The IKP Matlab Companion

Matlab exercises in TKP4105, TKP4110,TKP4120, TKP4140, TKP4160

and TMT4140

Tore Haug-Warberg

D C E, NTNUE-mail address: haugwarb@nt.ntnu.noURL: http://www.nt.ntnu/˜haugwarb

The Author wants to thank Ph.D. student Bjørn Tore Løvfall for his help with HYSYSand the diffusion problem, and Ph.D. student Antonio Carlos Brandao Araujo for hishelp with the control theory example and ODE description. They have also been very

helpful with the Matlab documentation in general. He also wants to thank prof. HallvardSvendsen for his kind interest in this little project.

Contents

List of Tables v

Preface vii

Chapter 1. Getting started 1

Chapter 2. Basic concepts 31. Homogeneous containers (vectors and matrices) 32. Heterogeneous containers (lists) 63. User-defined data types (structs) 74. Input/output issues 85. Tricks of the trade 8

Chapter 3. How to solve equations 91. Linear equations 92. Independent reactions 123. Linear parameter estimation 124. Polynomials 145. Single non-linear equations 156. Sets of many non-linear equations 16

Chapter 4. Write your own functions 191. The function concept 192. Vector functions 193. Iterative functions (algorithms) 204. Recursion — good, evil or maybe bad? 20

Chapter 5. Functionals or function(function)s 231. Integration 232. Ordinary differential equations 233. Unconstrained optimization 254. Optimization with constraints 255. Process control 26

Chapter 6. String parsing 291. Chemical formula arithmetics 292. Cool stuff (regular expressions) 30

Appendix A. Language elements 33

Appendix B. Examples 37

Appendix C. Matlab code 391. randolph 392. octane 413. figure 2 8 42

iii

iv CONTENTS

4. partialinv 435. fertilizer balance 436. acid base reaction matrix 457. water viscosity 1bar fit 468. water vapor pressure fit 479. water density 1bar fit 4910. acid base titration curve 5011. colebrook 5112. pressure drop at fixed flow 5113. friction and reynold 5214. flow at fixed pressure drop 5215. laminar 5316. colebrook vs laminar 5317. chon 5418. water viscosity 1bar 5419. water vapor pressure 5420. water density 1bar 5421. rachford rice 5522. diffusion 5523. pitot tube 5724. pitot integrand 5825. mingibb 5826. CSTR process control 5927. CSTR time derivatives 6128. ODE Jacobian 6229. mw from formula parser 6330. mw binop 6331. make formula matrix 6432. get formula vector 6633. greenhouse gases from HYSYS 67

Appendix D. Ruby code 691. greenhouse gases from HYSYS 69

Appendix E. Data files 711. pages from water vapor pressure data 712. pages from water density 1bar data 723. water viscosity 1bar data 734. pages from HYSYS report 73

List of Tables

1.1 Programming abstractions 1

3.1 Nitro-phosphate fertilizer plant 11

4.1 Recursive calculation of Fibonacci-numbers 22

6.1 Operator table 29

6.2 The most common regular expressions 31

A.1 File functions 33

A.2 Special variables 33

A.3 Special characters 33

A.4 Basic commands 33

A.5 Program flow instructions 34

A.6 Instantiation operators 34

A.7 Plotting commands 34

A.8 Relational operators 34

A.9 Algebraic operators 35

A.10 Set functions 35

A.11 Transcendental functions 35

A.12 String functions 35

A.13 Linear algebra functions 36

A.14 Advanced linear algebra functions 36

A.15 Functionals 36

v

Preface

My arguments for enrolling at the Chemistry Faculty in the late 1970s were subtle:I disliked mathematics and hated philosophy, but loved to experiment with (explosive)chemicals. It was therefore quite natural for me to study in Trondheim — just modestmathematics and no Exam Philosophicum. So, here I am, writing a philosophically in-clined preface to a companion in mathematical programming! My naive view lasted foronly one year, however, or more precisely till my first course in physical chemistry andthermodynamics. I then realized that the subject which had fascinated me as a young boycould be represented by thermodynamic state functions, and after a freshman course inFORTRAN programming I have since spent my professional life in the borderland be-tween chemistry, mathematics and computer science. In retrospect this has not been sucha bad fate because the field has expanded tremendously, which means I have had plentyof interesting things to do.

So what is studying, and why should we worry about programming? Basically, Ithink we have three qualities which make us stand out from other mammals, includingour cousins — the big apes:

Mental abstractionVerbal communication

Artistic inclination

To me it is important to master all three skill factors, at least to some degree. Itherefore enjoy to enjoy∗ mathematics (abstract thinking), computer programming (mindextension) and carpentry (artistic pleasure). However, communication is maybe the mostimportant skill factor these days. But, talking about communication there are at least fourlevels we need to worry about†:

Talking (99)Reading (80)Writing (25)

Programming (1)

The numbers indicate the percentage of the population that I expect master the sub-ject to such a degree that it help them in communicating with other people. I have noscientific background for my statements, and, before someone accuses me for havingfascistic motives, I want to say that it is not my intention (nor is it my profession) to cate-gorize people according to their skills — I simply want you to realize that learning aboutprogramming is likely to be a “painful” process in the beginning. However, the fractionof students who enroll to the technical studies at NTNU makes up about two percent ofthe juvenile vintage in Norway and provided it is true that our university recruits among

∗It is meant to be a recursive statement — think about it twice.†These subjects form, together with mathematics, the five basic skills that all children shall master

according to the 200X-reform of the Norwegian primary school system.

vii

viii PREFACE

the “science buffs”, it is likely that much more than one percent of you already havesome acquaintance with computers and computer programming. My hope is thereforethat many of you shall come to appreciate computer programming as a main productivitytool. Second, it is also my hope that some of you will join my research group working oncomputerized thermodynamics!

Before we get started I should mention that this little Matlab companion is intendedto follow the students at IKP over 2–3 years in several different subjects of engineeringdisciplines:

TKP4105 Separation technologyTKP4110 Chemical reaction engineeringTKP4120 Process engineeringTKP4140 Process controlTKP4160 Transport phenomenaTMT4140 Applied thermodynamics

Since I lack a first hand experience with each of the course syllabuses it is difficultfor me to give a chronological overview of the exercises. A thematic presentation is thenmore natural, especially when keeping maintainability in mind.

Even though this is a course in mathematical programming it will not give you anyformal training, neither in mathematics nor in programming. The focus is rather on prac-tical problem solving. Now, since most of the code is vectorized it means that linearalgebra is a prerequistite for reading the companion. Having said this, I still want toshow that math is done simpler, safer and faster in a mathematical programming lan-guage like Matlab, as compared to hand calculations, spread-sheet programming, or nocalculations at all∗. The exercises are taken from many different subjects, but I have putquite a lot of effort in keeping the programming style both as general and as uniform aspossible (which is the main reason this is a one-man show — it is hard to synchronizetwo programmers).

Throughout the course we shall make use of Matlab, but please understand me rightwhen I say that this is not a Matlab course. Matlab is excellent for prototyping issues, itis abundant, and it has a gentle learning curve, but it is unfortunately a rather inconsis-tent language. Without some critical sense you may therefore develop bad programminghabits, although less so than in e.g. Excel. This is not to say I want a crusade againstMatlab or Excel (well, perhaps), what I really want to tell is that the world is big and thatthere are hundreds of programming languages out there — all having their strong andweak points. The moral is: Pick the language that is best suited for your job! Take e.g.PostScript. Few people realize that this is anything but a relict from Adobe which makesartistic print-outs and fancy fonts available to the end user. However, PostScript is a pro-gramming language in its own right and you can write a complete simulation code in thislanguage if you really want to†. I have therefore avoided Matlab-specific programmingthe best I can, just to make sure you will be able to migrate to other (script) languageslike Python or Ruby without too much difficulty.

∗Plain belief or guts feeling are student favorites which seldom turn out to yield better results thanto-the-point calculations.

†Wouldn’t it be an intellectual challenge to let the printer do the number-crunching while the computercould concentrate on other things (the question is only what job that should be).

CHAPTER 1

Getting started

First of all I should let you know that the Matlab code which is included in AppendixC can also be found on my home page http://www.nt.ntnu/˜haugwarb/matlabcourse. Thismight save you same typing errors.

Before learning about programming (or any other scientific subject) you should def-initely ask yourself what the subject is about. In computer terminology program meansone out of three things: A systematic plan for the automatic solution of a problem by acomputer, or the precise sequence of instructions enabling a computer to solve a problem,or to prepare a program code (ref: Webster’s Encyclopedic Unabridged Dictionary). Allthree meanings are adequate for the current purpose.

Most programming languages are built around the idea that you shall be able to de-clare variables, write expressions of some kind, make assignments to the variables, andtransfer execution control to external procedures∗. External in this context means the pro-cedure has its own namespace where the variables are given a local meaning. In otherwords; to bring information from the outside world into the procedure it has to be copied,and vice versa; everything you want to be visible outside has to be copied back. Somelanguages got more abstraction mechanisms, some got less, but these four are the onesneeded in Matlab (see Table 1.1).

T 1.1: Programming abstractions. With this I mean thedifferent building blocks a language consists of.

Declaration function[lnp]=antoine(T,c) T and c hold copies ofthe arguments used inthe call to antoine

Expression c(1)+c(2)/(T+c(3)) The expression is eval-uated but not assigned

Assignment lnp=c(1)+c(2)/(T+c(3)) Variable lnp now holdsa copy of the evaluatedexpression

Procedure antoine(x,y) x and y are copied intoantoine, the calculatedresult is copied back

Matlab belongs to a family of languages called script languages (Python, Ruby, Perl,etc.). This is contrary to languages which has to be compiled before you get an executablecode† (Pascal, C++, FORTRAN, etc.). The big challenge in this respect is programming

∗In Matlab, the function abstraction is mixed up with the declaration keyword function which is usedto define functions. For this reason I write procedure when I am referring to the abstraction, and functionwhen I refer to the declaration, see Table 1.1.

†Nowadays often called an application. On the Windows platform this is a file having the file extension.exe

1

2 1. GETTING STARTED

speed versus calculation efficiency and code syntax versus semantics∗. Script languagesare very suitable for rapid prototyping and testing, while a compiled code is more securein the sense that you can ask for full control over its execution.

In Appendix A you will find a small subset of commands, operators, etc. taken fromthe standard Matlab implementation. The list is complete in the respect that nothing morewill be needed in this course. In Appendix B I have given a few examples on how theMatlab-code looks.

∗Whereas correct syntax only tells us that the program can be executed, a correct semantics tells usthat the program does exactly what it is intended to do — now that is a real challenge!

CHAPTER 2

Basic concepts

The expression syntax∗ in Matlab is not very different from what you have been taughtin mathematics, e.g. a=2; b=3; c=(a-b)ˆ2; c==1;. The central issue here is the statement.A statement is interpreted from the left and proceeds all the way to the first semicolonor the end-of-line (there are four statements in the previous example). If the statementcontains an equal sign it is said to be an assignment (there are three assignments in theprevious example). At this point things start to diverge from what you have seen before.To illustrate possible pit-falls let us rewrite the Matlab code into mathematical terms:

a =2

b =3

cdef= (a − b)2

c?=1

First, I assign the values 2 and 3 to the variables a and b, then I define c to be the completesquare in a and b, and in the fourth line I ask myself whether c is equal to 1 or not. Indaily speech most people just continue to use the = operator and say “equal to” in allfour cases. Now, this is something a computer cannot understand without some help!E.g. in Matlab, the first two lines are OK, but the third line gives trouble. Matlab doesnot know about symbolic algebra (contrary to e.g. Maple, but that is another story). Theclosest we come to a definition in Matlab is the function concept explained in Section 4.1.Therefore, issuing the statement c=(a-b)ˆ2; in line three, Matlab will simply evaluate thesquare using the current values of a and b. If a and b are not yet assigned you will get theerror message ??? Undefined function or variable ’a’.

The fourth line represents a boolean test (booleans are numbers that can take the val-ues true or false†). In Matlab there are no booleans so this feature has to be sneaked inusing ordinary numbers combined with the convention that 1 is true, and 0 is false. Totest the equality we have to write c==1; which in this particular case will return 1 (true).Hence, we can also write something like d=c==1; where the outcome from the test ex-pression is assigned to the variable d. The moral is: Beware of the difference betweendefinitions, assignments and boolean tests! You must also understand that Matlab exe-cutes one line at a time. Hence you must make sure that all statements (lines) comes inthe correct order.

1. Homogeneous containers (vectors and matrices)

The basic data type in Matlab is a matrix-like‡ container of numbers. The numbersthat Matlab operate on are so-called doubles. One double occupies 8 byte of information.

∗Rules for writing correct expressions, i.e. 2 + 3 ∗ 2 will be evaluated as 2 + (3 ∗ 2) and not (2 + 3) ∗ 2.†Don’t be too puzzled, this is not very different from an electric charge having the values plus or minus.‡What exactly is a matrix? There are at least four meanings of the term: A point in vector space, a

linear operator, a row or column basis for a vector space, or just a data “container”. The last interpretationfits the use of matrices in Matlab best.

3

4 2. BASIC CONCEPTS

There are 8 bit in one byte which means there are 63 bit available for storing numbers (thelast bit is the so-called sign-bit). In practise this means you can represent all integer valuesin the range

[−253, 253

]exactly. Outside this region the number field gets more and more

scattered until you reach the limit ±21024. Beyond this Matlab will use the notation infto signify that you are outside the number representation (over-flow). Creeping towardszero you can represent numbers down to ±2−1075. Inside the range

[2−1075, 2−1075

]you will

always get the result 0, i.e. there is no under-flow in Matlab.Let us go back to the container issue. In mathematical texts you will meet several

notations for vectors and matrices. The notation I tend to favor is this one:

A =(

1 23 4

)

a =(

5 6)

a = 3.141592 . . .

N = 13,

where capital boldface denotes a matrix, lower boldface denotes a vector, and normal font(either capital or lower case) is a scalar. In Matlab there are no bold fonts which meansvectors and scalars will look the same:

snippet from matlab matrix.m starts here

A = [1,2;3,4];

a = [5,6];

a = 3.141592;

N = 13;

matlab matrix.m ends hereOf importance for us is to realize that Matlab does not distinguish between matrices,

vectors and scalars. To Matlab everything is a matrix which means that executing thesize() command on A, a and N will tell they are all two-dimensional objects (do this).We could therefore forget about most of what we have learned in mathematics and startdoing things entirely the Matlab-way. This is not encouraged, however, and I shall do mybest to present code that is “mathematically correct” in this course.

Assume we want to calculate and plot the ideal gas pressure as a function of volumefor a given temperature. Mathematically, the operations can be written∗

R = 8.31453 J mol-1 K-1

T = 273.15 K

N = 7 mol

v =(0.1 m3, 0.3 m3, . . . , 1.5 m3

)e = (1, 1, . . . , 1)T

t = Te

p = NR diag (v)−1 t

Note the use of matrix/vector notation in the example. It certainly makes the code com-pact, but you must know how to handle array and matrix products! Translated to Matlabthe same operations look like this:

snippet from ideal gas isotherm.m starts here

∗Since there are no standard mathematical notation available for plotting vectors I cannot show thisoperation.

1. HOMOGENEOUS CONTAINERS (VECTORS AND MATRICES) 5

clear;

R = 8.31453;

T = 273.15;

N = 7;

v = [0.1:0.1:1.5];

e = ones(size(v))’;

t = T*e;

p = N*R*inv(diag(v))*t;

plot(v,p);

xlabel(’Volume [mˆ3]’);

ylabel(’Pressure [Pa]’);

title([’p(V)-plot at ’,num2str(T),’ K’]);

ideal gas isotherm.m ends hereThere are a couple of things we should note here. First of all Matlab has a vector con-

structor [from:step:to] which is very handy. Second, the matrix constructor ones(size(v))do two things at one time. Besides filling the vector e with 1’s, it also ensures that einherits the dimensions of v (or shape if you like). Now, this is a good programmingpractise! The opposite is also valid: Never, never, . . ., never enter the same definitiontwice in the code, and never let two dependent matrices (or two vectors) have indepen-dent size definitions! Finally, I should mention that in practise I would consider to changethe last line in the calculation to p=N*R*(t./v). This is both a matter of CPU-time andreadability, although I shall not claim that the compact code is a better alternative underall circumstances.

§ 1 Repeat the calculations in the last example for the isotherms T=[250:20:350].Use v=logspace(0.1,10,100)’ to construct the pV-diagram. Make your best at cal-culating everything as matrix products (no for-loops, index variables, etc.), andmake sure your results can be plotted using the command semilogx(v,P) where Pis a dim (v) × dim (T )-matrix. For your own understanding of the physical behavioryou should also try the command loglog(v,P).

I . Nothing really new here, but to become fluent in array program-ming you must exercise a lot. I want therefore to stress the importance of the compactnotation used in ideal gas isotherms. So, do not fake you have grasped the idea but makesure you understand! �

snippet from ideal gas isotherms.m starts here

clear;

R = 8.31453;

t = [250:20:350];

N = 7;

v = logspace(-1,1,100)’;

e = ones(size(v));

p = N*R*(e./v)*t;

semilogx(v,p);

xlabel(’Volume mˆ3’);

ylabel(’Pressure Pa’);

6 2. BASIC CONCEPTS

title(’p(V)-plot at several isotherms’);

legend(num2str(t’));

ideal gas isotherms.m ends hereEverything is a matrix in Matlab, and depending on your attitude to linear algebra this

may be a relief or a nuisance. However, take advantage of this feature and try to eliminatefor-loops, if-tests, and other spaghetti-ingredients. My own experience is that Matlab-programming may result in only 10% of the code compared to compiled languages likeFORTRAN and C++.

