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 Chapter 0 Preliminaries
The schematic description of a complex chemical process in terms of 
individual units and the principles which govern the distribution of material 
and energy flows among those units are the most fundamental tools of the 
chemical engineer. The analysis of chemical process flowsheets is what this 
electronic book is about.
Before beginning to analyze process flowsheets, we will review some basic 
concepts from physics and chemistry. These include dimensions and units, 
the mol concept, and the variables used to describe streams consisting of 
several chemical species.
0.1 Analysis vs Design
0.2 Dimensions and Units
0.3 The Mol Concept
0.4 Composition Variables
0.5 Flow Rates
0.1 Analysis vs Design
This section serves as an orientation to the goals of this course.
Role of the 
Chemical
Engineer
The primary functions of the chemical engineer are to design and safely 
operate processes that convert raw materials into desired products.
Design The design of a process involves the selection of an appropriate sequence 
of transformations and the specification of the equipment necessary to 
accomplish them.
Suppose it is known that a desired substance P results from the reaction of 
materials A and C via the reaction A + C = P. Reactant A is expensive 
but available in a mixture with substance B, which is inert. A schematic of 
a process to produce P then might be drawn as follows:
First the reactants are mixed. The resulting mixture is then fed to a vessel in 
which the conditions of temperature and pressure are selected to promote 
the formation of the desired product, which then must be separated from 
the other species in the reactor exit stream. It is assumed here that the 
reactant A is not completely converted.
Since the reactant A is expensive, it might be worthwhile to recycle part of 
the overhead stream from the separator so that more A can be converted.
We cannot recycle all of this stream because the substance B is inert. If no 
amount of B left the process, there would be no way to run this process at 
steady state. A steady state is one in which the conditions (temperatures, 
pressures, flow rates) are all independent of time. In this case, if no B was 
purged, the amount of B in the vessels composing the process would 
continually increase. The modified process can then be visualized as 
shown below.
In this figure we have numbered the streams. A new equipment item has 
been added: a splitter. It is worthwhile to specify more precisely what is 
meant by each of these equipment items.
A mixer is a device with more than one inlet stream and a single exit 
stream.
A splitter has a single inlet stream and multiple exit streams of the same 
composition.
A separator has multiple exit streams of different compositions. 
A reactor is a device with a single inlet and a single exit stream of different 
compositions.
The schematic diagram showing the numbered streams entering and leaving 
each equipment item and labeled with the species present is referred to as a 
flowsheet.
A Look Ahead To proceed further with the design for this process we would need to 
specify the conditions of temperature and pressure that minimize the cost of 
production. You will acquire the tools necessary to make these decisions 
in subsequent courses. For example, not shown in the flowsheet is the size 
of the pipes used to conduct fluids from one vessel to another. You will 
learn in a Fluid Mechanics course how to select pipe sizes and pumps to 
move fluids. Clearly the size of the separator must depend on the nature of 
the separation being accomplished. You will learn in Thermodynamics and 
Mass Transfer courses how to select the separation process and to specify 
the geometry of the vessels and their internal components. Similarly, you 
will learn in a Reactor Design course the principles upon which reactor 
selection and sizing are based.
Analysis This is a very different activity from design. When we do analysis we are 
given a flowsheet and asked to compute the flow rates, temperatures and 
pressures of all streams given some set of specifications.
For example, suppose in the flowsheet above the following specifications 
were given: the composition of the feed (stream 1), the desired flow rate 
of product (stream 8), the conversion of A in the reactor. Could we then 
compute the flow rates of all other streams? If not, what additional 
information is necessary?
Design and analysis are not disjoint activities. At many points in the design 
process it may be necessary to determine the flow rates of all streams in 
order to size certain equipment items before other design decisions can be 
made.
The Goals for
This Course
Since you are not yet prepared to design processes, your assignments in 
this course will be limited to analysis of given flowsheets. For each 
flowsheet to be analyzed some set of specifications will be given. You will 
learn how to apply material and energy balances to compute the flow rates 
and temperatures of all streams. For complex flowsheets the number of 
equations to be solved can be quite large. One of the major goals of this 
course is to develop the tools to quickly determine an efficient method of 
solution.
Questions to be answered for each flowsheet are: 
1. Is there enough information to proceed?
2. If there is enough information to proceed, where should calculations 
begin?
3. Can some minimum set of calculations be selected in advance of 
actually performing them?
4. What is the nature of the equations that must be solved?
5. Can the equations be solved sequentially, or must an entire set of 
equations be solved simultaneously?
6. Will non-linear equations arise? 
A Necessary
Limitation
In this course, it will not be necessary for us to be concerned with the 
internal details of the equipment items that compose the flowsheets under 
study. You will learn about those details in subsequent courses. Some 
students find this frustrating, but this is a necessary limitation in order to 
have a goal attainable by second- and third-semester students.
Whenever possible, class discussion will be supplemented by enrichment 
material which will include photographs, slides, and drawings of actual 
equipment.
Logical
Progression The nature of the mathematical problems which arise is directly related to
 -- the number of process units
 -- the number of reactions which occur
 -- the nature of the specifications
 -- the need to determine temperatures or energy demands.
If you glance at the chapter headings in the Table of Contents, you will see 
that the chapters are arranged in increasing order of difficulty as each of 
these issues is addressed.
Significant Figures0.2.6
Dimensional Homogeneity0.2.5
Unit Conversion0.2.4
Other Unit Systems0.2.3
The SI Unit System0.2.2
Dimensions and Units0.2.1
In this section we explore the most common unit systems, the issues 
surrounding the use of quantities with units in evaluationg mathematical 
expressions, and conventions regarding significant figures.
0.2 Dimensions and Units
0.2.1 Dimensions and Units
Fundamental
Dimension
A dimension is a fundamental concept from physics and is universal, that is, 
the concept would be recognized by a Martian as unerringly as an 
Earthling. Fundamental dimensions include mass (M), length (L), and time 
(T). Although it is not necessary, we will treat temperature (Θ) as a 
fundamental dimension also.
A dimension is a fundamental attribute of physical objects. All other 
attributes of such objects can be expressed in terms of these fundamental 
dimensions through the use of natural laws, that is, principles of physics.
Units A unit on theother hand is an arbitrary assignment of a "size" to some 
quantity of a particular dimension. Units are the result of some social 
agreement among living beings and are entirely unique to a culture. For 
example, the original definition of the kilogram was based on an actual 
metallic cylinder housed under glass in an inert atmosphere. Similarly, 
there was a rod, also preserved under glass, which served as the standard 
meter. The definition of the unit "ft" has an anthropomorphic origin. Take 
a guess at the body part to which it was initially related.
The value of a measured quantity is expressed in terms of a numerical 
magnitude and a unit. The property of the object we are describing is 
independent of the units in which we express that quantity. For example, a 
mass of 3.750 kg is still the same mass even if we describe it as 8.2673 lb.
All we are doing when we say that a mass is 3.75 kg is that when we 
determine the mass of the object in question, it was found to be 3.75 times 
the mass of our standard kilogram or 8.2673 times the mass of our standard 
pound.
Similarly, a length of 5.0 ft is simply 5.0 times the length of a standard ft or 
1.524 times that of a standard meter. It is still the same length.
0.2.2 The SI Unit System
A unit system is simply an agreement among people on units of measure for 
the fundamental dimensions. One such system is the SI (Systeme 
Internationale) system.
The SI
System
Fundamental SI
Dimension Unit
___________________
M kg
L m
T s
Θ K
It is assumed that you have encountered these units in high school chemistry 
and physics courses. What is unique about the SI system is an extensive 
and well-defined set of prefixes to represent multiples of the base units.
Some
Differences
in Mathcad
The symbols adopted in Mathcad for some units are not consistent with the 
SI definitions. For example, the SI symbols for second and gram are "s" 
and "g", while "sec" and "gm" are used in Mathcad. In the text regions, we 
will consistently use the SI symbols. The Mathcad built-in units will be 
used in equation regions.
