the sudden change of C4 at tp2 = 0.05 and tpl = 0.1, is clearly observed. In cases e and f, C2 and C3 are greater than unity because the volume of reactors 2 and 3 are reduced by increasing pi (for a constant flow rate), i.e. p2 = p3 = 100 by comparison to p1 = p4 = 1.

435 1

0.8

0.6

6-

0.4

0.2

0

./ /--

K- - - -

0

0.2

t

0.8

1

I

0.8

--

1

(9

-

-

I fI 4

0.6

u"

i

a

0.2 1,2,4! 1,2 I

0

0.05

---"1 1 /4

i

I

I

0

3 - - - -.- I

p i = 100 100-

\I

0.4

I

I

0.6

0.4

---

--

-

p.1

12

= -~

1-3 I

0.15

0.2

0

0.1

0.05

0.15

t

0.2

p = p = 1 p = p = 100 ( 1

2

4

12

0

0.2

0.4

t

0.6

3

11

a = 0.5

a =O

12

0.8

1

0

0.2 1 0.4

t

0.6

0.8

1

Fig4.4-6a. Ci versus t demonstrating the effect of pi and a12

4.4-6(11 A simplification of case 4.4-6 is demonstrated in Fig.4.4-6( 1) simulating the configuration in ref.[78, p.881.

436

5=4 Q13

934

Fig.4.4-6( 1). An open system with divided flow

1 P11 P12 p13 0 2 0 1 0 0 P= 3 0 0 P33 P34 4 0 0 0 1

1 P11 2 0

P=

3 4

0

0

P12

p13

0

P22

0

P24

0 0

P33 p34 0 1

Eqs.(4-12a) to (4.12-c) give that p24 = a12, p34 = a13 where a12+a13 =1. Eqs.(4- 14) to (4.17b) yield the following probabilities: p11= 1 - p1At PI2 = al2p2At p13 = al3p3At p22 = 1 for 0 c t < tp2 and p22 = 1 - a12p2At for t 2 tp2 p24 = a12p4At P33 = 1 - al3p3At P34 = a13p4At As seen, the parameters of the solutions are: pi = Q1/Vi (i = 1, ..., 4), a 1 2 = Q12/Q1, tpl and tp2. In the numerical solution it was assumed that tpl = 2, tp2 = 1, At = 0.01 as well as C(0) = [l, 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.4-6(1). The effect of a12 = Q12/Q1, 0, 0.5 and 1, is demonstrated in Fig.4.4-6(la); cases a to c correspond to pi = 10 and cases d to e for pi = 100. The effect of pi, 10 and 100, is demonstrated in cases a and d, b and e and c and f for various values of a12.

437

u-

-

a=O-

I

- f , F -7

p I t= l o -

-

0.2

0.4

t

I

0.6

0.8

1

-

'--> I

I

0

0.5

x 0.5

I

*

a12=

p i = 10

1

1.5

2

1 0.8 0.6

6-

0.4

0.2

0

1

'r

:::

-2 1-

4 -------

'i=l

6-

p

0.2 0 .

1.34 I

- -(0-

a = l

i I.,

12,4

0.4 -

p i = 10

I2

\

t

2

0

= 0.5 12

-4 [/':3\

I

a

I

-

=loo-

-

-

,

I

I

1

1.5

2

438

4.4-6(2) This case simulates the situation demonstrated in ref.[8 11, designated as 'partial mixing with piston flow and short-circuit' in continuous flow systems. It is an extension of case 4.4-6( 1) which is demonstrated in Fig.4.4-6(2) below.

Fig.4.4-6(2). An open system with plug flow and short-circuit The following transition matrices are applicable: 0 < t < tp2: 1 PI1 2 0

P= 3 4 5

0 0 0

P12

P13

p14

0

1

1

0

0

0

2

0

p33

0 0

0

0

0

p35 P44 p45 0 1

P= 3 4 5

Similar to case 4.4-6( l), the following probabilities are obtained: P11 = 1 - PlAt P12 = a12p2At PI3 = al3p3At P14 = al4p4dt p22 = 1 for 0 < t < tp2 and p22 = I - a12p2At for t 2 tp2 p25 = a12p5At P33 = 1 - al3p3At

P35 = al3p5At

P44 = 1 - a14p4At P45 = a14P5At where a 1 4 = 1- (a12 + (3113). The parameters of the solution are: tp, pi, a12 and a13 and the pulse was introduced into reactor 1 in Fig.4.4-6(2), i.e. C(0) = [l, 0, 0, 0, 01. In the numerical solution it was assumed that p1 = p2 = p4 = 15 = 30, i.e. the response of the reactors is relatively fast by comparison to the response of

439 reactor 3 for which p3 = 2. These conditions, approximately, simulate the model in ref."] where reactors 1, 2, 4 and 5 may be considered as measurement 'points' (very small reactors) of the concentraton of the pulse. Cases a, b and c in Fig.4.4-6(2a) demonstrate the effect of tp2 = 0.03, 0.3 and 3 for a12 = a13 = a14 = 1/3. Cases c and d demonstrate the effect of a12 and ~ 1 3 In . the later case a12 = 0.1 and a13 = 0.9 for $,2 = 3, i.e. 414 = 0. In the solution, At = 0.0003, 0.003 and 0.03 for $2 = 0.03,0.3and 3, respectively. t

= 0.03

12

14

t

t

0.8

= a = a = 1/3

12

14

13

I - -

-I t =3

-

PZ

1 ,

a = 0.1 a = 0.9

-,

12

13

-

t

t

Fig4.4-6(2a). Ci versus t demonstrating the effect of tp2 and ail

4.4-7

This example extends the previous one and simulates a contacting pattern of fluid elements of different ages [21, p.3341. The fluid flow Q1 is divided into flows 412, 4 1 3 and 4 1 4 as shown in Fig.4.4-7. The tracer, in a form of a pulse input, is introduced into reactor 1 whereas reactor 5 is the collector of the tracer.

440

Fig.4.4-7. An open system with divided flow between three plug flow reactors Noting that the residence time in each plug flow reactor is tpl, tp2 and tp3 and assuming that tpl > tp2> tp3, yields the following matrices: 0 < t < tp3: 1 2 3 4 1 P11 P12 p13 P14 2 0 1 0 0

P=

4

0 0

0 0

1 0

0 1

tp2 5 t < tpl: 1 2 3 4 5 1 P11 P12 P13 P14 0 2 0 P22 0 0 P25 P= 3 0 0 P33 0 P35 4 0 0 0 0 1 1 I 5 0 0 0 0

tp3 5 t < tp2 :

1 2 P= 3

4 5

1

2

3

4

5

P11

P12

P13

0

0

P22

0 0 0

0 0 0

0 1 0 0

P14

1

2

3

4

5

P12

P13

P14

0

0 0 1 0

P25

0 0 1

tpl I t :

1 P11 2 0 P= 3 0

4 5

0 0

P22

0 0 0

0

p33

0 0

0

P25

0

p35 p44 p45 0 1

Referring to Fig.4- 1, yields for the configuration in Fig.4.4-7 the following probabilities by considering reactor j (hence j = 1, a = 2,... and 6 = 5) and taking Q1 as reference flow. From Eqs.(4-12a) to (4-12c) one obtains that a 1 2 = p25, a13 = p35 and a 1 4 = p45 where ajk and Pis are defined in Eq.(4-2). In addition a 1 2 + a13 +a14 = 1 where Qs.(4-14) to (4-17b) yield:

441

As seen, the parameters of the solutions are: pi = QlNi (i = 1,

..., 5), a12, ~ 1 3 ,

$1, tp2 and tp3. In the numerical solution it was assumed that $1 = 0.4, $2 = 0.2 and 9= 0.1; pi = 200, At = 0.001 as well as C(0)= [l, 0, 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.4-7. The effect of a12 = 412141 = 1, a13 = Q13/Q1 = 1, a 1 4 = QldQl = 1 and a12 = 0.2 a13 = 0.3 and a14 = 0.5, is demonstrated in Fig.4.4-7a.

1

1=

0.8

0.6

'-0.4 0.2 0 -0.1

0

0.1

0.2

0.3

0.4

0.5

-0.1

0

0.1

Ot2

0.3

0.4

0.5

t

442

- - - - - - - - - -4- - - - - - - ~~

1 -

0.8

-

0.6

-

v-0.4

-

0.2

-

0 -

1 0.8

0.6

-I, I, I

S

I, L I 1

I 1

-

>- -

aI4-1 -

8\

I

i=ll

1.2.3.5 I

1

a

-

12

0.2 a

-

13

I

1-4

I

I

I

0.3 a

-

14

t I

0.5

“

u-0.4

1

0.2

0 -0.1

0

0.1

0.2

0.3

0.4

0.5

Fig4.4-7a. Ci versus t demonstrating the effect of aij

4.4-8

The following configuration of two interacting plug flow reactors was applied elsewhere [21, p.2981 for describing deviation from plug flow and long tails. Due to the interaction between the reactors, it is necessary to divide the reactor into perfectly mixed reactors. In the following example, the reactor was divided to five reactors. Generally, the number of perfectly mixed reactors needed, must be determined by comparing the response curve of the divided system to that of the plug flow reactor.

443

9

6

6

6

Fig.4.4-8. Two plug flow interacting reactors Referring to Fig.4-1, yields for the configuration in Fig.4.4-8 the following probabilities by considering reactor j and taking Q1 as reference flow. From Eqs.(4-12a) to (4-12c) one obtains a set of equations for the determination of aik. Assuming that: a = a67 = a85 = a58 = a 9 4 = a 4 9 = a10,3 = a3,lO = a1 1,2 p = a23 = a34 = a45 = a56 y = a76 = a2,11 6 = a87 = a 9 8 = a l 0 , 9 = a 1 1,lO as well as p+s =1

it follows that a+6=y The probability matrix for the above configuration reads:

(4.4-8a) (4.4-8b) (4.4-8~)

444 1 2 3 4 5 P= 6 7 8 9 10 11 12

1

2

3

p11

p12

0

0

'22

0

4

5

6

7

0

0

0

0

'23

0

0

0

0

0

p33

p34

0

0

0

0

0

0

p44

p45

0

0

0

0

0

0

'55

'56

0

0

0

0

0

0 ' 6 6

'67

0

0

0

0

0 ' 7 6

p77

0

0

0

0 ' 8 5 0

0

0

o p 9 4

0

0

'10.3

0

p11,2

0

0

'81

0

0

0

'10,9

0

0

0

0

0

0

0

0

0

0

0

0

0

Assuming that all pi's are equal, i.e. pi = p, the following probabilities are obtained from Eqs.(4-14) to (4-19): PI 1 = 1 - PAt p22 = 1 - (p + $pAt p33 = p44 = p55 = 1 - (a+ P)@t p66 = 1 - (1+ a)FAt P77 = P88 = P99 = PlO, 10 = PI 1,I 1 = 1 - WAt p12 = p6,12 = pAt P23 = p34 = P45 = p56 = PpAt P87 = P98 = ~ 1 0 = , spii,io = 6pAt p2,11 = P 7 6 = w t P3,lO = P49 = P58 = p67 = P11,2 = p10,3 = P94 = p85 = a w t The parameters of the solution are a and p related by Eqs.(4.4-8b) and (4.4-8c) as well as p. Fig.4.4-8a demonstrates the effect of the circulation intensity a between the reactors on the transient response in various reactors of the tracer, introduced in reactor 1. In case a, a = 0.05, p = 0.5, 6 = 0.5 and y = 0.55; in case b, a = 5, p = 0.5, 6 = 0.5, y = 5.5 and At = 0.001 where in both cases p = 50. As observed, by increasing a,the C-t curves for reactors 4 and 9 (see Fig.4.4-8 above), reactors 6 and 7 and reactors 2 and 11 become identical.

445 1

0.8

u

.-

0.6

0.4

/'

:i= 1

'

12

/ u = 0.05

-.

0.2

0 0

0.1

0.2

0.3

0.4

0.5

0.3

0.4

0.5

t

1

0.8

0.6 V'

0.4

0

0.1

0.2 t

Fig4.4-8a. Ci versus t demonstrating the effect of the circulation a The variation of the mean residence time tm in reactors 1 to 11, computed by Eq.4-26, was obtained from the response curve of reactor 6. The results are summarized in the following Table for the various operating parameters a to 6 listed in the Table and defined by Eq.4.4-8a. The following trends are observed: a) Increasing p (i.e. the flow rate), decreases tm. b) Taking a = 0 in case c, i.e. decreasing the number of effective reactors in Fig.4.4-8, decreases tm.

446 (a)

P

10

50 100

n

(c)

(b)

a = 0.05 a = 0.05

a =0

p=os

p=o

y = 0.55 6=0.5

y = 1.05 6=1

p=0 y= 1 6= 1

tm

tm

tm

1.096 0.219

1.080 0.219

0.797 0.159

0.109 11

0.108 11

0.079 8

4.4-9 The following configuration is a simulation (see also case 4.4-6(2)) of "partial mixing with piston flow and short-circuit" in continuous flow systems treated in ref. [8 11.

Fig.4.4-9. Mixing with plug flow and short-circuit The following transition matrices are applicable: O < t 2, the transient response of Ci remains unchanged; and b) for tp = 10, the effect of R is negligible because the response was transferred straight into reactor 4, bypassing reactors 3 and the plug flow. (b)

__------ -

6-

t

P

=2 R=O

-

\1

4-

c

-\

\

-

\

k

/

’

-

I ’

t

-- p -

0

I

2

4 t

I

I

6

8

100

P

= 2 R=0.2

-

- - $2 The following transition matrices are applicable: 0 < t < tp2:

1 P11 p12 p14 p = 2 P21 p22 0 1 4 0 0

tp2 I t < tpl: 1

2

4

5

1 P11 P12 P14 Pl5 P = 2 P21 P22 0 P25 4 0 0 P44 P45 0 0 0 1 5

tpl I t : 1 2

P= 3 4 5

Eqs.(4.4-10a) to (4.4-1Oc) are applicable also in this case. For 0 < t < tp2, the probabilities pli, p12, pi4 are given by Eq.(4.4-10d); p22 and p21 by Eq.(4.4-1&). For tp2 I t < tpl: PI1 = 1 - (a12 a14 + P15)PlAt where pi2 and pi4 are given in Eq.(4.4-10d), pi5 in Eq.(4.4-10f), p22 in Eq.(4.4log), p21 in Eq.(4.4-10e), p~ and p45 in Eq.(4.4-10h).

453 For tpl I t: p11 = 1 - (0112 4-a13 -+ a 1 4 4-P15)plAt p13 and Pi5 are given in Eq.(4.4-10f), pi2 and p14 in Eq.(4.4-10d), p22 and p25 by Eq.(4.4-10g) and p21 in Eq.(4.4-10e). Finally, p33, p34, p44 and p45 are give by Eq.(4.4-10h). The parameters of the solution are: pi = Q1/Vi (i = 1, ..., 5), tpl, tp2, 0112, 1x13, a14 and a 2 1 = R. The rest of the parameters can be obtained from Eqs.(4.41Oa) to (4.4-1Oc). Fig.4.4-lob demonstrates the effect of the residence time in the plug flow reactors for tpl = 2tp2 where tp2 = 0.01 and 0.1 (corresponding to At = 0.00005 and 0.0005, respectively) and pi = 25 and 250 (i = 1, ..., 5) for C(0) = [ l , 0, 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.4-10. Other unchanged parameters were: a12 = a 1 4 = 0.05, a13 = 0.9 and R = a21 = 0.025. The effect of tpl, tp2 is demonstrated in cases a, b and c, d, respectively. The effect of pi is demonstrated in cases a, c and b, d. Note the similarity between cases b and c although the time scale is different by a factor of ten. 1

t

i=l

=0.02 t = 0.01 (a)

0.8 -

0.6 va.4 -

-

0.2 0

4

I

I

I

3,

-

-

-

1 5

I

I

I

t

I

t

- - - - - - - - - - -cg ."

U

-

0

I

I

I

I

0.01

0.02

0.03

0.04

t

0.05 0

I

I

I

I

0.1

0.2

0.3

0.4

t

Fig4.4-lob. Ci versus t demonstrating the effect of

tpi

and pi

0.5

454 Case c: tp2 > t p l

The following transition matrices are applicable:

1 P11 Pl2 P14 p = 2 P21 p22 0 4 0 0 1

2

P= 3 4

P11

P12

P13

p14

P21

P22

0

0

0

0

P33

P34

0

0

0

1

tp2 I t: 1 2

P=

3 4 5

Eqsa(4.4-10a)to (4.4-1Oc) are applicable also in this case. For 0 < t < tpi, the probabilities pi 1, p12, pi4 are given by Eq.(4.4-10d); p22 and p21 by Eq.(4.4-10e). For tpl I t < tp2: p11 = 1 - (a12 -k a 1 3 4- a14)plAt where pi2 and pi4 are given in Eq.(4.4-10d), pi3 in Eq.(4.4-10f), p22 in Eq.(4.410e), p21 in Eq.(4.4-10e), p33 and p34 in Eq.(4.4-10h). For tp2 5 t: p11 = 1 (a12 4- a13 + a 1 4 -tP15)plAt pi3 and pi5 are given in Eq.(4.4-10f), pi2 and pi4 in Eq.(4.4-10d), p22 and p25 by Eq.(4.4-10g) and p21 in Eq.(4.4-10e). Finally, p33, p34, p44 and p45 are give by Eq.(4.4- 10h). The parameters of the solution are: pi = Ql/Vi (i = 1, ..., 5), tpl, tp2, a12, a13, a 1 4 and a21 = R. The rest of the parameters can be obtained from Eq~(4.41Oa) to (4.4-1Oc). Fig.4.4-10c demonstrates the effect of the residence time in the plug flow reactors for tp2 = 2tp1 where tpl = 0.01 and 0.1 (corresponding to At = 0.00005 and 0.0005, respectively) and pi = 25 and 250 (i = 1, ..., 5) for C(0) =

455 [l, 0, 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.4-10. Other unchanged parameters were: a12 = a14 = 0.05, a13 = 0.9 and R = a21 = 0.025.

The effect of $,I, $,2 is demonstrated in cases a, b and c, d, respectively. The effect of pj is demonstrated in cases a, c and b, d. Note the similarity, as in previous example, between cases b and c although the time scale is different by a factor of ten. 1 0.8 0.6

u'a.4

0.2 0 0

0.01

0.02

t

0.03

0.04

0.05 0

0.1

0.2

t

0.3

0.4

0.5

1

0.6

u"0.4 0.2 0 0

0.01

0.02

0.03 t

0.04

0.05 0

I

I

I

I

I

0.1

0.2

0.3

0.4

0.5

t

Fig4.4-10c. Ci versus t demonstrating the effect of tpiand pi

4.4-1 1 The codigurations in Fig.4.4- 11 demonstrate a model for small deviations from plug flow and long tails in continuous flow system described in ref.[21, p.2981. The deviation from plug flow, as well as adding time delay, is achieved by introducing plug flow reactors in parallel, as shown below. In case a the upper reactor is of plug flow type. In case b the plug flow reactor was divided into perfectly mixed reactors in order to obtain a solution for the following cases, 4.4 ll(1, 2, 3). In general, the perfectly mixed reactors are not of equal size, depending on the number of reactors in parallel and on their location.

