Calculus Checar

8/9/2019 Calculus Checar

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Process Calculations andReactor Calculations

For Environmental Engineering

Per Warfvinge


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Contents

Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1 Why do we perform process and reactor calculations? 8

1.1 Processes and reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.1 System boundaries and sub-systems . . . . . . . . . . . . . . . 111.1.2 Non-reaction and reaction systems . . . . . . . . . . . . . . . . 121.1.3 Element, components and inert substances . . . . . . . . . . . 121.1.4 Static and dynamic systems . . . . . . . . . . . . . . . . . . . . 13

1.2 The mass balance principle . . . . . . . . . . . . . . . . . . . . . . . . 131.3 Process calculations and reactor calculations . . . . . . . . . . . . . . . 151.4 Numerical methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4.1 Systems of linear algebraic equations . . . . . . . . . . . . . . . 151.4.2 Numerical solution of systems of algebraic equations . . . . . . 16

1.4.3 Solution of diff

erential equations . . . . . . . . . . . . . . . . . 172 Process calculations for non-reaction systems 22

2.1 Component mass balances . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Degrees of freedom analysis . . . . . . . . . . . . . . . . . . . . . . . . 242.3 Process calculations for a separation process . . . . . . . . . . . . . . . 26

3 Process calculations with multiple units 32

3.1 Calculation methodology . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2 Common types of units . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2.1 Mixer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2.2 Splitter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2.3 Recirculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.2.4 Bypass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.2.5 Purge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4 Process calculations for reaction systems 41

4.1 Mass balances mass or mole? . . . . . . . . . . . . . . . . . . . . . . 414.2 Important concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.2.1 Inert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.2.2 Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.2.3 Yield . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.2.4 Selectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

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Contents 4

4.3 The element mass balance method . . . . . . . . . . . . . . . . . . . . 44

4.3.1 The atomic matrix . . . . . . . . . . . . . . . . . . . . . . . . . 454.4 The component mass balance method . . . . . . . . . . . . . . . . . . 474.4.1 The reaction parameter . . . . . . . . . . . . . . . . . . . . . 474.4.2 The reaction matrix . . . . . . . . . . . . . . . . . . . . . . . . 484.4.3 Degree of freedom analysis . . . . . . . . . . . . . . . . . . . . 49

5 Reactor calculations 52

5.1 Kinetic rate equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.1.1 First order kinetics . . . . . . . . . . . . . . . . . . . . . . . . . 545.1.2 Second order kinetics . . . . . . . . . . . . . . . . . . . . . . . . 555.1.3 Overview of kinetic rate equation . . . . . . . . . . . . . . . . . 555.1.4 Reversible reactions . . . . . . . . . . . . . . . . . . . . . . . . 565.1.5 Consecutive reactions . . . . . . . . . . . . . . . . . . . . . . . 56

5.1.6 Parallel reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 575.2 Mean residence time and reaction time . . . . . . . . . . . . . . . . . . 575.3 Reactor models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5.3.1 The differential mass balance . . . . . . . . . . . . . . . . . . . 605.3.2 Methodology for reactor calculations . . . . . . . . . . . . . . . 62

5.4 The ideal completely stirred tank reactor, CSTR . . . . . . . . . . . . 625.5 The ideal batch reactor . . . . . . . . . . . . . . . . . . . . . . . . . . 665.6 The ideal plug-flow reactor, PFR . . . . . . . . . . . . . . . . . . . . . 71

5.6.1 PFR reactor modeling in terms of conversion . . . . . . . . . . 72

6 Non-ideal reactors 76

6.1 Examples of non-ideal mixing . . . . . . . . . . . . . . . . . . . . . . . 766.2 Residence-time distributions . . . . . . . . . . . . . . . . . . . . . . . . 77

6.2.1 The normalized RTD E(t) . . . . . . . . . . . . . . . . . . . . 786.2.2 c(t) for an ideal CSTR . . . . . . . . . . . . . . . . . . . . . . . 796.2.3 E(t)for an ideal CSTR . . . . . . . . . . . . . . . . . . . . . . 816.2.4 Mean residence time from E(t) . . . . . . . . . . . . . . . . . . 816.2.5 The F(t)-distribution . . . . . . . . . . . . . . . . . . . . . . . . 82

6.3 Non-ideal reactor models . . . . . . . . . . . . . . . . . . . . . . . . . . 856.3.1 The CSTR in series model . . . . . . . . . . . . . . . . . . . . . 85

6.4 The segregation model . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.5 Other reactor models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

7 Instationra CSTR 91

7.1 Svar p ndring i ingngskoncentration . . . . . . . . . . . . . . . . . . 92

7.2 Arbetsmetodik . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

8 Dispersion in porous media 99

8.1 Water flow- Darcys law . . . . . . . . . . . . . . . . . . . . . . . . . . 1008.1.1 Flow rate distribution - reactor modeling . . . . . . . . . . . . 1038.1.2 Advective transport . . . . . . . . . . . . . . . . . . . . . . . . 104

8.2 Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1048.2.1 Molecular diffusion - Ficks law . . . . . . . . . . . . . . . . . . 1048.2.2 Analogy dispersion - diffusion . . . . . . . . . . . . . . . . . . . 1058.2.3 Deduction of expression for spread . . . . . . . . . . . . . . . . 106


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Contents 5

8.2.4 Interpretation of the standard deviation . . . . . . . . . . . . . 108

8.2.5 Application to a advection-dispersion system . . . . . . . . . . 1098.3 Dispersion coefficients in groundwater aquifers . . . . . . . . . . . . . 1108.3.1 Dispersivitet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

9 Transport in porous media 113

9.1 The continuity equation - advection and dispersion . . . . . . . . . . . 1139.1.1 Continuous source . . . . . . . . . . . . . . . . . . . . . . . . . 114

9.2 Advection, dispersion and reaction . . . . . . . . . . . . . . . . . . . . 1169.2.1 Continuous source . . . . . . . . . . . . . . . . . . . . . . . . . 1179.2.2 Continuous addition during a limited time - step up and down 1179.2.3 Steadystate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

9.3 Adsorption and transport . . . . . . . . . . . . . . . . . . . . . . . . . 1199.3.1 Advection, dispersion, reaction and adsorption . . . . . . . . . 121

9.3.2 Determination ofKdand R for organic substances . . . . . . . 1239.4 The continuity equation in multiple dimensions . . . . . . . . . . . . . 123

10 Bioreaction engineering - biofilms 125

10.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12510.2 Kinetic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

10.2.1 Kinetic equations for conversion on cellular level . . . . . . . . 12710.2.2 Rate equations for conversion in reactors . . . . . . . . . . . . 13010.2.3 Rate equation for the microbial population . . . . . . . . . . . 130

10.3 Biofilms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13210.3.1 General rate equations for biofilms . . . . . . . . . . . . . . . . 13510.3.2 Concentration profile and flux in a thin film for 0th order reaction13510.3.3 Concentration profile and flux in a deep film for 0th order reaction13710.3.4 Reactor calculations . . . . . . . . . . . . . . . . . . . . . . . . 140

10.4 External mass transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . 14110.4.1 Mass transfer in packed beds . . . . . . . . . . . . . . . . . . . 14110.4.2 Coupled external mass transfer resistance and diffusion/reaction

in a biofilm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142


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Examples 7

Notations

Note that the choice of mass unit (g, kg, mol) often is arbitrary

Properties introduced in Chapters 1-7

cj Mole ormass concentration of substancej mol L1, kg L1

Fi,j Molar flux of substancej in stream i molor mol s1

k Kinetic rate constant VariesQ Volume flux m3 s1V Volume m3

Wi,j Mass flux ofj in stream i kgor kg s1

xi,j Mole fraction ofj in liquid stream iX Conversion with respect to substancej

yi,j Mole fraction ofj in gas stream i

i,j Stoichometric coeffficient forj in reaction ii,j Mass fraction of substancej in liquid stream ii,j Mass fraction of substancej in gas stream i Residence time in a reactor time1

Properties introduced in Chapters 8-10

B Penetration fractionD Diffusion coefficient m s1

E Dispersion coefficient m s1

J Diffusive flux mass m2 s1

km Mass transfer coefficient m s1KX Hydraulic conductivity m s1

Kd Distribution coefficient m3 kgsoil1

Koc Distribution coefficient m3 kgSOM1

L Length coordinate mLB Thichness of biological film mLP Penetration depth mR Retardation factorS Sorbed amount g kg1

u Linear flow velocity m s1

v Darcy velocity m s1

Dispersivity m Boundary layer thickness m Porositye Effective porosity Substrate turnover rate g s1

Bulk density kg m3

Volumetric water content


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C H A P T E R 1

Why do we perform process and

reactor calculations?

There are several situations when an engineer needs to gain an understanding of howdifferent substances flow and react in a system. The system in question may exist inan industry where a substance should be produced, destroyed or redistributed. Theanalyses of such processes is based on the principle of conservation of matter, andthe methods are built upon mass balance. In the chemical industry, and in chemicalengineering, process and reactor calculations based on mass balances are the singlemost important tool for the analysis and design of chemical processes.

