modeling notes complete
TRANSCRIPT
INTRODUCTION TO MATHEMATICAL
MODELING OF CHEMICAL PROCESSES
Emad Ali, AbdelHamid Ajbar and Khalid Alhumaizi
King Saud University College of Engineering
Chemical Engineering department
June 2006
i
Table of Contents 1. Chapter 1: Fundamentals 1
1.1. Introduction 1 1.2. Incentives for Process Modeling 1.3. Systems 3
1.3.1. Classification based on Thermodynamics 3 1.3.2. Classification based on number of phases 4
1.4. Classification of Models 4 1.5. State Variables and State Equations 5 1.6. Classification of theoretical models 6
1.6.1. Steady State vs Unsteady State 6 1.6.2. Lumped vs Distributed Parameters 6 1.6.3. Linear vs Non-Linear 7 1.6.4. Continuous vs Discrete 7 1.6.5. Deterministic vs Probabilistic 8
1.7. Building Steps for a Mathematical Model 1.8. Conservation Laws 9
1.8.1. Total Mass Balance 10 1.8.2. Component Balance 11 1.8.3. Momentum Balance 11 1.8.4. Energy Balance 12
1.9. Microscopic Balance 12 1.10. Macroscopic Balance 14 1.11. Transport rates 14
1.11.1. Mas Transport 15 1.11.2. Momentum Transport 16 1.11.3. Energy Transport 18
1.12. Thermodynamic relations 19 1.13. Phase Equilibrium 21 1.14. Chemical Kinetics 24 1.15. Control Laws 25 1.16. Degrees of Freedom 26 1.17. Model Solution 27 1.18. Model Validation 27
2. Chapter 2: Examples of Mathematical Models for Chemical processes 29 2.1. Examples of Lumped parameter Systems 29
2.1.1. Liquid Storage tank 29 2.1.2. Stirred Tank Heater 33 2.1.3. Isothermal CSTR 39 2.1.4. Gas-Phase Pressurized CSTR 43 2.1.5. Non-Isothermal CSTR 45 2.1.6. Mixing Processes 48 2.1.7. Heat Exchanger 53 2.1.8. Heat Exchanger with Steam 56
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2.1.9. Single Stage Heterogeneous Systems 58 2.1.10. Two-Phase Reactor 60 2.1.11. Reaction with Mass Transfer 64 2.1.12. Multistage Heterogeneous Systems 67 2.1.13. Binary Absorption Column 71 2.1.14. Multi-Component Distillation Column 73
2.2. Examples of Distributed Parameter Systems 82 2.2.1. Liquid Flow in Pipe 82 2.2.2. Velocity profile in a Pipe 84 2.2.3. Diffusion with Chemical Reaction in a Slab catalyst 86 2.2.4. Temperature Profile in a heated Cylindrical Rod 89 2.2.5. Isothermal Plug Flow Reactor 91 2.2.6. Non-Isothermal Plug-Flow Reactor 94 2.2.7. Heat Exchanger: Distributed Parameter Model 97 2.2.8. Mass Exchange in a Packed Column 99
3. Chapter 3: Equations of Change 103 3.1. Total Mass Balance 103 3.2. Component Balance Equation 106 3.3. Momentum balance Equation 109 3.4. Energy balance 113 3.5. Conversion between the Coordinates 116
3.5.1. Balance Equations in Cartesian Coordinates 117 3.5.2. Balance equation in Cylindrical Coordinates 118 3.5.3. Balance Equation in Spherical Coordinates 119
3.6. Examples of Application of equations of Change 121 3.6.1. Liquid Flow in a Pipe 121 3.6.2. Diffusion with Chemical Reaction in a Slab Catalyst 121 3.6.3. Plug Flow Reactor 122 3.6.4. Energy Transport with Heat Generation 123 3.6.5. Energy Transport in a Circular Tube 125 3.6.6. Unsteady State Heat Generation 127 3.6.7. Laminar Flow Heat Transfer with Constant Wall Temperature 127 3.6.8. Laminar Flow and Mass Transfer 129
Problems (Lumped Parameter Systems) 131 Problems (Distributed Parameter Systems) 139
1
Chapter 1 : Fundamentals
1.1 Introduction
The objective of mathematical modeling is the development of sets of
quantitative (mathematical) expressions that capture the essentials aspects of an existing
system. A mathematical model can assist in understanding the complex physical
interactions in the system and the causes and effects between the system variables
.Mathematical models are valuable tools since they are abstract equations that can be
solved and analyzed using computer calculations. It is therefore safer and cheaper to
perform tests on the model using computer simulations rather than to carry out
repetitive experimentations and observations on the real system. This becomes vital if
the real system is new, hazardous, or expensive to operate . Modeling, thus prevails the
field of science, engineering and business. It is used to assist in the design of equipment,
to predict behavior, to interpret data, to optimize resources and to communicate
information.
1.2 Incentives for process modeling
In the chemical engineering field, models can be useful in all the phases, from
research and development to plant operation. Models and their simulation are tools
utilized by the chemical engineer to help him analyze the process in the following ways:
• Better understanding of the process
Models can be used to study and investigate the effects of various process
parameters and operating conditions on the process behavior. It can also be used
to evaluate the interactions of different parts of the process. This analysis can be
carried out easily on a computer simulation without interrupting the actual
process, thus avoiding any delay or upsets for the process.
• Process synthesis and design
2
Model simulation can be utilized in the evaluation of equipment's size and
arrangements and in the study of alternative process flow-sheeting and strategies.
Furthermore, to verify the reliability and safety of the process design tests can be
carried out even prior to plant commissioning.
• Plant operators training
Models can be used to train plant personnel to simulate startup and shutdown
procedures, to operate complex processes and to handle emergency situations and
procedures .
• Controller design and tuning
Models help in developing and evaluating better controller structure and
configuration. Dynamic simulation of models is usually employed for testing and
assessing the effectiveness of various controller algorithms. It is worthwhile to
mention that models play a vital part in designing advanced model-based control
algorithms such as model predictive and internal model controllers .Moreover it
is a common practice of many control engineers to determine the optimum values
of the controller settings through dynamic simulation .
• Process optimization
It is desirable from economic standpoint to conduct process optimization before
plant operation to determine the optimum values of the process key parameters
or/and operating conditions that maximizes profit and reduces cost. Process
optimization is also performed during process operation to account for variations
in the feed-stock and utilities market and for changing environmental regulations.
It is worth mentioning that despite all their usefulness, models at their best are no more
than approximation of the real process since they do not necessarily incorporate all the
features of the real system . Therefore modeling can not eliminate completely the need
for some plant tests, especially to validate developed models or when some poorly
known parameters in the process need to be experimentally evaluated.
Models can be classified in a number of ways. But since mathematical models
are developed from applying the fundamental physical and chemical laws on a specific
system,
3
1
we review first the classification of systems since their nature affect the modeling
approach and the resulting model.
1.3 Systems
A system is a whole consisting of elements or subsystems. The system has
boundaries that distinguish it from the surrounding environment (external world) as
shown in Figure 1.1.The system may exchange matter and/or energy with the
surrounding through its boundary. Consequently, the state of a system can be defined or
understood via the interactions of its elements with the external world. A system may
be classified in different ways, some of which are as follows:
1.3.1 Classification based on thermodynamic principles
• Isolated system
This type of system does not exchange matter nor energy with the surrounding.
Adiabatic batch reactor is an example of such systems.
• Closed system
This type of system does not exchange matter with the surrounding but it does
exchange energy. Non-adiabatic batch reactor is an example of such systems.
• Open system
This system exchanges both matter and energy with the external environment.
An example of this system is the continuous stirred tank reactor (CSTR).
4
System
Boundary
Figure 1.1: system and its boundary
Suroundings
1.3.2 Classification based on number of phases
• Homogeneous system
This is a system that involves only one phase such as gas-phase or liquid-phase
chemical reaction processes.
• Heterogeneous system
This is a system that involves more than one phase. This kind of systems exists in
multi-phase reaction processes and in phase-based separation processes.
1.4 Classification of Models
Models can be classified according to how they are derived:
• Theoretical models.
These are models that are obtained from fundamental principles, such as the laws
of conservation of mass, energy, and momentum along with other chemical
principles such as chemical reaction kinetics and thermodynamic equilibrium,
etc. This first-principle model is capable of explaining the underlying physics of
the process and is often called `phenomenological model'. For this reason it is
particularly suitable for process design and optimization . Theoretical models are,
however, generally difficult to obtain and sometimes hard to solve.
5
• Empirical models
These models are based on experimental plant data. These models are developed
using data fitting techniques such as linear and non-linear regression. Models
obtained exclusively form experimental plant data are also known as black-box
models . Such models do not provide detailed description of the underlying
physics of the process. However, they do provide a description of the dynamic
relationship between inputs and outputs. Thus they are sometimes more adequate
for control design and implementation.
• Semi-empirical models
These models are somehow between the two previous models where uncertain or
poorly known process parameters are determined from plant data.
This book focuses only on developing theoretical models. The interested reader in
empirical modeling is referred to books listed in the references.
1.5 State variables and state equations
Once the system has been classified, developing a theoretical model for it
amounts to characterizing its behavior at any time and at any spatial position. For most
processing systems a number fundamental quantities are used to describe the natural
state of the system. These quantities are the mass, energy and momentum. Most often
these fundamental quantities can not be measured directly thus they are usually
represented by other variables that can be measured directly and conveniently. The most
common variables are density, concentration, temperature pressure and flow rate. There
are conveniently called 'state variables' since they characterize the state of the
processing system.
In order to describe the behavior of the system with time and position, the state
(dependent) variables should be linked to the independent variables (time, spatial
position) through sets of equations that are derived from writing mass, energy and
momentum balances. The set of equations describing these variables are called 'state
equations '.
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1.6 Classification of theoretical models
With this in mind, theoretical models may be further classified in more practical
ways as discussed in the following.
1.6.1 Steady state Vs. unsteady state
When the physical state of the processing system remains constant with the time,
the system is said to be at steady state. Models that describe steady state situations are
also called static, time-invariant or stationary models. Basically, almost all chemical
process unit designs are carried out on static models. On the other hand unsteady state
processes represent the situation when the process state (dependent variables) changes
with time. Models that describe unsteady-state situations are also called dynamic and
transient models .Such models are useful for process control design and development.
Process dynamics are encountered in practice during startup, shutdown, and upsets
(disturbances )
1.6.2 Lumped Vs. distributed parameters
Lumped parameters models are those in which the state variables and other
parameters have/or assumed to have no spatial dependence, i.e. they are considered to
be uniform over the entire system. In this case the time (for unsteady state models) is the
only independent variable. The chemical engineering examples for this case include the
perfectly mixed CSTR, distillation columns. etc. Conceptually, these models are
obtained through carrying out a macroscopic balance for the process as it will be
discussed in chapter 2 . On the other hand, distributed parameters models are those in
which states and other variables are function of both time and spatial position. In this
case, modeling takes into account the variation of these variables with time and from
point to point throughout the entire system. Some examples of such systems include
plug flow reactor, heat exchangers, and packed columns. These models are essentially
obtained through writing microscopic balances equations for the process as it will be
discussed later in chapter 2.
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1.6.3 Linear Vs non-linear
Linear models have the important property of superposition whereas nonlinear
models do not. Superposition means that the response of the system to a sum of inputs is
the same as the sum of responses to the individual inputs .In linear models all the
dependant variables or their derivatives appear in the model equations only to the first
power. These properties do not hold for nonlinear models . In this respect, it is
important to recognize the fact that most physical and chemical systems are nonlinear.
Linear models are commonly obtained through linearization of the nonlinear model
around a certain steady state. Linear models obtained this way are valid approximation
of the original non linear model only in the neighborhood of the selected state .
1.6.4 Continuous Vs discrete
When the dependant variables can assume any values within an interval the
model is called continuous .When on the other hand one or several variables are
assumed to take only discrete values, the model is called discrete. In chemical
engineering discrete models arise for example when some variables are required to take
only integer values, for example the number of stages in a distillation column, the
number of heat exchanger in a plant, etc.
1.6.5 Deterministic Vs probabilistic
Deterministic models are those in which each variable can be assigned a definite
fixed number, for a given set of conditions. On the other hand in probabilistic or
stochastic models some or all the variables used to describe the system are not precisely
known. They are considered as random variables. Probabilistic models are often
encountered when modeling systems that are subject to noise.
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1.7 Building steps for a mathematical model
Building a theoretical mathematical model of a processing system requires the
knowledge of the physical and chemical interactions taking place within the boundaries
of the system. With a given degree of fundamental knowledge of the system at a certain
stage one can build different models with different degree of complexity depending on
the purpose of the model building and the level of rigor and accuracy required .The
choice of the level of rigor and degree of sophistication is in itself an art that requires
much experience. Theoretical modeling of chemical processes may encounter
difficulties that can be classified as follows:
• Problems arising from processes that exhibit complicated physical and chemical
phenomena. Such problems appear for instance in multi-component interactive
processes .
• Problems arising from imprecisely known process parameters. This situation can
be handled by periodically estimating the unknown parameters from plant data.
• Problems arising from the size and complexity of the ensuing model. These can
be overcomed by proper use of simplifying assumptions.
Consequently a modeler should practice careful utilization of the simplifying
assumptions based on engineering sense and experience. Failure to do so, the modeler
may fall into one of the two extremes, i.e. creating rigorous but over complicated model
or creating oversimplified model that does not capture all the critical features of the true
process.
The general procedure for building up a mathematical model includes the following
steps:
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• Identification of the system configuration, its surrounding environment and the
modes of interaction between them, the identification of the relevant state
variables that describe the system and identification of the process taking place
within the boundaries of the system
.
• Introduction of the necessary simplifying assumptions.
• Formulation of the model equations based on principles of mass, energy and
momentum balances appropriate to the type of the system. This also requires the
determination of the fundamental quantitative laws (chemical kinetics,
thermodynamic relations…) that govern the rates of the process in terms of the
state variables .
• Determination of the solvability of the model using degree of freedom analysis.
• Development of the necessary numerical algorithms for the solution of the model
equations.
• Validation of the model against experimental results to ensure its reliability and
to re-evaluate the simplifying assumptions which may result in imposing new
simplifying assumptions or relaxing others.
1.8 Conservation Laws
As mentioned in previous section, the state equation forming the mathematical
model defines a relationship between the state variables (dependent variables) and the
independent variables, i.e., time and spatial variables of the system. These equations are
derived, from applying the conservation law for a specific fundamental quantity say S,
on a specific system with defined boundaries (see Figure 1.2) as follows:
10
System
Figure 1.2
Sin
S
Sout
Total flow
rate of (S)
into the
system
+
Generation
rate of (S)
within system
=
Total flow
rate of (S) out
of the system
+
Accumulation
rate of (S)
within system
±
amount of (S)
exchanged with
the surrounding
(1.1)
The quantity S can be any one of the following quantities:
• Total Mass
• Component Mass (Mole)
• Total Energy
• Momentum
1.8.1 Total mass balance
Since mass is always conserved, the balance equation for the total mass (m) of a given
system is:
Rate of mass in = rate of mass out + rate of mass accumulation (1.2)
11
The mass balance equation has the SI unit of kg/s.
1.8.2 Component balance
The mass balance for a component A is generally written in terms of number of moles
of A. Thus the component balance is
flow of moles
(A) in
+
Rate of Generation
of moles of (A)
=
Flow of moles
of (A) out
+
Rate of Accumulation
of moles of (A) (1.3)
The component balance has the unit of moles A/s. It should be noted that unlike the total
mass, the number of moles of species A is not conserved. The species A can be
generated or consumed by chemical reaction.
1.8.3 Momentum balance
The linear momentum (π) of a mass (m) moving with velocity (v) is defined as:
π = mv (1.4)
Since the velocity v is a vector, the momentum, unlike the mass is also a vector. The
momentum balance equation using (Eq 1.1) is:
Rate of
momentum in
+
Rate of Generation
of momentum
=
Rate of
momentum out
+
Rate of Accumulation
of momentum
(1.5)
The momentum balance has the unit of kg.m/s2. The momentum balance equation is
usually written using the Newton's second law. The law states that the time rate of
change of momentum of a system is equal to the sum of all forces F acting on the
system,
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∑= Fdtmvd )(
(1.6)
1.8.4 Energy balance
The energy balance for a given system is:
Rate of
energy
in
+
Rate of
Generation of
energy
=
Rate of
energy
out
+
Rate of
accumulation
of energy
± amount of
energy
exchanged
with the
surrounding
(1.7)
The energy generated within a system includes the rate of heat and the rate of work. The
rate of heat includes the heat of reaction (if a reaction occurs in the system), and the heat
exchanged with the surroundings. For the rate of work we will distinguish between the
work done against pressure forces (flow work) and the other work such as the work
done against the gravity force, against viscous forces and shaft work. The reason for this
distinction will appear clearly in the next chapter.
For a given system the general conservation law (Eq. 1.1) can be carried out either on
microscopic scale or macroscopic scale.
1.9 Microscopic balance
In the microscopic case, the balance equation is written over a differential
element within the system to account for the variation of the state variables from point
to point in the system, besides its variation with time. If we choose for example
cartesian coordinates, the differential element is a cube as shown in Figure 1.3. Each
state variable V of the system is assumed to depend on the three coordinates x,y and z
plus the time. i.e. V = V(x,y,z,t). The microscopic balance can be also written in
cylindrical coordinates (Figure 1.4) and in spherical coordinates (Figure 1.5). The
selection of the appropriate coordinates depends on the geometry of the system under
13
study . It is possible to convert from one coordinate system to an other as it will be
discussed in chapter 3.
∆x
∆z∆y
x
z
y
Figure 1.3: Cartesian coordinates
ry
z
x
θ
z
Figure 1.4: Cylindrical coordinates
14
r zθ
φ
y
x
z
Figure 1.5: Spherical coordinates
1.10 Macroscopic balance
In some cases the process state variables are uniform over the entire system, that
is each state variable does not depend on the spatial variables, i.e. x,y and z in cartesian
coordinates but only on time t. In this case the balance equation is written over the
whole system using macroscopic modeling . When modeling the process on
microscopic scale the resulting models consists usually of partial differential equation
(PDE) where time and one or more spatial position are the independent variables. At
steady state, the PDE becomes independent of t and the spatial positions are the only
independent variables .When, on the other hand, the modeling is based on macroscopic
scale the resulting model consists of sets of ordinary differential equations (ODE) . In
the next chapter we present examples of applying macroscopic and microscopic
balances to model various chemical processes.
The fundamental balance equations of mass, momentum and energy already
discussed are usually supplemented with a number of equations associated with
transport rates and thermodynamic relationships. In the following we present an
overview of some of these relations.
1.11 Transport rates
15
Transport of the fundamental quantities, mass, energy and momentum occur by two
mechanisms
• Transport due to convection or bulk flow
• Transport due to molecular diffusion or potential difference .
In many cases the two transport mechanism occur together. Therefore, the flux due to
the transport of any fundamental quantity is the sum of a flux due to convection and a
flux due to diffusion.