SI Prefixes Prefix Power Symbol
of 10
______________________
Giga 9 G
Mega 6 M
kilo 3 k
milli -3 m
micro -6 µ
nano -9 n
pico -12 p
Masses,
Lengths, and
Derived Units
The purpose of the prefixes is to permit the expression of physical 
quantities related to mass and length and derived units (defined below) 
with numerical values that are close to 1.0 while simultaneously avoiding 
power of ten notation. For example, 2.5 mg/s is preferable to 0.0025 
g/s. We refer to a 100 MW power plant rather than a 100,000,000 W
or a 1.0*108 W power plant. No other system of units in current use has 
such a consistent and flexible set of prefixes.
Note that only prefixes that represent powers of ten divisible by three are 
SI prefixes. The prefixes "deci" for 0.1 and "centi" for 0.01 of a base unit 
were in common use in the older literature but are discouraged now. The 
one exception is the cm, which is an accepted SI unit.
Temperature
and Time
Ordinarily, the prefixes are not used with temperatures. Typically, only the 
prefixes indicating negative powers of ten are used with time, such as ms, 
ns, or µs. Rather than using multiples of the second, other commonly 
accepted units are allowed such as the hour, day or year. Whenever a 
magnitude of a rate process is inconvenient with the second, either a prefix 
on the numerator or an alternate time unit can be used. For example, the 
mass flow rate 5,712,000 g/s is best expressed as either 5.712 Mg/s or 
1.5867 kg/h.
Derived Units In addition to the units associated with the fundamental dimensions, there 
are two types of derived units: compound units and derived equivalents.
Compound
Units
A compound unit is simply a product or ratio of the defined units. For 
example, the fundamental dimensions of acceleration are L/T2. Thus the 
SI "unit" of acceleration is m/s2. No new definition is involved.
Defined
Equivalents
Certain compound units are used so frequently, they are given special 
names for convenience. For example, from Newton's Law we know that 
the fundamental dimensions of force are ML/T2. The SI unit of force is 
then the kg m/ s2. This combination occurs in so many contexts it is given 
a special name: the Newton. Thus 1 N = 1 kg m /s2.
Other defined equivalents you should know are the Pascal ( Pa = N/m2), a 
unit of pressure, the Joule ( J = N m), a unit of energy, and the Watt (W = 
J/s) a unit of power.
Note that a mN is a milli-Newton, while a Nm is a Joule. To avoid such 
possible confusion, whenever units are multiplied, a multiplication symbol is 
employed between them. For example, a m*N is a Joule, while a mN is a 
milli-Newton. What are mg, gm, m*g?
The prefixes can be applied to the derived units just as they are applied 
with the fundamental units. A GJ is 109 J. What is a MPa?
0.2.3 Other Unit Systems
There are other unit systems still in general use. An effective engineer must 
be able to convert among all of these.
British
Gravitational
System
The British gravitational system selects force (F) as fundamental instead of 
mass.
Fundamental BG
Dimension Unit
________________
F lbf
L ft
T s
Θ R
In this system mass is a derived unit. From Newton's law, mass = 
force/acceleration. The derived unit for the mass is the slug.
slug = lbf s2/ ft
As long as all mass quantities are expressed in slugs no confusion should 
arise. However, an alternate choice of fundamental dimensions was 
made in the American engineering system.
American
Engineering
System
In the AE system, mass is fundamental, so force is once more a 
derived unit.
Fundamental AE
Dimension Unit
________________
M lbm
L ft
T s
Θ R
One reason for the confusion is that the unit of mass in the AE system was 
given the same name as the unit of force. Strike one! Since it is common 
to encounter situations in which both are used, it is necessary to 
distinguish the two uses of the word "pound" with subscripts. Strike two!
Recall that in the AE system force is a derived unit. But the choice made 
for the force unit has been a source of consternation for engineering 
students for two centuries. Instead of defining a unit that is consistent with 
Newton's Law, a unit was adopted that made the mass and weight of an 
object at sea level on earth exactly the same. It was simply a bad idea, 
considering that the convenience it offered is but a slight one. The defined 
unit of force in the AE system is 
lbf = 32.174 ft lbm/ s2
The only reasonable viewpoint to take today is to consider the above 
equation as the definition of a conversion factor. A conversion factor is 
any product or ratio of numbers and units that is identically equal to unity.
1 = 32.174 ft lbm / s2 lbf
Mass and
Weight
In the Newtonian view of the world, the mass of an object is one of its 
fundamental properties. The weight of an object is not an intrinsic 
property, but instead varies relative to its location in a gravitational field.
On earth, the sea level acceleration of gravity is 32.174 ft/s2, or 9.80665 
m/s2. Thus a mass of 1 kg weighs 9.80665 N. A mass of 1.0 lbm weighs 
exactly 1.0 lbf.
W = mg = (1.0 lbm) (32.174 ft/s2) = 1 lbf
Note that the last step involved the use of the conversion factor referred to 
earlier. Strike three! If the local acceleration of gravity is not 32.174 
ft/s2, the equality between mass and weight in pounds is no longer 
maintained.
0.2.4 Unit Conversion
Recall that the number 1 is called the multiplicative identity. Multiplying 
any quantity by this number results in the same quantity. A conversion 
factor is simply a special form of the numberone.
Suppose you have a value of time in days. To convert it to a time in hours 
you must find a conversion factor with hours in the numerator and days in 
the denominator. Since the expression 1 day = 24 h is an identity, the 
needed conversion factor is simply
1 = 24 h/d
The conversion process for the case where t = 1.5 day can be written:
t = (1.5 day) ( 24 h/day) = 36 h
Double click on the popup icon at the left to check your answer. When 
you are through examining the solution, just click anywhere outside the 
popup to return here.
Convert 57.2 mi/h to m/s.Example 0.2.4.1
W = J/s
hp = 550 ft lbf / s = 2544 Btu / h = 746 W
Power
J = Nm = 107 erg
cal = 4.1868 J
Btu = 778.2 ft lbf = 1055 J
kwh = 3.600 MJ
Energy
atm = 101.325 kPa = 1.01325 bar 
 = 760 mm Hg = 33.9 ft water
 = 14.696 psi
Pressure
N = 105 dynes
lbf = 4.4482 N
Force
m3 = 1000 liter = 35.3145 ft3
ft3 = 7.4805 gal
Volume
m = 39.37 in = 3.2808 ft
ft = 12 in = 30.48 cm
mi = 5280 ft
Length
kg = 2.20462 lb
lb = 453.593 g = 16 oz
Mass
As shown above, unit conversion is very simple process as long as the 
needed identities are available. Here is a short list that you should 
memorize.
Tables of
Identities
y = 1.5 m + (320 mi/h) (5280 ft/mi) (m/3.2808 ft)(h/3600 s) (3.7 s)
 - (1/2) 9.8 m/s2 (3.7s)2 = 463.7 m
It should be clear that the velocity must be converted to m/s.
y = 1.5 m + 320 mi/h (3.7 s) - (1/2) 9.8 m/s2 (3.7s)2
For example, substitute the values y0 = 1.5 m, v0 = 320 mi/h and t = 3.7s 
into the projectile motion equation above. Even though the equation is 
valid (both dimensionally homogeneous and consistent with physical laws), 
when the suggested values are substituted for each symbol, the result does 
not reduce to a single length as desired.
Only when we are in the process of evaluating symbolic mathematical 
expressions are we concerned with a related topic: unit consistency. Each 
additive term in an expression must have the same units.
Unit
Consistency
 L [=] L + (L/T) T - (L/T2) T2
When the fundamental dimensions of each term are written out explicitly, it 
is readily verified that each term has the dimensions of length, so that this 
equation is dimensionally homogeneous. In the expression below, the 
bracketed equal sign [=] may be read "has the dimensions of".
y = y0 + v0t -1/2 g t2
For example, a projectile fired vertically from an elevation y0, with a 
velocity v0, in a constant gravitational field with acceleration -g attains a 
height y after time t given by
In symbolic equations, the fundamental dimensions of every additive term 
must be the same. This is the requirement of dimensional homogeneity.
Any equation that is not dimensionally homogeneous is certainly invalid.
Dimensional
Homogeneity
Three topics are discussed here: dimensional homogeneity, unit 
consistency, and arguments of transcendental functions.
0.2.5 Dimensional Homogeneity
Example 0.2.5.1 Is the equation P/ρ + v2/2 + gz = c dimensionally homogeneous if
P is pressure, ρ is density, v is velocity, g is acceleration, z is elevation, 
and c is a constant with dimensions (L/T)2 ?
