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Table of Contents

TABLE OF CONTENTS...........................................................................................................................1

PREFACE................................................................................................................................................... 3

1 DEVELOPMENT OF LUMPED MODELS..................................................................................4

1.1 I NTRODUCTION ..........................................................................................................................4 1.2 MASS BALANCES .......................................................................................................................4 1.3 SYSTEMS OF MASS BALANCES...................................................................................................8 1.4 E NERGY BALANCES .................................................................................................................10 1.5 MOMENTUM BALANCES...........................................................................................................11 1.6 SUMMARY................................................................................................................................ 12

2 STEADY-STATE SIMULATION OF LUMPED MODELS .....................................................13

2.1 I NTRODUCTION ........................................................................................................................13 2.2 METHOD FOR LINEAR EQUATIONS ........................................................................................... 13 2.3 METHODS FOR NONLINEAR EQUATIONS .................................................................................. 16 2.4 NEWTON METHODS .................................................................................................................18

2.5 SUMMARY................................................................................................................................ 21 3 DYNAMIC SIMULATION OF LUMPED MODELS ................................................................22

3.1 I NTRODUCTION ........................................................................................................................22 3.2 EXPLICIT O NE-STEP SOLVERS..................................................................................................22 3.3 IMPLICIT O NE-STEP METHODS.................................................................................................28 3.4 MULTI-STEP METHODS............................................................................................................30 3.5 DIFFERENTIAL-ALGEBRAIC EQUATION SOLVERS..................................................................... 31 3.6 ODE SIMULATION IN MATLAB..............................................................................................33 3.7 SUMMARY................................................................................................................................ 33

4 DEVELOPMENT OF DISTRIBUTED MODELS......................................................................35

4.1 I NTRODUCTION ........................................................................................................................35 4.2 MASS BALANCES ..................................................................................................................... 35

4.3 E NERGY BALANCES .................................................................................................................40 4.4 MOMENTUM BALANCES...........................................................................................................43 4.5 BOUNDARY CONDITIONS ......................................................................................................... 44 4.6 SUMMARY................................................................................................................................ 45

5 STEADY-STATE SIMULATION OF DISTRIBUTED SYSTEM............................................ 46

5.1 I NTRODUCTION ........................................................................................................................46 5.2 SHOOTING METHOD.................................................................................................................46 5.3 FINITE DIFFERENCE METHOD ..................................................................................................53 5.4 SUMMARY................................................................................................................................ 59

6 DYNAMIC SIMULATION OF DISTRIBUTED SYSTEMS..................................................... 60

6.1 I NTRODUCTION ........................................................................................................................60 6.2 METHOD OF LINES ................................................................................................................... 60

6.3 SPACE DISCRETIZATION...........................................................................................................61 6.4 SUMMARY................................................................................................................................ 69

7 COMPUTER AIDED PROCESS ENGINEERING.................................................................... 70

7.1 I NTRODUCTION ........................................................................................................................70 7.2 FLOWSHEETING........................................................................................................................71 7.3 COMPUTATIONAL FLUID DYNAMICS........................................................................................76 7.4 HYSYS – A FLOWSHEETING PROGRAM ....................................................................................79 7.5 SUMMARY................................................................................................................................ 80

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8 PARAMETER ESTIMATION......................................................................................................81

8.1 I NTRODUCTION.........................................................................................................................81 8.2 LINEAR R EGRESSION ................................................................................................................81 8.3 NONLINEAR R EGRESSION .........................................................................................................85 8.4 R EGRESSION A NALYSIS............................................................................................................87 8.5 SUMMARY ................................................................................................................................92

9 INDEX..............................................................................................................................................93

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PrefaceProcess Si mul at i on usi ng MATLAB is a set of lecture notes for the courseProcess Simulation at the Department of Chemical Engineering, Lund Institute of

Technology, Lund, Sweden. Current version contains eigth chapters on modelling,

simulation methods, simulation tools and on parameter estimation. There are

corresponding exercises for the chapters and they are found on the course home page at

the site.

ht t p: / / www. chemeng. l t h. se

The course is inspired by the textbook, Computational Methods for Process Simulation,

by Fred Ramirez. This book serves as a reference in the course.

Simulation techniques presented in the course are frequently solved using MATLAB.

The solutions to examples and exercises sometimes use predefined MATLAB functions,

while others are programmed by the user in MATLAB M-files. There is no need for a

deep knowledge about MATLAB before entering the course. There is no need for

additional toolboxes to solve the exercises in the course.

The course uses some Internet sources with MATLAB M-files.

• Ramirez library of M-files to his book is found on this pageht t p: / / opt i mal . col orado. edu/ ~r ami r ez/ chen4580. ht ml

• A general textbook written by Mathews, Numerical Methods, describes a large set ofsimple numerical routines and some of them are used in the course in order to

illustrate fundamental properties.ht t p: / / t i t an. f ul l er t on. edu/ ~mat hews/ sof t war e/ sof t war e. ht m

• In a selection of M-files developed by Kelley there is a set of nonlinear solvers thatare very interesting. One of them, nsol , is modified into newsol and usedfrequently through out the course.ht t p: / / www. si am. or g/ books/ kel l ey/ kel l code. ht m

These sources are also available on the computer network.

Lund, January 10th, 2005.

Bernt Nilsson

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1 Development of Lumped Models

1.1 IntroductionThe basic properties in process systems are the conservation of mass, energy and

momentum. In process modelling this is formulised into the development of

mathematical expressions representing balance equations of mass, energy andmomentum. Conservation of a property follows the general balance expression below

In + Generation = Out + Consumption + Accumulation

Sometimes the generation and consumption terms are expressed as only one production

term as

In + Production = Out + Accumulation

This principle is illustrated in Figure 1.1 below. The definition of the system boundary

is important in order to develop the different terms in the balance equations. The system

boundary can be over a macroscopic object or over a microscopic fluid element.

mass

ener gy ener gy

mass

moment um

moment um

SYSTEM

In Out = + Accumulation

+Production

Figure 1.1: Illustration of the conservation principle.

The most convenient way of developing models is to ignore the spatial aspects of the

system that is going to be modelled. This means that all spatial properties are lumpedtogether and the model describes the integral averaged value of the properties. This is

the reason why this kind of models is called lumped models or models of lumped

systems. Models that describe spatial properties are called distributed models but they

are discussed in a later chapter.

1.2 Mass BalancesMass balances over a system can express the total mass in the system. A mass balance

can also be developed for a specific chemical component, a component mass balance or

just component balance. The only difference is that in chemical applications there is

never production of total mass while in reacting systems the component mass balance

can have production terms. In nonreacting systems there is no principle difference between total and component mass balances.

Example 1.1: Mixing Tank with two inlets

A tank has two inflows and one outflow and is seen in Figure 1.2. In to the tank

flows pure water from one inlet and pure ethanol from the other inlet. Out from

the mixing tank flows a mixture of the two components, the solution.

Model assumptions:

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• Total mass in the system is m. No limits on the amount of mass in the tank. m can be infinite large and it can be negative.

• The mass flow rates are expressed as w. In this particular case the mass flowrates are independent of the mass inside the system.

The total mass balance over the control volume now becomes

sewsewwww

dt

dm

dt

dmwww −+=⇒+=++

+=+

0

AccOutProdIn

Controlvolume

Water Ethanol

100 kg/hr 40 kg/hr

Solution

120 kg/hr

Figure 1.2: A mixing tank with two inlets.

The mass balances can often be rewritten into other variables. Below follows two

examples of such transformations.

Example 1.2: Mixing Tank with liquid volumetric inflow

Assume that the mixing tank in Example 1.1, which is seen in Figure 1.2, has

volumetric inflows and outflow.

Additional model assumption:• The density in the different fluids is ρ . Assume that the density in the solution

is constant (practical assumption in this case).

The mass balance can now be rewritten into an equation expressing the change of

tank volume.

( )

se

s

ew

s

w

sseewws

qqqdt

dV

qqqdt

V d

−+=

−+=

ρ

ρ

ρ

ρ

ρ ρ ρ ρ

Example 1.3: Mixing Tank with vapour inflow

Assume that the mixing tank in Example 1.1 has vapour inflows and outflow.Additional model assumption:

• Assume that the vapour follows the ideal gas law.• Isothermal conditions and constant volume in the vapour mixing tank.• Constant molar weight in the mixture (a practical assumption).The mass balance can now be rewritten into an equation expressing the change of

tank pressure.