1.11.1 Mass Transport
The total flux nAu (kg/m2s) of species A of density ρA (kg/m3) flowing with velocity vu
(m/s) in the u-direction is the sum of the two terms:
nAu = jAu + ρAvu (1.8)
total flux = diffusive flux + bulk flux (1.9)
The diffusive flux jAu (kg/m2s) for a binary mixture A-B is given by Fick's law:
dudwDj A
ABAu ρ−= (1.10)
where wA = ρA/ρ is the mass fraction of species Α, ρ (kg/m3) the density of the mixture
and DAB (m2/s) is the diffusivity coefficient of A in the mixture. In molar unit the flux
JAu (mol A/m2s) is given by:
dudxCDJ A
ABAu −= (1.11)
16
where C is the total concentration of A and B (Kg(A+B)/m3) and xA = CA/C is the mole
fraction of A in the mixture. For constant density the flux Eqs.( 1.10 and Eq. 1.11)
become:
dudDj A
ABAuρ
−= (1.12)
dudC
DJ AABAu −=
(1.13)
1.11.2 Momentum transport
Momentum is also transported by convection and diffusion. But unlike the mass,
the linear momentum π = mv is a vector. We have to consider then its transport in all
directions (x,y,z) of a given system. Let consider for instant the transport of the x-
component of the momentum. Similar analysis can be carried out for the transport of the
y and z component.
The flux due to convection of the x-component of the momentum in the y-direction, for
instant, is
(ρvx)vy (kg.m/s2) (1.14)
To determine the diffusion flux denoted by τyx of the x-component of the momentum in
the y-direction, consider a fluid flowing between two infinite parallel plates as shown in
Figure 1.6. At a certain time the lower plate is moved by applying a constant force Fx
while the upper plate is maintained constant. The force Fx is called a shear force since it
is tangential to the area Ay on which it is applied (Fig. 1.6).
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∆ Vx
∆ y F, force
x
y
Figure 1.6: Momentum transfer between two parallel plates
The force per unit area
)/.( 2smkgAF
y
x (1.15)
is called a stress and denoted τyx (kg.m/s). It is also a shear stress since it is tangential.
The force Fx imparts a constant velocity V = vx (y = 0 ) to the layer adjacent to the plate.
Because of molecular transport, the layer above it has a slightly slower velocity vx(y)
and so on as shown in Figure 1.6. Therefore, there is a transport by diffusion of the x-
component of the momentum in the y-direction. The flux of this diffusive transport is in
fact the shear stress τyx.
Therefore, the total flux πyx of the x-component in the y-direction is the sum of the
convection term (Eq. 1.14) and diffusive term τyx
πyx = τyx + (ρvx)vy (1.16)
momentum flux = diffusive flux + bulk flux (1.17)
For a Newtonian fluid the shear stress τyx is proportional to the velocity gradient:
yvx
yx ∂∂
−= µτ (1.18)
18
where µ (kg/m.s) is the viscosity of the fluid.
1.11.3 Energy transport
The total energy flux eu (J/s.m2) of a fluid at constant pressure flowing with a velocity vu
in the u-direction can be expressed as:
eu = qu + (ρCpT)vu (1.19)
energy flux = diffusive flux + bulk flux (1.20)
The heat flux by molecular diffusion, i.e. conduction in the u-direction is given by
Fourier's law:
uTkqu ∂
∂−=
(1.21)
where k (J/s.m.K) is the thermal conductivity.
The three relations (Eq. 1.10, 1.18, 1.21) show the analogy that exists between mass,
momentum and energy transport. The diffusive flux in each case is given by the
following form:
Flux = − transport property × potential difference
(gradient)
(1.22)
The flux represents the rate of transfer per area, the potential difference indicates the
driving force and the transport property is the proportionality constant . Table 1.1
summarizes the transport laws for molecular diffusion.
Table 1.1: One-dimensional Transport laws for molecular diffusion
Transport
Type
Law Flux Transport
property
gradient
19
Mass Fick's JAu D du
dC A
Heat Fourrier qu k dudT
Momentum Newton τux µ dudvx
When modeling a process on macroscopic level we can also express the flux by a
relation equivalent to (Eq. 1.22). In this case the gradient is the difference between the
bulk properties, i.e. concentration or temperature in two medium in contact, while the
transport property represents an overall transfer coefficient. For example for mass
transfer problems, the molar flux can be expressed as follows:
AA CKJ ∆×= (1.23)
where K an overall mass transfer coefficient.
The heat flux on the other hand is expressed as
TUq ∆×= (1.24)
where U is an overall heat transfer coefficient.
As for the momentum balance, the macroscopic description generally uses the pressure
drop as the gradient while the friction coefficient is used instead of the flux. The
following relation is for instance commonly used to describe the momentum laminar
transport in a pipe
PLv
Df ∆=ρ22
(1.25)
f is the Fanning friction factor, ∆P is the pressure drop due to friction, D is the diameter
of the pipe, L the length and v is the velocity of the fluid.
1.12 Thermodynamic relations
20
An equation that relates the volume (V) of a fluid to its temperature (T) and pressure
(P) is called an equation of state. Such equations are used to determine fluid densities
and enthalpies .
• Densities :
The simplest equation of state is the ideal gas law:
PV = nRT (1.26)
which can be used to determine the vapor (gas) density
ρ = MwP/RT (1.27)
where Mw is the molecular weight . As for liquids, tabulated values of density
can be used and can be considered invariant unless large changes in composition
or temperature occur.
• Enthalpies :
Liquid and vapor enthalpies for pure component can be computed from simple
formulas, based on neglecting the pressure effect as follows:
)(~refp TTCh −= (1.28)
λ+−= )(~refp TTCH (1.29)
Where h~ is the liquid specific enthalpy, H~ is its vapor specific enthalpy, pC
the average liquid heat capacity and λ is the latent heat of vaporization
Note that the reference condition is taken to be liquid at temperature Tref. If heat
of mixing is negligible, the enthalpy of a mixture can be taken as the sum of the
specific enthalpies of the pure components multiplied by their corresponding
21
mole fractions Note that the above enthalpy functions are valid for small
temperature variation and/or when the heat capacities of fluids are weak function
of temperature. In general cases the heat capacity Cp can be taken as function of
temperature such as:
Cp = a + bT + cT2 + dT3 (1.30)
The specific enthalpies for vapors and liquids are computed as integrals as
follows:
∫=T
Tp
ref
dTCh~ (1.31)
λ+= ∫T
Tp
ref
dTCH~ (1.32)
• Internal energy :
The internal energy U is a fundamental quantity that appears in energy balance
equation. For liquids and solids the internal energy can be approximated by
enthalpy .This can be also a good approximation for gases if the pressure change
is small.
1.13 Phase Equilibrium
A large number of chemical processes involve more than one phase. In many
cases the phases are brought in direct contact with each other such as in packed or tray
towers. When the transfer of either mass or energy occurs from a fluid phase to another
phase the interface between fluid phases is usually at equilibrium. In the case of heat
transfer between phase I and Phase II, the equilibrium dictates that the temperature is
the same at the interface. This is not the case for mass transfer. Figure 1.7 shows for
instant the mass transfer of species A from a liquid to a gas phase. The concentration of
22
the bulk gas phase yAG decreases at the interface. The liquid concentration increases, on
the other hand, from xAL to xAi. At the interface, an equilibrium exists and yAi and xAi are
related by a relation of the form of
yAi = F(xAi) (1.33)
The equilibrium relations are in general nonlinear. However, quite satisfactory results
can be obtained through the use of simpler relations that are derived form assumptions
of ideal behavior of the two phases:
Gas-phase mixtureof A in gas G
Liquid -phase solutionof A in liquid L
xAL
xAi
yAiyAG
NA
Interface
Figure 1.7: Equlibrium at the interface
• For vapor phases at low concentration, Henry's law provides the following
simple equilibrium relation:
pA = HxA (1.34)
Where pA (atm) is the partial pressure of species A in the vapor, xA is the mole
fraction of A in the liquid and H is the Henry's constant (atm/mol fraction)
Dividing by the total Pressure (P) we get another form of Henry's law:
yA = H~ xA (1.35)
23
The constant ( H~ ) depends on temperature and pressure.
• Raoult's law provides a suitable equilibrium law for ideal vapor-liquid mixtures
S
AAA PxPy = (1.36)
Where Ax is the liquid-phase mole fraction, Ay is the vapor-phase mole
fraction, PAs is the vapor pressure of pure A at the temperature of the system,
and P is the total pressure on gas-phase side.
The dependence of vapor pressure PAs on temperature can be approximated by
the Antoine equation
TCBAP S
A +−=)ln(
(1.37)
where A, B and C are characteristic parameters of the fluid.
• Raoult’s law can be modified to account for non-ideal liquid and vapor behavior
using the activity coefficients γi and φi of component (i) for liquid and vapor
phase respectively. The following equilibrium relation, also known as the
Gamma/Phi formulation, can be used:
S
iiiii PxPy γφ =
(1.38)
24
The activity coefficients can be determined using correlations found in standard
thermodynamics text books. Equation (1.38) is reduced to Raoult law for ideal
mixture i.e. ( iφ = iγ =1).
• In some cases the transfer occurs in liquid-liquid phases (such as liquid
extraction) or liquid-solid (such as ion exchange). In such cases an equilibrium
relation similar to Henry's law can be defined:
yA = KxA (1.39)
where K is the equilibrium distribution coefficient that depends on pressure,
temperature and concentration.
Phase equilibrium relations are commonly used in the following calculations:
• Bubble point calculations :
For a given molar liquid compositions (xi) and either T or P, the bubble point
calculations consist in finding the molar vapor composition (yi) and either P or
T.
• Dew point calculations :
For a given molar vapor compositions (yi) and either T or P, the dew point
calculations consist in finding the molar liquid compositions (xi) and either P or
T.
• Flash calculation :
For known mixture compositions (zi) and known (T) and (P), the flash
calculations consist in finding the liquid and vapor compositions via
simultaneous solution of component balance and energy balance equations.
1.14 Chemical kinetics
The overall rate R in moles/m3s of a chemical reaction is defined by:
25
dtdn
VvR i
i
1=
(1.40)
where
ni the number of moles
νi the stochiometric coefficient of component i
V is the volume due to the chemical reaction.
The rate expression R is generally a complex relation of the concentrations (or partial
pressures) of the reactants and products in addition to pressure and temperature.
For a general irreversible reaction
νr1R1 + νr2R2 + …+ νrrRrr νp1P1 + νp2P2 + …+ νppRpp (1.41)
The law of mass action stipulates that the reaction rate is a power law function of
temperature and concentration of reactants, i.e.
rrRRR CCkCR ζζζ= L2
211
(1.42)
The powers ξi are determined experimentally and their values are not necessarily
integers. The temperature dependence comes from the reaction rate constant k given by
Arrhenius law
RTE
oekk−
= (1.43)
where ko is the pre-exponential factor, E the activation energy, T the absolute
temperature and R is the ideal gas constant
1.15 Control Laws
Although the discussion of feedback control systems is beyond the scope of this
book, the existence of control loops in any processing system provides extra relations
26
for the process. Specifically in many cases some key process variables are required to
be controlled, i.e. maintained within desired range. This control objective can be
achieved by closing the control loop, i.e., relating the controlled variables to inputs
(manipulated variables). This introduces additional independent equations.
1.16 Degrees of Freedom
A key step in the model development and solution is checking its consistency or
solvability, i.e. the existence of exact solution. This is done by checking the degrees of
freedom of the model after the equations have written and before attempting to solve
them .
For a processing system described by a set of Ne independent equations and Nv
variables, the degree of freedom f is
F = Nv − Ne (1.44)
Depending on the value of f three cases can be distinguished:
• f = 0. The system is exactly determined (specified) system .Thus, the set of
balance equation has a finite number of solutions (one solution for linear
systems)
• f < 0. The system is over-determined (over-specified) by f equations. f
equations have to be removed for the system to have a solution.
• f > 0. The system is under-determined (under-specified) by f equations. The set
of equation, hence, has infinite number of solution .
To avoid the situations of over-specified or under-specified systems it is advised to
follow the following steps while checking the consistency of the model.
1. Determine known quantities of the model that can be fixed such as equipment
dimensions, constant physical properties, etc.
27
2. Determine other variables that can specified by the external world, for example,
variables that are the outcome of an upstream processing units, and/or variables
that can be used as forcing function or manipulated variables.
1.17 Model solution
After the solvability of the model has been checked, the next step is to solve the
model. The purpose of the solution of the model is be able to obtain the variations of
the state variables with the model independent variables (time, spatial positions..). The
solution of the model permits also a parametric investigation of the model, that is a
study of the effects of the changing the value of some parameters. It would be ideal to
be able to solve the model analytically, that is to get closed forms of the state variables
in term of the independent variables. Unfortunately this seldom occurs for chemical
processes . The reason is that the vast majority of chemical processes are nonlinear.
They may be a sets of nonlinear partial differential equations (PDE) as it is the case for
distributed parameter models, or sets of nonlinear ordinary differential equations
(ODE) or nonlinear algebraic equations as it is the case for lumped parameter models.
However most non linear problems can not be solved analytically. In fact the only class
of differential equations for which there is a well-developed framework are linear ODE .
However linearizing the original nonlinear model and solving it is not always
recommended (expect for control purposes) since the behavior of the linearized model
matches the original nonlinear model only around the state chosen for linearization .For
these reasons the solution of process models is usually carried out numerically, most
often through a computer programs.
1.18 Model validation
Model verification (validation) is the last and the most important step of model
building. Reliability of the obtained model depends heavily on faithfully passing this
test. Implementation of the model without validation may lead to erroneous and
28
misleading results. So, it is essential, as it is saves a lot of effort, time and frustration, to
verify the model against plant operating data, experimental data, or at least published
correlations. If the model failed in the test, then it might be necessary to adjust some of
the model parameters, which is believed to be poorly known, in order to minimize the
mismatch between the model and the true plant. In worst cases, a modeler may need to
reconsider some of the simplifying assumptions used or the neglected modeling parts.
However, it should be kept in mind that the model is no more than approximation of
the real world, thus some degree of mismatch will remain and could be overlooked.
29
Chapter 2: Examples of Mathematical Models for Chemical Processes
In this chapter we develop mathematical models for a number of elementary
chemical processes that are commonly encountered in practice. We will apply the
methodology discussed in the previous chapter to guide the reader through various
examples. The goal is to give the reader a methodology to tackle more complicated
processes that are not covered in this chapter and that can be found in books listed in
the reference. The organization of this chapter includes examples of systems that can be
described by ordinary differential equations (ODE), i.e. lumped parameter systems
followed by examples of distributed parameters systems, i..e those described by partial
differential equations (PDE). The examples cover both homogeneous and heterogeneous
systems. Ordinary differential equations (ODE) are easier to solve and are reduced to
simple algebraic equations at steady state. The solution of partial differential equations
(PDE) on the other hand is a more difficult task. But we will be interested in the cases
were PDE's are reduced to ODE's. This is naturally the case where under appropriate
assumptions, the PDE's is a one-dimensional equation at steady state conditions. It is
worth to recall, as noted in the previous chapters, that the distinction between lumped
and distributed parameter models depends sometimes on the assumptions put forward
by the modeler. Systems that are normally distributed parameter can be modeled under
appropriate assumptions as lumped parameter systems. This chapter includes some
examples of this situation.
30
2.1 Examples of Lumped Parameter Systems
2.1.1 Liquid Storage Tank
Consider the perfectly mixed storage tank shown in figure 2.1. Liquid stream
with volumetric rate Ff (m3/s) and density ρf flow into the tank. The outlet stream has
volumetric rate Fo and density ρο. Our objective is to develop a model for the variations
of the tank holdup, i.e. volume of the tank. The system is therefore the liquid in the
tank. We will assume that it is perfectly mixed and that the density of the effluent is the
same as that of tank content. We will also assume that the tank is isothermal, i.e. no
variations in the temperature. To model the tank we need only to write a mass balance
equation.
Figure 2-1 Liquid Storage Tank
Since the system is perfectly mixed, the system properties do not vary with position
inside the tank. The only variations are with time. The mass balance equation can be
written then on the whole system and not only on a differential element of it. This leads
to therefore to a macroscopic model.
We apply the general balance equation (Eq. 1.2), to the total mass m = ρV. This yields:
Mass flow in:
ρf Ff (2.1)
31
Mass flow out:
ρo Fo (2.2)
Accumulation:
dtVd
dtdm )(ρ
=
(2.3)
The generation term is zero since the mass is conserved. The balance equation yields:
dtVdFF ooff
)(ρ+ρ=ρ
(2.4)
For consistency we can check that all the terms in the equation have the SI unit of kg/s.
The resulting model (Eq. 2.4) is an ordinary differential equation (ODE) of first order
where time (t) is the only independent variable. This is therefore a lumped parameter
model. To solve it we need one initial condition that gives the value of the volume at
initial time ti, i.e.
V(ti) = Vi (2.5)
Under isothermal conditions we can further assume that the density of the liquid is
constant i.e. ρf = ρo=ρ. In this case Eq. 2.4 is reduced to:
of FFdtdV
−=
(2.6)
The volume V is related to the height of the tank L and to the cross sectional area A by:
V = AL (2.7)
32
Since (A) is constant then we obtain the equation in terms of the state variable L:
of FFdtdLA −=
(2.8)
with initial condition:
L(ti) = Li (2.9)
Degree of freedom analysis
For the system described by Eq. 2.8 we have the following information:
• Parameter of constant values: A
• Variables which values can be externally fixed (Forced variable): Ff
• Remaining variables: L and Fo
• Number of equations: 1 (Eq. 2.8)
Therefore the degree of freedom is:
Number of remaining variables – Number of equations = 2 – 1 = 1
For the system to be exactly specified we need therefore one more equations. This extra
relation is obtained from practical engineering considerations. If the system is operated
without control (at open loop) then the outlet flow rate Fo is a function of the liquid
level L. Generally a relation of the form:
LFo α= (2.10)
could be used, where α is the discharge coefficient.
If on the other hand the liquid level is under control, then its value is kept constant at
certain desired value Ls. If Fo is used to control the height then a control law relates Fo
to L and Ls:
33
Fo = Fo(L,Ls) (2.11)
For instant, if a proportional controller Kc is used then the control law is given by:
Fo = Kc(L − Ls) + Fob (2.12)
Where Fob the bias, i.e. the constant value of Fo when the level is at the desired value
i.e., L = Ls.
Note that at steady state, the accumulation term is zero (height does not change with
time), i.e., dL/dt = 0. The model of the tank is reduced to the simple algebraic equation:
F0 = Ff (2.13)
2.1.2 Stirred Tank Heater
We consider the liquid tank of the last example but at non-isothermal conditions.
The liquid enters the tank with a flow rate Ff (m3/s), density ρf (kg/m3) and temperature
Tf (K). It is heated with an external heat supply of temperature Tst (K), assumed
constant. The effluent stream is of flow rate Fo (m3/s), density ρo (kg/m3) and
temperature T(K) (Fig. 2.2). Our objective is to model both the variation of liquid level
and its temperature. As in the previous example we carry out a macroscopic model over
the whole system. Assuming that the variations of temperature are not as large as to
affect the density then the mass balance of Eq. 2.8 remains valid.
To describe the variations of the temperature we need to write an energy balance
equation. In the following we develop the energy balance for any macroscopic system
(Fig. 2.3) and then we apply it to our example of stirred tank heater.
The energy E(J) of any system of (Fig. 2.3) is the sum of its internal U(J), kinetic K(J)
and potential energy φ(J):
E = U + K + φ (2.14)
Consequently, the flow of energy into the system is:
34
ρf Ff ( fff K U φ++~ ~~ ) (2.15)
where the ( •~ ) denotes the specific energy (J/kg).