456

random location

n J

Fig.4.4-11. Schemes for describing deviations from a plug flow reactor

4.4-1 l ( 1 ) The simplest scheme corresponding to Fig.4.4- 11 is demonstrated in case 4.4-l(3). An additional scheme is shown in Fig.4.4-1l(1) below. Reactors 2 and 4 were added to account for the recycle streams 4 1 2 and 434. The response of these reactors can be controlled by the magnitudes of p2 and p.4. If reactor 2, for example, is a measurement "point" of the concentration, i.e. a very small reactor, p2 should be assigned a relatively large value.

Fig.4.4-ll(1). Simplified scheme for describing deviations from plug flow reactor Three cases will be treated below, depending on the magnitudes of tpl and tp2. Case a: tpl = tp2 The following transition matrices are applicable: O c t <tpl: tpl lt : 1 3 5 1 2 3 1 pll = 3 0 5 0

P13 0 P33 P35 0 1

4

5

1 pll P12 P13 0 0 2 P2l P22 0 0 0 p= 3 0 0 P33 P34 P35 4 0 0 P43 P44 0 5 0 0 0 0 1

From Eqs.(4- 12a) to (4- 12c) one obtains that: (4.4-1 l(1a)) P35 = a13 = 1, a21 = a 1 2 and a34 = a 4 3 The following probabilities are obtained from Eqs.(4-14) to (4-19) for O < t <tpl: p11= 1 - plAt P13 = pgAt p33 = 1 - p3At p35 = p ~ A t (4.4-1 l(1b)) For tpl I t : pi1 = 1 - (I+ ai2)plAt p12 = al2p2At P13 = p3At (4.4-1 l ( 1 ~ ) ) (4.4-1 l(1d)) P22 = 1 - a21p2At P21 = a21p1At P33 = 1 - (I+ a341C13At

P34 = a34~4At

P35 = I@

(4.4-1 l(1e))

458 p44= 1 - a43p4At P43 = a43p3At (4.4-11(1f)) The parameters of the solutions are: pi = QlNi (i = 1, ..., 5), tpl = tp2 , 0 < R1 = a 1 2 < 00 and 0 < R2 = a34 < 00. Fig.4.4-11( la) demonstrates the effect of the residence time in the plug flow reactors for $1 = tp2, 0.1 and 1 (corresponding to At = 0.0005 and 0.005, respectively), the effect of pi =10 and 100 (i = 1, ..., 5) as well as the effect of the recycle R1 = R2,O and 10, for C(0) = [ l , 0, 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.4-11(1). The effect of R1 = R2 is demonstrated in cases a and b, the effect of tpl = tp2 is demonstrated in cases b and c and the effect of pi is demonstrated in cases c and d. The following observation should be noted: a) The similarity between cases a and c although the time scale is different by a factor of ten. b) The effect of R1 = R2 in case c is negligible due to the relatively large value of tpl = tp2 = 1. In other words, no response is observed in reactors 2 and 4.

i 0

0.1

0.2

t

0.3

0.4

0.50

0.2

0.1

R =R =O I

2

t

t

0.3

p2*.

0.5

(d2.

=1

=t pl

0.4

'

.

'

. '

-

- -- -- -I

I

-I

-

._

I

-----I

I

I

--__ I

459 Case b: $1 > $2

The following transition matrices are applicable:

0 < t < tp2:

tp2 It < tpl: 1

1

p35 1

2

4

5

1 P11 P13 0 0 P = 3 0 P33 P34 P35 4 0 P43 P44 0 0 0 0 1

0

1 P11 p13 p = 3 0 p33 5 0 0

3

3

4

5

1 P11 P12 p13 0 0 2 P21 P22 0 0 0 P= 3 0 0 p33 p34 p35 4 0 0 p43 p44 0 5 0 0 0 0 1 Eq.(4.4-11(la)) is applicable also in this case. For 0 < t < tp2 the probabilities are given by Eq.(4.4-1l(1b)). For tp2 I t < tpl the probabilities pi1 and pi3 are given by Eqs(4,4-11(lb)), p33, p34 and p35 are given in Eq.(4.4-11(le)); p44 and p43 are given by Eq.(4.4-11(10). For tpi It the probabilities p11, pi2 and pi3 are given in Eq.(4.4-11(1~)); p33, p34 and p35 are given by Eq.(4.4-11( le)); p44 and p43 are given by Eq.(4.41l(1f)). The parameters of the solutions are: pi = Q l Ni (i = 1, ..., 5 ) , tpl, tp2,

0 < R1 = a12 < 00 and 0 < R2 = a34 < A numerical solution was obtained for tpl = 0.2 and tp2 = 0.1 (corresponding to At = 0.0005), pi = 10 and 2 (i = 1, ..., 00.

3,R1 = R2, 0, 10 and 100 for C(0) = [l, 0, 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.4- 11(1). The effect of R1 = R2 is demonstrated in cases a, b and c in Fig.4.4-11( lb) and the effect of pi in cases c and d.

460

u-0.4 -

. - - -_

0.2 -/

I

I

0

0.1

0.2

t

I

I

0.3

0.4

0.5 0

0.1

0.2

0.3

0.4

0.5

(C)

1

-i

0.8 0.6

u-

0.4 0.2

0 0

0.1

0.2

t

0.3

0.4

0.5

Fig.4.4-ll(lb). Ci versus t demonstrating the effect of hiand Ri

1 P11 P13 0 = 3 0 P33 p35 5 0 0 1

P=

1 P11 P12 p13 2 P21 P22 0 3 5

tp2 It:

1

2

3

4

5

1 PI1 P12 p13 0 0 2 P21 P22 0 0 0 P= 3 0 0 P33 p34 p35 4 0 0 P43 P44 0 5 0 0 0 0 1

0 0

0

0

P33

P35

0

0

0

1

46 1 Eq.(4.4-1 l(1a)) is applicable also in this case. For 0 < t < tpl the probabilities are given by Eq.(4.4-11(lb)). For tpl I t < tp2 the probabilities p11, pi2 and pi3 are given by Eq.(4.41l(lc)), p21 and p22 by Eq.(4.4-1 l(ld)), p33 and p35 by Eq.(4.4-11(1b)). For tp2 5 t the probabilities pi 1, pi2 and pi3 are given by Eq.(4.4-11( lc)); p21 and p22 by Eq.(4.4-11(1d)); p33, p34 and p35 are given by Eq.(4.4-11(1e)); p44 and p43 are given by Eq.(4.4-11( If)). The parameters of the solutions are: pi = QlNi (i = 1, ..., 3,tpl, tp2,

0 < R1= a12 < = and 0 < R2 = a34 < =. A particular solution was obtained for tpl = 0.1 and tp2 = 0.2 (corresponding to At = 0.0005), pi = 10 and 2 (i = 1, ..., 5 ) ,

R1 = R2, 0, 10 and 100 for C(0) = [l, 0, 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.4-11( 1). The effect of R1 = R2 is demonstrated in cases a, b and c in Fig.4.4-11( lc) and the effect of pi in cases c and d. One should compare the behavior in the above figure corresponding to tp2 > tpl with that in Fig.4.411(lb) above for tpl > tp2.

0

0.1

0.3

0.2 t

0.4

0.5 0

0.1

0.2

t

0.3

0.4

Fig.4.4-ll(lc). Ci versus t demonstrating the effect of piand Ri

0.5

462

4.5 IMPINGING-STREAM SYSTEMS Impinging streams [73] is a unique and multipurpose configuration of a twophase suspension for intensifying heat and mass transfer processes in the following heterogeneous systems: gas-solid, gas-liquid, liquid-liquid and solid-liquid. The essence of the method demonstrated in Fig.4.5, lies in bringing two streams of a suspension, flowing along the same axis in opposite directions, into collision. As a result of collisions between the opposed streams, a relatively narrow zone of high turbulent intensity is generated, which offers excellent conditions for intensifying the heat and the mass transfer rates. In this zone, too, the concentration of the particles is the highest [73], and continuously decreases towards the injection point of the system. The opposed flow configuration of the suspensions encourages multiple inter particle collisions as well as mutual penetration and multiple circulation of the particles from one stream into the other. The penetration of particles arises due to their inertia, whereas deceleration takes place due to the opposite flow of the gas stream as a result of drag forces. At the end of the deceleration path, the particle is accelerated and once again regains its original stream. After performing several such oscillatory motions, the particle velocity eventually vanishes until it is withdrawn from the system. The latter might occur even earlier due to inter particle collisions.

IMPINGEMENT PLANE

Fig.4.5. The principle of impinging streams The intensification of the transfer processes is due to the following effects: a) An increase of the relative velocity between the penetrating particles and the opposed gas stream. Under extreme conditions, where the particle attains the gas velocity at the point of entering the opposite stream, the relative velocity may

463 reach twice that of the gas velocity; the increase is thus significant. Consequently, the external resistance will decrease. b) An increase of the mean residence time of the particles in the system or their holdup due to their penetration into the opposed stream, consequently undergoing multiple circulation followed by damped oscillations. However, in a dense system of particles, inter particle collisions might reduce the effect of the increase of the mean residence time. Finally, an increase of the mean residence time also allows a decrease of the geometrical size of the system. c) In gas-liquid and liquid-liquid systems, shear forces exerted between the phases or inter droplet collisions, can result in a breakup of the droplets, leading to an increase in their surface area, their rejuvenation, and therefore an increase of the mass transfer rates. d) Collision of the continuous phase of the opposed streams, namely, jet-jet impingement, induces pressure pulsations or generates intense radial and axial velocity components in turbulent flow. Consequently, good mixing is created in the impingement zone of the streams. The mixing is also enhanced by the multiple circulation of the particles in this zone. As indicated previously, the key phenomenon in impinging-stream systems is the penetration of particles into opposite streams, causing an increase of their holdup in the reactor, of the relative velocity between the particles and the carrying stream as well as a longer mean residence time of the particles. Consequently, the mixing properties of the reactor may be significantly improved. A quantitative analysis of this phenomenon and others related to impinging streams is presented in the following examples, where the major aim is to determine the behavior of the particles by introducing a pulse of these into one of the reactors. In section 4.5, the following designations are made: Qi, Qij and qjj are the mass flow rates of the particles stream (kg particleshec) and Cj is the concentration of the tracer particles (kg tracer particleskg particles). In such systems, the tracer particles are, usually, those of the original ones. They are, however, made radioactive or are painted, in order to distinguish them from the original particles. The latter makes it possible to determine their concentration versus time in the RTD experiments [73, p.1761, thus their mean residence time tm in the system. tm =

464 Vj/Qj in a perfectly mixed reactor. An additional key quantity is the holdup of the particles in the reactor, Vi (kg particles); instead of the volume of the reactor. Thus, in consistent units, Eqs.(4-1) to (4-11) hold. An important quantity used in this section for comparing various effects is the mean residence time tm of the particles in the system.

4.5-1 The simplest model [73, p.1801 of a two impinging-stream reactor is shown schematically in Fig.4.5-1. On the LHS is demonstrated the actual configuration of the reactor and on the RHS the Markov-chain model. The latter employs the following considerations and assumptions: a) There are three main zones in which mixing of particles takes place: zones 1 , 2 and 3. These zones are considered to behave like perfectly mixed vessels, i.e. states. b) The lower part of the reactor behaves like a plug flow reactor in which no back mixing takes place. This is designated as tp in Fig.4.5-1. c) The time needed for a particle to pass from one vessel to the other is equal to zero. This is true since the actual borders of one vessel overlap with the neighboring one. On the other hand, a particle might stay in one of the above zones during a finite time. d) The holdups, Vj, of the particles in each model vessel are the same. The resulting model is shown on the RHS of Fig.4.5-1. Clearly each vessel represents a state in a Markov process; vessel 4 is the collector of the particles from which only the carrying stream is leaving at flow rate 2Q1. Q12,421,Q23 and Q 32 are recycle streams. This is due to the penetration of particles from one stream into the other where the impingement zone is designated by 2 in Fig.4.5-1 and is simulated as a perfectly mixed reactor. The effect of penetration is emphasized by the fact that movement of particles to vessel 4 is only possible from vessels 1 and 3. Whenever a particle reaches vessel 4,it remains there, i.e. the vessel is a trapping state.

465

q 4 = !

t 2Ql Fig.4.5-1. A model for a single stage two impinging-stream reactor Noting that the residence time in the plug flow reactor is tp yields the following matrices:

Considering reactor j in Fig.4-1, hence j = 1, a = 2, b = 3 and 5 = 4, as well as Eqs.(4-12a) to (4-12c), and taking Q1 as a reference flow, yields: 1 + a 3 = p i 4 + p34, 1 + a 2 1 = a 1 2 + p14, for i = 2: a 1 2 + a 3 2 = a 2 1 + a23, i = 3: a3 + a23 = a 3 2 + p34 The following assumption were made: a 1 2 = a 2 1 = a23 = a 3 2 = R, a3 = a1 = 1; this yields that p i 4 = p34 = 1 The holdups of particles in all reactors are the same, i.e. pi = p, i = 1, ..., 4 From Eqs.(4-14) to (4- 17b) the following probabilities were obtained: For 0 < t < tp: p11 = p33 = 1 - RpAt p12 = p21= p23 = p32 = RpAt

466 p22 = 1 - 2RMt FortpSt:pll=p33=1-(1+R)pAt PI4 = P34 = @t where the rest of the probabilities are as for 0 c t c tp. As seen, the parameters of the solutions are: p, R and tp. In the numerical solution it was assumed that tp = 1 corresponding to At = 0.005, p = 10 and that C(0) = [l, 0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.5-1, RHS.

The effect of the recycle R, 0, 0.1, 1 and 10 is demonstrated in Fig.Fig.4.5-la. It is observed that by increasing R, the central reactor 2 becomes active with respect to the response of the pulse input in reactor 1 and the system of reactors 1 , 2 and 3 behaves as a single perfectly-mixed reactor for R = 10.

t

t

Fig4.5-la. Ci versus t demonstrating the effect of R

467

4.5-2 The reactor [82; 73, p. 1881 depicted schematically in Fig.4.5-2 is a modified form of the original two impinging-stream reactor described in Fig.4.5-1 on the LHS with two additional air streams located below the upper streams where particles are introduced. A brief description of the vessels-flows model in Fig.4.52, is as follows. The inlet pipes to the reactor are simulated by two plug flow reactors 1 and 5. The entrance of the particles to the reactor are followed by three zones 2, 3 and 4, simulated by perfectly-mixed vessels. Particles leaving vessels 2 and 4 enter another mixing zone, designated as a perfectly mixed vessel 6, formed by the secondary air stream. The particles leave this reactor through a tubular reactor, where the time needed for a particle to pass from one vessel to the other is zero but it is finite for staying in the reactors. The recycle streams between vessels 2 , 3 , 4 simulate the harmonic motion of the particles.

t 2Ql Fig.4.5-2. An impinging-stream reactor with two pairs of tangential air feeds

468 Case a: tpl > 0

Noting that the residence time in the plug flow reactors is tp and tpl, yields the following matrices: O < t <tp: 1 P11 P12

p= 3 5 6

3 0

4 0

5 0

6 0

0

0

0

0

0

P22

p23

0

0

p26

0

P32

P33

p34

0

0

0

0

p43

p44

0

p46

0 0

p54

p55

0 1

0 0

0

0

0

0

7 0

1 P11 P12 2 0 P22 P23 0 0 P26 0 3 0 P32 P33 P34 0 0 0 P = 4 0 0 P43 P44 0 p46 0 5 0 0 0 P54 P55 0 0 6 0 0 0 0 0 P66 P67 7 0 0 0 0 0 0 1 Considering reactor j in Fig.4-1, hence j = 1, a = 2, b = 3, c = 4, d = 5, e = 6 and 6 = 7, as well as Eqs.(4-12a) to (4-12c), and taking Q1 as reference flow, yields: 1 + a5 = p67, a 1 2 = 1, for i = 2: a 1 2 + a 3 2 = a 2 3 + a26, i = 3: a 2 3 + a 4 3 = a 3 2 + a34, i = 4:a 5 4 + a 3 4 = a 4 3 + a46, i = 5 : a 5 = (3.54, i = 6: a 2 6 + a 4 6 = p67 The following assumption were made: a5 = a1 = 1, thus, a 5 4 = 1 and p67 = 2 a 2 3 = a 3 2 = a 4 3 = a 3 4 = R,thus, a 2 6 = a 4 6 = 1 yielding the following probabilities:

469 p11= 1 - PlAt P22 = 1 - ( 1 + p33 = 1 - 2Rp3At p~ = 1 - (1 + R)p4At

P12 = p2At P26 = p6A.t P23 = Rp3At P32 = Rp2At P34 = Rp4At p43 = Rp3At p46 = P55 = 1 - p5A.t P54 = MAt P66 = 1 - P6At P67 = CL?At In the numerical solution it has been assumed that pi = p, yielding the following parameters: p, R, tp and t p l . In addition, tp = 0.2, tpl = 0.4 corresponding to At = 0.001 and that C(0) = [ l , 0, 0,0,0, 0, 01, i.e. the pulse was introduced into reactor 1 in Fig.4.5-2. The effect of the recycle R, 0 and 10 is demonstrated in cases a, b and c, d for p = 1 and 50, respectively, in Fig.4.5-2a. The effect of p = 1,50 is demonstrated in cases a, c and b, d, respectively.

n.6

.-

uo14 0.2

t

.--_.-

i 0

t

L 0.1

t

Fig.4.5-2a. Ci versus t demonstrating the effect of R and p

470

Case b: tpl = 0 In this case, the plug flow reactor before the exit from the system in Fig.4.52, is absent. The results for identical parameters corresponding to case a are depicted in Fig.4.5-2b. It was observed that for p = 1, the results for the present case coincide with the results in case a above in the investigated range oft. The results in cases a and b below for i = 6, 7 differ from the results in cases c and d above due to the effect of $1.

0

I

I

I

0.1

0.2

0.3 t

I

0.4

I

0.5

0.6 t

Fig.4.5-2b. Ci versus t demonstrating the effect of R and p As indicated before, a possible increase in the mean residence time tm of the

particles might be anticipated due to penetration of particles into the opposed stream, consequently undergoing multiple circulation followed by damped oscillations. This behavior is controlled in the above model by the quantity R, i.e. the recycle stream, and is demonstrated as follows for extreme cases. R = 0: In this case the effective reactors are 1,2 and 6. Excluding reactor 7, yields the following equation for the total mean residence time in the reactors for a pulse introduced into reactor 1:

Assuming that the holdup V of the particles in the reactors is the same, thus p = QlN,yields that:

47 1

3 tm=tp+tp1+-

(4.5-2a)

P

R+-:

In this case all reactors are effective and reactors 2, 3 and 4 behave as a single reactor of holdup V2 + V3 + V4 , thus Vl h=Qi

v,+v,+v, '5'

24,

v, +-24,

+

tpl

or alternatively a J

tm=tp+tp1+

ji

(4.5-2b)

Eqs.(4.5-2a) and (4.5-2b) indicate that the relative increase in tm, due to the recycle R, with respect to its value for R = 0 is 16.67%.

4.5-3 As indicated at the beginning of this section, there is an increase of the

holdup of the particles at the impingement zone, resulting an increase in the mean residence time of the particles in the system. However, inter particle collisions might decrease the above effect as well as back flow of the particles. It is the aim of this case, depicted in Fig.4.5-3, to investigate these effects. Reactors 1 to 5 , demonstrate the impingement zone and reactor 3 with the highest concentration, demonstrates the impingement plane. Reactor 6 is the exit of the particles from the impingement zone and reactor 7 is the collector of the particles.