Show your reasoning on scratch paper and then check your work by 
clicking on the popup icon to the left.
Example 0.2.5.2 Using the equation discussed in the previous example, what is the pressure 
if the density is 0.967 g/cm3, the velocity is 18.2 m/s, the elevation is 9.1m 
and the constant c = 930 (m/s)2?
Try solving this problem both by hand and with Mathcad. Then check 
your work by clicking on the popup icon to the left.
Arguments of
Transcendental
Functions
Arguments of trigonometric and logarithmic functions must be 
dimensionless. One cannot evaluate ln(apple), sin(3.0 ft), or exp(37 s).
One often encounters situations in which this rule is apparently violated.
For example, suppose the concentration C (mol/liter) of a reactant in a 
vessel is exponentially decaying with time (h) as
 C = a e -bt
where it is stated that a, b are constants with a = 7.8 and b = 2.5. What is 
meant is that a = 7.8 mol/liter and b = 2.5 h-1, for this is the only way that 
the expression can be dimensionally homogeneous and have a 
dimensionless argument for the exponent.
0.2.6 Significant Figures
All engineers speak a common language when the topic is significant 
figures. You must learn this language in order to understand what you read 
and to report the results of computations.
The Rules To determine the significant figures in a number
1. Find the leftmost, nonzero digit.
234
0456.3
0078000003
2.0000001
^
Leftmost nonzero digit
2. The number of significant figures in a number without a decimal point is 
the number of digits from the leftmost nonzero digit to the rightmost 
nonzero digit.
Number Significant
Figures
_________________
2300 2
230001 6
02301 4
023010 4
The Rule
for Decimals
3. For a number with a decimal point, include all the digits from the 
leftmost nonzero digit to the last digit, including zeros.
Number Significant
Figures
_________________
2300. 4
230.001 6
0230.10 5
0.0103 3
0.01030 4
Use of 
Significant
Figures
There is an accepted convention regarding the reporting of computed 
results. It is simply that a final result should be listed to no more significant 
figures than the least accurate factor in the expression evaluated.
For example, the distance traveled by a vehicle moving with velocity 
16.223 m/s in 2.0s is 32m. The fact that your calculator or Mathcad may 
list 32.446000 m as the result of the computation is an electronic artifact.
It is your job to round the result to an appropriate number of significant 
figures. Intermediate calculations should be performed with the full 
precision of the tool you are using (slide rule, calculator, computer, abacus) 
but the final result must reflect the true precision.
Reporting a physical quantity to an unjustified number of significant figures 
is considered poor practice.
0.3 The Mol Concept
In this section we briefly review one fundamental principle of chemistry. 
Our goal is only to review the "why" and the "how" of the mol concept.
Some Simple
Facts
1. If we wanted to carry out the reaction A + 2B = AB2 so that all the 
A was converted, we would have to measure out some number of 
molecules of A and twice that many of B. That is, we are interested in the 
relative number of molecules.
2. What we can measure easily is the mass of some quantity of material, 
not the number of molecules in the given quantity of matter.
What we need is a method of converting an easily measured mass of 
substance into an equivalent number of molecules of that substance. But 
molecules are very small and the number of molecules in even a tiny mass 
of any substance is enormous, so we don't want to count by 1's. Instead, 
we will count by a large number. The value of that number is not 
important, only that it is big enough to give us a convenient measure of the 
number of molecules we are dealing with. That number we call one mol.
This is the "dozen" of chemistry. We count eggs in terms of "bunches", 
either by the dozen by the gross. We count molecules by bunches too, 
only we call the bunch a mol.
A Simple 
Conclusion
And this all comes together when we recall from chemistry that n, the 
number of molecules ( in units of mols ), in any mass m of a substance is 
obtained by dividing the mass by the molecular weight M of that substance.
n
m
M
=
Many Types 
of Mol
The molecular weight of methane is 16.042.
Then 16.042 g of CH4 is 1.00 gmol of methane, 
and 16.042 kg of CH4 is 1.00 kgmol of methane,and 16.042 lb of CH4 is 1.00 lbmol of methane,
and 16.042 oz of CH4 is 1.00 ozmol of methane, etc.
There can be no logical reason to ever use more than one "type" of mol in 
any single calculation, but if it is ever needed, 
1 kgmol = 1000 gmol
1 lbmol = 453.593 gmol
That is, the different "flavors" of mol are related exactly as the corresponding 
mass units.
Note that the Mathcad unit "mole" is equal to the gmol. The only distinction 
that this particular type of mol has is that the number of molecules in a gmol 
of any substance is exactly equal to Avogadro's number: 6.023 * 10 23. It 
should be clear by now that the number of molecules in a lbmol is simply 
453.593 times as many as this.
Worth
Remembering
Later in this course you will need to compute the molecular weights of many 
substances. With only a few exceptions, the substances that participate in 
the processes that we will consider consist of only a small number of 
elements. The following list of atomic weights is small enough to be worth 
memorizing.
Element At Wt
_______________
H 1.008
C 12.011
N 14.007
O 16.000
S 32.064
Cl 35.453
Example 0.3.1 What masses of nitrogen and hydrogen gas are required to make exactly 
2.76 lbmol of ammonia?
Solve this problem on your own worksheet. Check your answer by 
clicking on the popup icon at the left.
Example 0.3.2 What mass of chlorine gas is required to completely convert 33.73 kg of Al 
to aluminum chloride via the reaction Al + 3/2 Cl2 = AlCl3?
Solve this problem on your own worksheet. Check your answer by 
clicking on the popup icon at the left.
NaOH Composition0.4.6
Wet Sand Composition0.4.5
Flue Gas Composition0.4.4
Conversions Among Composition Variables0.4.3
Mass and Molar Concentrations0.4.2
Mass, Mol, and Volume Fractions0.4.1
A mixture is formed when quantities of two or more distinct substances are 
combined. An amount of substance can be measured in mass, molar or 
volumetric units and composition variables involving all of these can be 
defined and many are in common use. In this section we will explore the 
definitions and interconversions of composition variables.
0.4 Composition Variables
0.4.1 Mass, Mol, and Volume Fractions
The composition of a mixture gives information about the relative amounts 
of the component substances. These amounts can be measured in terms of 
mass, molar, or volume units. By dividing the amount of each substance 
by the total for the sample, an intensive property is obtained.
An intensive property is one which is independent of the amount of 
substance under consideration. Properties like composition, temperature, 
and pressure are intensive. They can be measured at a point. Properties 
like mass and volume are extensive, that is, their values depend on the size 
of the sample.
Mass Fraction We adopt the following notation:
 mi mass of substance i in the mixture
 m total mass of mixture
ωi mass fraction of substance i.
These quantities are related by the following expressions.
m
i
mi∑=
ωi
mi
m
=
As a consequence of their definition, the mass fractions sum to unity.
i
ωi∑ 1=
Mol Fraction We adopt the following notation:
 ni molar amount of substance i in the mixture
 n total molar abundance of mixture
 xi mol fraction of substance i.
These quantities are related by the following expressions.
n
i
ni∑=
xi
ni
n
=
As a consequence of their definition, the mol fractions sum to unity.
i
xi∑ 1=
Average
Molecular
Weight
In a mixture a useful quantity is the average molecular weight. In this 
course the desired average is a "number" average and is defined as the 
following weighted sum
M
i
xi Mi⋅∑=
in which Mi is the molecular weight of species i and M is the average 
molecular weight of the mixture. Note that it is the mol fraction xi that is 
used in this definition, not the mass fraction.
Click on the figure at the left to view a popup screen with the answer.
What is the mass composition of a mixture containing 23 g of A, 19 g of B 
and 74 g of C?
Ex 0.4.1.2
Click on the figure at the left to view a popup screen with the answer.
Click anywhere outside of the popup to return here.
What is the composition of an equimolar mixture of n-pentane, n-hexane 
and n-octane?
Ex 0.4.1.1
Volume fractions are used extensively in describing polymer solutions, but 
in general, volume fractions are of little use in material balance calculations 
because in reality, the volume of a mixture is not equal to the sum of the 
component volumes. We will not use volume fractions in this course.
i
φ i∑ 1=
As a consequence of their definition, the volume fractions sum to unity.
