Section 2: FR&B, Chapter 4

INTRODUCTION TO MATERIAL BALANCES

Most problems you will have to solve in this course will look something like this:
A liquid mixture containing 45.0 wt% benzene (B) and 55.0% toluene (T) is fed to a distillation
column operating at steady state. Product streams emerge from the top and bottom of the column.
The top product stream contains 95.0 wt% B, and the bottom product stream contains 8% of the
benzene fed to the column. The volumetric flow rate of the feed stream is 2000 liters/h and the
specific gravity of the feed mixture is 0.872. Determine the mass flow rate of the overhead
product stream and the mass flow rate and composition (mass fractions) of the bottom product
stream.

You can take two approaches to solving such problems:

1. Start writing and solving equations until all of the required quantities have been determined.

2. Draw and label a flow chart, do a degree of freedom analysis, lay out a solution strategy, and then
write equations and solve for the required quantities.

Students almost always initially lean toward the first approach, which looks more efficient to them. That
choice has several likely consequences that you should be aware of.

Feature of Course Writing and solving equations

Problem-solving time early in Slightly faster
Chapter 4
Problem-solving time, most of Slower, then much slower, then
Ch. 4 and Ch. 5–9 may never get the solution
Course grade C-F (as opposed to A-C for the
other approach)

Suggestion: Start using the second approach immediately, even on the simple problems, so that you’ll be
used to it when the problems start getting harder (which they will very soon). This handout guides you
through the basics of that approach. The labeled flow chart of the process described in the first paragraph
might appear as follows: m 2 (kg/h)
0.95 kg B/kg
0.05 kg T/kg
m 1 (kg/h) Distillation
0.45 kg B/kg Column
0.55 kg T/kg
2000 L/h (SG = 0.872)
m B 3 (kg B/h) (8% of B in feed)
m T 3 (kg T/h)

Figure 1. Flow Chart of a Distillation Process

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Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.
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 Section 2: FR&B, Chapter 4

We will refer back to Figure 1 throughout this handout.

Nomenclature:
 Batch process: Add contents, then remove them later

 Continuous process: Continuously add then remove contents

 Semi-batch process: Neither batch nor continuous; e.g. allow contents of gas tank to escape;
slowly blend liquids in a tank from which nothing is withdrawn

 Steady state process: Process variables (flowrate, T, P) are constant

 Transient process: Process variables change with time

Balance equation
We can write a balance on any quantity entering or leaving a system — mass, moles, energy, people:

I+G–O–C=A (4.2-1)

Input + Generation – Output – Consumption = Accumulation

What What is What What is Change in amount

comes produced leaves consumed contained in the system
in across within the across within the
system system system system
boundary (reaction) boundary (reaction)

We sometimes refer to this as one of the “tattoo equations” – i.e., the one you should tattoo on your arm,
or in your brain! We’ll be using it for the rest of this course (and you’ll be using it for the rest of your
career).

Simplifications

 Continuous process at steady state: A = 0

 Nonreactive system, or balance on nonreactive species: G = 0, C = 0
 Batch process: I = 0, O = 0, A = final value – initial value

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 Section 2: FR&B, Chapter 4

Example:
Referring back to Figure 1, write balance equations on benzene, toluene, and total mass.
Solution: The general balance equation is: I + G – O – C = A

In all balances on the process shown in Figure 1, we may drop the terms ___, ___, and ___.

Why? _______________________________________________________________________

The balance equation therefore reduces to __________________________________________

Balance on benzene (B): _______________________________________________________

Balance on toluene (T): ________________________________________________________

Balance on total mass: _________________________________________________________

Drawing and labeling flow charts

The first step in approaching a typical problem in this course is to draw and completely label the flow
chart. What does it mean to be completely labeled?

A stream on a flow chart is completely labeled when you can write an expression for the
amount (batch) or flow rate (continuous) of each species in the stream in terms of
numbers and variables written on the flow chart. A flow chart is completely labeled if
every stream on it is completely labeled.

For example, a continuous stream of natural gas containing 85 wt% methane and the balance ethane might
be labeled as:
0.85 kg CH4/kg
0.15 kg C2H6/kg

Is this stream completely labeled? (Hint: No!) Why not? ______________________________________

 (kg/s) denote the unknown mass flow rate of the stream. The labeling would then be
Suppose we let m
m (kg/s)
0.85 kg CH4/kg
0.15 kg C2H6/kg

The stream is now completely labeled, since we can express the mass flow rates of methane and ethane in
terms of what is written on the chart:

Methane flow rate: m CH 4 (kg CH 4 /s) = _____________________

Ethane flow rate: m C2 H 6 (kg C 2 H 6 /s) = _____________________

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 Section 2: FR&B, Chapter 4

Rules for labeling:

 You may label either a total flow rate (or amount) and component mass or mole fractions, or
individual component flow rates (amounts).
m(kg)
x1 (kg NaCl/kg) OR m1 (kg NaCl)
x 2 (kg KCl/kg) m2 (kg KCl)
(1  x1  x 2 )(kg H 2 O/kg) m3 (kg H 2 O)

Once you have one, you can always calculate the other:

In terms of m, x1, and x2, m1 = _____________________________

In terms of m1, m2, and m3, m = _______________ and x1 = __________________________

 Label in the way that builds in given information about the stream. (For instance, if you know
or can easily calculate either a total flow rate or values of component mass or mole fractions,
label the total and fractions; otherwise label individual component flow rates or amounts
(likely to get easier algebra that way).