2. Heterogeneous containers (lists)

I first decided not to talk about lists, but after joggling back and forth with the codeI wrote for this course I finally decided to trouble you with one extra abstraction level.Appearantly, we could have done without lists, but they make things easier, especiallyin program interfaces. The problem is that Matlab has a dirty implementation of the listconcept called a cell array which makes it difficult to remember the syntax from one dayto the other. The big difference between a list and a vector is that vectors hold numberswhile lists can hold strings, lists, matrices, structs, etc. In Matlab there are a number offunctions that let you manipulate the transition from a list to a vector and back:

listchar→ character matrix

abs→ matrixchar→ character matrix

cellstr→ list

Here, a character matrix is like any other matrix execpt that it contains numbers in therange 0–255. If you display this matrix on screen you will see letters and not numbers.However, you can still make use of the matrix as a numerical object, but once you manip-ulate its content Matlab forget about ASCII-decoding and display plain numbers (unlessyou explicitly tell something different). A short tour de lists follows:

snippet from matlab list.m starts here

clear;

format compact;

mylist = {’some’,’words’,’in’,’the’,’list’};

disp(’List syntax: mylist = {’’some’’,’’words’’,’’in’’,’’the’’,’’list’’}’);

disp(’This is how ’’mylist’’ looks:’);

pause;

disp(mylist);

disp(’Run size(mylist) to see the matrix-like size of the list:’);

pause;

disp([’size(mylist) = ’,num2str(size(mylist))]);

disp(’Run char(mylist) to convert to a character matrix:’);

pause;

disp(char(mylist));

disp(’The new size is:’);

pause;

disp([’size(char(mylist)) = ’,num2str(size(char(mylist)))]);

disp(’Convert to ASCII-numbers using A = abs(char(mylist))’);

pause;

3. USER-DEFINED DATA TYPES (STRUCTS) 7

A = abs(char(mylist));

disp(A);

disp(’Replace spaces (32) by periods (46)’)

pause;

i = find(A==32);

A(i) = A(i) + 14;

disp(A);

disp(’Convert back to a character matrix using char(A):’)

pause;

disp(char(A));

disp(’Convert into a new list using cellstr(char(A)):’);

pause;

disp(cellstr(char(A)));

matlab list.m ends here

3. User-defined data types (structs)

A struct is an extensible data type that Matlab has inherited from C. It is now ac-cepted as a standard data storage facility in most programming languages, not only C andMatlab. Compared to matrices, which are like dense tables, the struct is an open tree-likecontainer. The branches of the tree are given names by the programmer, and at the veryend of each branch sits a leaf which keeps a data object. An important feature of thestruct is its extensible and heterogenous nature which makes it ideally suited for objectoriented programming. This means the struct can hold several different data types at onetime; matrices, other structs, strings, etc. One example follows:

snippet from matlab struct.m starts here

book.author.name.given = ’Tore’;

book.author.name.family = ’Haug-Warberg’;

book.author.age = ’(now-datenum(’’13-Nov-1958’’))/365’;

book.title = ’The IKP Matlab companion’;

book.title.subtitle = ’Exercises in TMT4140 etc.’;

book.ISBN = 0;

matlab struct.m ends hereTo figure out my age (uh-oh!) you can issue the command eval(book.author.age).

This operation is known as string parsing, see chapter 6 for an introduction to this fasci-nating subject.

Structs can obviously be used to capture data in a systematic way, much like a bookwith chapters, sections, paragraphs, etc. One specific use is to make programs less sen-sitive to software upgrades. Imagine you have been asked to program the van der Waalequation function[p]=vdw(v,t,n,tc,pc). Months later, when you have forgotten what youdid the first time, you are asked to make the model more general, say function[p]=vdw(v,t,n,tc,pc,K), where K is an interaction parameter matrix. If the model was implemented asfunction[p]=vdw(v,t,n,par) in the first place, this update had been trivial since you wouldnot need to do any changes to the program interface, just include one new field par.K inthe input and change the model accordingly.

8 2. BASIC CONCEPTS

4. Input/output issues

To save the current work space simply issue the command save. To load it back useload. All variables in the workspace will then be written to/from a standard (binary)file named matlab.mat. This is OK for day-to-day use, but you should know about onepossible pit-fall — you need Matlab installed to view the information in matlab.mat! Tohave easier access to your data you must save the information in a text file rather than abinary file. Fortunately this is also very easy, just issue the command save yourFileName-ascii to save data with 8 digits, or save yourFileName -ascii -double to save data with 16digits. You can also write formatted data to a file using the fprintf command, see the files. . .fit.m for examples on how to generate a program from the inside of another program(this is a quite powerful technique).

Plotting data in Matlab is a snap once you have become familiar with vectors andmatrices. In its simplest version you enter the command plot(x,y), where x and y aretwo vectors of the same shape. There are numerous ways to produce labels, multipleplots, 3-dimensional plots, etc. Take a look at the files . . .fit.m if you want to see someexamples.

5. Tricks of the trade

• Always start your script with the clear command to be sure you get a fresh start(avoid overlapping workspaces).• Function variables should always be entered explicitly as in z = f(x,y).• Function parameters should be wrapped into a struct and given at the end of the

argument list as in z = f(x,y,s).• Use long variable names e.g. Hungarian notation longNamesLikeThis at the top

level of a program, and successively shorter names deeper down in the programhierarchy (i, j, x, y, etc. at the lowest level). Do not use underscore in variablenames.• Use underscore in long function (file) names (long function name like this). Do

not use Hungarian notation for files names (Windows does not support casesensitive naming).• Segment the code into readable blocks. Consider using whole-line comments in

front of each block to explain the gross action of the code.• Use right margin comments to explain complex expressions.• Align the assignment operators in each block (the = operator often plays a par-

ticularly important role in mathematical programs).

CHAPTER 3

How to solve equations

1. Linear equations

Solving linear equations is all what engineering is about, at least this is not far fromthe truth, and the importance of this chapter is so immense that you can hardly take it tooseriously. Matlab makes the solution of big equation systems a snap, but beware of thedangers if you do not know exactly how the solver works. In the worst case you may getfatally wrong results. This statement is quite harsh, so let us start out with one simpleexample:

§ 2 You are asked to make 98 octane gasoline containing 1% aromatics. Youhave three petroleum fractions at your disposal: Fraction one holds 90 octane with0.2% aromatics, fraction two holds 105 octane with 2% aromatics, and fraction threeholds 95 octane with 0.5% aromatics. The octane number is a mass linear functionof each petroleum fraction. Set up an equation system Ax=b that can be solved usingMatlab. Invert the equation system by hand (row elimination), and compare withthe Matlab solution x=inv(A)*b.

98 . We are given two pieces of information (octane number and aro-matic content) which expresses two equations in three unknowns. The total amounts ofeach fraction (in tons, kilograms or whatever) has no importance, which means we cando the calculations on a normalized basis. This gives us the third piece of informationneeded to invert the equations:

90 105 950.2 2 0.51 1 1

abc

=

980.96

1

The Matlab-script listed in Appendix C.2 takes you through the elimination process, stepby step. Notice that “hand-elimination” yields more accurate results than Matlab inver-sion (this is not a general result though). The difference is barely visible in this case, butin some practical problems like e.g. chemical process simulation there maybe thousandsof variables, and the differences between two numerical solutions can be like day andnight. Therefore, always check your result by back-calculating the right hand side! �

So far so good, but things are quite different when working on a computer comparedto doing hand calculations. In Section 2.1 I told you that the computer has only a limitedrange of numbers available. From time to time you will therefore experience that Matlabcomplains about singular matrices or bad condition numbers. Without any understandingof the possible causes of these messages you are totally lost and may easily producegarbage results. The next exercise gives you a clue of what appears to be a commonpitfall.

§ 3 Change the aromatic analysis in the last example to 1 − 10−15, 1, and 1% forthe three gasoline fractions. What happens to the solution now?

9

10 3. HOW TO SOLVE EQUATIONS

M 98 . The Matlab script is unchanged from the last exercise(your prompt answer decides what to do). The first thing you should notice is the neg-ative (nonphysical) element in the solution vector [−0.1583, 0.2208, 0.9375]. Next youshould not take too easy on Matlab’s warning message: “Matrix is close to singular orbadly scaled—Results may be inaccurate”. Finally, you should recognize that the back-calculated Matlab solution [51.25, 0.5, 0.5] is pure nonsense. �

By now we have solved one nice and clean equation system, and we have modified itinto something ill-conditioned. Next, let us take the full step and look at a truely singularproblem. This may look terrifying, but the important message is: Don’t panic∗! You aretold to solve the equations and should focus on your task without too much hesitation.

§ 4 Solve the component balances of this flowsheet as a simultaneous set of equa-tions Ax = b, where x = (n1, n2, n3, n4, n5) and ni[mol s-1] are the molar flow rates ofthe streams i ∈ {1, 2, . . . , 5}. Write the solution as

x = xp + Nxh

where xp is a particular solution and xh is the (homogeneous) solution Axh = 0. Plotthe equilibrium quotient Q = x2

B/xA in stream 3 as a function of the recirculationrate. Determine the smallest recirculation flow that is theoretical possible consider-ing that the equilibrium constant is K = 0.74.

F 2.8 S’ . You are given a flowsheet with 10 unknowns and 10equations, but the coefficient matrix is singular! Instead of one unique answer you haveinfinitly many possible answers, which in practise must be settled by knowing some extrainformation. In this case it is the conversion constraint Q = x2

B/xA ≤ K = 0.74 for thereaction A → 2B. This is a non-linear constraint which we shall not devote time onright now, but we can easily plot Q as function of the recirculation rate and determine anapproximate region for possible plant operation (Appendix C.3). �

In all the examples we have met so far there have one way of expressing the equations— the free variables have been unknown and the specifications have been fixed. Some-times, however, you will need to solve an partially inverted specification problem, i.e. anequation system

Ax = b

where some of the x’s are known and a similar number of the b’s are unknown. I havecoined the term “partially inverted problem” for this behavior, because in the ultimatecase all the x’s are known and you do not need to solve any equations at all (the invertedproblem is formulated explicitly).

§ 5 Write a function [x,b]=partialinv(A,x,b) which solves a partially invertedequation system. Let x(i) = b(j) = NaN denote the free variables x(i), and the spec-ifications b(j), that are considered unknown (on input). Update the vectors x and bsuch that Ax=b on output.

P . You may wonder why it is necessary to spend time onwriting a function in this case. After all, re-ordering the variables (and the specifications)will solve the problem. If it is a one-time problem this is definitely the right strategy,

∗Yes, I have admittedly taken this pun from the hilarious trilogy (in five volumes): A Hitch-hiker’sGuide to the Galaxy by Douglas Adams.

1. LINEAR EQUATIONS 11

T 3.1: Production data for a typical nitro-phosphatefertilizer plant.

Component Case I Case IIAmmonia 1300 ton d-1 1300 ton d-1

Mono-ammonium phosphate 200 ton d-1 200 ton d-1

Calcite − 0Calcium nitrate tetrahydrate 200 ton d-1 −Oxygen (output) 0 0Carbon dioxide (input) 0 0Water (input) 100 ton d-1 0

but if the equations are part of a more general problem setup, it can be easier to keepthe coefficient matrix unchanged, and flag the x(i)’s and b(j)’s as suggested above (seeAppendix C.4). �

In process simulation you will find that most of the unit operations are without chem-ical reactions. Hence, simulating a complete chemical plant will most of the time meanthat the mass balance and the component balance are equivalent concepts. However, ifyou view the plant from the outside and just look at the net reactions going on in itscomplex inner, you will frequently find that the atom balance is the key concept. Oneexample follows:

§ 6 The chemistry of a nitro-phosphate fertilizer plant is quite complex and in-volves several steps of leaching, evaporation, neutralization, etc. The main reactionsare

NH3 + 2O2 �HNO3 + H2O

2NH3 + Ca3 (PO4)2 + 6HNO3 �3Ca (NO3)2 + 2NH4H2PO4

3H2O + CaCO3 + 2HNO3 �Ca (NO3)2 × 4H2O + CO2

HNO3 + NH3 �NH4NO3

where NH3, O2, Ca3 (PO4)2, CaCO3, H2O and CO2 are considered to be (possible)reactants, and Ca (NO3)2 × 4H2O, NH4H2PO4, CO2, NH4NO3, O2 and H2O are (pos-sible) products. Note that H2O, O2 and CO2 appear as both reactants and products(the recipient act as a buffer). Your task is to calculate all input and output flowsgiven the production data in Table 3.1. Hint: Let x and y be vectors of reactant andproduct flows respectively, and A and B the corresponding formula matrices. Makesure your definitions are compliant to Ax = By. You can then choose to solve eitherB−1Ax = y or A−1By = x using the partial inversion algorithm from the last example.

F . Nothing really new here, except that you have to specify two coef-ficient matrices in order to solve one equation system (Appendix C.5). �

12 3. HOW TO SOLVE EQUATIONS

2. Independent reactions

The calculation of mass conservation due to chemical reactions is tightly coupled tolinear algebra. Consider the so-called formula matrix for the system CO–H2O–CO2–H2:

A = (atoms × species) =

0 2 0 21 0 1 01 1 2 0

Given a mole number vector n for the species mentioned above, we can calculate the molenumbers of the atoms from the standard matrix product b = An. For all reactions takingplace in a closed system we know that the atoms must be conserved (not talking aboutnuclear reactors here), and if the changes in the species mole numbers are represented by∆n this leads to

∆b = A∆n = 0.

The next step is to find a general solution to this problem. The full-blown theory belongsto a course in linear algebra, but from physical reasoning we can argue that ∆n must bea function of the independent reactions of the system. Hence, if we let x represent theextents of (independent) chemical reactions, and N the reaction stoichiometry matrix, wecan write

∆n = Nx.

Combining the last two equations, and realizing that x is by definition a free vector vari-able, we get the necessary condition

AN = 0.

By standard row elimination, and column permutation if this proves necessary, matrix Acan be factored into

A→(

I B0 0

)

Thus, by setting N =(

B −I)

we obviously have a stoichiometric matrix N whichfulfills the mass balance equation AN = 0. Neat, isn’t it?

§ 7 Find a set of independent reaction equations that fully describe monoproticacid–base titration:

HA + B� A− + BH+

The equilibrium reactions are assumed to take place in an aqueous environment andyou should therefore end up with 3 equations.

A- . The Matlab code listed in Appendix C.6 will assist you in get-ting the reaction stoichiometry right, but traditional hand calculations are perfectly OK.

�

3. Linear parameter estimation

The issue in this section is to find the minimum sum-of-squares of a function thatdepends linearly on the coefficients. Let the sum-of-squares function be

s =n∑

i=1

(yi − f (xi, c))2 ,

3. LINEAR PARAMETER ESTIMATION 13

where yi and xi = (xi,1, xi,2, . . . , xi,n) are the experimental observations, and c = (c1,c2, . . . , cn) are the parameters of the model f (xi, c). If the sum-of-square function shall bea minimum one with respect to the parameters c1, c2, . . . , cn we must show that

∂s∂c1=∂s∂c2= . . . =

∂s∂cn= 0.

Provided the model f (xi, c) is linear in the parameters we can write fi =∑

j ai j(xi)c j, andfor s we get the expression:

s =n∑

i=1

yi −n∑

j=1

ai j(xi)c j

2

= (y − A (x) c)T (y − A (x) c)

I have deliberately chosen a matrix definition of the function, because this is one of thekey issues in this course: Try to think in terms of matrices and vectors if at all possible!Differentating s with respect to ck and putting the result equal to zero yields

− ∂s∂ck= 2

n∑i=1

yi −n∑

j=1

ai j(xi)c j

aik = 2A (x)T (y − A (x) c) = 0,

or simply

c =(ATA

)−1ATy.

These are the so-called normal equations∗ of the system A(x)c = y. Note that the equationsystem is over-determined, i.e. there are more observations (equations) than parameters.Thus, there will be no solution except if the observations are in perfect agreement withthe experimental data. Such coincidence never happens in practise and we must thereforelook for an inexact solution. Among all the alternatives we shall go for the solution thatminimizes the Euclidean length of the observed deviations yi − f (xi, c).

§ 8 Fit the coefficients ci∈{1,2,3} in the expression log µ = c1 + c2T−1 + c3T−2 toµ, T -data found in water viscosity 1bar data.txt see Appendix E.3

L . You may think this is a non-linear problem, but no! Even though µis a non-linear function in T it will still be a linear function in the coefficient vector c.The formula matrix is

A (T ) =

1 T−1

1 T−21

......

...1 T−1

n T−2n

,

and the corresponding observation vector is y =(log µ1, . . . , log µn

). A Matlab script for

solving the least square problem is given in Appendix C.7. �

§ 9 Fit the coefficients ci∈{1,2,3,4} in Wagner’s vapor pressure equation to p, T -data found in water vapor pressure.txt see Appendix E.1. Wagner’s equation can be

∗Actually, Matlab is capable of solving such problems directly with the special “operator” c = A\c butthe backslash means several things depending on the arguments and we shall not pursue this alternative anyfurther.

14 3. HOW TO SOLVE EQUATIONS

written:

Tr log Pwagnerr

def=c1τ + c2τ

1.5 + c3τ3.0 + c4τ

6.0

τdef=1 − Tr

Trdef=

TTc

prdef=

ppc

The critical constants of water are Tc,water = 647.1 K and pc,water = 220.64 bar.

V . Don’t let this seemingly complex model scare you! Note carefullythat the function Tr log pwagner

r is linear in the coefficients c1 − c4. You can therefore usestandard linear parameter estimation theory if the observed T, p-data are converted toTr log pr and τ. This yields

A (T ) =

τ1 τ

1.51 τ3

1 τ61

............

τn τ1.5n τ3

n τ6n

,and

y =

Tr,1 log pr,1

...Tr,n log pr,n

The Matlab code is given in Appendix C.8. �

§ 10 Fit the coefficients c1 and c2 in the expression ρ = ρmaxec1τ1.3

ec2τ1.6

to ρ, T -datafound in water density 1bar data.txt see Appendix E.2. The definitions of τ and ρmax

are:

τ =

∣∣∣∣∣ T[K]− 273.15 − 3.98

∣∣∣∣∣ρmax = 999.97490 kg

−3m

L . As before, the trick is to transform the equation; this time to log(ρ/ρmax) = c1τ

1.3 + c2τ1.6. The rest is standard, see Appendix C.9. �

4. Polynomials

Easy to understand, easy to manipulate, and easy to calculate — there are many rea-sons why polynomials appear in engineering calculations, e.g. when an exact solutionis missing or for some reasons out of scope for numerical work. However, polynomialsmay also be the outcome of theoretical investigations as shown in the pH-calculation dis-cussed below. Dealing with such problems it is important to know that polynomial rootfinders are very stable, but that it may sometimes be difficult to interpret the physical rootamong the several false ones.

