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Chapter 4 Reactive Systems
When reactors are present in the flowsheet under consideration, there 
are more choices that must be made regarding how variables are 
counted and how the balances are formulated.
In previous chapters a pattern was established of beginning with a section 
on tools and techniques and then presenting collections of sample 
flowsheets in which the application of the new tools is illustrated. That 
pattern is continued here. Since the number of new tools and techniques 
required to deal with reactive systems is extensive, section 4.1 consists of 
five sub-sections.
Section 4.1.1 introduces the additional terms and tools necessary to 
analyze flowsheets with reactors. Terms defined here include 
stoichiometric coefficient, reaction rate, fractional conversion, 
limiting reactant, stoichiometric air, and excess air fraction. The 
formalism for dealing with more than one reaction is developed in 
section 4.1.2. When multiple reactions occur a fractional yield may 
be given as a specification. The issue of reaction independence also 
arises.
Element balances find use under special circumstances explained in 
section 4.1.3. The concept of the atom matrix is introduced along 
with the question of the independence of the element balances.
In section 4.1.4 a technique is introduced which permits the 
identification of an independent set of reactions given only the atom 
matrix for a system. This technique is referred to as the 
stoichiometric procedure. Lastly, the issue of dependence in 
reactor specifications is presented in section 4.1.5 along with a 
method for detecting combinations of reactor specifications which do 
not represent independent information.
Section 4.2 is a collection of flowsheets in which a single reaction 
occurs, with and without recycle. The flowsheets in section 4.3 
contain a single reactor but multiple reactions occur. The focus is on 
the formalism needed to deal with material balances for such systems.
Section 4.4 is a collection of problems in which element balances are 
effectively used. Some issues dealing with fossil fuels are also presented.
In section 4.5 several flowsheets are analyzed in which no reactions are 
given for reactors with a complex set of products in the exit stream. The 
stoichiometric procedure is applied in each of these systems to obtain an 
independent set of reactions. 
4.1 Fundamental Tools
4.2 Single Reactions
4.3 Multiple Reactions
4.4 Unspecified Stoichiometries
4.5 Formal Stoichiometries
4.1 Fundamental Tools
A generalized approach to stoichiometry makes available a set of tools 
that are useful in dealing with reacting systems. This section summarizes 
the techniques needed to count and to formulate meaningful species or 
element balances, and to interpret specifications that are unique to 
reacting systems.
4.1.1 Single Reactions
4.1.2 Multiple Reactions
4.1.3 Element Balances
4.1.4 Generating Independent Reactions
4.1.5 Dependence in Reactor Specifications
4.1.1 Single Reactions
The terms stoichiometric coefficient, reaction rate, conversion, excess 
fraction and limiting reactant are defined here. Topics related to the 
combustion of fuels are discussed, in particular the concepts of an 
ultimate analysis, stoichiometric air ratio and excess air fraction. The 
counting of variables is reviewed for the special case of a single 
reaction.
Stoichiometric
Coefficient
The coefficients in a balanced chemical reaction are referred to as 
stoichiometric coefficients. Any chemical reaction among S 
components can be symbolized as 
0
1
S
s
σs Cs⋅∑
=
=
where Cs is the chemical symbol for species s. For example, the 
reaction to form ammonia is N2 + 3H2 = 2NH3. It can be also be 
written
0 2 NH3⋅ N2− 3 H2⋅−=
You will note that the coefficients for products are positive, while 
those for reactants are negative. For the above reaction: σNH3 = 2,
σN2 = -1, σH2 = -3.
Reaction Rate Let the molar flowrates of species s into and out of a reactor be ns,0 and 
 ns,1 respectively. Then the reaction rate r is independent of the species 
selected and is defined as
r
ns 1, ns 0,−
σs
=
For example:
We can compute the rate of the reaction using any species for which 
information is known in both reactant and product streams.
Using hydrogen: r = [(28 - 40) mol/h] /(-3), nitrogen: r = [(8 - 
12) mol/h]/(-1) or ammonia: r = [(8 - 0) mol/h] /(+2) the same 
reaction rate r = 4 mol/h is obtained.
Limiting
Reactant
The reactant species for which the quantity [ ns,0 / (-σs) ] is a 
minimum is called the limiting reactant.
For the reactor above, there are two reactants. [ nH2,0 /(-σH2)] = 
40/3 while [ nN2,0 / (-σN2) ] = 12/1, so that for the feed shown, N2 is 
the limiting reactant.
Note that the term "limiting reactant" is only defined for a fully specified 
set of feed streams.
Fractional
Conversion
The term fractional conversion applies only to reactants. Using inlet 
and exit streams numbered 0 and 1 respectively, the fractional 
conversion of species s is defined as
Xs
ns 0, ns 1,−
ns 0,
=
For example, the fractional conversion of H2 is XH2
40 28−
40
0.3=:=
The fractional conversion of N2 is XN2
12 8−
12
1
3
=:=
Multiple Feed,
Product Streams
If a reactor has more than one feed stream and several product streams 
the definitions presented above can be unambiguously extended.
For example, let the ammonia formation reaction occur N2 + 3 H2 = 2 
NH3 in a reactor with I feed streams and J product streams numbered as 
in the figure above. Then the definitions of reaction rate, fractional 
conversion, and limiting reactant become:
r
I 1+
I J+
i
ns i,∑
= 1
I
i
ns i,∑
=
−
σs
= s = H2, N2, NH3
Xs
1
I
i
ns i,∑
= I 1+
I J+
i
ns i,∑
=
−
1
I
i
ns i,∑
=
= s = H2, N2
The limiting reactant is the reactant species for which the following 
quantity is a minimum.
1
I
i
ns i,∑
=
σs−
s = H2, N2
Combustion of
Fuels
Many industrial processes involve the combustion of a fuel, usually with air. 
Many different methods for designating the amount of air used have been 
devised. These are reviewed here, first for simple fuels consisting of a 
single, well-defined species and then for complex fuels such as the fossil 
fuels: coal, oil.
CH4 Combustion The stoichiometric amount of oxygen required to burn a fuel is defined to 
be that amount necessary to convert all carbon in the fuel to CO2, all 
hydrogen to H2O and all sulfur to SO2. If the fuel contains any nitrogen, it 
is assumed to leave as N2 in the product stream. Note, that this definition 
says nothing about the reactions that actually occur in the combustion 
process. The "combustion" reaction is that defined above. For 
example, the combustion reaction for methane is
CH4 + σ O2 = CO2 + 2 H2O
Notice that the coefficients for CO2 and H2O are determined by the 
molecular formula for methane. The stoichiometric coefficient for oxygen is 
determined by an O-balance. In this case σ = 2.
Stoichiometric
Oxygen
Now consider a fuel with an elemental composition given by a set of atomic 
coefficients α. Let the elements occur in the order C, H, S, O, N. So 
that the fuel has the formula Cα0 Hα1 Sα2 Oα3 Nα4.
The balanced combustion reaction for one "mol" of fuel is then 
Fuel + σ O2 = α0 CO2 + ( α1/2 ) H2O + α2 SO2 + (α4/2) N2
The coefficient for oxygen is obtained by an O-balance.
σ
2 α0 α2+( )⋅ α12+ α3−
2
=
Stoichiometric
Air Ratio
The molar amount of air required per mol of fuel is simply the 
stoichiometric coefficient of oxygen in the combustion reaction divided by 
the concentration of oxygen inair. The stoichiometric air/fuel ratio β is 
usually placed on a mass basis
β fair
stoich
ffuel
=
nair( )stoich
nfuel
Mair
Mfuel
⋅= σ
xO2 air,
Mair
Mfuel
⋅=
Ultimate
Analysis
Many substances are used in combustion equipment including fossil fuels 
such as coal, oil, and "natural" gas as well as many agricultural residues 
such as waste wood, corn stover, bagasse, peanut shells etc. Municipal 
solid waste is often burned to recover energy. These substances are 
typically characterized by their elemental analyses on a mass basis.
Chemical
Formula
To convert the elemental analysis to a chemical formula, simply divide the 
mass fraction of each element by its atomic weight. This yields the 
relative molar amounts of each element. Only the ratios of these 
quantities are significant. To obtain a chemical formula, it is customary to 
select one of two bases: (1) a fixed molecular weight or (2) a specified 
coefficient for carbon, often six. This simply means scaling all the 
coefficients to give carbon a coefficient of exactly six. Both of these 
bases are illustrated in the following examples.
Mair 28.97
gm
mol
⋅:=Molecular weight of air
xO2 0.21:=Oxygen concentration in air
αC 6:=Desired coefficient of carbon
A
12.011
1.008
32.064
16.00
14.007














gm
mol
⋅:=η
0.8400
0.1140
0.0320
0.0000
0.0140














:=
Atomic weightsUltimate analysisData:
αeCoefficient of element e in chemical
formula for oil
e 0 4..:=Element indices (C, H, S, O, N)Define:
A fuel oil has the chemical composition: C: 84.00 wt%, H:11.40%, S: 
3.20%, N: 1.40%. Determine a formula for the oil on a C-6 basis and 
the stoichiometric amount of air to burn this oil.
Example 4.1.1.1
Moil 100
gm
mol
=Moil
e
αe Ae⋅∑:=
α
6.994
11.31
0.1
0
0.1