φ i
Vi
V
=
(This equation is not quite "true", but
 it remains part of the definition.)
V
i
Vi∑=
These quantities are related by the following expressions.
We adopt the following notation:
 Vi volume of pure substance i in the mixture
 V total volume of mixture
φi volume fraction of substance i.
Volume Fraction
0.4.2 Mass and Molar Concentrations
A concentration is a ratio of an amount of substance to a specified volume.
This measure of composition is seldom used in material balance calculations 
but it arises frequently enough in other areas of Chemical Engineering to 
merit consideration.
Although it is not essential, ordinarily concentrations are used in cases 
where one substance is present in great excess and is usually referred to as 
the solvent. 
Mass
Concentration
The mass concentration ρi of a species i in a mixture is the ratio of the 
mass of that substance to the total volume of the mixture.
ρ i
mi
V
=
Molarity The molar concentration, or molarity Ci, of a species i in a mixture is the 
ratio of the moles of that substance to the total volume of solution.
Ci
ni
V
=
Molality This measure of composition is based on a fixed mass of solvent. The 
molality λi of species i is the number of moles of that species in exactly one 
kg of solvent. The arbitrary selection of one substance as a solvent makes 
this measure of compostion less useful to us than the others and we will not 
make use of this quantity in this course.
0.4.3 Conversions Among Composition Variables 
Process data usually comes from a variety of sources and it is often 
necessary to convert among the various composition scales. 
The fundamental idea behind all of these conversions is that composition 
is an intensive variable, and therefore independent of sample size. We 
are therefore free to select whatever basis is convenient.
1. If mol fractions are given, then use a basis of some number of moles
 of mixture. 
2. If mass fractions are given, then use a basis of some mass of mixture. 
3. If concentrations are given, it is usually convenient to select some
 volume of mixture as the basis for computation.
These ideas are explored in the examples in sections 0.4.4 - 0.4.6.
Points of
Confusion
One often hears compositions given in terms of percentages. For example, 
a particular gas is 27% ethylene, 33% butane, and the balance (40%) is 
nitrogen.
First, percentages are never used in Chemical Engineering computations.
The figure 27% is simply an alternate notation for the deciaml 0.27. In 
computations, a percentage is always converted to the corresponding 
decimal.
Secondly, it is not clear from the statement above whether the decimal 
values represent molar or mass compositions. In this course we adopt the 
convention that gas-phase compositions are always assumed to be molar 
compositions unless explicitly stated otherwise.
Another convention is that elemental compositionsare always given in mass 
fractions. So that the statement: "This dry coal sample is 67% carbon, 
29% hydrogen and 4% oxygen" is assumed to mean that the mass fraction 
of C in the sample is 0.67, that of H is 0.29, and that of O is 0.04.
0.4.4 Flue Gas Composition
Problem
Description
A flue-gas composition is 16.0% oxygen, 17.3% carbon dioxide, 2.80% 
carbon monoxide, and the balance nitrogen. Calculate the mass fraction of 
each species in the mixture.
Click on the popup icon at the left to view the solution.
0.4.5 Wet Sand Composition
Problem
Description
A mixture is 54 wt% sand (SiO2), 17% salt (NaCl), and 29% water.
What is the molar composition of the mixture?
Click on the popup icon at the left to view the solution.
0.4.6 Sodium Hydroxide Composition
An aqueous solution of sodium hydroxide has a concentration of 9.142 
mols of NaOH per liter, and the solution specific gravity is 0.9406.
Calculate the molar and mass fractions of NaOH.
Problem
Description
Click on the popup icon at the left to view the solution.
0.5 Flow Rates
In most process flowsheet analyses, the variables of interest are not 
masses, molar abundances, or volumes but flow rates. Just as quantities 
of substance are meassured in mass, molar, or volumetric units, flow rates 
also come in these "flavors". There is no fixed, accepted notation for 
flow rates in the Chemical Engineering field, so one purpose for this 
section is to establish some notational conventions that will be used 
throughout this course.
The definitions and nomenclature of total and component flow rates are 
presented in section 0.5.1. The following two sections are simply 
examples illustrating their use.
0.5.1 Component Flow Rates
0.5.2 Aqueous Alcohol Stream
0.5.3 Binary Fractionator
0.5.1 Component Flow Rates
For our purposes, a flow rate is a quantity of substance (mass, mol, 
volume) passing a fixed location in some conduit in a unit time. At times, 
we may wish to refer to a total flow rate. Since our streams will often be 
mixtures, we will often find it useful to refer to the flow rate of a particular 
component in that stream.
Mass Flow
Rates
We will adopt the following notation for mass flow rates.
Total stream flow rate: F
Component flow rate of species i: fi i = 1, ..., S
Number of species: S
These quantities are related through the mass composition.
fi F ωi⋅=
Since the mass fractions sum to unity, if the set of equations for the 
component flow rates are summed we obtain:
i
fi∑ F=
Note that the set of variables fi (i = 1,..., S where S is the number of 
species) is entirely equivalent to the set F, ωi. When performing material 
balance calculations, the nature of the specifications determines which set 
to use in any particular instance.
Note that the set of variables ni (i = 1, S where S is the number of 
species) is entirely equivalent to the set N, xi. When performing 
material balance calculations, the nature of the specifications determines 
which set to use in any particular instance.
i
ni∑ N=
As before, the component flow rates must sum to the total flow rate.
ni N xi⋅=
These are related through the molar composition.
Total stream flow rate: N
Component flow rate of species i: ni
The corresponding symbols for total and component flow rates expressed 
in molar units are:
Molar Flow
Rates
i
fi∑ 100
kg
sec
=
As required by their definition, the component flow rates sum to the total 
flow rate.
f
30
70






kg
sec
=fi F ωi⋅:=Component flow rates:
ω
0.3
0.7






:=Mass fractions:
F 100
kg
sec
⋅:=Total flow rate:
i 1 2..:=Species indices: methanol, water
For example, if a stream containing methanol (30 wt%) and water is 
flowing at a rate of 100 kg per second then, letting methanol be component 
1, the following expressions characterize this stream:
Aqueous
Methanol
ni j,Molar flow rate of species i in stream j
fi j,Mass flow rate of species i in stream j
NjTotal molar flow rate of stream j:
FjTotal mass flow rate of stream j :
Stream indices: j = 1, .., J
Species indices: i = 1, .., S
All process flowsheets have more than one stream. Our notation must 
permit us to refer to the flow rate of a given component in any stream. We 
will use a letter to refer to a concept and one index to specify a stream and 
another index to specify a species. For example, in a process with S 
species and J streams we would implement the following nomenclature:
Notation for
Multiple
Streams
These are seldom used in flowsheet analysis because the requisite density 
information is seldom available. For those occasions when a volumetric 
flow rate is given, we will use the symbol Φ. If the density of the stream is 
ρ, then the flow rate can be converted to a mass basis with the expression:
F = ρΦ.
Volumetric
Flow Rates
x
0.194
0.806






=xi
ni
N
:=
Compositions:
N 4.822
kmol
sec
=N
i
ni∑:=Total flow rate:
n
0.936
3.885






kmol
sec
=ni
fi
Mi
:=Component flow rates:
M
32.042
18.016






kg
kmol
⋅:=Molecular weights:
To continue with the previous example in molar units, let us convert the flow 
rates to a molar basis. For this we need the molecular weights.
Aqueous
Methanol
Revisited
Test
Yourself
If you are not accustomed to using indexed quantities, then the 
double-index notation can be challenging at first. It is so useful however, 
that we cannot avoid it. To get some practice using indexing and 
summations, expand the following sets of equations completely for the case 
of a process with three components (S = 3) and four streams (J = 4).
Fj
1
3
i
fi j,∑
=
= Nj
1
3
i
ni j,∑
=
=
Each of the above expressions stands for exactly four equations, one for 
each value of j = 1, .., 4.
ORIGIN 1≡ kmol 1000 mole⋅≡
0.5.2 Aqueous Alcohol Stream
Problem
Description
A stream has molar composition: 37% methanol (CH3OH), 32% ethanol 
(CH3CH2OH), and 31 % water. The stream flows at a rate of 1500 kg/h. 
 Compute the mass and molar component flow rates.