For natural systems such as lakes, streams, groundwater aquifers as well as entireecosystems, mass balances are used to describe and predict how matter is transportedand transformed. In nature, transformations of matter are often controlled by pro-cesses that man cannot influence, making the design aspect less important for naturalsystems.

Every environmental engineer needs to know how to perform mass balance cal-culations for different systems. Just as importantly, they also have great use of aprofound ability to thinkin terms of mass conservation and mass balances. Indeed,the ability to think abstractly in terms of the principles of conservation is a hallmarkof engineers.

Mass balances play an important role in environmental studies and environmentalscience. For example, atmospheric transport models, groundwater models and modelsto predict the water quality of lakes and streams are all based on mass balances forchemical components. The integrated models used to predict climate change also

include mass balance equations for different compartments and pools relevant for theglobal carbon cycle.

This compendium treats various techniques to carry apply mass balance calcula-tions to natural and engineered systems. The notations used throughout the text aregiven on the previous pages. The text is structured as follows:

This Introductory chapter provides a brief introduction to mass balance calcula-tions. The key concepts introduced are systems, system boundaries, the mass balanceprinciple,processandreactors, as well as the entitiesmass fractionandmolar fraction,and examples ofnumerical methods.

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1.1 Processes and reactors 10

solids

separationdenitrification nitrification P-precipitation

org-C

org-N

org-Psolids

org-C

org-N

org-P

solids

sludge sludge

sludge

CO2N2

org-C

NH4+

org-P

org-C

NO3-

P

O2

org-C

NO3-

P

Fe3+

org-C

NO3-

P

Figur 1.1: Examples of a wastewater treatment process which includes physical,chemical and biological sub-processes.

1.1 Processes and reactors

Figure 1.1 is a schematic illustration of a treatment plant for municipal wastewater.The goal of treatment is to remove contaminants so that the water may be displacedto a recipient in a safe and environmentally friendly way. The most importance

polluting substances are carbon (C), nitrogen (N) and phosphorus (P). In sewage,however, these are not present as pure elements or as easily identified ions, but in theform of organic substances, such as carbohydrates and proteins.

A water treatment plant contains many parts where various physical, biologicaland chemical processes take place. Many issues related to the design and the operationof such plants involve quantitative estimates. Process and reactor calculations basedon mass balance calculations provide powerful tools to address questions such as:

How much sludge is produced, and how high is the P content in the sludge?

How effective is the denitrification process?

Is enough carbon available in the first biological treatment step (denitrification)?

How large should the recirculation ofNO3 from the nitrification process to thedenitrification process be?

How much will the purification efficiency decrease if 25 % more households areconnected to the sewage treatment plant?

How will the system react when the temperature drops in winter?

Some of the above questions are about how the whole treatment plant operatesunder steady-stateconditions. Other questions arise when the system is exposed to adisturbance (e.g., increased load).


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solids

separationdenitrification nitrification P-precipitation

org-C

org-N

org-Psolids

org-C

org-N

org-P

solids

sludge sludge

sludge

CO2N2

org-C

NH4+

org-P

org-C

NO3-

P

O2

org-C

NO3-

P

Fe3+

org-C

NO3-

P

Figur 1.2: Example of how system boundaries can be drawn around the differentunits in a process.

1.1.1 System boundaries and sub-systems

All the above questions deal with how much material that flows at different pointsin the treatment plant. When one calculates the magnitude of these flows, one mustfirst define the systems boundaries. If one is only interested in the total flows in and

out, the whole plant is chosen when the system boundaries are drawn, as in Figure1.1. Then it is possible to calculate how the process works as a whole: How muchmaterial goes into treatment plant (as liquid), and how much leaves either as a gas,liquid or solid (sludge).

A general rule is that one can only calculate the mass flows in the streams crossingthe system boundary. Hence, one cannot say anything about the mass flows betweenthe various units within waste water treatment systems, in cases where the systemboundary is drawn outside all minor units.

If we want to be able to say something about the individual sub-systems, wemust develop our systems analysis and draw the system boundary in a different man-ner. If we, for example, want to calculate how effective the last sub-system (the P-precipitation) is, we must draw the system boundary as shown in Figure 1.2. Therethe system has been confined so that all components that enter or leave this unit arerepresented by streams across the system boundary.


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1.1.2 Non-reaction and reaction systems

The sub-systems mentioned in the previous section, the sand trap and the biologicalpurification step were selected to illustrate two fundamentally disparate systems. Thesand trap is an example of a non-reaction systemwhich characterized by the fact thatno chemical reaction takes place in the system. Input and output streams flowinginto the sub-system will thus be in exactly the same chemical form, but perhaps ina different phase or separated from other substances. All separation processesareexamples of non-reaction systems.

Process calculations based on mass balances for non-reaction system is usuallyrelatively simple. Chapters 2 and 3 deal with mass balances for non-reaction systems.

The biological purification step is an example of a reaction system. This is atypical topic for chemical engineering, which deals with how to design and operatereaction systems where a raw material is transformed into a chemical product of any

kind. In our example, it is microorganisms using C, N and P in wastewater for theirbiochemical metabolism.In chemical plants, it is obvious that one knows the exact chemical composition

of both the reactant and products. In ecosystems, however, it is difficult to deter-mine exactly which chemicals react and what products that are formed. One thecontrary, there are many examples of natural reaction systems that are very difficultto characterize, such as the atmosphere.

1.1.3 Element, components and inert substances

In the following chapters, the concepts of elements, components and inert substancesare used frequently. A characteristics of elements is that they are neither consumednor produced in a physical or chemical process. This means that elements are con-servativewithin a system. If the system is at steady state (i.e., nothing in the systemchanges with time) one can always assume that the mass flow of an element that goesinto a system, both non-reaction and reaction systems, is equal to the mass flow outof the system.

A componentis a chemical compound of any kind, for example ions such as NO3and SO24 , or compounds such as CO2 or benzene. In a steady-state, non-reactionsystem the mass flow of a component in is always equal to the mass flow out. In asteady-state, reaction system, however, this precondition never applies. This makesprocess balance calculations for reaction systems more difficult than those for non-reaction system.

Aninert substanceis one that does not react, even if it is present in a system whereother substances react. In many systems, water is an inert substance, especially if it

only serves as a solvent. In other cases, one can comfortably make the simplificationthat a substance is inert. This applies, for example, if it only reacts to a smallextent, or is present in such large surpluses that the change in the total amount ofthe substance in the system can be neglected.


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1.2 The mass balance principle 14

The term Input refers to the flow of the substance into the system. The flow

can have the unit kg s1

or similar. The term Output concerns, in the same way,the flow of the substance out of the system.The term Prod refers to the amounts of the substance that is produced inside

the system. It must however be noted that Prod can be either positive or negative.If Prod is negative, it means that the substance is consumed inside the system. Inthe chemical context, it is obvious that reactants are consumed (Prod< 0), while thedesired products are produced (Prod> 0).

The last term, Acc is conceptually the most difficult. Acc stands for the amountof a substance accumulated per unit time inside the system. That means that in asteady-state system Acc = 0 by definition. In a non-steady state system,Acc> 0 ifthe amount of the substance increases in the system. If it decreases, Acc < 0. As weshall see later, the term Acc introduces a time derivative in system models. That iswhy the mass balance model for non-steady-state systems always is composed of one

or more difference or differential equations.

Example 11: Balance calculation for a bank account

Problem 1

Each year, you deposit 50 000 SEK in your bank account, while you withdraw a totalof 45 000 SEK. The bank pays you 2 000 SEK in interest, but draws 500 SEK annuallyin incomprehensible fees. How much will your bank balance change during the year?

Solution 1

We apply the equation

Input + Prod = Output + Acc

Since the question dealt with the change during a time-interval, we are seeking theAcc-term. The equation i re-written as:

Acc = InputOutput + ProdWith the information given the solution becomes:

Acc = 50000SEK

year 45000 SEK

year+ (2000 500) SEK

year= 6500

SEK

year

Problem 2

Given the information about Input and Output, how large must the annual interest(i.e., the Prod term) be if the account balance should increase by 5 600 SEK a year?

Solution 2

We seek the Prod term and rewrite the equation as:

Prod = Output Input + Accand calculate the interest as:

Prod = 45000SEK

year 50000SEK

year+ 5600

SEK

year= 600

SEK

year


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1.4 Numerical methods 16

only A1 can be inverted, while A2 cannot. Furthermore, linear algebra has taught

us that the dimension of the largest (square) sub-matrix that is possible to invert iscalled the rank of the original matrix. Hence, the rank ofA1=2, while the rank ofA2=1.