=αe αe
M'oil
Moil
⋅:=
The elemental coefficients calculated need only be scaled to match the 
desired formula weight.
M'oil 100
gm
mol
⋅:=Data:
Repeat the above calculations but use a basis of a specified molecular 
weight for the oil.
Example 4.1.1.2
Thus 13.686 g-air is required to burn one g of fuel.
β 13.686=β σ
xO2
Mair
Moil
⋅:=Stoichiometric air/fuel ratio:
σ 8.511=σ
2 α0 α2+( )⋅ α12+ α3−
2
:=Stoichiometric oxygen:
Moil 85.793
gm
mol
=Moil
e
αe Ae⋅∑:=
Since the coefficient of C was set, the molecular weight of the oil is also 
determined by this choice.
That is, the chemical formula for the oil is C6 H9.703 S.086 N.086.
α
6
9.703
0.086
0
0.086














=αe
ηe
η0
A0
Ae
⋅ αC⋅:=
Chemical formula for oil:Example 4.1.1.1
(continued)
nair nair
stoich 1 Φ+( )⋅= σ
xO2
1 Φ+( )⋅=
Given an excess air fraction Φ, the actual amount of air used is then 
computed using the stoichiometric amount of air for that fuel: 
Note that an excess air fraction of 0.0 means that a stoichiometric 
amount of air was used, while an excess air fraction of 0.1 means "10% 
excess air", and a fraction 1.2 means "120% excess air".
Φ
nair
nair
stoich
1−=
A common specification for combustion processes is the "percent excess 
air". We will use the term excess air fraction defined as 
When burning fossil fuels or agricultural waste materials to generate 
energy, it is usual practice to use more than the stoichiometric amount of 
air. The excess oxidant promotes a more intense combustion reaction 
and makes it possible to achieve complete conversion of the fuel.
Excess Air 
Fractions
As before, 13.686 g-air is required to burn one g of fuel.
β 13.686=β σ
xO2
Mair
Moil
⋅:=Stoichiometric air/fuel ratio:
σ 9.921=σ
2 α0 α2+( )⋅ α12+ α3−
2
:=
Stoichiometric oxygen:
The stoichiometric coefficient of oxygen in the combustion reaction has 
changed because the formula for the oil has changed, but the air fuel ratio is 
the same.
n
40
12
0
30
8.667
6.667








mol
hr
=
ns 1, ns 0, σs r⋅+:=Species balances:
n1 0, 12
mol
hr
⋅:=
σ
3−
1−
2








:=
n2 0, 0
mol
hr
⋅:=n0 0, 40
mol
hr
⋅:=
s 0 2..:=Let the species be indexed in the order H2, N2, NH3:
Now all the product flowrates can be computed from the species balances.
r
10
3
mol
hr
⋅:=r
XH2 n0 H2,⋅
σH2−
:=
The DF analysis indicates that there is sufficient information to solve this 
problem. The definitions of reaction rate and conversion are combined to 
give
Item Reactor
__________________________
Variables
Stream 5
Rxn rate 1
--------------------------
Sp. Balances 3
Flowrates 2
Conversion 1
__________________________
DF 0
In the reactor shown below, the conversion of H2 is 0.25. Compute all 
the flow rates.
Each independent reaction adds one reaction rate variable to the total 
number of variables for the flowsheet. If a single reaction is occurring 
then of course, only one reaction rate is needed.
Counting
Variables
mol mole≡
4.1.2 Multiple Reactions
When multiple reactions occur there is a need for a formal approach to 
keep track of the stoichiometry. The stoichiometric matrix is defined 
for this purpose. When counting variables, a number of reaction rates 
equal to the number of independent reactions must be included. The 
number of independent reactions is simply the rank of the stoichiometric 
matrix. These issues are explored in this section. When multiple 
reactions occur, an additional type of specification arises: fractional 
yield. The definition of this quantity 
Stoichiometric
Matrix
The stoichiometric coefficient of species s in reaction k is σs,k. This 
two-index quantity is the stoichiometric matrix for the reacting system.
The extension of the species balances to multiple reactions is 
straightforward. Each species is augmented by its weighted 
participation in each reaction that occurs. For input stream 0, exit 
stream 1 of a reactor in which R reactions occur with rates rk, we have
ns 1, ns 0,
1
R
k
σs k, rk⋅∑
=
+=
The use of the stoichiometric matrix in performing material balances is 
illustrated in the next example.
Example 4.1.2.1 One hundred mols/hr of an equimolar mixture of A and B reacts with 
20% conversion of A to produce a mixture of A,B,C,E in which the 
concentration of C is 1.5 times that of E. There are only two reactions 
occurring:
 A + B = 2C and B + C = E.
Stream Vars: 6
Equipment Vars: 2
---------------------
Balances: 4
Flows: 2
Conversion 1
Comp. ratio 1
_____________________
DF 0
It is clear that we have sufficient information to proceed. Try doing the 
material balances on your own. Click on the text below to check your 
work.
View the material balances
Independent
Reactions
In many systems, particularly at high temperatures, the precise pathway 
by which reactants are converted to products is unknown or extremely 
complex. These are vital issues for reactor design. But for material 
balance purposes it is usually satisfactory to represent the overall 
chemistry by some list of simple reactions. This list must be complete 
enough to account for all products.
For the purposes of counting variables, the list of reactions must also be 
independent. That is, no reaction can be a sum or difference of any other 
subset of reactions. The precise number of such reactions needed to 
represent the stoichiometry of any system can be determined from the 
stoichiometric matrix. Let the number of nonzero columns obtained after 
exhaustive column reduction of σ be symbolized by ρ(σ). Wecall this the 
rank of σ. In performing a DF analysis for a multi-reactive system , we 
count exactly ρ(σ) reaction rate variables. 
Example 4.1.2.2 The steam reforming of methane results in the following reactions.
CH4 + CO2 = 2CO + 2H2
CO + H20 = CO2 + H2
CH4 + H20 = CO + 3H2
CH4 + 2H2O = CO2 + 4H2
There are five species (CH4,CO2,CO,H2O,H2) and four reactions, so the 
stoichiometric matrix is 5 x 4.
σ
1−
1−
2
0
2
0
1
1−
1−
1
1−
0
1
1−
3
1−
1
0
2−
4