Suppose you know that a gas stream containing sulfur dioxide and air (21 mole% O2, 79%
N2) flows at a rate of 125 mol/s. The stream might be labeled

125 mol/s Exercise: Prove that this stream is

x (mol SO2 / mol) completely labeled.
(1-x) (mol air/mol) Solution:
0.21 mol O2/mol air nSO (mol SO 2 / s)  ___________________
2

0.79 mol N2/mol air

__________________________________
__________________________________

 If volumes or volumetric flow rates of streams are either given or required, include labels
both for them and for mass or molar quantities.

 For practice, do the Test Yourself on p. 92.

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 Section 2: FR&B, Chapter 4

Exercise: Enriching Air with O2.

A stream of air (21.0 mole% O2, 79.0% N2), a stream of pure oxygen, and a stream of liquid water flowing
3
at a rate of 20.0 cm /min, are fed to a steady-state evaporation chamber in which all of the liquid
evaporates. The flow rate of the pure oxygen is 20% of the flow rate of the air. The emerging gas stream
contains 1.5 mole% water vapor.
20% of air flow rate

O2 (g)
Evaporator
Air
0.015 mol H2O(v)/mol
0.21 mol O2/mol
0.79 mol N2/mol
20.0 cm3 H2O(l)/min

 Completely label the flow chart (including the molar flow rate of the liquid water stream),
following the rules given above.

Evaporator
0.015 mol H2O(v)/mol

0.21 mol O2/mol

0.79 mol N2/mol

20.0 cm3 H2O(l)/min

 Convince yourself that your labeling is complete. (For example, what is the flow rate of oxygen
in the outlet stream in terms of labeled values and variable names?)

Basis of Calculation and Flowchart Scaling. At least one stream quantity or flow rate (mass, moles,
volume) should be specified before any calculations are done. If the problem statement doesn’t do it, you
should choose a convenient basis (an amount of a stream with known composition.) Having degrees of
freedom = 1 is often a reminder that you need to choose a basis.

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 Section 2: FR&B, Chapter 4

Example: A stream containing 30 wt% ethanol (E) and 70% water (W) is blended with a stream
containing 60% E and 40% W. The product contains 35% E. What is the mass ratio of Stream 1 to Stream
2 (= m1/m2 in the flow chart)?
m1 (g)
0.30 g E/g m3 (g)
0.70 g W/g Mixer 0.35 g E/g
0.65 g W/g
m2 (g)
0.60 g E/g
0.40 g W/g

Solution: The problem statement does not specify a stream amount or flow rate. There is enough
information to calculate the required stream ratio without specifying one, but arbitrarily choosing a basis
of calculation (an amount or flow rate of one of the streams) makes the required math easier. We might
proceed as follows:

Basis of calculation: 100 g of Stream 1 (m1 = 100 g):

100 g
0.30 g E/g
m3 (g)
0.70 g W/g
Mixer 0.35 g E/g
m2 (g) 0.65 g W/g
0.60 g E/g
0.40 g W/g

We could then write balance equations on ethanol and water, solve for m1 and m2, and determine the
requested ratio as 100/m2. The same result would be obtained if we chose any other basis of calculation:
the values of m1 and m2 might be different, but the ratio would be the same.

 Read Section 4.3b to see how to scale a flow chart from one basis to another, and work through the
Test Yourself on p. 104.

Degree-of-Freedom Analysis and Solution Strategy

Once a basis of calculation has been chosen and the flow chart is drawn and fully labeled, carry out a
degree-of-freedom analysis:

______ unknowns (the number of unknown variables on the flow chart)

– ______ balances (minus the number of independent balance equations)
– ______ other equations (minus the number of other equations relating the unknowns)
= DF degrees of freedom

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 DF = 0  number of equations equals number of unknowns  can solve for all unknowns
 DF > 0 (more unknowns than equations)  either you’ve overlooked some equations, problem is
underspecified, or you can solve for all requested quantities without knowing all the unknowns.
 DF < 0 (more equations than unknowns)  either flowchart is not fully labeled or problem is
overspecified.

Here are some simple examples using generic equations and variables to illustrate:

x + y = 10 ____ unknowns
- ____ independent equations
_______________________
____ DF

x + y = 10 ____ unknowns
x–y=0 - ____ independent equations
_______________________
____ DF

x + y = 10 ____ unknowns
2x + 2y = 20 - ____independent equations
_______________________
____ DF

x + y = 10 ____ unknowns
x–y=0 - ____independent equations
xy = 3 _______________________
____ DF

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 Section 2: FR&B, Chapter 4

In this course, when you have zero degrees of freedom, the problem can be solved.

