CHEE 221: Chemical Processes and Systems
Module 1.
Material Balances: Single Process Units
without Reaction
(Felder & Rousseau Ch 4.1‐4.3)
General Material Balance Equation (“GMBE”)
Accumulation = In – Out + Generation – Consumption
system boundary
Input streams to
system
System over which mass
balance is made
output streams
from system
Accumulation
within the system
(mass buildup)
=
Input through
system boundary
+
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Generation within
the system
Output through
system boundary
‐
Consumption
within the system
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What is the System?
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Some Basic Process Unit Functions
 Splitter – divides a single input into two
or more outputs of the same composition
(no reaction)
splitter
 Mixer – combines two or more inputs
(usually of different compositions) into a
single output) (no reaction)
mixer
 Separator – separates a single input into
two or more outputs of different
composition (no reaction)
separator
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Basic Process Unit Functions…cont’d
 Reactor – carries out a chemical reaction
that converts atomic or molecular species
in the input to different atomic or
molecular species in the output
 Heat exchanger – transfers heat from one
input to a second input (no reaction)
 Pump – changes the pressure of an input
to that of the corresponding output (no
reaction)
reactor
heat
exchanger
pump
Actual process units can combine these different functions into a single piece of
hardware, and are given different names, e.g. a separator can be a distillation
column, a filter press, a centrifuge, etc.
F&R Encyclopedia of Chemical Engineering Equipment ( textbook website)
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Steam Boiler
Steam Boiler
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Heat Exchanger
(no reaction)
+
Reactor (reaction)
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Distillation—A Very Common Separation Process
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Material (Mass) Balances (“MBs”)‐No Reaction
A material balance is simply an accounting of material. For a given system in
which no reaction is occurring (you will not be told this, and will need to
know this from the type of unit that is under consideration; crystallizer,
evaporator, filter, furnace, etc.), a material balance can be written in terms of
the following conserved quantities:
1.
2.
3.
Total mass (or moles)
Mass (or moles) of a chemical compound
Mass (or moles) of an atomic species
To apply a material balance, you need to define the system and the quantities of
interest (e.g. mass of a component, total mass, moles of an atomic species).
What is your system, and what are you keeping track of?
System – a region of space defined by a real or imaginary closed envelope
(envelope = system boundary)
– may be a single process unit, collection of process units or an
entire process
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Process Classification
Before writing a material balance (MB) you must first identify the type of
process in question.
 Batch – no material (mass) is transferred into or out of the system over the
time period of interest (e.g., heat a vessel of water)
 Continuous – material (mass) is transferred into and out of the system
continuously (e.g., pump liquid into a distillation column and remove the
product streams from top and bottom of column)
 Semibatch – any process that is neither batch nor continuous (e.g., slowly
blend two liquids in a tank)
 Steady‐State – process variables (i.e., T, P, V, flow rates) do not change with
time
 Transient – process variables change with time
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F&R Ch 4.1
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F&R Example 4.1
Classify the following processes as batch, continuous, or
semibatch, and transient or steady‐state.
1.
2.
3.
4.
A balloon is filled with air at a steady rate of 2 g/min.
A bottle of milk is taken from the refrigerator and left on the kitchen table.
Water is boiled in an open flask.
Carbon monoxide and steam are fed into a tubular reactor at a steady rate
and react to form carbon dioxide and hydrogen. Products and unused
reactants are withdrawn at the other end. The reactor contains air when
the process is started up. The temperature of the reactor is also constant,
and the composition and flow rate of the entering reactant stream are also
independent of time. Classify the process (a) initially and (b) after a long
period of time has elapsed.
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Material Balance Simplifications
The following rules may be used to simplify the material balance equation:
Accumulation = In – Out + Generation – Consumption
 If the system is at steady‐state, set accumulation = 0
 In – Out + Generation – Consumption = 0
 If the balanced quantity is total mass, set generation = 0 and consumption
= 0 (law of conservation of mass)
 Accumulation = In – Out
 If the balanced substance is a nonreactive species, (neither a reactant nor
a product) or for non‐reacting systems in general, set generation = 0 and
consumption = 0
 Accumulation = In – Out
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Problems Involving Material Balances
 Initial procedures will be outlined for solving single unit processes
– No reaction (consumption = generation = 0)
– Continuous steady‐state (accumulation = 0)
– And so the Conservation Equation becomes….. (what?)