In Ex.7 we discussed the independent reactions of monoprotic acid-base titration. Thenet reaction equation was

HA + B� A− + BH+

5. SINGLE NON-LINEAR EQUATIONS 15

Assume that the initial concentration of the base (in the flask) is C0B, and that the acid is

being added in arbitrary amounts. The nominal∗ concentration of the acid CHA is thena free variable. Eliminate the other variables in the equation system till you end upwithP4(CH+) = f (C0

B,NHA, k1, k2, k3), where ki∈{1,2,3} are the equilibrium constants and P4

stands for a 4th degree polynomial.

§ 11 Write a Matlab script that calculates pH as a function of the nominal acidconcentration using ca=linspace(0,2*cb,1000), where cb is 0.1 mol l-1. Neglect possi-ble volume changes during this “numerical experiment”.

H-. There are infinite many choices of independent reactions, but a nicechoice is the following one (H+ appears only once and this makes the elimination processeasier):

HA + Bx� BH+ + A−

H2O + By� BH+ + OH−

H+ + Bz� BH+

The extents of reaction x, y, z (in the right direction) can be used to express the massbalances as shown below:

CBH+COH−

CB=

(x + y + z) y

C0B − x − y − z

= CH2O10−pKb def= k1

CA−CH+

CHA=−xz

CA − x= 10−pKa def

= k2

COH−CH+ = −yz = CH2O10−pKwdef= k3

Elimination of x and y gives

−k1z4 +(k1C

0B + k3 + k1k2

)z3 +

(k1k2CHA + k1k3 − k1k2C0

B − k2k3

)z2

−(k2k3CHA + k2

3 + k1k2k3

)z + k2

3k2 = 0,

which can be solved repeatedly to yield the titration curve for the acid–base pair. At eachtitration point we have to choose among four possible solutions. There are no clues as towhich of the roots is the physical one, but in this case it appears to be the negative root(Appendix C.10). �

5. Single non-linear equations

There exist several “general-purpose” equation solvers in Matlab, but their mathe-matical methods are well beyond the current scope. Regretfully so because we are thenreduced to ordinary users of advanced code, which in turn makes it hard to tell what isclever and what is dumb. I mean, how could we possibly tell without knowing the under-lying theory? An important concept in this chapter will be the functional that is discussedmore carefully in Chapter 5. It is achieved in this context by the function handle operator@ which lets the user specify a run-time definition of the equation to be solved. Thenecessary actions are then taken by the solver through the built-in function feval whichis used to run a user specified function with solver specific variables. This is a goodexample on the “tool-box” idea of Matlab.

∗Nominal means measured before any reactions take place.

16 3. HOW TO SOLVE EQUATIONS

§ 12 Calculate the pressure drop over a L = 2000 m long pipe transportingQ = 0.5 m3 min-1 of water. The pipe (D = 5 cm) is made of commercial steel. Thetemperature of the water is θ = 8 ◦C. Use density and viscosity functions from Ex.19and Ex.17. Hint: Colebrook’s equation for the friction factor is:

1√f= −2 lg

ε

3.7 · D +2.51√

f Re

∆p =

f Lv2

2Dg[m H2O]

Use function fzero to solve this equation inside your pressure drop calculation. Hint:write a function colebrook(lnf,eod,lnRe) which returns the residual on the form∗:

12

log ( f ) + log

−2 lg

eD

3.7+

2.51

exp(

12 log( f ) + log (Re)

)

P-. The friction factor can be solved explicitly as lnf = fzero(@cole-brook,lnf,opt,eod,lnRe) where opt=[’display’,’iter’] can be used to display the iterationprogress. The Matlab code for the Colebrook residual is given in Appendix C.11, and forthe pressure drop calculations in Appendix C.12. �

This exercise illustrates an extremely important issue in non-linear equation solving.In theory any residual function that evaluates to zero at the solution point is sufficientfor finding the correct answer, but in practise only a few residual functions will be same.Finding a good residual is an art, but the following rules of thumb may be of value:

(1) Make sure the residual is comparable to 1, i.e. 10−6 should be considered a smallresidual and 106 a big residual. This can of course be achieved by pure scaling.

(2) Make sure the gradient has comparable magnitude in all directions (at least closeto the solution point).

(3) Residuals that are appealing to the human eye are not necessarily what makesyour algorithm effective.

Finding one single root of a function is usually not a big problem, but when the issue isto build a more general function these rules may become important (see Exercise 13).

6. Sets of many non-linear equations

§ 13 The inverted variant of Ex.12 is to calculate the volumetric flow rate for agiven pressure drop. This calculation is a little more involved because there are twonon-linear functions to be solved simultaneously (Re = ρvD/µ):

∆p = ρ fLD

v2

21√

f= −2 lg

ε

3.7 · D +2.51√

f Re

F-. The same procedure as in Ex.12, except that we must call fsolve ratherthan fzero. In the function call I have chosen to define a struct for copying the parameters.This is much easier than to remember explicit parameter ordering in the function call

∗It took me two full days to convince myself that there is no way to avoid the logarithmic form.However, feel free to try other residual forms and prove that I am wrong!

6. SETS OF MANY NON-LINEAR EQUATIONS 17

(just try to remember the order of six parameters for more than five minutes). I havealso taken the trouble to vectorize the calculations in order to be prepared for futuredemands. The solution is this time written x = fsolve(@friction and reynold,[lnRe,lnf],opt,@colebrook,s); where lnRe=x(:,1) and lnf=x(:,2) on output. The Matlab code forthe two flow residuals is listed in Appendix C.13, and for the flow rate calculations inAppendix C.14. �

§ 14 From fluid mechanics you know that turbulent flow develops at Re � 2300.Write a Matlab script that reveals the transition from laminar to turbulent flow.

T . Needs a f -relation for laminar flow which can becompared with Colebrook’s function in Ex.12, i.e. f = 64

Re . This relation is explicit in fand there is no need for a non-linear solver, but who cares since the point here is to learnabout solvers (see Appendix C.15)! The flow rate calculations comparing laminar withturbulent flow are found in Appendix C.16. �

§ 15 Hallvard Svendsen, Ex.7: Compressible friction flow.

C . Not implemented. �

CHAPTER 4

Write your own functions

1. The function concept

When you start Matlab you will enter (the topmost) workspace. From there, you canenter commands at the keyboard, or you can import several commands at once from ascript file, in order to build up the content of your workspace. You can also make calls toexternal procedures (called functions in Matlab). When a function has finished its job itwill return to the workspace, hopefully with some results. Functions may even call otherfunctions, but eventually the control will be returned to the topmost workspace. Note,however, that all variables passed back and forth in a function call is done “by copy”:

In fact, entering x=exp(3) into Matlab, copies the number 3 into the exp function,which returns a copy of the result to the variable x. The exp function does not changethe number 3 in any way! To illustrate this further let us try the command sequencex=3; x=exp(x). In this case the variable x is first asked to hold a copy of the number 3.Thereafter, the value of x (which is still 3) is copied into the exp function, and the resultis finally copied back into x. In this case x really did change, but only because we told itexplicitly to do so. This is called imperative programming.

§ 16 Write a function chon(x,y,z,w) for calculating the molecular weight of thefictitious compound CxHyOzNw.

M CHON-. See Appendix C.17. �

2. Vector functions

By vector function I mean a function that calculates more than one output value. Typ-ically, these functions should mimic the shape of their input arguments. As an exampletake the built-in function exp. If you enter something like x=[1 2; 3 4]; y=exp(x); fromthe keyboard, you will get an y-matrix with the same shape as x. This is veeery handyand is certainly something you should pursue when you write your own functions.

§ 17 Make a Matlab function that calculates log µH2O = c1 + c2T−1 + c3T−2 for anarbitrary temperature matrix t. The function syntax is water viscosity 1bar(t) andthe parameters are c = [-8.0003593134, -1470.7557834, 526396.01377] see also Ex.8

W . The trick is to inherit the shape of the temperature argument in thefunction body. The matrix e=ones(size(t)) will do this, as shown in Appendix C.18. �

§ 18 Make a Matlab function that calculates log pwagnerwater = f (T ) for an arbi-

trary temperature matrix t. The Wagner equation is described in Ex.9. The func-tion syntax is water vapor pressure(t) and the parameters are c = [-7.78664494819,1.50070473755, -2.82073715777, -1.20214640863]

19

20 4. WRITE YOUR OWN FUNCTIONS

W . Same trick as in Ex.17, see Appendix C.19. �

§ 19 Make a Matlab function that calculates ρwater = ρmaxec1τ1.3+c2τ

1.6for an ar-

bitrary temperature matrix t. The function syntax is water density 1bar(t) and theparameters are c = [0.00003932621, -0.00003868077], see also Ex.10.

W . Same trick as in Ex.17, see Appendix C.20. �

3. Iterative functions (algorithms)

The K-value method for solving 2-phase equilibria consists of solving the equationsyi = Kixi and (1 − α)yi + αxi = zi for all components i ∈ [1, n] in the mixture. Normally,yi and xi will be the mole fractions of component −i− in the vapor and liquid phases re-spectively, and α will be the mole fraction of the liquid phase (i.e. the number of molesin the liquid phase compared to the total numbers of mole present in both phases). Fid-dling around with the symbols it can be shown that everything boils down to solving onenon-linear equation in α:

f (α) =∑ zi (1 − Ki)α + (1 − α) Ki

= 0

After this equation has been solved the two mole fractions can be calculated explicitly:

xi =zi

α + (1 − α) Ki

yi =Ki

xi

§ 20 Write a function [x,y,l] = rachford rice(z,k) that calculates the componentmole fraction vectors x and y along with the phase mole fraction of the first phase(i.e. the phase from where the x is taken). On the input side z is the total compositionvector and k is the vector of K-values.

R-R. This problem is ideally suited for Newton’s method because it turnsout that the merit function is monotonically decreasing:

∂ f∂α= −

∑zi

(1 − Ki

α + (1 − α) Ki

)2

< 0

The implementation is therefore straight forward, see Appendix C.21. �

4. Recursion — good, evil or maybe bad?

A favorite example on recursive programming is the calculation of factorials. Thefactorial n! can obviously be programmed using a for-loop, but a “cleaner” alternative isthe recursive call procedure:

snippet from recursive factorial.m starts here

function [fac] = recursive_factorial(n)

if (n>0)

fac = n*recursive_factorial(n-1);

else

fac = 1;

end

4. RECURSION — GOOD, EVIL OR MAYBE BAD? 21

recursive factorial.m ends hereAnother favorite is the calculation of the Fibonacci sequence Fn

def= Fn−2 + Fn−1.

The Fibonacci numbers derived from F0 = 1 and F1 = 1, and other sequences builton the same principle, appears in surprisingly many fields of science like e.g. biology,number theory, architecture, optimization theory, chemistry, etc. The code listed in (re-cursive randmine) uses the Fibonacci sequence to generate random numbers given thetwo start values 0.8 and 0.0 and the imposed rule that all addition is modulo 1. Thismeans we shall not be interested in the magnitude of the number, just the fractional partof it.

snippet from recursive randmine.m starts here

function [x] = recursive_randmine(n)

if (n>2)

x = recursive_randmine(n-1) + recursive_randmine(n-2);

else

x = 0.6 + 0.2*n;

end

x = x - floor(x);

recursive randmine.m ends hereWhereas n! eats CPU-time in proportion to n, Fn is much more greedy. It is in fact

the worst kind of application you can possibly write because the CPU-time grows expo-nentially with n, look at Table 4.1 for details. Another interesting observation concerningthe modified Fibonacci sequence is its periodicity. Look carefully at the list generatedbelow,

n 1 2 3 4 5 6 · · · 19 20 21 22 23 24mod(Fn, 1) 0.8 0.0 0.8 0.8 0.6 0.4 · · · 0.6 0.2 0.8 0.0 0.8 0.8

and you will see that the sequence repeats itself after 20 calculations. In other words:F21 = F1, F22 = F2, and F23 = F3. Since the recursion needs only two members Fn−2

and Fn−1 to calculate Fn it is sufficient to observe the repetition of three numbers in or-der to prove that the entire sequence will repeat. Another feature is that the sequence isreproduced each time it is evaluated, irrespective of the programming language, the pro-grammer or the computer. This is the reason why random numbers produced by genericformulae are called pseudo-random. So, what has all this to do with chemistry students?A lot really, since random numbers are used heavily to:

(1) Replace observations of random nature, e.g. add “white” or “pink” noise to othersimulations in order to produce more realistic results.

(2) Test algorithms. I cannot possibly think of any better way of testing an algorithmthan to let it work on a massive set of random input data.

(3) In statistical theory like statistical mechanics, molecular dynamics and MonteCarlo methods.

(4) In optimization theory, i.e. “simulated annealing” techniques used in e.g. processoptimization.

(5) Make you work with individual problem solutions in this course.One comment about the last item: To make sure I will be able to verify your results(on midterm exams), we must use the same random number generator and I have tocontrol the “seed” you use to start the sequence. This is the reason you should implement“randolph” listed in Appendix C.1.

Not all programming languages open up for recursive definitions (e.g. FORTRAN-77), and the reason is very simple: If you forget to test the input arguments properly,the function will call itself forever! It is like executing the loop while(1>0) without

22 4. WRITE YOUR OWN FUNCTIONS

T 4.1: Calculation ofFibonacci-numbers using re-cursive function calls. Noteexponential growth in CPU-time.

n mod (Fn, 1) time[s]1 0.8 0.000...

......

14 0.2 0.02018 0.6 0.17122 0.0 1.10126 0.4 7.50030 0.8 51.38

breaks, but because things are less transparent when using function calls such ill-definedrecursiveness can be hard to debug.

§ 21 Solve the problem y = x2 using recursive programming. The function syn-tax is x = recursive sqrt(y,x0), where x0 is a user supplied start estimate for thealgorithm. Hint: The Newton method gives rise to 2xδx +

(x2 − y

)= 0. From this

scheme we can write an explicit update of xk+1 like this:

xk+1 =12

(yxk+ xk

)

S . One possible recursive update is x = recursive sqrt(y, 12

(yx + x

)),

and this really works! E.g. evaluating recursive sqrt(2,6.023e23) yields the correct an-swer 1.4142 . . . after 84 calls (the Avogadro number is a rather poor estimate of

√2). The

code from recursive sqrt is shown below: �

snippet from recursive sqrt.m starts here

function [x] = recursive_sqrt(y,x)

if (abs(x*x-y)>eps*y)

x = recursive_sqrt(y,(y/x+x)/2);

else

x = x;

end

recursive sqrt.m ends hereYou may wonder why you should bother about recursive definitions at all since the

applications we have met so far are rather trivial. My advice is: Recursive calculationis very nice if you are able to prove convergence a priori, or if you are interested in themathematical behavior of the solution, or if you are interested in compact definitions (thismay yield a tremendous increase in program readability if your algorithm is complex).Besides, recursiveness is in the heart of declarative (functional) programming. However,since function calls are much more expensive than arithmetic operations (+, i, ∗, /), mas-sive recursiveness should be avoided if the function overhead steals excessive CPU-time.

CHAPTER 5

Functionals or function(function)s

A functional in this context means a function that takes another function as input andproduces a number. Examples are: Definite integrals, minimization procedures, partialdiffentials, Taylor expansions, etc.

1. Integration

§ 22 The velocity profile in a circular duct is needed to calculate the averagevolumetric flow of gas. A series of pitot-tube measurements gave the following resultwhere the dynamic pressure is in mm fluid head:

hNS

hEW=

4.7 10.7 18.3 20.5 20.5 23.3 29.6 20.4 16.2 11.82.1 8.6 15.4 16.2 21.3 21.3 20.5 16.8 13.3 2.2

The gas has molecular weight 28 gmol−1, temperature 110 ◦C and pressure 1 bar.The density of the sealing fluid is 850 kgm−3. The cross-section is divided in fiveconcentric rings, all with the same area (four measurements in each ring). The totaldiameter of the duct is 1.6 m. Calculate the volumetric flow in m3/h using Simpson’sintegration rule, confer function quad.

P-. The determination of gas velocities in large ducts and chimneys is quiteregularly done with a pitot-tube. From several measurements at different radial positionsthe velocity profile is easily constructed. In this particular case we have measurementsin the North-South and East-West cross sections. If we average the measurements thevelocity profile can be based on polynomial approximation. We can e.g. assume laminarflow (very wrong!) and thereby ask for a parabolic velocity profile. Then a second orderpolynomial is correct and the function polyfit suffices. However, the velocities at theboundaries are not zero, see Appendix C.23! A better fit is obtained by increasing thenumber of parameters in the polynomial, but remember that higher order polynomials canoscillate violently (even inside the fitted region). The result may become terribly wrongalthough in the current example it is not too bad. Alternatively we could give up everyanalytical effort and go for a cubic spline interpolation. This time the velocity profilebecomes flat i.e. the flow is turbulent, but the spline will allways go through all the pointsso it is not possible to jugde how accurate the “fit” is. It may still be a useful tool to getan idea of shape of the profile, however. �

2. Ordinary differential equations

When Chemical Engineers face up to model and simulate a given system, it seemsthat a major concern is whether the ODE-solver will work. This turns into a nightmarewhen things go wrong and it is really disappointing when the computation crashes. Mostoften the problem is that we fully trust the “stupid” built-in code to do the whole job forus. In order to solve their simulation models, Chemical Engineers usually deal with ex-plicit first-order systems of differential equations (ODEs) with algebraic side constraints,

23

24 5. FUNCTIONALS OR FUNCTION(FUNCTION)S

turning the problem into a system of differential–algebraic equations (DAEs) of the form:f (dx/ dt, x, t, u, y) = 0 where x is the set of n differential variables, y is the set of m alge-braic variables, u is the set of k input variables, and f is the set of n + m DAE equations.The solution of such a problem requires two steps:

(1) Defining and solving for the initial state x(0) of the system at t = 0 (initial time).(2) Numerical integration of the DAEs from t = 0 to t (final time).