 For a non-reactive process, the number of independent material balance equations equals the number
of independent species involved in the process.
 Independent molecular species: We discussed this in the context of how many balances we can do
around a process. If two molecular species are in the same ratio everywhere in the process and the
ratio is incorporated in the flowchart labeling, the species are not independent, and neither are
balances on them.
100 mol CH4/s

Gas blender n2 (mol/s)

n1 (mol air/s) x1 (mol CH 4 / mol)
0.21 mol O2/mol (1  x1 )(mol air / mol)
0.79 mol N2/mol 0.21 mol O2 / mol air 
0.79 mol N / mol air 
 2 

Looks like 3 unknowns – 3 equations (balances on CH4, O2, N2) = 0 DF, but it won’t work. Write
balances on O2 & N2 and see what happens.

O2 balance:

N2 balance:

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Notes with Gaps to accompany Felder, Rousseau, & Bullard, Elementary Principles of Chemical Processes.
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 Section 2: FR&B, Chapter 4

Example. Reconsider the distillation process from Figure 1, shown again below. The basis of calculation
is 2000 L/h of the feed stream (exact, not 1 significant figure).

m 2 (kg/h)
0.95 kg B/kg
m 1 (kg/h)
Distillation 0.05 kg T/kg
0.45 kg B/kg
Column
0.55 kg T/kg
2000 L/h (SG = 0.872)
m B 3 (kg B/h) (8% of B in feed)
m T 3 (kg T/h)

Degree-of-freedom analysis (count only variables labeled on the chart):

__ unknowns
– __ balances (_______, ________)
– 1 ___________________
– 1 ___________________
= 0 DOF All unknowns can be calculated
Since the problem is solvable, we may proceed to write out the equations. All balances have the form
(_________).

2000 L _____________ kg
Feed density: m 1   m 1 (kg/h)  (2000)(0.872) (1)
h L

8%: m B 3  _____________________ (2)

Mass balance: m 1  m 2  m B 3  m T 3 (3)

Benzene balance: ________________________________________________ (4)

Either do the algebra and arithmetic manually (easy in this problem, not so easy in problems to come) or
use APEx or Excel’s Solver to calculate the unknown quantities.

Equations:

m1 = 2000*0.872 kg kg kg B kg T
 m 1  1744 , m 2  760  B 3 =62.8
, m  T 3 =921
, m
mB3 = 0.08*m1*0.45 h h h h
m B 3 kg B m kg overhead
m1 = m2 + mB3 + mT3  xB  = 0.0638 , ROF  2  0.436
m1*0.45 = m2*0.95 + mB3
m B 3  m T 3 kg m 1 kg feed

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Scale the flowchart: Now suppose we are asked to calculate the volumetric feed rate needed to produce
2500 lbm/day of overhead product. The scale factor (see Example 4.3-2) is the desired value of the
specified quantity (overhead product rate = 2500 lbm/day) divided by the value of the same quantity
corresponding to the original basis of calculation (= 760 kg/h).

2500 lbm /day lb /day

Scale Factor (SF) =  3.29 m
760 kg/h kg/h
 kg   lb /day  3 lb m feed
 (m 1 )new  (m 1 )old  SF  1744   3.29 m   5.74  10
 h  kg/h  day

Multiple-unit balances (Section 4.4)

We’ve looked at processes represented by single boxes—mixer, distillation column. Think about plants
you’ve seen from the highway—all those towers & stacks & pipes running all over the place.

How do you analyze something like that?

 Overall system: large box around system enclosing all units, recycle & bypass streams. Subsystems:
individual units or combinations of units, stream mixing points & stream splitting points. DOF
analyses as with single units.

 Go through Ex. 4.4-1 (next page)

 Work through Prob. 4.40 and check your answers in the back of the book.

 Go through Ex. 4.4-2 on your own or in a group. If you just read it you think you get it but you don’t.
Ask questions about each step, and don’t go on until you can answer them.

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Problem 4.4-1: A Two-Unit Process.

A labeled flowchart of a continuous steady-state two-unit process is shown below. Each stream contains
two components, A and B, in different proportions. Three streams whose flow rates and/or compositions
are not known are labeled 1, 2, and 3.

40 kg/h 30 kg/h
0.9 kg A/kg 0.6 kg A/kg
0.1 kg B/kg 0.4 kg B/kg

100 kg/h 1 2 3
0.5 kg A/kg
0.5 kg B/kg
30 kg/h
0.3 kg A/kg
0.7 kg B/kg

Calculate the unknown flow rates and compositions of streams 1, 2 and 3.

Solution. First, fully label the chart and mark systems about which balances might be written.