Figure 2-2 Stirred Tank Heater
Figure 2-3 General Macroscopic System
The flow of energy out of the system is:
ρo Fo( ooo K U φ++~ ~~ ) (2.16)
The rate of accumulation of energy is:
35
( )dt
K UVd )~ ~~( φ++ρ (2.17)
As for the rate of generation of energy, it was mentioned in Section 1.8.4, that the
energy exchanged between the system and the surroundings may include heat of
reaction Qr (J/s), heat exchanged with surroundings Qe (J/s) and the rate of work done
against pressure forces (flow work) Wpv (J/s), in addition to any other work Wo.
The flow of work Wpv done by the system is given by:
ffoopv PFPFW −= (2.18)
where Po and Pf are the inlet and outlet pressure, respectively.
In this case, the rate of energy generation is:
)( ffooore PFPFWQQ −+−+ (2.19)
Substituting all these terms in the general balance equation (Eq. 1.7) yields:
( ) ( ) ( ))(
~ ~~ ~ ~~ )~ ~~(
ffooore
ooooofffff
PFPFWQQ
K UF K UFdt
K UVd
−+−++
φ++ρ−φ++ρ=φ++ρ
(2.20)
We can check that all terms of this equation have the SI unit of (J/s). Equation (2.20)
can be also written as:
( ) ( ) ( )
f
fff
o
oooore
ooooofffff
PFPFWQQ
K UF K UFdt
K UVd
ρρ+
ρρ−−++
φ++ρ−φ++ρ=φ++ρ ~ ~~ ~ ~~ )~ ~~(
(2.21)
36
The term ρ= /1~V is the specific volume (m3/kg). Thus Eq. 2.21 can be written as:
( ) ( ) ( )ore
ooooooofffffff
WQQ
KV P UF KV P UFdt
K UVd
−++
+++−+++=++ φρφρφρ ~ ~~~ ~ ~~~ )~ ~~(
(2.22)
The term V P U ~~ + that appears in the equation is the specific enthalpy h~ . Therefore,
the general energy balance equation for a macroscopic system can be written as:
( ) ( ) ( ) oreooooofffff WQQ KhF K hFdt
K UVd−++φ++ρ−φ++ρ=
φ++ρ ~ ~~ ~ ~~ )~ ~~( (2.23)
We return now to the liquid stirred tank heater. A number of simplifying assumptions
can be introduced:
• We can neglect kinetic energy unless the flow velocities are high.
• We can neglect the potential energy unless the flow difference between the inlet
and outlet elevation is large.
• All the work other than flow work is neglected, i.e. Wo = 0.
• There is no reaction involved, i.e. Qr = 0.
The energy balance (Eq. 2.23) is reduced to:
( )eooofff QhFhF
dtUVd
+ρ−ρ=ρ ~ ~
~
(2.24)
Here Qe is the heat (J/s) supplied by the external source. Furthermore, as mentioned in
Section 1.12, the internal energy U~ for liquids can be approximated by enthalpy, h~ .
The enthalpy is generally a function of temperature, pressure and composition.
However, it can be safely estimated from heat capacity relations as follows:
37
)(~~refTTpCh −= (2.25)
where pC~ is the average heat capacity.
Furthermore since the tank is well mixed the effluent temperature To is equal to process
temperature T. The energy balance equation can be written, assuming constant density
ρf = ρo = ρ , as follows:
( )erefporeffpf
refp QTTCFTTCF
dtTTVd
C +−ρ−−ρ=−
ρ )(~ )(~ )(~
(2.26)
Taking Tref = 0 for simplicity and since V = AL result in:
( )epofpfp QTCFTCF
dtLTdAC +ρ−ρ=ρ
~ ~ ~ (2.27)
or equivalently:
( )p
eoff C
QTFTFdtLTdA ~
ρ+−=
(2.28)
Since
( ) ( ) ( )dtTdAL
dtLdAT
dtLTdA +=
(2.29)
and using the mass balance (Eq. 2.8) we get:
p
eoffof C
QTFTFFFT
dtdTAL ~ )(
ρ+−=−+
(2.30)
or equivalently:
38
p
eff C
QTTFdtdTAL ~)(
ρ+−=
(2.31)
The stirred tank heater is modeled, then by the following coupled ODE's:
of FFdtdLA −=
(2.32)
p
eff C
QTTFdtdTAL ~)(
ρ+−=
(2.33)
This system of ODE's can be solved if it is exactly specified and if conditions at initial
time are known,
L(ti) = Li and T(ti) = Ti (2.34)
Degree of freedoms analysis
For this system we can make the following simple analysis:
• Parameter of constant values: A, ρ and Cp
• (Forced variable): Ff and Tf
• Remaining variables: L, Fo, T, Qe
• Number of equations: 2 (Eq. 2.32 and Eq. 2.33)
The degree of freedom is therefore, 4 − 2 = 2. We still need two relations for our
problem to be exactly specified. Similarly to the previous example, if the system is
operated without control then Fo is related to L through (Eq. 2.10). One additional
relation is obtained from the heat transfer relation that specifies the amount of heat
supplied:
Qe = UAH (Tst−T ) (2.35)
39
U and AH are heat transfer coefficient and heat transfer area. The source temperature Tst
was assumed to be known. If on the other hand both the height and temperature are
under control, i.e. kept constant at desired values of Ls and Ts then there are two control
laws that relate respectively Fo to L and Ls and Qe to T and Ts:
Fo = Fo(L, Ls), and Qe = Qe (T, Ts) (2.36)
2.1.3 Isothermal CSTR
We revisit the perfectly mixed tank of the first example but where a liquid phase
chemical reactions taking place:
BA k⎯→⎯ (2.37)
The reaction is assumed to be irreversible and of first order. As shown in figure 2.4, the
feed enters the reactor with volumetric rate Ff (m3/s), density ρf (kg/m3) and
concentration CAf (mole/m3). The output comes out of the reactor at volumetric rate Fo,
density ρ0 and concentration CAo (mole/m3) and CBo (mole/m3). We assume isothermal
conditions.
Our objective is to develop a model for the variation of the volume of the reactor
and the concentration of species A and B. The assumptions of example 2.1.1 still hold
and the total mass balance equation (Eq. 2.6) is therefore unchanged
Figure 2.4 Isothermal CSTR
40
The component balance on species A is obtained by the application of (Eq. 1.3)
to the number of moles (nA = CAV ). Since the system is well mixed the effluent
concentration CAo and CBo are equal to the process concentration CA and CB.
Flow of moles of A in:
Ff CAf (2.38)
Flow of moles of A out:
Fo CAo (2.39)
Rate of accumulation:
dtVCd
dtdn A)(
=
(2.40)
Rate of generation: -rV
where r (moles/m3s) is the rate of reaction.
Substituting these terms in the general equation (Eq. 1.3) yields:
rV CF CFdtVCd
AoAffA −−=
)( (2.41)
We can check that all terms in the equation have the unit (mole/s).
We could write a similar component balance on species B but it is not needed
since it will not represent an independent equation. In fact, as a general rule, a system of
n species is exactly specified by n independent equations. We can write either the total
mass balance along with (n −1) component balance equations, or we can write n
component balance equations.
41
Using the differential principles, equation (2.41) can be written as follows:
rV CF CFdtVdC
dtCdV
dtVCd
AoAffAAA −−=+=
)()()( (2.42)
Substituting Equation (2.6) into (2.42) and with some algebraic manipulations we
obtain:
rVC CFdtCdV AAff
A −−= )()( (2.43)
In order to fully define the model, we need to define the reaction rate which is for a
first-order irreversible reaction:
r = k CA (2.44)
Equations 2.6 and 2.43 define the dynamic behavior of the reactor. They can be solved
if the system is exactly specified and if the initial conditions are given:
V(ti) = Vi and CA(ti) = CAi (2.45)
Degrees of freedom analysis
• Parameter of constant values: A
• (Forced variable): Ff and CAf
• Remaining variables: V, Fo, and CA
• Number of equations: 2 (Eq. 2.6 and Eq. 2.43)
The degree of freedom is therefore 3 − 2 =1. The extra relation is obtained by the
relation between the effluent flow Fo and the level in open loop operation (Eq. 2.10) or
in closed loop operation (Eq. 2.11).
The steady state behavior can be simply obtained by setting the accumulation terms to
zero. Equation 2.6 and 2.43 become:
42
F0 = Ff (2.46)
rVC CF AAff =− )( (2.47)
More complex situations can also be modeled in the same fashion. Consider the
catalytic hydrogenation of ethylene:
A + B P (2.48)
where A represents hydrogen, B represents ethylene and P is the product (ethane). The
reaction takes place in the CSTR shown in figure 2.5. Two streams are feeding the
reactor. One concentrated feed with flow rate F1 (m3/s) and concentration CB1 (mole/m3)
and another dilute stream with flow rate F2 (m3/s) and concentration CB2 (mole/m3). The
effluent has flow rate Fo (m3/s) and concentration CB (mole/m3). The reactant A is
assumed to be in excess.
VFo, CB
F1, CB1 F2, CB2
Figure 2-5 Reaction in a CSTR
The reaction rate is assumed to be:
)./()1(
32
2
1 smmoleCk
CkrB
B
+=
(2.49)
43
where k1 is the reaction rate constant and k2 is the adsorption equilibrium constant.
Assuming the operation to be isothermal and the density is constant, and following the
same procedure of the previous example we get the following model:
Total mass balance:
oFFFdtdLA −+= 21
(2.50)
Component B balance:
rVC CFC CFdtCdV BBBB
A −−+−= )()()(2211
(2.51)
Degrees of freedom analysis
• Parameter of constant values: A, k1 and k1
• (Forced variable): F1 F2 CB1 and CB2
• Remaining variables: V, Fo, and CB
• Number of equations: 2 (Eq. 2.50 and Eq. 2.51)
The degree of freedom is therefore 3 − 2 =1. The extra relation is between the effluent
flow Fo and the level L as in the previous example.
2.1.4 Gas-Phase Pressurized CSTR
So far we have considered only liquid-phase reaction where density can be taken
constant. To illustrate the effect of gas-phase chemical reaction on mass balance
equation, we consider the following elementary reversible reaction:
BA 2↔ (2.52)
taking place in perfectly mixed vessel sketched in figure 2.6. The influent to the vessel
has volumetric rate Ff (m3/s), density ρf (kg/m3), and mole fraction yf. Product comes out
of the reactor with volumetric rate Fo, density ρo, and mole fraction yo. The temperature
44
and volume inside the vessel are constant. The reactor effluent passes through control
valve which regulate the gas pressure at constant pressure Pg.
P, T, V ,y Fo, yoFf, f , yf
Pg
Figure 2-6 Gas Pressurized Reactor
Writing the macroscopic total mass balance around the vessel gives:
ooff FFdt
Vd ρρρ−=
)(
(2.53)
Since V is constant we have:
ooff FFdtdV ρρρ
−=
(2.54)
Writing the component balance, for fixed V, results in:
VrVrCFCFdt
dCV AoAffA
210+−−=
(2.55)
The reaction rates for the reversible reaction are assumed to be:
r1 = k1 CA (2.56)
2
22 BCkr = (2.57)
Equations (2.54) and (2.55) define the variations of density and molar concentration.
One can also rewrite the equation to define the behavior of the pressure (P) and mole
45
fraction (y). The concentration can be expressed in term of the density through ideal gas
law:
CA = yP/RT (2.58)
CB = (1 − y)P/RT (2.59)
Similarly, the density can be related to the pressure using ideal gas law:
ρ = MP/RT = [MAy + MB (1 − y)]P/RT (2.60)
Where MA and MB are the molecular weight of A and B respectively. Therefore one can
substitute equations (2.58) to (2.60) into equations (2.54 & 2.55) in order to explicitly
write the latter two equations in terms of y and P. Or, alternatively, one can solve all
equations simultaneously.
Degrees of freedom analysis:
• Parameters: V, k1, k2, R, T, MA and MB
• Forcing function: Ff, CAf, yf
• Variables: CA, CB, y, P, ρ, F
• Number of equations: 5 (Eqs. 2.54, 2.55, 2.58, 2.59, 2.60)
The degree of freedom is therefore 6 − 5 =1. The extra relation relates the outlet flow to
the pressure as follows:
ρg
vo
PPCF
−=
(2.61)
where Cv is the valve-sizing coefficient. Recall also that Pg is assumed to be constant.
2.1.5 Non-Isothermal CSTR
We reconsider the previous CSTR example (Sec 2.1.3), but for non-isothermal
conditions. The reaction A B is exothermic and the heat generated in the reactor is
46
removed via a cooling system as shown in figure 2.7. The effluent temperature is
different from the inlet temperature due to heat generation by the exothermic reaction.
Figure 2-7 Non-isothermal CSTR
Assuming constant density, the macroscopic total mass balance (Eq. 2.6) and mass
component balance (Eq. 2.43) remain the same as before. However, one more ODE will
be produced from the applying the conservation law (equation 2.23) for total energy
balance. The dependence of the rate constant on the temperature:
k = koe-E/RT (2.62)
should be emphasized.
The general energy balance (Eq. 2.23) for macroscopic systems applied to the CSTR
yields, assuming constant density and average heat capacity:
( )errefporeffpf
refp QQTTCFTTCF
dtTTVd
C −+−ρ−−ρ=−
ρ )(~ )(~ )(~
(2.63)
where Qr (J/s) is the heat generated by the reaction, and Qe (J/s) the rate of heat
removed by the cooling system. Assuming Tref = 0 for simplicity and using the
differentiation principles, equation 2.63 can be written as follows:
erpofpfpp QQTCFTCFdtdVTC
dtdTVC −+ρ−ρ=ρ+ρ
~ ~ ~~ (2.64)
47
Substituting Equation 2.6 into the last equation and rearranging yields:
erfpfp QQTTCFdtdTVC −+−ρ=ρ )(~ ~
(2.65)
The rate of heat exchanged Qr due to reaction is given by:
Qr = −(∆Hr)Vr (2.66)
where ∆Hr (J/mole) is the heat of reaction (has negative value for exothermic reaction
and positive value for endothermic reaction). The non-isothermal CSTR is therefore
modeled by three ODE's:
of FFdtdV
−= (2.67)
rVC CFdtCdV AAff
A −−= )()( (2.68)
erfpfp QVrHTTCFdtdTVC −−+−ρ=ρ )()(~ ~
∆ (2.69)
where the rate (r) is given by:
r = koe-E/RTCA (2.70)
The system can be solved if the system is exactly specified and if the initial conditions
are given:
V(ti) = Vi T(ti) = Ti and CA(ti) = CAi (2.71)
Degrees of freedom analysis
• Parameter of constant values: ρ, E, R, Cp, ∆Hr and ko
• (Forced variable): Ff , CAf and Tf
48
• Remaining variables: V, Fo, T, CA and Qe
• Number of equations: 3 (Eq. 2.67. 2.68 and 2.69)
The degree of freedom is 5−3 = 2. Following the analysis of example 2.1.3, the two
extra relations are between the effluent stream (Fo) and the volume (V) on one hand and
between the rate of heat exchanged (Qe) and temperature (T) on the other hand, in either
open loop or closed loop operations.
A more elaborate model of the CSTR would include the dynamic of the cooling
jacket (Fig. 2.8). Assuming the jacket to be perfectly mixed with constant volume Vj,
density ρj and constant average thermal capacity Cpj, the dynamic of the cooling jacket
temperature can be modeled by simply applying the macroscopic energy balance on the
whole jacket:
ejjfpjjj
jpj QTTCFdt
dTVC
jj+−ρ=ρ )(~ ~
(2.72)
Since Vj, ρj, Cpj and Tjf are constant or known, the addition of this equation introduces
only one variable (Tj). The system is still exactly specified.
V
Ff , CAf , Tf
Fo , CA , T
Fj , Tjf
Fj , Tj
Figure 2-8 Jacketed Non-isothermal CSTR
2.1.6 Mixing Process
Consider the tank of figure 2.9 where two solutions 1 and 2 containing materials
A and B are being mixed. Stream 1 has flow rate F1 (m3/s), density ρ1 (kg/m3), T1 (K),
49
concentration CA1 (mole/m3) and CB1 (mole/m3) of material A and B. Similarly stream 2
has flow rate F2 (m3/s), density ρ2 (kg/m3), T2 (K), concentration CA2 (mole/m3) and CB2
(mole/m3) of material A and B. The effluent stream has flow rate Fo (m3/s), density ρo
(kg/m3), To (K), concentration CAo (mole/m3) and CBo (mole/m3) of material A and B. We
assume that the mixing releases heat of rate Q (J/s) which is absorbed by a cooling fluid
flowing in a jacket or a coil.
Our objective is to develop a model for the mixing process. We will assume that
the tank is well mixed. In this case all the effluent properties are equal to the process
properties. We also assume for simplicity that the densities and heat capacities of the
streams are constant and equal:
ρ = ρ1 = ρ2 = ρo (2.73)
Cp = Cp1 = Cp2 = Cp3 (2.74)
Fo, To, CAo, CBo
F1, T1, CA1, CB1 F2, T2, CA2, CB2
Q
Figure 2-9 Mixing Process
Total mass balance
The mass balance equation yields
( ) ooFFFdt
Vd ρρρρ−+= 2211
)(
(2.75)
50
Since the densities are equals we have:
( ) oFFFdtdV
−+= 21
(2.76)
Component balance
The component balance for species A for instant yields:
( ) AoooAAAoo CFCFCF
dtVCd ρρρρ
−+= 222111)(
(2.77)
Expanding Eq. 2.77 yields:
( ) AoooAAAoAo
o CFCFCFdtdVC
dtdCV ρρρρ −+=⎟
⎠⎞
⎜⎝⎛ + 222111
(2.78)
Substituting Eq. 2.76 into 2.78 yields after some manipulation:
( ) ( )AoAAoAAo CCFCCF
dtdCV −+−= 2211
(2.79)
A similar equation holds for component B,
( ) BoooBBBoo CFCFCF
dtVCd ρρρρ
−+= 222111)(
(2.80)
Energy balance
The general energy balance equation (Eq. 2.23) yields, assuming negligible kinetic,
potential energy and since no reaction or shaft work occurs:
51
( )QhFhFhF
dthVd
mixooomixmixmixoo −−+= ,,222,111
, ~~ ~ ~
ρρρρ
(2.81)
where mixih ,~ (J/kg) is the specific enthalpy of mixture i.
The specific enthalpy mixh~ of n components can be written as:
)()(~),(~, refmixrefmixmix TTCpPThPTh −+= (2.82)
where
),,(ˆ),,(ˆ),(~1
PTChCPTChCPTh iskrefi
n
iiirefmixmix ∆+= ∑
=
ρ
(2.83)
with ih (J/mole) being the molar enthalpy of component i and sh∆ is the heat of
solution per mole of a key component k. Assume constant process pressure (P) then
we can write the enthalpy of each stream as follows, taking component (A) as the key
component,
)(ˆˆˆ)(~1,11,11111,11 refmixsABBAAmix TTCphChChCTh −+∆++= ρρ (2.84)
)(ˆˆˆ)(~2,22,22222,22 refmixsABBAAmix TTCphChChCTh −+∆++= ρρ (2.85)
)(ˆˆˆ)(~,,, refmixoosoAoBBoAAomixoo TTCphChChCTh −+∆++= ρρ (2.86)
Substituting Eq. 2.86 into the left hand side of Eq. 2.81 and expanding yields:
( )dt
hVd mixoo ,~ρ
=dtdVCpT
dtdTCpV
dtVCdh
dtVCdh
dtVCdh Ao
soBo
BAo
A ρρ ++∆++)(ˆ)(ˆ)(ˆ
,
(2.87)
52
The right hand side of equation (2.81) is equal after some manipulation to:
QTFTFTFCphFChFChFC
FCFCFChFCFCFCh
oo
sooAosAsA
oBoBBBoAoAAA
−−++
∆−∆+∆+
−++−+
)()ˆˆˆ(
)(ˆ)(ˆ
2211
,,222,111
22112211
ρ
(2.88)
Substituting into the right hand side of Eq. 2.87, the total mass balance equation (Eq.
