英文版化学工程设计技术：08Material Balance Notes

MATERIAL BALANCE NOTES
Revision 3
Irven Rinard
Department of Chemical Engineering
City College of CUNY
and
Project ECSEL
October 1999
1999 Irven Rinard
i
CONTENTS
INTRODUCTION 1
A,Types of Material Balance Problems
B,Historical Perspective
I,CONSERVATION OF MASS 5
A,Control Volumes
B,Holdup or Inventory
C,Material Balance Basis
D,Material Balances
II,PROCESSES 13
A,The Concept of a Process
B,Basic Processing Functions
C,Unit Operations
D,Modes of Process Operations
III,PROCESS MATERIAL BALANCES 21
A,The Stream Summary
B,Equipment Characterization
IV,STEADY-STATE PROCESS MODELING 29
A,Linear Input-Output Models
B,Rigorous Models
V,STEADY-ST ATE MATERIAL BALANCE CALCULATIONS 33
A,Sequential Modular
B,Simultaneous
C,Design Specifications
D,Optimization
E,Ad Hoc Methods
VI,RECYCLE STREAMS AND TEAR SETS 37
A,The Node Incidence Matrix
B,Enumeration of Tear Sets
VII,SOLUTIO N OF LINEAR MATERIAL BALANCE MODELS 45
A,Use of Linear Equation Solvers
B,Reduction to the Tear Set Variables
C,Design Specifications
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VIII,SEQUENTIAL MODULAR SOLUTION OF NONLINEAR 53
MATERIAL BALANCE MODELS
A,Convergence by Direct Itera tion
B,Convergence Acceleration
C,The Method of Wegstein
IX,MIXING AND BLENDING PROBLEMS 61
A,Mixing
B,Blending
X,PLANT DATA ANALYSIS AND RECONCILIATION 67
A,Plant Data
B,Data Reconciliation
XI,THE ELEMENTS OF DYNAMIC PROCESS MOD ELING 75
A,Conservation of Mass for Dynamic Systems
B,Surge and Mixing Tanks
C,Gas Holders
XII,PROCESS SIMULATORS 87
A,Steady State
B,Dynamic
BILIOGRAPHY 89
APPENDICES
A,Reaction Stoichiometry 91
B,Evaluation of Equipment Model Parameters 93
C,Complex Equipment Models 96
D,Linear Material Balance by Spreadsheet - Example 97
E,The Kremser Model of Gas Absorbers 98
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INTRODUCTION
Revised October 2,1999
The material balance is the fundamental tool of chemical engineering,It is the basis for
the analysis and design of chemical processes,So it goes without saying that chemical engineers
must thoroughly master its use in the formu lation and solution of chemical processing problems.
In chemical processing we deal with the transformation of raw materials of lower value
into products of higher value and,in many,cases unwanted byproducts that must be disposed of.
In addition many of these chemical compounds may be hazardous,The material balance is the
chemical engineer's tool for keeping track of what is entering and leaving the process as well as
what goes on internally,Without accurate material balances,it is impossible to design or operate
a chemical plant safely and economically.
The purpose of these notes is to provide a guide to the use of material balances in
chemical engineering,Why one might ask? Aren't there already enough books on the subject,
books such as those by Felder and Rouseau,by Himmelblau,and by Reklaitis? To answer that
question we need to look briefly at the history of the prob lem.
A,Types of Material Balance Problems
First let us look at the types of material balance problems that arise in chemical
engineering,There are four basic types of problems:
(1) Flow sheet material balance models for continuous processes operating in the steady
state,
(2) Mixing and blending material balances,
(3) Flow sheet material balances for non-steady state processes,either continuous or
batch,and
(4 ) Process data analysis and reconciliation
A flow sheet is a schematic diagram of a process which shows at various levels of detail
the equipment involved and how it is interconnected by the process piping (See,for instance
Figures II-1 and II-2 in Chapter II),A flow sheet material balance shows the flow rates and
compositions of all streams entering and leaving each item of equipment,
Most of the emphasis on material balance problems has been on continuous processes
operating in the steady state,Again one might ask why,The reason is simple,Of the total
tonnage of chemi cals produced,the vast majority is produced using con tin uous steady-state
processes,This includes oil refin eries as well as chemical plants producing large ton nage prod ucts
such as sulfu ric acid,ethylene,and most of the other commod ity chemicals,petro chemi cals and
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polymers,It has been found that the most econom ical and efficient way to produce such
chemicals on a large scale is via the continuous process operat ing in t he steady-state,This is the
reason for the emphasis on this type of material balance problem.
Another class of material balance problems is those in volving blending and mixing,A sub-
stan tial number of the products pro duced by the chem ical pro ces sing industries are blends or mix-
tures of various constitu ents or ingredients,Exam ples of blends are gasoline and animal feeds ; of
precise mixtures,prescription drugs and polymeric resins.
Dynamic material balance problems arise in the opera tion an d control of continuous
processes,Also,batch pro cesses,by their very nature,are dynamic,In either case we must
consider how the state of the process varies as a function of time,In addition to determining the
flow rates and composi tions of the i nterconnecting streams,we must also follow the chang es in
inventory within the process itself.
In the three types of problems just discussed,we are interested in predicting the
performance of the process or equipment,Our models start by assuming that the law of
conservation of mass is obeyed,A fourth type of problem,which en countered by engineers in the
plant,starts with actual operating data,generally flow rates and compositions of various streams,
The problem is to determine the actual performance of the plant from the available data,
This,in many ways,is a much more difficult problem than the first three,Why? Simple,
The data may in error for one reason or another,A flow meter may be out of calibration or
broken entirely,A composition measurement is not only subject to calibration errors but sampling
errors as well,Thus the first thing one must do when dealing with plant data is to determine,if
possible,whether or not it is accurate,If it is,then we can proceed to use it to analyze it to
determine process performance,If not,we must try to determine what measurements are in error,
by how much,and make the appropriate corrections to the data,This is known as data
reconciliation and is possible only if we have redundant measurements,
B,Historical Perspective
The solution of material balance problems for continuous steady-state processes of any
complexity used to be very diffi cult,By its nature,the problem is one of solv ing a large number
of simul ta ne ous algebraic equa tions,many of which are highly nonlinear,Before the avail ability
of comput ers and the appro priate soft ware,the solution of the material balance model for a
chemical process typically took a team of chemical engi neers using slide rules and adding
machines days or weeks,if not months,And given the complexity of the problem,errors were
common.
The methods used in those days to solve material balance problems days are best described
as ad hoc,Typically an engineer start ed with the pro cess specifications such as the production
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rate and product quality and calculated backwards through the pro cess,Interme diate specifi-
cations would be used as addition al starting points for calcula tions,As will be seen later in these
notes,such an approach goes against the output-from-input structure of the process and can lead
to severe numerical insta bilities,
The growing availability of digital computers in the late 1950's led to the development of
the first material balance programs such as IBM's GIFS,Dartmouth's PACER and Shell’s
CHEOPS,Almost every major oil and chemical company soon developed in-house programs of
which Monsanto's Flowtran is the best-known example,By the 1970's several companies
specializing in flow sheet programs had come into existence,Today companies such as
Simulation Scienc es,Aspen Technology and Hyprotech provide third-generation versions of
steady-state flow sheet simulation programs that provide a wide range of capabilities and are
relatively easy to use compared to earlier versions,
Dynamic simulation is less advanced than steady-state simulation,This is due,in part,to
the lack of emphasis until recently on the dynamic aspects of chemical engineering operations,
This situa tio n is chang ing rapid ly due to demands for improved process control and for simulators
for training operating personnel,The companies mentioned in the previous paragraph have all
recently added dynamic simulators to their product lines,In addition several companies such as
ABB Simcon offer training simulators for the process industries.
C,Material Balance Methodology
There are two major steps involved in applying the principle of conservation of mass to
chemical processing prob lems,The first i s the formu la tion of the problem ; the second,its solu-
tion,
By formulation of the problem is meant determining the appropriate mathematical
description of the system based on the applicable principles of chemistry and physics,In the case
of material balances,the appropriate physical law is the con serva tion of mass,The resulting set of
equations is some times referred as a mathematical model of the system.
What a mathematical model means will be made clearer by the examples contained in these
notes,However,some general comments are in order,First,there may be a number of mathe-
matical models of varying levels of detail that can apply to the same system,Which we use
depends upon what aspects of the process we wish to study,This will also become clearer as we
proceed,Second,for many systems of practical interest,the number of equations involved in the
model can be quite large,on the order of several hundred or even several thousand,
Thus,the process engineer must have a clear of how to formulate the model to insure that
it is a correct and adequate representation of the process for the purposes for which it is intended.
This is the subject of Sections I - IV of these notes.
Today,using process simulation program such as PRO-II,ASPEN,and HYSIM,a single
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engineer can solve significant flow sheeting problems in as little as a day or two and,moreover,
do it much more accurately and in much more de tail than was previously possible,The process
engineer can now con cen trate on th e pro cess model and the results rather than con coct ing a
scheme to solve the model equations themselves,The simu lation program will do that,at least
most of the time,Howev er,things do go wrong at times,Either the problem is very diffi cult for
the simulator to solve or a mistake has been made in describing the process to the program,Thus,
in order to fix what is wrong,the engi neer does need to know something about how the simu la-
tion pro gram attempts to solve the prob lem,This is the su bject of Sections V,VIII,and IX of
these notes,Steady-state simulation programs are described briefly in Section XII,
Simple material balance problems involving only a few variables can still be solved
manually,However,it is general ly more ef ficient to use a computer program such as a spread-
sheet,Both approaches are discussed in Section VII,
In order to achieve high levels of mass and energy utiliza tion effi ciency,most processes
involve the use of recycle,As will be seen,this creates recycle loops within the process which
complicate the solution of material balances models for the process,A systematic procedure for
identifying recycle loops is presented in Section VI.
An introduction to problems encountered in determining plant performance from plant is
given in Section X,
There are two basic process operating modes that are of inter est to chemical engineers,
dynamic and steady state,All pro cesses are dynamic in that some or all of the process vari ables
change with time,Many processes are deliberately run dynami cally; batch processes being the
prime example,However,many large-scale processes such as oil refineries and petrochem ical
plants are run in what is called the con tin u ous or steady state opera tion,The a ppropriate model
for dynamic processes are dif feren tial equations with respect to time,In general,con tin uous pro-
cesses operating in the steady state are mod eled by algebraic equations,Dynamic process
modeling is discussed briefly in Section XI and dynamic process simulators in Sec tion XII.
Many topics in process modeling are not covered in these notes,The most serious
omission is the companion to the mate rial balance,namely,the energy balance,Also,little atten-
tion is paid to what are known as first-principle or rigorous equipment models,Such modeling is
more properly covered in texts and courses on unit operations and chemical reaction engineering.
However,a few of the simpler and more useful models are given in Appendix E,
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I,CONSERVATION OF MASS
Revised October 2,1999
The principle of conservation of mass is fundamental to all chemical engineering analysis,
The basic idea is relatively easy to understand since it is fact of our everyday life,
Let us consider a simple example,Suppose we are required to prepare one kilogram of a
solution of ethanol in water such that the solution will contain 40% ethanol by weight,So,we
weigh out 400 grams of ethanol and 600 grams of water and mix the two together in a large
beaker,If we weigh the resulting mixture (making appropriate allowance for the weight of the
beaker),experience says it will weigh 1000 grams or one kilogram,And it will,This is a
manifestation of the conservation of mass,
That,in the absence of nuclear reactions,mass is
conserved is a fundamental law of nature,
This law is used throughout these notes and throughout all chemical engineering.
Suppose we happened to measure the volumes involved in making up our alcohol
solution,Assuming that we do this at 20 C,we would find that we added 598.9 ml of water to
315.7 ml of ethanol to obtain 935.2 ml of solution,However,the sum of the volumes of the pure
components is 914.6 ml,We conclude that volume is not conserved,
Let us take note of one other fact about our solution,If we were to separate it back into
its pure components (something we could do,for instance,by azeotropic distillation) and did this
with extreme care to avoid any inadvertent losses,we would obtain 400 gm of ethanol and 600
gm of water,Thus,in this case,not only was total mass conserved but the mass of each of the
components was also,
This is not always true,Suppose that instead of adding ethanol and water,we added
(carefully and slowly) sodium hydroxide to sulfuric acid,Suppose that the H 2 SO 4 solution
contains exactly 98.08 pounds of H 2 SO 4 and that we add exactly 80.00 pounds of NaOH,A
chemical reaction will take place as follows:
H 2 SO 4 + 2 NaOH à Na 2 SO 4 + 2 H 2 O.
Notice that the amount of H 2 SO 4 in the original solution is 1.0 lb-mol and that the amount of
NaOH added is exactly 2.0 lb-mols,What we are left with is 1.0 lb-mol of Na 2 SO 4 or 142.05 lbs
and 2.0 lb-mols of H 2 O or 36.03 lbs,No individual component is conserved; the H 2 SO 4 and the
NaOH have disappeared and in their place we have Na 2 SO 4 and H 2 O,However,if we look at the
atomic species H,O,S,and Na,we will find that these are all con-served,That is exactly what
the reaction equation expresses.
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Thus we have to be careful to identify the appropriate conserved species for the system we
are analyzing,If no chemical reactions are involved,then each of the molecular species is
conserved,If chemical reactions are involved,then only atomic species are conserved,There will
be a mass balance for each of the conserved species,In the example above it does not make much
difference since there are four conserved atomic species and four molecular species,But,if
additional reactions take place involving,say,Na 2 S and NaHSO 4,then the number of molecular
species exceeds the number of conserved atomic species,This will generally be the case.
A,Control Volumes
We apply the principle of the conservation of mass to systems to determine changes in the
state of the system that result from adding or removing mass from the system or from chemical
reactions taking place within the system,The system will generally be the volume contained
within a precisely defined section of a piece of equipment,We refer to this precisely defined
volume as a control volume.
It may be the entire volume of the equipment,This would be the case if the system is a
cylinder containing a gas or gas mixture,Or it may be the volume associated with a particular
phase of the material held within the system,For instance,a flash drum is used to allow a mixture
of vapor and liquid to separate into separate vapor and liquid phases,The liquid phase will
occupy part of the total volume of the drum ; the vapor,the remainder of the volume,If we are
interested only in what happens to the liquid phase,then we would specify the volume occupied
by the liquid as our control volume,
Note that the control volume can change over the course of an operation,Suppose we are
adding liquid to a tank that contains 100 Kg of water to start with and that we add another 50 Kg.
The tank would originally contain 100 liters of water but would contain 150 liters after the
addition,On the other hand,if our interest is in the entire contents of the tank - both the liquid
and the vapor in the space above it - then we would take the volume of the tank itself as our
control volume,This volume,of course,will not change.
B,Holdup or inventory
Another concept that we will need to make precise is that of holdup,also known as
inventory or accumulation,Holdup refers to the amount of a conserved species contained within
a control volume,We can refer to the total holdup as simply the total mass of material contained
within the control volume,Or we can refer to the holdup of a particular component,sodium
chloride say,which is contained within the control volume,Needless to say,the sum of the
holdups of all of the individual components within the control volume must equal the total hold-
up,
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C,Material Balance Basis
Whenever we apply the principle of conservation of mass to define a material balance,we
will want to specify the basis for it,Generally,the basis is either the quantity of total mass or the
mass of a particular component or conserved species for which the material balances will be
defined,Or,for continuous processes,it might be the mass flow rate of a component or
conserved species,
Quite often the basis will be set by the specification of the problem to be solved,For
instance,if we are told that a tank contains 5,000 pounds of a particular mixture about which
certain questions are to be answered,then a natural basis for the problem would the 5,000 pounds
of the mixture,Or,if we are looking at a continuous process to make 10,000 Kg/hr of ethanol,a
reasonable choice for a basis would be this production rate,
Some problems,however,do not have a naturally defined basis so we must choose one,
For instance,if we are asked what is the mass ratio of NaOH to H 2 SO 4 required to produce a
neutral solution of NaCl in water,we would have to specify a basis for doing the calculations,
We might choose 98.08 Kg of H 2 SO 4 (1.0 Kg-mol) as a basis,Or we could chose 1.0 lb of
H 2 SO 4,Either is acceptable,One basis may make the calculations simpler than another,but in
this day of personal computers the choice is less critical than it might have been years ago,
Whatever the choice of basis,it is mandatory that all material balances are defined to be consistent
with it.
D,Material Balances
We are now in a position to define material balances for some simple systems,(Note,
Material balances are sometimes referred to as mass balances.) There are three basic situations
for which we will want to do this:
1) A discrete process in which one or more steps are carried out over a finite but
indefinite period of time.
An example of such a process is the dissolving of a specified quantity of salt in a quantity
of water contained in a tank,We are only interested in the concentration in weight % of the salt
in the water after it is completely dissolved and not how long it takes for the salt to dissolve.
2) A continuous process operating in the steady state,
By definition,continuous process operating in the steady state undergoes no changes in its
internal state variables such as temperatures,pressures,compositions,and liquid levels,In
addition,all the flow rates of all streams entering and leaving each item of equipment are constant.
What this means from the standpoint of material balances is that there is no change in any of the
holdups in the system.
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3) Dynamic processes.
This is the general case in which both the holdups within equipment as well as the flow
rates and compositions of the input and output streams can vary with time,This type of mass
balance is discussed in more detail in Section XI.
Examples
a,Discrete process
Let us now consider the application of the principle of the conservation of mass to a
discrete process,For each of the conserved species,it can be stated as follows:
(I-1) Change in holdup = additions to the control volume
- withdrawals from the control volume
Consider the following example:
We have a tank that initially holds 100 Kg of a solution containing 40% by weight of salt
in water,
(1) We add 20 Kg of salt to the tank and allow it to dissolve.
What do we now have in the tank?
First we have to identify the conserved species,Since there are no chemical reactions
involved,both salt and water are conserved species,Next we have to define the control volume,
It seems natural to choose the salt solution in the tank,Our basis is the amount of solution
originally contained in the tank.
Now we can define a material balance for each conserved species as follows:
Water
Initial holdup of water = (1 - 0.4)(100) = 60 Kg
Change in holdup of water = additions of water to the control volume
- withdrawals of water from the control volume
Since no water is either added or withdrawn,the change in this holdup is zero,Therefore the
holdup of water after the salt addition is still 60 Kg.
Salt
Initial holdup of salt = (0.4)(100) = 40 Kg
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Change in holdup of salt = additions of salt to the control volume
- withdrawals of salt from the control volume
We add 20 Kg of salt and withdraw no salt,Therefore the change in the holdup of salt = +20
Kg and the holdup of salt after the addition is 40 + 20 = 60 Kg,A simple calculation shows that
the concentration of salt in the tank is now 50 weight %.
(2) We withdraw 40 Kg of the solution now in the tank.
Since the solution in the tank is 50 weight % salt and 50 weight % water,in withdrawing
40 Kg of solution,we will withdraw 20 Kg of salt and 20 Kg of water,This will leave 60 - 20 =
40 Kg of each component in the tank,The composition has not changed from Step 1.
b,Continuous steady-state process
Let us consider a continuous mixer which has two input streams and,of course,one
output stream (See Figure 6 for a diagram),Suppose the first input stream has a flow rate of
10000 lb/hr of a 40 wt,% solution of salt in water while the second input stream has a flow rate
of 20000 lb/hr of a 70 wt,% solution of salt in water,What is the flow rate and composition of
the output stream?
Since the system is now characterized in terms of rates of flow into the control volume
(additions) and rates of flow out of the system (withdrawals),we need to restate the principle of
conservation of mass as follows:
(I-2) Rate of change of holdup =
rate of additions to the control volume
- rate of withdrawals from the control volume
For a continuous system operating in the steady state,the holdup does not change with time,
Therefore,the rate of change of holdup is zero and Eqn,I-2 becomes
(I-3) Rate of withdrawals from the control volume =
Rate of additions to the control volume
Let us apply this to the mixer problem,The control volume is the contents of the mixer (even
though these do change) and the basis is the total rate of flow to the mixer,As in the previous
example,the conserved species are salt and water,
Salt
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F in
F out
z
Rate of withdrawal of salt =
rate of additions of salt to the mixer
Rate of additions = (10000)(0.4) + (20000)(0.7)
= 18000 lb/hr of salt
Water
Rate of withdrawal of water =
rate of additions of water to the mixer
Rate of additions = (10000)(0.6) + (20000)(0.3)
= 12000 lb/hr
Thus the stream leaving the mixer has a flow rate of 18000 lb/hr of salt and 12000 lb/hr of water,
for a total of 30000 lb/hr,This is exactly the total flow rate of the mixer output we get by adding
up the total flow rates to the mixer,Also,the composition of salt of the stream leaving the mixer
=
(100)(18000)/(30000) = 60 wt %.
c,Dynamic Process
Consider the surge tank shown in Figure I-1,
Water flows into the tank with a flow rate F in lb/hr,It
flows out at a rate F out lb/hr,The flow rates in and
out can be adjusted by means of the valves in the inlet
and outlet piping,The tank has the form of an
upright cylinder that has a cross section area of S ft 2,
The liquid level in the tank is z ft.