Example 12: Solving linear equation systems

Problem

Solve the following system of equations with Matlab:

3x1+ 4x2 = 10

x1 x2 = 2

SolutionThe system of equations can be written as a matrix in the form:

AX=Y

where:

A=

3 41 1

, X=

x1x2

, Y =

102

WithMatlabthe system of equations is solved as:

1.4.2 Numerical solution of systems of algebraic equations

Matlabcan also solve the system of equations AX = Yby finding the roots to theequation 0 = AX Y. In contrast to the above method, this calls for some morework by the computer. However, in this case one is not limited to linear systems ofequations, non-linear systems can also be solved easily. WhatMatlabactually doesis to seek F as follows:

F =AX Y, so that F<

where is so small that it, in practice, equals 0. The only thing one has rememberas a user is to write the equations in a correct form and then submit a guess ofX isthat is not completely out of line.


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Example 13: Solving linear equations with fsolve

Problem

Solve the above systems of equations in 12 using fsolve in Matlab after havingdefined the equations in matrix notation.Solution

The system of equations can be written as a matrix on the form:

0 =AX Y

where

3 41 1 , Y = 102

Write an m-file: where f0is a guess at the solution needed to initialize fsolve. TheMatlabfunction fsolveis run from workspace with: Of course, the solution is thesame as in the above example; x1= 2.5714, x2= 0.5714.

Example 14: Solving non-linear equations with fsolve

Problem

Solve the following system of equations with the routine fsolvein Matlab:

3x1+ 4x22 = 10

x31 x2 = 2Solution

The system of equations can be written as a matrix in the form:

0 = 3x1+ 4x22 10

0 =x31 x2 2The two functions are the defined in an m-file exfsolve1.m: which is executedfromworkspacewith: Solution: x1= 1.4708, x2= 1.1819. The vector [1 ; 1], abovecalledf0, is an initial guess of what the solution vector X is.

1.4.3 Solution of differential equations

Dynamics in all types of systems is described by differential equations, ordinary orpartial. It is only for models of rather idealized systems that the resulting differentialequations are analytically soluble. For many classes of problems it is possible to lin-earize the equations, and thereby obtain analytical mathematical solution. However,this is a trick, which is most often used in Automatic Control and it will not befurther discussed here.


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In making the mass balance calculations for non-steady state systems, the starting

point is normally one or more ordinary non-linear diff

erential equations. In thiscompendium we will almost exclusively solve differential equations numerically. Theexceptions are the simplest cases.

How do we solve a differential equation numerically? First we will take a quicklook at the simplest (and worst) method, Eulers solution method.

An ordinary differential equation can be written in the form:

dX

dt =f(t, X)

This differential equation states that ifX = Xt at time t, then X = Xt+dt at timet + dt:

Xt+dt = Xt+ dtf(t, Xt)

We can interpret this graphically if we follow the derivative of the function X(t)a rectilinear piece dt, starting at the point (t, Xt). The approximately estimatedsubsequent value of the function is thus a small distance away, in the direction of thederivative. This value is defined as(t + dt,Xt+dt). After this value of the function hasbeen calculated, another value is calculated in the same way. The whole procedure isrepeated to create a vector of estimates function value.

As we see in Figure 1.4 the approximation may differ significantly from the realfunctionX. With Eulers method, the calculation error is relatively large, but if onereduces the dt, the calculation error is also reduced. However, the smaller the dt,the longer the calculations take. For all practical purposes it is better to use moresophisticated solution methods such as the ode45routine in Matlab.


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1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3-1

0

1

2

3

4

5

6

t+dtt

Xt

Xt+dt

approximering av Xt+dt

X

dX/dt i (t,Xt)

Figur 1.4: Illustration of Eulers method.

Example 15: Numerical integration with Eulers method

Problem

Integrate (simulate) the ordinary non-linear differential equation:

dc

dt = 0.5c Initial value:

t= 0

c= 1

using Eulers method. The simulations shall be done over the ranget = 0to t = 4insteps (dt) of 0.1 units of time.


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Solution

Write an m-file exEuler.m: which is executed from workspace with exEuler. Inthe m-file, one can see how the time interval (from 0 to 4) has been divided into 40pieces, each representing 0.1 units of time. The results are given below, together withthe exact solution. The conclusion is: As long as the solution is monotonous, thissimple integration method works quite well!

0 0.5 1 1.5 2 2.5 3 3.5 40.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

exakt

euler


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Example 16: Numerical integration with ode45

Problem

Integrate (simulate) the ordinary non-linear differential equation withode45inMatlab:

dX1dt

= k1X1 k1= 0.5dX2

dt = k2X1X2 k2= 0.1

Solution

Write an m-file exODE45.m: and run it fromworkspacewith: where[0 10]denotes

the time interval that should be covered, and the vector [1 1] represents the initialvalues for r X1andX2 respectively. The solution is presented as a diagram:

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

X1

X2

Vrdetp

tid

svariabelttillstnd

X

tiden t


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C H A P T E R 2

Process calculations for

non-reaction systems

In this chapter we shall go through the methods for resolving integral mass balancesfor the very simplest type of system. As mentioned in the previous chapter, processcalculations allow us to describe systems with flows of various chemical substances.The process calculations my be used to quantify fluxes, but may also be extended toprocess design and, partly, to explain what is happening within a process system.

Process calculations have played an enormously important role in industry, butalso in environmental research. Without mass balances, we would not have hadany quantitative understanding of flows of mass and matter in any one of earthsecosystems.

Mass balances are known under different names. Often one uses the terms budgetcalculationsor material balances.

Here we will use a somewhat formal approach to process calculations. They willinvolve models with several equations. Solving these equations involves the collectionof as many equations as that there are unknown variables. By working with degree of

freedom analysesas a tool we will systematically examine wether it is possible or notto solve a certain system of equations. We will also see how they can become solvedby doing certain tricks.

In a series of very simple examples, different types of equations and informationwill be slowly introduced and will illustrate key concepts in the mass balance calcu-lations.

The methodology will probably seem pretty tedious, and give the impression of

being overly formal. If so, the impression is correct: the idea is to show a formaland systematic approach that always leads to a solution if a solution exists, and thatreveals why a solutions sometimes cannot be provided.

The mass balance for element/component A will be denoted as MB A etc.

22


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2.1 Component mass balances 23

W1,A

W1,B

W2,A

W3,B

W4,A W4,B

1

2

3

4

Process

systemgrns

Figur 2.1: Illustration of the concepts of process, system and stream variable.

2.1 Component mass balances

Let us consider a process where substances flow into the system at one end, and out

of the system at the other end. The mass flux of the substance is denoted as Wi,j ,wherei refers to streami and j to the substances. For each substance, we can alwayswrite the general mass balance:


If the system is a non-reaction system and is at steady state, we recognize thatProd = Acc = 0 and we get:

Input = Output

or, expressed in terms of mass fluxes:

in

Wi,j = out

Wi,j

Figure 2.1 shows a process where two physical streams, W1 and W2 are inputfluxes, while W3 and W4 are output fluxes. W1 and W4 both contain the chemicalcomponents A and B, while W2 only contains A and W3 only contains B.

The units for mass fluxWmay be mass (kg) or mass flux (kg/time). Under theseconditions, the integral component mass balances for A and B become:

MB A : W1,A+ W2,A =W4,A

MB B : W1,B =W3,B+ W4,B

We can also see that these component mass balances show linear independence(i.e.,they are not identical). If they were independent, one of the balances would be

worthless. The solution would be trivial, of the type 0=0, or 1=1.The variables W1,A, etc., are called stream variables. A stream variable refers to

the flux of a specific substance in a specific physical stream.The maximum number of stream variables in a system always equals the number

of streams multiplied by the number of components. If there are 4 streams with 2components in each, there is a maximum of 8 stream variables. This means that weneed 8 equations in order to allow the system to be unambiguously defined. This isalso true in the example above, although only 6 stream variables are explicitly writtenout. The reason is that implicitly W2,B =W3,A= 0.


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2.2 Degrees of freedom analysis 24

2.2 Degrees of freedom analysis

The difference between the number of stream variables and the number of equationsavailable is called the number of degrees of freedom. We can now distinguish threefundamentally different cases:

Degrees of freedom >0 The system is not completely defined.

Degrees of freedom = 0 The system is defined.

Degrees of freedom


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2.2 Degrees of freedom analysis 25

1,AW

1

1,BW1

1

2

3

4

Process

systemgrns

2,AW2

4,AW4

4,BW4

3,AW

3

Figur 2.2: Steam variables expressed in terms of mass fractionsi,j and the totalflux Wi.

component fluxes (quantified by the stream variables) must add up to the total flux.Hence:

Wi =

nj=1 Wi,j

Win

j=1i,j

The corresponding relationships also hold true when the stream variables are ex-pressed in molar units:

Fi =

nj=1 Fi,j

Fi

nj=1 xi,j

The equations are illustrated in Figure 2.2 which shows how the stream variablesin Figure 2.1 may be expressed in terms of the mass fraction i,j and the total fluxWi.