472

5

P11

p12

0

0

0

p16

2 P21 P22 p23 0 0 0 3 0 P32 p33 p34 0 p36 p= 4 0 0 0 p43 p44 p45 5

0

6 P61 7 0

0 0 0 0

0 0

0 0

p54

p55

p56

0

0

P65

P66

P67

0

0

0

0

0

1

Considering reactor j in Fig.4-1, hence j = 1, a = 2, b = 3, c = 4, d = 5, e = 6 and 6 = 7, as well as Eqs.(4-l2a) to (4-12c), and taking Q1 as reference flow, yields: 1 + a5 = p67

1 + a61 + a21 = a 1 2 + (3116 for i = 2: a 1 2 + a 3 2 = a 2 1 + a23, i = 3: a 2 3 + a 4 3 = a32 + a 3 4 + a36, i = 4: a 3 4 + a 5 4 = a 4 3 + a45, i = 5: a5 + a45 + a 6 5 = a 5 4 + a56, i = 6: a 1 6 + (336 + a 5 6 = a61 + a 6 5 + p67 Assuming that: a5 = a1 = 1, a 3 2 = a 3 4 = a 2 1 = (145, a12 = a 5 4 = a 2 3 = a43,

473 a16 = a569 a61 = a 6 5

and taking the following quantities as known, designated by 0.21 = R, 0136 = a,a16 = y yields that: a12 = a54 = a23 = a43 = R + a / 2 a61 = 0165 = y + a/2 -1 a56 = y a32 = a 3 4 = a45 = R p67 = 2

Considering the above coefficients and Eqs.(4-14) to (4-19), yields the following probabilities: p11= 1 - (R + W2 + y)plAt pl2 = (R + a/2)plAt p16 = w t p22 = 1 - (2R + a/2)~2At p21= RplAt p23 = (R + a/2)p3At p33 = 1 - (2R + a)p3At P32 = Rp2At P34 = Rp4At P36 = a p d t p44 = 1 - (2R + a/2)p4At p43 = (R + a/2)p36t p45 = Rp5At p55 = 1 - (R + a/2 + 9 ~ 5 A t p54 = (R + a/2)w4At p56 = '@& p6i=(y+a/2- 1)piAt i = 1,5 P66 = 1 - (2'Y+ a)p6At P67 = 2p7At Because of symmetry, pi = p5 = p, p2 = p4 = p', yielding the following parameters: pi (i = 3, 6, 7), p, p', R, a and y. The mean residence time in the system (Fig.4.5-3), for a pulse introduced into reactor 1 and considering reactors 1 to 6, is determined below for R = 0 and R + 00. R=O: Ignoring recycles 461 and 465, assuming that the holdups V1 = v6, thus p = Q1N1, yields vl

v6

3

t"=Q1+24,=%

474

R+-:

where p' = Q 1 N 2 = Q1N4 and p3 = Q1N3. The above equation indicates that if the holdup of the particles V3 in the impingement zone is increased, p3 is decreased and hence tm is increased. The latter is an important property of the impingingstream configuration [73, p.3 and 1381. It has also been observed in the calculations that increasing a = a36 and y = a16 = a56 has a negligible effect on tm. Finally, it should be noted that values of tm calculated by the above equations, coincide with numerical values obtained by numerical integration according to Eq.(4-26) for a pulse introduced into reactor 1 and collected in reactor 6 . Fig.4.5-3a demonstrates the effect of R = a32 = a34 = a45 = 0, 5, 25, 100 on the concentration profile for y = a56 = 1 and constant pi = 1 (i = 1, ..., 7) for a unit pulse introduced into reactor 1.

475

R = 25

R = 100

- -

7

2!

--_

-

I t

I

I

t

Fig.4.5-3a. Ci versus t demonstrating the effect of R

4.5-4 The major characteristic of impinging streams is the penetration of particles from one stream into the opposite one through the impingement plane (Fig.4.5), thus, increasing their mean residence time in the reactor as well as their relative velocity with respect to the air. The scheme in Fig.4.5-4 demonstrates this effect in the following way. Reactors 1 to 10 simulate the impingement zone of the particles and in each reactor particles reside for some time. A pulse of particles introduced in reactor 1 may be divided into three streams. One stream occupies reactors 3 to 5, the other, reactors 6 to 8, and the third one will occupy reactors 9 and 10. Eventually, the pulse accumulates in reactor 12 while passing reactor 11.

476

ir

QlQ1-

Q5

1

+p+Q5 k = 12

Fig.4.5-4. A scheme for demonstrating the effect of different penetration distances in impinging streams The following matrix is applicable for the configuration in Fig.4.5-4:

1 2 3 4 5 6 7 8

P= 9 1c 11 12

The following simplifying assumptions were made: pi = p (i = 1, ..., 12). For reactors 1,2, 3 , 4 , 5 all interactions are equal, i.e., R1 = aij. For reactors 1,6,7,8, 5 all interactions are equal, designated as R2. For reactors 1,9, 10,5 all interactions are equal, designated as R3.

477 Considering Eqs.(4-12a) to (4-12c) and Eqs.(4-14) to (4-19), yields the following probabilities: p11 =p55 = 1 - (1 +R1+ R2 + R3)pAt P12 = p21= P23 = P32 = p34 = p43 = p45 = p54 = RipAt P16 = P61 = P58 = PI35 = p67 = p76 = p78 = p87 = R2pAt P19 = P91= P5,lO = P10,5 = p9,io = pio,g = R3pAt PI,11 = P5,ll = ClAt ~ 2 =2 p33 = ~ 4 =4 1 - 2R1PAt p99= Pl0,lO = 1 - 2R3CLAt p11,11= 1 - 2pAt p11,12 = 2pAt As seen the parameters of the solution are: p and R1, R2 and R3. The mean residence time in the system (Fig.4.5-4), for a pulse introduced into reactor 1 and considering reactors 1 to 11, is determined below for Ri = 0 and Ri + 00. Ri = 0: In this case only reactors 1 and 11 are effective for a unit pulse introduced into reactor 1. Assuming that the holdups V1 = V11, thus p = Q1N1, yields

Rj + 00: In this case reactors 1 to 11 are effective. Assuming the same holdup of particles in all reactors, Vi = V, where p = Q l N , yields

The above equations indicate that the relative increase of tm due to the recycles Ri, with respect to tm for Ri = 0, amounts to 266.7%. Fig.4.5-4a demonstrates typical response curves to a unit pulse input introduced into reactor 1 for R1 = 10, R2 = 5 , R3 = 1, p = 10 and At = 0.00005.

478

1

0.8

0.6

Li' 0.4

0.2 0

. .

0

0.01

.....

0.02

.... .

0.03

.

..

..

0.04

-

.

.

0.05

t

Fig.4.5-4a. Ci versus t

4.5-5 The following schemes demonstrate configurations comprising of 2 , 3 and 4 impinging streams. The effect of the number of impinging streams on the mean residence time of the particles in the system will be investigated below.

479 case a: 2-impinging streams

case b: 3-impinging streams

case c: 4-impinging streams

case d: 2-impinging streams

p5%

case e: 2-impinging streams

rl

5=4

py 5=4

Fig.4.5-5. The effect of the number of impinging streams

480

Case a: 2-impinging streams The following matrix applicable for case a in Fig.4.5-5 is: 1

P=

2

3

4

5

1 PI1 PI2 0 p14 0 2 P2l P22 P23 0 0 3 0 P32 P33 P34 0 4 0 0 0 p44 p45 5 0 0 0 0 1

Considering Eqs.(4-12a) to (4-12c) and assuming that pi = p (i = 1, ..., 5), a12 = a 2 1 = a 2 3 = a 3 2 = R and that a 1 4 = a34, yields from Eqs.(4-14) to (4-19), the following probabilities: p11= p33 = 1 - (1 + R)yAt p22 = 1 - 2RpAt P14 = P34 = VAt p21 = p12 = p23 = P32 = RpAt p44 = 1 - 2 M t p45 = 2pAt Fig.4.5-5a demonstrates the effect of the recycle R for y = 1 and At = 0.001 for a unit pulse introduced into reactor 1. It is observed that for R = 50, reactors 1, 2 and 3 behave as a single reactor due to the relatively high recycle.

-

t

t

Fig.4.5-5a. Ci versus t demonstrating the effect of R

481

Case b: 3-impinging streams The following matrix is applicable for case b depicted in Fig.4.5-5:

1 2 3

P= 4 5 6 Considering Eqs.(4-12a) to (41%) and assuming that = p (i = 1, ..., 6), a12 = a21 = a23 = a32 = a 2 4 = a42 = R, a15 = a35 = a45 yields from Eqs.(414) to (4-19), the following probabilities: PI 1 = P33 = PM = 1 - (1 + R)pAt p22 = 1 - 3RpAt

p12 = p21= pu = ~ 3 =2 p24 = p42 = RpAt pls = ~ 3 =5p45 = pAt p55 = 1 - 3pAt ~ 5 =6 3pAt Fig.4.5-5b demonstrates the effect of the recycle R for p = 1 and At = 0.005 for a unit pulse introduced into reactor 1. It is observed that for R = 50, reactors 1

to 4 behave as a single reactor due to the relatively high recycle.

-

-

t

t

Fig.4.5-Sb. Ci versus t demonstrating the effect of R

482

Case c: 4-impinging streams The following matrix is applicable for case c depicted in Fig.4.5-5:

1 2 3 P= 4 5

6 7

As before,the following probabilities are obtained: P11=P33=P44=P55=1-(1+R)pAt ~22=1-4RpAt P12 = P21= P23 = P32 = p24 = p42 = p25 = p52 = RpAt P16 = P36 = P46 = P56 = PAt p66 = 1 - 4@t p67 = 4pAt where R = a12 = a21 = a23 = a 3 2 = a 2 4 = a 4 2 = a25 = 0152 Fig.4.5-5c demonstrates the effect of the recycle R for p = 1 and At = 0.005 for a unit pulse introduced into reactor 1. It is observed that for R = 50, reactors 1 to 5 behave as a single perfectly-mixed reactor due to the relatively high recycle.

0.8

0.6

v-0.4

c

\.;

0.2 O - 1 0

1

2 t

3

4

5 t

Fig.4.5-5c. Ci versus t demonstrating the effect of R

483

Case d: 2-impinging streams This is another scheme of 2-impinging streams depicted in Fig.4.5-5 case d where part of the impingement zone is a plug flow reactor. Noting that the residence time in the plug flow reactor is tp yields the following matrices:

p= 3 4

P11 0 0

p13 p33 0

0 p34 1

1 2 P= 3 4

2

3

P11

P12

P13

0 0 0

P22

P23

0 0

P33

0

4

0 0 p34 1

4 0

0 P34

1

Considering Eqs.(4-12a) to (4-12c) and assuming that pi = p (i = 1, ..., 4), a 1 2 = a 2 1 = R, a 1 3 = a 2 3 , yields from Eqs.(4-14) to (4-19) the following probabilities: For 0 < t < tp: p11= 1 -@t p13 = PAt p33 = 1 - 2@t p34 = 2pAt For tp I t: p11= P22 = 1 - (1 + R)@t P12 = P21 = RCldt P13 = P23 = @t p33 = 1 - 2@t p34 = 2pAt Fig.4.5-5d demonstrates the effect of R, p and tp on the distributions Cj - t for At = 0.005. Cases a and b show the effect of R; if R > 0 reactor 2 becomes active after some time. Cases b and c depict the effect of p and cases c, d

484

demonstrate the effect of tp. In the latter case it is interested to note that reactor 2 is inactive all the time due to the relatively high value of tp.

(4

p = 5 R = O t =0.1 P

-

0

I

I

I

I

0.1

0.2

0.3

0.4

0.5 0

I

I

0.1

0.2

t

I

t

p=10 R=5

0.3 t

=1

p---

I

0.4

0.5

- - - -(Q -

c

--i---

0

0.1

I

I

I

0.2

0.3

0.4

t

0.50

I

I

I

0.1

0.2

0.3 t

I

0.4

0.5

Fig.4.5-5d. Ci versus t demonstrating the effect of R, p. and t, Case e: 2-impinging streams In this scheme depicted in Fig.4.5-5 case e, an additional tubular reactor is added, simulating the plug flow at the lower part of the reactor (Fig.4.5-1, LHS) in which no back mixing takes place. The residence time in the plug flow reactor simulating the impingement zone is tp and of the lower part it is tpl. In the following we consider the interesting case where tpl > tp; a unit pulse introduced into reactor 1, yielding the following information: For 0 < t <tp, and because tpl > tp, no change in concentration takes place in the system. For tp It < 2tp (c tpl), there is a transfer of the tracer from reactor 1 to 2 only through the tubular reactor. As a result of this situation Q1 = 0 and there are only

485

circulation flows, i.e. 4 1 2 = 421. However, when t > tpl, Q1 > 0. We define R with respect to this flow, i.e. R = Q12/41 = a 1 2 = a21. Under this condition, the following matrix holds:

p=

:I

1

2

P11 0

P12 1

For 2tp 5 t < 3tp (< tpl), the following matrix holds: 1

2

For tpl I t ,the following matrix holds:

P=

3

4

PI1 P21 0 0

Pl2 p22

p13 p23

0 0

P33

0 0 p34

0

1

where p11 =p22 = 1 - (1 +R)pAt P12 = P21 = RpAt P13 = P23 = PAt p33 = 1 - 2pAt p34 = 2pAt As seen, the parameters of the solution are: R, p, tp and tpl. Fig.4.5-5e demonstrates the effect of the above parameters on the distributions Ci - t for At = 0.00005 and 0.005. The following quantities were assigned for the parameters: tpl = 5tp; tp = 0.01 and 1, R = 0, 1 and 10 whereas p = 50, 100 and 500. Cases a, b

486

and b, e show the effect of p; cases a, c depict the effect of tp and tpl; cases d, e, f demonstrate the effect of R. 1

0.8

0.6 .-

u 0.4 0.2 0

I

I

-

0.6

-

0.4

-

6-

I I I t =0.01 t = 0.05 P Pi

I

i=l/ ;'.$ .

1;

I,'

\R=1

I

-4

I

p=500.

: :

-

0.2 3,4 I

0

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0 t

I

I

I

I

I

0.01 0.02 0.03 0.04 0.05 0.06 0.07 t

487

R = 10

0.8

p = 500 .

0.6

0

0.01 0.02 0.03t0.04 0.05 0.06 0.07

Fig.4.5-5e. Ci versus t demonstrating the effect of R, p, t, and tpl Mean residence time tm of the impinging-stream systems in Fig 4.5-5 The comparison will be made for extreme cases, i.e. R = 0 (part of the reactors are not interacting) and R + 00 (appropriate reactors are interacting and considered as a reactor of volume equal to the total volume of the single reactors). Note that R = Q12/Q1= Q21/Q1. The following examples elaborate the above for a unit pulse introduced into reactor 1 and assuming that the holdups of the particles in each reactor are the same, i.e. Vi = V, thus, pi = p = Q1N Case c R = 0: tm = tml+ tm6 = V1/Q1 + Vd(4Q1) = 1/p + 1/(4p)= 5/(4p)

tm = (5 + R)/(2p(l+ R)) + tp where tp = Vd(2Q12) = 1/(2Rpp) and pp= QINp.

488

Case e ($1 > tP) R = 0: For t Itpl tm = tpl Fort 1 tpl tm = tml 342P) + tp1 R + 00: Fort Itpl tm = tpl

+ tpl+ tm3 = \. r/Q1 + tpl + V3/(2Q1) =

Fort 2 tpl tm = tml+ tp + tm2 + tp + tm3 = (5 + R)/(2p( 1+ R)) + tp + tpl where tp is given above in case d, and tpl = Vpl/(2Q1) = 1/(2pp1) and ppl = QlNpl

Similarly, the following Table may be obtained for a pulse input introduced into reactor 1: Table 4.5-5. Expressions for the mean residence time tm in the reactor-system in Fig.4.5-5 (excluding the collector reactor 6 )

tp and tp,l are given in cases d and e above. It should be emphasized that the value of tm calculated by the above equations, coincides with numerical values obtained by numerical integration according to Eq.(4-26) for a pulse introduced into reactor 1 and collected in reactor 6-1.

489 The following ratios may be obtained from Table 4.5-5:

a1 demonstrates the effect of the recycle R on increasing tm and it approaches 1 for large n. Maximum a1 = 4/3 for a 2-impinging-stream reactor (n = 2).

a2 and a3 show the effect of the number of streams n on the mean residence time ratio for R = 0 and R + 00. It may be concluded that increasing n causes the ratios to approach unity. The maximum ratio is obtained for n = 2, yielding 012 = 9/8 = 1.125 and a3 = 12/10 = 1.2

4.5-6 Fig.4.5-6 shows a single stage four impinging-stream reactor [73, p. 1861. On the RHS is schematic of the reactor and on the LHS is the model of reactors and flows. It is assumed to comprise eight perfectly-mixed reactors designated as 1, ..., 8. All reactors are assumed to have equal holdups. The model assumes also tangential feed of the streams and that more mixing zones are generated by the direct impingement of the streams. Recycle occurs through intermediate perfectlymixed reactors 2,4,6, 8 where impingement is expected.

490

-

Fig.4.5-6. A model for a single stage four impinging-stream reactor For the configuration in Fig.4.5-6, the following matrix holds:

1 2

p=

3 4 5 6

7 8 9 10 Considering reactor j in Fig.4-1 and making the following designations: j = 9, a = 1, b = 2, ..., Z = 8 and 6 = 10. Applying Eqs.(C12a) to (4-12c), taking Q1 as reference flow and making the following assumptions, i.e. a 7 = a5 = a3 = a1 = 1

49 1 R = a 1 2 = a 2 1 = a 2 3 = a32 = a34 = a 4 3 = a45 = a54 = a56 = a65 = a67 = 0176 = a 7 8 = a 8 7 = a 1 8 = a81 yields that a19 = a39 = a59 = a79 as well as P9,lo = 4 Assuming also that all reactors have the same holdup of particles, i.e. Vi = V (i = 1, ..., 10) or pi = p = Q i N , and applying Eqs.(4-14) to (4-19), yields the following probabilities: PI 1 = ~ 3 =3 PSS = p77 = 1 - (1 + 2R)pAt p22 = p~ = p66 = p88 = 1 - 2RpAt

P12 = P21 = P23 = P32 = P34 = P43 = P45 = P54 = P56 = P65 = P67 = p76 = p78

= P87 = PlS = p81= RpAt

P19 = P39 = P59 = P79 = @t ~ 9 =9 1 - 4 W PSJO = 4 W As seen the parameters of the solution are p and R. The mean residence time in the system (Fig.4.5-6), for a pulse introduced into reactor 1 and considering reactors 1 to 9, is determined below for R = 0 and R + -. R = 0: Assuming that the holdups V1= Vg, thus p = Q1N1, yields

R+-:

The above equations indicate that the relative increase in the mean residence time with respect to R = 0, due to the recycles, is 80%. In the numerical solution the effect of the recycle R, 0,2 and 50 and of p, 1 and 0.1, is demonstrated in Fig.4.5-6a for At = 0.005 and a unit pulse introduced into reactor 1 in Fig.4.5-6. Note that due to symmetry C2 = Cs,C3 = C7 and C4 = c6. The effect of R is demonstrated in cases a, b and c; the effect of p in cases c and d.