We know from experience that if the flow
rates in and out are not exactly equal,the level in the
tank will change with time,If the inlet flow rate
exceeds the outlet flow rate,then the level will rise
and vice versa,Now,the purpose of a surge tank is
to absorb changes in the inlet flow rate while
maintaining a relatively constant outlet flow rate,
(The reservoirs that supply water to a town or city
are surge tanks where the inlet flow is the run-off from rainstorms and the outlet flow Figure I-
1,Surge Tank
is the daily consumption by the town or city.) Thus,a question that designers of surge tanks must
ask is,given an estimate of the variations of inlet and outlet flow rates as functions of time,how
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big must the surge tank be so that it never runs dry (town loses its water supply) or never
overflows (area surrounding the reservoir is flooded),In general this is a complex design problem
but let us look at a simple example to at least illustrate the concept,
Suppose that under normal conditions the level in the tank is to be half the height of the
tank,If the flow rate into the tank becomes zero for a period of time (no rain),how long will it
take for the tank to run dry if the outlet flow rate is maintained at its usual value? Specifically,
suppose that the cross section area of the tank is 10 ft 2 and its height is 10 ft and the normal outlet
flow rate is 12,480 lb/hr,
First,let us take the volume of liquid in the tank as the control volume,The holdup of
water in the control volume will be
Holdup = Sz ρ,where ρ is the density of water (62.4 lb/ft 3 ).
Consider an interval of time?t,Suppose that over that time interval the inlet and outlet flow
rates are constant but not necessarily equal,Then,by the conservation of mass the change in the
holdup will be given by
(I-1) Holdup| t= t - Holdup| t=0 = F in?t - F out?t
Now,if we divide both sides of the mass balance equation (Eqn,I-1) by?t and take the limit as
t -> 0,we get the differential form of the mass balance,to wit,
(I-2) d[ Sz ρ]/ dt = F in - F out
If we assume that is constant (a reasonable assumption if the temperature is also reasonably
constant),then our mass balance equation becomes
(I-3) dz/ dt = (F in - F out )/S ρ
For our problem F in = 0 and F out = 12,480 lb/hr,both constant,We can calculate dz/ dt,that is
dz/ dt = (0 - 12480)/(10)(62.4) = -20 ft/hr.
Since the nominal level is 5.0 ft (half the tank height of 10ft),it will take 0.25 hour or 15 minutes
for the tank to run dry.
Further examples of the application of the principle of conservation of mass,particularly
for reacting systems,will be found in the subsequent sections of these notes,
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II,PROCESSES
Revised October 2,1999
A,The Concept of a Process
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Processes are the main concern of chemical engineers,A process is a system that converts
feedstocks of lower intrinsic value to products of higher value,For instance,the block diagram of
a pro cess to manufacture a mmonia from natural gas is shown in F i g ure II-1,This process has
several sections,each one of which car ries out a specific task and is,in effect,a mini-process,
Natural gas,steam and air are fed to the Reformer Section that converts these feeds into a mixture
of H2,CO,
Fig,II-1,Block Diagram for Ammonia Process
CO2,N2,and H2O,The overall chemical reactions involved are:
CH4 + H2O à CO + 3 H2
CH4 + O2 à CO + 2 H2
CH4 + 2 O2 à CO2 + 2 H2O.
Since H2 is the desired raw material from which to make ammonia,this gas mixture is sent
to the CO Shift Section where addition al steam is added to improve conversion by the water gas
shift reaction:
CH4
Steam
Air Steam
Methane
Reformer
Water Gas
Shift
Converter
CO2
Removal
CO2
H2O
Argon
Purge
Methanation
CO
CO2
H2
H2O
N2
CH4
Ar
CO
H2
N2
CH4
Ar
SYN GAS
H2
N2
CH4
ArNH3Product Ammonia
Synthesis
Loop
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CO + H2O à H2 + CO2.
The CO2 and H2O present in the gas mixture leaving the CO Shift Section are removed in
the CO2 Removal Section,The gas mixture leaving the CO2 Removal Section contains primarily
a 3/1 mixture of H2 and N2 (the N2 coming from the air fed to the Reformer Section),It also
contains small amounts of CO (the water gas shift reaction is equilibrium limited) as well as argon
from the air feed to the Reformer Section,
The CO must be removed from this mixture because it will deacti vate the cata lyst used in
the ammonia converter,This is done in the Methan ation Section vi a the reaction
CO + 3 H2 à CH4 + H20.
The gas mixture leaving the Methanation Section contains a 3/1 mixture of H2 and N2 and
trace amounts of CH4 and Ar,This is sent to the NH3 Synthesis Loop where the ammonia is
made via the well-known reaction
N2 + 3 H2 à 2 NH3.
So far each section of the process has been considered as a black box,Let us look at the
Ammonia Synthesis Loop in more detail,A process flow diagram (PFD) is shown in Fig,II-2.
The ammonia synthesis reaction is equilibrium limited to the point where it must be run at
rather high pres sures (3000 PSIA or 20 mPa) in order to achieve a reasonable conversion of Syn
Gas to ammo nia across the Convert er,Since the front end of the pro cess is best run at much
lower pres sures,a Feed Compres sor is needed to compress the Syn Gas to the operating pressure
of the Synthesis Loop,
Due to the unfavorable reaction equilibrium,only part of the Syn Gas is converted to ammonia on
a single pass through the Converter,Since the unconverted Syn Gas is valuable,the majority of it
is recycled back to the Converter,Due to pres sure drop through the Synthesis Loop equipment,a
Recycle Com pressor is required to make up this pressure drop,The recycled Syn Gas is mixed
with fresh Syn Gas to provide the feed to the Synthesis Converter.
The effluent from the Synthesis Converter contains product ammonia as well as unreacted
Syn Gas,These must be separated,At the pressure of the synthesis loop ammonia can be con-
densed at reasonable temperatures,This is done in the Ammonia Con denser where the Syn Gas is
cooled to approximately ambient temperature using cooling water (CW),The liquid ammonia is
then
- 15 -
Fig,II-2,Ammonia Synthesis Loop Process Flow Diagram
,
separated from the cycle gas in the Ammonia Knockout (KO) Drum,The liquid ammonia is
removed from the bottom of the drum and the cycle gas leaves the top,
Some of the cycle gas must be purged from the Synthesis Loop,Otherwise,the argon that
enters the loop in the Syn Gas has no way to leave and will build up in concentration,This will
reduce the rate of the ammonia synthesis reac tion to an unacceptable level,To prevent this from
happening,a small amount of the cycle gas must be purged,the amount being deter mined by the
amount of argon in the feed and its acceptable level in the Synthesis Converter feed (generally
about 10 mol %).
This description of the Ammonia Synthesis Loop covers only the most important aspects.
Modern ammonia plants are much more complex due to attention paid to maximizing the amount
of ammo nia produced per mol of Syn Gas fed to the loop,However,the process shown in Figure
II-2 is typical of many chemical processes.
C-1
Feed
Compressor
M-1
Feed
Mixer
R-1
Synthesis
Converter
E-1
Ammonia
Condenser
F-1
Ammonia
KO Drum
P-1
Argon
Purge
C-2
Recycle
Compressor
ST1
Syn
Gas
ST2
ST3
ST4 ST5
ST6
ST7
Liquid
Ammonia
ST8
Purge
ST10
ST9
- 16 -
B,Ba sic Pr ocessing Functions
Several basic processing activities are required by the Ammonia Synthesis Loop in order
to convert the hydrogen and nitrogen to ammonia product,These activities are common to
almost all chemical processes; the functionality of each can be con sidered independently of any
specific process,
There are five of these processing activities that are of major interest in chemical
engineering:
(1 ) Chemical Reaction
(2 ) Mixing
(3 ) Separation
(4) Materials Transfer (Fluid flow)
(5) Energy (Heat) Transfer
Of these the first three are involved in the process material balance,The last two are necessary
adjuncts to operation of chemical processes,Material must be transported from one piece of
equipment to another,For the large number of chemical plants processing only liquids and
vapors,this involves fluid flow,Also,streams must be heated to or cooled to specified
temperatures as dictated by the needs of the process,For example,reactors in general are
operated at temperatures higher than those that are acceptable for most separation operations,
Thus,streams must be heated to reaction temperature and then cooled back down for subsequent
processing.
1,Chemical Reaction
What distinguishes the chemical process industries from almost all others is the use of
chemical reactions to convert less valuable raw materials to more valuable products,In other
words,chemical reaction is the heart and soul of almost all pro cesses.
In the Ammonia Synthesis Loop example,R-1,the Synthesis Converter is where the
ammonia synthesis reaction takes place,
2,Mixing
Many chemical reactions involve two or more reactants,In order for these reactants to
react,these must be brought into contact at the molecular level,i.e.,mixed,before the desired
reactions can proceed properly,
Mixing is also required if several substances are to be blend ed to create a product mixture
with the desired properties.
- 17 -
For our example process,mixing of fresh synthesis gas and recyc led syn the sis gas takes
place in the Feed Mixer,M-1,before being sent to the Synthesis Converter.
3,Separation
In an ideal chemical process,exactly the right amounts of reactants would be mixed and
reacted completely to the desired product,Unfortunately,this is seldom the case,Many reac tions
cannot be carried to completion for various reasons,Seldom do the reactants react only to the
desired product,Unwanted byproducts are formed in addition to the target prod uct,Finally,the
reactants are seldom 100% pure,again for many reasons,
The result is that the material leaving the reactor is a mixture containing the desired
product,by-prod ucts,unreacted raw materials,and impu ri ties,It may also contain other compo-
nents delib er ately intro duced for one reason or another,The use of a homo geneous catalyst is
just one example of this.
This mixture must separated into its various constituents,The product must be separated
from almost everything else and brought to an acceptable level of purity such that it can be sold,
The reactants,being valuable,must be recovered and recycled back to the reactor,The impurities
and by-products must be separated out for disposal in a suitable manner,
Thus,the activity of separation is fundamental to the operation of almost any process,
Some separation systems are relatively simple,Others constitute the major part of the process,
As will be seen,separation takes on many forms.
Separation of ammonia from cycle gas takes place in the Ammonia KO Drum,F-1,The
cycle gas leaving the Ammonia Con denser,E- 1,contains droplets of liquid ammonia,This mix-
ture enters the middle of the KO Drum,The vapor,being less dense,flows upward while the
liquid ammonia falls to the bottom of the drum.
4,Materials Transfer
The Ammonia Synthesis Loop (Figure II-2) consists of several items of equipment,each
of which has material flowing in and material flowing out,These flows take places though
process piping connecting the various items of equipment,If the pres sure in the up stream item of
equip ment is sufficient ly higher than that in t he down stream item,then material will flow from the
upstream equipment to that downstream without the need for any additional equipment,This
pressure difference is necessary to overcome the friction due to fluid flow through the piping,
As a result,the pressure will decrease in the direction of flow through the process,
Thus,as synthesis gas flows from the inlet to the reactor (Stream 3) through the reactor,
condenser and knockout drum,the pressure decreases significantly,In order to be able to recycle
the unreacted hydrogen and nitrogen back to the reactor,some means is required to increase the
- 18 -
pressure of this stream back to that of the reactor inlet,Hence the presence of the compressor in
the recycle loop.
C,Unit Operations
One of the major contributions to the practice of chemical engineering is the concept of
the unit operations,This concept was developed by Arthur D,Little and Warren K,Lewis in the
early 1900's,Prior to its development,chemical engineering was,to a large extent,practiced
along the lines of specific process technologies,For instance,if distillation was re quired in the
manufacture of acetic acid,it became a problem in acetic acid distillation,The fact that a similar
distillation might be required for the manufacture of,say,acetaldehyde was largely ignored,What
Little and Lewis did was to show that the principles of distillation (as well as many other
processing operations) were the same regardless of the materials being processed,So,if one
knew how to design distillation columns,one could do so for acetic acid,acetaldehyde,or any
other mixture of reasonable volatility with equal facility,The same proved to be true of other
operations such as heat transfer by two-fluid heat exchangers,gas compression,liquid pumping,
gas absorption,liquid-liquid extraction,fluid mixing,and many other operations common to the
chemical industry,
The case for chemical reactors is less clear,Each reac tion system tends to be somewhat
unique in terms of its reac tion conditions (temperature,pressure,type of catalyst,feed compo-
sition,residence time,heat effects,and equilibrium limita tions),Thus,each reaction system must
be approached on its own merits with regard to the choice of reactor type and design,However,
the field of chemical reaction engineering has under gone substantial development over the past
few decades,the result being that much of what is required for the choice and design of reactors is
subject to a rational and quantitative approach.
D,Modes of Process Opera tion
As mentioned previously,there are several modes of process operation,The one that has
been most widely studied by chem ical engineers is that of the steady-state operation of contin uous
processes,The reasons have also been discussed,In this mode of operation we assume that the
process is subject to such good control that as feed materials flow into the process as constant
flow rates,the necessary reactions,separations,and other operations all take place at conditions
that do not vary with time,The amount of material in each item of equipment (its inventory) does
not vary with time; pressures and liquid levels are constant,Nor do temperatures,compositions,
and flow rates at each point in the process vary with time,From the standpoint of the casual
observer,nothing is happening,However,finished product is flowing out of the other end of the
process into the product storage tanks.
The next most common mode of process operation is known as batch operation,Here,
every processing operation is carried out in a discrete step,Reactants are pumped into a reactor,
- 19 -
mixed,and heated up to reaction temperature,After a suitable length of time,the reactor turned
off by cooling it down,It now contains a mixture of products,by-products,and unreacted
reactants,These are pumped out of the reactor to the first of the various separation steps,
possibly a batch distillation or a filtration if one of the products or byproducts is a solid,Each
batch step has a beginning,a time duration,and an end,(Most activities around the home are
batch in nature - cooking,washing clothes,etc.),
A third mode is cyclic operation,From the standpoint of flows in and out of both the
process and individual items of equipment,operation is continuous,However,in one or more
items of equipment,operating conditions vary in time in a cyclical manner,A typical example is
reactor whose catalyst deactivates fairly rapidly with time due to,say,coke formation on the
catalyst,To recover the catalyst activity,it must be regenerated by being taken out of reaction
operation,The coke is removed either by stripping by blowing an inert gas over the catalyst or,in
the more difficult cases,by burning the coke off with dilute oxygen in an inert gas carrier,
Two characteristics of cyclical operation become apparent,First off,if the process is to be
operated continuously but it one or more items of equipment must be taken off line for regen-
eration of one sort or another,we must have at least two of such items available in parallel,One
is on line while another is being regenerated,The second characteristic is the is that operation
conditions in the cyclically operated equipment must vary with time,If catalyst activity decrease
with time,then something must be done to maintain the productivity of the reactor,Usually this
is achieved by raising the reactor tem perature,Thus,a freshly regenerated reactor will start off at
a relatively low temperature; the temperature will be raised during the cycle; and the reactor will
be taken off line when no further benefit is to be obtained by raising the temperature any further,
(The catalyst may melt,for instance.)
- 20 -
III,PROCESS MATERIAL BALANCES
Revised October 10,1999
- 21 -
Material balances result from the application of the law of conservation of mass to
individual items of equip ment and to entire plants (or subsections thereof),When the mass
conserva tio n equations are combined with enough other equations (energy balances,equi librium
relationships,reaction ki netics,etc.) for an individual item of equipment (such as a reactor or a
distillation column),the result is a mathematical model of the performance of that equipment item.
The model can be dynamic or steady state,depending upon how it is formulated,
For the present let us confine our attention to steady-state models,Such equipment
models are generally nonlinear and must be solved by iterative procedures,usually with the help
of a computer,However,if we make enough simplifying assumptions,linear models will result,
While these will not be as accurate as the more rigorous nonlinear models,they are a good
starting point for many applications,one which is pursued in subsequent sections of these notes.
Individual models can be combined to represent the performance of an entire chemical
plant or sections thereof,For instance,one might start by modeling the performance of the
reaction section,When this is in hand,one can then add other parts of the plant such as the
separation and purification sections,As will be seen,if we limit ourselves to simple,linear
equipment models,then the overall process or flowsheet model will also be linear,This we can
solve using a spreadsheet for instance,Not only that,we can solve rather large flowsheet models
with a relatively modest amount of effort,This is one motivation for using linear models.
If the equipment models are nonlinear,we almost always require the use of a computer,
Indeed,there are special-purpose programs that have been developed just for solving process
flowsheet models,These programs are generally known as steady-state process simulators or
flowsheeting programs,
A,The Stream Summary
Before discussing equipment characterization in de tail,it is necessary to consider the
character iza tion of the streams entering and leaving each item of equipment,While there is no
unique way of doing this,the following character ization is typical,Each stream is represented by
a vector of various quanti ties,If there are nc components of signifi cance in the stream,then the
first nc entries in the vector will be
f i,n - the molar flow rate of the component i in stream n.
Additional quantities that may be included in this vector are:
F n - the total molar flow rate of stream n,
T n - the temperature of stream n,
- 22 -
P n - the pressure of stream n,
R n - V/F,(the mols of stream n that are vapor)/(the total mols of stream n) [V/F = 0,
stream is liquid; V/F = 1,stream is vapor; 0 V/F 1,stream is a two-phase mixture of liquid
and vapor]
n - mass density of stream n,
Mw n - average molecular weight of stream n,
W n - the total mass flow rate of stream n,
H n - the total enthalpy of stream n,
S n - the total entropy of stream n,and
G n - the total free energy of stream n,
Not all of these quantities have to be included in the stream vector for all problems,Only the f i,n
are needed for linear material bal ance calculations,However,most process simulati on pro grams
include all of the variables listed above in the stream char ac ter iza tion vector as well as mol
fractions and weight fractions of the individual components,Since the total mass balance around
each item of equipment is easy to check,we will include W n in the stream vector (and will need
Mw n to calculate it from F n ).
A typical summary is shown in the Table III-1 below,It is for the Ammonia Synthesis
Loop whose PID was presented in Fig,II-1 of the previous chapter,It was computed using the
EXCEL spreadsheet program based on the linear material solution technique developed in
Chapter VII,The details of the spreadsheet solution are given in Appendix D.
B,Equipment Characteriza tion
1,Reactors
The reactor is the heart of almost any chemical process,A sim ple re ac tor is shown
schematically in Fig,III-1,The input or feed stream STin con tains the re actants along any inerts,
feed impuri ties,and dil uents (all of which have usu al ly been pre mixed in a mix er),One or mo re
chemi cal reac tions take place in the reac tor and the re ac tion pro ducts,any remain ing unreac ted
re actants,and the in er ts,dilu ents,and feed impu rities leave in the reactor output or efflu ent
stream ST out,
Stream Summary
Comp ST1 ST3 ST4 ST6 ST7 ST8 ST9
- 23 -
f out,if in i
f fout i in i i= +,?
STin STout
Lb-mol/hr
H2 750.00 2894.03 2193.37 2191.18 2.19 47.15 2144.03
N2 250.00 934.20 700.65 699.25 1.40 15.05 684.20
Argon 10.00 425.95 425.95 425.10 0.85 9.15 415.95
Ammonia 0.00 4.62 471.72 4.72 467.00 0.10 4.62
Total mol/hr 1010.00 4258.79 3791.69 3320.25 471.45 71.45 3248.79
Mol percent
H2 74.26% 67.95% 57.85% 65.99% 0.47% 65.99% 65.99%
N2 24.75% 21.94% 18.48% 21.06% 0.30% 21.06% 21.06%
Argon 0.99% 10.00% 11.23% 12.80% 0.18% 12.80% 12.80%
Ammonia 0.00% 0.11% 12.44% 0.14% 99.06% 0.14% 0.14%
Total 100.00% 100.00% 100.00% 100.00% 100.00% 100.00% 100.00%
Average MW 8.83 11.53 12.95 12.37 17.04 12.37 12.37
Total lb/hr 8916.4 49101.78 49101.78 41069.19 8032.59 883.81 40185.38
Table III-1,Stream Summary for Ammonia Synthesis Loop
There are many ways of charac terizing reactors,In general,what we want to know is the
extent of reac tion?i
for each component i.
We de fine the extent
of reaction of compo-
nent i as the mols of i
that are made or used
up in the reactor,For
a con tinuous steady-
state reaction,the
extent of reaction is
really a rate,namely,
the mols of i formed
or consumed per unit
time,
Figure III-1,Reactor
The extent of reaction can be determined in var i ous ways,Regardless o f how determined,
the?i must satisfy the reaction stoichiometry,Let us consider a simple example,Let us sup pose
that methane (CH4) is being oxidized using air to a mixture of CO and CO2 according to the
following chemistry:
1) CH4 + 3/2 O2 ---> CO + 2 H2O
2) CH4 + 2 O2 ---> CO2 + 2 H2O
- 24 -
This reaction system involves two reactions,There are five components involved the reactions as
well as one inert - the nitrogen that comes with the air,(Let us assume that for present purposes,
the small amount of argon also present in the air can be lumped with the nitrogen.)