40 kg/h 30 kg/h
0.9 kg A/kg 0.6 kg A/kg
0.1 kg B/kg 0.4 kg B/kg

100 kg/h m 1 (kg/h) m 2 (kg/h) m 3 (kg/h)

0.5 kg A/kg x1 (kg A/kg) x2 (kg A/kg) x3 (kg A/kg)
0.5 kg B/kg (1 - x1)(kg B/kg) (1 – x2)(kg B/kg) (1 – x3)(kg B/kg)

30 kg/h
0.3 kg A/kg
0.7 kg B/kg

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 Section 2: FR&B, Chapter 4

Degree-of-freedom analysis:

Overall system: Mixing point:

_______ Unknowns: _______ Unknowns:

- _______ Independent balances: - _______ Independent balances:
_____________________________ _______________________________
_______ Degrees of freedom _______ Degrees of freedom

Unit 1: Mixing point (after solving unit 1):

_______ Unknowns: _______ Unknowns:

- _______ Independent balances: - _______ Independent balances:
_____________________________ _______________________________
_______ Degrees of freedom _______ Degrees of freedom

Our strategy for solving the problem:

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 Section 2: FR&B, Chapter 4

Overall mass balance:

Overall balance on A:

Mass Balance around Unit 1:

Balance on A around Unit 1:

Mass balance around Mixing Point:

Balance of A around Stream Mixing Point:

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 Section 2: FR&B, Chapter 4

Recycle & Bypass (Section 4.5).

Sell
100 kg B/min
10 kg A/min
200 kg A/min AB 100 kg A/min
100 kg B/min Dump
90 kg A/min

What’s wrong with this picture? How would you redraw it to be more cost efficient and environmentally
conscious?

Notes:
 Real reaction and real separation are always less than perfect. There is always some
unreacted raw material, and some unwanted material in the product.
 Recycling is not free (requires pumping, pipes, etc.) but is only a one-time cost and usually
pays for itself quickly
 Reasons to recycle:
o Recovering a catalyst
o Diluting a process stream
o Controlling a process variable
o Circulating a working fluid
o Increasing the overall percent conversion of a reaction

Sometimes a stream might bypass a process unit, where a fraction of the stream is diverted around the
unit and combined with the output stream. This allows us to vary the composition and properties of the
product stream. A stream that bypasses a unit introduces a splitting point in which a stream is split into
two streams.

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 Section 2: FR&B, Chapter 4

Splitting point: A stream is split into two streams.

100 mol/s 60 mol/s

0.70 mol A/mol x1 (mol A/mol)
0.30 mol B/mol (1  x1 )(mol B/mol)
m 2 (mol/s)
x2 (mol A/mol)
(1  x2 )(mol B/mol)

Do a degree-of-freedom analysis:

Did you get 1 DOF? Think about it…if all we’re doing is splitting a stream, the compositions of the
outlet streams must be the same as that of the feed stream.

OK, so here’s the revised flowchart.

100 mol/s 60 mol/s
0.70 mol A/mol 0.70 mol A/mol
0.30 mol B/mol 0.30 mol B/mol
m 2 (mol/s)
0.70 mol A/mol
0.30 mol B/mol

Before we do the formal D.F. analysis, how many degrees-of-freedom do we have?

Q: We know there should be 0. What went wrong?

A: We are only allowed 1 balance.

Q: Why?
A: Because # balances = # independent species, which A and B are not in this case. Try writing the
possible balances:

Overall Mass Balance:

A Balance:

B Balance:
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 Go through Example 4.5-1.

 Note reasons for recycling in Section 4.5.
 Go through Ex. 4.5-2. If you can do that without looking at solution, you can handle recycle
problems.
 Further practice: Problems 4.43 and 4.48.

Example 4.5-2: An Evaporative Crystallization Process.

The flowchart of a steady-state process to recover crystalline potassium chromate (K2CrO4) from an
aqueous solution of this salt is shown below.

H2O
Crystals
4500 kg/h Evaporator Crystallizer
0.333 kg K/kg 0.494 kg K/kg and filter Solution
0.667 kg W/kg 0.506 kg W/kg
0.364 kg K/kg
0.636 kg W/kg
Filtrate
0.364 kg K/kg
0.636 kg W/kg

Forty-five hundred kilograms per hour of a solution that is one-third K2CrO4 by mass is joined by a
recycle stream containing 36.4% K2CrO4, and the combined stream is fed into an evaporator. The
concentrated stream leaving the evaporator contains 49.4% K2CrO4; this stream is fed into a crystallizer in
which it is cooled (causing crystals of K2CrO4 to come out of solution) and then filtered. The filter cake
consists of K2CrO4 crystals and a solution that contains 36.4% K2CrO4 by mass; the crystals account for
95% of the total mass of the filter cake. The solution that passes through the filter, also 36.4% K2CrO4, is
the recycle stream.

1. Calculate the rate of evaporation, the rate of production of crystalline K2CrO4, the feed rates that the
evaporator and the crystallizer must be designed to handle, and the recycle ratio (mass of
recycle)/(mass of fresh feed)
2. Suppose that the filtrate were discarded instead of being recycled. Calculate the production rate of
crystals. What are the benefits and costs of the recycling?

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 Section 2: FR&B, Chapter 4

Solution.