A basis of calculationconsists of an arbitrary definition of a flux. This number isused as an equation in the calculations. A basis of calculation may be (and has tobe!) introduced if no other values of fluxes (i.e. no values of stream variables) areknown or given. This occurs when only concentrations or mass/molar fractions arespecified. For example, a basis of calculation can be set to 1 kg, 100 kg/hour, etc.,but may onlybe introduced of it does not contradict other information.


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2.3 Process calculations for a separation process 26

2.3 Process calculations for a separation process

In this section we shall see how we methodicallycan solve a process calculation prob-lem for a non-reaction system at steady state. The starting point is a very simpleseparation process where a mixture of ethanol, C2H5OHand H2Oare separated intotwo streams with another composition than the input stream. By varying the waywe look at the problem, we will see how different types of information can be usedfor problem solution. All examples are solved withMatlab, but it is valuable to solveequation systems by hand now and then, for sake of training.

The solution methodology is:

Draw a picture, a process chartwith the system boundary and all the physicalstreams.

Enter all stream variables and information in the process chart.

Make a degree of freedom analysis.

Set up and solve the equation system.

Answer the question(s) asked.

The first examples dealing with separation of the two components is schematicallyillustrated below. In the process, C2H5OH and H2O flow into a process with oneinput stream and leave the system in the other end via two output streams. Themass flux of each component is denoted as Wi,j , whereidenotes streami, whilej = 1denotesC2H5OHand j = 2 denotes H2O:

W1,1kg C2H5OH

W1,2kg H2O

W2,1kg C2H5OH

W2,2kg H2O

W3,1kg C2H5OH

W3,2kg H2O

1

2

3

separations-process

Example 21: Separation of ethanol and water

Problem:The input stream (the feed) to a separation process contains 500 kg C2H5OH and500 kg H2O. There are two output streams from the process, one containing 460 kgC2H5OHand 60 kgH2O. How muchC2H5OHand H2Oleave the system in the otherstream?

Solution:

The first thing to do is to draw a process chart, where the streams are numbered1-3 and the components 1-2. The stream variables and the information given are allintroduced into the process chart:


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Example 22: Separation of ethanol and water - 6 unknowns

Problem:

Solve the problem above while considering every stream variable as an unknown, andall of the information given as indepenent information.

Solution:

First, the process chart is drawn with the streams numbered 1-3 and the components1-2.

W1,1

kg C2H

5OH

W1,2kg H2O

W2,1kg C2H5OH

W2,2kg H2O

W3,1kg C2H5OH

W3,2kg H2O

1

2

3

separations-process

There are six unknown stream variable, and the independent information are W1,1=500, W1,2= 500, W2,1= 460and W2,2= 60.

Unknown stream variables 6Independent component mass balances -2Independent information -4

Degrees of freedom 0The equations are:

MB C2H5OH : 500 = 460 + W3,1

MB H2O : 500 = 60 + W3,2

Info 1 : 500 =W1,1

Info 2 : 500 =W1,2

Info 3 : 460 =W2,1

Info 4 : 60 =W2,2

Or, with matrix notation AX=Y:

A=

1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

, X=

W3,1W3,2W1,1W1,2W2,1W2,2

, Y =

500 460500 60

50050046060

Ais a unity matrix, which can be created with the Matlabcommand eye(6):


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Example 23: Ethanol and water concentration constraints

Problem:

A separation process is fed with an input stream consisting of 500 kg C2H5OHand500 kg H2O. The output consists of two streams. One amounts to a total of 400kg and contains 96% C2H5OH. Calculate the mass fractions of C2H5OH and H2Orespectively in the other stream.

Solution:

First, the process chart is drawn with the streams numbered 1-3 and the components

1-2.

500 kg C2H5OH

500 kg H2O

96%, W2,1 C2H5OH

W2,2H2O

W3,1kg C2H5OH

W3,2kg H2O

1

W2=400kg

3

The next step is to carry out a degree of freedom analysis. They way the process chartis drawn, there are four unknowns,W2,1and W2,2as well as W3,1and W3,2. However,

we have two information: the concentration constraint 0.96 = W2,1/(W2,1+W2,2)and the total constraint 400 =W2,1+ W2,2.


Degrees of freedom 0

The four equations are:

MB C2H5OH : 500 =W2,1+ W3,1

MB H2O : 500 =W2,2+ W3,2

Info 1 : 400 =W2,1+ W2,2

Info 2 : 0.96 =W2,1/(W2,1+ W2,2)

0 = 0.04W2,1+ 0.96W2,2In matrix notation, AX=Ywe have:

A=

1 0 1 00 1 0 11 1 0 0

0.04 0.96 0 0

, X=

W2,1W2,2W3,1W3,2

, Y =

500500400

0


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and the solution becomes: The mass fractions in stream 3 are thus3,1= 116/(116+

484) = 0.1933, and 3,2= 484/(116 + 484) = 0.8067.

Example 24: Separation of ethanol and water moved constraint

Problem:

A separation process is fed with an input stream consisting of 500 kg C2H5OHand500 kg H2O. The output consists of two streams. The first of these output streamscontains 96%C2H5OH, while the second amounts to a total of 600 kg. Calculate themass fractions ofC2H5OHand H2Orespectively in the second stream.

Solution:

First, the process chart is drawn with the streams numbered 1-3 and the components1-2.

500 kg C2H5OH

500 kg H2O

96%, W2,1 C2H5OH

W2,2H2O

W3,1kg C2H5OH

W3,2kg H2O

1

2

W3=600kg

Just as in the previous example, there are four unknown stream variables. In addition,the concentration constraint 0.96 = W2,1/(W2,1+ W2,2) is the same as above, whilethe total constraint is moved so that 600 =W3,1+ W3,2.



The four equations are:

MB C2H5OH : 500 =W2,1+ W3,1

MB H2O : 500 =W2,2+ W3,2

Info 1 : 600 =W3,1+ W3,2

Info 2 : 0.96 =W2,1/(W2,1+ W2,2)

0 = 0.04W2,1+ 0.96W2,2In matrix notation, AX=Ywe have:

A=

1 0 1 00 1 0 10 0 1 1

0.04 0.96 0 0

, X=

W2,1W2,2W3,1W3,2

, Y =

500500600

0


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and the solution is: The mass fractions of all constituents are, of course, the same as in

the previous example: 3,1= 116/(116+484) = 0.1933, and3,2= 484/(116+ 484) =0.8067.


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C H A P T E R 3

Process calculations with

multiple units

In practice, it is rare for natural or engineered systems to be described with onlyone unit. As examples of this we have . Instead, it is normally several streams andunits with different characteristics, which together form a complex process, such asthe treatment plant in Figure 1.1.

In this chapter, we will see how to manage process calculations for non-reactionsystem with multiple units. An important part is to implement the degree of freedomanalysis of multiple-unit systems. The chapter also introduces typical units that areimportant to recognize, especially the mixer and the splitter. Some fairly comprehen-sive examples, such as 32, are included.

3.1 Calculation methodology

One important aspect that distinguishes multiple unit systems from those discussedin the previous chapter is that the system boundaries can be drawn in several ways.For example, consider the system:

Here, we can distinguish three levels:

1. Degree of freedom analysis and equation system for a specific unit.

2. Degree of freedom analysis and equation system relating to the outer systemboundaries, from now on referred to as the Process.

32


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3.2 Common types of units 34

one splitter constraint is that:

W1,AW1,A+ W1,B

= W2,A

W2,A+ W2,B

If one composition inW1is known, one can use this information to re-write the aboveexpression as:

1,A= W2,A

W2,A+ W2,B

However, it is not possible to define any number of splitter constraints. Instead,the number of independent splitter constraints is limited. This fact origins from thenature of the splitter: All stream do have the same composition! If too many splitterconstraints are added to each other, the result will be the component mass balance.

Thus, one can only define a limited number of independentsplitter constraints.If the input stream to a splitter is divided into S output streams, and there

are Ncomponents in the input, one can define (N 1)(S 1) independent splitterconstraints. This holds true if the components mass balances are used in the processcalculation.

For example, if a stream with two components is split into two streams, we mayform exactly (2-1)(2-1)=1 independent splitter constraint. However, for a splitterreceiving four components, split into four output streams as many as (4-1)(4-1)=9splitter constraints can be included in the equation system.

The splitter constraints are not only a whole lot of fun; they give rise to non-linearalgebraic equations that cannot be solved by simple numerical methods. Therefore,it is advised to make a short-cut to overcome this obstacle. If one knows that:

1,A= W2,A

W2,A+ W2,Bwhere 1,A= 0.2

one may directly write the linear equation:

0.2 = W2,A

W2,A+ W2,Bor 0 = 0.8W2,A 0.2W2,B


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Example 31: Limitation of the number of splitter constraints

Problem:

Show that only 1 independent splitter constraint exists if one at the same time setsup the component mass balances of A and B.