Open up your own worksheet and try to solve this problem on your own. 
Click on the popop icon at the left to check your solution.
0.5.3 Binary Fractionator
Problem
Description
An equimolar mixture of ethylene and butane (stream 0) is separated in a 
distillation column into an overhead stream (stream 1) with an ethylene mol 
fraction of 0.9764 and a bottom stream (stream 2) with an ethylene 
composition of 0.0875. If the feed flow rate is 1000 mol/s, determine the 
flow rates of the overhead and bottom streams by applying total and 
ethylene balances to the column.
Open your own worksheet and solve this problem. Click on the popup 
icon at the left to check your answer.
Chapter 1 Fundamentals
This chapter is a summary of some of the concepts involved in 
performing material and energy balances on chemical process 
flowsheets. It focuses on the degree-of-freedom (DF) analysis as a 
tool in organizing the balance calculations.
1.1 What is a Degree-of-Freedom (DF) Analysis?
1.2 Material -Balance Variables
1.3 Material-Balance Principles
1.4 Material-Balance Specifications
1.5 What About Pressure?
1.1 What is a DF Analysis?
A degree-of-freedom (DF) analysis is simply an accounting of the 
variables associated with a problem and the facts and relations 
available to determine values for those variables. This chapter 
introduces the classes of variables, relations, andtypes of information 
that are used in the analysis of material balance problems.
Variables Any problem in engineering has associated with it some set of physical 
quantities. If we assign a symbol to each of those quantities, then we 
can abstract the problem into a mathematical form which involves only 
those symbols. We may call each of these symbols a variable, even 
though in many applications the value assigned to the symbol never 
changes.
Double click on the example icon at the left.
Definition The degree of freedom of a problem is simply the difference between 
the number of symbols that characterize a problem and the number 
of equations that are available to solve for those symbols ( variables). 
Double click on this square for an example:
Equations Equations are mathematical statements that involve only the symbols for the 
problem at hand or constant values. The equations used in DF analyses 
are classified into two types.
The first type are physical laws and mathematical identities that apply 
to all systems.
Physical laws: Newton's laws, Conservation of 
Mass, Conservation of Energy, etc.
Mathematical identities: trigonometric relations,
mensuration formulas from plane geometry,
Pythagorean theorem, etc. 
Specifications The second type of equation is referred to as a specification and this 
type applies only to the problem at hand. Specifications can be 
simply assignment of a constant value to a problem variable, or some 
statement that can be translated into an equation relating problem 
variables.
Recall: the degree of freedom of a problem is simply the difference 
between the number of variables and the number of equations available 
to assign values to those variables.
Double-click on the square at the left to see an example.
Things
to Note
In the farm example discussed in the popups you may have noted 
that a DF of zero is a characteristic of a solvable problem. We say 
in such cases that the problem is well-posed. This merely means 
that sufficient information is supplied to solve for the values of all 
symbols.
When performing a DF analysis, no effort is expended in figuring out 
how to solve the equations that we list. We merely have to count 
them.
For problems as simple as the farm example, a "DF analysis" 
seems trivial. The conclusions are obvious. However, material 
balances on chemical process flowsheets typically involve 
hundreds of symbols and the results of a DF analysis are much 
more useful in such cases.
Benefits 1. The most important result of a DF analysis is the determination 
that a problem statement contains enough information to solve the 
problem. Most chemical process flowsheets involve hundreds of 
variables, are extremely tedious to solve, and the specifications are 
often hidden in technical jargon. Identifying a well-posed problem 
is then a non-trivial undertaking.
2. The DF analysis techniques to be presented in subsequent 
sections not only tell us if a problem is well-posed, they can also 
yield a very precise calculation order. A typical flowsheet may 
require that hundreds of equations be solved. The DF analysis can 
determine the order in which these equations should be solved.
Summary A DF analysis is simply a process of checking that the number of
symbols (variables) associated with a problem matches the number 
of equations we have to solve for them. It forces us to answer the 
question: "Have I missed anything?"
 As part of the analysis, we also determine a feasible order in which 
to solve the equations that model our system. In other words, it 
answers the question: "How do I proceed?"
1.2 Material Balance Variables
Recall that a degree-of-freedom analysis involves counting the
variables associated with a problem and then counting the number of 
equations that apply. In this section we explore the types of 
variables that will arise in analyzing chemical-process flowsheets.
1.2.1 Flowsheets & Processing Equipment
1.2.2 Stream Variables
1.2.3 Equipment Variables
1.2.4 Rules for Counting Variables
1.2.1 Flowsheets and Processing Equipment
Flowsheets A chemical-process flowsheet is a diagram containing boxes 
representing equipment items and numbered, directed lines representing 
material streams. Each line is labeled with the species present in the 
stream that it represents. Each box is labeled with the type of operation 
that it carries out.
Double-click the example icon at the left to see a sample flowsheet.
Equipment
Items
The boxes in the flowsheet represent processing equipment. The 
flowsheet in the previous popup incorporated nearly all the equipment 
types that we will encounter in this course.
The three equipment items that appear in virtually all flowsheets are:
Mixer: two or more streams combine to form a single exit stream.
Splitter: a stream is divided into two or more streams of identical 
composition.
Separator: a stream is divided into two or more streams. The 
streams differ in either temperature, composition or both. If the 
stream temperatures are not problem variables, then the separator 
is said to operate adiabatically.
If a flowsheet is composed of only mixers, splitters, or separators 
and stream temperatures are not considered, the resultant analysis 
is referred to as an elementary material balance. 
Reactors If one or more reactors are present, then the problem is a 
reactive-system material balance.
Reactor: a single entering stream is converted into a single exit 
stream with a different composition. A reactor can operate 
adiabatically, or it can be heated or cooled. 
Notes If a reactor has more than one entering stream, then it is sometimes 
convenient to replace the reactor (even if only temporarily) with a 
mixer-reactor combination. Similarly, if a reactor has more than one 
exit stream, it may be convenient (while thinking about the reactions 
occurring) to replace the reactor with a reactor-separator combination. 
 In this way, any reactor can be treated as having a single feed stream 
and a single product stream.
Energy
 Balances
If temperatures or heating/cooling rates are considered as system
variables, then the problem involves combined material and energy 
balances. In this case, three other process items may appear.
Heater/Cooler: a single stream enters at one temperature and leaves at 
another temperature. This is the only device which operates 
non-adiabatically in all cases.
Heat Exchanger: two streams enter and two streams leave the device.
Only the temperatures of the streams are changed. The material in one 
entering stream is not mixed with that of the other entering stream. This 
device operates adiabatically.
Compressor/Turbine: in these devices a single gaseous stream enters 
and one or more streams leave at different pressures. If the exit 
stream is at a lower pressure, the device is a turbine. If the exit stream 
is at a higher pressure than the entering stream, then the device is a 
compressor. We will employ only the most elementary analysis of such 
devices, and not until chapter eight. Typically only the power required 
(developed) in compressing (expanding) gases is considered in a 
preliminary flowsheet analysis. The power required to compress 
liquids is usually quite small and can be ignored.
Summary It is important to develop the skill to rapidly classify the problem 
that is being analyzed.
If there are no reactors, and stream temperatures are not
considered, then the problem is an elementary material balance.
The other cases are summarized in the following table.
Flowsheet Type Equipment Items
__________________________________________-
________
Elementary Mixer, Splitter, Separator
Reactive All above + Reactor
Mat'l & EnergyAbove+Heater+Heat
Exch+Compressor
The only devices that can be associated with duties (heating or 
cooling loads) are the reactor, separator, and heater.
V 29=V
1
J
j
ST( ) j∑
=
:=
The total number of variables (V) for the flowsheet is then
S 3 3 3 4 3 3 3 2 2 3( ):=Number of species in each stream:
J 10:=Number of streams:
For some flowsheet let there be ten streams, and let the numbers of 
species in each stream be the elements of vector S below.
The number of stream variables for a flowsheet is simply the sum 
over all streams of the number of species in each stream.
Counting
Variables
For the stream numbered j above, the number of variables is simply 
the number of species. Sj = 4
Associated with each stream on a flowsheet is a set of component
flowrates. These are the only variables in a non-reactive, 
material balance.
The number of variables associated with each stream is simply the 
number of distinct chemical species present in that stream.