:= σ
1
0
1−
1
3−
0
1
1−
1−
1
0
0
0
0
0
0
0
0
0
0














:=
The second form is obtained from the first by elementary column 
operations. These include multiplying by a constant, or adding a multiple 
of one column to another. The final form shows clearly that there are 
only two independent reactions for this system, that is, r(s) = 2.
The reactions represented by the second form of the stoichiometric 
matrix are: CO + 3H2 = CH4 + H2O and C0 + H2O = CO2 + 
H2.
For many systems it is easy to determine by inspection that a set of 
reactions is independent. If each reaction contains one unique species 
then each reaction is necessarily independent. In more complex cases, 
the matrix reduction procedure may be required.
When a feed substance is converted to multiple products by more than one 
reaction, it is common to specify the relative amount of some desired 
product obtained. This new type of specification is called the yield.
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 s'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.
4.1.3 Element Balances
For most systems, the species balances give rise to algebraically 
simpler equation sets than do the element balances. The material 
balances for a special class of problems, however, can be simpler 
when phrased in terms of element balances. This section explores the 
issues involved in setting up and solving element balances.
Elements are 
Conserved
In reacting systems, species are not conserved and the balance 
equations must reflect this. Recall that the species balances for reacting 
systems have terms involving the stoichiometric matrix and reaction 
rates.
In contrast, elements are conserved. But the balances do not reduce to the 
unrestricted form of the non-reactive species balances. The elements are 
"packaged" into distinct species, and our material balance variables are the 
flowrates of these species. The balances must be expressed in terms of 
species flowrates but reflect element conservation.
Atom Matrix The formal method of dealing with this is to define an atom matrix. The 
element αes of this matrix is simply the coefficient of element e in the 
molecular formula for species s. The element balances merely tally the 
amount of each element accompanying each species entering and leaving 
the system. If there are I input streams, and J exit streams in a system 
containing S species consisting of E elements, then the balances are
1
J
j 1
S
s
αe s, ns j,⋅∑
=
∑
= 1
I
i 1
S
s
αe s, ns i,⋅∑
=
∑
=
= e = 1 ... E
There is one such equation for each element. The formidable double-sum 
form of these balances hides an underlying simplicity: each element is 
conserved. In practice, the equations are very simple to formulate, as 
shown in the next example.
First, notice how many terms are involved in the element balances even 
for a unit with only one inlet and one exit stream. The balances are very 
easy to write down, because element conservation is a simple concept.
But the equations are usually algebraically less tractable than the 
corresponding species balances because typically many terms appear in 
each equation.
Things to 
Note About 
Ex4.1.3.1
nH2O
41
15
mol
hr
⋅:=nCH4
2
15
mol
hr
⋅:=nCO2
21
15
mol
hr
⋅:=nCO
7
15
mol
hr
⋅:=
Substituting nCO2 = 3 nCO and solving the resulting system of three 
unknowns by elementary methods we obtain the solution: 
2 nCO2⋅ nCO+ nH2O+ 2 nO2⋅=Oxygen
nH2O 2 nCH4⋅+ 3 nC2H6⋅=Hydrogen
nCO2 nCO+ nCH4+ 2 nC2H6⋅=Carbon
DF = 6 - (3 + 2 + 1) = 0 where we have three element balances, two 
flowrates, and one compostion ratio. The problem is well posed. Note 
that we do not count any reaction rate variables when doing element 
balances. The element balances are
Ethane is burned with oxygen to produce carbon dioxide, carbon 
monoxide, methane, and water. When a 3:1 ratio of oxygen to ethane is 
used, the ratio CO2/CO in the product is 3:1.
Example 4.1.3.1
Independent
Element
Balances
If there are E elements in a system, there will be E element balances that 
can be written. Ordinarily, these will be independent. But for some 
systems this isn't true. This occurs frequently enough that a method of 
counting independent balances is needed.
The number of independent element balances for a system with atom 
matrix α is simply ρ(α), the rank of the atom matrix. The rank of the 
atom matrix is the number of nonzero rows remaining after exhaustive 
row reduction. The technique is illustrated in the next example.
Example 4.1.3.2 Urea is produced by reacting ammonia with carbon dioxide. When a 
feed consisting of 1/3 CO2 is reacted with 99% conversion of the carbon 
dioxide, the product stream contains the species shown in the figure. Is 
this problem well-posed?
The problem can be well-posed only if there are four independent 
element balances. DF = 7 - [ ρ(α) + 1 + 1 + 1] Numbering the 
species in the order listed for the product stream, and numbering 
elements in the order (C,O,H,N) the atom matrix is
α
1
1
4
2
0
1
2
0
1
2
6
2
1
2
0
0
0
0
3
1










:=
Elementary row operations can be used to reduce the atom matrix to a 
form in which there is a zero row. Then ρ(α) = 3 and the problem is 
underspecified from this viewpoint.
α
1
0
0
0
0
1
0
0
1
1
0
0
1
1
2
0
0
0
1−
0










:=
Summary In general, a good rule of thumb is to always use species balances for 
non-reactive systems. The number of independent species balances is 
always S, the number of species, and S >= ρ(α) for any system. Also, 
the species balances are usually easier to solve.
For reactive systems, when ρ(σ) = S - ρ(α), it is feasible to use either 
species or element balances. When this condition does not hold, use 
species balances. In other words, use element balances only when 
they offer the possibility of an obvious simplification. 
Element balances are only essential when a reactant is a complex 
mixture such as a fossil fuel or agricultural product such as wood, corn 
stover, or bagasse.
4.1.4 Generating Independent Reactions
Species balances are preferable to element balances in most cases. But 
what if there are no reactions given? Suppose that all we know about 
the system is the set of species present. The species balances require 
that we have a set of independent reactionsin order to formulate 
meaningful expressions for the product flow rates. This section 
discusses how to generate such a set of reactions knowing only the atom 
matrix. The resulting technique is called the stoichiometric procedure.
Stoichiometric Procedure: α ==> σ
Stoichiometric
Procedure
With knowledge only of the species present in the system we can write 
the atom matrix α for the system. This is an array with E rows and S 
columns, where E is the number of elements present and S is the number 
of species.
To determine a set of independent reactions for this system, use 
elementary row operations to put α in the form 
α
I
0
β
0






=
I is an identity matrix of size ρ(α), and β has dimensions ρ(α) x [S - 
ρ(α)]. The stoichiometric matrix is then an S x [S - ρ(α)] matrix of the 
form
σ
β−
I






=
The next example illustrates the procedure.
Example 4.1.4.1 This "alphabet soup" system consists of five species A,B,C,D,E and four 
elements w,x,y,z. The molecular formulas of the species are reflected in 
the atom matrix shown below.
A = w x y3
B = y2 z
C = w y
D = w2 y4 z
E = x y2
The system consists of the species:
It is represented by the atom matrix: α
1
1
3
0
0
0
2
1
1
0
1
0
2
0
4
1
0
1
2
0










=
The atom matrix can be row-reduced to
the form:
α
1
0
0
0
0
1
0
0
0
0
1
0
0
1
2
0
1
0
1−
0










=
We identify a 3x3 identity matrix in the
upper left, and the matrix β is: β
0
1
2
1
0
1−








=
The stoichiometric matrix is obtained by
taking the negative of b and appending an
identity matrix of size [S - ρ(α)] or 2.
σ
0
1−
2−
1
0
1−
0
1
0
1














=
This matrix corresponds to the two reactions: B + 2C = D and A = C + 
E.
Additional examples are given by the ammonia oxidation system of 
section 4.5.1 and the alcohols from syngas system of section 4.5.2.
Summary When a system is characterized by knowledge only of the species 
present, a set of independent reactions can easily be found by the 
stoichiometric procedure.
4.1.5 Dependence in Reactor Specifications
Once a set of R independent reactions has been specified for a reactor, 
information given in the specifications is then used to fix the reaction 
rates. Fixing a reaction rate requires that at least one species be 
completely specified: that is, both its input and output flow rates must be 
known. Some combinations of completely specified species do not 
represent independent information. In this section a test is presented to 
detect when this situation occurs. 
Example Let the combustion of methane with pure oxygen be modeled with the 
following two reactions:
 CH4 + 2 O2 = CO2 + 2 H2O
2CH4 + 3 O2 = 2CO + 4 H2O
The methane is completely burned to produce an oxygen-free stack gas 
consisting of CO2, CO, and H2O.
The two reactions are clearly independent since each involves a unique 
species. There are five balances and a basis can be selected. We will 
use the methane flow rate. This means that one additional specification 
can be made. The same conclusion is drawn from a DF analysis.
DF = [5 + 2] - [5 + 1] = 1
Let this additional specification be the water flow rate. The two species 
that are completely specified are methane and water. Let the methane 
flow rate be 1.0 mol/h and that of water be 2.0 mol/h. Their balances are 
CH4: 0 = 1 - r1 - 2 r2
H2O: 2 = 2 r1 + 4 r2
By an obvious rearrangement, the water balance is seen to be a multiple 
of the methane balance, and is therefore not an independent specification.
Criterion for 
Independence
For species balances, a specification is independent if the rows of the 
stoichiometric matrix corresponding to the completely specified species are 
linearly independent.
In the above example, the stoichiometric matrix with species ordering 
methane, oxygen, water, carbon dioxide, carbon monoxide is:
σ
1−
2−
2
1
0
2−
3−
4
0
2