Let r j = the rate of the jth reaction in,say,lb-mol/hr and
a i,j = the mols of component i that are formed or consumed by one mol of reaction j,a i,j is
known as the stoic hiometric coefficient of component i with respect to reaction j.
Then if there are nr reactions in the system,
For the CH4 oxidation reactions,the stoichiometric coefficient matrix is:
Stoichiometric Coefficient for:
Reaction CH4 O2 CO CO2 H2O N2
1 -1 -3/ 2 1 0 2 0
2 -1 -2 0 1 2 0
Note that the stoichiometric coefficients for reactants are negative ; those for products,positive.
Let us suppose that r 1 = 20 lb-mol/hr and r 2 = 10 lb-mol/hr.
Then?CH4 = (-1)(20) + (-1)(10) = -30 lb-mol/hr
O2 = (-3/2)(20) + (-2)(10) = -50 lb-mol/hr
CO = (1)(20) + (0)(10) = 20 lb-mol/hr
CO2 = (0)(20) + (1)(10) = 10 lb-mol/hr
H2O = (2)(20) + (2)(10) = 60 lb-mol/hr
N2 = (0)(20) + (0)(10 ) = 0 lb-mol/hr
Note,For a reactor to operate at these reaction rates,the feed will have to contain at least 30 lb-
mol/hr of CH4 and 50 lb-mol/hr of O2,If this is not the case,the reactant that is in short supply
is termed the limiting reactant,For instance,suppose that the feed contains 60 lb-mol/hr of O2
but only 15 lb-mol/hr of CH4,CH4 will then the limiting reactant,The best we can expect to do
is 50% of the assumed reaction rates.
( ),III a ri i j j
j
nr
=
=
∑1
1
- 25 -
Often,the overall reactor performance is characterized in terms of conversion and
selectivity,We pick a key component,usually either the more valuable reactant or the limiting
reactant.
We define the conversion C k with respect to the key component k as follows:
C k = (Mols of key component converted by all reactions)
(Total mols of key component in the reactor feed)
For our reaction system above,suppose we feed 50 mols/hr of CH4 to the reactor,The
conversion of CH4 is then
C CH4 = 30/50 = 0.6 or 60%
We define the selectivity S k,j of the key component with respect to the jth reaction as
follows:
S k,j = (Mols of key component converted by reaction j)
(Mols of key component converted by all reactions)
For our reaction system above,the selectivity of CH4 to CO (Reaction 1) is
S CH4,1 = 20/30 = 0.667 or 66.7%,
and the selectivity to CO2 (Reaction 2) is
S CH4,2 = 10/30 = 0.333 or 33.3%.
Note that the selectivities over all reactions must sum to unity,i.e.,
One should be aware that there are other definitions of selectivity that are used in the literature,
However,the one given above is the most commonly used and the only one that has the property
expressed in Eqn,III-2.
Let f k,in = the feed rate of the key component to the reactor.
Then,
(III-3) r j = C k S k,j f k,in,j = 1,...,nr.
( ),III S j k
j
nr
=
=
∑2 1
1
- 26 -
f in i,
f s f
f s f
out i i in i
out i i in i
,,
,,( )
1 =
=?2 1
f out i,1
f out i,2
STin
STout1
STout2
The extents of reaction i can be calculated using Eqn,II-1.
2,Separators
to any chemical process,A simple separator is shown in Figure III-2,It has one in put
stream,ST in,and two output streams,ST 1out and ST 2out,This separa tor can be used to rep resent
flash drums,simple distil la tion col umns and other separa tors that do not re quire a mass separating
agent.
More complex separators are
shown in Fi g ure III-3,Sepa rator A is
typical of gas ab sorbers and liq uid-liq-
uid ex tra c tors where ST 1in is the input
stream which is to be sepa rat ed and
ST 2in is the lean mass sepa rating agent.
ST 1out cor re sponds to ST 1in from
which the compo nents of inter est have
been removed,ST 2out is a mass sep a-
rating agent en ri ch ed with the
components of inter est th at were to be
sepa rated from ST 1in,
Sep ara tor B is typical of a
com plex distil lation column with side
stream (ST 2out ) as well as the distil late
or ove rhead (ST 2out ) and the bot toms
(ST 3out ),
There are two indices of how
well a separator does its job,The first
is the frac tion of a given component
that is recov ered from a sp ec ified feed
stream,The sec ond is the purity of one Figure III-2,Simple Separator
or more output streams from the separa tor,
The frac tional recovery is im portant from an econom ic point of view,Purity spe cifications have
to be met in or der to sat isfy the feed pu rity re qui re men ts of dow n-stream equ ip ment or,if the
output stream is a final product,the sales purity spe cifi cations,
- 27 -
f out,i
f in i,1
f in i,2
f in i,3
f in i,nin
f out i f in i,j
j
nin
= ∑
= 1
STin1
STin2
STin3
STnin
STout
Fig,III-3,Complex Separators
Simple separators can be charac terized at the simplest level in terms of a separation coeffi cient.
This is defined in Chapter
IV,Evaluation of
equipment model
parameters such as the
separation coefficient is
covered in Appendix B,
Complex separators can be
characterized using the
basic material balance
models as is described in
Appendix C.
3,Mixers
A mixer is a device that
brings together two or more
input streams of different
compositions and pro duces an Fig,III-4,Mixer
A
STout1
STin1
STin1
STin2
STin2
STout1
STout2
STout3STout2
B
- 28 -
f out i,1
f in i,
( )
f P f
f P f
out i in i
out i in i
,,
,,
1
2 1
=
=?
f out i,2
STin
STout1
STout2
output stream that is a uniform mix ture of the in put streams,A typical mixer is shown in Figure
III-4,As will be seen in the next chap ter,cha rac teri zation of mix ers from the sta nd point of con-
ser va tion of mass is quite straig ht for ward.
4,Flow Splitters
A flow splitter divides a
stream into two or more
output streams,each having
the composition of the feed
stream,The sum of all of the
stream flow rates leaving the
splitter must equal the flow
rate of the feed stream,A
simple flow splitter is shown
in Figure III-5.
Figure III-5,Flow Splitter
- 29 -
STin1
STin2
STout1
STout2
P(arameters)
IV,STEADY-STATE PROCESS MODELING
Revised October 10,1999
A,Unit Models -- General Considerations
Material balance calculations begin with the charac teriza tion of the individual unit
operations by math e mat ical models,These are known as unit models or process blocks.
Consider the generalized pr o cess
block shown in Figure IV-1,Here,in1
and in2 are the numbers or names of the
input streams to the process block and
out1 and out2 are the numbers or names
of the output streams,A block can have
more or less than the two input streams
shown; the same is true of output
streams.
Also,the quantities P 1,P 2,...,P m
are a vector of parameters re quired to
characterize the process block,For
instance,if the block represents an
isothermal flash,two parameters,namely
the flash pressure and tempera ture,would
have to be specified as part of the Figure IV-1,Generalized Unit Equipment Model
characterization of the block.
A process unit operation block can now be characterized as follows:
Let X n = the stream characterization vector for stream n.
Then
(IV-1a) X out1 = G J,1 [X in1,X in2,...,X inn ; P 1,P 2,...,P m ]
(IV- 1b) X
out2 = G J,2 [X in1,X in2,...,X inn ; P 1,P 2,...,P m ]
,
,
,
(IV-1c) X outn = G J,n [X in1,X in2,...,X inn ; P 1,P 2,...,P m ]
- 30 -
In general,these models are nonlinear and difficult to solve without using a computer.
A,Linear Input-Output Models
For many purposes such as preliminary material balance calculations for scoping out a
d e sign,simple linear unit models are perfectly adequate,These allow the material balance to be
done by manual calculations (or by a spreadsheet program),This direct participation in the
calculations at the early stages is recommended since in general the engineer will develop more
insight into the workings of the process than if the calcula tions are done at one remove b y a
process simulation program.
Four simple models are all that are needed in order to do linear material balances,
Actually only two are needed,The remaining two are special cases of the others.
1,Mixer (MIX)
In this model two or more streams are added together to produce a single output stream
that is a mixture of all the input streams (See Figure III-4),
A mass balance on the ith component gives the equation describing the mixer,i.e.,
Note that no parameters are required to characterize a mixer,This is not the case for any of the
other equipment models,
2,Reactor (REACT)
A Reactor takes a feed stream,and by chemical reactions,con verts some into other
comp o nents (See Figure III-1).
A mass balance on the ith component again gives the performance equation for the reactor,i.e.,
(IV-3) f i,out = f i,in + i
where i = the extent of reaction of component i,i.e.,the net number of mols of component i
pr o duced by reaction.
Note,i will be negative for reactants,positive for prod ucts.
( ),,IV f out i f in i j
j
nin
=
=
∑2
1
- 31 -
Note also that REACT is a special case of MIX where i can be considered the flow rate of
comp o nent i in the second input stream to MIX,
The procedure for evaluating the?i given conve rsion and selectivity information is
discussed in Section II,Or the?i may be given as constant values based on a specified production
rate and the reaction stoichiometry,If a mathemati cal model for the reaction kinetics is available
and the reactor type has been chosen,the?i can be estimated from a separate reactor calculation.
3,Separator (SEPAR)
A Separator is used to model process units in which each component in the feed is
separated into two output streams (See Figure III-2).
The mass balance equations describing the performance of the Separator for the ith component
are:
(IV-4a) f i,out1 = s i f i,in
(IV-4b) f i,out2 = (1 - s i ) f i,in
where s i = the separation coefficient for component i,
Note that conservation of mass dictates that 0 s i 1.
As with the i for REACT,the s i for SEPAR must be known or estimat ed by other
means,For conceptual design material bal anc es for which the separation equipment generally has
not yet been se lected,let alone designed,one will generally assume reasonable values,say 99%
recovery of one key component in the overhead (ST 1out ) and a similar recovery of the other key
compo nent in the bottoms (ST 2out ),
Complex separators,such as those shown in Figure III-3,can be modeled by a
combination of the models for MIX and SEPAR,
4,Flow Splitter (SPLIT)
In many processes,a stream is split into two smaller streams,each having the same composition
as the input stream (See Figure III-5).
The governing equation for the Stream Splitter is an overall or total mass balance,i.e.,
- 32 -
(IV-5) F out1 = S F in
The same relationship holds for each component,So,
(IV-6a) f i,out1 = S f i,in and
(IV-6b) f i,out2 = (1 - S) f i,in
Note that SPLIT is a special case of SEPAR for which s i = S for all i,Also,0 S 1,A flow
splitter having more than two output streams can be written along similar lines,
The major difference between SEPAR and SPLIT is that the s i are dic tat ed by physi cal
consider a tions such as relative volatili ties and how the equipme nt is operated,However,S is can
be assigned any value between 0 and 1.
B,Rigorous Models
The models described in the previous have two advantages,They are both simple and
linear,Thus they are well suited to calculating the material balances for entire processes,Howev-
er,beyond conserving mass,these models are not very realistic,Their model parameters must be
known from other sources,This is discussed in Appendix B.
This presents a problem in doing accurate flowsheet materi al balances,In order to get the
model parameters for the linear material balance calculations,we must do separate rigorous
calcul a tions for the individual items of equipment,To do this we must know the feed streams to
each item of equip ment for which we must d o the flowsheet material balances,Thus we have a
situation in which we must assume values of all of the parameters,do the flowsheet material
balances using the linear models,then evaluate the parameters from the rigorous models for the
individual items of equipment,It the assumed and calculated values of the parameters do no
agree within some reaso n able tolerance,we must repeat the procedure,Indeed,we may have to
do it many times,Clearly,there should be a better way.
The better way is to use rigorous equipment models directly in the flowsheet material
balance calculations,Since most rigorous models are nonlinear and difficult to solve in their own,
this approach is not amenable to hand calculations,It is possible with a spreadsheet but
considerable effort,skill,and time are required,
The way that has been adopted for doing rigorous flowsheet calculations is to use a
steady-state flowsheet simulation program,Solution of the flowsheet materials using rigorous
equipment models is discussed in the next section,A brief introduction to commercially available
programs is given in Section X.
V,STEADY-STATE MATERIAL BALANCE CALCULATIONS
- 33 -
Revised October 10,1999
There are a number of techniques that have been developed for the solution of steady-state
flowsheet performance equations,In general,this is an exercise in the numerical solution of a set
of algebraic (generally nonlinear) equations for which there are many algorithms and computer
codes available,One approach is to write out all the equations,specify enough parameters so that
the number of unknown variables equals the number of equations,and use the equation solver of
choice,Indeed,this is the approach taken in many of the texts on material and energy balances,
There is nothing wrong with it,However,it tends to obscure the underlying physical significance
of the problem,particularly where recycles are involved,Instead we will look at some of the
techniques that have been developed specifically for solving the flowsheeting problem.
A,Sequential Modular
One technique for solving the material balances for an entire process (although not the
only one) is called Sequential Modular,In this technique the material balances for a entire process
are solved one module (process block) at a time,
Let us first consider the process shown schematically in Figure V-1,There are three
process operations or process blocks,The exact nature of each is not important at the moment,
How ever,it is assumed that block equati ons ( Eqns,IV-1a,IV-1b,...,IV-1c) can be solved for
each block in the process,In other w ords,if all of the input stream vectors and the parameter
vector are known,then the output stream vectors for the block can be computed via a well
defined procedure,(For the linear models of Eqns,IV-2 thro ugh IV-6,the computations are
simple and direct,For the nonlinear models used in more realistic block characterizations,the
com putational procedure may require a trial-and-error or iterative algorithm.)
So,the ground
rule for direct sequential
modular material
balance calculations
(a.k.a,pro cess
simulation),the out put
streams can be calcu-
lated if the input streams
and the block
parameters Figure V-1,Sequential Process Flow Diagram
are known,
- 34 -
Thus,in the three-block process shown,one can start by calculating Block A to determine
Streams 3 and 4,Then Block B can be calculated to give Streams 5 and 6,Finally Block C can
be calculated to give Streams 7 and 8,
Now let us
consider a second
process,one with recycle,
as shown in Figure V-2,
If the calculations are to
be started at Block A,
there is a prob lem,
Stream 2 is not known
since it is an output
stream from Block B that
has not been calculated
yet,Suppose we decide to start with Block B instead,Again there is a problem; Streams 3 and 4
are unknown,A similar prob - Figure V-2,Recycle Process Flow Diagram
lem occurs if we attempt to
start with Block C,
So how can we possibly calculate this process? Only by agreeing to guess values for the
unknown input streams to each block in the process,Suppose now we start with Block A,
Stream 2 must be guessed,To calculate Block B another stream,Stream 4,must be guessed,
Block 5 can be calculated without having to guess any further streams.
Once all the blocks have been calculated,we will have computed values for all the streams
we originally guessed,If the computed values agree with the guessed values within some
acceptable tolerance,then the material balance for the process has been solved,If not,the whole
procedure has to be repeated with new guesses,One strategy is to use the previously comput ed
values for the next round of guesses.
This technique for solving the overall f low sheet material balance pr ob lem is known as the
sequential modular ap proach,We calculate each unit operations block or mod ule in the process
sequence,providing initial guesses of unknown recycle streams where necessary,New values of
the guessed (or tear) streams are produced as a result of each pass through the process sequence
calculations,Methods for doing so are discussed in Section VII of these notes,When the
differences between successive guessed values become sufficiently small,the procedure is
considered to be converged,The conditions under which it converges are Section VIII,We will
see that because of the conservation of mass,convergence is guaranteed for linear flowsheet
models and strongly favored for nonlinear models.
The large majority of commercially available steady-state process simulators use the
sequential modular approach.
- 35 -
B,Simultaneous
The sequential modular approach can generally be made to converge,even for difficult
problems,However,it tends to be inefficient,Made of the unit operations models have internal
iterative procedures just to solve for their output streams as functions of their input streams and
operating parameters,Embed these iterative calculations within the sequential modular flowsheet
calculations and one has a massive loops-within-loops calculation with the potential to be very
inefficient,So,why not solve all of the equations simultaneously?
As we will see,this procedure is straightforward if all of the equations are linear,If they
are not,an appropriate algorithm such as Newton- Raphson must be employed to solve large sets
of nonlinear equations,This approach has been taken in several simulators developed in
academia,SPEEDUP,developed at Imperial College,and ASCEND,developed at Carnegie-
Mellon,are two of the more advanced of this type of simulator,SPEEDUP is currently available
commercially through Aspen Technology,Inc.
It is not the intent of these notes to go into the pros and cons of the simultaneous
approach versus the sequential modular,Suffice it to say that neither approach is without its
drawbacks and difficulties,We note,however,that the sequential modular approach has been
much widely used in commercial simulators than the simultaneous,And it tends to be fairly
robust for reasons that are discussed in Section VIII.
C,Design Specifications
The calculations discussed so far are,in process control terminology,for processes
operated in open loop,All the input streams and block model parameters are specified and all of
the block output streams are then calculated,We have no way of knowing beforehand what the
values of the output stream vectors will be,
Usually,the specification of the performance that the process must achieve will involve
selected variables in the output stream vectors,The production rate,i.e.,the flow rate of the
product stream must meet the capacity specification,This stream must also meet the product
purity specifications,The composition of streams being discharged to the environmental must
meet emission specifications,Many other specifications such as reactor temperature and pH,flash
drum vapor flow rates,and fractional recoveries in separators must be met,This can only be done
by adjusting feed stream flow rates and block model operating parameters,In some process
simulators,these are known as design specifications,In fact,they have the form of feedback
control loops,
There are many techniques for incorporating design specifications into the flowsheet
calculations,All commercially available flowsheet simulators provide the means to do so,In
sequential modular simulators,the design specifications are handled as control loops around the
open loop simulation,which adjust selected parameters to satisfy the design specifications,One
- 36 -
of the advantages of simulators using the simultaneous approach is that the design specifications
can be added directly to the other equations being solved,
D,Optimization
Quite often one is not just interested in a solution to a flowsheet material balance problem
but one that is best in some sense,usually economic,Finding such a solution by adjusting selected
input and block model parameters is known as optimization,Most commercially available
simulators have optimization capability,Many spreadsheets also have some optimization
capability such as a linear programming solver,
Optimization is beyond the scope of these notes,We will,however,look at a specific type
of optimization that arises in blending problems,The optimization technique we will use is linear
programming.
E,Ad Hoc Methods
We term the computational method employed as ad hoc when one attempts to solve a
flowsheet material balance by starting with design specifications and working backwards through
the various block models and even the entire process,For flowsheets of any complexity,this a
difficult approach since flowsheet calculations tend to be quite stable and well-behaved if one
calculates block output streams from block input streams but notoriously unstable if one attempts
to do the reverse,
Another characteristic of ad hoc methods is the use of overall material balances around
two or more items of equipment,either on a component or total flow rate basis,The use of
overall material balances is to be avoided,Use component material balances around individual
items of equipment,This is the rule followed in Section IV of these notes.
For simple flowsheets,particularly when one is just interested in the input-output
structure,ad hoc methods can be used to estimate raw material requirements based on reactor
selectivities and separator efficiencies,The problem is that each material balance solution is a
special case,Change the form of a design specification and the entire procedure must be revised.
One is advised to use the linear material balance approach,at least initially,since it is
straightforward in its application to almost any problem,If the calculations are done using a
spreadsheet,then adjusting inputs and parameters to meet design specifications can be done by
trial-and-error if there are not too many,The inclusion of design specs is discussed in Section
VII-C.
VI,RECYCLE STREAMS AND TEAR SETS
- 37 -
Revised October 11,1999
In the previous section,the sequential modular method of solving process material balance
was described,the first step of which was to choose a set of streams which,if their values are
known,allow us to calculate the process material balance unit by unit,It is presumed that for
each unit the input streams and operating parameters are known and that the unit calculations
involve calculating the output streams from that unit,If the process contains no recycle streams,
then the choice is obvious,namely,the input streams to the process that originate from outside
the process,These are often referred to as the feed streams to the process,For example,in Fig.
VI-1,there is only one such stream,namely,Stream 1.
M-1 M-2S-1 S-2 S-3
1 2
3
4 5
6
7 9
8
Figure VI-1,Process Flow Diagram for Example #1.