1. First, fully label the chart and mark systems about which balances might be written (next page):

m 2 (kg W/h)
m 4 (kg K/h)
4500 kg/h m 1 (kg/h) Evaporator m 3 (kg/h) Crystallizer
0.333 kg K/kg x1 (kg K/kg) 0.494 kg K/kg and filter
m 5 (kg/h)
0.667 kg W/kg (1 - x1) (kg W/kg) 0.506 kg W/kg
0.364 kg K/kg
0.636 kg W/kg
m 6 (kg/h)
 4 = 0.95 m 5 ]
[ 0.05 m
0.364 kg K/kg
0.636 kg W/kg

Degree-of-freedom analysis:
Overall system: Recycle-fresh feed mixing point:

_______ Unknowns: _______ Unknowns:

- ____ Independent balances: - ____ Independent balances:

- ____ Additional relations: - ____ Additional relations:

_______________________________ _______________________________
_______ Degrees of freedom _______ Degrees of freedom

Evaporator: Crystallizer/filter:

_______ Unknowns: _______ Unknowns:

- ____ Independent balances: - _____Independent balances:

- ____ Additional relations: - _____Additional relations:

_______________________________ _______________________________
_______ Degrees of freedom _______ Degrees of freedom

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Here’s our strategy for solving the problem:

Overall K2CrO4 balance:

Overall total mass balance:

95% specification:

Mass balance around crystallizer:

Water balance around crystallizer:

Mass balance around recycle-fresh feed mixing point:

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 Section 2: FR&B, Chapter 4

4.6. Chemical Reaction Stoichiometry

 Stoichiometry (4.6a) — the proportions in which chemicals combine in reaction. Stoichiometric

equation:
1 mol O2 consumed 2 mol SO3 generated
2SO2 + O2  2SO3  , , etc.
2 mol SO2 consumed 2 mol SO2 consumed

The stoichiometric coefficients define conversion factors from moles of one species consumed or
formed in the reaction to moles of another species consumed or formed. For instant, if 50 mol/s of
SO3 are produced in the reaction

50 mol SO3 formed 1 mol O2 consumed mol O2 consumed

oxygen consumed   25
s 2 mol SO3 formed s

Work through Test Yourself on p. 117.

 Limiting and excess reactants (4.6b)

Feed reactants in stoichiometric proportion (e.g., 2 mol SO2/1 mol O2)  all reactants run out at the
same time. What if they’re not in stoichiometric proportion?

2SO2 + O2  2SO3 (Sulfuric acid production, acid rain)

100 mol SO2/s n1 (mol SO3/s)

90 mol O2/s n2 (mol SO2/s)
n3 (mol O2/s)

Limiting reactant: reactant that runs out first, fed in less than stoichiometric proportion to all other
reactants. In example, SO2 limits. (Stoichiometric proportion = 2 SO2 : 1 O2, actual feed proportion =
10 SO2 : 9 O2)

Excess reactants: all other reactants but limiting one. In example, O2 is in excess.

Theoretical (stoichiometric) requirement of an excess reactant: How much would be required to react
completely with the limiting reactant.

100 mol SO2 fed 1 mol O2

Theoretical oxygen =  50 mol O2 (theoretical)
s 2 mol SO2

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Percent excess of a reactant: (Amount fed – theoretical amount)/(theoretical amount) x 100%

90 mol O2  50 mol O2
%XS O2 =  100%  80% excess O2 (=80%XS air if O2 comes from air)
50 mol O2

Suppose you are told that oxygen is now fed in 60% excess (instead of at a rate of 90 mol/s). Then

(nO2 )fed  (nO2 )theoretical  (nO2 )excess  (nO2 )theoretical  0.60(nO2 )theoretical
 (nO2 )theoretical (1  0.60)  1.60  (50 mol O 2 /s)  80 mol O 2 /s

 % XS 
or in general, if A is fed in excess, (n A )fed  (n A )theoretical 1  
 100 

Note: The theoretical and excess quantities depend only on the feed amounts and the stoichiometric
reaction, not on what actually happens in the reactor.
moles reacted moles in  moles out
 Fractional conversion of a reactant: 
mole fed moles in

2SO2 + O2  2SO3

100 mol SO2/s n1 80(mol SO3/s)

90 mol O2/s n2 20 mol SO2/s
n3 50 mol O2/s

100 mol in/s  20 mol out/s

SO2: fSO2   0.80 mol SO 2 react/mol fed
100 mol in/s
(Percentage conversion of SO2 =100 fSO2 = 80%)

90 mol in/s  50 mol out/s

O2: fO2   0.44 mol O 2 react/mol fed
90 mol in/s
(Percentage conversion of O2 =100 f O2 = 44%)

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 Section 2: FR&B, Chapter 4

Exercise:

N2 + 3H2  2NH3

200 mol N2 n1 (mol N2)

300 mol H2 n2 (mol H2)
n3 (mol NH3)

(a) The limiting reactant is ____________________________

(b) The percentage excess of the other reactant is ___________________________

(c) The theoretical amount of NH3 produced is

____________________________________________________

Now suppose the percentage conversion of nitrogen is 25%

(d) n1 = _____________________ , n2 = _____________________ , n3 = _________________

(e) The percentage conversion of hydrogen is ________________

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 Section 2: FR&B, Chapter 4

Balances on Reactive Processes (4.7). There are three approaches:

1. Molecular species balances. Look again at the flowchart:

2SO2 + O2  2SO3

100 mol SO2/s n1 (mol SO3/s)

90 mol O2/s n2 (mol SO2/s)
n3 (mol O2/s)

We have three unknowns and three species ( 3 balances), so it looks like we should have zero
degrees of freedom. We don’t, however—no matter how hard you try, you can’t calculate any of
those unknown variables without getting more information. Try using 0 = I – O + G - C

However, let’s say we are now told that n2 = 20 mol SO2/s.

