CHEE 221: Chemical Processes and Systems

Module 1. 

Material Balances: Single Process Units without Reaction 

(Felder & Rousseau Ch 4.1‐4.3)

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General Material Balance Equation (“GMBE”)

Accumulation = In – Out + Generation – Consumption

System over which mass balance is made

Input streams to system

output streams from system

system boundary

Accumulation   within the system (mass buildup)

=Input through 

system boundaryOutput through system boundary‐

Generation within the system+ ‐ Consumption 

within the system

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)

Mixer – combines two or more inputs (usually of different compositions) into a single output) (no reaction)

Separator – separates a single input into two or more outputs of different composition (no reaction)

splitter

mixer

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

pump

heat exchanger

Actual process units can combine these different functions into a single piece ofhardware, and are given different names, e.g. a separator can be a distillationcolumn, a filter press, a centrifuge, etc.F&R Encyclopedia of Chemical Engineering Equipment ( textbook website)

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Steam Boiler

Steam Boiler

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 inwhich no reaction is occurring (you will not be told this, and will need toknow this from the type of unit that is under consideration; crystallizer,evaporator, filter, furnace, etc.), a material balance can be written in terms ofthe following conserved quantities:

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

To apply a material balance, you need to define the system and the quantities ofinterest (e.g. mass of a component, total mass, moles of an atomic species).What is your system, and what are you keeping track of?

1. Total mass (or moles)2. Mass (or moles) of a chemical compound3. Mass (or moles) of an atomic species

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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

F&R Ch 4.1

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F&R Example 4.1

Classify the following processes as batch, continuous, orsemibatch, and transient or steady‐state.

1. A balloon is filled with air at a steady rate of 2 g/min.2. A bottle of milk is taken from the refrigerator and left on the kitchen table. 3. Water is boiled in an open flask.4. Carbon monoxide and steam are fed into a tubular reactor at a steady rate

and react to form carbon dioxide and hydrogen. Products and unusedreactants are withdrawn at the other end. The reactor contains air whenthe process is started up. The temperature of the reactor is also constant,and the composition and flow rate of the entering reactant stream are alsoindependent of time. Classify the process (a) initially and (b) after a longperiod 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

If the balanced quantity is total mass, set generation = 0 and consumption = 0 (law of conservation of mass)

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

In – Out + Generation – Consumption = 0

Accumulation = In – Out

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 30mole % n‐butanol is separated by distillation into two fractions. The flow rateof n‐butanol in the overhead stream is 120 kmol/h and that of iso‐butanol inthe bottom stream is 300 kmol/h. The operation is at steady‐state.

Calculate the unknown component flow rates in the output streams. What isthe mole fraction of n‐butanol in the bottom stream? What is the massfraction?

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Fractionation of Oil

Wk 2 pre‐tutorial exercise:Draw a schematic of a plate(or tray) distillation column(continuous operation), andbriefly explain how separationoccurs

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:  

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

Given values of some input and output stream variables (e.g. flowrates,compositions), derive and solve equations for the others

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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 subsequentcalculations.

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.

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:  

mVV

xy

m – mass n – moles

– mass flow rate – molar flow rate

– volume – volumetric flow rate

– component fractions (mass or mole) in liquid streams

– component fractions in gas streams

n

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Example 2

A spent sulfuric acid solution is brought up to strength for a pickling processin 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% acidby weight. All are aqueous solutions. Determine all flowrates on the basis of100 lbm/h of product. If the actual flow of the spent stream is 300 lbm/h, whatmust 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 flowrates by a proportional amount while leaving the stream compositions andconditions 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/min70 mol B/min

40C, 1 atm

60 mol A/min140 mol B/min

40C, 1 atm

200 mol/min0.30 mol A/mol0.70 mol B/mol40C, 1 atm

100 mol/min0.30 mol A/mol0.70 mol B/mol40C, 1 atm

Scale up process by a factor of 2

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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 calculated7. 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 betested on “Understanding the Concepts”. You will be tested on yourability 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 adistillation column at a flowrate of 100 kg/min. The product stream leaving thetop of the column (the overhead product) contains 90 wt% benzene, and 85wt% of the total benzene fed to the column exits in this overhead productstream.Calculate the mass flowrate and mass composition of the product streamleaving the bottom of the column. Calculate the volumetric flowrate of theoverhead product, assuming that it exits the distillation column as a vapourstream 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 productstreams: an overhead fraction, a middle cut and a bottom fraction, whosemole fraction compositions are shown below. Seventy per cent of the C8entering the column in the feed is recovered in the overhead. On the basisof 100 lb‐moles/h of feed determine the molar flow rates of the 3 productstreams.