The second step is generally straightforward for simple systems. The challenge isto consistently initialize the DAE system which means to solve step 1 above. If notperformed well it may cause the system to diverge in step 2. However, a DAE systemmight be easier to deal with than a pure ODE system because the extra equations givedegrees of freedom we can use in the initialization. E.g. if the temperature evolution isconstrained by an energy balance it is easier to initialize T such that the energy balanceis fulfilled, than it is to solve the non-linear energy equation with respect to T as wouldbe required in a pure ODE set-up.

§ 23 Consider a gas A that dissolves into liquid B and diffuses isothermally in theliquid phase. During the diffusion A undergoes an irreversible first order homoge-nous reaction A + B → AB. If the mixture is treated as a binary solution, ignoringAB, the mass balance can be written:

DABd cA

d z2− k1cA = 0

This equation can be solved with the boudary conditions z = 0, cA = cA0 and z = L,d cA/ d z = 0. The equation can be rewritten into dimensionless variables

(5.1)d2Γ

dζ2− φ2Γ = 0

with boundary conditions ζ = 0, Γ = 1 and ζ = 1, dΓdζ = 0, and ζ = 0, Γ = 1 and ζ = 1,

Γ = 0. Solve the equation numerically and compare the solution you get to the exactsolution given by

(5.2) Γ = C1 cosh(φζ) + C2 sinh(φζ)

D. Any ODE function in matlab will produce the solution to this problemwith no problems. ODE functions can only solve (systems of) first order ordinary dif-ferential equations. This problem is of second order. This can easily be rewritten to asystem of two first order equations. y′′ + y′ + y = 1 can be written

u′1 = u2

u′2 = 1 + u1 + u2

Another consideration of this problem is the fact that the ODE solvers only support initialconditions. Here, one initial condition is given, and one boundary condition. In orderto satisfy the initial conditions one have to iterate as shown in Appendix C.22. Thediscrepancy between the numerical and analytical solution comes as a consequence ofthe boundary conditions not being satisfied completely. �

4. OPTIMIZATION WITH CONSTRAINTS 25

3. Unconstrained optimization

§ 24 Terje Hertzberg: Non-linear parameter estimation.

N- . Not implemented. �

4. Optimization with constraints

A thermodynamic equilibrium state is always optimal, i.e. it is either at a minimum ora maximum of the pertinent thermodynamic state function. Exactly which state functionwe are talking about depends on the constraints, and e.g. for ideal gas combustion it iscorrect to minimize Gibbs energy:

minn

G(T, p, n)

An = b.

Note that the minimization is subject to n only, keeping T and p constant during thecalculation. There are several methods available for solving linearly constrained minima,but my favorite is the Newton-Lagrange scheme which can be derived from the so-calledLagrange function

L = G −m∑

i=1

λi

n∑

j=1

Ai jNj − bi

= G − λT (An − b) .

By differentiation L with respect to Nk and λl at constant T, p, and putting the partialderivatives equal to zero, we end up with the following equations describing the thermo-dynamic equilibrium:

∂L∂n= µ − ATλ = 0

∂L∂λ= An − b = 0

These stationary conditions are finally solved by Newton’s method:

(5.3)

(∂µ∂nT AT

A 0

) (δn−δλ

)=

(ATλ − µb − An

)

§ 25 Write a function [n] = mingibb(n0, T, p, mu0, A) which solves Eq.5.3 itera-tively for an ideal gas system. The chemical potential vector of an ideal gas is

µig = µ◦ + RT log( PP◦

N−1n)

n =(

N1 N2 . . . Nn

)N =

∑Ni,

and the Hessian matrix:

(∂µ

∂nT

)igdef= Hig = RT

N−1

1. . .

N−1n

−RTN

I

26 5. FUNCTIONALS OR FUNCTION(FUNCTION)S

G . Once you master vector/matrix notation, the calculationsare straight forward, but you must remember to test for negative mole numbers. Thealgorithm will quite frequently take you into the unphysical region and the question iswhat to do then. Many things can be said about this topic, but it appears that in thisspecial case it will be sufficient to shorten the step to, say, a fraction 0.9 of the maximumallowable step, see Appendix C.25. �

5. Process control

Process control is an engineering discipline that deals with architectures, mecha-nisms, and algorithms for controlling the output of a specific process (from WikipediaThe Free Encyclopedia). Applications are found almost everywhere e.g. in the chemicalprocess industry, in living organisms and in robotics. Here we shall deal with one briefexample taken from the control of chemical reactors. The essential steps in developing acontrol structure include the following:

(1) Objectives: Study the physical behavior of the system and decide on the controlobjectives.

(2) Modeling: Build a mathematical model and simplify if necessary (either fromexperimental data using identification techniques, or from physical equationsdescribing the plant dynamics).

(3) Controllability: Decide which variables are to be controlled (identify the manip-ulated and measured variables and which links should be made between them).

(4) Structure: Decide on the measurements and manipulated variables (what sensorsand actuators will be used and where shall they be placed).

(5) Design: Select the control configuration and decide on the type of controller tobe used (formulate a mathematical design problem which captures the engineer-ing design and synthesize a corresponding controller).

(6) Analysis: Decide on performance specifications, based on the overall controlobjectives (assess the control system by analysis and simulation against the per-formance specifications or the designers expectations).

(7) Implementation: Simulate the resulting controlled system (either on a computeror a pilot plant).

In the following example we use steps one to four to study a simple reactor controller.

§ 26 A continuous stirred tank reactor (CSTR) has variable feed qin and variableheat load Q. In the stationary state the incoming fluid has temperature T�in = 350 K,concentration C�A,in = 2 kmol m−3 and volume flow q�in = 1 m3 min−1. The correspond-ing heat load and reactor temperature are Q� = 5000 kJ min−1 and T� = 400 Krespectively. The (only) first order reaction taking place is A→ B. The reaction rater = kCA follows the Arrhenius law

k = k◦e−ER

(1T − 1

T◦)

where k◦ = 0.5 min−1, E = 20 kJ mol−1, R = 8.31453 J mol−1 K−1 and T◦ = 400 K. Thetask is to control the output concentration CA using qin and Q as control variables.The temperature Tin and concentration CA,in are disturbances to the process. Thereis no B in the feed. Other (constant) data are: reactor volume V = 4 m3 fluid densityρ = 1000 kg m−3 and heat capacity CP = 0.15 kJ kg−3 K−1.

5. PROCESS CONTROL 27

CSTR. The Matlab code in Appendices C.26, Appendices C.27 and Appendices C.28produces four plots. The steady state is first verified (plot 1) before a series of dynamicresponses to some step changes and disturbances are calculated (plot 2). The model isthen linearized and the state-space matrices A, B, C, D, E and F are calculated alongwith the controllability, the observability and the stability of the model. Finally, the codegenerates the transfer function and calculates its poles and zeros, and the RGA in differentfrequencies (plot 3 and 4).

�

CHAPTER 6

String parsing

1. Chemical formula arithmetics

Consider a simple chemical formula like H2O. What does it tell us? Of course youknow the answer; take two atoms of hydrogen and one atom of oxygen, combine andyou have water. However, I am quite sure you have never reflected over how close thechemical formula is an arithmetic expression! E.g. the expansion

H2O→ H ∗ 2 + O

is a valid expression for the molecular weight of water (assuming that H and O are con-stants defined as the molecular weight of hydrogen and oxygen respectively). This ideacan be carried further, e.g. the expansion

Ca (NO3)2 (H2O)4 → Ca + (N + O ∗ 3) ∗ 2 + (H ∗ 2 + O) ∗ 4

is a valid expression for the molecular weight of calcium nitrate. The whole idea isto identify tokens in the formula and insert arithmetic operators when required. In thesimple examples above we have to distinguish five tokens

0 − 9 a − z A − Z ) (

and two operators+ ∗

In addition we must obey three rules which are specific to chemical formulas: Elementnames contain at most one lower case letter, which cannot be the first letter in the name,and empty parentheses are not allowed in the formula. The full operator table is shownin Table 6.1. It tells you which operator to insert given a left and right operand. Parsinge.g. the input string ’H2O’ from left to right yields the following sequence:

H2Ole f t=H,right=2→ H ∗ 2O

le f t=2,right=O→ H ∗ 2 + O

Note that the two operators + and ∗ are what we call binary operators, i.e. they are of theform x ◦ y where x is the left operand and y is the right operand. Hence, it is sufficient toanalyse just two characters at a time!

T 6.1: Operator table for converting chemical formulasinto arithmetic expressions.

rightleft

0 − 9 a − z A − Z ) (

0 − 9 none + + none +

a − z ∗ error + none +

A − Z ∗ none + none +

) ∗ error + none +

( + + none error none

29

30 6. STRING PARSING

§ 27 Write a function expr = mw from formula parser(formula) which returns avalid arithmetic expression from a given input string (chemical formula). E.g. thefunction call expr = mw from formula parser(’H2O’) shall return the string expr =’H*2+O’. Make use of the function op = mw binop(x,y) in your code.

C . No big surprises here, except that I for some reason(don’t remember exactly why) implemented the parser as RL and not LR. Anyway, theMatlab code is nice and clean (Appendix C.29). �

§ 28 Write a function op =mw binop(x,y) that analyses the two operands x (left)and y (right) and returns the correct operator from Table 6.1 (as a string).

C . There are many ways to implement this operatortable and the one shown in Appendix C.30 is definitely not the most compact. However,legibility is also an important issue! �

In the next exercise you will benefit from knowing the function [s,b] = get formulavector(s, b, e, f). This function takes an input string s, a vector of stoichiometric factors bwhich accompanies the list of known element symbols e, and an external multiplicativefactor f . The function syntax you should use is get formula vector(s, zeros(length(n),1), e, 1.0). It will parse s from right to left and try to resolve the stoichiometric factorscorresponding to the supplied list of element symbols. The parsing continues till a rightparanthesis is found, or till the string is exhausted. If a right paranthesis is found getformula vector calls itself recursively till the string s is exhausted, so in your case s shallallways be an empty string on output.

§ 29 Write a function [A,e] = make formula matrix(formula1,formula2,. . .) thatanalyses an arbitrary number of chemical formulas and calculates the corespondingformula matrix A, and a list of chemical symbols e representing the rows of A.

F . I have elaborated the interface a little in Appendix C.31 to makethe function more flexible, but the biggest job was to get the chemical symbols correct(it turned out that the chemical symbol of Lawrencium had changed from Lw to Lr sincethe time I was a student). �

§ 30 Last but not least, try to program the function [s,b] = get formula vector(s,b, e, f) as it was introduced above.

F . This is a fairly big programming job where you will need recursiveprogramming and ASCII-encoding. To make the code more versatile I have in AppendixC.32 opened up for decimal numbers in the chemical formula. E.g. ’FeO0.977’ is a legalinput string. �

2. Cool stuff (regular expressions)

Consider the situation where you have a huge amount of information with some(known) structure but are only interested in a small part of the data. Typically you maybe interested in extracting information from an HTML file (garbage filtering) or to sum-marise the output from a big simulation run into a small table in Excel (report generation).

2. COOL STUFF (REGULAR EXPRESSIONS) 31

In order to retain the data of interest, regular expressions are very powerful. Table 6.2shows a list of the most common expressions.

T 6.2: The most common regularexpressions

Symbol Meaningˆ Start of string$ end of string. any character\ quote next character∗ match zero or more+ match one or more? match zero or one[] match set of characters[ˆ] exclude set of characters() group subexpression\w match word [a-z A-Z0-9]\W not a word [ˆa-z A-Z0-9]\d match digit [0-9]\D not a digit [ˆ0-9]\s match white space\S not a white space| logical “or”

Assume we want to count the number of sentences in a text document. This can bedone using the regular expression [A-Z][ˆ.|ˆ!|ˆ?]+[.|!|?] on a string representation of thedocument. The search will match everything which starts with a capital letter and endswith . or ! or ? The end of the matching sub-string is used to define the start of the nextsearch till the entire string is fully explored.

§ 31 Calculate the CO2 equivalents for the vapor streams V1, V2 and V3 re-ported in the data file HYSYS report.txt. The report format is shown in AppendixE.4. Note that the HYSYS file contains much more data than what is needed tosolve the task. Regular expressions are useful to do the filtering job, but the Matlabimplementation of RE is rather ackward. It would of course be smarter to filter thedata in another language with gives a better support for RE, but we shall stick toMatlab here because “to know a language you must learn about its limits”.

G . First, the vapour streams in the output file from HYSYSmust be located. It is then possible to filter out the stream data needed and calculatethe CO2 equivalents. But, as you can see from Appendix C.33 the code is quite involved.The job is still worth the efforts if there are lots of data to be processed. Ideally, however,you should consider a language which is better suited for string handling and regular ex-pressions like Ruby or Python. They yield less code which at the same time is easier towrite and maintain, see Appendix D.1 for an implementation in Ruby. �

APPENDIX A

Language elements

T A.1: File functions.

fopen open filefclose close file

T A.2: Special variables of mathe-matical character.

eps machine precision εans last calculated result −pi pi πinf infinity ∞NaN Not-a-Number −

T A.3: Special characters.

% comment (rest of line is ignored). . . statement continues on next line

T A.4: Basic commands that makes life easier, buthas little to do with programming.

pwd print working directorywhos list all variables in the work spacewhat list all functions in the local directorydisp display text on screenexit exit the Matlab sessionhelp give help topicsload read table from filesave save table to fileformat change output formatcd change local directory

33

34 A. LANGUAGE ELEMENTS

T A.5: Program flow control instructions.

if;elseif;else;end declare conditional execution block(s)while;end repeat execution block while logical argument is truefor start:step:stop; end repeat statements a specific number of timesbreak break innermost execution blockreturn break function executionpause suspend executionpersistent save variables in local namespacefunction declare functionfeval evaluate functionalkeyboard halt execution while reading from keyboard

T A.6: Instantiation operators. Used when shaping or filling a ma-trix, a vector, etc.

[] vector (matrix) constructor, row separator; column separator= assignment operator′ vector or matrix transpositionend last row or column elementfind evaluate logical argument for all vector/matrix elements

T A.7: Plotting commands.

loglog logarithmic plottitle graph titleylabel y-axis labelxlabel x-axis labelaxis control axes rangesplot linear plothold hold current graph for additional plottinggrid add grid lineslegend place graph legends on the current plot

T A.8: Relational operators (logical operations).

eq;ne;gt;lt;ge;le numerical test operators =,�, >, <,≥,≤and;or;not logical (boolean) test operators ∨,∧,¬any equivalent to sum(arg > 0) > 0 ∃all equivalent to prod(arg > 0) > 0 ∀isnan true if argument is NaN −isinf true if argument is∞ −isempty true if empty arrayiscell true if cell arraysort sort in ascending order

A. LANGUAGE ELEMENTS 35

T A.9: Algebraic operators.

+;− vector or matrix addition (subtraction)∗ vector or matrix multiplication/ right matrix division (dangerous)\ left matrix division (dangerous).∗ element-by-element product./ element-by-element division.ˆ element-by-element exponentiation

T A.10: Set functions.

mod modulus divisionmax column maxiummin column minimumround round towards nearest integerfloor round towards −∞ceil round towards +∞abs absolute valuerand random (uniform distribution) number

T A.11: Transcendental functions.

exp exponentiationlog natural logarithmsin sinecos cosinesqrt square root

T A.12: String functions.

char create character array (string)num2str convert number to stringstrcat catenationstrcmp compare stringsupper convert to upper caselower convert to lower casestrtok find token in stringfindstr find substring in string

36 A. LANGUAGE ELEMENTS

T A.13: Linear algebra functions.

eye identity matrixdiag diagonal matrixsize vector/matrix dimensionlinspace vector of equidistant pointslogspace vector of logarithmic equidistant pointsones vector/matrix filled with 1’szeros vector/matrix filled with 0’sreshape reshape vector/matrixsum column sum of vector/matrixprod column product of vector/matric

T A.14: Advanced linear algebra functions.

inv matrix inversionlu LU-factorizationqr QR-factorizationrref matrix on row reduced echelon formnorm vector normeig matrix eigenvaluesrank matrix rank

T A.15: Functionals i.e. functions that take other functions asarguments and produce a number.

fzero solve non-linear functionsquad Simpson integrationode45 4’th order Runge-Kutta integrationrbvp4c 4’th order boundary value integrationoptimset optimization data structurefsolve nonlinear solver using the method of least squares

APPENDIX B

Examples

In Matlab a command can be entered at the system prompt without any left assigment.It will then return the numerical value of the operation in the default variable ans (so thereis an assigment after all). Some trivial examples follow:

>> mod(6, 4)

ans = 2

>> floor(3.95)

ans = 3

>> ceil(3.95)

ans = 4

>> isempty([])

ans = 1

>> char([′a′, ′b′, ′c′])ans = abc

>> num2str([1, 2, 3])

ans = 1 2 3

>> sort([2, 1,−4, 0])

ans = −4 0 1 2

Matrix inversion is simple in Matlab, but note that direct inversion is not much used —there are more efficient ways to solve a system of linear equations:

A = inv([2, 1; 1, 2])

A =

(0.6667 −0.3333−0.3333 0.6667

)

LU-factorization is typically used in the first stage of a solution. It holds the outcome ofan ordinary Gauss elimination of the rows of the matrix:

[A, B] = lu([2, 1; 1, 2])

A =

(1.0000 0.00000.5000 1.0000

)

B =

(2.0000 1.00000.0000 1.5000

)

37

38 B. EXAMPLES

QR-factorization is similar to LU but has some nice additional properties (orthogonalQ-matrix):

[A, B] = qr([2, 1; 1, 2])

A =

( −0.8944 −0.4472−0.4472 0.8944

)

B =

( −2.2361 −1.78890.0000 1.3416

)

The row reduced echelon (staircase) form of a matrix properly identifies linear depen-dencies:

[A, B] = rref([2, 1; 1, 2])

A =

(1.0000 0.00000.0000 1.0000

)

B =(

1 2)

The 2-norm is a measure for the Euclidian length of a vector. It is defined as the squareroot of the inner product of the vector with itself:

a = norm ([3, 4])

a = 5

Matrix eigenvalues and eigenvectors are important in the stability analyses in processcontrol, differential equation solving and thermodynamic equilibrium theory:

[A, B] = eig([2, 1; 1, 2])

A =

( −0.7071 0.70710.7071 0.7071

)

B =

(1.0000 0.00000.0000 3.0000

)

The matrix rank is important for the solvability of an equation system. Matrices with fullranks yield unique solutions while rank deficient matrices do not:

a = rank ([2, 1; 1, 2])

a = 2

APPENDIX C

Matlab code

1. randolph

Randolph is a simple, yet effective random number generator for internal uses in thiscourse. The rand function in Matlab yields far better randomness, but is more difficultto control by simple means (it used to be, but MathWorks has changed it — grrrr). Withemphrandolph you can set the initial state of the generator by simply issuing the com-mand randolph(n), where n ≥ 1 is a random number (called the seed) of your choice.The seed is much like a password, you must remember it in order to reproduce your cal-culations! After randolph has been initialized you can issue the command randolph(n,m)to generate n × m matrices of random numbers.