However,if the process contains recycle streams,the choice of a set of streams with
which to start the calculations is not so simple,Just knowing the feed streams is not sufficient,
Referring again to Fig,VI-1,suppose we want to start the unit calculation sequence with the
mixer M-1,Stream 1 is a feed stream and is presumed known,But we also need to know
Streams 3 and 8,But Stream 3 is an output stream from the separator S-1 while Stream 8 is an
output of S-3,At the outset these are not known,What to do? We could,as previously
described,guess values for these two streams,Then we could calculate M-1,getting a value for
Stream 2 as a result,This would then allow us to calculate S-1 getting Streams 3 and 4 as a
result,Next we would like to calculate M-2 but we do not know Stream 6,Proceeding as we did
for M-1,we could guess Stream 6 and thereby calculate S-2 and S-3 in turn as well as the
remaining streams for the process,Does this complete out material balance calculations? The
answer probably is no,Unless we were very clever (or very lucky),the values we guessed for
Streams 3,8 and 6 will differ from those we subsequently calculated,If any of these differ by
more than an acceptable amount,we will have to guess new values of these streams and repeat
the calculations,This procedure will have to be repeated as many times as is necessary to achieve
an acceptable agreement between the guessed and calculated values of the three streams.
But this is not the subject of this section of these notes,Our topic is more limited,namely,
how to chose the a set of streams and the associated sequence of unit models to begin the
- 38 -
calculations,Now we could do this by just starting at the upstream end of the process and
guessing as many streams as we need to get the calculations started,But,as we will see,this may
require guessing more streams than are absolutely necessary which,in turn,may cause the
calculations to be more difficult or inefficient than would otherwise be the case,So,what is
presented in this section is a simple technique for determining all sets of streams that minimize the
number to be guessed,Further,this technique will show us what the recycle structure of the
process is,something that is important not only for the calculations but also for the behavior and
operability of the process.
The procedure of choosing streams to guess is called tearing or cutting,The process flow
diagram can be considered as a directed graph where each process block is a node and the nodes
are con necte d by streams,each of which has direction from one node toward another,The
terminology "cutting" or "tearing" comes from graph theory (See Mah [1990] for more details).
The question is,for a given process graph or diagram,what and how many streams have
to be cut or torn so that the process can be calcu lated sequentially? Each set of streams which
allow this to be done is called a cut or tear set,For each choice of sequence,there will be a tear
set,For some choices the tear set will be smaller than for others,Those tear sets which require
the fewest number of streams to be torn are called minimal tear sets.
There are several ways to determine the minimal tears sets for a process graph,The first is
by determining the tear set for each possible choice of sequence and then selecting the minimal
tear sets from among all the tear sets,This is what we did above,However,since the number of
sequences for N process units is N !,this method becomes imprac tical for processes having more
than a few process units.
A second method one which much less time-consuming,is based on the node-incidence
matrix,The method proceeds in four steps,The first step is to construct the matrix,The second
is to deter mine all of the closed cycles within the gra ph that the matrix repre s ents,The third is to
identify the minimal tear sets and the fourth is to list the computational sequences corresponding
to each minimal tear set.
A,Construction of the Node Incidence Matrix
The node-incidence matrix contains a row for each process block and a column for each
process stream that connects two process blocks,Feed streams to the process from the outside
and product streams to the outside are not included.
As an example,let us construct the node-incidence matrix for the process shown in Figure
VI-1,For each stream,enter a 0 in the row for the process block from which the stream comes,
Enter a 1 in the row for the process block at which the stream terminates,(Some authors use -1
instead of 0,Since no numerical computations are involved,the basis for choice is clarity.)
- 39 -
The node-incidence matrix for our example is:
Block Streams:
2 3 4 5 6 7 8
M-1 0 1 1
S-1 1 0 0
M-2 1 0 1
S-2 1 0 0
S-3 1 0
Note that Streams 1 and 9 have not been included in the node-incidence matrix,Since these two
streams either enter or leave the process flow diagram,they cannot be part of recycle loop,For a
stream to be a candidate for a recycle loop,it must both originate and terminate within the
process flow diagram.
B,Determination of all Closed Cycles
In the second step,each cyclical path through the graph is determined,First,determine all
the cycles formed by pairs of streams (2-loops),Start with the first stream,in this case Stream 2.
It originates at M-1,as indicated by the 0,and terminates at S-1,as indicated by the 1,Now,try
to find a stream originating at S-1 that terminates at M-1,In this case there is one,Stream 3,
Repeat the procedure for each stream,ignoring duplicate cycles,Two 2-loopss are found,namely
(2,3) and (5,6),
Next,repeat the process,this time looking for three-stream cycles ( 3-loops),There are
none,Proceed to look for four-stream cycles ( 4-loops),Again,there are none,Go on to look
for 5-loops,Here,there is one,namely (2,4,5,7,8),This terminates our search since there can be
no cycles having more than five streams since there are only five nodes (blocks) in the process
graph.
C,Identification of all Minimal Tear Sets
In the third step,examine all the cycles from Step B to identify those sets of streams that
qualify as tear sets,A tear set is a set of streams such that at least one member of this set is also a
member of at least one cycle (or loop),Those tear sets that contain the smallest number of
streams are the minimal tear sets.
For our example,there are three loops,namely,(2,3),(5,6) and (2,4,5,7,8),There are
many candidate tear sets,The set [2,5] is a tear set in that either Stream 2 or Stream 5 is a
member of all three loops,So are the sets [2,6] and [3,5],There are no more tear sets of only
two streams,It makes no sense to look for tear sets containing three streams (such as [3,6,7] for
instance) since we are only interested in minimal tears sets,So,the question remains are there any
tear sets containing only one stream? The answer is clearly no,So,as a result we have identified
three minimal tear sets.
- 40 -
D,Listing of Computational Sequences
If we are to calculate the process material balance based on guessed values for a given
minimal tear set,we must determine what computational sequences can be used,i.e.,what
sequences allow us to compute a given process block for which all of the input streams are
known,This means that each input stream must be (1) an external feed stream to the process,(2)
a tear stream for which a guessed value has been made,or (3) the output stream from a process
block that has already been calculated.
A computational sequence for the tear set [2,5] is [S-1,S- 2,S-3,M-1,M-2],There are
others,all of which are minor variations on this one,For instance we could calculate S-2 before
S-1,We must always calculate S-3 before we can calculate M-1 but we can calculate M-2 any
time after we have calculated S-1 and S-2,The question is whether or not one sequence is better
than the others,The answer in general is no unless the initial guess of one or more tear streams
happens to be zero,If the input streams to a process block are all zero,then the calculations for
that block are meaningless and must be bypassed until a later iteration has produced a non-zero
value for at least one input stream,
We can similarly determine the computational sequences for the other minimal tear sets,
The final question to be considered concerns the choice tear set,Is one better than another? If
we have some basis for making reasonable guesses of some recycle streams but not others,then
we should chose the tear for which we can make the best guesses,Otherwise,there is not much
of a basis for choosing one over another,
After all this,one might ask why do we need to construct a node-incidence matrix? Why
can’t we just identify the various cycles by inspection of the process flow diagram? The answer is
that,in principle,one can,However,for more complex processes involving more process blocks
and streams than the simple examples contained herein,determining all the cycles is more difficult
and it is easy to overlook some,The use of the node-incidence matrix makes the procedure
somewhat more straightforward,but does not guarantee a complete enumeration either,(There
are computer algorithms for doing this which,of course,can be expected to identify all the cycles
and the minimal tear sets.) But,for anything but the simplest processes,use of the node-
incidence matrix is recommended.
For most processes,there is a rule of thumb that will allow one to identify at least one
minimal tear set within going through the entire node-incidence procedure,Choose as a tear set
those streams that are the outputs of all mixers that have one or more recycle streams as inputs.
For Example #1,these would be Streams 2 and 5 which are one of the three minimal tear sets we
identified using the node-incidence method.
E,Further Examples
- 41 -
Let us look at two further examples,each of which will tell us something more about the
recycle structure of processes,Example #2 (shown in Fig,VI-2) looks to be very similar to
Example #1 but is distinctly different in one important respect,Note that Stream #8 recycles
M-1 M-2S-1 S-2 S-3
1 2
3
4 5
6
7 9
8
Fig,VI-2,Process Flow Diagram for Example #2.
back to mixer M-2 rather than mixer M-1,We can determine all the cycles by inspection,They
are (2,3),(5,6) [the same as in Example #1] and (5,7,8),Now,however,there are only two
minimal tear sets,namely,[2,5] and [3,5],The important difference between this example and the
previous one is in the structure of the loops,In Example #1 both 2-loops contained streams
which were common to the 5-loop,namely,Streams 2 and 5,However,in Example #2,this is not
the case,The 2-loop (2,3) can be torn by either Steam 2 or Stream 3 alone while both the 2-loop
(5,6) and the 3-loop (5,7.8) can be torn by Stream 5 alone,Put another way,as is obvious from
the diagram,there are two separate recycle structures in Example #2,This is not the case for
Example #1,In Example #2 we can solve the sequence [M-1,S-1] as though the rest of the
process is not there,Then,having a value for Stream 4,we can solve the sequence [M-2,S-2,S-
3] as a separate problem,Indeed,this is exactly how a process flowsheeting program such as
ASPEN would do it,The same is not true of Example #1; the entire sequence must be solved
simultaneously,Thus,we must be careful to look for separate subproblems so as to minimize the
computational work involved,
Our third example (Fig,VI-3) is a four-tray gas absorber that is to be solved as a process
flow sheet problem,(There are more efficient specialized algorithms for gas absorbers,but the
purpose of this example is to illustrate the recycle structure of multi-stage countercurrent
separators.)
- 42 -
S-4
S-3
S-2
S-1
1
3
5
7
9 10
8
6
4
2
Figure VI-3,Example #3
The node-incidence matrix is:
Block Streams:
3 4 5 6 7 8
S-1 0 1
S-2 1 0 0 1
S-3 1 0 0 1
S-4 1 0
The cycles are:
- 43 -
2-loops,(3,4),(5,6),and (7,8)
3-loops,(3,5,6,4) and (5,7,8,6)
4-loops,(3,5,7,8,6,4)
There are many minimal tear sets,[3,5,7],[4,6,8],[3,6,8],[4,6,8],and [3,5,8],The
computational sequence for [3,5,7] is [S-4,S-3,S-2,S-1] (top down) while that for [4,6,8] is
[S-1,S-2,S-3,S-4] (bottom up),The computational sequences for the other tear sets are
mixed.
The important point to be made here is that multi-staged separator is highly recycled if we
look at it from the standpoint of a process consisting of a number of separation stages arranged in
a countercurrent sequence,Looked at from the standpoint of a single piece of equipment,the
recycle structure does not show,It is internal to that piece of equipment,If we use one of the
many algorithms for solving the steady-state problem for distillation or gas absorption,then the
internal recycle structure does not enter into the solution of a process flow sheet for which the
separator is merely one process block,But the internal recycle structure does have important
consequences,particularly when it comes to the dynamics of both the separator and the process.
- 44 -
- 45 -
VII,SOLUTION OF LINEAR MATERIAL BALANCE MODELS
Revised October 11,1999
In this section we consider the solution of steady-state process material balances when the
equations are linear,First we look at the solution of performance material balances,i.e.,those
for which the parameters for each equipment model and the external input streams to the process
are all specified,In the terminology of the control engineer,this is the solution of the open-loop
problem.
It is often the case however,that we want some internal stream variable,say the mol
fraction of a given component or a total flow rate,to achieve a specified value,This is known as
a design specification,or design spec for short,In the terminology of the control engineer,this is
the solution of the closed-loop or control problem.
Achieving a design spec is accomplished by varying either one of the model parameters or
the flow rate of one of the external input streams,The problem that arises from adding a design
spec to a process material balance may or may not be linear,Fortunately,the most commonly
imposed design specs generally lead to linear problems,Even if this is not the case,clever use of
a spreadsheet can still allow us to solve a material balance with a limited number design specs in
an efficient manner,
A,Use of Linear Equation Solvers
The material balance for any process can be modeled,albe it at very simple level,by the
unit models MIX,REACT,SEPAR,and SPLIT,Further,if the model parameters are specified,
each model is linear so that the overall system of equations will be linear also,Since the methods
for solving sets of linear alge braic equations,even large sets,are well developed,the use of these
four models to perform process material balanc es has obvi ous advantages,The disadvantage is
that the engi neer must evaluate all of the parameter sets before being able to do the material
balance,This is not a serious problem once the engi neer has some idea of what performance is
expected of each pro cess unit.
The simple form of these equations confers another advan tage,Each equation for the ith
component involves only the ith component,Thus,once the parameters for the ith component are
known,its material balance can be solved independently of all the other components,This
reduces the material balance problem in NS streams and NC components from one large NS x NC
set of equations to NC sets,each of dimension NS,For instance if a process has 6 components
and 21 streams,we have only to solve 21 equations 6 times rather than 6 x 21 = 126 equations at
one time.
B,Reduction to the Tear Set Variables
- 46 -
A further reduction in the size of the problem can be accomplished by exploiting the
structure of the process as fol lows:
a,Enumerate the tear sets for the flowsheet,Choose a minimal tear set that is convenient,
A minimal tear set that includes all the reactor feed streams is generally a good choice.
b,Write out all the equations for a typical component,using the appropriate model for each
process unit,Each equation will have the form:
(VII-1a) f i,out = αi,1 f i,in1 + αi,2 f i,in2
(VII-1b) f i,out2 = αi,3 f i,i n1 + αi,4 f i,in2
Note,A reactor is really a Stream Add where the component flows in the second stream are the
αi and αi,2 = 1.
For each component there will be one equation that is the output of a block,Some of these
streams are outputs from the process,i.e.,are inputs to other blocks,The equations for these
streams can be set aside until the last step.
c,Substitute for all of the non-cut set stream variables,Of the remaining equations,there
will be one for each of the cut set streams which its f i,n appears on the left-hand side of the = sign.
Start with this set of equations,as many as there are cut set streams,Substitute from the
remaining equations for all of the non-cut set flow rates,Continue back substituting until all of
the flow rate variables which are neither cut set or input variables have been eliminated,
d,Solve the remaining equat ions for the tear set variables one component at a time,If there
are only one or two tear,this can be done analytically,If there are more,it can be done
numerically by Gaussian elimination or by using a linear equation solver,(Most of the flowsheets
we will encounter will have only a few streams in their minimal tear sets.)
e,Compute all of the other stream flow rates by direct of the process sequence that
corresponds to the cut set,The process output streams that were set aside in Step b should also
be evaluated at this point.
Suggestions for avoiding pitfalls and mistakes.
1,Convert all problem data to molar quantities before doing anything else,Remember:
(VII-2) F mix = W mix / MW avg
where W mix = the flow rate of a mixture in lb/hr (kg/hr)
- 47 -
F mix = the flow rate of that mixture in lb-mol/hr
( kg-mol/hr)
and MW avg = the average molecular weight of the stream.
(VII-3) MW avg = Σ y i MW i
where y i = the mol fraction of component i,and
MW i = the molecular weight of component i,
2,Do all material balance calculations in mols (unless,per haps,the system is a single non-
reacting component).
3,In writing out the equations in Step b and doing the reduc tion to cut set variables in Step
c,do not substitute numbers for the parameters (coefficients in the equations),Do this in Step d,
It is easier to check for mistakes and you can reuse the equations for all the components without
having to repeat Steps b and c.
Example
The ammonia loop shown in Figure 2 will be used as an example,There is only a single
recycle loop so let us choose Stream 3 as the tear stream,Since we are only interested in the
material balance in this example,items of equipment which do not affect either the flowrate or
composition of their output streams can be ignored,This includes the Feed Compressor C-1,the
Recycle Compressor C-2,and the Ammonia Condenser E-1,Their output streams,ST2,ST10,
and ST5 can be dropped from the stream list.
The process is specified as follows:
(1) The Synthesis Gas feed (ST1) provides 750 lb-mol/hr of H2,250 of N2,and 10 of Argon,
(2) The Synthesis Converter R-1 operates at 25% conversion per pass of the N2 in its feed (ST3).
(3) The Ammonia Condenser condenses 99% of the NH3 in Synthesis C onverter effluent
(ST4) which is removed as liquid in the Ammonia KO Drum F-1,The other gases are slightly
soluble in the liquid ammonia resulting in a 0.1% removal of H2 and 0.2% of both N2 and Argon.
(4) 5% of Stream 6 is purged.
We can write the linear material balance equations for the Synthesis Loop as follows:
- 48 -
Feed Mixer M-1
(E-1) f i,3 = f i,1 + f i,9
Synthesis Converter R-1
(E-2) f i,4 = f i,3 +?i
Ammonia KO Drum F-1
(E-3) f i,6 = s i f i,4
f i,7 = (1 - s i )f i,4
Argon Purge
(E-4) f i,8 = P f i,6
(E-5) f i,9 = (1 - P)f i,6
The Synthesis Converter performance is given in terms of the conversion of N2,C N2,
Thus
(E-6)?N2 = - C N2 f N2,3
We can now solve for f N2,3 by back-substituting Eqns,E-5,
E-3,E-2,and E-6 into Eqn,E-1 as follows:
(E-7) f N2,3 = f N2,1 + (1 - P)f N2,6
= f N2,1 + (1 - P)s N2 f N2,4
= f N2,1 + (1 - P)s N2 (f N2,3 +?i )
= f N2,1 + (1 - P)s N2 (1 - C N2 )f N2,3
Solving for f N2,3 gives
(E-8) f N2,3 = f N2,1 /[1 - (1 - P)s N2 (1 - C N2 )]
We can now calculate?N2 from Eqn,E-6,Knowing N2,we can,by the reaction stoichiometry,
calculate the other extents of reaction as follows:
(E-9)?H2 = 3?N2,
- 49 -
NH3 = -2?N2,and,of course,
Ar = 0.
Knowing the extents of reaction for the non-key components,we can now solve for their
component molar flow rates in the tear stream,again by back substitution,This gives
(E-10) f i,3 = [f i,1 + (1 - P) s i?i ]/[1 - (1 - P) s i ],i ≠ N2.
Once the tear stream component flow rates are known,the flow rates for the remaining streams
can be calculated directly starting with the Synthesis Converter,These calculations can be done
by spreadsheet as shown in the following summary:
Stream Summary:
Comp ST1 ST3 ST4 ST6 ST7 ST8 ST9
lb -mol/hr:
H2 750.00 2632.13 1983.17 1981.19 1.98 99.06 1882.13
N2 250.00 865.28 648.96 647.66 1.30 32.38 615.28
Ar 10.00 192.68 192.68 192.29 0.39 9.61 182.68
NH3 4.15 436.79 4.37 432.42 0.22 4.15
Total
mol/hr
1010.00 3694.23 3261.60 2825.51 436.09 141.28 2684.23
MWavg 8.83 10.10 11.44 10.58 17.02 10.58 10.58
Total lb/hr 8914.98 37314.01 37314.01 29893.71 7420.29 1494.69 28399.03
We note in passing that this stream summary was produced using a spreadsheet and imported
directly into this document,The details are given in the appendix,Also,the stream summary
shown in Chapter III was produced using the same spreadsheet program,The only differences
are that it includes more information and that the purge fraction was adjusted so that the Argon
concentration in the reactor feed would be 10%,This is an example of a design specification,the
subject of the next section.
C,Incorporation of Design Specifications
A performance material balance such that given in the example above allows us to
calculate the calculate the component molar flow rates f i,j of all of the output streams from all of
the equipment blocks,However,we may want one or more of the stream variables to meet a
performance (or design) specification,
For instance,if we look at Stream 3 in the stream summary above,we can calculate that
the mol fraction of argon in the stream is approximately 5.9%,On the other hand,if we look at
the purge stream,we see that we are losing 99.06 mol/hr of hydrogen,Since this amounts to
- 50 -
13.2% of the hydrogen fed to the synthesis loop,it represents a substantial yield loss,Suppose
we can operate the loop with up to 10% argon in the reactor feed,This means that we can reduce
the purge fraction from its nominal value of 5%,thereby reducing the amount of valuable
hydrogen that is lost in the purge,The question is how to determine the purge fraction that will
just meet the design spec of 10% argon in the reactor feed,
The most typical material balance design specs are:
1) Total flow rate of a stream
nc
F j = Σ f i,j = φFj
i=1
where φFj = the value for the flow rate design spec for the jth stream.
2) Mol fraction of a component in a stream
f i,j / F j = φMij
where φMij = the value for the mol fraction design spec for the ith component in the jth
stream,and
3) Molar ratio of component a to component b
f a,j / f b,j = φRabj
To meet a design spec we must have a process parameter available which can be varied,
Typical parameters are
1) Purge fraction
2) Flow split fraction
3) Total fresh feed flow rate,i.e.,the flow rate of stream which originates outside
battery limits (OSBL).
4) Reactor conversion,or
5) A separator recovery parameter.
Let us now consider a simple example that leads to a linear solution for a design spec,
Consider the hypo thetical process to make methanol (MeOH) via the partial oxidation of methane
(CH4),The proposed process is shown in Figure VII-1,
- 51 -
1
2
3
4
5
6
7
8
98% CH4
2% C2H6
99.5% O2
0.5% Ar
Purge
Methanol
Fresh CH4 (which contains 2% C2H6,an inert) is mixed with fresh O2 and recycle gas
and fed to a reactor,The
reaction chemistry is:
CH4 + 1/2 O2 --> CH3OH
The reactor is run at a conver-
sion of 20% of the methane in
the reactor feed,The reactor
effluent is sent to a separator
where all of the MeOH is
removed but none of the other
components,The overhead is
recycled to the reactor after a
purge is taken to maintain the
C2H6 concen tration at 10
mol%,
Figure VII-1,Hypothetical
Methanol Process
Taking the reactor feed stream (Stream 3) as the tear stream and solving the linear
material balance equations gives for CH4
where P = the fraction purged to Stream 7 and
C = the fractional conversion across the reactor.