=
The rows corresponding to the completely specified species ( methane and 
water ) are clearly not linearly independent. The row for water is just (-2) 
times the row for methane.
σcs
1−
2
2−
4






:=
Note that the original problem can be solved if another specification is 
selected. For example, a flowrate of CO2 or CO, or a CO2/CO ratio 
would be valid specifications.
4.2 Single Reactions
Each flowsheet in this section involves a single reaction. The sample 
problems are intended to give you practice in setting up a 
stoichiometric matrix and using an assortment of specifications to find 
the product composition.
4.2.1 Chlorine Dioxide Synthesis
4.2.2 Hydration of Ethylene
4.2.3 Acetic Acid Synthesis
4.2.1 Chlorine Dioxide Synthesis
Problem
Statement
Chlorine dioxide is a bleaching agent produced by the following 
reaction. 6 NaClO3 + 6 H2SO4 + CH3OH = 6 NaHSO4 + 6 ClO2
+ CO2 + 5 H2O
If 14 mols of the chlorate stream are fed for every mol of methanol and a 
conversion of 90% is achieved, what is the composition of the product?
Click on the icon at the left to see the DF analysis for this problem.
Click on the underlined text below to see the material balances.
View the material balances for the chlorine dioxide synthesis.
4.2.2 Hydration of Ethylene
Problem
Statement
The catalytic hydration of ethylene is carried out with a single-pass 
conversion of only 4.5%. The separator removes all the alcohol and 
the recycle gas contains 6.5 mol% steam. The molar ratio of steam to 
ethylene in the mixed feed to the reactor is 0.55. Calculate the 
compositions of all streams, and the ratio of steam to ethylene fed to the 
process.
On Your Own Click on the underlined text below to check your DF analysis.
View the DF analysis for the hydration of ethylene.
Complete the material balances for this problem, then check your 
work by clicking on the underlined text below.
View the material balances for the hydration of ethylene.
4.2.3 Acetic Acid Synthesis
Problem
Statement
The recycle reactor shown below oxidizes ethanol to acetic acid with a 
mixture of dichromate and sulfuric acid. A 90% overall conversion is 
achieved with a recycle to ethanol feed ratio of 1:1, a 10% excess of 
sulfuric acid, and a 20% excess of sodium dichromate. The recycle 
stream contains only ethanol and sulfuric acid. The concentration of 
acid in the recycle is 94 mol%. A perfect separation of acetic is 
achieved. The reaction occurring is
3C2H5OH + 2NaCr2O7+ 8H2SO4 =
 3CH3COOH + 2Cr2(SO4)3 + 2Na2SO4 + 
11H2O
On Your Own Perform a DF analysis on the acetic acid synthesis flowsheet. Check 
your work by clicking on the underlined text below.
View the DF analysis for the acetic acid synthesis
Now do the material balances in your worksheet. Check your work 
by clicking on the underlined text below.
View the material balances for the acetic acid synthesis.
4.3 Multiple Reactions
The flowsheets in this section all involve multiple reactions. These 
problems focus on calculating the reaction rates from given specifications 
and then using those rates to compute final product compositions.
4.3.1 Acetaldehyde Synthesis
4.3.2 Carbon Disulfide Synthesis
4.3.3 Reformate Hydrodealkylation
4.3.4 Toluene Dealkylation
4.3.1 Acetaldehyde Synthesis
Problem
Statement
Acetaldehydecan be produced by the catalytic dehydrogenation of 
ethanol. There is however, a parallel reaction that produces ethyl 
acetate.
 C2H5OH = CH3CHO + H2
 2C2H5OH = CH3COOC2H5 + 2H2
In a particular reactor, an ethanol conversion of 95% is obtained with 
an 80% yield of acetaldehyde. Calculate the product composition 
assuming the feed consists of pure ethanol.
On Your Own Try performing the DF analysis for this problem. Check your work by 
clicking on the exercise icon at the left.
Now compute the product composition. Click on the underlined text to 
check your work.
View the Acetaldehyde Synthesis material balances.
4.3.2 Carbon Disulfide Synthesis
Problem
Statement
Vaporized sulfur is reacted with methane to produce CS2. There are 
three reactions known to occur.
 CH4 + 4S = CS2 + 2H2S
 CH4 + 2S = CS2 + 2H2
 CH4 + 2H2S = CS2 + 4H2
A feed containing four mols of sulfur per mol of methane results in a 
product mix characterized by 90% methane conversion and 70% sulfur 
conversion. Calculate the product composition.
On Your Own Complete a DF analysis for this problem. Click on the underlined text 
below to check your work.
View the DF analysis for the Carbon Disulfide synthesis.
Do the material balances for this problem. Click on the underlined text 
below to check your work.
View the material balances for the Carbon Disulfide synthesis.
Click here to check your DF analysis for the hydrodealkylation 
Determine if the problem is well-posed.On Your Own
What is the composition of the product stream?
(4)2C6H6 = (C6H5)2 + H2
(3)C6H3(CH3)3 + H2 = C6H4(CH3)2 + CH4
(2)C6H4(CH3)2 + H2 = C6H5CH3 + CH4
(1)C6H5CH3 + H2 = C6H6 + CH4
The reactions occurring are:
The process of removing alkyl groups from aromatic compounds with 
hydrogen is called hydrodealkylation. In the process shown below, a 
refinery reformate stream consisting of 5 mol% benzene (B), 20% 
toluene (T), 35% xylene (X), and 40% mesitylene (Y) reacts with 
hydrogen. When 5.0 mols of hydrogen is used per mole of 
hydrocarbon feed, the following conversions are obtained: T:80%, 
X:74%, Y:13%. The product stream also contains 0.1% biphenyl (P).
Problem
Statement
4.3.3 Reformate Hydrodealkylation
Compute the composition of the product stream.
Click here to see the material balances for the 
hydrodealkylation
4.3.4 Toluene Dealkylation
Problem
Statement
The dealkylation of toluene to produce benzene is accompanied by a 
side reaction which produces biphenyl.
 C6H5CH3 + H2 = C6H6 + CH4 2C6H5CH3 + H2 = (C6H5)2 + 
2CH4
The side reaction requires that conversions be kept low in the reactor, 
and this complicates the separation sequence. The first stage of 
separation recovers pure benzene and produces two recycle streams.
The overhead consists of hydrogen and methane, part of which must 
be purged to remove the methane. The bottom product of the first 
separation stage is a liquid stream which undergoes further separation 
into a pure toluene recycle stream, and a pure diphenyl byproduct.
When the reactor feed has a 5:1 ratio of hydrogen to toluene, a 75% 
toluene conversion is achieved.
a. If the reactor outlet stream is 5% benzene, and 2% toluene, 
calculate the fractional yield of benzene and the make-up (stream 2) 
hydrogen required per mol of toluene fed to the process.
b. Repeat for a reactor outlet specification of 2% toluene and 58% 
methane.
On Your Own In your worksheet, perform a degree of freedom analysis for part a.
Click on this text to check your DF analysis.
Continuing in your worksheet, compute all the material flows.
Click on this text to check your material balances.
Now repeat the DF analysis using the specifications for part b.
Click on this text to check your DF analysis for part b.
4.4 Unspecified Stoichiometries
The problems in this section focus on the application of element
balances to process flowsheets. Species balances are usually preferred 
over element balances. Element balances can be useful, however, in 
processes involving natural products or in cases where the reaction 
stoichiometry is unspecified.
Natural products such as fossil fuels or agricultural residues are 
complex mixtures of materials with a wide range of molecular weights.
Their composition as a mixture of distinct species is too complex to deal 
with effectively for material balance purposes. Instead, balances are 
done at the element rather than the species level. 
The reactions involved in some processes, even those involving only 
simple, distinct species, are often unknown or very complex. Element 
balances are sometimes useful for these systems.
4.4.1 Fuel Oil Combustion
4.4.2 Coal Gasification
4.4.3 Fuel Oil Desulfurization
4.4.4 Formaldehyde Synthesis
4.4.1 Fuel Oil Combustion
Problem
Statement
A fuel oil with analysis C:84.0%, H:11.4%, N:1.4%, S:3.2% is burned 
with air. The stack is found to contain 2130 ppm by wt (dry basis) of 
sulfur dioxide. Determine a chemical formula for the oil on a C-6 basis, 
and compute the percent excess air that was used in the burner.
On Your Own Perform a DF analysis for the oil burner.
Click on the this text to see the DF analysis for the oil burner.
On your worksheet, solve for the percent excess air.
Click on this text to view the material balances for the oil 
burner.
4.4.2 Coal Gasification
Problem
Statement
A dry coal consisting of ash:3%, N:1.5%, S:0.6% and unknown 
amounts of C, H, and O is gasified with steam and air to produce a fuel 
gas. The dry-basis composition of the fuel gas is H2:23%, CH4:3.2%,
CO:16.2%, CO2:12%, H2S:1%, and N2:44.6%. In addition to the fuel 
gas, a solid residue is produced consisting of 10% C and the rest ash.
If a 1:1mass ratio of steam/coal is fed, calculate the air rate and 
moisture content of the fuel gas. 
On Your Own Perform the DF analysis for this system. 
Click on this text to check your DF analysis for the coal gasifier.
Now complete the material balances.
Click on this text to see the material balances for the coal 
gasifier.
4.4.3 Fuel Oil Desulfurization
Problem
Statement
A fuel oil with analysis C:84.0%, H:11.4%, N:1.4%, S:3.2% is 
desulfurized by contact with a hydrogen-rich gas ( 44 mol% H2, 56% 
CH4) in a catalytic reactor. The desulfurized oil contains only 1 ppm N, 
9 ppm S. The gas stream exiting the reactor contains H2, CH4, NH3,
H2S. The methane to hydrogen ratio in the off-gas is 7:3, while the 
ammonia:hydrogen ratio is 1:4.
The reactor off-gas is subjected to two stages of purification. The first 
stage removes 90% of the hydrogen sulfide and all of the ammonia. The 
second stage removes all the remaining H2S. The recycle stream is 40% 
H2. It is mixed with fresh gas containing 48% H2. All gas compositions 
are in mol %, while the oil compositions are in wt%.
If 60 lb oil are fed to the reactor per mol of gas, calculate the product oil 
composition.
On Your Own Perform a DF analysis for this system.
Click here to check your DF analysis of the oil desulfurization
Now complete all the material balances.
Click here to check your material balances for this process.
4.4.4 Formaldehyde Synthesis
Problem
Statement
Formaldehyde can be produced by partial oxidation of methanol over a 
silver catalyst. Formic acid, hydrogen, carbon monoxide, carbon 
dioxide and water are also produced. An overall yield of formaldehyde 
of 2/3 is achieved in the process shown below. The reactor product is 
scrubbed to separate the light gases, and the residue is distilled to 
produce pure methanol for recycle.The scrubber and distillation units 
are combined into a single separator in the flowsheet. The reactor inlet 
stream is 35 mol% methanol and the off-gas composition is H2: 20.2%,
CO2:4.8%, CO:0.2%, N2:74.5%, O2:0.3%. Calculate the 
compositions of all streams.
On Your Own Perform a DF analysis for this flowsheet.
Click here to see the DF analysis for the formaldehyde process.
Complete the material balances for the formaldehyde process.
Click here to view the material balances for this process.
4.5 Formal Stoichiometries
This section deals with problems in which species balances are used in 
preference to element balances even if no reactions are given.
We have two reasons for preferring species balances to element 
balances. There are always as many or more species balances as there 