2.76) and the component balance equation for species (A) (Eq. 2.77) and that of B (Eq.
2.80), and equating Eq. 2.87 to Eq. 2.88 yields, after some manipulations:
( ) QTTFTTFCp
hhFChhFCdtdTVCp sosAsosA
−−+−+
∆−∆+∆−∆=
)()(
)ˆˆ()ˆˆ(
2211
,,222,,111
ρ
ρ
(2.89)
The mixer is then described by three ODE's Equations (2.76, 2.79, 2.89). To these
relations we should add the relations that give the heats of mixing:
),,(ˆ1111,1 TCCfh BAs =∆ (2.90)
),,(ˆ2222,2 TCCfh BAs =∆ (2.91)
),,(ˆ3, oBoAoso TCCfh =∆ (2.92)
Degrees of freedom analysis
• Parameter of constant values: ρ, A, Tref, Cp, Ah and Bh
• (Forced variable): F1 , F2, CA1, CA2, T1 and T2
• Remaining variables: V, Fo, To, CAo, Q, sh ,1∆ , sh ,2ˆ∆ , soh ,
ˆ∆
• Number of equations: 6 (Eq. 2.76, 2.79, 2.89 and 2.90-2.92)
53
The degree of freedoms is therefore 8 – 6 = 2. The two needed relations are the relation
between effluent stream Fo and height L and the relation between the heat Q and
temperature To in either open loop or closed-loop operations.
2.1.7 Heat Exchanger
Consider the shell and tube heat exchanger shown in figure 2.10. Liquid A of
density ρA is flowing through the inner tube and is being heated from temperature TA1
to TA2 by liquid B of density ρB flowing counter-currently around the tube. Liquid B
sees its temperature decreasing from TB1 to TB2. Clearly the temperature of both liquids
varies not only with time but also along the tubes (i.e. axial direction) and possibly with
the radial direction too. Tubular heat exchangers are therefore typical examples of
distributed parameters systems. A rigorous model would require writing a microscopic
balance around a differential element of the system. This would lead to a set of partial
differential equations. However, in many practical situations we would like to model the
tubular heat exchanger using simple ordinary differential equations. This can be
possible if we think about the heat exchanger within the unit as being an exchanger
between two perfect mixed tanks. Each one of them contains a liquid.
Liquid, ATA1 TA2
Liquid, BTB1
Liquid, BTB2
Tw
Figure 2-10 Heat Exchanger
For the time being we neglect the thermal capacity of the metal wall separating the two
liquids. This means that the dynamics of the metal wall are not included in the model.
We will also assume constant densities and constant average heat capacities.
54
One way to model the heat exchanger is to take as state variable the exit temperatures
TA2 and TB2 of each liquid. A better way would be to take as state variable not the exit
temperature but the average temperature between the inlet and outlet:
221 AA
ATTT +
=
(2.93)
221 BB
BTTT +
= (2.94)
For liquid A, a macroscopic energy balance yields:
QTTCFdt
dTVC AApAAA
ApA AA+−= )( 21ρρ
(2.95)
where Q (J/s) is the rate of heat gained by liquid A. Similarly for liquid B:
QTTCFdt
dTVC BBpBBB
BpB BB−−= )( 21ρρ
(2.96)
The amount of heat Q exchanged is:
Q = UAH (TB – TA) (2.97)
Or using the log mean temperature difference:
Q = UAH ∆Tlm (2.98)
where
)()(ln
)()(
21
12
2112
BA
BA
BABAlm
TTTT
TTTTT
−−
−−−=∆
(2.99)
55
with U (J/m2s) and AH (m2) being respectively the overall heat transfer coefficient and
heat transfer area. The heat exchanger is therefore describe by the two simple ODE's
(Eq. 2.95) and (Eq. 2.96) and the algebraic equation (Eq. 2.97).
Degrees of freedom analysis
• Parameter of constant values: ρΑ, CpA, VA, ρΒ, CpB, VB, U, AH
• (Forced variable): TA1, TB1, FA, FB
• Remaining variables: TA2, TB2, Q
• Number of equations: 3 (Eq. 2.95, 2.96, 2.97)
The degree of freedom is 5 − 3 = 2. The two extra relations are obtained by noting that
the flows FA and FB are generally regulated through valves to avoid fluctuations in their
values.
So far we have neglected the thermal capacity of the metal wall separating the
two liquids. A more elaborated model would include the energy balance on the metal
wall as well. We assume that the metal wall is of volume Vw, density ρw and constant
heat capacity Cpw. We also assume that the wall is at constant temperature Tw, not a bad
assumption if the metal is assumed to have large conductivity and if the metal is not
very thick. The heat transfer depends on the heat transfer coefficient ho,t on the outside
and on the heat transfer coefficient hi,t on the inside. Writing the energy balance for
liquid B yields:
)()( ,,21 WBtotoBBpBBB
BpB TTAhTTCFdt
dTVCBB
−−−= ρρ (2.100)
where Ao,t is the outside heat transfer area. The energy balance for the metal yields:
)()( ,,,, AwtitiwBtotow
wpw TTAhTTAhdt
dTVCw
−−−=ρ (2.101)
where Ai,t is the inside heat transfer area. . The energy balance for liquid A yields:
56
)()( ,,21 AwtitiAApAAA
ApA TTAhTTCFdt
dTVCAA
−+−= ρρ (2.102)
Note that the introduction of equation (Eq. 2.101) does not change the degree of
freedom of the system.
2.1.8 Heat Exchanger with Steam
A common case in heat exchange is when a liquid L is heated with steam (Figure 2.11).
If the pressure of the steam changes then we need to write both mass and energy balance
equations on the steam side.
Liquid, LTL1 TL2
SteamTs(t)
condensate, Ts
Tw
Figure 2-11 Heat Exchanger with Heating Steam
The energy balance on the tube side gives:
sLLpLLL
LpL QTTCFdt
dTVCLL
+−ρ=ρ )( 21 (2.103)
where
221 LL
LTTT +
= (2.104)
Qs = UAs (Ts – TL) (2.105)
57
The steam saturated temperature Ts is also related to the pressure Ps:
Ts = Ts (P) (2.106)
Assuming ideal gas law, then the mass flow of steam is:
s
ssss RT
VPMm = (2.107)
where Ms is the molecular weight and R is the ideal gas constant. The mass balance for
the steam yields:
ccsss
ss FFdtdP
RTVM
ρ−ρ= (2.108)
where Fc and ρc are the condensate flow rate and density. The heat losses at the steam
side are related to the flow of the condensate by:
Qs = Fc λs (2.109)
Where λs is the latent heat.
Degrees of freedom analysis
• Parameter of constant values: ρL, CpL, Ms, As, U , Ms, R
• (Forced variable): TL1
• Remaining variables: TL2, FL, Ts, Fs, Ps, Qs, Fc
• Number of equations: 5 (Eq. 2.103, 2.105, 2.106, 2.108, 2.109)
The degrees of freedom is therefore 7 – 5 = 2. The extra relations are given by the
relation between the steam flow rate Fs with the pressure Ps either in open-loop or
closed-loop operations. The liquid flow rate F1 is usually regulated by a valve.
58
2.1.9 Single Stage Heterogeneous Systems: Multi-component flash drum
The previous treated examples have discussed processes that occur in one single
phase. There are several chemical unit operations that are characterized with more than
one phase. These processes are known as heterogeneous systems. In the following we
cover some examples of these processes. Under suitable simplifying assumptions, each
phase can be modeled individually by a macroscopic balance.
A multi-component liquid-vapor separator is shown in figure 2.12. The feed
consists of Nc components with the molar fraction zi (i=1,2… Nc). The feed at high
temperature and pressure passes through a throttling valve where its pressure is reduced
substantially. As a result, part of the liquid feed vaporizes. The two phases are assumed
to be in phase equilibrium. xi and yi represent the mole fraction of component i in the
liquid and vapor phase respectively. The formed vapor is drawn off the top of the
vessel while the liquid comes off the bottom of the tank. Taking the whole tank as our
system of interest, a model of the system would consist in writing separate balances for
vapor and liquid phase. However since the vapor volume is generally small we could
neglect the dynamics of the vapor phase and concentrate only on the liquid phase.
Fvyi
P, T, Vv
VL LFLxi
FoziToPo
Figure 2-12 Multicomponent Flash Drum
For liquid phase:
Total mass balance:
59
vvLLffLL FFF
dtVd ρρρρ
−−=)(
(2.110)
Component balance:
ivviLLiffiLL yFxFzF
dtxVd
ρρρρ
−−=)(
(i=1,2,….,Nc-1)
(2.111)
Energy balance:
HFhFhFdt
hVdvvLLfff
LL ~~~)~( ρρρρ−−=
(2.112)
where h~ and H~ are the specific enthalpies of liquid and vapor phase respectively.
In addition to the balance equations, the following supporting thermodynamic
relations can be written:
• Liquid-vapor Equilibrium:
Raoult's law can be assumed for the phase equilibrium
PPx
ys
iii = (i=1,2,….,Nc)
(2.113)
Together with the consistency relationships:
11
=∑=
Nc
iiy
(2.114)
11
=∑=
Nc
iix
(2.115)
• Physical Properties:
The densities and enthalpies are related to the mole fractions, temperature and pressure
through the following relations:
60
ρL = f(xi,T,P) (2.116)
ρv = f(yi,T,P) ≈ MvaveP/R T (2.117)
Mvave = ∑
=
Nc
iiiMy
1
(2.118)
h = f(xi,T) ≈ )(1
ref
Nc
iii TTCpx −∑
=
(2.119)
H = f(yi,T) ≈ mref
Nc
iii TTCpy λ+−∑
=
)(1
(2.120)
λm = ∑=
Nc
iiiy
1
λ (2.121)
Degrees of freedom analysis:
• Forcing variables: Ff, Tf, Pf , zi (i=1,2..Nc),
• Remaining variables:2Nc+5: VL, FL, FV, P, T, xi (i=1,2..Nc), yi(i=1,2,…Nc)
• Number of equations: 2Nc+3: (Eq. 2.110, 2.111, 2.112, 2.113, 2.114, 2.115)
Note that physical properties are not included in the degrees of freedom since they are
specified through given relations. The degrees of freedom is therefore (2Nc+5)-
(2Nc+3)=2. Generally the liquid holdup (VL) is controlled by the liquid outlet flow rate
(FL) while the pressure is controlled by FV. In this case, the problem becomes well
defined for a solution.
2-1-10 Two-phase Reactor
Consider the two phase reactor shown in figure 2.13. Gaseous A and liquid B
enters the reactor at molar flow rates FA and FB respectively. Reactant A diffuses into
the liquid phase with molar flux (NA) where it reacts with B producing C. The latter
diffuses into the vapor phase with molar flux (NC). Reactant B is nonvolatile. The
product C is withdrawn with the vapor leaving the reactor. The objective is to write the
mathematical equations that describe the dynamic behavior of the process. We consider
all flows to be in molar rates.
61
Figure 2-4 Two Phase Reactor
Assumptions:
• The individual phases are well mixed and they are in physical equilibrium at
pressure P and temperature T.
• The physical properties such as molar heat capacity Cp, density ρ , and latent heat
of vaporization λ are constant and equal for all the species.
• The reaction mechanism is: A+B C and its rate has the form: Rc = k CA CB VL
• The two phases are in equilibrium and follows the Raoult’s law.
• Total enthalpy for the system is given as: H = NL HL + Nv Hv where HL and Hv are
molar enthalpies in the liquid and vapor phases respectively, and NL and Nv are
their corresponding molar holdups.
The assumption of well mixing allows writing the following macroscopic balances:
Vapor phase:
Total mass balance:
vcAAv FNNF
dtdN
−+−=
(2.122)
62
Component balance for A:
AvAAAv yFNF
dtyNd
−−=)(
(2.123)
Since d(NvyA)/dt = Nv dyA/dt + yA dNv/dt, and using equation (2.122), equation (2.123)
can be written as follows:
AcAAAAA
v yNyNyFdt
dyN −−−−= )1()1(
(2.124)
Liquid phase:
Total mass balance:
cLcABL RFNNF
dtdN
−−−+=
(2.125)
Component balance for A:
cALAAL RxFN
dtxNd
−−=)(
(2.126)
Since d(NL xA)/dt = NL dxA/dt + xA dNL/dt, and using equation (2.125), equation (2.126)
can be written as follows:
AcAcABAAA
L xNxRxFxNdt
dxN +−−−−= )1()1(
(2.127)
Component balance for B:
Repeating the same reasoning used for component A, we can write:
63
BcBcBBBAB
L xNxRxFxNdt
dxN +−−+−= )1()1(
(2.128)
Energy balance, assuming Tref = 0:
QHRCpTFCpTFCpTFCpTFdt
HNHNdrcvLAABB
vvLL +−−−λ++=+
∆)()(
(2.129)
Note that:
dtNdCpT
dtTdCpN
dtNdH
dtHdN
dtHNd L
LL
LL
LLL )()()()()(
+=+=
(2.130)
dtNdCpT
dtTdCpN
dtNdH
dtHdN
dtHNd v
vv
vv
vvv )()()()()()(
λ++=+=
(2.131)
Substituting the last two equations, and using the definition of dNL/dt and dNv/dt from
equations (2.122) and (2.125), in equation (2.131) yields:
CpQNN
CpCpHTRTTFTTF
dtTdNN cA
rcBBAAvL +−
λ+−+−+−=+ )()()()()( ∆
(2.132)
The following additional equations are needed:
Vapor-liquid equilibrium relations:
yAP − xA PAs = 0 (2.133)
yAP − (1 − xA − xB) Pcs = 0 (2.134)
Total volume constraint:
V = VL + Vv (2.135)
64
Or, using ideal gas law for vapor volume and total volume and knowing that VL = NL/ρ ,
we can write:
nRT = NvRT + NLP/ρ (2.136)
or
V = NvRT/P + NL/ρ (2.137)
Degrees of freedom analysis:
• Forcing variables: FA, FB, TA, TB, Q, P
• Physical properties and parameters: ∆Hr, Cp, λ, R, ρ, V, sAP , s
CP
• Remaining variables: NA, Nc, NL, Nv, FL, T, xA , xB, yA
• Number of equations: (Eq. 2.122, 2.124, 2.125, 2.127, 2.128, 2.132, 2.133,2.134,
2.137)
The degree is freedom is 9-9=0 and the problem is exactly specified. Note that the
reaction rate Rc is defined and that the outlet flow Fv can be determined from the overall
mass balance.
2-1-11 Reaction with Mass Transfer
Figure 2.14 shows a chemical reaction that takes place in a gas-liquid
environment. The reactant A enters the reactor as a gas and the reactant B enters as a
liquid. The gas dissolves in the liquid where it chemically reacts to produce a liquid C.
The product is drawn off the reactor with the effluent FL. The un-reacted gas vents of
the top of the vessel. The reaction mechanism is given as follows:
A + B C (2.138)
Assumptions:
• Perfectly mixed reactor
• Isothermal operation
• Constant pressure, density, and holdup.
65
• Negligible vapor holdup.
Figure 2-14 Reaction with Mass Transfer
In such cases, when the two chemical phenomena, i.e., mass transfer and chemical
reaction, occur together, the reaction process may become mass transfer dominant or
reaction-rate dominant. If the mass transfer is slower reaction rate, then mass transfer
prevail and vise versa.
Due to the perfectly mixing assumption, macroscopic mass transfer of
component A from the bulk gas to the bulk liquid is approximated by the following
molar flux:
NA = KL (C*A − CA) (2.139)
where
KL is mass transfer coefficient
CA* is gas concentration at gas-liquid interface
CA is gas concentration in bulk liquid
To fully describe the process, we derive the macroscopic balance of the liquid phase
where the chemical reaction takes place. This results in:
66
Liquid phase:
Total mass balance:
LAmABB FNAMFdt
Vdρ−+ρ=
ρ (2.140)
Component balance on A:
rVCFNAdt
dCV ALAmA −−=
(2.141)
Component balance on B:
rVCFCFdt
dCV BLBoBB −−=
(2.142)
Vapor phase:
Here, since vapor holdup is negligible, we can write a steady state total continuity
equation as follows:
Fv = FA − MA Am NA/ρA (2.143)
where
Am total mass transfer area of the gas bubble
MA molecular weight of component A
ρ density
V liquid volume
Degrees of freedom analysis:
• Forcing variables: FA, FB, CB0,
• Parameters of constant values: KL, MA, Am, ρ, ρA, ρB,
67
• Remaining variables: CB, NA, CA, Fv, V
• Number of equations: (Eq. 2.139-2.143)
Note that the liquid flow rate, FL can be determined from the overall mass balance and
that the reaction rate r should be defined.
2.1.12 Multistage Heterogeneous Systems: Liquid-liquid extraction
There are many chemical processes which consist in a number of consecutive
stages in series. In each stage two streams are brought in contact for separating
materials due to mass transfer. The two streams could be flow in co-current or counter
current patterns. Counter-current flow pattern is known to have higher separation
performance Examples of these processes are distillation columns, absorption towers,
extraction towers and multi-stage flash evaporator where distillate water is produced
from brine by evaporation.
The same modeling approach used for single stage processes will be used for the
staged processes, where the conservation law will be written for one stage and then
repeated for the next stage and so on. This procedure will result in large number of state
equation depending on the number of stages and number of components.
The separation process generally takes place in plate, packed or spray-type
towers. In tray or spay-type columns the contact and the transfer between phases occur
at the plates. Generally, we can always assume good mixing of phases at the plates, and
therefore macroscopic balances can be carried out to model these type of towers.
Packed towers on the other hand are used for continuous contacting of the two phases
along the packing. The concentrations of the species in the phases vary obviously along
the tower. Packed towers are therefore typical examples of distributed parameters
systems that need to be modeled by microscopic balances.
In the following we present some examples of mass separation units that can be
modeled by simple ODE's, and we start with liquid-liquid extraction process.
68
Liquid-liquid extraction is used to move a solute from one liquid phase to another.
Consider the single stage countercurrent extractor, shown in Figure 2.15, where it is
desired to separate a solute (A) from a mixture (W) using a solvent (S). The stream
mixture with flow rate W (kg/s) enters the stage containing XAf weight fraction of solute
(A). The solvent with a flow rate (S) (kg/s) enters the stage containing YAf weight
fraction of species (A). As the solvent flows through the stage it retains more of (A) thus
extracting (A) from the stream (W). Our objective is to model the variations of the
concentration of the solute. A number of simplifying assumptions can be used:
• The solvent is immiscible in the other phase.
• The concentration XA and YA are so small that they do not affect the mass flow
rates. Therefore, we can assume that the flow rates W and S are constant. A total
mass balance is therefore not needed.