2SO2 + O2  2SO3

100 mol SO2/s n1 (mol SO3/s)

90 mol O2/s 20 mol SO2/s
n3 (mol O2/s)

If molecular species balances are used (there are alternatives), then

DOF = #unknowns (2) – #independent molecular balances (3) – #other equations (0)

+ #independent reactions (1) = 0

 Independent reactions (Section 4.7b). Chemical reactions are not independent if we can get one in
terms of the other by adding, subtracting, and multiplying them.
AB (1)
2A  2B (2) (1) and (2) are not independent [(2) = 2  (1)]

A  2B (1)
BC (2) (1) and (2) are independent, but

A  2C (3) (1), (2), and (3) are not [(3) = (1) + 2  (2)]

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 Section 2: FR&B, Chapter 4

mol SO 2 consumed
SO 2 balance : I  G  O  C  A  C  I  O  80
s
SO3 balance : I  G  O  C  A  O  G
80 mol SO 2 consumed ___ mol SO3 generated mol SO3
 n1   _____
s ___ mol SO 2 consumed s

O 2 balance : I  G  O  C  A  O  I  C
mol O 2 80 mol SO 2 consumed ___ mol O 2 consumed
 n3  90   50 mol O 2 / s
s s ___ mol SO 2 consumed

The original system must therefore have had ________ degree(s) of freedom.

2. Extent of reaction
For 2SO2 + O2  2SO3, if 10 mol SO3/s are formed, 10 mol SO2/s and 5 mol O2/s must react.
Moles formed & reacted are always proportional to stoichiometric coefficients.

mol SO 3 generated mol SO 2 consumed mol O 2 consumed

  =  (pronounced zi)
2 mol SO 3 2 mol SO 2 1 mol O 2

where  is the extent of reaction (same for all species, must be positive) and can be expressed as

(ni  ni 0 )(moles i )

 i (moles i )
 starts at zero when the reaction begins, increases as the reaction proceeds, and has the same value
for all species.

It follows that

(nSO2 )final  (nSO2 )initial  2

(nO )final  (nO )initial  
2 2

(nSO3 )final  ( nSO3 )initial  2

For any single reaction,

ni  ni 0   i (Batch)
ni  ni 0  i (Continuous, steady-state) (4.6-3)

where i (or i ) is the stoichiometric coefficient of species i (+ for products, – for reactants) and has
units of moles (or moles/time) of species i. One of these equations may be written for every
independent molecular species involved in the process. For nonreactive species ( = 0), the equation
reduces to ni  ni 0 ).

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 Section 2: FR&B, Chapter 4

Q: What is ?
A. It’s a dummy variable that reflects how far each reaction proceeds. It reflects the stoichiometry
of the reaction and helps us keep up with generation and consumption.
2SO2 + O2  2SO3

100 mol SO2/s n1 (mol SO3/s)

90 mol O2/s 20 mol SO2/s
n3 (mol O2/s)

DOF = #unknowns (2) – #extent of reaction equations (3) – #other equations (0)
+ #independent reactions (1) = 0

Apply extent of reaction equation ( ni  ni 0  i ), starting with the species we know most about:

SO2: 20 mol/s = 100 mol/s – 2    = 40

O2: n3 = ___________________________________________= 50 mol O2/s

SO3: n1 = __________________________________________ = 80 mol SO3/s

3. Atomic species balances. Same process once more.

2SO2 + O2  2SO3

100 mol SO2/s n1 (mol SO3/s)

90 mol O2/s 20 mol SO2/s
n3 (mol O2/s)

Instead of writing balances on the molecular species involved in the process (SO2, O2, SO3), let’s do it
on the atomic species (S, O). Since atomic species are neither generated nor consumed (except in
nuclear reactions, which we don’t consider), all balances reduce to I = O.

With atomic species balances,

DOF = #unknowns (2) – #independent atomic balances (2) – #other equations (0)

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 Section 2: FR&B, Chapter 4

100 mol SO 2 ___ mol S 20 mol SO 2 ___ mol S n1 (mol SO3 ) ___ mol S
S balance :  
s 1 mol SO 2 s 1 mol SO2 s 1 mol SO3
 n1  80 mol SO 3 / s

100 mol SO 2 ___ mol O 90 mol O 2 ___ mol O n1 (mol SO3 ) ___ mol O
O balance :  =
s 1 mol SO 2 s 1 mol O 2 s 1 mol SO3

+ 

 n3  50 mol O 2 / s

 Independent atomic species: If two atomic species are in the same ratio everywhere in the process,
they are not independent. Look at the Test Yourself on p. 143.
Q: How many independent atomic balances are there?