are element balances. Also, the species balances are usually 
algebraically simpler. If the species present in the system are known, 
then the atom matrix for the system can be constructed. The 
technique of generating a feasible stoichiometric matrix from the atom 
matrix, and thus an independent set of reactions, was discussed in 
section 4.1.4. The problems in this section illustrate the use of this 
technique.
4.5.1 Ammonia Oxidation
4.5.2 Alcohols from Syngas
4.5.3 Formaldehyde Synthesis
4.5.1 Oxidation of Ammonia
Problem
Statement
A feed stream formed from air and ammonia (20 mol air/mol NH3)
reacts to produce a complex mixture consisting of H2O, O2, N2, NH3,
NO, and NO2. When an 80% conversion of ammonia is achieved, the 
product stream contains 3% O2 and 6% NO. Calculate the 
compositions of the other species. 
On Your Own Perform a DF analysis on the ammonia oxidation reactor.
Click here to check your DF analysis.
Continuing with your worksheet, do the material balances.
Click here to check your material balances.
4.5.2 Alcohols from Syngas
Problem
Statement
When a mixture of CO and H2 (syngas) with a H2:CO ratio of 3.2 is 
reacted over a nickel catalyst, a mixture is produced that consists of 
hydrogen, carbon monoxide, water, methane, ethane, propane, 
methanol, ethanol, and carbon dioxide. When the conversion of CO is 
97.9% and that of hydrogen is 73.8%, the product gas is characterized 
as follows: CO2/CO = 2.0, C2H5OH/CH3OH = 0.05, C3H8/CH4 = 
0.01, C2H6/CH4 = 0.05. Find the composition of the product gas using 
only element balances. Repeat using only species balances.
On Your Own Perform a DF analysis for this process.
Click here to see a DF analysis of the alcohols from syngas 
process.
In your worksheet, solve for the composition of the product gas using 
only the element balances.
Click here to see the element balances for this process.
In your worksheet, solve for the composition of the product gas using 
only species balances.
Click here to see the species balances for this process.
4.5.3 Formaldehyde Synthesis
Problem
Statement
Formaldehyde can be produced by partial oxidation of methanol over a 
silver catalyst. Formic acid, hydrogen, carbon monoxide, carbon 
dioxide and water are also produced. An overall yield of formaldehyde 
of 2/3 is achieved in the process shown below. The reactor product is 
scrubbed to separate the light gases, and the residue is distilled to 
produce pure methanol for recycle. The scrubber and distillation units 
are combined into a single separator in the flowsheet. The reactor inlet 
stream is 35 mol% methanol and the off-gas composition is H2: 20.2%,
CO2:4.8%, CO:0.2%, N2:74.5%, O2:0.3%. Calculate the 
compositions of all streams using species balances.
On Your Own Perform a DF analysis for this flowsheet.
Click here to check your DF analysis for the formaldehyde 
process.
Complete the material balances for the formaldehyde process.
Click here to view the material balances for this process.
5.6.2 Multiple Non-Linear Equations
5.6.1 Numerical Techniques for Single Equations
Solving Non-Linear Equations5.6
Acid-Gas Removal5.5
Methyl Iodide Synthesis5.4
Perchloric Acid Synthesis5.3
Simple Recycle5.2
Extending the Sequential-Unit Approach5.1
We follow the pattern established in previous chapters. The first 
section defines relevant terms and explores the special issues associated 
with this new technique. Subsequent sections apply the technique to 
selected process flowsheets.
In all the flowsheets considered thus far we have implemented a 
sequential-unit solution strategy. We were able to solve all the balances 
for a selected process unit, then we moved on to the next unit. This 
chapter deals with flowsheets for which this procedure breaks down.
Rather than abandoning the sequential-unit approach, we will augment 
our strategy by proceeding with partial solution of one or more sets of 
balances.
Chapter 5 Sequencing with Partial Solution
5.1 Extending the Sequential-Unit Approach
For all the flowsheets considered thus far, our calculation sequencing 
procedure has resulted in a partitioning of the material balances into 
ordered subsets that were completely and separately solvable. We 
simply identified a process unit with a degree-of-freedom of zero, 
solved the balances for that unit, and then used the new information thus 
obtained to reduce the DF for subsequent units to zero. This process 
was repeated until all the streams were fully characterized.
Sequencing
with Partial 
Solution
This chapter deals with flowsheets whose specifications do not permit 
the balances for some unit(s) to be solved completely. Rather than 
abandoning the sequential-unit approach, we augment it with the 
following technique.
1. We first solve the balances for all units that have a zero DF, if there 
are any.
2. Then we consider the balances for that unit with the lowest DF. For 
this discussion, let the DF for this unit be '1'. The positive value of DF 
means that we will not be able to solve all the balances completely.
Instead, we will have to estimate the value of one carry-forward 
variable, and express the solutions for all other computed quantities in 
terms of this variable. 
3. We then continue with the balances for the remaining process units.
4. Eventually, since the DF for the process is zero, we will have a 
redundant balance that can serve as an "objective function". We can 
vary the value of the estimated variable in order to satisfy this objective 
function.
Before giving more details about implementing the procedure of 
sequencing with partial solution, let us deal with some preliminary issues 
that sometimes cause confusion.
An Alternative
Strategy
Instead of solving the material balances sequentially, unit by unit, we 
could write down all the balance equations at one time and solve them 
simultaneously. The fact that the DF for the process is zero means that 
this procedure can be made to work. Modern calculators and tools like 
Mathcad make the computational burden a manageable one. However, 
there are some good reasons why this is not a good approach for hand 
calculations.
1. There is the practical problem of selecting an independent set of 
balances. Verifying that a set of three or four balances for one unit is an 
independent set is much easier than trying to make this determination for a 
much larger set of equations.
2. It is easier to locate an error in one equation (or in a small set of 
equations) than it is to locate such an error in a large set of equations.
A complex flowsheet with a dozen process units and several species 
may involve hundreds of variables. The sequential-unit approach is 
predicated on the principle that a small set of equations is easier for a 
human being to deal with than a large set of equations. This ability to 
"debug" a solution is a compellingreason to favor the sequential-unit 
strategy over the simultaneous-equation approach.
3. By dealing with the balances on a unit-by-unit basis, intuitive 
approaches for a particular application are more likely to be employed.
These "obvious simplifications" are more likely to be recognized in the 
sequential approach than in the simultaneous-equation approach.
Although it would be nice to believe that a solution to the general 
material balance problem is available in some sense, it is still true that 
leaps of intuition can often arrive at a solution with far greater efficiency 
than a "general" method. Although we are trying to develop a 
framework of tools to lessen the need for such intuitive approaches, 
many flowsheets display unique features which can be profitably 
exploited by an engineer with the skill to recognize them.
Hand
Calculation
The reader may have wondered why we continue to employ he term 
"hand calculation" when we are clearly gearing our solution methods for 
implementation by computing tools such as Mathcad, or similar 
packages.
We define a hand calculation as one in which we formulate the balances 
ourselves. Whether we use an abacus, slide rule, calculator, or 
Mathcad to actually solve those balances is irrelevant. These tools only 
handle the tedious mathematical manipulations required to solve the 
equations. We still have to tell the tool what equations to solve.
There are software tools called process simulators which (with some 
restrictions on the specifications) could formulate as well as solve all the 
balances for us. All the user has to do is enter the data. Such a tool 
permits fully automated solution of all the material balances. This 
electronic book does not concern itself with such tools.
We return now to the issue of determining the value of a carry-forward 
variable.
Non-Linear
Equations
The final determination of the estimated variable is made by satisfying 
some objective function. The objective function is usually one of the 
redundant balances that involves the process unit for which the stream 
variable estimate was made. In one sense, the problem has been 
reduced to solving f(x) = 0, where f(x) is some non-linear expression in 
x.
Evaluating the
Objective
Function
Unlike the situation in which we have a single, equation to "plug into", it is 
usually the case that evaluating the objective function may involve 
performing many intermediate balances. In the figure below, let x be the 
carry-forward variable.
There is no distinction in a mathematical sense between a closed-form 
expression, like f(x) = 3 x2 - sin(x) + 5, and a procedure that can 
uniquely assign a value to our objective function, given a value of the 
variable x. All that matters is that we can indeed evaluate the function.
Converging to
a Solution
Once we have estimated the value of the carry-forward variable, say x, 
and selected an objective function f(x), we must determine the value of x 
that satisfies f(x) = 0. Distinguish clearly the process of evaluating the 
objective function and the process of solving the objective function.
One way to solve the equation is to simply guess values of x until we get 
lucky enough to find one that makes f(x) =0. This method is called 
solving by "trial and error". It is usually impractical to proceed by this 
fashion.
A better method is to guess only once, and then use an updating rule to 
successively refine the value of x, that is, converge to a solution. Many 
such updating rules exist.
The Newton, Secant and Wegstein methods are examples of such 
convergence techniques. They all employ some rule for updating the 
current estimate of the solution. The Secant method is illustrated below. 
 Other iterative methods are discussed in section 5.6.
The sample flowsheet analyses in sections 5.2 - 5.5 illustrate the 
technique with far greater clarity than any word description can afford.
The essence of the technique of carry-forward variables is simple.
When the DF for a unit is positive, introduce one or more carry-forward 
variables. Values for these variables must be estimated.
To determine the actual values of the variables, complete sufficient 
additional balances to permit evaluation of some objective function. If the 
objective function is not satisfied within some tolerance, then vary the 
assumed values of the variables by some convergence technique, and 
re-evaluate the objective function. Repeat this procedure until the 
objective function is satisfied to within the desired tolerance.
Summary
Err 6.381− 10 10−×=Err x6 2
1
3
−:=
The last value of x is very close to the exact solution.
x
1
1.5
1.21053
1.25141
1.26027
1.25992
1.25992




