1.2.2 Stream Variables
For those just learning to use Mathcad: a column vector is expected as 
the argument of the sum function. Since S was entered as a row vector 
(to save space) the transpose function was applied to S before indexing 
it. The transpose button is on the Vector & Matrices Palette in 
Mathcad 6.0.
Flowrate
Symbols
The component flowrates are the variables we associate with each 
stream. If there are S species in a stream, there are S such flowrates.
We will use different symbols to represent component mass flowrates, 
and component molar flowrates.
The mass flowrate of species s in stream j is represented by: fsj
The molar flowrate of species s in stream j is represented by: nsj
We will adopt the convention that doubly-indexed quantities referring 
to the components of a stream use the first index for the species and 
the second for the stream. 
Compositions Stream information is often given in terms of compositions, either as 
mole fractions or mass fractions.
Mass fraction of species s in stream j: ωsj
Mol fraction of species s in stream j: xsj
For any stream j, the set of variables fsj is entirely equivalent to the set 
 consisting of the total mass flow rate Fj, and the compositions ωsj .
For any stream j, the set of variables nsj is entirely equivalent to the 
set consisting of the total molar flow rate Nj, and the compositions
xsj.
These are related by the definitions: fsj = Fj ωsj and nsj = Nj xsj
One Step at
a Time
Remember: when counting variables for a flowsheet, we count only S 
variables for each stream, where S is the number of species in that 
stream.
We may later use any combination of component flowrates, total 
flowrates, and compositions to actually solve the balances for the 
flowsheet, but this is not an issue for a DF analysis. Solving 
equations is a task that is not initiated until the DF analysis is 
completed.
This is an important distinction: while doing the DF analysis, we are 
counting equations, not solving them.
Keeping this distinction in focus is a difficult aspect of learning to do 
a DF analysis. One must resist the temptation to start solving 
equations until this analysis is completed.
Temperature If the flowsheet involves combined material and energy balances, then 
with each stream is also associated one temperature. Thus the 
number of variables associated with a stream containing S species is 
S+1.
We will not be considering energy balances until chapter six.
Temperature will not be a flowsheet variable in any of the flowsheets 
considered in chapters one through five.
Summary A stream with S species is characterized by S variables if only material
balances are of concern. If energy balances are also an issue, then 
we must count S+1 variables for each stream.
1.2.3 Equipment Variables
In this course, we are not concerned with designing equipment. Each 
 equipment item for a flowsheet is assumed to fulfill its function. The 
internal details of such items are not of interest in this context. This 
means that all quantities such as lengths, diameters, capacities,
stages, etc. are not of any concern to us.
The only variables we associate with equipment items are:
reaction rates for reactors 
duties (heating/cooling rates) for heaters/coolers, and 
non-adiabatic reactors and separators
powers for compressors and turbines
Only reaction rates will be encountered until chapter six.
Reactors We associate one reaction rate variable with each independent reaction 
that can be written for the reaction system of a particular reactor. We 
will explore later the question of how to determine the number of 
independent reactions. The reaction rate has units of mol/time. 
Heaters Equipment items are assumed to operate adiabatically, except for
heaters. For some flowsheets, reactors and occasionally separators 
will be specified as operating non-adiabatically. We associate one 
duty with every heater/cooler and each separator or reactor which is 
identified as operating non-adiabatically. The duty has units of power: 
energy/time.
Note Heat exchangers operate adiabatically and have no duties associated 
with them for flowsheet energy balance purposes.
Compressors Although the power required to pump liquids may be ignored in a 
preliminary analysis of a flowsheet, the power required (developed) on 
compressing (expanding) a gas stream can never be ignored. The units 
normally used are hp or kW.
1.2.4 Rules for Counting Variables
Summary The number of variables for a non-reactive, material balance is simply 
the sum of the numbers of species in each stream.
If there are reactors present then we add one reaction rate variable for 
each independent reaction in each reactor.
If energy balances are also a consideration, then we must add one 
temperature for each stream, and one duty for each heater and for each 
separator or reactor that is heated/cooled. Corresponding to each 
compressor or turbine, a power variable is added.
Problem 1.2.4.1 Consider the following flowsheet for which we are interested only in 
material balances. How many variables are involved?
Double-click on the exercise icon at the left to see the answer.
Problem 1.2.4.2 Both material and energy balances are needed for the flowsheet below.
In the reactor, which operates adiabatically, the following reactions 
occur:
 CO + 3H2 = CH4 + H2O and CO + H2O = CO2 +H2
Count the number of variables associated with this flowsheet.
Double-click on the exercise icon at the left to see the answer.
1.3 Material Balance Principles
A degree-of-freedom analysis explores the match between the 
number of variables associated with a flowsheet and the number of 
equations available to determine values for those variables. In the 
last section, you learned how to count variables. This section 
explores how to count equations for a flowsheet. The discussion 
is restricted to the kinds of equations that apply to all flowsheets, 
that is, equations that are statements of the fundamental principles 
that mass and energy are conserved.
1.3.1 Species Balances
1.3.2 Element Balances
1.3.3 Energy Balance
1.3.4 Properties of the Balance Equations
1.3.1 Species Balances
There are two issues to be explored: how to count material balance 
equations, and how to formulate them. For the DF analysis one only 
needs to know how to count the equations. In practice, one must also 
know how to formulate these balances for different types of flowsheets.
Counting We perform species balances on each process unit. If a process unit 
has S species associated with it, then there are exactly S independent 
balances that can be writtenfor that unit.
In a multiunit flowsheet, we simply count the number of balances that 
can be written for each process unit and sum them to determine the 
total number of balances E for the flowsheet. If there are U process 
units, and Su species associated with unit u, then
E
1
U
u
Su∑
=
:=
Problem 1.3.1.1 How many species balances can be written for the flowsheet below?
Double-click on the exercise icon to the left to see the answer.
Independent
Balances
When solving the material balances for some flowsheet, we may 
actually employ balances around two or more (or all) process units.
But these balances are not independent of the E balances found by 
the procedure just described.
In the flowsheet above, it may be advantageous, when actually solving 
the material balances, to use not just balances around individual units, 
but balances around some set of units. For example, the dashed lines 
in the figure above indicate balances around two subunits each 
containing four process units.
Overall
Balance
One often employs an overall balance: a balance around the set of all
process units. This particular balance is useful when more is known 
about external streams than about internal streams. An external 
stream is one that has either a head or a tail that is not attached to a 
process unit. The only streams involved in an overall balance are 
external streams.
Even though such balances are useful when solving the material 
balances, these additional balances are not independent of the E 
balances around the individual process units.
Distinguish between counting material balance equations and solving 
them.
Formulating
the Balances
The step of formulating the species balances is actually not needed for
the DF analysis. However, an awareness of the alternative formulations 
is needed to make intelligent choices in subsequent steps, so we will 
explore this topic here.
The first choice is that between mass or mol balances.
Mass
Balances
Let some steady-state, process unit have I input streams, and J output 
streams. If there are S species involved, the principle of mass 
conservation can be expressed in terms of the component flow rates as: 
1
I
i
f si∑
= 1
J
j
f sj∑
=
= s 1 S..:=
If you have difficulty "seeing" the meaning of equations with such sums in 
them, you should write out the equations for some sample flowsheet.
Note that there is one equation for each species. Usually the set of 
equations is very simple, because there are seldom more than a few 
streams entering or leaving any single device.
Sample
Mass Balance
The following material balances can be written for this system.
f s 0, f s 1, f s 2,+=
This is three equations, one for each species: A,B, C.
If the unit pictured is not a reactor, then it is also true that
ns 0, ns 1, ns 2,+=
Ordinarily, the reaction rates are determined from specifications such as 
conversions or product compositions.
nD1 0 r2+=
nC1 0 r1+ r2−=
nB1 nB0 2 r1⋅−=
nA1 nA0 r1− r2−=
The molar material balances for this process unit, in which it is assumed that 
no C or D are present in the feed stream, are:
In the reactor shown below, two reactions occur. These are numbered 1, 
2.
Sample Mol
Balance
If there are no reactions, then the second term on the right contributes 
nothing and the set of equations to the same form as those for the mass 
balances. That is, in nonreacting systems, moles are conserved.