Splitter1

W1,A

W1,B

2

3

W2,A

W2,B

W3,A

W3,B

Solution:We try to formulate 4 equations:

MB A W1,A= W2,A+ W3,A

MB B W1,B =W2,B+ W3,B

Splitter constraint 1 W1,AW1,A+W1,B

= W2,AW2,A+W2,B

Splitter constraint 2 W3,AW3,A+W3,B

= W2,AW2,A+W2,B

Re-write the mass balances:

MB A W3,A =W2,A W1,AMB B W3,B =W2,B

W1,B

and substitute W3,A andW3,B into splitter constraint 2, which is then simplifies as:

W3,AW3,A+W3,B

= W2,AW2,A+W2,B

(W1,A W2,A)(W2,AW2,B) = W2,A(W1,A W2,A+ W1,B W2,B)(W1,AW2,A W22,A+ W1,AW2,B+ W2,AW2,B) =

W2,AW1,A W22,A+ W2,AW1,B W2,AW2,BW1,AW2,B =W2,AW1,B

Now, we re-write and simplify splitter constraint 1:

Splitter constraint 1 W1,A(W2,A+ W2,B) = W2,A(W1,A+ W1,B)W1,AW2,B =W2,AW1,B

Splitter constraint 1 is identical to a combination of splitter constraint 2 and the twocomponent mass balances. Hence, only 3 out of 4 equations are independent!


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Example 32: Process calculation based on total system analysis

Problem:

From a manufacturing process, a sewage output stream contains 10 weight-% of theenvironmentally harmful component A. The substance A exists in a liquid streamwhich mainly contains water, here referred to as B.

A government requirement exists that says the concentration of A may not bemore than 3 % in the outgoing stream to the recipient. To meet this requirement aseparation plant in which A is separated to 90% has been introduced.

The separation requires expensive additional chemicals. Therefore, it is economi-cally favourable to clean as small a part of the sewage flow as possible. Thus, a sewageflow bypass is introduced.

Calculate all the flows in the system, and determine how large should the part ofthe sewage flow stream that is led through the process should be as compared to thefraction that is bypassed? Base the calculation on simultaneous solution of the totalsystem.

The process can be described by the following chart:

Process chart and stream variables:First we draw a process chart where the streams have been numbered 16 and thestream variables necessary are introduced.

I fact, the problem only states that the ratio between stream 2 and 3 needs to be

calculated. But to do this, all other stream variables have to be calculated.Degree of freedom analysis:

First, a degree of freedom analysis is carried out. For an analysis based on the totalsystem, including all sub-systems, the system boundaries are drawn as follows:


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We can now make the degree of freedom analysis to investigate if the problem ispossible to solve. Calling the basis of calculation BoC we get:

TotaltStream variables 11MB (A and B, Splitter) - 2MB (A and B, Separation) - 2MB (A and B, Mixer) - 2Info conc. of A in W1 - 1Info conc. of A in W5 - 1Info separation of A -1Splitter constraint -1Basis of calc. -1Degrees of freedom 0

Equations:

Here, we will utilize the fact that the system can be solved by solving for all the 11equations at once.

MB A Splitter: W1,A = W2,A+ W3,AMB B Splitter: W1,B =W2,B+ W3,BMB A Separation: W2,A = W4,A+ W6,AMB B Separation: W2,B =W4,B

MB A Mixer: W3,A+ W4,A= W5,AMB B Mixer: W3,B+ W4,B =W5,B

Conc. of A in stream 1: W1,A = 0.1(W1,A+ W1,B)

Conc. of A in stream 5: W5,A = 0.03(W5,A+ W5,B)

Separation of A: W4,A= 0.1W2,ASplitter constraint (trick): W3,A = 0.1(W3,A+ W3,B)

Basis of Calc.: W1,A+ W1,B = 1000


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Then, all equations are re-written so that all the stream variables end up on the

left side:MB A Splitter: W1,A W2,A W3,A = 0MB B Splitter: W1,B W2,B W3,B = 0

MB A Separation: W2,A W4,A W6,A = 0MB B Separation: W2,B W4,B = 0

MB A Mixer: W3,A+ W4,A W5,A = 0MB B Mixer: W3,B+ W4,B W5,B = 0

Conc. of A in stream 1: 0.9W1,A 0.1(W1,B = 0Conc. of A in stream 5: 0.97W5,A 0.03W5,B = 0

Separation of A: W4,A 0.1W2,A = 0Splitter constraint (trick): 0.9W3,A 0.1(W3,B = 0

Basis of Calc.: W1,A+ W1,B = 1000

This system of equations can also be written in matrix form AX=Y whereA is thematrix of coefficients, Xis a column vector containing all the stream variables, whileY is the right hand side of the equations, only containing numbers. We then get thefollowing matrixes:

A=

1 0 1 0 1 0 0 0 0 0 00 1 0 1 0 1 0 0 0 0 00 0 1 0 0 0 1 0 0 0 10 0 0 1 0 0 0 1 0 0 00 0 0 0 1 0 1 0

1 0 0

0 0 0 0 0 1 0 1 0 1 00.9 0.1 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0.97 0.03 00 0 1 0 0 0 1 0 0 0 00 0 0.9 0.1 0 0 0 0 0 0 01 1 0 0 0 0 0 0 0 0 0

, X=

W1,AW1,BW2,AW2,BW3,AW3,BW4,AW4,BW5,AW5,BW6,A

, Y =

00000

00000

1000

It is easy to solve this system of equations with Matlab. The Matlabsolution m-fileis: and the final results are: We now recall that the initial aim was to calculatethe fraction of the stream W1 that could bypass the separation process while stillpreventing the concentration of A in the output stream W5 from getting too high.The result is that the ratio between the steam can be:

W3,A+ W3,BW2,A+ W2,B

= 20.6979 + 186.280779.3021 + 713.7193

= 0.2610


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Example 33: Degree of freedom analysis based on sub-systems

Problem:

Make a degree of freedom analysis for Example 32 for the Process and the threesub-system (units) that form the system.

Solution:

In this case, the degree of freedom analysis is:

Process Splitter Separation MixerStream variables 5 6 5 6MB (A and B, Process) - 2MB (A and B, Splitter) - 2

MB (A and B, Separation) - 2MB (A and B, Mixer) - 2Info conc. of A in W1 - 1 - 1Info conc. of A in W5 - 1 - 1Info Separation of A -1Splitter constraint - 1Degr. of freed.w/o BoC 1 2 2 3Basis of Calc. - 1 -1 -1 -1Degr. of freed. w BoC 0 1 1 2

Thus, the Process may be solved for, but this does not allow us to calculate all fluxesin the system. If the Process is solved for, W1,A, W2,A, W2,B, W5,A and W5,B arequantified. If we should continue to with the other sub-systems, we must take into

consideration that: We may not introduce an additional basis of calculation, this has already been

done.

The information regarding W1 and W5 have already been used.

The information regarding the separation has already been used.

Since no additional basis of calculation may be introduced at this stage, the degreeof freedom analysis for the remaining units becomes:

Splitter Separation MixerStream variables 6 5 6MB (A and B, Splitter) - 2MB (A and B, Separation) - 2MB (A and B, Mixer) - 2Known stream variables -2 -1 -2Splitter constraint - 1Degrees of freedom 1 2 2

Conclusion:

We may now conclude that we cannot get any further. Thus, only the Total and theProcess may be solved for.


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C H A P T E R 4

Process calculations for reaction

systems

In many natural and engineered systems chemical reactions take place in which reac-tants are converted into products. Products and reactants are the usual nomenclaturein chemistry and chemical engineering. Regarding biological systems one says thata substrate is metabolized to a metabolite. The treatment plant in Figure 1.1 is anexample of a technological system that is dominated by metabolic chemical processes.

It is not as easy to describe a system with chemical reactions as it is to describenon-reaction systems. One reason is that new components are created in the system;another reason is that we may not know exactly which components are actually inthe system.

This chapter describes different techniques to make process calculations in thesystem with one or more chemical reactions. Some important new concepts are definedto help when independent information is to be interpreted.

4.1 Mass balances mass or mole?

Suppose that the following chemical reaction occurs in a system:

N2+ 3H2 2NH3Here, it is important to note that a chemical reaction formula is always written interms of molar units. Indeed, we all know that it is nottrue that 1 kg N2 and 3 kgH2 form 2 kg NH3.

We can also note two important things happening for the above reaction system:

1. N2 och H2 are consumed while NH3 is produced.

2. The chemical form of the elements N and H change.

These simple observations have two implications with respect to the mass balancesthat can be used to define the system in which the reaction takes place. If oneformulates mass balances for the componentsN2, H2, NH3 they will be in a different

41


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4.2 Important concepts 42

form compared to if mass balance equations are set up for the elementsN and H:

Element mass balance or N, H : Input = Output (4.1)Component mass balance for N2, H2, NH3 : Input + Prod = Output (4.2)

The difference, obviously, is caused by the fact that elements are indivisible whilecomponents may react and change their nature in chemical reactions. In terms ofconstraints related to stream variables, this can be expressed more formally as:

For components:in

Wi,j=out

Wi,jandin

Fi,j=out

Fi,j

For elements:in

Wi,j =out

Wi,jandin

Fi,j =out

Fi,j

When one solves mass balance problems for reaction systems one should therefore:1. Never make mass balances in terms of mass units, but always use molar units.