 These procedures will form the foundation for more complex problems
involving multiple units and processes with reaction
 Following a standard methodology to solve problems is the key to success.
This standard methodology will be illustrated via many examples in class,
and is the one used by F&R.
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Example 1
450 kmol per hour of a mixture of n‐butanol and iso‐butanol containing 30
mole % n‐butanol is separated by distillation into two fractions. The flow rate
of n‐butanol in the overhead stream is 120 kmol/h and that of iso‐butanol in
the bottom stream is 300 kmol/h. The operation is at steady‐state.
Calculate the unknown component flow rates in the output streams. What is
the mole fraction of n‐butanol in the bottom stream? What is the mass
fraction?
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Fractionation of Oil
Wk 2 pre‐tutorial exercise:
Draw a schematic of a plate
(or tray) distillation column
(continuous operation), and
briefly explain how separation
occurs
F&R Encyclopedia of Chemical
Engineering Equipment
(textbook website)
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Material Balance Procedures
All material balance calculations are variations on a single theme:
Given values of some input and output stream variables (e.g. flowrates,
compositions), derive and solve equations for the others
Solving the equations is a matter of simple algebra (the math is easy!),
however, you first need to:
 convert the problem statement into a process flow diagram; what are the
streams in/out and what components are in each stream?
 label the PFD with the ‘knowns” (flows, compositions, etc.), assign variables
to the unknowns (remaining flows, compositions), identify the system on
which you are doing the MB, and decide on your basis
(mass/moles/input/output….)
 derive the necessary equations from the component and/or overall MB
equations, and process constraint (PC) equations
 follow the standard methodology to solve the problem
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Process Flow Diagrams
A flowchart, or process flow diagram (PFD), is a convenient (actually,
necessary) way of organizing process information for subsequent
calculations.
To obtain maximum benefit from the PFD in material balance calculations,
you must:
1. Write the values and units of all known stream variables (flows and
compositions) at the locations of the streams on the chart.
2. Assign algebraic symbols to unknown stream variables (flows and
compositions) and write these variable names and their associated units
on the chart.
Your PFD is an essential part of the problem solution,
and will be assigned marks for completeness.
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F&R Ch 4.3a, Example 4.3‐1
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Note on Notation
The use of consistent notation is generally advantageous. For the purposes of
this course, the notation adopted in Felder and Rousseau will be followed. For
example:
n – moles
m – mass
n – molar flow rate
m – mass flow rate
V – volume
V – volumetric flow rate
x – component fractions (mass or mole) in liquid streams
y – component fractions in gas streams
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Example 2
A spent sulfuric acid solution is brought up to strength for a pickling process
in a mixer. Spent solution at 3% sulfuric acid (by weight) is mixed with a 50%
solution (by weight) to obtain the desired product concentration of 40% acid
by weight. All are aqueous solutions. Determine all flowrates on the basis of
100 lbm/h of product. If the actual flow of the spent stream is 300 lbm/h, what
must the flowrates of the streams be?
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Basis of Calculation
Basis of calculation – is an amount or flow rate of one of the process streams
on a mass or mole basis
 If a stream amount or flow rate is given in the problem statement, use
this as the basis of calculation (usually)
 If no stream amounts or flow rates are known, you can assume one,
preferably a stream of known composition
– if mass fractions are known, choose a total mass or mass flow
rate of that stream (e.g., 100 kg or 100 kg/h) as a basis
– if mole fractions are known, choose a total number of moles or
a molar flow rate
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Flowchart Scaling
Scaling – the process of changing the values of all stream amounts or flow
rates by a proportional amount while leaving the stream compositions and
conditions unchanged.
 Scaling up – final stream quantities are larger than the original
quantities
 Scaling down – final stream quantities are smaller than the
original quantities
30 mol A/min
70 mol B/min
40C, 1 atm
100 mol/min
0.30 mol A/mol
0.70 mol B/mol
40C, 1 atm
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60 mol A/min
140 mol B/min
Scale up process
by a factor of 2
40C, 1 atm
200 mol/min
0.30 mol A/mol
0.70 mol B/mol
40C, 1 atm
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Methodology for Solving Material Balance Problems
1. Choose a basis of calculation (input, output, mass, moles)
2. Draw and fully label a flowchart with all the known and unknown process
variables (flows, compositions) as well as the basis of calculation. Be sure
to include units.