The equation above makes it clear that while "mass in = mass out", a 
similar statement cannot in general be made for molar balances. Mols 
are not conserved. The second term on the right hand side accounts 
for the change in the amount of species s via the R reactions occurring.
s 1 S..:=
1
I
i
nsi∑
= 1
J
j
nsj∑
= 1
R
k
σsk rk⋅∑
=
−=
Using the same process unit, let there be R independent reactions 
occurring, and the corresponding reaction rates be rk, with 
stoichiometric coefficients σsk. Reaction rates are discussed at greater 
length in chapter four. At this point we merely state that stoichiometric 
coefficients are defined to be positive for products and negative for 
reactants. The mole balances are:
Mole
Balances
Equivalence Note that for all process units except reactors, the mole balances have 
the same form as the mass balances. One may use either set of 
equations.
For reactors, it is usually more convenient to use mol balances, but this 
is still a matter of choice. The number of equations is the same in either 
case.
One serious mistake to avoid is that of using the form of the mass 
balance equations with molar flow rates for a reactor. Mols are not 
conserved in a reactor. Mols are conserved in all other processing
units.
1.3.2 Element Balances
In general, species balances are preferable to element balances.
However, there are some situations for which element balances offer 
great simplification. It is worthwhile to consider them in order to be be 
able to recognize when their use will be advantageous.
Element balances are often used when the stoichiometry of a reaction is 
either unknown or complex.
For example, suppose ultimate analyses are available for the substances 
wood, mayonnaise, tar, and char. The following reaction is most easily 
handled using element balances.
wood + mayo + air = char + tar + lots of other substances
The formulation of the element balances requires the use of the atom 
matrix for which we will use the symbol α.
Atom Matrix The stoichiometric coefficient of element e in species s is αes.
Problem 1.3.2.1 For a particular reactor, the species involved are CO, H2, and 
CO2. What is the atom matrix for this system if the species are 
numbered in the order listed and the elements are indexed in the 
order H,C,O?
Double-click the exercise icon at the left to see the answer.
Element
Balances
Consider the processing unit shown below with I input streams and J 
output streams. In contrast to species, elements are conserved in 
reactors.
If there are E elements present in the system then (usually) there are E 
element balances that can be written.
1
J
j 1
S
s
αes nsj⋅∑
=
∑
= 1
I
i 1
S
s
αes nsi⋅∑
=
∑
=
=
Note that there is no reference to any reaction stoichiometry.
Each term accounts for the flow of element e in a stream that 
accompanies species s.
Double Sums If you find the double sums in the element balances formidable, just 
remember that there are only as many terms in the I sum as there are 
inlet streams, and there are only as many terms in the J sum as there 
are exit streams. Each term simply accounts for the amount of element 
e that accompanies species s in stream i.
The reason for the complexity is that while we are balancing on 
elements, the variables of interest are species flowrates. We must 
express the very simple idea that elements are conserved even though 
they are "packaged" into a set of species.
An example is worth a thousand words here. Examine the flowsheet 
below.
Problem 1.3.2.2 Write out the element balances for the flowsheet shown below.
Double-click on the exercise icon at the left to see the answer.
Independent
Element
Balances
When using species balances, if there are S species present, then there 
are always exactly S independent balances. If there are E elements 
present, there are E balances that can be written. However, for some 
systems, some of the balances are not independent.
The number of independent element balances is equal to the rank of 
the atom matrix for the system. The system is the set of species 
present. The rank of the atom matrix is the number of non-zero rows 
remaining after exhaustiverow reduction. We will use the expression 
ρ(α) to represent the rank of the matrix α.
Problem 1.3.2.3 Determine the number of independent element balances for the 
system (H2O, SO3, H2SO4).
Double-click on the exercise icon at the left to see the answer.
Summary The element balances are conceptually easier than species balances
because no reaction stoichiometry appears in their formulation. There 
is no need to distinguish between reacting and non-reacting systems. 
The element balances are algebraically more complex than most 
species balances because they typically involve more unknowns per 
equation.
1.3.3 Energy Balance
We will not need to formulate energy balances until chapter six. For 
completeness, this section focuses only on the issue of counting energy 
balances. The answer is simple, the details are not. There is exactly 
one energy balance for each process unit. If you are not interested in 
details right now, you are invited to skip the rest of this section on a 
first reading.
For the flowsheets considered in this course, we will make some 
simplifying assumptions. These restrictions are not essential. They 
merely make the process of applying an energy balance more convenient.
Assumptions 1. All processes operate at steady-state. This means that properties are 
independent of time.
2. Equipment and piping are sized such that fluid velocity changes are 
negligible.
3. There are no net changes in elevation for any fluid stream.
4. Fluid properties are not functions of pressure.
5. Mixture properties may be computed assuming the ideal solution model.
Energy Balance With the restrictions outlined above, the energy balance for a process 
unit with I input streams and J output streams is 
1
J
j
Fj h j⋅∑
= 1
I
i
Fi hi⋅∑
=
− q w−=
The net rate of heating is q, and net shaft power output is w. The 
expression above is phrased in terms of mass flow rates and specific 
enthalpies. The corresponding expression for molar quantities would 
use the total molar flowrates Nj in place of Fj. The same symbol is 
used for molar and specific enthalpy, because we will never have 
occasion to need both in the same expression.
If you don't know what enthalpy is yet, do not be concerned. This new 
property will be introduced more formally in chapter six. Enthalpies, 
like all material properties, are never considered variables. Instead, 
they are computed from system variables like temperature, pressure, 
and composition. For now, just notice where stream variables occur in 
the energy balance. This equation will be derived for you later.
Ideal Solution The ideal-solution assumption makes it possible for the enthalpy of a 
stream to be computed solely from the enthalpies of its components.
For any stream j containing S species, the enthalpy is computed from 
the pure-component enthalpies hsj, and the compositions xsj.
h j
1
S
s
xs j, hs j,⋅∑
=
:=
Evaluating the pure-component enthalpies requires data on heats of 
formation, heats of phase-change, and heat capacity data. The 
techniques for doing this are discussed in chapter 6.
Counting There is exactly one energy balance for each process unit. If there are 
U process units in a given flowsheet, then exactly U independent energy 
balances can be written. 
1.3.4 Properties of the Balance Equations
The balance equations (species, element, energy) all share the property 
of homogeneity. If only component flowrates are used, they are also 
linear in all unknowns. If instead, total flowrates and compositions are 
used, some of the balances may be non-linear.
Linearity An equation is said to be linear in a set of variables if each member of 
the set appears only to the first power and no products of variables 
occur.
Consider the balance on species s for a process unit with one entering 
stream (indexed by 1) and two exit streams.
A balance phrased solely in terms of component flowrates is inherently 
linear because products of such flow rates have no meaning and never 
occur.
fs1 = fs2 + fs3
Balances expressed in terms of total flowrates and compositions are 
inherently non-linear since the species flowrates are products by 
definition.
 F1xs1 = F2 xs2 + F3 xs3
Homogeneity A function f(x,y) is said to be homogeneous in y if f(x,cy) = c f(x,y).
That is, if the variable y is replaced by some multiple cy, where c is some 
constant, then the function value is similarly a multiple of the original.
The balance equations are homogeneous in the flowrates. This means 
that if a set of flowrates is found that satisfies all constraints (except for 
capacity) for some flowsheet, then if each flowrate in that set is multiplied 
by some constant, the new flowrates would also be a solution.
It is this property of homogeneity that allows us to select any basis for 
the flowsheet, solve the balances, and then scale the solution to satisfy 
some given capacity. The basis is the flow rate of some selected stream 
or component in a stream.
Scaling Let all the flowrates for a given flowsheet be determined for some 
arbitrary basis. Suppose that a particular product stream flowrate is 
350 kg/h using that basis. In obtaining this solution suppose that a 
capacity constraint was ignored, and that the desired product flowrate is 
700 kg/h. If all stream flowrates are simply doubled, then the new set 
of flowrates is a solution to the original problem that also satisfies the 
capacity constraint.
In general, let the flowrates determined for an arbitrary basis be Fj', and 
the actual basis be Fb for stream b. Define a scale factor ζ = Fb/Fb'.
Then the actual flowrates are Fj = ζ Fj'. 