A: Two. Write balances on C and H and see what happens.

For more practice, do the Test Yourself on p. 134.

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 Section 2: FR&B, Chapter 4

Summary: Balances on reactive systems

 Three ways to write balances on reactive processes:

Form #DF
Molecular (# unknowns on chart)
species + (# independent reactions)
a
balances – (# independent molecular species balances )
– (# other equations)
Atomic species (# unknowns on chart)
b
balances -- (# independent atomic species balances )
– (# other equations)
Extents of (# unknowns on chart)
reaction + [# independent reactions: (1  per reaction)]
a
– (# independent species balances )
– (# other equations)

a
If two molecular species are in the same ratio everywhere in the process
and the ratio is incorporated in the flowchart labeling, the species are not
independent
b
If two atomic species are in the same ratio everywhere in the process,
they are not independent

 Which approach to use?

 DOF: If you’re confident about independence of atomic species, use the atomic species balance
approach, & if you’re more confident about independence of reactions, use the extent of reaction
approach. Note that all DOF approaches should give the same answer, so if you’re unsure, use
more than one to confirm.

 Balances:
 Don’t use molecular species balances for reactive species (especially for multiple reactions –
it gets hairy fast)
 Hand calculations—atomic balances
 Solver or chemical equilibrium—extents of reaction. (Be sure reactions are independent)

 Note: You can use one method for the DOF analysis & another for the calculations.

 Work through Example 4.7-1. If you fully understand the solution, you can solve any single-unit
reactive balance problem you are likely to encounter on homework and tests.

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 Section 2: FR&B, Chapter 4

Chemical Equilibrium Calculations (Section 4.6c).

A + 2B  C

If you are given an initial composition of a mixture of A, B, and C and an expression for the equilibrium
constant K(T), and you are told that the reaction proceeds to equilibrium at a final temperature T, you can
calculate the final composition using extents of reaction. See Example 4.6-2 for an illustration.

Multiple reactions (4.6d)

C2H6  C2H4 + H2 (Ethylene is the desired product, hydrogen is a byproduct)

C2H6 + H2  2CH4 (Methane is an undesired byproduct)

C2H4 + C2H6  C3H6 + CH4 (Propylene is an undesired byproduct)

 The second and third reactions both consume ethylene and so are undesired. Sometimes byproducts
can be sold, and at other times they are worthless and possibly hazardous and have to be disposed of
(another cost).

 Yield = mole of desired product / [moles that would have been formed if there were no side reactions
and the reaction went to completion (the limited reactant was completely consumed)]

 Selectivity = moles of desired product / moles of undesired product

(note, selectivity always refers to selectivity of desired A with respect to undesired B)

 If yield and selectivity are high, then we have successfully suppressed the undesired reactions.

 Note: in the ethylene example, H2 is not an undesired byproduct. An undesired byproduct is the
result of a competing reaction that results in less of the desired product.

A  2B + C
BD
B: Desired product, D: Unwanted byproduct

20 mol A
100 mol A 120 mol B
80 mol C
40 mol D

Maximum possible B produced = _______ mol (All A fed reacts, no side reaction)

Yield (Eq. 4.6-4): YB = _____________________

Selectivity B/D (Eq. 4.6-5): SBD = _____________________________________

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 Section 2: FR&B, Chapter 4

 A process engineer might take two different approaches to this reaction system:

 Maximize yield: Get most B you can, even if it means producing more D.
 Maximize selectivity: Hold down production of D, even if it means producing less B.

Q: Why would you want to suppress production of a byproduct if it means getting less of the product
you’re selling? (Think of several possible reasons.)

A:

Extents of reaction for systems with multiple reactions

Batch: ni  ni 0   
j
ij j [ j , i = species, j = independent reactions]
(4.6-7)
Continous, Steady State: ni  ni 0  ij j
j
[ j , i = species, j = independent reactions]

Recall or  is positive for products, negative for reactants, and ZERO for inerts (N2, etc.) that go
through the process without reacting.

Example:
A  2B + C [Rxn 1]  A1 = –1 mol/s,  B1 = +2 mol/s,  C1 = +1 mol/s
BD [Rxn 2]  B2 = –1 mol/s,  D2 = +1 mol/s (All others = 0)

100 mol A/s n A (mol A/s)

20 mol B/s n B (mol B/s)
10 mol I/s n C (mol C/s)
n D (mol D/s)
nI (mol I/s)
n A  100  1 , nB  ____________________, nC  ____________________

nD  ____________________, nI  ____________________

DOF: 5 unknowns ( n A , n B , nC , n D , n I ) – 5 equations + 2 reactions = 2 DOF.

Specify any 2 of the 7 variable values, fractional conversion of A, yield of B, or selectivity of
B/D, & calculate the others. (Write equations, solve with APEx or Excel’s Solver.)