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seww

sew

w

wwwdt

dp

RT

VM

wwwdt

RT

pVM d

−+=

−+=⎟

⎠

⎞⎜⎝

⎛

Component Mass Balances

It is possible to put up component balances, one for each chemical component. Note

that all component mass balances sum to the total mass balance. In order to set up a

consistent equation system it is recommended to take away one mass balance, often one

component mass balance, for instance the solvent balance. Sometimes this knowledge

can be used as redundant information to check the correctness of the model.

Example 1.4: Mixing Tank with component mass balances

Component mass balances over the mixing tank in Example 1.1, which is seen in

Figure 1.2 becomes as follows

Additional model assumptions:

• The inflows have pure chemical components. The solution has the massfraction of ethanol expressed as ω . This means that the mass fraction of water

is ω w=(1-ω ) because the sum of the fractions must be equal to 1.

( )

( )

dt

dmwwwtotal

dt

md wwwater

dt

md wwethanol

swe

sw

se

+=+

−+−=+

+=+

+=+

:

)1()1(0:

0:

AccOutProdIn

ω ω

ω ω

We see that the two component mass balances sum to form the total mass balance.

The ethanol component mass balance become

( )sse

wwdt

md ω

ω −=

The conservation of mass is valid for each chemical component. It is often more

convenient to express it in composition instead of mass. This means that the component

mass balance must be rewritten using the rule for the derivative of a product.

( )dt

dm

dt

d m

dt

md ω

ω ω +=

This manipulation is just pure mathematics. Remember that the left hand side is the

component balance and the second part of the second term on the right hand side is the

total mass balance. Let us rewrite it as

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( )dt

dm

mdt

md

mdt

d ω ω ω −=

1

This means that both the total mass balance and the component balance are used to

derive an expression of the composition change.

Example 1.5: Mixing Tank with component mass balancesThe change in composition in mixing tank in Example 1.1, which is seen in

Figure 1.2, can easily be found using the above expansion.

Additional model assumptions:

• The mass must be nonzero.

The total mass balance from Example 1.1 and the component balance from

Example 1.4 are used together to derive the change of composition.

sewwww

dt

dm−+=

( ) sse wwdt md ω ω −=

These two balances are used in

( )dt

dm

dt

d m

dt

md ω

ω +=

to form a composition equation

( ) ( )

( ) ω ω

ω ω

ω

m

w

m

w

www

m

ww

mdt

d

we

sewse

−−=

−+−−=

1

1

Note that this expression is only valid when the total mass is nonzero (positive).

Component mass balances are often rewritten into component mole balances, using m =

M n. It can further be rewritten using concentrations, using m =M n = M V c. This

means that a component mass balance can be transformed into a component mole

balance just by dividing all terms with the molar weight of the chemical component.

Example 1.6 : Mixing Tank expressed in concentration changesA tank with constant volume has an inflow with a volumetric flow rate, q, and an

inflow concentration, cin. Develop a mathematical expression for the

concentration in the tank.

Model assumptions:

• Constant volume in the tank.• Ideal mixing and that the concentration in the tank is equal every where and

that the outflow concentration is the same as the tank concentration.

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( )ccV

q

dt

dc

dt

dc MV qcqc

dt

MVcd c Mqc Mq

dt

dmww

in

out in

out out inin

out in

−=

⇒+=

+=

+=

+=

)(

AccOutIn

qVc

in c

out

Figure 1.3: A Mixing Tank

1.3 Systems of Mass BalancesModels of process systems with sets of subsystems generate equation systems of mass

balances. This means that the equation set is coupled and the equations must be solved

simultaneously.

Example 1.7 : Two Tanks in Series

Two well-mixed tanks are connected in series. They have constant but different

volumes. They are seen in the figure below.

Model assumptions:

1. Constant volumes and the same volumetric flow through the tank series.2. Well-mixed tanks, which means that the outlet concentration is the same as

the tank concentration.

Figure 1.4: Two-tank series .

Material balances over one chemical component, expressed in moles, in the two

tanks becomes

( )

( )

In Out Acc= +

= + ⇒ = −

= + ⇒ = −

qC qC V dC

dt

dC

dt

q

V C C

qC qC V dC

dt

dC

dt

q

V C C

in in1 1

1 1

1

1

1 2 2

2 2

2

1 2

and rewritten on state-space form using matrices it becomes

d

dt

C

C

q

V

q

V

q

V

C

C

q

V C in1

2

1

2 2

1

21

0

0

⎡

⎣⎢

⎤

⎦⎥ =

−

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣⎢

⎤

⎦⎥ +

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

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At steady state the accumulation terms are zero and in this case this is the same as

the concentration derivatives are zero. Instead we get the following algebraic

equation system (linear in this case). Note that the solution is trivial.

inC V

q

C

C

V

q

V

q

V

q

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−=⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢

⎢

⎣

⎡

−

−

0

0

12

1

22

1

Example 1.8: Six-Plate Absorption Column

An absorption column with six plates is used to absorb a chemical component in a

vapour into a liquid. The liquid, L, enters at the top and the vapour, V, enters at

the bottom. It is assumed that the liquid and vapour flow rates do not change

through the column.

1

2

3

4

5

6

L, x0

L, x6

V, y1

V, y7

Tray i

xi

yi

xi-1

yi+1

Figure 1.5: Six-plate adsorption column. left) column configuration, right) column tray

A steady-state component material balance over a tray givesIn Out=

+ +− + Lx Vy = Lx Vyi i i i1 1

Assume a linear equilibrium relation between liquid and vapour composition.

y = ax bi i

+

Let us put up the six component balances together with the equilibrium relation

In Out=

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + +

1

2

3

4

5

6

0 2 1 1

1 3 2 2

2 4 3 3

3 5 4 4

4 6 5 5

5 7 6 6

: ( ) ( )

: ( ) ( )

: ( ) ( )

: ( ) ( )

: ( ) ( )

: ( )

Lx V ax b = Lx V ax b





Lx Vy = Lx V ax b

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1.4 Energy BalancesIn process systems chemical components are transformed into other chemicals under

energy changes. Energy is released or consumed in the reaction step and it is often

consumed in the separation step. Energy balances are important parts of the process

model in analysis and design. Total energy can be decomposed into different energy

forms, similar to the decomposition of total mass into component masses. The most

common decomposition of total energy is into thermal energy and mechanical energy.There are other energy forms like electrical and magnetically energy.

A general total energy balance can be written as follows

( )( ) ( )( ) ( )( )mPK U dt

d QPK U W W QQPK U out out out out out pV Pininininin ++++++=++++++

+=+

ω ω

AccOutProdIn

U is the internal energy, K is the kinetic energy and P is the potential energy, all three

per mass unit. The three energy forms enter and leave the system in the process streams.

Energy can also be transferred into and from the system, Qin and Qout and it can be produced in the system through reaction and phase change, Q p. The work done by the

system on the surroundings is shaft work, W , and pressure/volume work, W pV . The most

common situation is when the internal energy dominates the total energy balance. Then

kinetic and potential energy are neglected and we get

( ) ( )

W QQQ p pU U

dt

mU d

dt

dmU QU W W QQU

out inPout

out

out

in

in

in

out out inin

out out out pV Pininin

−−++−+−=

⇒

++=−−++

+=+

ρ

ω

ρ

ω ω ω

ω ω

)(

AccOutProdIn

The enthalpy of a system can be expressed as the sum of the internal energy and the

pressure-volume

ρ

pU H +=

If we use this and assume that the change of the pressure-volume is neglectable the

internal energy balance can be expressed as an enthalpy balance

W QQQ H H dt

mH d

W QQQ p p

p

H

p

H dt

p H md

out inPout out inin

out inPout

out

out

in

in

in

out

out

in

in

−−++−=

−−++−+⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

−−⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

−=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

ω ω

ρ

ω

ρ

ω

ρ ω ρ ω

ρ

)(

Example 1.9: Mixing Tank with one inlet

A tank with constant volume has one inflow with a volumetric flow rate, q, and an

inflow temperature, Tin. Develop a mathematical expression for the enthalpy in

the tank. Model assumptions are constant volume in the tank and ideal mixing.

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The enthalpy in the tank is equal every where and that the outflow enthalpy is the

same as the tank enthalpy. The density is constant.