%The program implements a lagged Fibonacci pseudo-random number generator

%using subtraction [0]. A sequence of random numbers X(n) is obtained from

%the formula:

%

% X(i) = X(i-q) - X(i-p) mod M

%

%where M is 1 if the X’s are taken to be floating point reals in [0,1) as

%they are here. The period of this generator is (2ˆp-1)*2ˆ(b-1),where p is

%the largest lag and b is the number of significant bits in the numbers

%X(n).The lag needs to be chosen large enough to give a period much longer

%than the number of random numbers to be generated.Only certain lags (p,q)

%will give the maximal period of the generator [4, 5]. Some of these are

%listed below:

%

% p q

% 9689 4187

% 4423 2098

% 2281 1029

% 1279 418

% 607 273

% 521 168

% 250 103

% 127 63

% 97 33

% 55 24

% 31 13

% 17 5

%

%Small lags can cause problems with this generator [1, 2, 5]. Larger lags

%give larger periods and empirically seem to give "better" random numbers

%[2, 4, 5]). I recommend using at least (1279,418),and ideally much larger

%lags such as (9689,4187). Larger lags give higher quality random numbers

%and longer period, with the only tradeoff being increased memory use and

%initialization time. However initialization is very fast, and the memory

39

40 C. MATLAB CODE

%used is only of order p words, so p=2281 will add less than 10 Kbytes to

%the memory required for the user program. This generator performs well in

%empirical statistical tests [2, 4] as long as a large lag is used. This

%does not necessarily mean it will perform well for your application.It is

%recommended that you check your results by also running your program

%using another good random number generator (see [1] for recommendations).

%

%0. http://www.npac.syr.edu/projects/random/generators/fibadd/fibadd.html

%1. P. D. Coddington, Random number generators for parallel computers,

% NHSE Review, Volume 2, 1996.

%2. P. D. Coddington, Analysis of Random Number Generators Using Monte

% Carlo Simulation, Int. J. Mod. Phys. C 5, 547 (1994). NPAC technical

% report SCCS-526.

%3. F. James, A Review of Pseudo-random Number Generators,

% Comput.Phys.Comm. 60, 329 (1990).

%4. D. E. Knuth, The Art of Computer Programming Vol. 2: Seminumerical

% Methods, Addison-Wesley, Reading, Mass., 1981.

%5. G. A. Marsaglia, A current view of random number generators, in

% Computational Science and Statistics: The Interface,

% ed. L. Balliard, Elsevier, Amsterdam, 1985.

function [x] = randolph(n,m)

persistent x0 o p q;

if (nargin==1)

if (n<1)

disp(’Your seed number must be >= 1, try again!’);

return;

end

x0 = [0.52189398762154 0.22586125989448 0.78962235300517 ...

0.73889796980862 0.61383531911974 0.13010525023880 ...

0.64626222776828 0.22126297192155 0.92287218663571 ...

0.58354708643659 0.36750946939959 0.31286745921603 ...

0.59021783397155 0.05315767833273 0.37021959254501 ...

0.88726820502120 0.97633988457881 0.76793407983075 ...

0.24525730723827 0.91975917519955];

x0 = x0*n; % start modifying the seed vector

x0 = x0 - floor(x0); % calculate the new seed

o = max(size(x0)); % dim(x0)

p = 17; % 1st random parameter. Note that dim(x0) >= p

q = 5; % 2nd random parameter

else

if (isempty(x0))

disp(’You must initialize first, use randolph(n) where n>=1’);

return;

end

x = [x0,zeros(1,n*m)]; % initialize random vector

for i=1:n*m % repeat for all elements

x(o+i) = x(o+i-p) - x(o+i-q); % Fibonacci sequence

x(o+i) = x(o+i) - floor(x(o+i)); % Calculate random numbers

end

2. OCTANE 41

x0 = x(end-o+1:end); % update the seed vector

x = reshape(x(o+1:end),n,m); % reshape to requested format

end

2. octane

clear;

format short;

prompt = input(’Normal or bad case? [n/b]: ’,’s’);

if (strcmp(prompt,’b’))

A = [90 105 95; 1-1e-15 1 1; 1 1 1];

disp(’Bad coefficient matrix:’)

else

A = [90 105 95; 0.2 2 0.5; 1 1 1];

disp(’Normal coefficient matrix:’)

end

disp(A);

pause;

b = [98; 1; 1];

disp(’Right hand side vector (specifications):’)

disp(b);

pause;

x = zeros(size(b));

B = [A,b];

disp(’Adjoint matrix:’)

disp(B);

pause;

B(2,:) = B(2,:) - (B(2,1)/B(1,1))*B(1,:);

disp(’Element 2,1 eliminated:’)

disp(B);

pause;

B(3,:) = B(3,:) - (B(3,1)/B(1,1))*B(1,:);

disp(’Element 3,1 eliminated:’);

disp(B);

pause;

B(3,:) = B(3,:) - (B(3,2)/B(2,2))*B(2,:);

disp(’Element 3,2 eliminated:’)

disp(B);

pause;

x(3) = B(3,4)/B(3,3);

disp(’Solution variable 3:’)

disp(x(3));

pause;

x(2) = (B(2,4)-B(2,3)*x(3))/B(2,2);

disp(’Solution variable 2:’)

42 C. MATLAB CODE

disp(x(2));

pause;

x(1) = (B(1,4)-B(1,3)*x(3)-B(1,2)*x(2))/B(1,1);

disp(’Solution variable 1:’)

disp(x(1));

pause;

format long e;

disp(’The solution vector compared to Matlab:’)

disp([x,inv(A)*b]);

pause;

disp(’Back calculated results compared to Matlab:’)

disp([A*x,A*(inv(A)*b)]);

3. figure 2 8

clear;

I = eye(2,2);

O = zeros(2,2);

a = [2 , 1];

s1 = [99, -1];

s2 = [10,-90];

o = O(1,:);

e = I(1,:);

b = [101; 0; 2; 0; 0; o’; 0; o’];

A = [I, O, O, O , O ; ... % specification of stream 1

o, o, o, o , e ; ... % specification of A in stream 5

o, o, o, o , s1; ... % specification of top product

o, o, o, s2, o ; ... % specification of bottom product

I,-I, O, I , O ; ... % component balance (mixer)

o, a,-a, o , o ; ... % atom balance (reactor)

O, O, I,-I ,-I ]; % component balance (separator)

[B,j] = rref([A,b]); % stair case (exchelon form)

xp = zeros(size(b)); % initialize

xp(j) = B(1:length(j),end); % particular solution

xh = null(A); % homogenous solution

n = 200; % number of plot points

X = xp*ones(1,n) - xh*linspace(0,2000,n); % n solution vectors

s3 = [5,6]; % stream 3

s4 = [7,8]; % stream 4

n3 = sum(X(s3,:),1); % total flow of stream 3

n4 = sum(X(s4,:),1); % total flow of stream 4

Q = X(s3(2),:).ˆ2./(X(s3(1),:).*n3); % equilibrium quotient

loglog(n4,Q,’r’);

5. FERTILIZER BALANCE 43

title(’Equilibrium restricted reaction’)

ylabel(’x_{B}ˆ{2}/x_{A}’)

xlabel(’Resirculation flow [mol sˆ{-1}]’)

axis([10 1000 0.05 10]);

hold on;

grid;

K = 0.74; % equilibrium constant

i = find(Q<K); % possible states

loglog(n4(i),Q(i),’g’);

legend(’Q>K’,’Q<K’);

hold off;

disp(round(xp-1568.44*xh));

4. partialinv

%Purpose : Solve the partial inversion problem Ax = b where some of the

% elements in b are unknown, while an equal number of elements in

% x are known (a total of rank(A) elements in x and b must be

% known all over)

%Author : Tore Haug-Warberg

%Date : October 16 1999

%Syntax : [x,b] = partialinv(A,x,b)

%Input : A = coefficient matrix

% x = left hand side vector

% b = right hand side vector

%Output : x = updated left hand side vector

% b = updated right hand side vector

%External: eye inv isnan not size

function [x,b] = partialinv(A,x,b)

i = isnan(x);

j = isnan(b);

I = eye(size(A));

c = inv([I(not(i),:);A(not(j),:)])*[x(not(i));b(not(j))];

x(i) = I(i,:)*c;

b(j) = A(j,:)*c;

5. fertilizer balance

clear;

% Reactant formula matrix (in)

% NH3 O2 Ca3(Po4)2 CaCO3 H2O CO2

A = [1 0 0 0 0 0; ... % N

3 0 0 0 2 0; ... % H

0 2 8 3 1 2; ... % O

0 0 0 1 0 1; ... % C

0 0 3 1 0 0; ... % Ca

44 C. MATLAB CODE

0 0 2 0 0 0]; ... % P

NH3 = 1; % column number in A belonging to NH3

O2_in = 2; % column number in A belonging to O2

phosphate = 3; % column number in A belonging to Ca3(PO4)2

calcite = 4; % column number in A belonging to CaCO3

H2O_in = 5; % column number in A belonging to H2O

CO2_in = 6; % column number in A belonging to CO2

% Product formula matrix (out)

% CN MAP CO2 AN O2 H2O

B = [2 1 0 2 0 0; ... % N

8 6 0 4 0 2; ... % H

10 4 2 3 2 1; ... % O

0 0 1 0 0 0; ... % C

1 0 0 0 0 0; ... % Ca

0 1 0 0 0 0]; ... % P

CN = 1; % column number in B belonging to Ca(NO3)2

MAP = 2; % column number in B belonging to NH4H2PO4

CO2_out = 3; % column number in B belonging to CO2

AN = 4; % column number in B belonging to NH4NO3

O2_out = 5; % column number in B belonging to O2

H2O_out = 6; % column number in B belonging to H2O

%Molecular weights

w = [14.0067 1.00797 15.9994 12.01115 40.08, 30.9738]; % atoms

wa = w*A; % reactants

wb = w*B; % products

x = zeros(size(wa))’;

y = zeros(size(wb))’;

x(NH3) = 1300/wa(NH3); % NH3 capacity is 1300 ton/day

x(O2_in) = NaN; % unknown O2 consumption

x(phosphate) = NaN; % unknown phosphate consumption

x(calcite) = NaN; % minimize calcite input

x(H2O_in) = 100/wa(H2O_in); % minimize H2O input

x(CO2_in) = 0; % minimize CO2 input

y(CN) = 1000/wb(CN); % 1600 ton/day CN production

y(MAP) = 200/wb(MAP); % MAP capacity is 500 ton/day

y(CO2_out) = NaN; % unknown CO2 production

y(AN) = NaN; % unknown AN production

y(O2_out) = 0; % zero O2 production

y(H2O_out) = NaN; % unknown H2O production

[x,y] = partialinv(inv(B)*A,x,y); % solve the mole balance

disp(’Case I’); % show title

disp([diag(wa)*x,diag(wb)*y]); % show mass flows in ton/day

x(NH3) = 1300/wa(NH3); % NH3 capacity is 1300 ton/day

6. ACID BASE REACTION MATRIX 45

x(O2_in) = NaN; % unknown O2 consumption

x(phosphate) = NaN; % unknown phosphate consumption

x(calcite) = 0; % minimize calcite input

x(H2O_in) = 0; % minimize H2O input

x(CO2_in) = 0; % minimize CO2 input

y(CN) = NaN; % unknown CN production

y(MAP) = 200/wb(MAP); % MAP capacity is 500 ton/day

y(CO2_out) = NaN; % unknown CO2 production

y(AN) = NaN; % unknown AN production

y(O2_out) = 0; % zero O2 production

y(H2O_out) = NaN; % unknown H2O production

[x,y] = partialinv(inv(B)*A,x,y); % solve the mole balance

disp(’Case II’); % show title

disp([diag(wa)*x,diag(wb)*y]); % show mass flows in ton/day

6. acid base reaction matrix

clear;

format short;

format compact;

%Let’s first put up the formula matrix (identical to the atom matrix if

%the B, H, A, OH symbols are interpreted differently).

% B BH+ HA A- H2O OH- H+

A = [1 1 0 0 0 0 0; ... % B

0 1 1 0 1 0 1; ... % H

0 0 1 1 0 0 0; ... % A

0 0 0 0 1 1 0]; % OH

name = {’B’,’BH+’,’HA’,’A-’,’H2O’,’OH-’,’H+ ’}’;

disp(’Formula matrix:’)

disp(A);

pause;

A(1,:) = A(1,:) - (A(1,2)/A(2,2))*A(2,:);

disp(’Element 1,2 eliminated:’)

disp(A);

pause;

A(1,:) = A(1,:) - (A(1,3)/A(3,3))*A(3,:);

disp(’Element 1,3 eliminated:’)

disp(A);

pause;

A(2,:) = A(2,:) - (A(2,3)/A(3,3))*A(3,:);

disp(’Element 2,3 eliminated:’)

disp(A);

pause;

46 C. MATLAB CODE

A(1,:) = A(1,:) - (A(1,5)/A(4,5))*A(4,:);

disp(’Element 1,5 eliminated:’)

disp(A);

pause;

A(2,:) = A(2,:) - (A(2,5)/A(4,5))*A(4,:);

disp(’Element 2,5 eliminated:’)

disp(A);

pause;

tmp = A(:,5);

A(:,5) = A(:,4);

A(:,4) = tmp;

tmp = name{5};

name{5}= name{4};

name{4}= tmp;

disp([’Swap components ’,name{4},’ and ’,name{5}])

disp(A);

pause;

N(:,1) = [A(:,5);-1;0;0];

disp(’First set of reaction coefficients:’)

disp([char(name),abs(zeros(length(name),3)),num2str(N)]);

pause;

N(:,2) = [A(:,6);0;-1;0];

disp(’Second set of reaction coefficients:’)

disp([char(name),abs(zeros(length(name),3)),num2str(N)]);

pause;

N(:,3) = [A(:,7);0;0;-1];

disp(’Third set of reaction coefficients:’)

disp([char(name),abs(zeros(length(name),3)),num2str(N)]);

pause;

N(:,2) = -N(:,2);

N(:,1) = -N(:,1);

disp(’Change direction of reaction 1 and 2:’)

disp([char(name),abs(zeros(length(name),3)),num2str(N)]);

7. water viscosity 1bar fit

clear;

K2C = 273.15;

load -ascii ..\data\water_viscosity_1bar_data.txt

t = water_viscosity_1bar_data(:,1) + K2C;% experimental temperatures

visc = water_viscosity_1bar_data(:,3); % experimental viscosities

e = ones(size(t)); % unity vector

[t,i] = sort(t); % sort in ascending order

visc = visc(i); % re-order viscosities as well

A = [e e./t e./(t.ˆ2)]; % coefficient matrix

8. WATER VAPOR PRESSURE FIT 47

c = inv(A’*A)*A’*log(visc); % least square solution

%Generate (overwrite) the viscosity function "on the fly"

ans = input(’Overwrite old ’’water_viscosity_1bar.m’’ y/n [n]: ’,’s’);

if ans==’y’

out = 0;

func = ’water_viscosity_1bar’;

mfil = [func,’.m’];

format = ’%s\r\n’;

fid = fopen(mfil,’w’);

out = out + fprintf(fid,format, ...

[’%Viscosity of water calculated from ’, ...

’log(visc)=c(1)+c(2)/T+c(3)/Tˆ2’]);

out = out + fprintf(fid,format, ...

’%Viscosity in [kg/m*s], temperature in [K].’);

out = out + fprintf(fid,format, ...

[’ function [visc] = ’,func,’(t);’]);

out = out + fprintf(fid,format, ...

[’ c = [’,num2str(c’,’%15.11g’),’];’]);

out = out + fprintf(fid,format, ’ e = ones(size(t));’);

out = out + fprintf(fid,format, ...

’ visc = exp(c(1)*e+c(2)*e./t+c(3)*e./(t.ˆ2));’);

fclose(fid);

disp([num2str(out),’ bytes of code written to ’,mfil]);

end

%Show results from parameter fitting

plot(t-K2C,visc,’*’,t-K2C,exp(A*c));

title(’Viscosity of water at 1 bar’);

xlabel(’Temperature [C]’);

ylabel(’Viscosity [kg/m*s]’);

grid;

8. water vapor pressure fit

clear;

load -ascii ..\data\water_vapor_pressure_data.txt; % load ASCII-file

%Define critical constants and fit experimental data to Wagner’s equation

tc = 647.1; % critical temperature [K]

pc = 220.64e5; % critical pressure [Pa]

t = water_vapor_pressure_data(:,1); % observed temperatures [K]

p = water_vapor_pressure_data(:,2)*1e6; % observed pressures [Pa]

[t,i] = sort(t); % temperatures in increasing order

p = p(i); % rearrange p in the same order

i = find(t>tc); % find temperatures outside range

t(i) = []; % remove them (avoid trouble)

p(i) = []; % rearrange p in the same order

e = ones(size(t)); % unity vector

tr = t/tc; % reduced temperature

pr = p/pc; % reduced pressure

A = [e-tr,(e-tr).ˆ1.5,(1-tr).ˆ3,(e-tr).ˆ6]; % coefficient matrix

b = tr.*log(pr); % right hand side

48 C. MATLAB CODE

%c = inv(A’*A)*A’*b; % solve normal equations

[Q,R] = qr(A); % QR-decomposition

c = inv(R’*R)*A’*b; % more stable solution

%Generate (overwrite) the vapor pressure function "on the fly"

ans = input(’Overwrite old ’’water_vapor_pressure.m’’ y/n [n]: ’,’s’);

if ans==’y’

out = 0;

func = ’water_vapor_pressure’;

mfil = [func,’.m’];

format = ’%s\r\n’;

fid = fopen(mfil,’w’);

out = out + fprintf(fid,format,[’%Vapor pressure of water’, ...