Keep in mind that we are assuming that the only component removed in the separator is MeOH,
Thus,the separation coefficient s i with respect to Stream 5 for the all the other components is 1.0.
The tear stream solution for all the components other than CH4 and methanol (which is
not recycled) is
(E - 11) f = f (1 - P) CCH4,3 CH4,11 1( )
- 52 -
Now,the design spec φ is
Substituting Eqn,E-12 into E-13 and simplifying gives
This equation is linear in P so its solution is straightforward,If the flow rate of Stream 1 (F1) is
100 mol/hr and that of Stream 2 is 50 mol/hr,then solving for P gives P = 0.02549.
Unfortunately,not all design specs lead to linear equations for the associated model
parameters,This is the case for the NH3 syn thesis loop if we want to adjust its purge rate to keep
the argon concentration in the reactor feed to 10%,This is because not all the s i are either 0 or
1.0,In this case it is easier to solve the problem using a spreadsheet,This was done by
programming a cell to display the mol % of argon in the reactor feed and then adjusting the purge
parameter until the spec is met,The following are the results that were obtained:
P Mol% Ar
0.05 5.216
0.04 6.208
0.03 7.765
0.021525 10.000
0.02 10.562
Note that at the design spec purge rate,the yield is reduced to 6.29 %.
(E - 12) f = f + f + (1 - P) P
where = - C f
= - 12 C f
= C f
= 0
i,3
i,1 i,2 i
CH4 CH4,3
O2 CH4,3
MeOH CH4,3
C2H6
(E - 13) = f F = 0.1
where F = f
C2H6,3
3
3
i = 1
nc
i,3
f
Σ
(E - 14) = ( f )[P + (1 - P)C]
[P - C 2 (1 - P)] f + [P + (1 - P)C]( f + f )
CH4,1
CH4,1 O2,1 C2H6,1
f
- 53 -
VIII,SEQUENTIAL MODULAR SOLUTION OF
NONLINEAR MATERIAL BALANCE MODELS
Revised September 23,1998
In this section we look at some commonly used procedures for converging a flowsheet
model using the sequential modular method of solution when some or all of the unit operations
models are nonlinear,First we will look at the method direct iteration,This method is simple to
use,highly reliable,and not sensitive to initial guesses of the tear streams,But,for tightly re-
cycled processes it is quite inefficient,(By tightly recycled we mean processes for which the total
flow rates of the tear streams are large compared to the fresh feed streams.)
The inefficiency of direct iteration was recognized early on and methods to accelerate it
were developed,The best known of these is the method of Wegstein,Its convergence
capabilities are explored in Part B of this section.
A,Convergence by Direct Iteration
As mentioned earlier the sequential modular method starts with an initial guess of the
values of the tear stream vectors,The process is then calculated unit by unit in the sequence
determined by the choice of tear streams,The result is a set of calculated values for the tear
streams,If these do not agree with guessed values within a specified tolerance,a new set of
values have to be g uessed (or estimated) for the tear streams,If the previously calculated set are
used as the new guesses,the procedure is referred to as direct iteration.
In order to learn something
about the convergence
characteri s tics of direct iteration as
applied to flowsheets cal cul a tions,
let us look at the simplest of all
problems,that of the single-
component,single-recycle process
as shown in Figure VIII-1.
Stream 1 is mixed with
Stream 3 in Mixer M-1 and sent to
the Purge Splitter P-1 where it is
separated into Streams 3 and 4,
Stream 3 is obviously a recycle
stream,Let us assume that the
model for the Purge Splitter is non - Fig,VIII-1,Purge Splitter R e cycle Process
linear,
- 54 -
Let X i be the molar flow rate in Stream i of a typical component present in this process,
Then,the material balance equations for the process are:
For the Mixer M-1
(VIII-1) X 2 = X 1 + X 3
For the Purge Splitter P-1
(VIII-2) X 3 = F[X 2,p],
where p is an operating parameter,
Now,let us solve Eqns,VII-1 and VII-2 iteratively,We start by assuming a value for X 3,
call it X 3 (0) where X 3 (n) is the valve of X 3 after n iterations,Then,
X 2 (1) = X 1 + X 3 (0),and
X 3 (1) = F[X 2 (1),p]
Let E(n) = the relative error in X 3 after the nth iteration,Then,if
E(1) = |X 3 (1) - X 3 (0)|/X 3 (1) < ε
where ε is a pre-assigned tolerance,the solution is finished,Otherwise,we must repeat the
proc e dure,i.e.,
X 2 (2) = X 1 + X 3 (1),and
X 3 (2) = F[X 2 (2),p]
If E(2) < ε,the solution is now finished,Otherwise,repeat the procedure until E(n) < ε or n >
Nmax,where Nmax is the maximum number of iterations for which we are willing to continue the
iterative procedure.
The question is,under what conditions does this procedure converge? The answer is,
from numerical analysis (See Ostrowski,Chapter 4,or any other good text on numerical analysis),
if |dF/dX 2 | < 1,then the procedure will converge.
Let us examine this in a little more detail,Suppose dF/dX 2 = 1-p = r,where p = the
fraction of Stream 2 which is sent to Stream 4,Then,combining Eqns,VII-1 and VII-2 gives
- 55 -
(VIII-3) X 2 (n+1) = X 1 + r X 2 (n)
If we start the iterative procedure with X 2 (0),we have for the first few iterations
Iteration 1,X 2 (1) = X 1 + r X 2 (0)
Iteration 2,X 2 (2) = X 1 + r [X 1 + r X 2 (0)]
= (1+r) X 1 + r 2 X 2 (0)
Iteration 3,X 2 (3) = X 1 + r [(1+r)X1 + r 2 X 2 (0)]
= (1+r+r 2 ) X 1 + r 3 X 2 (0)
It can be shown that after n iterations,
(VIII-4) X 2 (n) = (1-r r+1 )/(1-r) X 1 + r n X 2 (0)
Now,if an algorithm is to converge to a unique solution,the value of that solution should not
depend on the initial assumed value X 2 (0),For this to be the case,
(VIII-5) lim r n -- > 0
n-> ∞
This will only happen if r < 1,which is the condition for convergence of the nonlinear procedure,
Also,if r < 1,the first term on the right hand side of Eqn,VII-4 becomes 1/(1-r),so the steady-
state solution X 2,ss by iteration is
(VIII-6) X 2,ss = 1/(1-r) X 1
We get the same solution if we assume that X 2 (n+1) = X 2 (n) in Eqn,VII-3,which is reassuring,
Of course,in the nonlinear case,we cannot solve directly for X 2,ss,We can only estimate its value
by the iteration.
We can make the following observations about the physical nature of the solution,First,r
is the fraction of the component under consideration that is transferred from Stream 2 to Stream
3,If this component is a conserved quantity,then from physical considerations 0 < r < 1.0,In
other words,if only conserved quantities are used in modeling the various unit operations in the
process,then the input-output coefficients for all the process units must lie between 0 and 1.0,It
can be shown that because of this,the direct iteration solution of a linear process always
converges.
- 56 -
However,if in our simple example,r --> 1.0,the rate of convergence can be very slow,
For instance,if r = 0.9,and we want r n < 0.0001,88 iterations will be required,If r = 0.95,180
are necessary,Since values of r = 0.9 - 0.95 are not untypical in tightly recycled processes,it is
obvious that convergence by direct iteration can require a lot of computer time for realistic
process models.
Recycle has another effect on the process flowsheet other than making convergence slow.
From Eqn,VII-6 we also see that it increases the internal stream flow rates by a factor of
1/(1-r) with respect to the feed streams to the process,This,in turn,will increase all the
equipment sizes in the recycle loop for a fixed throughput,
In sequential modular simulation,the standard method for checking for convergence is to
monitor the change from iteration to iteration of the tear set variables,For our simple example,
the relative error Er is given by
(VIII-7) Er = | X 2 (n+1) - X 2 (n)|/X 2 (n+1)
However,a chemical process engineer is generally interested in how well the overall process
mat e rial balance is converged,Let Eo be the relative error in the overall material balance,For
the process of Figure 10,it is given by
(VIII-8) Eo = | X 4 (n+1) - X 1 |/X 1
The question is,if Er =,what does Eo =? If the solution is close to convergence,
(VIII-9) X 2 (n+1) ≈ [1/(1-r)] X 1
So,
(VIII-10) Er = (1-r) | X 2 (n+1) - X 2 (n)|/X1
Also,
(VIII-11) Eo = |(1-r) X 2 (n) - X 1 |/X 1
= | X 2 (n) - {r X 2 (n) + X 1 }|/X 1
= | X 2 (n) - X 2 (n+1)|/X 1
So,
- 57 -
(VIII-12) Eo/Er = 1/(1-r)
The relative error in the overall material balance Eo is a factor of 1/(1-r) larger than the relative
error between iterations Er,For instance,if R = 0.9,Eo = 10 Er,It must be kept in mind that in
tightly recycled processes,the relative error in the overall material balance can be an order of
magnitude or more larger than the relative error from iteration to iteration Er,It is Er that is the
quantity usually monitored by the convergence checking routine in most simulation programs.
It is interesting to estimate how many iterations it will take to achieve convergence to a
given tolerance,Suppose that for the system of Figure 11,we want to converge the simulation to
a tole r ance on Er of 0.0001,The question is how many iterations will it take as a function of r?
This obviously depends upon how close the initial guess of X 2 (0) is to the converged value,To
keep things simple and straightforward,let us take this initial guess to be zero,i.e.,X 2 (0) = 0,
Then,substituting Eqn,VIII-4 into Eqn,VIII-7 and solving for n gives
(VIII-13) n = ln[ ε/(1-r(1- ε))]/ ln(r) - 1
This is demonstrated numerically as follows:
r n Eo/ Er
0.2 5 1.25
0.5 12 2.0
0.7 22 3.33
0.9 65 10.0
0.95 121 20.0
0.99 459 100.0
We can see that the number of iterations required for a relative convergence to four significant
figures grows quite rapidly as the system becomes more tightly recycled (r --> 1.0),Also,the
error in the overall material balance Eo increases substantially,If we want to converge the
simulation to a overall material balance tolerance of 0.0001,then it is easy to show that for r =
0.99,the number of iterations is 916 rather than 459,an increase of a factor of 2.0.
B,Convergence Acceleration
One can ask whether or not anything can be done to improve the rate of convergence of
s equential modular simulations? The answer is yes,Instead of using the previously calculated
value X 2 (n) for the (n+1) th iteration,let us develop a better estimate,call it Y 2 (n),A very simple
strategy for doing this is
(VIII-14) Y 2 (n) = α Y 2 (n-1) + (1- α) X 2 (n)
- 58 -
Note that for α = 0,this is equivalent to direct iteration and if α = 1,no acceleration at all will
take place,If 0 < α < 1,the effect is to damp the iterative procedure,For instance,if α = 0.5,
this amounts to averaging the previous guess and the current calculated value to produce the next
guess,If the convergence is to be accelerated α must be negative.
Now,α is a parameter whose value must be determined in one manner or another,Let us
see how this might be done,If an estimated value is to be used for each iteration,Eqn,VIII-3
becomes
(VIII-15) X 2 (n) = X 1 + r Y 2 (n)
So,from Eqn,VIII-14,
(VIII-16) Y 2 (n) = α Y 2 (n-1) + (1- α)[X 1 + r Y 2 (n-1)]
Or,
(VIII-17) Y 2 (n) = λ Y 2 (n-1) + Φ
where λ = α + (1- α)r and
Φ = (1- α)X 1
Eqn,VIII-17 is similar in form to Eqn,VIII-3 where now r becomes and X 1 becomes (1- )X 1,
Thus,the condition for convergence is | λ| < 1.0,Since α is a parameter at our disposal,we can
choose it to make λ take on any value we want,An obvious value is zero ; convergence will then
take place in one iteration,Solving for α for λ = 0 gives
(VIII-18) α0 = -r/(1-r)
It is also possible to choose α such that | λ| > 1 in which case the sequential modular iterative
proc e dure will diverge,What are the limits on α so that this will not happen,Since λ is real if α
is,we need to determine what values of α make λ = +1 and λ = -1,For λ = +1,we get
(VIII-18a) α+1 = 1
which,as was previously established,is the value of which results in no change from iteration
to iteration,For λ = -1 we get
(VIII-18b) α-1 = -(1+r)/(1-r)
- 59 -
Let us examine this numerically:
r α0 α-1
___ _____ _____
0.2 -0.25 -1.50
0.5 -1.0 -4.00
0.7 -2.33 -5.67
0.9 -9.0 -19.0
0.95 -19.0 -39.0
0.99 -99.0 -199.0
We see that the value required for α0 increases rapidly as r approaches 1.0 and that value of α-1 is
asymptotically only twice that of α0,This,as will be seen,limits the extent to which this form of
acceleration can be applied to real problems.
C,The Method of Wegstein
One of the first methods used to accelerate the convergence of sequential modular
simul a tions was that of Wegstein,It is still used today; it is,for instance,the default method in
ASPEN,It is based on the procedure outlined in Eqns,VIII-16 and VIII- 18 which assumes that
the value of r is known,In real problems,it is not,Furthermore,if the problem is nonlinear (Why
else would one do sequential modular simulation if the problem were not?),r is not known,
Furthermore,its value varies from iteration to iteration,Therefore,in order to use this
acceleration procedure,one must have a means of estimating r.
This can be done as follows,Suppose we execute two direct iterations of the simulation,
This can be expressed as follows:
(VIII-19a) X 2 (1) = X 1 + r X 2 (0) and
(VIII-19b) X 2 (2) = X 1 + r X 2 (1).
Solving for r gives
(VIII-20) r = [ X 2 (2) - X 2 (1)]/[X 2 (1) - X 2 (0)]
This is all well and good but what we are estimating in effect is a slope,The way in which
Wegstein procedure is applied is component by component and tear stream by tear stream,
ignoring all intera c tions,The effect of these interactions is to introduce errors into the estimation
- 60 -
of r for each comp o nent and each tear stream,As can be seen from the previous table,small
errors in the estimation of r,particularly when r --> 1.0,can result in choosing a value for that
will make the procedure unstable,This has led in practice to placing a bound on ; that used in
ASPEN is -5.0,As can be seen,this will reduce the efficiency of the procedure,For instance,for
r = 0.99,is only 0.94 instead of 0,However,it still helps since the number of iterations
required for our example problem is reduced from 459 to about 120,a factor of almost 4.0,
Our investigation of convergence and convergence acceleration is quite limited,It is
i n tended only to point out some of the difficulties that can be encountered in using present-day
sequential modular simulators,Many of these simulators,such as ASPEN,have alternative
alg o rithms available as well,These generally involve using Newton- Raphson or pseudo-Newton
met h ods such as that of Broyden applied simultaneously to all of the tear set variables,These
methods can also be used to solve simultaneously for the design specs and tear stream variables,
This can result in a considerable saving of computer time for complex problems with many design
specs,The drawback is that these methods are much more sensitive to initial guesses than those
based on direct iteration or Wegstein,As a practical matter one has to experiment to determine
which is best in a given situation.
- 61 -
IX,MIXING AND BLENDING PROBLEMS
Revised September 23,1998
Many of the products produced by the chemical processing industries are mixtures or
blends of various constituents,One example is gasoline,Today's gasoline must meet a number of
specifications including octane rating,volatility,and oxygenate content,Another example is
pharmaceutical products,For instance,each capsule in a bottle of Extra Strength Excedrin,an
analgesic,contains 250 mg each of aspirin and acetaminophen and 65 mg of caffeine in addition to
the binder,A third example is animal feed,The feed for,say,chickens must contain protein,fat,
and carbohydrates in the proper ratios as well as the appropriate amounts of vitamins,minerals,
and other nutrients.
In each example,the product is the result of the blending and mixing together of a number
of individual constituents,many of which are themselves mixtures,We will refer to this process
as mixing when the composition of the end product must meet exact composition specifications,
For instance,each Excedrin capsule must contain 250 mg of aspirin,no more and no less,If the
end product must merely meet its specifications in term of ranges,then we will refer to it as a
blend,For instance,our chicken feed might be required to contain no less than 20% protein by
weight and no more than 25%.
A mixture can be a mixture of gases (Air,for instance,is a mixture of oxygen,nitrogen,
argon,carbon dioxide,and other trace gases.),liquids (Rubbing alcohol,for instance,is a mixture
of isopropyl alcohol in water.),or solids (Premixed concrete,for instance,is a mixture of sand,
cement,and lime.),A mixture that is liquid is quite often referred to as a solution,Mixtures of
solids or gases have no special terminology,
A,Mixing
Mixtures can be specified on a weight basis,a molar basis,or a volume basis,A weight
basis is commonly used for many large-scale commodities such as solutions containing one or
more components that solids in their pure form,Many mixtures that are a solution of two liquids
are characterized in terms of volume,Rubbing alcohol,for instance,which is generally 50 to 70%
isopropyl alcohol by volume,Mixtures,particularly solutions for use in a chemical laboratory,are
quite often characterized in terms of the number mols of a given constituent contained (usually) in
a given volume of solution.
Let start with a simple example,namely,to make up 100 Kg of a 40% solution by weight
of NaOH in water given available supplies of pure NaOH and pure water,All we would do is
weight out 40 Kg of NaOH and 60 Kg of water and then dissolve the NaOH in the water (with
due attention to safe handling).
However,if we only happen to have a solution of 60% NaOH in water and pure water
available,then we have to do some material balance calculations,
- 62 -
Let F 1 = the Kg of NaOH solution which must be used and
let F 2 = the Kg of water to be used,Then,a overall material balance for mixture is
(IX-1) F 1 + F 2 = 100 Kg
while a material balance on NaOH gives
(IX-2) 0.6 F 1 + 0.0 F 2 = (0.4)(100) = 40 Kg.
These two equations are easy enough to solve giving
F 1 = 40/0.6 = 66.67 Kg and
F 2 = 100 - F 2 = 100 - 66.67 = 33.33 Kg.
As a second example,consider making up a solution that approximates the composition of
seawater,namely,22,000 ppm by weight of NaCl and 13,000 ppm of MgCl2,Let us suppose that
in addition to a supply of pure water,we have available a solution containing 5.0 wt% of NaCl
and a second solution containing 4.0 wt% of MgCl2,
Since a target amount of solution has not been specified,let us choose 1000 Kg as a basis,In this
1000 Kg of solution we will need:
(0.022)(1000) = 22 Kg of NaCl,
(0.013)(1000) = 13 Kg of MgCl2,
and by difference,965 Kg of H2O.
Let W1 = the Kg of NaCl solution to be used,
W2 = the Kg of MgCl2 solution to be used,and
W3 = the Kg of pure H2O to be used.
We can write the following mass balances:
W1 + W2 + W3 = 1000 (overall mass balance)
0.05 W1 = 22 (NaCl balance)
0.04 W2 = 13 (MgCl2 balance)
This is an easy problem to solve,Doing so gives
W1 = 440 Kg,
W2 = 325 Kg,and
W3 = 275 Kg.
Suppose instead that the two solutions have the following com positions:
Solution 1,5% NaCl and 1% MgCl2 in H2O
- 63 -
Solution 2,1% NaCl and 4% MgCl2 in H2O
The overall mass balance remains the same but the NaCl and MgCl2 balances become:
0.05 W1 + 0.01 W2 = 22
0.01 W1 + 0.04 W2 = 13
These two equations must be solved simultaneously,Doing so gives
W1 = 394.74 Kg,
W2 = 226.32 Kg,and
W3 = 378.94 Kg.
However,if the solutions have the compositions:
Solution 1,5% NaCl and 3% MgCl2 in H2O and
Solution 2,3% NaCl and 4% MgCl2 in H2O,
then W2 < 0,For this case there is no physically meaningful solution.
The important fact to note about exact mixing problems is that for a solution to exist at
all,the number of independent variables that can be adjusted must be exactly equal the number of
specifications for the mixture,In the last example,there were three specifications ( ppp NaCl,
ppm MgCl2,and total weight of the solution) and three independent variables (W1,W2,and W3).
B,Blending
Blending problems arise when either there are more inde pendent variables than there are
specifications to be met or if some of the specifications are in the form of inequalities,In either
case there is not a unique solution but rather a range of solutions,One can introduce,then,
another consideration,namely,cost,
Let us look at the previous mixing problem as a blending problem to see how cost
considerations can arise,Suppose that instead of having two solutions available,we have three,
all of different compositions,If the problem is feasible at all,there range of combinations of the
three solutions (plus the pure H2O) which will meet the specifications,Suppose that each
solution has a different cost per Kg,Then the question is,which of all the possible combinations
yields the lowest cost? This is what is traditionally known as an optimization problem.