( )hhV

q

dt

dh

dt

dhV qhqh

dt

hV d hqhq

in

in

out out out ininin

−=

⇒

+=

+=

+=

)(

AccOutIn

ρ ρ ρ

q VTin

Tout

Figure 1.6 : A Mixing Tank

Assume also that the enthalpy can be expressed as proportional to the temperature

difference from reference temperature as

T C H m p ∆= ,

The enthalpy balance can be rewritten into a temperature expression

( ))( ) ( )( )

( )T T V

q

dt

dT

T T C T T C V

q

dt

T T C d

in

m pininm p

m p

−=

−−−=−

0,0,

0,

This is only valid under the assumption that the heat capacity is (or almost)

constant.

Mechanical Energy BalancesThe neglected kinetic and potential energies in the total energy balance can sometimes

be expressed almost independent of the enthalpy balance. For incompressible flows this

is described in the famous Bernoulli equation, which is a mechanical energy balance.

Mechanical energy balances are not discussed here.

1.5 Momentum BalancesThe conservation of momentum is not so important in lumped model development of

process systems. The reason is that it is unusual to have moving lumped models.

Dynamic momentum balances also occur in mechanics. Steady-state momentum

balances are also called force balances.

( )dt

mvd vF v out out lossinin +=−

+=+

ω ω

AccOutProdIn

One major difference between mass and energy balances compared to momentum

balance is that momentum has direction. This means that we can put up one momentum

balance for each dimension in the modelling problem. Momentum balances becomes

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much more important in distributed models and is the fundamental issue in

Computational Fluid Dynamics, which will be discussed more in a later chapter.

1.6 SummaryLumped models are based on the assumption that spatial effects can be neglected.

Lumped models are often composed of a set of mass, component and thermal energy balances. Under the assumption that we are studing steady-state behaviour the lumped

model results in a algebraic equation system, which is discussed in the next chapter.

Unsteady-state or transient behaviour studies results in a set of ordinary differential

equaitons in time. Simulation of unsteady-state models is discussed in chapter three.

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2 Steady-State Simulation of Lumped Models

2.1 IntroductionA steady-state lumped model results in a set of algebraic equations. These algebraic equations

describe static relations between a set of variables. The equation system can be solved if

the number of equations is the same as the number of variables. For computationalreasons one often differs between linear and nonlinear equation systems. Nonlinear

steady-state lumped models can be written on a general form as

f (x , x , ..,x ) =

f (x , x , ..,x ) =

f (x , x , ..,x ) =

n

n

n n

1 1 2

2 1 2

1 2

0

0

0

...

or on a more compact representation on vector form

f x 0( ) =

If the algebraic equation system is linear it can have the following representation

a x a x a x u =

a x a x a x u =

a x a x a x u =

n n

n n

n n nn n n

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

0

0

0

+ + + −

+ + + −

+ + + −

...

...

...

...

or

A ux =

using matrix notation. First we discuss one numerical method for the solution of linear

algebraic equation system, namely Gaussian elimination, and then different nonlinear

equation solvers are discussed.

2.2 Method for Linear EquationsThe general method for simultaneously solving linear equation systems are based on

Gaussian elimination. This is done by performing a decomposition of the A-matrix into

one lower and one upper triangular matrix, a so called LU-decomposition.

LU-decomposition

In MATLAB LU-decomposition is done by the command l u( A) . The l u-commandreturns the two triangular matrices, L and U, and a partitioning matrix, P. Partitioning isdone if the diagonal element, the pivots, becomes zero or very small and it is done by

reordering the rows. To solve the linear equation system, A x = b, in MATLAB one can

write x = A\ b. Note the backslash notation on the left hand side which is interpreted

as left division of vector b by matrix A. This command implicitly makes a lu-

decomposition with pivoting.

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An absorption column with six plates is used to absorb a chemical component in a

vapour into a liquid. The liquid, L, enters at the top and the vapour, V, enters at

the bottom. It is assumed that the liquid and vapour flow rates do not change

through the column.

1

2

3

4

5

6

L, x0

L, x6

V, y1

V, y7

Tray i

xi

yixi-1

yi+1

Figure 2.2: Six-plate adsorption column. left) column configuration, right) column tray

A steady-state component material balance over a tray gives

In Out=

+ +− + Lx Vy = Lx Vyi i i i1 1

Assume a linear equilibrium relation between liquid and vapour composition.

y = ax bi i +

Let us put up the six component balances together with the equilibrium relation

In Out=

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + +

1

2

3

4

5

6

0 2 1 1

1 3 2 2

2 4 3 3

3 5 4 4

4 6 5 5

5 7 6 6

: ( ) ( )

: ( ) ( )

: ( ) ( )

: ( ) ( )

: ( ) ( )

: ( )






Lx Vy = Lx V ax b

This can now be rewritten with unknowns on the left and remember that x0 and y7

are known values.

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( )

( )

( )

( )

( )

( )

− + + −

− + +

− + +

− + +

− + +

− + −

L Va x Vax = Lx

Lx L Va x Vax =

Lx L Va x Vax =

Lx L Va x Vax =

Lx L Va x Vax =

Lx L Va x = Vy

1 2 0

1 2 3

2 3 4

3 4 5

4 5 6

5 6 7

0

0

0

0

This can now be put up on matrix form as A x = b.


A linear model of a six-plate absorption column is developed in Example 1.1

above. The mathematical problem becomes a linear equation system and it can be

solved by MATLAB in the following way:

a = 0. 72;b = 0;

V = 66. 7;L = 40. 8;x0 = 0;y7 = 0. 3;eyel ow = di ag( ones( 1, 5) , - 1) ;A = - ( L+V*a) *eye(6) + L*eyel ow+ V*a*eyel ow' ;u = [ - L*x0 0 0 0 0 - G*a*( y7- b) / a] ' ;x = A\ u

x =0. 09210. 17030. 23680. 29330. 3413

0. 3820

Note the creation of a matrix with ones on the lower off diagonal, eyel ow. It isalso used to create the elements in the upper off diagonal and this is done by

transposing the matrix, eyel ow' . Note also that the u-vector is defined as a row

but it is transposed at the end.

Analysis of Linear Equation

Analysis of the equation system gives insight regarding the error in the solution. There

are some tools available in MATLAB for this. The rank of a matrix is calculated by the

r ank-command. The rank measurement indicates if there are singularities in the problem formulation, like redundant or contradictory equations. This is not so

interesting in this stage of the problem solving. A more interesting measurement is a

numerical value of how close we are a singular equation system. The condition number

is a measurement of the quality of the solution. The reciprocal condition number is

found by the r cond-command. The rcond value is between 1 (good) and 0 (bad).Values of the r cond down to 0.0001 are common.

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Example 2.3: Analysis of the Six-Plate Adsorption Column

The rank and the reciprocal condition number of the problem are calculated by the

following commands:

r = r ank( A)

r =6

c = rcond(A)

c =0. 0508

This condition number is satisfactory. Bad condition numbers are at least 3

magnitudes lower. Make it a good habit to always calculate the rcond-value for

linear problems.

2.3 Methods for Nonlinear EquationsIt is much easier to solve linear than nonlinear equations. In solving linear equationssystem there are one dominating method, Gaussian elimination. For nonlinear equations

there are a number of different methods that are good for different problems. A number

of different nonlinear solvers are discussed in Ramirez, section 2.3. In this text we

present three different methods. First we illustrate the use of a predefined equation

solver in MATLAB for one-dimensional problems called f zer o. For many nonlinear problems the Newton method are used. A multidimensional Newton solver, called

newsol , is presented .

One-dimensional Nonlinear Problems

There are a number of well known equation solvers for nonlinear one-dimensional

problems. Three similar methods are presented below. The first is the bisection method.The idea is very simple. Choose an interval where the function has different values.

Calculate the value of the function in the middle of the interval. The zero is now in the

interval where the function has different sign. The procedure can now be repeated until

the interval is small enough. The procedure is illustrated on the left in Figure 1.

a

b

c

(a)

(b)

(c)

a

b

c

(a)

(b)

Figure 2.2 - right) Bisection method, left) Regula Falsi method.

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Another simple method is the Regula Falsi or linear inverse interpolation. The idea is to

make an interpolation between two function values with different sign. The zero of the

interpolation is now used as a new function value. The method is illustrated on the right

in Figure 1.

fzero

The fzero method in MATLAB can be use for nonlinear problems in one unknownvariable. It is a combination of bisection and secant methods (using inverse quadratic

interpolation). The secant method is a simplified Newton method.