3. Write any Process Constraint (PC) equations that relate variables.
4. Determine the number of unknowns and the number of equations that
can be written to relate them. That is, does the number of equations
equal the number of unknowns?
5. Solve the equations
6. Check your solution – does it make sense? Calculate the quantities
requested in the problem statement if not already calculated
7. Cleary present your solution with the proper units and the correct
number of significant figures
“Understanding the Concepts” is not good enough. You will not be
tested on “Understanding the Concepts”. You will be tested on your
ability to set up and solve problems, and to get the correct answer.
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Example 3: Quiz 1 2007
A mixture containing 42 wt% benzene (B) and 58 wt% toluene (T) is fed to a
distillation column at a flowrate of 100 kg/min. The product stream leaving the
top of the column (the overhead product) contains 90 wt% benzene, and 85
wt% of the total benzene fed to the column exits in this overhead product
stream.
Calculate the mass flowrate and mass composition of the product stream
leaving the bottom of the column. Calculate the volumetric flowrate of the
overhead product, assuming that it exits the distillation column as a vapour
stream at 82 ºC and 1 atm (abs).
Physical Property Data (S.G.=specific gravity) from Table B1:
• Benzene
S.G.=0.879
MW=78.11 g/mol
• Toluene
S.G.=0.866
MW=92.13 g/mol
• Water
density = 1.00 kg/L
MW=18.02 g/mol
R = 0.08206 L∙atm/(mol∙K)
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Degrees of Freedom Analysis: A Motivating Example
A feed stream containing C8 and C10 hydrocarbons is split into 3 product
streams: an overhead fraction, a middle cut and a bottom fraction, whose
mole fraction compositions are shown below. Seventy per cent of the C8
entering the column in the feed is recovered in the overhead. On the basis
of 100 lb‐moles/h of feed determine the molar flow rates of the 3 product
streams.
C8: 0.516
C10: 0.484
100 lb‐moles/h
C8: 0.300
C10: 0.700
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C8: 0.352
C10: 0.648
C8: 0.146
C10: 0.854
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Independent Equations
A set of equations are independent if you cannot derive one by adding and
subtracting combinations of the others.
x  2y  z 1
Is this set of equations independent?
2x  y  z  2
y  2z  5
1 2 1   x   1 
2 1  1  y   2
   

0 1 2   z  5
1 0 0  x   6 
0 1 0  y    5

   
0 0 1  z   5 
row
reduce
Rank = 3. No non‐zero
rows in reduced form
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Independent Equations… cont’d
Are these sets of equations independent?
x  2y  z 1
2 y  4 z  10
y  2z  5
x  2y  z 1
2x  y  z  2
3x  3 y  3
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Degree‐of‐Freedom Analysis
A degree‐of‐freedom analysis (DFA) is a determination of the number of
unknowns in a problem, and the number of independent equations that can
be written. The difference between the number of unknowns and the number
of independent equations is the number of degrees‐of‐freedom, DF or ndf, of
the process.
ndf  nunknowns  nindependent equations
Possible outcomes of a DFA:
– ndf = 0, there are n independent equations and n unknowns. The
problem can be solved.
– ndf > 0, there are more unknowns that independent equations. The
problem is underspecified. ndf more independent equations or
specifications are needed to solve the problem.
– ndf < 0, there are more independent equations than unknowns. The
problem is overspecified with redundant and possibly inconsistent
relations.
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DFA: Sources of Equations
Sources of equations that relate unknown process variables include:
1. Material balances – for a nonreactive process, usually but not always, the
maximum number of independent equations that can be written equals
the number of chemical species in the process
2. Process constraints– given in the problem statement
3. Physical constraints – e.g., mass or mole fractions must add to 1 (usually
taken care of when setting up PFD)
4. Stoichiometric relations – systems with reaction (later)
5. Energy balances – 2nd half of course
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Notes on DFA: Dependent Material Balances
There are two common situations where you will find fewer independent
equations than species, and they are:
1.