’ using Wagner’’s equation.’]);

out = out + fprintf(fid,format,[’%Pressure in [Pa],’, ...

’ temperature in [K].’]);

out = out + fprintf(fid,format,[’ function [lnp] = ’,func,’(t);’]);

out = out + fprintf(fid,format,[’ tc = ’,num2str(tc),’;’]);

out = out + fprintf(fid,format,[’ pc = ’,num2str(pc),’;’]);

out = out + fprintf(fid,format,[’ c = [’,num2str(c’,’%15.11f’),’];’]);

out = out + fprintf(fid,format, ’ e = ones(size(t));’);

out = out + fprintf(fid,format, ’ tr = t/tc;’);

out = out + fprintf(fid,format,[’ tmp = c(1)*(e-tr) + c(2)*(e-tr).ˆ1.5’,...

’ + c(3)*(e-tr).ˆ3 + c(4)*(e-tr).ˆ6;’]);

out = out + fprintf(fid,format, ’ lnp = tmp./tr + log(pc)*e;’);

fclose(fid);

disp([num2str(out),’ bytes of code written to ’,mfil]);

end

%The experimental data span the entire range from the triple point to the

%critical point, which means it will be quite pointless to show the fit in

%an ordinary 1/T,lnP-plot. The data are therefore transformed into the

%deviation plot 1/T,dlnP, where dlnp = log(p) - [a + b(1/Tmin-1/T)], and

%a = log(Pmin), and b = (log(Pmax)-log(Pmin))/(1/Tmin-1/Tmax).

tmin = min(t); % mininum temperature

tmax = max(t); % maximum temperature

lnpmin = min(log(pr)); % minimum log(reduced pressure)

lnpmax = max(log(pr)); % maximum log(reduced pressure)

a = lnpmin; % 1 parameter in linearized plot

b = (lnpmax-lnpmin)/(1/tmin-1/tmax);% 2 parameter in linearized plot

lnplin = a*e + b*(e/tmin-e./t); % linearized 1/T,lnP-plot

dpobs = log(pr) - lnplin; % observed deviations

dtnorm = (e./t - e/tmin)/(1/tmax-1/tmin); % normalized 1/T data

dpcalc = (A*c)./tr - lnplin; % calculated pressure devations

plot(dtnorm,dpobs,’o’,dtnorm,dpcalc);

title(’Vapor pressure of water (Wagner’’s equation)’)

xlabel(’Normalized 1/T (from triple to critical point)’)

ylabel(’Transformed log(p) - [a+b(1/Tmin-1/T)]’)

grid;

pause;

9. WATER DENSITY 1BAR FIT 49

plot(t,exp((A*c)./tr)./pr-e);

title(’Vapor pressure of water (Wagner’’s equation)’)

xlabel(’Temperature [K] (from triple to critical point)’)

ylabel(’Relative pressure deviation’)

grid;

9. water density 1bar fit

clear;

%Maximum density of air-free water is approx. 999.97490+/-0.0001 kg mˆ{-3}

%at 3.98 degree Celsius. Interpolated from Takenaka,M.,Masui,R.Metrologia,

%27, 165-171, (1990).

K2C = 273.15;

dmax = 999.97490;

tmax = K2C + 3.98;

load -ascii ..\data\water_density_1bar_data.txt

t = water_density_1bar_data(:,1); % experimental temperatures

d = water_density_1bar_data(:,3); % experimental densities

e = ones(size(t)); % unity vector

[t,i] = sort(t); % sort in ascending order

d = d(i); % re-order the densities as well

dt = t - tmax*e; % make sure max. temperature is right

A = [abs(dt).ˆ1.3,abs(dt).ˆ1.6]; % coefficient matrix

c = inv(A’*A)*A’*log(d/dmax); % least square solution

%Generate (overwrite) the density function "on the fly"

ans = input(’Overwrite old ’’water_density_1bar.m’’ y/n [n]: ’,’s’);

if ans==’y’

out = 0;

func = ’water_density_1bar’;

mfil = [func,’.m’];

format = ’%s\r\n’;

fid = fopen(mfil,’w’);

out = out + fprintf(fid,format, ’%Density of water (1 bar) calculated’);

out = out + fprintf(fid,format, ’%from log(rho) = tauˆ1.3 + tauˆ1.6,’);

out = out + fprintf(fid,format, ’%where rho is d/dmax and tau is ’);

out = out + fprintf(fid,format, ’%t-tmax. Densities in [kg/mˆ3] and’);

out = out + fprintf(fid,format, ’%temperatures in [K].’);

out = out + fprintf(fid,format,[’ function [d] = ’,func,’(t);’]);

out = out + fprintf(fid,format,[’ c = [’,num2str(c’,’%15.11f’),’];’]);

out = out + fprintf(fid,format, ’ e = ones(size(t));’);

out = out + fprintf(fid,format,[’ tmax = ’,num2str(tmax,’%15.11f’),’;’]);

out = out + fprintf(fid,format,[’ dmax = ’,num2str(dmax,’%15.11f’),’;’]);

out = out + fprintf(fid,format, ’ dt = abs(t-tmax*e);’);

out = out + fprintf(fid,format, ’ d = dmax*exp(c(1)*dt.ˆ1.3+c(2)*dt.ˆ1.6);’);

fclose(fid);

disp([num2str(out),’ bytes of code written to ’,mfil]);

end

%The model looks quite OK in absolute terms

plot(t-K2C,d,’*’,t-K2C,dmax*exp(A*c));

50 C. MATLAB CODE

title(’Density of water at 1 bar’);

xlabel(’Temperature [C]’);

ylabel(’Density [kg/mˆ3]’);

pause;

%But the deviation plot shows oscillations (typical for heuristic models)

plot(t-K2C,d-dmax*exp(A*c),’o’);

title(’Residual density of water at 1 bar’);

xlabel(’Temperature [C]’);

ylabel(’Density [kg/mˆ3]’);

grid;

10. acid base titration curve

clear;

%Solve the titration problem B+HA = BH(+) + A(-) using direct substitution

%of variables to obtain a fourth order polynomial in n(H+), i.e the number

%of moles of protons at equilibrium.

k1 = 10ˆ(-4.76); % pKb NH3

%k1 = 10ˆ(-2.35); % pKb H2O2

%k1 = 10ˆ(+2.00); % pKb NaOH

k2 = 10ˆ(-4.76); % pKa CH3COOH

%k2 = 10ˆ(-3.74); % pKa HCOOH

%k2 = 10ˆ(+1.37); % pKa HNO3

k3 = 10ˆ(-14); % pKw

cb = 0.1; % molar concentration of base

ca = linspace(0,2*cb,1000)’; % moles of acid added

e = ones(size(ca)); % unity vector with same shape as n

ph = zeros(size(ca)); % negative log10(molar H+ concentration)

c = [-k1*e, ... % 4th order coefficient

k1*k2*e + k1*cb + k3, ... % 3rd order coefficient

k1*k2*ca + k1*k3 - k1*k2*cb - k2*k3, ... % 2nd order coefficient

-k2*k3*ca - k3*k3 - k1*k2*k3, ... % 1st order coefficient

k3*k3*k2*e]; % 0th order coefficient

%Solve fourth order polynomial repeatedly (i.e. all titration points)

for k=1:max(size(ca))

x = roots(c(k,:));

ph(k) =-log10(-max(x(find(x<0))));

end

plot(ca,ph);

title(’Titration curve for NH3 - CH3COOH’);

xlabel(’Acid added [mol]’);

ylabel(’pH’);

hold on;

grid;

%Find inflexion points (quite simple because the x-axis values are mono-

%tonic and equidistant)

12. PRESSURE DROP AT FIXED FLOW 51

curvature = diff(diff(ph)); % numerical second derivative

inflexion = find(curvature(1:end-1).*curvature(2:end)<0); % sign shift

plot(ca(inflexion) ,ph(inflexion) ,’or’); % plot lower limit

plot(ca(inflexion+1),ph(inflexion+1),’or’); % plot upper limit

hold off;

11. colebrook

function [r] = colebrook(lnf,lnRe,eod)

r = lnf/2 + log(-2*log10(eod/3.7+2.51./(exp(lnf/2).*exp(lnRe))));

12. pressure drop at fixed flow

T = 8 + 273.15; % temperature [K]

L = 2000; % pipe length [m]

Q = 0.5; % volumetric flow rate [m3/min]

D = 0.05; % pipe diameter [m]

eod = 1e-3; % relative pipe surface roughness (commercial steel)

mu = water_viscosity_1bar(T); % viscosity [kg/m*s]

rho = water_density_1bar(T); % density [kg/m3]

v = Q/(60*pi*(D/2)ˆ2); % flow velocity [m/s]

lnRe = log(rho*v*D) - log(mu); % log(Reynold number)

lnf = -5; % initial estimate log(Fanning friction)

opt = optimset(’Display’,’iter’); % display iteration progress

lnf = fzero(@colebrook,lnf,opt,lnRe,eod);

g = 9.80665; % standard acceleration of free fall [m/s2]

hs = exp(lnf)*L*vˆ2/(2*D*g); % m H2O at T [K]

dp = hs*rho*g/1e5; % bar

disp([’ ’]);

disp([’Case I’]);

disp([’Diameter : ’,num2str(D ,4),’ [m]’]);

disp([’Pressure drop : ’,num2str(hs,4), ...

’ [m H2O at ’,num2str(round(T-273.15)),’ C]’]);

disp([’Pressure drop : ’,num2str(dp,4),’ [bar]’]);

disp([’ ’]);

disp([’Press any key to continue’]);

pause;

D = 0.10; % new pipe diameter [m]

v = Q/(60*pi*(D/2)ˆ2); % new flow velocity [m/s]

lnRe = log(rho*v*D) - log(mu); % new log(Reynold) number

lnf = fzero(@colebrook,lnf,opt,lnRe,eod);

hs = exp(lnf)*L*vˆ2/(2*D*g); % m H2O at T [K]

dp = hs*rho*g/1e5; % bar

disp([’ ’]);

disp([’Case II’]);

52 C. MATLAB CODE

disp([’Diameter : ’,num2str(D ,4),’ [m]’]);

disp([’Pressure drop : ’,num2str(hs,4), ...

’ [m H2O at ’,num2str(round(T-273.15)),’ C]’]);

disp([’Pressure drop : ’,num2str(dp,4),’ [bar]’]);

13. friction and reynold

function [r] = friction_and_reynold(x,fun,s)

lnRe = x(:,1);

lnf = x(:,2);

r = zeros(size(x));

dp = s.dp;

D = s.D;

rho = s.rho;

L = s.L;

eod = s.eod;

mu = s.mu;

r(:,1) = log(2*dp.*rho.*D.ˆ3) - log(L.*mu.ˆ2) - lnf - 2*lnRe;

r(:,2) = feval(fun,lnf,lnRe,eod);

14. flow at fixed pressure drop

clear;

T = 8 + 273.15; % temperature [K]

s.L = 2000; % pipe length [m]

s.D = 0.05; % pipe diameter [m]

s.eod = 1e-3; % relative pipe surface roughness (commercial steel)

s.mu = water_viscosity_1bar(T); % viscosity [kg/m*s]

s.rho = water_density_1bar(T); % density [kg/m3]

s.dp = 3e5; % pressure drop [Pa]

opt = optimset(’Display’,’iter’); % display iteration progress

lnRe = 10; % initial estimate of log(Reynold number)

lnf = -8; % initial estimate of log(Fanning friction)

x = fsolve(@friction_and_reynold,[lnRe,lnf],opt,@colebrook,s);% solve

Q = 60*pi*exp(x(:,1)).*s.mu.*s.D./(4*s.rho); % flow rate [mˆ3/min]

f = exp(x(:,2)); % friction factor

disp([’ ’]);

disp([’Case I’]);

disp([’Diameter : ’,num2str(s.D ,4),’ [m]’]);

disp([’Pressure drop : ’,num2str(s.dp,4),’ [bar]’]);

disp([’Friction factor : ’,num2str(f ,4),’ [-]’]);

disp([’Flow rate : ’,num2str(Q ,4),’ [mˆ3/min]’]);

disp([’ ’]);

disp([’Press any key to continue’]);

pause;

16. COLEBROOK VS LAMINAR 53

s.D = 0.10; % new pipe diameter [m]

x = fsolve(@friction_and_reynold,[lnRe,lnf],opt,@colebrook,s); % solve

Q = 60*pi*exp(x(:,1)).*s.mu.*s.D./(4*s.rho); % flow rate [mˆ3/min]

f = exp(x(:,2)); % friction factor

disp([’ ’]);

disp([’Case II’]);

disp([’Diameter : ’,num2str(s.D ,4),’ [m]’]);

disp([’Pressure drop : ’,num2str(s.dp,4),’ [bar]’]);

disp([’Friction factor : ’,num2str(f ,4), ’[-]’]);

disp([’Flow rate : ’,num2str(Q ,4),’ [mˆ3/min]’]);

15. laminar

function [r] = laminar(lnf,lnRe,dummy)

r = lnf + lnRe - log(64);

16. colebrook vs laminar

clear;

n = 100; % number of plot points

T = 20 + 273.15; % temperature [K]

s.L = 2000; % pipe length [m]

s.D = logspace(-3,1,n)’; % pipe diameter [m]

s.eod = 1e-3; % relative pipe surface roughness (commercial steel)

s.mu = water_viscosity_1bar(T); % viscosity [kg/m*s]

s.rho = water_density_1bar(T); % density [kg/m3]

s.dp = 3e5; % pressure drop [Pa]

opt = optimset(’Display’,’iter’); % display iteration progress

lnRe = 10*ones(n,1); % initial estimate of log(Reynold number)

lnf = -8*ones(n,1); % initial estimate of log(Fanning friction)

x = fsolve(@friction_and_reynold,[lnRe,lnf],opt,@laminar,s); % solve

Rel = exp(x(:,1)); % laminar Reynold number

Ql = 60*pi*Rel.*s.mu.*s.D./(4*s.rho); % laminar flow rate [mˆ3/min]

fl = exp(x(:,2)); % laminar friction factor

x = fsolve(@friction_and_reynold,[lnRe,lnf],opt,@colebrook,s); % solve

Ret = exp(x(:,1)); % turbulent Reynold number

Qt = 60*pi*exp(x(:,1)).*s.mu.*s.D./(4*s.rho); % flow rate [mˆ3/min]

ft = exp(x(:,2)); % turbulent friction factor

loglog(s.D,Ql,’b’,s.D,Qt,’r’);

title(’Turbulent & laminar flow at constant \Delta P’);

xlabel(’Pipe diameter [m]’);

ylabel(’Flow rate [mˆ3/min]’);

legend(’Laminar’,’Turbulent’);

grid;

disp(’ ’)

54 C. MATLAB CODE

disp(’Press any key to continue’)

pause;

loglog(s.D,Rel,’b’,s.D,Ret,’r’)

title(’Turbulent & laminar Re at constant \Delta P’);

xlabel(’Pipe diameter [m]’);

ylabel(’Re’);

legend(’Laminar’,’Turbulent’);

grid;

pause;

loglog(Rel,fl,’b’,Ret,ft,’r’);

title(’Turbulent & laminar Fanning friction at constant \Delta P’);

xlabel(’Re’);

ylabel(’Friction factor f’);

legend(’Laminar’,’Turbulent’);

grid;

pause;

17. chon

%Calculates molecular weight of CHON-compounds.

function [mw] = chon(x,y,z,w)

mw = x*12.01115 + y*1.00797 + z*15.9994 + w*14.0067;

18. water viscosity 1bar

%Viscosity of water calculated from log(visc)=c(1)+c(2)/T+c(3)/Tˆ2

%Viscosity in [kg/m*s], temperature in [K].

function [visc] = water_viscosity_1bar(t);

c = [-8.0003593134 -1470.7557834 526396.01377];

e = ones(size(t));

visc = exp(c(1)*e+c(2)*e./t+c(3)*e./(t.ˆ2));

19. water vapor pressure

%Vapor pressure of water using Wagner’s equation.