Example 2.4: Heat Capacity for Carbon Dioxide

Calculate the temperature where the heat capacity for carbon dioxide is equal to a

given value, in this case C p = 1. The temperature dependency can be expressed

with the following equation (T in Kelvin, C p in kJ/kg K)

C T T

p = − −1716 4 257 10

15046

. . *.

Nonlinear problems in general are represented as functions in MATLAB. This

means the creation of a M-file. Our small example becomes a M-file with thesame name as the function name, Cpf un. This function requires one inputvariable, the unknown variable, in this case T. The function returns the residual,i.e., the deviation from zero.

f uncti on r es = Cpf un( T)r es = 0. 716 - 4. 257e- 6*T - 15. 04. / sqr t ( T) ;

The Cpf un can now be used in the nonlinear equation solver fzero. The firstinput variable to f zero is a string with name of the function (M-file). Thesecond is the starting guess of the unknown variable, in this case T.

T = fzero('Cpfun',600)

T =443. 5712

Note that fzero can only find the zero for a nonlinear function in one unknown.

In nonlinear problems with more than one unknown other types of solvers are needed.

One simple and often used method is direct substitution. It can be used in multi

dimensional problems but it is more used in problems with one unknown. The most

dominating method is a multidimensional version of the Newton method.

Direct substitutionA third simple method is called direct substitution and this is very common method.

The algebraic problem is rewritten as

x xg x f −== )()(0and put on the following form

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The procedure starts with a guess of x0. The initial guess is used to calculate g(x0) and

the answer is a new value of x, i.e. x1, and the procedure can now be repeated until x

converges. Direct substitution has problems for equation systems with derivatives

greater than one, as seen in Figure 1.3. The problem on the right converges quite rapidly

while the problem on the left diverges.

x0x1x2

d i a g

o n a l

g(x0)

g(x1)g(x2)

x0 x1 x2

d i a g

o n a l

g(x0)

g(x1)

Figure 2.3 - right) Converging direct substitution , left) Diverging direct substitution.

2.4 Newton MethodsThe most common method for nonlinear systems is the Newton method or Newton-

Raphson method. The Newton method is based on Taylor expansion of the function

with the truncation of all terms with order higher than one, HOT.

HOT x x x f x f x f +−+= ))((')()( 0100

At the solution the function f is zero and let us through away higher order terms.

))((')(0 0100 x x x f x f −+=

The approximation can now be rewritten where a new x-value are calculated based on a

function evaluation, f , and its derivative, f' . The calculations are now repeated in order

to find a new even better value. A graphical interpretation is seen in Figure 2.4.

)('

)())((')(

0

0010100

x f

x f x x x x x f x f −=⇒−=−

x0x1x2

f'(x0)

Figure 2.4 – The Newton-Raphson method.

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The Newton method converges to a solution if we start close enough. For some

problems it is hard to find good starting guesses in order to get a desired Newton

solution. Other problems can be slow converging rates and some times the derivative

goes to zero and the problem becomes singular.

Problems are very often multidimensional, i.e., contains more than one unknown.

Sometimes it is possible to rewrite multidimensional problems as a one-dimensional

problem by making symbolic manipulations. Even when it is possible it may beundesirable.

A multidimensional nonlinear problem is written as iteration of a linear equation

system. Define J as the Jacobian matrix, which is a matrix with the partial derivatives as

elements

f x J x' ( ) ( ) x x x

x x

f

x

f

x

f

x

f

x

0

0 0

0 0

0

1

1

1

2

2

1

2

2

= =

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

∂

∂

∂

∂

∂

∂

∂

∂

Note that the notation x1

and x2

is the definition of the two unknowns. Make amultidimensional Newton method and rewrite it as a linear equation system

)()()())(()( 000100100 xf xxJxxJxxxJxf −=⇒−=−

To find a new x-vector, x1, one have to solve A x1 = b where A = J and b = J x0 + f . In

other words the Newton method for nonlinear equation systems solves a linear equation

system in each iteration. Below are two different Newton methods briefly presented,

namely new2di m and newsol . The routine new2di m is a Newton solver fornonlinear problems in 2 unknowns and it is developed by Mathews. It requires one

function describing the nonlinear problem and one function describing the Jacobian.

The second method,newsol

, is described in the following subsection.

Example 2.5: Exothermic CSTR

An exothermic reaction occurs in a continuous stirred tank reactor, CSTR.

Assume constant volume, V , and flow rates, F . The CSTR is cooled by a coolant

and it is also assumed that the coolant temperature is constant, T C .

F, C A0

, T0 F, C A, T

TC

Figure 2.5 - Continuous stirred tank reactor.

A steady-state model with one component material balance and one energy

balance is as follows

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( )

In Prod Out+ =

+ =

+ − = + −

FC VR FC

FC C H VR FC C kA T T

A A

p A p A C

0

0 ρ ρ ∆ ( )

Assume also a first-order reaction and an Arrhenius temperature dependency as

R k C e A

E

RT

a

= −−

0

The model can be rewritten in dimensionless form as follows

0 1

0

501

2

1

0 1

501

2

1

= e

= u ( ) ec

− − ⋅

− − − + ⋅

−

−

ξ ξ

η η η η ξ

η

η

( )

( )

( )

where ξ and η are the unknown dimensionless concentration and temperature.This is implemented in the following MATLAB functions, cstrss for the

problem and cstrssJ for the Jacobian. Note that the input variable is a vector, x

= [ ξ; η ] and the output variable is also a vector containing two values.

f uncti on [ res]=cst rss( x)% CSTRSS i s a di mensi onl ess st eady- st at e% l umped model of an exot her mi c cont i nuous% st i r r ed t ank r eact or . See Rami r ez ex. 2. 3

% par amet ers u1=2;nyc=1. 75;ny0=1. 75;% unknown var i abl esksi =x(1) ;ny=x( 2) ;% equat i ons P=ksi . *exp( 50*( 0. 5- 1. / ny)) ;Us=u1. *( ny- nyc) ;r es=[ 1- ksi - P; (ny0- ny)- Us+P];

f uncti on [ J ] =cst rss J (x)% CSTRSSJ cal cul at es t he J acobi an f or t he% di mensi onl ess st eady- st ate l umped model% f or an exot her mi c cont i nuous st i r r ed% t ank r eact or. See Rami r ez ex. 2. 3

% par amet ers

u1=2;nyc=1. 75;ny0=1. 75;% equat i ons ksi =x(1) ;ny=x( 2) ;% equat i ons Pksi =exp( 50*( 0. 5- 1. / ny)) ;Pny=( 50. / ny. 2̂) *ksi *exp( 50*( 0. 5- 1. / ny)) ;Usny=u1; J =[ - 1- Pksi , - Pny; Pksi , - ny- Usny+Pny] ;

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This is now solved by the new2di mfunction

sol = new2dim('cstrss','cstrssJ',[1 1],1e-6,1e-6,40)

sol =0. 9676 1. 7608

newsol

Advanced Newton solvers have routines for numerical differentiation. The newsol MATLAB M-file is a method of this kind and it is a modification of a solver developed

by C.T.Kelley, called nsol . The user only have to specify the problem on residualform as r = f(x), i.e. a set of algebraic equations. The numerical Jacobian is calculated

by the diffjac function and it is possible to change the updating mechanism of the

numerical Jacobian. There is no Newton solver in the basic MATLAB program but

there is one solver in Optimization toolbox called f sol ve which is similar to

newsol .The newsol Newton solver is used in the following ways. The solver only needs thename of the user defined function and a column vector of initial guesses and it returns

the solution column vector as

sol = newsol ( ' cstr ss' , [ 1 1] ' ) ;

There are some optional input arguments to newsol which makes it possible to settolerances, t ol =[r t ol ato l ] (relative/absolute tolerance), and to specify the Jacobian

update mechanism, par am=[ 40 1 0 0] , for more information use the help command. It

is possible to parameterise the function with the mpar input argument. This means that it

is possible to set parameters from the outside in the user defined function. The method

returns the solution, sol , the solution history, sol hi st , and an error indicator for a

successful iteration, i er r . It is also possible to return variables from the user defined

function to the outside through a fourth output variable, like mvar . A complete function

call to newsol look as follows

[ sol , sol hi st, i err , mvar] =newsol ( ' cst rss' , [ 1 1] ' , t ol , par am, mpar ) ;

Example 2.6 : Exothermic CSTR, continue

The nonlinear problem in Example 2.5 can be solved by the Newton solver

newsol . The solver starts at [ 1 1] . Note that newsol only needs the function

cstrss.

sol = newsol ( ' cst rss ' , [ 1 1] ' )sol =

0. 96761. 7608

2.5 SummaryIn the course we solve linear equation systems with Gaussian elimination as x = A\ b.

Nonlinear equations can be solved by fzero, the predefined MATLAB function. Thenonlinear equation system solver, newsol , is used frequently throughout the course.The solver is available on the course home page on the web.