492

1

c

I

50

1

I

I

I

(q

3

4

5

0.8 0.6

u 0.4 .-

0.2

0 0 1

1

2

I

I

t

3

4

I

I

(C

I

i=l

0.8

2

t

p = 0.1 R=50

F = 1-

i

R=50 -

0.6

.-

0.4 0.2

0 -0.5

0

0.5

t

1

1.5

2-1

0

1

2

t

3

4

5

Fig.4.5-6a. Ci versus t demonstrating the effect of R and

4.5-7 An extension of the single-stage two impinging-stream reactor to a 3-stage

reactor [73, p.1951, and similarly to a multi-stage reactor, is demonstrated in Fig.4.5-7. The reactor is composed of sections separated by plates a, with appropriate openings c, d etc. Two successive plates and their top views A-A and B-B' are shown. Between two plates there is a partition designated by b. The partition is dividing the gas-particle stream so that the impinging-stream effect is maintained at each of the reactor's stages. In order to operate the multi-stage reactor, two gas-solid streams are fed to the top of the reactor. The streams collide in the impingement zone 2 (Fig.4.5-7, top view). The combined gas-solid stream formed after the impingement enters through opening c and flows downwards where it is divided into equal horizontal streams by the partition b. The streams impinge above opening d and enter the next stage where they are divided again by

493 partition b. Eventually, the solid particles leave at point e and the gas exits through pipe f located at the reactor's exit. The vessel-flow arrangement which is modeled, is shown on the RHS of Fig.4.5-7. The n stages may be envisioned as n identical reactor segments, shown schematically in Fig.4.5-1, LHS. Each segment consists of four mixed vessels where the multi-stage reactor is terminated by a plug flow reactor. The inlet pipes are also simulated by plug flow reactor. It is assumed that the transition time from one stage to the other is negligible as compared to the time spent in the mixed vessel. top view gas-blm 4

i

n

Fig.4.5-7. Structure and model of a 3-stage two impinging-stream reactor

494 The transition probability matrix for the 3-stage reactor is given in the following and can easily be extended to a multi-stage reactor. For tp < t < $1 the matrix is of 12 by 12 since reactor 13 is yet inactive whereas for tpl I t, it reads: 1

2

1 2

'11

p12

3

0

4

0

0

0

5

0

0

0

0

6

0

0

0

0

P= 7

3

4

0

p14 0

'21 p22 p23

p32 p33 p34

5

6

7

8

9

10

11

12

13

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

p47

0

0

0

0

0

0

p55 p56

0

p58

0

0

0

0

0

p65 p66

p67

0

0

0

0

0

0

0

0

0

0

0

P44 P45

0

0

0

0

o p 7 6 p77 p78

8

0

0

0

0

0

0

0

p88

p89

0

p8,ll

0

0

9 10 11 12 13

0

0

0

0

0

0

0

0

p99

p9,lO

0

p9,12

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

p10,9 P10,lO P 1 0 , l l

P 1 1 , I O P 1 1 , l l p11,12

0

p12,12 p12.13 0

1

The derivation of the probabilities is similar to case 4.5-1. The following probabilities were obtained assuming symmetry in the flows with respect to reactors 2, 4, 6, 8, 10 and 12, and that all reactors have the same holdup of particles, thus, pi = p, i = 1, ..., 13: P I I = P33 = PSS=p77 = p99 = p11,11= 1 - (1 + R)pAt P22 = P66 = Pl0,lO = 1 - 2RPAt p44 = p88 = 1 - 2pAt

P12 = P21 = P23 = P32 = P56 = P65 = P67 = p76 = P9,lO = PlO,9 = Pl0,ll = Pl1,lO = RpAt P45 = P47 = P89 = P8,ll = FAt P14 = P34 = P58 = P78 = p9,n = p11,12 = pAt whereas for tp < t < tpl: p12,12 = 1 and for tpl I t : p12,12 = 1 - 2pAt ~ 1 2 =~2pAt 3

495

As seen, the parameters of the solutions are: p, R, tp and tpl. In the numerical solution it was assumed that tp = 0, tpl = 1, 5 corresponding to At = 0.001 and 0.005, respectively. Other parameters are: R = 0, 1, 20, p = 1, 5 and the unit pulse was introduced into reactor 1 in Fig.4.5-7, RHS. Fig.4.5-7a demonstrates the relationship Ci-t in which the effect of R is demonstrated in cases a, b and c, the effect of p is depicted in cases c and d whereas the effect of tpl in cases d and e. Note that due to symmetry C1 = C3, C5 = C7 and Cg = C11.

1

0.8 0.6 U-

0.4 0.2 0

0

1

2

3

I

I

I

4

5

6

I

I

I

7

t

1 -

PI

i=l

(b)

-

p=1 R = l t =5

-

0.8

U-

0

1

2

3

4 t

5

6

7

496 p = l R=20 t = 5 PI

=1

0.6 .-

u

0.4

12 ..

13.’

1

1

0.2 0

I

I

I

1 ,’

0.8 -

I

1

p = 5 R=20 t = 5

13

PI

(4

. . . . . .-

-

-

-

0.4 -

-

-2

-

0.6

u-

0.2

0 -

.-._.-__. _ _ ”._

-

497 1

0.8

0.6 U'

0.4 5.6

I

-0.5

0

I

,'

0.5

I

I

I

I

1

1.5

2

2.5

3

t

Fig.4.5-7a. Ci versus t demonstrating the effect of R, IJ. and tpl The mean residence time in the system (Fig.4.5-7), for a pulse introduced into reactor 1 and considering reactors 1 to 12, is determined below for R = 0 and

R + -, R=O: In this case reactors 1, 4, 5, 7, 8, 9, 11, and 12 are effective for the pulse introduced into reactor 1. Assuming that the holdups of the reactors are identical, i.e. Vi = V, thus p = Q N , yields:

R+-: In this case reactors 1 to 12 are effective, and reactors 1-2-3, 5-6-7 and 9-10-1 1 may be considered as single reactors due to the recycle R. The above assumption is valid also here, thus it follows that: tm = tp + tpl

+ (15/(21~.)

The above equations indicate that the relative increase of tm due to the recycles R, with respect to tm for R = 0, and ignoring tp and tpl is 15.4%.

498

Chapter 5

APPLICATIONS OF MARKOV CHAINS IN CHEMICAL PROCESSES The major objective of this chapter is to demonstrate how Markov chains can be applied to determine the transient behavior of complicated open systems undergoing simultaneously heat and mass transfer processes as well as chemical reactions. It is an extension of chapter 4 in which the RTD of a complicated system was investigated.

5.1 MODELING OF THE PROBABILITIES The model. A multi-component and multi-reactor system, arranged according to the general model depicted in Fig.5-1, is considered, which extends the scheme in Fig.4-1. It covers numerous flow arrangements and processes encountered in Chemical Engineering. The general scheme consists of a central reactor designated by j and peripheral reactors a, b, ..., Z. A process may terminate at the exit of the peripheral reactors, but the model makes it possible also to collect in reactor 5 the streams leaving the peripheral reactors. In reactor 5 no chemical or physical processes take

place. - flows The flow system comprise the following flows: Qj,Qa, Qb, ..., Q'b, ..., QqZ - flows leaving from feed vessels to reactors j, a, b, ..., Z. Q;, reactors j, a, b, ..., Z, outside. There are also interacting flows between the reactors, i.e. each reactor is feeding all the others. Finally, flows qs are from each reactor to the collector 5. Qla,

Each of the streams and the reactors may contain species f where f = 1,2, .. ., F; the total number of species is F. The concentrations of the species at the exit of the reactors, i.e., C'G, C'fi (i = a, ..., z) and C'fs are not, in general, equal to the concentrations inside the reactors, CG, Cfi (i = a, ..., z) and Cfs. If the

499 concentrations C'fj, C'fi (i = a, ..., z) are equal to zero, such a flow configuration simulates, for example, a concentration process. If the concentration of the species at the exit of reactor 5 equals zero, Cfs = 0, the species are completely accumulated in reactor 5 which is considered as "total collector" or "dead state" for the species, If C'g= Cg, the species are not accumulated in reactor 6. If 0 c C'gc Cg,the species are partially accumulated in reactor 5 which is considered as a "partial collector" of the species. The basic element in the flow system is the perfectly-mixed reactor. In the multi-reactor system heat and mass transfer operation (absorption, desorption, dissolution of solids, heat generation or absorption as well as heat interaction between the reactor and the surroundings etc.) as well as chemical reactions may occur simultaneously, or not. The processes are governed by Eqs.(5-8), (5-12), (5-16), (5-19), (5-23) and (5-25) in the following, on the basis of which transition probabilities are derived as well as the single step transition matrix. A specific configuration is determined by appropriate selection of the interacting flows and the number of reactors as well as the operating conditions, When the central reactor j is considered, usually, j = 1 and a = 2, b = 3, etc. As indicated, the collector is designated by 5. When the central reactor is not considered, usually, a = 1, b = 2, etc. Each reactor designates a state, thus, the total number of reactors is the number of states.

500

a) Overall scheme

b) Collector

Fig.5-1. Scheme of a continuous flow system

50 1 Definitions. A,

fk designates a system with respect to its chemical formula f

and location (reactor) k in the flow system. The system is a fluid element containing F chemical species for which f = 1, 2, 3, ..., F where each figure designates a certain species with its chemical formula. In addition, k = a, b, ...,Z, where each letter designates a reactor in the flow system composed of Z + 1 perfectly mixed reactors, including reactor 6. The states of the system are the following concentrations and enthalpies (or temperatures) of the different species inside the various reactors: a) The concentrations Cf,in,jof species f at the peripheral feed vessels to reactor j in Fig.51 as well as the specific enthalpy hinj (or Tinj) of the fluid in the feed vessel to j; b) The concentrations Cf,in,k of species f at the feed vessels to reactors k and the specific enthalpy hin,k (or Tin,k) of the fluid in the feed vessels to k; c) The concentrations C, of species fin reactors k as well as the specific enthalpies hk, (or Tk, Tg ), respectively, of the fluid in reactors k and in collector 5. Thus, the state space SS for the system will read:

where the total number of states, only if all above elements do exist, is (F+1)(2Z+3). Z is the number of perfectly-mixed reactors excluding the collector 6 and F is the total number of species. A compact form of the above equation for f = 1 , 2 , 3 , ..., F reads:

502

(5-la)

If the system occupies a state, it means that the concentrations of species f i n all reactors are prescribed as well as the enthalpies of the fluid in the reactors. Finally, the movement of a fluid element from a state to another is the transition between the states. The initial state vector is given by Eq.(2-22) ,i.e.,

For the case under consideration it reads:

The above equation is expressed compactly for f = 1,2,3, ..., F by:

503 The state vector at step n, or time t is, similarly, given by:

A compact representation for f = 1,2,3, ..., F is given by:

The transition from step n to n+l is carried out according to: S(n+l) = S(n)P + L(n)

(5-4)

where P is the one-step transition probability matrix given by Eq.(5-27). S(n) and S(n+l) are state vectors at times t and t+At, or step n to n+l, given by Eq.(5-3). L(n) is an expression corresponding to time t or step n which is spelled out in the appropriate equations below. Specific elements of the state vector, corresponding to Fig.5-1 are derived in the following and are given by Eqs.(5-9a), (5-13a), (517a), (5-21a), (5-24a) and (5-26a).

504 Scheme of the chemical reactions. In the derivations below, chemical reactions are involved. To remain consistent with the nomenclature in chapter 3.1

and the modified nomenclature in this chapter, we repeat part of the material as follows. Consider a chemically reacting system containing species A1, A2, A3, ... ., AF. It is assumed that a certain species Af can react simultaneously in several reactions, designated in the following by superscripts (m), m = 1, 2, ..., R where R is the total number of reactions in which Af is involved. The following scheme of irreversible reactions by which Af is converted to products is assumed, where each set of reactions involving reversible reactions can always be written according to the scheme below in order to apply the following derivations. (1)

‘f

... + af(1 1Af + ... + products

1st reaction:

r(2) f

... + af(2)A,

2nd reaction:

+ ... +

products

0 ‘f

@’A~ + ... + @)A,

mth mction:

+ ... + ... + %(m)% + ... 02) ‘f

... + af(R)Af + ... -.,

Rth d o n :

products

@’ is the stoichiometriccoefficients of species f in the rnth reaction.

(5-5)

The rate of

conversion of species f in the rnth reaction based on volume of the fluid in the reactor V, i.e.

p),is defined by:

where Nf are the number of moles of species f and Cf is its concentration in moles

of flm3. kp), in consistent units, is the reaction rate constant with respect to the

conversion of species f in reaction m. In the case of a plus sign before kp), this

means (moles of f formed)/(s m3) in reaction m. The discrete form of Eq.(5-9) reads: 0

$“(n)

= - kp)C>

0

0

613

(n)C> (n). .. C) (n). .. C$ (n)

505

where the reaction rate and the concentrations CAn) refer to step n. In addition, the conservation of the molar rates for all reacting species in reaction m in Eq.(5-5) is given by:

This makes it possible to compute the reaction rates of all species on the basis of given by Eq.(5-6).

Derivation of probabilities from mass balances Reactor j. A mass balance on species f undergoing various mass transfer processes as well as chemical reactions due to changes in the operating conditions in the central reactor j in Fig.5-1, gives:

+ x k fPJ.aPJ. A C f p , j + x $ % j

k = a, b,

..., Z

(5-8)

m

P

where

k#j,

5

f = 1, 2, ..., F

p = 1, 2, ..., P. The last

summation with respect to m, is on all reactions in Eq.(5-6 ) that species f is involved with, i.e. m = 1, ..., R. Cf,i,,j is the concentration of species fat the inlet stream Qj flowing from a feed vessel into reactor j; C e and CQare, respectively, the concentrations of species fin reactors k and j. C'fj is the concentration off in the stream Q j leaving reactor

j.

kfp,japjACfp,j= pfp,jACfp,jVjis the mass transfer rate of species f into (or from)

reactor j by some transfer mechanism designated by p (such as absorption, dissolution, etc., or simultaneously by several processes). kfp,j(ds) is the mass transfer coefficient for process p with respect to species f corresponding to conditions in reactor j; apjis the mass transfer area for process p corresponding to conditions prevailing in reactor j. If the mass transfer area apj is not known, the

506

volumetric mass transfer coefficient p .(l/s) defined in Eq.(5-10) is used. ACfp,j fP J

is the driving force for the transfer process p with respect to species fat conditions prevailing in reactor j. It should be noted that a positive sign before kfpj indicates mass transfer into the reactor whereas a negative sign indicates mass transfer from the reactor outside.

im) is the reaction rate of species f by reaction m per unit fj

volume of reactor (or fluid in reactor) j corresponding to the conditions in this reactor. A plus sign before kp' in Eq.(5-7) means (moles of f formed in reaction m)/(s m3). Integration of Eq.(5-8) between the times t and t+At, or step n to n+l, assuming that Vj remains constant, yields:

k = a , b ,..., Z w h e r e k # j , c

f = 1 , 2,..., F

p = l , 2 ,..., P

m =1,

..., R

An alternative form of Eq.(5-9) in terms of transition probabilities reads

(5-9a) where detailed expressions for the probabilities are summarized in Eq.(5-28a). Cg(n+l) is the concentration at time t+At, or step n+l, of species f in reactor j; Cf,inj is the concentration of species f at the inlet stream Qj flowing from a feed vessel into reactor j;Co(n) and Cfi(n) are, respectively, the concentrations at time t or step n of species f i n reactors j and k. C'fj(n)is the concentration of f in the stream Q j leaving reactor j. ACfpj(n)and r'"'(n) are, respectively, at time t or step fj

n, the driving force for the transfer process p and the reaction rate per unit volume of species f by reaction m, corresponding to the conditions in reactor j. Pinj is the single step transition probability from the state of the feed reactor to the state of reactor j; p i is the probability of remaining in the state of reactor j and pkj is the transition probability from the state of reactor k to the state of reactor j. The complete expressions for the probabilities in Eq.(5-9a) are given, respectively, in

507 Eq.(5-9). Lfj(n) is the last term on the RHS of the equation corresponding to species f (= 1,2,3, ...,F) in reactor j and at time t. Other definitions are: (5-10) where pfpj, the volumetric mass transfer coefficient (Us), indicates that this quantity corresponds to species f for process p in reactor j. In addition, (5-1 1) Reactors i. A mass balance on species f undergoing various mass transfer processes as well as chemical reactions due to changes in the operating conditions in reactor i in Fig.5-1, gives:

(5-12) P

m

i , k = a , b ,..., Z where k # i , j , t f = 1 , 2,..., F p = l , 2 ,..., P m = l , ..., R. Cf,in,iis the concentration of species f in the stream Qi flowing from the feed vessel into reactor i, Cfj is the concentration of species f in reactor j, Cfi is the concentration of species f in reactor k and Cfi is the concentration of species f in reactor i. C'fi is the concentration of species f in stream Q'i leaving reactor i. kfp,iapiACfp,i= pfp,iACf p J.v. is the mass transfer rate of species f into (or from) J reactor i by some transfer mechanism designated by p (such as absorption, dissolution, etc., or simultaneously by several processes). kfp,i(m/s) is the mass transfer coefficient with respect to species f for the process p corresponding to conditions in reactor i; api is the mass transfer area for process p corresponding to conditions in reactor i. If the mass transfer area api is not known, the volumetric mass transfer coefficient p ,(Us) defined in Eq.(5-14) is used. ACfp,iis the fPJ

driving force with respect to species f for the transfer process p at conditions of reactor i. It should be noted that a positive sign before kg,i indicates mass transfer

508

into the reactor whereas a negative sign indicates mass transfer from the reactor outside. $'I is the reaction rate of species f by reaction m per unit volume of reactor i corresponding to conditions in this reactor. In the case of a plus sign before kp)in Eq.(5-6), this means (moles off formed in reaction m)/(s m3).