• An equilibrium relationship exists between the weight fraction YA of the solute in
the solvent (S) and its weight fraction XA in the mixture (W). The relationship can
be of the form:
YA = K XA (2.144)
Here K is assumed constant. Since both phases are assumed perfectly mixed a
macroscopic balance can be carried out on the solute in each phase. A component
balance on the solute in the solvent-free phase of volume V1 and density ρ1 gives:
AAAfA NWXWX
dtdXV −−=ρ 11
(2.145)
whereas NA (kg/s) is the flow rate due to transfer flow between the two phases. A similar
component balance on the solvent phase of volume V2 and density ρ2 gives:
AAAfA NSYSY
dtdYV +−=ρ 22
(2.146)
Since YA = K XA and K is constant, the last equation is equivalent to:
69
AAAfA NSKXSY
dtdXKV +−=22ρ
(2.147)
Adding Eq. 2.145 and 2.147 yields:
AAfAfA XKSWSYWX
dtdXKVV )()( 2211 +−+=+ ρρ
(2.148)
The latter is a simple linear ODE with unknown XA. With the volume V1, V2 and flow
rates W, S known the system is exactly specified and it can be solved if the initial
concentration is known:
XA(ti) = XAi (2.149)
Note that we did not have to express explicitly the transferred flux NA.
Figure 2-15 Single Stage Liquid-Liquid Extraction Unit
The same analysis can be extended to the multistage liquid/liquid extraction
units as shown in Figure 2.16. The assumptions of the previous example are kept and
we also assume that all the units are identical, i.e. have the same volume. They are also
assumed to operate at the same temperature.
Figure 2-16 Multi-Stage Liquid-Liquid Extraction Unit
70
A component balance in the ith stge (excluding the first and last stage) gives:
• Solvent-free phase, of volume V1i and density ρ1i
AiAiAiAi
ii NWXWXdt
dXV −−=ρ −111 (i=2…,N-1) (2.150)
where NAi is the flow rate due to transfer between the two phases at stage i.
• Solvent phase of volume V2i and density ρ2i:
AiAiAiAi
ii NSYSYdt
dYV +−= +122ρ … (i = 2…,N-1) (2.151)
Writing the equilibrium equation (Eq. 2.144) for each component YAi = K XAi
(i=1,.,.N) and adding the last two equations yield:
AiAiAiAi
iiii XKSWSYWXdt
dXKVV )()( 112211 +−+=ρ+ρ +− (i=2…,N-1) (2.152)
Since the volume and densities are equal, i.e.:
V1i = V1 and V2i = V (2.153)
ρ1i = ρ1 and ρ2i = ρ2 (2.154)
Equation 2.152 is therefore equivalent to:
AiAiAiAi XKSWSYWX
dtdXKVV )()( 112211 +−+=ρ+ρ +− (i=2…,N-1)
(2.155)
The component balance in the first stage is:
71
111
2211 )()( AAAfA XKSWSYWX
dtdXKVV +−+=ρ+ρ
(2.156)
And that for the last stage is:
ANAfANAN XKSWSYWX
dtdXKVV )()( 12211 +−+=ρ+ρ −
(2.157)
The model is thus formed by a system of linear ODE's (Eq. 2.155, 2.156, 2.157)
which can be integrated if the initial conditions are known:
XA(ti) = XAi (i=1,2…,N) (2.158)
Degrees of freedom analysis
• Parameter of constant values: ρ1, ρ2, K, V1, V2, W and S
• (Forced variable): XAf, YAf
• Remaining variables: XAi (2N variables): (i=1,2…,N) and YAi (i=1,2…,N)
• Number of equations: 2N [2.144 (N equations, one for each component), Eq.
2.155 (N-2 eqs), 2.156(1 eq), 2.157(1 eq)].
The problem is therefore is exactly specified.
2.1.13 Binary Absorption Column
Consider a N stages binary absorption tower as shown in figure 2.17. A Liquid
stream flows downward with molar flow rate (L) and feed composition (xf). A Vapor
stream flows upward with molar flow rate (G) and feed composition (yf). We are
interested in deriving an unsteady state model for the absorber. A simple vapor-liquid
equilibrium relation of the form of:
yi = a xi + b (2.159)
can be used for each stage i (i=1,2,…,N).
Assumptions:
72
• Isothermal Operation
• Negligible vapor holdup
• Constant liquid holdup in each stage
• Perfect mixing in each stage
According to the second and third assumptions, the molar rates can be considered
constants, i.e. not changing from one stage to another, thus, total mass balance need not
be written. The last assumption allows us writing a macroscopic balance on each stage
as follows:
Component balance on stage i:
)()( 11 iiiii xxLyyG
dtdxH −+−= +− (i=2…,N-1)
(2.160)
where H is the liquid holdup, i.e., the mass of liquid in each stage. The last equation is
repeated for each stage with the following exceptions for the last and the first stages:
Figure 2-17 N-stages Absorbtion Tower
In the last stage, xi+1 is replaced by xf
In the first stage, yi-1 is replaced by yf
73
Degrees of freedom analysis
• Parameter of constant values: Η, a, b
• (Forced variable): G, L, xf, yf
• Remaining variables: xi (i=1,2…,N), yi (i=1,2…,N)
• Number of equations:2N (Eqs.2.159, 2.160)
The problem is therefore is exactly specified.
2.1.14 Multi-component Distillation Column
Distillation columns are important units in petrochemical industries. These units
process their feed, which is a mixture of many components, into two valuable fractions
namely the top product which rich in the light components and bottom product which is
rich in the heavier components. A typical distillation column is shown in Figure 2.18.
The column consists of n trays excluding the re-boiler and the total condenser. The
convention is to number the stages from the bottom upward starting with the re-boiler as
the 0 stage and the condenser as the n+1 stage.
Description of the process:
The feed containing nc components is fed at specific location known as the feed tray
(labeled f) where it mixes with the vapor and liquid in that tray. The vapor produced
from the re-boiler flows upward. While flowing up, the vapor gains more fraction of the
light component and loses fraction of the heavy components. The vapor leaves the
column at the top where it condenses and is split into the product (distillate) and reflux
which returned into the column as liquid. The liquid flows down gaining more fraction
of the heavy component and loses fraction of the light components. The liquid leaves
the column at the bottom where it is evaporated in the re-boiler. Part of the liquid is
drawn as bottom product and the rest is recycled to the column. The loss and gain of
materials occur at each stage where the two phases are brought into intimate phase
equilibrium.
74
Bxb
Fz
Dxd
Cw
steam
Figure 2-18 Distillation Column
Modeling the unit:
We are interested in developing the unsteady state model for the unit using the flowing
assumptions:
• 100% tray efficiency
• Well mixed condenser drum and re-boiler.
• Liquids are well mixed in each tray.
• Negligible vapor holdups.
• liquid-vapor thermal equilibrium
Since the vapor-phase has negligible holdups, then conservation laws will only be
written for the liquid phase as follows:
Stage n+1 (Condenser), Figure 2.19a:
Total mass balance:
75
)( DRVdt
dMn
D +−=
(2.161)
Component balance:
1,1)()(
,,, −=+−= ncjxDRyV
dtxMd
jDjnnjDD
(2.162)
Energy balance:
cDnnDD QhDRhV
dthMd
−+−= )()( (2.163)
Note that R = Ln+1 and the subscript D denotes n+1
Stage n, Figure fig2.19b
Total Mass balance:
nnnn LRVV
dtdM
−+−= −1 (2.164)
Component balance:
1,1)(
,,,,11, −=−+−= −− ncjxLRxyVyV
dtxMd
jnnjDjnnjnnjnn
(2.165)
Energy balance:
nnDnnnnnn hLRhHVHV
dthMd
−+−= −− 11)(
(2.166)
Stage i, Figure 2.19c
Total Mass balance:
76
iiiii LLVV
dtdM
−+−= +− 11 (2.167)
Component balance:
1,1)(
,,11,,11, −=−+−= ++−− ncjxLxLyVyV
dtxMd
jiijiijiijiijii
(2.168)
Energy balance:
iiiiiiiiii hLhLHVHV
dthMd
−+−= ++−− 1111)(
(2.169)
Stage f (Feed stage), Figure 2.19d
Total Mass balance:
)())1(( 11 qFLLFqVVdt
dMffff
f +−+−+−= +− (2.170)
Component balance:
1,1
)())1(()(
,,11,,11,
−=
+−+−+−= ++−−
ncj
qFzxLxLFzqyVyVdt
xMdjjffjffjjffjff
jff
(2.171)
Energy balance:
)())1(()(
1111 ffffffffffff qFhhLhLFhqHVHV
dthMd
+−+−+−= ++−− (2.172)
Stage 1, Figure 2.19e
Total Mass balance:
77
1211 LLVV
dtdM
B −+−= (2.173)
Component balance:
1,1)(
,11,22,11,,11 −=−+−= ncjxLxLyVyV
dtxMd
jjjjBBj
(2.174)
Energy balance:
11221111 )( hLhLHVHV
dthMd
BB −+−= (2.175)
Stage 0 (Re-boiler), Figure 2.19f
Total Mass balance:
BLVdt
dMB
B −+−= 1 (2.176)
Component balance:
1,1)(
,,11,, −=−+−= ncjBxxLyV
dtxMd
jBjjBBjBB
(2.177)
Energy balance:
rBBBBB QBhhLHV
dthMd
+−+−= 11)(
(2.178)
Note that L0 = B and B denotes the subscript 0
Additional given relations:
Phase equilibrium: yj = f (xj, T,P)
Liquid holdup: Mi = f (Li)
78
Enthalpies: Hi = f (Ti, yi,j), hi = f (Ti, xi,j)
Vapor rates: Vi = f (P)
Notation:
Li, Vi Liquid and vapor molar rates
Hi, hi Vapor and liquid specific enthalpies
xi, yi Liquid and vapor molar fractions
Mi Liquid holdup
Q Liquid fraction of the feed
Z Molar fractions of the feed
F Feed molar rate
Degrees of freedom analysis
Variables
Mi n
MB, MD 2
Li n
B,R,D 3
xi,j n(nc − 1)
xB,j,xD,j 2(nc − 1)
yi,j n(nc − 1)
yB,j nc − 1
hi n
hB, hD 2
Hi n
HB 1
Vi n
VB 1
Ti n
TD, TB 2
Total 11+6n+2n(nc−1)+3(nc−1)
79
Equations:
Total Mass n + 2
Energy n + 2
Component (n + 2)(nc − 1)
Equilibrium n(nc − 1)
Liquid holdup n
Enthalpies 2n+2
Vapor rate n
hB = h1 1
yB = xB (nc − 1)
Total 7+6n+2n(nc-1)+3(nc-1)
Constants: P, F, Z
Therefore; the degree of freedom is 4
To well define the model for solution we include four relations imported from inclusion
of four feedback control loops as follows:
• Use B, and D to control the liquid level in the condenser drum and in the re-
boiler.
• Use VB and R to control the end compositions i.e., xB, xD
80
R, xd
Ln, xn
Vn, yn
Vn-1, yn-1
stage n
Li, xi Vi-1, yi-1
stage i
Li+1, xi+1 Vi, yi
(a)(b)
Lf+1, xf+1
Lf, xf Vf-1, yf-1
stage f
Vf , yf
(c)
L2, x2
L1, x1
V1, y1
stage 1
(d)
VB, yB
VB, yB
L1, x1
B, xB
(e)
Vn, yn
R, xd D, xd
Qc
Qr
(f)
Figure 2-19 Distillation Column Stages
Simplified Model
One can further simplify the foregoing model by the following assumptions:
(a) Equi-molar flow rates, i.e. whenever one mole of liquid vaporizes a tantamount
of vapor condenses. This occur when the molar heat of vaporization of all
components are about the same. This assumption leads to further idealization
that implies constant temperature over the entire column, thus neglecting the
energy balance. In addition, the vapor rate through the column is constant and
equal to:
VB = V1 = V2 =… = Vn (2.179)
81
(b) Constant relative volatility, thus a simpler formula for the phase equilibrium can be
used:
yj = αj xj/(1+(αj − 1) xj) (2.180)
Degrees of Freedom:
Variables:
Mi, MB, MD n + 2
Li, B,R,D n + 3
xi,xB,xD (n + 2)(nc − 1)
yj, yB (n + 1)(nc − 1)
V 1
Total 2 + 2n + (2n + 3)(nc − 1)
Equations:
Total Mass n + 2
Component (n + 2)(nc − 1)
Equilibrium n(nc − 1)
Liquid holdup n
yB = xB 1
Total 2+2n+(2n+3)(nc-1)
It is obvious that the degrees of freedom is still 4.
82
2.2 Examples of Distributed Parameter Systems
2.2.1 Liquid Flow in a Pipe
Consider a fluid flowing inside a pipe of constant cross sectional area (A) as
shown in Figure 2.20. We would like to develop a mathematical model for the change
in the fluid mass inside the pipe. Let v be the velocity of the fluid. Clearly the velocity
changes with time (t), along the pipe length (z) and also with the radial direction (r). In
order to simplify the problem, we assume that there are no changes in the radial
direction. We also assume isothermal conditions, so only the mass balance is needed.
Since the velocity changes with both time and space, the mass balance is to be carried
out on microscopic scale. We consider therefore a shell element of width ∆z and
constant cross section area (A) as shown in Fig. 2.20.
Figure 2-20 Liquid flow in a pipe
Mass into the shell:
ρvA∆t|z (2.181)
where the subscript (.|z) indicates that the quantity (.) is evaluated at the distance z.
Mass out of the shell:
ρvA∆t|z+∆z (2.182)
Accumulation:
83
ρA∆z|t+∆t − ρA∆z|t (2.183)
Similarly the subscript (.|t) indicates that the quantity (.) is evaluated at the time t.
The mass balance equation is therefore:
ρvA∆t|z = ρvA∆t|z+∆z + ρA∆z|t+∆t − ρA∆z|t (2.184)
We can check for consistency that the units in each term are in (kg). Dividing Eq. 2.184
by ∆t ∆z and rearranging yields:
zvAvA
tAA
zzzttt
∆∆∆∆ ++
ρ−ρ=
ρ−ρ )()()()(
(2.185)
Taking the limit as ∆t 0 and ∆z 0 gives:
zvA
tA
∂ρ∂
−=∂ρ∂ )()(
(2.186)
Since the cross section area (A) is constant, Eq. 2.186 yields:
zv
t ∂∂
−=∂∂ )(ρρ
(2.187)
or
0)(=
∂∂
+∂∂
zv
tρρ
(2.188)
The ensuing equation is a partial differential equation (PDE) that defines the variation
of ρ and v with the two independent variables t and z. This equation is known as the
one-dimensional continuity equation. For incompressible fluids for which the density is
constant, the last equation can also be written as:
84
0=∂∂
zv
(2.189)
This indicates that the velocity is independent of axial direction for one dimensional
incompressible flow.
2.2.2 Velocity profile inside a pipe
We reconsider the flow inside the pipe of the previous example. Our objective is
to find the velocity profile in the pipe at steady state. For this purpose a momentum
balance is needed. To simplify the problem we also assume that the fluid is
incompressible. We will carry out a microscopic momentum balance on a shell with
radius r, thickness ∆r and length ∆z as shown in Fig 2.21.
Figure 2-21 Velocity profile for a laminar flow in a pipe
Momentum in:
(τrz2πr∆z)|r (2.190)
where τrz is the shear stress acting in the z-direction and perpendicular to the radius r.
Momentum out:
85
(τrz2πr∆z)|r+∆r (2.191)
As for the momentum generation we have mentioned earlier in section 1.8.3 that the
generation term corresponds to the sum of forces acting on the volume which in this
example are the pressure forces, i.e.
(PA)|z – (PA)|z+∆z = P(2πr∆r)|z − P(2πr∆r)|z+∆z (2.192)
There is no accumulation term since the system is assumed at steady state. Substituting
this term in the balance equation (Eq. 1.6) and rearranging yields,
zPPr
rrr zzzrrzrrrz
∆∆∆∆ )||(|| ++ −
=τ−τ
(2.193)
We can check for consistency that all the terms in this equation have the SI unit of
(N/m2). Taking the limit of (Eq. 2.193) as ∆z and ∆r go to zero yields:
dzdPr
drrd rz −=τ )(
(2.194)
Here we will make the assumption that the flow is fully developed, i.e. it is not
influenced by the entrance effects. In this case the term dP/dz is constant and we have:
LP
LPP
dzdP ∆
=−
= 12 (2.195)
where L is the length of the tube. Note that equation (2. 195) is a function of the shear
stress τrz, but shear stress is a function of velocity. We make here the assumption that
the fluid is Newtonian, that is the shear stress is proportional to the velocity gradient:
drdvz
rz µτ −= (2.196)
86
Substituting this relation in Eq. 2.194 yields:
LPr
drrdv
drd z ∆
=)(µ (2.197)
or by expanding the derivative:
LP
drdv
rdrvd zz ∆
=+ )1( 2
2
µ (2.198)
The system is described by the second order ODE (Eq. 2.198). This ODE can be
integrated with the following conditions:
• The velocity is zero at the wall of the tube
vz = 0 at r = R (2.199)
• Due to symmetry, the velocity profile reaches a maximum at the center of the
tube:
0=drdvz at r = 0
(2.200)
Note that the one-dimensional distributed parameter system has been reduced to a
lumped parameter system at steady state.
2.2.3 Diffusion with chemical reaction in a slab catalyst
We consider the diffusion of a component A coupled with the following
chemical reaction A B in a slab of catalyst shown in figure 2.22. Our objective is to
determine the variation of the concentration at steady state. The concentration inside the
87
slab varies with both the position z and time t. The differential element is a shell
element of thickness ∆z.
Flow of moles A in:
(SNA)|z (2.201)
where S (m2) is the surface area and NA (moles A/s m2) is the molar flux.
Flow of moles A out:
(SNA)|z+∆z (2.202)
Rate of generation of A:
−(S∆z)r (2.203)
where r = kCA is the rate of reaction, assumed to be of first order. There is no
accumulation term since the system is assumed at steady state. The mass balance
equation is therefore,
(SNA)|z − (SNA)|z+∆z − (S∆z)kCA =0 (2.204)
z = 0 z = L
Porous catalyst particle
Exterior surface
z
Figure 2-22 diffusion with chemical reaction inside a slab catalyst
88
Dividing equation (2.204) by S∆z results in:
0|)(|)(=−
− +A
zzAzA kCzNN
∆∆
(2.205)
Taking the limit when ∆z 0, the last equation becomes:
0=− AA kC
dzdN
(2.206)
The molar flux is given by Fick's law as follows:
dzdC
DN AAA −=
(2.207)
where DA is diffusivity coefficient of (A) inside the catalyst particle. Equation (2.206)
can be then written as follows:
02
2
=− AA
A kCdz
CdD
(2.208)
This is also another example where a one-dimensional distributed system is reduced to a
lumped parameter system at steady state. In order to solve this second-order ODE, the
following boundary conditions could be used:
at z = L, CA = CAo (2.209)
at z = 0, dCA/dr = 0 (2.210)
The first condition imposes the bulk flow concentration CAo at the end length of the
slab. The second condition implies that the concentration is finite at the center of the
slab.
89
2.2.4 Temperature profile in a heated cylindrical Rod
Consider a cylindrical metallic rod of radius R and length L, initially at a
uniform temperature of To. Suppose that one end of the rod is brought to contact with a
hot fluid of temperature Tm while the surface area of the rod is exposed to ambient
temperature of Ta. We are interested in developing the mathematical equation that
describes the variation of the rod temperature with the position. The metal has high
thermal conductivity that makes the heat transfer by conduction significant. In addition,
the rod diameter is assumed to be large enough such that thermal distribution in radial
direction is not to be neglected. The system is depicted by figure 2.23. For modeling
we take an annular ring of width ∆z and radius ∆r as shown in the figure. The following
transport equation can be written:
Figure 2-23 Temperature Distribution In a cylindrical rod
Heat flow in by conduction at z:
qz(2π r ∆r)∆t (2.211)
Heat flow in by conduction at r:
qr(2π r ∆z)∆t (2.212)
90
where qz and qr are the heat flux by conduction in the z and r directions.