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3 Dynamic Simulation of Lumped Models

3.1 IntroductionAn unsteady-state lumped model results in a set of ordinary differential equations, ODEs. These

equations describe time-dependent relations between a set of variables. Nonlinear

unsteady-state lumped models can often be written as ODEs on state-space form withderivatives of the states explicitly on the RHS like

dx

dt f x x u u

dx

dt f x x u u

n n

n

n n n

1

1 1 1

1 1

=

=

( ,..., , ,..., )

...

( ,..., , ,..., )

or on a more compact representation on vector form

d

dt

xf x u= ( , )

(the general form, F(d x /dt, x ,u) = 0, is not discussed in this course).

If the ODE system is linear (and has one input u) it has the following representation

dx

dt a x a x a x b u

dx

dt a x a x a x b u

n n

n

n n nn n n

1

11 1 12 2 1 1

1 1 2 2

= + + + +

= + + + +

...

...

...

or using matrix notation it becomes

d

dt A B

xx u= +

The difference between linear and nonlinear ODEs is not an important issue here as it

were in the steady-state case. Solvers of ODE systems are categorised into one-

step/multi-step methods and into explicit/implicit methods. The different methods and

their properties are briefly discussed in this chapter. First, explicit one-step solvers are

discussed, as they are simple and commonly used, followed by a short discussion about

implicit one-step solvers. Finally, the multi-step methods are discussed.

3.2 Explic it One-Step SolversThe simplest ODE solver uses the explicit Euler method. More accurate explicit

methods are called Runge-Kutta methods which is a family of methods. These methods

are the most common methods for ODE systems. A one-step method uses only one

previous step to calculate a new step. An explicit method uses an explicit formula to

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calculate a new step, i.e. a direct calculation without internal iteration or equation

solving.

Euler method

The simplest ODE solver uses the Euler method. Assume an ODE expressed on state-

space form as,

dx

dt f x= ( )

Make a Taylor expansion of x(t+h) where h t t t i i= = −+∆ 1 and truncate theexpansion after the first derivative. The first derivative is equal to the function value.

x t h x t dx

dt t h t h x t h x t f x t h( ) ( ) ( ) ( ) ( ) ( ) ( ( ))+ = + + ⇒ + ≈ +φ 2

This approximation gives us the following explicit one-step method.

x x f x

x x

( ) ( ) ( ( ))

( )

t t h t

t

i i i+ = +

=1

0 0

One drawback with this method is that the error decreases proportional to the step-size.

This results in small step-sizes in order to keep the error low. A related problem is that

the solution can become unstable if the step-size is to large, see Ramirez section 3.1.

Assume a linear ODE written on state-space form and that it is stable, i.e. K > 0.d

dt

xKx= −

An Euler approximation gives the following expression

x x Kx K xi i i ih h+ = − = −1 1( )

The stability of this discrete system is determined by the size of the coefficient in front

of the state variable, x. This coefficient is the eigenvalue for this simple system. If the

coefficient is less than one then the solution decreases and the solution is stable and if it

is larger than one then the solution x increases for each calculation and the solution is

unstable.

1 1 1 0

1 1 1

1

1

− < ⇒ = − < ⇒ →

− > ⇒ = − > ⇒ → ∞

+ ∞

+ ∞

h x h x x x

h x h x x x

i i i

i i i

K K

K K

( )

( )

We get the following step-size criteria on h for the stability limit.

1 1 22

− = ⇒ − = − ⇒ =h h hK K K

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The step-size for the Euler method for a linear ODE must be less than a given value in

order to give a stable solution.

h <2

K

Example 3.1: Concentration Dynamics in a Mixing Tank Using the Euler MethodA tank with constant volume, V=1, and inflow, q=1 and inflow concentration

cin= 1, has the initial concentration equal to zero.

( )

( )ccV

q

dt

dc

ccqdt

dcV

cqcqdt

Vcd

in

in

out out inin

−=

−=

−=)(

Figure 3.1: A Mixing Tank

This is expressed in the following M-file.

f uncti on [ dc] =onet ank(t , c)% ONETANK i s a model of t he concent r at i on var i at i on % i n a cont i nuous t ank. % par amet ers ci n = 1;q = 1;V = 1;% equat i ons dc = - q/ V*c + q/ V*ci n;

An explicit Euler method called eul er s , is found in Mathews, Numerical Methods. The one tank model can be simulated by the Euler method with the

following settings, from 0 to 10, with initial concentration 0 and with 20 steps:

[ t , y] =eul ers( ' onet ank' , 0, 10, 0, 20) ; TY = [ t ' y' ]

TY =0 0

0. 5000 0. 50001. 0000 0. 75001. 5000 0. 87502. 0000 0. 93752. 5000 0. 96883. 0000 0. 98443. 5000 0. 99224. 0000 0. 99614. 5000 0. 9980

5. 0000 0. 99905. 5000 0. 99956. 0000 0. 99986. 5000 0. 99997. 0000 0. 99997. 5000 1. 00008. 0000 1. 00008. 5000 1. 00009. 0000 1. 00009. 5000 1. 0000

10. 0000 1. 0000

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0 2 4 6 8 10

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

time

c

o

n

c

e

n

tr

a

ti

o

n

Figure 3.2: Simulation of the concentration in a tank using Eulers method (5, 10, 20 and 100 steps).

The solution is shown in Figure 3.2. The plot illustrates the solutions with 5, 10, 20 and100 steps. Figure 3.2 indicates that the solutions depend on the choice of step-size. The

solution goes unstable for step-sizes larger then 2. The solution based on 100 steps

(step-size = 0.1) are close to the exact solution.

Runge-Kutta methods

To increase the step-size the derivative approximation must include higher order terms.

In Runge-Kutta methods this is done by taking the average of a number of function

derivatives in the current step. The local error decrease with the number of derivatives

used. A fourth order Runge-Kutta solver called rk4, found in Mathews Numerical

Methods, is used to simulate the mixing tank example below.

0 2 4 6 8 10

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time

c

o

n

c

e

n

tr

a

ti

o

n

Figure 3.3: Simulation of a tank using Eulers method (1000 steps) and RK4 method (10 steps)

Example 3.2: Concentration Dynamics in a Continuous Tank, continue.

Two solutions of the mixing tank concentration dynamics are shown in Figure 3.3.

The solid line solution is based on the Euler method with 1000 steps and the

dashed line is based on 4-order Runge-Kutta, r k4, with 10 steps. As can be seen

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in this case that the Runge-Kutta method performs very well. It uses only 1% of

the steps compared with Euler. On the other hand it uses 4 function evaluations in

each step, i.e. 40 function evaluations which is 4% compared to the 1000 function

evaluations in the Euler method used in this case. An Euler solution based on 10

steps is seen in Figure 3.2.