On the other hand suppose that only solutions are available but that we relax the
specification on NaCl,Instead of being exactly 22,000 ppm,suppose it only has to be between
18,000 and 24,000 ppm,Let us consider the situation where the two solutions have the
compositions:
- 64 -
Solution 1,5% NaCl and 1% MgCl2 in H2O
Solution 2,1% NaCl and 4% MgCl2 in H2O
Further Solution 1 costs $0.05/Kg while Solution 2 costs $0.10/Kg,The cost of the pure water is
negligible,What is the minimum cost recipe for this blending problem?
First,let us define an objective function which,in this case will be the total cost of the target
solution:
Γ = 0.05 W1 + 0.10 W2 (dollars)
The material balance specifications now become
0.05 W1 + 0.01 W2 ≥ 18
0.05 W1 + 0.01 W2 ≤ 24
0.01 W1 + 0.04 W2 = 13
These three equations are known in optimization jargon as con straints,There are two additional
constraints that must be satisfied to guarantee an physically meaningful solution to the
optimization problem,namely,
W1 ≥ 0 and
W2 ≥ 0.
Note that both the objective function and the constraints are linear in the problem variables
W1 and W2,When this is the case,the optimization problem has a very specific known a linear
program,Linear programming or LP was developed over forty years ago and has been applied to
many problems in planning,operations,and design,Very effective algorithms such as the Simplex
Method have been developed for the solution of LP programs and problems involving thousands
of con straints and thousands of variables are solved daily,These algorithms are widely available,
Most of the popular spread sheet programs have LP solvers as part of their repertoire.
Since we have only two variables,namely W1 and W2,we can explore the nature of the
optimization problem graphically,This is shown in Figure IX-1,The MgCl2 exact constraint is
the almost horizontal line and the two NaCl inequality constraints are the other two lines,The
solution must lie to the right of the lower NaCl constraint and to the left of the upper one,It must
also lie on the MgCl2 constraint,If we solve for the in tersections of the two NaCl constraints and
the MgCl2 constraint,we get,for the lower NaCl constraint,W1 = 310.53 and W2 = 247.37,
while for the upper NaCl constraint,W1 = 436.84 and W2 215.78,If we evaluate the objective
function at these two intersections,we get for the lower NaCl constraint
Γ = (0.05)(310.53) + (0.1)(247.37) = $40.26
- 65 -
and for the upper NaCl constraint
Γ = (0.05)(436.84) + (0.1)(215.78) = $43.42.
Any choice in between of W1 and W2 that lies on the MgCl2 con straint will have a value of be-
tween these two values,This is al-
ways the situation with an LP solu-
tion,namely,that the optimum lies
at the intersection of constraint
bound aries,The Simplex Method
exploits this fact in finding the
optimum so lution for problems
much larger than this one,
0
250
500
750
1000
1250
1500
1750
2000
2250
2500
W2
0 50 100 150 200 250 300 350 400 450 500
W1
Blending Example,Constraint Map
- 66 -
X,PLANT DATA ANALYSIS AND RECONCILIATION
Revised September 23,1998
- 67 -
Up to this point we have been considering hypothetical processes,processes that exist
only on paper,Equipment models have been chosen on the assumption that these describe the
behavior of the process with sufficient accuracy for the purposes at hand,Flowsheet calculations
have then been done,the results being presented in some form of a stream summary,If these
calculations have been properly done,the stream summary satisfies the conservation of mass to as
many significant figures as our computer will support.
Once a plant is built and operating,the situation changes,The chemical engineer's job is
now to determine the performance of the actual,as-built plant rather than that of the hypothetical
one on which the design is based,And this can only be done by the analysis of operating plant
data.
A,Plant Data
Before proceeding to some examples,let us examine the nature of plant data,Data are the
results of measurements,And measurements are subject to a number of potential difficulties.
1) The nature of measurements
Some measurements are direct but most are inferential,The determination of the weight
of a sample using an old-fashioned swing-type gravimetric balance is direct,One puts the sample
in one pan and adds weights of known denominations to the other pan until a balance is achieved.
The two weights must be equal; the sample weight must be equal to the sum of the known
weights added to the other pan of the balance.
Almost all other measurements are inferred,Temperature as measured by a mercury-in-
glass thermometer is inferred from the length of the mercury column,The accuracy of the
measurement depends on the uniformity of the bore of the thermometer and how well the
dependence of the coefficients of thermal expansion of both the mercury and the glass are known.
If a thermocouple is used,then the accuracy depends upon whether or not there is contamination
of the junction weld and how well the dependence of the emf on temperature is known,
The two types of measurements most commonly used in plant material balance
calculations are flow rate and stream composition,There are a variety of instruments used for
each,all of which infer flow rate or composition from some other direct measurement,In the case
of flow rate,the most widely used instrument is the orifice meter; mass flow rate is inferred from
the measured pressure drop across the orifice,There are many other types of flow meters but
their description is beyond the scope of these notes,
There are two ways in which composition is measured,One is on-line using an instrument
called an analyzer,A small sample stream is continuously diverted to the analyzer,The
measurement itself may be continuous in time,examples being infrared and ultra-violet
Spectrome ters,Or it may be periodic,a new analysis becoming available every so often,say
- 68 -
every ten seconds or every five minutes depending upon the requirements of the specific analysis.
The most common example of this type of instrument is the gas chromatograph.
Some composition measurements are too difficult or too expensive to do on-line,In this
case a discrete sample is taken in a sample flask and sent the plant analytical laboratory to be
processed there,This is generally done on a frequency of one an hour to once per shift,
Depending upon the complexity of the analysis,the results may not be available for anywhere
from an hour to a day or more.
It is not necessary to know the specifics of every measurement to use plant data for
material balances and other purposes,However,there are a few general characteristics of plant
that should be kept in mind,Above,never assume (even if the on-line analyzer cost a half million
dollars) that the data is correct,Beyond that one should know the following:
a) Zero and span
Every on-line measurement is good only over some predetermined range of values,The
transducer that converts the primary signal into one that is compatible with the rest of the plant
instrumentation is designed with a compromise between accuracy and range in mind,For
example,suppose a reactor is to be run at temperature of 500 C with allowable excursions of 1.0
C,If a transducer is chosen with a range of 0 to 1000 C and an accuracy of 0.1% of span,then
the best resolution it will give is 1.0 C,Since we must control to less than 1.0 C,this choice of
transducer would not be acceptable,Choosing one with a range from 400 to 600 C with the small
accuracy specification would give a of 0.2 C which is at least within the realm of possibility,It
might be necessary to narrow the range even more (or buy a more accurate but more expensive
transducer).
The lowest value for which the transducer provides an output is called the zero,For the
second temperature transducer,this is 400 C,The range over which the transducer provides a
useful output is called the span,This is 200 C for the second transducer,The important thing to
keep in mind is that if the measurement is outside the range of the transducer,it is not valid,The
best that can be said if the second temperature transducer indicates a temperature of 600 C is that
the temperature is at least 600 C,We should know how the transducer fails,Some fail high while
others fail low,Suppose our temperature transducer fails low,Then,if it reads 400 C,we do not
know whether the temperature is 400 C or less or whether the transducer has failed and we have
no idea what the temperature is.
b) Bias and error
There are two major sources of inaccuracy in an on-line measurement,One is error,and
by this is meant random error,Suppose that the value of the variable being measured has not
changed for the last hour but that when we read the transducer output every five minutes,we get
a significantly different value,some higher,some lower than the others,In this case the error is
- 69 -
probably random and a good estimate of the underlying value would be to take the average of the
20 readings.
On the other hand,suppose that the instrument technician incorrectly calibrated the
transducer so that its real zero is 390 C but its span is still 200 C,Or,suppose that the transducer
has not been calibrated for a long time and its zero has drifted to the lower value,This means
that every temp erature measurement will be 10 C lower than the actual value,This is called bias,
c) Sampled versus continuous
Sampled measurements such as those described above are generally subject to much more
random error than continuous on-line measurements,The opposite tends to hold for on-line
measurements,namely that these subject to more bias but less random error than sampled
measurements,There are many reasons for this,An important one is that most laboratory
instruments are recalibrated much more often that plant instruments,
2) Plant material balances
Let us now consider the problem of determining plant performance from plant data,Two
questions arise,The first is whether sufficient measurements are available to permit the
calculation of those aspects of plant performance that are of interest,The second concerns the
accuracy of the data,
It should be kept in mind that only a small fraction of all of the variables which
characterize the complete performance of a plant are actually measured,For instance the complete
p erformance of a distillation column distillation column of thirty trays separating a ternary mixture
is characterized by approximately 130 variables,In practice only about 15 would be measured,
Also,measurements are expensive,(Typi cally,10% of the total investment in a plant is for
instrumentation.) And some are very difficult and potentially unreliable,As a result,the number
of measurements available represents a compromise between cost,feasibility,and suf ficiency.
a) Surge Tank
Now let us consider a few simple examples,A surge tank is shown in Figure X-1,The i-
nstru menta tion consists of flow meter on the inlet stream (FM1),another on the outlet stream
(FM2),and a level meter (LM),S uppose the system is at steady state,This will be the case none
of the measurements change with time,If this is the case,the inlet flow rate F1 as measured by
FM1 should equal the outlet flow rate F2 as measured by FM2.
- 70 -
Suppose F1 = 10,050 lb/hr as measured and F2
= 9,975 lb/hr,The two measurements differ by less
than 1 % which is well within the expected accuracy of
most flow meters,Thus,we might conclude that the
two measurements are fundamentally correct and
agree to use the average of the two as the most
correct value,On the other hand if F1 = 11,050 lb/hr
and F2 = 8,250 lb/hr,we recognize that we have a
problem,
Figure X-1,Surge Tank PID
Further,with only this data,there is no way to
determine which,if either is correct,Our only
recourse is to call the instrument technician to check
and recalibrate the two flow meters,
Let us note in passing that if we have only flow meter,we have no way of knowing
whether or not it is even remotely correct,Having two allows us to check one against the other,
which is an improvement,If one can calculate the same aspect of plant performance from two
independent sets of data,the data are said to be redundant,As we will see,redundancy is the key
to the effective use of plant data.
Our next example is shown in Figure X-2,
The only difference between this surge tank and the
previous one is the additional flow meter FM3 on
the inlet stream,Suppose the measurements are F1
= 10,050 lb/hr,F2 = 9,975 and F3 = 7,200,Clearly,
some-thing is amiss,However,if we,
Fig.
XI-2,Surge Tank PID #2
chose not to think about it,we might just use the
average over the three values which is 9075 lb/hr,
F1 and F2 each differ from this value by about 10%
while F3 is more than 20% low,Thus,this way of
adjusting the data does not appear to be satisfactory,On the other hand,if we use the average of
just F1 and F2,which is 10012.5 lb/hr,then F1 and F2 differ from this value be less than 1%,But
F3 is low by 28%,Thus it would appear that F3 is in serious error and should be disregarded,
The value of redundant data is obvious,If we only had flow meters FM2 and FM3,there
would be no way to identify which was in error,But with three,there is at least the possibility of
determining which one is in error.
- 71 -
c) Flash Drum
Let us look next at the flash drum shown in Figure X-3,A feed stream containing
nitrogen and water is flashed,Most of the nitrogen goes overhead while most of the water leaves
as liquid in the bottom stream,All three streams
are equipped with a flow meter,The feed stream
has an analyzer AM1 which measures the amount
of nitrogen in the stream,The overhead analyzer
AM2 measures the amount of water in that
stream while AM3 measures the amount of
nitrogen dissolved in the liquid leaving the flash
drum,Fig,XI-3,Flash Drum
PID
The following data have been taken at
steady-state conditions:
F1 (measured by FM1) = 25,015 Kg/hr
F2 (measured by FM2) = 20,300 Kg/hr
F3 (measured by FM3) = 4,920 Kg/hr
N2 in feed = 70 mol%
H2O in overhead = 4 mol%
N2 in bottoms = 0.24 mol%
We would like to know the following:
i ) What is the % of N2 in the feed which leaves in the overhead vapor? From a
separation operations point of view,what is the recovery of N2?
ii) What is the separation factor for the flash dru m?
Digression,
The separation factor is measure of the completeness of separation produced by any item
of separation equipment,Suppose that a feed stream,such as that to the flash drum,is separated
into two output streams,If a separation occurs,then the composition of the two output streams
must differ,The separation factor between two components is defined as follows:
Chose the component of most interest,generally a product or other valuable component,
Call this component A and the other component B,Identify the stream in which that component
A s concentration should be higher than it is in the feed,Call this Stream 1 and the other output
stream from the separator Stream 2,Then the separation factor between A and B (SF AB ) is:
- 72 -
where X I,J is the composition or molar flow rate of component I in stream J,Properly defined,the
separation factor is always greater than 1.0 if any separation takes place,For high purity
separations,separation factors of 10,000 or more are not uncommon,Thus,the separation factor
is useful measure of the effectiveness of a given separation.
End of Digression
Returning to our flash drum,suppose that we want to determine the fractional recovery of
nitrogen and the separation factor,First,following good practice,we should examine the the
quality of the data,We can see immediately that the feed rate is 25,015 Kg/hr while the total
output is 25,220 Kg /hr,The discrepancy is not particularly significant compared to the expected
errors in flow rate measurements.
The overall mass balance is not the only thing that needs to be checked,We must also
check the individual component balances,The first thing to check is the validity of the analytical
data,if possible,Some on-line analyzers such as IR and UV spectrometers only analyze for a
specific component,If we know that the mixture is binary,then the amount of the second com-
ponent can be inferred by difference,
Other analyzers such as gas chromatographs and mass spectrometers have the capability of
determining the amount of each component present in a mixture,In this case,we can check the
creditability of the analysis by summing up the amounts,If these are reported as mol or weight
percents,the sum should equal 100%,Generally it will not do so exactly,If the sum is
reasonable,say between 99% and 101%,we might simply renormalize the by dividing each
component analysis by the sum,
For the example at hand we have only partial analyses,so no creditability check is
possible,Let us do a mass balance on nitrogen,Before doing that,we must the measured mass
flow rates into molar flow rates which requires computing the average molecular weight for each
of the three streams,These are:
MWfeed = (0.7)(28.014) + (0,3)(18.016) = 25.015
MWvapor = (0.96)(28.014) + (0.04)(18.016) = 27.614
MWbottoms = (0.0024)(28.014) + (0.9976)(18.016) = 18.040
The molar flow rates of the three streams are:
F1molar = (25015/25.015) = 1000.00 Kmol/hr
F2molar = (20300/27.614) = 735.13 Kmol/hr
F3molar = (4920/18.040) = 272.73 Kmol/hr
(X - 1) SF = x / xx / xAB A,1 A,2
B,1 B,2
- 73 -
The N2 balance is:
N2in = (1000.)(0.7) = 700.00 Kmol/hr
N2out = (735.13)(0.96)
+ (272.73)(0.0024)
= 705.72 + 0.654 = 706.37 Kmol/hr
(Vapor) (Liquid)
The balance is not in serious error (less than 10%) but notice that the amount of N2 leaving in the
vapor alone exceeds the amount in the feed,Moisture analyses can somewhat unreliable unless
precautions are taken to prevent condensation and adsorp tion,If the overhead vapor ana lysis
were 4.9% H2O instead 4.0%,the balance would be almost exact,Since this correction is in the
right direction if there were condensation of H2O in the sample line,we will accept it.
B,Data Reconciliation
Data reconciliation is the adjustment of the raw plant data so that the material balances
computed the adjusted data are exact,The previous examples have employed what we might
term ad hoc data reconciliation,i.e.,adjustments based on technical judgment as to what
measurements are most likely to be in error,One might ask if such reconciliation is necessary if
the errors associated with the raw data are not excessive,The answer should be clear from the
second example,Without even a small adjustment,the calculation of the fractional recovery of
N2 is meaningless.
The ad hoc approach is fine if the problem is relatively small,Different choices of
adjustments can be made,The one that leads to exact balances with the most reasonable
adjustments of the data can then be selected,If the problem is large involving,say,an entire
section of a plant,the ad hoc procedure becomes unwieldy since the number of choices grows fac-
torially,
A more structured approach is required,This has been the subject of extensive research
over the past thirty years,What one wants is a procedure that,for one,identifies measurements
that are in gross error and,for the other,determines what are the most logical adjustments to be
made to the remaining data,There are many approaches to this problem,some of which are
outlined in the book by Mah (1990),It is beyond the scope of these notes to go into any of these.
Our purpose here has been to introduce the reader to the nature of the problems that arise in
dealing with actual plant data.
- 74 -
We are now in a position to calculate the required quan tities,The fractional recovery fr of
N2 in the overhead vapor is
fr = (0.951)(735.13)/700,= 0.9987
Note that if we had used the unadjusted analysis,fr would be greater than 1.0 which is not a
useful result,The separation factor Sf is
Sf = (0.951/0.0024)/(0.049/0.9976) = 8067
XI,THE ELEMENTS OF DYNAMIC PROCESS MODELING
Revised September 23,1998
- 75 -
Up to this point almost all of our attention has been focused on processes operating in the
continuous steady state,While this is a useful idealization for many purposes,in reality processes
do not operate in the steady state,Their true behavior is dynamic,their state changing with time
due to disturbances of various kinds and deliberate changes in the required operating conditions,
Furthermore,many processes are operated in the unsteady-state mode for one reasons or another.
This includes batch processes such as are commonly used in the manufacture of fine chemicals
and pharmaceuticals,Some processes or parts thereof,are operated in the cyclic mode,examples
being pressure swing absorbers (PSA) and fixed bed reactors with a rapid catalyst de-
cay/regeneration cycle.
Thus a full understanding of the behavior of any process requires that we look at its
dynamics,Dynamic process modeling,when done fully,is very a very demanding undertaking
( Denn 1985,Franks 1972,Ramirez 1989,Silebi and Schiesser 1992),Our intent here is to provide
an introduction to such modeling and the solution of simple dynamic process problems.
We will look at two systems that are typical of various processing operations,namely,the
liquid-filled continuous stirred tank,and the well-mixed vapor drum or gasholder,In each case
we are interested in both the changes in total inventory as a function of the total flow rates into
and out of the system and the changes in inventory of individual components as functions of both
total flow rates and feed compositions.
In the case of the liquid-filled tank,one concern is its total inventory,If the flow rates into
the tank exceed those leaving it for a long enough period of time,the tank will overflow,If the
tank happens to be our bathtub,the result will be a wet floor or perhaps damage to the ceiling of
the room below,If the tank happens to be in a chemical plant and contains a corrosive or toxic
solution,the results could be far more serious,If,on the other hand,our tank is a drinking-water
supply tank and the flow rates out exceed the flow rates in for long enough,the tank will run dry
with unpleasant consequences.
In the case of a gasholder,pressure is the measure of total inventory,If the flow rate of
gas into the holder exceeds that leaving,the pressure will increase,If the supply pressure exceeds
the design pressure of the holder,eventually the pressure in the holder will exceed its design
pressure,As a result,the holder may rupture or burst with potentially disastrous con-sequences,
Thus,there are several reasons for developing a dynamic model of a process,If the
process is inherently dynamic,being either batch on cyclic in operation,then only a dynamic
model makes sense,Even if the process is intended to be operated in the continuous steady state,
its underlying behavior is dynamic,Steady state operation can only be achieved by good control,
In turn,good control can only be achieved through a proper understanding of the process
dynamics,
We are also interested in what happens when things fail,Suppose the control valve that
controls the level in a tank fails? What are the consequences as a function of time? How long
does the operator have to take remedial action? What sort of safety features should be included in
- 76 -
the process design so this catastrophic event will be prevented? Again,a dynamic model is
required to predict the course of events both without and with the safety system,The use of
dynamic models for hazard evaluation is a very important part of chemical engineering de-sign
and operations analysis,
A,Conservation of Mass for Dynamic Systems
In what follows the principle of conservation of mass will be applied to the two systems in
which we are interested,In order to carry our analysis,some assumptions are required:
1) Contr ol volume is well mixed
In dynamic modeling it is important to know how the physical quantities that describe the
state of system (such as composition and temperature) vary throughout the control volume,In
general these will vary over the three spatial coordinates and the model equations will be partial
differential equations ( PDE's) in time and the three spatial dimensions,While such models have
the most potential to accurately represent the dynamic performance of the system,they are
relatively difficult to solve,For many systems some simplifying assumptions are reasonably valid
and allow us to reduce the number of spatial dimensions that must be considered,If the system
involves flow through a cylindrical vessel or pipe (often the case in chemical processing),we can
assume that the spatial distribution is axisymmetrical and thereby reduce the number of spatial
dimensions to two,If the flow is highly turbulent,we can assume that the turbulent mixing is
intense enough that at any point along axis,the fluid is well mixed,(This is the so-called plug-flow
idealization.) It reduces the number of spatial dimensions to one,Still we are faced with the
solution of partial differential equations in time and one spatial dimension,a more manageable
prospect than the general case but one that can present some difficulties,
The assumption that leads to the simplest of dynamic models is that of well mixedness,It
is assumed that within the control volume the fluid is intensely mixed,either by turbulence
generated by the flow through the control volume or by the use of mechanical mixing device such
as a motor-driven agitator,Variables such as temperature and composition are assumed to have
the same value over the entire control volume at any in point in time,Therefore,no spatial
dimension is involved and the model equations become ordinary differential equations ( ODE's),
The control volume for which this assumption is most commonly made is an agitated tank through
which fluid continually flows (hence the terminology continuous stirred tank or CST).