Heat flow out by conduction at z+∆z:
qz+∆z(2π r ∆r)∆t (2.213)
Heat flow out by conduction at r+∆r:
qr+∆r(2π (r+∆r)∆z)∆t (2.214)
Heat accumulation:
ρ(2π r ∆r ∆z)( h~ t+∆t − h~ t) (2.215)
where h~ is the specific enthalpy. Summing the above equation according to the
conservation law and dividing by (2π ∆r ∆z ∆t), considering constant density, gives:
rrqrq
zqqr
thhr rrrzzzttt
∆−
+∆
−=
∆− ∆+∆+∆+
~~ρ
(2.216)
Taking the limit of ∆t, ∆z, and ∆r go to zero yield:
rrq
zqr
thr rz
∂∂
−∂∂
−=∂∂ )(~
ρ (2.217)
Dividing by r and replacing h~ by pC (T − Tref), where pC is the average heat
capacity, and substituting the heat fluxes q with their corresponding relations (Fourier
law):
zTkq zz ∂
∂−=
(2.218)
91
and
rTkq rr ∂
∂−=
(2.219)
Equation (2.217) is then equivalent to:
)(2
2
rTr
rrk
zTk
tTpC r
z ∂∂
∂∂
+∂∂
=∂∂ρ
(2.220)
This is a PDE where the temperature depends on three variables: t, z, and r. If we
assume steady state conditions then the PDE becomes:
)(0 2
2
rTr
rrk
zTk r
z ∂∂
∂∂
+∂∂
= (2.221)
If in addition to steady state conditions, the radius of the rod is assumed to be small
so that the radial temperature gradient can be neglected then the PDE (Eq. 2.221) can
be further simplified. In this case, the differential element upon which the balance
equation is derived is a disk of thickness ∆z and radius R. The heat conduction in the
radial direction is omitted and replaced by the heat transfer through the surface area
which is defined as follows:
Q = U(2πRL)(Ta – T) (2.222)
Consequently, the above energy balance equation (Eq. 2.221) is reduced to the
following ODE
))(2(0 2
2
TTRLUdz
Tdk az −−= π (2.223)
2.2.5 Isothermal Plug Flow Reactor
92
Let consider a first-order reaction occurring in an isothermal tubular reactor as
shown in figure 2.24. We assume plug flow conditions i.e. the density, concentration
and velocity change with the axial direction only. Our aim is to develop a model for the
reaction process in the tube.
z
z = L z z+ z
v(t,z)
(t,z), CA(t,z)
CAo CA
Figure 2-5 Isothermal Plug flow reactor
In the following we derive the microscopic component balance for species (A) around
differential slice of width ∆z and constant cross-section area (S).
Flow of moles of A in:
As has been indicated in section 1.11.1 mass transfer occurs by two mechanism;
convection and diffusion. The flow of moles of species A into the shell is therefore the
sum of two terms:
(vCA S ∆t) |z + (NA S ∆t)|z (2.224)
where NA is the diffusive flux of A ( moles of A/m2 s).
Flow of moles of A out:
(vCA S ∆t) |z+∆z + (NA S ∆t)|z+∆z (2.225)
Accumulation:
(CA S ∆z) |t+∆t − (CA S ∆z) |t (2.226)
93
Generation due to reaction inside the shell:
− r(S∆z∆t) (2.227)
where r = k CA is the rate of reaction.
Substituting all the terms in the mass balance equation (Eq. 1.3) and dividing by ∆t and
∆z gives:
SkCz
SNSvCSNSvCt
SCSCA
zzAAzAAtAttA −+−+
=− ++
∆∆∆∆ |)(|)(|)(|)(
(2.228)
Taking the limit of ∆t 0 and ∆z 0 and omitting S from both sides give the
following PDE:
AAAA kC
zN
zvC
tC
−∂
∂−
∂∂
−=∂
∂ (2.229)
where NA is the molar flux given by Fick’s law as follows:
dzdCDN A
ABA −= (2.230)
where DAB is the binary diffusion coefficient. Equation 2.229 can be then written as
follows:
AA
AbAA kC
zCD
zvC
tC
−∂
∂+
∂∂
−=∂
∂2
2)( (2.231)
Expanding the derivatives, the last equation can be reduced to:
AA
AbAAA kC
zCD
zvC
zCv
tC
−∂
∂+
∂∂
−∂
∂−=
∂∂
2
2
(2.232)
94
This equation can be further simplified by using the mass balance equation for
incompressible fluids (Eq. 2.189). We get then:
AA
AbAA kC
zCD
zCv
tC
−∂
∂+
∂∂
−=∂
∂2
2
(2.233)
The equation is a PDE for which the state variable (CA) depends on both t and z.
The PDE is reduced at steady state to the following second order ODE,
AA
AbA kC
dzCdD
dzdCv −+−= 2
2
0 (2.234)
The ODE can be solved with the following boundary conditions (BC):
BC1: at z = 0 CA(0) = CA0 (2.235)
BC2: at z = L 0)(
=dz
zdCA (2.236)
The first condition gives the concentration at the entrance of the reactor while the
second condition indicates that there is no flux at the exit length of the reactor.
2.2.6 Non-Isothermal Plug-Flow reactor
The tubular reactor discussed earlier is revisited here to investigate its behavior
under non-isothermal conditions. The heat of reaction is removed via a cooling jacket
surrounding the reactor as shown in figure 2.25. Our objective is to develop a model
for the temperature profile along the axial length of the tube. For this purpose we will
need to write an energy balance around an element of the tubular reactor, as shown in
Fig.2.25. The following assumptions are made for the energy balance:
95
Figure 2-25 Non-isothermal plug flow reactor
Assumptions:
• Kinetic and potential energies are neglected.
• No Shaft work
• Internal energy is approximated by enthalpy
• Energy flow will be due to bulk flow (convection) and conduction.
Under these conditions, the microscopic balance around infinitesimal element of width
∆z with fixed cross-section area is written as follows:
Energy flow into the shell:
As mentioned in Section 1.11.3 the flow of energy is composed of a term due to
convection and another term due to molecular conduction with a flux qz.
(qzA + v Aρ h~ )∆t|z (2.237)
Energy flow out of the shell:
(qzA + v Aρ h~ )∆t|z+∆z (2.238)
Accumulation of energy:
(ρA h~ ∆z)|t+∆t − (ρA h~ ∆z)|t (2.239)
96
Heat generation by reaction:
(−∆Hr )kCA A ∆z ∆t (2.240)
Heat transfer to the wall:
ht(πD∆z)(T − Tw)∆t (2.241)
where ht is film heat transfer coefficient.
Substituting these equations in the conservation law (equation 1.7) and dividing by
Α∆t ∆z give:
))((
|||)~(|)~(|)~(|)~(
wtAr
zzzzzzzzttt
TTADhkCH
zqq
zhvhv
thh
−−∆−
∆−
+∆−
=∆
− ∆+∆+∆+
π
ρρρρ
(2.242)
Taking the limit as ∆t and ∆z go to zero yields:
))(()~()~(wtAr
z TTADhkCH
zq
zhv
th
−−∆−∂∂
−∂
∂−=
∂∂ πρρ
(2.243)
The heat flux is defined by Fourier’s law as follows:
zTkq tz ∂
∂−=
(2.244)
where kt is the thermal conductivity. The specific enthalpy ( h~ ) can be approximated by:
)(~refTTpCh −= (2.245)
Since the fluid is incompressible it satisfies the equation of continuity (Eq. 2.189).
Substituting these expressions in Eq. 2.243 and expanding gives:
97
))((/2
2
wtARTE
ort TTADhCekH
zTk
zTpvC
tTpC −−∆−
∂∂
+∂∂
−=∂∂ − πρρ
(2.246)
At steady state this PDE becomes the following ODE,
))((0 /2
2
wtARTE
ort TTADhCekH
dzTdk
dzdTpvC −−∆−+−= − πρ
(2.247)
Similarly to Eq. 2.234 we could impose the following boundary conditions:
B.C1:
at z = 0 T(z) = To (2.248)
B.C2:
at z = L 0)(
=dz
zdT (2.249)
The first condition gives the temperature at the entrance of the reactor and the second
condition indicates that there is no flux at the exit length of the reactor.
2.2.7 Heat Exchanger: Distributed parameter model
We revisit the shell-and-tube heat exchanger already discussed in example 2.1.6.
Steam of known temperature Ts flowing around the tube is heating a liquid L of density
ρL and constant velocity v from temperature TL1 to TL2. The temperature in the tube
varies obviously with axial direction z, radial direction r and time t. To simplify the
problem we will assume that there are no change in the radial direction. This
assumption is valid if the radius is small and no large amount of heat is transferred. The
heat transfer from the steam to the liquid depends on the heat transfer coefficient on the
steam side, hto and on the transfer on the liquid side hti. We also neglect the thermal
capacity of the metal wall separating the steam and the liquid and assume that the
exchange between the steam and liquid occurs with an overall heat transfer coefficient
U. We also assume constant heat capacity for the liquid. An energy balance on a
differential element of the exchanger of length ∆z and cross-sectional area A, yields:
98
Flow of energy in:
(vAρ pC )∆t|z (2.250)
Flow of energy out:
(vAρ pC )∆t|z+∆z (2.251)
Energy accumulation:
(Aρ h~ ∆z)|t+∆t − (Aρ h~ ∆z)|t (2.252)
Energy generated:
U(πD∆z)(T – Ts)∆t (2.253)
Using the expression for specific enthalpy (Eq. 2.245) and dividing by A∆t∆z, the
energy balance yields:
( ) ( )))((|)(|)(|)(|)(
szzzrefttref TT
ADU
zpTCvApTCvA
ttzTTpCAzTTpCA
−−∆
−=
∆
∆−−∆− ∆+∆+ πρρρρ (2.254)
Taking the limit as ∆t an ∆z goes to zero gives:
))(()()(sTT
ADU
zpTCvA
tpTCA
−−∂
∂−=
∂∂ πρρ
(2.255)
Since A = πD2/4 and dividing by ρ pC Eq. (2.255) is equivalent to:
))(4( sTTpDCU
zTv
tT
−−∂∂
−=∂∂
ρ
(2.256)
99
At steady state the PDE becomes the following ODE,
))(4(0 sTTpDCU
dzdTv −−−=
ρ
(2.257
With the following condition:
0)0( TzT == (2.258
2.2.8 Mass exchange in packed column
In previous section (section 2.1.12) we presented some examples of mass
transfer units that can be described by simple ODE's. This includes all the operations
that can occur in tray or spray-tray towers. In this section we present an example of
modeling a mass transfer operation that occurs in packed tower. Absorption is a mass
transfer process in which a vapor solute (A) in a gas mixture is absorbed by contact with
a liquid phase in which the solute is more or less soluble. The gas phase consists usually
of an inert gas and the solute. This process involves flow transfer of the solute A
through a stagnant non diffusive gas B into a stagnant liquid C. The liquid is mainly
immiscible in the gas phase. An example is the absorption of ammonia (A) from air (B)
by liquid water (C). The operation can be carried out either in tray (plate) towers or in
packed towers. The operation in tray towers can be modeled similarly to the liquid-
liquid extraction process in Example 2.1.12 and it is left as an exercise. We consider
here the absorption taking place in a packed tower.
Consider the binary absorption tower shown in Figure 2.26. A liquid stream flow
downward with molar flow rate L and feed composition (XAf). Vapor stream flows
upward with molar flow rate (G) and feed composition (YAf). A simple vapor-liquid
equilibrium relation of the form of:
YA = HXA (2.265)
is used, where H (mole fraction gas/mole fraction liquid) is the Henry's law constant.
This assumption is valid for dilute streams. The molar rates can be considered constants,
100
i.e. not changing from one stage to another, thus the total mass balance need not be
written. To establish the model equations we need to write equations for liquid and
vapor phase. To simplify the problem we assume constant liquid and vapor holdup in
each stage. We also assume isothermal conditions. An energy balance therefore is not
needed.
The flux NA transferred from bulk liquid to bulk gas is given by:
NA = KY (YA – YA*) (2.266)
Where KY is the overall mass transfer in the gas-phase (kgmole/m2s mole fraction) and
YA* is the value that would be in equilibrium with XA. The flux can also be expressed as:
NA = KX (XA – XA*) (2.267)
Where KX is the overall mass transfer coefficient in the liquid-phase and XA* is the value
that would be in equilibrium with YA.
A mass balance on the liquid phase for a differential volume (Fig. 2.26) of the column
length z and cross sectional area S yields:
Flow of mole in:
[(SLXA)∆t|z + (NAS∆t) |z]∆z (2.268)
Flow of moles out:
(SLXA)∆t|z+∆z (2.269)
Rate of accumulation:
(SHLXA∆z)|t+∆t - (SHLXA∆z)|t (2.270)
where XA is the liquid fraction of A and HL the liquid holdup (mole/m3).
101
Figure 2-26 Packed column
The balance equation yields:
(SHLXA∆z)|t+∆t - (SHLXA∆z)|t = (SLXA)∆t|z+∆z - [(SLXA)∆t|z + (NAS∆t)∆t|z]∆z (2.271)
Dividing by S∆t∆z and taking the limits as ∆z and ∆t goes to zero yield:
AAA
L Nz
XL
tX
H +∂
∂=
∂∂
(2.272)
which is equivalent to
)( *AAY
AAL YYK
zXL
tXH −+
∂∂
=∂
∂ (2.273)
We could also use the expression of flux (Eq. 2.267):
102
)( *AaX
AAL XXK
zXL
tXH −+
∂∂
=∂
∂ (2.274)
We can develop material balances for the gas phase that are similar to Eq. 2.274. This
gives:
)( *AAY
AAG YYK
zYG
tXH −+
∂∂
−=∂
∂ (2.275)
or alternatively:
)( *AaX
AAG XXK
zYG
tYH −+
∂∂
−=∂
∂ (2.276)
It should be noted that the analysis carried here can be used for a number of operations
where packed columns are used. This includes liquid-liquid extraction, gas-liquid
absorption and gas-solid drying. In each of these operations an equilibrium relation of
the type:
YA = f (XA) (2.277)
is generally available.
103
Chapter 3: Equations of Change
In the last chapter, we presented examples of microscopic balances in one or two
dimensions for various elementary examples. In this chapter we present the general balance
equations in multidimensional case. The balances, also called equations of changes can be
written in cartesian, cylindrical or spherical coordinates. We will explicitly derive the balance
equations in cartesian coordinates and present the corresponding equations in cylindrical and
spherical coordinates. The reader can consult the books in reference for more details. Once
the equations are presented we show through various examples how they can be used in a
systematic way to model distributed parameter models.
3.1 Total Mass balance
Our control volume is the elementary volume ∆x∆y∆z shown in Figure 3.1. The
volume is assumed to be fixed in space. To write the mass balance around the volume we
need to consider the mass entering in the three directions x,y, and z.
104
Figure 0-1 Total Mass balance in Cartesian coordinates Mass in:
The mass entering in the x-direction at the cross sectional area (∆y∆z) is
(ρvx)|x ∆y∆z∆t (3.1)
The mass entering in the y-direction at the cross sectional area (∆x∆z) is
(ρvy)|y ∆x∆z∆t (3.2)
The mass entering in the z-direction at the cross sectional area (∆x∆y) is
(ρvz)|z ∆x ∆y ∆t (3.3)
Mass out:
The mass exiting in the x-direction is:
(ρvx)|x+∆x ∆y∆z∆t (3.4)
105
The mass exiting in the y-direction is:
(ρvy)|y+∆y ∆x∆z∆t (3.5)
The mass exiting in the z-direction is:
(ρvz)|z+∆z ∆x∆y ∆t (3.6)
Rate of accumulation:
The rate of accumulation of mass in the elementary volume is:
(ρ)|t+∆t ∆x ∆y ∆z - (ρ)|t ∆x ∆y ∆z (3.7)
Since there is no generation of mass, applying the general balance equation Eq. 1.2 and
rearranging gives:
(ρ|t+∆t − ρ|t) ∆x ∆y ∆z = (ρvx|x − ρvx|x+∆x)∆y∆z∆t + (ρvy|y − ρvy|y+∆y)∆x∆z∆t
+ (ρvz|z − ρvz|z+∆z) ∆x ∆y ∆t (3.8)
Dividing the equation by ∆x ∆y ∆z∆t results in:
zvv
yvv
xvv
tzzzzzyyyyyxxxxxttt
∆∆∆∆∆∆∆∆ ++++ ρ−ρ
+ρ−ρ
+ρ−ρ
=ρ−ρ ||||||||
(3.9)
By taking the limits as ∆y,∆x,∆,z and ∆t goes to zero, we obtain the following equation of
change:
zv
yv
xv
tzyx
∂ρ∂
−∂ρ∂
−∂ρ∂
−=∂ρ∂
(3.10)
Expanding the partial derivative of each term yields after some rearrangement:
106
)(zv
yv
xv
zv
yv
xv
tzyx
zyx ∂∂
+∂∂
+∂∂
ρ−=∂ρ∂
+∂ρ∂
+∂ρ∂
+∂ρ∂
(3.11)
This is the general form of the mass balance in cartesian coordinates. The equation is also
known as the continuity equation. If the fluid is incompressible then the density is assumed
constant, both in time and position. That means the partial derivatives of ρ are all zero. The
total continuity equation (Eq. 3.11) is equivalent to:
)(0zv
yv
xv zyx
∂∂
+∂∂
+∂∂
ρ−= (3.12)
or simply:
zv
yv
xv zyx
∂∂
+∂∂
+∂∂
=0 (3.13)
3.2 Component Balance Equation
We consider a fluid consisting of species A, B … , and where a chemical reaction is
generating the species A at a rate rA (kg/m3s). The fluid is in motion with mass-average
velocity v = nt/ρ (m/s) where nt = nA + nB + … (kg/m2s) is the total mass flux and ρ (kg/m3s) is
the density of the mixture. Our objective is to establish the component balance equation of A
as it diffuses in all directions x,y,z (Figure 3.2).