Below follows a brief presentation of a second order Runge-Kutta method. The basicRunge-Kutta formula is as follows.

x t h x t h x t e( ) ( ) ( ( ))+ = + φ

A new step is explicitly expressed as a function of the previous step. The choice of the

φ e

x t ( ( )) function is different for different RK methods. A second order RK method has

the following expression

φ = +ak bk 1 2

where the k -coefficients are different function evaluations

k f t xi i1 = ( , ) k f t ph x q x f t ph x qhk i i i i i2 1= + + = + +( , ) ( , )∆

Make a Taylor expansion of the second function evaluation, k 2. ( HOT represents higher

order terms.)

k f t x phf t x qhf t x f t x HOT i i t i i i i x i i2 = + + +( , ) ( , ) ( , ) ( , )

Now use the two function evaluation expressions in the Runge-Kutta formula

( ) x t h x t h ak bk ( ) ( )+ = + +1 2

and rewrite the expression as

( )

( )

x x h af t x bf t x

h bpf t x bqf t x f t x h

i i i i i i

t i i i i x i i

+ = + +

+ + +

1

2 3

( ) ( )

( ) ( ) ( ) ( )

, ,

, , , Ο

In order to find the parameters, a, b, p and q the RK formula is compared with a Taylor

expansion of the original ODE

( )

x t x t hf t x t

h f t x t f t x t f t x t h

i i i i

t i i x i i i i

( ) ( ) ( , ( ) )

( , ( ) ) ( , ( ) ) ( , ( ) ) ( )

+ = +

+ + +

1

2

3

2Ο

We now have two expressions, one is the RK approximation and the other is the Taylor

expansion of the original equation. If the different terms in the both expressions are

assumed to be equal then we get the following expressions

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( )

( )

( )

x t x

f t x t a b f t x

f t x t f t x t f t x t

bpf t x t bqf t x t f t x t

i i

i i i i

t i i x i i i i

t i i x i i i i

( )

( , ( ) ) ( , )

( , ( ) ) ( , ( ) ) ( , ( ) )

( , ( ) ) ( , ( ) ) ( , ( ) )

+ =

= +

+ =

+

1

1

2

These expressions give

a b bp bq+ = = =11

2

1

2; ;

This means that three RK parameters are given when the fourth parameter is assumed. If

we choose b=0 then it gives the Euler method. If we choose b=1 instead then it gives

a=0 and p=q=0.5 which is a midpoint approximation. The choice b=0.5 results in

a=0.5 and p=q=1. This RK method uses two function evaluations, one on each side of

the interval.

The Runge-Kutta method discussed above is a second order method based on two

function evaluations on the interval. Other RK methods like fourth order RK4 methodsuses four function evaluations on the interval. The error decreases with the order, i.e.

the number of function evaluation on the interval.

Variable Step-Size

To keep the step-size fixed at a given value is not common practice. Instead almost all

professional methods have variable step-sizes based on some kind of step-size control.

The step-size control is often based on the required error tolerance of the problem.

Runge-Kutta solvers have error estimators included in the code. These estimators must

have higher order than the actual solver. Therefore a second-order Runge-Kutta is

equipped with a third-order error estimator and a solver like this is called rk23 or as in

MATLAB ode23. In MATLAB there is also a fourth-order Runge-Kutta solver called ode45. (In MATLAB 4 and older versions there are two kinds of ODE-solvers, ode23 and ode45. These are not used in this course. Instead there are new ode23 and

ode45 in the MATLAB ODE Suite. From a user point of view the only difference is

the set of input variables.)

RK methods in MATLAB

ODE-solvers in MATLAB have the following simplified syntax:

[ T, Y] = ODE23( ' F' , TSPAN, Y0) ;

TSPAN is a vector with the start and stop time, e.g. [0 10], and Y0 is the initial values.

Example 3.3: Concentration Dynamics in a Mixing Tank

The simple mixing tank is solved by the ode23 method in the following way. Notice that the solution has many steps in the beginning of the solution. At the

end there is only a few steps and in total 22 step with guaranteed error estimates.

The default tolerances are a relative error tolerance of 1e-3 and an absolute error

tolerance of 1e-6. The ode45 solver works exactly in the same way.

[ t , y]=ode23( ' onet ank' , [ 0 10] , 0) ;

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TY=[ t y]

TY =0 0

0. 0001 0. 00010. 0005 0. 00050. 0025 0. 00250. 0125 0. 01240. 0625 0. 06060. 1791 0. 16400. 3485 0. 29430. 5668 0. 43280. 8350 0. 56631. 1574 0. 68601. 5431 0. 78682. 0077 0. 86642. 5790 0. 92513. 3139 0. 96494. 3139 0. 98835. 3139 0. 99616. 3139 0. 99877. 3139 0. 99968. 3139 0. 99999. 3139 1. 0000

10. 0000 1. 0000

3.3 Implici t One-Step Methods Explicit solvers uses current function evaluations to predict a new step, like

xi+1 = xi + h f(xi).

Implicit solvers are based on the idea to use the new step in the function evaluation to

generate the new step, like

xi+1 = xi + h f(xi+1),

which in turn generates an implicit equation system. This means that an implicit method

solves a nonlinear equation system in each step. This results in very good stability properties. In MATLAB, ode23s is an implicit second-order Rosenbrock solver.

Example 3.4: Two Tank Series and Stiff Problems

Two tanks connected in series are simulated with different volumes. The total

volume is constant. In the first case the volumes are the same and in the second

case the first tank has 0.1% of the volume of the second tank. This is clearly

illustrated in the figure below.

Figure 3.4: Two-tank series left) same volumes right) different volumes.

Material balances over one chemical component in the two tanks becomes

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( )

( )

In Out Acc= +

= + ⇒ = −

= + ⇒ = −

qC qC V dC

dt

dC

dt

q

V C C

qC qC V dC

dt

dC

dt

q

V C C

in in1 1

1 1

1

1

1 2 2

2 2

2

1 2


d

dt

C

C

q

V

q

V

q

V

C

C

q

V C in1

2

1

2 2

1

21

0

0

⎡

⎣⎢

⎤

⎦⎥ =

−

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

⎡

⎣⎢

⎤

⎦⎥ +

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

The first case solved by ode23 is seen on the left in Figure 3.5. The solution by

ode23s of the second case is seen on the right in the same figure.

f uncti on [ dc] =t wot anks(t , c)

% TWOTANKS i s a model of t he concent r at i on var i at i on % i n a t ank ser i es of 2 t anks.

% par amet ers c0 = 1;q = 1;V1 = 0. 5; Vt ot = 1;V2 = Vt ot - V1;% equat i ons dc = [ - q/ V1 0; q/ V2 - q/ V2] *c + [ q/ V1; 0] *c0;

Numerical statistics is clearly illustrated in the table below. In the first case the

two solvers take almost the same amount of steps but the nonlinear solver in the

implicit ode23s makes almost twice as much floating point operations, called flops, compared with the explicit Runge-Kutta ode23. In the second case theimplicit ode23s takes some more steps and require 20% more flops comparedwith the first case. The explicit ode23 needs almost 15 times more flops andneeds an enormous amount of steps.

Table 3.1: Number of steps and floting point operations during the solution of a two-tank problem

with stiffness properies.

V1 = 0.5 V1 =0.001

ode23 33 steps, 4968 flops 3995 steps, 63932 flops

ode23s 35 steps, 9453 flops 41 steps, 11143 flops

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0 2 4 6 8 10

0

0 2

0 4

0.6

0.8

1

1.2

1 4

time 0 2 4 6 8 10

0

0 2

0.4

0.6

0 8

1

1 2

1.4

time Figure 3.5: Simulation of a two-tank series left) tanks with equal sizes

right)volume of first tank is 0.1% of the second

The implicit solver is superior in cases when fast dynamics die out and the interesting

dynamics is slow. Problems with these kinds of characteristics are called stiff problems. Note that a simulation of the second case between 0 and 0.01 is not a stiff problem.

3.4 Multi-Step MethodsTo increase the accuracy in a RK-method we have to increase the order of the method

which increases the number of function evaluations needed in each step. One way to

increase the order without increasing the function calls is to reuse previous steps. This

means using one new function evaluation together with a number of old ones. There are

both explicit and implicit multi-step methods.

Adams Methods

Explicit multi-step methods are often called Adams-Bashforth methods. These methodsare rarely used because they have very small stability regions and they require very

small step-sizes. Methods like this are often modified. The explicit multi-step formula is

used as a prediction of the new step. This predicted value is then used in an implicit

multi-step formula, which correct the predicted value. This method uses direct

substitution to solve the nonlinear equation. In this way we get a semi-implicit multi-

step method with good performance. Methods like this are called predictor-corrector

methods or Adams-Bashforth-Moulton methods or just Adams methods. These methods

can also include variable step-size and variable order, both controlled by error

tolerances. In MATLAB there is a variable step-size variable order multi-step method

(Adams-Bashforth-Moulton) called ode113.

Example 3.5: Two Tank Series and High Accuracy

If the error tolerances are decreased higher order methods are favourable. The

simple two tank problem from Example 3.4 illustrates this in the table below. If

the default tolerances are used the lowest order Runge-Kutta is the most economic

method. If the tolerances are halved the situation becomes the opposite. Adams

methods have good performance for non-stiff problems with high accuracy

demands.