Integration of Eq.(5-12) between the times t and t+At, or step n to n+l, assuming that Vj remains constant, yields:

i, k = a, by ..., Z where k f i, j, 5 f = 1, 2, ..., F p = 1, 2, ..., P m = 1, ..., R. An alternative form of 1245-13)

in terms of transition probabilities reads

Cfi(n+l) = ~ ~ , + ~ cfl(n)pji ~ , ~+ C,(n)pii p ~ ~ +CC,(n)pki , ~ + Ln(n>

(5-13a)

k

where detailed expressions are summarized in Eq.(5-29a). Pin,i is the single step transition probability from the state of the feed reactor to the state of reactor i; pji is the transition probability from the state of reactor j to the state of reactor i; pii is the probability of remaining in the state of reactor i and p h is the transition probability from the state of reactor k to the state of reactor i . The complete expressions for the probabilities in Eq.(5-13a) are given, respectively, in Eq.(5-13) where Lfi(n) is the last term on the RHS of Eq.(5-13) corresponding to species f (= 1, 2, 3, ...,F) in reactors i (= a, by..., Z) and at time t or step n. Other definitions appear after Eq.(5-12) whereas the (n) or (n+l) in Eq.(5-13a) stands for t and t+At, or step n to n+l. In addition, the following definitions are applicable: (5-14)

509

pfp,i, the volumetric mass transfer coefficient ( U s ) , indicates that this quantity corresponds to species f for process p (for example: absorption, p = 1; desorption, p = 2; dissolution, p = 3; etc.) in reactors i or j. In addition, Qij a..= 'J

Qj

Qik

Qji =

Uik = Qj

Qj

Qki Uki

=

pi,

=

915 7

(5-15)

Reactor 5. A mass balance on species f i n reactor 5 in Fig.5-1 for k = a, b,

..., Z where k f j, 5 reads:

- (qjk + ~ q k ~ ) C ) i v cdC9 d t -- qjscfj +cqkccfk k

(5-16)

It is assumed that the volume of the fluid in the reactor remains unchanged and that no chemical reactions take place in the reactor as well as other mass transfer processes. Cfg is the concentration of species f in reactor 6 , C'fg is the concentration of species leaving reactor 6 (not necessarily equal to Cfs), Cfi is the concentration of species f in reactor k and CQis the concentration of species f i n reactor j. If the concentration C'fs = 0, the species are completely accumulated in reactor 5 which is considered as "total collector" or "dead state" for the species. If C ' g = Cfs, the species are not accumulated in reactor 5. If 0 < C'fg < C g , the species are partially accumulated in reactor 5 which is considered as a "partial collector" of the species. Integration of Eq.(5-16) between the times t and t+At, or step n to n+l ,yields:

An alternative form of Eq.(5-17) in terms of transition probabilities reads:

(5-17a)

5 10

Detailed expressions for the probabilities are summarized in Eq.(5-28a). pjc is the transition probability from the state of reactor j to the state of reactor 6 and pk{ is the transition probability from the state of reactor k to the state of reactor 6. The complete expressions for the probabilities in Eq.(5-9a) are given, respectively, in Eq.(5-9). In addition, the following definition are applicable:

(5-18) Finally, it should be noted that the determination of the parameters of the type aij is carried out by the following mass balances which extend Eqs.(4-12a) to (4-

12c). An overall mass balance on the flow system in F i g 5 1, i.e. on reactors j and

k (k = a, b, ..., Z where k f j) as well as 6, gives:

(5- 18a) ak,

aAt Pin,l, pi1 above and pi2 below are given by by Eq.(5.2-l(ld)). In addition, for reactor 2:

where p22 = 1 - p2At is obtained from Eq.(5-30a) noting that C'Q = C a . In the numerical solution it was assumed that the reactors are of an identical volume, i.e. pi = Q1/V1 = p2 = Q1/V2 = p. The initial state vector C(0) = [Cf,in,l(O), Cfi(O), Ca(O)] reads: [l, 0, 01 for f = 1, [0.2, 0, 01 for f = 2, 6 as well as [0, 0, 01 for f = 3,4, 5, 7. Additional parameters were: kl = 1, kl = 104, k2 = 40, k 2 = 0, k3 = 100, k-3 = 20, = 1000, k-4 = 100, k5 = 1, k5 = 0, = 1 and At = 0.015. The transient response of C11, C21, C31 and C71, i.e. the concentrations in reactor 1 where reactions take place, is depicted in Fig.5.2-l(3). The effect of p = 0,00001, 0.0001,0.0005 as well as g = 0 and 0.5 is demonstrated.

532 p = 0.00001, g = 0

0.0005

fl = 11

0.0003

c

-0m . 1-

0

0.004

/

4.0002

0.000lL

p = 0.0002, g = 0

/

10

20

30

40

50

-0.001

0

1

I

I

I

10

20

30

40

t

50

t

p = 0.0005,g = 0

0.025

I

p = 0.0005,g = 0.5 I

I

0.02

0.015

I

(d)

.......

,

-0.005 0

10

20

30

t

40

50

-0.005-

0

10

30

20

40

50

t

Fig.5.2-l(3). C11, C21, C31 and C71 versus t demonstrating the effect of p and g

5.2-1(4) This is an extension of case 3.13-3 for an open system comprising of two reactors; in the first one chemical reaction takes place. The flow scheme is shown in Fig.5.2-1(1). The configuration in Fig.5-1 is reduced to the present one by choosing reactors j and 6 designated as j = 1 and 5 = 2, respectively. Considering Eq.(5-18a) for a'k = 0 = 0, yields that a1 = Q1/Q1= pi2 = q12/Q1 = 1 while taking Q1 as reference flow. Eqs.(5.2-l(lb)) to (5.2-1(1h)) in case 5.2-l(1) are applicable also in the present case where f = 1, ..., 7. The following model [65],appearing in case 3.13-3, was applied also for the open system:

533 k2

kl

i=1:A1+Az4A3 c

i = 2 A 3 -.A2

k3

i=3:2A2-+A4

k-1

rl = r(ll)= - k,ClC2 + k-lC,

r3

- rt)+

r4 = I-(:) r5 = I-(:)=

r(32)+ r$)

= k,ClC2

- k-1C3 - $C3 - k4C,C4

+ r t ) + rt) = &Ci - k4C3C4- k5C4 k5C4

Considering Eqs.(5.2-l(le)) to (5.2-1(1g)) yields for f = 1, ..., 5 the following equations for reactor 1 where the different rf(n)'s are given above:

Pin.1, pi1 above and pi2 below are given by by Eq.(5.2-l(ld)). In addition, for reactor 2

where p22 = 1 - p2At is obtained from Eq.(5-3Oa) assuming that C'n= Ca. In the numerical solution it was assumed that the reactors are of an identical volume, i.e. p i = Q l N l = p2 = QlN2 = p. The initial state vector C(0) = [Cfjn,l(0), Cfl(O),

534

Cfi(O)]reads: [20, 1, 11 for f = 1, 2 and [0, 0, 01 for f = 3, 4 and 5. Additional

parameters were: kl = 2.5, k-1 = 0.1, k2 = 1, k3 = 0.03, k4 = 1, k5 = 1 and At = 0.005. The transient response of C11,C21, C31 and C41 in reactor 1 shown in

2 1.5

p = 0.01

IJ.=O

1

I

I

I

I

2

I

...........

n c

I

0-

c

I

I

2

0

14 12-

I

I

4 I

I

6 I

8

-

10

-

/.,..-

I

I

0 14

I

10-

8u-6 420-2.

!

2 1,. ........

-

f l =2! ...............

12-

10-

86-

4

6

8

1

0

, --------I

I

I

I

31

1 I 1 ................. ?.!. ..............:. .. -

*

4-31- - - - 2- I: 011 I -21 I -

2

-

11

-

I

I

I

I

5.2-1 ( 5 ) This is an extension of case 5.2-l(1) for an open system comprising of three reactors; in the first two ones chemical reaction takes place, and the third reactor is a "total collector" of the reactants and products. If this reactor is not a total collector, QE.(5.2-l(a)) is applicable, i.e. in the following matrix p33 = 1 - p3 At. Note that in previous cases, 5.2-l(1) to 5.2-1(4), chemical reaction took place only in one reactor. The flow scheme is shown in Fig.5.2-1(5),which is slightly different than the scheme in case 4.3-4.

535 Ql

~

cb q13 1 -

Ql+

3

-

Q,

The general configuration in Fig.5-1 is reduced to the present one by choosing reactors j, a and 6 designated as j = 1, a = 2 and 5 = 3, respectively. Considering Eqs.(5-18a) to (5-18c) for a ' k = 0, yields that p i 3 = 1 + a2, 1 + a 2 1 = a 1 2 + p i 3 and a2 + a12 = a21, where a2 = 42/41, pi3 = qi3/Qi, a12 = 412141 and a21 = 421141. A numerical solution was obtained in the following for a2 = 1; thus p i 3 = 2 and a 2 1 = a12 + 1. From Eq.(5-la) the state space in the present case reads:

noting that no heat and mass transfer take place, i.e. pfpj = phj = 0 which are given by Eqs.(5-10) and (5-22). From Eq.(5-3a) the state vector reads: (5.2- 1(5b))

0

0

1

0

Pin,l

0

1

0

0

0

P11

P12

PI3

0

0

P2 1

P22

0

0

0

0

0

1

~in,2

0

5.2-1(5~))

From Eqs.(5-28a) for j = 1, (5-29a) for a = 2 and (5-30a) for follows, noting that pi3 = 2 and a 2 1 = a12 + 1, for reactor 1 that:

6 = 3, it

536 for reactor 2:

for reactor 3: (5.2- 1( 1d))

P33 = 1 In addition, for all species, i.e. f = 1, ..., 4:

for reactor 3: Cn(n+l) = Cfl(n)pl3 + CD(n)

(5.2- 1(le))

where

Subscripts fl, f2, f3 designate species fin reactors 1, 2, 3, respectively. The following reactions, as in cases 5.2-l(1) and 3.13-4, known as the Brusselator model [60],are assumed: kl

i = 1: A1 + A3

k2

i = 2: 2A3 + A2 +, 3A3

k3

i = 3: A3 + A2

k4

i=4:A3-,A4 thus, m = 1, ..., 4 in the equations for Lfl(n) and Lf2(n) above. Considering the derivations in case 5.2-1( l), the following equations are obtained which enables one to compute Lfl(n) and Lfz(n):

537 f = 1:

for reactor 2: r12(n)= - klC12(n)

for reactor 1: rll(n) = - k,C,,(n) f=2:

for reactor 1: r21(n)= - k2C2,(n)C;,(n) + k3C31(n) for reactor 2: r2,(n) = - k2C,,(n)Ci2(n) + k3C3,(n> f=3: for reactor 1: r31(n) = klCll(n) + k,C,,C;,(n)

- (k3+ k4)C31(n)

for reactor 2: r32(n) = k,C12(n)+ k2C2,Ci,(n) - (k3+ k4)c32(n)

f=4: for reactor 1: r41(n) = k4C31(n)

for reactor 2: r42(n) = k4C3,(n)

Thus, the above equations as well as Eqs.(5.2-l(le)) to (5.2-1(1g)), makes it possible to calculate the concentration distributions in reactors 1,2 and 3 of species f = 1, ..., 4. In the numerical solution it was assumed that the reactors are of an identical volume, thus, p1 = Q l N l = p2 = QlN2 = p3 = Q1N3 = p. The initial state vector C(0) = [Cf,in,l(O),Cf,in,2(0),Cfl(O), Cn(O), Cn(O)] reads: [loo, 10, 1, 1, 01 for f = 1 as well as [0, 0, 0, 0, 01 for f = 2, 3, 4. Other parameters were: kl = 10, k2 = 0.1, k3 = 2, = 1, a12 = 0, 1, 5, 10 and 50, p = 0, 0.02, 0.03 and 0.05 and At = 0.05. The transient response of C21 and C31 versus t, i.e. the concentrations in reactor 1 in Fig.5.2-1(1) and of C22,the concentrations in reactor 2, as well as the attractor C31 against C21, are depicted in Fig.5.2-1(5a) where the effect of 0112 is demonstrated for p = 0.02. In Fig.5.2-1(5b) the effect of p is shown for a 1 2 = 5.

538 al2=0

6

5

-4 rn

u3

2 1 100

150

200

0 0

2

4

6

t

8

10

12

I

I

14

c2 1

a 1 2= 1

a 1 2= 1

6

I

I

I

2

4

6

I

5

-4 um3 2 1 0

50

100 t

t

150

200

0

8 1 0 1 2 1 4 c2 1

L

21

Fig.5.2-1(5a). C21,C22 versus t and C31 versus C21 demonstrating the effect of a 12

539 0.7

0.6 0.5

0.2 0.1 0 -0.1 L

21

5

I

I

I

I

I

I

I

-

4-

-3u"2 -

-

-1

1

I

I

I

1-

0-1

I

I

I

I

I

I

I

540

5.2-1(61 IMPOSSIBLE PRODUCTS' BEHAVIOR IN

BELOUSOV-ZHABOTINSKI MODEL CREATING A HUMOROUS PATTERN This case terminates the examples of chapter 5.2-1 on reacting systems. We apply here the Belousov-Zhabotinski model [57], presented in case 5.2-1(3), for operating conditions which generate non-realistic results, on the one hand, but a humorous unexpected behavior, on the other, due to a certain presentation of the results. The equations are those appearing in case 5.2-l(3) for the following data. loq5,01 The initial state vector C(0) = [Cf,i,,l(O), Cfl(O), Ca(O)] reads: [5.0 for f = 1, [3.5 lO-5,0,0] for f = 2, [0, 0, 01 for f = 3,4, 6, 7 and for f = 5 it reads [1.24 10-4, 10-11,0] . Additional parameters were: kl = 0.084, kl = 1.0 1069 k2 = 4.0 108, k2 = 0, k3 = 2000, k 3 = 2.0 107, = 1.3 105, k4 = 2.4 107, k5 = 4.0 106, k5 = 4.010-11, = 1.0 106, g = 1, p = 0.03 and At = 0.01, These data are slightly similar to those appearing in [57]. The transient response of C21 and C31, i.e. the concentrations in reactor 1 where reactions take place, is depicted in Fig.5.2-1(6a) as well as the plot of C21 against C31 is demonstrated. As observed, the C21-t and C31-t behavior is unrealistic after 25 time units since negative concentrations are obtained and somehow chaotic behavior. However, the plot of C21 against C31 creates a combined eyes and nose which were complimented by the author to a face.

54 1 5 4

3

u

2

z

1

-1

-2 10.71 -5

I

I

I

I

I

I

I

0

5

10

15

20

25

30

35

t

Fig.5.2-1(6a). C21 and C31 versus t

-2 1 0 7

-1 l o 7

0

1

2

Fig.5.2-1(6b). C21 versus C3 1

3 lo7

542

5.2-2 ABSORPTION SYSTEMS In the following cases absorption processes with and without chemical reaction will be demonstrated.

5.2-2(1) The flow system shown below comprises of three reactors. In the first two reactors absorption of a single component takes place, whereas the third reactor is assumed as "total collector" for the absorbed gas. If this is not the case, Eq.(5.2l(c)) is applicable and in the following matrix p33 = 1- p3 At. The fluid in which the species are absorbed, enters the first reactor and leaves the third one at a rate Qi. I

q&M" Q12

1

2

Ql '1

q13

Fig.5.2-2( 1). The flow system The configuration in Fig.5-1 is reduced to that in Fig.5.2-2(1) by choosing reactors j, a and 6 designated as j = 1, a = 2 and 6 = 3, respectively. Considering Eqs.(5-18a) to (5-18c) for a'k = 0, yields that p i 3 + p23 = 1, a 2 1 + p23 = a 1 2 where ajk and pjk are given by Eq.5-11. From Eq.(5-la), the state space for species f = 1 reads: (5.2-2( 1a)) where from Eq.(5-3a) the state vector reads: (5.2-2(1b)) The probability matrix given by Eq.( 5-27) is reduced to:

543

Cl,in,1

P=

c11

c12

5.2-2( Ic))

c13

From Eqs.(5-28a) for j = 1, (5-29a) for a = 2 and (5-30a) for 5 = 3, noting that there is a single mass transfer process, i.e. absorption and hence p = 1, it follows that:

(5.2-2( Id))

(5.2where from Eq.(5-28a) Lll(n) = Pii,iAC11,1At

(5.2-2( If))

and from Eq.(5-14) k1,l

=

kl l , l % 1 Vl

ACll,l = CY, - Cll(n)

Cy is the equilibrium concentration of the species 1 absorbed on the surface of the liquid in reactor 1 corresponding to its partial pressure in the gas phase above the liquid. For reactor 2:

544 (5.2-2( lh)) and p11,2 =

k11,2a12 v2

AC,,,, = Cy,

- C12(n)

Cy, is the equilibrium concentration of species 1 absorbed on the surface of the liquid in reactor 2 corresponding to its partial pressure in the gas phase above the liquid. For reactor 3:

In the numerical solution it was assumed that the reactors are of an identical volume, thus, p1 = Q1/V1 = p2 = Q1/V2 = p3 = Q1/V3 = p, and that the third

reactor behaves as "total collector" for the reactants and products, namely, p33 = 1.The transient response of C11 , C12 and C13, i.e. the concentrations of species 1 in reactors 1, 2 and 3 is depicted in Fig.5.2-2(la) where the effect of p = 1, 10 (cases a, b in the figure) and the mass transfer coefficient for absorption (for which p = 1 in Eq.5-28a) of species lin reactor 2, i.e. p11,2 = 1, 50 (cases d, e) as well as Cy2 = 3.10-69 6.10-6 (cases c, d) is demonstrated. The initial state vector C(0)= [Cf,i,,l(O), Cf1(0), Cfz(O), Cn(O)] = [0, 0, 0, 01 for f = 1. Other parameters were: = 0, p11,1 = 50, Cy, = 3.10-6and At = 0.0001.

a 1 2 = 1, p i 3

545 u=

p= 1

3 lo6-

1

(a)

-

2.5 l o 6 -

5 107

--

i=? I

0.05 0.1

'0

10

0.2

0.15

0.25

t

0.3 0

0.05

0.2

0.15

0.1

0.25

t

0.3

1.5 /

1 0 5 1

/

3/

-

/

.-

u

L

:

I

/

i=l

/

0 0

0.05

I

1

I

1

0.1

0.15

0.2

0.25

.

0.3 0

0.05

0.1

0.15 t

t

1.5 l o 5

I

0.2 0.25

0.3

c*]*=3 l o 6 , p,r,2= 1 I

I

I

0.15

0.2

I

(e)

5 i =1 ........

0

0.05

0.1

t

0.25

0.3

Fig.5.2-2(la). C11, C12 and C13 versus t demonstrating the effect of CL11,2 and c;, CL9

A simpler case which can be treated by applying the above model is the absorption of one component, f = 1, into a single reactor, i.e. reactor 1 where reactor

546 3 is the "total collector". Under these conditions, a12 = 0 and p i 3 = 1 in Fig.5.2-

2( 1). In the numerical solution it was assumed that the reactors are of an identical volume, i.e. a constant p. Other parameters kept unchanged were: Cy = 3.10-6 and Cy2 = 0. The initial state vector C(0) = [Cl,in,l(O),Cll(O), C13(0)] = [0, 0, 01 and At = 0.0001. The transient response of C11 and C13, i.e. the concentrations of species 1 in reactors 1 and 3 is depicted in Fig.5.2-2(lb) where the effect of p = 0.1, 10 (cases b, c in the figure) and p11,1= 10,50 (cases a, c) is demonstrated. 106, 3 106

2.5

-

II I

b = 0.1, 1 I I I

I

I

11.1

= 10 I

I

I

I

I

I

I

(b) -

l i = 11

-

-

l o 6-

-

l o 6-

-

-

_ _ _ 1_3 . _ _ _ .___ ____._.....-. _ _ .. -..... 0

0.05 0.1 0.15 t

0.2 0.25 0.3

I

I

,.I

I

I

I

(4

-

I

I

I

-

547

5.2-2(2) The flow system is shown in Fig.5.2-1(1). It comprises of two reactors and only in the first one the chemical reaction given below takes place, i.e.: kl

i = l:Ai+A2+Ag (5.2-2(2a)) The carrying fluid of reactant 1 (f = 1) dissolved in it enters the first reactor and leaves the second one at rate Q1. Reactant 2 (f = 2) is absorbed in reactor 1 and reacts there with reactant 1 yielding product 3 (f = 3). The configuration in Fig.5-1 is reduced to that in Fig.5.2-1(1) by choosing reactors j and 5 designated as j = 1 and 5 = 2, respectively. Considering Eqs.(5-18a) to (5-18c) for a'k = 0, yields that a1 = Q1/Q1= p12 = q12/Q1 = 1 while taking Q1 as reference flow. From Eq.(S-la) the state space reads:

From Eq.(5-3a) the state vector reads:

The probability matrix given by Eq.( 5-27) is reduced, for f = 1, ..., 3, to:

(5.2-2(2d)) From Eqs.(5-28a), for j = 1, and (5-30a) for 5 = 2, noting that p12 = 1 and C'a = C n , it follows that: Pin,l= PI At PI 1 = 1 - PI At ~ 1 =2~ 2 A t p22 = 1 - ~ 2 A t Considering Eq.(5.2-2(2a)), yields the following relationship:

(5.2-2(2e))

548 where kl1 indicates the rate constant for reaction i = 1 in reactor 1. However, since the chemical reaction takes place only in reactor 1, kl should replace kl1. For reactor 1, it follows from Eq.(5-28a) that:

where Lll(n) = rll(n)At = rl(n)At = - k,C,,(n~C,,(n)At

L31(n)= rjl(n)At = r3(n)At = k,C,,(n)C2,(n)At For reactor 2: Cfz(n+l) = Cfl(n)pl2 + Cfz(n)p22

f

= 1, ..., 3

(5.2-2(2h))

In the numerical solution it was assumed that the reactors are of an identical volume, thus, p1 = Q l N l = p2 = Q1N2 = p. The transient response of C11, C21 and C31, i.e. the concentrations in reactor 1 of species 1, 2 and 3 in Fig.5.2-2(1), is depicted in Fig.5.2-2(2). The effect of p is demonstrated in cases a, b as well as in cases d, e, and f; the effect of the mass transfer coefficient for absorption, i.e. p21,1(for which p = 1 in Eq.5-28a), is depicted in cases by c. The initial state vector C(0) = [Cf,in,l(O),Cfl(O), Cn(O)] was [0, 0, 01 for f = 2, 3 in all cases and [3.10-5, 0, 01 as well as [0, 3.10-5, 01 for f = 1 when exploring the effect of p in cases a, b and d, e, f. Other common parameters were: kl = lo6,C;, = 3.10-5 and At = lom4.