2) System is isothermal at all times
This is an assumption of conve nience for our present analysis necessitated by the fact that
our only tool at this point is the mass balance,It is easily removed by applying the First Law of
- 77 -
Thermodynamics (conservation of
energy) but this is beyond the scope of
these notes,So,we will as sume that the
temperature of the system is constant
with respect to time.
Fig,XI-1,Surge and Mixing
Tank
B,Surge and Mixing Tanks
Consider the agitated tank shown
in Figure XI-1,There is a liquid flow in of flow rate F1 and a liquid flow out of flow rate F2,
There can,of course,be more than one input stream to the tank; inclu sion of additional feed
streams is straightforward,So is the inclusion of additional output streams.
First,let us chose a control volume,We could chose the entire volume of the tank but this
is not directly related to how much liquid there is in the tank,Let us chose instead the actual
volume of liquid in the tank at time t,This means if the height h of liquid in the tank varies with
time,so will the control volume,If the tank has a constant cross section area S,then the control
volume at any time is Sh.
Let Ψ = the holdup of a conserved quantity of in the control volume,If quantity of
interest is the total mass of liquid in the tank,then Ψ = Sh ρ where ρ = the fluid density in,say,lb-
mols/ ft 3,If we are interested instead in amount of the ith component,then Ψi = Sh ρx i where x i =
the mol fraction of the ith component in the tank,(Under the assumption of well mixedness,this
will be the same everywhere throughout the control volume.)
Let us calculate the change in inventory over a very small time interval?t starting at time
t,This will be given by
Where the superscript (a) denotes the average over the time interval?t,Also,x i,1 and x i,2 are the
mol fractions of the ith component in Streams 1 and 2 respectively and F 1 and F 2 are the
corresponding total molar flow rates,Also,if there are a total of nc components in the mixture,
Eqn,XI-1 represents a set of nc equations,one for each component.
(XI - 1) (t + t) - (t) = ( F x ) t - ( F x ) ti i 1 i,1 (a) 2 i,2 (a)y y
- 78 -
Note,Unless specifically mentioned,all dynamic mass balances in this section will be derived on a
molar basis,In non-reacting systems,mols are conserved,In reacting systems,care must be
taken to include terms to account for the extent of reaction for each component,This will be
illustrated in this example.
Let us divide Eqn,XI-1 by?t which gives
Taking the limit of the left-hand side as?t -> 0 gives
The limit of the left-hand side is merely the derivative and,in the limit,the average values in right-
hand side become the point values at time t,
Earlier,we related Ψi to the mol fraction x i of the ith component in the control volume,
One of the physical consequences of withdrawing fluid from a well-mixed control volume is that it
has to be some of the same fluid that is in th control volume,i.e.,its temperature and composition
must be equal to those in the control volume,Therefore,x i,2 must equal x i,(For clarity,we will
use x i,2 to represent x i,) Applying this fact and the definition of Ψi to Eqn,XI-3 gives
Let us take stock of the situation,We have nc ODE's,But we have nc+1 variables,namely the
nc mol fractions x i,2 and the liquid height h,Clearly,we need one more equation,This can be
obtained by invoking the closure condition on mol fractions,namely,
where j refers to any stream or control volume,Now,Eqn,XI-5 represents a problem,It is an
algebraic equation which states that the sum of the mol fractions of the jth stream or control
volume must equal unity at all times,While Eqns,XI-4 and XI-5 comprise a set of nc+1
equations to go with the set of nc+1 variables for the system,this set of nc+1 equations is of
mixed type,They are what is known as a set of differential and algebraic equations or DAE's for
short,
(XI - 2) (t + t) - (t)t = ( F x ) - ( F x ),i = 1,.,,,nci i 1 i,1 (a) 2 i,2 (a)y y
(XI - 3) d dt = F x - F x,i = 1,.,,,nci 1 i,1 2 i,2y
(XI - 4) d[Sh x ]dt = F x - F x,i = 1,.,,,nci,2 1 i,1 2 i,2r
(XI - 5) x
i = 1
nc
i,j = 1.0Σ
- 79 -
For a number of reasons,we would like the equations representing the dynamic behavior
of a system to be a pure set of ODE's of the form
where the y i are the state variables (i.e.,those variables which describe the internal state of the
system) and the z j are the input variables which drive the system,For our CST,h and the x i,2
while F 1,F 2,and the x i,1 are the input variables,There are excellent software packages (such as
ODEPACK) available for solving sets of ODE's of the form of Eqn,XI-6,And,for the purposes
of control system design,the dynamic equations for the system also need to be in that form.
So let us see how to convert Eqns,XI-4 and XI-5 from a set of DAE's to ODE's,First,
expand the derivative in Eqn,XI-4 using the chain rule of differentiation,Assume for the moment
that the density ρ is constant,We get
Now let us sum Eqn,XI-7 over I which gives
If Eqn,XI-5 is differentiated with respect to time for Stream 2,we get
Substituting Eqns,XI-5 and XI-9 into Eqn,XI-8 gives
(XI - 6) dy dt = f ( y,y,.,,,y,z,.,,,z )
dy dt = f ( y,y,.,,,y,z,.,,,z )
,
,
,
dy dt = f ( y,y,.,,,y,z,.,,,z )
1
1 1 2 n 1 m
2
2 1 2 n 1 m
n
n 1 2 n 1 m
(XI - 7) Sh d xdt + S x dh dt = F x - F xi,2 i,2 1 i,1 2 i,2r r
(XI - 8) Sh dx dt + S ( x F x F x
i = 1
nc i,2
i = 1
nc
i,2 1
i = 1
nc
i,1 2
i = 1
nc
i,2)
dh
dt = - r rΣ Σ Σ Σ
(XI - 9) ddt x d xdt = 0
i = 1
nc
i,2
i = 1
nc i,2 = Σ Σ
- 80 -
Eqn,XI-10 is known the overall mass balance and could be derived directly,However,the
approach outlined avoids confusion with regard to whether or not all of the equations are
independent.
If the density ρ is not constant but is a function of composition (and more generally
temperature as well),this does not change how to reduce the equations to a set of pure ODE's but
it does introduce some additional terms into Eqn,XI-10.
Now let us substitute Eqn,XI-10 into Eqn,XI-7,The component mass balance becomes
Except for d i v iding t h rough by some constants,Eqns,XI-10 and XI-11 are a set of nc+1
equations in the form required by Eqn,XI-6.
Now that we have this dynamic model,what can we do with it? One thing is dynamic
simulation,We can write a computer program that uses a standard set of routines for integrating
ODE's (IMSL and ODEPACK are two such) to solve these equations for specified variations in
the input variables,The computed responses of the state variables then tell us something about
the behavior of the system,This is generally what must be done since the model equations are
nonlinear,However,for some simple situations the model equations can be solved analytically,
Let us look at some.
Suppose we are only interested in how the liquid height h (also called the liquid level) in
the tank varies with time as the flow rates in and out vary,For this we only need to solve Eqn,XI-
10,Let us suppose that prior to time t = 0,F 1 and F 2 are equal but at time t = 0 one or the other
is changed to a new,constant value,Let this difference be?F = F 1 - F 2,Suppose also that at t=
0,h(t) = h 0,Then,direct integration of Eqn,XI-10 gives
We see that if?F > 0,then h increases linearly with t for as long as the imbalance in flow rates
continues,A new steady state value of h will never be reached,at least not until the tank
(XI - 10) S dh dt = F - F1 2r
Sh dx dt + x ( F - F ) = F x - F x,or
(XI - 11) Sh dx dt = F ( x - x ),i = 1,.,,,nc
i,2
i,2 1 2 1 i,1 2 i,2
i,2
1 i,1 i,2
r
r
(XI - 12) h(t ) = h + 1S F t0 r?
- 81 -
overflows,at which point our model in its present form no longer applies,there being two flows
out of the system,Conversely,if?F < 0,h will decrease until the tank runs dry.
One might ask how this can be prevented,One is by manually adjusting one of the flow
rates to exactly match the other,Now,if this is to be done by manual control,a number of
problems arise,Suppose that the flow in is truly constant,i.e,,does not vary with time,Then we
must set the output flow rate exactly equal to it using the available plant instrumentation,The
lesson of Chapter X should be that,with well-maintained instrumentation,we can set the output
flow rate almost equal to the input flow rate but very likely there will be a small error in one
direction or the other,Therefore h will change with time,very slowly perhaps,but change it will.
Occasional readjustments will be necessary,The situation will be worse if the input flow rate
varies with time,Frequent readjustments may be necessary making manual control a labor-
intensive activity.
Instead,we might want to use an automatic controller to maintain h at some desired value
or setpoint,Let this value be h s,The simplest control law we can use is proportional control,If
the output flow rate is to be adjusted to compensate for variations in the input flow rate,then the
control law will have the form
where F 2 0 and K c are adjustable controller parameters,In particular,K c is controller proportional
gain,Let us assume that prior to turning on the controller at t = 0,the system has been lined out
so that F 2 0 = F 1 and h = h s,Also at t = 0 let there be a change in net flow rate?F,Then,Eqn,XI-
11 becomes
Solving by the standard method for a 1st order ODE and keeping in mind that h(o) = h s gives
We see that the controller will maintain the level at the set point h s except for an offset that
increases with t to a final value of?F/ K c,This offset can be made smaller by increasing K c,
However,a more sophisticated control law might do better but this gets into the subject of
process control,The important point is that our model can be used to show how a control system
can be used to improve system performance.
(XI - 13) F = F - K ( h - h)2 20 c s
(XI - 14) dh dt = ( FK + h ) 1 - h
where = S K
c
s
c c
c
c
t t
t r
(XI - 15) h(t ) = h + FK (1 - e )s
c
-t
c
t
- 82 -
Suppose we are interested in the composition dynamics of the well-mixed tank,There are
a number of questions we might like to answer about its dynamic behavior,One concerns the
time it will take the output of the tank to come to a new steady state following a sudden change in
the input composition,If this change takes place instantaneously,it is commonly referred to as a
step change,Let us suppose that prior to t = 0 the tank has been operated with a input of
constant composition,Then the output will have the same composition,(This is the steady-state
solution of Eqn,XI-11.) Then,at t = 0,the input composition is instantly increased by a fixed
amount,This leads to a problem of the form
The solution to this equation is
Note that at t = 0 the output is equal to the initial value,As t increases,the exponential term goes
to zero and the output approaches
When t = τm the solution will have reached 63.2% of its final value,Thus the time constant tells
us how fast the system will respond to a sudden change in input,In the case of the well-mixed
tank,the time constant is the residence time or the ratio of the holdup in the tank to the flow rate
leaving it,For a given flow rate,this residence time can be varied by varying the active volume of
the tank,Thus the designer can adjust the transient behavior of the tank by choosing the
appropriate volume.
It is also instructive to look at the behavior of the tank when the input varies sinusoidally
with respect to time,As in the previous case,suppose the tank is at steady state at t = 0 at which
time the input concentration begins to vary sinusoidally,The mathematical expression of this
situation is
(XI - 16) d xdt = 1 (( x - x )
where x (t) = x for t < 0
and = (1 + ) x for t 0
also = Sh F (the res idence tim e or mixin g time con stant)
i,2
m
i,1 i,2
i,1 i,1
0
i,1
0
m
1
t
b
t r
≥
(XI - 17) x (t) = x [1 + (1 - e )]i,2 i,10 -t / mb t
i,2 i,1
0x (t) (1 + ) x→ b
- 83 -
The solution is
Note the solution consists of two terms,The first is multiplied by a decaying exponential in time
(the same exponential that appears in the step response transient) and therefore goes to zero after
enough time,The second term continues as a function of time and represents,in effect,a
sinusoidal steady state,Applying some trigonometric identities,we can rewrite the steady-state
part of Eqn,XX-19 as Note that the output is also a sinusoid and of the same period at the input
sinusoid,However the output lags behind the input by a factor which is called the phase lag or
phase angle,This phase angle is essentially zero at low frequencies,i,e,,frequencies for which
τm ω << 1,At high frequencies,the phase angle approaches - pi/2 (or -90 degrees),Also,the
amplitude of the output sinusoid compared to the input sinusoid decreases as the frequency
increases,What this means physically is that variations in the input which are extremely rapid
compared to the residence time will be strongly attenuated (or filtered out) by the tank while
those which are r elatively slow will pass through with little or no attenuation,This relationship of
the amplitude ratio (or gain) and the phase angle of a physical system is known as its frequency
response,The frequency response is a powerful tool for the dynamic analysis of systems and for
the design their control systems.
Let us now look at how chemical reaction can be included in our mathematical model of
the well-mixed tank,Consider the case of single reaction whose rate per unit volume is given by
r,Let a i be the stoichiometric coefficient for the ith component for this reaction,R i,the rate at
which the ith component is generated by reaction in the tank,is given by
(XI - 18) d xdt = 1 (( x - x )
where x (t) = x for t < 0
and = x [1 + ( t)] for t 0
where = the frequ ency at wh ich the in put is var ied
i,2
m
i,1 i,2
i,1 i,1
0
i,1
0
t
a w
w
sin ≥
(XI - 19) x (t) = x ( )( ) + 1 e
+ x (1 + ( ) + 1 [ ( t) - ( t)])
i,2 i,1
0 m
2
m
2
- t /
i,1
0
m
2 m
m
a t w
t w
a
t w w t w w
t
sin cos
(XI - 20) x (t) = x [1 + 1
( ) + 1
( t + )]
where = - ( )
i,2 i,1
0
m
2
- 1
m
a
t w
w f
f t w
sin
tan
- 84 -
Now,R i is just another input term in the component mass balance,Including this in Eqn,XI-4
gives
To get the overall mass balance,we do as we did before,namely sum Eqn,XI-22 over i,Doing
so and collecting terms gives the reacting version of Eqn,XI-10,namely,
Again,using Eqn,XI-23 to put Eqn,XI-22 in standard form gives,for the component mass
balance
C,Gas Holders
Let now look briefly at another version of
the well-mixed tank,namely,a vapor drum
or gas holder,The major dif -
Fig,XI-2,Gas Holder
ferences between this system and the well
mixed tank previously analyzed are:
1) The gas or vapor completely fills the available volume,(Needless to say,a vapor drum
must be closed while a mixing tank can be open at the top.)
(XI - 21) R = a S h ri i
(XI - 22) d[Sh x ]dt = F x - F x + a S h r,i = 1,.,,,nci,2 1 i,1 2 i,2 ir
(XI - 23) S dh dt = F - F + S h r
where = a
1 2
i = 1
nc
i
r s
s Σ
(XI - 24) Sh dx dt = F ( x - x ) + ( a - x ) ) S h r,i = 1,.,,,nci,2 1 i,1 i,2 i i,2r s
- 85 -
2) While assuming constant density for a liquid is not a bad first approximation,it
definitely is for a gas,
Thus,the major the major difference between the two systems will be in variable used to
track changes in inventory,For the mixing or surge tank,it is the liquid height h,For the vapor
drum it is the gas pressure P inside the drum.
Consider the vapor drum shown in Figure XI-2,Except that it must be a c losed volume,it
is very similar to the liquid-filled,well-mixed tank of Figure XI-1,Thus,we would expect the
mass balance equa tions to be similar and indeed they are,The only difference is in the overall
mass balance,In Eqn,XI-10 (non-reacting system ) or Eqn,XI-23 (reacting system),it was
assumed that is constant and that h varies with time,In the case of the flash drum,the situation
is reversed and the overall mass balance becomes
However,ρ is not a state variable but is a function of several state variables including pressure,
temperature,and composition,At the beginning of this section,we stipulated that our c-
onsiderations would be limited to isothermal systems,Also,to keep the analysis within hand,let
us assume that ρ is described by the ideal gas law,If a molar basis is used,then ρ is also
independent of composition,This leaves only pressure,A pplying the chain rule of differentiation
to Eqn,XI-25 gives
The ideal gas law states that ρ = P/RT,Carrying out the partial differentiation indicated in Eqn.
XI-26 provides a useable form of the overall mass balance,namely,
The reader should verify that the component mass balances are the same for both the mixing tank
and the gasholder.
(XI - 25) Sh d dt = F - F1 2r
(XI - 26) Sh ( P ) ( dP dt ) = F - F1 2r
(XI - 27) Sh P dP dt = F - F1 2r
- 86 -
XII,PROCESS SIMULATORS
- 87 -
Revised October 12,1999
Process simulators are computer programs for performing the kinds of process flowsheet
calculations described in these notes,These are of particular utility for simulating processes using
the more accurate,rigorous unit operations models,A good process simulator offers a broad
selection of such models,In addition,present day simulators have comprehensive physical
properties systems offering a wide choice of equations of state,activity coefficient models,etc.,as
well as extensive databases of physical properties data for both pure components and binary
mixtures,They also have extensive capabilities for flowsheet convergence and optimization,
Many also provide an elementary,though useful,equipment sizing and cost estimation capability.
The following are some commercially available process simulators,
A,Steady State
Two of the most widely used steady-state process simula tors are ASPEN PLUS (Aspen
Technology,Inc.,Cambridge,MA) and PRO-IV (Simulation Sciences,Fullterton,CA),Both of
these pro grams offer a large variety of unit operations models,a sub stantial physical properties
system and database,and robust methods for flowsheet convergence and optimization,ASPEN
PLUS also provides a conceptual level equipment sizing and cost est imation capability.
Hyprotech and COADE also offer steady-state simulators particularly suited for use on
high-end desktop computers,Aspen Technology also provides desktop capability.
B,Dynamic
For many years about the only dynamic simulator that was readily available was DYFLO
developed by Roger Franks of DuPont,It is still used in many quarters as attested to by the fact
that Franks's book,originally published in 1972,is still in print,DYFLO,while quite useful,
requires considerable effort and skill on the part of the user,
Dynamic process simulators of reasonable reliability are just beginning to become
commercially available,One is SPEEDUP that was originally developed by Roger Sargent and
John Perkins at Imperial College and is now offered commercially by Aspen Technology,
Another is OTIS which is currently be commercial ized by Simulation Sciences,Hyprotech has
recently come to the market with HYSIS,The leading academic dynamic simulators are
ASCEND (developed by A,Westerberg and colleagues at Carnegie-Mellon) and DIVA
(developed by E,Gilles and colleagues at the University of Stuttgart),Another,more recent,
simulator that originated at Imperial College but is now commercially available is gPROMS,
- 88 -
The use of real-time dynamic simulators for personnel training has grown considerably in
recent years,Several companies,such as ABB Simcon,offer a complete of services in this area
from process modeling to development of a set of training exercises for use with the simulator.
Dynamic simulation is undergoing rapid development,Its use for both engineering
simulation and personnel training has lead to a demand for simulators that are easy to use and that
can provide realistic simulations of real processes,Heretofore,the development of dynamic
simulator models has been a time-consuming and error-prone activity,The most recent trend is
the development of modeling environments that allow the user to configure new equipment
models with a minimum of effort,Two such systems are MODEL.LA developed by
Stephanopoulos at MIT and ForeSee being developed at CCNY.
89
BIBLIOGRAPHY
Revised October 12,1999
Brickman,L.,Mathematical Introduction to Linear Programming and Game Theory,Springer-
Verlag,New York,1989.
Denn,M.M.,Process Modeling,Longman,New York,1986.
Douglas,J.M.,Conceptual Design of Chemical Processes,Mc-Graw Hill,New York,1988.
Felder,R.,and Rousseau,Elementary Principles of Chemical Processes,2nd ed.,Wiley,New
York,1984.
Franks,R.G.E.,Modeling and Simulation in Chemical Engineering,Wiley,New York,1972.
Himmelblau,D.,Basic Principles and Calculations in Chemical Engineering,4th ed.,Prentice
Hall,Englewood Cliffs,New Jersey,1983,
Mah,R.S.H.,Chemical Process Structures and Information Flows,Butterworths,Boston,1990.
Ostrowski,Solutions of Equations and Systems of Equations,Academic Press,New York,1960.
Ramirez,W.F.,Computational Methods for Process Simulation,Butterworths,Boston,1989.
Reklaitis,G.V.,Introduction to Material and Energy Balances,Wiley,New York,1980.
Silebi,C.A.,and Schiesser,Dynamic Modeling of Transport Pro cess Systems,Academic Press,
New York,1992.