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Table 3.2: Number of steps and floting point operations during the solution of a two-tank problem

with high accuracy.

default tolerances halved tolerancesode23 33 steps, 6463 flops 251 steps, 40180 flopsode45 81 steps, 13539 flops 201 steps, 31539 flopsode113 46 steps, 13056 flops 95 steps, 25576 flops

Gear methods

True implicit multi-step methods are often based on backward difference formula, so

called BDF-methods. These methods were developed by Gear and are also called Gear

methods. The methods use a number of previous steps to approximate the derivatives,

backward differences. In MATLAB there is one BDF-method called ode15s. Thismethod is a variable order method between 1 and 5. BDF-methods with order higher

then 6 become unstable. ode15s has quasi-constant step-size control which means thatthe step-size is fixed as long as possible. When the step-size is changed the solver must

recompute a number of previous steps. The ode15s method is very good on stiff problems. Specially when the number of states increase then it performs better than

ode23s. (ode15s is actually a NDF-method which means numerical difference formulaand it uses numerical calculated derivatives instead of using previous function calls

direct.)

Example 3.6 : Two Tank Series and Multiple Components

The tank series in Example 3.4 requires stiff ODE solvers if the tank volumes

differs in magnitude. A comparison between the two stiff solvers ode23s and

ode15s is illustrated in the table below. If the tank series only has one chemicalcomponent then the ODE-system have 2 states, i.e. 2 differential equations. If the

tanks have 4 chemical components the ODE-system have 8 states. When the order

of the ODE-system increases the high order method becomes more efficient

compared to the low order method. This is seen below. For the tank series from

Example 4 the ode23s is slightly better then ode15s. When the same tank serieshas 4 chemical components ode15s is much more efficient. Note that ode15stakes more steps due to the quasi-constant step-size control.

Table 3.3: Number of steps and floting point operations during the solution of a multi component

tank problem.

2 state system 8 state systemode23s 35 steps, 9453 flops 41 steps, 170647 flopsode15s 49 steps, 10701 flops 65 steps, 59228 flops

3.5 Differential-Algebraic Equation SolversWhen fast dynamics becomes so fast compared to the dynamics of interest it can be

approximated to be instantaneous. Some physical phenomena can also be so fast that it

is hared to measure the dynamic behaviour. In chemical engineering we assume that the

phenomenon is at equilibrium. A differential-algebraic equation system, DAE, can be

expresses as

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),(0

),(

21

211

x xg

x x f dt

dx

=

=

The problem is a set of differential equations, f , and a set of algebraic equations, g. The

algebraic equations have to be fulfilled in each time step that the differential equationsas solved. These kinds of problems can be solved by implicit ODE-methods with

modified step-size control. In MATLAB the methods ode23t and ode15s can solve

DAE problems.

)()(00

01

),(*0

),(*1

2122

2111

x f dt

dx M x f

dt

dx

x x f dt

dx

x x f dt

dx

=⇒=⎥⎦

⎤⎢⎣

⎡⇒

⎪⎭

⎪⎬

⎫

=

=

The matrix on the left hand side, M, is called the mass matrix. For DAE problems the

mass matrix is singular. To solve DAE problems in MATLAB the right hand side isdescribed in an m-file as usual and the mass matrix is defined in the options using

odeset .

M=[ . . . ] ;opt i ons=odeset ( ‘ Mass’ , M)[ t , y] =ode15s( @f unc, [ 0, 1] , x0, opt i ons) ;

Example 3.6 : Two Tank Series again

Two tanks connected in series are simulated with different volumes. The total

volume is constant, see axample 3.4. In this case we assume that the first tank is

so small that the dynamics is neglected.

Material balances over one chemical component in the two tanks becomes

( )

( )212

22221

11 0

AccOutIn

C C V

q

dt

dC

dt

dC V qC qC

C C qqC qC inin

−=⇒+=

−=⇒=

+=


inC q

C

C

V

q

V

q

q

C

C

dt

d ⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−

−=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡0

0

10

00

2

1

222

1

This problem is solved using ode15s as follows. Define the right hand side of the

DAE syetm, i.e., the differential and algebric equations.

f unct i on dx=DynTankSer i es( t , x)c1=x( 1) ;c2=x( 2) ;ci n=1;q=1;V2=1;

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r es=q*( ci n- c1) ;dc2=q/ V2*( c1- c2) ;dx=[ r es; dc2] ;

Then define the mass matrix and call the DAE solver.

M=[ 0 0; 0 1] ;

opt i ons=odeset ( ' Mass' , M) ;t span=[ 0 10] ;xi ni t =[ 0; 0] ;[ t , x]=ode15s( @DynTankSer i es, t span, xi ni t , opt i ons) ;

Figure 3.6: Simulation of a two-tank series using a DAE solver.

The DAE solvers in MATLAB can not solve all kinds of DAE problems. They can only

solve index 1 problems. The index of a DAE problem is the number of timed the

equation system have to be differentiated in order to be a set of ODE:s. In the example

above the first equation only have to be differentiated once in order to be an ODE. The

derivative of the first concentration is equal to the derivative of the secondconcentration. In general this is not the case and the solvers fail to solve the problems

indicating in the error message that the index is higher than one.

3.6 ODE Simulation in MATLABAbove we have discussed five different methods that are available in MATLAB. There

are two additional ones, ode23t and ode23tb. These methods can handle moderate

stiff problems and stiff problems respectively. ode23t is an implementation of thetrapezoidal rule. ode23t b is an implementation of an implicit Runge-Kutta formulawith a first stage that is a trapezoidal rule step and a second stage that is a backward

differentiation formula of order two. All solvers have a number of options for setting

different solver parameters in order to increase solver performances. These option parameters are handles through odeset and odeget . MATLAB also advise the user

to write its ODE-file in a predefined way. This is well described in the help text of

odef i l e or in the manual Using MATLAB. When you become a more skilled user use

these guidelines to develop simulation problems in MATLAB.

3.7 Summary

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The ODE solvers in MATLAB6 are categorised in the table below. In ordinary

simulation problems the low order methods can be used, ode23 and ode23s. In

process simulation, high order stiff problems often occur in which ode15s is preferable. These three methods are probably the ones that are used in the Process

Simulation course. The ode15s and ode23t can handle differential-algebraicequation systems, DAE, but this is not discussed futher in the course.

Table 3.4: Selection of ODE-methods in MATLAB.

Nonstiff Stiff DAELow order ode23 ode23s

ode23t bode23t

High order ode113,ode45

ode15s ode15s

High accuracy ode113 ( ode15s) ( ode15s)

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4 Development of Distributed Models

4.1 IntroductionIn the previous chapters process models are based on balance equations of mass, energy

and momentum over a lumped system. Inside the system the properties are uniform with

no spatial change. Lumped systems are well mixed and they have no spatial dependent behaviour. In this chapter we are going to develop models for systems with spatial

change. The behaviour is distributed in the space dimension and therefore they are

called distributed systems or distributed parameter models. The development is similar

to lumped model development. The difference is that the balance equations are set up

for arbitrary small fluid elements or infinite small control volumes.

A general balance equation is expressed as follows

In + Production = Out + Accumulation

This is set up for the volume in Figure 4.1.

Figure 4.1: A general fluid element

Below we are discussing distributed models based on mass, energy and momentum

balances first in one space dimension, in this case x-dimension, and second very briefly

present balances in three dimensions.

4.2 Mass BalancesMass balances over a distributed system are very similar to mass balances over lumped

systems. The major difference is the special change. Assume that the fluid in the control

volume in Figure 4.1 has a given density, than the accumulation of mass is simply the

time change of the mass.

For simplicity let us assume a convective flow of mass only in the positive x-direction.

The inflow of mass can be expressed as the density times the volume flow rate. In this

case the element surface, ∆y∆z, times the velocity in x-dimension.

y z

∆x

∆y

∆z

x

( ) z y xt d

d ∆∆∆ ρ

∆ ∆ y z v x x( ) ρ

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The outflow at x+∆ x becomes

There is no generation of mass and therefore the total mass balance is as follows

Assume that the volume boundaries are time independent. This means that the control

volume derivative is zero and it is possible to divide all terms with the volume. Then let

the volume size goes to zero, i.e. take the limit when ∆V goes to zero, and use thedefinition of derivation. In the simple example with flow only in the x-direction the

total mass balance becomes

An equation, like the one on the right above, with derivatives in more the one

independent variable is called a partial differential equation, shortly just called PDE.