%Pressure in [Pa], temperature in [K].

function [lnp] = water_vapor_pressure(t);

tc = 647.1;

pc = 22064000;

c = [-7.78664494819 1.50070473755 -2.82073715777 -1.20214640864];

e = ones(size(t));

tr = t/tc;

tmp = c(1)*(e-tr) + c(2)*(e-tr).ˆ1.5 + c(3)*(e-tr).ˆ3 + c(4)*(e-tr).ˆ6;

lnp = tmp./tr + log(pc)*e;

20. water density 1bar

%Density of water (1 bar) calculated

%from log(rho) = tauˆ1.3 + tauˆ1.6,

%where rho is d/dmax and tau is

%t-tmax. Densities in [kg/mˆ3] and

%temperatures in [K].

function [d] = water_density_1bar(t);

c = [0.00003932621 -0.00003868077];

22. DIFFUSION 55

e = ones(size(t));

tmax = 277.13000000000;

dmax = 999.97490000000;

dt = abs(t-tmax*e);

d = dmax*exp(c(1)*dt.ˆ1.3+c(2)*dt.ˆ1.6);

21. rachford rice

%Purpose : Rachford-Rice 2-phase flash calculation

%Author : Tore Haug-Warberg

%Date : March 14 1993

%Syntax : [x,y,l] = rachford_rice(z,k)

%Input : z = feed mole fractions vector

% k = K-value vector

%Output : x = liquid mole fraction vector

% y = vapor mole fraction vector

% l = Liquid fraction

%External: abs length min ones sum

function [x,y,l] = rachford_rice(z,k)

%Initializations

tol = 1.e-6; %Convergence tolerance

e = ones(length(k),1); %Unity vector

l = 0.5; %Initial estimate of liquid fraction

dl = 1.0; %Liquid fraction increment

z = z/sum(z); %Normalize feed composition

%Verifications

if length(z)˜= length(k); return; end

if min (z) < 0.0 ; return; end

if min (k) < 0.0 ; return; end

if min (k) > 1.0 ; l=0.0; x=0.0*z; y=z; return; end

if max (k) < 1.0 ; l=1.0; y=0.0*z; x=z; return; end

%Rachford-Rice iteration

while abs(dl) > tol

f = e’*((z.*(e-k))./(l*e+(1-l)*k));

dfdl = -e’*((z.*(e-k).ˆ2)./((l*e+(1-l)*k).ˆ2));

dl = -f/dfdl;

l = l + dl;

end

x = z./(l*e+(1-l)*k);

y = k.*x;

if l > 1.0; l=1.0; x=z; y=0.0*e; end

if l < 0.0; l=0.0; y=z; x=0.0*e; end

return

22. diffusion

function dummy

56 C. MATLAB CODE

clear all

close all

% Solution of the differential equation with no reaction on the surface D

zeta = [0 1]; % Starting value for zeta

% The starting values of gamma is chosen to satisfy the boundary conditio-

% ns given for the problem. Here with a better starting value, the numeri-

% cal solution could be closer to the analytical.

gamma = [1 -0.26]; % Starting value for gamma and its derivative

[a,b] = ode23(@equations,zeta,gamma); % Solving the differential equation

plot(a,b(:,1),’:’)

hold on

% Solution of the differential equation with reaction on the surface D

zeta = [0 1]; % starting value for zeta

gamma = [1 -1.1]; % Starting value for gamma and its derivative

[a,b] = ode23(@equations,zeta,gamma); %solving the differential equation

plot(a,b(:,1),’-.’)

% To compare the numeric solution with the analytical the analytical solu-

% tion is found

zetad = [0:.001:1];

kodab = 0.25;

l = 1;

th = sqrt(kodab*lˆ2);

% Analytical solution with no reaction on surface D

g1 = (cosh(th).*cosh(th.*zetad) - sinh(th).*sinh(th.*zetad))./cosh(th);

plot(zetad,g1,’-.g’)

% Analytical solution with reaction on surface D

g2 = (sinh(th).*cosh(th.*zetad)-cosh(th).*sinh(th.*zetad))./sinh(th);

plot(zetad,g2,’:m’)

legend(’No reaction on D, numeric sol’,’No reaction on D, analytic sol’,...

’Reaction on D analytic sol’,’Reaction on D numeric sol’)

xlabel(’\zeta’)

ylabel(’\Gamma’)

hold off

% Function containing the equations to be solved by the ODE solver

function d = equations(z,g)

kodab = 0.25;

l = 1;

th = sqrt(kodab*lˆ2);

dz = g(2);

ddz = thˆ2*g(1);

d = [dz;ddz];

23. PITOT TUBE 57

23. pitot tube

clear all

close all

%Initial data

g = 9.8065; % m/sˆ2

R = 8.31453; % J/(mol K)

D = 1.6; % pipe diamerter in m

Ptot = 1.01325e5; % Pa

Tg = 273.15 + 110; % K

Mgas = 28; % molecular weight g/mol

rholiq = 850; % kg/mˆ3

Cp = 1; % assume incompressible flow

%Velocity profiles North-South and West-East

hNS = [4.7 10.7 18.3 20.5 20.5 23.3 29.6 20.4 16.2 11.8]; % North-South

hEW = [2.1 8.6 15.4 16.2 21.3 21.3 20.5 16.8 13.3 2.2]; % East-West

%Now we calculate the velocity from the the measured heights. In order to

%do this, we first need to know the radial position of the measurement. We

%know there are five concentric rings, with the same area.

N = 5;

n = [1 2 3 4 5];

r = (D/2)*sqrt((2*n-1)/(2*N)); % position of the area center in each ring

%In order to plot the velocity profiles, we need an axis from wall to wall.

z = [fliplr(D/2-r) D/2+r];

%Next we need the preassure drop

delpNS = hNS*rholiq*g/1000;

delpEW = hEW*rholiq*g/1000;

rhogas = Ptot*Mgas/(R*Tg*1000);

vNS = Cp*sqrt(2*delpNS/rhogas);

vEW = Cp*sqrt(2*delpEW/rhogas);

%We add the boundary conditions

vNSw = [0 vNS 0];

vEWw = [0 vEW 0];

z = [0 z 1.6];

vmid = (vNSw+vEWw)./2; % Calculate the average velocity

z1 = [0:D/30:D]; % Adding a few extra points gives a smooth plot

%Do polynomial approximation

p2 = polyfit(z,vmid,2); % 2nd degree polynomial

p4 = polyfit(z,vmid,4); % 4th degree polynomial

sp = spline(z,vmid); % spline approximation

t2 = polyval(p2,z1);

t4 = polyval(p4,z1);

fsp = ppval(z1,sp);

plot(z,vmid,’*’,z1,t2)

title(’2nd degree polynomial’)

ylabel(’Average velocity, m/s’)

58 C. MATLAB CODE

xlabel(’Distance from left wall, m’)

axis([min(z),max(z),0,30])

legend(’Datapoints’,’Fitted profile’)

pause

plot(z,vmid,’*’,z1,t4)

title(’4th degree polynomial’)

ylabel(’Average velocity, m/s’)

xlabel(’Distance from left wall, m’)

axis([min(z),max(z),0,30])

legend(’Datapoints’,’Fitted profile’)

pause

plot(z,vmid,’*’,z1,fsp)

title(’Spline approximation’)

ylabel(’Average velocity, m/s’)

xlabel(’Distance from left wall, m’)

axis([min(z),max(z),0,30])

legend(’Datapoints’,’Fitted profile’)

%Compute the volumetric flow in mˆ3/h for the different cases.

Q_sp = num2str(round(quad(@pitot_integrand,0,D,[],[],sp)*3600));

Q_p2 = num2str(round(quad(@pitot_integrand,0,D,[],[],p2)*3600));

Q_p4 = num2str(round(quad(@pitot_integrand,0,D,[],[],p4)*3600));

disp([’Volumetric flow (cubic spline) : ’,Q_sp,’ mˆ3/h’])

disp([’Volumetric flow (2nd order polynomial): ’,Q_p2,’ mˆ3/h’])

disp([’Volumetric flow (4th order polynomial): ’,Q_p4,’ mˆ3/h’])

24. pitot integrand

function f = pitot_integrand(r,p)

%Function to calculate the integrand of the pitot-tube experiment.

%

% p = transferred polynomial coefficients for the velocity

% r = The position in the tube.

%

%The velocity is calculated from the polynomial coefficients. Since spline

%and polyfit uses different formats, we need to do a test and choose the

%appropriate function.

if isstruct(p) == 1

v = ppval(r,p);

else

v = polyval(p,r);

end

f = pi.*r.*v./2;

25. mingibb

%Equilibrium calculation n = minGibb(n0,t,p,m0,A) at a given temperature,

%pressure and total composition in ideal mixture phases.

%

%n0 = initial mole number vector [mol]

%t = temperature [K]

26. CSTR PROCESS CONTROL 59

%p = pressure [Pa]

%m0 = standard chemical potential (ideal gas at T)

%A = formula matrix (atoms x substances)

%n = final mole number vector [mol]

function[n] = minGibb(n0,t,p,m0,A)

%Parameters and physical quantities

r = 8.31453; % universal gas constant

p0 = 1e5; % standard state pressure

tol = 1e-6; % convergence tolerance

maxiter = 100; % maximum number of iterations

maxstep = 0.9; % maximum relative step length

epx = 1e-6; % arbitrarily small mole fraction

[nc,ns] = size(A); % number of atoms and substance

%Initialization

n = n0;

i = find(n==0);

n(i) = zeros(length(i),1) + sum(n)*epx;

e = ones(ns,1);

%Equilibrium calculation

for iter=1:maxiter

m = m0 + r*t*log(p/p0*n/sum(n));

D = diag(n)/(r*t);

lhs = [A*D*A’, A*n; (A*n)’, 0];

rhs = [A*D*m-A*(n-n0); n’*m];

x = inv(lhs)*rhs;

dn =-D*(m-A’*x(1:nc)) + n*x(nc+1);

step =-maxstep/min([-maxstep,min(dn./n)]);

n = n + step*dn;

if max(abs(dn./n))<tol; return; end

end

%Not converged

n = NaN*n;

return

26. CSTR process control

clear all

warning off

func = @cstr_time_derivatives; % make function handle

ca=2/3; T =400; x0=[ca T]; % equilibrium point: steady-state values

Q =5000; q =1; u0=[Q q]; % inputs

T1=350; caf=2; d0=[T1 caf]; % disturbances

t = [0 100];

[t0,y0] = ode15s(func,t,x0,[],u0,d0); % steady-state check

figure(1)

60 C. MATLAB CODE

subplot(2,1,1);

plot(t0,y0(:,1)), grid; title(’Steady-state behavior of the system’);

ylabel(’C_a [kmol/mˆ3]’);

subplot(2,1,2);

plot(t0,y0(:,2)), grid; xlabel(’t [s]’); ylabel(’T [K]’)

%Transient response to step changes

st = 2; % step order

u = u0; d=d0; d(1)=st*d(1); [t1,y1]=ode15s(func,t,x0,[],u,d); %T1

u = u0; d=d0; d(2)=st*d(2); [t2,y2]=ode15s(func,t,x0,[],u,d); %C_a final

u = u0; d=d0; u(1)=st*u(1); [t3,y3]=ode15s(func,t,x0,[],u,d); %Q

u = u0; d=d0; u(2)=st*u(2); [t4,y4]=ode15s(func,t,x0,[],u,d); %q

figure(2)

subplot(4,2,1); plot(t1,y1(:,1)), ylabel(’T1 [K]’), grid; title(’ca [kmol/mˆ3]’);

subplot(4,2,2); plot(t1,y1(:,2)), grid; title(’T [K]’);

subplot(4,2,3); plot(t2,y2(:,1)), ylabel(’caf [kmol/mˆ3]’), grid;

subplot(4,2,4); plot(t2,y2(:,2)), grid;

subplot(4,2,5); plot(t3,y3(:,1)), ylabel(’Q [kJ/min]’), grid;

subplot(4,2,6); plot(t3,y3(:,2)), grid;

subplot(4,2,7); plot(t4,y4(:,1)), ylabel(’q [mˆ3/min]’), grid; xlabel(’t [s]’)

subplot(4,2,8); plot(t4,y4(:,2)), grid; xlabel(’t [s]’)

%Linearization

[A,B,E] = ode_jacobian(func,[],x0,u0,d0);

C = eye(2);

D = zeros(2);

F = zeros(2);

%Stability, controllability, and observability

Co = ctrb(A,B);

Ob = obsv(A,B);

disp(’Eigenvalues of A’)

disp(eig(A))

disp(’Size Co, Rank Co’)

disp([size(Co,1), rank(Co)])

disp(’Size Ob, Rank Ob’)

disp([size(Ob,2), rank(Ob)])

Gss = ss(A,B,C,D);

Gdss = ss(A,E,C,D);

G = tf(Gss);

Gd = tf(Gdss);

[num,den] = tfdata(G);

[numd,dend] = tfdata(Gd);

sz = size(G);

i = 1:sz(1);

j = 1:sz(2);

%The data below are transferred to the Simulink file

[num11 ,den11 ] = tfdata(tf(num(1,1),den(1,1)),’v’); %g11

27. CSTR TIME DERIVATIVES 61

[num12 ,den12 ] = tfdata(tf(num(1,2),den(1,2)),’v’); %g12

[num21 ,den21 ] = tfdata(tf(num(2,1),den(2,1)),’v’); %g21

[num22 ,den22 ] = tfdata(tf(num(2,2),den(2,2)),’v’); %g22

[numd11,dend11] = tfdata(tf(numd(1,1),dend(1,1)),’v’); %gd11

[numd12,dend12] = tfdata(tf(numd(1,2),dend(1,2)),’v’); %gd12

[numd21,dend21] = tfdata(tf(numd(2,1),dend(2,1)),’v’); %gd21

[numd22,dend22] = tfdata(tf(numd(2,2),dend(2,2)),’v’); %gd22

%Poles and zeros

Poles_of_G = pole(Gss);

Zeros_of_G = zero(Gss);

w = logspace(-2,2);

figure(3)

sigma(G,w)

grid

%RGA

sz = size(G);

i = 1:sz(1);

j = 1:sz(2);

w1 = logspace(-3,1,200);

H = freqresp(G,w1);

for k=1:length(w1)

Hinv = inv(H(:,:,k));

RGA(i,j,k) = H(i,j,k).*Hinv’;

end

rga_diag = abs(RGA(1,1,:));

rga_offdiag = abs(RGA(1,2,:));

figure(4)

plot(w1,rga_diag(:),w1,rga_offdiag(:)); grid

xlabel(’Frequency [rad/s]’), ylabel(’Magnitude of the RGA Elements’)

legend(’\Lambda_{11} = \Lambda_{22}’,’\Lambda_{12} = \Lambda_{21}’)

27. CSTR time derivatives

function xdot = cstr_time_derivatives(t,x,u,d)

To = 400; % K

V = 4; % reactor volume [m3]

cp = 0.150; % heat capcity [kJ/kg*K]

E = 20000; % activation energy [kJ/kmol]

rho = 1000; % density [kg/m3]

ko = 0.5; % rate constant [minˆ-1]

R = 8.31453; % universal gas constant [kJ/K*kmol]

H = 1875; % reaction enthalpy [kJ/kmol]

ca = x(1); % concentration of A

T = x(2); % temperature

Q = u(1); % heat (input)

q = u(2); % volumetric flow (input)

T1 = d(1); % temperature in (disturbance)

caf = d(2); % concentration (disturbance)

k = ko*exp(-E/R*(1/T-1/To)); % Arrhenius rate

62 C. MATLAB CODE

dca_dt = (caf-ca)*q/V-k*ca; % concentration derivative

dT_dt = (T1-T)*q/V+H*k*ca/(rho*cp)+Q/(rho*cp*V); % temperature deriv

xdot = [dca_dt dT_dt]’;

28. ODE Jacobian

function [J_x,J_u,J_d] = ODE_Jacobian(fun,t,x,u,d)

% Evaluates the Jacobian of "fun" by the first order difference method.

%

% t = time

% x, u, d = input variables (arrays)

%

% The Jacobian is calculated by pertubing these variables, one at a time.

del = 1.0e-10;

%Calculation of matrix A (states x)

n_x = length(x);

J_x = [];

F = feval(fun,t,x,u,d);

for i=1:n_x

x_new = x;

pertub = x(i)*del + eps*100;

x_new(i) = x(i) + pertub;

F_new = feval(fun,t,x_new,u,d);

J_x = [J_x,((F_new-F)./pertub)];

end

% Calculation of matrix B (inputs u)

n_u = length(u);

J_u = [];

F = feval(fun,t,x,u,d);

for i=1:n_u

u_new = u;

pertub = u(i)*del + eps*100;

u_new(i) = u(i) + pertub;

F_new = feval(fun,t,x,u_new,d);

J_u = [J_u,((F_new-F)./pertub)];

end

% Calculation of matrix E (disturbances d)

n_d = length(d);

J_d = [];

F = feval(fun,t,x,u,d);

for i=1:n_d

d_new = d;

pertub = d(i)*del + eps*100;

d_new(i) = d(i) + pertub;

F_new = feval(fun,t,x,u,d_new);

J_d = [J_d,((F_new-F)./pertub)];

end

30. MW BINOP 63

29. mw from formula parser

%Chemical formula to molecular weight parser. Example on use:

%formula_to_mw_parser(’Ca(NO3)2(H2O)6’) returns the string

%’Ca+(N+O*3)*2+(H*2+O)*6’ which can be evaluated by Matlab if Ca, N, O and

%H are assigned Matlab variables. I.e. eval(’Ca+(N+O*3)*2+(H*2+O)*6’)

%should yield 272.1815.

%

%Author : Tore Haug-Warberg

%Version : April 30 2002 (THW)

%

%function [mwe] = mw_from_formula_parser(formula)

%

%formula = Chemical formula [string].

%mwe = Molecular weight expression [string].

function [mwe] = mw_from_formula_parser(formula)

%Parse the formula from right to left

mwe = formula(end);

right = mwe;

for left=formula(end-1:-1:1)

mwe = strcat(left,mw_binop(left,right),mwe);

right = left;

end

30. mw binop

%Binary operator (x o y) table for molecular

%weight parsing of chemical formulas.

%

%Author : Tore Haug-Warberg

%Version : April 30 2002 (THW)

%

%function [o] = mw_binop(x,y)

%

%x = Left operand [char].

%y = Right operand [char].