=
The sequence of values generated is:
xk 1+
xk 1− f xk( )⋅ xk f xk 1−( )⋅−
f xk( ) f xk 1−( )−:=k 1 5..:=
Currently, we have two argument values x1 = 1, x2 = 1.5 and two 
function values: f1 = -1, f2 = 1.375. Let us generate some new 
estimates for x. We will use index k to keep track of them. The 
updating rule for the secant method is
f x1( ) 1.375=x1 1.5:=
f x0( ) 1−=x0 1:=
f x( ) x3 2−:=
To illustrate the technique, let us solve f(x) = x3 - 2 = 0. The answer is 
the cube root of two. But let us converge to that value by starting with 
an initial estimate of 1.0.
Secant Method
5.2 Simple Recycle
Problem
Description
A feed of 1000 mol/h consisting of 1/3 A, 2/3 B is mixed with a recycle 
stream and reacts with the stoichiometry A + B --> D. The conversion 
of A per pass is 20%. The split fractions from the separator feed to the 
recycle stream are A:0.80, B:0.90, D:0.10. Calculate the flow rates of 
all streams.
Click on the diamond icon to copy the figure and the problem 
statement to your worksheet.
On Your Own Show that the problem is specified correctly and determine a calculation 
order. If any carry-forward variables are required, be sure to note 
when they will be needed, and how they will be determined.
Click on this text to view the DF analysis for this problem.
Set up and solve all the material balances.
Click on this text to view the material balances for this problem.
5.3 Perchloric Acid Synthesis
Problem
Description
Perchloric acid is produced by reacting barium chlorate with sulfuric acid.
Ba(ClO4)2 + H2SO4 = BaSO4 + 2 HClO4
In the process shown below, an 80% conversion of barium chlorate is 
achieved when the ratio of sulfuric acid to barium chlorate in the 
combined reactor feed is 1:1.2. The feed (stream 1) contains 90 wt% 
barium chlorate in perchloric acid. The recycle stream contains only 
Ba(ClO4)2. Calculate all flowrates.
Click on the diamond icon to copy the figure and problem statement to your
worksheet.
On Your Own Show that the problem is specified correctly and determine a calculation 
order. Note if any carry-forward variables are required.
Click on this text to view the DF analysis for this problem.
In your own worksheet, set up and solve all the material balances for this 
problem.
Click on this text to check your material balances.
5.4 Methyl Iodide Synthesis
Problem
Description
Two thousand lb/day of HI is added to an excess of methanol in the 
reactor shown below. The compositions in the product stream are 
given as mass fractions. The recycle stream is pure HI. What is the 
recycle ratio (stream 5: stream 1) if a single-pass conversion of 30% 
is achieved via the reaction
HI + CH3OH = CH3I + H2O
Click on the diamond icon to copy the problem statement and the figure to
your worksheet.
On Your Own Determine if the problem is specified correctly. Specify a calculation 
sequence, and determine if anycarry-forward variables are required.
Click on this text to see the DF analysis for this problem.
Solve for the required recycle ratio.
Click on this text to see the material balances for this problem.
5.5 Acid-Gas Removal
The absorber-stripper system shown in the figure removes CO2 and H2S
from a feed consisting of 30% CO2 and 10% H2S in an inert gas. A 
proprietary solvent is used to selectively absorb the acid gases. The 
absorber bottoms are flashed. The ratio of the concentration of CO2 in 
the exit vapor (stream 6) to that in the liquid leaving the flash unit is 
13.33. A similar ratio of 6.9 is measured for H2S. The liquid leaving 
the flash contains 5% CO2, and the overhead contains 20% solvent. The 
stripper overhead contains 30% solvent. The bottoms stream leaving the 
stripper is pure solvent.
The feed is cleaned to the extent that stream 2 contains only 1% CO2
and no H2S. The splitter beneath the flash unit operates to produce 
equal flowrates in streams 3 and 8. Compute the compositions of all 
streams and the amount of solvent per mole of feed required.
Problem
Description
On Your Own Is there sufficient data supplied to solve this problem?
Click on this text to see the DF analysis for this problem.
On your own worksheet, determine the flowrates of all streams.
Click on this text to view the material balances for this problem.
5.6 Solving Non-Linear Equations
Whenever a carry-forward variable, or tear variable, is required to 
proceed with the balances for a flowsheet, some technique must be 
implemented to converge on the correct value of the assumed quantity.
If only a single tear variable is needed, then there are several techniques 
which can reliably converge to a solution. The Newton, Secant, 
Successive Substitution, and Wegstein methods are discussed in the first 
section below.
The Wegstein convergence method is easily extended to deal with the 
case of multiple tear variables. This technique is discussed in section 
5.6.2.
5.6.1 Numerical Techniques for Single Equations
5.6.2 Multiple Non-Linear Equations
5.6.1 Numerical Techniques for Single Equations
The techniques in this section involve a very general class of equations.
Any problem that can be phrased in the form f(x) = 0 can be treated by 
the methods discussed here. Whether a numerical method actually 
converges to a root, of course, depends heavily on the nature of the 
function whose root is sought.
Closed-Form
Objective
Function
We first consider the case when the function f(x) can be written in closed 
form. Such functions arise frequently when physical systems are 
simulated.
An Example Suppose we are asked to find the value of the argument x when the 
functions y(x) = exp(x) and z(x) = 5 - 4(x-3/2)2 intersect? First, let's 
plot these functions to see what they look like.
M 30:= i 0 M..:= Xi 2
i
M
⋅:=
y x( ) exp x( ):= z x( ) 5 4 x
3
2
−