This particular PDE, with first order derivative in space and time, is called convective

type PDE or first order hyperbolic PDE .

In the general case there is a flow field in all directions giving changes in all directions.

Convection in three dimensions is just the sum of the individual convection terms.

Note the velocity vector v and the divergence of the velocity, the nabla operator. Below

follows a tube example based on an alternative development of the total mass balance. It

is based in the integral form of the ideas presented above.

Example 4.1: Tube

A medium is flowing through a pipe. The pipe has cylindrical geometry with a

constant cross section. The accumulation of mass in the finite volume is the

volume integral of the density. Mass enters the finite volume at position z and

leaves at position z+dz.

Figure 4.1: Tube.

ρ L

z=0 z z+dz z=L

ρ 0

∆ ∆∆

y z v x x x( ) ρ +

( ) ( )Acc In Out= −

= −+

d x y z

d t y z v v x x x x x

ρ ρ ρ

∆ ∆ ∆∆ ∆

∆( ) ( )

( ) xv

t xvv

t d d x x x x x x

∂ ρ ∂

∂ ρ ∂ ρ ρ ρ −=⇒⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛

∆−= ∆+

( )∂ρ

∂

∂

∂ ρ

∂

∂ ρ

∂

∂ ρ ρ t x v y v z v x y z= − + +

⎛

⎝ ⎜

⎞

⎠⎟ = −∇ ⋅ v

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The accumulation of mass in the control volume is the time derivative of the

volume integral of the density.

∫+

=dz z

z

d Adt

d ζ ρ )(massof onAccumulati

The inlet and outlet of mass is in this case only convection dependent

z z Av ρ =massof Inlet

dz zdz z Av ++= ρ massof Outlet

The three terms are put together in a mass balance giving

∫

+

++ +=

+=+dz z

z

dz zdz z z z d A

dt

d Av Av ζ ρ ρ ρ )(

AccOutProdIn

The inlet and outlet terms are put on the same side and can be rewritten as the

primitive function of the integral of its derivative. We see that there is the same

integral on both sides of the equation.

[ ] ( )

∫∫+

+++

+

∂

∂−=−=−=

dz z

z

dz z

zdz zdz z z z

dz z

z

d Av

Av Av Avd Adt

d ζ

ζ

ρ ρ ρ ρ ζ ρ

ζ ζ

ζ ζ )(

If the boundaries, z and z+dz, are time independent then the time derivative in theaccumulation term can be moved inside the integral.

( )∫∫++

∂∂

−=∂∂ dz z

z

dz z

z

d v

Ad t

A ζ ζ

ρ ζ

ρ

Let us now remove the integral on the both sides and the result is a partial

differential equation.

( ) z

v

t ∂∂

−=∂∂ ρ ρ

We see that the development in Example 4.1 results in the same equation.

Diffusion

In systems with molecular diffusion effects there will be additional terms in the mass

balance. Often one can assume that the molecular flux is proportional to the

concentration gradient described in Fick's first law.

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This expression assumes a binary mixture where component A diffuse in a solvent B.

Component Mass BalancesIt is possible to put up component balances, one for each chemical component. Note

that all component mass balances sum to the total mass balance as in the lumped

parameter case. In order to set up a consistent equation system it is recommended to

take away one mass balance, often one component mass balance, for instance the

solvent balance. Sometimes this knowledge can be used as redundant information to

check the correctness of the model.

Let us derive a component balance with only molecular diffusion as diving force for the

fluid control volume in Figure 4.1. The mass of component A in the control volume is

the product of mole weight and concentration.

The inflow of component A becomes

and the outflow

Let us now put these three terms together in a component mass balance of component

A.

Divide all terms with the control volume and the mole weight. Let the volume size go to

zero.

J Ddc

dx A AB A= −

( ) z y xc M t d

d A A ∆∆∆

( ) x

A AB A x A A

dx

dc D M z y J M z y ⎟

⎠

⎞⎜⎝

⎛ −∆∆=∆∆

( ) x x

A AB A x x A A

dx

dc D M z y J M z y

∆+∆+ ⎟

⎠

⎞⎜⎝

⎛ −∆∆=∆∆

( ) ( ) ( ) x x A A x A A A A

J M z y J M z y z y xc M t d

d ∆+∆∆−∆∆=∆∆∆

( ) ( )( ) x

J

t

c J J

xt d

dc A A

x x A x A A

∂∂

−=∂

∂⇒−

∆= ∆+

1

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The flux J can be expressed as Fick's first law which results in Fick's second law.

A partial differential equation with a second order derivative in space and first order in

time is called a parabolic PDE . It is also some times called diffusion type PDE or heat

equation.

Dispersion

In flow systems there appear a phenomenon called dispersion. This occurs in problems

with turbulent flow, flow in complicated passages such as packed beds or porous media.

Dispersion is mathematically expressed similar to diffusion.

The major difference is that the dispersion coefficient, Dax, is flow dependent while the

diffusion coefficient is not. Below follows an example with dispersion in a tubular

reactor.

Example 4.2: Tubular reactor with dispersion

A medium is flowing through a tubular reactor. A chemical component has the

concentration c. The reactor has cylindrical geometry with a constant diameter.

The accumulation of mass in the finite volume is the volume integral of the

density. Mass enters the finite volume at position z and leaves at position z+dz.

Figure 4.2: Tubular reactor.

∫+

=dz z

z

d Acdt d ζ )(moleof onAccumulati

z

ax z zd

dc A D Acv

ζ −=moleof Inlet

c L

z=0 z z+dz z=L

c0

2

2

x

c D

x

c D

x x

J

t

c A AB

A AB

A A

∂∂

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂∂

−∂∂

−=∂

∂−=

∂∂

dx

dc D J Aaxax −=

2

2

x

c D

t

c A AB

A

∂

∂=

∂

∂

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dzz

axdz zdz zd

dc A D Acv

+

++ −=ζ

moleof Outlet

( )∫+

=dz z

z

d rA ζ moleof Production

The three terms are put together in a mass balance

( ) ∫∫+

+

++

+

++=++

+=+dz z

zdzz

axdz zdz z

dz z

z z

ax z z d Acdt

d

d

dc A D Acvd rA

d

dc A D Acv ζ

ζ ζ

ζ )(

AccOutProdIn

Rearrage the component mass balance

( )

[ ] ( )

( )( )∫∫∫

∫∫∫

+++

+++

+

+

++

+

+∂

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ∂∂

∂

+∂

∂−=

=+⎥⎦

⎤⎢⎣

⎡

∂∂

+−=

=+

∂

∂+−

∂

∂−=

∂

∂

dz z

z

dz z

z

axdz z

z

dz z

z

dz z

z

ax

dz z

z

dz z

zdzz

axdz zdz z

z

ax z z

dz z

z

d rAd

c A D

d Acv

d rAc

A D Acv

d rAc

A D Acvc

A D Acvd Ac

t

ζ ζ ζ

ζ ζ

ζ

ζ ζ

ζ

ζ ζ

ζ

ζ ζ

ζ ζ

)(

Remove all integrals and we get a partial differential equation with a second order

space derivative term.

( )r

z

c D

z

vc

t

cax +∂

∂+

∂∂

−=∂∂

2

2

This kind of partial differential equation has one convective term (hyperbolic),

one dispersion term (parabolic) and one reactive term. It is a hyperbolic-parabolic

PDE.

4.3 Energy BalancesThe development of distributed energy balance is identical to the development of

distributed mass balance. In microscopic energy balances there occur different termscompared to lumped models.

Assume that the total energy is composed of internal energy and kinetic energy.

Accumulation of potential energy is neglected together with other energy forms.

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ∆∆∆⎟⎟

⎠

⎞⎜⎜⎝

⎛ + z y x

vU

dt

d

2

2

ρ ρ

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U represents the internal energy per mass unit. The difference between the convection

of energy in the x-direction is

The energy conduction terms are

Work done by the fluid element against the surroundings consists of work against

volume forces (gravity) and work against surface forces, such as pressure

and viscous forces.

Put all terms together and taking the limit of the volume that goes to zero gives the

following energy balance

This is a general distributed energy balance in on space dimension. Some special cases

are discussed below before the general

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