C8: 0.300C10: 0.700

C8: 0.516C10: 0.484

C8: 0.352C10: 0.648

C8: 0.146C10: 0.854

100 lb‐moles/h

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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.

Is this set of equations independent? 

522212

zyzyxzyx

521

210112

121

zyx

55

6

100010001

zyx

rowreduce

Rank = 3.  No non‐zero rows in reduced form

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Independent Equations… cont’d

52104212

zyzyzyx

3332212

yxzyxzyx

Are these sets of equations independent?

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Degree‐of‐Freedom Analysis

A degree‐of‐freedom analysis (DFA) is a determination of the number ofunknowns in a problem, and the number of independent equations that canbe written. The difference between the number of unknowns and the numberof independent equations is the number of degrees‐of‐freedom, DF or ndf, ofthe process.

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.  

equationstindependenunknownsdf nnn

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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

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

Splitter

m3 kg/hx3 kg A/kg

(1‐x3) kg B/kg

m1 kg/hx1 kg A/kg

(1‐x1) kg B/kg

m2 kg/hx2 kg A/kg

(1‐x2) kg B/kg

There are two common situations where you will find fewer independent equations than species, and they are:

CHEE 221 29

Distillation Column with Total Condenser

Notes on DFA: Dependent Material Balances… cont’d

CHEE 221 30

2. If two species are in the same ratio to each other wherever they appear ina process and this ratio is incorporated in the flowchart labeling, balanceson 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

n1 mol O2/s3.76n1 mol N2/s

n5 mol CCl4(L)/sn2 mol CCl4(L)/s

n3 mol O2/s3.76n3mol N2/sn4 mol CCl4(v)/s Best to treat air 

as a single species in this situation

Notes on DFA: Dependent Material Balances… cont’d

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• 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)– selecting a basis of calculation (4.3b)– setting up material balances (4.3c)– performing a degree of freedom analysis (4.3d)

Summary: MB applied to single process units without reaction

These are critical sectionsof the text and

form the basis for Quiz 1

CHEE 221 32

Example 4

Hot soap is chilled on a roller and scraped continuously from the roller onto a movingconveyor belt which carries the soap into a dryer (see below). The entering soapcontains 25% water by weight. It is desired to reduce the water content to 15% byweight and to produce 1200 lb/h of nearly dry soap chips. The entering air contains0.3 mole % water vapour. The dryer manufacturer suggests that the dryer operatesefficiently when the nearly dry air/wet soap flow ratio is 3.0. Calculate the unknownflowrates and compositions. Air is 21% oxygen and 79% nitrogen (mole basis) and hasa molecular weight of 29.0 g/mol.

Soap Dryer

moist air

wet soap chips

hot, nearly dry air

dried soap chips

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Condensers and Evaporators

Wk 3 pre‐tutorial exercise:Dryers and dehumidifiers aretwo examples of a generalclass of separators known ascondensers.

Look into the differencebetween condensers andevaporators, and explain theirindustrial usage. Can you thinkof a household example of anevaporator?

F&R Encyclopedia of Chemical Engineering Equipment(textbook website)

CHEE 221 34

Example F&R 4.3‐1

An experiment on the growth rate of certain organisms requires anenvironment of humid air enriched in oxygen. Three input streams are fedinto an evaporation chamber to produce an output stream with the desiredcomposition.

A:  Liquid water, fed at a rate of 20.0 cm3/minB:  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. Drawand label a PFD, and calculate all unknown stream variables (i.e. flows andcompositions).

Work through on your own, then check with solution in the textbook

CHEE 221 35

Example: Quiz 1 2009

A continuous distillation column is to be used to separate a 3‐component mixture ofacetic acid (AA), water (W) and Benzene (B), and a trial run gave the data below (massbasis). The data for the benzene in the feed (which consists of AA, W and B) was nottaken because of an instrument malfunction. Use a degree of freedom analysis, andthen calculate the benzene flow in the feed in kg/h.

Aqueous Solution(containing 80% AA + 20% W)+B (data not available)

Waste10.9% AA21.7% W67.4% B

Product350 kg/h pure AA

Answer: B in feed = 311 kg/h