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 Section 2: FR&B, Chapter 4

Product Separation and Recycle (Section 4.7f)

 Step through flowchart on p. 151

 Identify “fresh feed”, “recycle”, “feed to the reactor”

 What is the percentage conversion of A? Depends on the definition of “conversion”:

 Single pass conversion: based on what goes into and comes out of the reactor. In this case
it is 75%.
A fed to reactor  A leaving reactor
Single-pass conversion of A =
A fed to reactor
=

 Overall conversion: Based on what comes in and out of the overall process.

A fed to process  A leaving process

Overall conversion of A =
A fed to process
=

We must specify whether a given conversion is overall or single-pass.

 Do Test Yourself on p. 151. Work through Example 4.7-2.

 Note: only 10% of the propane entering the reactor is converted to propylene in a single pass.
99% of the unconsumed propane is recovered in the separation unit and recycled back, where it
gets another chance to react. Net result = 95% of propane entering the process is converted to
propylene and 5% leaves with final product.

 To achieve high conversion:

1. Design the reactor to achieve a high single-pass conversion
o Requires very large residence time in reactor, large reactor volume, expensive

2. Design the reactor to achieve a low single-pass conversion, follow with a separation unit to
recover and recycle unconsumed reactant.
– Decreased reactor cost
– Incur cost of separation process and cost of recycle line

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 Section 2: FR&B, Chapter 4

Purge

 Inert gases (e.g. nitrogen, argon) are used in processes because they undergo chemical reactions
under given conditions

 Using an inert gas in a system prevents undesirable chemical reactions from occurring (e.g. oxidation,
hydrolysis, combustion)

 Purging with nitrogen minimizes fire hazards of residual solvents or process fluids

 If an inert gas is introduced to the system and it doesn’t react, it has to exit the system somewhere –
otherwise, it builds up and would shut down the process

 Look at Fig. 4.7-2 — production of ethylene oxide from ethylene. N2 is an inert. How did it get in
the process?

 Purge stream: note it also contains ethylene and O2. Why are we throwing away product and
reactant?

 Note there is a cost to discard. But there is also a cost to separate and recycle the last little bit. Gas
separation is hard and expensive. Engineers must evaluate the options based on economics,
environmental regulations, etc. Note that purge stream take-off is a splitter (we can only write one
balance one equation in the DOF).

 Work through Example 4.7-3. If you understand Example 4.7-3, you know how to do recycle
problems, which are the hardest problems to solve. The key is flowcharting and systematic DOF
analysis.

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 Section 2: FR&B, Chapter 4

Combustion Processes (Section 4.8)

Combustion: Rapid reaction of a fuel with oxygen (usually but not always in air)

Complete combustion: All C in fuel forms CO2, all H forms H2O, all S forms SO2

Partial combustion: Some C forms CO

Air: Actual composition given on p. 161. For most of our purposes, assume 21.0 mole% O2, 79.0% N2 
3.76 mol N2/mol O2, 4.76 mol air/mol O2.

Percent excess oxygen (= percent excess air): Based on complete combustion of fuel, regardless of
whether all fuel actually reacts and whether some CO is formed.

Wet-basis and dry-basis product gas compositions: Mole fractions of components with water included
and not included, respectively.

Example
20 mol CH4/s
100 mol CH4/s 95 mol O2/s
940 mol N2/s
1190 mol air/s 70 mol CO2/s
0.21 mol O2/mol 10 mol CO/s
0.79 mol N2/mol 160 mol H2O/s

Q: Is the flow chart balanced?

Q: What reaction(s) are taking place?

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 Section 2: FR&B, Chapter 4

Q: What is the percent excess air fed to the reactor?

Q: What about the fact that not all the methane reacted and some CO was formed?

Q: What are the wet-basis and dry-basis compositions of the product gas?

 Do Test Yourself on p. 164

 Most material balance problems on combustion reactors are no different from those on any other
reactor. Go through Example 4.8-3 for an illustration.
 Sometimes composition of fuel may be unknown, but if you know the atomic constituents of the fuel
you can determine their ratio. Example 4.8-4 illustrates such a computation.
 Read Section 4.9, especially noting p. 171. It outlines the differences between this course and the real
world.

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 Section 2: FR&B, Chapter 4

Summary: General Strategies and Tips for Problem Solving

1. Break the problem down into small sub-problems when possible.

2. Even if the DOF analysis is not asked for, do it. That’s your roadmap and confirmation that the
problem is solvable. (5 minutes on the DOF analysis can save hours of trying to solve an
impossible problem.)
3. For multiple unit processes or system with split, mix, recycle, purge, etc., do the DOF for each
subprocess. Having 10 equations and 10 unknowns for the entire process is literally correct, but
if it doesn’t converge you don’t have a clue as to where to start looking for the error.
4. Use standard nomenclature: m for mass, n for moles, and x or y for composition. Avoid A-F, all
Xi, all Mi. Someone may have to decipher your solution later. You may have to decipher it later.
Save your creativity for solving the problem, not labeling it.
5. Clearly label your equations and the basis for the DOF analysis. This helps remind you and
anyone checking your work what you have done.
6. Reality check: negative flowrates, negative compositions are a signal. Very small or very large
numbers relative to others are also suspicious.
7. Always show complete units on the flow chart and in calculations.

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