%o = Operator [char].

function [o] = mw_binop(x,y)

o = ’’;

if (isdigit(x))

if (islower(y)|isupper(y)|isleft(y))

o = ’+’;

end

elseif(islower(x))

if(isdigit(y))

o = ’*’;

elseif(islower(y))

o = ’<multiple_lower_case_letters>’;

elseif(isupper(y)|isleft(y))

o = ’+’;

64 C. MATLAB CODE

end

elseif(isupper(x))

if(isdigit(y))

o = ’*’;

elseif(isupper(y)|isleft(y))

o = ’+’;

end

elseif(isright(x))

if(isdigit(y))

o = ’*’;

elseif(islower(y))

o = ’<name_starts_with_lower_case>’;

elseif(isupper(y)|isleft(y))

o = ’+’;

end

elseif(isleft(x))

if(isdigit(y))

o = ’<name_starts_with_digit>’;

elseif(islower(y))

o = ’<name_starts_with_lower_case>’;

elseif(isright(y))

o = ’<empty_name>’;

end

end

%Verify that input string is a digit 0-9

function [i] = isdigit(s)

i = and(ge(abs(s),48),le(abs(s),57));

%Verify that input string is a left paranthesis (

function [i] = isleft(s)

i = eq(abs(s),40);

%Verify that input string is a right paranthesis )

function [i] = isright(s)

i = eq(abs(s),41);

%Verify that the input string is a lower case letter a-z

function [i] = islower(s)

i = and(ge(abs(s),97),le(abs(s),122));

%Verify that the input string is a lower case letter A-Z

function [i] = isupper(s)

i = and(ge(abs(s),65),le(abs(s),90));

31. make formula matrix

%Calculates the formula matrix for a specified list of chemical species.

%The complete list of 103 chemical elements is recognized by the parser.

%To add some flexibility a mixed list of strings and cell arrrays are

%allowed as input arguments:

%

%make_formula_matrix(’NaCl’,’KCl’,’MgCl2’,’CaCl2(H2O)6’);

%make_formula_matrix(’CH3(CH2)3OH’,{’H2’,’O2’,’H2O’},’CH3(CH2)COOH’);

31. MAKE FORMULA MATRIX 65

%make_formula_matrix({’CH3OH’,{’H2’,’O2’,’H2O’},’CO’,’CO2’});

%

%Note, however, that the output variables do not mimic the input cell

%structure, i.e. A will be a straightforward matrix and b will be a list.

%

%Author : Tore Haug-Warberg

%Version : April 30 2002 (THW)

%

%function [A,b] = make_formula_matrix(arg1,arg2,...,argN)

%

% A = Formula matrix [elements x species].

% b = Element name cell array [elements].

% arg1 = species formula (or cell array)

% arg2 = species formula (or cell array)

% ...

% argN = species formula (or cell array)

function [A,b] = make_formula_matrix(varargin);

persistent name;

if (isempty(name))

name = {’H’ ,’He’,’Li’,’Be’,’B’ ,’C’ ,’N’ ,’O’ ,’F’ ,’Ne’,’Na’,’Mg’,...

’Al’,’Si’,’P’ ,’S’ ,’Cl’,’Ar’,’K’ ,’Ca’,’Sc’,’Ti’,’V’ ,’Cr’,...

’Mn’,’Fe’,’Co’,’Ni’,’Cu’,’Zn’,’Ga’,’Ge’,’As’,’Se’,’Br’,’Kr’,...

’Rb’,’Sr’,’Y’ ,’Zr’,’Nb’,’Mo’,’Tc’,’Ru’,’Rh’,’Pd’,’Ag’,’Cd’,...

’In’,’Sn’,’Sb’,’Te’,’I’ ,’Xe’,’Cs’,’Ba’,’La’,’Ce’,’Pr’,’Nd’,...

’Pm’,’Sm’,’Eu’,’Gd’,’Tb’,’Dy’,’Ho’,’Er’,’Tm’,’Yb’,’Lu’,’Hf’,...

’Ta’,’W’ ,’Re’,’Os’,’Ir’,’Pt’,’Au’,’Hg’,’Tl’,’Pb’,’Bi’,’Po’,...

’At’,’Rn’,’Fr’,’Ra’,’Ac’,’Th’,’Pa’,’U’ ,’Np’,’Pu’,’Am’,’Cm’,...

’Bk’,’Cf’,’Es’,’Fm’,’Md’,’No’,’Lr’}’;

end

varargin = unpack(varargin); % make straight list

A = zeros(length(name),length(varargin)); % init formula matrix

for i=1:length(varargin)

[s,A(:,i)] = get_formula_vector(varargin{i},A(:,i),name,1.0);

end

i = find(sum(A,2)>0); % find non-zero rows in matrix A

A = A(i,:); % shrink matrix A to include non-zero rows only

b = {name{i}}’; % shrink the element list accordingly

%Recursive unpacking of input argument list

function [out] = unpack(in);

out = {}; % empty cell array

for i=1:length(in); % do for all members of the list

if (iscell(in{i})) % check if this is a new cell

tmp = unpack(in{i}); % recursive unpacking

out = {out{:},tmp{:}}; % augment output list

else %

out = {out{:},in{i}}; % augment output list

66 C. MATLAB CODE

end

end

32. get formula vector

%Adds stoichiometric coefficients for a specified chemical species using

%recursive string parsing.Eg [s,b]=get_formula_vector(’(CH3(CH2)2COOH)1.5’

%,[0 0 0],{’C’,’H’,’O’},1) returns s=’’ and b=[6 12 3]. The chemical ele-

%ments need not be specified in any particular order, but their order must

%coincide with the stoichiometric coefficients also specified in the input

%

%Author : Tore Haug-Warberg

%Version : March 10 2000 (THW)

%

%function [s,b] = get_formula_vector(s,b,name,factor)

%

% s = Chemical formula [string].

% b = Stoichiometric coefficients [vector].

% name = Recognized formula symbols [cell array].

% factor = External multiplier.

%

function [s,b] = get_formula_vector(s,b,name,factor)

coef = ’’; % initialize stoichiometric coefficient

sym = ’’; % initialize chemical symbol

while not(isempty(s)) % continue while string is non-empty

next = s(end); % pick next token from end of string

s = s(1:end-1); % decrement the formula string

if isletter(next) % token is a letter a-z or A-Z

sym = strcat(next,sym); % build a chemical symbol (He, Mg, etc.)

i = strmatch(sym,name,’exact’); % test against known symbols

if any(i) % SYM is a recognized symbol

b(i) = b(i) + factor*stoichiometry(coef); % increment stoich. coeff

coef = ’’; % reset the stoichiometric coefficient

sym = ’’; % reset the chemical symbol

end %

elseif isnumber(next) % token is a period or a number 0-9

coef = strcat(next,coef); % build a stoichiometric coefficient

sym = emptystring(sym); % test that SYM is empty

elseif isright(next) % token is a right paranthesis )

[s,b] = get_formula_vector(s,b,name, ... % call myself recursively

factor*stoichiometry(coef)); % increase premultiplic. factor

coef = ’’; % reset the stoichiometric coefficient

sym = emptystring(sym); % test that SYM is empty

elseif isleft(next) % token a left paranthesis (

sym = emptystring(sym); % test that SYM is empty

return; % return (from recursive call only)

end %

end %

sym = emptystring(sym); % test that SYM is empty

33. GREENHOUSE GASES FROM HYSYS 67

%Verify that input string is empty

function [s] = emptystring(s)

if (not(isempty(s)))

disp([’Unrecognized string: ’,s]);

end

sym = ’’;

%Verify that input string is a period or a number in the range 0-9

function [i] = isnumber(s)

i = or(and(ge(abs(s),48),le(abs(s),57)),eq(abs(s),46));

%Verify that input string is a left paranthesis (

function [i] = isleft(s)

i = eq(abs(s),40);

%Verify that input string is a right paranthesis )

function [i] = isright(s)

i = eq(abs(s),41);

%Convert input string to number (empty input string is treated as 1)

function [n] = stoichiometry(s)

if isempty(s)

n = 1;

else

n = str2num(s);

end

33. greenhouse gases from HYSYS

clear all % clear work space

warning(’off’) % avoid warning when the while loop terminates

component = {’Methane’,’Ethane’,’Propane’,’n-Butane’,’n-Pentane’}; % names

cnum = [1 2 3 4 5]; % Relative amount of carbon in each component

co2 = cell(4,2);

co2{1,1} = ’Stream’;

co2{1,2} = ’Co2 equivalent’;

fid = fopen(’..\data\HYSYS_report.txt’);

t = 1;

c = 1;

st = ’’;

% Read file into one long string.

while t>0;

t = fgets(fid);

st = strcat(st,t);

end

fclose(’all’); % close file

a = regexp(st,’(Material Stream:)’); % find start position of the streams

a(end+1) = length(st); % add end positions of the string to a vector

for i=1:length(a)-1

68 C. MATLAB CODE

stream = st(a(i):a(i+1)); % Look through the string one stream at a time

if ˜isempty(regexp(stream,’Material Stream: V’)) % Find wanted data

[tmp,tmp1] = regexp(stream,’(Overall Phase).*(Vapour Phase)’);

c = c + 1;

co2{c,2}(1) = 0; % initialize the container for CO2 equivalents

vap = stream(tmp:tmp1); % copy the interesting part

[tb,te] = regexp(vap,’Total ,\d*\.\d*’); % find the total stream

tmp2 = vap(tb:te);

[tb1,te1] = regexp(tmp2,’\d*\.\d*’); % pick out the numbers

total = str2num(tmp2(tb1:te1)); % convert from text to numbers

for j=1:length(component)

% find the position of each component, and take out the

% amount from the string. afterward the CO2 equivalent is

% calculated and added to the container.

[start,stop] = regexp(vap,strcat(component{j},’,\d*\.\d*’));

comp = vap(start:stop);

[start1,stop1] = regexp(comp,’\d+\.\d+’);

co2{c,2}(1) = co2{c,2}(1) + ...

str2num(comp(start1:stop1))*cnum(j)/total*100;

co2{c,1} = str2num(stream(19));

end

else

% do nothing, we are only seeking vapour data

end

end

disp(co2); %display the final result

APPENDIX D

Ruby code

1. greenhouse gases from HYSYS

#read output file and remove end-of-line character

f = File.open(’..\data\HYSYS_report.txt’).read.gsub!(/\n/,’’)

#Split file into sections of stream data. Remove non-vapour streams

arr = f.split(/(?=Material Stream: V)/).each{|i|

i.gsub!(/Material Stream: [ˆV].*/,’’)

}

#Create hash with key="component name" and value="relative amount of C"

co2 = {"Methane",1.0,"Ethane",2.0,"Propane",3.0,

"n-Butane",4.0,"n-Pentane",5.0}

#Create empty Hash to store the final result. Default flow is zero.

hsh = Hash.new(0.0)

#Define some practical regular expressions

streams = Regexp.new(’Material Stream: V(\d)’)

totflow = Regexp.new(’COMPONENTS,MOLAR.*Total ,(\d*\.\d*).*Vapour Phase’)

vapflow = Regexp.new(’COMPONENTS,MOLAR.*Vapour Phase’)

cmpflow = Regexp.new(’([a-z]\-[A-Z][a-z]*|[A-Z][a-z]*),(\d*\.\d*)’)

#Scan the vapor stream sections of the output file

arr.each{|i|

n = i.scan(streams).to_s # vapor stream number

t = i.scan(totflow).to_s.to_f # total stream flow

i.scan(vapflow).to_s.scan(cmpflow).each{|j| # each component flow

hsh[n] += co2[j.first]*j.last.to_f/t*100 # % CO2 equivalent

}

}

#Make a report

hsh.each{|n,p|puts "Stream V#{n}: #{p.to_i}% CO2 of total molar flow"}

69

APPENDIX E

Data files

1. pages from water vapor pressure data

%PDOUS71

%Douslin, D.R.

%J. Chem. Thermodynamics, 3, 187 (1971).

% T90(K) P(MPa)

270.651 .0005088

273.150 .0006114

273.160 .0006121

274.150 .0006573

275.150 .0007061

276.149 .0007587

277.149 .0008137

278.149 .0008730

280.649 .0010370

283.148 .0012291

288.147 .0017066

285.647 .0014485

290.646 .0020007

293.145 .0023395

%

%PHANA84

%Hanafusa,H.,Tsuchida,T.,Kawai.,Sato,H.,Uematsu,M.,and Watanabe, K.

%Proc. 10th ICPS, Moscow 1984, 180;High Temp.-High Pres., 15, 311(1983).

% T90(K) P(MPa)

643.174 21.0502000

645.607 21.6715000

645.607 21.6705000

646.106 21.7997000

646.356 21.8645000

646.607 21.9309000

646.607 21.9303000

646.857 21.9974000

%

%PKELL85

%Kell, G.S., McLaurin, G.E., and Whalley, E.

%Phil. Trans. R. Soc. London, 315A, 235 (1985).

% T90(K) P(MPa)

423.133 .4759000

423.134 .4759000

423.133 .4758000

423.108 .4752000

423.112 .4757000

448.114 .8918000

448.114 .8918000

71

72 E. DATA FILES

473.109 1.5536000

473.109 1.5535000

498.107 2.5474000

498.118 2.5479000

523.168 3.9780000

523.168 3.9772000

548.108 5.9416000

548.108 5.9416000

573.111 8.5821000

573.110 8.5822000

598.110 12.0446000

598.111 12.0443000

623.183 16.5370000

623.183 16.5370000

623.109 16.5196000

2. pages from water density 1bar data

%Kell, G.S.

%J. Chem. Eng. Data, 20(1) 97 (1975).

% T90(K) P(MPa) d(kg/m3)

273.150 .101325 999.84260

278.147 .101325 999.96690

283.144 .101325 999.70311

288.141 .101325 999.10362

293.138 .101325 998.20874

298.136 .101325 997.05027

303.134 .101325 995.65331

308.132 .101325 994.03845

313.130 .101325 992.22289

318.128 .101325 990.22074

323.137 .101325 988.03930

328.125 .101325 985.70316

333.134 .101325 983.20192

338.124 .101325 980.56279

343.132 .101325 977.77266

348.124 .101325 974.85613

353.123 .101325 971.80450

358.123 .101325 968.62628

363.123 .101325 965.32527

373.124 .101325 958.36664

383.126 .101325 950.94723

393.127 .101325 943.07942

403.130 .101325 934.76863

413.132 .101325 926.01485

423.134 .101325 916.81308

%

%Owen, B.B., White, J.R., and Smith, J.S.

%J. Am. Chem. Soc., 78, 3561 (1956).

% T90(K) P(MPa) d(kg/m3)

313.130 .101325 992.22250

318.128 .101325 990.21900

323.127 .101325 988.04030

328.125 .101325 985.69720

4. PAGES FROM HYSYS REPORT 73

333.124 .101325 983.19870

338.124 .101325 980.55310

343.124 .101325 977.76690

348.123 .101325 974.84530

353.123 .101325 971.79200

358.123 .101325 968.61060

3. water viscosity 1bar data

%Barrow, G. M., Physical Chemistry, 3rd ed.,

%McGraw-Hill, 535 (1973).

% T(C) P(MPa) visc(kg/m*s)

0 .101325 0.001792

20 .101325 0.001005

40 .101325 0.000656

60 .101325 0.000469

80 .101325 0.000356

100 .101325 0.000284

4. pages from HYSYS report

Material Stream: V3,Fluid Package: Basis-1

,

,Property Package: Peng Robinson CONDITIONS

,Overall,Vapour Phase,Liquid Phase,,

Vapour / Phase Fraction ,1.0000,1.0000,0.0000,,

Temperature: (C),45.00,45.00,45.00,,

Pressure: (kPa),3924,3924,3924,,

Molar Flow (kgmole/h),42.19,42.19,0.0000,,

Mass Flow (kg/h),1223,1223,0.0000,,

Std Ideal Liq Vol Flow (m3/h),3.090,3.090,0.0000,,

Molar Enthalpy (kJ/kgmole),-8.852e+004,-8.852e+004,-1.179e+005,,

Molar Entropy (kJ/kgmole-C),158.3,158.3,114.1,,

Heat Flow (kJ/h),-3.735e+006,-3.735e+006,0.0000,,

Liq Vol Flow @Std Cond (m3/h),,,0.0000,,

COMPOSITION

Overall Phase ,Vapour Fraction 1.0000

,

COMPONENTS,MOLAR FLOW,MOLE FRACTION,MASS FLOW,MASS FRACTION,

LIQUID VOLUME,LIQUID VOLUME

, (kgmole/h), , (kg/h), ,FLOW (m3/h),FRACTION

Methane,17.7435,0.4206,284.6569,0.2328,0.9508,0.3077

Ethane,13.4946,0.3199,405.7803,0.3319,1.1408,0.3692

Propane,7.8865,0.1869,347.7726,0.2845,0.6864,0.2221

n-Butane,2.6142,0.0620,151.9480,0.1243,0.2605,0.0843

n-Pentane,0.4487,0.0106,32.3749,0.0265,0.0514,0.0166

Total ,42.1875,1.0000 ,1222.5327,1.0000 ,3.0899,1.0000

Vapour Phase,Phase Fraction 1.000

,

COMPONENTS,MOLAR FLOW,MOLE FRACTION,MASS FLOW,MASS FRACTION,

LIQUID VOLUME,LIQUID VOLUME

, (kgmole/h), , (kg/h), ,FLOW (m3/h),FRACTION

Methane,17.7435,0.4206,284.6569,0.2328,0.9508,0.3077

Ethane,13.4946,0.3199,405.7803,0.3319,1.1408,0.3692

74 E. DATA FILES

Propane,7.8865,0.1869,347.7726,0.2845,0.6864,0.2221

n-Butane,2.6142,0.0620,151.9480,0.1243,0.2605,0.0843

n-Pentane,0.4487,0.0106,32.3749,0.0265,0.0514,0.0166

Total ,42.1875,1.0000 ,1222.5327,1.0000 ,3.0899,1.0000

Liquid Phase,Phase Fraction 0.0000

,

COMPONENTS,MOLAR FLOW,MOLE FRACTION,MASS FLOW,MASS FRACTION,

LIQUID VOLUME,LIQUID VOLUME

, (kgmole/h), , (kg/h), ,FLOW (m3/h),FRACTION

Methane,0.0000,0.1091,0.0000,0.0404,0.0000,0.0664

Ethane,0.0000,0.2463,0.0000,0.1710,0.0000,0.2363

Propane,0.0000,0.3223,0.0000,0.3282,0.0000,0.3184

n-Butane,0.0000,0.2364,0.0000,0.3172,0.0000,0.2673

n-Pentane,0.0000,0.0859,0.0000,0.1431,0.0000,0.1117

Total ,0.0000,1.0000 ,0.0000,1.0000 ,0.0000,1.0000