Balance around a splitter
– Single input – two or more outputs with same composition
– Only 1 independent balance equation, since:
m1 = m2 + m3
(Overall Balance)
and x1m1 = x2m2 + x3 m3
(Balance on A)
but since x1 = x2 = x3, these balances are not independent
– Splitters are used for:
• Purge streams (reactor systems with recycle)
• Total condensers at the top of distillation columns
m1 kg/h
x1 kg A/kg
(1‐x1) kg B/kg
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m2 kg/h
x2 kg A/kg
(1‐x2) kg B/kg
Splitter
m3 kg/h
x3 kg A/kg
(1‐x3) kg B/kg
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Notes on DFA: Dependent Material Balances… cont’d
Distillation Column with Total Condenser
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Notes on DFA: Dependent Material Balances… cont’d
2. If two species are in the same ratio to each other wherever they appear in
a process and this ratio is incorporated in the flowchart labeling, balances
on those species will not be independent equations.
– Situation occurs frequently when air is present in a nonreactive
process (21 mol% O2; 79 mol% N2)
– E.g., vapourization of liquid carbon tetrachloride into an air stream
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n1 mol O2/s
3.76n1 mol N2/s
n3 mol O2/s
3.76n3 mol N2/s
n4 mol CCl4(v)/s
n2 mol CCl4(L)/s
n5 mol CCl4(L)/s
Best to treat air
as a single
species in this
situation
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Summary: MB applied to single process units without reaction
•
Standard procedures was developed for single‐unit processes (F&R 4.3)
– No reaction (Consumption=Generation=0)
– Continuous steady‐state (Accumulation=0)
•
Develop good habits now, and practice. Problems will get more complex as we
extend the procedures to multiple‐unit processes (starting in ≈Week 3) and
processes with reaction (starting in Week 4/5)
•
Standard procedures are summarized in F&R Section 4.3 and include:
– drawing/labeling a process flow diagram (4.3a)
These are critical sections
– selecting a basis of calculation (4.3b)
of the text and
– setting up material balances (4.3c)
form the basis for Quiz 1
– performing a degree of freedom analysis (4.3d)
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Example 4
Hot soap is chilled on a roller and scraped continuously from the roller onto a moving
conveyor belt which carries the soap into a dryer (see below). The entering soap
contains 25% water by weight. It is desired to reduce the water content to 15% by
weight and to produce 1200 lb/h of nearly dry soap chips. The entering air contains
0.3 mole % water vapour. The dryer manufacturer suggests that the dryer operates
efficiently when the nearly dry air/wet soap flow ratio is 3.0. Calculate the unknown
flowrates and compositions. Air is 21% oxygen and 79% nitrogen (mole basis) and has
a molecular weight of 29.0 g/mol.
moist air
hot, nearly dry air
Soap Dryer
wet soap chips
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dried soap chips
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Condensers and Evaporators
Wk 3 pre‐tutorial exercise:
Dryers and dehumidifiers are
two examples of a general
class of separators known as
condensers.
Look into the difference
between condensers and
evaporators, and explain their
industrial usage. Can you think
of a household example of an
evaporator?
F&R Encyclopedia of Chemical
Engineering Equipment
(textbook website)
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Example F&R 4.3‐1
An experiment on the growth rate of certain organisms requires an
environment of humid air enriched in oxygen. Three input streams are fed
into an evaporation chamber to produce an output stream with the desired
composition.
A: Liquid water, fed at a rate of 20.0 cm3/min
B: Air (21 mole% O2, the balance N2)
C: Pure oxygen, with a molar flow rate one‐fifth of the molar flow rate
of stream B.
The output gas is analyzed and is found to contain 1.5 mole% water. Draw
and label a PFD, and calculate all unknown stream variables (i.e. flows and
compositions).
Work through on your own, then check with solution in the textbook
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Example: Quiz 1 2009
A continuous distillation column is to be used to separate a 3‐component mixture of
acetic acid (AA), water (W) and Benzene (B), and a trial run gave the data below (mass
basis). The data for the benzene in the feed (which consists of AA, W and B) was not
taken because of an instrument malfunction. Use a degree of freedom analysis, and
then calculate the benzene flow in the feed in kg/h.
Waste
10.9% AA
21.7% W
67.4% B
Aqueous Solution
(containing 80% AA +
20% W)
+
B (data not available)
Product
350 kg/h pure AA
Answer: B in feed = 311 kg/h
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