107
x
z
y
nAx|xnAx|x+ x
x
y
znAz|z
nAz|z+ znAy|y
nAy|y+ y
Figure 0-2 Mass balance of component A
Mass of A in:
The mass of species A entering the x-direction at the cross sectional (∆y∆z) is:
(nAx)|x ∆y∆z∆t (3.14)
where nAx kg/m2 is the flux transferred in the x-direction
Similarly the mass of A entering the y and z direction are respectively:
(nAy)|y ∆x∆z∆t (3.15)
(nAz)|z ∆x ∆y ∆t (3.16)
Mass of A out:
The mass of species A exiting the x, y and z direction are respectively
(nAx)|x+∆x ∆y∆z∆t (3.17)
(nAy)|y+∆y ∆x∆z∆t (3.18)
108
(nAz)|z+∆z ∆x ∆y ∆t (3.19)
The rate of accumulation is:
ρA|t+∆t ∆x ∆y ∆z − ρA|t ∆x ∆y ∆z (3.20)
The rate of generation is:
-rA∆x ∆y ∆z∆t (3.21)
Applying the general balance equation (Eq. 1.3) yields:
(ρA|t+∆t − ρA|t) ∆x ∆y ∆z = (nAx|x+∆x − nAx|x)∆y∆z∆t + (nAy|y+∆y − nAy|y)∆x∆z∆t
+ (nAz|z+∆z − nAz|z)∆x∆y∆t +rA∆x ∆y ∆z∆t (3.22)
Dividing each term by ∆x∆y∆z∆t and letting each of these terms goes to zero yields:
AAzAyAxA rt
nt
nt
nt
=∂
∂+
∂
∂+
∂∂
+∂ρ∂
(3.23)
We know from Section 1.11.1, that the flux nA is the sum of a term due to convection (ρAv)
and a term due to diffusion jA (kg/m2s):
nA = ρAv + jA (3.24)
Substituting the different flux in Eq. 3.23 gives:
AAzAyAxzAyAxAA rz
jy
jx
jzv
yv
xv
t=
∂∂
+∂
∂+
∂∂
+∂
∂+
∂
∂+
∂∂
+∂
∂ )()()( ρρρρ
(3.25)
For a binary mixture (A,B), Fick’s law gives the flux in the u-direction as :
uwDj A
ABAu ∂∂
ρ−= (3.26)
109
where wA = ρA/ρ. Expanding Eq. 3.25 and substituting for the fluxes yield:
AAABAABAAB
Az
Ay
Ax
zyxA
A
rz
wDzy
wDyx
wDx
zv
yv
xv
zv
yv
xv
t
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
ρ∂∂∂
+∂
ρ∂∂∂
+∂
ρ∂∂∂
−
⎟⎟⎠
⎞⎜⎜⎝
⎛∂ρ∂
+∂ρ∂
+∂ρ∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂
∂+
∂∂
ρ+∂ρ∂
)()()(
(3.27)
This is the general component balance or equation of continuity for species A. This equation
can be further reduced according to the nature of properties of the fluid involved. If the binary
mixture is a dilute liquid and can be considered incompressible, then density ρ and diffusivity
DAB are constant. Substituting the continuity equation (Eq. 3.13) in the last equation gives:
AAAA
ABA
zA
yA
xA r
zyxD
zv
yv
xv
t=⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂ρ∂
+∂
ρ∂+
∂ρ∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂ρ∂
+∂ρ∂
+∂ρ∂
+∂ρ∂ )2
2
2
2
2
2
(3.28)
This equation can also be written in molar units by dividing it by the molecular weight MA to
yield:
{reaction
A
Diffusion
AAAAB
Convection
Az
Ay
Ax
onaccumulati
A RzC
yC
xC
Dz
Cv
yC
vx
Cv
tC
=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂−⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
∂+
∂∂
+∂
∂+
∂∂
44444 344444 21444444 3444444 213212
2
2
2
2
2
(3.29)
The component balance equation is composed then of a transient term, a convective term, a
diffusive term and a reaction term.
3.3 Momentum Balance
We consider a fluid flowing with a velocity v(t,x,y,z) in the cube of Figure 3.3. The
flow is assumed laminar. We know from Section 1.11.2 that the momentum is transferred
through convection (bulk flow) and by molecular transfer (velocity gradient).
110
xx |x+ xxx |x
zx |z+ zyx |y
yx |y+ y zx |z
y
x
z
Figure 0-3 Balance of the x-component of the momentum
Since, unlike the mass or the energy, the momentum is a vector that has three components,
we will present the derivation of the equation for the conservation of the x-component of the
momentum. The balance equations for the y-component and the z-component are obtained in
a similar way. To establish the momentum balance for its x-component we need to consider
its transfer in the x-direction, y-direction, and z-direction. Momentum in:
The x-component of momentum entering the boundary at x-direction, by convection
is:
(ρvxvx)|x ∆y∆z∆t (3.30)
The x-component of momentum entering the boundary at y-direction, by convection
is:
(ρvyvx)|y ∆x∆z∆t (3.31)
and it enters the z-direction by convection with a momentum:
(ρvzvx)|z ∆x∆y∆t (3.32)
111
The x-component of momentum entering the boundary at x-direction, by molecular
diffusion is:
(τxx)|x ∆y∆z∆t (3.33)
The x-component of momentum entering the boundary at y-direction, by molecular
diffusion is:
(τyx)|y ∆x∆z∆t (3.34)
and it enters the z-direction by molecular diffusion with a momentum:
(τzx)|z ∆x∆y∆t (3.35)
Momentum out:
The rate of momentum leaving the boundary at x+∆x, by convection is:
(ρvxvx)|x+∆x ∆y∆z∆t (3.36)
and at boundary y+∆y,:
(ρvyvx)|y+∆y ∆x∆z∆t (3.37)
and at boundary z+∆z:
(ρvzvx)|z+∆z ∆x∆y∆t (3.38)
The x-component of momentum exiting the boundary x+∆x, by molecular diffusion is:
(τxx)|x+∆x ∆y∆z∆t (3.39)
and at boundary y+∆y:
112
(τyx)|y+∆y ∆x∆z∆t (3.40)
and at boundary z+∆z::
(τzx)|z+∆z ∆x∆y∆t (3.41)
Forces acting on the volume:
The net fluid pressure force acting on the volume element in the x-direction is:
(P|x – P|x+∆x) ∆y∆z∆t (3.42)
The net gravitational force in the x-direction is:
ρg|x ∆x ∆y∆z ∆t (3.43)
Accumulation is:
(ρvx|t+∆t − ρvx|t) ∆x ∆y∆z (3.44)
Substituting all these equations in Eq. 1.5, dividing by ∆x ∆y ∆z∆t and taking the limit of each
term goes zero gives:
xzxyxxxzxyxxxx g
xP
zyxzvv
yvv
xvv
tv ρτττρρρρ
+∂∂
−∂
∂+
∂∂
+∂
∂−=
∂∂
+∂
∂+
∂∂
+∂
∂ )()()()()(
(3.45)
Expanding the partial derivative and rearranging:
xgxP
zyx
zv
vy
vv
xv
vt
vzv
yv
xv
tv
zxyxxx
xz
xy
xx
xzyxx
ρτττ
ρρρρρ
+∂∂
−∂
∂+
∂
∂+
∂∂
−
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂+
∂∂
+∂
∂+
∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂
∂+
∂∂
+∂∂
)(
(3.46)
113
Using the equation of continuity (Eq. 3.10) for incompressible fluid, Equation (3.46) is
reduced to:
xzxyxxxx
zx
yx
xx g
xP
zyxzvv
yvv
xvv
tv ρτττρ +
∂∂
−∂
∂+
∂∂
+∂
∂−=⎟
⎠
⎞⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂
∂ )( (3.47)
Using the assumption of Newtonian fluid, i.e.
zv
yv
xv x
zxx
yxx
xx ∂∂
−=∂∂
−=∂∂
−= µτµτµτ ,,
(3.48)
Equation 3.47 yields:
434214444 34444 2144444 344444 21321generationforces by viscoustransport
2
2
2
2
2
2
flowbulk by transport onaccumulati
xxxxx
zx
yx
xx g
xP
zv
yv
xv
zvv
yvv
xvv
tv ρµρρ +
∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
=⎟⎠
⎞⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂
∂
(3.49)
The momentum balances in the y-direction and z-direction can be obtained in a similar
fashion:
434214444 34444 2144444 344444 21321generationforces by viscoustransport
2
2
2
2
2
2
flowbulk by transport onaccumulati
yyyyy
zy
yy
xy g
yP
zv
yv
xv
zv
vyv
vxv
vt
vρµρρ +
∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
+∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂
∂
(3.50)
434214444 34444 2144444 344444 21321generationforces by viscoustransport
2
2
2
2
2
2
flowbulk by transport onaccumulati
zzzzz
zz
yz
xz g
zP
zv
yv
xv
zvv
yvv
xvv
tv ρµρρ +
∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
=⎟⎠
⎞⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂
(3.51)
These equations constitute the Navier-Stock’s equation.
3.4 Energy balance
114
In deriving the equation for energy balance we will be guided by the analogy that
exists between mass and energy transport mentioned in Section 1.11.3 We will assume
constant density, heat capacity and thermal conductivity for the incompressible fluid. The
fluid is assumed at constant pressure (Fig 3.4).
The total energy flux is the sum of heat flux and bulk flux:
e = q + ρCpTv (3.52)
Therefore, the energy coming by convection in the x-direction at boundary x is:
(qx + ρCpTvx)∆y∆z∆t (3.53)
Similarly the energy entering the y and z directions are
(qy + ρCpTvy)∆x∆z∆t (3.54)
(qz + ρCpTvz) ∆x ∆y ∆t (3.55)
The energy leaving the x,y and z directions are:
(qx + ρCpTvx)|x+∆x ∆y∆z∆t (3.56)
(qy + ρCpTvy)|y+∆y ∆x∆z∆t (3.57)
(qz + ρCpTvz)|z+∆z ∆x ∆y ∆t (3.58)
The energy accumulated is approximated by:
(ρCpT|t+∆t − ρCpT|t) ∆x ∆y∆z (3.59)
115
y
x
z
qz |z
qz |z+ z
qx |x+ x
qy |y+ y
qx |x
qy |y
Figure 0-4 Energy Balance in Cartesian coordinates
The rate of generation is ΦH where ΦH includes all the sources of heat generation, i.e.
reaction, pressure forces, gravity forces, fluid friction, etc. Substituting all these terms in the
general energy equation (Eq. 1.7) and dividing the equation by the term ∆x∆y∆z∆t and letting
each of these terms approach zero yield:
Hzyxzyx
zq
yq
xq
zCpTv
yCpTv
xCpTv
tCpT
Φ=∂∂
+∂∂
+∂∂
+∂
∂+
∂∂
+∂
∂+
∂∂ )()()()( ρρρρ
(3.60)
Expanding the partial derivative yields:
Hzyxzyx
zyx zq
yq
xq
zv
yv
xv
tCpT
zTv
yTv
xTv
tTCp Φ=
∂∂
+∂
∂+
∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂+
∂
∂+
∂∂
+∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ ρρρρρρ
(3.61)
Using the equation of continuity (Eq. 3.10) for incompressible fluids the equation is reduced
to:
Hzyx
zyx zq
yq
xq
zTv
yTv
xTv
tTCp Φ=
∂∂
+∂∂
+∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂
ρ (3.62)
Using Fourier's law:
116
dudTkqu −=
(3.63)
into the last equation gives:
{generation
Hzyx
onaccumulati
zT
yT
xTk
zTv
yTv
xTvCp
tTCp Φ+⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
ρ+∂∂
ρ444 3444 2144444 344444 2143421
diffusion by thermalTransport
2
2
2
2
2
2
flowbulk by Transport
(3.64)
The energy balance includes as before a transient term, a convection term, a diffusion term,
and generation term. For solids, the density is constant and with no velocity, i.e. v = 0, the
equation is reduced to:
{generation
H
onaccumulati
zT
yT
xTk
tTCp Φ+⎟⎟
⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
=∂∂
ρ444 3444 2143421
diffusion by thermalTransport
2
2
2
2
2
2
(3.65)
3.5 Conversion between the coordinates
So far we have shown how to derive the equation of change in Cartesian coordinates.
In the same way, the equations of change can be written other coordinate systems such as the
cylindrical or spherical coordinates. Alternatively, one can transform the equation of change
written in the Cartesian coordinates to the others through the following transformation
expressions.
The relations between Cartesian coordinates (x,y,z) and cylindrical coordinates (r,z,θ)
(Figures 1.3 and 1.4) are the following:
x = rcos(θ), y = rsin(θ), z = z (3.66)
117
Therefore;
)(tan, 122
xyyxr −=θ+=
(3.67)
The relations between Cartesian coordinates (x,y,z) and spherical coordinates (r,θ,φ) (Figures
1.3 and 1.5) are:
)cos(),sin()sin(),cos()sin( θ=φθ=φθ= rzryrx (3.68)
Therefore;
)(tan),(tan, 122
1222
xy
zyx
zyxr −− =φ+
=θ++= (3.69)
Accordingly, we list the following general balance equations in the three coordinates. These
equations are written under the assumptions mentioned in previous sections. For the more
general case, where density is considered variable the reader can consult the books listed in
the references.
3.5.1 Balance Equations in Cartesian Coordinates
Mass Balance
0)( =∂∂
+∂∂
+∂∂
ρzv
yv
xv zyx
(3.70)
Component balance for component A in binary mixture with chemical reaction rate RA:
AAAA
ABA
zA
yA
xA R
zC
yC
xCD
zCv
yCv
xCv
tC
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂+
∂∂
+∂
∂=⎟⎟
⎠
⎞⎜⎜⎝
⎛∂
∂+
∂∂
+∂
∂+
∂∂ )2
2
2
2
2
2
(3.71)
118
Energy balance
Hzyx zT
yT
xTk
zTv
yTv
xTvCp
tTCp Φ+
∂∂
+∂∂
+∂∂
=∂∂
+∂∂
+∂∂
+∂∂ )()( 2
2
2
2
2
2
ρρ
(3.72)
Momentum balance
• x component
xxxxx
zx
yx
xx g
xP
zv
yv
xv
zvv
yvv
xvv
tv
ρ+∂∂
−∂∂
+∂∂
+∂∂
µ=∂∂
+∂∂
+∂∂
ρ+∂
∂ρ )()( 2
2
2
2
2
2
(3.73)
• y component:
yyyyy
zy
yy
xy g
yP
zv
yv
xv
zv
vyv
vxv
vt
vρ+
∂∂
−∂∂
+∂∂
+∂∂
µ=∂∂
+∂∂
+∂∂
ρ+∂
∂ρ )()( 2
2
2
2
2
2
(3.74)
• z component
zzzzz
zz
yz
xz g
zP
zv
yv
xv
zvv
yvv
xvv
tv
ρ+∂∂
−∂∂
+∂∂
+∂∂
µ=∂∂
+∂∂
+∂∂
ρ+∂∂
ρ )()( 2
2
2
2
2
2
(3.75)
3.5.2 Balance Equations in Cylindrical Coordinates
Mass balance
0)11( =∂∂
+θ∂
∂+
∂∂
ρ θ
zvv
rrrv
rzr
(3.76)
Component balance for component A in binary mixture (A-B)with reaction rate RA:
119
AAAA
ABA
zAA
rA R
zCC
rrC
rrr
Dz
Cv
Cr
vr
Cv
tC
+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂∂
∂∂
=⎟⎠
⎞⎜⎝
⎛∂
∂+
∂∂
+∂
∂+
∂∂
)1)(112
2
2
2
2 θθθ
(3.77)
Energy balance
Hzr zTT
rrTr
rrk
zTvT
rv
rTvCp
tTCp Φ+
∂∂
+∂∂
+∂∂
∂∂
=∂∂
+∂∂
+∂∂
+∂∂ )1)(1()1( 2
2
2
2
2 θθρρ θ
(3.78)
Momentum balance
• r component
rrrr
rz
rrr
r
grP
zvv
rv
rrrv
rr
zv
vr
vvr
vr
vv
tv
ρθθ
µ
θρρ
θ
θθ
+∂∂
−∂
∂+
∂∂
−∂
∂+
∂∂
∂∂
=∂
∂+−
∂∂
+∂
∂+
∂∂
)21)1((
)(
2
2
22
2
2
2
(3.79)
• θ component:
θθθθ
θθθθθθ
ρθθθ
µ
θρρ
gPzvv
rv
rrrv
rr
zvv
rvvv
rv
rvv
tv
r
zr
r
+∂∂
−∂∂
+∂∂
+∂∂
+∂
∂∂∂
=∂∂
++∂∂
+∂∂
+∂
∂
)21)1((
)(
2
2
22
2
2
(3.80)
• z component
zzzzz
zzz
rz g
zP
zvv
rrvr
rrzvvv
rv
rvv
tv
ρθ
µθ
ρρ θ +∂∂
−∂∂
+∂∂
+∂∂
∂∂
=∂∂
+∂∂
+∂∂
+∂
∂ )1)(1()( 2
2
2
2
2 (3.81)
3.5.3 Balance Equations in Spherical Coordinates
Mass Balance
120
0)sin(
1))sin(()sin(
1)(1 2
2 =⎟⎟⎠
⎞⎜⎜⎝
⎛φ∂
∂
θ+
θ∂θ∂
θ+
∂∂
ρ φθv
rv
rrvr
rr
(3.82)
Component balance for component A in binary mixture (A-B) with reaction rate RA :
AAAA
AB
AAAr
A
RCr
Crr
Crrr
D
Cr
vCr
vr
Cvt
C
+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂
∂∂∂
+∂
∂∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂
∂+
∂∂
+∂
∂+
∂∂
))(sin
1))(sin()sin(
1)(1
)sin(
2
2
2222
2 φθθθ
θθ
φθθφθ
(3.83)
Energy balance
H
r
Tr
Trr
Trrr
k
Tr
vTr
vrTvCp
tTCp
Φ+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
∂∂
+∂∂
∂∂
=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂
2
2
2222
2 )(sin1))(sin(
)sin(1)(1
)sin(
φθθθ
θθ
φθθρρ φθ
(3.84)
Momentum balance
• r component
rrr
r
rrrr
r
grPv
rv
rvr
rr
rvvv
rvv
rv
rvv
tv
ρφθθ
θθθ
µ
φθθρρ φθφθ
+∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
∂∂
+∂∂
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ +−
∂∂
+∂∂
+∂∂
+∂
∂
2
2
2222
2
2
2
22
)(sin1))(sin(
)sin(1)(1
)sin(
(3.85)
• θ component:
θφ
θθθ
φθθφθθθθ
ρθφθ
θθ
µ
φθθθ
θθ
µ
θφθθ
ρρ
gPr
vvr
vr
vrr
vrrr
rv
rvvv
rvv
rv
rvv
tv
r
rr
+∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂−
∂∂
+
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂
∂∂∂
+∂∂
∂∂
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−+
∂∂
+∂∂
+∂∂
+∂
∂
1)(sin)cos(2
)(sin1))sin()(sin(1)(1
)cot()sin(
22
2
2
2222
2
2
(3.86)
121
• Φ component
φθ
φφφ
φθφφφφθφφ
ρφθφθ
θφθ
µ
φθθθ
θµ
θφθθ
ρρ
gPr
vr
vr
v
r
vθrr
vr
rr
rvv
rvvv
rvv
rv
rv
vt
v
r
rr
+∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛
∂∂
+∂∂
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂+
∂
∂
∂∂
+∂
∂
∂∂
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛++
∂
∂+
∂
∂+
∂
∂+
∂
∂
)sin(1
)(sin)cos(2
)sin(2
)(sin1)
)sin()sin(
1(1)(1
)cot()sin(
222
2
2
2222
2
(3.87)
3.6 Examples of Application of Equations of change
Practically all the microscopic balance examples treated in the previous chapter can be
treated using the equations of change presented in this chapter. In this section we review some
of the previous examples and present additional applications.
3.6.1 Liquid flow in a Pipe
To model the one dimensional flow through the pipe of an incompressible fluid (Example
2.2.1) we may use the continuity balance. For constant density we have the continuity
equation in cylindrical coordinates (Eq. 3.76).
01=
∂∂
+θ∂
∂+
∂∂ θ
zv
rv
rrv
rzr
(3.88)
The plug flow assumptions imply that vr = vθ = 0, and the continuity balance is reduced to:
0=∂∂
zvz
(3.89)
3.6.2 Diffusion with Chemical Reaction in a Slab Catalyst
122
To model the steady state diffusion with chemical reaction of species A in a slab
catalyst, (Example 2.2.3) we use the equation of change (Eq. 3.71). The fluid properties are
assumed constant,
AAAA
AA
zA
yA
xA R
zC
yC
xCD
zCv
yCv
xCv
tC
=∂
∂+
∂∂
+∂
∂−
∂∂
+∂
∂+
∂∂
+∂
∂ )( 2
2
2
2
2
2
(3.90)
Since the system is at steady state we have 0=∂
∂t
CA . If we assume that there is no bulk flow
then vx = vy = vz = 0. For diffusion in the z-direction only, the following holds: =∂∂x
0=∂∂y
.