2. Only use component mass balances if there are good reasons, otherwise elementmass balances should be used (still in molar units).

Unfortunately, in most natural systems one can never know exactly which compo-nents are present, how they react and what they form. However, by use of chemicalanalytical methods it is quite possible to track the elements that are present in vari-ous phases, and how they are transported within the system. In conclusion, systemswhere biological transformations are important should normally be handled be meansof element mass balances.

Another exciting possibility, not further discussed here, is the use of isotopes.For example 18Oand 16O both have element characteristics, but do in fact behavesomewhat differently in nature.

4.2 Important concepts

4.2.1 Inert

An inert substance is one that does not react chemically. In many processes thesynthesis ofNH3 described above is a rare exception N2 is an inert. N2 stays inertas long as the temperature is below 1200 oC. An inert substance can be consideredas an element, and handled just like any other element.


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4.2 Important concepts 43

4.2.2 Conversion

The conversion is a number that represents the extent to which a reactant takes partin a chemical process. The conversion Xfor a certain reactant is defined as:

X=

Converted amountInput flux

Input flux - Output amountInput flux

1

output

F

inputF

Hence, if the output flux (

output F) is equal to the input flux (

input F) the conver-sion is obviously X= 0(i.e., the reactant has not reacted at all within the system).On the contrary, if the output flux (

output F) is equal to 0, all of the reactant has

been converted (i.e., X= 1, or as percentage in the range 0-100%).

4.2.3 Yield

The yield quantifies the amount of a substance formed, in relation towhat could havebeen formedif the limiting reactant were converted to a desired product. With thisdefinition, the yield is 100% if:

1. All of the reactant converted is converted to the desired product (i.e. no-sidereactions take place).

2. The conversion of the reactant in question is 100%.

For example, let us define the yield if the limiting reactant R is converted to themain product P. The stoichiometry of the reaction is:

RR PPand the yield is thus defined as:

Yield =

output

Fi,P/P

input

Fi,R/R

4.2.4 Selectivity

The selectivity tells the amount of the main product that is formed in relation to theamount of (undesired) bi-products that are formed. If the molar fluxes of all productshave been determined, the selectivity is calculated as:

Selectivity =

output

Fmain product

output

Fbi-products


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4.3 The element mass balance method 45

Hence the system is uniquely defined. UsingFto denote molar flux, we can set

up mass balances that describe the number of moles of C and H that are associatedwith the input and output streams. For example, a flux of 1 mole ofC2H6 obviouslycarries 2 moles of C and 6 moles of H. The system of equations then becomes:

MB C : 2F1,C2H6 = 2F2,C2H6 + 2F2,C2H4MB H : 6F1,C2H6 = 2F2,H2+ 6F2,C2H6+ 4F2,C2H4Info 1 : F1,C2H6 = 100

Info 2 : F2,C2H6 = (1 0.6)F1,C2H6In matrix notation AX=Ywe get:

A=

2 0 2 26 2 6 41 0 0 0

(1 0.6) 0 1 0

, X=

F1,C2H6F2,H2

F2,C2H6F2,C2H4

, Y =

00

1000

and the Matlabsolution becomes:

4.3.1 The atomic matrix

In example 41 it was simply assumed that the element mass balances for C andH were independent. Normally, this assumption is valid but there are reasons to becautious.

The number of independent element mass balances is given by the rank of theatomic matrix. The columns of the atomic matrix contains the stoichiometric com-position of all components in the system, while the rows refer to each element. This

means that if any rows are linearly dependent, one of the rows is redundant. Theconsequence is that two elements are dependent.For example, let us consider the process above (example 41), which includes the

componentsC2H6, C2H4andH2:

C2H6 H2 C2H4

CH

2 0 26 2 4

We can now calculate the rank of this matrix. If the rank is 2, then the elementsC and H are independent. And if they are independent, it is possible to form twoindependent element mass balances, one for C and one for H. In Matlabthe rank iscalculated as: As we see, the rank of the atomic matrix is 2 which means that we can

form the corresponding independent element mass balances which both add uniqueinformation.

Example 42: Atomic matrix with linearly dependent rows

Problem

In a process, the conversion:

2C2H4catalysisC4H8


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4.3 The element mass balance method 46

is carried out over a catalytic bed.

C2H41 Process

systemgrns

C4H82

C2H4

Make a degree of freedom analysis provided that the feed contains 100 mol C2H6andthe conversion is 60%.

Solution

The atomic matrix is:

C2H4 C4H8

CH

2 44 8

Obviously, the rank for the atomic matrix is 1, which means that only one of theelement mass balances for C or H is dependent on the other. Hence, the degree offreedom analysis is:

Stream variables 3- Independent element MB - 1- Independent information - 2


We can thus conclude that the reaction may actually be viewed as if:

2Acatalysis

A2

where A = C2H4. Obviously, the elements C and H will always appear in the sameproportions. They are thus not independent, but dependent on each other.


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4.4 The component mass balance method 47

4.4 The component mass balance method

In 4.3 the method to solve all fluxes in a system was based on i) element mass balancesand ii) stream variables for components.

In this section, we will introduce a method to base the calculations on componentmass balances instead. We then have to take into consideration that componentsreact, while elements do not.

The reason why an alternative to element mass balance calculation can be usefulis that sometimes, one needs to know the extent to which chemical reactions occurin a system. With the element balances method, the chemical reactions were notconsidered at all. With the component mass balance calculations, quantification ofthe chemical reactions are essential.

When applying a component mass balance method, the degree of freedom analysisis somewhat different compared to the element mass balance method. The reason

is that the chemical reactions need to be quantified by some kind of Prod term asoutlined in equation 4.2, which introduces mathematical constraints in the calculation.

4.4.1 The reaction parameter

The reaction parameteris a calculation helper. It is defined as:

= The number of moles converted (per unit time) in a system via a certain reaction .

For the most simple example, whereA Band there is only one input and oneoutput stream, the component mass balances become:

Input + Prod = Output

Fin,A = Fout,A

Fin,B+ = Fout,B

As we see,will always be a positive number since Fin,A> Fout,Aand Fin,B < Fout,B.In order to generalize, let us consider more complex situation where one chemical

reaction takes place:N2+ 3H2 2NH3

For the two reactants N2 and H2, has a negative sign in front of it since Prod


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In many (most) systems, more than one chemical reaction takes place. In that case

it is necessary to define one reaction parameter for each reaction. This is indeed veryuseful since the ratio between calculated values of the different reaction parametersis the same as the ratio between the extent to which the chemical reactions occur.

If there are many chemical reactions i in the system, the general component massbalance for component j will thus become:

Finput,j +

all

recationsi

i,ji = Foutput,j

4.4.2 The reaction matrix

One cannot introduce any number of reaction parameters when component massbalance problems are to be solved. The maximum number of reaction parameterscorresponds to the maximum number of independent chemical reactions. An inde-pendent chemical reaction is one that cannot be expressed in terms of other chemicalreactions used to characterize the system.

The concept of independent chemical reactions is also important in chemical equi-librium calculations; if too many equilibrium equations are formed for a chemicalsystem, the calculations will not result in an unambiguous solution.

The number of independent chemical reactions is determined by means of a reac-tion matrix. This is formed by the the chemical reactions (as rows) and the stoichio-metric coefficients associated with the different chemical reactions (as columns).

The number of independent chemical reactions is equal to the rank of the reactionmatrix. If the rank is lower than the number of rows, one of the rows should beremoved as it is not linearly independent of the others.

Example 43: The rank of a reaction matrix

Problem

Investigate the number of independent chemical reactions in a system where the fol-lowing chemical reactions occur:

Reaction 1 A 2B

Reaction 2 B CReaction 3 A 2C

Solution

Form the reaction matrix by setting up the chemical reactions as rows and the chemicalcomponents involved as columns:

A B C

Reaction 1Reaction 2Reaction 3

1 2 00 1 11 0 2


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We now useMatlabto calculate the rank of the reaction matrix: Obviously, the rank

of the matrix is 2, not 3. Hence only two of the rows and reactions are linearly inde-pendent. These two can, in this case, be chosen freely among the three reactions.

4.4.3 Degree of freedom analysis

Example 43 indicates how the Prod terms have to be quantified when reaction sys-tems are treated by means of mass balances. One way to look at this is:

1. Input terms are quantified by stream variables representing fluxes into the sys-tem.

2. Output terms are quantified by stream variables representing fluxes out of the

system.3. Prod terms are quantified by means of reaction parameters, one for each reac-

tion.