549

I

I

I

2.5 10-5 - ....................

I

2 105 1

j

uY.5 10-5

3,0

0

/

1 105L

/

-

3

2.5 E

I

J

!

-

- - - -3 - - - -

-

21.1

'

(b)

-

p = 100, 1 = 1000, c

105

I

2.. .............T

--,.....................

I

I

-

/

I

I

-

I

f= 1

I

I

~

/

/

I

(a)

I

2................

I.in.1

I

l

=3

'

I

I

I

Cl(0)= 0

(C) ...... ...... ...... !..2..! ...... ...... f1 -- 11

I

2

u1.5

1

5 O

0

X

0.05 0.1 0.15 0.2 0.25 0.3

=o, cl(o)= 3 10" P = 10, 121,l = 100,c l,in,l

3.5 3 2.5 i:

I

I

I

I

2

u1.5 1 5

.....................

0

0.01 0.02 0.03 0.04 0.05 P = 0, P ~ ~ 100, , ~ =c

3.5 1 o - ~ .,-5 3 2.5 1 o - ~ c2 1.5 1 5 10-6 01 0

yyl

I

~ , 0,~cl(o) ~ , =~3=10.~

I

...

-

(01

I

--

2 ........ '2........ .-. ..... 0

.-.

/

"

;, /

I

0.05

0.1 1

I

0.15

0.2

Fig.5.2-2(2). C11,C12 and C13 versus t the effect of p and p21,1

550

5.2-2(3)

The flow system is shown in Fig.5.2-1(1). It comprises of two reactors and the chemical reaction given below takes place only in the first reactor. The following reactions were considered in case 3.13- 1 for a closed system: kl

i = 1: A1 +A2

k2

i = 2: A2 +A3

(5.2-2(3a))

The carrying fluid, entering reactor 1 at rate Q1, may or may not contain the reactants or the products. However, in reactor 1 absorption of reactant 1 (f = l), i.e. Al, takes place with the formation of products 2 and 3 (f = 2 and 3). The configuration in Fig.5-1 is reduced to that in Fig.5.2-1(1) by choosing reactors j and 6 designated as j = 1 and 6 = 2, respectively. Considering Eq.(5-18a) for a ' k = 0, yields that a1 = Q1/Q1= pi2 = q12/Q1 = 1 while taking Q1 as reference flow. Thus, Eqs.(5.2-2(2b)) to (5.2-2(2e)) of the previous case are also applicable here and with respect to Eq.(5.2-2(3a)) the following are the rate equations [55]: rl = - klC1C2

r2 = klC1C2 - k2C2

r3 = k2C2

(5.2-2(3b))

For reactor 1, it follows from Eq.(5-28a) that:

where

L,,(n) = r3,(n)At = r3(n)At= k,C,,(n)At For reactor 2: f = 1, ..., 3

(5.2-2(3d))

55 1 where Pin,l, pi1 and pi2 are given in Eq.(5.2-2(2e)); p22 = 1- p2 At. In the numerical solution it was assumed that the reactors are of an identical volume, thus, pi = Qi/Vi = p2 = Q1/V2 = p. The transient response of C11 and C21,i.e. the concentrations of species 1 and 2 in reactor 1 in Fig.Fig.5.2-1(1), is depicted in Fig.5.2-2(3) where the effect of p (cases a, b and c) and k2 (cases d, e and f) is

demonstrated. In case g the operating conditions are identical to case e where C21 is plotted versus C11. The initial state vector C(0) = [Cf,jn,l(O), Cfl(O),Ca(O)] reads: [0,0,01 for f = 1 , 3 and [0,3.10-*0,0]for f = 2. Other common parameters were: kl = 106, At = 5.10-4, Cy,= 1 and p l l , l = 10-3. As seen, the oscillatory behavior of C11 depends on p and k2.

552 p = 10, k = 50 2

-

-

...-. . . 2 ................

0

0.4

0.2

t

0.6

0

I

0.8

0.4

0.2

0.0005

I

p = 1, k = 1

0.0004-

0.0007

0.0006 -

0.0005 -

'

2 / . . ......

........... 0.00035 (d) - 0.0003-

.!

.'

....2.......

0.0004-

.,

03 u

.

.

.: -

I

I

6 10.' -

; I

I

.

, .:

I

0 ,'

A

-

U20.0003 -

0.0002 - ,:. 0.0001-/

1

u = 1.' k 2= 150

'

2

N

:I

0.8

p = 1, k = 150

-

I

0.6

i

2 10-61

0 - . 2 ................................. 0 0.2 0.4 0.6 0.8

t

- 2 10.~ -

-

0

A I

0.0001

0.0002

c 0.0003 11

Fig.5.2-2(3). C11 and C21 versus t for the effect of p and k,

1

553

5.2-2(4) The flow system shown in Fig.5.2-1(1) comprises of two reactors. The following reactions known as the Brusselator model [60], taking place in the first reactor, were also considered in case 3.13-4 for a closed system. In reactor 1 absorption of reactant 1 (f = l), i.e. Al, takes place with formation of products 2, 3 and 4 (f = 2 , 3 and 4). kl

i=l:Al+A3

k3

k2

i = 2 : 2 A 3 + A 2 +3A3

i=3:A3+A2

k4

i=4:A3+A4 (5.2-2(4a))

The following are the rate equations: rl = - klCl

r2 = - k2C2C: + k3C3

r3 = klCl - 2k2C2Ci+ 3k2C2Ci - (k3 + k4)C3

r4 = k4C3

(5.2-2(4b))

For reactor 1, it follows from Eq.(5-28a) that:

L41(n) = r4,(n>At = r4(n)At= k4C3,(n)At For reactor 2: Ca(n+l) = Cfi(n)pi2 + Cn(n)p22

f

= 1, ...,

(5.2-2(41

where Pin,l, pi1 and pi2 are given in Eq.(5.2-2(2e)); p22 = 1- p2At. In the numerical solution it was assumed that the reactors are of an identical volume, thus, PI = QiN1= p2 = Q1N2= p. The transient response of C11, C21 and C31, i.e.

554 the concentrations of species 1 , 2 and 3 in reactor 1 in Fig.5.2-1(1), is depicted in Figs.5.2-2(4a), 5.2-2(4b), 5.2-2(4c). In addition the variation of C31 versus C21 is depicted. The data applied were: the initial state vector C(0) = [Cf,i,,l(O), = 30,p = 0.06, kl = 1, k2 = 1, Cf1(0), Cf2(0)] = [0, 0, 01 for f = 1, ..., 4, pll,l k 3 = 2 , k q = l , c ; , = 1 andAt=0.05. The effect of p = 0,0.06, 0.1 and 0.2 is demonstrated in Fig.5.2-2(4a). The effect of p1 = 30, 5 and 0.5 is demonstrated in Fig.5.2-2(4a), cases c and d for which p l l , l = 30, as well as in Fig95.2-2(4b),cases a to d. The effect of c;,= 1.5, 1 and 0.15 is demonstrated in Fig.5.2-2(4a), cases c and d for which C;,= 1, 9

as well as in Fig.5.2-2(4b), cases e to f. Finally, the effect of the reaction rate constant k2 = 10, 1,0.7, 0.5 and 0.3 is demonstrated in Fig.5.2-2(4a), cases c and d for which k2 = 1, as well as in Fig.5.2-2(4c). As observed, the oscillatory behavior of the concentrations depends on the above operating parameters.

555 p=O

2

I

I

I

I

r\

I

(b)

1.5 (c.

" 1

0.5

0

20

10

t

30

50

40

-0

0.5

2

1.5

1

cz

2.5

3

2.5

3

1

1.2 1

-0.8 um0.6

0.4

0.2

O

0

0.8

t r;-u=2t i 2s

OQ 0

2

0

i

0.5

2

1.5

1

cz p = 0.1

1

I

I

I

0.6 -

L

I

I

10

20

I

t

I

30

40

I

50

0

0

1

0.5

2

1.5

1

2.5

3

2.5

3

c2 1

p = 0.2 I

I

I

I

0.5

1

1.5

2

t

i

0.5 I

2.5

(g)

3

0

0

I

I

0.5

1

p = 0.2

Fig.5.2-2(4a). The effect of p

I

1.5

1

2 c2 1

556 p =5

3.5

3

I

I

'F-.;'

2.5

I

I

:

, 3, ,

,

0.5

0' 0

11.1

f= 1

1 (a)

1 o*o;l

11.1

=5

I

I

I

I

I

I

I

I

I

I

,

0.5

1

1.5

2

2.5

3

I

0.02

.

20

40

t

60

80

100

0

C*],= 1.5

I

(b)

4

3.5

2

I

I

1.5

-

G1 0.5 - 0

0

0.5

1.5

1 1

I

I

I

I

(g)

1.5

1:/: -

-

U'0.5

or. '

c

f=l

3

11.

1.

0

2

2.5

557 1.6 1.4 1.2.-

k2= 10

I

1-

(4

I

-

i

i

0.8-

-

f= 1

k2= 10

i

-0.6rn

0

1

2

t

-

0

3

1

0.3

0.4

3

4

0.8

3 E

u

k-= 0.7

L2 1

k-= 0.7

4

-

0.2

0.1

-0.6 c,

2

0.4

0.2

- 0 0

1

...--I

(el

2

3L

4-

1

-

1.1

u=2

0 0

0.8

...__... I....__. 4 _.......

2

1

1

1

k2= 0.5

c2 I

1

1

1

--0.61 0.6-

c1 ci

u

-

I y

3 20

10

t

1 30

I

40

50

- 0 0

1

2

3

4

5

6

c2 1

k2= 0.3

." I

21.. 0,' 0

'

,

f=l

t

, ,3 100

150

O OO

2e

Fig.5.2-2(4c). The effect of k2

4

l c2 1

6

l8

558

5.2-2(51 IMPOSSIBLE REACTIONS CREATING AESTHETIC PATTERNS The flow system is shown in Fig.5.2-1(1) and comprises of two reactors. The following "reactions", only partially possible, are based on the Lorenz equations [85, p.6971. The reactions in case 3.14-3 occur in the present example only in the first reactor. In this reactor, j = 1, absorption of reactant 2 (f = 2), i.e. A2, takes place with the formation of "products" 1 and 3 (f = 1 and 3) according to the following Lorenz "reactions": 10

r l = 10C2 - lOC1

for which

-+ A1 A2 may be written t 10

r2 = 28C1 - C2 - C1C3

for which no reaction may be realized 813

r3 = C1C2 - (8/3)C3

for which

+ A3 t 1

A1 + A2 may be written

For reactor 1, it follows from Eq.(5-28a) that:

where

For reactor 2: Cf2(n+l) = Cfl(n)Pl2 + Cf200P22

f = 1, ..., 3

(5.2-2(5b))

where Pin,l, p i i , pi2 and p22 are given in Eqe(5.2-2(2e)). In the numerical solution it was assumed that the reactors are of an identical volume, thus, ~1 = Qi/Vi = p2 = Q1/V2 = p. The transient response of (211, C21 and C31, i.e. the

559 "concentrations" of species 1, 2 and 3 in reactor 1 in Fig.5.2-1(1), is depicted in Figs.5.2-2(5a), 5.2-2(5b), 5.2-2(5c). In addition the variation of C31 versus C21 and C31 versus C11 are depicted, which reveal very nice patterns from the artistic point of view. This is, by the way, a mean to generate art from scientific models, which has been demonstrated also in case 5.2-l(6) above. The data applied in the computations were: the initial state vector C(0) = [Cf,j,,l(O), Cfi(O), Cn(O)] = [0, 0.01, 01 for f = 1, 2, 3, p21,1= 0, Cll = 0 and At = 0.01. The effect of p = 0, 1, 2, 2.07, 3 and 8 is demonstrated in the following in Figs.5.2-2(5a), 5.2-2(5b), 5.2-2(5c). As seen, for p = 0 the "concentrations",

which are positive and negative, reveal chaotic behavior as expected from the Lorenz equations. By increasing p, i.e. reducing the mean residence time of the fluid in the reactor, the "concentrations" become positive all the time, and nonchaotic for p > 2.

5 60

Fig.5.2-2(5a). The effect of p = 0 and 1

56 1

Fig.5.2-2(5b). C11, C21 and C31 versus t, C31 versus C21 and C31 versus C11 demonstrating the effect of p = 2 and 2.07

5 62

Fig.5.2-2(5c). C11,C21 and C31 versus t, C31 versus C21 a n d C31 versus C11 demonstrating the effect of p = 3 and 8

563

5.2-3 COMBINED PROCESSES In the following, combined processes are presented which incorporate several chemical engineering operations acting simultaneously.

5.2-3(1]

-

Chemical reaction and heat transfer

The flow system shown below comprises two reactors. In the first one the chemical reaction Al+ A, takes place and simultaneously the heat of reaction is removed by a cooler. The carrying fluid in which the reacting species are dissolved, enters the first reactor and leaves the second one at rate Q1.

T0,l TO,l

Fig.5.2-3(1). The flow system The configuration in F i g 5 1 is reduced to that in Fig.5.2-3(1) by choosing reactors j and 6,designated as j = 1 and 6 = 2, respectively. From Eq.(5-18a) for a ' k = 0, it follows that a1 = Q1/Q1 = p12 = q12/Q1 = 1 while taking Q1 as reference flow, From Eq.(S-la) the state space reads: (5.2-3( 1a)) and from Eq.(5-3a) the state vector reads:

where S(n+l) is given by Eq.(5-4). The probability matrix given by Eq.(5-27) is reduced to:

564

(5.2-3( Ic)) Noting that the process in reactor 1 incorporates the chemical reaction Al+ A, for which rl = - klC1 as well as a heat transfer process, it follows from Eqs.(5-28a) and (5-30a) for j = 1 and 5 = 2 that:

The last probability indicates that the second reactor behaves as a "total collector" of the reactants and products. If C'n = C n , i.e. the reactants and products are not accumulated in the reactor, Eqs.(5-30a) gives p22 = 1 - p2At. From Eq.(5-28b) it follows for j = 1 that: (5.2-3( 1e)) From Eq.(5-30b) it follows for j = 1 and 6 = 2 that:

(5.2-3( If)) From Eqs.(5-28a) and (5-28b) it follows that:

Thus, from Eqs.(5-28a) and (5-30a) it follows for j = 1, 6 = 2 and f = 1 , 2 that for the 1st reactor: cfl(n+l) = Cf,in,lPin,l + Cfl(n)P11 + h ( n )

565 (5.2-3(1h))

In the case under consideration the following expressions are applicable for f = 1, 2: Lll(n) = rll(n)At = rl(n)At- - klCllAt L21(n) = - Lll(n)

where AHrl is negative for an exothermic reaction. It is also assumed that

yielding from Eqs.(5.2-3(le)) and (5.2-3(If)) that

If all reactors are of the same volume pi = p2 = p. Also for Ah = CphT it follows that

Substitution of the above expression into Eq.(5.2-3(lh)), considering Eq(5.23( lj) and the above probabilities, yields:

where AHrl is negative for an exothermic reaction. Similarly, Eq.(5.2-3(li)) yields:

566 T2(n+l) = Tl(n)pzAt + T2(n)(1 - p2At )

(5.2-3(1m)

Eqs.(5.2-3(lh)), (5.2-3(10) for Cfl(n+l) and CE(n+l) and (5.2-3(11), (5.23( lm) make it possible to calculate the concentration and temperature distributions versus time in reactors 1 and 2 for species f = 1,2. In the numerical solution it was assumed that the reactors are of an identical volume, thus, p1 = Q1/V1 = p2 =

Q1N2 = p. The initial state vector S(0) = [Cf,in,l(O),Cfl(O), Cn(O), hin,l, h1(0), h2(0)], in terms of temperatures, reads:

In the computations kl(l/min) = exp( 17.2-11600/[1.987T(K)]), AHrl = -18000 cal/gr-mole A, p = lgr/cc and C, = lcal/gr OC. Other common parameters were: Tin,l = Ti(0) = T2(0) = 25OC and At = 0.0005. The transient response of T1(OC) and T2(OC), i.e. the temperatures in reactors 1 and 2, as well as the concentrations C11, C21, C12 and C22 versus t, i.e. the concentrations of A1 and A2 in reactors 1 and 2 are demonstrated in Figs.5.2-3( la) and 5.2-3( lb). The effect of p(l/min) = 0, 10 for a heat transfer coefficient ph(l/min) = 100 is depicted in cases a to d; other common parameters are given in the figure. The effect p h = 10, 100 for p = 10 is given in cases c to f. The effect of Cll(0) = 0.2, 0.5, i.e the initial concentration of A1 in reactor 1, is shown in cases c, d, e, h. The effect of C1,in (0) = 0, 0.2, i.e. the inlet concentration of A1 into reactor 1, is shown in cases g to j.

567 30 1

I

If 26

I

I

I

I

c

i=l

"

0.1

f l = 11

I >

2 ...................... ......

240

0. p = 0, p = 100, C,JO) = 0.2, Cl,in=0

p = 0, pLh=100, CI1(O) = 0.2, c 1 . h= 0

0.2

t

0.3

I

21

" . ....................... . ~

f2 = 12,22

O t

0.4

0

0.1

0.2

I

I

I

t

0.3

0.4

I

0.5

I

-

c

-

(4 -

-

f2= 12

........................... 21, 22

4

2

45

0.25

0

I

I

0

0.5

t

I

I

'

(el

0.25-

0.1

I

I

0.2

I

0.3

I

. . . . . . .