2
⋅−:=
0 0.5 1 1.5 2
0
2
4
6
y Xi( )
z Xi( )
Xi
The function z(x) is an inverted parabola centered on x = 1.5. There 
are two roots, i.e. there are two values of x that satisfy the equation 
y(x) = z(x). These two roots lie near x = 0.6 and x = 1.6.
General Form The problem can be restated in general as f(x) = 0, in which the function is
f x( ) exp x( ) 4 x
3
2
−






2
⋅+ 5−:=
0 1 2
2
0
2
f Xi( )
0
Xi
The function f(x) crosses the axis near x = 0.6 and x = 1.6. Suppose we 
wish to find the exact value of the root near x = 1.6? We already have a 
good estimate for the root. What we need is a technique to update and 
improve this estimate. Below, we develop such a technique. Later we 
will come back to this example and find the desired root.
Newton's
Method
Illustrated above is an arbitrary function f of a single variable x. We seek 
to find the root x* given the estimate of the root: x0. Newton first 
suggested that an improved estimate of the root could be found by 
following the tangent to the curve back to where it meets the axis at x1.
Two expressions can be written for the slope of the tangent: one is the 
derivative f'(x), the other is the divided difference representing the slope of 
the curve at x = x0.
f' x0( ) f x0( ) 0−x1 x0−=
This expression can be solved for the new estimate.
x1 x0
f x0( )
f' x0( )−=
Note that both the function f and its derivative f' are evaluated at x0. That 
is the entire right-hand side is a function only of x0. The calculated value x1
will in general be closer to x* than the starting value x0.
The updating expression, or iteration formula can be applied multiple times 
to generate the sequence x2, x3, x4, .... which under certain conditions will 
converge as closely as desired to the root. That is, after we have 
calculated x1, we can use x1 on the right to generate x2, etc. We stop 
when the new value of x differs from the last value by some desired 
tolerance.
Iteration Scheme
Characteristics
All iteration methods have the following characteristics:
0. An initial estimate must be supplied. This estimate is usually based 
on intuition about a physical system, or analysis of the characteristics of the 
mathematical expression. This is the place where "engineering problem 
solving" skills are applied.
1. An iteration formula must be available to generate new values of the 
argument. Newton's formula is an example. We will see others shortly.
2. Some convergence criterion must be selected. This is also an issue 
which depends on the nature of the analysis being carried out. In essence 
one must decide how many places of accuracy are reasonable for the 
argument of interest.
xr 1.601209=xr root f ξ( ) ξ,( ):=ξ 1.6:=
Of course, we could just as easily have let Mathcad solve for this root 
directly:
f x3( ) 0=... and it is.
It is clear that the third and fourth values are identical to six places. As a 
check, the function value should be near zero ...
x
1.6
1.601211
1.601209
1.601209












=The sequence of values is:
xk 1+ xk
f xk( )
fp xk( )−:= ... by applying Newton's iteration formula:
k 0 2..:=Generate a sequence of values ... 
x0 1.6:=The initial estimate of the root:
fp x( ) exp x( ) 8 x
3
2
−






⋅+:=
We need the derivative of f:
f x( ) exp x( ) 4 x
3
2
−






2
⋅+ 5−:=
Newton's formula is applied below to the problem of finding the root of 
f(x) near x = 1.6. 
Newton Method
Example
Extension to
Implicitly-Defined
Functions
Whenever an analytical expression for a function is available, it is clear that 
the Mathcad root function will solve for the desired root. However, 
Mathcad cannot deal directly with an implicitly defined function such as 
those which arise in the analysis of process flowsheets. The "function" can 
be visualized in the following context:
The argument "x" is some flowsheet variable, such as a species 
composition, flow rate, or a dimensionless equivalent of such a quantity.
The objective function is some material balance, an expression involving 
other flowsheet variables. Values for these other flowsheet variables are 
computed from other balances, independent of the balance that is used as 
the objective function. The values of the compositions and flow rates that 
appear in the objective function depend on the value of the argument, even 
if the argument itself does not appear explicitly in this balance.
A critical distinction must be made between the process of evaluating this 
function and the method used to solve it, that is, to find the value of the 
argument that satisfies the balance. The evaluation consists simply in 
performingall the necessary intermediate balances. The "solving" consists 
in the application of an iterative scheme that converges to the desired root.
The Newton method, which requires an explicit derivative, is not suitable 
for this task. Instead, a related method in which the derivative is estimated 
numerically can be employed. This method is termed the secant method.
x1 x0 0.05+:=Another value of the argument: 
x0 1.6:=Initial estimate of the root:
f x( ) exp x( ) 4 x
3
2
−






2
⋅+ 5−:=
We can apply this technique to the function defined above.Secant
Application
This is the secant updating rule.
x2
x0 f x1( )⋅ x1 f x0( )⋅−
f x1( ) f x0( )−=
This expression can then be solved for the next value of x in the sequence.
f x1( ) f x0( )−
x1 x0−
f x1( ) 0−
x2 x1−
=
We equate two expressions for the slope of the secant at x1.
The Newton method was based on following the tangent to the curve at x0
back to the x-axis. The tangent can be approximated by the secant to the 
curve through two closely-spaced points, say x0 and x1.
Secant
Method
k 0 7..:=x0 0.5:=Initial estimate:
g x( )
1 2 x2⋅+ x2.2−
2.8
:=One reformulation is
f x( ) x2.2 2 x2⋅− 2.8 x⋅+ 1−:=
A simpler technique for solving f(x) = 0 is to reformulate the objective 
function (when possible) in the form x = g(x). The right-hand-side is then 
an updating rule. For the function
Successive
Substitution
If the convergence tolerance δ = 0.001 then we could have stopped with 
the fourth value (the second iteration) in the sequence. With δ = 0.000001 
we needed six values (four iterations).
xk xk 1−− δ<
In general, iteration is stopped when a convergence criterion like
x
1.6
1.65
1.6011462
1.6012062
1.6012095
1.6012095


