This means that the total number of unknowns in the equation system is increasedby the number of reaction parameters. Consequently, the degree of freedom analysisbecomes:

Number of (unknown) stream variables+ Number of reaction parameters- Number of independent component mass balances- Number of independent information

= Number of degrees of freedom


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Example 44: Application of the reaction parameter method

Problem

In a reactor, two chemical reactions occur simultainuously. In one of those, ethane(C2H6) is dehydrated to ethene (C2H4) and H2. In the other, ethane reacts withH2 to form the biproduct methane (CH4):

Reaction 1 C2H6 C2H4+ H2Reaction 2 C2H6+ H2 2CH4

The feed (i.e., the input flow of reactant gas) consist of 85% C2H6 while the rest isinert gas (N2). Calculate the magnitude of all output fluxes, if the total conversion

ofC2H6 is 50.1%, while the yield with respect to formed ethene C2H4 is 47.1%. Inaddition, calculate the selectivity with respect to ethene relative to formed CH4.

Solution

The process chart becomes::

F1,C2H6

F1,N2

1 2dehydro-genering

F2,C2H6 F2,N2

F2,C2H4 F2,H2F2,CH4

The next step is to form a reaction matrix. Although it is obvious that the reactionsare linearly independent (CH4 only appears in one of the reactions), this can beanalyzed formally by calculating the rank of:

Reaction 1Reaction 2

C2H6 C2H4 CH4 H21 1 0 11 0 2 1

Obviously, the rank of the reaction matrix is 2, there are 2 independent reactionsand we need to introduce 2 reaction parameters in our calculations. The degree offreedom analysis, combined with the information given results in:

Stream variables 7+ Reaction parameters 2

- Independent MB - 5- Independent info - 3


We now introduce the basis of calculation F1 = 100 mole in order to eliminate thelast degree of freedom. This is allowed since no other flux is given. The equations


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become:

MB C2H6: F1,C2H6 1 2 =F2,C2H6MB C2H4: 1 =F2,C2H4

MB H2: 1 2 =F2,H2MB CH4: 22 =F2,CH4

MB Inert : F1,N2 =F2,N2Conc. inF1: F1,C2H6 = 0.85(F1,N2+ F1,C2H6)

Conversion : F2,C2H6 = (1 0.501)F1,C2H6Yield : F2,C2H4 = 0.471F1,C2H6

Basis of calc.: F1,N2+ F1,C2H6 = 100

As usual, the equations are written in matrix notation AX=Y:

A=

1 0 1 0 0 0 0 110 0 0 1 0 0 0 1 00 0 0 0 0 1 0 1 10 0 0 0 1 0 0 0 20 1 0 0 0 01 0 0

0.15 0.85 0 0 0 0 0 0 00.499 0 1 0 0 0 0 0 00.471 0 0 1 0 0 0 0 0

1 1 0 0 0 0 0 0 0

,X=

F1,C2H6F1,N2

F2,C2H6F2,C2H4F2,CH4F2,H2F2,N212

,Y =

00000000

100

and theMatlabsolution becomes: Thus, the selectivity is F2,C2H4/F2,CH4 = 40.03/5.1 =7.849.


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C H A P T E R 5

Reactor calculations

In the previous chapters, we have mainly dealt with mass balance problems to calcu-late certain material flows over system boundaries, utilizing other flows, informationand balances. For reaction systems, we have also defined valuable quantities such asconversion, yield and selectivity.

However, we have notgiven the chemical reactions much attention. The integralmass balances can, at the most, provide a possibility to calculate the reaction param-eter which is a measure of the total production/consumption of a substance in asystem. The integral mass balances cannot, however, explain why a reaction occursto a certain extent, and certainly cannot be used to describe the dynamics of thereaction system.

In this chapter, we will study the chemical reactorby which we mean simply a

volume in which occur one or several chemical reactions. Figure 5.1 shows somechemical reactors: a lake, a denitrification basin in a sewage treatment plant anddough which is in the process of becoming bread in a bakery. Despite the differencesbetween reactors, they have two important features in common: they have a clearsystem boundary, and they contain reactants that participate in chemical reactions.

In several aspects, the three reactors shown are very different. The lake is largelymixed, although it has periodically stable stratification between the epilimnion andthe hypolimnion. On the contrary, the denitrification basin is horizontally but notvertically mixed. If it were, oxygen would be mixed into the water and the denitrifica-tion process, which occurs under anaerobic, reducing conditions would slow down oreven stop. In terms of system boundaries, both the lake and the basin have inflow andoutflow streams. Flow rates vary with time, as well as the concentrations of dissolved

substances in the inputs and outputs as well as within the system itself.The bread dough undergoing fermentation, however, has no inflows or outflows.The chemical process occurring in the dough systems is primarily a transformationof carbohydrates (sugar) to CO2 and H2O. However, these products do not leavethe dough; they stay within the system. We see this as the dough rises. When thefermentation has progressed to a certain level, the dough is placed on a moving beltthat will convey it into a continuous baking oven.

In this chapter, we limit ourselves to treating chemical reactions in liquid systems.These are called homogeneous liquid phase systems. We will not treat gas phase

52


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Reactor calculations 53

Figur 5.1: Three examples of reactors; a lake, a basin in a treatment plant anda piece of dough at Lockarps in Malm.


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5.1 Kinetic rate equations 54

reaction systems.1

We will focus on three fundamental concepts that are very important for chemicalreactor calculations:

Kinetic models used to calculate the rate of chemical reactions, and rate con-stants.

The concept of residence time which describes how much that that flows in andout of the system.

The mixing models orreactor modelsthat are used to describe how the reactantscome in contact with each other within the reactor.

5.1 Kinetic rate equations

The kinetic rate equation describes how fast a reaction proceeds at a given momentper unit volume and time. It has units of the type

mol

volume time.

Kinetic rate equations are not simply made up; they are deduced on theoreticalgrounds, while the rate coefficients are determined experimentally. Table 5.1 gives anoverview of different common rate equations for different reaction types.

5.1.1 First order kinetics

The most simple is the first order rate equation. It is valid for irreversible reactionsincluding only one reactant:

A CThe rate equation describing how fast the reaction proceeds from left to right is:

r= kcAmole

volumeunit time

wherer is the reaction rate,k is the rate coefficientand cAis the concentration of thereactant A.

The theoretical basis is simple: the probability that a certain molecule of A willreact within the system is directly proportional to the concentration of A!

One must note that the rate equation refers to the rate by which the reactionproceeds from left to right. For the rate of consumption of A and the production ofC, one must take the stoichiometric coefficients into consideration. For A C, thestoichiometric coefficient of A, A is -1 (it is negative since A is consumed) while Cequals 1.

To get the rate equations for A and C, one should multiply the general rateequationr by the stoichiometric coefficient:

r = kcA

rA = Ar = 1kcA= kcArC = Br= 1kcA= kcA

The reaction is called first orderbecause the exponent on cA is 1, ascA= c1A.1Gas phase reaction systems differ from liquid phase reaction systems in two ways: as the total

number of moles change due to the reaction, the pressure and the flow rate will also change.


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5.1 Kinetic rate equations 55

Tabell 5.1: Examples of kinetic rate equations. The abbreviations rev. and

irrev. denote reversible and irreversible reactions respectively. The unitsare based on the use of mole, liters (L) and hours (hr) but any measure of volumeand time may be used.

ReactionType Example rA Unitsorder

0 irrev. A B rA= k k: molLhr1 irrev. A B rA= kcA k: 1hr1 rev. A B rA= kfcA+ kbcB k: 1hr

Pseudo 1 irrev. A + H2O B r= kcA k: 1hr2 irrev. 2A

B r

A=

kc2A

k: Lmolhr

2 irrev. A + B C rA = kcAcB k: Lmolhr2 rev. A + B C rA = kfcAcB+ kbcC k: Lmolhr

Monod irrev. AEnzymecatalysis

B r= kcAK+ cA

k: molLhr ; K :

molL

5.1.2 Second order kinetics

An example of a simple second order reaction is:

A + B CThe theoretical basis is that, in order for A and B to react they must meet, andthe probability that a certain molecule of A meets B within the system is directlyproportional to the product of the concentrations of A and B. Thus, the general rateequation becomes:

r= kcAcB

Since the stoichiometric coefficients are -1 or 1, we get the following rate equationsfor A,B and C:

rA = kcAcBrB = kcAcBrC = kcAcB

The reaction is called second orderbecause the sum of exponents oncAand cB in therate equations is 1+1=2.

5.1.3 Overview of kinetic rate equation

Table 5.1 gives an overview of common rate equations for common stoichiometries.


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5.2 Mean residence time and reaction time 57

As we see, the stoichiometric coefficients are either -1 (for reactants) or 1 (for prod-

ucts). Hence the rate equations for A, B and C become:rA =r1 = k1cArB =r1 r2 =k1cA k2cBrC =r2 =k2cB

It is thus possible to express the equilibrium constant in terms of a ratio between tworate coefficients, or cA and cB.