-

0.4

I

0.5

(0

.f2.=.12 .

T

-

30 25

*-.

I

I

-

-

-

- - 22 - -

.-.__. I

I

-0.05

I

I

I

I

I

Fig.5.2-3(la). Ti(0C) and T2(OC), C11, C21, C12 and C22 versus t, demonstrating the effect of p and ph

568 36

p = 10, kh= 100, C1,(0)= 0.5, Cl,in=0 I

(€9

-

22

-0.1 0.25

0

I

0

0.5

t

21

0.2

0.4

t

- -

I

I

0.6

0.8

1

F = 10,P = 100, CI1(O) = 0.5, Cl,in=0.2

f2= 12,

1.5

'

c

u 0.5 241

1.

1

0

1

I

I

I

0.2

0.4

0.6

0.8

t

1

0

0.2

0.4

0.6

0.8

1

Fig.5.2-3(lb). Ti(0C) and T2(OC), C11, C21, C12 and C22 versus t, demonstrating the effect of Cl(0) and C1,in

5.2-3(21

- Cooling heat transfer

The flow system shown below comprises of three reactors. Cooling or heating of the entering fluid takes place in reactors 1 and 2.

-

Q21

-

I I Q12 * cJo a=2

Ql

Tim1

j=1

4 t

k t

*O,l

To,,

To,l

q25Q1

ob-Q,_

5=3

T0,2

Fig.5.2-3(2). The flow system

569

The configuration in Fig.5-1 is reduced to that in Fig.5.2-3(2) by choosing reactors j, a and 5, designated as j = 1, a = 2 and 6 = 3, respectively. From Eqs.(5-18a), (5-18b) for a'k = 0, it follows that p23 = 1, a12 = 1 + a21; a1 = Q1/Q1 taking Q1 as reference flow. From Eq.(S-la) the state space reads:

and from Eq.(5-3a) the state vector reads:

where S(n+l) is given by Eq.(5-4). The probability matrix given by Eq.(5-27) is reduced to:

P =

hl

h2

hin, 1

hin,l 1

Pin,l,h

0

h3 0

hl

0

Pll,h

P12,h

0

P21,h

P22,h

P23,h

0

0

P33,h

h2 h3

0

(5.2-3(2 ~ ) )

It follows from Eqs.(5-28b), (5-29b) and (5-30b) for j = 1, i = a = 2 and 6 =

3, assuming that the density of the fluid flowing in the system remains constant, that: Pin,l,h = p1 At Pl1,h = 1 - a12klAt P12,h = al2p2At

(5.2-3(2d))

(5.2-3(2e))

In addition:

570

The following substitution is made for replacing enthalpies by temperatures, i.e. Ah = CpA thus:

Substitution of the above expression into Eq.(5.2-3(2f)), assuming that Cp,l = Cp,2I C, and considering the above probabilities, yields:

T3(n+l) = Tdn)pgAt + Tj(n)[ 1 - p3At]

(5.2-3(28)

where a12 = 1 + a21. Eqs.(5.2-3(2i)) make it possible to calculate the temperature distributions versus time in reactors 1, 2 and 3. In the numerical solution it was assumed that pi = p2 = p3 = p. The initial state vector S(0)= [hin,l, h1(0), h2(0), h3(0)], in terms of temperatures, reads S(0) = [Tin,l, T1(0), T2(0), T3(0)] = [50OC, 25OC, 250C, 25OCl. Other common parameters were T1,o = T2,o = lOOC and At = 0.005. The transient response of TI(OC), T2(OC) and T3(OC), i.e. the temperatures in reactors 1, 2 and 3, is depicted in Fig.5.2-3(2a). The effect p = 0, 1, 10, 100 is demonstrated in cases a to d; the effect of a 1 2 = 0,25 in cases b and e and of phi= ph2 = 0,10 in cases e and h.

57 1 p = l , p = p =10,a = O

p=O,p = k =10,a = O

30 I

I

hl I

h2

2

t

h2

hl

12

I

I

3

4

I

I

I

12

I

I

3

4

(b

25 20 "v,

r+ 15 10 5

0

1

5

0

1

30

t

5

p = 100, p = p = 10, a = 0

50

32

2

hl I

h2

12

1

I

I

I

I

I

I

0

0.02

0.04

0.06

0.08

28

22 20 18 0

0.1

0.2

I

I

t

0.3

0.4

0.5

t

0.1

p = l , p = p =O,a = 2 5 I

I

hl

h2

12

(el -

-

0

I

I

I

I

1

2

3

4

t

5

2oo

;

I

4

t

I

I

6

8

10

Fig.5.2-3(2a). Tl(oC), T2(OC) and T3(OC) versus t, demonstrating the effect of p, phi = ph2 and a12

572

5.2-3(31 - Heat transfer in impinging streams

Impinging streams were thoroughly treated in chapter 4.5 when studying the RTD of such systems. The flow system below comprises of four reactors three of which are equipped with heat exchangers.

'14

T0,l T0,l

T0,2

T0,2

'0,3

T0,3

q3,

I

ob

5=4-

The configuration in Fig.5-1 is reduced to that in Fig.5.2-3(3) by choosing reactors j, a, b and 5, designated as j = 1, a = 2, b = 3 and 5 = 4, respectively. From Eqs.(518a) to (5-18c) for a'k = 0,it follows: 1 + a3 = p14 + p34; 1 + a21 = p i 4 + a 1 2 ; a 1 2 + a 3 2 = a 2 1 + a 2 3 ; a3 + a 2 3 = a 3 2 + p34. From symmetry considerations a 1 2 = a21 = a 2 3 = 0132 a ; thus, p i 4 = p34 = a3 = a1 = 1 while taking Q1 as reference flow.

From Eq.(S-la) the state space reads: (5.2-3(3a)) and from Eq.(5-3a) the state vector reads:

where S(n+l) is given by Eq.(5-4). The probability matrix given by Eq.(5-27) is reduced to:

573

(5.2-3(3~))

From Eqs.(528b), (529b) and (5-30b) forj = 1, i = a = 2, i = b = 3 and 5 =

4, assuming that the density of the fluid flowing in the system remains constant,

follows that: Pin,lb=

v

At

P I I J ~ =1 - (1 + a12)piAt

~ 2 i =b a2ipiAt

The following substitution is made for replacing enthalpies by temperatures, i.e. Ah = C$; thus, hi(n+l) = hin.1 + Cpfli(n) - Tin.1) i = 1, ..., 4

(5.2-3(3g))

574

Substitution of the above expression into Eq.(5.2-3(2f)), assuming that Cp,i = Cp, hin,l = hin,3 and applying the probabilities in Eq.(5.2-3(3d)) as well as that a 1 2 = a 2 1 = a 2 3 = a 3 2 a, yields:

T4(n+l) = Tl(n)pdt + T3(n)bAt + T4(n)(l - p4)At

(5.2-3(3h))

To obtain the last expression, one should take hin,l = CpTin,l stemming from Eq.(5.2-3(3g)). Eqs.(5.2-3(3h)) make it possible to calculate the temperature distributions versus time in reactors 1,2,3 and 4. In the numerical solution it was assumed that pi = p2 = p3 = p4 = p. The initial state vector S(0)= [hin,l, h1(0), h2(0), h3(O), h4(0)], in terms of temperatures, reads S(0) = [Tin,l, T1(0), T2(0), T3(O), T4(0)] = [50OC, 25OC, 25OC, 25OC, 25OCl. Other common parameters were: T1,o = T2,o = T3,o = 10°C and At = 0,0005. The transient response of Tl(oC), T2(OC), T3(OC) and T4(OC), i.e. the temperatures in reactors 1, 2, 3 and 4, is depicted in Fig.5.2-3(3a). The effect of p = 0, 10, 100 is demonstrated in cases b, c and d; the effect of a = 0, 10 in cases a and b and of phl = j . 4 ~= ph3 = 0, lo, 100 in cases b, e and f.

575

;

I

60 -

. ' .

- .

. (aL

-

-

-

-

-

50 340"v,

10 -

-

I

1, 3 .......................... 2

-

I

a = 10, p I

I

I

-

I

I

I

= 10, p = 100,

ph= 10

t

= 0,p = 10 h

a I

I

25

80

c

60

,

1,Z3

40

10

5

-

I

I

t

30

(U

-

-

...._.-...2...

I

4 , , _ _ _ - . - . * .

~.--.-'-..

20 L....

I

I

-

i=l,3

b30-'

I

0

0.1

0.2 t

100 -

I

0.3

I

0.4

0.02

.

,

1 . .

.

.

__

(e)

25

c:

'

.

_

0.08

0.1

.4.

I

t

0.06

0.04 t

I

i=4.

0.5 2o0

0

2

I

0.2

0.1 t

Fig.5.2-3(3a). Ti (OC), T2(OC),T3(OC) and T4(OC) versus t, demonstrating the effect of p, a and phl = ph2 = F h 3

0.3

576

5.2-3(4)

-

Concentration of solutions

The concentrator is a 4-stage system where evaporation of the solution takes place in reactors 1 , 2 and 3. The inlet concentration of the solution is Cl,in,l.

-

j=1

a=2

'

*

b=3

m

'

* 5=4-

The configuration in Fig.5-1 is reduced to that in Fig.5.2-3(4) by choosing reactors j, a, b and 6, designated as j = 1, a = 2, b = 3 and 6 = 4, respectively. From Eqs.(5-18a) to (5-18c), for a'k > 0,it follows: 1 = p34 + a ' l + a'2 + a ' 3 1 = a 1 2 + all a 1 2 = a 2 3 + a ' 2 a 2 3 = p34 + a ' 3 where a'i= Q'i/Q1 and taking Q1 as reference flow. It assumed that a'i = a'(i = 1, 2, 3), thus:

From Eq.(5-la) the state space reads:

From Eq.(5-3a) the state vector reads:

where S(n+l) is given by Eq.(5-4). The probability matrix given by Eq.(5-27) is reduced to:

577

Cl,in,l

1

Pin,l

0

0

0

c11

0

P11

P12

0

0

P = c12

0

0

P22

P23

0

c13

0

0

0

P33

P34

c14

0

0

0

0

P44

(5.2-3(4d))

From Eqs.(5-28a) to (5-30a), assuming C'14 = C14, i.e. no accumulation in reactor 4, noting that Lfi(n) = 0, i = 1,2,3, it follows: Cll(n+l) = Cl,in,lPin,l+ Cll(n)~11

C12(n+l) = Cll(n1~12+ C12(n)p22

Eqs.(5.2-3(4e)) and (5.2-3(40) make it possible to calculate the concentration distributions of the salt versus time in reactors 1 to 4. In the numerical solution it was assumed that pi = p2 = p3 = p4 = p. The initial state vector S(0)= [Cl,in,l, C11(0), C12(0), C13(O), C14(0)] = [l, 0, 0, 0, 01 in cases a to c in Fig.5.2-3(4a) and [0, 1, 0, 0, 01 in case d; At = 0.005. The transient response of Cli (i = 1,2, 3, 4),i.e. the salt concentration in reactors 1 to 4,is depicted in Fig.5.2-3(4a). The effect p = 1, 10 is demonstrated in cases a and b; the effect of a = 0.1,O in cases b and c. The effect of the initial concentration in reactor 1 is demonstrated in case d.

578 p = 1, a = 0.1, Cl,in,l= 1

Cli(0)= 0 (i = 1,2,3,4)

1.5

I

I

I

.- - I

-I ~ - - -J --. - - -

I

/. . 2 .............................

: ,

I

1

(b)

I

#'

,

,

,

I .

.

I

I

I

t l = l O , a = O , Cl,in,l = 1 Cli(0) = 0 (i = 1,2,3,4)

0

0.5

t

1

1.5

t

Fig.5.2-3(4a). Cli (i = 1, 2, 3, 4) versus t, demonstrating the effect of p and a

5.2-3(5)

- Electrolysis of solutions-model A

[86] The following scheme was suggested as a possible network model to describe a real electrolytic processes. Reactors 1 and 2 are continuous-flow stirredtank electrolytic reactors (CSTER), reactor 3 is a reactor for the recycling electrolyte and reactor 4 is collector in which no electrolytic process takes place.

579

a2

Ql

*

j=1 CSTER

924

*

a=2 CSTER

o$

Ql

5=4-

I 4

Q31

b = 3 .a

Q23

Fig.5.2-3(5). The electrolyser flow system The configuration in Fig.5-1 is reduced to that in Fig.5.2-3(5) by choosing reactors j, a, b and 5, designated as j = 1, a = 2, b = 3 and 5 = 4, respectively. From Eqs.(5-18a)-(5-18~),for a'k = 0 and taking Q1 as reference flow, it follows: p34 =

1, a 2 3 = a31 = a as well as that a 1 2 = 1 + a 2 3

(5.2-3(5a))

where a = Q23/Q1 is the recycle. From Eq.(S-la) the state space reads:

ss = [Cl,in,l, c11, c12,

7

(5.2-3(5b))

c13, CI4I

where Cl,in,l is the concentration at the inlet to reactor 1 of the species designated by 1 undergoing electrolysis in reactors 1 and 2. From Eq.(5-3a) the state vector reads: (5.2-3(5~))

S(n>= [Cl,in,l, cll(n>,C12(n), C13(n), cl4(n)I

where S(n+l) is given by Eq.(5-4). The probability matrix given by Eq.(5-27) is reduced to: c11

c12

c13

c14

C1,in,1

1

Pin.1

0

0

0

c11

0

P11

P12

0

0

c12

0

0

P22

P23

P24

c13

0

P3 1

0

P33

0

c14

0

0

0

0

P44

Cl,in,l

P=

(5.2-3(5d))

From Eqs.(5-2&)-(5-30a), assuming C'14 = C14, i.e. no accumulation in reactor 4 of species 1, it follows: i C13(n)p31+ L i i W Cll(n+l) = Cljn,l(n)Pin,l+ C i i ( n ) ~ i +

pfp,i (l/sec) are mass transfer coefficients for the transfer of solute f in process p (=

1 to designate an electrolytic p m s s ) from the bulk of the solution to the electrode in reactor i. f = 1, p = 1 and i = 1 means the mass transfer efficient for the transfer of solute 1 in process 1 in reactor 1; f = 1, p = 1 and j = 2 stands for the mass transfer coefficient for the transfer of solute 1 in process 1 in reactor 2. kM is the mass transfer coefficient in mlsec, Ai and Vi are, respectively, the electrode area and the effective electrolyser volume. It is assumed [86] that the electrolyhc process takes place under limiting current conditions, i.e. the solute concentrations on the surface of the electrode, C*ll= C*12 = 0. The probabilities are as follows:

where a31 = a23 = @/Q1= a is the recycle. Eqs.(5.2-3(%))to (5.2-3(5g))make it possible to calculate the concentration distributions of Cli versus time in reactors i = 1 , 2 , 3 and 4. In the numerical solution, fully described by a,pi, p11,1, p11.2 and dt, it was assumed that p i = p2 = p3 = w = p. The data in the calculations were based on ref.[%].

The initial statevector S(0)= [Cljn,l, Cll(O), C12(0),

58 1

C13(O), C14(0)] = [0,0.04,0, 0, 01 in cases a to e in Fig.5.2-3(5a), [0.04,0,0, 0, 01 in cases f and g and [0.04,0.04,0, 0,01 in case h; At = 0.1. The transient response of Cli (i = 1, 2, 3,4)in reactors 1 to 4,is depicted in Fig.5.2-3(5a). The effect of the recycle a = 0.5, 5 is demonstrated in cases a and b ; the effect of p = 0.002, 0.02, 0.2 in cases a, c and d; the effect of the mass transfer coefficient p11,1 = p11,2 = 0.0094,0.2 in cases b and e as well as in cases f and g in which pi1,i = p11,2 = 0.0094,0,respectively. Note that cases g and h demonstrate absence of an electrolytic process. The effect of the location of the introduction of species A1 into the system is shown in cases g and f. In cases a to e, A1 was introduced initially into reactor 1; in cases f and g it was introduced into the inlet of reactor 1 where the initial concentration in this reactor as well as in the others was zero. Only in case h, A1 was introduced both in reactor 1 and continuously at the inlet to it. 0.04-

'

a = 0.5,' p = 0.02, p = 0.0094 11,l

(9

Cl,h,l= 0,Cl1(O) = 0.04,c l p= 0

a = 5,'p = 0.02, p

-

I= 0.0094

11.1

(bJ

(= i 2,3,4)

-

0

50

t

100

150 0

50

'

I

a = 5,'p = 0.002, p 11.1 = O.O094(C) = 0,ClI(O) = 0.04,ClJ0) = 0 (i = 2,3,4)

150

100

(4 ~

I

a = 5 , p = 0.2, p 11,1= 0.0094

-

Cl,h,l = 0, ClI(O) = 0.04,Cl,(0) = 0 (i=2,3,4)

-

-

.__.. - . -. . _ _ 0Lr.-

t

. ( _ . . . . . . . . . . - . -.

,: 113f.

'

. - . _ _ \

.--..-

.

14

.

. . .

,

I

I

,

-

582 I

a = 5 , 1 = 0.02, 1

I

11.1

(4

= 0.2

0.025

I

I

I

I

(0

0.015 -

(i=2,3,4)

-

I

Cl,in,l = 0.04,Cli(0) = 0 (i = 14) -

,

02

0

5

15

10

0

I

I

100

200

t

I

t

300

I

400

500

-

0

100

200

300

400

500

600

-100 0

t

100 200 300 400 500 600 t

Fig.5.2-3(5a). Clj (i = 1, 2, 3, 4) versus t, demonstrating the effect of a, p, p11,1 and the introduction location of species A 1

5.2-30

-

Electrolysis of solutions-model B [86]

In this configuration, a CSTER is imbedded between two perfectly-mixed reactors 1 and 3 in the forward loop. As in model A above, electrolyte recycling is represented by a perfectly-mixed reactor 4 in the feedback loop shown in Fig.5.23(6). Electrolysis takes place in reactor 2 and the collector is reactor 5.