Westerberg,A.W.,et al.,Process Flowsheeting,Cambridge University Press,New York,1979.
90
91
APPENDIX A,REACTION STOICHIOMETRY
October 12,1999
The proper handling of reaction stoichiometry in material balance calculations is of the
utmost importance,It must be done correctly,otherwise the calculations are at best valueless
and,if the errors are not caught,downright dangerous,The purpose here is to provide some
guidelines to avoiding the most types of errors.
The recommended procedure for any reactor calculations for which a set of reactions is
known is as follows:
1) Convert any series reactions to parallel reactions,
While series reactions may be correct from the point of view of chemical kinetic
mechanisms,we are only interested here in getting the material balance bookkeeping straight,
This is more readily done if all the reactions are in parallel form.
2) Divide all reactions through by the stoichiometric coefficient of the key component(s)
Conversion and selectivity are based on one or more key components,The correct
interpretation of these is easier if all the key components have a stoichiometric coefficient of -1,
For most simple reaction systems there will only be one key component,However,there are
more complex systems for which there may be more than one key component,This will depend
on the reaction chemistry and the conversion and selectivity definitions,Note that each reaction
can have only one key component.
3) Convert all reaction system performance specifications to a molar basis
Reactor performance specifications are generally made on a weight basis,e.g.,(1) convert
a specified number of pounds of a key raw material to products and byproducts or (2) produce a
given number of kilograms of a specified product,Although raw materials are bought and
product sold by weight,reactions take place in mols,
4) Calculate the extents of reaction of all key components first
Since the selectivity structure is based on the extents of reaction of the key components,
these should obviously be calculated first,This is straightforward since in 2) we have insured that
all key components have unity stoichiometric coefficients.
5) Calculate the extents of reaction of all other components based on their stoichiometric
coefficients and the selectivity structure
92
Keep in mind that the selectivity of a key component to a particular product or byproduct
is the ratio of the mols of the key component converted to the target product or byproduct to the
totals mols of key component converted via all applicable reactions.
6) Calculate the stream summary for both the reactor feed and the effluent
Do this on both a molar and a weight basis.
7) Check the overall and conserved species balances
Do pounds in equal pounds out? This is a simple check if the calculations have been done
by spreadsheet,If there is a significant error,something has done wrong in the calculations,If it
is small,the error probably stems from not using consistent molecular weights,(Note,The
molecular weights given in Perry's Handbook are not consistent due to round-off.) The conserved
species balances are most readily checked on a molar basis.
93
APPENDIX B,EVALUATION OF EQUIPMENT MODEL PARAMETERS
October 12,1999
In order to utilize the linear material balance technique,one must be able to estimate
reasonable values for all the equipment model parameters,Let us look at the parameters required
for each of the four models described in Chapter IV:
1) Mixer (MIX)
No parameters are required for this model.
2) Reactor (REACT)
We require?i (extent of reaction) for each component,This is generally known in one of
three ways:
a,The rate of each reaction r j is specified in,say,kmol/hr,
Then extent of reaction of the ith component is given by
b,The conversion C k and selectivities S kj for each of a set of key components are specified,
Then,for the kth key component,its extent of reaction is given by
For the other components in the reaction set assigned to the kth key component
3) Separator (SEPAR)
SEPAR can be applied directly to simple separators such as a flash drum or simple
distillation column,For more complex separators (more than one feed or more than two output
streams),an appropriate model will have to be developed as demonstrated in Appendix C,For
SEPAR we require s I for each component in the feed to the separator,
( )B a ij
j
nr
ri j? =
=
∑1
1
( ),B C fk k in k? =?2?
( ),,B S ai k j i j k? =?3
94
Most separators in use in large-scale continuous plants work on the principle of the
equilibrium stage,The physical quantity that best characterizes the performance of this type of
equipment is the relative volatility,Thus,before beginning a material balance model,one should
estimate the relative volatilities of the components involved,These estimates do not have to be
highly accurate,What one want to know is the ranking of the components by relative volatility
and a reasonable estimate of the numerical values,particularly if any of these are less than,say,
1.2,
a,Flash Drum
If the feed to the flash drum consists of a number of components with very high relative
volatilities compared to the others,then one can assume that these components will be almost
entirely in the vapor leaving the drum while the low volatility components leave almost entirely in
the liquid,Thus,s I will be approximately 1.0 for the high volatility components and 0.0 for the
low.
The most general way to specify the performance of a flash drum is to specify the fraction
of the feed that is to leave as vapor,Let γ = V/F,Then we can solve the Rachford-Rice Equation
for γ,If k R is the k-value of the reference component,then we must solve
for γ,Then the component flow rates for the liquid are given by
So,if s i is defined with respect to the vapor,then
b,Simple Distillation Column
In general we know the specs on the key components,either purity specs or fractional
recovery specs,If the separation between the keys essentially complete,then a good first
approximation is to assume that all the components lighter than the light appear in the distillate
while all the components heavier than the heavy key leave in the bottoms,
( ) ( )B k zk i R i
i R
+? =∑1 11 0aa g g
( ) ( ),,B f k fL i
i R
F i? =
+?2
1
1
g
ga g
( ),
,
B s f fi L i
F i
=?3 1
95
If there are components whose volatilities are intermediate between the key components
or the separation between the keys is only partially complete,then we can estimate the distribution
of the key components using the Fenske Equation.
1) Calculate Nmin based on the specification s of the key components.
2) Choose one of the key components as the reference component,
3) Calculate the distribution of the non-key components using
where all the relative volatilities are with respect to the reference component.
4) Flow Split (FSPLIT)
The flow split parameter P is specified by the process engineer,It may be adjusted to
achieve a specified flow rate or composition spec in a recycle stream.
( ) ( / ) ( / )minB d b d bi i N R? =4 a
96
APPENDIX C,COMPLEX EQUIPMENT MODELS
October 13,1999
Complex equipment models can be developed using a combination of the four basic
equipment models developed in Chapter IV,We give a typical example,that for an absorber that
has two feeds,the vapor entering at the bottom and the absorption liquid entering at the top,This
is shown in Figure C-1.
Figure C-1,Representation of an Absorber as Complex Separator
The absorber is modeled as two simple separators,S-11 for the vapor phase and S-12 for the
liquid phase,The components absorbed from the vapor are added to the components not
stripped from the liquid in mixer M-2,M-1 does the same for the non-absorbed vapor-phase
components and the stripped liquid-phase components,Evaluation of the separation coefficients
for both phases can be done for dilute systems using the Kremser Equation,The details are given
in Appendix E.
A similar approach can be used for constructing models for other complex equipment items.
S-11 S-12
M-1
M-2
STin1
STin2
STin1 STin2
STout1
STout1
STout2 STout2
Complex Separator
(Gas Absorber)
Linear Model
Representation
S-1
97
APPENDIX D,LINEAR MATERIAL BALANCE SPREADSHEET
Revised October 13,1999
The following is the spreadsheet for the Ammonia Synthesis Loop example presented in Section VII,The
upper part of the spreadsheet contains the problem input data,One column,the reactor deltas,is not input but is
computed based on the molar flow rate of N2 in Stream 3 and the specified conversion,
The lower part of the spreadsheet is the stream summary that is shown in Section VII,Streams 1 and 2 are
input data,Each molar flow rate for Stream 3 is calculated based on the tear stream solution ( Eqns,E-8 and E-10),The
other streams are then calculated sequentially based on Eqns,E-1 through E-5.
Also shown in the upper part of the spreadsheet is the mol% of Argon in Stream 3,This is calculated directly
from the stream summary,Every time a new value for P is entered,the entire flowsheet is recalculated including the
mol% of Argon,This provides the mechanism for adjusting P to achieve the design spec of 10 Mol% Argon by trial and
error,
Ammonia SynLoop 7/12/95
Example for MB Notes
Problem Data:
Comp si ai Delta MW
H2 0.999 -3.0 -648.96 2.016 Conv = 0.25
N2 0.998 -1.0 -216.32 28.014 P = 0.05
Ar 0.998 0.0 0.00 39.948
NH3 0.010 2.0 432.64 17.031 %Argon 5.216%
Stream Summary:
Comp ST1 ST3 ST4 ST6 ST7 ST8 ST9
lb -mol/hr:
H2 750.00 2632.13 1983.17 1981.19 1.98 99.06 1882.13
N2 250.00 865.28 648.96 647.66 1.30 32.38 615.28
Ar 10.00 192.68 192.68 192.29 0.39 9.61 182.68
NH3 4.15 436.79 4.37 432.42 0.22 4.15
Total mol/hr 1010.00 3694.23 3261.60 2825.51 436.09 141.28 2684.23
MWavg 8.83 10.10 11.44 10.58 17.02 10.58 10.58
Total lb/hr 8914.98 37314.01 37314.01 29893.71 7420.29 1494.69 28399.03
98
APPENDIX E,THE KREMSER MODEL OF GAS ABSORBERS
Revised September 26,1999
The preliminary design of staged gas absorption systems and liquid-liquid extractors can
generally be accomplished with a modest amount of effort using the Kremser Equation,This is
particularly the case if the components to be absorbed or extracted from a process stream are
"dilute",i.e.,present in the stream at relatively low concentrations,The system must also be
essentially isothermal which is generally the case for liquid-liquid extractors but not always so
with gas absorbers,Also,the vapor-liquid or liquid-liquid equilibrium should not be a strong
function of composition,This requirement is satisfied for dilute solutions due to the small
concentration range of either an absorbed or extracted component.
Consider the staged absorber shown in Figure E-1,The column contains N stages (6 are
shown) that,for the moment,we will take to be 100% efficient,Vapor feed enters below the
bottom tray of the column (Tray #1) and liquid absorbent enters on the top or Nth tray,Yin is the
mol fraction in the feed of a typical component being absorbed; Yout is the mol fraction of that
component leaving the column,Similarly,Xin and Xout are the mol fractions of that component
in the liquid entering and
leaving the column.
There are two cases of
interest,For the first,the
rating case,we are told
how many trays there are
in the column (N),what
the vapor and liquid feed
rates (V and L) are,and
what the inlet compositions
Yin and Xin are,The
problem is to determine the
separation performance of
the column,i.e.,what are
Yout and Xout? For the
second Figure E-1,Staged Gas Absorber
case,that of design,we are
told what V,Yin,and Xin are as well as the value of Yout that the column is to achieve,The
problem is to determine suitable values of L and N,
A,The Kremser Model
Let us start with a mass balance around the nth tray for a typical component,This gives
99
If the tray is at thermodynamic equilibrium,then
Substituting Eqn,2 into Eqn,1 give
In general we would also need an energy balance and an overall mass balance to complete the
performance equations for the tray,We eliminate this requirement by making some simplifying
assumptions:
a) y n << 1.0 and x n << 1.0,i.e.,the feeds to the column are dilute with respect to the
components to be transferred from one phase to the other,Therefore there will be a
negligible change in V and L from tray to tray,As a result,we have
i.e.,V and L are constant from tray to tray.
b) The column is isothermal and isobaric,Th is means that k n does not vary from tray
to tray,i.e.,k n = k.
With these assumptions,Eqn,3 becomes
Where S = Vk/L,This quantity is known as the stripping factor for the component in question.
Eqn,5 is a difference equation,Before we can solve it,we must establish the boundary
conditions which are,for the vapor feed to the bottom of the column (n=0),
x 0 = y in /k
And,for the liquid feed to the top of the column (n=N+1),
(1) V y V y L x L x = 0n 1 n 1 n n n n n 1 n + ++ + + 1
(2) y = k xn n n
(3) V k x ( V k + L ) x + L x = 0n 1 n 1 n 1 n n n n n 1 n + 1 +
(4) V = V = V and L = L = Ln n 1 n n + 1?
(5) S x (S + 1) x + x = 0n n n + 1?1
100
x N +1 = x in,
To solve Eqn,5 we assume a solution of the form
Substituting Eqn,6 into Eqn,5 and collecting terms gives a characteristic equation of the form
Solving Eqn,7 for its characteristic values gives 1 = 1 and 2 = S,Our assumed solution (Eqn.
6) now has the form
Applying the boundary conditions,solving for c 1 and c 2,and collecting terms gives
We are generally interested how much is absorbed,namely the difference between y out (= y N ) and
y in,This is given by
Now,let us define a fractional removal factor as
Rearranging Eqn,10 gives
(6) x = c n nl
(7) (S + 1) + S = 02l l?
(8a) x = c + c or
(8b c + c S
n 1 1
n
2 2
n
1 2
n
l l
) =
(9) x = S SS 1 y k + S 1S 1 xn
N + 1 n
N + 1
in
n
N + 1 in
(10) y = S S 1S 1 y + S 1S 1 k xout N N + 1 in
N
N + 1 in
(11) = y yk x yout in
in in
Φ
101
First,let us look at some limiting cases for Eqn,10 as we let the number of trays increase to
infinity,For S < 1,we have
For S > 1,we have
What can we learn from these limiting cases? From Eqn,13 we see that if we want to
reduce the amount of material in the vapor feed to zero,S must be < 1,In this case,the amount
of the component in question in the vapor leaving the column will be equal to the amount in
equilibrium with the liquid feed to the top of the column,If we are using a "clean" absorbent,
then x in = 0 and we can theoretically have a clean vapor leaving the column,However,economics
and environmental considerations dictate that we reuse the absorbent,This means separating the
absorbed material from it and recycling it back to the absorber,Since no separation using finite
means can be 100% complete,x in will generally not be zero,
If,instead of absorbing material from a vapor,we want to remove it from a liquid using a
vapor (a procedure known as stripping),then Eqn,14 provides guidance,This shows that if we
have a clean stripping vapor (y in = 0),then we can theoretically reduce the amount of material in
the liquid leaving the column to zero,Otherwise,x out is in equilibrium with y in,
B,Absorber Design
1) Selection of Oper ating Conditions
The value of S can be set by the designer by adjusting any of a number of variables,
Usually V is set process considerations,So if we want to enhance absorption,we must make S <
1 by either making k sufficiently small or L sufficiently large,L can be made large up to a point,
For a given V there is a limit on how large L can be without causing mal-operation of the column.
(12) = S 1S 1
N
N + 1Φ
(13) lim y x
N out in
→∞
→ 0 + k
(14) x
N out
+ lim
→∞
→ iny k 0
102
k can be made small by raising the pressure and in some cases by lowering the
temperature,If the component being absorbed is subcritical and obeys Raoult's Law with respect
to the absorbent,then
where P 0 (T) = the vapor of the absorbed component,It is obvious that k can be reduced by either
increasing the column operating pressure P col or by reducing the temperature T of the absorbent,
Generally the available feed pressure will determine the column operating pressure; it is seldom
economic to compress the feed to a higher pressure,The absorbent temperature cannot generally
be reduced below a the available cooling water temperature plus a suitable approach temperature.
Otherwise refrigeration will be required which is expensive,Thus,it is clear that design of an
absorber system is requires the evaluation of alternatives and some design optimization.
2) Tray Requirements
If the absorber performance is specified,then fractional removal factor is known and
Eqn,12 can be solved for N,This gives a familiar form of the Kremser Equation
Let us examine how the choice of S affects the tray requirements as a function,Table 1 gives
and idea of the relationship.
Table 1.
Number of trays N for φ =
S 0.9 0.95 0.99 0.999
0.95 7.2 13.0 34.8 76.6
0.90 6.1 10.1 22.7 43.8
(15) k = P (T)P
0
col
(16) N =
[ 1 S 1 ]
S
ln
ln
f
f
103
0.80 4.6 7.0 13.6 23.8
0.70 3.7 5.3 9.6 16.0
We see that for high fractional removals ( φ > 0.99),the tray requirements are quite high for S >
0.7 to 0.8,Since tray efficiencies for absorption are lower than those for distillation (25 to 40% is
typical),ideal tray requirements should not exceed 15 to 25 if the column height is not to be
excessive (> 100 ft at a tray spacing of 2.0 ft.),Thus adjusting L to give S ≤ 0.8 is useful rule of
thumb for choosing a reasonable starting point for a design.
The effect of tray efficiency can also be included in the Kremser model,If we use the
Murphree Vapor Phase efficiency E MV defined as
we can solve for the actual tray requirement N ACTUAL and the overall column efficiency E O which
gives
where N Ideal is given by Eqn,16,We see that the actual tray requirement is independent of and
is a function only S and E MV,The overall column efficiency E O as a function of these last two
variables is shown in Table 2.
Table 2.
E O for S =
E MV 0.95 0.9 0.8 0.7
1.0 1.0 1.0 1.0 1.0
0.9 0.898 0.895 0.889 0.882
0.8 0.79 6 0.791 0.781 0.769
0.5 0.494 0.456 0.472 0.456
0.3 0.295 0.289 0.277 0.264
We see that there is not much difference between E MV and E O over most of this table,The
maximum error (S=0.6,E MV = 0.3) is 12%,Thus,for preliminary design,setting E O = E MV will
not introduce a significant error.
C,Absorber-Stripper Systems
(17) E = y yk x yMV n n1
n n1
(18) E = NN = [(1 E ) + S E ] SO ideal
actual
MV MVln
ln
104
Absorbing a component from a gas stream into a liquid (mass separating agent or MSA)
merely transfers a separation problem from one stream to another if our goal is to recover the
absorbed component in a relatively pure form,Even if all we want to do is throw it away or burn
it,we still have the problem of dealing with the MSA,The flow rate of this stream will generally
be several times larger than the molar flow rate of the component being absorbed,If it is an
expensive solvent,we must reuse it for our absorption system to be economic,This means
separating most of the absorbed component from it before recycling it back to the absorber,Even
if the MSA is water and relatively inexpensive,we may still have to reuse it for environmental
reasons if the material being absorbed is toxic,carcinogenic,or merely smells bad,The question
arises as to what is the most economical separation system for this task.
F
igur
e E-
2,
Abs
orbe
r-
Stri
pper
Syst
em
The two most common
choices are stripping and
distillation,Since distillation is a
separate subject,we will only consider stripper in these notes,A schematic diagram of an
absorber-stripper system is shown in Figure E-2,The absorber is similar to that shown in Fig,E-1
except that the liquid stream leaving the bottom of the column goes to the stripper,A stripping
vapor of flow rate V S is used to strip the component of interest from the liquid feed to the
stripper.
Now,having a stripper will do little good if the concentration of the absorbate in the vapor
leaving the stripper is not considerably higher than it is in the feed to the absorber,We have
already established that the absorber must be designed so that its stripping factor S A < 1,This
sets a value for L,Therefore,the first design decision for the stripper is the value of V S,
Let us assume that the Kremser model applies to the stripper as well as the absorber,As
in the case of the absorber itself,let us look at the limiting case of infinite trays,From Eqn,14,
X Sout = 0 for clean stripping vapor ( y Sin = 0) if S S > 1,Now,if we use the same rule of thumb for
S S as we do for S A,S S 1/0.8 = 1.25,Therefore S S /S A = 1.25/0.8 = 1.6,
105
This means that
Thus,for V S to be much less than V A,K S must be much greater than K A,How can this be
accomplished? Two of the most commonly used ways are by pressure swing and by thermal
swing,quite often a combination of both,
Let us consider pressure swing first,Suppose that the absorber operates at a pressure of
10 atm and we choose to operate the stripper at 1 atm,Then,from Eqn,15,K S = 10 K A so V S =
0.16 V A,This is an improvement but not an overwhelming one,We have only increased the
concentration of the absorbate by a factor of 6.25,It is obvious that,unless the absorber pressure
is very high,stripping by pressure swing only is not very effective,
Suppose that the absorber is run at a temperature just above that of cooling water,say 30
C,and that we heat the bottoms of the absorber to a much higher temperature before feeding it to
the stripper,If the absorbate is subcritical and follows Raoult's Law,we would expect a
substantial increase in K going from the absorber to the stripper,As an example,suppose we are
absorbing methanol in water and we operate the absorber at 30 C and the stripper at 85 C,Then
K in the absorber will be about 0.25 atm while that in the stripper will be about 2.0 atm,an
improvement of a factor of 8.0,If we operate the stripper almost at the atmospheric boiling point
of water,then the improvement will be closer to a factor of 15.0.
If the absorber also operates at 10 atm,then the overall improvement by the combination of
pressure and temperature swing is on the order of 50 to 100 depending upon the operating
temperature of the stripper.
Recovering the absorbate from the stripping vapor may still be a problem,If the amount
of vapor is small enough,the absorbate can be recovered by condensation if that is our purpose,
On the other hand,if the absorber is used for pollution control,the overhead from the stripper can
be sent directly to an incinerator to recover the heating value of the absorbate,There are many
possibilities.
It should be kept in mind that the above analysis is based on the limiting behavior of
absorbers and strippers for infinite trays,Since the number of trays must be finite,there will be
some unstripped absorbate in the bottoms of the stripper,some of which will,on recycle,appear
in the overhead of the absorber (refer to Eqn,10),The amount left in the stripper bottoms
becomes another design decision subject to optimization.
(19) V KV K = 1.6S S
A A