The equation is then reduces to:
AA
A Rdz
CdD =− 2
2
(3.91)
3.6.3 Plug Flow Reactor
The isothermal plug flow reactor (Example 2.2.5) can be modeled using the
component balance equation (3.77). The plug flow conditions imply that vr = vθ = 0 and
=∂∂r
0=θ∂∂ . Equation (3.77) is reduced to:
AA
ABA
zA R
zCD
zCv
tC
−∂
∂+
∂∂
−=∂
∂2
2
(3.92)
For the non-isothermal plug flow reactor (Example 2.2.6), the energy balance is obtained by
using Eq. 3.78. For fluid with constant properties and at constant pressure, we have:
Hzr zTT
rrT
rrTk
zTvT
rv
rTvCp
tTCp Φ+
∂∂
+∂∂
+∂∂
+∂∂
=∂∂
+∂∂
+∂∂
+∂∂ )11()( 2
2
2
2
22
2
θθρρ θ
(3.93)
123
Using the plug-flow assumptions, vr = vθ = 0 and =∂∂r
0=θ∂∂ , and neglecting the viscous
forces, the term ΦH includes the heat generation by reaction rate RA and heat exchanged with
the cooling jacket, htA(T –Tw). Equation (3.93) is reduced to:
)(/2
2
wtARTE
or TTADhCekH
zTk
zTCpv
tTCp −
π−−
∂∂
+∂∂
ρ−=∂∂
ρ −∆
(3.94)
3.6.4 Energy Transport with Heat Generation
Consider the example of a solid cylinder of radius R in which heat is being generated
due to some reaction at a uniform rate of ΦH (J/m2s). A cooling system is used to remove heat
from the system and maintain its surface temperature at the constant value Tw (Figure 3.5).
Our objective is to derive the temperature variations in the cylinder. We assume that the solid
is of constant density, thermal conductivity and heat capacity. Clearly this is a distributed
parameter system since the temperature can vary with time and with all positions in the
cylinder. We will use then the equation of change (Eq. 3.78) in cylindrical coordinates for a
solid:
CpzTT
rrT
rrT
Cpk
tT H
ρ+
∂∂
+θ∂
∂+
∂∂
+∂∂
ρ=
∂∂ Φ)11( 2
2
2
2
22
2
(3.95)
Figure 0-5 Cylindrical solid rod
A number of assumptions can be made:
124
• The system is at steady state i.e. 0=∂∂
tT
• The variation of temperature is only allowed in radial directions. Therefore, the terms
2
2
zT
∂∂ and 2
2
θ∂∂ T are zero.
The energy balance is reduced to:
CpdrdT
rdrTd
Cpk H
ρρΦ
++= )1(0 2
2
(3.96)
Or equivalently:
kdrdT
rdrTd HΦ
−=+ )1( 2
2
(3.97)
with the following boundary conditions:
• The temperature at the wall is constant:
wTRrT == )( (3.98)
• The maximum temperature will be reached at the center ( r = 0), therefore:
0 0 == ratdrdT
(3.99)
Note that Equation (3.97) can also be written as follows:
kdrdTr
drd
rHΦ
−=)(1
(3.100)
since qr = −k dT/dr, this equation is equivalent to:
125
Hrrqdrd
rΦ−=)(1
(3.101)
The left hand side is the rate of diffusion of heat per unit volume while the right hand side is
the rate of heat production per unit volume.
3.6.5 Momentum Transport in a Circular Tube
We revisit example 2.2.2 where we derived the steady state equations for the laminar
flow inside a horizontal circular tube. We will see how the model can be obtained using the
momentum equation of change. We assume as previously that the fluid is incompressible and
Newtonian. The momentum equations of change in cylindrical coordinates are given by Eqs.
(3.79-3.81). A number of simplifications are used:
• The flow has only the direction z, i.e. vr = vθ = 0.
• The flow is at steady state, 0=∂
∂t
vz
The momentum equation in cylindrical coordinates Eq. 3.81 is reduced to:
zzzzzz
z gzP
zvv
rrv
rrv
zvv ρ+
∂∂
−⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+θ∂
∂+
∂∂
+∂∂
µ=∂∂
ρ 2
2
2
2
22
2 11
(3.102)
Using the continuity equation (Eq. 3.76), and since vr = vθ = 0 gives:
0=dzdvz
(3.103)
We also note that because the flow is symmetrical around the z-axis we have necessarily no
variation of the velocity with θ, i.e.
02
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛θ∂
∂ zv
(3.104)
Equation 3.102 is then reduced to
126
dzdP
drdv
rdrvd zz =⎟⎟
⎠
⎞⎜⎜⎝
⎛+
12
2
µ
(3.105)
Since the left hand side depends only on r, this equation suggests that dzdP is constant.
Therefore:
LP
dzdp ∆
=
(3.106)
where ∆P is the pressure drop across the tube. Equation 3.105 is equivalent to:
LP
drdv
rdrvd zz ∆
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
12
2
µ
(3.107)
with the following conditions identical to those in Example 2.2.2. Note also that Eq. 3.107 can
also be written as:
LP
drrv
drd z ∆
=)(µ
(3.108)
which is also equivalent to:
LP
rdrrd rz ∆
=)( τ
(3.109)
where τrz is the shear stress. The left term is the rate of momentum diffusion per unit volume
and the right hand side is in fact the rate of production of momentum (due to pressure drop).
Note then the similarity between Eq. 3.109 for momentum transfer with Eq. 3.101 for heat
transfer.
127
3.6.6 Unsteady state Heat Generation
We reconsider Example 3.6.4 but we are interested in the variations of the temperature
of the reactor with time as well. This may be needed to compute the heat transferred during
start-up or shut-down operations. Keeping the same assumptions as Example 3.3.4 (except the
steady state assumption), the energy balance in cylindrical coordinates yields:
kdrdTr
drd
rk
tTCp HΦ
+=∂∂
ρ )(1
(3.110)
with the initial and boundary conditions:
w
r
w
TrTdrdT
TtRT
=
=
=
=
)0,(
0|
),(
0
(3.111)
(3.112)
(3.113)
3.6.7 Laminar Flow Heat Transfer with Constant Wall Temperature
We consider a fluid flowing at constant velocity vz into a horizontal cylindrical tube.
The fluid enters with uniform temperature Ti. The wall is assumed at constant temperature Tw.
We would like to model the variations of the fluid temperature inside the tube. To apply the
energy equation of change (Eq. 3.78) we will assume that the fluid is incompressible,
Newtonian and of constant thermal conductivity. Since the system is at steady state d/dt = 0
and the flow is one-dimensional vr = vθ = 0, the energy equation in cylindrical coordinates
Eq. 3.78 is reduced to:
)11( 2
2
2
2
22
2
zTT
rrT
rrTk
tTCpvz ∂
∂+
θ∂∂
+∂∂
+∂∂
=∂∂
ρ
(3.114)
128
Figure 0-6 heat transfer with constant wall temperature
Since the temperature is symmetrical then θ∂
∂T = 2
2
θ∂
∂ T = 0. In some cases we can neglect the
conduction term 2
2
zT
∂
∂ compared to the convective term zTvz ∂
∂ . The system is then described
by the following energy and momentum equations:
)1( 2
2
rT
rrTk
tTCpvz ∂
∂+
∂∂
=∂∂
ρ (3.115)
LP
drrv
drd z ∆
=)(µ
(3.116)
These two equations are therefore coupled by vz, with the previous boundary conditions for vz,
At r = R, vz = 0 (3.117)
At r = 0, dvz/dr = 0 (3.118)
At z = 0, T = Ti (3.119)
At r = 0, dT/dr = 0 (3.120)
At r = R, T = Tw (3.121)
129
3.6.8 Laminar Flow and Mass Transfer
We consider the example of a fluid flowing through a horizontal pipe with constant
velocity vz. The pipe wall is made of a solute of constant concentration CAw that dissolved in
the fluid. The concentration of the fluid at the entrance z = 0 is CAo. The regime is assumed
laminar and at steady state. The fluid properties are assumed constant. We would like to
model the variation of the concentration of A along the axis in the pipe. The component
balance for A (Eq. 3.77) is:
))(
( 2
2
23
2
zC
rC
rrr
Cr
Dz
CvCvr
Cv AA
A
ABA
zAA
r ∂∂
+θ∂
∂+
∂∂
∂∂
=∂
∂+
θ∂∂
+∂
∂θ
(3.122)
Since vr = vθ = 0, the balance equation becomes:
))(
( 2
2
zC
rrr
Cr
Dz
Cv A
A
ABA
z ∂∂
+∂∂
∂∂
=∂
∂
(3.123)
If the diffusion term 2
2
zCA
∂
∂ in the z-direction is negligible compared to convection term
zC
v Az ∂
∂ then the last equation reduces to:
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
∂∂
∂∂
=∂
∂rrr
Cr
Dz
Cv
A
ABA
z
)(
(3.124)
The equation (Eq. 3.116) is unchanged.
The two equations are coupled through vz. The boundary conditions are analogous to the
previous example:
130
At r = R, vz = 0 (3.125)
At r = 0, dvz/dr = 0 (3.126)
At z = 0, CA= CAo (3.127)
At r = 0, dCA/dr =0 (3.128)
At r = R, CA= CAw (3.129)
Note the similarity between this example and the heat transfer case of the previous example.
131
Problems: Lumped parameter systems
Problem 1
Consider the three storage tanks in series shown by Figure X-1.
1. Write the mathematical model that describes the dynamic behavior of the process.
Assume the density of the fluid constant.
2. Identify the states of the process and the degrees of freedom.
h1
F0
h2
F1 F2
F3h3
Fr
Figure X-1
Problem 2
Consider the three cooling storage tanks in series shown in Figure X-2.
h1
F0, T0
h2 h3
F1, T1 F3, T3
F4, T4
Ff , Tf
wc1, Tc1 wc2, Tc2 wc3, Tc3 Figure X-2
1. Assuming constant fluid properties, write the mathematical model for the process.
Identify the states and the degrees of freedom.
132
2. Assume now the following non-isothermal reversible first-order liquid-phase reaction:
A B is taking place in the three above CSTR in series. Assuming all CSTRs are
adiabatic, write down the unsteady state model for the process if the holdups is kept
constant in all reactors.
Problem 3
Consider the well-stirred non-isothermal reactor shown in Figure X-3 below. Write down the
mathematical equations that describe the dynamic behavior of the fundamental quantities of
the process. Consider an exothermic liquid-phase second order reaction rate in the form of A
B is taking place. The density is assumed to be constant, develop the necessary equations
describing the process dynamic behavior.
h1
F1, CA1
F3, CA3
Figure X-3
Problem 4
Consider a two CSTRs in series with an intermediate mixer introducing a second feed as
shown in Figure X-4. A first order irreversible exothermic reaction: A B is carrier out in
the process. Water at ambient temperature (Tc1i and Tc2i) is used to cool the reactors. The
densities and heat capacities are assumed to be constant and independent of temperature and
concentration. Develop the necessary equations describing the process dynamic behavior.
Note that the mixer has negligible dynamics and that the inlet feed to CSTR2 has the same
temperature as that of the outlet of CSTR1.
133
Mixer
Q1, C1f, T1f
QC1, TC1i
Q4, C4, T4
QC2, TC2i
Q2, C2, T2
Q3, C3, T3
CSTR 1
CSTR 2
QC1, TC1
QC2, TC2
Figure X-4
Problem 5:
Consider a thermometer bulb with pocket is immersed in hot fluid of temperature T1 as shown
in Figure X-5 below. The pocket temperature is T2 (assumed uniform through the thickness)
and that for the bulb is T3. Write the model equation that describes how T2 and T3 varies with
time.
T1
T2
T3
Figure X-5
Problem 6
Consider the single-effect steam evaporator shown in Figure X-6. A salty water (brine) with
mass fraction Cb0 and mass flow rate B0 is fed to the evaporator where it is heated with
saturated steam with mass flow rate W. The concentrated product comes out of the evaporator
134
with mass flow rate B1 and mass fraction Cb1 while the vapor is withdrawn from the top with
mass flow rate V. Develop the dynamic model that describe the process behavior. Assume the
heat supplied by the steam is mainly equal to the heat used for vaporizing the brine.
Steam
Feed
Vapor
Condensate
Concentratedliquid
Figure x-6
Problem 7
Consider the process shown in Figure X-7. A stream of pure component A is mixed with
another stream of a mixture of component A and B in an adiabatic well stirred mixing tank.
The effluent is fed into an adiabatic CSTR where the following reaction takes place: A + B
C. Assume the process is isothermal. Develop the dynamic model for the process and
determine the degrees of freedom assuming constant fluid properties.
Q1, CA1
Q3, CA3
Q2, CA2
Q4, CA4
Figure X-7
135
Problem 8:
Consider the following irreversible first-order liquid-phase reaction:
CBA kk ⎯→⎯⎯→⎯ 21
where k1 and k2 are the reaction rate constants in sec-1.
1. Write down the unsteady state model for the process if the reaction takes place in an
isothermal well-mixed CSTR.
2. Write down the unsteady state model for the process if the reaction takes place in a
non-isothermal well-stirred batch reactor, where the heat needed for the endothermic
reaction is supplied through electrical coil.
3. Repeat part 2 but assuming that the reactor is cooled by cold fluid flowing in a jacket
and that the reactor wall has high resistance to heat transfer.
Problem 9:
A perfectly mixed non-isothermal adiabatic reactor carries out a simple first-order exothermic
reaction, A B in the liquid phase. The product from the reactor is cooled from the output
temperature T to Tc and then introduced to a separation unit where the un-reacted A is
separated from the product B. The feed to the separation unit is split into two equal parts top
product and bottom product. The bottom product from the separation unit contains 95% of the
un-reacted A in the effluent of the reactor and 1% of B in the same stream. The bottom
product which is at Temperature Tc (since the separation unit is isothermal) is recycled and
mixed with the fresh feed of the reactor and the mixed stream is heated to temperature Tf
before being introduced to the reactor. Write the steady state mass and energy balances for the
whole process assuming constant physical properties and heat of reaction. Discuss also the
degree of freedom of the resulting model. The process is depicted in figure X-8
136
CA, CB , Tc
Fo, To ,CAo
F, Tf
F
L = 0.5 F
V= 0.5 FCSTR
Separator
Figure X-8
Problem 10:
Consider a biological reactor with recycle usually used for wastewater treatment as depicted
in figure X-9 below. Substrate and biomass are fed to the reactor with concentrations Sf and
Xf. The effluent form the well-mixed reactor is settled in a clarifier and a portion of the
concentrated sludge is returned to the reactor with flow rate Qr and Xr. If the reaction of
substrate in the clarifier is negligible, the recycle stream would contain the same substrate
concentration as the effluent from the reactor. Sludge is withdrawn directly from the reactor
with a fraction W. Assuming constant holdup develop the dynamic model for the process.
Assume the rate of disappearance of S is given by: r = mS/(K + S) and the rate of generation
of X by r/Y where Y is constant.
ReactorV
WXS
XS
QXfSf
Q-WXtS
Qr , Xr , S
Settler
Figure X-9
Problem 11
137
Consider the two phase reactor and a condenser with recycle. Gaseous A and liquid B enters
the reactor at flow rates FA and FB respectively. Gas A diffuses into the liquid phase where it
reacts with B producing C. The latter diffuses into the vapor phase where B is nonvolatile.
The vapor phase is fed to the condenser where the un-reacted A is cooled and the condensate
is recycled back to the reactor. The product C is withdrawn with the vapor leaving the
condenser. For the given information develop the dynamic model for the process shown in
Figure X-10. Consider all flows are in moles.
FB, TB
FA, TA
F1L, xA1, xB1, T1
F2L,xA2
F1v, yA1, yc1
T1, P1
P2
T2
NA
NA
NC
Q1
A + B = 2C
Figure X-10
Problem 12:
Develop the mathematical model for the triple-effect evaporator system shown in figure X-11
below. Assume boiling point elevations are negligible and that the effect of composition on
liquid enthalpy is neglected.
138
Vo
V1=F-L1
L1
V2=L1-L2
V1
L2
V2
L3
V3=L2-L3
Vo
Thick Liquor Thick Liquor Thick Liquor
F (feed)
Vapor Vapor Vapor
Figure X-11
Problem 13
Liquid vaporizer as depicted in Figure X-12 below is one of the important processing units in
a chemical plant. A liquid feed, enters the vaporizer at specific flow rate F0 density ρ0 and
temperature T0. Inside the vessel, the liquid is vaporized by continuous heating using hot oil.
The mass rate of vaporization is wn. The formed vapor is withdrawn continuously from the
top of the vessel. We would like to develop a mathematical model to describe the process.
Unlike the adiabatic flash operation, the temperature and pressure in the two phases are
different. Correspondingly, the volume of the two phases varies with time.
Fo , To , ρo
Fv
Q
PL, VL, ρL
Pv , Vv , ρv
wv
Figure X-12
139
Problems: Distributed Parameter Systems
Problem 14
At time t = 0 a billet of mass M, surface area A, and Temperature To is dropped into a tank of
water at Temperature Tw. Assume that the heat-transfer coefficient between the billet and
water is h. If the mass of water is large enough that its temperature is virtually constant,
determine the equation describing the variation of billet temperature with time for the
following two cases:
1. The thermal conductivity of the metal in the billet is sufficiently high that the billet is
of uniform temperature.
2. The thermal conductivity of the metal is low and the billet radius is large enough that
the temperature inside the billet is not uniform.
Problem 15
(a)Consider a wall made up of stacked layers of various materials each with different thermal
conductivity usually used for insulation as shown by figure X-13a. Assume the inner side is at
high temperature Tb and the outer side is at the ambient temperature Ta. Derive the
temperature profile through the wall.
x
Ta
Tb
Figure X-13a
(b) Repeat the above development for a composite cylindrical wall shown by figure X-13b.
140
Ta
Tb
r
Figure X-13b
Problem 16
A chemical reaction is being carried out in a fixed bed flow reactor. The reaction zone is
filled with catalyst pellets. Assume plug flow and the reactor wall is well insulated such that
the temperature is uniform in the radial direction. If the fluid enters the reactor with
temperature T1 and superficial velocity v and that thermal energy per unit volume (S) is
produced inside the reactor due to chemical reaction, derive the steady-state temperature axial
distribution. The unit is depicted in figure X-14.
z = 0 z = LReaction zone
Reactant Product
Insulating wall Catalyst particles
zr
Figure X-14
Problem 17
A first order irreversible reaction is taking place in a tubular reactor. The reactor can be
considered to be isothermal and radial dispersion is not to be neglected. Using Fick’s law with
effective diffusion coefficient to describe the axial and radial dispersion, derive the steady
state mass balance equation and its boundary conditions.
Problem 18
141
(a) Derive the unsteady state mass balance equation for the diffusion of species A through a
spherical porous pellet.
(b) Derive the steady state mass balance equation for the diffusion of species A through a
spherical porous pellet where it converts to B according to the first-order reaction: A B.
Problem 19
Species A dissolves in liquid B and diffuses into stagnant liquid phase contained in a
cylindrical tank where it undergoes a homogeneous irreversible second-order chemical
reaction: A+B C. Derive the mass balance equation for component A. Assume that the
depth of liquid is so large compared to the radius of the tank, thus, the radial distribution can
be neglected. The process is shown schematically in figure X-15.
Gas A
Liquid B∆z
z = 0
z = L
Figure X-15