=
The method converged to the same
root as before, albeit more slowly.
xk 2+
xk f xk 1+( )⋅ xk 1+ f xk( )⋅−
f xk 1+( ) f xk( )−:=
k 0 3..:=Let's generate a sequence of values ... 
Generate the sequence: xk 1+ g xk( ):=
The sequence converges slowly. x
0.5
0.45799
0.44289
0.43773
0.43599
0.43542
0.43523
0.43516
0.43514


























=
Convergence
Problem
Although successive substitution is temptingly simple, it often fails to 
converge. We can try solving the function used in the Newton and Secant 
examples by this technique to see this behavior.
f x( ) exp x( ) 4 x
3
2
−






2
⋅+ 5−:=
g x( )
5 exp x( )−
4
3
2
+:=
Initial estimate: x0 1.601:=
Generate the sequence: xk 1+ g xk( ):=
The sequence fails to 
converge even when supplied
with an initial estimate that is
very close to the root.
x
1.601
1.602
1.593
1.642
1.5 0.205i+
1.962 0.246i−
1.79 0.748i+
2.269 0.662i−
2.196 1.068i+


























=
x
1.601
1.6024842
1.5930696
1.5997952
1.6013528
1.6011789
1.6012104
1.6012095
1.6012095
1.6012095






























=The sequence converges.
Sk 1+
xk 2+






g xk 1+( ) g xk( )−
xk 1+ xk−
xk 1+ 1 θ Sk( )−( )⋅ θ Sk( ) g xk 1+( )⋅+










:=
Generate a sequence:
S0
x1






0
g x0( )






:=k 0 7..:=x0 1.601:=Initialization:
g x( )
5 exp x( )−
4
3
2
+:=Auxiliary functions: θ η( )
1
1 η−
:=
f x( ) exp x( ) 4 x
3
2
−






2
⋅+ 5−:=We apply the Wegstein method to:Example
2. Update: x = (1 - θ) x1 + θ g(x1)
1. Evaluate x1 = g(x0), g(x1), S = ∆g/∆x, and θ = 1/(1-S)
0. Estimate the argument value x0, and supply a tolerance δ.
This step-limited version of successive substitution displays better 
convergence properties. This technique also requires that the objective 
function f(x) = 0 be reformuled as x = g(x). The algorithm is:
Wegstein
Iteration
Back to
Reality
The secant method is geared to solving f(x) = 0 while the successive 
substitution and Wegstein techniques are applied to problems that can be 
phrased as x = g(x). Let's recall why we are interested in solving such 
equations.
In the simple recycle system shown below, let us assume that the DF 
analysis indicates that a carry-forward variable in stream 2 is required to 
initiate the material balances on the reactor. Let this variable be the flow 
rate of species A. That is, the argument of our objective function is x = nA2.
We proceed by performing the balances on the reactor, the separator, and 
the overall subunit. The mixer balance is our objective function.
Secant
Formulation
The objective function, the mixer balance on A, can be written as an error 
function:
f x( ) nA1 nA5+ nA2−= 0=
This formulation is tailor-made for the secant algorithm.
Wegstein In this case we formulate the mixer balance as the function g(x):
g x( ) nA1 nA5+=
Summary Whenever a tear variable is required to proceed with the material 
balances in a process flowsheeet, it is necessary to implement some 
technique for converging to the correct value of the assumed variable. In 
essence, the problem has been reduced to the solution of a single equation 
in a single unknown.
The Newton, Secant, Successive Substitution, and Wegstein techniques 
were presented for the case of an analytic function of a single variable.
The only difficulty in applying these ideas to the solution of the material 
balances for a process flowsheet is that the evaluation of the objective 
function may involve the evaluation of many intermediate balances.
y0 0.5:=x0 1:=Initial estimate of the roots:
g2 x y,( )
2
x 7 y⋅+
1
2
+:=y g2 x y,( )=
g1 x y,( )
x2
5
y
10
+ 0.8+:=x g1 x y,( )=
As was the case for a single equation, the system must first be reformulated 
for updating. One such reformulation is:
Multidimensional
Successive
Substitution
The desired root is at (1.139,0.798).
f2 x y,( )
2
x 7 y⋅+
1
2
+ y−:=
f1 x y,( )
x2
5
y
10
+ 0.8+ x−:=
The system to be considered consists of two non-linear equations in two 
unknowns: x, y.
Sample System
All the methods discussed in the previous section can be extended to the 
case of finding roots to multiple, non-linear equations. However, these 
methods differ markedly in the relative amount of computation involved in 
the iterative algorithm. Each iteration may involve multiple evaluations of 
the objective functions. Due to the implicit nature of the objective 
functions involved in flowsheet analysis, and the subsequently heavy 
computational load involved simply in function evaluation, only the two 
simplest algorithms are discussed below. These are Successive 
Substitution and the Wegstein method.
If the degree-of-freedom analysis for a given flowsheet indicates that 
more than one tear variable is required to proceed with the material 
balances then one is faced with the task of converging simultaneously on 
two or more arguments. 
5.6.2 Multiple Non-Linear Equations
Sxk 1+
xk 2+
Syk 1+
yk 2+














g1 xk 1+ yk,( ) g1 xk yk,( )−
xk 1+ xk−
1 θ Sxk( )−( ) xk 1+⋅ θ Sxk( ) g1 xk 1+ yk 1+,( )⋅+
g2 xk 1+ yk 1+,( ) g2 xk 1+ yk,( )−
yk 1+ yk−
1 θ Syk( )−( ) yk 1+⋅ θ Syk( ) g2 xk 1+ yk 1+,( )⋅+




















:=
Iterate:
θ s( )
1
1 s−
:=Auxiliary function:
Sx0
x1
Sy0
y1














0
g1 x0 y0,( )
0
g2 x0 y0,( )











:=
y0 0.5:=x0 1:=Initialize:
This algorithm is applied to the same system used to illustrate the 
multidimensional successive substitution method.
0. Estimate the array of argument values x0
1. Calculate x1 = g(x0), the slopes Si = ∆gi/∆xi, and θi = 1/(1-Si)
2. Update with x2 = (1 - θ) x1 + θ g(x1)
The algorithm is similar to that used in the one-dimensional case: Multidimensional
Wegstein Iteration
y
0.5
0.944
0.761
0.81
0.794
0.799
0.797




















=x
1
1.05
1.115
1.125
1.134
1.137
1.138




















=
The sequence converges to
the desired multidimensional root.
xk 1+
yk 1+






g1 xk yk,( )
g2 xk yk,( )






:=k 0 5..:=Generate the sequence:
The convergence is more 
rapid in this case.
x
1
1.05
1.115
1.132
1.139
1.139
1.139
1.139
























= y
0.5
0.944
0.761
0.796
0.798
0.798
0.798
0.798
























=
Relation to
Flowsheet
Analysis
Let us recall once more the context in which multiple non-linear equations 
arise in process flowsheet analysis. It may occur that a degree-of-freedom 
analysis indicates that several tear variables may be required to initiate 
balances. For example, assume that the flow rates of all species in stream 
two below must be estimated in order to proceed with the balances on the 
reactor.
The unknowns then are the species flow rates in stream two ni2. One 
possible scenario is that the reactor, separator, and overall subunit balances 
can then be completed, leaving the mixer balances as an objective function. 
 The formulation for S species is: 
ni2 ni 1, ni 5,+= i 1 S..:=
This form is precisely that needed to implement successive substitution or 
Wegstein iteration x = g(x).
Summary The extension of the successive substitution and Wegstein methods to 
multiple non-linear equations involves no new principles. The 
convergence of systems of more than two equations is much slower than 
for the case of a single equation.