Selecting a
Basis
For some flowsheets, the flow rate of no stream may be given. In that 
case, we are free to select any stream as a basis. Only the 
compositions of the streams and the ratios of the flow rates will be 
significant.
If the flow rate of at least one stream is given, we may refer to that flow 
rate as a capacity constraint. This flow rate may be used directly as a 
basis. However, it may not be the best basis to use for intermediate 
computation.
Even if a capacity constraint is given, it may be more convenient to 
temporarily relocate the basis. 
Double-click the example icon on the left to see an example where such 
a relocation might be useful.
Scaling
Example
Determine the flow rates and compositions of all streams in the flowsheet 
shown below.
Click on the popup icon to view the solution.
For a given flowsheet, the most convenient basis may be used, and then 
all flow rates can be scaled to match the actual capacity constraint. It 
is the homogeneity of the balance equations that makes this possible.
Summary
Summary1.4.6
Splitter Restrictions1.4.5
Equilibrium Relations1.4.4
Flow and Composition Ratios1.4.3
Flowrates and Compositions1.4.2
Understanding Specifications1.4.1
This section explores the different types of specifications that are often 
encountered in flowsheets for chemical processes.
Recall that a DF analysis simply examines the match between the 
number of variables associated with a given system and the number of 
equations available to set values for those variables. Section 1.2 
explained how to count variables, while the previous section explored 
how general conservation principles are used to formulate balance 
equations involving the variables.
1.4 Material Balance Specifications
1.4.1 Understanding Specifications
The number of balance equations for any flowsheet is always less than 
the number of variables. A flowsheet can represent a well-posed 
problem only if a precise number of specifications is also supplied.
Unlike the balance equations, specifications arenot derived from 
general principles. Instead, they are simply statements of fact about a 
specific flowsheet.
Many Possible 
Variations
There are only a small number of choices in phrasing the balance 
equations: species vs. element, mass vs. mol, flowrate vs. compostion.
The situation is very different for specifications. There are hundreds of 
ways of expressing constraints among the variables for a flowsheet.
Sections 1.4.2 through 1.4.6 attempt to categorize some of the most 
common of these.
In spite of the large variety of specifications possible, there are some 
common features.
Requirements 1. Specifications must be phrased as equations relating system variables.
You must develop the skill of translating declarative statements about a 
flowsheet into a mathematical form involving flowsheet variables. For 
example:
"The recovery of CO in the flash overhead (stream 5) is 78% of that 
in the feed (stream 1)" translates to fCO,5 = 0.78 fCO,1.
2. The specifications must be independent. That is, each specification 
must be a new piece of information. Suppose it is stated that a stream 
containing only CO and H2 is 43% CO and 57% H2. This is really just 
one specification. Once it is recorded that xCO = 0.43, then writing 
xH2 = 0.57 is certainly a true statement, but it is not independent of the 
first expression, therefore it is not a specification.
1.4.2 Flowrates and Compositions
Flowrates and compositions are the simplest types of specifications 
because they translate directly into expressions involving flowsheet 
variables.
Flowrates come in three "flavors":
1. total: the flow rate of an entire stream
2. component: the flow rate of one component in a stream
3. pseudo-component: the flow rate of some combination of 
components in a stream.
Flowrates Total flowrate specifications appear in statements like "The flowrate of 
the feed is 260 mol/h". If the feed is stream 1, then this translates to
N1 = 260 mol/h.
A sample component flowrate specification is "The flowrate of CO in 
the flash overhead is 38 mols/h". If the flash overhead is stream 7, this 
translates to
 nCO,7 = 38 mol/h
Alternatively, a statement may refer to some combination of 
components as "The total flowrate of all sulfur-containing species in the 
absorber bottoms is 5.6 mols/h". If the bottoms stream of the 
absorber is stream 9 and it contains H2S, SO2, and CS2 among other 
species which do not contain sulfur, then the statement gets translated to 
 nH2S,9 + nSO2,9 + nCS2,9 = 5.6 mol/h.
Such combinations of species are encountered frequently and the general 
term pseudo-component is used to describe this artifice.
Composition This is the most common type of specification yet it is a source of great 
confusion. Because of the summation constraint, the complete set of 
mole fractions for a stream containing S species is only S-1 
specifications.
If a stream is pure, that is, if it consists of a single species, then the 
statement that xs = 1 is not a specificaton. By counting only one 
variable for that stream we have already taken into account its trivial 
"composition".
Example 1.4.2.1
Double-click the icon at the left to see an example that illustrates 
some of these issues.
1.4.3 Flow and Composition Ratios
In addition to specifying flowrates or compositions directly, one 
frequently encounters statements which establish a relation between 
two or more flowrates or compositions, usually expressed as a ratio.
The quantities involved may be for
i. a single stream
ii. two streams on the same unit 
iii. or two streams on that have no units in common
Terminology Various terms are applied to such ratios. With splitters one refers to 
"split ratios". The term "recovery" is often used with separators.
"Conversion", "recycle ratio", and "yield" apply to reactors. They are 
fundamentally the same kind of information but they require different 
treatments in translating word descriptions of the ratios to mathematical 
expressions involving only material balance variables.
In all cases, be sure to distinguish mass ratios from molar ratios.
Split Ratio This kind of ratio always applies to a single equipment item. Let 
stream 5 enter a splitter and let streams 6,7 exit the splitter. A split 
ratio usually refers to an exit stream as some fraction (less than one) of 
the inlet stream. For example, in terms of mass flow ratio:
 F6/F5 = 0.73
However, ratios of exit steams are also encountered. Such ratios are
typically greater than one. For example: "Streams 6 and 7 exit the 
splitter in a molar ratio of 3:2" means
N6/N7 = 1.5
Recovery A recovery specification is phrased in language like this: "78% of the 
CO2 in the feed (stream 1) is recovered in the flash overhead (stream 
6)". It can be expressed as:
 fCO2,6 = 0.78 fCO2,1
A recovery specification is ambiguous unless the component and the two 
streams involved are clearly identified.
Recycle Ratio This term applies exclusively to splitters. It expresses the ratio of the 
flow of one of the exit streams of the splitter to some other stream.
The other stream is usually the feed to a mixer, but may be any stream 
that is feeding or is part of a recycle loop. In the figure below the flow 
rate ratios of streams 4:3, 4:0, 4:2 are used, depending on the 
particular emphasis of the specification.
This is another case in which the terminology is often confusing. Unless 
the identities of the two streams involved are clearly stated, this kind of 
specification is ambiguous. There is no accepted convention with regard 
to recycle ratios. The two streams involved in a given recycle ratio are 
whatever the author intends for them to be.
Conversion This type of specification has the distinction of being the only ratio that is 
unambiguously defined. Let the feed stream to a reactor be indexed with 
'0', and the exit stream be indexed by '1'. The conversion of reactant 
species s is simply the amount of s that reacts expressed as a fraction of 
the amount in the feed. 
Xs
ns0 ns1−
ns0
=
If the species s is not explicitly identified, then by convention it is 
assumed to refer to the limiting reactant. See section 4.1 for a precise 
method of distinguishing this reactant.
Single-Pass
Conversion vs
Overall
Conversion
If the streams in the definition above refer to reactor inlet and exit 
streams, then the conversion is called a single-pass conversion. If the 
reactor is part of a recycle loop, then stream 0 may instead be 
interpreted as the raw feed to a mixer upstream of the reactor and the 
other stream is the non-recycle, exit stream of some splitter. In that 
case, X is referred to as an overall conversion.
Using the stream numbers in the figure above, a conversion relating molar 
flow rates in streams 1 and 2 is a single-pass conversion, while a 
conversion applying to molar flow rates in streams 0 and 3 is an overall 
conversion.
Yield This ratio applies only to reactors with at least two competing reactions.
Let reactant A form desired product C by one reaction and one or more 
other products (but no C) by competing reactions. The yield of C is 
simply the actual amount of C formed divided by the maximum amount 
of C that would have been formed if all the A converted went to 
producing C. Again using streams 0, 1 as inlet and exit streams:
YC
σA nC1 nC0−( )⋅
σC nA1 nA0−( )⋅=
The σ's are the stoichiometric coefficients in the balanced reaction that 
produces C. Recall that such coefficients are defined to be negative 
for reactants, so that the ratio above is always positive.
Problem 1.4.3.1 Translate each of