5.1.6 Parallel reactions

In parallel reactions, a reactant may participate in several reactions. For example, ifthe reactant A may react either to B or to C, the reaction system includes:

A k1 B

A k2 C

Then, if:

r1= k1cA

r2= k2cA

we get:

rA = r1 r2 = (k1+ k2)cArB =r1 =k1cA

rC =r2 =k2cA

One example of parallel reactions in nature, is the reaction of nitrous oxide, NO2, inthe formation of nitric acid, HNO3, from reactions either with OH orNO3:

NO2 + OH HNO3NO2 + NO3 N2O5+ H2O 2HNO3

5.2 Mean residence time and reaction time

In all systems in which chemical reactions occur, the conversion depends on how longthe reaction takes place inside the system. If the chemical reaction proceeds for a

short time, the conversion from reactants to products will be lower than if it takeslonger.In Figure 5.1, three examples of chemical reactors were shown. One of the differ-

ences between them was that while two (the lake and the basin) both has inflow andoutflow, the third (the dough) has neither. Thus, the lake and the basin are examplesof opensystems while the dough is an example of a closedsystem.

We can now introduce three very important quantities relevant to opensystems:

The volumeof a reactor: V (liters, m3, etc.)

The volumetric flow rateinto or out of a reactor: Q(liters/s, m3/hour, etc.)


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5.2 Mean residence time and reaction time 59

Example 51: Average residence time of a lake

Problem

Lake Kvarnsjn in the province of Vsterbotten receives its inflow water from thestream Aborrbcken. In the environmental monitor program, the flow rate is mea-sured the 15th of each month. Calculate the average residence time of the water inKvarnsjn, if the lake area is A = 15 ha and the average lake depth is h = 3.5 m,expressed in years. Use the data for 1999 given below.

Month Flow

(m3 day1)January 222February 609March 3327April 2730May 2870June 693July 33August 93September 441October 1461November 1147December 435

Solution

The following calculations are made in Matlab, resulting in an average residence timeof water in the lake of = 1.2 years:

1 2 3 4 5 6 7 8 9 10 11 120

500

1000

1500

2000

2500

3000

3500

Uppmttflde(m3/dygn)

Mnad

Medelflde


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5.3 Reactor models 60

5.3 Reactor models

In this section we deal with the basics of reactor calculations by studying ideal reactormodels. They always are the starting point when trying to describe the conditionsin a reactor system, especially with respect to mixing. Mixing is of fundamentalimportance because it affects the concentrations in the system, which in turn affectthe rate of chemical reactions. Here, we will only consider macroscopicmixing, notdiffusive transport at the molecular level.

It is also of fundamental importance whether the reactor is an open or closedsystem.

Given these to dimension, mixing and open/closed, there are three ideal reactormodels to consider:

1. The ideal tank reactor

2. The ideal batch reactor

3. The ideal plug-flow reactor

As the drawings illustrate, these types of chemical reactor all differ since:

1. The ideal tank reactor is open and perfectly mixed.2. The ideal batch reactor is closed and perfectly mixed.

3. The ideal plug-flow reactor is open and not mixed.

5.3.1 The differential mass balance

All of the ideal reactor models can be characterized by the same mass balance equa-tion, the differential (component) mass balance2. As reaction systems are of primeinterest, the component mass balance will always be expressed in terms of molar fluxesof components, denoted FA, FB, etc. Furthermore, the fluxes will be given in molarfluxes per unit time (i.e., mol/hour etc.).

The general mass balance for a given component, valid for all ideal reactor modelsis:


Fin+ rV = Fout+d(cV)

dt

mole

unit time

where Fis the molar flux (mole/unit time),c is the molar concentration (mole/unit vol-ume),Qis the volumetric flow rate (volume/unit time) and V is the reactor volume(volume).

2From now on it will be understood that mass balances refer to components, not elements.


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5.3 Reactor models 61

Although based on Input + Prod = Output + Acc, there are differences relative to

how the mass balance principle was used in Chapters 2-4: the Prod-term is expressedin terms of kinetic rate equations, and the Acc-term deals with possible changes withtime.

The Input and Output-terms: The terms representing the fluxes across the re-actor system boundaries,Finand Fout, are given by the product of the concentrationcand the volumetric flow rates Qinand Qoutrespectively. If the flow rate is constant,Qin = Qout= Q.

The Prod-term: The Prod-term quantifies how much of a substance that is pro-duced in the reactor per unit of time. The Prod-term is formed by multiplying thekinetic rate equation (mole/unit time and unit volume) by the volume. This is logicalsince scaling up will lead to a higher Prod while scaling down will decrease theamount converted in the chemical reaction.

If, for example, only the reaction A Boccurs with rA= kcA, we get:Prod = kcA V

In more general terms, we can express the Prod-term for component j reacting inseveral chemical reactions i, as the sum of the contributions for all reactions:

Prodj=allii=1

i,jriV

Thus, in terms of calculations there are three actions related to the Prod-term:

1. Identify all reactions.

2. Define the corresponding kinetic rate equations.

3. Include the kinetic rate equations in the Prod-term with the right stoichiometriccoefficients and signs (+ or -).

The Acc-term: The Acc-term quantifies the change in the number of moles perunit of time (i.e., the timederivative of the molar amount). The molar amount in thereactor is given by the product of the concentration c and the volume V (i.e., c V).By taking the differential ofcV, we get the Acc-term as:

Acc = d

dt

cV =

Vdc

dt+ c

dV

dt ifVvaries with time

Vdcdt ifV is constant

In summary: For a component A which reacts in one single reaction in a reactorsystem where the flow rate and the volume is constant, the differential mass balancebecomes:

Qincin,A+ rAV =Qincout,A+d(cAV)

dt

mole

unit time (5.2)


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5.4 The ideal completely stirred tank reactor, CSTR 62

Figur 5.2: Many environmental systems can be modeled by means of a CSTR.

5.3.2 Methodology for reactor calculations

Regardless of rector model, the chemical reactor calculations will follow the procedure:

1. Define the flow and mixing conditions in terms of an ideal reactor model.

2. Determine the reactor-specific parameters.

3. Define the kinetic equation(s) and rate constant(s).

4. Combine the reactor model and the kinetic equations(s) in the differential massbalance.

5. Do the calculations, analytically or numerically.

5.4 The ideal completely stirred tank reactor, CSTR

Almost all industrial and natural, open systems can be described in terms of one ormore ideal tank reactors, often called Continuously Stirred Tank Reactor, CSTR.

Since the CSTR is perfectly mixed, the concentrations (and temperature) are thesame everywhere in the entire reactor. Thus, there are no gradients whatsoever in thereactor. The important implication of perfect mixing is that the reaction rates arethe same everywhere in the reactor. The general mass balance for a CSTR becomes:

cutQut

cinQin cr V

In + Prod = Out + Acc

Qincin+ rV = Qoutcout+ d

dtcV

At steady state, when Acc=0 och Qin = Qout= Q the simplified model is:

In + Prod = Ut (5.3)

Qcin+ rV = Qcout (5.4)

Final remark: Never forget that the concentration in the reactorc is the same ascout!


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Example 52: CSTR with a 1st order irreversible reaction

Problem

Calculate output stream fluxes from a steady-state CSTR in which a 1st order reactionoccurs:

AB r= kcA= 0.5cA mole m3 minprovided that the volumetric flow rate is 1.25 m3 min1, the reactor volume is 5 m3,and the input stream concentration of the reactant A is 2 mole m3 while the inputconcentration of B is 0.2. mole m3.

Solution

First we select the appropriate model: the steady-state CSTR. For a CSTR the

differential mass balances are given by:Qincin,A+ rAV = Qoutcout,A

Qincin,B+ rBV = Qoutcout,B

Since Qin =Qout = Q, cout,A = cA and cout,B =cB the equations can be simplifiedto:

Qcin,A+ rAV = QcA

Qcin,B+ rBV = QcB

All of the reactorspecific parameters are given in the problem definition. The kineticrate equations for first order reaction with respect to the reactant A and the productB are:

rA= Ar= 1r = kcArB =Br= 1r =kcA

We can now combine the kinetic equations and the differential mass balance model.The combined equations are:

Qcin,A = +kcAV + QcA

Qcin,B = kcAV + QcB

or, if we introduce the residence time =V

Q:

cin,A= cA(1 + k) , cA =

cin,A

1 + kcin,B = kcA+ cB , cB =cin,B+ kcA (E52.5)

Equation E52.5 allows us to solve directly for cA. Before inserting numbers, we

recognize that = 5

1.25= 4minutes, so that:

cA = 2

1 + 0.54= 0.667 mole m3

and thus, cB = 0.2 + 0.5 4 0.667 = 1.533 mole m3.


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Example 53: Example 52 with matrix notation

Problem

Utilize the fact that the above reaction calculations can be seen as a system of linearequations, and solve this system with Matlab.

Solution

The first step is to re-write E52.5 :cin,A = cA(1 + k)

cin,B = kcA+ cB

with all terms including cA and cB on the left-hand side:(1 + k)cA+ 0 cB =cin,A

kcA+ 1 cB =cin,B

This system of equations can be written in the form AX=Y:

A=

1 + k 0k 1

, X=

cAcB

, Y =

c

Calculus Checar

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