Ql

a 2

j=1

935

Q*3 a=2

CSTER

I

b=3

I

do {=5-

Ql

583 From Eqs.(5-18a) to (5-18c), for a'k = 0 and taking Q1 as reference flow, it follows: p35 = 1, a 1 2 = a23, a34 = ~

1

a =and a12 = 1 + a34

(5.2-3(6a))

where a = Q34/Q1 is the recycle. From Eq.(S-la) the state space read:

ss = [Cl,in,l, c11, c12,

9

(5.2-3(6b))

c13, c14, CIS]

where Cl,in,l is the concentration at the inlet to reactor 1 of the species designated by 1 undergoing electrolysis in reactor 2. From Eq.(5-3a) the state vector reads: S(n) = [Cl,in,l, Cll(n>,C12(n), C13(n), c14(n), Cldn)I

(5.2-3(6~))

where S(n+l) is given by Eq.(5-4). The probability matrix given by Eq.(5-27) is reduced to:

P =

Cl,in,l

c11

c12

c13

c14

c15

Cl,in,l

1

Pin,l

0

0

0

0

c11

0

P11

P12

0

0

0

c12

0

0

P22

p23

0

0

c13

0

0

0

P33

P34

P35

c14

0

P41

0

0

P44

0

cl5

0

0

0

0

0

P55

(5.2-3(6d))

From Eqs.(5-28a)-(5-30a), assuming C'14 = C14, i.e. no accumulation in reactor 5 of species 1, it follows:

112,l = kMA2N2

(5.2-3(60)

p11,2(l/sec) is the mass transfer coefficient for the transfer of solute

1 in process 1 (electrolysis) in reactor 2. kM is the mass transfer coefficient in d s e c , A1 and V1 are, respectively, the electrode area and the effective electrolyser volume. It is assumed [86] that the electrolytic process takes place under limiting current conditions, i.e. the solute concentrations on the surface of the electrode, C*12 = 0. The probabilities are as follows:

noting that a 1 2 = a 2 3 = 1 + a34 and a 4 1 = a 3 4 = Q34/Q1 = a is the recycle. Eqs.(5.2-3(6e)) to (5.2-3(6g)) make it possible to calculate the concentration distributions of Cli versus time in reactors i = 1, ..., 5 . In the numerical solution, fully described by a,pi, p11,2and At, it was assumed that Pi = p. The initial state vector S(0) = [Cl,in,l, Cll(O), ..., C15(0)] = [0, 0.04, 0, 0, 0, 01 in cases a to d in Fig.5.2-3(6a) and [0.04,0,0,0,0,0] in case e and f; At = 0.1,0.5in cases a to d and e, f, respectively. The transient response of Cli (i = 1, 2, 3) in reactors 1, 2, 3, is depicted in FigS5.2-3(6a). The effect of the recycle a = 0.05, 5 is demonstrated in cases a and b; the effect p = 0.02,0.2 in cases b and c; the effect of the mass transfer coefficient pi 1,2 = 0.0094 and 0.2 in cases c and d as well as e and f. The effect of the introduction location of species A l , i.e. the inlet concentration to reactor 1, Cl,in,l = 0.04, while the initial concentration in all reactors is zero, is demonstrated in cases e and f for p11,2 = 0.0094, 0 which demonstrate again the effect of pi 1.2. Note that in case f no electrolysis takes place in reactor 2 indicated by pi 1,2 = 0.

585

-

I

(9

I

a = 5, p = 0.02,pll,p 0.0094

(i = 2-5)

(i = 2-5)

-

-

0

5

cl,id

o

fl

100

t

150

I

0

= 0, CI1(O) = 0.04.Clf 0

"0.02

100

5 I

0.04-

10

t I

I

(i = 2-5)

15 I

0.03

-

150

= 0, C,l(0) = 0.04,Cli=0

C

(i = 2-5)

0

-

I

(el

0

100

~~

50

I

t

I

I

I

150 I

01

i=l=

-

c 0

-

200

400,

600

800

lo00 0

= 0.04, Cli=0 (i = 1-5)

200

400

t

600

800

lo00

Fig.5.2-3(6a). C l i (i = 1, 2, 3) versus t, demonstrating the effect of a, p, ~ 1 1 . 2and the introduction location of species A1

-

-

Simultaneous chemical reaction

5.2 3 (7)

dissolution,

absorption

and

In the following, simulation is carried out of a combined process incorporating dissolution, absorption of species f = 1 takes place into the solution as well as reacting with species 2 arriving from reactor 2 at flow rate 412 according to

586 kl

A,

+ A, + A,

for which - rl = - r2 = r3 = k2C1C2

(5.2-3(7a))

The feed to reactors 1 and 2 may contain species 1 and 2 at concentrations Cf,in,l and Cf,in,2 where f = 1, 2.

'Y j =1 absorption

5=3

+ reaction

I dissolution I Fig.5.2-3(7). Flow system for the simultaneous dissolution, absorption and chemical reaction It is assumed that the quantities dissolved and absorbed do not change the flow rate Q1 + 4 2 . From Eqs.(518a) to (5-18c), for d k = 0 and taking Q1 as reference flow, it follows: pi3 = 1 and a21 = a2 = Q2/41. where a = 42/41. From Eq.(S-la) the state space reads:

From Eq.(5-3a) the state vector reads:

where S(n+l) is given by Eq.(5-4). The probability matrix for f = 1, 2, given by Eq.(5-27), is reduced to:

587

(5.2-3(7d))

where

In pfp,i the following designations are applicable: f = 1, 2 indicate species 1 and 2,

respectively; p = 1, 2 stand for processes of absorption and dissolution, respectively; i = 1 , 2 indicate reactors 1 and 2, respectively. Thus,

The probabilities are as follows:

588

noting that a21 = a2. Eqs.(5.2-3(7e)) to (5.2-3(7h)) make it possible to calculate the concentration distributions of Cfi versus time in reactors i. In the numerical , C*ll,C*22 and At, it was solution, fully described by a 2 , pi, p11,1,~ 2 2 . 2 k2, assumed that pi = p, p l ~=, ~~2 2 . 2= 1. In addition, C*l1 = C*22 = 1 in cases a to e and C * l l = C*22 = 0 in case f. The initial state vector S(0) = [Cf,in,l, Cf,in,2, Cfl(O), Cn(O), Ca(O)] = [0, 0, 0, 0, 01 for f = 1, 2 in cases a to e in Fig.5.2-3(7a) and [0, 0, 1, 0, 01 for f = 1 and [0, 0, 0, 1,0] for f = 2 in case f; At = 0.002. The effect of p = 0, 1, 10 is demonstrated in cases a, b and c; the effect of k2 = 1, 10 in cases b and d; the effect of a 2 = 1, 10 in cases b and e. In case f, no dissolution or absorption take place; however chemical reaction occurs since the initial concentrations of A1 in rector 1 and of A2 in reactor 2 were unity. The latter species was transferred by the flow to reactor 1 undergoing there a chemical reaction.

589 L

t

c I l = C22 = 1, Cfi (0) = 0 (i = 1-3, f = 1.2)

C

*

I

II

= C 22 = 1 , Cri(0) = 0 (i = 1-3, f = 1,2)

p=O,a =1, k =1 2

2

I

I

I

(b

1

I-

1

0

3

2 t

4

Cell=CoZ2= 1, Cfi = 0 (i = 1-3, f = 1,2)

0

2

1

t

3

4

C'I1 =C*22= 1, Cri(0)= 0 (i = 1-3, f = 1,2)

p = 10, a = 1, k 2 = 1

p = l , a2 = 1 , k2 =10 I

I

I

(41

t

0.8 u*:6k0.4

I

0

I

22

1

2

3

t

4

toll = C*22= 1, C ,(O)= 0 (i = 1-3, f = 1, 2) I

p = 1 , a =10,k = 1 1.4 1.2

2

I

2

I

I

.--

/

c

- - -(el -

i

i

0.6b\,

0

1

3

3

4

Fig.5.2-3(7a). C11, C21 and C22 versus t, demonstrating the effect of p, k2, and a2

590 NOMENCLATURE designation of peripheral reactors in Figs.4-1 and 4-la. stoichiometric coefficients of species j in the ith reaction. stoichiometric coefficients of species j. heat transfer area to reactors i and j, m2. mass transfer area for process p corresponding to conditions prevailing in reactors i or j, m2. ZxZ square matrices given by Eq.(2-34). designates a system with respect to its chemical formula f and

location k in the flow system. designates the state of the system (a molecule), i.e. a specific chemical formula (Chapter 3). concentration of a chemical species i at time t (Chapter 2). initial concentration of species i. elements of the matrices A, B. concentration of A at time t and t = 0, respectively. Chapter 4: concentration of the stimulating input in reactor j in moles/(m3 reactor); concentration of species j, moles of j/m3; also designating the state of the system. Cj(n), Cj(n+l) concentration of species j in the mixture at time interval t and t + At or step n and n+l, respectively. concentration vector at step n+ 1. C(n+l) concentration of speciesj in reaction i. Cii) equilibrium concentration of species 1 absorbed on the surface of the c;,

CY 1 Cfi, cfj C'fi, C'fj

liquid in reactor 2 corresponding to its partial pressure in the gas phase above the liquid. as above, but for species 1 in reactor 1. concentration of species f in reactors i or j, kg or kg-moUm3. concentration of species f leaving reactors i or j, kg or kg-moUm3. These concentrations are not, in general, equal to the concentrations Cfi, Cfj in the reactor.

concentration of species fin reactor i at time t or at step n, kg or kg-moYm3. concentration of species f leaving reactor i at time t or at step n, kg or kg-mol/m3. concentration of species fin reactor 6, kg or kg-moVm3. concentration of species f leaving reactor 6,kg or kg-moVm3.

This concentration is not, in general, equal to the concentrations

CfS. concentration of species f leaving reactor 5 at time t or at step n, kg or kg-moVm3. concentration of the tracer in reactor 5, kg or kg-moVm3. concentration of the tracer leaving reactor 6, kg or kg-moVm3. This concentration is not, in general, equal to the concentrations

c5.

CSTER Di(%)

concentration of species fat the inlet to reactor j, kg or kg-moVm3. concentration of species fat the inlet to reactor i, kg or kg-moVm3. specific heat of the fluid mixture in reactor j, kcaV(kg K); similarly for reactor i. continuous-flow stirred-tank electrolytic reactor deviation in the concentration for the ith pair of data between the exact solution and Markov chain solution , i.e., Di(%)= 1wlCexactj - CMarkov,il/Cexact,i.

mean deviation defined by

2

D,,,(%) = ( 1 ~ ) Di(%) where H is the number of pairs. i= 1

considered in the comparison. maximum value of Di(%). age distribution defined by Eq.(4-20). at equilibrium. exact solution running index, 1, 2, ..., F, for designating the different species. subscript; designating species fin reactors i or j.

592

fjk

probability that, starting from state Sj, the system will ever pass through Sj. Defined in Eqs.(2-95). probability that, starting from Sj, the system will ever pass through s k . Defined in Eqs.(2-98). probability that state Sj is avoided at steps (times) 1, 2, ..., n - 1 where re-occupied at step n. Defined in Eqs.(2-94). fjk(n) indicates that Sk is avoided at steps (times) 1, ...,n-1 and occupied exactly at step n, given that state Sj is occupied initially. Defined in Eqs.(2-97). total number of species or reactants. specific enthalpy of the fluid in the feed vessels to reactor i, kcal/kg. specific enthalpies of the fluid in reactors i, j and in collector 6, kcavkg or kg-mole. specific enthalpies of the fluid in reactors i and in collector 6 at time t

h'i, h'j, h'c

H

khi, khj kM LHS

or step n, kcavkg or kg-mole. specific enthalpies of the fluid leaving reactors i, j and 5, kcal/kg or kg-mole. the rate of supply of A1 in moles/sec from the vapor phase ( A I ) ~ ~ into the condensed phase. integers designating states j and k, respectively. total number of cities in the external circle and in the internal city, respectively. reaction rate constant with respect to the conversion of species j in the ith reaction (in consistent units). reaction rate constant (Chapter 3). reaction rate constant for the conversion from state i (species Ai) to state j (species Aj). mass transfer coefficients for process p with respect to species f corresponding to conditions in reactors i and j, d s . heat transfer coefficient corresponding to the conditions in reactors i and j, kcal/(s m* K). mass transfer coefficient, d s . left hand side.

593 a mathematical expression corresponding to time t or step n.

P Pinj, Pinj

Pinj.h

designates step n in discrete Markov chains. number of moles of A at time t and t = 0, respectively. total number of inhabitants in the state. number of inhabitants occupying statej (an external city j) at time t. number of inhabitants occupying state k (an internal city k) at time t. number of reacting species in reaction i. probability density function, i.e. probability per unit length. transition probability function defined by Eq.(2-185). constant one-step transition probabilities. subscript designating some transfer mechanism such as absorption, dissolution, etc., or simultaneously several processes. It is assigned arbitrarily numerical values such as absorption - 1, dissolution - 2. total number of transfer mechanisms. single step transition probabilities from the state of the feed reactors (to reactors i and j) to the state of reactors i and j. single step transition probability with respect to enthalpy (or temperature) from the state of the inlet reactor (to reactor i) to the state of reactor i; similarly for reactor j, i.e., pinj,h. single step transition probability to remain in the state of reactor j

%jI

with respect to enthalpy (or temperature). single step transition probability with respect to enthalpy (or

Pki.h

temperature) from the state of reactor k to the state of reactor j. single step transition probability with respect to enthalpy (or

pj i,h

temperature) from the state of reactor k to the state of reactor i. single step transition probability with respect to enthalpy (or

Pii,h

tempemture) from state of reactor j to state of reactor i. single step transition probability to remain in the state of reactor i

P k 9 qk

Pjk, Pkj

with respect to enthalpy (or temperature). one-step transition probabilities which depend on the state k. one-step probability or the transition probability from state j to state k (or the opposite) in one step (one time interval) for each j and k

594 probability of occupying state k after one step given that the system occupied statej before. Defined in Eqs.(2-13, 13a) and (4-4,4-4a). transition probabilities from reactor j to 6, from k to 6,respectively. the probability of the transition Aj + A, for the ith reaction. probability of remaining in step j during one step; probability of remaining in reactor 6.

probability of occupying Sj after n steps (or at time n) while initially occupying also this state. n-step transition probability function designating the conditional probability of occupying Sk at the nth step given that the system initially occupied Sj. probability of occupying state k at step r given that state j was occupied at step n. transition probability of a system to occupy state k at time t subjected to the fact that the system occupied state j at time 2.

probability that the number N(t) of events occurred (customers arrived) is equal to i, given that the service time is t; defined in Eq.(2-89). Probability of the system to occupy Sj at time t; defined in Eq.(2-112). defined in Eq.(2-119). probability that the system remains at x = 0, i.e. state So, until time t. one-step transition probability matrix defined by Eq.(2-16). a probability matrix containing the elements Pjk(n). Defined in Eq.(2-31). probability of observing event Sj. Probability of occupying state Sj. prob{ s k I Sj } conditional probability. Probability of observing an event Sk under the condition that event Sj has already been observed. Probability of occupying state Sk under the condition that state Sj has already been occupied. prob{sksj} probability for the intersection of events Sk and Sj or probability of observing, not simultaneously, both Sk and Sj. rate or intensity function indicating the rate at which inhabitants qj(t) leave state Sj (external city j), Us.

595

Qi, Qj Q'i, Q'j Qji

and Qij

QP

Q 'lit

rj = dCj/dt

transition probability of the inhabitants from Sj to occupy Sk at time t. volumetric flow rate of the fluid entering reactors i and j, Figs.4-1 and 5-1, m3/s. volumetric flow rate of the fluid leaving reactors i and j, Fig.5-1, m3/s, This flows do not contain any dissolved material. interacting flows between reactors (states)j and i or i and j, respectively, Fig.4-1, m3/s. Similarly for Qk and Qik. external flow into reactor P , m3/s. external flows into the reactors, r = j, a, b, ..., Z, m3/s. volumetric flow rate from reactor (state) i to 5 ,Fig.4-1, m3/s. Similarly, qjk, qkk and qpt. rate of change by reaction of the concentration of species j per unit volume of fluid in reactor. rate of change by reaction at time t of the concentration of species j in the ith reaction per unit volume of fluid in reactor. rate of change by reaction at step n of the concentration of speciesj in the ith reaction per unit volume of fluid in reactor. reaction rate of species f by reaction m per unit volume of reactor (or fluid in reactor) j corresponding to the conditions in this reactor; similarly for reactor i. as above, at time t or step n; similarly for reactor i. rate of change at step n by all reaction of the concentration of species f i n reactor j which equals

R RHS RTD Ri sj, s k

ss

&,,T(n) where m = 1, ..., R.

total number of reactions. right hand side residence time distribution. recycle ratio defined in Eq.(4-22). designate events or states j and k,respectively. S stands for state and the subscriptj designates the number of the state. state space. Set of all states a system can occupy.

596

tm

Ti, Tj Tinj V 'j,

X

i'

occupation probability of state i at time n by the system. Defined in Eq.(2-21a). initial occupation probability of state i by the system. state vector of the system at time n (step n). Defined in Eq.(2-22a). initial state vector. Defined in Eq.(2-22). designates generally time where in a discrete process t designates the number of steps from time zero, s. mean residence time of the fluid in the reactor, s. Defined in Eq.(426). mean residence time of the fluid in reactor j, s. residence time of the fluid in a plug-flow reactor, s. residence time of the fluid in a plug-flow reactor j, s. source temperature from which (to which) heat is transferred into (from) reactor j, K. Similarly for Toj temperatures in reactors i and j, K. temperature at inlet to reactor j, K. volume of reactor, m3. volume of reactor (or fluid in reactor) j or i, m3. a prescribed value in Eq.(2-119) indicating the number of events occurring during the time interval (0,t) = t; x indicates also a numerical value corresponding to the state of a system, i.e. x = 0, 1, 2, ... initial magnitude of the state. a random variable describing the states of the system with respect to time and referring to Eq.(2-8). It also designates the fact that the system has occupied some state at time or step t. Referring to Eq.(2-119) and the following ones, X(t) which is a random variable, designates the number of events occurring during the time interval (0,t). the position at time or step n of a moving particle (n = 0, 1,2, ... ). number of customers in the queue immediately after the nth customer has completed his service.

597

yn Z

the number of customers arriving during the service time of the nth customer. total number of events, states, chemical species or reactors that a system can occupy. size of the jump of the particle at the nth step.

Greek letters ratio between the flow from reactor j to reactor i and the total flow. rate Qj. Similarly, aij, ski, a i k defined in Eqs.(4-2), (4-6) as well as aji, akj, ajk, Similarly defined. ratio between the flow entering reactor i or k to the total flow rate Qj. Defined in Eq.(4-12d). defined in Eqs.(5-18), (5-18d). the mean number of customers being serviced per unit time. defined in Eqs.(4-2), (4-6). Similarly, pr{, r = i, j and k. driving force for the transfer process p with respect to species fat conditions prevailing in reactor j; similarly for reactor i replacing subscript j; kg or kg-mol/m3. as above at time t or step n, kg or kg-moVm3.

= To,i - Ti on reactor i; similarly for reactor j, K. = To,j - Ti(") at time t or step n, K. heat of reaction m at conditions in reactor j, kcavkg or kg-mol. heat of reaction at conditions in reactor j, kcavkg or kg-mol. rate factor; rate of arrival of customers or the rates eventdtime or birthshime. birth rate which is a function of the state Si. mean occurrence rate of the events which is a function of the actual state x; mean birth rate. time interval, s. volumetric heat transfer coefficients defined in Eq.(5-22) for

reactors i and j, l/s. mean recurrence time defined in Eqs.(2-96). pj (l/sec) is a measure of the transition rate of the system defined in Eqs.(4-3).

598

h PX

defined in Eq.(4-11). death rate in Eq.(2-158). Mean occurrence rate of the events which is a function of the actual state x. volumetric mass transfer coefficient for species f, for mass transfer

V

5 xk n

process of type p (for example: absorption, p = 1; desorption, p = 2; dissolution, p =3; etc.) corresponding to reactors i or j; defined in Eq.(5-14), I/s. period. subscript designating the collection reactor for the tracer. The final reactor in the flow system symbolized as a "dead" or an "absorbing state" for the tracer for which p g = 1. limiting probabilities. Defined in Eqs.(2-105). stationary distribution of the limiting state vector. Defined in Eqs.(2105a). density of the content of reactors i, j or 6,kg/m3. density of the streams leaving reactors i, j or 5, kg/m3.

density of the stream entering reactors i or j, kg/m3.

599

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